## Noble frequency ratios as prime-count vectors in ℚ(√5)

- Dave Keenan
- Site Admin
**Posts:**2180**Joined:**Tue Sep 01, 2015 2:59 pm**Location:**Brisbane, Queensland, Australia-
**Contact:**

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

Yes. I think one post a day would be plenty to take in. Thank you for the correction. I have made the edit.

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

Very interesting thread, I just have one little pedantic mathematical correction. Note that this pattern you mentioned

This also means that your GSRCF method

Of course this should not bother you at all in the context of this thread, because the largest ordinary prime we need to factor here is 79, but it may become important if people choose to consider nobles coming from more levels of the Stern-Brocot tree.

is only true for ordinary primes < 100, as 101 factors into (9ϕ - 4) * (9ϕ - 5), and not some numbers with golden part 10.Dave Keenan wrote: ↑Wed Sep 21, 2022 12:33 am The golden parts of the prime factors are always the same, and are the floor of the square root of the ordinary prime.

This also means that your GSRCF method

doesn't work onDave Keenan wrote: ↑Wed Sep 21, 2022 11:02 pm Call this numberp. Find the greatest square number that is not greater thanp, call its. Its square root is the golden part of both factors, call itb. Sob² =sandb= ⌊√p⌋. Subtractsfrompto obtain a remainder. Factorise this remainderp-sinto two factors (not negative, but may include 0 or 1) that sum tob, call these factorsaandA, whereais the smaller factor. So we needa×A=p-sanda+A=b= √s. Then fp= -A+bϕ and Fp= -a+bϕ.

*p*= 101, as you'll get*b*= 10 instead of 9, and then need to factor 1 into two numbers that sum to 10. You can mitigate this problem to make the method work on primes up to 200 by choosing the largest*b*so that*b*² +*b*- 1 ≤*p*, or equivalently so that (2*b*+ 1)² ≤ 4*p*+ 5, i.e.*b*= ⌊(√(4*p*+ 5) - 1)/2⌋, but this breaks again because 211 = (13ϕ - 6) * (13ϕ - 7) gives*b*= 14 using this method, and in general there is no clear cutoff between ordinary primes that are factored to feudal primes with different golden parts, as (15ϕ - 1) * (15ϕ - 14) = 239 < 241 = (14ϕ - 5) * (14ϕ - 9).Of course this should not bother you at all in the context of this thread, because the largest ordinary prime we need to factor here is 79, but it may become important if people choose to consider nobles coming from more levels of the Stern-Brocot tree.

- Dave Keenan
- Site Admin
**Posts:**2180**Joined:**Tue Sep 01, 2015 2:59 pm**Location:**Brisbane, Queensland, Australia-
**Contact:**

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

Thank you! I have now added asterisk notes to both of the posts you mentioned, which link to your post above.

- Dave Keenan
- Site Admin
**Posts:**2180**Joined:**Tue Sep 01, 2015 2:59 pm**Location:**Brisbane, Queensland, Australia-
**Contact:**

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

[I've pasted the fully-expanded current contents of the related facebook thread below (as 5 posts), because I'm tired of having to click "See more" and "View more replies" all the time to find the recent gems (proofs and counterexamples) by רועיסיני (Roee Sinai), hilted below.]

Dave Keenan

The Feudal Manifesto

While procrastinating from paid work recently, I followed up on some tuning math that I've been meaning to investigate for years.

The executive summary is below.

The detailed explanation is here: viewtopic.php?f=21&t=555

An effective cadence can be made by playing a chord having high-harmonic entropy followed by a chord having low harmonic-entropy (HE).

Regular Temperament Theory (RTT) is, at present, designed to target intervals having low HE, while users rely on the fact that any octave-repeating scale of more than 5 notes is guaranteed to have some intervals with high HE.

But what if we could target specific high-HE intervals as well as low-HE intervals?

Minima of HE are at simple rational-number ratios.

Many maxima of HE are close to simple noble-number ratios.

Noble numbers are irrational and so cannot be be expressed as (finite) prime-count vectors (PC-vectors or monzos) whose basis is rational (e.g. the ordinary prime numbers).

But it turns out noble-numbers can be expressed as a quotient of two "feudal" primes or units, where "feudal" means "belonging to the quadratic field ℚ(√5)" (my coinage).

Feudal units are numbers of the form ±ϕⁿ where ϕ is the golden ratio and n is an integer.

Feudal primes are a subset of the feudal integers which are numbers of the form a+bϕ where a and b are integers.

By appending small feudal primes (and the unit ϕ) to the basis of our PC-vectors, it may be possible to find feudal commas and therefore feudal temperaments that generate a large number of both simple rationals and simple nobles using a small number of generators.

forum.sagittal.org

Noble frequency ratios as prime-count vectors in ℚ(√5) [superseded] - The Sagittal forum

439 comments

Dave KeenanAuthor

Units and primes of ℚ(√5), by Prof. Dirk Dekker, annotated by me.

Reply

32 wEdited

Mike BattagliaAdmin

Great stuff Dave Keenan and I will read when I get the chance, but I was also just thinking about this - that if you add "anti-JI" intervals to JI, you get the number field Q√5.

Reply

32 w

Mike BattagliaAdmin

I had some results on this a while ago I never posted which I worked on with Ryan Avella. Of course, Q[√5] is a field, so everything is invertible and divides 1 (thus every element of a field is a "unit"). I think that there is some subtlety to what you are calling "primes" and "units" - the units you are talking about in your post are those things that are units in the ring of integers derived from this number field (which I think corresponds to Z[√5] (EDIT: oops, it really is Z[phi] for this.)). That is one method, but it is also interesting to look at units in Z[phi] (note this is not the same ring). You also get some interesting results for "primes" if you look at the monoid N[phi] of elements a+b*phi with non-negative a and b. I had an interesting method of looking at all elements of the form a + b*phi for non-negative a and b, and iterating upward starting at 0 + 0*phi, and for each element checking if it was divisible by any of the elements with smaller a and b; those which don't are also prime in a certain sense.

Reply

32 wEdited

Dave KeenanAuthor

Mike Battaglia This is all dealt with in the full series of articles.

Reply

32 w

Mike BattagliaAdmin

Dave Keenan yes I see it. What I'm saying is phi is a unit in the ring of integers Z[phi], but in the monoid of "positive" integers N[phi], it doesn't divide 1, so I think it'd be just another "prime." I'm not 100% sure on this though.

Reply

32 w

Paul ErlichAdmin

RTT relies on no such fact as far as I am aware; the 9-limit Otonality and 5-tET are perfectly fine counterexamples and are not ruled out by any RTT considerations.

Reply

32 w

Dave KeenanAuthor

Paul Erlich Why the uncharitable reading? It's only a brief summary. I didn't mean that RTT relies on it, but that /users/ of RTT often rely on it, to ensure they have discordance as well as concordance in their RTT-generated scales. The 9-limit otonality and 5-tET are /not/ counterexamples because they do not contain "more than 5 notes".

Reply

32 w

Paul ErlichAdmin

Dave Keenan sorry I misread!

Reply

32 w

Paul ErlichAdmin

Of course phi itself is almost never near a local, let alone global, maximum of HE, and repeated stacks of phi are more concordant than repeated stacks of any similar-sized interval. This gives another interesting reason to want to include phi along with conventional primes. Aggelos Boshidis would be interested.

Reply

32 w

Dave KeenanAuthor

Paul Erlich I never suggested it was near a global maximum. But I don't know why you claim that ϕ (≈ 833 ¢) is almost never near a local maximum of HE. 829 ¢ is marked as a local maximum on the second HE chart here: http://tonalsoft.com/enc/h/harmonic-entropy.aspx

harmonic entropy - accordance model of musical intervals

TONALSOFT.COM

harmonic entropy - accordance model of musical intervals

harmonic entropy - accordance model of musical intervals

Reply

Remove Preview

32 wEdited

Paul ErlichAdmin

Dave Keenan ha you're right! I wasn't thinking about the cases where 8:5 was an inflection point. Thanks for the correction! It's just often claimed that if simple ratios are maximally concordant then phi must be maximally discordant.

Reply

32 w

Dave KeenanAuthor

Paul Erlich Sorry about that last post. Facebook failed to show me your previous until I switched from my desktop to my phone.

Reply

32 w

Dave KeenanAuthor

Surely there's almost always a peak between 8/5 (814 ¢) and 13/8 (841 ¢), but closer to 13/8?

Reply

32 w

Paul ErlichAdmin

Dave Keenanso to answer your question, no, my original or default 2HE curves would probably have no peak between 8/5 and 13/8, and in fact none 746 and 845 cents

Reply

31 w

Mike BattagliaAdmin

Dave Keenan and Paul Erlich , I was playing around with some of these theorems about phi to see if we can relate Dirichlet's approximation theorem to what we are doing with HE. Some of these results suggest one may tend to get local maxima that are closer to phi if one a) uses the min-entropy rather than the regular Shannon entropy, and b) the Laplace distribution rather than the Gaussian. Kind of curious how well that does.

Reply

31 w

Mike BattagliaAdmin

Oh, and with the Farey weighting instead of Tenney as well.

Reply

31 w

Dave KeenanAuthor

Paul Erlich Mike Battaglia We agree that nobles are not necessarily the most discordant ratios in their vicinity, but there may still be a sense in which they are audibly the most "metastable" with respect to cadences. Or audibly the most "ambiguous" with respect to approximating just intervals. In any case, the fact that they exist in a unique factorisation domain that (hopefully) doesn't have too much extra "junk", is hard to resist.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan I'm not sure it matters; it's clearly good enough and tends to give sensible results at least for varying values of s.

Reply

31 w

Paul ErlichAdmin

Mike Battagliavarying results for s where? Good enough for what, how?

Reply

31 w

Paul ErlichAdmin

Mike Battaglia are you looking at Steve Martin 's graphs above?

Reply

31 w

Mike BattagliaAdmin

Paul Erlich: "repeated stacks of phi are more concordant than repeated stacks of any similar-sized interval" - is this true? Would be kind of interested to see if Steve Martin's HE script can compute this - suppose we sweep from all intervals from 0-1200 cents, and for each one, compute (let's say) the 5-HE of a stack of five of those intervals on top of one another. Is there really a local minimum at exactly phi?

Reply

31 w

Paul ErlichAdmin

Mike Battagliayes that's exactly what we found for 3HE/4HE. Not sure about 5HE.

Reply

31 w

Mike BattagliaAdmin

Paul Erlich that's really interesting. Is there a graph of the results somewhere?

Reply

31 w

Paul ErlichAdmin

Mike BattagliaI don't recall if Steve Martin graphed them but he reported the other local minima which were pretty far away (3:2 was the closest on one side). Maybe he has a graph he could share. I made a big deal about this for a few years because it arose despite the fact that HE doesn't even account for combinational tones which are main the reason one might think stacks of phi would be more concordant than stacks of an interval a bit off from phi.

Reply

31 w

Paul ErlichAdmin

*the main reason

Reply

31 w

Mike BattagliaAdmin

Well I don't think it's worth throwing the baby out with the bathwater re what Dave is saying about phi, but I do think what you are saying is also important in that it tells you that much of this phi business may fall apart once you leave the realm of dyads. I

There are some interesting papers on simultaneous Diophantine approximation of multidimensional real vectors, and I'm not sure if the golden ratio has quite as special of a role there.

Reply

31 w

Paul ErlichAdmin

Mike Battaglia maybe the tribonacci constant etc. would . . .

Reply

31 w

Paul ErlichAdmin

Mike Battagliaalso I see this as a positive, not negative, reason to want to have phi as a generator or at least temperamentally simple, and I know Aggelos Boshidis is interested in more possibilities along those lines. I mentioned good old Sqrtphi temperament, highly complex mappings for most ratios other than 11:7 but with remarkably low 'badness' overall; it fit with his observation that (pi/2 ~) 11/7 ~ 2/sqrt(phi), which means one period up and one generator down in Sqrtphi. But I'm sure other interesting possibilities emerge when explicitly targeting phi along with the first several primes, and in no way would I want to put the kibosh on such research -- quite the contrary! But FWIW 845 cents is the peak I'm seeing in all of these graphs Dave Keenan . . . http://tonalsoft.com/.../harmonic-entro ... commentary...

Paul Erlich's older ideas on Harmonic Entropy, comments by Monzo

TONALSOFT.COM

Paul Erlich's older ideas on Harmonic Entropy, comments by Monzo

Paul Erlich's older ideas on Harmonic Entropy, comments by Monzo

Reply

31 w

Steve Martin

Mike Battaglia, Paul Erlich, I can make graphs but for now just the minima extracted:

3HE: 702, 600, 832, 952 ...

4HE: 833, 600, 45, 107 ...

5HE: 600, 833, 43, 1077 ...

Reply

31 wEdited

Paul ErlichAdmin

Steve MartinI'm really looking forward to seeing the graph

Reply

31 w

Paul ErlichAdmin

Aggelos Boshidisnot sure I understand. Can you break down what you're trying to demonstrate, step-by-step through each calculation, with a narrative to guide the reader?

Reply

31 w

Paul ErlichAdmin

Aggelos Boshidislooks like you're describing a nearly equal tuning, where the step size is approximately 48.3 cents?

Reply

31 w

Steve Martin

Paul, Mike 3HE for stacks

Reply

31 wEdited

Steve Martin

... 4HE for stacks ...

Reply

31 w

Steve Martin

... 5HE for stacks ...

Reply

31 w

Steve Martin

I suspect noise appearing in the 5HE one at larger cent values

Reply

31 w

Aggelos Boshidis

Paul Erlich I will reply asap, maybe this helps (with some 'quantizing' : https://sevish.com/scaleworkshop/?name= ... penv=organ

Scale Workshop

SEVISH.COM

Scale Workshop

Scale Workshop

Reply

31 w

Aggelos Boshidis

Paul Erlich So I used sevish workshop to generate a rank2 temperament.

I used sqrtphi as a generator which is apprx 1272/1000

I used phi inversion as a period (367c.) apprx 1236/1000, scale size was seven with six generators up 1/1. What I found interesting is that the intervals fall onto almost the exact 1200 cents of one octave, 2400 of second and 3600 of third octave

:

~50,99,149,199,249,298,367,

416(pi/2 inversion interval~sqrtphi ), 466,516,566,615,665,733

783(pi/2),833(phi), 883(8/5),932,982,1031, 1100,1150

1199.85~1200

....1516 (6/5*2),...2399.7~2400(4/1),

3599.5~3600 (8/1)

Reply

31 wEdited

Paul ErlichAdmin

Aggelos Boshidis six generators up from 1/1? I see six approximate quartertones up from 1/1. 1272/1236 = 106/103 would be period minus generator according to what you said, but that's 49.7 cents, so I suspect what you actually did was construct six of your generators *down*, not *up*, from 1/1. Is that correct so far?

Reply

31 w

Aggelos Boshidis

Paul Erlich

Reply

31 w

Aggelos Boshidis

Paul Erlich I forgot to add 1031 and 1100 cents on my previous message regarding the tuning. Edited it. Also sent you a screenshot with the exact figures

Reply

31 w

Paul ErlichAdmin

Aggelos Boshidisthanks. well my question still stands. It looks like you're building the generator *downward* (not upward) from 1/1; the math I presented seems to corroborate that -- no?

Reply

31 w

Paul ErlichAdmin

Steve Martinmaybe make a new post with your graphs (and maybe try a couple different s values)? Some people might be missing them buried in a subthread here, and I'd like to tag others.

Reply

31 w

Steve Martin

Paul Erlich I missed this, will do but not for a week or so.

Reply

30 w

רועי סיני

Paul Erlich If you are interested, there are other temperaments that can be tuned so that the golden ratio has a low complexity. The simplest examples, that also come from the Fibonacci series if you try to equate the ratios between successive elements, are gamelismic temperaments that also include a half-octave (which happens to be another interval whose stacks have a low harmonic entropy). Some examples are lemba, hendec, semimiracle, semivalentine and semimothra (where the last, like the others, tempers out 169/168).

Reply

1 d

Paul ErlichAdmin

The "Vos" HE curves used the Laplace distribution (and shannon), but they look like they might be "fractal" with unendingly many local minima and maxima on ever-finer scales . . .

Reply

31 w

Mike BattagliaAdmin

Paul Erlich yeah I think they may be "continuous nowhere differentiable" or something like that.

Reply

31 w

Paul ErlichAdmin

https://m.facebook.com/photo.php?fbid=1 ... 2912705601

Reply

31 w

Paul ErlichAdmin

https://m.facebook.com/photo.php?fbid=1 ... 1370594482

Reply

31 w

Aggelos Boshidis

Paul Erlich but did you see the screenshot I sent you? According to the program its the opposite direction

Reply

31 w

Paul ErlichAdmin

Aggelos Boshidis wow, how is that? Maybe Sean Archibald could explain. AFAICT the 49.7-cent interval is period minus generator, not the generator itself.

Reply

31 w

Sean Archibald

What I'm seeing is that the generator is greater than period, and generators are being stacked up, resulting in the 49.7c interval which looks expected to me

Reply

31 w

Paul ErlichAdmin

Sean Archibaldthanks! I never imagined someone would specify a tuning using a generator larger than the period, so I inadvertently mentally swapped Aggelos Boshidis 's generator and period. Sorry for the confusion guys! Anyway, this means most of the "~" in Aggelos' original list of relationships were not approximate, they were exact, since the scale simply repeats the same 106:103 step interval 6 times. So obviously one interval in the scale is 6/5 times the size of another when measured in cents, etc. (This has nothing to do with the consonant frequency ratio 6:5 of course).

Reply

31 w

Aggelos Boshidis

Paul Erlich the thing that is interesting for me is that after three repetitions plus two 49.7c you get 1200, 2400, 3600 (then it deviates) Try other rank2 temperaments with the same logic and it's not easy to get the same result. 6/5 appears as 1516 c. In the (1200-3600 cent octave)

Reply

31 w

Paul ErlichAdmin

Aggelos Boshidis12/5, yes.

You can use various intervals of repetition such that neither the period nor the generator is an octave or 1/N octave even for rank-2 temperaments featuring prime 2 (in Graham's app). Just use the interval of repetition as the first element of the JI basis you specify, and alter the other basis elements accordingly if necessary (and click "subgroup TE" for the results). So if we want e.g. the usual meantone temperament, but repeating at say 5:3, we can specify the basis as 5/3.2.3, giving

http://x31eq.com/cgi-bin/rt.cgi?limit=5%2F3_2_3...

where the first Scala file is:

http://x31eq.com/cgi-bin/scala.cgi?limit=5%2F3_2_3...

Repeating this up and down over many repetitions of the ~5:3 period, one will find many multiples of ~1201 cents.

Obviously there are infinite variations on this for any rank-2 temperament featuring prime 2 in its JI basis.

X31EQ.COM

5/3.2.3-limit Regular Temperament

5/3.2.3-limit Regular Temperament

Reply

31 w

Aggelos Boshidis

Paul Erlich yes agreed, but I was talking in reference to the phi "business", using sqrtphi and phi inverted.

Reply

31 w

Paul ErlichAdmin

רועי סיניthanks, I'm familiar with most of those of course; would love to see some more explicit examples of equating the ratios between successive elements of the Fibonacci sequence & understanding how you always seem to get rank-2 temperaments from the process (?)

Reply

1 d

רועי סיני

Paul Erlich If you set ϕ ~ 13/8 ~ 21/13 ~ 34/21 ~ ... up to infinity, the √2.ϕ.5 group gives you a rank-3 temperament up to 31-limit. Its map is the following:

[⟨2 -1 0 7 6 6 4 -4 -9 0 -10]

⟨0 3 0 -1 4 1 3 9 13 7 11]

⟨0 0 1 0 -1 0 0 0 0 0 1]⟩

For example, 13 is 8 * ϕ = √2⁶ϕ, and 3 = √(144 / 16) is √(8 * ϕ⁶ / 16) = ϕ³/√2.

Now, if you want it to be rank-2, you essentially need to represent 5 in terms of the other generators, and each of the temperaments I provided comes from another way to do so:

Lemba comes from the most naïve way, which is to say that 8/5, the fibonacci ratio before the ones we considered, should also be ϕ. It gives you low complexity in exchange for a high error for 5, 11 and 31.

Hendec comes from noting that √2 is also approximated by a simple continued fraction, and in fact, is already represented by 17/12 and 41/29, which are both present in this continued fraction (41 is larger than 31 but we can still get a value for it using F₂₀ / F₁₀ = 3*41), and therefore it makes sense to also require the next ones. 99/70 gives you a mapping for the 5, and the more complex ones seems to not affect small primes and only give much larger primes than is musically interesting.

Semimiracle and semivalentine just come from the fact that both miracle and valentine are good 11-limit temperaments that are consistent with this rank-3 one, and they already have these known 13-limit extensions that split the octave in half and are also consistent with this temperament.

Semimothra comes from the fact that the terms in the Lucas sequence are also all represented in this temperament, as Lₙ = F₂ₙ / Fₙ, which from the sixth one is always represented by ϕⁿ, and the ratio between them and the Fibonacci numbers, that are again represented from the sixth one formulaically, this time by Fₙ ~ 8ϕⁿ⁻⁶, approaches √5 as n approaches infinity, and therefore it makes sense to equate the ratio between these two representations, that equals ϕ⁶/8, to the square root of 5. This ratio also represents 9/4, and therefore we get a meantone temperament, which I call semimothra as it is identical to mothra in the 11-limit and then cuts the octave in half.

Reply

17 h

Paul ErlichAdmin

Dave Keenan Cmloegcmluin Xenharmonic Feisbeuk

Reply

17 h

Dave KeenanAuthor

רועי סיני This is mind-blowing stuff.

Reply

16 h

Paul ErlichAdmin

רועי סיני very interesting. However we don't describe any weak extensions of meantone as itself "a meantone temperament".

Reply

17 h

Dave KeenanAuthor

רועי סיני This deserves to be the topic of its own thread, not buried here. Perhaps with some introduction for those who haven't been following. And perhaps with a link to this thread. And perhaps spelled out in more detail.

Reply

16 h

Paul ErlichAdmin

Steve Martin may be busy but hope he gets back to this eventually

Reply

1 d

Paul ErlichAdmin

PC-vectors as in academic musical set theory? Or is this just confusingly similar terminology?

Reply

32 w

Dave KeenanAuthor

Paul Erlich I wrote: "prime-count vectors (PC-vectors or monzos)" so it should have been clear that I did not mean "pitch class vectors". But thanks for the heads-up. I was not previously aware of "pitch class vectors".

Reply

32 w

Ralph Hutchison

"feudal" because the knight in chess moves a square root of 5 in any direction.

Reply

32 w

Dave KeenanAuthor

Ralph Hutchison That's so cool. Thanks. No, I was not aware of that. I've added it, with credit to you, to my main article. As you'll read there, it is because ℚ(√5) contains both nobles and non-nobles. And since we care about the difference, we must be in a feudal system. Also because it relates to Margo Schulter's neo-medieval harmony. In fact I dedicate the series of articles to Margo, with whom I have previous collaborated on this topic.

Reply

32 w

Paul ErlichAdmin

Thanks for the great work Dave Keenan !

To what extent does 72-equal emerge (more).favorably in the context of targeting both rational and feudal ratios?

Reply

32 w

Dave KeenanAuthor

Thanks Paul. I haven't applied any of this to 72edo. I fact I haven't attempted to apply it to anything yet. I'm hoping others will take up the challenge. I'm content to have opened up the field. I'm too busy writing a textbook on RTT with Douglas Blumeyer (as you know, because you'll be the first to review a draft). Feudal RTT may even turn out to be a dead end. But you're right to suggest that 72 is fertile ground in which to look for feudal commas.

Reply

32 w

Dave KeenanAuthor

I note that we would not be targeting feudal ratios in general, but only the noble subset, and of those, only the highest (smallest-numbered) orders (the dukes, marquises, counts, viscounts and maybe the barons). I expect the knights and lords will rarely get a look in.

Reply

32 w

Dave KeenanAuthor

I note also that "feudal" includes both noble and rational (and a bunch of irrational ignobles we don't care about).

Reply

32 wEdited

Mike BattagliaAdmin

Some thoughts as I think about some of these memories from years ago, for Dave Keenan and Paul Erlich:

The main challenge with this, if memory serves, was that the "noble mediants" don't form any kind of group. The product of two noble mediants isn't a noble mediant. Even worse, the product of a noble mediant with a JI ratio is also not a noble mediant. So let's say you have, for instance, the noble mediant between 6/5 and 5/4. This is a very good candidate for a canonical "blue note," FWIW, to my ears - and I will even call it the "noble blue note" for the moment and suggest it as a great "prime" to start looking at. But anyway, even though this is a noble mediant, when you multiply it by 3/2, the thing you get is not a noble mediant; it is also not the same as the noble mediant between 15/8 and 9/5.

This so far isn't a problem; it just tells us that noble mediants aren't the only intervals we naturally care about when looking at this kind of thing. In your post on the Sagittal forum, you make a distinction between "noble numbers" and "ignoble numbers" for this reason. But it is important to note here the difference between a "noble" and "ignoble" number is in some sense entirely relative and due to the choice of tonic. For instance, the "noble blue note" is a "noble number," and its transposition up by a fifth is not. But if we change the tonic and modulate up by 3/2, this pitch now becomes "noble" relative to the new tonic. I am not sure if every "ignoble number" can be built in this way as a transposed noble number, but I would not be surprised if so; this would basically mean the "ignoble numbers" are really just "second-order," "third-order," etc noble numbers relative to whatever choice of tonic. This is all pretty different from JI, where rational numbers are always rational no matter how much you modulate around within JI.

So instead, we can look at the group generated by these numbers. As you mention, if we look at every possible just, noble, ignoble and every other kind of number, we get the quadratic field Q[√5]. However, we already know that we mostly care about multiplication, not addition - and in particular, we really don't care about negative frequencies - so we really only care about the multiplicative group of strictly positive elements here. We also would like some kind of "integers" for which we can talk about something like a prime factorization. If we do this, we get basically the structure that you have laid out in your post, with the exception that it may be kind of tenuous to call the powers of phi a "unit" given that we don't care about the entire ring of integers Z[phi], only the strictly positive elements. But we're in the ballpark either way.

The next challenge is to note that there are basically infinitely many "noble primes" one can care about which are relevant, even to a subgroup of finite rank. For instance, suppose we care about all JI intervals one can get using only factors of 2, 3, and 5. This is a rank-3 group; the 2.3.5 subgroup. But if we are now inspired to explore this "anti-JI" idea, we may think it'd be a good start if we just add every possible "anti-JI" noble number one can get by taking noble mediants of every 2.3.5 interval - except we end up with an enormous group of countably infinite rank. This is kind of a shame. The answer is to note that, in some sense all of the elements in this group are still derived from the 2.3.5 subgroup - we've just added a new binary operation, which is "noble mediant", alongside the usual multiplication. This should give us a very interesting structure - and the structure we are really studying, whether we identify it as such or not. I am not quite sure how it all shakes out, if it's a ring or a semiring or something else. I note that if anyone wants to pursue this, Minkowski's question mark function gives us a way to generalize the mediant of two rational numbers to arbitrary real numbers which may be of use.

Lastly, one thing I didn't see in your post, although I may have missed it, is that the number field Q[√5] has an associated "modulus", which gives us a natural way to define the complexity of any noble number (or "feudal number", I guess) that agrees with and generalizes the usual Tenney Height on rationals.

EDIT: one afterthought above regards the number of generators needed to express all of the "feudal numbers" one can generate from the 2.3.5 subgroup. I claimed that if you try to take noble mediants of everything in the 2.3.5 subgroup, and add those to the 2.3.5 subgroup itself along with all of the "ignoble" numbers thus generated, you end up with a group of countably infinite rank. However, typically we only take noble mediants of ratios a/b and c/d that have the property that ad-bc = 1; let's call these "adjacent ratios." I am not actually sure what you get if you enrich the 2.3.5 subgroup with *only* those noble mediants you can get by starting with some pair of *only* adjacent ratios drawn from the 2.3.5 subgroup itself. I am not even sure how many pairs of adjacent ratios there *are* in the 2.3.5 subgroup, to be honest. Does anyone know this?

Reply

32 wEdited

Mike BattagliaAdmin

Just a quick note this above post may have been edited while people were reading it...

Reply

32 w

Mike BattagliaAdmin

As another addendum, I notice that not all of the "feudal primes" listed - e.g. those of the form a+b*phi, for positive a, b - are actually "noble." For instance, 3+2*phi is called a feudal prime, but it's really (3+2*phi)/(1+0*phi), and 3*0 - 2*1 = -2, which is not ±1, thus this is not the noble mediant of any two intervals.

It would perhaps be a very good idea to determine which of the feudal primes really are "noble" to begin with, and see how the "ignoble" ones are produced. How many "noble primes" are there of the form a+b*phi w/ a and b both in the 2.3.5 subgroup, for instance? And so on.

Reply

32 w

Dave KeenanAuthor

Mike Battaglia That's good stuff. Thanks. I have not come across any "modulus" with the properties you describe. Neither Dekker nor Dodd have it. You wouldn't be thinking of the "arithmetic norm" would you? For a number of the form a+bϕ, a, b rational, N(a+bϕ) = a²+ab-b².

I note that the "noble mediant" is only /defined/ when |ad-bc| = 1, otherwise it wouldn't deserve the name "noble". I guess that could still be called the "ϕ-mediant" or something, but it isn't really of any interest except when it produces nobles.

Reply

32 wEdited

Dave KeenanAuthor

Yes, the concept of noble pitches is meaningless. Only noble intervals. And yes, nobility is fragile. It rarely survives multiplication. Very few pairs of nobles can breed successfully.

Reply

32 wEdited

Mike BattagliaAdmin

Dave Keenan I am going from memory but that a^2 + ab - b^2 looks correct. I forget why the name modulus was in my head but we can call it a norm. So for the general complexity of any feudal number, you can take the product of norms of the numerator and denominator. Try it with the noble mediant associated to 5/4 and 6/5, for instance. There are several ways to express this - the noble mediant of 5/4 and 6/5, or of 1/1 and 5/4, etc - but you should always get the same complexity no matter what. I don't remember what meaning the sign of the norm had.

Reply

32 w

Mike BattagliaAdmin

Dave Keenan: I am sort of puzzled by these "ignoble primes" like 3+2*phi. The noble ones will all be of the form a+1*phi. Something like 3+2*phi is not noble. So what is it and is it somehow generated in any musical way?

Is there some clever way that it's a transposition of some noble interval, or something like that, or is it just irreducibly generated as a "phi-mediant" of two non-adjacent ratios?

Reply

32 w

Dave KeenanAuthor

I never expected the primes (or any feudal integers other than ϕ and ϕ²) to be noble. Thanks for pointing out that all of the form a+1ϕ are. I expected all nobles to be a quotient of feudal integers. I was surprised to discover they were all quotients of primes-or-units. I'm not sure what qualifies as "generated in a musical way" for you here. As shown in various of my tables, 3+2ϕ is the numerator or denominator of 6 different nobles down to level 6 of the SB-tree, and it is one of the prime factors of 11.

Reply

32 w

Mike BattagliaAdmin

I guess it's also the transposition of the noble mediant (3+2*phi)/(1+1*phi) up by the other noble interval (1+1*phi). I am still curious if every ignoble interval is just a transposition of a noble interval in this way.

Reply

32 w

Dave KeenanAuthor

Re the monoid of ℤ⁺[ϕ] under multiplication, which I agree is all we actually need: I assume that what we mean by a "positive" feudal integer is one whose real-number value is positive, i.e. the value obtained when you treat the formal addition as real addition, substituting 1.618... for ϕ. In that case, all integer powers of ϕ, i.e. ϕⁿ for all integers n, including n<0, are included in the monoid. Therefore the monoid has integers other than 1 that have reciprocals within the monoid, therefore the monoid has "units" other than 1. And you can't describe ϕ as just another prime if you want to preserve a version of the fundamental theorem of arithmetic. But as far as our vectors go, ϕ is just another basis element, along with the feudal primes.

Reply

32 w

Mike BattagliaAdmin

Dave Keenan: I sorted through a bunch of this and I think I have better clarity on what I was trying to say. Had an earlier post but replaced it with this instead after sorting through some stuff... This is pretty subtle!

The big idea is that we would like some kind of monzo decomposition of a feudal number, similarly to the way we have monzo decompositions of rationals. To do this, we need something like "primes."

If I understand the basic theorem involved here correctly, we know that every single element in Q[sqrt(5)] is a product of feudal primes and powers of phi. We know that, "up to equivalence of powers of phi and -1," we have unique factorization.

So one preliminary question is: if we want to just treat phi as another basis element in our monzo, and get rid of -1, *do we still have unique factorization?* In other words, for any positive feudal number, is there always one unique monzo representation in terms of feudal primes *and* phi, even though phi is a "unit?" Again, I am going from memory, but I think this is true - it would be good to prove this.

Given that we do, then the next step is to ask what our generalized "natural numbers" are. In general, given the prime factorization of an arbitrary feudal number, some exponents will be positive and some negative, so we can look at the set of all feudal numbers with only positive exponents. Let's call these the "feudal natural numbers." Then every positive element of Q[sqrt(5)] is the quotient of two feudal natural numbers. If the above unique representation theorem is true, these feudal naturals will form a *free* commutative monoid, meaning it has a unique choice of basis. In some sense, this monoid of feudal naturals is almost the main thing we care about.

It turns out that the set of feudal naturals is *not* the set N[phi] I was talking about before. For instance, the element "5" is not a feudal natural number. It is, rather, a quotient of the two feudal natural numbers (2+phi)^2 = 5 + 5*phi and phi^2 = 1 + phi. Put another way, using the quotient representation, we'd really write "5" as "(5+5*phi)/(1+phi)" as a quotient of two of feudal naturals, derived from the positive and negative exponents of the monzo representation for 5. This is something I think is super interesting.

Where I got thrown off is this notion of "feudal integer" that you have for all elements of the form a+b*phi. You also show that every feudal number is a quotient of two feudal integers. But there is this subtlety with this notion of feudal integers that threw me off. We know that we don't care about negative numbers, for instance. With the regular integers, we know that if we look at only the positive numbers, we get the naturals, and coincidentally this is also our free multiplicative monoid that generates the positive rationals Q+ as quotients. But this isn't how it works with feudal integers: if we just naively do this, we get that the set of positive feudal integers is "too large." It isn't a free monoid, because we have, as you say, unlimited powers of 1/phi thrown in the mix. The result is kind of like looking at the regular natural numbers, but with all the numbers also multiplied by every possible power of 1/2, or something, and treating 2 as a "unit." What we really want is to go one step further.

This has some pretty profound implications when we treat an arbitrary feudal number as a quotient of two feudal integers. For instance, even some positive feudal integers, like 13 + 16*phi, are actually nontrivial *quotients* of positive feudal naturals, in this situation (10+3*phi)/(1+phi). This kind of thing never happens with the regular integers. So we may want to ask which feudal integers we should choose in our quotient representation of an arbitrary feudal number. You suggest a normal form in your post which involves doing this Fibonacci-esque thing in reverse until we have a+b*phi with a>b. However, there is another natural normal form which is to look at the unique reduced representation of a feudal number as a quotient of two feudal naturals with no common prime factors (including positive powers of phi), which can be plainly seen by looking at positive and negative exponents of the monzo representation. These two are not in general the same due to the weirdness involving feudal integers; (5+0*phi)/(1+0*phi) is in normal form in only the first sense, whereas the second sense would have it as (5+5*phi)/(1+1*phi).

Anyway, I guess that answers that for now...

Reply

31 wEdited

Mike BattagliaAdmin

It would also be interesting to see if there is some easy way to identify if an arbitrary feudal integer is a feudal natural using only the coefficients a and b. Is there some pattern that lets us easily determine if it's a product of only feudal primes and positive powers of phi?

Reply

31 w

Dave KeenanAuthor

Mike Battaglia: That stuff you wrote, about using the product of the norms of the numerator and denominator as a complexity measure, checks out so far. Thanks.

I'm worried that you're overthinking this monoid stuff. I believe I've already described how to generalise "monzos" (prime count vectors or PC-vectors) to give a unique representation for every positive feudal number ℚ⁺(√5) where a positive feudal number is defined as one whose "real value" is positive. The "real value" of a feudal number is defined in the obvious way as Re(a+bϕ) = a+bϕ, a, b rational, where the "+" and the juxtaposition of b and ϕ on the left are the /formal/ addition and multiplication used in representing feudal numbers, and the "+" and the juxtaposition of b and ϕ on the right are real number addition and multiplication. And for any feudals α, β, Re(αβ) = Re(α)Re(β), and Re(α/β) = Re(α)/Re(β) when Re(β)≠0.

I believe we can avoid representing any negative feudals simply by not having -1 (or any other negative feudal) in our basis. And prime count vectors are incapable of representing zero, as their entries are constrained to be integers (which do not include negative infinity).

I don't understand why I should care whether the positive feudal integers ℤ⁺[ϕ], or your feudal naturals ℕ[ϕ]\{0}, constitute the /smallest/ monoid that will give us the numerators and denominators we need for our nobles and rationals, because we have no choice in the matter. There is no way to limit the integer entries in the vectors in order to restrict them to representing any set smaller than ℤ⁺[ϕ], while still being able to avail ourselves of the power and convenience of all the usual linear algebra methods of RTT.

Yes, it would be good to formally prove that "Up to powers of phi (with ±integer exponents), we have unique factorisation of the positive feudals (as defined above), simply by treating ϕ (and not -1) as a basis element for our PC-vectors. But I'm so sure of it, I'm happy to proceed on that basis (double meaning) until someone comes up with a counterexample. You can tell I'm more of a physicist and engineer than a mathematician.

I note that your phrase "positive powers of phi" is ambiguous, given that "power" is sometimes treated as synonymous with the exponent alone, and sometimes (more correctly) as referring to the whole expression consisting of base and exponent. The two cases can be distinguished as "powers of phi with positive exponents" versus simply "powers of phi" (which are necessarily positive). Their opposites are "powers of phi with negative exponents" versus "the negations of powers of phi".

You wrote: "For instance, even some positive feudal integers, like 13 + 16*phi, are actually nontrivial *quotients* of positive feudal naturals, in this situation (10+3*phi)/(1+phi). This kind of thing never happens with the regular integers."

The denominator in your example is a unit, namely 1+1ϕ = ϕ². So while I agree it's non-trivial, I think it's fair to say that the analogous thing /does/ happen with the ordinary integers, e.g. 2 = 2/1.

But you might be onto something with your proposed normal form. Consider the feudal-prime-factorisation of 5. In my current normal form we have 5 = (2+1ϕ)²/(1+1ϕ) = (2+1ϕ)²/ϕ². With your proposed normal form I believe it would be 5 = (-1+2ϕ)²/(1+0ϕ) = (-1+2ϕ)²/1 = (-1+2ϕ)². I believe this form requires doing the Fibonacci-esque thing in reverse for one more step beyond where I stopped it, i.e. continuing until we have a+bϕ with a<0. (You can normalise primes but not units). It's nice to get rid of those ϕ² from the factorisations of the ordinary primes, but it's annoying to be guaranteed to have a minus sign in every prime. It makes one want to write them with the golden part on the left, e.g. 2ϕ-1.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan I think we are talking past each other. There is a very simple way to look at all of this: if you take this monzo decomposition, what I am calling the "feudal naturals" are those with only positive coefficients. Every feudal number is uniquely a quotient of two of those which have no prime factors in common, which is the "reduced form" I talked about. This parallels the way that every rational is a quotient of two natural numbers with no common prime factors. Given any feudal number, the reduced form can easily be read straight from the monzo: zeroing out all of the negative coefficients gives the numerator, and vice versa for the denominator.

I'm not sure how you got that normal form for 5 in your last paragraph but it is not the same as this. Using the monzo decomposition, the numerator of 5 is (2+1*phi)^2 = (5+5*phi), and the denominator is (1+phi). This is what you gave in your post as *your* normal form, but I get something different with your method: if we take the Fibonacci thing one step further then we get (5 + 0*phi)/(1 + 0*phi) without any negative numbers.

Some of the things you are saying seem to be misconceptions that I addressed directly in the previous post, for instance "There is no way to limit the integer entries in the vectors in order to restrict them to representing any set smaller than ℤ⁺[ϕ]" -- this is incorrect, the feudal naturals are strictly smaller than this. Feudal naturals have only positive monzo coefficients, whereas feudal integers allow for the phi coordinate (and I think only the phi coordinate) to be an arbitrary integer.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia You misquoted me above, by leaving off the second half of my sentence. I wrote: "There is no way to limit the integer entries in the vectors in order to restrict them to representing any set smaller than ℤ⁺[ϕ], while still being able to avail ourselves of the power and convenience of all the usual linear algebra methods of RTT."

At least I now understand how your "feudal naturals" differ from the positive feudal integers, namely choose a specific normal form for the primes in the vector basis, and omit ϕ from the vector basis (and hence omit all units other than 1). But I don't see how that is at all useful for our RTT application. ϕ and ϕ² are two of the nobles we want to represent.

I have no problem with calling those the feudal naturals. But if you're claiming we can actually omit ϕ from our basis, I'm not seeing it.

My reverse-fibonacci normalisation method, as described in my forum post, is only shown there as being performed simultaneously on the numerator and denominator of nobles (which seem to be always primes or units). I've since realised, it is applicable also to primes individually, but not to units or composites individually. That's why it doesn't work on unfactorised 5. So it's purely a way to obtain a normal form for primes.

And now I find it gives a simpler result for our prime-count vectors (monzos) if I take it one step further and end up with a negative wooden part for all the primes. Because that eliminates all the ϕ²'s from the factorisations of ordinary primes. For example, it's better to have the simplest prime (the single prime factor of 5) as 2ϕ-1 = √5 rather than my existing 2+1ϕ = ϕ√5.

I understand now, that this new normal form is not one you proposed.

I'm considering doing a wholesale find and replace on my forum articles, using my new normal form for the primes, aϕ-c (rather than -c+aϕ), where a and c are positive integers with c<a. The relationship to the previous choice of a+bϕ, where a and b are positive integers with b<a, is simply c = a-b.

Does anyone have any thoughts on why that might be a good or bad idea?

Reply

31 w

Mike BattagliaAdmin

Dave Keenan I don't see how linear algebra has any bearing either way on the feudal natural numbers. You already have a basis of phi + feudal primes, and the set of all vectors w positive integer coefficients is feudal natural. This is already a set smaller than the positive members of Z[phi]. What is the problem?

The thing about omitting phi from the basis is very far from anything I am talking about. I've said explicitly that phi is to be treated as just another basis element again and again. I'm not sure where we got off track but maybe try reading my post again...

Reply

31 w

Mike BattagliaAdmin

I guess that in the last bit, you are saying that the choice of "feudal prime" is somewhat arbitrary - in Z[phi], feudal primes come in equivalence classes which are only equivalent up to a power of phi. So depending on which one we choose, we get an entirely different basis, and thus a different set of "feudal naturals" as the set of vectors with non negative coefficients relative to that bsis. This is a good point. I would be curious if the arithmetic norm can make it easier to choose a representative from each feudal prime class.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia Yes, we do seem to be talking past each other. Can we perhaps reframe the discussion as trying to agree on the most practically-useful choice for the normal form of the primes to be used in our monzo basis?

Reply

31 w

Mike BattagliaAdmin

Dave Keenan sure. A "noble" goal.

I guess one suggestion is that the prime should certainly be larger than 1/1, but this is kind of weak.

Here is an important question: given some prime, suppose we generate the two-sided "fibonacci" sequence it generates. Is the number 1 or -1 guaranteed to appear exactly once in this sequence? Because then we could (and should) choose the representation in which the ±1 is the phi coefficient, because then this is the unique representative in the feudal prime class which is "noble." This would also affirmatively answer the question I posed earlier, if every "ignoble ratio" is a transposed noble ratio.

EDIT: sadly, turns out this line of reasoning doesn't work. 7 + 4*phi doesn't have any noble representative; the associated Fibonacci sequence goes ...-13, 10, -3, 7, 4, 11, ... and diverges from there. We could try to choose the "most noble" representative, perhaps, which would be 10-3*phi, although sometimes this isn't unique, in that 2+phi and 3-phi are in the same equivalence class and both equally noble. So I'm not sure.

But a bigger question I have about this is what the point of intervals like 7+4*phi is. I thought the goal here was to formalize the structure of these noble mediants. If 7+4*phi isn't a noble mediant, or some product of noble mediants and JI ratios, does it have any musical relevance? I mean, I get that it's technically a "prime" in the ring Z[phi], and that you can factor noble mediants according to this set of primes, but is that all?

Reply

31 wEdited

Mike BattagliaAdmin

I think we may get that there's usually a unique "most noble" representative, except for the rare sporadic prime where one would need to choose a tiebreaker. Either way, I'm really not sure if this is the best way to do it.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia Here are the first two counterexamples. The factor of 5 and numerator of both 3rd-order nobles:

..., -11, 7, -4, 3, -1, 2, 1, 3, 4, 7, 11, ...

It has both -1 and +1.

A factor of 31 and the numerator of two 5th-order and denominator of two 6th-order nobles:

..., -30, 19, -11, 8, -3, 5, 2, 7, 9, 16, 25, ...

It has neither -1 nor +1.

I agree with >1. One possibility is to make them as close to 1 as possible while still being greater. But my earlier suggestion is not equivalent to that, as it is equivalent to making them as close as possible to the square root of the ordinary prime that they are a factor of.

Reply

31 w

Mike BattagliaAdmin

Dave Keenan you may have missed my later edit about the musical point of some of these primes, see the last paragraph. Basically my view is - if we view all of this as deriving from the noble mediant idea, which has a clear musical use, then the intervals we care about, at most, are:

1. JI ratios

2. Noble mediants

3. Products of noble mediants and other noble mediants and JI ratios

Given those criteria, does an interval like 7+4*phi even matter, and does it have any musical use? it's technically a prime in Z[phi] and we can factor noble mediants in those terms if we want, but is it "irreducibly ignoble" in some sense, and if so why does it matter?

Reply

31 w

Dave KeenanAuthor

I don't see any particular value in making the primes themselves noble (when possible), as this may lead to them being quite large numbers. And nobles wider than say a diatonic 13th, say 3.25, are not of much interest. They are too wide (>2041 ¢) for accurate tuning to matter.

Making them as close as possible to 1 while being greater, means constraining them to be between 1 and ϕ. I don't see any particular value in this either.

Reply

31 w

Mike BattagliaAdmin

At this point the main question I have is if there are simply "too many" feudal primes. Noble mediants matter and transposed noble mediants matter. But when we factor these according to the primes of Z[phi], are we really factoring relative to a somewhat arbitrary set containing non-noble junk intervals without direct musical relevance? Or is there some other way to look at it?

Reply

31 w

Dave KeenanAuthor

Mike Battaglia As an interval, 7+4ϕ is of no musical interest whatsoever. But as a prime, we do need it, or one of its associates (that's a technical term in the literature), to encode the noble mediant of 11/8 and 15/11 (as the numerator), and possibly others. And yes, it is irreducibly ignoble.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia: Yes I missed the edit. You wrote: "I mean, I get that it's technically a "prime" in the ring Z[phi], and that you can factor noble mediants according to this set of primes, but is that all?"

Yes. That's all. Isn't it enough? However we might find a more useful associate than 7+4ϕ. I think 7ϕ-3 is more useful, but it might not be the most useful.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan:

1) How do we know that it is irreducibly ignoble?

2) Suppose that instead of doing it this way, we just took all of the noble mediants we care about, along with the JI ratios, and we look at all possible products and quotients of those. Which feudal primes does this generate? Apparently not all of them?

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia: I'd qualify that to say that we care about:

1. JI ratios

2. Simple noble mediants

3. Products of simple noble mediants and other simple noble mediants and JI ratios

I note that, to me, JI means "simple rational", not just any rational. So the counterpart of JI is not noble but rather "anti-JI" or "MI (merciful intonation)".

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia Taking all of the noble mediants we care about, along with the JI ratios, and listing the feudal primes this generates is exactly what I did in my series of articles on the Sagittal forum! Haven't you read it? And this automatically includes all possible products and quotients of those.

Reply

31 w

Dave KeenanAuthor

How do we know 7+4ϕ is irreducibly ignoble? I was assuming that a prime can only be noble if it is of the form a+1ϕ. Is that not so? If it is so, then we only have to look at its fibo-like sequence to see that it will never generate a 1 or -1.

... 59 -36 23 -13 10 -3 7 4 11 15 26 41 ...

Reply

31 w

Mike BattagliaAdmin

Dave Keenan you looked at which "primes" generate the noble mediants, not which primes are generated *by* the noble mediants as a product of noble mediants and JI ratios.

Reply

31 wEdited

Mike BattagliaAdmin

Let's put it this way: forget noble intervals for now. Suppose we only care about JI. The regular prime numbers are a useful basis for JI intervals. But, we could always add other new elements to our group such that our prime numbers become composite. For instance, if we added the elements (3+sqrt(2)) and (3-sqrt(2)), then 7 can be factored into a product of those. But we don't do this because there is no point to introducing those intervals to begin with (if we only care about JI).

Are we perhaps doing something similar with these feudal primes; factoring noble mediants into extraneous, unnecessary, "unmusical" intervals that fall outside our criteria? Or is there some mathematical reason why all of these feudal primes are somehow necessary?

Reply

31 w

Dave KeenanAuthor

Mike Battaglia But products of nobles and rationals will not generate any new primes. You will still only have the primes that were contained in either the nobles or the rationals to begin with.

Reply

31 wEdited

Dave KeenanAuthor

I assume we're talking prime equivalence classes here, i.e. ignoring powers of ϕ.

Reply

31 w

Mike BattagliaAdmin

No primes are inherently "contained" in any of these things; we can factor it according to whatever basis we want. From before: "Are we perhaps doing something similar with these feudal primes; factoring noble mediants into extraneous, unnecessary, "unmusical" intervals that fall outside our criteria? Or is there some mathematical reason why all of these feudal primes are somehow necessary?"

Reply

31 wEdited

Mike BattagliaAdmin

I think you get what I'm asking at this point. Do you know the answer to this question?

Reply

31 w

Mike BattagliaAdmin

Dave Keenan: Here is a great example. This will hopefully make clear what I am talking about, in a way that I am sure is impossible to misunderstand.

Suppose we want 11-limit JI, and the only two noble mediants we care about are 1) phi, and 2) the noble mediant between 11/8 and 15/11, call it "N." We can add those two as basis elements to the 2.3.5.7.11 subgroup to get a 7D group, which I'll call 2.3.5.7.11.phi.N.

"N" can be written, in your lowest terms form, as (7 + 4*phi)/(5 + 3*phi). The numerator and denominator are both feudal primes which are not noble, and as far as I know, not a transposed version of any noble interval. Thus, they don't have any direct musical use, at least not given our criteria for before.

So we could take our nice, musically valid interval "N" and further factor it into a quotient of these other two musically useless intervals which appear in the numerator and denominator.

Why do we want to do this?

Reply

31 wEdited

Dave KeenanAuthor

I think it's just wonderful that ℚ(√5) is a unique factorisation domain (UFD) and that it contains the rationals and the nobles and amazingly little junk (compared to say, some imagined set that needed to also contain square roots of ordinary primes other than 5 before it became a UFD). We're never going to do any better.

I don't think it matters that the primes themselves are unmusical. I think there's an argument that the prime 5 is unmusical. 2786 ¢. How often does such a wide interval occur in a chord, and when it does, who cares about tuning it accurately? If not prime 5, then 7, or 11. While we value simple ratios containing those primes, we don't really care to hear the primes themselves.

Reply

31 wEdited

Dave Keenan

The Feudal Manifesto

While procrastinating from paid work recently, I followed up on some tuning math that I've been meaning to investigate for years.

The executive summary is below.

The detailed explanation is here: viewtopic.php?f=21&t=555

An effective cadence can be made by playing a chord having high-harmonic entropy followed by a chord having low harmonic-entropy (HE).

Regular Temperament Theory (RTT) is, at present, designed to target intervals having low HE, while users rely on the fact that any octave-repeating scale of more than 5 notes is guaranteed to have some intervals with high HE.

But what if we could target specific high-HE intervals as well as low-HE intervals?

Minima of HE are at simple rational-number ratios.

Many maxima of HE are close to simple noble-number ratios.

Noble numbers are irrational and so cannot be be expressed as (finite) prime-count vectors (PC-vectors or monzos) whose basis is rational (e.g. the ordinary prime numbers).

But it turns out noble-numbers can be expressed as a quotient of two "feudal" primes or units, where "feudal" means "belonging to the quadratic field ℚ(√5)" (my coinage).

Feudal units are numbers of the form ±ϕⁿ where ϕ is the golden ratio and n is an integer.

Feudal primes are a subset of the feudal integers which are numbers of the form a+bϕ where a and b are integers.

By appending small feudal primes (and the unit ϕ) to the basis of our PC-vectors, it may be possible to find feudal commas and therefore feudal temperaments that generate a large number of both simple rationals and simple nobles using a small number of generators.

forum.sagittal.org

Noble frequency ratios as prime-count vectors in ℚ(√5) [superseded] - The Sagittal forum

439 comments

Dave KeenanAuthor

Units and primes of ℚ(√5), by Prof. Dirk Dekker, annotated by me.

Reply

32 wEdited

Mike BattagliaAdmin

Great stuff Dave Keenan and I will read when I get the chance, but I was also just thinking about this - that if you add "anti-JI" intervals to JI, you get the number field Q√5.

Reply

32 w

Mike BattagliaAdmin

I had some results on this a while ago I never posted which I worked on with Ryan Avella. Of course, Q[√5] is a field, so everything is invertible and divides 1 (thus every element of a field is a "unit"). I think that there is some subtlety to what you are calling "primes" and "units" - the units you are talking about in your post are those things that are units in the ring of integers derived from this number field (which I think corresponds to Z[√5] (EDIT: oops, it really is Z[phi] for this.)). That is one method, but it is also interesting to look at units in Z[phi] (note this is not the same ring). You also get some interesting results for "primes" if you look at the monoid N[phi] of elements a+b*phi with non-negative a and b. I had an interesting method of looking at all elements of the form a + b*phi for non-negative a and b, and iterating upward starting at 0 + 0*phi, and for each element checking if it was divisible by any of the elements with smaller a and b; those which don't are also prime in a certain sense.

Reply

32 wEdited

Dave KeenanAuthor

Mike Battaglia This is all dealt with in the full series of articles.

Reply

32 w

Mike BattagliaAdmin

Dave Keenan yes I see it. What I'm saying is phi is a unit in the ring of integers Z[phi], but in the monoid of "positive" integers N[phi], it doesn't divide 1, so I think it'd be just another "prime." I'm not 100% sure on this though.

Reply

32 w

Paul ErlichAdmin

RTT relies on no such fact as far as I am aware; the 9-limit Otonality and 5-tET are perfectly fine counterexamples and are not ruled out by any RTT considerations.

Reply

32 w

Dave KeenanAuthor

Paul Erlich Why the uncharitable reading? It's only a brief summary. I didn't mean that RTT relies on it, but that /users/ of RTT often rely on it, to ensure they have discordance as well as concordance in their RTT-generated scales. The 9-limit otonality and 5-tET are /not/ counterexamples because they do not contain "more than 5 notes".

Reply

32 w

Paul ErlichAdmin

Dave Keenan sorry I misread!

Reply

32 w

Paul ErlichAdmin

Of course phi itself is almost never near a local, let alone global, maximum of HE, and repeated stacks of phi are more concordant than repeated stacks of any similar-sized interval. This gives another interesting reason to want to include phi along with conventional primes. Aggelos Boshidis would be interested.

Reply

32 w

Dave KeenanAuthor

Paul Erlich I never suggested it was near a global maximum. But I don't know why you claim that ϕ (≈ 833 ¢) is almost never near a local maximum of HE. 829 ¢ is marked as a local maximum on the second HE chart here: http://tonalsoft.com/enc/h/harmonic-entropy.aspx

harmonic entropy - accordance model of musical intervals

TONALSOFT.COM

harmonic entropy - accordance model of musical intervals

harmonic entropy - accordance model of musical intervals

Reply

Remove Preview

32 wEdited

Paul ErlichAdmin

Dave Keenan ha you're right! I wasn't thinking about the cases where 8:5 was an inflection point. Thanks for the correction! It's just often claimed that if simple ratios are maximally concordant then phi must be maximally discordant.

Reply

32 w

Dave KeenanAuthor

Paul Erlich Sorry about that last post. Facebook failed to show me your previous until I switched from my desktop to my phone.

Reply

32 w

Dave KeenanAuthor

Surely there's almost always a peak between 8/5 (814 ¢) and 13/8 (841 ¢), but closer to 13/8?

Reply

32 w

Paul ErlichAdmin

Dave Keenanso to answer your question, no, my original or default 2HE curves would probably have no peak between 8/5 and 13/8, and in fact none 746 and 845 cents

Reply

31 w

Mike BattagliaAdmin

Dave Keenan and Paul Erlich , I was playing around with some of these theorems about phi to see if we can relate Dirichlet's approximation theorem to what we are doing with HE. Some of these results suggest one may tend to get local maxima that are closer to phi if one a) uses the min-entropy rather than the regular Shannon entropy, and b) the Laplace distribution rather than the Gaussian. Kind of curious how well that does.

Reply

31 w

Mike BattagliaAdmin

Oh, and with the Farey weighting instead of Tenney as well.

Reply

31 w

Dave KeenanAuthor

Paul Erlich Mike Battaglia We agree that nobles are not necessarily the most discordant ratios in their vicinity, but there may still be a sense in which they are audibly the most "metastable" with respect to cadences. Or audibly the most "ambiguous" with respect to approximating just intervals. In any case, the fact that they exist in a unique factorisation domain that (hopefully) doesn't have too much extra "junk", is hard to resist.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan I'm not sure it matters; it's clearly good enough and tends to give sensible results at least for varying values of s.

Reply

31 w

Paul ErlichAdmin

Mike Battagliavarying results for s where? Good enough for what, how?

Reply

31 w

Paul ErlichAdmin

Mike Battaglia are you looking at Steve Martin 's graphs above?

Reply

31 w

Mike BattagliaAdmin

Paul Erlich: "repeated stacks of phi are more concordant than repeated stacks of any similar-sized interval" - is this true? Would be kind of interested to see if Steve Martin's HE script can compute this - suppose we sweep from all intervals from 0-1200 cents, and for each one, compute (let's say) the 5-HE of a stack of five of those intervals on top of one another. Is there really a local minimum at exactly phi?

Reply

31 w

Paul ErlichAdmin

Mike Battagliayes that's exactly what we found for 3HE/4HE. Not sure about 5HE.

Reply

31 w

Mike BattagliaAdmin

Paul Erlich that's really interesting. Is there a graph of the results somewhere?

Reply

31 w

Paul ErlichAdmin

Mike BattagliaI don't recall if Steve Martin graphed them but he reported the other local minima which were pretty far away (3:2 was the closest on one side). Maybe he has a graph he could share. I made a big deal about this for a few years because it arose despite the fact that HE doesn't even account for combinational tones which are main the reason one might think stacks of phi would be more concordant than stacks of an interval a bit off from phi.

Reply

31 w

Paul ErlichAdmin

*the main reason

Reply

31 w

Mike BattagliaAdmin

Well I don't think it's worth throwing the baby out with the bathwater re what Dave is saying about phi, but I do think what you are saying is also important in that it tells you that much of this phi business may fall apart once you leave the realm of dyads. I

There are some interesting papers on simultaneous Diophantine approximation of multidimensional real vectors, and I'm not sure if the golden ratio has quite as special of a role there.

Reply

31 w

Paul ErlichAdmin

Mike Battaglia maybe the tribonacci constant etc. would . . .

Reply

31 w

Paul ErlichAdmin

Mike Battagliaalso I see this as a positive, not negative, reason to want to have phi as a generator or at least temperamentally simple, and I know Aggelos Boshidis is interested in more possibilities along those lines. I mentioned good old Sqrtphi temperament, highly complex mappings for most ratios other than 11:7 but with remarkably low 'badness' overall; it fit with his observation that (pi/2 ~) 11/7 ~ 2/sqrt(phi), which means one period up and one generator down in Sqrtphi. But I'm sure other interesting possibilities emerge when explicitly targeting phi along with the first several primes, and in no way would I want to put the kibosh on such research -- quite the contrary! But FWIW 845 cents is the peak I'm seeing in all of these graphs Dave Keenan . . . http://tonalsoft.com/.../harmonic-entro ... commentary...

Paul Erlich's older ideas on Harmonic Entropy, comments by Monzo

TONALSOFT.COM

Paul Erlich's older ideas on Harmonic Entropy, comments by Monzo

Paul Erlich's older ideas on Harmonic Entropy, comments by Monzo

Reply

31 w

Steve Martin

Mike Battaglia, Paul Erlich, I can make graphs but for now just the minima extracted:

3HE: 702, 600, 832, 952 ...

4HE: 833, 600, 45, 107 ...

5HE: 600, 833, 43, 1077 ...

Reply

31 wEdited

Paul ErlichAdmin

Steve MartinI'm really looking forward to seeing the graph

Reply

31 w

Paul ErlichAdmin

Aggelos Boshidisnot sure I understand. Can you break down what you're trying to demonstrate, step-by-step through each calculation, with a narrative to guide the reader?

Reply

31 w

Paul ErlichAdmin

Aggelos Boshidislooks like you're describing a nearly equal tuning, where the step size is approximately 48.3 cents?

Reply

31 w

Steve Martin

Paul, Mike 3HE for stacks

Reply

31 wEdited

Steve Martin

... 4HE for stacks ...

Reply

31 w

Steve Martin

... 5HE for stacks ...

Reply

31 w

Steve Martin

I suspect noise appearing in the 5HE one at larger cent values

Reply

31 w

Aggelos Boshidis

Paul Erlich I will reply asap, maybe this helps (with some 'quantizing' : https://sevish.com/scaleworkshop/?name= ... penv=organ

Scale Workshop

SEVISH.COM

Scale Workshop

Scale Workshop

Reply

31 w

Aggelos Boshidis

Paul Erlich So I used sevish workshop to generate a rank2 temperament.

I used sqrtphi as a generator which is apprx 1272/1000

I used phi inversion as a period (367c.) apprx 1236/1000, scale size was seven with six generators up 1/1. What I found interesting is that the intervals fall onto almost the exact 1200 cents of one octave, 2400 of second and 3600 of third octave

:

~50,99,149,199,249,298,367,

416(pi/2 inversion interval~sqrtphi ), 466,516,566,615,665,733

783(pi/2),833(phi), 883(8/5),932,982,1031, 1100,1150

1199.85~1200

....1516 (6/5*2),...2399.7~2400(4/1),

3599.5~3600 (8/1)

Reply

31 wEdited

Paul ErlichAdmin

Aggelos Boshidis six generators up from 1/1? I see six approximate quartertones up from 1/1. 1272/1236 = 106/103 would be period minus generator according to what you said, but that's 49.7 cents, so I suspect what you actually did was construct six of your generators *down*, not *up*, from 1/1. Is that correct so far?

Reply

31 w

Aggelos Boshidis

Paul Erlich

Reply

31 w

Aggelos Boshidis

Paul Erlich I forgot to add 1031 and 1100 cents on my previous message regarding the tuning. Edited it. Also sent you a screenshot with the exact figures

Reply

31 w

Paul ErlichAdmin

Aggelos Boshidisthanks. well my question still stands. It looks like you're building the generator *downward* (not upward) from 1/1; the math I presented seems to corroborate that -- no?

Reply

31 w

Paul ErlichAdmin

Steve Martinmaybe make a new post with your graphs (and maybe try a couple different s values)? Some people might be missing them buried in a subthread here, and I'd like to tag others.

Reply

31 w

Steve Martin

Paul Erlich I missed this, will do but not for a week or so.

Reply

30 w

רועי סיני

Paul Erlich If you are interested, there are other temperaments that can be tuned so that the golden ratio has a low complexity. The simplest examples, that also come from the Fibonacci series if you try to equate the ratios between successive elements, are gamelismic temperaments that also include a half-octave (which happens to be another interval whose stacks have a low harmonic entropy). Some examples are lemba, hendec, semimiracle, semivalentine and semimothra (where the last, like the others, tempers out 169/168).

Reply

1 d

Paul ErlichAdmin

The "Vos" HE curves used the Laplace distribution (and shannon), but they look like they might be "fractal" with unendingly many local minima and maxima on ever-finer scales . . .

Reply

31 w

Mike BattagliaAdmin

Paul Erlich yeah I think they may be "continuous nowhere differentiable" or something like that.

Reply

31 w

Paul ErlichAdmin

https://m.facebook.com/photo.php?fbid=1 ... 2912705601

Reply

31 w

Paul ErlichAdmin

https://m.facebook.com/photo.php?fbid=1 ... 1370594482

Reply

31 w

Aggelos Boshidis

Paul Erlich but did you see the screenshot I sent you? According to the program its the opposite direction

Reply

31 w

Paul ErlichAdmin

Aggelos Boshidis wow, how is that? Maybe Sean Archibald could explain. AFAICT the 49.7-cent interval is period minus generator, not the generator itself.

Reply

31 w

Sean Archibald

What I'm seeing is that the generator is greater than period, and generators are being stacked up, resulting in the 49.7c interval which looks expected to me

Reply

31 w

Paul ErlichAdmin

Sean Archibaldthanks! I never imagined someone would specify a tuning using a generator larger than the period, so I inadvertently mentally swapped Aggelos Boshidis 's generator and period. Sorry for the confusion guys! Anyway, this means most of the "~" in Aggelos' original list of relationships were not approximate, they were exact, since the scale simply repeats the same 106:103 step interval 6 times. So obviously one interval in the scale is 6/5 times the size of another when measured in cents, etc. (This has nothing to do with the consonant frequency ratio 6:5 of course).

Reply

31 w

Aggelos Boshidis

Paul Erlich the thing that is interesting for me is that after three repetitions plus two 49.7c you get 1200, 2400, 3600 (then it deviates) Try other rank2 temperaments with the same logic and it's not easy to get the same result. 6/5 appears as 1516 c. In the (1200-3600 cent octave)

Reply

31 w

Paul ErlichAdmin

Aggelos Boshidis12/5, yes.

You can use various intervals of repetition such that neither the period nor the generator is an octave or 1/N octave even for rank-2 temperaments featuring prime 2 (in Graham's app). Just use the interval of repetition as the first element of the JI basis you specify, and alter the other basis elements accordingly if necessary (and click "subgroup TE" for the results). So if we want e.g. the usual meantone temperament, but repeating at say 5:3, we can specify the basis as 5/3.2.3, giving

http://x31eq.com/cgi-bin/rt.cgi?limit=5%2F3_2_3...

where the first Scala file is:

http://x31eq.com/cgi-bin/scala.cgi?limit=5%2F3_2_3...

Repeating this up and down over many repetitions of the ~5:3 period, one will find many multiples of ~1201 cents.

Obviously there are infinite variations on this for any rank-2 temperament featuring prime 2 in its JI basis.

X31EQ.COM

5/3.2.3-limit Regular Temperament

5/3.2.3-limit Regular Temperament

Reply

31 w

Aggelos Boshidis

Paul Erlich yes agreed, but I was talking in reference to the phi "business", using sqrtphi and phi inverted.

Reply

31 w

Paul ErlichAdmin

רועי סיניthanks, I'm familiar with most of those of course; would love to see some more explicit examples of equating the ratios between successive elements of the Fibonacci sequence & understanding how you always seem to get rank-2 temperaments from the process (?)

Reply

1 d

רועי סיני

Paul Erlich If you set ϕ ~ 13/8 ~ 21/13 ~ 34/21 ~ ... up to infinity, the √2.ϕ.5 group gives you a rank-3 temperament up to 31-limit. Its map is the following:

[⟨2 -1 0 7 6 6 4 -4 -9 0 -10]

⟨0 3 0 -1 4 1 3 9 13 7 11]

⟨0 0 1 0 -1 0 0 0 0 0 1]⟩

For example, 13 is 8 * ϕ = √2⁶ϕ, and 3 = √(144 / 16) is √(8 * ϕ⁶ / 16) = ϕ³/√2.

Now, if you want it to be rank-2, you essentially need to represent 5 in terms of the other generators, and each of the temperaments I provided comes from another way to do so:

Lemba comes from the most naïve way, which is to say that 8/5, the fibonacci ratio before the ones we considered, should also be ϕ. It gives you low complexity in exchange for a high error for 5, 11 and 31.

Hendec comes from noting that √2 is also approximated by a simple continued fraction, and in fact, is already represented by 17/12 and 41/29, which are both present in this continued fraction (41 is larger than 31 but we can still get a value for it using F₂₀ / F₁₀ = 3*41), and therefore it makes sense to also require the next ones. 99/70 gives you a mapping for the 5, and the more complex ones seems to not affect small primes and only give much larger primes than is musically interesting.

Semimiracle and semivalentine just come from the fact that both miracle and valentine are good 11-limit temperaments that are consistent with this rank-3 one, and they already have these known 13-limit extensions that split the octave in half and are also consistent with this temperament.

Semimothra comes from the fact that the terms in the Lucas sequence are also all represented in this temperament, as Lₙ = F₂ₙ / Fₙ, which from the sixth one is always represented by ϕⁿ, and the ratio between them and the Fibonacci numbers, that are again represented from the sixth one formulaically, this time by Fₙ ~ 8ϕⁿ⁻⁶, approaches √5 as n approaches infinity, and therefore it makes sense to equate the ratio between these two representations, that equals ϕ⁶/8, to the square root of 5. This ratio also represents 9/4, and therefore we get a meantone temperament, which I call semimothra as it is identical to mothra in the 11-limit and then cuts the octave in half.

Reply

17 h

Paul ErlichAdmin

Dave Keenan Cmloegcmluin Xenharmonic Feisbeuk

Reply

17 h

Dave KeenanAuthor

רועי סיני This is mind-blowing stuff.

Reply

16 h

Paul ErlichAdmin

רועי סיני very interesting. However we don't describe any weak extensions of meantone as itself "a meantone temperament".

Reply

17 h

Dave KeenanAuthor

רועי סיני This deserves to be the topic of its own thread, not buried here. Perhaps with some introduction for those who haven't been following. And perhaps with a link to this thread. And perhaps spelled out in more detail.

Reply

16 h

Paul ErlichAdmin

Steve Martin may be busy but hope he gets back to this eventually

Reply

1 d

Paul ErlichAdmin

PC-vectors as in academic musical set theory? Or is this just confusingly similar terminology?

Reply

32 w

Dave KeenanAuthor

Paul Erlich I wrote: "prime-count vectors (PC-vectors or monzos)" so it should have been clear that I did not mean "pitch class vectors". But thanks for the heads-up. I was not previously aware of "pitch class vectors".

Reply

32 w

Ralph Hutchison

"feudal" because the knight in chess moves a square root of 5 in any direction.

Reply

32 w

Dave KeenanAuthor

Ralph Hutchison That's so cool. Thanks. No, I was not aware of that. I've added it, with credit to you, to my main article. As you'll read there, it is because ℚ(√5) contains both nobles and non-nobles. And since we care about the difference, we must be in a feudal system. Also because it relates to Margo Schulter's neo-medieval harmony. In fact I dedicate the series of articles to Margo, with whom I have previous collaborated on this topic.

Reply

32 w

Paul ErlichAdmin

Thanks for the great work Dave Keenan !

To what extent does 72-equal emerge (more).favorably in the context of targeting both rational and feudal ratios?

Reply

32 w

Dave KeenanAuthor

Thanks Paul. I haven't applied any of this to 72edo. I fact I haven't attempted to apply it to anything yet. I'm hoping others will take up the challenge. I'm content to have opened up the field. I'm too busy writing a textbook on RTT with Douglas Blumeyer (as you know, because you'll be the first to review a draft). Feudal RTT may even turn out to be a dead end. But you're right to suggest that 72 is fertile ground in which to look for feudal commas.

Reply

32 w

Dave KeenanAuthor

I note that we would not be targeting feudal ratios in general, but only the noble subset, and of those, only the highest (smallest-numbered) orders (the dukes, marquises, counts, viscounts and maybe the barons). I expect the knights and lords will rarely get a look in.

Reply

32 w

Dave KeenanAuthor

I note also that "feudal" includes both noble and rational (and a bunch of irrational ignobles we don't care about).

Reply

32 wEdited

Mike BattagliaAdmin

Some thoughts as I think about some of these memories from years ago, for Dave Keenan and Paul Erlich:

The main challenge with this, if memory serves, was that the "noble mediants" don't form any kind of group. The product of two noble mediants isn't a noble mediant. Even worse, the product of a noble mediant with a JI ratio is also not a noble mediant. So let's say you have, for instance, the noble mediant between 6/5 and 5/4. This is a very good candidate for a canonical "blue note," FWIW, to my ears - and I will even call it the "noble blue note" for the moment and suggest it as a great "prime" to start looking at. But anyway, even though this is a noble mediant, when you multiply it by 3/2, the thing you get is not a noble mediant; it is also not the same as the noble mediant between 15/8 and 9/5.

This so far isn't a problem; it just tells us that noble mediants aren't the only intervals we naturally care about when looking at this kind of thing. In your post on the Sagittal forum, you make a distinction between "noble numbers" and "ignoble numbers" for this reason. But it is important to note here the difference between a "noble" and "ignoble" number is in some sense entirely relative and due to the choice of tonic. For instance, the "noble blue note" is a "noble number," and its transposition up by a fifth is not. But if we change the tonic and modulate up by 3/2, this pitch now becomes "noble" relative to the new tonic. I am not sure if every "ignoble number" can be built in this way as a transposed noble number, but I would not be surprised if so; this would basically mean the "ignoble numbers" are really just "second-order," "third-order," etc noble numbers relative to whatever choice of tonic. This is all pretty different from JI, where rational numbers are always rational no matter how much you modulate around within JI.

So instead, we can look at the group generated by these numbers. As you mention, if we look at every possible just, noble, ignoble and every other kind of number, we get the quadratic field Q[√5]. However, we already know that we mostly care about multiplication, not addition - and in particular, we really don't care about negative frequencies - so we really only care about the multiplicative group of strictly positive elements here. We also would like some kind of "integers" for which we can talk about something like a prime factorization. If we do this, we get basically the structure that you have laid out in your post, with the exception that it may be kind of tenuous to call the powers of phi a "unit" given that we don't care about the entire ring of integers Z[phi], only the strictly positive elements. But we're in the ballpark either way.

The next challenge is to note that there are basically infinitely many "noble primes" one can care about which are relevant, even to a subgroup of finite rank. For instance, suppose we care about all JI intervals one can get using only factors of 2, 3, and 5. This is a rank-3 group; the 2.3.5 subgroup. But if we are now inspired to explore this "anti-JI" idea, we may think it'd be a good start if we just add every possible "anti-JI" noble number one can get by taking noble mediants of every 2.3.5 interval - except we end up with an enormous group of countably infinite rank. This is kind of a shame. The answer is to note that, in some sense all of the elements in this group are still derived from the 2.3.5 subgroup - we've just added a new binary operation, which is "noble mediant", alongside the usual multiplication. This should give us a very interesting structure - and the structure we are really studying, whether we identify it as such or not. I am not quite sure how it all shakes out, if it's a ring or a semiring or something else. I note that if anyone wants to pursue this, Minkowski's question mark function gives us a way to generalize the mediant of two rational numbers to arbitrary real numbers which may be of use.

Lastly, one thing I didn't see in your post, although I may have missed it, is that the number field Q[√5] has an associated "modulus", which gives us a natural way to define the complexity of any noble number (or "feudal number", I guess) that agrees with and generalizes the usual Tenney Height on rationals.

EDIT: one afterthought above regards the number of generators needed to express all of the "feudal numbers" one can generate from the 2.3.5 subgroup. I claimed that if you try to take noble mediants of everything in the 2.3.5 subgroup, and add those to the 2.3.5 subgroup itself along with all of the "ignoble" numbers thus generated, you end up with a group of countably infinite rank. However, typically we only take noble mediants of ratios a/b and c/d that have the property that ad-bc = 1; let's call these "adjacent ratios." I am not actually sure what you get if you enrich the 2.3.5 subgroup with *only* those noble mediants you can get by starting with some pair of *only* adjacent ratios drawn from the 2.3.5 subgroup itself. I am not even sure how many pairs of adjacent ratios there *are* in the 2.3.5 subgroup, to be honest. Does anyone know this?

Reply

32 wEdited

Mike BattagliaAdmin

Just a quick note this above post may have been edited while people were reading it...

Reply

32 w

Mike BattagliaAdmin

As another addendum, I notice that not all of the "feudal primes" listed - e.g. those of the form a+b*phi, for positive a, b - are actually "noble." For instance, 3+2*phi is called a feudal prime, but it's really (3+2*phi)/(1+0*phi), and 3*0 - 2*1 = -2, which is not ±1, thus this is not the noble mediant of any two intervals.

It would perhaps be a very good idea to determine which of the feudal primes really are "noble" to begin with, and see how the "ignoble" ones are produced. How many "noble primes" are there of the form a+b*phi w/ a and b both in the 2.3.5 subgroup, for instance? And so on.

Reply

32 w

Dave KeenanAuthor

Mike Battaglia That's good stuff. Thanks. I have not come across any "modulus" with the properties you describe. Neither Dekker nor Dodd have it. You wouldn't be thinking of the "arithmetic norm" would you? For a number of the form a+bϕ, a, b rational, N(a+bϕ) = a²+ab-b².

I note that the "noble mediant" is only /defined/ when |ad-bc| = 1, otherwise it wouldn't deserve the name "noble". I guess that could still be called the "ϕ-mediant" or something, but it isn't really of any interest except when it produces nobles.

Reply

32 wEdited

Dave KeenanAuthor

Yes, the concept of noble pitches is meaningless. Only noble intervals. And yes, nobility is fragile. It rarely survives multiplication. Very few pairs of nobles can breed successfully.

Reply

32 wEdited

Mike BattagliaAdmin

Dave Keenan I am going from memory but that a^2 + ab - b^2 looks correct. I forget why the name modulus was in my head but we can call it a norm. So for the general complexity of any feudal number, you can take the product of norms of the numerator and denominator. Try it with the noble mediant associated to 5/4 and 6/5, for instance. There are several ways to express this - the noble mediant of 5/4 and 6/5, or of 1/1 and 5/4, etc - but you should always get the same complexity no matter what. I don't remember what meaning the sign of the norm had.

Reply

32 w

Mike BattagliaAdmin

Dave Keenan: I am sort of puzzled by these "ignoble primes" like 3+2*phi. The noble ones will all be of the form a+1*phi. Something like 3+2*phi is not noble. So what is it and is it somehow generated in any musical way?

Is there some clever way that it's a transposition of some noble interval, or something like that, or is it just irreducibly generated as a "phi-mediant" of two non-adjacent ratios?

Reply

32 w

Dave KeenanAuthor

I never expected the primes (or any feudal integers other than ϕ and ϕ²) to be noble. Thanks for pointing out that all of the form a+1ϕ are. I expected all nobles to be a quotient of feudal integers. I was surprised to discover they were all quotients of primes-or-units. I'm not sure what qualifies as "generated in a musical way" for you here. As shown in various of my tables, 3+2ϕ is the numerator or denominator of 6 different nobles down to level 6 of the SB-tree, and it is one of the prime factors of 11.

Reply

32 w

Mike BattagliaAdmin

I guess it's also the transposition of the noble mediant (3+2*phi)/(1+1*phi) up by the other noble interval (1+1*phi). I am still curious if every ignoble interval is just a transposition of a noble interval in this way.

Reply

32 w

Dave KeenanAuthor

Re the monoid of ℤ⁺[ϕ] under multiplication, which I agree is all we actually need: I assume that what we mean by a "positive" feudal integer is one whose real-number value is positive, i.e. the value obtained when you treat the formal addition as real addition, substituting 1.618... for ϕ. In that case, all integer powers of ϕ, i.e. ϕⁿ for all integers n, including n<0, are included in the monoid. Therefore the monoid has integers other than 1 that have reciprocals within the monoid, therefore the monoid has "units" other than 1. And you can't describe ϕ as just another prime if you want to preserve a version of the fundamental theorem of arithmetic. But as far as our vectors go, ϕ is just another basis element, along with the feudal primes.

Reply

32 w

Mike BattagliaAdmin

Dave Keenan: I sorted through a bunch of this and I think I have better clarity on what I was trying to say. Had an earlier post but replaced it with this instead after sorting through some stuff... This is pretty subtle!

The big idea is that we would like some kind of monzo decomposition of a feudal number, similarly to the way we have monzo decompositions of rationals. To do this, we need something like "primes."

If I understand the basic theorem involved here correctly, we know that every single element in Q[sqrt(5)] is a product of feudal primes and powers of phi. We know that, "up to equivalence of powers of phi and -1," we have unique factorization.

So one preliminary question is: if we want to just treat phi as another basis element in our monzo, and get rid of -1, *do we still have unique factorization?* In other words, for any positive feudal number, is there always one unique monzo representation in terms of feudal primes *and* phi, even though phi is a "unit?" Again, I am going from memory, but I think this is true - it would be good to prove this.

Given that we do, then the next step is to ask what our generalized "natural numbers" are. In general, given the prime factorization of an arbitrary feudal number, some exponents will be positive and some negative, so we can look at the set of all feudal numbers with only positive exponents. Let's call these the "feudal natural numbers." Then every positive element of Q[sqrt(5)] is the quotient of two feudal natural numbers. If the above unique representation theorem is true, these feudal naturals will form a *free* commutative monoid, meaning it has a unique choice of basis. In some sense, this monoid of feudal naturals is almost the main thing we care about.

It turns out that the set of feudal naturals is *not* the set N[phi] I was talking about before. For instance, the element "5" is not a feudal natural number. It is, rather, a quotient of the two feudal natural numbers (2+phi)^2 = 5 + 5*phi and phi^2 = 1 + phi. Put another way, using the quotient representation, we'd really write "5" as "(5+5*phi)/(1+phi)" as a quotient of two of feudal naturals, derived from the positive and negative exponents of the monzo representation for 5. This is something I think is super interesting.

Where I got thrown off is this notion of "feudal integer" that you have for all elements of the form a+b*phi. You also show that every feudal number is a quotient of two feudal integers. But there is this subtlety with this notion of feudal integers that threw me off. We know that we don't care about negative numbers, for instance. With the regular integers, we know that if we look at only the positive numbers, we get the naturals, and coincidentally this is also our free multiplicative monoid that generates the positive rationals Q+ as quotients. But this isn't how it works with feudal integers: if we just naively do this, we get that the set of positive feudal integers is "too large." It isn't a free monoid, because we have, as you say, unlimited powers of 1/phi thrown in the mix. The result is kind of like looking at the regular natural numbers, but with all the numbers also multiplied by every possible power of 1/2, or something, and treating 2 as a "unit." What we really want is to go one step further.

This has some pretty profound implications when we treat an arbitrary feudal number as a quotient of two feudal integers. For instance, even some positive feudal integers, like 13 + 16*phi, are actually nontrivial *quotients* of positive feudal naturals, in this situation (10+3*phi)/(1+phi). This kind of thing never happens with the regular integers. So we may want to ask which feudal integers we should choose in our quotient representation of an arbitrary feudal number. You suggest a normal form in your post which involves doing this Fibonacci-esque thing in reverse until we have a+b*phi with a>b. However, there is another natural normal form which is to look at the unique reduced representation of a feudal number as a quotient of two feudal naturals with no common prime factors (including positive powers of phi), which can be plainly seen by looking at positive and negative exponents of the monzo representation. These two are not in general the same due to the weirdness involving feudal integers; (5+0*phi)/(1+0*phi) is in normal form in only the first sense, whereas the second sense would have it as (5+5*phi)/(1+1*phi).

Anyway, I guess that answers that for now...

Reply

31 wEdited

Mike BattagliaAdmin

It would also be interesting to see if there is some easy way to identify if an arbitrary feudal integer is a feudal natural using only the coefficients a and b. Is there some pattern that lets us easily determine if it's a product of only feudal primes and positive powers of phi?

Reply

31 w

Dave KeenanAuthor

Mike Battaglia: That stuff you wrote, about using the product of the norms of the numerator and denominator as a complexity measure, checks out so far. Thanks.

I'm worried that you're overthinking this monoid stuff. I believe I've already described how to generalise "monzos" (prime count vectors or PC-vectors) to give a unique representation for every positive feudal number ℚ⁺(√5) where a positive feudal number is defined as one whose "real value" is positive. The "real value" of a feudal number is defined in the obvious way as Re(a+bϕ) = a+bϕ, a, b rational, where the "+" and the juxtaposition of b and ϕ on the left are the /formal/ addition and multiplication used in representing feudal numbers, and the "+" and the juxtaposition of b and ϕ on the right are real number addition and multiplication. And for any feudals α, β, Re(αβ) = Re(α)Re(β), and Re(α/β) = Re(α)/Re(β) when Re(β)≠0.

I believe we can avoid representing any negative feudals simply by not having -1 (or any other negative feudal) in our basis. And prime count vectors are incapable of representing zero, as their entries are constrained to be integers (which do not include negative infinity).

I don't understand why I should care whether the positive feudal integers ℤ⁺[ϕ], or your feudal naturals ℕ[ϕ]\{0}, constitute the /smallest/ monoid that will give us the numerators and denominators we need for our nobles and rationals, because we have no choice in the matter. There is no way to limit the integer entries in the vectors in order to restrict them to representing any set smaller than ℤ⁺[ϕ], while still being able to avail ourselves of the power and convenience of all the usual linear algebra methods of RTT.

Yes, it would be good to formally prove that "Up to powers of phi (with ±integer exponents), we have unique factorisation of the positive feudals (as defined above), simply by treating ϕ (and not -1) as a basis element for our PC-vectors. But I'm so sure of it, I'm happy to proceed on that basis (double meaning) until someone comes up with a counterexample. You can tell I'm more of a physicist and engineer than a mathematician.

I note that your phrase "positive powers of phi" is ambiguous, given that "power" is sometimes treated as synonymous with the exponent alone, and sometimes (more correctly) as referring to the whole expression consisting of base and exponent. The two cases can be distinguished as "powers of phi with positive exponents" versus simply "powers of phi" (which are necessarily positive). Their opposites are "powers of phi with negative exponents" versus "the negations of powers of phi".

You wrote: "For instance, even some positive feudal integers, like 13 + 16*phi, are actually nontrivial *quotients* of positive feudal naturals, in this situation (10+3*phi)/(1+phi). This kind of thing never happens with the regular integers."

The denominator in your example is a unit, namely 1+1ϕ = ϕ². So while I agree it's non-trivial, I think it's fair to say that the analogous thing /does/ happen with the ordinary integers, e.g. 2 = 2/1.

But you might be onto something with your proposed normal form. Consider the feudal-prime-factorisation of 5. In my current normal form we have 5 = (2+1ϕ)²/(1+1ϕ) = (2+1ϕ)²/ϕ². With your proposed normal form I believe it would be 5 = (-1+2ϕ)²/(1+0ϕ) = (-1+2ϕ)²/1 = (-1+2ϕ)². I believe this form requires doing the Fibonacci-esque thing in reverse for one more step beyond where I stopped it, i.e. continuing until we have a+bϕ with a<0. (You can normalise primes but not units). It's nice to get rid of those ϕ² from the factorisations of the ordinary primes, but it's annoying to be guaranteed to have a minus sign in every prime. It makes one want to write them with the golden part on the left, e.g. 2ϕ-1.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan I think we are talking past each other. There is a very simple way to look at all of this: if you take this monzo decomposition, what I am calling the "feudal naturals" are those with only positive coefficients. Every feudal number is uniquely a quotient of two of those which have no prime factors in common, which is the "reduced form" I talked about. This parallels the way that every rational is a quotient of two natural numbers with no common prime factors. Given any feudal number, the reduced form can easily be read straight from the monzo: zeroing out all of the negative coefficients gives the numerator, and vice versa for the denominator.

I'm not sure how you got that normal form for 5 in your last paragraph but it is not the same as this. Using the monzo decomposition, the numerator of 5 is (2+1*phi)^2 = (5+5*phi), and the denominator is (1+phi). This is what you gave in your post as *your* normal form, but I get something different with your method: if we take the Fibonacci thing one step further then we get (5 + 0*phi)/(1 + 0*phi) without any negative numbers.

Some of the things you are saying seem to be misconceptions that I addressed directly in the previous post, for instance "There is no way to limit the integer entries in the vectors in order to restrict them to representing any set smaller than ℤ⁺[ϕ]" -- this is incorrect, the feudal naturals are strictly smaller than this. Feudal naturals have only positive monzo coefficients, whereas feudal integers allow for the phi coordinate (and I think only the phi coordinate) to be an arbitrary integer.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia You misquoted me above, by leaving off the second half of my sentence. I wrote: "There is no way to limit the integer entries in the vectors in order to restrict them to representing any set smaller than ℤ⁺[ϕ], while still being able to avail ourselves of the power and convenience of all the usual linear algebra methods of RTT."

At least I now understand how your "feudal naturals" differ from the positive feudal integers, namely choose a specific normal form for the primes in the vector basis, and omit ϕ from the vector basis (and hence omit all units other than 1). But I don't see how that is at all useful for our RTT application. ϕ and ϕ² are two of the nobles we want to represent.

I have no problem with calling those the feudal naturals. But if you're claiming we can actually omit ϕ from our basis, I'm not seeing it.

My reverse-fibonacci normalisation method, as described in my forum post, is only shown there as being performed simultaneously on the numerator and denominator of nobles (which seem to be always primes or units). I've since realised, it is applicable also to primes individually, but not to units or composites individually. That's why it doesn't work on unfactorised 5. So it's purely a way to obtain a normal form for primes.

And now I find it gives a simpler result for our prime-count vectors (monzos) if I take it one step further and end up with a negative wooden part for all the primes. Because that eliminates all the ϕ²'s from the factorisations of ordinary primes. For example, it's better to have the simplest prime (the single prime factor of 5) as 2ϕ-1 = √5 rather than my existing 2+1ϕ = ϕ√5.

I understand now, that this new normal form is not one you proposed.

I'm considering doing a wholesale find and replace on my forum articles, using my new normal form for the primes, aϕ-c (rather than -c+aϕ), where a and c are positive integers with c<a. The relationship to the previous choice of a+bϕ, where a and b are positive integers with b<a, is simply c = a-b.

Does anyone have any thoughts on why that might be a good or bad idea?

Reply

31 w

Mike BattagliaAdmin

Dave Keenan I don't see how linear algebra has any bearing either way on the feudal natural numbers. You already have a basis of phi + feudal primes, and the set of all vectors w positive integer coefficients is feudal natural. This is already a set smaller than the positive members of Z[phi]. What is the problem?

The thing about omitting phi from the basis is very far from anything I am talking about. I've said explicitly that phi is to be treated as just another basis element again and again. I'm not sure where we got off track but maybe try reading my post again...

Reply

31 w

Mike BattagliaAdmin

I guess that in the last bit, you are saying that the choice of "feudal prime" is somewhat arbitrary - in Z[phi], feudal primes come in equivalence classes which are only equivalent up to a power of phi. So depending on which one we choose, we get an entirely different basis, and thus a different set of "feudal naturals" as the set of vectors with non negative coefficients relative to that bsis. This is a good point. I would be curious if the arithmetic norm can make it easier to choose a representative from each feudal prime class.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia Yes, we do seem to be talking past each other. Can we perhaps reframe the discussion as trying to agree on the most practically-useful choice for the normal form of the primes to be used in our monzo basis?

Reply

31 w

Mike BattagliaAdmin

Dave Keenan sure. A "noble" goal.

I guess one suggestion is that the prime should certainly be larger than 1/1, but this is kind of weak.

Here is an important question: given some prime, suppose we generate the two-sided "fibonacci" sequence it generates. Is the number 1 or -1 guaranteed to appear exactly once in this sequence? Because then we could (and should) choose the representation in which the ±1 is the phi coefficient, because then this is the unique representative in the feudal prime class which is "noble." This would also affirmatively answer the question I posed earlier, if every "ignoble ratio" is a transposed noble ratio.

EDIT: sadly, turns out this line of reasoning doesn't work. 7 + 4*phi doesn't have any noble representative; the associated Fibonacci sequence goes ...-13, 10, -3, 7, 4, 11, ... and diverges from there. We could try to choose the "most noble" representative, perhaps, which would be 10-3*phi, although sometimes this isn't unique, in that 2+phi and 3-phi are in the same equivalence class and both equally noble. So I'm not sure.

But a bigger question I have about this is what the point of intervals like 7+4*phi is. I thought the goal here was to formalize the structure of these noble mediants. If 7+4*phi isn't a noble mediant, or some product of noble mediants and JI ratios, does it have any musical relevance? I mean, I get that it's technically a "prime" in the ring Z[phi], and that you can factor noble mediants according to this set of primes, but is that all?

Reply

31 wEdited

Mike BattagliaAdmin

I think we may get that there's usually a unique "most noble" representative, except for the rare sporadic prime where one would need to choose a tiebreaker. Either way, I'm really not sure if this is the best way to do it.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia Here are the first two counterexamples. The factor of 5 and numerator of both 3rd-order nobles:

..., -11, 7, -4, 3, -1, 2, 1, 3, 4, 7, 11, ...

It has both -1 and +1.

A factor of 31 and the numerator of two 5th-order and denominator of two 6th-order nobles:

..., -30, 19, -11, 8, -3, 5, 2, 7, 9, 16, 25, ...

It has neither -1 nor +1.

I agree with >1. One possibility is to make them as close to 1 as possible while still being greater. But my earlier suggestion is not equivalent to that, as it is equivalent to making them as close as possible to the square root of the ordinary prime that they are a factor of.

Reply

31 w

Mike BattagliaAdmin

Dave Keenan you may have missed my later edit about the musical point of some of these primes, see the last paragraph. Basically my view is - if we view all of this as deriving from the noble mediant idea, which has a clear musical use, then the intervals we care about, at most, are:

1. JI ratios

2. Noble mediants

3. Products of noble mediants and other noble mediants and JI ratios

Given those criteria, does an interval like 7+4*phi even matter, and does it have any musical use? it's technically a prime in Z[phi] and we can factor noble mediants in those terms if we want, but is it "irreducibly ignoble" in some sense, and if so why does it matter?

Reply

31 w

Dave KeenanAuthor

I don't see any particular value in making the primes themselves noble (when possible), as this may lead to them being quite large numbers. And nobles wider than say a diatonic 13th, say 3.25, are not of much interest. They are too wide (>2041 ¢) for accurate tuning to matter.

Making them as close as possible to 1 while being greater, means constraining them to be between 1 and ϕ. I don't see any particular value in this either.

Reply

31 w

Mike BattagliaAdmin

At this point the main question I have is if there are simply "too many" feudal primes. Noble mediants matter and transposed noble mediants matter. But when we factor these according to the primes of Z[phi], are we really factoring relative to a somewhat arbitrary set containing non-noble junk intervals without direct musical relevance? Or is there some other way to look at it?

Reply

31 w

Dave KeenanAuthor

Mike Battaglia As an interval, 7+4ϕ is of no musical interest whatsoever. But as a prime, we do need it, or one of its associates (that's a technical term in the literature), to encode the noble mediant of 11/8 and 15/11 (as the numerator), and possibly others. And yes, it is irreducibly ignoble.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia: Yes I missed the edit. You wrote: "I mean, I get that it's technically a "prime" in the ring Z[phi], and that you can factor noble mediants according to this set of primes, but is that all?"

Yes. That's all. Isn't it enough? However we might find a more useful associate than 7+4ϕ. I think 7ϕ-3 is more useful, but it might not be the most useful.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan:

1) How do we know that it is irreducibly ignoble?

2) Suppose that instead of doing it this way, we just took all of the noble mediants we care about, along with the JI ratios, and we look at all possible products and quotients of those. Which feudal primes does this generate? Apparently not all of them?

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia: I'd qualify that to say that we care about:

1. JI ratios

2. Simple noble mediants

3. Products of simple noble mediants and other simple noble mediants and JI ratios

I note that, to me, JI means "simple rational", not just any rational. So the counterpart of JI is not noble but rather "anti-JI" or "MI (merciful intonation)".

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia Taking all of the noble mediants we care about, along with the JI ratios, and listing the feudal primes this generates is exactly what I did in my series of articles on the Sagittal forum! Haven't you read it? And this automatically includes all possible products and quotients of those.

Reply

31 w

Dave KeenanAuthor

How do we know 7+4ϕ is irreducibly ignoble? I was assuming that a prime can only be noble if it is of the form a+1ϕ. Is that not so? If it is so, then we only have to look at its fibo-like sequence to see that it will never generate a 1 or -1.

... 59 -36 23 -13 10 -3 7 4 11 15 26 41 ...

Reply

31 w

Mike BattagliaAdmin

Dave Keenan you looked at which "primes" generate the noble mediants, not which primes are generated *by* the noble mediants as a product of noble mediants and JI ratios.

Reply

31 wEdited

Mike BattagliaAdmin

Let's put it this way: forget noble intervals for now. Suppose we only care about JI. The regular prime numbers are a useful basis for JI intervals. But, we could always add other new elements to our group such that our prime numbers become composite. For instance, if we added the elements (3+sqrt(2)) and (3-sqrt(2)), then 7 can be factored into a product of those. But we don't do this because there is no point to introducing those intervals to begin with (if we only care about JI).

Are we perhaps doing something similar with these feudal primes; factoring noble mediants into extraneous, unnecessary, "unmusical" intervals that fall outside our criteria? Or is there some mathematical reason why all of these feudal primes are somehow necessary?

Reply

31 w

Dave KeenanAuthor

Mike Battaglia But products of nobles and rationals will not generate any new primes. You will still only have the primes that were contained in either the nobles or the rationals to begin with.

Reply

31 wEdited

Dave KeenanAuthor

I assume we're talking prime equivalence classes here, i.e. ignoring powers of ϕ.

Reply

31 w

Mike BattagliaAdmin

No primes are inherently "contained" in any of these things; we can factor it according to whatever basis we want. From before: "Are we perhaps doing something similar with these feudal primes; factoring noble mediants into extraneous, unnecessary, "unmusical" intervals that fall outside our criteria? Or is there some mathematical reason why all of these feudal primes are somehow necessary?"

Reply

31 wEdited

Mike BattagliaAdmin

I think you get what I'm asking at this point. Do you know the answer to this question?

Reply

31 w

Mike BattagliaAdmin

Dave Keenan: Here is a great example. This will hopefully make clear what I am talking about, in a way that I am sure is impossible to misunderstand.

Suppose we want 11-limit JI, and the only two noble mediants we care about are 1) phi, and 2) the noble mediant between 11/8 and 15/11, call it "N." We can add those two as basis elements to the 2.3.5.7.11 subgroup to get a 7D group, which I'll call 2.3.5.7.11.phi.N.

"N" can be written, in your lowest terms form, as (7 + 4*phi)/(5 + 3*phi). The numerator and denominator are both feudal primes which are not noble, and as far as I know, not a transposed version of any noble interval. Thus, they don't have any direct musical use, at least not given our criteria for before.

So we could take our nice, musically valid interval "N" and further factor it into a quotient of these other two musically useless intervals which appear in the numerator and denominator.

Why do we want to do this?

Reply

31 wEdited

Dave KeenanAuthor

I think it's just wonderful that ℚ(√5) is a unique factorisation domain (UFD) and that it contains the rationals and the nobles and amazingly little junk (compared to say, some imagined set that needed to also contain square roots of ordinary primes other than 5 before it became a UFD). We're never going to do any better.

I don't think it matters that the primes themselves are unmusical. I think there's an argument that the prime 5 is unmusical. 2786 ¢. How often does such a wide interval occur in a chord, and when it does, who cares about tuning it accurately? If not prime 5, then 7, or 11. While we value simple ratios containing those primes, we don't really care to hear the primes themselves.

Reply

31 wEdited

- Dave Keenan
- Site Admin
**Posts:**2180**Joined:**Tue Sep 01, 2015 2:59 pm**Location:**Brisbane, Queensland, Australia-
**Contact:**

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

Dave KeenanAuthor

We need a UFD to make the whole monzo thing work. Right?

Reply

31 wEdited

Mike BattagliaAdmin

OK, well, I'm not sure what to say - it seems like you are mostly just defending what you already have... I think it's all interesting and was trying to explore a little bit further. All I am really asking are some basic questions, which I think are immediately apparent from all of this:

1) Given some p-limit, what noble mediants/feudal numbers/etc can we actually generate from just the intervals in that p-limit?

2) If we take all of the JI ratios and noble mediants and throw them into a blender, do we actually get the feudal numbers or just some of them?

3) Do the other feudal numbers have some musical use?

I don't know how to answer them, but I certainly think these are interesting questions that are worth looking into.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia: Ah. Thanks for the specific example question. In that specific case, and at this point in time, there is almost certainly no advantage in factoring that N. But if N was instead the noble mediant between 4/3 and 5/4, which can be written as (3ϕ-2)/(2ϕ-1), then it would be worth factoring. Why? Because a feudal monzo would replace 5 in the basis with 2ϕ-1 = √5, which is also the numerator of N, and it would replace 11 in the basis with its two factors 3ϕ-2 and 3ϕ-1, one of which is also the denominator of N. So the basis encompassing 11-limit JI, ϕ and N would be 2.3.(2ϕ-1).7.(3ϕ-2).(3ϕ-1).ϕ. Still only 7 entries.

And if we already had a list of low badness noble/rational commas to choose from, having standardised the choice of associates for the feudal primes, there might be an advantage in factoring even your original N, and we would still replace 5 and 11 in the basis with their feudal factors.

Reply

31 wEdited

Dave KeenanAuthor

If we didn't replace 5 and 11 in the basis with their feudal factors, we wouldn't have a UFD. There would be more than one way to make 5 or 11.

Reply

31 w

Mike BattagliaAdmin

Dave Keenan that's an interesting example. I guess it's the same as something like the 2.9.5/3 subgroup. In that case, "9" happens to factor as a square of the denominator of 5/3. So you would make it 2.3.5. But that doesn't make 2.9.5/3 somehow not a UFD (or rather not a free abelian group, since we aren't talking about rings). It just means that 2.9.5/3 is a subgroup of finite index within the 2.3.5 group, meaning it is basically a dilation of it. The 2.3.5.7.11.phi.N subgroup is similar and I guess is a subgroup of 2.3.(2phi-1).7.(3phi-2).(3phi-1).phi.

Reply

31 w

Mike BattagliaAdmin

2.3.5 is an even better example. 5, treated as a feudal number, is now composite. Do we really need to factor it into 2.3.√5, though? Of course not, and similarly we don't need to always factor noble mediants. Of course, if we add several noble mediants to a subgroup and doing this somehow automatically causes various primes to split even before we've done the factoring into numerator and denominator, then you just get that for free.

Reply

31 wEdited

Dave KeenanAuthor

Right. You have to check if any noble you're adding has what I call f5 or f11 or f11′ as factors, and split 5 and/or 11 as the case may be. The next splitter doesn't come until ordinary prime 19 splits into f19 and f19′.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan well what I'm saying is I don't think it is really required that you split it, just like you don't have to split 9 in the 2.9.5/3 subgroup. In your example you also don't have to split it, and if you don't, you just get some subgroup which isn't "saturated" relative to the entire group of feudal numbers, or which is "enfactored" or "impure" or whatever you are calling it these days. The saturation of the non-split version will be the one which is split.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia I just want you to know that I really appreciate you wrestling with this stuff with me. You're probably the only person who could grasp it so fully and so quickly. It's great that you did some investigation previously.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan Thanks man. I originally got interested in this stuff from your noble mediant paper many years ago, so I'm really just continuing your work, I guess. I similarly appreciate the wrestling, probably more than you do 🙂

Reply

31 w

Paul ErlichAdmin

The major seventeenth occurs quite often on chords on the piano or organ let alone any multi-instrument combo, where there's usually a bass player several octaves below the melody (let alone any upper vocal harmonies, string arrangements, etc.) or at least say a guitar accompanying a higher instrument or voice. So e.g. 5:10:15:20:24 and 4:8:12:16:19 are competing (in terms of minimizing 5HE) tunings for a very common voicing of the minor triad in real music. And 1:2:3:4:5 dominates for a very common voicing of the major triad. As a pianist these are the voicings I instinctively grab on the keyboard. Oddly 1:5's tuning may matter most when the upper note's fundamental is quiet enough to just cancel out the 5th harmonic of the lower note once per cycle as they beat against one another.

Reply

31 wEdited

Paul ErlichAdmin

I know Gene wrote a lot about manipulating noble numbers and other aspects of this topic, but wouldn't know where to find that in the archives . . .

Reply

31 w

Paul ErlichAdmin

For example discussing the golden meantone generator

v = 2^[(3Φ+1)/(5Φ+2)]

he explained that the exponent of 2 can be simplified further.

Gene Ward Smith wrote (Yahoo tuning-math group message 2507 (Mon Jan 7, 2002 7:31 pm):

"Ratios of the sort (a+br)/(c+dr) define an algebraic number field, which can always be put into the form of a sum of rational numbers times powers of a single algebraic number r. In this case, that results in

(a+br)/(c+dr) = (ac+ad-bd + (bc-ad)r)/(c^2+cd-d^2)

This form of the algebraic numbers in the field Q(r) is unique, since {1, r} are a basis for a vector space over the rationals Q; hence we can determine if two elements of Q(r) are the same by putting them both into this form."

If we rearrange the exponent of 2 (in the ratio for v) so that it reads (1+3Φ)/(2+5Φ), and plug the values a=1, b=3, c=2, d=5, and Φ=r, into Gene's equation, the resulting answer is (-8+Φ)/-11, which can be simplified to (8-Φ)/11.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia: This empirical statistic may be relevant to your recent questions: At any given depth (order-limit) of the SB-tree, the number of feudal primes needed to generate all the nobles is about half the number of nobles. Every noble needs at most 2 primes, but every prime is used by about 4 nobles (typically 2 uses as numerator and 2 uses as denominator). That seems to me like a reason to factor the nobles.

Reply

31 wEdited

Dave KeenanAuthor

Here's another thought. Sure you don't need primes in your monzo basis, but you need to know that your basis is linearly independent. How do you check that? You express each vector of your non-prime basis as a vector in a prime basis, because we know the primes are independent.

So whether you have composite nobles in your basis or not, you'll still need a prime basis to check their independence. And it would be good to standardise those feudal primes.

Reply

31 w

Dave KeenanAuthor

Some thoughts on your 3 questions:

1) Given some p-limit, what noble mediants/feudal numbers/etc can we actually generate from just the intervals in that p-limit?

It is mathematically interesting if the set of noble mediants of p-limit ratios is finite, despite the fact that the number of p-limit ratios is infinite. I didn't expect that. But in musical terms, I'd be more interested in the noble mediants of some kind of integer-limited or complexity-limited subset of the p-limit.

It may be educational to take a copy of Erv's scale tree, which has more levels than the sideways SB-tree in my feudal posts, and circle all the 5-limit ratios. Then embolden the line between any two that are adjacent, and circle the noble that this corresponds to. Putting it another way: If a zigzag path has two consecutive 5-limit ratios anywhere along it, the noble corresponding to that zigzag lives. Then see if there's any pattern to the nobles that survive.

2) If we take all of the JI ratios and noble mediants and throw them into a blender, do we actually get the feudal numbers or just some of them?

I take this to mean: Can we get all the feudal numbers as products and quotients of rationals and nobles? i.e. Can we get all feudal numbers as sums and differences of the feudal-prime-count-vectors of rationals and nobles. Sorry I was so slow to understand this question. I guess it's because I don't understand why it matters (except as an interesting math question). I'm just happy that it works the other way round, namely that by being able to represent all feudal numbers, we get all nobles and rationals (and their products and quotients).

But I will give this some thought, and see if I can find a proof or counterexample to answer your question.

3) Do the other feudal numbers have some musical use?

If there are no feudal numbers other than those that can be obtained as products and quotients of rationals and nobles, the question is moot. If there are any, I can't conceive of any musical use for them, but I'd be willing to tolerate them in order to have a prime basis for vectors that can encode both rationals and nobles.

Reply

31 w

Dave KeenanAuthor

Here is a sketch of a proof that all feudal numbers can be generated as products or quotients of nobles and rationals. It suffices to prove that all feudal primes can be so generated. If we can obtain all feudal primes, we know we can can generate all feudal numbers as products and quotients of those.

Some statements below are merely observations from the nobles examined so far, and require their own proofs in general. Hence calling this a "sketch".

From Dekker we know that feudal primes fall into 3 classes. (a) Those which are ordinary primes (2 and those ending in 3 or 7), (b) the prime factor that's squared to make 5, (c) those primes which are one of a pair of complementary prime factors of ordinary primes ending in 1 or 9.

The rationals directly supply us with the primes of class (a). The nobles supply us with the primes of classes (b) and (c) as follows.

It will help to look at this table: viewtopic.php?p=4592#p4592

The nobles of the top two levels of the SB-tree consist only of units, no primes. All noble numbers beyond level 2 have a prime as their numerator, and a prime-or-unit as their denominator. As one descends the SB-tree, each new level of nobles has, as its denominators, only those primes-or-units that have appeared as numerators on previous levels. New primes always appear first as numerators. All units (red) can be treated as equivalent.

Therefore every noble can be reduced to only the prime in its numerator (and possibly some unit), by multiplication by an ascending series of nobles from levels above it. i.e. Cancel its denominator with the numerator of a noble from a previous level, and then cancel /its/ denominator with the numerator of a noble from a level above /that/, and so on, until reaching a noble with a unit denominator.

e.g. to obtain the prime 7+5ϕ we can multiply the following series of ascending nobles.

(7+5ϕ)/(4+3ϕ) × (4+3ϕ)/(3+2ϕ) × (3+2ϕ)/(2+1ϕ) × (2+1ϕ)/(1+0ϕ)

Here's a shorter path

(7+5ϕ)/(3+2ϕ) × (3+2ϕ)/(1+1ϕ) x (1+1ϕ)/(1+0ϕ)

The prime of class (b) appears as a numerator on level 3. What remains is to show that the Stern-Brocot tree will eventually include every feudal prime in class (c) as the numerator of some noble. I don't have a proof for that, but it seems to be on track so far, generating primes of increasing complexity as we descend. Any gap left on one level seems to be filled in on the next level, such that it seems likely there will be some ordering of the class (c) primes that is generated in sequence, without gaps, by some traversal of the SB-tree.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan wow! Ok, that makes sense. Great stuff, thank you!

I guess an immediate follow up question is: can we somehow stratify the feudal numbers into classes based on how "close" to noble they are?

I am thinking of this primarily for musical reasons. Basically noble numbers are important. "Justly transposed noble numbers," which are a product of a rational and a noble number, are probably of equal importance to the nobles themselves, unless your music never changes chords or tonics at all. In fact for now let's grant them honorary "noble enough" status. Then there are those numbers which are a product of two noble numbers and perhaps a rational, and then those which are a product of three noble numbers and perhaps a rational, etc. These numbers will be less musically frequent unless you like to do very strange modulations directly by a noble number or something like that.

So in general we can ask about the shortest such decomposition. Is there some easy way for us to determine this "compound nobility rank" for any feudal number, or at least for the feudal integers or primes?

I have this strange feeling that perhaps the determinant is involved here: for (a+b*phi)/(c+d*phi), it's noble iff ad-bc = ±1, so maybe the other determinants mean something relevant to this question. For feudal integers we have (a+b*phi)/(1+0*phi), so the determinant is just -b; again that b coefficient seems very relevant.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia No worries mate. (I think that translates into American as "You're welcome"). 😉 I agree that determinant seems like it should represent some kind of nearness to nobility. There is a concept called near-noble numbers, defined in terms of continued fractions, here:

https://archive.lib.msu.edu/crcmath/mat ... n/n039.htm

I don't think it's related.

Near Noble Number

ARCHIVE.LIB.MSU.EDU

Near Noble Number

Near Noble Number

Reply

Remove Preview

31 wEdited

Mike BattagliaAdmin

Dave Keenan here is an interesting thought to perhaps supply the missing part of the proof.

Suppose we have an arbitrary feudal integer of the form a+b*phi in which a and b are coprime. We would like to determine some sequence of nobles whose product is our integer. To generate the first noble, we look for c, d such that the 2x2 matrix [a b;c d] has determinant 1 (or -1). This means we want

ad - bc = 1

We can easily locate the values for c and d using the modular inverse. If we take the above expression mod a and b, we get the two following expressions

ad = 1 mod b

-bc = 1 mod a

Since a is known, and a is coprime to b, d is the unique modular inverse of a mod b, and c the unique modular inverse of -b mod a.

As a first step, we have turned (a+b*phi) into (a+b*phi)/(c+d*phi) * (c+d*phi). The key thing is that we get two values of c and d which are strictly smaller than b and a. So now we can repeat the same thing with (c+d*phi), decomposing it as (c+d*phi)/(e+f*phi) * (e+f*phi) in which e and f are smaller than d and c, and then repeat again.

We can formalize this as a table. The first row are our numbers a and b, and then we keep taking modular inverses in the way above for the next row. Then we repeat, building each time the next new row from the last previous row. If we start with 23+29*phi, we get the following table:

23 29

19 24

15 19

11 14

7 9

3 4

2 3

1 2

0 1

This is the entire decomposition: (23+29*phi)/(19+24*phi) * (19+24*phi)/(15+19*phi) * ... * (1+2*phi)/(0+1*phi) * (0+1*phi). All noble, because each 2x2 submatrix of the above table has determinant 1.

The main thing is that this expression terminated at 0+1*phi. Are we guaranteed to get this, or something like it, for all initial a+b*phi as long as a and b are coprime?

Reply

31 w

Dave KeenanAuthor

Mike Battaglia Well done! I can't answer your final question, but my engineering attitude says, I've pushed down the SB-tree a little beyond where I think it's musically relevant, and it all works, so that's good enough for me. 🙂

Reply

31 w

רועי סיני

Mike Battaglia Very cool method! If you're still interested in this question now, and for future readers, the process indeed always ends in either 1 or phi if a and b are coprime. This is just because if it doesn't end there then you can continue the process more, and the gcd always stays 1 because of the determinants of the 2x2 submatrices.

Also, this option to choose one pair of modular reciprocal and a negative of a reciprocal over another gives you noble triads, which Dave Keenan has expressed interest in, because if the matrix

a b

c d

has determinant 1, then so does

a b

a - c b - d

and also

c d

a - c b - d

which means that (a-c+(b-d)ϕ):(c+dϕ):(a+bϕ) is a noble triad.

Also, this process seems to disprove the conjecture stated in viewtopic.php?p=4620#p4620 about the noble mediants always being a quotient of two primes, because if you take (-1+7ϕ)*ϕ³ = 13+19ϕ, which is not a prime but an associate of f5*f11, you still get a noble number with it as a numerator, namely (13+19ϕ)/(11+16ϕ).

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Reply

20 hEdited

Dave KeenanAuthor

I'm concerned to standardise a normal form for the split primes as soon as is decent, because the later we do it, the more people will have read my articles, and the more will have to unlearn the old ones and learn the new ones, in the case that I decide to change them.

My current form, where f5 is 2+1ϕ and f11 and f11′ are 3+1ϕ and 3+2ϕ, does the following when you split the primes for a 13-limit vector:

[ a b c d e f ⟩

goes to

[ a b 2c d e e f; -2(c+e) 0 0 ... ⟩

where the entry after the semicolon is the ϕ-count.

I figure it would be better if it went like this:

[ a b c d e f ⟩

goes to

[ a b 2c d e e f; 0 0 0 ... ⟩

i.e. no power of phi is introduced merely by splitting primes.

To achieve that, we need f5 to be -1+2ϕ and f11 and f11′ to be -2+3ϕ and -1+3ϕ.

But they are more compactly written if flipped horizontally so f5 is 2ϕ-1 and f11 and f11′ are 3ϕ-2 and 3ϕ-1.

What do you think?

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan I would always recommend a+b*phi even if a is negative; I almost misread one of your earlier posts making that mistake when you wrote it as b*phi+a.

Beyond that I think the cart is way ahead of the horse when it comes to standardized conventions and etc; we are just now starting to get the absolute basics of a structure theorem together and that will probably make clear what the best representative of each feudal prime should be.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia IIUYC, the feudal naturals ℕ[ϕ] are those feudal numbers whose feudal-prime-count-vector has no negative counts and zero for the ϕ-count. But that depends on which of each set of associated primes we choose for our prime basis.

Is it too early then to decide that we'd like all ordinary naturals 1, 2, 3, 4, 5, ... to also be feudal naturals?

Reply

31 w

Mike BattagliaAdmin

Yeah, exactly. We get different feudal naturals based on different choices of prime, which I missed when writing this idea for the first time. There is a "maximal" possible set of naturals we could get based on if we choose the unique representative between 1 and phi for each prime, and all other possible choices for the primes give some subset of this maximal set of naturals. But as you mention, this is probably not right, at least not unless we want 3/phi^2 to be the representative for 3. But the idea to make the ordinary naturals also feudal naturals is intriguing. Does that lead to some particular choice of basis?

Of course if we choose any particular basis such that the true naturals all appear with positive exponents, then we can multiply everything in that basis by 1/phi and get another basis where the naturals still have all positive exponents, just with some extra factors of phi thrown in. So we can ask about the phi-coordinate for each natural. Is there some basis we can choose so that the phi coordinate is 0 for each of the ordinary naturals?

I guess this requirement would correspond, for instance, to your choice of sqrt(5) = (-1 + 2*phi) for that particular feudal prime. But is there some way we can use this to choose, for instance, what you are calling f11 and f11'?

Reply

31 w

Mike BattagliaAdmin

I think the only question is basically the primes in your third class stated previously "those primes which are one of a pair of complementary prime factors of ordinary primes ending in 1 or 9." So let's call these two primes, whatever they are f11 and f11'. It would be nice if we had f11 * f11' = 11, rather than f11 * f11' * phi = 11 or anything like that. The only problem is that given any choice of f11 and f11' which have that property, we can replace f11 with f11*phi and f11' with f11'/phi to get another pair of representatives which multiply to 11. Is it possible to choose some kind of unique pair which are, perhaps, conjugates of one another, either in the (1, phi) basis or the (1, sqrt(5)) basis?

Reply

31 w

Mike BattagliaAdmin

Dave Keenan Here is a great example - in your original post about this, you suggest f11 = (-2*phi + 3) and f11' = (-1*phi + 3).

However, we could also choose f11 = (3+phi) and f11' = (4-phi), and we still have f11*f11' = 11 without any extra phi coordinate.

This is a very natural choice: since you are deriving all of this from the algebraic number theory view, we have that

3+phi = 7/2+sqrt(5)/2

4-phi = 7/2-sqrt(5)/2

so that these two are conjugates of one another in Q[sqrt(5)]. I think that's about as good of a pair as you could possibly want. In this case, these two representatives also have the benefit of both being noble.

Do these always come in conjugate pairs for each prime? If so, these pairs will always be of the form a + b*phi and (a+1) - b*phi. And if things are that neat, and we are guaranteed some pair like that for each relevant prime, I think that's about all she wrote! That would be a perfect basis, probably as good and standardized as you can get from an algebraic number theory view. And I would be very curious if the "b" coefficient has any interesting interpretation - clearly if it's 1 it's noble, and what otherwise?

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia I don't think so, because you can always multiply any choice for f11 by ϕ and multiply f11′ by ϕ⁻¹ and keep the overall ϕ-exponent for 11 being zero. We'd need to introduce some other constraint, preferably one that gives us some other desirable property. One such is that nobles should also have a zero ϕ-count, whenever possible. And since f11/f5 and f11′/f5 are nobles, and we want f5 to be -1+2ϕ, I believe that constrains f11 and f11′ to be -1+3ϕ and -2+3ϕ, not necessarily in that order.

Reply

31 w

Mike BattagliaAdmin

Dave Keenan I'm not sure what you're talking about... Did you see my post above about the primes for f11 and f11' being in conjugate pairs?

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia No. I did not see your previous two in this subthread until now. I was responding to the question at the end of your post that begins "Yeah, exactly".

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia I totally agree we want f11×f11′ = 11, and the same for every other split prime. Yes, the split primes can always be written as a conjugate pair in the (1, √5) basis, which, in the (1, ϕ) basis will be of the form a+bϕ and (a+b)-bϕ [not (a+1-bϕ as you have above]. Dodd defines a constant ϕ̅ (pronounced phi bar - the bar is meant to be above the phi) such that ϕ̅ = -ϕ⁻¹ ≈ -0.618, so that he can write the (1,√5)-conjugates as e.g. f11 = 3+1ϕ and f11′ = 3+1ϕ̅. (Or at least he does the equivalent, using ω instead of ϕ). I don't like it. Why have two irrational constants when one will do.

But what is best for number theory isn't necessarily best for RTT scale engineering. Dodd never once mentions noble numbers. We've agreed we don't want powers of ϕ in our rationals. I think there is a similar advantage in not having powers of ϕ in our nobles (whenever possible).

The desire to have no powers of ϕ in our nobles, combined with the fact that f11/f5 and f11′/f5 are noble, and f5 = √5 = -1+2ϕ, forces f11 and f11′ to be -1+3ϕ and -2+3ϕ, which could be called (√5,1)-conjugates (as opposed to (1,√5)-conjugates) since they are +1/2 + 3/2×√5 and -1/2 + 3/2×√5. They are generally of the form -a+bϕ and -(b-a)+bϕ. We could call them "complements" rather than conjugates. And treat the prime symbol ′ as the complement operator, so that f11 = -1+3ϕ and f11′ = (-1+3ϕ)′ = -2+3ϕ.

I believe they are also the pair that are closest to each other in their real values. Or in other words, they are the pair that are closest to √11. One smaller one greater. But which should be called f11 and which f11′?

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia I'm still keen to discuss the best RTT normal form for feudal primes. Did you see my post beginning "I totally agree we want f11×f11′ = 11" above?

I've started a parallel version of my original Sagittal forum thread, using the -a+bϕ form that I propose above.

viewtopic.php?f=21&t=557

It eliminates powers of ϕ from the rationals without introducing them into the nobles as the conjugate pairs do.

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

Reply

Remove Preview

31 w

Dave KeenanAuthor

Those negative wooden-parts on my new normal-form split primes were driving me crazy. e.g. √5 = -1+2ϕ. Particularly since all the unsplit primes have positive wooden parts e.g. 2 = 2+0ϕ.

Then I remembered being puzzled by this MathWorld article on noble numbers: https://mathworld.wolfram.com/NobleNumber.html

where they say any nobles can be written as (a+bϕ⁻¹)/(c+dϕ⁻¹). Why do they use ϕ⁻¹ when it would be just as true if they had written it using ϕ?

Then I thought: What does √5 look like when we use the basis (1, ϕ⁻¹) instead of (1, ϕ)? Answer: 1+2ϕ⁻¹. It has both parts positive! As do all the other split primes whose form doesn't give powers of phi in the rationals or most nobles.

But I'm not going to type ϕ⁻¹ all the time. [Incidentally, I use WinCompose to type this stuff. It's great.] But I can't find a good single symbol for ϕ⁻¹. I mentioned that phi bar ϕ̅ is the conjugate of ϕ, i.e. -ϕ⁻¹. Unfortunately uppercase phi Φ (or "Phi" with a capital P) has also been used for the conjugate -ϕ⁻¹. I guess we could just declare that we are using uppercase phi Φ for the reciprocal ϕ⁻¹. Then we can write √5 = 1+2Φ.

f11 and f11′ are then 1+3Φ and 2+3Φ. They multiply to give 11 without any factors of Φ.

And the noble n5/4 (the noble mediant of 5/4 with its Stern-Brocot parent 4/3) is f11/f5 = (1+3Φ)/(1+2Φ) with no factors of Φ.

Reply

31 wEdited

Mike BattagliaAdmin

The only reason we care about this basis of "feudal primes" at all, rather than a basis of noble numbers, is because of this algebraic number theory perspective that, as you mention, is already neatly worked out for us. So given that, I think the conjugate pair method is the cleanest way to do it. It has several nice properties - it gives us all the naturals, the conjugate pairs are decently close in size, and we get lots of nice mathematical properties for free.

"The desire to have no powers of ϕ in our nobles, combined with the fact that f11/f5 and f11′/f5 are noble, and f5 = √5 = -1+2ϕ, forces f11 and f11′ to be -1+3ϕ and -2+3ϕ"

I don't see what you mean by this. Having f11 and f11' be 3+phi and 4-phi also has f11/f5 being noble and so on. As you already know, there is no absolute way to measure "how many powers of phi" some particular representative of a feudal prime class has, or at least not that I can see. Rather, we are arbitrarily picking a representative that we are *declaring* to have no powers of phi, or rather a zero "phi-coordinate," to use in our basis.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia We agree f5 = -1+2ϕ = √5. But you prefer f11 = 3+1ϕ and f11′= 4-1ϕ? What then is your prime exponent vector for the noble-mediant of 4/3 and 5/4 which has a real value of approximately 1.2764. We expect it to have +1 for its f11 coordinate and -1 for its f5 coordinate. But f11/f5 = (3+1ϕ)/(-1+2ϕ) ≈ 2.0652. It's too large by a factor of ϕ. So you're forced to have a phi-coordinate of -1 to make it right.

If instead we use f11 = -2+3ϕ and f11′ = -1+3ϕ, we find that (-2+3ϕ)/(-1+2ϕ) ≈1.2764. Zero phi-coordinate.

And if we want, we can rewrite those primes with Φ = ϕ⁻¹ instead of ϕ, to eliminate the minus signs. f5 = 1+2Φ, f11 = 1+3Φ, f11′ = 2+3Φ. We'll still have a zero phi-coordinate.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia Once we decide that f5 must be √5 = -1+2ϕ, so that 5 has no phi-coordinate, if we want those nobles that are the ratio of two primes to have no phi-coordinate, then that choice for f5 propagates to all the other split primes, all the way down the Stern-Brocot tree. The normal form of f5, and 2 nobles on level 4 of the SB-tree, determine the normal forms of f11 and f11′. Then the normal forms of f11 and f11′, and 4 nobles on level 5, determine the normal forms of f19, f19′, f31 and f31′, and so on.

BTW, Thanks to Douglas Blumeyer, I've realised it's bad to use the prime symbol ′ for these, since we will eventually want to write powers of these primes, like f11′³. So I'm switching to "f11" and "F11", pronounced "small eff eleven" and "big eff eleven". "Small" and "big" relate to their real values as well as their letters. The "f"s of both cases can be thought of as an abbreviation of both "factor of" or "feudal prime".

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan hm. That is quite interesting. Do you have a table of primes in this form? Are there other noble numbers that instead have no phi coordinate with the conjugate basis?

Reply

31 w

Mike BattagliaAdmin

Dave Keenan btw, the use of uppercase and lowercase phi makes it almost impossible to read in certain fonts, for instance the standard font that is used on FB Android. See picture

No photo description available.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan I think this is an artefact of the particular Unicode phi you are using. If you go with the standard greek Φ and φ it's much easier to read. See picture:

No photo description available.

Reply

31 w

Dave KeenanAuthor

You're right about Capital phi and phi symbol being nearly indistinguishable in many fonts. This is not a problem for \Phi and \phi in LaTeX. But the curly phi (\varphi in LaTeX) is not the standard math symbol for the golden ratio.

Fortunately I've stopped worrying about minus signs now, since I expect we'll mostly use the fF form of the non-wooden primes. So I have no need for capital phi and can continue to use the unicode that is supposed to be straight phi. But many fonts have straight phi and curly phi swapped, because that's how unicode had them up until version 3. This is all in my "Rant about ϕ" in my first feudal Sagittal forum article.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia You wrote: "Do you have a table of primes in this form?"

I assume you've seen the table in this thread by now:

viewtopic.php?p=4605#p4605

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

Reply

Remove Preview

31 wEdited

Dave KeenanAuthor

Mike Battaglia You wrote: "Are there other noble numbers that instead have no phi coordinate with the conjugate basis?"

Yes. If you use the (1, √5)-conjugate basis, only those nobles involving f5 = √5 = -1+2ϕ have a phi in them. You could even get rid of those if you split 5 into the conjugate pair f5 = 3-1ϕ and F5 = 2+1ϕ. But it makes no sense to split 5 like that because that f5 and F5 are associates. i.e. they are really the same prime, differing only by the unit ϕ².

It's purely the fact that we want f5 = F5 = √5 = -1+2ϕ (i.e. a single non-conjugate factor of 5) that propagates the non-conjugate form down the SB-tree to all the other primes.

Reply

31 w

Mike BattagliaAdmin

Dave Keenan, OK, I'm signing off for now, but a last thought for the night. Given that the algebraic number theory perspective seems decent at this point, here is a totally different method of analysis which I think is also useful, and possibly better, at least for my purposes:

We have agreed that the things we primarily care about are the rationals and noble numbers, or at least the simple ones. These are really the true "primitives" from which we are building arbitrary musical intervals. Other intervals can be formed as compounds of these.

Thus, for very clear and direct musical reasons, I should very much like to see the structure of how arbitrary feudal numbers decompose as a product of nobles and rationals, learning how these things all fit together in the same way I learned about JI. If you were going to hand me this theory and have me start actually using it to make music, figuring out this kind of decomposition, and how to move around, is the very first thing I would do.

The most basic question possible is this: what is *any* choice of basis for the noble numbers, which is comprised entirely of other noble numbers? This basis, plus the usual primes (at least those which are still prime in Z[phi]), will give us, let's call it an "Arthurian basis" which is entirely rational and noble and generates all of the feudal nobles.

It turns out that there is an easy way to compute such a basis. This only requires some way to list all of the noble numbers in terms of increasing "complexity", using pretty much any method you want. Your method of doing this using the Stern-Brocot tree is good enough to get started at least. So we start with phi, which is a noble number, and it goes into the basis. Then we go to 1+phi, which is not independent of phi, and thus it is not a new basis element. Then we have (2+phi)/(1+phi), which is independent, so it goes in. (2+phi) itself is next, and is not independent of the last, so it's not a new basis element. And so on. This gives us at least one choice of basis.

There is probably some way to generate a slightly cleaner list. After all, it is kind of weird that (2+phi)/(1+phi) makes it into the basis instead of (2+phi). This is only because it appears first in the way that we are ordering nobles. You could probably do it directly in terms of feudal primes - add them one by one if they are noble, and if not, try to divide somehow by the simplest possible denominator that does make it noble. Either way though, this all just serves to show that this method is possible.

There is also a very nice notation for the elements of this basis. This basis has the property that the continued fraction expansion of every element is either finite (if rational), or ends in a stream of 1's (if noble). In the first situation, the notation is to just write the rational. In the second, we look at the continued fraction and truncate it immediately *after* the first "1" in the tail. We then write that as a rational, followed by the symbol "~". So for instance, the noble mediant between 5/4 and 6/5 would be 6/5~. This works for everything except for phi, since phi and 1/phi would both be 1~, so we just use the symbol phi for it.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia As you know, at this stage in development I'm not particularly interested in any basis that isn't all primes, and I don't even see much advantage in those primes being noble, but I'm still keen to see where you take this idea.

I like your postfix tilde operator, which can be pronounced as "ennobled". I suggested a similar thing to Douglas back in 2020, namely to add a boolean to his computer implementation of prime-count vectors (monzos) to let them also represent nobles. I suggested that it should work so that 2/1~ = ϕ and 1/2~ = 1/ϕ, which has the intuitive reading that the octave ennobled is the phitave. I believe that is consistent with you saying that the noble mediant between 5/4 and 6/5 would be 6/5~. i.e. all 3 examples imply that we should use the ϕ-weighted one of the two mediends that I give in my tables.

viewtopic.php?p=4592#p4592

This also makes it the mediend that is closer in size to the noble.

But it's not the rational at the top of the zigzag. It's the second one down.

Noble frequency ratios as prime-count vectors in ℚ(√5) [a+bϕ] - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) [a+bϕ] - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) [a+bϕ] - The Sagittal forum

Reply

Remove Preview

31 wEdited

Mike BattagliaAdmin

Dave Keenan you know, I think that's technically right given what I wrote, but turns out that it also it isn't what I intended. I should clarify a bit...

Initially I had suggested looking at this in terms of continued fraction expansions, but it turns out to be slightly easier for our purposes to instead look at the Stern Brocot tree. Every noble number can be represented as an initial segment, followed by an infinite zig-zag pattern. The idea was to represent the noble number as a/b~, where a/b is where it *starts* zig-zagging, but there are a few different places where one can determine what we are calling the "start." I think the thing I originally suggested above is not exactly what was going for, so I will slightly modify my previous proposal to give the larger picture of what I was after.

We would like, given any rational number a/b, the two rationals ~a/b and a/b~ to be the noble numbers obtained by first going to a/b on the Stern-Brocot tree, and then starting a zig-zag pattern to the left and the right respectively. Every noble number will, in general, have several representations using this method: for instance, the noble mediant of 5/4 and 6/5 is equally ~5/4, 6/5~, ~11/9, 17/14~, and so on. Having the ~ on the left, in the case of 5/4, tells you that this is the unique noble number associated to 5/4 which is *flat* of 5/4, and having it on the right as in 6/5~ tells you it's the unique noble number which is *sharp*. I'll call this a "noble flat" and "noble sharp" respectively.

We can use this to develop two unique representations for any noble number: we can look at the *simplest* representation of it as a noble flat, and the simplest as a noble sharp. So the two representations ~5/4 and 6/5~ are the unique representatives for that blue noble number from before. For phi, we get the two representations 1/1~ and ~2/1, and for 1/phi we have ~1/1 and 1/2~. We also have that for any noble representation a/b~, its reciprocal is ~b/a.

If we want a unique representative for the entire thing, we could, for instance, take the simpler of both. So the unique representative of that noble blue note would then be ~5/4. This is the absolute first point where the zig zag occurs on the Stern-Brocot tree. For whatever reason, the convention last night that I suggested would have us take the more complex of the two instead, which is the first rational in the zig zag that changes direction. I don't remember why I thought this was better last night.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia Isn't "two unique representations" a contradiction in terms? 🙂

I liked your right or postfix tilde. I don't like the left or prefix tilde because that already has at least two meanings in RTT, namely "approximate" and "Grassmann complement".

I'd much prefer to name nobles using a single one-to-one and onto mapping (bijection) between rationals (excluding 1/1) and nobles. That's what I proposed last night, thinking I was agreeing with you.

That would greatly reduce the cognitive load in figuring out what noble someone is referring to, and would allow people to memorise the cents values of the simpler nobles given as ennobled ratios.

The SB-tree is a binary tree*, so every node (every rational) has a left and a right child, but it only has one parent (except for 1/1 which has none). So the way to obtain a bijection is to define the ennoblement of a rational as the noble mediant it makes with its /parent/. Yes, this will be the same as the one it makes with one of its children, but we don't need to worry about that, or whether it is the right or the left.

Yes, that makes it the second rational from the top of the zigzag corresponding to a given noble. Or the first corner, as you say. But that does not give the more complex of the two noble as you claim above. It gives the /simpler/.

[EDIT: * When you strip away all the dashed or dotted lines people like to decorate it with.]

Reply

31 wEdited

Dave KeenanAuthor

Some possible abbreviations or symbols for "the noble that 6/5 makes with its SB-parent" are:

6/5~

n6/5

6n5

The postfix tilde has the advantage that it can be treated as an operator and used with a rational variable or expression. But the "n"s could be associated with a function called "noble()".

What's the simplest way to compute the rational that is the SB-parent of a given rational?

Reply

31 w

Mike BattagliaAdmin

Dave Keenan I like the way that I was doing it, because it's easy to see if the noble number we are talking about is located to the left or the right of the rational in question. It isn't difficult to define a bijection if that's what you want - I already wrote two different ways to do it. I'd like some notation for my "noble sharp" and "noble flat" thing but I'm not really all that married to the ~ sign. We could do 6/5< and 6/5> or something.

Reply

31 w

Mike BattagliaAdmin

Dave Keenan I'll write more later, but a quick note here that there is more to what we were calling the "arithmetic norm" than meets the eye - in particular the thing Wolfram calls AlgebraicNumberNorm[x] is not the same as the thing I was talking about before. I will look at my notes and see if I can remember how I was extending Tenney height. That expression a^2 + ab - b^2 works well for any feudal integer but can be kind of subtle.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan I'll just make some notes as I read them:

The thing we were calling the "arithmetic norm" turns out to be a^2 + ab - b^2 for the feudal integers, and we may take the absolute value to get the unsigned version. In fact, it is slightly more useful if we take the square root to get sqrt(|a^2 + ab - b^2|).

Using this expression, the norm for any standard prime is the prime itself, the norm for sqrt(5) is sqrt(5), and the norm for f11 and f11' is sqrt(11). And so on. The norm for phi is 1 (actually the unsigned version would be i), which is kind of weird, but we'll ignore this for now.

So right away, the big question then is: *is the norm of any feudal integer the product of the norms of the primes?* Put another way: is the norm of any product of feudal integer the product of the norms, given that we ignore the sign and normalize properly? EDIT: Turns out that indeed it is.

If so, then we have something generalizing Tenney height on the feudal integers, which we can easily extend to all feudal numbers, minus the quirk of phi having a norm of 1 which is easily adjusted.

Lastly, I note Wolfram has some of this implemented as AlgebraicNumberNorm[...], but it is very quirky. You need to take the square root of AlgebraicNumberNorm[a+b*GoldenRatio] whenever b is nonzero, and just naively doing AlgebraicNumberNorm[p/q] also just gives p/q, rather than max(p,q) or p*q or etc. sqrt(|a^2+ab-b^2|) seems to give decent results, though.

Reply

31 wEdited

Mike BattagliaAdmin

Lastly I note that there is also this notion of a "height function" on an arbitrary algebraic number field. Some of these height functions I'm looking at are very strange and I'm not sure if they make sense here. They also seem to generalize the max(n,d) height rather than n*d height, but maybe some of them are useful (or even agree with the above, except for the weird quirk involving phi).

Reply

31 w

Mike BattagliaAdmin

Dave Keenan, after thinking about this, there turns out to be a much simpler and possibly better generalization of Tenney height for our purposes.

Once some basis of feudal primes is chosen, one can simply take the "weight" of each prime to be the log of the prime itself. We can then put things into "weighted coordinates," the same way that we do with regular monzos. Then everything derives naturally from that: the L1 norm of this weighted monzo gives us a natural measure of the complexity of each feudal number, and the JIP is <1 1 1 1 1 1 ...| in the weighted basis, and so on.

This gives very similar results to the sqrt(|a^2 + ab - b^2|) method with one quirk. sqrt(5) has the same complexity in both, as do the regular primes. The two factors of 11, using the first method, both have complexity equal to log2(sqrt(11)) ≈ 1.730, whereas with this method we have f11 and F11 are two close values which average to 1.730. For instance, using your basis we have log2(-2+3*Phi) ≈ 1.513 and log2(-1+3*Phi) = 1.946.

The only quirk is that with the original method, we had that phi has a weight of 0 (which we don't have here). We could perhaps grandfather in a weight of sqrt(5) for phi as well, treating sqrt(5) as its "unofficial" prime pair (it seems reasonable that phi and sqrt(5) would be pairs in Q[sqrt(5)] in some sense). But with this newer method you get that all for free.

Lastly, I note that the "conjugate basis" I suggested gives us values which are slightly further apart in value for f11 and F11. We have log2(3+Phi) ≈ 2.207 and log2(4-Phi) ≈ 1.252. These still average to ~1.730, but the two primes that you choose (-2+3*Phi) and (-1+3*Phi) are slightly closer. Does that uniquely characterize the choice of primes in your preferred basis? Will they always be the two closest in size?

Reply

31 w

Dave KeenanAuthor

Mike Battaglia Yes, I believe the choice of primes in my preferred basis will always be the two closest in size, and therefore closest to the square root. I mentioned this several times previously.

Reply

31 w

Mike BattagliaAdmin

There is no need to be rude, but thanks, I guess.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia I apologise.

Reply

31 wEdited

Dave KeenanAuthor

I spent most of the day converting all my articles, with all their tables and diagrams, over to the new choice of fundamental primes. Very tedious, and not yet complete.

Now that I've had a chance to digest your work above. I agree that the log of the real value of the prime has just as much claim, if not more, than the log of the square root of the absolute value of the arithmetic norm, as the generalisation of our usual log of primes, for computing complexities, provided we use the basis where fp and Fp are as close as possible to √p, because p is the absolute value of the arithmetic norm of fp and Fp. As you say, it's not clear what it means psychoacoustically to compare complexities between rationals and nobles, but we can go with log-of-real-value for now.

BTW, when we need to distinguish this norm, a²+ab-b², from the various p-norms we use on our vectors, we could just call it the "feudal norm". That's 2 less syllables than "arithMETic norm", and a whole lot less than "quadratic field norm" or "algebraic number norm", which, as you say, may not even be the same thing. I thought of yet another reason why "feudal" is a good name for ℚ(√5) and ℤ[ϕ]: It sounds a bit like "ϕ-dal".

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan sounds good to me. So we can also use this to generalize the TE norm, and thus we get the entire theory powering Graham's temperament search thus generalized to feudal numbers. This will let us search for good feudal temperaments and let us answer your original question about what are some good feudal commas. The only thing is that I think this choice of weighting may give very strange results: for instance, f11 and F11 are both weighted more simply than 5/1 and 7/1, and the simplest element of all is phi, simpler even than 2/1. This way of prioritizing intervals may give us an "optimal" tuning for some given temperament that is less than optimal, as far as our ears are concerned, and thus it will also skew the results of the temperament search. But I will try it when I get a second and give the results.

I will just add the only real thing left is to figure out this complexity function; I need to know what weighting matrix to put into the temperament search which will give sensible results. Note the weighting matrix need not be diagonal, which in plain terms is equivalent to saying that we could run the search relative to a different basis internally, even if we keep using the feudal prime basis for notation. We just need some L2 norm whose elliptical unit sphere approximates whatever we view as the true complexity function. I will give some thought to this.

Reply

31 w

רועי סיני

Dave Keenan In fact, your choice of primes is not always the two closest in size. For example, F29/f29 is 1.733..., while (f29ϕ)/(F29/ϕ) is 1.510...

However, they are always either the best or the second best choice, because they are both between bϕ-b = b/ϕ and bϕ, where b is their equal golden part, which means the quotient between them is less than ϕ², and any time you multiply one of them by ϕ and divide the other by ϕ you either multiply or divide the directed quotient by ϕ². Anyway, I don't think this is a large enough issue to use the other choice instead in this case, because of the consistent normal forms they have.

Reply

19 h

Mike BattagliaAdmin

Also a quick note that this also easily generalizes to things like "Kees Expressibility" and etc of feudal numbers.

However, I would also add that it is a little bit strange that we are going to have an interval like (-1+3*Phi) have a lower complexity than 11 itself. Part of this is because naively putting an L1 norm on a basis of "feudal primes" may not really represent perceived interval complexity (which is one reason I was interested in the basis of primes and noble mediants). It is a pretty good question to ask what criteria we should even use to compare noble and just intervals at all.

So I guess for now, my point is that the log(prime) weighting is "at least as good" as the sqrt(|b²+ab-a²|) weighting, being approximately equivalent but slightly better behaved, with a clearer choice of weight for phi, but also with the same shortcomings for both.

That being said, some of this is unavoidable if we want even the basic bare minimum of linear algebra w/ normed vector spaces to work - we must have that 5/1 has double the complexity of sqrt(5/1), for instance. But I don't think that particular choice of weight will affect the results of any temperament search much. The better question is what to do with things like f11 and F11.

Reply

31 wEdited

Mike BattagliaAdmin

I note that weighting phi as log(phi) gives some extremely weird results; phi has a lower complexity than any rational number, and using this weighting temperament searches tend to rank 30 higher than 31 for this reason (due to the extreme emphasis on phi).

Reply

31 w

Paul ErlichAdmin

Mike Battaglia I believe that's how this search is carried out. I do see a 30 and no 31, but the higher-ranked results look fine, make sense, and represent some appealing solutions. It all depends on how you want to use phi, with what timbres, volume/distortion, etc. how much importance it and its precise tuning should get IMO.

http://x31eq.com/cgi-bin/more.cgi?r=1...

X31EQ.COM

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 1 Temperaments

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 1 Temperaments

Reply

31 w

Paul ErlichAdmin

Mike Battagliaa difference with the results you posted is that 27 is not showing up for me here for some reason. Any idea why that is?

Reply

31 w

Paul ErlichAdmin

(Also worth noting is that 62 = 31×2 shows up pretty strongly)

Reply

31 w

Mike BattagliaAdmin

Paul Erlich the weighting on phi is so extreme that phi is weighted stronger than any rational number, so I decided to try some other weightings. Graham's algorithm just weights each of these basis elements by their size, so one simple way to get a different weighting is just to replace phi in the basis with 2*phi, so that it is weighted somewhere between 2 and 3. So if you do that, replacing 833 cents with 2033 cents, 27 is at the top: http://x31eq.com/cgi-bin/more.cgi?r=1...

X31EQ.COM

1200.000, 2033.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 1 Temperaments

1200.000, 2033.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 1 Temperaments

Reply

31 w

Steve Martin

Just to say I have a book on quadratic fields, reading up now so maybe I can help, if only by reaching the point where I can say I agree with you guys ...

Reply

31 w

Steve Martin

Hi Dave, I have read your article, blog, this thread and the 2.5 chapters of Hardy&Wright that cover quadratic fields. Everything looks correct, I like the way ℚ(√5) is a UFD and has its own units (countably many of them!) and primes. Your canonical choice of primes is neat and is not in H&W. The annotated diagram is very helpful.

I'd need to think more to see how to use this, and I'll look at Zest24 too. One question to start: is every noble number a quotient of two primes? Would the quotient of two composites (or a composite and a prime) not be a noble? I've not understood this.

Reply

30 w

Dave KeenanAuthor

Steve Martin, Thanks for your interest, and thanks for checking my work. At this time it remains a conjecture on my part, that every noble number is a quotient of two feudal primes-or-units. Apart from the prince of nobles (ϕ itself), I've found nobles to always be ratios between two feudal integers taken from the set of non-wooden canonical primes (the fF-primes) plus the units f1 and F1 (ϕ⁻¹ and ϕ¹). And beyond the prince and the duke (ϕ and ϕ²) I've only seen ratios between two big F's or two small f's (where √5 counts as either f5 or F5). I conjecture that this will continue ad infinitum. But it's more constrained than that. You can't just choose any pair of f's or F's and assume you'll get a noble. This is due to the "unimodular" requirement — cross-multiply and subtract on the (a+bϕ)/(n+mϕ) form and you must get am-bn = ±1.

Reply

30 wEdited

Steve Martin

Dave Keenan thx, it's good to be reminded of the unimodular requirement - which is not mentioned in the Wolfram page unless I missed it - but I guess is easily proved to be equivalent to the continued fraction definition?

You are right, Hardy & Wright do not discuss noble numbers.

Reply

30 w

Paul ErlichAdmin

Steve Martinit is implied on the Wolfram page in that the 4 coefficients are required to be numerators and denominators of successive convergents.

Reply

30 w

Steve Martin

Paul Erlich ah yes I see

Reply

30 w

Steve Martin

Paul Erlich what is the syntax to use more.cgi ? I think there is no UI page for it, could you re-post an example please (as I am stuck on phone where search history is hard)

Reply

30 w

Paul ErlichAdmin

Steve Martinlike this?

http://x31eq.com/cgi-bin/more.cgi?r=1...

X31EQ.COM

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 1 Temperaments

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 1 Temperaments

Reply

30 w

Paul ErlichAdmin

Steve Martin you can get to it in the UI by clicking "show more of these " e.g. here:

http://x31eq.com/cgi-bin/pregular.cgi...

which you get to from here:

http://x31eq.com/temper/pregular.html

by typing the relevant numbers into the boxes.

X31EQ.COM

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Regular Temperaments

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Regular Temperaments

Reply

30 w

Dave KeenanAuthor

Steve Martin Thanks for checking Hardy & Wright. I take it they don't discuss noble numbers as a subset of ℚ(√5). The only place I've seen anything about that is the MathWorld article: https://mathworld.wolfram.com/NobleNumber.html

I've ordered a copy of the Manfred Schroeder book referenced there.

Wolfram MathWorld: The Web's Most Extensive Mathematics Resource

MATHWORLD.WOLFRAM.COM

Wolfram MathWorld: The Web's Most Extensive Mathematics Resource

Wolfram MathWorld: The Web's Most Extensive Mathematics Resource

Reply

Remove Preview

30 w

Paul ErlichAdmin

Dave Keenangreat book!!!!!!!!

Reply

30 w

Dave KeenanAuthor

Paul Erlich If you have the Schroeder book, perhaps you can tell us if it sheds any light on my conjecture that every noble number is a quotient of two primes or units of ℚ(√5)?

But only /some/ quotients of two primes or units are noble numbers.

Reply

30 wEdited

Paul ErlichAdmin

Dave Keenanit's deep in storage! 🤣

Reply

30 w

Steve Martin

Yes, thank you!

Reply

30 w

Paul ErlichAdmin

Steve Martin rank-2

http://x31eq.com/cgi-bin/more.cgi?r=2...

X31EQ.COM

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 2 Temperaments

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 2 Temperaments

Reply

30 w

Steve Martin

Thanks again, Paul, I'd not used that one before

Reply

30 w

Steve Martin

Here's an example using the parameters I wanted to try:

http://x31eq.com/cgi-bin/rt.cgi?ets=17_43...

X31EQ.COM

1200.000, 1901.955, 1393.157, 833.090, 3368.826, 1815.643, 2335.673-limit Regular Temperament

1200.000, 1901.955, 1393.157, 833.090, 3368.826, 1815.643, 2335.673-limit Regular Temperament

Reply

30 w

We need a UFD to make the whole monzo thing work. Right?

Reply

31 wEdited

Mike BattagliaAdmin

OK, well, I'm not sure what to say - it seems like you are mostly just defending what you already have... I think it's all interesting and was trying to explore a little bit further. All I am really asking are some basic questions, which I think are immediately apparent from all of this:

1) Given some p-limit, what noble mediants/feudal numbers/etc can we actually generate from just the intervals in that p-limit?

2) If we take all of the JI ratios and noble mediants and throw them into a blender, do we actually get the feudal numbers or just some of them?

3) Do the other feudal numbers have some musical use?

I don't know how to answer them, but I certainly think these are interesting questions that are worth looking into.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia: Ah. Thanks for the specific example question. In that specific case, and at this point in time, there is almost certainly no advantage in factoring that N. But if N was instead the noble mediant between 4/3 and 5/4, which can be written as (3ϕ-2)/(2ϕ-1), then it would be worth factoring. Why? Because a feudal monzo would replace 5 in the basis with 2ϕ-1 = √5, which is also the numerator of N, and it would replace 11 in the basis with its two factors 3ϕ-2 and 3ϕ-1, one of which is also the denominator of N. So the basis encompassing 11-limit JI, ϕ and N would be 2.3.(2ϕ-1).7.(3ϕ-2).(3ϕ-1).ϕ. Still only 7 entries.

And if we already had a list of low badness noble/rational commas to choose from, having standardised the choice of associates for the feudal primes, there might be an advantage in factoring even your original N, and we would still replace 5 and 11 in the basis with their feudal factors.

Reply

31 wEdited

Dave KeenanAuthor

If we didn't replace 5 and 11 in the basis with their feudal factors, we wouldn't have a UFD. There would be more than one way to make 5 or 11.

Reply

31 w

Mike BattagliaAdmin

Dave Keenan that's an interesting example. I guess it's the same as something like the 2.9.5/3 subgroup. In that case, "9" happens to factor as a square of the denominator of 5/3. So you would make it 2.3.5. But that doesn't make 2.9.5/3 somehow not a UFD (or rather not a free abelian group, since we aren't talking about rings). It just means that 2.9.5/3 is a subgroup of finite index within the 2.3.5 group, meaning it is basically a dilation of it. The 2.3.5.7.11.phi.N subgroup is similar and I guess is a subgroup of 2.3.(2phi-1).7.(3phi-2).(3phi-1).phi.

Reply

31 w

Mike BattagliaAdmin

2.3.5 is an even better example. 5, treated as a feudal number, is now composite. Do we really need to factor it into 2.3.√5, though? Of course not, and similarly we don't need to always factor noble mediants. Of course, if we add several noble mediants to a subgroup and doing this somehow automatically causes various primes to split even before we've done the factoring into numerator and denominator, then you just get that for free.

Reply

31 wEdited

Dave KeenanAuthor

Right. You have to check if any noble you're adding has what I call f5 or f11 or f11′ as factors, and split 5 and/or 11 as the case may be. The next splitter doesn't come until ordinary prime 19 splits into f19 and f19′.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan well what I'm saying is I don't think it is really required that you split it, just like you don't have to split 9 in the 2.9.5/3 subgroup. In your example you also don't have to split it, and if you don't, you just get some subgroup which isn't "saturated" relative to the entire group of feudal numbers, or which is "enfactored" or "impure" or whatever you are calling it these days. The saturation of the non-split version will be the one which is split.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia I just want you to know that I really appreciate you wrestling with this stuff with me. You're probably the only person who could grasp it so fully and so quickly. It's great that you did some investigation previously.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan Thanks man. I originally got interested in this stuff from your noble mediant paper many years ago, so I'm really just continuing your work, I guess. I similarly appreciate the wrestling, probably more than you do 🙂

Reply

31 w

Paul ErlichAdmin

The major seventeenth occurs quite often on chords on the piano or organ let alone any multi-instrument combo, where there's usually a bass player several octaves below the melody (let alone any upper vocal harmonies, string arrangements, etc.) or at least say a guitar accompanying a higher instrument or voice. So e.g. 5:10:15:20:24 and 4:8:12:16:19 are competing (in terms of minimizing 5HE) tunings for a very common voicing of the minor triad in real music. And 1:2:3:4:5 dominates for a very common voicing of the major triad. As a pianist these are the voicings I instinctively grab on the keyboard. Oddly 1:5's tuning may matter most when the upper note's fundamental is quiet enough to just cancel out the 5th harmonic of the lower note once per cycle as they beat against one another.

Reply

31 wEdited

Paul ErlichAdmin

I know Gene wrote a lot about manipulating noble numbers and other aspects of this topic, but wouldn't know where to find that in the archives . . .

Reply

31 w

Paul ErlichAdmin

For example discussing the golden meantone generator

v = 2^[(3Φ+1)/(5Φ+2)]

he explained that the exponent of 2 can be simplified further.

Gene Ward Smith wrote (Yahoo tuning-math group message 2507 (Mon Jan 7, 2002 7:31 pm):

"Ratios of the sort (a+br)/(c+dr) define an algebraic number field, which can always be put into the form of a sum of rational numbers times powers of a single algebraic number r. In this case, that results in

(a+br)/(c+dr) = (ac+ad-bd + (bc-ad)r)/(c^2+cd-d^2)

This form of the algebraic numbers in the field Q(r) is unique, since {1, r} are a basis for a vector space over the rationals Q; hence we can determine if two elements of Q(r) are the same by putting them both into this form."

If we rearrange the exponent of 2 (in the ratio for v) so that it reads (1+3Φ)/(2+5Φ), and plug the values a=1, b=3, c=2, d=5, and Φ=r, into Gene's equation, the resulting answer is (-8+Φ)/-11, which can be simplified to (8-Φ)/11.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia: This empirical statistic may be relevant to your recent questions: At any given depth (order-limit) of the SB-tree, the number of feudal primes needed to generate all the nobles is about half the number of nobles. Every noble needs at most 2 primes, but every prime is used by about 4 nobles (typically 2 uses as numerator and 2 uses as denominator). That seems to me like a reason to factor the nobles.

Reply

31 wEdited

Dave KeenanAuthor

Here's another thought. Sure you don't need primes in your monzo basis, but you need to know that your basis is linearly independent. How do you check that? You express each vector of your non-prime basis as a vector in a prime basis, because we know the primes are independent.

So whether you have composite nobles in your basis or not, you'll still need a prime basis to check their independence. And it would be good to standardise those feudal primes.

Reply

31 w

Dave KeenanAuthor

Some thoughts on your 3 questions:

1) Given some p-limit, what noble mediants/feudal numbers/etc can we actually generate from just the intervals in that p-limit?

It is mathematically interesting if the set of noble mediants of p-limit ratios is finite, despite the fact that the number of p-limit ratios is infinite. I didn't expect that. But in musical terms, I'd be more interested in the noble mediants of some kind of integer-limited or complexity-limited subset of the p-limit.

It may be educational to take a copy of Erv's scale tree, which has more levels than the sideways SB-tree in my feudal posts, and circle all the 5-limit ratios. Then embolden the line between any two that are adjacent, and circle the noble that this corresponds to. Putting it another way: If a zigzag path has two consecutive 5-limit ratios anywhere along it, the noble corresponding to that zigzag lives. Then see if there's any pattern to the nobles that survive.

2) If we take all of the JI ratios and noble mediants and throw them into a blender, do we actually get the feudal numbers or just some of them?

I take this to mean: Can we get all the feudal numbers as products and quotients of rationals and nobles? i.e. Can we get all feudal numbers as sums and differences of the feudal-prime-count-vectors of rationals and nobles. Sorry I was so slow to understand this question. I guess it's because I don't understand why it matters (except as an interesting math question). I'm just happy that it works the other way round, namely that by being able to represent all feudal numbers, we get all nobles and rationals (and their products and quotients).

But I will give this some thought, and see if I can find a proof or counterexample to answer your question.

3) Do the other feudal numbers have some musical use?

If there are no feudal numbers other than those that can be obtained as products and quotients of rationals and nobles, the question is moot. If there are any, I can't conceive of any musical use for them, but I'd be willing to tolerate them in order to have a prime basis for vectors that can encode both rationals and nobles.

Reply

31 w

Dave KeenanAuthor

Here is a sketch of a proof that all feudal numbers can be generated as products or quotients of nobles and rationals. It suffices to prove that all feudal primes can be so generated. If we can obtain all feudal primes, we know we can can generate all feudal numbers as products and quotients of those.

Some statements below are merely observations from the nobles examined so far, and require their own proofs in general. Hence calling this a "sketch".

From Dekker we know that feudal primes fall into 3 classes. (a) Those which are ordinary primes (2 and those ending in 3 or 7), (b) the prime factor that's squared to make 5, (c) those primes which are one of a pair of complementary prime factors of ordinary primes ending in 1 or 9.

The rationals directly supply us with the primes of class (a). The nobles supply us with the primes of classes (b) and (c) as follows.

It will help to look at this table: viewtopic.php?p=4592#p4592

The nobles of the top two levels of the SB-tree consist only of units, no primes. All noble numbers beyond level 2 have a prime as their numerator, and a prime-or-unit as their denominator. As one descends the SB-tree, each new level of nobles has, as its denominators, only those primes-or-units that have appeared as numerators on previous levels. New primes always appear first as numerators. All units (red) can be treated as equivalent.

Therefore every noble can be reduced to only the prime in its numerator (and possibly some unit), by multiplication by an ascending series of nobles from levels above it. i.e. Cancel its denominator with the numerator of a noble from a previous level, and then cancel /its/ denominator with the numerator of a noble from a level above /that/, and so on, until reaching a noble with a unit denominator.

e.g. to obtain the prime 7+5ϕ we can multiply the following series of ascending nobles.

(7+5ϕ)/(4+3ϕ) × (4+3ϕ)/(3+2ϕ) × (3+2ϕ)/(2+1ϕ) × (2+1ϕ)/(1+0ϕ)

Here's a shorter path

(7+5ϕ)/(3+2ϕ) × (3+2ϕ)/(1+1ϕ) x (1+1ϕ)/(1+0ϕ)

The prime of class (b) appears as a numerator on level 3. What remains is to show that the Stern-Brocot tree will eventually include every feudal prime in class (c) as the numerator of some noble. I don't have a proof for that, but it seems to be on track so far, generating primes of increasing complexity as we descend. Any gap left on one level seems to be filled in on the next level, such that it seems likely there will be some ordering of the class (c) primes that is generated in sequence, without gaps, by some traversal of the SB-tree.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan wow! Ok, that makes sense. Great stuff, thank you!

I guess an immediate follow up question is: can we somehow stratify the feudal numbers into classes based on how "close" to noble they are?

I am thinking of this primarily for musical reasons. Basically noble numbers are important. "Justly transposed noble numbers," which are a product of a rational and a noble number, are probably of equal importance to the nobles themselves, unless your music never changes chords or tonics at all. In fact for now let's grant them honorary "noble enough" status. Then there are those numbers which are a product of two noble numbers and perhaps a rational, and then those which are a product of three noble numbers and perhaps a rational, etc. These numbers will be less musically frequent unless you like to do very strange modulations directly by a noble number or something like that.

So in general we can ask about the shortest such decomposition. Is there some easy way for us to determine this "compound nobility rank" for any feudal number, or at least for the feudal integers or primes?

I have this strange feeling that perhaps the determinant is involved here: for (a+b*phi)/(c+d*phi), it's noble iff ad-bc = ±1, so maybe the other determinants mean something relevant to this question. For feudal integers we have (a+b*phi)/(1+0*phi), so the determinant is just -b; again that b coefficient seems very relevant.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia No worries mate. (I think that translates into American as "You're welcome"). 😉 I agree that determinant seems like it should represent some kind of nearness to nobility. There is a concept called near-noble numbers, defined in terms of continued fractions, here:

https://archive.lib.msu.edu/crcmath/mat ... n/n039.htm

I don't think it's related.

Near Noble Number

ARCHIVE.LIB.MSU.EDU

Near Noble Number

Near Noble Number

Reply

Remove Preview

31 wEdited

Mike BattagliaAdmin

Dave Keenan here is an interesting thought to perhaps supply the missing part of the proof.

Suppose we have an arbitrary feudal integer of the form a+b*phi in which a and b are coprime. We would like to determine some sequence of nobles whose product is our integer. To generate the first noble, we look for c, d such that the 2x2 matrix [a b;c d] has determinant 1 (or -1). This means we want

ad - bc = 1

We can easily locate the values for c and d using the modular inverse. If we take the above expression mod a and b, we get the two following expressions

ad = 1 mod b

-bc = 1 mod a

Since a is known, and a is coprime to b, d is the unique modular inverse of a mod b, and c the unique modular inverse of -b mod a.

As a first step, we have turned (a+b*phi) into (a+b*phi)/(c+d*phi) * (c+d*phi). The key thing is that we get two values of c and d which are strictly smaller than b and a. So now we can repeat the same thing with (c+d*phi), decomposing it as (c+d*phi)/(e+f*phi) * (e+f*phi) in which e and f are smaller than d and c, and then repeat again.

We can formalize this as a table. The first row are our numbers a and b, and then we keep taking modular inverses in the way above for the next row. Then we repeat, building each time the next new row from the last previous row. If we start with 23+29*phi, we get the following table:

23 29

19 24

15 19

11 14

7 9

3 4

2 3

1 2

0 1

This is the entire decomposition: (23+29*phi)/(19+24*phi) * (19+24*phi)/(15+19*phi) * ... * (1+2*phi)/(0+1*phi) * (0+1*phi). All noble, because each 2x2 submatrix of the above table has determinant 1.

The main thing is that this expression terminated at 0+1*phi. Are we guaranteed to get this, or something like it, for all initial a+b*phi as long as a and b are coprime?

Reply

31 w

Dave KeenanAuthor

Mike Battaglia Well done! I can't answer your final question, but my engineering attitude says, I've pushed down the SB-tree a little beyond where I think it's musically relevant, and it all works, so that's good enough for me. 🙂

Reply

31 w

רועי סיני

Mike Battaglia Very cool method! If you're still interested in this question now, and for future readers, the process indeed always ends in either 1 or phi if a and b are coprime. This is just because if it doesn't end there then you can continue the process more, and the gcd always stays 1 because of the determinants of the 2x2 submatrices.

Also, this option to choose one pair of modular reciprocal and a negative of a reciprocal over another gives you noble triads, which Dave Keenan has expressed interest in, because if the matrix

a b

c d

has determinant 1, then so does

a b

a - c b - d

and also

c d

a - c b - d

which means that (a-c+(b-d)ϕ):(c+dϕ):(a+bϕ) is a noble triad.

Also, this process seems to disprove the conjecture stated in viewtopic.php?p=4620#p4620 about the noble mediants always being a quotient of two primes, because if you take (-1+7ϕ)*ϕ³ = 13+19ϕ, which is not a prime but an associate of f5*f11, you still get a noble number with it as a numerator, namely (13+19ϕ)/(11+16ϕ).

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Reply

20 hEdited

Dave KeenanAuthor

I'm concerned to standardise a normal form for the split primes as soon as is decent, because the later we do it, the more people will have read my articles, and the more will have to unlearn the old ones and learn the new ones, in the case that I decide to change them.

My current form, where f5 is 2+1ϕ and f11 and f11′ are 3+1ϕ and 3+2ϕ, does the following when you split the primes for a 13-limit vector:

[ a b c d e f ⟩

goes to

[ a b 2c d e e f; -2(c+e) 0 0 ... ⟩

where the entry after the semicolon is the ϕ-count.

I figure it would be better if it went like this:

[ a b c d e f ⟩

goes to

[ a b 2c d e e f; 0 0 0 ... ⟩

i.e. no power of phi is introduced merely by splitting primes.

To achieve that, we need f5 to be -1+2ϕ and f11 and f11′ to be -2+3ϕ and -1+3ϕ.

But they are more compactly written if flipped horizontally so f5 is 2ϕ-1 and f11 and f11′ are 3ϕ-2 and 3ϕ-1.

What do you think?

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan I would always recommend a+b*phi even if a is negative; I almost misread one of your earlier posts making that mistake when you wrote it as b*phi+a.

Beyond that I think the cart is way ahead of the horse when it comes to standardized conventions and etc; we are just now starting to get the absolute basics of a structure theorem together and that will probably make clear what the best representative of each feudal prime should be.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia IIUYC, the feudal naturals ℕ[ϕ] are those feudal numbers whose feudal-prime-count-vector has no negative counts and zero for the ϕ-count. But that depends on which of each set of associated primes we choose for our prime basis.

Is it too early then to decide that we'd like all ordinary naturals 1, 2, 3, 4, 5, ... to also be feudal naturals?

Reply

31 w

Mike BattagliaAdmin

Yeah, exactly. We get different feudal naturals based on different choices of prime, which I missed when writing this idea for the first time. There is a "maximal" possible set of naturals we could get based on if we choose the unique representative between 1 and phi for each prime, and all other possible choices for the primes give some subset of this maximal set of naturals. But as you mention, this is probably not right, at least not unless we want 3/phi^2 to be the representative for 3. But the idea to make the ordinary naturals also feudal naturals is intriguing. Does that lead to some particular choice of basis?

Of course if we choose any particular basis such that the true naturals all appear with positive exponents, then we can multiply everything in that basis by 1/phi and get another basis where the naturals still have all positive exponents, just with some extra factors of phi thrown in. So we can ask about the phi-coordinate for each natural. Is there some basis we can choose so that the phi coordinate is 0 for each of the ordinary naturals?

I guess this requirement would correspond, for instance, to your choice of sqrt(5) = (-1 + 2*phi) for that particular feudal prime. But is there some way we can use this to choose, for instance, what you are calling f11 and f11'?

Reply

31 w

Mike BattagliaAdmin

I think the only question is basically the primes in your third class stated previously "those primes which are one of a pair of complementary prime factors of ordinary primes ending in 1 or 9." So let's call these two primes, whatever they are f11 and f11'. It would be nice if we had f11 * f11' = 11, rather than f11 * f11' * phi = 11 or anything like that. The only problem is that given any choice of f11 and f11' which have that property, we can replace f11 with f11*phi and f11' with f11'/phi to get another pair of representatives which multiply to 11. Is it possible to choose some kind of unique pair which are, perhaps, conjugates of one another, either in the (1, phi) basis or the (1, sqrt(5)) basis?

Reply

31 w

Mike BattagliaAdmin

Dave Keenan Here is a great example - in your original post about this, you suggest f11 = (-2*phi + 3) and f11' = (-1*phi + 3).

However, we could also choose f11 = (3+phi) and f11' = (4-phi), and we still have f11*f11' = 11 without any extra phi coordinate.

This is a very natural choice: since you are deriving all of this from the algebraic number theory view, we have that

3+phi = 7/2+sqrt(5)/2

4-phi = 7/2-sqrt(5)/2

so that these two are conjugates of one another in Q[sqrt(5)]. I think that's about as good of a pair as you could possibly want. In this case, these two representatives also have the benefit of both being noble.

Do these always come in conjugate pairs for each prime? If so, these pairs will always be of the form a + b*phi and (a+1) - b*phi. And if things are that neat, and we are guaranteed some pair like that for each relevant prime, I think that's about all she wrote! That would be a perfect basis, probably as good and standardized as you can get from an algebraic number theory view. And I would be very curious if the "b" coefficient has any interesting interpretation - clearly if it's 1 it's noble, and what otherwise?

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia I don't think so, because you can always multiply any choice for f11 by ϕ and multiply f11′ by ϕ⁻¹ and keep the overall ϕ-exponent for 11 being zero. We'd need to introduce some other constraint, preferably one that gives us some other desirable property. One such is that nobles should also have a zero ϕ-count, whenever possible. And since f11/f5 and f11′/f5 are nobles, and we want f5 to be -1+2ϕ, I believe that constrains f11 and f11′ to be -1+3ϕ and -2+3ϕ, not necessarily in that order.

Reply

31 w

Mike BattagliaAdmin

Dave Keenan I'm not sure what you're talking about... Did you see my post above about the primes for f11 and f11' being in conjugate pairs?

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia No. I did not see your previous two in this subthread until now. I was responding to the question at the end of your post that begins "Yeah, exactly".

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia I totally agree we want f11×f11′ = 11, and the same for every other split prime. Yes, the split primes can always be written as a conjugate pair in the (1, √5) basis, which, in the (1, ϕ) basis will be of the form a+bϕ and (a+b)-bϕ [not (a+1-bϕ as you have above]. Dodd defines a constant ϕ̅ (pronounced phi bar - the bar is meant to be above the phi) such that ϕ̅ = -ϕ⁻¹ ≈ -0.618, so that he can write the (1,√5)-conjugates as e.g. f11 = 3+1ϕ and f11′ = 3+1ϕ̅. (Or at least he does the equivalent, using ω instead of ϕ). I don't like it. Why have two irrational constants when one will do.

But what is best for number theory isn't necessarily best for RTT scale engineering. Dodd never once mentions noble numbers. We've agreed we don't want powers of ϕ in our rationals. I think there is a similar advantage in not having powers of ϕ in our nobles (whenever possible).

The desire to have no powers of ϕ in our nobles, combined with the fact that f11/f5 and f11′/f5 are noble, and f5 = √5 = -1+2ϕ, forces f11 and f11′ to be -1+3ϕ and -2+3ϕ, which could be called (√5,1)-conjugates (as opposed to (1,√5)-conjugates) since they are +1/2 + 3/2×√5 and -1/2 + 3/2×√5. They are generally of the form -a+bϕ and -(b-a)+bϕ. We could call them "complements" rather than conjugates. And treat the prime symbol ′ as the complement operator, so that f11 = -1+3ϕ and f11′ = (-1+3ϕ)′ = -2+3ϕ.

I believe they are also the pair that are closest to each other in their real values. Or in other words, they are the pair that are closest to √11. One smaller one greater. But which should be called f11 and which f11′?

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia I'm still keen to discuss the best RTT normal form for feudal primes. Did you see my post beginning "I totally agree we want f11×f11′ = 11" above?

I've started a parallel version of my original Sagittal forum thread, using the -a+bϕ form that I propose above.

viewtopic.php?f=21&t=557

It eliminates powers of ϕ from the rationals without introducing them into the nobles as the conjugate pairs do.

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

Reply

Remove Preview

31 w

Dave KeenanAuthor

Those negative wooden-parts on my new normal-form split primes were driving me crazy. e.g. √5 = -1+2ϕ. Particularly since all the unsplit primes have positive wooden parts e.g. 2 = 2+0ϕ.

Then I remembered being puzzled by this MathWorld article on noble numbers: https://mathworld.wolfram.com/NobleNumber.html

where they say any nobles can be written as (a+bϕ⁻¹)/(c+dϕ⁻¹). Why do they use ϕ⁻¹ when it would be just as true if they had written it using ϕ?

Then I thought: What does √5 look like when we use the basis (1, ϕ⁻¹) instead of (1, ϕ)? Answer: 1+2ϕ⁻¹. It has both parts positive! As do all the other split primes whose form doesn't give powers of phi in the rationals or most nobles.

But I'm not going to type ϕ⁻¹ all the time. [Incidentally, I use WinCompose to type this stuff. It's great.] But I can't find a good single symbol for ϕ⁻¹. I mentioned that phi bar ϕ̅ is the conjugate of ϕ, i.e. -ϕ⁻¹. Unfortunately uppercase phi Φ (or "Phi" with a capital P) has also been used for the conjugate -ϕ⁻¹. I guess we could just declare that we are using uppercase phi Φ for the reciprocal ϕ⁻¹. Then we can write √5 = 1+2Φ.

f11 and f11′ are then 1+3Φ and 2+3Φ. They multiply to give 11 without any factors of Φ.

And the noble n5/4 (the noble mediant of 5/4 with its Stern-Brocot parent 4/3) is f11/f5 = (1+3Φ)/(1+2Φ) with no factors of Φ.

Reply

31 wEdited

Mike BattagliaAdmin

The only reason we care about this basis of "feudal primes" at all, rather than a basis of noble numbers, is because of this algebraic number theory perspective that, as you mention, is already neatly worked out for us. So given that, I think the conjugate pair method is the cleanest way to do it. It has several nice properties - it gives us all the naturals, the conjugate pairs are decently close in size, and we get lots of nice mathematical properties for free.

"The desire to have no powers of ϕ in our nobles, combined with the fact that f11/f5 and f11′/f5 are noble, and f5 = √5 = -1+2ϕ, forces f11 and f11′ to be -1+3ϕ and -2+3ϕ"

I don't see what you mean by this. Having f11 and f11' be 3+phi and 4-phi also has f11/f5 being noble and so on. As you already know, there is no absolute way to measure "how many powers of phi" some particular representative of a feudal prime class has, or at least not that I can see. Rather, we are arbitrarily picking a representative that we are *declaring* to have no powers of phi, or rather a zero "phi-coordinate," to use in our basis.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia We agree f5 = -1+2ϕ = √5. But you prefer f11 = 3+1ϕ and f11′= 4-1ϕ? What then is your prime exponent vector for the noble-mediant of 4/3 and 5/4 which has a real value of approximately 1.2764. We expect it to have +1 for its f11 coordinate and -1 for its f5 coordinate. But f11/f5 = (3+1ϕ)/(-1+2ϕ) ≈ 2.0652. It's too large by a factor of ϕ. So you're forced to have a phi-coordinate of -1 to make it right.

If instead we use f11 = -2+3ϕ and f11′ = -1+3ϕ, we find that (-2+3ϕ)/(-1+2ϕ) ≈1.2764. Zero phi-coordinate.

And if we want, we can rewrite those primes with Φ = ϕ⁻¹ instead of ϕ, to eliminate the minus signs. f5 = 1+2Φ, f11 = 1+3Φ, f11′ = 2+3Φ. We'll still have a zero phi-coordinate.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia Once we decide that f5 must be √5 = -1+2ϕ, so that 5 has no phi-coordinate, if we want those nobles that are the ratio of two primes to have no phi-coordinate, then that choice for f5 propagates to all the other split primes, all the way down the Stern-Brocot tree. The normal form of f5, and 2 nobles on level 4 of the SB-tree, determine the normal forms of f11 and f11′. Then the normal forms of f11 and f11′, and 4 nobles on level 5, determine the normal forms of f19, f19′, f31 and f31′, and so on.

BTW, Thanks to Douglas Blumeyer, I've realised it's bad to use the prime symbol ′ for these, since we will eventually want to write powers of these primes, like f11′³. So I'm switching to "f11" and "F11", pronounced "small eff eleven" and "big eff eleven". "Small" and "big" relate to their real values as well as their letters. The "f"s of both cases can be thought of as an abbreviation of both "factor of" or "feudal prime".

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan hm. That is quite interesting. Do you have a table of primes in this form? Are there other noble numbers that instead have no phi coordinate with the conjugate basis?

Reply

31 w

Mike BattagliaAdmin

Dave Keenan btw, the use of uppercase and lowercase phi makes it almost impossible to read in certain fonts, for instance the standard font that is used on FB Android. See picture

No photo description available.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan I think this is an artefact of the particular Unicode phi you are using. If you go with the standard greek Φ and φ it's much easier to read. See picture:

No photo description available.

Reply

31 w

Dave KeenanAuthor

You're right about Capital phi and phi symbol being nearly indistinguishable in many fonts. This is not a problem for \Phi and \phi in LaTeX. But the curly phi (\varphi in LaTeX) is not the standard math symbol for the golden ratio.

Fortunately I've stopped worrying about minus signs now, since I expect we'll mostly use the fF form of the non-wooden primes. So I have no need for capital phi and can continue to use the unicode that is supposed to be straight phi. But many fonts have straight phi and curly phi swapped, because that's how unicode had them up until version 3. This is all in my "Rant about ϕ" in my first feudal Sagittal forum article.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia You wrote: "Do you have a table of primes in this form?"

I assume you've seen the table in this thread by now:

viewtopic.php?p=4605#p4605

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

Reply

Remove Preview

31 wEdited

Dave KeenanAuthor

Mike Battaglia You wrote: "Are there other noble numbers that instead have no phi coordinate with the conjugate basis?"

Yes. If you use the (1, √5)-conjugate basis, only those nobles involving f5 = √5 = -1+2ϕ have a phi in them. You could even get rid of those if you split 5 into the conjugate pair f5 = 3-1ϕ and F5 = 2+1ϕ. But it makes no sense to split 5 like that because that f5 and F5 are associates. i.e. they are really the same prime, differing only by the unit ϕ².

It's purely the fact that we want f5 = F5 = √5 = -1+2ϕ (i.e. a single non-conjugate factor of 5) that propagates the non-conjugate form down the SB-tree to all the other primes.

Reply

31 w

Mike BattagliaAdmin

Dave Keenan, OK, I'm signing off for now, but a last thought for the night. Given that the algebraic number theory perspective seems decent at this point, here is a totally different method of analysis which I think is also useful, and possibly better, at least for my purposes:

We have agreed that the things we primarily care about are the rationals and noble numbers, or at least the simple ones. These are really the true "primitives" from which we are building arbitrary musical intervals. Other intervals can be formed as compounds of these.

Thus, for very clear and direct musical reasons, I should very much like to see the structure of how arbitrary feudal numbers decompose as a product of nobles and rationals, learning how these things all fit together in the same way I learned about JI. If you were going to hand me this theory and have me start actually using it to make music, figuring out this kind of decomposition, and how to move around, is the very first thing I would do.

The most basic question possible is this: what is *any* choice of basis for the noble numbers, which is comprised entirely of other noble numbers? This basis, plus the usual primes (at least those which are still prime in Z[phi]), will give us, let's call it an "Arthurian basis" which is entirely rational and noble and generates all of the feudal nobles.

It turns out that there is an easy way to compute such a basis. This only requires some way to list all of the noble numbers in terms of increasing "complexity", using pretty much any method you want. Your method of doing this using the Stern-Brocot tree is good enough to get started at least. So we start with phi, which is a noble number, and it goes into the basis. Then we go to 1+phi, which is not independent of phi, and thus it is not a new basis element. Then we have (2+phi)/(1+phi), which is independent, so it goes in. (2+phi) itself is next, and is not independent of the last, so it's not a new basis element. And so on. This gives us at least one choice of basis.

There is probably some way to generate a slightly cleaner list. After all, it is kind of weird that (2+phi)/(1+phi) makes it into the basis instead of (2+phi). This is only because it appears first in the way that we are ordering nobles. You could probably do it directly in terms of feudal primes - add them one by one if they are noble, and if not, try to divide somehow by the simplest possible denominator that does make it noble. Either way though, this all just serves to show that this method is possible.

There is also a very nice notation for the elements of this basis. This basis has the property that the continued fraction expansion of every element is either finite (if rational), or ends in a stream of 1's (if noble). In the first situation, the notation is to just write the rational. In the second, we look at the continued fraction and truncate it immediately *after* the first "1" in the tail. We then write that as a rational, followed by the symbol "~". So for instance, the noble mediant between 5/4 and 6/5 would be 6/5~. This works for everything except for phi, since phi and 1/phi would both be 1~, so we just use the symbol phi for it.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia As you know, at this stage in development I'm not particularly interested in any basis that isn't all primes, and I don't even see much advantage in those primes being noble, but I'm still keen to see where you take this idea.

I like your postfix tilde operator, which can be pronounced as "ennobled". I suggested a similar thing to Douglas back in 2020, namely to add a boolean to his computer implementation of prime-count vectors (monzos) to let them also represent nobles. I suggested that it should work so that 2/1~ = ϕ and 1/2~ = 1/ϕ, which has the intuitive reading that the octave ennobled is the phitave. I believe that is consistent with you saying that the noble mediant between 5/4 and 6/5 would be 6/5~. i.e. all 3 examples imply that we should use the ϕ-weighted one of the two mediends that I give in my tables.

viewtopic.php?p=4592#p4592

This also makes it the mediend that is closer in size to the noble.

But it's not the rational at the top of the zigzag. It's the second one down.

Noble frequency ratios as prime-count vectors in ℚ(√5) [a+bϕ] - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) [a+bϕ] - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) [a+bϕ] - The Sagittal forum

Reply

Remove Preview

31 wEdited

Mike BattagliaAdmin

Dave Keenan you know, I think that's technically right given what I wrote, but turns out that it also it isn't what I intended. I should clarify a bit...

Initially I had suggested looking at this in terms of continued fraction expansions, but it turns out to be slightly easier for our purposes to instead look at the Stern Brocot tree. Every noble number can be represented as an initial segment, followed by an infinite zig-zag pattern. The idea was to represent the noble number as a/b~, where a/b is where it *starts* zig-zagging, but there are a few different places where one can determine what we are calling the "start." I think the thing I originally suggested above is not exactly what was going for, so I will slightly modify my previous proposal to give the larger picture of what I was after.

We would like, given any rational number a/b, the two rationals ~a/b and a/b~ to be the noble numbers obtained by first going to a/b on the Stern-Brocot tree, and then starting a zig-zag pattern to the left and the right respectively. Every noble number will, in general, have several representations using this method: for instance, the noble mediant of 5/4 and 6/5 is equally ~5/4, 6/5~, ~11/9, 17/14~, and so on. Having the ~ on the left, in the case of 5/4, tells you that this is the unique noble number associated to 5/4 which is *flat* of 5/4, and having it on the right as in 6/5~ tells you it's the unique noble number which is *sharp*. I'll call this a "noble flat" and "noble sharp" respectively.

We can use this to develop two unique representations for any noble number: we can look at the *simplest* representation of it as a noble flat, and the simplest as a noble sharp. So the two representations ~5/4 and 6/5~ are the unique representatives for that blue noble number from before. For phi, we get the two representations 1/1~ and ~2/1, and for 1/phi we have ~1/1 and 1/2~. We also have that for any noble representation a/b~, its reciprocal is ~b/a.

If we want a unique representative for the entire thing, we could, for instance, take the simpler of both. So the unique representative of that noble blue note would then be ~5/4. This is the absolute first point where the zig zag occurs on the Stern-Brocot tree. For whatever reason, the convention last night that I suggested would have us take the more complex of the two instead, which is the first rational in the zig zag that changes direction. I don't remember why I thought this was better last night.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia Isn't "two unique representations" a contradiction in terms? 🙂

I liked your right or postfix tilde. I don't like the left or prefix tilde because that already has at least two meanings in RTT, namely "approximate" and "Grassmann complement".

I'd much prefer to name nobles using a single one-to-one and onto mapping (bijection) between rationals (excluding 1/1) and nobles. That's what I proposed last night, thinking I was agreeing with you.

That would greatly reduce the cognitive load in figuring out what noble someone is referring to, and would allow people to memorise the cents values of the simpler nobles given as ennobled ratios.

The SB-tree is a binary tree*, so every node (every rational) has a left and a right child, but it only has one parent (except for 1/1 which has none). So the way to obtain a bijection is to define the ennoblement of a rational as the noble mediant it makes with its /parent/. Yes, this will be the same as the one it makes with one of its children, but we don't need to worry about that, or whether it is the right or the left.

Yes, that makes it the second rational from the top of the zigzag corresponding to a given noble. Or the first corner, as you say. But that does not give the more complex of the two noble as you claim above. It gives the /simpler/.

[EDIT: * When you strip away all the dashed or dotted lines people like to decorate it with.]

Reply

31 wEdited

Dave KeenanAuthor

Some possible abbreviations or symbols for "the noble that 6/5 makes with its SB-parent" are:

6/5~

n6/5

6n5

The postfix tilde has the advantage that it can be treated as an operator and used with a rational variable or expression. But the "n"s could be associated with a function called "noble()".

What's the simplest way to compute the rational that is the SB-parent of a given rational?

Reply

31 w

Mike BattagliaAdmin

Dave Keenan I like the way that I was doing it, because it's easy to see if the noble number we are talking about is located to the left or the right of the rational in question. It isn't difficult to define a bijection if that's what you want - I already wrote two different ways to do it. I'd like some notation for my "noble sharp" and "noble flat" thing but I'm not really all that married to the ~ sign. We could do 6/5< and 6/5> or something.

Reply

31 w

Mike BattagliaAdmin

Dave Keenan I'll write more later, but a quick note here that there is more to what we were calling the "arithmetic norm" than meets the eye - in particular the thing Wolfram calls AlgebraicNumberNorm[x] is not the same as the thing I was talking about before. I will look at my notes and see if I can remember how I was extending Tenney height. That expression a^2 + ab - b^2 works well for any feudal integer but can be kind of subtle.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan I'll just make some notes as I read them:

The thing we were calling the "arithmetic norm" turns out to be a^2 + ab - b^2 for the feudal integers, and we may take the absolute value to get the unsigned version. In fact, it is slightly more useful if we take the square root to get sqrt(|a^2 + ab - b^2|).

Using this expression, the norm for any standard prime is the prime itself, the norm for sqrt(5) is sqrt(5), and the norm for f11 and f11' is sqrt(11). And so on. The norm for phi is 1 (actually the unsigned version would be i), which is kind of weird, but we'll ignore this for now.

So right away, the big question then is: *is the norm of any feudal integer the product of the norms of the primes?* Put another way: is the norm of any product of feudal integer the product of the norms, given that we ignore the sign and normalize properly? EDIT: Turns out that indeed it is.

If so, then we have something generalizing Tenney height on the feudal integers, which we can easily extend to all feudal numbers, minus the quirk of phi having a norm of 1 which is easily adjusted.

Lastly, I note Wolfram has some of this implemented as AlgebraicNumberNorm[...], but it is very quirky. You need to take the square root of AlgebraicNumberNorm[a+b*GoldenRatio] whenever b is nonzero, and just naively doing AlgebraicNumberNorm[p/q] also just gives p/q, rather than max(p,q) or p*q or etc. sqrt(|a^2+ab-b^2|) seems to give decent results, though.

Reply

31 wEdited

Mike BattagliaAdmin

Lastly I note that there is also this notion of a "height function" on an arbitrary algebraic number field. Some of these height functions I'm looking at are very strange and I'm not sure if they make sense here. They also seem to generalize the max(n,d) height rather than n*d height, but maybe some of them are useful (or even agree with the above, except for the weird quirk involving phi).

Reply

31 w

Mike BattagliaAdmin

Dave Keenan, after thinking about this, there turns out to be a much simpler and possibly better generalization of Tenney height for our purposes.

Once some basis of feudal primes is chosen, one can simply take the "weight" of each prime to be the log of the prime itself. We can then put things into "weighted coordinates," the same way that we do with regular monzos. Then everything derives naturally from that: the L1 norm of this weighted monzo gives us a natural measure of the complexity of each feudal number, and the JIP is <1 1 1 1 1 1 ...| in the weighted basis, and so on.

This gives very similar results to the sqrt(|a^2 + ab - b^2|) method with one quirk. sqrt(5) has the same complexity in both, as do the regular primes. The two factors of 11, using the first method, both have complexity equal to log2(sqrt(11)) ≈ 1.730, whereas with this method we have f11 and F11 are two close values which average to 1.730. For instance, using your basis we have log2(-2+3*Phi) ≈ 1.513 and log2(-1+3*Phi) = 1.946.

The only quirk is that with the original method, we had that phi has a weight of 0 (which we don't have here). We could perhaps grandfather in a weight of sqrt(5) for phi as well, treating sqrt(5) as its "unofficial" prime pair (it seems reasonable that phi and sqrt(5) would be pairs in Q[sqrt(5)] in some sense). But with this newer method you get that all for free.

Lastly, I note that the "conjugate basis" I suggested gives us values which are slightly further apart in value for f11 and F11. We have log2(3+Phi) ≈ 2.207 and log2(4-Phi) ≈ 1.252. These still average to ~1.730, but the two primes that you choose (-2+3*Phi) and (-1+3*Phi) are slightly closer. Does that uniquely characterize the choice of primes in your preferred basis? Will they always be the two closest in size?

Reply

31 w

Dave KeenanAuthor

Mike Battaglia Yes, I believe the choice of primes in my preferred basis will always be the two closest in size, and therefore closest to the square root. I mentioned this several times previously.

Reply

31 w

Mike BattagliaAdmin

There is no need to be rude, but thanks, I guess.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia I apologise.

Reply

31 wEdited

Dave KeenanAuthor

I spent most of the day converting all my articles, with all their tables and diagrams, over to the new choice of fundamental primes. Very tedious, and not yet complete.

Now that I've had a chance to digest your work above. I agree that the log of the real value of the prime has just as much claim, if not more, than the log of the square root of the absolute value of the arithmetic norm, as the generalisation of our usual log of primes, for computing complexities, provided we use the basis where fp and Fp are as close as possible to √p, because p is the absolute value of the arithmetic norm of fp and Fp. As you say, it's not clear what it means psychoacoustically to compare complexities between rationals and nobles, but we can go with log-of-real-value for now.

BTW, when we need to distinguish this norm, a²+ab-b², from the various p-norms we use on our vectors, we could just call it the "feudal norm". That's 2 less syllables than "arithMETic norm", and a whole lot less than "quadratic field norm" or "algebraic number norm", which, as you say, may not even be the same thing. I thought of yet another reason why "feudal" is a good name for ℚ(√5) and ℤ[ϕ]: It sounds a bit like "ϕ-dal".

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan sounds good to me. So we can also use this to generalize the TE norm, and thus we get the entire theory powering Graham's temperament search thus generalized to feudal numbers. This will let us search for good feudal temperaments and let us answer your original question about what are some good feudal commas. The only thing is that I think this choice of weighting may give very strange results: for instance, f11 and F11 are both weighted more simply than 5/1 and 7/1, and the simplest element of all is phi, simpler even than 2/1. This way of prioritizing intervals may give us an "optimal" tuning for some given temperament that is less than optimal, as far as our ears are concerned, and thus it will also skew the results of the temperament search. But I will try it when I get a second and give the results.

I will just add the only real thing left is to figure out this complexity function; I need to know what weighting matrix to put into the temperament search which will give sensible results. Note the weighting matrix need not be diagonal, which in plain terms is equivalent to saying that we could run the search relative to a different basis internally, even if we keep using the feudal prime basis for notation. We just need some L2 norm whose elliptical unit sphere approximates whatever we view as the true complexity function. I will give some thought to this.

Reply

31 w

רועי סיני

Dave Keenan In fact, your choice of primes is not always the two closest in size. For example, F29/f29 is 1.733..., while (f29ϕ)/(F29/ϕ) is 1.510...

However, they are always either the best or the second best choice, because they are both between bϕ-b = b/ϕ and bϕ, where b is their equal golden part, which means the quotient between them is less than ϕ², and any time you multiply one of them by ϕ and divide the other by ϕ you either multiply or divide the directed quotient by ϕ². Anyway, I don't think this is a large enough issue to use the other choice instead in this case, because of the consistent normal forms they have.

Reply

19 h

Mike BattagliaAdmin

Also a quick note that this also easily generalizes to things like "Kees Expressibility" and etc of feudal numbers.

However, I would also add that it is a little bit strange that we are going to have an interval like (-1+3*Phi) have a lower complexity than 11 itself. Part of this is because naively putting an L1 norm on a basis of "feudal primes" may not really represent perceived interval complexity (which is one reason I was interested in the basis of primes and noble mediants). It is a pretty good question to ask what criteria we should even use to compare noble and just intervals at all.

So I guess for now, my point is that the log(prime) weighting is "at least as good" as the sqrt(|b²+ab-a²|) weighting, being approximately equivalent but slightly better behaved, with a clearer choice of weight for phi, but also with the same shortcomings for both.

That being said, some of this is unavoidable if we want even the basic bare minimum of linear algebra w/ normed vector spaces to work - we must have that 5/1 has double the complexity of sqrt(5/1), for instance. But I don't think that particular choice of weight will affect the results of any temperament search much. The better question is what to do with things like f11 and F11.

Reply

31 wEdited

Mike BattagliaAdmin

I note that weighting phi as log(phi) gives some extremely weird results; phi has a lower complexity than any rational number, and using this weighting temperament searches tend to rank 30 higher than 31 for this reason (due to the extreme emphasis on phi).

Reply

31 w

Paul ErlichAdmin

Mike Battaglia I believe that's how this search is carried out. I do see a 30 and no 31, but the higher-ranked results look fine, make sense, and represent some appealing solutions. It all depends on how you want to use phi, with what timbres, volume/distortion, etc. how much importance it and its precise tuning should get IMO.

http://x31eq.com/cgi-bin/more.cgi?r=1...

X31EQ.COM

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 1 Temperaments

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 1 Temperaments

Reply

31 w

Paul ErlichAdmin

Mike Battagliaa difference with the results you posted is that 27 is not showing up for me here for some reason. Any idea why that is?

Reply

31 w

Paul ErlichAdmin

(Also worth noting is that 62 = 31×2 shows up pretty strongly)

Reply

31 w

Mike BattagliaAdmin

Paul Erlich the weighting on phi is so extreme that phi is weighted stronger than any rational number, so I decided to try some other weightings. Graham's algorithm just weights each of these basis elements by their size, so one simple way to get a different weighting is just to replace phi in the basis with 2*phi, so that it is weighted somewhere between 2 and 3. So if you do that, replacing 833 cents with 2033 cents, 27 is at the top: http://x31eq.com/cgi-bin/more.cgi?r=1...

X31EQ.COM

1200.000, 2033.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 1 Temperaments

1200.000, 2033.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 1 Temperaments

Reply

31 w

Steve Martin

Just to say I have a book on quadratic fields, reading up now so maybe I can help, if only by reaching the point where I can say I agree with you guys ...

Reply

31 w

Steve Martin

Hi Dave, I have read your article, blog, this thread and the 2.5 chapters of Hardy&Wright that cover quadratic fields. Everything looks correct, I like the way ℚ(√5) is a UFD and has its own units (countably many of them!) and primes. Your canonical choice of primes is neat and is not in H&W. The annotated diagram is very helpful.

I'd need to think more to see how to use this, and I'll look at Zest24 too. One question to start: is every noble number a quotient of two primes? Would the quotient of two composites (or a composite and a prime) not be a noble? I've not understood this.

Reply

30 w

Dave KeenanAuthor

Steve Martin, Thanks for your interest, and thanks for checking my work. At this time it remains a conjecture on my part, that every noble number is a quotient of two feudal primes-or-units. Apart from the prince of nobles (ϕ itself), I've found nobles to always be ratios between two feudal integers taken from the set of non-wooden canonical primes (the fF-primes) plus the units f1 and F1 (ϕ⁻¹ and ϕ¹). And beyond the prince and the duke (ϕ and ϕ²) I've only seen ratios between two big F's or two small f's (where √5 counts as either f5 or F5). I conjecture that this will continue ad infinitum. But it's more constrained than that. You can't just choose any pair of f's or F's and assume you'll get a noble. This is due to the "unimodular" requirement — cross-multiply and subtract on the (a+bϕ)/(n+mϕ) form and you must get am-bn = ±1.

Reply

30 wEdited

Steve Martin

Dave Keenan thx, it's good to be reminded of the unimodular requirement - which is not mentioned in the Wolfram page unless I missed it - but I guess is easily proved to be equivalent to the continued fraction definition?

You are right, Hardy & Wright do not discuss noble numbers.

Reply

30 w

Paul ErlichAdmin

Steve Martinit is implied on the Wolfram page in that the 4 coefficients are required to be numerators and denominators of successive convergents.

Reply

30 w

Steve Martin

Paul Erlich ah yes I see

Reply

30 w

Steve Martin

Paul Erlich what is the syntax to use more.cgi ? I think there is no UI page for it, could you re-post an example please (as I am stuck on phone where search history is hard)

Reply

30 w

Paul ErlichAdmin

Steve Martinlike this?

http://x31eq.com/cgi-bin/more.cgi?r=1...

X31EQ.COM

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 1 Temperaments

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 1 Temperaments

Reply

30 w

Paul ErlichAdmin

Steve Martin you can get to it in the UI by clicking "show more of these " e.g. here:

http://x31eq.com/cgi-bin/pregular.cgi...

which you get to from here:

http://x31eq.com/temper/pregular.html

by typing the relevant numbers into the boxes.

X31EQ.COM

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Regular Temperaments

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Regular Temperaments

Reply

30 w

Dave KeenanAuthor

Steve Martin Thanks for checking Hardy & Wright. I take it they don't discuss noble numbers as a subset of ℚ(√5). The only place I've seen anything about that is the MathWorld article: https://mathworld.wolfram.com/NobleNumber.html

I've ordered a copy of the Manfred Schroeder book referenced there.

Wolfram MathWorld: The Web's Most Extensive Mathematics Resource

MATHWORLD.WOLFRAM.COM

Wolfram MathWorld: The Web's Most Extensive Mathematics Resource

Wolfram MathWorld: The Web's Most Extensive Mathematics Resource

Reply

Remove Preview

30 w

Paul ErlichAdmin

Dave Keenangreat book!!!!!!!!

Reply

30 w

Dave KeenanAuthor

Paul Erlich If you have the Schroeder book, perhaps you can tell us if it sheds any light on my conjecture that every noble number is a quotient of two primes or units of ℚ(√5)?

But only /some/ quotients of two primes or units are noble numbers.

Reply

30 wEdited

Paul ErlichAdmin

Dave Keenanit's deep in storage! 🤣

Reply

30 w

Steve Martin

Yes, thank you!

Reply

30 w

Paul ErlichAdmin

Steve Martin rank-2

http://x31eq.com/cgi-bin/more.cgi?r=2...

X31EQ.COM

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 2 Temperaments

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Rank 2 Temperaments

Reply

30 w

Steve Martin

Thanks again, Paul, I'd not used that one before

Reply

30 w

Steve Martin

Here's an example using the parameters I wanted to try:

http://x31eq.com/cgi-bin/rt.cgi?ets=17_43...

X31EQ.COM

1200.000, 1901.955, 1393.157, 833.090, 3368.826, 1815.643, 2335.673-limit Regular Temperament

1200.000, 1901.955, 1393.157, 833.090, 3368.826, 1815.643, 2335.673-limit Regular Temperament

Reply

30 w

- Dave Keenan
- Site Admin
**Posts:**2180**Joined:**Tue Sep 01, 2015 2:59 pm**Location:**Brisbane, Queensland, Australia-
**Contact:**

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

Dave KeenanAuthor

Mike Battaglia Paul Erlich Cmloegcmluin Xenharmonic Feisbeuk Steve Martin Here is a table/plot of my new preferred feudal primes — those that ensure that all rationals and the maximum number of nobles (all but 2 on each level of the Stern-Brocot tree) can be factorised without need of a unit, a power of ϕ, other than its zeroth power, i.e. 1.

viewtopic.php?p=4607#p4607

Be careful. The ink's still wet. 🙂

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

Reply

Remove Preview

31 wEdited

Mike BattagliaAdmin

Thanks Dave Keenan, one question: the first 32 noble numbers seem to be a quotient of exactly two feudal primes (counting phi). Three questions

1. Do we know if a noble number can *ever* have a strictly "wooden prime" in its prime factorization, or will the factorization always be that of feudal primes and phi?

2. Do we know if a noble number can ever have more than two factors in its prime factorization?

3. Can we *ever* have any nobles which have, in the prime factorization, two complementary prime pairs, like fp and Fp for some p?

I think it's easy to see that no such noble number can be of the form fp/Fp given that your "square root" analysis is correct, because the two pairs will be of the form a+b*phi and a'+b*phi with the same b and thus the determinant of the matrix [a b;a' b] will be a multiple of b. But the question, I guess, if is we can ever multiply fp/Fp by something else to get a noble number.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia You wrote: "1. Do we know if a noble number can *ever* have a strictly "wooden prime" in its prime factorization, or will the factorization always be that of feudal primes and phi?"

No proof, but I conjecture that all nobles have at most one non-wooden prime in the numerator and at most one non-wooden prime in the denominator and never a wooden prime anywhere.

And further that, apart from ϕ and ϕ², all nobles have numerator and denominator either both small-f or both big-F, where f5 = F5 = -1+2ϕ = √5, and f1 = -1+1ϕ = ϕ⁻¹ and F1 = 0+1ϕ = ϕ¹.

"2. Do we know if a noble number can ever have more than two factors in its prime factorization?"

Not proven. But conjectured that never more than 2.

"3. Can we *ever* have any nobles which have, in the prime factorization, two complementary prime pairs, like fp and Fp for some p?

I think it's easy to see that no such noble number can be of the form fp/Fp given that your "square root" analysis is correct, because the two pairs will be of the form a+b*phi and a'+b*phi with the same b and thus the determinant of the matrix [a b;a' b] will be a multiple of b. But the question, I guess, if is we can ever multiply fp/Fp by something else to get a noble number."

No. Never the same p in numerator and denominator*. And other limitations on which pairs of p's make a noble. They all come back to the "unimodular" requirement for nobles. i.e. the determinant must be ±1.

*Unless you count my degenerate use of f1 and F1 as ϕ⁻¹ and ϕ in naming nobles. In that case you only get the same "p" top and bottom in the case of the noble ϕ².

Reply

31 wEdited

רועי סיני

Mike Battaglia For the case that you are still interested, here are answers to these quesrions:

1. They can't, because if we have a wooden prime that divides the numerator but not the denominator (or vice versa) this means that the numerator is another feudal integer times this prime, which means that both its wooden part and its golden part are divisible by this prime, and therefore the determinant of the matrix divisible by this prime and not ±1. This also means that f5² or any pair of (opposite) conjugate primes together can't be found in either the numerator or the denominator.

2. This is essentially equivalent to Dave Keenan's conjecture about the numerator and denominator always being prime or unit, which I disproved. Specifically, (13+19ϕ)/(11+16ϕ) = (-1+2ϕ)*(-2+3ϕ)*ϕ/(-1+6ϕ) = f5*f11*ϕ/F41.

3. Yes we can. The key to finding them is realizing that (a+bϕ)*conj(c+dϕ) = (a+bϕ)*(c+d-dϕ) = ac+ad-bd + (-ad+bc+bd-bd)ϕ = ac+ad-bd - (ad-bc)ϕ with a golden part of -(ad-bc), which means that if we want to find a golden mediant that has F11 on the top and f11 on the bottom, for example, we need to find a multiple of F11*conj(f11) = -F11² which has a golden part of ±1. F11² = 10+3ϕ, which can be multiplied by 4-ϕ = F11/ϕ to give a number with golden part of 1, which means that F11²/f11ϕ = (10+3ϕ)/(3+ϕ) = (13+16ϕ)/(4+5ϕ) is a noble that has both F11 and f11 in its factorization.

Reply

19 h

Dave KeenanAuthor

I have the new primes on the Stern-Brocot tree now, in both a+bϕ and fp or Fp form:

viewtopic.php?p=4609#p4609

I think it will be useful to have noble-flat and noble-sharp operators as you describe. And I no longer think that human-friendly unique identifiers for the nobles need to be based on their mediends, or any rationals. I'm getting quite used to treating the f11/f5, F31/F19 form as an identifier for the noble.

I describe how to mentally extract the a+bϕ form from the fp or Fp form, in the fifth paragraph of this post:

viewtopic.php?p=4607#p4607

It's possible that there is a mental, or simple pen-and-paper, method for extracting the two ratios at the start of the SB-tree-zigzag, from the fp₁/fp₂, e.g. F31/F11, form.

Reply

31 wEdited

Dave KeenanAuthor

It turns out: All you have to do to get the two simplest Stern-Brocot-adjacent ratios whose noble-mediant is given as e.g. F31/F11 is to first use the above-linked greatest-square-remainder-complementary-factorisation method on the numerator and denominator to get it to the form (-2+5ϕ)/(-1+3ϕ), then "fibonacci" the numerator and denominator forward exactly 3 steps. It's always 3 steps, irrespective of the noble. In this case we get:

-2 5 3 8 11

-- -- -- -- --

-1 3 2 5 7

So it's the noble mediant of 8/5 and 11/7. Of course it's also the noble mediant of any other pair of succesive ratios in that sequence (and its continuation). But 8/5 and 11/7 are the pair at the start of the zigzag. Prior to them there was a non-adjacency, a discontinuity, a "teleport". 3/2 and 8/5 are not directly connected on the SB-tree.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan this is great stuff. Is this proven or just a conjecture?

Reply

31 w

Dave KeenanAuthor

Just conjecture. I know I'm bad. The way I state all these thing in my articles that I don't have proofs for. Particularly when I've watched those Matt Parker and/or Numberphile videos where, just when you think you've got some sequence nailed down, it goes off and does something crazy. But I keep rationalising it to myself with: Well it works for the "musically-significant" stuff so who cares if it breaks somewhere along the road to infinity. 😮

Reply

31 w

רועי סיני

Dave Keenan This is in fact true. If in the beginning you have a/b, c/d s.t. a and b are negative and c and d are positive and larger than their respective absolute values, then in the next step you'll get c/d, e/f s.t. 0 < e < c and 0 < f < d, which means, because of the ±1 determinant, that e/f is an ancestor of c/d in the SB tree. After the second step you get e/f, g/h, where now the first is an ancestor of the second, but not its direct parent because that is c/d. After the third step you indeed get i/j which is a direct child of g/h, and is its simpler child because e/f is not its parent.

You can run this process in reverse too - from a pair of parent and simpler child you first get the parent and an ancestor of it, secondly some other node descending from the same ancestor and finally the two numbers become negative and you get your primes in the normal form.

Reply

18 hEdited

Paul Johnson

Here’s a more practical question: 72edo gives fantastic primes and a near perfect phi. Do you think you could find something better using this method?

Reply

31 w

Paul ErlichAdmin

Paul Johnsonthe idea is to find a minimal basis of "primes" that will allow an array of noble mediants -- not just phi -- to be represented. Then the search for temperaments of any rank would proceed much as usual. If it were just a matter of phi or any set of already-known feudals (with the log weighting mentioned above) and some primes, you can do that with Graham's app already.

Reply

31 w

Paul Johnson

Paul Erlich ok here’s a dumb question, can all the noble mediants be found in Q(root(5))? Obviously phi can

Reply

31 w

Dave KeenanAuthor

Paul Johnson Yes. See: https://mathworld.wolfram.com/NobleNumber.html

Wolfram MathWorld: The Web's Most Extensive Mathematics Resource

MATHWORLD.WOLFRAM.COM

Wolfram MathWorld: The Web's Most Extensive Mathematics Resource

Wolfram MathWorld: The Web's Most Extensive Mathematics Resource

Reply

Remove Preview

31 w

Paul Johnson

Dave Keenan thank you

Reply

31 w

Paul ErlichAdmin

e.g., http://x31eq.com/cgi-bin/pregular.cgi...

X31EQ.COM

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Regular Temperaments

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Regular Temperaments

Reply

31 w

Paul ErlichAdmin

Mike Battaglia Dave Keenan

Reply

31 w

Mike BattagliaAdmin

Dave Keenan one simple idea, which may improve the weighting somewhat, is to say that f11 and F11 should have the same complexity as 11, f31 and F31 should have the same complexity as 31, and so on. This gives us an "L1-norm-of-Linf-norms" rather than an L1 norm; still easy to compute and can easily be done on the back of a napkin given the monzo decomposition of some feudal number.

For each of these prime factor pairs like f11 and F11, we weight them as log(11) rather than log(√11). But then, instead of just taking an L1 norm of everything, we first take the max of the absolute value of the two f11 and F11 coordinates, and likewise with the other prime factor pairs, and then take the L1 norm of all those intermediate results along with that of the normal primes. This gives f11, F11 and 11/1 the same complexity, and in general a very interesting choice of complexity for everything, with the only quirk being that again, -1+2*Phi and Phi itself are the two sporadic basis elements with different behavior (which I have some ideas about and will deal with later).

If any of these ideas sound good, I have a whole bunch of related results showing how we can "TE-ify" this to locate the best-fit approximate L2 norm for the temperament search, as well as a way to extend the above to a generalized complexity with one free parameter interpolating between the L1 version and this L1-of-Linf version. The question is if this way of thinking is sensible, and gives sensible results for nobles. Do you have any thoughts on that?

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan: One thought about all of this is that what I'm suggesting is really just kind of just a "hack" to try to do things relative to this basis of feudal primes. I'm still not quite sure if starting from that basis is the best way to measure the complexity of an arbitrary feudal number although maybe there's some way to get it to work.

Two criteria that we may want any decent complexity measure to meet, at least approximately, are the following:

1. We'd like the complexity of each rational to be the same as the usual log(n*d).

2. We'd like the complexity of each noble number to at least somewhat agree with its "inherent" complexity of the simplest rational generating that noble as the start of its zig zag pattern on the scale tree.

As previously mentioned there are a few different conventions for saying where the zig zag "starts"; I don't think it really matters as long as the complexity we have is at least somewhat approximately in agreement with just about any sensible way to do it. But we would like, for instance, the noble between 6/5 and 5/4 to be lower in complexity than the one between 6/5 and 7/6. In fact a pretty easy way to try this would be to see what we get for the nobles between two superparticular ratios.

I guess the question is, then, given some feudal number, how do we obtain the unique prime factorization? Is there some kind of generalization of the Euclidean algorithm?

Reply

31 w

Dave KeenanAuthor

Mike Battaglia You wrote: "I guess the question is, then, given some feudal number, how do we obtain the unique prime factorization? Is there some kind of generalization of the Euclidean algorithm?"

I'm pretty sure there is. I thought it would be in either Dekker or Dodd, but I can't see it.

https://staff.fnwi.uva.nl/t.j.dekker/Pr ... Primes.pdf

https://archive.org/det.../numbertheory ... d/mode/2up

But do we really need it? Factoring rationals is easy. You just do the ordinary prime factorisation, then split the primes that end in 1, 5, or 9. My conjecture is that the nobles never have more than one prime in the denominator and one in the numerator, so you can factorise them by fibonacciing them backwards until you get a negative number.

Reply

31 w

רועי סיני

Mike Battaglia If you are still interested, there is a way to factor a feudal number into primes, but I don't know of a way as simple as the Euclidean algorithm.

First, you find the gcd of the wooden and golden part, this the greatest ordinary natural number that divides this feudal number. Factor it, and divide by it to get a number with no ordinary integer factors.

Then, factor the new number's arithmetic norm. Because of the multiplicativity of the norm, the factors will be the norms of the primes that divide it. To find the feudal primes this primes factor into, you need to solve the equation 1+x-x² = 0 mod p, which means find (1±√5)/2 mod p (where √5 is a number that when squared becomes equivalent to 5 mod p). Each solution x will have that 1+xϕ is divisible by another feudal prime that divides p. Use the Euclidean algorithm in ℤ[ϕ] (which is an Euclidean domain) to find one of fp, Fp, where the other is minus its conjugate. To check which of the primes divide your number, multiply it by one of them. If the result is divisible by p (i.e. both the wooden and golden parts are divisible by p), this means that the *other one* divides your number, and the quotient is the product you got divided by p. If one of the two feudal primes doesn't work, the other one should. In the case of 5, if you multiply a number whose norm is divisible by 5 with f5 you'll always get a feudal number divisible by 5 and it divided by 5 will be the original number divided by f5.

Reply

18 h

Dave KeenanAuthor

Mike Battaglia I think we are free to adjust the complexity of the phi-coordinate in whatever way seems to make sense psychoacoustically. I don't think there is any mathematics that will tell us what that weighting should be. Only our ears. But as you say, we want to preserve the usual complexity of the rationals, so that severely constrains what we can do with the non-wooden feudal primes. The complexity of f5 has to be the square root (or half if we're talking logs) of the complexity that 5 used to be. And otherwise the complexities of fp and Fp have to multiply (add in log space) to what p used to be. And as you say, we want to apportion it between the small-f and big-F factors in a way that reflects the complexity of the "simplest adjacent mediends" (SAMs) [to coin a term] of the noble number. Simply using the real value (or its log) of the non-wooden prime seems to lean in the right direction. Nobles that are the ratio of big-Fp's seem to have more complex SAMs than the ratio of the correspond little-fp's.

So I think it's only the complexity of the phi-coordinate (the unit rather than the primes) that we're free to fiddle with to our hearts (or ears) content.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia Do you think that "f" and "F" is a good notation for these complementary primes? Or is posterity going to curse us for the lack of visual distinctiveness and the two-syllable pronunciation "small eff" and "big eff"?

An alternative would be to use "f" and "g" which are quite distinct visually and only one syllable ("eff" and "gee"). They make me think of the orange and blue buttons on the first "computer" I bought with my first pay-check back in 1976.

https://thimet.de/CalcC.../Calculators/ ... 5C-2-M.JPG

Problems with "f" and "g" are that it's not so obvious which one is bigger, and "g" is not an abbreviation of anything useful that I can think of.

So I still prefer "f", "F" over "f", "g". But I just thought I should ask.

thimet.de

THIMET.DE

thimet.de

thimet.de

Reply

Remove Preview

31 wEdited

Mike BattagliaAdmin

Dave Keenan I think f and F are alright. But note my previous post about how the complexities of f11 and F11 don't need to multiply to the complexity of 11 if we do the L1 of Linf norms thing. I think the better way forward is just to see what gives sensible results for nobles though.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia That all sounds reasonable to me. You're a smart guy. I trust you will figure out something that is a good compromise between keeping the math simple and modelling the psychoacoustics.

I'm going to have to take a back seat for a while, to catch up on other projects. As well as the RTT "textbook" I'm writing with Douglas Blumeyer, I'm designing and building a movable-fret 7-string electric guitar. It's an evolution of the choob, but oval rather than circular, and made of wood, not carbon fibre, and with stainless-steel (movable) frets, not nylon.

Reply

31 w

Mike BattagliaAdmin

Some preliminary results from Graham's temperament search suggest that 72, 46, 27e, 43 are very good in the 2.3.5.7.11.phi subgroup, and to a lesser extent 26 and 19. Part of the problem, again, is that the algorithm is with the default log(p) weighting we're even prioritizing phi over 2, which is one reason why 26 does so well.

We can also do something similar by adding a factor of phi to the Riemann zeta function, though the results are really weird (if we just use a default weighting of phi, which again is stronger than that of any rational number). Results:

EDIT: A typo in these links made all of these results wrong! See post below

https://www.wolframalpha.com/input?i=pl ... +y%3D0..40

https://www.wolframalpha.com/input?i=pl ... +y%3D0..40

https://www.wolframalpha.com/input?i=pl ... +y%3D0..80

https://www.wolframalpha.com/input?i=pl ... y%3D0..100

Note 30 is even better than 31 using this, again because of the extreme emphasis on phi.

This is probably not the best way to do it. The Dedekind zeta function, however, is a generalization of the Riemann zeta function to arbitrary number fields, including Q[sqrt(5)], so I'm kind of curious what that looks like.

Wolfram|Alpha: Making the world’s knowledge computable

WOLFRAMALPHA.COM

Wolfram|Alpha: Making the world’s knowledge computable

Wolfram|Alpha: Making the world’s knowledge computable

Reply

31 wEdited

Paul ErlichAdmin

Mike Battagliathat's the same subgroup I had searched earlier as well as more recently for the Graham Breed's app results I posted here (with error parameters 1.25 and 4.27, respectively).

Reply

31 w

Mike BattagliaAdmin

Paul Erlich, FWIW, I tried the Dedekind zeta function in Mathematica, and got this. It likes 71 better than 72. Not sure what on earth to make of it. I'm not sure how this is weighting phi, really.

No photo description available.

Reply

31 w

Mike BattagliaAdmin

Actually, I think that there's no real weight for phi at all; instead there's just one single term for the entire phi-equivalent ideal for each feudal prime, so that's that.

Reply

31 w

Mike BattagliaAdmin

I guess one question is if this generalized zeta function has any use at all. I mean, algebraic number theory has gotten us this far, and it's the "zeta function" associated with Q[sqrt(5)], but some of these results are weird. I wonder if they make sense, though, and if there's some reason 71-EDO is better than 72-EDO with the feudal numbers. Maybe 71 approximates all of the noble numbers plus JI better than 72, for instance. Who knows.

Reply

31 w

Mike BattagliaAdmin

It also likes 28 better than 27. I dunno, this is pretty out there. Maybe it's another situation where the "natural" weighting that it's using isn't the one we really care about.

No photo description available.

Reply

31 w

Mike BattagliaAdmin

Just a quick note that the original posts about adding phi to zeta had a typo and were wrong. The correct version is here:

https://www.wolframalpha.com/input?i=pl ... +y%3D0..40

It really likes 26, 36, 46, and 72. 46 seems like a standout. Surprised 27 didn't turn up quite as much.

Wolfram|Alpha: Making the world’s knowledge computable

WOLFRAMALPHA.COM

Wolfram|Alpha: Making the world’s knowledge computable

Wolfram|Alpha: Making the world’s knowledge computable

Reply

30 w

Paul ErlichAdmin

10, 26, 62, 72?

Reply

30 w

Mike BattagliaAdmin

Yeah, also 10 and 62.

Reply

30 w

Paul ErlichAdmin

Mike Battaglia you mean they are repeated?

Reply

30 w

Mike BattagliaAdmin

Er, what do you mean?

Reply

30 w

Paul ErlichAdmin

Mike BattagliaI listed 10 and 62, and you said also 10 and 62, so I thought maybe you meant two closely-spaced peaks at 10 and 26. But, I guess that doesn't happen with zeta . . .

Reply

30 w

Aggelos Boshidis

David Guillot

Reply

31 w

Dave KeenanAuthor

I have finished the new version of my series of articles, updated to use the superior choice of fundamental primes:

viewtopic.php?f=21&t=557

Noble frequency ratios as prime-count vectors in ℚ(√5) - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) - The Sagittal forum

Reply

Remove Preview

31 w

Dave KeenanAuthor

I just posted another article in the Feudal Manifesto series, in which I clarify a bunch of things, such as the (non) relationship between noble numbers and combination tones, and I give unique identifiers for noble numbers using an "ennoblement function" n, where e.g.

n7/4 = (7+9ϕ)/(4+5ϕ) = f31/f11 ≈ 1002 ¢

viewtopic.php?p=4620#p4620

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Reply

Remove Preview

30 w

Mike BattagliaAdmin

Dave Keenan it looks like the ennoblement function is the same as my "~" from before, or rather the first version of it?

You claim in this that noble numbers are not sensitive to mistuning in the sense that there are no partials to try to get to cohere, but I don't really agree with that. In particular the thing you are now calling n6/5 seems to be pretty fragile. There is a real sweet spot somewhere in the 330-340 cent range where things are maximally "stanky," let's call it, and then things kind of fall off pretty steeply, particularly as you tune sharp.

Reply

30 w

Dave KeenanAuthor

Mike Battaglia Thanks for this feedback. I have updated my post accordingly. Search on your name within it. Yes, I think my prefix "n" is the same as your first definition of your postfix "~". However you confused the issue at the time, by giving as an example, that (6/5)~ was the noble at 339c. It seems you're still confused about that. Or one of us is. Because you imply above, that I am now calling the noble near 339c "n6/5" when in fact I am calling it "n5/4". "n6/5" is the noble near 284c.

Reply

30 w

Mike BattagliaAdmin

Ah, I'm sorry, you're right. So it's the other variant then.

Reply

30 w

Dave KeenanAuthor

Mike Can you tell me how to get the simplest child of a given ratio, without having to look up the SB-tree, perhaps related to that trick you did using ModularInverse?

Reply

30 wEdited

רועי סיני

Dave Keenan If you still need this, there is indeed a way to do that – given a/b use the modular inverse trick to find numbers c, d s.t. ad-bc = ±1. If c < a/2 then (a+c)/(b+d) is your simplest child, otherwise it's (2a-c)/(2b-d).

Reply

18 h

Cmloegcmluin Xenharmonic Feisbeuk

I was having trouble visualizing it otherwise, so I put this diagram together to help me understand how Dave and Mike's ennoblement function gives a one-to-one correspondence between non-subunison rationals and noble numbers. Hope it helps others too:

No photo description available.

Reply

30 w

Mike BattagliaAdmin

Yes, basically. You can also get it by truncating the continued fraction expansion.

Reply

29 w

Cmloegcmluin Xenharmonic Feisbeuk

Mike hmm. I see you've mentioned this above. so given n(n/d), you can find the other mediend by appending a single 1 to the continued fraction form of n/d, and you can find the noble by appending infinite 1's. that's pretty handy.

Reply

29 w

Mike BattagliaAdmin

The < and > "noble flat" and "noble sharp" operators also have a nice visual representation here, being just the noble mediant of the ratio and its left or right child.

Reply

29 w

Cmloegcmluin Xenharmonic Feisbeuk

Mike yes, I can see a separate use for those, because n(n/d) gives you the simpler of the two children, which 50% of the time is the left child and 50% of the time is the right child, based on n/d's location in the tree.

Reply

29 w

Mike BattagliaAdmin

Right, although note that you also have 11/9< is the same as 6/5> which is the same as 5/4<, all of which are n(5/4). Kind of an interesting property.

Reply

29 w

Steve Martin

Phi rhymes with sky, or with key? I agree with Dave but have heard it the other way (on you tube), is it a Aus/UK vs US thing?

Reply

30 w

Dave KeenanAuthor

Steve Martin I understand that when Greeks say their alphabet it's pronounced like the English word "fee". So no one can fault you for pronouncing it that way. But in all the maths and physics lectures I attended in Australia, the math symbol was pronounced like the "Fi" in "Wi-Fi". I can't speak for any other jurisdictions, but I read somewhere that this was generally the case in English-speaking maths and science usage.

Same thing with π. Greek "pee", English "pie".

Reply

30 wEdited

Steve Martin

Notes on where I have got with this, below

Reply

30 w

Steve Martin

As a nerdy mathematician I found the quadratic fields in Hardy and Wright very interesting; here a few general notes that you probably know already: Q(root(m)) is a field for any nonzero integer m; if negative then we have a subfield of the complex numbers, if positive then a subfield of the real numbers; the norm for an element X=c+d*root(m) where c and d are rational is N(X)=(c+d*root(m))*(c-d*root(m)), this is always positive if m is negative, not necessarily so if m is positive, note it yields the formula we have been using in Q(root(5)) when re-expressed in terms of phi. The norm obeys the rule N(X*Y)=N(X)*N(Y); units are elements e having N(e)=+/-1. Any two elements related by X=Y*e where e is a unit are called associates. In the field "integers" are defined in terms of polynomials that they satisfy. The field is sometimes a UFD and sometimes has a Euclidean algorithm. If a non-UFD then you can have irreducibles that are not primes, you have to define each carefully.

But Q(root(5)) is a UFD and has Euclidean algo; it turns out that the units are phi^n where n is a (usual kind of) integer, and that the "integers" in this field are X=c+d*(phi) where c and d are (usual kind of) integers. Note, in this field any associate of an integer is an integer, and of a prime is a prime. It is found that any (usual kind of) prime of form 5*n +/- 2 is irreducible in this field so is a prime here, but those of form 5*n +/- 1 can be factorised and their irreducible factors are primes. Finally, root(5) =-1+2*phi is irreducible hence a prime (and 5 is not).

A final general note, in terms of Dave's diagram, you can find units arbitrarily far from the origin, and the same for associates of f5, f11, etc.

Reply

30 wEdited

Dave KeenanAuthor

Steve Martin Not root(phi), just phi. Phi already contains root(5), so root(phi) would be like a fourth root.

Units are elements e such that N(e) = ±1, not only 1.

N(ϕⁿ) = (-1)ⁿ for all ordinary integers n.

Thanks for reminding me to add that to my forum article.

Reply

30 w

Steve Martin

For RTT use, i was fairly certain early on that we should be using the primes as defined above; and indeed your table confirms it. Adding phi alone to the usual basis 2.3.5.7... does not get many noble ratios, because the group is multiplicative and only the first two noble numbers are simple multiples of phi; but with f5, f11 etc we can reach more as shown in the table, and maybe all as you conjecture. However, we cannot do without phi as well, as you already observed; a simple reason is that f1 and F1 feature often in the table; more explicatively we can say that although it is a unit, it is not 0 cents, or that phi allows us to move between associates cheaply, so that our choice of canonical primes is less significant.

My first look included all primes up to F11, but to simplify things I focused on 560c (n3/2) and 422c (n4/3) which only need f11 and f5 hence my basis 2.3.f5.F1.f11. I like the look of 26&43 for this (very close to 69edo). Generators now c1201.5c, c139.3c. (When 7 and F11 were included, the octave went to c1198...c) Not the only possibility of course, but it's a start.

http://x31eq.com/cgi-bin/more.cgi?r=2...

X31EQ.COM

1200.000, 1901.955, 1393.157, 833.090, 1815.643-limit Rank 2 Temperaments

1200.000, 1901.955, 1393.157, 833.090, 1815.643-limit Rank 2 Temperaments

Reply

30 wEdited

Dave KeenanAuthor

Steve Martin It's great that Graham built his temperament finder to work with inharmonic primes, so we can just plug these things in, although as Mike has noted, it might benefit from weightings other than the log of their real value being applied to the feudal units and primes.

Reply

30 w

Steve Martin

Commas: sqrtphi: can we have the square root of phi in our setup? Not sure, but plough ahead anyway: even without x31eq's help if we assume that the generator is being n4/3 i.e. f11/f5 as well as sqrtphi, then:

(f11/f5)^2 = F1.

This comma is given for my 26&43 alongside:

3/2 = phi^3/f11

9/4 = f5 (of course!)

f5^3 = phi^5

et al.

Reply

30 w

Dave KeenanAuthor

Steve Martin I don't understand why we'd want √ϕ. The integers of ℚ(√5) are not c+d√ϕ as you say above, but rather c+dϕ, where c and d are ordinary integers. ϕ = ½+½√5

Reply

30 wEdited

Steve Martin

Dave Keenan yes, crikey, that's a typo re root(phi), will fix - done; but my mention of it in connection with a comma was intended, because Paul mentioned the temperament of that name and indeed it does come up in the answers from Graham's app.

Reply

29 wEdited

Dave KeenanAuthor

Steve Martin Then I'll take your question to be, "Can we have the square root of phi [as a generator] in our setup?"

Of course we can. And we don't need it (or want it) in our vector basis in order to do so. Generators are not constrained to have integer entries in their vectors.

Reply

29 w

Mike BattagliaAdmin

One interesting thing I've been thinking about recently, which maybe Paul Erlich and Dave Keenan would have some thoughts on:

The mediant can be thought of as a weighted arithmetic mean, where the weights are proportional to the denominators of the ratios. For instance, the mediant of 5/4 and 6/5 is equal to (5/4)*(4/9) + (6/5)*(5/9) = 11/9. This tells us that the mediant will be in between the two original ratios, and weighted closer to the more complex of the two. These are both useful properties for the same reason that the Lagrange point between the Earth and Moon is slightly closer to the Moon.

We may ask what were to happen if we were to derive something similar, but with the geometric mean replacing the arithmetic mean above. We would like to derive something like a "geometric mediant," or at least a "geometric noble mediant."

The main snag is that a geometric mean of two ratios is not, in general, another ratio, so it is an interesting question on how we should recurse to get our noble mediant. There are several interesting ways to answer this question, and though I'm still trying to figure out which is the "right" way, all of them basically lead to the same big picture of everything: instead of adding new "feudal primes," we instead change our monzos so the *coordinates* can be feudal numbers rather than regular integers. This is a fairly useful property as we don't need to add any new feudal primes or expand the dimensionality of our vector space at all to add these new phi-based intervals; for the 5-limit we get the same 3D space but with generalized coordinates that can be feudal numbers. You can also think of this as a 6D space in which the basis elements are 2, 2^phi, 3, 3^phi, 5, and 5^phi.

The same space of intervals is generated by all of the various candidates for "geometric noble mediant" that I've looked at so far, with slight differences on which particular interval in this space, for instance, we are identifying as the "geometric noble mediant" of 5/4 and 6/5. The results are all typically pretty close to one another, and also tend to be pretty close to the regular noble mediant, at least for intervals which are mostly similar in size. "Geometric phi", on the other hand, is very different from regular phi and turns out to be about 742 cents. I'll write some details once I have it together, but has anyone thought about this kind of thing?

Reply

29 wEdited

Paul ErlichAdmin

I'm familiar with 742 cents as 1200/Phi or 1200*phi cents. Not yet grasping how to derive it without the octave, but sounds interesting!

Reply

29 w

Cmloegcmluin Xenharmonic Feisbeuk

right that’s apparently another way to describe logarithmic phi: https://en.xen.wiki/w/Logarithmic_phi

Logarithmic phi - Xenharmonic Wiki

EN.XEN.WIKI

Logarithmic phi - Xenharmonic Wiki

Logarithmic phi - Xenharmonic Wiki

Reply

29 w

Paul ErlichAdmin

Cmloegcmluin Xenharmonic Feisbeuksure but the octave is right there in the definition.

Reply

29 w

Cmloegcmluin Xenharmonic Feisbeuk

right, just sharing in case any links there give Mike additional insight he was asking for

Reply

29 w

Dave KeenanAuthor

Mike Battaglia You probably should have started a new post for this. I note that the geometric mean is just the exponential of the arithmetic mean of the logs, hence the appearance of "logarithmic phi" or 742 cents.

I note some misnomers: What you're calling a "noble geometric mean" should instead be called a "phi-weighted geometric mean" or similar. "Noble" is not synonymous with "phi-weighted". "Noble" means the results are noble numbers, which already have a well-accepted mathematical definition.

When you wrote that the coordinates can be "feudal numbers" instead of regular integers, you should have said they can be "feudal integers". Feudal numbers include quotients of feudal integers.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan Thanks, glad you found it interesting!

Reply

29 w

Mike BattagliaAdmin

Paul Erlich Dave Keenan OK, turns out there really is one "best" way to do the thing I was looking for. Although I derived it from this standpoint of replacing the arithmetic with the geometric mean, it turns out to simplify to something so simple it's almost absurd. However, because it was derived in a rigorous way, this simple idea turns out to have a number of interesting properties.

It all simplifies to this: let's say that you have two ratios a/b and c/d, and you want the noble mediant from a/b to c/d, which is (a+b*phi)/(c+d*phi). Then the "geometric noble mediant" of those two is simply (a/b)^(1-p) * (c/d)^(p), where p = phi-1 ≈ 0.618. In other words, you draw a ruler in cents space going from a/b and c/d, and then go 61.8% of the way there. This can be viewed as an approximation of the original which is better as the two ratios get closer together in size (and I guess complexity).

To derive this formally, I had this entire thing worked out with "radical numbers," which are rational numbers raised to a rational power - this structure naturally arises when looking at geometric means of rationals. Each radical number is just a monzo with rational coefficients. Then, for each radical number, we can say the "geometric mediant" is the monzo you get taking the mediants of each coordinate separately, then you can recurse to get a "geometric noble mediant," although it becomes an interesting question what we call "noble" in this situation. You get an infinite set of independent Stern-Brocot trees and a sort of "continued fraction matrix" and all kinds of weird stuff like this.

*HOWEVER*, we don't really care about arbitrary radical numbers, really, but only rational numbers which have integer monzo coefficients. Thus, in this situation, the entire thing almost becomes trivial - basically as trivial as taking the "noble mediant" of two integers. All of the "denominators" of our coordinates are 1 and everything simplifies to the nice simple property above.

However, because it was derived in such a rigorous way, it turns out to have some really amazing properties:

Because the geometric and arithmetic mean of two ratios are approximately equal if the two ratios are not too far apart, this turns out to approximate the noble mediant very well for pairs of ratios which are "reasonably" close together, and I will post a table of results about that. I am quite curious to see if the geometric or additive noble mediant is a better approximation of local maxima of HE.

The geometric noble mediant is transpositionally invariant, in that the geometric noble mediant of 2/1 and 3/1 is the same as the geometric noble mediant of 1/1 and 3/2 transposed up an octave. This parallels how the regular noble mediant is "additively invariant", in the sense that NobleMediant(z+a/b, z+c/d) = z*NobleMediant(a/b, c/d) for any integer z. In this situation, we have GeometricNobleMediant(q*a/b, q*c/d) = q*GeometricNobleMediant(a/b, c/d) for any *rational* number q.

Lastly, it doesn't require we add any new feudal primes or anything like that, just let our monzo coordinates be feudal integers rather than regular integers. Or we could say our basis now becomes 2, 2^phi, 3, 3^phi, 5, and 5^phi and the coordinates remain regular integers.

If for any reason people later care about these "radical numbers" - kind of interesting as they are basically equal divisions of some rational number, which are certainly perceptually relevant for a very different set of melodic reasons - we can extend these results to them and you get some interesting structure.

I will post a table of comparisons between regular and geometric noble mediants for some superparticular ratios, keeping in mind that it's still kind of tricky to call anything "noble" in this situation.

Reply

29 wEdited

Paul ErlichAdmin

Mike BattagliaiCYMI above, FWIW, the nearest HE local maximum to phi in most of the original graphs was at 845 cents.

Reply

29 w

Mike BattagliaAdmin

Paul Erlich to which phi?

Reply

29 w

Paul ErlichAdmin

Mike Battaglia to "acoustical" phi at 833 cents.

Reply

29 w

Mike BattagliaAdmin

Paul Erlich how would that be relevant in this situation?

Reply

29 w

Paul ErlichAdmin

Mike Battaglia I don't know if it would be. That's why it was labeled "FWIW". You said something about "a better approximation of local maxima of HE" and I was just pointing out the 845-cent local maximum since I pointed that out in response to Dave Keenan much earlier in this thread. So if you had something which predicted that rather than 833 cents, instead something more like 845 cents is near a local maximum, that would appear to be along the lines of what you said you'd be curious about. But perhaps I'm missing something that makes this inapplicable to the thing you're curious about somehow.

Reply

29 w

Paul ErlichAdmin

Maybe Dave Keenan and/or Cmloegcmluin Xenharmonic Feisbeuk are grasping more of this at this juncture and could help explain why my observation wouldn't be pertinent here.

Reply

29 w

Mike BattagliaAdmin

Paul Erlich Rather than having Dave and Douglas explain to you why your observation isn't pertinent, I suggest trying to explain to me why it is. Given any arbitrary interval, one can look for the nearest local maximum, and I'm sure there are many noble mediants, as well as geometric noble mediants, that are approximately equal to 845 cents. Which one am I supposed to be looking for?

Reply

29 w

Mike BattagliaAdmin

Paul Erlich I'm not sure if your Facebook client is glitching out but you're responding in all kinds of different subthreads... I think your comment was meant to be here. The geometric noble mediant between 5/3 and 8/5 is 840.68 cents.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia no I don't see anything glitching out. I respond to things I see in different subthreads each in their own subthreads as usual, though obviously some subthreads spun off from others initially and so the conversations end up being related. My comment was a reply to your list of pairs of ratios with the different mediants for each pair. I looked at that and then replied with another such pair.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia I don't know if refreshing the thread would help show more of what I wrote, but it often does.

Again I said "FWIW"; if the answer is that it's worth nothing, then that's fine . . . in which case clearly I am missing something but don't know what it is.

Again, all I know was that

(1) in response to the OP, suggesting that phi itself (833 cents, the prototypical noble mediant, and certainly the one between what are normally the two nearest local minima of HE, 8:5 and 5:3) is near a local maximum of HE, I pointed out that in most of the original graphs, the only local maximum in that region was at 845 cents;

(2) you said something about curiosity whether your approach might offer "a better approximation of local maxima of HE";

(3) I asked whether it might, correspondingly, better match this data point about a local maximum of HE.

Perhaps that's a poorly thought-out question, and I have no reason to suppose it isn't, but I didn't see the harm in throwing it out there. But clearly I don't understand something about the idea, which is why I asked for help. Perhaps someone can see where I'm coming from and point out why it doesn't apply in this situation. And I didn't see the harm in asking that either.

Reply

29 w

Mike BattagliaAdmin

Paul Erlich perhaps I misunderstood; I was interpreting this in the context of our original conversation about "logarithmic phi," which is about 742 cents and very far from 845 cents. See also my note about how it depends on which pair of ratios you ask for; the geometric noble mediant between 5/3 and 8/5 is about 841 cents and closer to your local maximum, but you get different results for other pairs of starting ratios (like 3/2 and 5/3), for instance. In general, it seems like you start to get sensible results as the two ratios get about a half step apart.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia I think the pair of ratios Dave and Margo start with in their article are the two flanking consonances or local minima.

Reply

29 w

Mike BattagliaAdmin

Paul Erlich it may be a good idea to try this with neighboring ratios that are local minima of HE and see what the local maxima are. Do you have a list of of local minima and maxima for different values of s?

Reply

29 w

Paul ErlichAdmin

Mike Battaglia I don't have that handy, and though I've posted a lot of relevant numbers and graphs over the years and Steve could calculate them again, I imagine your own HE calculator could readily be called into service? The graphs with local maxima 845 I posted earlier in the thread appeared with both Farey and more "straight" versions, but with a consistent s value of I believe 1%. We also saw that when s is increased all the way to 1.5% so that 8:5 becomes a hillock, more like an inflection point than much of a visible minimum, the nearest maximum shifts way over (to 829 cents -- but only a hair higher entropy than 8:5).

Obviously the global maximum that might be near 40-70 cents might be a great test case, though it's hard to necessarily say what the next local minimum is (very sensitive to s), and even the global maximum depends a lot on N (and you apparently found a way to find the global maximum in the limit as N->infinity for a given value of s, which is incredibly impressive and interesting, but it seems to me that there *might* be some psychoacoustical equivalent of an actual maximum complexity N of ratios in the "template" for fundamental-finding for an interval in a given register, given that our hearing frequency range is finite, etc.).

Reply

29 w

Paul ErlichAdmin

I imagine any local maximum can be pushed right up to the "edge" of (arbitrarily close to) the next local minimum (the more complex of the two flanking it) just by tweaking s to just below where the local minimum disappears entirely Mike Battaglia.

Reply

29 w

Mike BattagliaAdmin

I think that 40-70 cent maximum will be the hardest to get right with this, as it'll need to be "phi%" of the way between 1/1 and the next local minimum, meaning there'd need to be a minimum at like, 60-100 cents or so.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia yeah for sure! but since it seems a very shallow local minimum can have the next local maximum on one side arbitrarily close to it, there are liable to be cases even harder to get right (in terms of quotient of the two distances of the local maximum from its flanking local minima shooting off to infinity or zero rather than any kind of average or mean) for many values of s.

Reply

29 w

Mike BattagliaAdmin

Paul Erlich well I'm pretty sure all of these results are within the margin of error of how good HE is supposed to be, so if things are a few cents off I don't think it matters much. Actually, I think the true local maximum between 6/5 and 5/4 is somewhere closer to maybe 333 cents, tbh. Either way, I think this is all doing pretty well.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia well in the scenario I described, there can be more than a few cents from being right up against a flanking local minimum (say 820 cents, right up against 8:5) to being some kind of average or golden mean or whatever 845 cents is. But since the local minimum in question (here 8:5) is on the verge of disappearing in the scenario I described, somehow this is still within some margin of error of something somehow I suppose?(?idk?)

Reply

29 wEdited

Paul ErlichAdmin

^edited

Reply

29 w

Mike BattagliaAdmin

Paul Erlich: right, but if we're only viewing the results as valid for neighboring minima (relative to some s), then this problem goes away. Or rather, we could say that for any noble mediant - regular or geometric - there is some choice of "s" such that this is approximately the local maximum between two neighboring local minima.

Reply

29 w

Paul ErlichAdmin

Mike Battagliayes, "some choice of s" is probably right.

Reply

29 w

Mike BattagliaAdmin

BTW, I sometimes think that maybe the min-entropy is what we should really be looking at... I mean, it's about as simple as it gets. For instance, with the laplace distribution it's literally just a piecewise-linear graph (see picture). For the Gaussian it's piecewise-parabolic and so on.

No photo description available.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia FWIW I don't find any aural "landmarks" that make it easy to tune to local maxima by ear, i.e., I think the minima should look a good bit pointer than the maxima (though still round if it's just HE and not roughness etc.). It seems like a=2 already makes them equally pointy, and beyond 2 the maxima are pointier than the minima, based on playing around on your calculator, though I was using the Gaussian and not Laplacian distribution.

Reply

29 w

Mike BattagliaAdmin

There is clearly a point between 6/5 and 5/4 that is maximally "stanky." Have you considered consulting your internal stankiness barometer for measuring this? As a guitarist I would imagine it is highly well developed

Reply

29 w

Paul ErlichAdmin

Mike Battaglia I've spent hours just jamming in 7-equal because of how stanky it is! But I've also spent hours trying to identify some kind of local maximum of discordance and nothing seems to hold up long-term. Just the other day on a SoundCloud stream it sounded to me like Joseph McDermott used a 7-note mode found within Miracle (maybe within Blackjack, don't remember) featuring a 333-cent interval over his tonic and a lot of 333-cent harmonic intervals. I kept going back and forth and trying to decide whether it sounded consonant or not. The more I dialed in the treble so as to hear the higher harmonics, the more it seemed to stand out as dissonant, but still in some kind of medial unfully-realized way, and that's probably due to roughness rather than HE, IMHO.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia I don't know what you mean -- the minima *were* neighboring in my example.

Reply

29 w

Mike BattagliaAdmin

I thought you were saying that if you decrease "s" enough, you can make the nearest maximum larger than some local minimum arbitrarily close to that minimum.

Reply

29 w

Paul ErlichAdmin

Mike Battagliaright, or it might be the nearest maximum smaller than that local minimum. But the same minima still neighbor that maximum, so I don't know what you meant by "if we're only viewing the results as valid for neighboring minima". For the case of 8:5 and 5:3, they are consistently the neighboring minima of the maximum which takes on values from 845 cents to <820 etc. cents as s is increased from 1% to ~1.6%.

Reply

29 w

Mike BattagliaAdmin

I was just saying that as s decreases, then there will be new minima that appear in between 8/5 and 5/3. Once that happens, it no longer makes sense to take the noble mediant of 8/5 and 5/3 at all, but rather 8/5 and whatever the next local minimum is, and of course this will be closer to 8/5, and thus so will the local maximum between the two (and the noble mediant between the two and so on).

Reply

29 wEdited

Paul ErlichAdmin

Mike Battaglia in my example there were no other local minima between 8/5 and 5/3 anywhere in the range of s values mentioned

Reply

29 w

Mike BattagliaAdmin

Paul Erlich Oh, you're saying *increasing* s brings the maximum to 1.6%. Yes, for the Gaussian that's right, although again I think much of this probably depends on what probability distribution you're using...

Reply

29 w

Paul ErlichAdmin

Mike Battaglia well when the Laplace distribution is used, the HE curves become "fractal", and there seems to be an infinite density of minima and maxima regardless of s, and 8:5 and 5:3 never become neighboring minima . . . right?

Reply

29 w

Dave KeenanAuthor

Mike Battaglia I think you should replace "GeometricNobleMediant" with "GeometricPhiMediant".

I don't understand in what sense it is more "clean" to have numbers like 2ᵠ, 3ᵠ, 5ᵠ in our basis as opposed to -1+2ϕ, -2+3ϕ, -1+3ϕ.

Our ears should be the ultimate arbiter, not any particular HE curve. As Paul has noted, the local maxima of HE are very unstable wrt the parameter s. I only mentioned HE maxima in my OP because I knew it was something people reading this group would be familiar with. Noble number frequency ratios at least have the guaranteed property that they are maximally distant from all nearby simple ratios, with their distance being a monotonically-increasing function of their simplicity. Other kinds of phi-weighted mean do not have this property.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan: the thing about ears vs HE is quite a strawman. Did you read my entire post?

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Which entire post? When I read posts, I read entire posts. But there are so many, and facebook seems so arbitrary in what it reveals when. Why don't you just explain why you think ears vs HE is a strawman.

Reply

29 w

Mike BattagliaAdmin

It's a strawman because it isn't a response to anything I've written. I gave a giant list of noble mediants and showed that in most situations, the "geometric" version is usually only off by a few cents from the regular version. Are you saying that you listened to all of these and your ears like the regular version better? Or are you just nitpicking random stuff or what? It seems unrelated.

Reply

29 w

Dave KeenanAuthor

Or ignore the strawman and address the other thing, where noble number frequency ratios at least have the guaranteed property that they are maximally distant from all nearby simple ratios, with their distance being a monotonically-increasing function of their simplicity, while other kinds of phi-weighted mean do not have this property.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan they do have this property, since I keep telling you it's usually only a few cents off.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia There's no need to be rude.

At that time, I had not yet seen the giant list of noble mediants that showed that in most situations, the geometric version is usually only off by a few cents from the regular version.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan My response wasn't intended to be rude at all; apologies if it came off that way. It was just to suggest slowing down and catching up on some of these posts. Most of the points you brought up (the use of "feudal numbers" rather than "feudal integers," the questions about how to best generalize the word "noble," the quality of the approximation) are things that are mentioned in the very post you are responding to and posts I thought you'd already seen.

It seemed like you aren't really interested in this idea and were mostly looking to debate things. That may have been a misguided interpretation, but anyway I would just suggest to give it a chance. It's a simple idea: a mathematical approximation that has a lot of nice properties, including transpositional invariance. It exists in a *much* lower dimensional vector space than the original and is thus computationally much easier to deal with, and for the most important pairs of ratios, it's only a few cents off, often in the direction that's closer to the HE maximum. I thought that may be useful to some people.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia When you say it exists in a much lower dimensional vector space than the original, that's the first thing I've read that makes me understand why it might be a better basis. I understand you to be saying, in effect, that the 32 simple nobles that I've listed (the only ones I consider musically relevant), can be approximated well enough, with many fewer entries in the vectors, when the usual prime basis is augmented with a few pᵠ for a few low (ordinary) primes p, compared to the large number of feudal primes required (which I noted as a problem in my first forum article).

If so, that's great. I'd love to see a table similar to my table 14 or 15, with the vectors replaced with those using your new basis for the approximations of those nobles.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan it means that *all* of the noble mediants that you can generate from some p-limit fit into a Z-module of twice the dimension, where we add one corresponding p^phi for each prime p. So every possible 11-limit noble mediant fits into a 10-dimensional vector space. How large of a vector space do we need for your 32 nobles?

Reply

29 w

Mike Battaglia Paul Erlich Cmloegcmluin Xenharmonic Feisbeuk Steve Martin Here is a table/plot of my new preferred feudal primes — those that ensure that all rationals and the maximum number of nobles (all but 2 on each level of the Stern-Brocot tree) can be factorised without need of a unit, a power of ϕ, other than its zeroth power, i.e. 1.

viewtopic.php?p=4607#p4607

Be careful. The ink's still wet. 🙂

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) [-a+bϕ] - The Sagittal forum

Reply

Remove Preview

31 wEdited

Mike BattagliaAdmin

Thanks Dave Keenan, one question: the first 32 noble numbers seem to be a quotient of exactly two feudal primes (counting phi). Three questions

1. Do we know if a noble number can *ever* have a strictly "wooden prime" in its prime factorization, or will the factorization always be that of feudal primes and phi?

2. Do we know if a noble number can ever have more than two factors in its prime factorization?

3. Can we *ever* have any nobles which have, in the prime factorization, two complementary prime pairs, like fp and Fp for some p?

I think it's easy to see that no such noble number can be of the form fp/Fp given that your "square root" analysis is correct, because the two pairs will be of the form a+b*phi and a'+b*phi with the same b and thus the determinant of the matrix [a b;a' b] will be a multiple of b. But the question, I guess, if is we can ever multiply fp/Fp by something else to get a noble number.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia You wrote: "1. Do we know if a noble number can *ever* have a strictly "wooden prime" in its prime factorization, or will the factorization always be that of feudal primes and phi?"

No proof, but I conjecture that all nobles have at most one non-wooden prime in the numerator and at most one non-wooden prime in the denominator and never a wooden prime anywhere.

And further that, apart from ϕ and ϕ², all nobles have numerator and denominator either both small-f or both big-F, where f5 = F5 = -1+2ϕ = √5, and f1 = -1+1ϕ = ϕ⁻¹ and F1 = 0+1ϕ = ϕ¹.

"2. Do we know if a noble number can ever have more than two factors in its prime factorization?"

Not proven. But conjectured that never more than 2.

"3. Can we *ever* have any nobles which have, in the prime factorization, two complementary prime pairs, like fp and Fp for some p?

I think it's easy to see that no such noble number can be of the form fp/Fp given that your "square root" analysis is correct, because the two pairs will be of the form a+b*phi and a'+b*phi with the same b and thus the determinant of the matrix [a b;a' b] will be a multiple of b. But the question, I guess, if is we can ever multiply fp/Fp by something else to get a noble number."

No. Never the same p in numerator and denominator*. And other limitations on which pairs of p's make a noble. They all come back to the "unimodular" requirement for nobles. i.e. the determinant must be ±1.

*Unless you count my degenerate use of f1 and F1 as ϕ⁻¹ and ϕ in naming nobles. In that case you only get the same "p" top and bottom in the case of the noble ϕ².

Reply

31 wEdited

רועי סיני

Mike Battaglia For the case that you are still interested, here are answers to these quesrions:

1. They can't, because if we have a wooden prime that divides the numerator but not the denominator (or vice versa) this means that the numerator is another feudal integer times this prime, which means that both its wooden part and its golden part are divisible by this prime, and therefore the determinant of the matrix divisible by this prime and not ±1. This also means that f5² or any pair of (opposite) conjugate primes together can't be found in either the numerator or the denominator.

2. This is essentially equivalent to Dave Keenan's conjecture about the numerator and denominator always being prime or unit, which I disproved. Specifically, (13+19ϕ)/(11+16ϕ) = (-1+2ϕ)*(-2+3ϕ)*ϕ/(-1+6ϕ) = f5*f11*ϕ/F41.

3. Yes we can. The key to finding them is realizing that (a+bϕ)*conj(c+dϕ) = (a+bϕ)*(c+d-dϕ) = ac+ad-bd + (-ad+bc+bd-bd)ϕ = ac+ad-bd - (ad-bc)ϕ with a golden part of -(ad-bc), which means that if we want to find a golden mediant that has F11 on the top and f11 on the bottom, for example, we need to find a multiple of F11*conj(f11) = -F11² which has a golden part of ±1. F11² = 10+3ϕ, which can be multiplied by 4-ϕ = F11/ϕ to give a number with golden part of 1, which means that F11²/f11ϕ = (10+3ϕ)/(3+ϕ) = (13+16ϕ)/(4+5ϕ) is a noble that has both F11 and f11 in its factorization.

Reply

19 h

Dave KeenanAuthor

I have the new primes on the Stern-Brocot tree now, in both a+bϕ and fp or Fp form:

viewtopic.php?p=4609#p4609

I think it will be useful to have noble-flat and noble-sharp operators as you describe. And I no longer think that human-friendly unique identifiers for the nobles need to be based on their mediends, or any rationals. I'm getting quite used to treating the f11/f5, F31/F19 form as an identifier for the noble.

I describe how to mentally extract the a+bϕ form from the fp or Fp form, in the fifth paragraph of this post:

viewtopic.php?p=4607#p4607

It's possible that there is a mental, or simple pen-and-paper, method for extracting the two ratios at the start of the SB-tree-zigzag, from the fp₁/fp₂, e.g. F31/F11, form.

Reply

31 wEdited

Dave KeenanAuthor

It turns out: All you have to do to get the two simplest Stern-Brocot-adjacent ratios whose noble-mediant is given as e.g. F31/F11 is to first use the above-linked greatest-square-remainder-complementary-factorisation method on the numerator and denominator to get it to the form (-2+5ϕ)/(-1+3ϕ), then "fibonacci" the numerator and denominator forward exactly 3 steps. It's always 3 steps, irrespective of the noble. In this case we get:

-2 5 3 8 11

-- -- -- -- --

-1 3 2 5 7

So it's the noble mediant of 8/5 and 11/7. Of course it's also the noble mediant of any other pair of succesive ratios in that sequence (and its continuation). But 8/5 and 11/7 are the pair at the start of the zigzag. Prior to them there was a non-adjacency, a discontinuity, a "teleport". 3/2 and 8/5 are not directly connected on the SB-tree.

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan this is great stuff. Is this proven or just a conjecture?

Reply

31 w

Dave KeenanAuthor

Just conjecture. I know I'm bad. The way I state all these thing in my articles that I don't have proofs for. Particularly when I've watched those Matt Parker and/or Numberphile videos where, just when you think you've got some sequence nailed down, it goes off and does something crazy. But I keep rationalising it to myself with: Well it works for the "musically-significant" stuff so who cares if it breaks somewhere along the road to infinity. 😮

Reply

31 w

רועי סיני

Dave Keenan This is in fact true. If in the beginning you have a/b, c/d s.t. a and b are negative and c and d are positive and larger than their respective absolute values, then in the next step you'll get c/d, e/f s.t. 0 < e < c and 0 < f < d, which means, because of the ±1 determinant, that e/f is an ancestor of c/d in the SB tree. After the second step you get e/f, g/h, where now the first is an ancestor of the second, but not its direct parent because that is c/d. After the third step you indeed get i/j which is a direct child of g/h, and is its simpler child because e/f is not its parent.

You can run this process in reverse too - from a pair of parent and simpler child you first get the parent and an ancestor of it, secondly some other node descending from the same ancestor and finally the two numbers become negative and you get your primes in the normal form.

Reply

18 hEdited

Paul Johnson

Here’s a more practical question: 72edo gives fantastic primes and a near perfect phi. Do you think you could find something better using this method?

Reply

31 w

Paul ErlichAdmin

Paul Johnsonthe idea is to find a minimal basis of "primes" that will allow an array of noble mediants -- not just phi -- to be represented. Then the search for temperaments of any rank would proceed much as usual. If it were just a matter of phi or any set of already-known feudals (with the log weighting mentioned above) and some primes, you can do that with Graham's app already.

Reply

31 w

Paul Johnson

Paul Erlich ok here’s a dumb question, can all the noble mediants be found in Q(root(5))? Obviously phi can

Reply

31 w

Dave KeenanAuthor

Paul Johnson Yes. See: https://mathworld.wolfram.com/NobleNumber.html

Wolfram MathWorld: The Web's Most Extensive Mathematics Resource

MATHWORLD.WOLFRAM.COM

Wolfram MathWorld: The Web's Most Extensive Mathematics Resource

Wolfram MathWorld: The Web's Most Extensive Mathematics Resource

Reply

Remove Preview

31 w

Paul Johnson

Dave Keenan thank you

Reply

31 w

Paul ErlichAdmin

e.g., http://x31eq.com/cgi-bin/pregular.cgi...

X31EQ.COM

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Regular Temperaments

1200.000, 833.090, 1901.955, 2786.314, 3368.826, 4151.318-limit Regular Temperaments

Reply

31 w

Paul ErlichAdmin

Mike Battaglia Dave Keenan

Reply

31 w

Mike BattagliaAdmin

Dave Keenan one simple idea, which may improve the weighting somewhat, is to say that f11 and F11 should have the same complexity as 11, f31 and F31 should have the same complexity as 31, and so on. This gives us an "L1-norm-of-Linf-norms" rather than an L1 norm; still easy to compute and can easily be done on the back of a napkin given the monzo decomposition of some feudal number.

For each of these prime factor pairs like f11 and F11, we weight them as log(11) rather than log(√11). But then, instead of just taking an L1 norm of everything, we first take the max of the absolute value of the two f11 and F11 coordinates, and likewise with the other prime factor pairs, and then take the L1 norm of all those intermediate results along with that of the normal primes. This gives f11, F11 and 11/1 the same complexity, and in general a very interesting choice of complexity for everything, with the only quirk being that again, -1+2*Phi and Phi itself are the two sporadic basis elements with different behavior (which I have some ideas about and will deal with later).

If any of these ideas sound good, I have a whole bunch of related results showing how we can "TE-ify" this to locate the best-fit approximate L2 norm for the temperament search, as well as a way to extend the above to a generalized complexity with one free parameter interpolating between the L1 version and this L1-of-Linf version. The question is if this way of thinking is sensible, and gives sensible results for nobles. Do you have any thoughts on that?

Reply

31 wEdited

Mike BattagliaAdmin

Dave Keenan: One thought about all of this is that what I'm suggesting is really just kind of just a "hack" to try to do things relative to this basis of feudal primes. I'm still not quite sure if starting from that basis is the best way to measure the complexity of an arbitrary feudal number although maybe there's some way to get it to work.

Two criteria that we may want any decent complexity measure to meet, at least approximately, are the following:

1. We'd like the complexity of each rational to be the same as the usual log(n*d).

2. We'd like the complexity of each noble number to at least somewhat agree with its "inherent" complexity of the simplest rational generating that noble as the start of its zig zag pattern on the scale tree.

As previously mentioned there are a few different conventions for saying where the zig zag "starts"; I don't think it really matters as long as the complexity we have is at least somewhat approximately in agreement with just about any sensible way to do it. But we would like, for instance, the noble between 6/5 and 5/4 to be lower in complexity than the one between 6/5 and 7/6. In fact a pretty easy way to try this would be to see what we get for the nobles between two superparticular ratios.

I guess the question is, then, given some feudal number, how do we obtain the unique prime factorization? Is there some kind of generalization of the Euclidean algorithm?

Reply

31 w

Dave KeenanAuthor

Mike Battaglia You wrote: "I guess the question is, then, given some feudal number, how do we obtain the unique prime factorization? Is there some kind of generalization of the Euclidean algorithm?"

I'm pretty sure there is. I thought it would be in either Dekker or Dodd, but I can't see it.

https://staff.fnwi.uva.nl/t.j.dekker/Pr ... Primes.pdf

https://archive.org/det.../numbertheory ... d/mode/2up

But do we really need it? Factoring rationals is easy. You just do the ordinary prime factorisation, then split the primes that end in 1, 5, or 9. My conjecture is that the nobles never have more than one prime in the denominator and one in the numerator, so you can factorise them by fibonacciing them backwards until you get a negative number.

Reply

31 w

רועי סיני

Mike Battaglia If you are still interested, there is a way to factor a feudal number into primes, but I don't know of a way as simple as the Euclidean algorithm.

First, you find the gcd of the wooden and golden part, this the greatest ordinary natural number that divides this feudal number. Factor it, and divide by it to get a number with no ordinary integer factors.

Then, factor the new number's arithmetic norm. Because of the multiplicativity of the norm, the factors will be the norms of the primes that divide it. To find the feudal primes this primes factor into, you need to solve the equation 1+x-x² = 0 mod p, which means find (1±√5)/2 mod p (where √5 is a number that when squared becomes equivalent to 5 mod p). Each solution x will have that 1+xϕ is divisible by another feudal prime that divides p. Use the Euclidean algorithm in ℤ[ϕ] (which is an Euclidean domain) to find one of fp, Fp, where the other is minus its conjugate. To check which of the primes divide your number, multiply it by one of them. If the result is divisible by p (i.e. both the wooden and golden parts are divisible by p), this means that the *other one* divides your number, and the quotient is the product you got divided by p. If one of the two feudal primes doesn't work, the other one should. In the case of 5, if you multiply a number whose norm is divisible by 5 with f5 you'll always get a feudal number divisible by 5 and it divided by 5 will be the original number divided by f5.

Reply

18 h

Dave KeenanAuthor

Mike Battaglia I think we are free to adjust the complexity of the phi-coordinate in whatever way seems to make sense psychoacoustically. I don't think there is any mathematics that will tell us what that weighting should be. Only our ears. But as you say, we want to preserve the usual complexity of the rationals, so that severely constrains what we can do with the non-wooden feudal primes. The complexity of f5 has to be the square root (or half if we're talking logs) of the complexity that 5 used to be. And otherwise the complexities of fp and Fp have to multiply (add in log space) to what p used to be. And as you say, we want to apportion it between the small-f and big-F factors in a way that reflects the complexity of the "simplest adjacent mediends" (SAMs) [to coin a term] of the noble number. Simply using the real value (or its log) of the non-wooden prime seems to lean in the right direction. Nobles that are the ratio of big-Fp's seem to have more complex SAMs than the ratio of the correspond little-fp's.

So I think it's only the complexity of the phi-coordinate (the unit rather than the primes) that we're free to fiddle with to our hearts (or ears) content.

Reply

31 wEdited

Dave KeenanAuthor

Mike Battaglia Do you think that "f" and "F" is a good notation for these complementary primes? Or is posterity going to curse us for the lack of visual distinctiveness and the two-syllable pronunciation "small eff" and "big eff"?

An alternative would be to use "f" and "g" which are quite distinct visually and only one syllable ("eff" and "gee"). They make me think of the orange and blue buttons on the first "computer" I bought with my first pay-check back in 1976.

https://thimet.de/CalcC.../Calculators/ ... 5C-2-M.JPG

Problems with "f" and "g" are that it's not so obvious which one is bigger, and "g" is not an abbreviation of anything useful that I can think of.

So I still prefer "f", "F" over "f", "g". But I just thought I should ask.

thimet.de

THIMET.DE

thimet.de

thimet.de

Reply

Remove Preview

31 wEdited

Mike BattagliaAdmin

Dave Keenan I think f and F are alright. But note my previous post about how the complexities of f11 and F11 don't need to multiply to the complexity of 11 if we do the L1 of Linf norms thing. I think the better way forward is just to see what gives sensible results for nobles though.

Reply

31 w

Dave KeenanAuthor

Mike Battaglia That all sounds reasonable to me. You're a smart guy. I trust you will figure out something that is a good compromise between keeping the math simple and modelling the psychoacoustics.

I'm going to have to take a back seat for a while, to catch up on other projects. As well as the RTT "textbook" I'm writing with Douglas Blumeyer, I'm designing and building a movable-fret 7-string electric guitar. It's an evolution of the choob, but oval rather than circular, and made of wood, not carbon fibre, and with stainless-steel (movable) frets, not nylon.

Reply

31 w

Mike BattagliaAdmin

Some preliminary results from Graham's temperament search suggest that 72, 46, 27e, 43 are very good in the 2.3.5.7.11.phi subgroup, and to a lesser extent 26 and 19. Part of the problem, again, is that the algorithm is with the default log(p) weighting we're even prioritizing phi over 2, which is one reason why 26 does so well.

We can also do something similar by adding a factor of phi to the Riemann zeta function, though the results are really weird (if we just use a default weighting of phi, which again is stronger than that of any rational number). Results:

EDIT: A typo in these links made all of these results wrong! See post below

https://www.wolframalpha.com/input?i=pl ... +y%3D0..40

https://www.wolframalpha.com/input?i=pl ... +y%3D0..40

https://www.wolframalpha.com/input?i=pl ... +y%3D0..80

https://www.wolframalpha.com/input?i=pl ... y%3D0..100

Note 30 is even better than 31 using this, again because of the extreme emphasis on phi.

This is probably not the best way to do it. The Dedekind zeta function, however, is a generalization of the Riemann zeta function to arbitrary number fields, including Q[sqrt(5)], so I'm kind of curious what that looks like.

Wolfram|Alpha: Making the world’s knowledge computable

WOLFRAMALPHA.COM

Wolfram|Alpha: Making the world’s knowledge computable

Wolfram|Alpha: Making the world’s knowledge computable

Reply

31 wEdited

Paul ErlichAdmin

Mike Battagliathat's the same subgroup I had searched earlier as well as more recently for the Graham Breed's app results I posted here (with error parameters 1.25 and 4.27, respectively).

Reply

31 w

Mike BattagliaAdmin

Paul Erlich, FWIW, I tried the Dedekind zeta function in Mathematica, and got this. It likes 71 better than 72. Not sure what on earth to make of it. I'm not sure how this is weighting phi, really.

No photo description available.

Reply

31 w

Mike BattagliaAdmin

Actually, I think that there's no real weight for phi at all; instead there's just one single term for the entire phi-equivalent ideal for each feudal prime, so that's that.

Reply

31 w

Mike BattagliaAdmin

I guess one question is if this generalized zeta function has any use at all. I mean, algebraic number theory has gotten us this far, and it's the "zeta function" associated with Q[sqrt(5)], but some of these results are weird. I wonder if they make sense, though, and if there's some reason 71-EDO is better than 72-EDO with the feudal numbers. Maybe 71 approximates all of the noble numbers plus JI better than 72, for instance. Who knows.

Reply

31 w

Mike BattagliaAdmin

It also likes 28 better than 27. I dunno, this is pretty out there. Maybe it's another situation where the "natural" weighting that it's using isn't the one we really care about.

No photo description available.

Reply

31 w

Mike BattagliaAdmin

Just a quick note that the original posts about adding phi to zeta had a typo and were wrong. The correct version is here:

https://www.wolframalpha.com/input?i=pl ... +y%3D0..40

It really likes 26, 36, 46, and 72. 46 seems like a standout. Surprised 27 didn't turn up quite as much.

Wolfram|Alpha: Making the world’s knowledge computable

WOLFRAMALPHA.COM

Wolfram|Alpha: Making the world’s knowledge computable

Wolfram|Alpha: Making the world’s knowledge computable

Reply

30 w

Paul ErlichAdmin

10, 26, 62, 72?

Reply

30 w

Mike BattagliaAdmin

Yeah, also 10 and 62.

Reply

30 w

Paul ErlichAdmin

Mike Battaglia you mean they are repeated?

Reply

30 w

Mike BattagliaAdmin

Er, what do you mean?

Reply

30 w

Paul ErlichAdmin

Mike BattagliaI listed 10 and 62, and you said also 10 and 62, so I thought maybe you meant two closely-spaced peaks at 10 and 26. But, I guess that doesn't happen with zeta . . .

Reply

30 w

Aggelos Boshidis

David Guillot

Reply

31 w

Dave KeenanAuthor

I have finished the new version of my series of articles, updated to use the superior choice of fundamental primes:

viewtopic.php?f=21&t=557

Noble frequency ratios as prime-count vectors in ℚ(√5) - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) - The Sagittal forum

Reply

Remove Preview

31 w

Dave KeenanAuthor

I just posted another article in the Feudal Manifesto series, in which I clarify a bunch of things, such as the (non) relationship between noble numbers and combination tones, and I give unique identifiers for noble numbers using an "ennoblement function" n, where e.g.

n7/4 = (7+9ϕ)/(4+5ϕ) = f31/f11 ≈ 1002 ¢

viewtopic.php?p=4620#p4620

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Reply

Remove Preview

30 w

Mike BattagliaAdmin

Dave Keenan it looks like the ennoblement function is the same as my "~" from before, or rather the first version of it?

You claim in this that noble numbers are not sensitive to mistuning in the sense that there are no partials to try to get to cohere, but I don't really agree with that. In particular the thing you are now calling n6/5 seems to be pretty fragile. There is a real sweet spot somewhere in the 330-340 cent range where things are maximally "stanky," let's call it, and then things kind of fall off pretty steeply, particularly as you tune sharp.

Reply

30 w

Dave KeenanAuthor

Mike Battaglia Thanks for this feedback. I have updated my post accordingly. Search on your name within it. Yes, I think my prefix "n" is the same as your first definition of your postfix "~". However you confused the issue at the time, by giving as an example, that (6/5)~ was the noble at 339c. It seems you're still confused about that. Or one of us is. Because you imply above, that I am now calling the noble near 339c "n6/5" when in fact I am calling it "n5/4". "n6/5" is the noble near 284c.

Reply

30 w

Mike BattagliaAdmin

Ah, I'm sorry, you're right. So it's the other variant then.

Reply

30 w

Dave KeenanAuthor

Mike Can you tell me how to get the simplest child of a given ratio, without having to look up the SB-tree, perhaps related to that trick you did using ModularInverse?

Reply

30 wEdited

רועי סיני

Dave Keenan If you still need this, there is indeed a way to do that – given a/b use the modular inverse trick to find numbers c, d s.t. ad-bc = ±1. If c < a/2 then (a+c)/(b+d) is your simplest child, otherwise it's (2a-c)/(2b-d).

Reply

18 h

Cmloegcmluin Xenharmonic Feisbeuk

I was having trouble visualizing it otherwise, so I put this diagram together to help me understand how Dave and Mike's ennoblement function gives a one-to-one correspondence between non-subunison rationals and noble numbers. Hope it helps others too:

No photo description available.

Reply

30 w

Mike BattagliaAdmin

Yes, basically. You can also get it by truncating the continued fraction expansion.

Reply

29 w

Cmloegcmluin Xenharmonic Feisbeuk

Mike hmm. I see you've mentioned this above. so given n(n/d), you can find the other mediend by appending a single 1 to the continued fraction form of n/d, and you can find the noble by appending infinite 1's. that's pretty handy.

Reply

29 w

Mike BattagliaAdmin

The < and > "noble flat" and "noble sharp" operators also have a nice visual representation here, being just the noble mediant of the ratio and its left or right child.

Reply

29 w

Cmloegcmluin Xenharmonic Feisbeuk

Mike yes, I can see a separate use for those, because n(n/d) gives you the simpler of the two children, which 50% of the time is the left child and 50% of the time is the right child, based on n/d's location in the tree.

Reply

29 w

Mike BattagliaAdmin

Right, although note that you also have 11/9< is the same as 6/5> which is the same as 5/4<, all of which are n(5/4). Kind of an interesting property.

Reply

29 w

Steve Martin

Phi rhymes with sky, or with key? I agree with Dave but have heard it the other way (on you tube), is it a Aus/UK vs US thing?

Reply

30 w

Dave KeenanAuthor

Steve Martin I understand that when Greeks say their alphabet it's pronounced like the English word "fee". So no one can fault you for pronouncing it that way. But in all the maths and physics lectures I attended in Australia, the math symbol was pronounced like the "Fi" in "Wi-Fi". I can't speak for any other jurisdictions, but I read somewhere that this was generally the case in English-speaking maths and science usage.

Same thing with π. Greek "pee", English "pie".

Reply

30 wEdited

Steve Martin

Notes on where I have got with this, below

Reply

30 w

Steve Martin

As a nerdy mathematician I found the quadratic fields in Hardy and Wright very interesting; here a few general notes that you probably know already: Q(root(m)) is a field for any nonzero integer m; if negative then we have a subfield of the complex numbers, if positive then a subfield of the real numbers; the norm for an element X=c+d*root(m) where c and d are rational is N(X)=(c+d*root(m))*(c-d*root(m)), this is always positive if m is negative, not necessarily so if m is positive, note it yields the formula we have been using in Q(root(5)) when re-expressed in terms of phi. The norm obeys the rule N(X*Y)=N(X)*N(Y); units are elements e having N(e)=+/-1. Any two elements related by X=Y*e where e is a unit are called associates. In the field "integers" are defined in terms of polynomials that they satisfy. The field is sometimes a UFD and sometimes has a Euclidean algorithm. If a non-UFD then you can have irreducibles that are not primes, you have to define each carefully.

But Q(root(5)) is a UFD and has Euclidean algo; it turns out that the units are phi^n where n is a (usual kind of) integer, and that the "integers" in this field are X=c+d*(phi) where c and d are (usual kind of) integers. Note, in this field any associate of an integer is an integer, and of a prime is a prime. It is found that any (usual kind of) prime of form 5*n +/- 2 is irreducible in this field so is a prime here, but those of form 5*n +/- 1 can be factorised and their irreducible factors are primes. Finally, root(5) =-1+2*phi is irreducible hence a prime (and 5 is not).

A final general note, in terms of Dave's diagram, you can find units arbitrarily far from the origin, and the same for associates of f5, f11, etc.

Reply

30 wEdited

Dave KeenanAuthor

Steve Martin Not root(phi), just phi. Phi already contains root(5), so root(phi) would be like a fourth root.

Units are elements e such that N(e) = ±1, not only 1.

N(ϕⁿ) = (-1)ⁿ for all ordinary integers n.

Thanks for reminding me to add that to my forum article.

Reply

30 w

Steve Martin

For RTT use, i was fairly certain early on that we should be using the primes as defined above; and indeed your table confirms it. Adding phi alone to the usual basis 2.3.5.7... does not get many noble ratios, because the group is multiplicative and only the first two noble numbers are simple multiples of phi; but with f5, f11 etc we can reach more as shown in the table, and maybe all as you conjecture. However, we cannot do without phi as well, as you already observed; a simple reason is that f1 and F1 feature often in the table; more explicatively we can say that although it is a unit, it is not 0 cents, or that phi allows us to move between associates cheaply, so that our choice of canonical primes is less significant.

My first look included all primes up to F11, but to simplify things I focused on 560c (n3/2) and 422c (n4/3) which only need f11 and f5 hence my basis 2.3.f5.F1.f11. I like the look of 26&43 for this (very close to 69edo). Generators now c1201.5c, c139.3c. (When 7 and F11 were included, the octave went to c1198...c) Not the only possibility of course, but it's a start.

http://x31eq.com/cgi-bin/more.cgi?r=2...

X31EQ.COM

1200.000, 1901.955, 1393.157, 833.090, 1815.643-limit Rank 2 Temperaments

1200.000, 1901.955, 1393.157, 833.090, 1815.643-limit Rank 2 Temperaments

Reply

30 wEdited

Dave KeenanAuthor

Steve Martin It's great that Graham built his temperament finder to work with inharmonic primes, so we can just plug these things in, although as Mike has noted, it might benefit from weightings other than the log of their real value being applied to the feudal units and primes.

Reply

30 w

Steve Martin

Commas: sqrtphi: can we have the square root of phi in our setup? Not sure, but plough ahead anyway: even without x31eq's help if we assume that the generator is being n4/3 i.e. f11/f5 as well as sqrtphi, then:

(f11/f5)^2 = F1.

This comma is given for my 26&43 alongside:

3/2 = phi^3/f11

9/4 = f5 (of course!)

f5^3 = phi^5

et al.

Reply

30 w

Dave KeenanAuthor

Steve Martin I don't understand why we'd want √ϕ. The integers of ℚ(√5) are not c+d√ϕ as you say above, but rather c+dϕ, where c and d are ordinary integers. ϕ = ½+½√5

Reply

30 wEdited

Steve Martin

Dave Keenan yes, crikey, that's a typo re root(phi), will fix - done; but my mention of it in connection with a comma was intended, because Paul mentioned the temperament of that name and indeed it does come up in the answers from Graham's app.

Reply

29 wEdited

Dave KeenanAuthor

Steve Martin Then I'll take your question to be, "Can we have the square root of phi [as a generator] in our setup?"

Of course we can. And we don't need it (or want it) in our vector basis in order to do so. Generators are not constrained to have integer entries in their vectors.

Reply

29 w

Mike BattagliaAdmin

One interesting thing I've been thinking about recently, which maybe Paul Erlich and Dave Keenan would have some thoughts on:

The mediant can be thought of as a weighted arithmetic mean, where the weights are proportional to the denominators of the ratios. For instance, the mediant of 5/4 and 6/5 is equal to (5/4)*(4/9) + (6/5)*(5/9) = 11/9. This tells us that the mediant will be in between the two original ratios, and weighted closer to the more complex of the two. These are both useful properties for the same reason that the Lagrange point between the Earth and Moon is slightly closer to the Moon.

We may ask what were to happen if we were to derive something similar, but with the geometric mean replacing the arithmetic mean above. We would like to derive something like a "geometric mediant," or at least a "geometric noble mediant."

The main snag is that a geometric mean of two ratios is not, in general, another ratio, so it is an interesting question on how we should recurse to get our noble mediant. There are several interesting ways to answer this question, and though I'm still trying to figure out which is the "right" way, all of them basically lead to the same big picture of everything: instead of adding new "feudal primes," we instead change our monzos so the *coordinates* can be feudal numbers rather than regular integers. This is a fairly useful property as we don't need to add any new feudal primes or expand the dimensionality of our vector space at all to add these new phi-based intervals; for the 5-limit we get the same 3D space but with generalized coordinates that can be feudal numbers. You can also think of this as a 6D space in which the basis elements are 2, 2^phi, 3, 3^phi, 5, and 5^phi.

The same space of intervals is generated by all of the various candidates for "geometric noble mediant" that I've looked at so far, with slight differences on which particular interval in this space, for instance, we are identifying as the "geometric noble mediant" of 5/4 and 6/5. The results are all typically pretty close to one another, and also tend to be pretty close to the regular noble mediant, at least for intervals which are mostly similar in size. "Geometric phi", on the other hand, is very different from regular phi and turns out to be about 742 cents. I'll write some details once I have it together, but has anyone thought about this kind of thing?

Reply

29 wEdited

Paul ErlichAdmin

I'm familiar with 742 cents as 1200/Phi or 1200*phi cents. Not yet grasping how to derive it without the octave, but sounds interesting!

Reply

29 w

Cmloegcmluin Xenharmonic Feisbeuk

right that’s apparently another way to describe logarithmic phi: https://en.xen.wiki/w/Logarithmic_phi

Logarithmic phi - Xenharmonic Wiki

EN.XEN.WIKI

Logarithmic phi - Xenharmonic Wiki

Logarithmic phi - Xenharmonic Wiki

Reply

29 w

Paul ErlichAdmin

Cmloegcmluin Xenharmonic Feisbeuksure but the octave is right there in the definition.

Reply

29 w

Cmloegcmluin Xenharmonic Feisbeuk

right, just sharing in case any links there give Mike additional insight he was asking for

Reply

29 w

Dave KeenanAuthor

Mike Battaglia You probably should have started a new post for this. I note that the geometric mean is just the exponential of the arithmetic mean of the logs, hence the appearance of "logarithmic phi" or 742 cents.

I note some misnomers: What you're calling a "noble geometric mean" should instead be called a "phi-weighted geometric mean" or similar. "Noble" is not synonymous with "phi-weighted". "Noble" means the results are noble numbers, which already have a well-accepted mathematical definition.

When you wrote that the coordinates can be "feudal numbers" instead of regular integers, you should have said they can be "feudal integers". Feudal numbers include quotients of feudal integers.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan Thanks, glad you found it interesting!

Reply

29 w

Mike BattagliaAdmin

Paul Erlich Dave Keenan OK, turns out there really is one "best" way to do the thing I was looking for. Although I derived it from this standpoint of replacing the arithmetic with the geometric mean, it turns out to simplify to something so simple it's almost absurd. However, because it was derived in a rigorous way, this simple idea turns out to have a number of interesting properties.

It all simplifies to this: let's say that you have two ratios a/b and c/d, and you want the noble mediant from a/b to c/d, which is (a+b*phi)/(c+d*phi). Then the "geometric noble mediant" of those two is simply (a/b)^(1-p) * (c/d)^(p), where p = phi-1 ≈ 0.618. In other words, you draw a ruler in cents space going from a/b and c/d, and then go 61.8% of the way there. This can be viewed as an approximation of the original which is better as the two ratios get closer together in size (and I guess complexity).

To derive this formally, I had this entire thing worked out with "radical numbers," which are rational numbers raised to a rational power - this structure naturally arises when looking at geometric means of rationals. Each radical number is just a monzo with rational coefficients. Then, for each radical number, we can say the "geometric mediant" is the monzo you get taking the mediants of each coordinate separately, then you can recurse to get a "geometric noble mediant," although it becomes an interesting question what we call "noble" in this situation. You get an infinite set of independent Stern-Brocot trees and a sort of "continued fraction matrix" and all kinds of weird stuff like this.

*HOWEVER*, we don't really care about arbitrary radical numbers, really, but only rational numbers which have integer monzo coefficients. Thus, in this situation, the entire thing almost becomes trivial - basically as trivial as taking the "noble mediant" of two integers. All of the "denominators" of our coordinates are 1 and everything simplifies to the nice simple property above.

However, because it was derived in such a rigorous way, it turns out to have some really amazing properties:

Because the geometric and arithmetic mean of two ratios are approximately equal if the two ratios are not too far apart, this turns out to approximate the noble mediant very well for pairs of ratios which are "reasonably" close together, and I will post a table of results about that. I am quite curious to see if the geometric or additive noble mediant is a better approximation of local maxima of HE.

The geometric noble mediant is transpositionally invariant, in that the geometric noble mediant of 2/1 and 3/1 is the same as the geometric noble mediant of 1/1 and 3/2 transposed up an octave. This parallels how the regular noble mediant is "additively invariant", in the sense that NobleMediant(z+a/b, z+c/d) = z*NobleMediant(a/b, c/d) for any integer z. In this situation, we have GeometricNobleMediant(q*a/b, q*c/d) = q*GeometricNobleMediant(a/b, c/d) for any *rational* number q.

Lastly, it doesn't require we add any new feudal primes or anything like that, just let our monzo coordinates be feudal integers rather than regular integers. Or we could say our basis now becomes 2, 2^phi, 3, 3^phi, 5, and 5^phi and the coordinates remain regular integers.

If for any reason people later care about these "radical numbers" - kind of interesting as they are basically equal divisions of some rational number, which are certainly perceptually relevant for a very different set of melodic reasons - we can extend these results to them and you get some interesting structure.

I will post a table of comparisons between regular and geometric noble mediants for some superparticular ratios, keeping in mind that it's still kind of tricky to call anything "noble" in this situation.

Reply

29 wEdited

Paul ErlichAdmin

Mike BattagliaiCYMI above, FWIW, the nearest HE local maximum to phi in most of the original graphs was at 845 cents.

Reply

29 w

Mike BattagliaAdmin

Paul Erlich to which phi?

Reply

29 w

Paul ErlichAdmin

Mike Battaglia to "acoustical" phi at 833 cents.

Reply

29 w

Mike BattagliaAdmin

Paul Erlich how would that be relevant in this situation?

Reply

29 w

Paul ErlichAdmin

Mike Battaglia I don't know if it would be. That's why it was labeled "FWIW". You said something about "a better approximation of local maxima of HE" and I was just pointing out the 845-cent local maximum since I pointed that out in response to Dave Keenan much earlier in this thread. So if you had something which predicted that rather than 833 cents, instead something more like 845 cents is near a local maximum, that would appear to be along the lines of what you said you'd be curious about. But perhaps I'm missing something that makes this inapplicable to the thing you're curious about somehow.

Reply

29 w

Paul ErlichAdmin

Maybe Dave Keenan and/or Cmloegcmluin Xenharmonic Feisbeuk are grasping more of this at this juncture and could help explain why my observation wouldn't be pertinent here.

Reply

29 w

Mike BattagliaAdmin

Paul Erlich Rather than having Dave and Douglas explain to you why your observation isn't pertinent, I suggest trying to explain to me why it is. Given any arbitrary interval, one can look for the nearest local maximum, and I'm sure there are many noble mediants, as well as geometric noble mediants, that are approximately equal to 845 cents. Which one am I supposed to be looking for?

Reply

29 w

Mike BattagliaAdmin

Paul Erlich I'm not sure if your Facebook client is glitching out but you're responding in all kinds of different subthreads... I think your comment was meant to be here. The geometric noble mediant between 5/3 and 8/5 is 840.68 cents.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia no I don't see anything glitching out. I respond to things I see in different subthreads each in their own subthreads as usual, though obviously some subthreads spun off from others initially and so the conversations end up being related. My comment was a reply to your list of pairs of ratios with the different mediants for each pair. I looked at that and then replied with another such pair.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia I don't know if refreshing the thread would help show more of what I wrote, but it often does.

Again I said "FWIW"; if the answer is that it's worth nothing, then that's fine . . . in which case clearly I am missing something but don't know what it is.

Again, all I know was that

(1) in response to the OP, suggesting that phi itself (833 cents, the prototypical noble mediant, and certainly the one between what are normally the two nearest local minima of HE, 8:5 and 5:3) is near a local maximum of HE, I pointed out that in most of the original graphs, the only local maximum in that region was at 845 cents;

(2) you said something about curiosity whether your approach might offer "a better approximation of local maxima of HE";

(3) I asked whether it might, correspondingly, better match this data point about a local maximum of HE.

Perhaps that's a poorly thought-out question, and I have no reason to suppose it isn't, but I didn't see the harm in throwing it out there. But clearly I don't understand something about the idea, which is why I asked for help. Perhaps someone can see where I'm coming from and point out why it doesn't apply in this situation. And I didn't see the harm in asking that either.

Reply

29 w

Mike BattagliaAdmin

Paul Erlich perhaps I misunderstood; I was interpreting this in the context of our original conversation about "logarithmic phi," which is about 742 cents and very far from 845 cents. See also my note about how it depends on which pair of ratios you ask for; the geometric noble mediant between 5/3 and 8/5 is about 841 cents and closer to your local maximum, but you get different results for other pairs of starting ratios (like 3/2 and 5/3), for instance. In general, it seems like you start to get sensible results as the two ratios get about a half step apart.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia I think the pair of ratios Dave and Margo start with in their article are the two flanking consonances or local minima.

Reply

29 w

Mike BattagliaAdmin

Paul Erlich it may be a good idea to try this with neighboring ratios that are local minima of HE and see what the local maxima are. Do you have a list of of local minima and maxima for different values of s?

Reply

29 w

Paul ErlichAdmin

Mike Battaglia I don't have that handy, and though I've posted a lot of relevant numbers and graphs over the years and Steve could calculate them again, I imagine your own HE calculator could readily be called into service? The graphs with local maxima 845 I posted earlier in the thread appeared with both Farey and more "straight" versions, but with a consistent s value of I believe 1%. We also saw that when s is increased all the way to 1.5% so that 8:5 becomes a hillock, more like an inflection point than much of a visible minimum, the nearest maximum shifts way over (to 829 cents -- but only a hair higher entropy than 8:5).

Obviously the global maximum that might be near 40-70 cents might be a great test case, though it's hard to necessarily say what the next local minimum is (very sensitive to s), and even the global maximum depends a lot on N (and you apparently found a way to find the global maximum in the limit as N->infinity for a given value of s, which is incredibly impressive and interesting, but it seems to me that there *might* be some psychoacoustical equivalent of an actual maximum complexity N of ratios in the "template" for fundamental-finding for an interval in a given register, given that our hearing frequency range is finite, etc.).

Reply

29 w

Paul ErlichAdmin

I imagine any local maximum can be pushed right up to the "edge" of (arbitrarily close to) the next local minimum (the more complex of the two flanking it) just by tweaking s to just below where the local minimum disappears entirely Mike Battaglia.

Reply

29 w

Mike BattagliaAdmin

I think that 40-70 cent maximum will be the hardest to get right with this, as it'll need to be "phi%" of the way between 1/1 and the next local minimum, meaning there'd need to be a minimum at like, 60-100 cents or so.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia yeah for sure! but since it seems a very shallow local minimum can have the next local maximum on one side arbitrarily close to it, there are liable to be cases even harder to get right (in terms of quotient of the two distances of the local maximum from its flanking local minima shooting off to infinity or zero rather than any kind of average or mean) for many values of s.

Reply

29 w

Mike BattagliaAdmin

Paul Erlich well I'm pretty sure all of these results are within the margin of error of how good HE is supposed to be, so if things are a few cents off I don't think it matters much. Actually, I think the true local maximum between 6/5 and 5/4 is somewhere closer to maybe 333 cents, tbh. Either way, I think this is all doing pretty well.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia well in the scenario I described, there can be more than a few cents from being right up against a flanking local minimum (say 820 cents, right up against 8:5) to being some kind of average or golden mean or whatever 845 cents is. But since the local minimum in question (here 8:5) is on the verge of disappearing in the scenario I described, somehow this is still within some margin of error of something somehow I suppose?(?idk?)

Reply

29 wEdited

Paul ErlichAdmin

^edited

Reply

29 w

Mike BattagliaAdmin

Paul Erlich: right, but if we're only viewing the results as valid for neighboring minima (relative to some s), then this problem goes away. Or rather, we could say that for any noble mediant - regular or geometric - there is some choice of "s" such that this is approximately the local maximum between two neighboring local minima.

Reply

29 w

Paul ErlichAdmin

Mike Battagliayes, "some choice of s" is probably right.

Reply

29 w

Mike BattagliaAdmin

BTW, I sometimes think that maybe the min-entropy is what we should really be looking at... I mean, it's about as simple as it gets. For instance, with the laplace distribution it's literally just a piecewise-linear graph (see picture). For the Gaussian it's piecewise-parabolic and so on.

No photo description available.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia FWIW I don't find any aural "landmarks" that make it easy to tune to local maxima by ear, i.e., I think the minima should look a good bit pointer than the maxima (though still round if it's just HE and not roughness etc.). It seems like a=2 already makes them equally pointy, and beyond 2 the maxima are pointier than the minima, based on playing around on your calculator, though I was using the Gaussian and not Laplacian distribution.

Reply

29 w

Mike BattagliaAdmin

There is clearly a point between 6/5 and 5/4 that is maximally "stanky." Have you considered consulting your internal stankiness barometer for measuring this? As a guitarist I would imagine it is highly well developed

Reply

29 w

Paul ErlichAdmin

Mike Battaglia I've spent hours just jamming in 7-equal because of how stanky it is! But I've also spent hours trying to identify some kind of local maximum of discordance and nothing seems to hold up long-term. Just the other day on a SoundCloud stream it sounded to me like Joseph McDermott used a 7-note mode found within Miracle (maybe within Blackjack, don't remember) featuring a 333-cent interval over his tonic and a lot of 333-cent harmonic intervals. I kept going back and forth and trying to decide whether it sounded consonant or not. The more I dialed in the treble so as to hear the higher harmonics, the more it seemed to stand out as dissonant, but still in some kind of medial unfully-realized way, and that's probably due to roughness rather than HE, IMHO.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia I don't know what you mean -- the minima *were* neighboring in my example.

Reply

29 w

Mike BattagliaAdmin

I thought you were saying that if you decrease "s" enough, you can make the nearest maximum larger than some local minimum arbitrarily close to that minimum.

Reply

29 w

Paul ErlichAdmin

Mike Battagliaright, or it might be the nearest maximum smaller than that local minimum. But the same minima still neighbor that maximum, so I don't know what you meant by "if we're only viewing the results as valid for neighboring minima". For the case of 8:5 and 5:3, they are consistently the neighboring minima of the maximum which takes on values from 845 cents to <820 etc. cents as s is increased from 1% to ~1.6%.

Reply

29 w

Mike BattagliaAdmin

I was just saying that as s decreases, then there will be new minima that appear in between 8/5 and 5/3. Once that happens, it no longer makes sense to take the noble mediant of 8/5 and 5/3 at all, but rather 8/5 and whatever the next local minimum is, and of course this will be closer to 8/5, and thus so will the local maximum between the two (and the noble mediant between the two and so on).

Reply

29 wEdited

Paul ErlichAdmin

Mike Battaglia in my example there were no other local minima between 8/5 and 5/3 anywhere in the range of s values mentioned

Reply

29 w

Mike BattagliaAdmin

Paul Erlich Oh, you're saying *increasing* s brings the maximum to 1.6%. Yes, for the Gaussian that's right, although again I think much of this probably depends on what probability distribution you're using...

Reply

29 w

Paul ErlichAdmin

Mike Battaglia well when the Laplace distribution is used, the HE curves become "fractal", and there seems to be an infinite density of minima and maxima regardless of s, and 8:5 and 5:3 never become neighboring minima . . . right?

Reply

29 w

Dave KeenanAuthor

Mike Battaglia I think you should replace "GeometricNobleMediant" with "GeometricPhiMediant".

I don't understand in what sense it is more "clean" to have numbers like 2ᵠ, 3ᵠ, 5ᵠ in our basis as opposed to -1+2ϕ, -2+3ϕ, -1+3ϕ.

Our ears should be the ultimate arbiter, not any particular HE curve. As Paul has noted, the local maxima of HE are very unstable wrt the parameter s. I only mentioned HE maxima in my OP because I knew it was something people reading this group would be familiar with. Noble number frequency ratios at least have the guaranteed property that they are maximally distant from all nearby simple ratios, with their distance being a monotonically-increasing function of their simplicity. Other kinds of phi-weighted mean do not have this property.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan: the thing about ears vs HE is quite a strawman. Did you read my entire post?

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Which entire post? When I read posts, I read entire posts. But there are so many, and facebook seems so arbitrary in what it reveals when. Why don't you just explain why you think ears vs HE is a strawman.

Reply

29 w

Mike BattagliaAdmin

It's a strawman because it isn't a response to anything I've written. I gave a giant list of noble mediants and showed that in most situations, the "geometric" version is usually only off by a few cents from the regular version. Are you saying that you listened to all of these and your ears like the regular version better? Or are you just nitpicking random stuff or what? It seems unrelated.

Reply

29 w

Dave KeenanAuthor

Or ignore the strawman and address the other thing, where noble number frequency ratios at least have the guaranteed property that they are maximally distant from all nearby simple ratios, with their distance being a monotonically-increasing function of their simplicity, while other kinds of phi-weighted mean do not have this property.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan they do have this property, since I keep telling you it's usually only a few cents off.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia There's no need to be rude.

At that time, I had not yet seen the giant list of noble mediants that showed that in most situations, the geometric version is usually only off by a few cents from the regular version.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan My response wasn't intended to be rude at all; apologies if it came off that way. It was just to suggest slowing down and catching up on some of these posts. Most of the points you brought up (the use of "feudal numbers" rather than "feudal integers," the questions about how to best generalize the word "noble," the quality of the approximation) are things that are mentioned in the very post you are responding to and posts I thought you'd already seen.

It seemed like you aren't really interested in this idea and were mostly looking to debate things. That may have been a misguided interpretation, but anyway I would just suggest to give it a chance. It's a simple idea: a mathematical approximation that has a lot of nice properties, including transpositional invariance. It exists in a *much* lower dimensional vector space than the original and is thus computationally much easier to deal with, and for the most important pairs of ratios, it's only a few cents off, often in the direction that's closer to the HE maximum. I thought that may be useful to some people.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia When you say it exists in a much lower dimensional vector space than the original, that's the first thing I've read that makes me understand why it might be a better basis. I understand you to be saying, in effect, that the 32 simple nobles that I've listed (the only ones I consider musically relevant), can be approximated well enough, with many fewer entries in the vectors, when the usual prime basis is augmented with a few pᵠ for a few low (ordinary) primes p, compared to the large number of feudal primes required (which I noted as a problem in my first forum article).

If so, that's great. I'd love to see a table similar to my table 14 or 15, with the vectors replaced with those using your new basis for the approximations of those nobles.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan it means that *all* of the noble mediants that you can generate from some p-limit fit into a Z-module of twice the dimension, where we add one corresponding p^phi for each prime p. So every possible 11-limit noble mediant fits into a 10-dimensional vector space. How large of a vector space do we need for your 32 nobles?

Reply

29 w

- Dave Keenan
- Site Admin
**Posts:**2180**Joined:**Tue Sep 01, 2015 2:59 pm**Location:**Brisbane, Queensland, Australia-
**Contact:**

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

Mike BattagliaAdmin

FWIW, I was originally thinking about all of the nobles you can generate from the 5-limit being this very large dimensional space - what was it, like 150-dimensions or something, just for the 5-limit? I'm not sure how it compares if you only want to add one or two nobles. But then again, if you only want to add one or two nobles, I'm not sure why you wouldn't just add those nobles directly - what was the point of all of these feudal primes and etc then?

Reply

29 w

Mike BattagliaAdmin

Looks like the first 32 nobles fit into a 16-dimensional vector space and go up to the 17-limit, so the geometric version would give you a 14-dimensional vector space instead - not much better. So yeah, if you are only interested in using a few nobles, you don't get such a huge vector space to begin with. In that situation, the main benefit of this system would be that you don't have the weirdness of having to factor 5, 11, etc into smaller primes.

But again, I would ask what the point of the feudal prime decomposition is to begin with if you are just adding a small hand-selected set of nobles - why go through the hassle of having to factor 11 and etc? Just add the nobles themselves as basis intervals.

The other advantage of the geometric version is that you get very nice transpositional invariance properties, a nice "geometric phi mediant" operation that is closed on any two ratios within the space, etc. I'm not sure that the closeness to HE maxima matters all that much since they're so close, but FWIW the geometric version is also slightly closer, to the extent that matters. So one question is why the regular feudal numbers should be the gold standard to begin with.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia Of course, I've never seen the musical relevance of *all* the noble mediants that you can generate from some p-limit. And of course, to be mathematically rigorous you should have said "sufficiently good approximations of" *all* the noble mediants, since it can't actually generate the nobles. Not that that matters psychoacoustically.

For the dimensionality, see my table 14 or 15.

viewtopic.php?p=4611#p4611

16 dimensions. But that's only for the nobles. I consider these nobles to be adjoined to the 13-limit rationals. When we include those, that's another 6 dimensions for a total of 22 [EDIT: Actually only 4 more for a total of 20. I forgot you get 5 and 11 for free with the nobles]. The simplest mediends of these nobles require primes up to 17, so I assume that means with your basis it would only require 14 dimensions. That's definitely a worthwhile reduction.

I look forward to seeing these vectors, and how close they are to the nobles in cents.

Noble frequency ratios as prime-count vectors in ℚ(√5) - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) - The Sagittal forum

Reply

Remove Preview

29 wEdited

Mike BattagliaAdmin

Dave Keenan 'And of course, to be mathematically rigorous you should have said "sufficiently good approximations of" *all* the noble mediants'

Like I said, there is a sense in which "geometric nobility" exists; in that for some pairs of ratios, you can get a monzo in which the coefficients are themselves noble feudal integers. This happens if the quotient of two ratios has Linf norm = 1. Thus, I view these "geometric nobles" as useful entities in their own right, not just as approximations to the regular noble numbers.

But again, I would ask why the regular feudal numbers should be the gold standard to begin with when there are things like this...

Reply

29 w

Dave KeenanAuthor

I forgot you get primes 5 and 11 for free with the nobles (as f5² and f11.F11), so it's only 4 extra primes for a total of 20. But that doesn't alter the conclusion that 14 is better.

Reply

29 w

Mike BattagliaAdmin

I note earlier you said

"Or ignore the strawman and address the other thing, where noble number frequency ratios at least have the guaranteed property that they are maximally distant from all nearby simple ratios, with their distance being a monotonically-increasing function of their simplicity, while other kinds of phi-weighted mean do not have this property."

Can you clarify this? I don't see how this is true. Given some noble number, like phi, there will be plenty of simple ratios that are closer to it than more complex ratios, and vice versa. For instance, phi is closer to 21/13 than 23/14, or in fact anything/14.

Reply

29 wEdited

Dave KeenanAuthor

You're right. I spoke wrong. What I have in mind is the idea that phi, and to decreasing degrees the other nobles, are "the most irrational numbers" because their continued fractions end in all 1's. This is visualised on the SB-tree in the way that the nobles zoom down their zigzags, like Luke on the Death Star, staying as far as possible from the converging rationals on both sides. And yes, this is the basis of their claim to being the gold standard. Any approximation is going to "clip a wing" at some stage.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia I agree that in some cases you might as well just add the nobles instead of the feudal primes. But as Steve Martin noted, the case where you only want the 8 simplest nobles, along with the 11-limit, is interesting, since you only need a basis of 2.3.f5.7.f11.F11.ϕ. Only 7-dimensional, versus 10-dimensional with your (otherwise clever) approximations.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan the important thing is that when we say that phi is the "most irrational number," we are typically referring to a very particular notion of "distance" to each rational. In particular, we are talking about the linear absolute distance, e.g. |r - n/d| for some real number r, not the logarithmic absolute distance, e.g. |log(r) - log(n/d)|, which is almost certainly what we really care about.

The golden ratio and all of the noble numbers are "badly approximable": https://mathworld.wolfram.com/BadlyApproximable.html

This means that, for some rational p/q, we have that |x - p/q|*q^2 gets closer to some particular value as q -> inf, for the "best possible choice" of p/q for each choice of q. This is the "Markov constant" of the number. Good reading here: https://perso.univ-rennes1.fr/.../serie ... ry-markoff...

We note that |x - p/q|*q^2 is basically equal to err(x, p/q) * complexity(p,q), where we are using the absolute linear distance for error, and q^2 as the complexity.

In our situation, many of these things are different. Most importantly, the metric we really care about is the logarithmic distance. I would very much like to know how much any of these results hold if we're doing things logarithmically instead. Then we also use squared error sometimes, and sqrt(n*d) weighting instead of d^2, and so on. This is the main reason that all of these different models (noble numbers, HE, whatever) don't give exactly the same results.

The point is that all of these are approximations to whatever is happening in the brain, so there is some benefit to not taking them too seriously and going with the simplest thing that is in the ballpark and yields sensible results.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan regarding the 11-limit, I would say that if you only want a few noble numbers, then the computational advantages of using a 7 vs 10 vs 14 vs whatever dimensional space may be negligible as far as performance in temperament searches goes.

Mostly I just thought the geometric version was really neat, and had a neat structure with this basis of |2 2^phi 3 3^phi ... etc> that is easy to understand, whereas I'm still kind of thrown off by what is happening with f5, f11 and so on. But it may well be that if you do things in the feudal numbers with a basis of nobles things get simpler and are also about as easy to grasp.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia That's a good analysis of the badness of irrationals in terms of error and complexity. I agree we care more about log absolute distance than linear absolute distance. And I agree we care more about n×d than d². I agree with going with the simplest thing in the ballpark that yields sensible results psycho-acoustically.

I was never worried about how long the computations might take. Reasons to prefer lower dimensionality are rather: (a) ease of grasping by us humans, (b) increased likelihood that good temperaments are likely to be found at all, where "good" temperaments combine low rank, low complexity and low error.

I too, think your geometric-phi-mediant version is really neat, now that I understand it can reduce dimensionality. You should do a proper write-up of it somewhere more permanent than facebook, with a table like I suggested earlier. I'd like to link to it from my forum articles.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan yes that is fair. If I can get some time I'll write it up properly.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia I'm clearly not as thrown as you by what happens with f5, f11 etc. It seems easy enough to me, to grasp that 5 is replaced by its square root, and primes ending in 1 or 9 split into two almost-square-roots, and nobles are ratios of an f on an f or an F on an F. But I agree that once you go beyond the 8 simplest nobles, the dimensionality grows too fast and it's hard to keep track of which f/f's and F/F's are actually nobles.

Those Stern-Brocot zig-zag funnels of simple ratios exist in their linear arrangements, no matter that we'd prefer them to be logarithmic. So if you don't use an actual noble, you're going to blast through the side of the funnel at some point. But you're right that that doesn't matter, provided that point is far enough down the funnel that it only passes by ratios that are too complex to be considered JI.

So your geometric-phi-mediant scheme suggests yet another scheme. One that doesn't require /any/ irrationals in our vector basis. Just choose a complex rational far enough down each zigzag, to approximate each noble. For example, for approximating ϕ, 13/8 is too simple, too JI, but 21/13 (830¢) is probably fine, and no one could argue with 34/21 (834¢). This might only require 23-limit, i.e. 9 dimensions, with primes 17, 19 and 23 being used only for approximating nobles. Or maybe we have to go to 31-limit. That would still only be 11-dimensions for the 32 simplest nobles plus 13-limit JI.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan I'm not sure I get the significance of the zig zag pattern. It tells you that the best approximations to the number alternate between sharp and flat. But even if you blast through the funnel at some simple ratio, all that tells you is that you may have two successive best approximations that are both sharp, for instance. What does that mean for our "logarithmically inapproximable" numbers?

Reply

29 w

Dave KeenanAuthor

Mike Battaglia What I think it means, is that there may be no simple definition of logarithmically-inapproximable numbers. Maybe nobles are the best you can do. But this is from a pure math perspective, and doesn't necessarily matter when you bring in the psychoacoustics. So what do you think of the idea of using complex rationals (and hence higher ordinary primes) to approximate nobles? That's effectively what Margo Schulter has been doing, right from the beginning of her neo-medieval work.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan I think it isn't a terrible idea; in fact starting with nobles should yield these kinds of rational approximations as useful temperaments thereof! I mean, in some sense, the "geometric noble" thing was basically a midway point between the true nobles and the thing you are suggesting.

Reply

29 w

Mike BattagliaAdmin

Here are some results regarding neighboring pairs of superparticular ratios. The second column is the regular noble mediant, and the third is the geometric noble mediant, and the fourth is the difference between them. The approximation involved says that the geometric noble mediant ≈ the regular noble mediant as the difference between the two intervals gets smaller in size and complexity. We can see for instance that when we get to around 5/4, 6/5, etc, the two are only a few cents off.

"2/1 -> 3/2" "833.0903" "892.1913" "59.101"

"3/2 -> 4/3" "560.0666" "575.9317" "15.8651"

"4/3 -> 5/4" "422.4874" "428.9913" "6.5039"

"5/4 -> 6/5" "339.3439" "342.6358" "3.2919"

"6/5 -> 7/6" "283.6048" "285.4995" "1.8947"

"7/6 -> 8/7" "243.6192" "244.8091" "1.1898"

"8/7 -> 9/8" "213.5281" "214.324" "0.79588"

"9/8 -> 10/9" "190.0599" "190.6184" "0.5585"

We can also look at pairs of neighboring integers to get:

"1/1 -> 2/1" "833.0903" "741.6408" "91.4495"

"2/1 -> 3/1" "1666.1806" "1633.832" "32.3485"

"3/1 -> 4/1" "2226.2472" "2209.7637" "16.4834"

"4/1 -> 5/1" "2648.7345" "2638.755" "9.9795"

"5/1 -> 6/1" "2988.0784" "2981.3908" "6.6877"

"6/1 -> 7/1" "3271.6832" "3266.8903" "4.793"

"7/1 -> 8/1" "3515.3025" "3511.6994" "3.6031"

"8/1 -> 9/1" "3728.8305" "3726.0233" "2.8072"

"9/1 -> 10/1" "3918.8904" "3916.6417" "2.2487"

So again this approximation starts to get better around the 6/1 ish range.

It would be nice to try this with some other noble numbers.

Reply

29 wEdited

Paul ErlichAdmin

Mike Battaglia maybe my observation would be relevant if we start with 5:3 and 8:5, the two nearest local minima?

Reply

29 w

Mike BattagliaAdmin

Paul Erlich the geometric noble mediant is then (5/3)^(1-p) * (8/5)^(p) = about 840.68 cents.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia so maybe my idea did make sense after all????

Reply

29 w

Mike BattagliaAdmin

Paul Erlich yes, if you're specifically talking about 5/3 and 8/5 then it does make sense and the geometric version is closer to the local maximum of HE. Again I thought you were referencing our prior conversation about the noble mediant of 1/1 and 2/1...

Reply

29 w

Paul ErlichAdmin

Mike Battaglia repeating just for completion (not to be repetitive, and I know you already saw this): in Dave and Margo's original article, I believe they explicitly start from the two nearest (flanking) local minima or consonances.

Reply

29 w

Mike BattagliaAdmin

I guess it is noteworthy for the conversation with Paul Erlich to note that geometric noble mediants differ from regular noble mediants in that, for instance, some of the pairs of ratios that give the same noble mediant can give different geometric noble mediants. For instance:

"1/1 -> 2/1" "833.0903" "741.6408" "91.4495"

"2/1 -> 3/2" "833.0903" "892.1913" "59.101"

"3/2 -> 5/3" "833.0903" "814.6867" "18.4036"

"5/3 -> 8/5" "833.0903" "840.6808" "7.5905"

"8/5 -> 13/8" "833.0903" "830.2752" "2.8151"

These all have a noble mediant of "acoustical phi" whereas the geometric noble mediants are all different. Note again the results only tend to approximate the regular noble mediant well as the difference between the two intervals starts to get to about a half step or so.

That being said, even with regular noble mediants I'm not sure that you get sensible results for pairs of ratios that are hugely spaced apart.

Reply

29 wEdited

Mike BattagliaAdmin

Paul Erlich: OK, for s = 1% and using the zeta function HE for N=Inf, we get the following between neighboring minima, always going from the simpler to the more complex ratio:

ratios: regular geometric maximum

1/1>8/7: 213.528076 142.873447 48.600000

7/6>8/7: 243.619228 244.809062 238.230000

6/5>7/6: 283.604818 285.499534 287.550000

5/4>6/5: 339.343874 342.635752 349.800000

5/4>9/7: 422.487396 416.455467 423.030000

4/3>9/7: 448.459360 459.133021 458.580000

4/3>7/5: 560.066561 550.248596 538.770000

3/2>7/5: 606.851200 628.135286 658.590000

3/2>8/5: 792.105295 771.008733 745.290000

5/3>8/5: 833.090296 840.680751 846.120000

5/3>7/4: 942.516832 936.562309 924.840000

7/4>9/5: 1001.612335 998.967660 998.370000

2/1>9/5: 1038.629251 1087.268306 1153.050000

It looks like the geometric one is *usually* closer than the noble one, except interestingly for 5/4 > 9/7. The maxima at 48.6 and 1153 cents are both pretty far off in both cases. Sometimes I think we may have gotten better results if we went from the more complex to the simple ratio...

Reply

29 wEdited

Mike BattagliaAdmin

Paul Erlich here are similar results for the min-entropy at s=1% (actually approximate min-entropy, w/ a=10), rounded to the nearest cent:

ratios: regular geometric maximum

1/1>11/10: 155.819149 101.978222 46.000000

10/9>11/10: 171.243282 171.650240 171.000000

9/8>10/9: 190.059883 190.618384 191.000000

8/7>9/8: 213.528076 214.323958 216.000000

7/6>8/7: 243.619228 244.809062 248.000000

6/5>7/6: 283.604818 285.499534 291.000000

6/5>11/9: 339.343874 335.274159 338.000000

5/4>11/9: 355.865983 362.268624 360.000000

5/4>9/7: 422.487396 416.455467 413.000000

4/3>9/7: 448.459360 459.133021 468.000000

4/3>11/8: 541.418757 530.969489 529.000000

7/5>11/8: 560.066561 563.233086 562.000000

7/5>10/7: 606.851200 604.128311 603.000000

10/7>13/9: 630.430476 629.310707 634.000000

3/2>13/9: 644.639704 661.574303 667.000000

3/2>11/7: 770.641415 751.729626 737.000000

8/5>11/7: 792.105295 794.407179 794.000000

8/5>13/8: 833.090296 830.275169 834.000000

5/3>13/8: 848.859910 857.269634 856.000000

5/3>12/7: 923.029392 914.500466 913.000000

7/4>12/7: 942.516832 946.764063 946.000000

7/4>9/5: 1001.612335 998.967660 994.000000

9/5>11/6: 1038.629251 1037.229160 1036.000000

11/6>13/7: 1063.998505 1063.169088 1063.000000

13/7>15/8: 1082.472117 1081.940699 1085.000000

2/1>15/8: 1096.525910 1130.946268 1158.000000

Again, the two results are typically very close, but the geometric version is often just a little bit closer. Either way, it's certainly close enough to be useful, particularly since we get a nice clean vector space...

Reply

29 w

Mike BattagliaAdmin

Another unrelated thought: suppose you think the mediant doesn't go far enough towards the more complex rational. So instead of the mediant of 5/4 and 6/5, you take the "double mediant" (5+6+6)/(4+5+5) = 17/14. Then you repeat again, taking the double mediant of 6/5 and 17/14, and so on. Then you get a bunch of "silver noble numbers" instead, whatever they are called, whose continued fractions end in all 2's rather than all 1's. These may also be useful. In general you can replace with the n-fold mediant to get generalized metallic noble numbers.

Reply

29 w

Mike BattagliaAdmin

Actually, quick correction, you would have to go from the simpler to the more complex ratio to get a silver noble number.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia They have been called silver aristocratic numbers. Noble = gold aristocratic. I'm guessing that a tree like the Stern-Brocot tree but based on double-mediants would not cover all rationals. Is there a requirement for a determinant of +-1 or +-2 for the limit of a double-mediant series to be a silver aristocratic?

Reply

29 w

Dave KeenanAuthor

Answering my own questions: The determinant still has to be ±1 for silver aristocrats, same as for nobles. A zigzag of double-mediants, whose limit is a silver aristocratic, skips every other level of the Stern-Brocot tree, and is closer to its convergents (in logarithmic terms) than the corresponding noble. For example, the silver ratio itself has zigzag convergents 2/1 5/2 12/5 29/12. It is 10 cents from 12/5 and 1.8 cents from 29/12. The corresponding noble, n12/5, has zigzag convergents 12/5 17/7 29/12 and is 14 cents from 12/5, 6 cents from 17/7 and 2.3 cents from 29/12.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan I looked at this yesterday and got that the determinant must indeed be ±2. What pair of ratios are you talking about? If you look at consecutive "zig-zag" points, for instance, like (6+5+5)/(5+4+4) = 16/13, being the "double mediant" from 6/5 to 5/4, then the determinant of 6/5 and 16/13 is indeed 2.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia The recurrence relation you're using, namely

(2nᵢ₋₂ + nᵢ₋₁ ) /(2dᵢ₋₂ + dᵢ₋₁ ) = nᵢ/dᵢ

is not the silver aristocratic (silvercratic) one. In fact it doesn't even converge on irrational numbers. It converges on rational numbers, namely the ordinary mediants. In the case of 5/4 6/5 16/13 28/23 60/49 ... it converges on 11/9. Absolute values of successive determinants for that sequence are 1 2 4 8 16 32 ... i.e. increasing powers of 2. You just happened to pick the pair of successive ratios that gave an absolute determinant of 2.

The silvercratic recurrence relation is

(nᵢ₋₂ + 2nᵢ₋₁ ) /(dᵢ₋₂ + 2dᵢ₋₁ ) = nᵢ/dᵢ

Your first example was correctly silvercratic, namely 5/4 6/5 17/14 40/33 97/80 ... Notice successive pairs here are unimodular.

I didn't understand the full import of your "correction", where you said you would have to go from the simpler to the more complex ratio to get a silver noble number, because that's what I see the above sequence as doing, and in any case, the following sequence is also silvercratic: 6/5 5/4 16/13 37/30 90/73 ... Again unimodular.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan That first thing isn't the recurrence relation I was using. I'm just zig-zagging 2 to the right, 2 to the left, etc on the Stern-Brocot tree. After 16/13, the thing you'd be taking the double mediant of is 11/9, so you'd get 38/31 and the determinant of 16/13 and 38/31 is 2. I had it so you're always taking the double mediant of the parent of the current element in the tree, not the double mediant of the last output emitted by the algorithm. If you want it as a recurrence relation in terms of outputs then I think the second recurrence relation you have is correct.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia That's weird. You start with 5/4 and 6/5 which have a determinant of 1. But I agree that from then on your procedure gives 6/5 16/13 38/31 92/75 ... which do indeed have determinants of 2 and do obey the silvercratic recurrence relation (nᵢ₋₂ + 2nᵢ₋₁ ) /(dᵢ₋₂ + 2dᵢ₋₁ ) = nᵢ/dᵢ and they do indeed converge on a silvercratic number. But I note that the same silvercratic number can also be generated as the limit of a sequence having determinants of 1, using the same recurrence relation, namely: 1/1 5/4 11/9 27/22 65/53 ...

Reply

29 w

Mike BattagliaAdmin

Dave Keenan yes, I guess you would say the zig zag properly starts at the pair of 6/5 and 16/13.

Reply

29 w

Mike BattagliaAdmin

That is kind of interesting and I guess it depends on how you are describing the zig zag. Clearly if you use the "endpoints" of the zig-zag you'll get a determinant of 2, or of n for the n'th metallic mean. But I guess there are other nontrivial descriptions with different determinants.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia You could also back up to 4/3 and 6/5 to start it. Or, as I say you can start it from 1/1 and 5/4 and end up converging on the same place. So it can work with determinants of 1 or determinants of 2.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia It turns out, the series with a determinant of 1 are the convergents of the silvercratic. The series with a determinant of 2 are semiconvergents.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan yes, that makes sense. Curious how this works for other metallic means. You will always get semiconvergents as the endpoints of the zigzag and convergents as one prior to that.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia For the one bronzocratic example I tried, I find three equivalent series — the convergents with determinant 1:

1/1 5/4 16/13 53/43 ...

and two different semiconvergent series, both with determinant 3.

3/2 6/5 21/17 69/56 ...

4/3 11/9 37/30 122/99 ...

Reply

29 w

Mike BattagliaAdmin

BTW to Dave Keenan, answering an earlier question about approximating nobles with rationals... kind of a failed idea, but just making a note anyway.

As I mentioned, instead of monzos w/ integer coefficients in the basis 2, 2^phi, 3, 3^phi, etc, we can treat them as just monzos in the basis 2, 3, etc but with coefficients that are feudal integers. E.g. each monzo is of the form |(a1+a2*phi) (b1+b2*phi) ...>. I actually prefer that we go with "little phi" here, which I will write p = 1/phi, so the coefficients are of the form x+y*p.

To make this very simple, we can simply look for a rational approximation to phi within the monzo coordinates. For instance, let's say that p ≈ 2/3. Then our monzos become very simple and are of the form |a/3 b/3 ...>, e.g. we simply get monzo coordinates w/ denominators of at most 3.

This is the same basic idea as before, except we are now going 2/3 of the way from the simpler to the more complex ratio rather than 1/phi of the way (and I am kind of curious if that gives even better results).

The only problem, though, is that our "nobles" can't be tuned independently from the rationals like they can with independent basis vectors. For instance, the approximate noble from 5/4 to 6/5 is now 339.198 cents, which is always 6/5 * (25/24)^(1/3). This means that the relevant interval doesn't even *exist* in any EDO which maps 25/24 to only 1 or 2 steps. This also turns out to be an issue in a slightly different way if we just approximate nobles with rational numbers; we can't give nobles their own independent complexities and tunings and etc.

One answer is to look at several different possible rational approximations for phi at the same time and see what you get, but really the goal of finding these good rational approximations is the same as just looking for good "commas" in the larger original space of feudal numbers.

I think, though, I'm mostly happy with the "logarithmic feudal numbers" or whatever we're going to call them as some kind of sweet spot; probably the regular feudal numbers are also alright although I still wish we had a better choice of basis...

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan man, another really great property of the "logarithmic feudal numbers"!

Suppose after doing some listening tests, your ears tell you that 61.8% is not the best amount to go from ratio a to ratio b, on average, to get to some approximate local maximum between them. Suppose you think something like 70% is better, or whatever. Then this is as simple as changing 2^phi, 3^phi, etc to 2^(1.7), 3^(1.7) etc instead. So we can introduce an adjustment term which one can dial in empirically. In general we can write this as 2^X for some magic term X.

We can take it even further. Suppose that you like the double mediant better than the mediant, and you want to build things on the silver ratio and the silver aristocratic numbers and etc. Then doing the same geometric approximation as before, you just get 2^(1+sqrt(2)) instead. And likewise with all of the metallic means.

So really there is just one model. We start with the primes 2, 3, 5, ..., and then we add one more basis element 2^X, 3^X, 5^X, ... for whatever value of X. Different values of X correspond to different approximations of aristocratic numbers, and in fact we can choose X to be anything in between that we like empirically. If we wanted to do this with true silver numbers in general, we'd have to look at the field Q[sqrt(2)] and so on.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan just a quick note that I tried some different values of X for a bunch of pairs of intervals.

It really does seem to be true that, as the intervals you are looking at have nd -> inf, X=phi gives the best approximation. I guess this kind of makes sense given the geometric ≈ arithmetic mean approximation I was using but it is still a little bit surprising to me.

However, I note that there may be better values of X than phi if you only care about a few nobles.

One interesting question is that it's kind of nontrivial how to measure this. For instance, as you mentioned before, pairs of ratios that have the same noble mediant need not have the same approximation. So we are not actually approximating nobles, but "phi-weighted mediants of pairs of simple ratios," with it being kind of a fortuitous happenstance that the true noble mediant happens to have a bunch of these pairs generating the same noble number.

Of course, any noble can be approximated to arbitrary closeness if one goes far enough into the SB tree and gets some pair of rationals arbitrarily close to the noble in question, which generate that noble. But for each noble, there does seem to be a "sweet spot" where there are two relatively simple ratios, typically about a half step apart, which generate that noble and also give a decent phi-weighted approximation. I also note that this "sweet spot" changes as you vary the value of X.

Anyway, just some thoughts.

Reply

29 w

Mike BattagliaAdmin

In light of recent results, I would also just add that instead of saying the "sweet spot" changes as you vary the value of X, you can also say that the value of X can change as you would like to dial in the sweet spot.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan lastly, and perhaps most importantly, this also reveals another challenge with the regular feudal numbers. With those, you have to split 5 into sqrt(5), since f5 is a pretty important prime when looking at noble numbers. Thus when we search for "temperaments" thereof, there won't ever be any EDOs that map 5 to an odd number of steps.

If we try something with silver numbers instead, I expect 2 will split into sqrt(2), so we will get only even-numbered EDOs.

I'm not sure if there is some way around this using the true feudal numbers; we could try using a different basis to some extent but in general 5 will tend to be square.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia I'm not sure if you're saying the idea of approximating nobles with rationals is a failed idea, or just referring to the idea of using [a/3 b/3 ...⟩ or similar, based on a rational approximation of 1/ϕ, which you noted fails because you can't tune the nobles independently. I note that [a/3 b/3 ...⟩ with an ordinary prime basis, is equivalent to [a b ...⟩ with a and b integers, when the basis consists of the cube roots of the ordinary primes.

You claim that this is also a problem for approximating nobles with rational numbers. You say "we can't give nobles their own independent complexities and tunings and etc.". But that is not so if, as I suggested, we use the higher primes only for approximating nobles. Say we want the 13-limit rationals plus some simple nobles. Then we use the primes 17 19 23 (and maybe 29 and 31 if necessary), in combination with lower primes, to approximate the nobles. In this case we are not interested in using primes above 13 for JI, so they are free to be used to tune the anti-JI intervals, and to give them complexities independent of the rationals.

I think it's incredibly unlikely that listening tests would shift all the anti-JI points in the same /direction/ away from the phi-mediants, let alone to similar ratio-X-mediants, where "same direction" means all towards simplicity or all towards towards complexity. And so I think the probability that these would be consistently nearer to silvercrats than to nobles is close to zero.

I don't understand why it's a problem, that there won't ever be EDOs with good approximations of certain simple nobles, that that map 5 to an odd number of steps. To me, that's just a very useful observation of a mathematical fact.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan I may have misunderstood the original idea. If you are adding noble numbers as new basis elements using only new primes, why not just add the new nobles directly? Adding new primes means the problem is that the "complexity" of each noble is determined by the complexity of the primes, so the weighting may not be what you want.

Re listening tests shifting all of the anti-JI points away from the phi mediants on average, are you talking about the regular phi mediant or the geometric version?

Reply

29 w

Mike BattagliaAdmin

Regarding this: "I don't understand why it's a problem, that there won't ever be EDOs with good approximations of certain simple nobles, that that map 5 to an odd number of steps. To me, that's just a very useful observation of a mathematical fact."

There won't ever be EDOs with *any* approximations of certain simple nobles that map 5 to an odd numbers of steps, good or bad, as long as we are talking about "regular temperaments" derived from the ambient space of feudal numbers.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Right. Why is that a problem?

I don't understand why such EDOs* should exist with /any/ basis.

*EDOs with good approximations of certain simple nobles, that map 5 to an odd number of steps.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia I'm talking about both the regular phi-mediant and the geometric one in so far as it approximates the regular one. I'm not sure what the geometric values actually are. I'm not sure whether the geometric idea is an entirely coherent one. Don't you get slightly different values close together, depending on what ratios you feed into it, even when those values are all successive convergents of the same noble? Do the geometric mediants converge in any sense?

Reply

29 wEdited

Dave KeenanAuthor

The reason it's better to add new (ordinary) primes rather than just adding the nobles themselves to the basis, is lower dimensionality. Each new prime will allow for the approximation of multiple new nobles. And there is no reason why these noble-only higher primes need to have their complexity calculated in the same way as the lower primes that are used for rationals.

Reply

29 w

Mike BattagliaAdmin

The problem is simply that EDOs with 5 mapped to an odd number of steps can still have perfectly good approximations of noble numbers and regular primes. This method will rule those out.

I'm not sure what you're asking about geometric mediants or what you think is incoherent about it, but yes, different convergents give different values. That is one property you lose when going from the regular nobles to this thing. It's still a good approximation though, for any pair of ratios in a reasonably similar size range.

I don't think that last point about dimensionality is true - if you have a N-dimensional vector space and want to add M new linearly independent vectors, then the new vector space is of dimension M+N. You can't just change the basis and make the vector space of lower dimension.

Reply

29 w

Dave KeenanAuthor

I guess you're right re EDOs. But to set my mind at rest, can you find an example of an EDO that maps 5 to an odd number of steps and has good approximations (say ±5 cents) of the four simplest nobles: n1/1, n2/1, n3/2, n3/1 (833c, 1666c, 560c, 2226c)?

You have sufficiently answered my question about geometric mediants. Thanks.

We seem to be talking past each other re approximating nobles using higher primes. Of course, what you say is true, but I don't see how it argues against my proposal. Let's leave that question until I come up with a specific proposal for how to approximate the 32 simplest nobles using higher primes.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan that's a good question and I'm not sure offhand about that level of accuracy with that many nobles at once. I would guess that the more accuracy you want, and for more nobles simultaneously, it gets harder for that to be possible. If you relax the criteria somewhat you have that 34-EDO has a half-decent version of most of those, say within 10-ish cents rather than 5, with the exception of phi. It also has 5 mapped to an odd number of steps and so the corresponding val doesn't even exist. I'm sure there's some better example but it's late and I need to sleep, lol.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan another thought is that probably the simplest way around this would be to introduce an artificial "prime 5" basis element that is independent of sqrt(5)^2. The ideal tunings of 5 and sqrt(5)^2 will be identical, but they are treated as different elements. This parallels how sometimes we'll add a "prime 9" to 5-limit JI so we can model things like 16-EDO, which has a decent approximation to 9/8 at 225 cents, good for use in chords like 0-975-1425 ≈ 4:7:9 that is different from the mapping you get for 3^2.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia I just discovered something that suggests noble numbers are ideal in terms of /logarithmic/ distance from their convergents. If you multiply any noble's absolute-distance-in-cents from any of its convergents, by the product-complexity (n×d) of that convergent, you always get a badness figure between 680 and 1020, and the badness converges to approx 774.23 as the complexity of the convergent increases. How do the geometric mediants fare using this badness measure, which is the product of what I believe we previously agreed were desirable error and complexity measures?

This doesn't diminish the possible utility of the "logarithmic feudal integers" or "phith-roots-of-primes" in approximating nobles while reducing the number of dimensions required, but it helps to explain why the nobles are the "gold standard".

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan that is super interesting - how do you prove that property? About convergence to 774.23. What do other real numbers converge to - the silver ratio, or even just random real numbers?

Reply

29 w

Cmloegcmluin Xenharmonic Feisbeuk

hey, just wanted to say that I’ve been trying to keep up to date with you two across all these various comment sub-threads. Liking posts when they have cool thoughts and/or good questions. I don’t think I have much original thought to contribute here, so don’t let me slow y’all down, but if there’s anything either of you think I could try to help out with, I’d be happy to give it a go. In any case, I’m a big fan of metallic means and stanky sounds, so this is really exciting to me that y’all are making headway with this!

Reply

29 w

Mike BattagliaAdmin

Cmloegcmluin Xenharmonic Feisbeuk good stuff. I was going to make a Wiki page called "anti-JI" about this at some point, meaning any system which has intervals that are intended to be maximally dissonant, which summarizes a few different systems for modeling anti-JI. The feudal numbers are one such system - perhaps the prototypical system. So all of the stuff about feudal primes and noble mediants and Q[sqrt(5)] and etc would go there, as well as stuff about the silver ratio and etc. Then the "geometric feudal numbers," or whatever they are called, would be a different approach that models things in a different way but also gives similar results. There may be other systems beyond this which would be on that page. It may just make sense for anti-JI to be a page that links to separate pages on feudal numbers and etc since the results have become so extensive.

Reply

29 w

Cmloegcmluin Xenharmonic Feisbeuk

Mike word. the ink is pretty wet on all the feudal stuff... Dave's still editing those Sagittal forum posts on a daily basis, methinks. I can't imagine any good arguments against adding an "Anti-JI" page on the wiki at this point, for the reasons as you describe them (you may recall that *I* pushed back against the idea of "anti-JI" a week ago or so, but Dave has since disabused me of that position!) I agree the results are extensive (and the frontier is still open), but maybe let's start with that page for now and see where it goes. It's super easy to break pages out once stuff gets big enough.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia You wrote: "How do you prove that property about convergence to 774.23? What do other real numbers converge to - the silver ratio, or even just random real numbers?"

I don't know how to prove it. I just took the standard 32 nobles in a spreadsheet and ran off about 10 convergents for each and computed how badly the noble approximated each convergent according to the badness formula

|log₂(n/d)-log₂(α)|×1200×n×d, where α is a noble and n/d is a convergent of α.

And I saw that these badnesses never went below 680 and converged to about 774.23.

It's funny to say that the badness converges, because what that means is that there is a sense in which the convergents of a noble number never converge on the noble. The noble remains an equally bad approximation of them for their complexity, all the way to infinity. That quantifies the thing I was always led to believe about nobles, and tried to state, somewhat ineptly, earlier in our discussion.

I expect this to be true of all metallic means and their aristocrats, except that none of them will stay as far from their convergents as the nobles do. Those snooty nobles really stay as far as they can from those rational commoners.

I tried the silver ratio itself, and a few other random silvercrats. Their badness in approximating their convergents converged to about 612.1.

For bronzocrats the magic number is 480.2.

I can't help thinking a different choice of log base (and dropping the ×1200) might reveal some simple pattern to those numbers. But I haven't found it yet.

And I expect that all non-metallic non-aristocratic irrationals will pass arbitrarily close to one of their convergents eventually. Consider π and its famous convergent 355/113. Its badness by the above formula is only 5.9. In other words π is a damn good approximation of 355/113 (although we don't usually look at it that way).

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia I've got it! When you use the natural log, the magic badness numbers are 1/√5 for nobles, 1/√8 for silvercrats and 1/√13 for bronzocrats. i.e. when the badness of approximation of their convergents is given as:

|ln(n/d)-ln(α)|×n×d,

where α is an aristocratic number and n/d is one of its convergents,

then the badness converges to those reciprocal square roots above. In general, for aristocrats of the n-th metallic mean, the limit badness is 1/√(n²+4). As usual, no proof, just conjecture based on experimental results.

The nobles /are/ the gold standard.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia Putting it another way, those cents-based values are:

1200/ln(2)/√5 ≈ 774.2314037

1200/ln(2)/√8 ≈ 612.083668

1200/ln(2)/√13 ≈ 480.1579334

Reply

29 w

Cmloegcmluin Xenharmonic Feisbeuk

It’s cute how the radicands of those roots overlap with the Fibonacci series (but only for those three; the next one is 20 instead of 21, and the previous one would be 4 instead of 3). And yes I have been waiting for an opportunity to use that word “radicand” haha

Reply

29 w

Mike BattagliaAdmin

Dave Keenan that 1/sqrt(5) is the famous Markov constant of the golden ratio, but the usual expression is |n/d - a|*d. How are you showing that you get the same thing for |ln(n/d)-ln(a)|*nd?

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Like I said, I just calculated 32 x 10 of them in a spreadsheet and saw what they did. When I try the same thing with |n/d - a|×d it converges to zero. However |n/d - a|×d² converges to 1/√5 for nobles. Is that what you meant to write?

Reply

29 w

Mike BattagliaAdmin

Oops, yes, d^2 is it, sorry. You're saying you have just empirically determined you get the same thing using logs and nd? That's nuts. What happens if you use |ln(n/d)-ln(a)|*d^2 instead?

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Yes. That's what I was saying. But now I have a proof that |n/d - α|×d² and |ln(n/d)-ln(α)|×nd converge on the same value. It depends on the fact that ln(x) → x-1 as x → 1.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia |n/d - α|×d²

= |α - n/d|×d²

= |α - n/d|÷(n/d) × (n/d)×d²

= |α÷(n/d) - n/d÷(n/d)|×nd

= |α÷(n/d) - 1|×nd

→ |ln(α÷(n/d))|×nd as n/d → α

= |ln(α) - ln(n/d)|×nd

= |ln(n/d) - ln(α)|×nd

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia As you probably expect by now, |ln(n/d)-ln(α)|×d² doesn't work. It converges on a different value for each noble, namely 1/(α√5).

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan that is really amazing! OK, wow. So any result usual results still hold even with this metric.

Here is an interesting thought. Suppose that for some bizarre reason you decide to use a Cauchy distribution for HE rather than a Gaussian. Also, you give all of the rationals 1/(nd)^2 weighting rather than 1/sqrt(nd) weighting. And also use the min-entropy rather than the Shannon entropy (and maybe the Shannon entropy as well). Then I am thinking that, as the "s" of the Cauchy distribution->0, we may get that the maxima will all be noble numbers.

This is because as s->0, the Cauchy distribution just starts to look like k/x^2 for some tiny value of k. Thus for some arbitrary dyad, the strength of the match to each rational is going to basically be proportional to 1/err^2 * 1/nd^2, which is the reciprocal of the square of the quantity you had before.

With the normal distance, the numbers which have maximum the best-possible weighted error on all rationals are the noble numbers. *If* your proof suggests this result generalizes to this logarithmic distance instead, we would get the same thing and thus the maxima would be noble numbers. (However I'm not quite sure if we can say that as your proof only holds for rationals arbitrarily close to the real number.)

Anyway, something to think about.

Reply

29 w

Mike BattagliaAdmin

Paul Erlich may be interested in that...

Reply

29 w

FWIW, I was originally thinking about all of the nobles you can generate from the 5-limit being this very large dimensional space - what was it, like 150-dimensions or something, just for the 5-limit? I'm not sure how it compares if you only want to add one or two nobles. But then again, if you only want to add one or two nobles, I'm not sure why you wouldn't just add those nobles directly - what was the point of all of these feudal primes and etc then?

Reply

29 w

Mike BattagliaAdmin

Looks like the first 32 nobles fit into a 16-dimensional vector space and go up to the 17-limit, so the geometric version would give you a 14-dimensional vector space instead - not much better. So yeah, if you are only interested in using a few nobles, you don't get such a huge vector space to begin with. In that situation, the main benefit of this system would be that you don't have the weirdness of having to factor 5, 11, etc into smaller primes.

But again, I would ask what the point of the feudal prime decomposition is to begin with if you are just adding a small hand-selected set of nobles - why go through the hassle of having to factor 11 and etc? Just add the nobles themselves as basis intervals.

The other advantage of the geometric version is that you get very nice transpositional invariance properties, a nice "geometric phi mediant" operation that is closed on any two ratios within the space, etc. I'm not sure that the closeness to HE maxima matters all that much since they're so close, but FWIW the geometric version is also slightly closer, to the extent that matters. So one question is why the regular feudal numbers should be the gold standard to begin with.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia Of course, I've never seen the musical relevance of *all* the noble mediants that you can generate from some p-limit. And of course, to be mathematically rigorous you should have said "sufficiently good approximations of" *all* the noble mediants, since it can't actually generate the nobles. Not that that matters psychoacoustically.

For the dimensionality, see my table 14 or 15.

viewtopic.php?p=4611#p4611

16 dimensions. But that's only for the nobles. I consider these nobles to be adjoined to the 13-limit rationals. When we include those, that's another 6 dimensions for a total of 22 [EDIT: Actually only 4 more for a total of 20. I forgot you get 5 and 11 for free with the nobles]. The simplest mediends of these nobles require primes up to 17, so I assume that means with your basis it would only require 14 dimensions. That's definitely a worthwhile reduction.

I look forward to seeing these vectors, and how close they are to the nobles in cents.

Noble frequency ratios as prime-count vectors in ℚ(√5) - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) - The Sagittal forum

Reply

Remove Preview

29 wEdited

Mike BattagliaAdmin

Dave Keenan 'And of course, to be mathematically rigorous you should have said "sufficiently good approximations of" *all* the noble mediants'

Like I said, there is a sense in which "geometric nobility" exists; in that for some pairs of ratios, you can get a monzo in which the coefficients are themselves noble feudal integers. This happens if the quotient of two ratios has Linf norm = 1. Thus, I view these "geometric nobles" as useful entities in their own right, not just as approximations to the regular noble numbers.

But again, I would ask why the regular feudal numbers should be the gold standard to begin with when there are things like this...

Reply

29 w

Dave KeenanAuthor

I forgot you get primes 5 and 11 for free with the nobles (as f5² and f11.F11), so it's only 4 extra primes for a total of 20. But that doesn't alter the conclusion that 14 is better.

Reply

29 w

Mike BattagliaAdmin

I note earlier you said

"Or ignore the strawman and address the other thing, where noble number frequency ratios at least have the guaranteed property that they are maximally distant from all nearby simple ratios, with their distance being a monotonically-increasing function of their simplicity, while other kinds of phi-weighted mean do not have this property."

Can you clarify this? I don't see how this is true. Given some noble number, like phi, there will be plenty of simple ratios that are closer to it than more complex ratios, and vice versa. For instance, phi is closer to 21/13 than 23/14, or in fact anything/14.

Reply

29 wEdited

Dave KeenanAuthor

You're right. I spoke wrong. What I have in mind is the idea that phi, and to decreasing degrees the other nobles, are "the most irrational numbers" because their continued fractions end in all 1's. This is visualised on the SB-tree in the way that the nobles zoom down their zigzags, like Luke on the Death Star, staying as far as possible from the converging rationals on both sides. And yes, this is the basis of their claim to being the gold standard. Any approximation is going to "clip a wing" at some stage.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia I agree that in some cases you might as well just add the nobles instead of the feudal primes. But as Steve Martin noted, the case where you only want the 8 simplest nobles, along with the 11-limit, is interesting, since you only need a basis of 2.3.f5.7.f11.F11.ϕ. Only 7-dimensional, versus 10-dimensional with your (otherwise clever) approximations.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan the important thing is that when we say that phi is the "most irrational number," we are typically referring to a very particular notion of "distance" to each rational. In particular, we are talking about the linear absolute distance, e.g. |r - n/d| for some real number r, not the logarithmic absolute distance, e.g. |log(r) - log(n/d)|, which is almost certainly what we really care about.

The golden ratio and all of the noble numbers are "badly approximable": https://mathworld.wolfram.com/BadlyApproximable.html

This means that, for some rational p/q, we have that |x - p/q|*q^2 gets closer to some particular value as q -> inf, for the "best possible choice" of p/q for each choice of q. This is the "Markov constant" of the number. Good reading here: https://perso.univ-rennes1.fr/.../serie ... ry-markoff...

We note that |x - p/q|*q^2 is basically equal to err(x, p/q) * complexity(p,q), where we are using the absolute linear distance for error, and q^2 as the complexity.

In our situation, many of these things are different. Most importantly, the metric we really care about is the logarithmic distance. I would very much like to know how much any of these results hold if we're doing things logarithmically instead. Then we also use squared error sometimes, and sqrt(n*d) weighting instead of d^2, and so on. This is the main reason that all of these different models (noble numbers, HE, whatever) don't give exactly the same results.

The point is that all of these are approximations to whatever is happening in the brain, so there is some benefit to not taking them too seriously and going with the simplest thing that is in the ballpark and yields sensible results.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan regarding the 11-limit, I would say that if you only want a few noble numbers, then the computational advantages of using a 7 vs 10 vs 14 vs whatever dimensional space may be negligible as far as performance in temperament searches goes.

Mostly I just thought the geometric version was really neat, and had a neat structure with this basis of |2 2^phi 3 3^phi ... etc> that is easy to understand, whereas I'm still kind of thrown off by what is happening with f5, f11 and so on. But it may well be that if you do things in the feudal numbers with a basis of nobles things get simpler and are also about as easy to grasp.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia That's a good analysis of the badness of irrationals in terms of error and complexity. I agree we care more about log absolute distance than linear absolute distance. And I agree we care more about n×d than d². I agree with going with the simplest thing in the ballpark that yields sensible results psycho-acoustically.

I was never worried about how long the computations might take. Reasons to prefer lower dimensionality are rather: (a) ease of grasping by us humans, (b) increased likelihood that good temperaments are likely to be found at all, where "good" temperaments combine low rank, low complexity and low error.

I too, think your geometric-phi-mediant version is really neat, now that I understand it can reduce dimensionality. You should do a proper write-up of it somewhere more permanent than facebook, with a table like I suggested earlier. I'd like to link to it from my forum articles.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan yes that is fair. If I can get some time I'll write it up properly.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia I'm clearly not as thrown as you by what happens with f5, f11 etc. It seems easy enough to me, to grasp that 5 is replaced by its square root, and primes ending in 1 or 9 split into two almost-square-roots, and nobles are ratios of an f on an f or an F on an F. But I agree that once you go beyond the 8 simplest nobles, the dimensionality grows too fast and it's hard to keep track of which f/f's and F/F's are actually nobles.

Those Stern-Brocot zig-zag funnels of simple ratios exist in their linear arrangements, no matter that we'd prefer them to be logarithmic. So if you don't use an actual noble, you're going to blast through the side of the funnel at some point. But you're right that that doesn't matter, provided that point is far enough down the funnel that it only passes by ratios that are too complex to be considered JI.

So your geometric-phi-mediant scheme suggests yet another scheme. One that doesn't require /any/ irrationals in our vector basis. Just choose a complex rational far enough down each zigzag, to approximate each noble. For example, for approximating ϕ, 13/8 is too simple, too JI, but 21/13 (830¢) is probably fine, and no one could argue with 34/21 (834¢). This might only require 23-limit, i.e. 9 dimensions, with primes 17, 19 and 23 being used only for approximating nobles. Or maybe we have to go to 31-limit. That would still only be 11-dimensions for the 32 simplest nobles plus 13-limit JI.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan I'm not sure I get the significance of the zig zag pattern. It tells you that the best approximations to the number alternate between sharp and flat. But even if you blast through the funnel at some simple ratio, all that tells you is that you may have two successive best approximations that are both sharp, for instance. What does that mean for our "logarithmically inapproximable" numbers?

Reply

29 w

Dave KeenanAuthor

Mike Battaglia What I think it means, is that there may be no simple definition of logarithmically-inapproximable numbers. Maybe nobles are the best you can do. But this is from a pure math perspective, and doesn't necessarily matter when you bring in the psychoacoustics. So what do you think of the idea of using complex rationals (and hence higher ordinary primes) to approximate nobles? That's effectively what Margo Schulter has been doing, right from the beginning of her neo-medieval work.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan I think it isn't a terrible idea; in fact starting with nobles should yield these kinds of rational approximations as useful temperaments thereof! I mean, in some sense, the "geometric noble" thing was basically a midway point between the true nobles and the thing you are suggesting.

Reply

29 w

Mike BattagliaAdmin

Here are some results regarding neighboring pairs of superparticular ratios. The second column is the regular noble mediant, and the third is the geometric noble mediant, and the fourth is the difference between them. The approximation involved says that the geometric noble mediant ≈ the regular noble mediant as the difference between the two intervals gets smaller in size and complexity. We can see for instance that when we get to around 5/4, 6/5, etc, the two are only a few cents off.

"2/1 -> 3/2" "833.0903" "892.1913" "59.101"

"3/2 -> 4/3" "560.0666" "575.9317" "15.8651"

"4/3 -> 5/4" "422.4874" "428.9913" "6.5039"

"5/4 -> 6/5" "339.3439" "342.6358" "3.2919"

"6/5 -> 7/6" "283.6048" "285.4995" "1.8947"

"7/6 -> 8/7" "243.6192" "244.8091" "1.1898"

"8/7 -> 9/8" "213.5281" "214.324" "0.79588"

"9/8 -> 10/9" "190.0599" "190.6184" "0.5585"

We can also look at pairs of neighboring integers to get:

"1/1 -> 2/1" "833.0903" "741.6408" "91.4495"

"2/1 -> 3/1" "1666.1806" "1633.832" "32.3485"

"3/1 -> 4/1" "2226.2472" "2209.7637" "16.4834"

"4/1 -> 5/1" "2648.7345" "2638.755" "9.9795"

"5/1 -> 6/1" "2988.0784" "2981.3908" "6.6877"

"6/1 -> 7/1" "3271.6832" "3266.8903" "4.793"

"7/1 -> 8/1" "3515.3025" "3511.6994" "3.6031"

"8/1 -> 9/1" "3728.8305" "3726.0233" "2.8072"

"9/1 -> 10/1" "3918.8904" "3916.6417" "2.2487"

So again this approximation starts to get better around the 6/1 ish range.

It would be nice to try this with some other noble numbers.

Reply

29 wEdited

Paul ErlichAdmin

Mike Battaglia maybe my observation would be relevant if we start with 5:3 and 8:5, the two nearest local minima?

Reply

29 w

Mike BattagliaAdmin

Paul Erlich the geometric noble mediant is then (5/3)^(1-p) * (8/5)^(p) = about 840.68 cents.

Reply

29 w

Paul ErlichAdmin

Mike Battaglia so maybe my idea did make sense after all????

Reply

29 w

Mike BattagliaAdmin

Paul Erlich yes, if you're specifically talking about 5/3 and 8/5 then it does make sense and the geometric version is closer to the local maximum of HE. Again I thought you were referencing our prior conversation about the noble mediant of 1/1 and 2/1...

Reply

29 w

Paul ErlichAdmin

Mike Battaglia repeating just for completion (not to be repetitive, and I know you already saw this): in Dave and Margo's original article, I believe they explicitly start from the two nearest (flanking) local minima or consonances.

Reply

29 w

Mike BattagliaAdmin

I guess it is noteworthy for the conversation with Paul Erlich to note that geometric noble mediants differ from regular noble mediants in that, for instance, some of the pairs of ratios that give the same noble mediant can give different geometric noble mediants. For instance:

"1/1 -> 2/1" "833.0903" "741.6408" "91.4495"

"2/1 -> 3/2" "833.0903" "892.1913" "59.101"

"3/2 -> 5/3" "833.0903" "814.6867" "18.4036"

"5/3 -> 8/5" "833.0903" "840.6808" "7.5905"

"8/5 -> 13/8" "833.0903" "830.2752" "2.8151"

These all have a noble mediant of "acoustical phi" whereas the geometric noble mediants are all different. Note again the results only tend to approximate the regular noble mediant well as the difference between the two intervals starts to get to about a half step or so.

That being said, even with regular noble mediants I'm not sure that you get sensible results for pairs of ratios that are hugely spaced apart.

Reply

29 wEdited

Mike BattagliaAdmin

Paul Erlich: OK, for s = 1% and using the zeta function HE for N=Inf, we get the following between neighboring minima, always going from the simpler to the more complex ratio:

ratios: regular geometric maximum

1/1>8/7: 213.528076 142.873447 48.600000

7/6>8/7: 243.619228 244.809062 238.230000

6/5>7/6: 283.604818 285.499534 287.550000

5/4>6/5: 339.343874 342.635752 349.800000

5/4>9/7: 422.487396 416.455467 423.030000

4/3>9/7: 448.459360 459.133021 458.580000

4/3>7/5: 560.066561 550.248596 538.770000

3/2>7/5: 606.851200 628.135286 658.590000

3/2>8/5: 792.105295 771.008733 745.290000

5/3>8/5: 833.090296 840.680751 846.120000

5/3>7/4: 942.516832 936.562309 924.840000

7/4>9/5: 1001.612335 998.967660 998.370000

2/1>9/5: 1038.629251 1087.268306 1153.050000

It looks like the geometric one is *usually* closer than the noble one, except interestingly for 5/4 > 9/7. The maxima at 48.6 and 1153 cents are both pretty far off in both cases. Sometimes I think we may have gotten better results if we went from the more complex to the simple ratio...

Reply

29 wEdited

Mike BattagliaAdmin

Paul Erlich here are similar results for the min-entropy at s=1% (actually approximate min-entropy, w/ a=10), rounded to the nearest cent:

ratios: regular geometric maximum

1/1>11/10: 155.819149 101.978222 46.000000

10/9>11/10: 171.243282 171.650240 171.000000

9/8>10/9: 190.059883 190.618384 191.000000

8/7>9/8: 213.528076 214.323958 216.000000

7/6>8/7: 243.619228 244.809062 248.000000

6/5>7/6: 283.604818 285.499534 291.000000

6/5>11/9: 339.343874 335.274159 338.000000

5/4>11/9: 355.865983 362.268624 360.000000

5/4>9/7: 422.487396 416.455467 413.000000

4/3>9/7: 448.459360 459.133021 468.000000

4/3>11/8: 541.418757 530.969489 529.000000

7/5>11/8: 560.066561 563.233086 562.000000

7/5>10/7: 606.851200 604.128311 603.000000

10/7>13/9: 630.430476 629.310707 634.000000

3/2>13/9: 644.639704 661.574303 667.000000

3/2>11/7: 770.641415 751.729626 737.000000

8/5>11/7: 792.105295 794.407179 794.000000

8/5>13/8: 833.090296 830.275169 834.000000

5/3>13/8: 848.859910 857.269634 856.000000

5/3>12/7: 923.029392 914.500466 913.000000

7/4>12/7: 942.516832 946.764063 946.000000

7/4>9/5: 1001.612335 998.967660 994.000000

9/5>11/6: 1038.629251 1037.229160 1036.000000

11/6>13/7: 1063.998505 1063.169088 1063.000000

13/7>15/8: 1082.472117 1081.940699 1085.000000

2/1>15/8: 1096.525910 1130.946268 1158.000000

Again, the two results are typically very close, but the geometric version is often just a little bit closer. Either way, it's certainly close enough to be useful, particularly since we get a nice clean vector space...

Reply

29 w

Mike BattagliaAdmin

Another unrelated thought: suppose you think the mediant doesn't go far enough towards the more complex rational. So instead of the mediant of 5/4 and 6/5, you take the "double mediant" (5+6+6)/(4+5+5) = 17/14. Then you repeat again, taking the double mediant of 6/5 and 17/14, and so on. Then you get a bunch of "silver noble numbers" instead, whatever they are called, whose continued fractions end in all 2's rather than all 1's. These may also be useful. In general you can replace with the n-fold mediant to get generalized metallic noble numbers.

Reply

29 w

Mike BattagliaAdmin

Actually, quick correction, you would have to go from the simpler to the more complex ratio to get a silver noble number.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia They have been called silver aristocratic numbers. Noble = gold aristocratic. I'm guessing that a tree like the Stern-Brocot tree but based on double-mediants would not cover all rationals. Is there a requirement for a determinant of +-1 or +-2 for the limit of a double-mediant series to be a silver aristocratic?

Reply

29 w

Dave KeenanAuthor

Answering my own questions: The determinant still has to be ±1 for silver aristocrats, same as for nobles. A zigzag of double-mediants, whose limit is a silver aristocratic, skips every other level of the Stern-Brocot tree, and is closer to its convergents (in logarithmic terms) than the corresponding noble. For example, the silver ratio itself has zigzag convergents 2/1 5/2 12/5 29/12. It is 10 cents from 12/5 and 1.8 cents from 29/12. The corresponding noble, n12/5, has zigzag convergents 12/5 17/7 29/12 and is 14 cents from 12/5, 6 cents from 17/7 and 2.3 cents from 29/12.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan I looked at this yesterday and got that the determinant must indeed be ±2. What pair of ratios are you talking about? If you look at consecutive "zig-zag" points, for instance, like (6+5+5)/(5+4+4) = 16/13, being the "double mediant" from 6/5 to 5/4, then the determinant of 6/5 and 16/13 is indeed 2.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia The recurrence relation you're using, namely

(2nᵢ₋₂ + nᵢ₋₁ ) /(2dᵢ₋₂ + dᵢ₋₁ ) = nᵢ/dᵢ

is not the silver aristocratic (silvercratic) one. In fact it doesn't even converge on irrational numbers. It converges on rational numbers, namely the ordinary mediants. In the case of 5/4 6/5 16/13 28/23 60/49 ... it converges on 11/9. Absolute values of successive determinants for that sequence are 1 2 4 8 16 32 ... i.e. increasing powers of 2. You just happened to pick the pair of successive ratios that gave an absolute determinant of 2.

The silvercratic recurrence relation is

(nᵢ₋₂ + 2nᵢ₋₁ ) /(dᵢ₋₂ + 2dᵢ₋₁ ) = nᵢ/dᵢ

Your first example was correctly silvercratic, namely 5/4 6/5 17/14 40/33 97/80 ... Notice successive pairs here are unimodular.

I didn't understand the full import of your "correction", where you said you would have to go from the simpler to the more complex ratio to get a silver noble number, because that's what I see the above sequence as doing, and in any case, the following sequence is also silvercratic: 6/5 5/4 16/13 37/30 90/73 ... Again unimodular.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan That first thing isn't the recurrence relation I was using. I'm just zig-zagging 2 to the right, 2 to the left, etc on the Stern-Brocot tree. After 16/13, the thing you'd be taking the double mediant of is 11/9, so you'd get 38/31 and the determinant of 16/13 and 38/31 is 2. I had it so you're always taking the double mediant of the parent of the current element in the tree, not the double mediant of the last output emitted by the algorithm. If you want it as a recurrence relation in terms of outputs then I think the second recurrence relation you have is correct.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia That's weird. You start with 5/4 and 6/5 which have a determinant of 1. But I agree that from then on your procedure gives 6/5 16/13 38/31 92/75 ... which do indeed have determinants of 2 and do obey the silvercratic recurrence relation (nᵢ₋₂ + 2nᵢ₋₁ ) /(dᵢ₋₂ + 2dᵢ₋₁ ) = nᵢ/dᵢ and they do indeed converge on a silvercratic number. But I note that the same silvercratic number can also be generated as the limit of a sequence having determinants of 1, using the same recurrence relation, namely: 1/1 5/4 11/9 27/22 65/53 ...

Reply

29 w

Mike BattagliaAdmin

Dave Keenan yes, I guess you would say the zig zag properly starts at the pair of 6/5 and 16/13.

Reply

29 w

Mike BattagliaAdmin

That is kind of interesting and I guess it depends on how you are describing the zig zag. Clearly if you use the "endpoints" of the zig-zag you'll get a determinant of 2, or of n for the n'th metallic mean. But I guess there are other nontrivial descriptions with different determinants.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia You could also back up to 4/3 and 6/5 to start it. Or, as I say you can start it from 1/1 and 5/4 and end up converging on the same place. So it can work with determinants of 1 or determinants of 2.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia It turns out, the series with a determinant of 1 are the convergents of the silvercratic. The series with a determinant of 2 are semiconvergents.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan yes, that makes sense. Curious how this works for other metallic means. You will always get semiconvergents as the endpoints of the zigzag and convergents as one prior to that.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia For the one bronzocratic example I tried, I find three equivalent series — the convergents with determinant 1:

1/1 5/4 16/13 53/43 ...

and two different semiconvergent series, both with determinant 3.

3/2 6/5 21/17 69/56 ...

4/3 11/9 37/30 122/99 ...

Reply

29 w

Mike BattagliaAdmin

BTW to Dave Keenan, answering an earlier question about approximating nobles with rationals... kind of a failed idea, but just making a note anyway.

As I mentioned, instead of monzos w/ integer coefficients in the basis 2, 2^phi, 3, 3^phi, etc, we can treat them as just monzos in the basis 2, 3, etc but with coefficients that are feudal integers. E.g. each monzo is of the form |(a1+a2*phi) (b1+b2*phi) ...>. I actually prefer that we go with "little phi" here, which I will write p = 1/phi, so the coefficients are of the form x+y*p.

To make this very simple, we can simply look for a rational approximation to phi within the monzo coordinates. For instance, let's say that p ≈ 2/3. Then our monzos become very simple and are of the form |a/3 b/3 ...>, e.g. we simply get monzo coordinates w/ denominators of at most 3.

This is the same basic idea as before, except we are now going 2/3 of the way from the simpler to the more complex ratio rather than 1/phi of the way (and I am kind of curious if that gives even better results).

The only problem, though, is that our "nobles" can't be tuned independently from the rationals like they can with independent basis vectors. For instance, the approximate noble from 5/4 to 6/5 is now 339.198 cents, which is always 6/5 * (25/24)^(1/3). This means that the relevant interval doesn't even *exist* in any EDO which maps 25/24 to only 1 or 2 steps. This also turns out to be an issue in a slightly different way if we just approximate nobles with rational numbers; we can't give nobles their own independent complexities and tunings and etc.

One answer is to look at several different possible rational approximations for phi at the same time and see what you get, but really the goal of finding these good rational approximations is the same as just looking for good "commas" in the larger original space of feudal numbers.

I think, though, I'm mostly happy with the "logarithmic feudal numbers" or whatever we're going to call them as some kind of sweet spot; probably the regular feudal numbers are also alright although I still wish we had a better choice of basis...

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan man, another really great property of the "logarithmic feudal numbers"!

Suppose after doing some listening tests, your ears tell you that 61.8% is not the best amount to go from ratio a to ratio b, on average, to get to some approximate local maximum between them. Suppose you think something like 70% is better, or whatever. Then this is as simple as changing 2^phi, 3^phi, etc to 2^(1.7), 3^(1.7) etc instead. So we can introduce an adjustment term which one can dial in empirically. In general we can write this as 2^X for some magic term X.

We can take it even further. Suppose that you like the double mediant better than the mediant, and you want to build things on the silver ratio and the silver aristocratic numbers and etc. Then doing the same geometric approximation as before, you just get 2^(1+sqrt(2)) instead. And likewise with all of the metallic means.

So really there is just one model. We start with the primes 2, 3, 5, ..., and then we add one more basis element 2^X, 3^X, 5^X, ... for whatever value of X. Different values of X correspond to different approximations of aristocratic numbers, and in fact we can choose X to be anything in between that we like empirically. If we wanted to do this with true silver numbers in general, we'd have to look at the field Q[sqrt(2)] and so on.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan just a quick note that I tried some different values of X for a bunch of pairs of intervals.

It really does seem to be true that, as the intervals you are looking at have nd -> inf, X=phi gives the best approximation. I guess this kind of makes sense given the geometric ≈ arithmetic mean approximation I was using but it is still a little bit surprising to me.

However, I note that there may be better values of X than phi if you only care about a few nobles.

One interesting question is that it's kind of nontrivial how to measure this. For instance, as you mentioned before, pairs of ratios that have the same noble mediant need not have the same approximation. So we are not actually approximating nobles, but "phi-weighted mediants of pairs of simple ratios," with it being kind of a fortuitous happenstance that the true noble mediant happens to have a bunch of these pairs generating the same noble number.

Of course, any noble can be approximated to arbitrary closeness if one goes far enough into the SB tree and gets some pair of rationals arbitrarily close to the noble in question, which generate that noble. But for each noble, there does seem to be a "sweet spot" where there are two relatively simple ratios, typically about a half step apart, which generate that noble and also give a decent phi-weighted approximation. I also note that this "sweet spot" changes as you vary the value of X.

Anyway, just some thoughts.

Reply

29 w

Mike BattagliaAdmin

In light of recent results, I would also just add that instead of saying the "sweet spot" changes as you vary the value of X, you can also say that the value of X can change as you would like to dial in the sweet spot.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan lastly, and perhaps most importantly, this also reveals another challenge with the regular feudal numbers. With those, you have to split 5 into sqrt(5), since f5 is a pretty important prime when looking at noble numbers. Thus when we search for "temperaments" thereof, there won't ever be any EDOs that map 5 to an odd number of steps.

If we try something with silver numbers instead, I expect 2 will split into sqrt(2), so we will get only even-numbered EDOs.

I'm not sure if there is some way around this using the true feudal numbers; we could try using a different basis to some extent but in general 5 will tend to be square.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia I'm not sure if you're saying the idea of approximating nobles with rationals is a failed idea, or just referring to the idea of using [a/3 b/3 ...⟩ or similar, based on a rational approximation of 1/ϕ, which you noted fails because you can't tune the nobles independently. I note that [a/3 b/3 ...⟩ with an ordinary prime basis, is equivalent to [a b ...⟩ with a and b integers, when the basis consists of the cube roots of the ordinary primes.

You claim that this is also a problem for approximating nobles with rational numbers. You say "we can't give nobles their own independent complexities and tunings and etc.". But that is not so if, as I suggested, we use the higher primes only for approximating nobles. Say we want the 13-limit rationals plus some simple nobles. Then we use the primes 17 19 23 (and maybe 29 and 31 if necessary), in combination with lower primes, to approximate the nobles. In this case we are not interested in using primes above 13 for JI, so they are free to be used to tune the anti-JI intervals, and to give them complexities independent of the rationals.

I think it's incredibly unlikely that listening tests would shift all the anti-JI points in the same /direction/ away from the phi-mediants, let alone to similar ratio-X-mediants, where "same direction" means all towards simplicity or all towards towards complexity. And so I think the probability that these would be consistently nearer to silvercrats than to nobles is close to zero.

I don't understand why it's a problem, that there won't ever be EDOs with good approximations of certain simple nobles, that that map 5 to an odd number of steps. To me, that's just a very useful observation of a mathematical fact.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan I may have misunderstood the original idea. If you are adding noble numbers as new basis elements using only new primes, why not just add the new nobles directly? Adding new primes means the problem is that the "complexity" of each noble is determined by the complexity of the primes, so the weighting may not be what you want.

Re listening tests shifting all of the anti-JI points away from the phi mediants on average, are you talking about the regular phi mediant or the geometric version?

Reply

29 w

Mike BattagliaAdmin

Regarding this: "I don't understand why it's a problem, that there won't ever be EDOs with good approximations of certain simple nobles, that that map 5 to an odd number of steps. To me, that's just a very useful observation of a mathematical fact."

There won't ever be EDOs with *any* approximations of certain simple nobles that map 5 to an odd numbers of steps, good or bad, as long as we are talking about "regular temperaments" derived from the ambient space of feudal numbers.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Right. Why is that a problem?

I don't understand why such EDOs* should exist with /any/ basis.

*EDOs with good approximations of certain simple nobles, that map 5 to an odd number of steps.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia I'm talking about both the regular phi-mediant and the geometric one in so far as it approximates the regular one. I'm not sure what the geometric values actually are. I'm not sure whether the geometric idea is an entirely coherent one. Don't you get slightly different values close together, depending on what ratios you feed into it, even when those values are all successive convergents of the same noble? Do the geometric mediants converge in any sense?

Reply

29 wEdited

Dave KeenanAuthor

The reason it's better to add new (ordinary) primes rather than just adding the nobles themselves to the basis, is lower dimensionality. Each new prime will allow for the approximation of multiple new nobles. And there is no reason why these noble-only higher primes need to have their complexity calculated in the same way as the lower primes that are used for rationals.

Reply

29 w

Mike BattagliaAdmin

The problem is simply that EDOs with 5 mapped to an odd number of steps can still have perfectly good approximations of noble numbers and regular primes. This method will rule those out.

I'm not sure what you're asking about geometric mediants or what you think is incoherent about it, but yes, different convergents give different values. That is one property you lose when going from the regular nobles to this thing. It's still a good approximation though, for any pair of ratios in a reasonably similar size range.

I don't think that last point about dimensionality is true - if you have a N-dimensional vector space and want to add M new linearly independent vectors, then the new vector space is of dimension M+N. You can't just change the basis and make the vector space of lower dimension.

Reply

29 w

Dave KeenanAuthor

I guess you're right re EDOs. But to set my mind at rest, can you find an example of an EDO that maps 5 to an odd number of steps and has good approximations (say ±5 cents) of the four simplest nobles: n1/1, n2/1, n3/2, n3/1 (833c, 1666c, 560c, 2226c)?

You have sufficiently answered my question about geometric mediants. Thanks.

We seem to be talking past each other re approximating nobles using higher primes. Of course, what you say is true, but I don't see how it argues against my proposal. Let's leave that question until I come up with a specific proposal for how to approximate the 32 simplest nobles using higher primes.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan that's a good question and I'm not sure offhand about that level of accuracy with that many nobles at once. I would guess that the more accuracy you want, and for more nobles simultaneously, it gets harder for that to be possible. If you relax the criteria somewhat you have that 34-EDO has a half-decent version of most of those, say within 10-ish cents rather than 5, with the exception of phi. It also has 5 mapped to an odd number of steps and so the corresponding val doesn't even exist. I'm sure there's some better example but it's late and I need to sleep, lol.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan another thought is that probably the simplest way around this would be to introduce an artificial "prime 5" basis element that is independent of sqrt(5)^2. The ideal tunings of 5 and sqrt(5)^2 will be identical, but they are treated as different elements. This parallels how sometimes we'll add a "prime 9" to 5-limit JI so we can model things like 16-EDO, which has a decent approximation to 9/8 at 225 cents, good for use in chords like 0-975-1425 ≈ 4:7:9 that is different from the mapping you get for 3^2.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia I just discovered something that suggests noble numbers are ideal in terms of /logarithmic/ distance from their convergents. If you multiply any noble's absolute-distance-in-cents from any of its convergents, by the product-complexity (n×d) of that convergent, you always get a badness figure between 680 and 1020, and the badness converges to approx 774.23 as the complexity of the convergent increases. How do the geometric mediants fare using this badness measure, which is the product of what I believe we previously agreed were desirable error and complexity measures?

This doesn't diminish the possible utility of the "logarithmic feudal integers" or "phith-roots-of-primes" in approximating nobles while reducing the number of dimensions required, but it helps to explain why the nobles are the "gold standard".

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan that is super interesting - how do you prove that property? About convergence to 774.23. What do other real numbers converge to - the silver ratio, or even just random real numbers?

Reply

29 w

Cmloegcmluin Xenharmonic Feisbeuk

hey, just wanted to say that I’ve been trying to keep up to date with you two across all these various comment sub-threads. Liking posts when they have cool thoughts and/or good questions. I don’t think I have much original thought to contribute here, so don’t let me slow y’all down, but if there’s anything either of you think I could try to help out with, I’d be happy to give it a go. In any case, I’m a big fan of metallic means and stanky sounds, so this is really exciting to me that y’all are making headway with this!

Reply

29 w

Mike BattagliaAdmin

Cmloegcmluin Xenharmonic Feisbeuk good stuff. I was going to make a Wiki page called "anti-JI" about this at some point, meaning any system which has intervals that are intended to be maximally dissonant, which summarizes a few different systems for modeling anti-JI. The feudal numbers are one such system - perhaps the prototypical system. So all of the stuff about feudal primes and noble mediants and Q[sqrt(5)] and etc would go there, as well as stuff about the silver ratio and etc. Then the "geometric feudal numbers," or whatever they are called, would be a different approach that models things in a different way but also gives similar results. There may be other systems beyond this which would be on that page. It may just make sense for anti-JI to be a page that links to separate pages on feudal numbers and etc since the results have become so extensive.

Reply

29 w

Cmloegcmluin Xenharmonic Feisbeuk

Mike word. the ink is pretty wet on all the feudal stuff... Dave's still editing those Sagittal forum posts on a daily basis, methinks. I can't imagine any good arguments against adding an "Anti-JI" page on the wiki at this point, for the reasons as you describe them (you may recall that *I* pushed back against the idea of "anti-JI" a week ago or so, but Dave has since disabused me of that position!) I agree the results are extensive (and the frontier is still open), but maybe let's start with that page for now and see where it goes. It's super easy to break pages out once stuff gets big enough.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia You wrote: "How do you prove that property about convergence to 774.23? What do other real numbers converge to - the silver ratio, or even just random real numbers?"

I don't know how to prove it. I just took the standard 32 nobles in a spreadsheet and ran off about 10 convergents for each and computed how badly the noble approximated each convergent according to the badness formula

|log₂(n/d)-log₂(α)|×1200×n×d, where α is a noble and n/d is a convergent of α.

And I saw that these badnesses never went below 680 and converged to about 774.23.

It's funny to say that the badness converges, because what that means is that there is a sense in which the convergents of a noble number never converge on the noble. The noble remains an equally bad approximation of them for their complexity, all the way to infinity. That quantifies the thing I was always led to believe about nobles, and tried to state, somewhat ineptly, earlier in our discussion.

I expect this to be true of all metallic means and their aristocrats, except that none of them will stay as far from their convergents as the nobles do. Those snooty nobles really stay as far as they can from those rational commoners.

I tried the silver ratio itself, and a few other random silvercrats. Their badness in approximating their convergents converged to about 612.1.

For bronzocrats the magic number is 480.2.

I can't help thinking a different choice of log base (and dropping the ×1200) might reveal some simple pattern to those numbers. But I haven't found it yet.

And I expect that all non-metallic non-aristocratic irrationals will pass arbitrarily close to one of their convergents eventually. Consider π and its famous convergent 355/113. Its badness by the above formula is only 5.9. In other words π is a damn good approximation of 355/113 (although we don't usually look at it that way).

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia I've got it! When you use the natural log, the magic badness numbers are 1/√5 for nobles, 1/√8 for silvercrats and 1/√13 for bronzocrats. i.e. when the badness of approximation of their convergents is given as:

|ln(n/d)-ln(α)|×n×d,

where α is an aristocratic number and n/d is one of its convergents,

then the badness converges to those reciprocal square roots above. In general, for aristocrats of the n-th metallic mean, the limit badness is 1/√(n²+4). As usual, no proof, just conjecture based on experimental results.

The nobles /are/ the gold standard.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia Putting it another way, those cents-based values are:

1200/ln(2)/√5 ≈ 774.2314037

1200/ln(2)/√8 ≈ 612.083668

1200/ln(2)/√13 ≈ 480.1579334

Reply

29 w

Cmloegcmluin Xenharmonic Feisbeuk

It’s cute how the radicands of those roots overlap with the Fibonacci series (but only for those three; the next one is 20 instead of 21, and the previous one would be 4 instead of 3). And yes I have been waiting for an opportunity to use that word “radicand” haha

Reply

29 w

Mike BattagliaAdmin

Dave Keenan that 1/sqrt(5) is the famous Markov constant of the golden ratio, but the usual expression is |n/d - a|*d. How are you showing that you get the same thing for |ln(n/d)-ln(a)|*nd?

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Like I said, I just calculated 32 x 10 of them in a spreadsheet and saw what they did. When I try the same thing with |n/d - a|×d it converges to zero. However |n/d - a|×d² converges to 1/√5 for nobles. Is that what you meant to write?

Reply

29 w

Mike BattagliaAdmin

Oops, yes, d^2 is it, sorry. You're saying you have just empirically determined you get the same thing using logs and nd? That's nuts. What happens if you use |ln(n/d)-ln(a)|*d^2 instead?

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Yes. That's what I was saying. But now I have a proof that |n/d - α|×d² and |ln(n/d)-ln(α)|×nd converge on the same value. It depends on the fact that ln(x) → x-1 as x → 1.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia |n/d - α|×d²

= |α - n/d|×d²

= |α - n/d|÷(n/d) × (n/d)×d²

= |α÷(n/d) - n/d÷(n/d)|×nd

= |α÷(n/d) - 1|×nd

→ |ln(α÷(n/d))|×nd as n/d → α

= |ln(α) - ln(n/d)|×nd

= |ln(n/d) - ln(α)|×nd

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia As you probably expect by now, |ln(n/d)-ln(α)|×d² doesn't work. It converges on a different value for each noble, namely 1/(α√5).

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan that is really amazing! OK, wow. So any result usual results still hold even with this metric.

Here is an interesting thought. Suppose that for some bizarre reason you decide to use a Cauchy distribution for HE rather than a Gaussian. Also, you give all of the rationals 1/(nd)^2 weighting rather than 1/sqrt(nd) weighting. And also use the min-entropy rather than the Shannon entropy (and maybe the Shannon entropy as well). Then I am thinking that, as the "s" of the Cauchy distribution->0, we may get that the maxima will all be noble numbers.

This is because as s->0, the Cauchy distribution just starts to look like k/x^2 for some tiny value of k. Thus for some arbitrary dyad, the strength of the match to each rational is going to basically be proportional to 1/err^2 * 1/nd^2, which is the reciprocal of the square of the quantity you had before.

With the normal distance, the numbers which have maximum the best-possible weighted error on all rationals are the noble numbers. *If* your proof suggests this result generalizes to this logarithmic distance instead, we would get the same thing and thus the maxima would be noble numbers. (However I'm not quite sure if we can say that as your proof only holds for rationals arbitrarily close to the real number.)

Anyway, something to think about.

Reply

29 w

Mike BattagliaAdmin

Paul Erlich may be interested in that...

Reply

29 w

- Dave Keenan
- Site Admin
**Posts:**2180**Joined:**Tue Sep 01, 2015 2:59 pm**Location:**Brisbane, Queensland, Australia-
**Contact:**

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

Dave KeenanAuthor

Here's a specific proposal for how to approximate the 32 simplest nobles using higher primes.

viewtopic.php?p=4625#p4625

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Reply

Remove Preview

29 w

Mike BattagliaAdmin

Great stuff!

Reply

29 w

Mike BattagliaAdmin

Dave Keenan the only issue with the current proposal, that I see, is that the complexity is not necessarily what we want. For instance, I think you will have that phi is much more complex than n7/2. This is because although you've found decent rational approximations from a tuning standpoint, they are place on the lattice in sort of a random haphazard way, and as a result the natural complexity on that lattice places essentially a random weighting on the nobles themselves. However I think the basic idea is sound and there is probably some way to do it so that things all fit together nicely.

The question is what basis we can choose for our nobles such that the complexities and accuracies all are all put together in a nice way. I have this feeling that trying to derive this rigorously may once again lead to 2^phi 3^phi etc, or something close to it. I'll see if I can figure out how it all works.

Reply

29 w

Dave KeenanAuthor

I agree about the haphazardness and the resulting complexities. One way to make those irrelevant is to use a finite list of target intervals and minimax unweighted damage (absolute error).

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan another interesting result: suppose you take the noble mediant of two neighboring superparticular ratios. You would like to see where the result is, as some percentage of the way from the simpler toward the more complex ratio. Supposing our two intervals are n/(n-1) and (n+1)/n, then as n->inf, we indeed get precisely the same "logarithmic phi" from before, meaning about 61.8% of the way. In other words, you again get the same 2^phi 3^phi etc structure. So this is another way to derive that.

You seem to get the same thing if you take the noble mediant of two successive harmonics, e.g. two intervals of the form n/1 and (n+1)/n. The percentage will again be logarithmic phi.

On the other hand, this mainly seems to work when the two intervals are of reasonably similar complexity. If we instead look at the noble mediant from 1/1 to some arbitrary superparticular (n+1)/n, then as n->inf, the noble mediant is actually placed 100% of the way from 1/1 to (n+1)/n.

But at least if those are the interval pairs we care about, then this does show that in some sense phi really is the best value to put in for 2^X, 3^X etc. Although if we only care about approximating relatively simple nobles, then sometimes other values of X can be useful.

I think it's quite interesting that logarithmic phi really is intimately related to the structure of feudal numbers, in ways beyond just being a neat approximation...

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan perhaps more interestingly, we get something different if we look at where some noble fits within its own zig-zag pattern.

For instance, suppose we have the noble between 5/4 and 6/5. We can ask what percentage of the way that interval is located from 5/4 to 6/5. If we do this for two arbitrary superparticulars we get logarithmic phi as n->inf, as before.

But let's say we want to stay with the *same* noble and look at other rationals which generate that noble. For instance, the one between 5/4 and 6/5 is generated also by 6/5 and 11/9. So we can ask what percentage of the way this noble is located between those two. And then again from 11/9 to 17/14 and so on.

It looks like this does again converge to something nontrivial, but this time it isn't 1/phi. In fact, it seems to converge to 0.723606797749979. In addition, this magic number seems to be the result you get if you start with *any* pair of rationals, regardless of the determinant, and compute the phi-weighted mediant between them, and then do the same thing.

Any idea what on Earth this value may be?

Reply

29 w

Mike BattagliaAdmin

Dave Keenan According to Mathematica, anyway, the magic value is

1/10 (5 + Sqrt[5])

= 1/2 + Sqrt[5]/10

= 0.72360679774997896964091736687312762354406183596115257242708972454105209256378048994144144083787822749695081...

WTF number is this? This can also be rewritten as

(2 + phi)/5

Anyway, also may be interesting to put this into X for 2^X, 3^X, etc. This magic value, whatever it is, is the percentage of the way that every noble number sits between the n'th pair of convergents as n->inf. Also seem to be getting something similar for arbitrary non-noble feudal numbers w/ nontrivial golden part.

Reply

29 wEdited

Mike BattagliaAdmin

It looks like using that number gives us slightly better results for generating nobles from the simplest pairs of rationals. The first 10 superparticulars, w columns as "ratios" "true noble" "approx noble" "error"

"2/1 -> 3/2" "833.0903" "839.6113" "6.521"

"3/2 -> 4/3" "560.0666" "554.4043" "5.6622"

"4/3 -> 5/4" "422.4874" "417.1955" "5.2919"

"5/4 -> 6/5" "339.3439" "335.1747" "4.1692"

"6/5 -> 7/6" "283.6048" "280.3507" "3.2541"

"7/6 -> 8/7" "243.6192" "241.0404" "2.5788"

"8/7 -> 9/8" "213.5281" "211.4456" "2.0825"

"9/8 -> 10/9" "190.0599" "188.3479" "1.712"

"10/9 -> 11/10" "171.2433" "169.8133" "1.43"

The other one was

"2/1 -> 3/2" "833.0903" "892.1913" "59.101"

"3/2 -> 4/3" "560.0666" "575.9317" "15.8651"

"4/3 -> 5/4" "422.4874" "428.9913" "6.5039"

"5/4 -> 6/5" "339.3439" "342.6358" "3.2919"

"6/5 -> 7/6" "283.6048" "285.4995" "1.8947"

"7/6 -> 8/7" "243.6192" "244.8091" "1.1898"

"8/7 -> 9/8" "213.5281" "214.324" "0.79588"

"9/8 -> 10/9" "190.0599" "190.6184" "0.5585"

The second one quickly becomes a better approximation, although the first one seems to perform better for the very simplest pairs of nobles, which may be what matters.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Interesting! You get a simpler number if instead of looking at the ratio of the larger part to the whole, you look at the ratio between the two parts. This is simply 1:ϕ², and so your magic number is ϕ²/(1+ϕ²). Yes, that may be a better power to use for 2^X, 3^X, etc, but I haven't fully wrapped my head around that. That means the idea of an equivalence with feudal integer entries in the vectors goes out the window, and this is purely an approximation scheme. Nothing wrong with that.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia As you go from one convergent of a noble to the next more complex one, both the numerator and denominator increase by approximately a factor of ϕ, so the product complexity increases by approx ϕ². And that result I just proved says the logarithmic distance to each convergent is (approx) inversely proportional to the convergent's product complexity, so that's why the distances from the noble to the two convergents are in a ratio of 1:ϕ², and so the distance from the simple convergent to the noble, divided by the distance between the two convergents, is ϕ²/(1+ϕ²) ≈ 0.7236.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan that is very interesting!

OK, so here is another interesting note. Suppose that instead of looking at two neighboring superparticulars, we instead take the mediant of two neighboring superparticulars, and then take the noble mediant of the originally more complex superparticular and the mediant of both. These will also be unimodular.

In other words, we have the following sequence:

2/1, 3/1 (derived from 1/0, 2/1)

3/2, 5/3 (derived from 2/1, 3/2)

4/3, 7/5 (derived from 3/2, 4/3)

5/4, 9/7 (derived from 4/3, 5/4)

6/5, 11/9 (derived from 5/4, 6/5)

...

(N+1)/N, (2N-1)/(2N-3)

Then we also get something different than the usual phi thing. It looks like the proportion of the large to the whole is (2 phi)/(1 + 2 phi) ≈ 76.4%, and the proportion of the large to the small step is 2 phi.

I think I have an idea of how this all fits together...

Reply

29 w

Mike BattagliaAdmin

Dave Keenan

OK, some idea of the larger theory. Let's introduce some basic language. Suppose that we are taking the phi-weighted mediant from ratio r1 -> r2; call this result "m". Then we will say

a = log(m/r1)

b = log(r2/m)

c = log(r2/r1)

We can use whatever base log we want; octaves or cents or base e or whatever.

Thus if we have the most simple situation of r1 < r2, and r1 is less complex than r2, we have something like

|-----a-----|---b---|

with a+b = c.

Note that if we *don't* have r1 < r2, like when we go from 5/4 to 6/5, you still get a+b = c, but all of these quantities are negative. Also note that if the complexity of r2 is higher than r1 - which is the typical situation with noble mediants, we will always have the size of a > b. However, if we do very strange "improper noble mediants" going from some very complex ratio to a simpler one, we may have a < b.

Then there are three quantities of interest:

a/c: my original quantity, typically "large" to "whole"

a/b: your quantity, typically "large" to "small

b/c: 1 - a/c

I will just talk about a/c and a/b because b/c is derived from them.

So with this, we have

Successive superparticular ratios, e.g. 5/4 to 6/5:

a/c = 1/phi ≈ .618

a/b = phi ≈ 1.618

Successive harmonics, e.g. 5/1 to 6/1:

a/c = 1/phi ≈ .618

a/b = phi ≈ 1.618

They are the same. We also get the same thing if we take the noble mediant "the wrong way", e.g. from 6/5 to 5/4 or from 6/1 to 5/1: still a/b = phi.

We also have:

Successive convergents of the same noble mediant:

a/c = phi/(phi+1/phi) = 1/2+sqrt(5)/10 ≈ .723

a/b = phi^2 ≈ 2.618

Then for this "superparticular + mediant of two superparticulars" sequence above, e.g. from 6/5 to 11/9, we get

a/c = 2phi/(1+2phi) ≈ 0.764

a/b = 2phi ≈ 3.236

My thinking is that in general, for some arbitrary "pattern" of rational numbers we want to iterate to infinity, we can look at the "complexity quotient" q = (n1d1)/(n2d2) where r1 = n1/d1 and r2 = n2/r2. Then just playing around with it, we seem to get:

a/b = phi * sqrt(q)

I don't know why that sqrt is there but it seems to work. I'll think about how to prove this.

Reply

29 wEdited

Mike BattagliaAdmin

Another cute way to look at this: when we talk about these noble mediants of successive superparticulars, etc, we are in a certain sense talking about noble mediants of different "rational patterns." These patterns can be of the form

N/1 -> (N+1)/1 (harmonics)

N/(N-1) -> (N+1)/N (superparticulars)

N/(N-1) -> (2N-1)/(2N-3) (superparticular + mediant thing)

N/round(N/phi) -> round(N*phi)/N (successive convergents)

or whatever.

Now, we can treat each of these patterns as just a real function of N. Then, as N->inf, we get some value q, and it seems we have a/b = sqrt(q) from before.

But perhaps an even cuter way to look at this is to let N be an infinite number. If we do so, these patterns become idealized "infinite rationals" which are located "at infinity" on the SB tree. The things we are studying, then, are just the phi-weighted mediants of different infinite rationals. The regular real numbers are also "points at infinity" but these are different; they are all infinitesimally close to the real numbers but are true rational numbers with numerators and denominators that are generalized infinite integers. We can use nonstandard analysis to make this all rigorous if we want, but for now I'll just note it's a nice visual way to think of things.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia It seems to me, that what we are trying to get to here is to decide what pair of ratios to use to approximate each noble via the weighted geometric mediant scheme (or is it a weighted geometric mean?). If so, I think that instead of looking at all these different "rational patterns" of superparticulars, harmonics, etc, we only need to look at the convergents for each noble (the zigzags on the SB tree) and choose a pair of successive convergents that give a good compromise between simplicity and accuracy. We want them to be within the same prime-limit as the rationals that we are also targeting.

One option is to use only 13-limit convergents. These could be the two convergents just prior to the first non-13-limit convergent, which is the one I chose for my rational approximations of the nobles. But maybe some earlier simpler pair will be suitable in some cases.

Once such a pair of convergents is chosen for each of the 32 simplest nobles, then the 2.2ˣ.3.3ˣ.5.5ˣ... vectors can be constructed and the optimal value of X can be found to minimise the errors over the 32 nobles. In fact we might find a different value of X for each prime. Some trial and error may be required, with different pairs of successive convergents for some nobles, before settling on them all.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan that's one question this all raises, yes. Of course the simple method is just to search empirically for the best X as you suggest. But I am interested in peering deeper into the structure of this.

Using this same idea, you can derive different geometric approximations that are "correct" in different ways. For instance, with successive convergents to some noble number, we get the a/b ratio is something different. In general, if we go with that "q-value" from my earlier post, it seems we would get that the a/b ratio = 1/(1 + 1/(phi*sqrt(q))), for which some value of X can be derived. I will need to think about it.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan Of course there is also a simpler way to look at it. *Given* some value of X, like phi, the pair of "best" generators for some noble will differ. We can choose some X and look for whichever pair of ratios has the best tradeoff between simplicity and accuracy.

If we know in advance we have some kind of ratios in mind for this, e.g. in some small odd limit or prime limit or whatever, we can look for the X which best gives this result. Same with different values of X for different primes.

We could try to just to calculate this for each odd-limit, where we are only taking noble mediants of pairs of neighboring ratios. I think, though, that we will get the best results if neighboring ratios are "uniform" in some sense that they tend to have a similar complexity ratio to one another. I'm not sure if odd-limit gives us this or if something else will do.

Reply

29 wEdited

Mike BattagliaAdmin

Lastly, I also remember in some of my original posts about this that the approximation between 5/4 and 9/7 was a little worse than the rest. I am wondering if this is because one of the other approximations may have been a better choice for this pair of ratios.

This would suggest some method where the approximate geometric phi-mediant of two ratios isn't just some simple % of the distance between them for all such pairs, but that the % depends on the ratios themselves in some way.

This would to be a more detailed approximation than the ones we've been talking about, but I wonder if there is still some way to do it which is equally simple, in that you still just get some kind of 2.2^x.3.3^y.etc structure. It's difficult to do this perfectly if we want integer coordinates though.

Reply

29 wEdited

Dave KeenanAuthor

I have created a table that shows the options for pairs of rationals to be used in the geometric approximation scheme. In 12 of the 32 cases (shown green) there is no real choice in the matter, if you want to stay within the 13-prime-limit. See

viewtopic.php?p=4627#p4627

This shows, for example, that if 5/4 and 9/7 don't work so well, you can use 9/7 and 14/11.

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Reply

Remove Preview

29 wEdited

Dave KeenanAuthor

Mike Battaglia I've been doing some of the optimisations I described above, but only with a single value of X for all primes. I decided it's too messy to do otherwise. When they are all the same, I don't need to prime-factorise, i.e. I don't need to actually generate the vectors. Optimum values of X range from 0.71 to 0.75 with different sets of pairs of convergents. In other words, your 0.7236 is the right exponent, not 0.618. It turns out that exponent can also be described as n2/3 = F1/F5 = (0+1ϕ)/(-1+2ϕ) = ϕ/√5.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan I think if you are using generating rationals which are some decent length of the way into the zigzag then that'll be right. I still am trying to grasp what it all means though.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan I am still kind of curious what we get if we take the n-odd-limit and look for the best X value for all nobles generated by pairs of neighboring ratios, for each n. And also curious what we get if instead of the n-odd-limit, we instead go with all ratios of depth ≤ n in the Stern-Brocot tree. I have this feeling the SB tree version may have a more "uniform" error and thus may be better.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Odd limits per se, make no sense to me in this context since there's nothing special about prime 2. But I guess you mean the odd limit /diamond/. In this context, I prefer to think of that as the 22-integer-limit 13-prime-limit triangle. Yes, it turns out we don't need to use any noble convergents outside that set, to get good geometric mean approximations.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia It doesn't matter how close to the top of the zigzags the convergents are, the optimum X stays in the range 0.71 to 0.75. The optimum is constrained by a few nobles that only have two 13-prime-limit convergents, no matter whether we limit it to the 16 simplest nobles or go for the full 32.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan what difference is there between what you are calling the odd limit and odd limit diamond? Are you doing a minimax optimization or an rms or something else?

Reply

29 w

Dave KeenanAuthor

Mike Battaglia The odd limit has unlimited powers of 2. The diamond stays within the first octave. I'm being a pedant. Sorry.

I start with miniRMS, then I increase the power from 2 to 4 to 8 to 16 so I approximate minimax. X stays in the aforementioned range for all of these. I'm now allowing simpler nobles to have more error, by weighting the error by the complexity of the noble, but it doesn't make any difference. I'm taking the complexity of a noble to be the product of the feudal norms, N(), of its numerator and denominator, as you suggested way back. When the noble is expressed in the form f11/f5 etc, the norm is the number after the f. e.g. N(f11) = 11. So f11/f5 and F11/F5 have complexity 55.

So I believe we should just use X = ϕ/√5 and be done with it.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan ok, that makes sense. I am curious if that phi/(sqrt(5)) term is really related or if it is a coincidence that the best X happens to be near this value. If you increase the number of nobles you are looking at, do things seem like they tend toward that X?

Reply

29 w

Mike BattagliaAdmin

Tbh, maybe that really is the best value. The thing that threw me off before is that any pair of rationals can be viewed as more than one "pattern," and different patterns suggest different idealized approximations. For instance if we look at 5/4 and 6/5, we get a different idealized approximation if we view those as "two successive super particular ratios" than if we view them as "two successive convergents to some noble number." But the latter is perhaps what we really want when we are viewing pairs of ratios as seeds for some noble mediant, and then, as you say, phi/√5 seems to be the result.

Reply

29 w

Mike BattagliaAdmin

Last thought of the night.

Suppose we have two ratios for which r2 is much more complex than r1, then the phi-weighted mediant from r1 to r2 is clearly going to be much more than 72% of the way there. It could be close to 100% of the way there. So clearly this approximation is best if the two ratios are closest in complexity.

I am curious if there is a slight generalization of the approximation in which the distance from ratio r1 to r2 is, rather than just being the same percentage for all ratios, is somehow also a function of the complexity of r1 and r2. As r2 gets more complex, the mediant is closer to r2 and if not it's closer to r1. This would only be useful if we don't have to increase the dimensionality or add weird fractional coefficients though.

Reply

29 w

Nate BeDell

Woah, just skimming over this for the first time, this sounds super interesting. Time to break out the Dummit and Foote!

Reply

29 w

Mike BattagliaAdmin

Dave Keenan OK, I think I have managed to figure the entire thing out. The idea is we start with the expression for the noble mediant as a four-variable function N(n1,d1,n2,d2) = (n1+n2*phi)/(d1+d2*phi). Then, we get the expression for the ratio "a/b" of the small to large partition between the two ratios, which is a ratio of two logarithms derived from the same four variables. Then, playing around w/ Taylor series in all four variables, I somehow managed to get this:

phi * (n2/n1 - d2/d1) * 1/c

This expression is great, and is the "enhanced approximation" I was looking for. It's an expanded version of the original approximation with extra terms. The (n2/n1 - d2/d1) term can be viewed as kind of measuring the difference in relative complexity of the two rational numbers, and the "c" term is the chroma.

My main problem: I don't know how I got it! I was playing around with Mathematica and was trying to get the Taylor series for the ratio "a/c" of the large step to the entire thing, but somehow it gave this instead, and somehow the results were basically perfect as a Taylor series for the ratio "a/b" instead. I'm pretty stumped and think maybe this could be derived as a multivariate Padé approximant of some kind? I don't get it but it works great.

We can use this to determine the best approximations to things. Given some set of important intervals, like a tonality diamond or whatever, we can use this method to determine the "right" multiplier for phi by just computing some kind of average of (n2/n1 - d2/d1) * 1/c for all adjacent pairs of intervals. I think that if instead of the tonality diamond we use the ≤ n'th level of the Stern-Brocot tree, then as n -> inf we will indeed get the phi/sqrt(5) thing, because successive ratios in this set will be successive mediants, and I think the majority of them will have a complexity ratio of phi again. It would be interesting to try this and see.

We also note that previous results can be derived from this equation. If we let n1 -> n2 and d1 -> d2, this expression tends toward phi, which is the original approximation I was using. If we instead say that n2 -> n1*phi and d2 -> d1*phi, this instead tends to phi/sqrt(5), which we've been talking about.

Lastly, we can also write this expression as:

phi * (n1*d2 - n2*d1) * 1/(n1*d1) * (-1/c)

which is in terms of three quantities of interest. That -(n1*d2 - n2*d1) is the negative of the determinant (which for noble mediants should be +/- 1), the 1/(n1*d1) is a measure of complexity, and 1/c is the chroma size. I guess if the chroma is positive then the determinant will be negative and so on. But anyway, if we know that our ratios are going to have a determinant of +/- 1, then we can rewrite this as

phi * 1/(n1*d1) * |-1/c|

Anyway, I think this answers a bunch of questions, but sheesh, how does one even get this magical expression? For whatever reason, actually looking at the Taylor series for "a/b" didn't give this, but rather that for "a/c" did, even though it's a much better approximation for "a/b". Well, whatever...

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia I was stuck at the second paragraph for a long time. I had no idea what you were referring to as the "chroma" here. I thought at first you must mean the quotient of the two ratios, i.e. d₁n₂/n₁d₂. But when I tried that, it didn't give sensible numbers for the ratio of the two parts in logarithmic terms. I eventually reverse-engineered it and figured you must mean ln(d₁n₂/n₁d₂). In this case, it matters what log base you use.

If that's correct, then, as a pedagogical note: Why you thought "the natural log of the quotient of the two ratios" or "the difference between the natural logs of the two ratios" could be unpacked from the name "chroma" is beyond me. It would have been good if you had just given a formula for "c".

Yes, it does give much better approximations than simply assuming a ratio of ϕ² between parts, but I don't see how it has any practical value. A vector basis that made use of it would have more dimensions than the feudal prime basis that gives /exact/ results.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan I just explained all of this a few posts up, what I was calling "a" "b" and "c". I explained why it has practical value I'm the paragraph about using it to compute the best approximations.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Yes. 18 posts back, in a different subthread, where you explicitly say the log base doesn't matter. That and the fact that you hadn't previously used the term "chroma" led me to believe that this must be a different use of "c". Never mind.

Re practical use: I assume you're referring to where you write:

"We can use this to determine the best approximations to things. Given some set of important intervals, like a tonality diamond or whatever, we can use this method to determine the "right" multiplier for phi by just computing some kind of average of (n2/n1 - d2/d1) * 1/c for all adjacent pairs of intervals."

I don't see any point in trying to determine the best approximation for the set by computing the average of the approximations for all pair of adjacent elements, because we can more easily determine the best approximation for the set by computing the average of the /exact/ values for all pair of adjacent elements. Or better still, by doing a minimax or miniRMS optimisation of some suitable damage measure that weights the absolute cents errors by some kind of complexity measure.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia I also don't see any point in using pairs of ratios that are not successive convergents of the nobles being approximated. But I do see your point that these should be further limited to (what I prefer to describe as) some combination of low prime limit and low integer limit.

It appears at present, that for the first 32 nobles on the SB-tree we can stay within the 20-integer limit (maybe even the 18-integer limit), and all but two can stay within the 13-prime-limit. Those two (923c and 1530c) have pretty bad damage unless we let them go to the 17-prime limit.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan sorry, you're right, I should have mentioned that in this particular formula the log does need to be base e. I wrote this pretty late last night.

Of course you can just do some kind of minimax optimization and determine the values empirically, and if you do there will likely be various seemingly random values things seem to converge to, such as the 72% or whatever it is. Looking at some of these series gives us some way to see what we are really doing and why these approximations work so well, and seems to give us simple closed-form values for some of the empirically-determined values in terms of phi, which so far have agreed with the empirical estimates to within some reasonable margin of error. I think that's certainly interesting, and what's really interesting to me is that all of this stuff is just another part of the inherent structure of the SB tree.

I think there are several valid ways to approach this. We could say we want to approximate nobles, and then as you say, if we look for convergents we get the phi/sqrt(5) thing. But I also think it's useful to say that we want some binary operator that takes some arbitrary pair of ratios and returns some kind of idealized maximally dissonant interval between them which they "generate." To some extent the structure we are talking about gives you both of those things simultaneously and that's what I've been looking at.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia Well you set yourself a fun puzzle, and you solved it. But for my money, the breakthrough was when you figured out that the magic number for successive convergents was that constant that can be written ϕ/√5. I suggest we give that constant the Greek small letter chi χ as it looks a bit like the X you've been using for the variable, and it's traditional to use lowercase Greek letters for non-(quadratic)-integer values in quadratic fields, and I don't know of any other use of χ in RTT or related math. Then we can write the basis as e.g. 2.3.5.7.13.2ᵡ.3ᵡ.5ᵡ.7ᵡ.11ᵡ.13ᵡ.17ᵡ (yes, a superscript chi exists in unicode, U+1D61).

I think it better to order the basis in that way, rather than interleaving chi powers with 1st powers, because the sub-vector over the first powers of the primes is the vector for the simplest of the two chosen convergents, and the subvector over the chi powers is the vector for the quotient of the two convergents (note: not the vector for the second convergent).

You've called this ratio between the two generating ratios, the "chroma" (I'm assuming you didn't mean for the "chroma" to be strictly the natural log of that ratio). That's not bad, but "chroma" seems to have several connotations in addition to being a ratio between ratios, such as being a comma-sized interval that is not tempered out, and being a "side" of a periodicity block or MOS scale. Neither of which seem relevant here.

Douglas Blumeyer = Cmloegcmluin Xenharmonic Feisbeuk once asked me if I knew of a term that functioned multiplicatively in the way that the term "delta" and its symbol Δ function additively. In other words a term for a small ratio that is the quotient of two other ratios. I did not know of one, but suggested "qoppa" since it sounds a bit like a contraction of "quotient delta". "quotta" might be better in that regard, however "qoppa" happens to be the name of a Greek letter, albeit an archaic one, that looks a bit like a Q. In fact it looks like this: Ϙ.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan Regarding the basis, I got in the habit of writing it like this

|a b c d e> + |f g h i j>*X

So we explicitly write the monzo in terms of a "just part" and a "golden part". This is a little bit nicer, I thought, than either |(a+fX) (b+gX) ... > or something like |a b c d e ; f g h i j> or whatever.

I was just using the term "chroma" loosely here; I wasn't proposing it be the official name of anything. In my own writings, I've used the term "chroma" extensively when talking about MODMOS, as well as for epimorphic scales more generally, as the difference between some pair of steps (or thirds, or etc). But I don't really have much problem if people want to use the term "chroma" in a generalized sense to refer to the quotient of any two ratios or intervals in general, not just two steps in an MOS. In fact I haven't really thought much about it before and have probably done that myself plenty of times without thinking about it, as I did here.

FWIW I had different names for all of this stuff back in the day, although we've gone way further than I ever did. Looking at some of my old notes on this stuff, I apparently used to call feudal primes "phrimes", heh.

Reply

29 w

Mike BattagliaAdmin

BTW, how are you typing all of these ridiculous Unicode characters so easily? I mean damn, superscript phi? Are you just constantly Googling unicode letters and copying and pasting them? I'll probably just keep typing "X" but sure, I don't mind if you want to make it chi for when things are typeset nicely...

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia You wrote: "But I also think it's useful to say that we want some binary operator that takes some arbitrary pair of ratios and returns some kind of idealized maximally dissonant interval between them which they "generate.""

I think we already know that any "kind of idealized maximally dissonant interval" will always be a noble number. Of course there are intervals like the dissonance maximum near 70 cents, and others near the octave and fifth. But those are extremely dependent on properties of the human ear/brain, or on the spectrum of the timbre being used, and therefore can't be considered "idealized".

That's why I don't have any interest in a binary operator that takes an arbitrary pair of ratios. But I wish you luck.

Reply

29 wEdited

Mike BattagliaAdmin

I'm not sure I follow - yes, we're agreeing those are idealized, and now we're talking about approximating them within this other 2.2^X.3.3^X.etc space...

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia "Phrimes" is cute. And [EDIT]

[a b c d e⟩ + [f g h i j⟩×χ

or

[a b c d e⟩ + χ×[f g h i j⟩

is excellent.

I'm certainly not searching, copying and pasting.Or at least I only had to do that once per codepoint, long ago. The secret weapon is an insanely great utility called WinCompose.

http://wincompose.info/

If you're not using a Windows machine, I think there are similar utilities for Linux and Mac. I don't know of any equivalent for Android or iPhone. But if you are using Windows, as well as installing WinCompose, you'll need my custom sequences.

https://dkeenan.com/WinComposeSequences.txt

Don't bother trying to find the right place to put this file, just choose "Show sequences→Edit" then copy and paste its contents to overwrite the existing examples.

Then scan through it to learn the common prefixes, like ⎄8 for Greek, ⎄6 for superscript, ⎄- for subscript. ⎄ here stands for the right Alt key. You don't hold them down simultaneously, you just type them in sequence. The most often used characters get sequences with double characters like ⎄pp for pi, ⎄22 for superscript 2, ⎄88 for infinity. Some are pictorial, or like overstrikes, such as ⎄v/ for square root sign. Superscript phi is ⎄68f. Superscript chi is ⎄68c.

WinCompose | Easy typing of special characters on Windows™

WINCOMPOSE.INFO

WinCompose | Easy typing of special characters on Windows™

WinCompose | Easy typing of special characters on Windows™

Reply

Remove Preview

29 wEdited

Dave KeenanAuthor

Mike Battaglia Agreed. But then what am /I/ missing? Why do we need, or want, anything other than pairs of successive convergents, for approximating nobles in the 2.2^X.3.3^X.etc space?

Reply

29 w

Dave KeenanAuthor

Edit above, because you were right to have an explicit (non-matrix, non dot product ) multiply sign there, which can be assumed to be mapped over the vector. i.e.

[a b c d e⟩ + [f g h i j⟩×χ

or

[a b c d e⟩ + χ×[f g h i j⟩

Note that I'm using (upright) χ = ϕ/√5 ≈ 0.7236 to represent that specific constant. You should still use (italic) 𝑋 or 𝑥 in typeset material when it's a variable.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan Thanks, I'm on Mac but I'll check that program out and see if I can locate a port of some kind. It looks like there are some "Mac Compose key" programs out there so hopefully that'll be good. I also wish I had something on Android as I am often on mobile when doing stuff.

Regarding why, it's just that I'm trying to get the structure of what's going on. Right now, if you take any two monzos m1 and m2, the approximate noble mediant from m1 -> m2 = m1 + (m2-m1)*X. That's pretty neat. So for 5/4 -> 6/5, we literally get (5/4) * (24/25)^(X). Basically, what we have done is added, for each JI ratio n/d, a new element of the form (n/d)^X. It's really simple and gives great results.

What we've done is now point to some element in the space and say, this is the best representation of n6/5 or n5/4 or whatever, which is equivalent to pointing to some pair of two ratios and saying that's the right pair to generate the noble. However, these other elements still exist. Some of them are approximate phi-weighted mediants that aren't noble, and some of which are approximate noble mediants using other convergents, and so on. And since they're all there I'm trying to figure out what they are. Whatever they are, they all flow naturally from the new "primes" we've added and inherit this complexity function from them and everything.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia AFAIK we haven't yet pointed to pairs of ratios and said those are the right pairs to generate the nobles. But that's certainly what I'm working towards, and I'm very close to making a specific proposal. If you've already chosen a set for the usual 32 nobles, I'd like to see it, with errors.

It seems I'm viewing this in a different way from the way you are viewing it. Of course, this basis supports elements which are not simple rationals and are not the standard approximations of the simple nobles (whatever we decide those to be). Such elements are either (a) useful approximations of simple rationals, or (b) useful approximations of the standard approximations of the nobles, or (c) junk. In cases (a) and (b) what we have are commas that can be made use of by temperaments.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia When you write: "... and inherit this complexity function from them and everything", what complexity function are you referring to?

Reply

29 w

Mike BattagliaAdmin

Dave Keenan there is also option (d), which is transposed versions of other nobles which are not junk. One of the reasons I was so interested in this space is because I think that nothing in it is really junk at all.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia I'm viewing these vectors as representing intervals, not pitches, so I'm automatically including transpositions of pitch. To my way of thinking, an interval is not changed by being transposed. Do you have some meaning for a "transposed interval" that actually changes the size of the interval? I assume you don't mean things like octave extensions of intervals, because I assume you're aware that when you multiply a noble by a rational other than 1, it is no longer noble, because you break its unimodularity. And so it is no longer anti-JI.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan I view them interchangeably, but I just mean the product of various approximate nobles and just ratios (and maybe other approximate nobles). I don't think it really matters from a musical standpoint that, for instance, 3/2 * n5/4 isn't technically noble anymore; it's still a very important interval (or note or however you like to view things) and it still very much is part of what I would call "anti-JI."

Reply

29 wEdited

Mike BattagliaAdmin

Hm, here's an interesting theorem: the elements of this algebra which are approximations of noble numbers all have "golden part" equal to some superparticular ratio. That is, if we look at |a b c d e> + |f g h i j>*phi, it is required that |f g h i j> be either superparticular or the inverse of a superparticular in order for the result to be an approximate noble.

It goes the other way as well. If |f g h i j> is superparticular or inverse superparticular, then the entire thing is either an approx noble, or the product of an approx noble and a just ratio. Sometimes there are multiple |a b c d e> making |f g h i j> noble: for instance, if |f g h i j> is the monzo for 24/25, then we have that this is approx. noble when |a b c d e> is 5/4 or 5/3.

To see this we simply note that the ratio between any two adjacent elements of the SB tree is a superparticular ratio, and that every superparticular ratio appears somewhere in the SB tree.

Reply

29 w

Mike BattagliaAdmin

If the golden part is not a superparticular ratio, then the resulting ratio can simultaneously be viewed either as a phi-weighted mediant of non-adjacent ratios, or a product of multiple nobles and JI ratios. So I guess this property from the feudal numbers is also true here.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia I don't understand why the interval 3/2 * n5/4 (as opposed to its components) is of any musical interest. I would never include it in a target set for optimisation of the tuning of some temperament. I would include 3/2 and n5/4, but if the tuning was such that the errors in the 3/2 and n5/4 added to make a large error in 3/2 * n5/4, I would see this as being of no consequence, since there's nothing audibly special about the interval 3/2 * n5/4.

Reply

29 wEdited

Mike BattagliaAdmin

It's of interest to me because I play it all the time. n5/4, or something in the ballpark, is basically one of the most important intervals (or notes, as I prefer to think of it) that there is. Moving it up a fifth is also pretty important. In general, if some note or interval is important, moving it around by a fifth tends to get you to some other note or interval of interest.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia OK. We're talking about different things. There's seems to be conflation of pitches (notes) and intervals here. Moving it around doesn't change it. If we target the interval n5/4 for optimisation (in a tuning of a regular temperament) then it will be good no matter how you move it around. You don't need to also target 3/2 * n5/4.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan well, my goal with all of this is to build something that we can just put into the existing temperament finder infrastructure and get good results from it. Currently we need to give a list of "primes" and "weights" and it will do a least squares-based search on those primes, and we have theorems showing how such an optimization also optimizes on all intervals. It isn't set up to do some kind of minimax optimization on arbitrary intervals, but my hope with this is that this structure makes everything magically work out because the new "primes" we're adding are very basic and simple and more complex nobles are built up from them in a simple way.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Sure. That's a laudable goal. But you still don't need to worry about how well it approximates products of nobles and rationals or products of nobles and nobles. Approximating the nobles and rationals well is sufficient.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia I note that the dual-norms-based optimisations only optimise a specific kind of damage over all intervals, namely a damage which is simplicity-weighted absolute error. i.e. they have to allow the absolute error to increase as the complexity increases. But you'd really like the absolute error to decrease as complexity increases, because the more complex a noble is (or a rational) the more easily it loses its anti-JI (or JI) character by mistuning.

But we put up with this aspect of all-interval tunings in exchange for the mathematical convenience.

Reply

29 wEdited

Dave KeenanAuthor

Here's a possible set of pairs of convergents for the 32 nobles, listed with the errors they produce when used in your geometric-mean approximation scheme (using an exponent of 0.7236). Before you look at them, it would be good if you could arrive at your own list independently, so we'd have more confidence where they agree.

viewtopic.php?p=4628#p4628

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Reply

Remove Preview

29 wEdited

Mike BattagliaAdmin

Thanks Dave, will look at it when I get the chance. I'm sure I will get the same result; still trying to make sense of these other things...

Reply

29 w

Mike BattagliaAdmin

Dave Keenan I thought about this tonight and I think the way it should work is as follows:

1. Choose some basis set of "good" intervals which are deemed to "exist"

2. Take noble mediants of every pair of neighboring ratios in the set

I think starting with nobles and choosing convergents is backwards. First we should see what intervals we even have as generators, and then generate whatever nobles we can from them. Sometimes a noble can be generated in more than one way and I think that's alright for different basis sets.

I also think it's alright if the approximations we get are different for the "same noble" with different basis sets. For instance, if 11/9 doesn't "exist" to us, then we don't really need to worry about avoiding it as some kind of landmine in between 5/4 and 6/5, in the way that we would if we're playing it all over the place in 11-limit chords and training the listener's ear to follow that interval as a harmonic, consonant entity. Thus, it wouldn't matter in this situation if the tuning were a few cents closer to 11/9.

So I can imagine different answers to your question based on the music and the ratios we are trying to represent (and avoid, for nobles).

I have some ideas about systematic ways to do this and will write them up when I get a second...

Reply

29 w

Mike BattagliaAdmin

Dave Keenan FWIW, I played around with a bunch of different series approximations and here is what I found:

As two possiblities, we can choose, as our basis set, either the set of all ratios such that max(n,d) <= N, or the set of all ratios such that sqrt(n*d) <= N. The choice will not matter all that much. We will also only look at ratios between 1/1 and 2/1, although it turns out not matter all that much what we choose.

For almost any value of N, and for pretty much any of these choices, we will typically get that the best possible X is somewhere in that 70-75% range. In general, it seems X is slightly higher for lower values of N and then quickly settles in. It is really robust for a broad range of parameters and series approximations.

Probably the max(n,d) version is the best. So my current suggestion is this:

1. Pick whatever N-odd-limit you want, and then use all adjacent pairs as convergents, and see what nobles you get. These are, a priori, the only convergents we care about.

2. Just use something in the ballpark of that X = phi/sqrt(5). There is probably no point in messing with it for each different N; it seems to be good enough for basically every situation.

3. As you keep increasing N, new, better approximations will pop up for the same noble.

Note that the old approximations will also still "exist" in the space alongside the new ones as they pop up. That is, ApproxNoble(5/4, 6/5) still exists even when we get to N=11, where it is more properly viewed as ApproxNoble(6/5, 11/9). I don't think this is that much of a problem. The first one can be viewed as an approximation of the max-dissonance point between 5/4 and 6/5 which doesn't view 11/9 as "existing", and the second one does. And so on with ApproxNoble(11/9, 17/14). I can make this rigorous with some more endless series approximations but that's the idea and I think it works well.

This viewpoint gives pretty much everything we could ever want, I think, and I'm mostly happy with it.

Reply

29 w

Mike BattagliaAdmin

Some other thoughts, mostly to myself for later:

Noble numbers minimize the two quantities

lim inf d->∞ |r - n/d|*d^2

lim inf d->∞ |log(r) - log(n/d)|*nd

where n is chosen for each d so that n/d is closest to r. This quantity is the reciprocal of what is called the "Markov constant." It tells you that as you get better and better rational approximations, the weighted error asymptotically approaches this value. In other words, we are talking about the relative error of rational approximations "at infinity." Noble numbers maximize this relative error.

It is often true that some convergent early on in the sequence can perform better than the asymptotic performance of all convergents at infinity. In general, we can ask: what is the best rational approximation, and how good is it? So we can instead look at these related quantities, again with the best "n" for each "d":

min_d |r - n/d|*d^2

min_d |log(r) - log(n/d)|*nd

so we have replaced the "lim inf" with a "min" on all rationals. This evaluates each r as the weighted error of its best possible rational approximation on all rationals. This is closer to how models like HE work, as well as what our ears care about; it is perhaps a better measure of the "irrationality" of a number than the aforementioned quantity.

This metric will rank different nobles differently and I am curious to see how it does it. Also, it turns out that it isn't necessarily the noble numbers which maximize this quantity. For instance, the noble between 5/4 and 6/5 ranks as better approximable than 1+sqrt(2). I haven't found anything scoring higher than phi itself.

Reply

29 wEdited

Mike BattagliaAdmin

I asked about this at MSE: https://math.stackexchange.com/.../meas ... w-good-the...

One epiphany that makes this somewhat easy to see is that, for instance, there are noble numbers arbitrarily close to 3/2, such as all of those of the form [1;1,1,N,1,1,1,1,1,1,...] for some huge N.

The Markov constant will evaluate these numbers as being maximally inapproximable, attaining the maximum value of sqrt(5). Basically, the Markov constant only cares about how large the numbers get within the "tail" of the continued fraction, and since the tail is an infinite stream of 1's, which is the minimum possible, it is maximally inapproximable.

The metric that I'm talking about would care very much about these huge numbers at the *beginning* of the continued fraction, which are points where the number has an extremely good simple rational approximation. This is also related to how models like HE work, and is possibly a good launching point for a number-theoretic view of HE.

If we use this metric as some kind of "irrationality measure" that is perhaps more sophisticated than the Markov constant, then this metric will give us some kind of interesting measure of the complexity of a noble number, which I think will be related to the point on the SB tree where it first starts zig-zagging, e.g. should be related to the complexity of the numbers "n_/_" in that notation you are using. If we mix in nobles and silver aristocratic numbers, for instance, I'm not quite sure how things will be ranked, but some of the silver ones will be ranked higher than some of the nobles and so on.

We also have that for HE, with the Laplace distribution and 1/(nd) weighting (rather than 1/sqrt(nd)), as s->0 the maxima will be... well, maxima of this metric, whatever they are, which again will be some interesting thing that is looking at the coefficients in the beginning of the continued fraction rather than the tail.

Measuring how good the best possible rational approximation of a real number is

MATH.STACKEXCHANGE.COM

Measuring how good the best possible rational approximation of a real number is

Measuring how good the best possible rational approximation of a real number is

Reply

29 w

Mike BattagliaAdmin

Dave Keenan may think this is interesting so I will tag him, although this is only tangentially related to our other discussion about looking for convergents to nobles.

Reply

29 w

רועי סיני

Mike Battaglia Very interesting. If you are still interested in this now, it seems like both numbers ending in infinite 1s and numbers ending in infinite 2s (and probably numbers ending in infinite repetition of any positive integer) can be local minima of this function. For example, the number that gets the lowest value between 1 and 4/3 is 2 - 1/√2 = [1; 3, 2, 2, 2, ...] = 1.29289...

However, numbers that switch between 1 and 2, e.g. (1+√3)/2 = [1; 2, 1, 2, ...] can always be improved (or worsened, depending on your perspective) in any neighborhood you choose.

Reply

2 d

[End of facebook thread (as of 1-May-2023).]

Here's a specific proposal for how to approximate the 32 simplest nobles using higher primes.

viewtopic.php?p=4625#p4625

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Reply

Remove Preview

29 w

Mike BattagliaAdmin

Great stuff!

Reply

29 w

Mike BattagliaAdmin

Dave Keenan the only issue with the current proposal, that I see, is that the complexity is not necessarily what we want. For instance, I think you will have that phi is much more complex than n7/2. This is because although you've found decent rational approximations from a tuning standpoint, they are place on the lattice in sort of a random haphazard way, and as a result the natural complexity on that lattice places essentially a random weighting on the nobles themselves. However I think the basic idea is sound and there is probably some way to do it so that things all fit together nicely.

The question is what basis we can choose for our nobles such that the complexities and accuracies all are all put together in a nice way. I have this feeling that trying to derive this rigorously may once again lead to 2^phi 3^phi etc, or something close to it. I'll see if I can figure out how it all works.

Reply

29 w

Dave KeenanAuthor

I agree about the haphazardness and the resulting complexities. One way to make those irrelevant is to use a finite list of target intervals and minimax unweighted damage (absolute error).

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan another interesting result: suppose you take the noble mediant of two neighboring superparticular ratios. You would like to see where the result is, as some percentage of the way from the simpler toward the more complex ratio. Supposing our two intervals are n/(n-1) and (n+1)/n, then as n->inf, we indeed get precisely the same "logarithmic phi" from before, meaning about 61.8% of the way. In other words, you again get the same 2^phi 3^phi etc structure. So this is another way to derive that.

You seem to get the same thing if you take the noble mediant of two successive harmonics, e.g. two intervals of the form n/1 and (n+1)/n. The percentage will again be logarithmic phi.

On the other hand, this mainly seems to work when the two intervals are of reasonably similar complexity. If we instead look at the noble mediant from 1/1 to some arbitrary superparticular (n+1)/n, then as n->inf, the noble mediant is actually placed 100% of the way from 1/1 to (n+1)/n.

But at least if those are the interval pairs we care about, then this does show that in some sense phi really is the best value to put in for 2^X, 3^X etc. Although if we only care about approximating relatively simple nobles, then sometimes other values of X can be useful.

I think it's quite interesting that logarithmic phi really is intimately related to the structure of feudal numbers, in ways beyond just being a neat approximation...

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan perhaps more interestingly, we get something different if we look at where some noble fits within its own zig-zag pattern.

For instance, suppose we have the noble between 5/4 and 6/5. We can ask what percentage of the way that interval is located from 5/4 to 6/5. If we do this for two arbitrary superparticulars we get logarithmic phi as n->inf, as before.

But let's say we want to stay with the *same* noble and look at other rationals which generate that noble. For instance, the one between 5/4 and 6/5 is generated also by 6/5 and 11/9. So we can ask what percentage of the way this noble is located between those two. And then again from 11/9 to 17/14 and so on.

It looks like this does again converge to something nontrivial, but this time it isn't 1/phi. In fact, it seems to converge to 0.723606797749979. In addition, this magic number seems to be the result you get if you start with *any* pair of rationals, regardless of the determinant, and compute the phi-weighted mediant between them, and then do the same thing.

Any idea what on Earth this value may be?

Reply

29 w

Mike BattagliaAdmin

Dave Keenan According to Mathematica, anyway, the magic value is

1/10 (5 + Sqrt[5])

= 1/2 + Sqrt[5]/10

= 0.72360679774997896964091736687312762354406183596115257242708972454105209256378048994144144083787822749695081...

WTF number is this? This can also be rewritten as

(2 + phi)/5

Anyway, also may be interesting to put this into X for 2^X, 3^X, etc. This magic value, whatever it is, is the percentage of the way that every noble number sits between the n'th pair of convergents as n->inf. Also seem to be getting something similar for arbitrary non-noble feudal numbers w/ nontrivial golden part.

Reply

29 wEdited

Mike BattagliaAdmin

It looks like using that number gives us slightly better results for generating nobles from the simplest pairs of rationals. The first 10 superparticulars, w columns as "ratios" "true noble" "approx noble" "error"

"2/1 -> 3/2" "833.0903" "839.6113" "6.521"

"3/2 -> 4/3" "560.0666" "554.4043" "5.6622"

"4/3 -> 5/4" "422.4874" "417.1955" "5.2919"

"5/4 -> 6/5" "339.3439" "335.1747" "4.1692"

"6/5 -> 7/6" "283.6048" "280.3507" "3.2541"

"7/6 -> 8/7" "243.6192" "241.0404" "2.5788"

"8/7 -> 9/8" "213.5281" "211.4456" "2.0825"

"9/8 -> 10/9" "190.0599" "188.3479" "1.712"

"10/9 -> 11/10" "171.2433" "169.8133" "1.43"

The other one was

"2/1 -> 3/2" "833.0903" "892.1913" "59.101"

"3/2 -> 4/3" "560.0666" "575.9317" "15.8651"

"4/3 -> 5/4" "422.4874" "428.9913" "6.5039"

"5/4 -> 6/5" "339.3439" "342.6358" "3.2919"

"6/5 -> 7/6" "283.6048" "285.4995" "1.8947"

"7/6 -> 8/7" "243.6192" "244.8091" "1.1898"

"8/7 -> 9/8" "213.5281" "214.324" "0.79588"

"9/8 -> 10/9" "190.0599" "190.6184" "0.5585"

The second one quickly becomes a better approximation, although the first one seems to perform better for the very simplest pairs of nobles, which may be what matters.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Interesting! You get a simpler number if instead of looking at the ratio of the larger part to the whole, you look at the ratio between the two parts. This is simply 1:ϕ², and so your magic number is ϕ²/(1+ϕ²). Yes, that may be a better power to use for 2^X, 3^X, etc, but I haven't fully wrapped my head around that. That means the idea of an equivalence with feudal integer entries in the vectors goes out the window, and this is purely an approximation scheme. Nothing wrong with that.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia As you go from one convergent of a noble to the next more complex one, both the numerator and denominator increase by approximately a factor of ϕ, so the product complexity increases by approx ϕ². And that result I just proved says the logarithmic distance to each convergent is (approx) inversely proportional to the convergent's product complexity, so that's why the distances from the noble to the two convergents are in a ratio of 1:ϕ², and so the distance from the simple convergent to the noble, divided by the distance between the two convergents, is ϕ²/(1+ϕ²) ≈ 0.7236.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan that is very interesting!

OK, so here is another interesting note. Suppose that instead of looking at two neighboring superparticulars, we instead take the mediant of two neighboring superparticulars, and then take the noble mediant of the originally more complex superparticular and the mediant of both. These will also be unimodular.

In other words, we have the following sequence:

2/1, 3/1 (derived from 1/0, 2/1)

3/2, 5/3 (derived from 2/1, 3/2)

4/3, 7/5 (derived from 3/2, 4/3)

5/4, 9/7 (derived from 4/3, 5/4)

6/5, 11/9 (derived from 5/4, 6/5)

...

(N+1)/N, (2N-1)/(2N-3)

Then we also get something different than the usual phi thing. It looks like the proportion of the large to the whole is (2 phi)/(1 + 2 phi) ≈ 76.4%, and the proportion of the large to the small step is 2 phi.

I think I have an idea of how this all fits together...

Reply

29 w

Mike BattagliaAdmin

Dave Keenan

OK, some idea of the larger theory. Let's introduce some basic language. Suppose that we are taking the phi-weighted mediant from ratio r1 -> r2; call this result "m". Then we will say

a = log(m/r1)

b = log(r2/m)

c = log(r2/r1)

We can use whatever base log we want; octaves or cents or base e or whatever.

Thus if we have the most simple situation of r1 < r2, and r1 is less complex than r2, we have something like

|-----a-----|---b---|

with a+b = c.

Note that if we *don't* have r1 < r2, like when we go from 5/4 to 6/5, you still get a+b = c, but all of these quantities are negative. Also note that if the complexity of r2 is higher than r1 - which is the typical situation with noble mediants, we will always have the size of a > b. However, if we do very strange "improper noble mediants" going from some very complex ratio to a simpler one, we may have a < b.

Then there are three quantities of interest:

a/c: my original quantity, typically "large" to "whole"

a/b: your quantity, typically "large" to "small

b/c: 1 - a/c

I will just talk about a/c and a/b because b/c is derived from them.

So with this, we have

Successive superparticular ratios, e.g. 5/4 to 6/5:

a/c = 1/phi ≈ .618

a/b = phi ≈ 1.618

Successive harmonics, e.g. 5/1 to 6/1:

a/c = 1/phi ≈ .618

a/b = phi ≈ 1.618

They are the same. We also get the same thing if we take the noble mediant "the wrong way", e.g. from 6/5 to 5/4 or from 6/1 to 5/1: still a/b = phi.

We also have:

Successive convergents of the same noble mediant:

a/c = phi/(phi+1/phi) = 1/2+sqrt(5)/10 ≈ .723

a/b = phi^2 ≈ 2.618

Then for this "superparticular + mediant of two superparticulars" sequence above, e.g. from 6/5 to 11/9, we get

a/c = 2phi/(1+2phi) ≈ 0.764

a/b = 2phi ≈ 3.236

My thinking is that in general, for some arbitrary "pattern" of rational numbers we want to iterate to infinity, we can look at the "complexity quotient" q = (n1d1)/(n2d2) where r1 = n1/d1 and r2 = n2/r2. Then just playing around with it, we seem to get:

a/b = phi * sqrt(q)

I don't know why that sqrt is there but it seems to work. I'll think about how to prove this.

Reply

29 wEdited

Mike BattagliaAdmin

Another cute way to look at this: when we talk about these noble mediants of successive superparticulars, etc, we are in a certain sense talking about noble mediants of different "rational patterns." These patterns can be of the form

N/1 -> (N+1)/1 (harmonics)

N/(N-1) -> (N+1)/N (superparticulars)

N/(N-1) -> (2N-1)/(2N-3) (superparticular + mediant thing)

N/round(N/phi) -> round(N*phi)/N (successive convergents)

or whatever.

Now, we can treat each of these patterns as just a real function of N. Then, as N->inf, we get some value q, and it seems we have a/b = sqrt(q) from before.

But perhaps an even cuter way to look at this is to let N be an infinite number. If we do so, these patterns become idealized "infinite rationals" which are located "at infinity" on the SB tree. The things we are studying, then, are just the phi-weighted mediants of different infinite rationals. The regular real numbers are also "points at infinity" but these are different; they are all infinitesimally close to the real numbers but are true rational numbers with numerators and denominators that are generalized infinite integers. We can use nonstandard analysis to make this all rigorous if we want, but for now I'll just note it's a nice visual way to think of things.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia It seems to me, that what we are trying to get to here is to decide what pair of ratios to use to approximate each noble via the weighted geometric mediant scheme (or is it a weighted geometric mean?). If so, I think that instead of looking at all these different "rational patterns" of superparticulars, harmonics, etc, we only need to look at the convergents for each noble (the zigzags on the SB tree) and choose a pair of successive convergents that give a good compromise between simplicity and accuracy. We want them to be within the same prime-limit as the rationals that we are also targeting.

One option is to use only 13-limit convergents. These could be the two convergents just prior to the first non-13-limit convergent, which is the one I chose for my rational approximations of the nobles. But maybe some earlier simpler pair will be suitable in some cases.

Once such a pair of convergents is chosen for each of the 32 simplest nobles, then the 2.2ˣ.3.3ˣ.5.5ˣ... vectors can be constructed and the optimal value of X can be found to minimise the errors over the 32 nobles. In fact we might find a different value of X for each prime. Some trial and error may be required, with different pairs of successive convergents for some nobles, before settling on them all.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan that's one question this all raises, yes. Of course the simple method is just to search empirically for the best X as you suggest. But I am interested in peering deeper into the structure of this.

Using this same idea, you can derive different geometric approximations that are "correct" in different ways. For instance, with successive convergents to some noble number, we get the a/b ratio is something different. In general, if we go with that "q-value" from my earlier post, it seems we would get that the a/b ratio = 1/(1 + 1/(phi*sqrt(q))), for which some value of X can be derived. I will need to think about it.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan Of course there is also a simpler way to look at it. *Given* some value of X, like phi, the pair of "best" generators for some noble will differ. We can choose some X and look for whichever pair of ratios has the best tradeoff between simplicity and accuracy.

If we know in advance we have some kind of ratios in mind for this, e.g. in some small odd limit or prime limit or whatever, we can look for the X which best gives this result. Same with different values of X for different primes.

We could try to just to calculate this for each odd-limit, where we are only taking noble mediants of pairs of neighboring ratios. I think, though, that we will get the best results if neighboring ratios are "uniform" in some sense that they tend to have a similar complexity ratio to one another. I'm not sure if odd-limit gives us this or if something else will do.

Reply

29 wEdited

Mike BattagliaAdmin

Lastly, I also remember in some of my original posts about this that the approximation between 5/4 and 9/7 was a little worse than the rest. I am wondering if this is because one of the other approximations may have been a better choice for this pair of ratios.

This would suggest some method where the approximate geometric phi-mediant of two ratios isn't just some simple % of the distance between them for all such pairs, but that the % depends on the ratios themselves in some way.

This would to be a more detailed approximation than the ones we've been talking about, but I wonder if there is still some way to do it which is equally simple, in that you still just get some kind of 2.2^x.3.3^y.etc structure. It's difficult to do this perfectly if we want integer coordinates though.

Reply

29 wEdited

Dave KeenanAuthor

I have created a table that shows the options for pairs of rationals to be used in the geometric approximation scheme. In 12 of the 32 cases (shown green) there is no real choice in the matter, if you want to stay within the 13-prime-limit. See

viewtopic.php?p=4627#p4627

This shows, for example, that if 5/4 and 9/7 don't work so well, you can use 9/7 and 14/11.

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Reply

Remove Preview

29 wEdited

Dave KeenanAuthor

Mike Battaglia I've been doing some of the optimisations I described above, but only with a single value of X for all primes. I decided it's too messy to do otherwise. When they are all the same, I don't need to prime-factorise, i.e. I don't need to actually generate the vectors. Optimum values of X range from 0.71 to 0.75 with different sets of pairs of convergents. In other words, your 0.7236 is the right exponent, not 0.618. It turns out that exponent can also be described as n2/3 = F1/F5 = (0+1ϕ)/(-1+2ϕ) = ϕ/√5.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan I think if you are using generating rationals which are some decent length of the way into the zigzag then that'll be right. I still am trying to grasp what it all means though.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan I am still kind of curious what we get if we take the n-odd-limit and look for the best X value for all nobles generated by pairs of neighboring ratios, for each n. And also curious what we get if instead of the n-odd-limit, we instead go with all ratios of depth ≤ n in the Stern-Brocot tree. I have this feeling the SB tree version may have a more "uniform" error and thus may be better.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Odd limits per se, make no sense to me in this context since there's nothing special about prime 2. But I guess you mean the odd limit /diamond/. In this context, I prefer to think of that as the 22-integer-limit 13-prime-limit triangle. Yes, it turns out we don't need to use any noble convergents outside that set, to get good geometric mean approximations.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia It doesn't matter how close to the top of the zigzags the convergents are, the optimum X stays in the range 0.71 to 0.75. The optimum is constrained by a few nobles that only have two 13-prime-limit convergents, no matter whether we limit it to the 16 simplest nobles or go for the full 32.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan what difference is there between what you are calling the odd limit and odd limit diamond? Are you doing a minimax optimization or an rms or something else?

Reply

29 w

Dave KeenanAuthor

Mike Battaglia The odd limit has unlimited powers of 2. The diamond stays within the first octave. I'm being a pedant. Sorry.

I start with miniRMS, then I increase the power from 2 to 4 to 8 to 16 so I approximate minimax. X stays in the aforementioned range for all of these. I'm now allowing simpler nobles to have more error, by weighting the error by the complexity of the noble, but it doesn't make any difference. I'm taking the complexity of a noble to be the product of the feudal norms, N(), of its numerator and denominator, as you suggested way back. When the noble is expressed in the form f11/f5 etc, the norm is the number after the f. e.g. N(f11) = 11. So f11/f5 and F11/F5 have complexity 55.

So I believe we should just use X = ϕ/√5 and be done with it.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan ok, that makes sense. I am curious if that phi/(sqrt(5)) term is really related or if it is a coincidence that the best X happens to be near this value. If you increase the number of nobles you are looking at, do things seem like they tend toward that X?

Reply

29 w

Mike BattagliaAdmin

Tbh, maybe that really is the best value. The thing that threw me off before is that any pair of rationals can be viewed as more than one "pattern," and different patterns suggest different idealized approximations. For instance if we look at 5/4 and 6/5, we get a different idealized approximation if we view those as "two successive super particular ratios" than if we view them as "two successive convergents to some noble number." But the latter is perhaps what we really want when we are viewing pairs of ratios as seeds for some noble mediant, and then, as you say, phi/√5 seems to be the result.

Reply

29 w

Mike BattagliaAdmin

Last thought of the night.

Suppose we have two ratios for which r2 is much more complex than r1, then the phi-weighted mediant from r1 to r2 is clearly going to be much more than 72% of the way there. It could be close to 100% of the way there. So clearly this approximation is best if the two ratios are closest in complexity.

I am curious if there is a slight generalization of the approximation in which the distance from ratio r1 to r2 is, rather than just being the same percentage for all ratios, is somehow also a function of the complexity of r1 and r2. As r2 gets more complex, the mediant is closer to r2 and if not it's closer to r1. This would only be useful if we don't have to increase the dimensionality or add weird fractional coefficients though.

Reply

29 w

Nate BeDell

Woah, just skimming over this for the first time, this sounds super interesting. Time to break out the Dummit and Foote!

Reply

29 w

Mike BattagliaAdmin

Dave Keenan OK, I think I have managed to figure the entire thing out. The idea is we start with the expression for the noble mediant as a four-variable function N(n1,d1,n2,d2) = (n1+n2*phi)/(d1+d2*phi). Then, we get the expression for the ratio "a/b" of the small to large partition between the two ratios, which is a ratio of two logarithms derived from the same four variables. Then, playing around w/ Taylor series in all four variables, I somehow managed to get this:

phi * (n2/n1 - d2/d1) * 1/c

This expression is great, and is the "enhanced approximation" I was looking for. It's an expanded version of the original approximation with extra terms. The (n2/n1 - d2/d1) term can be viewed as kind of measuring the difference in relative complexity of the two rational numbers, and the "c" term is the chroma.

My main problem: I don't know how I got it! I was playing around with Mathematica and was trying to get the Taylor series for the ratio "a/c" of the large step to the entire thing, but somehow it gave this instead, and somehow the results were basically perfect as a Taylor series for the ratio "a/b" instead. I'm pretty stumped and think maybe this could be derived as a multivariate Padé approximant of some kind? I don't get it but it works great.

We can use this to determine the best approximations to things. Given some set of important intervals, like a tonality diamond or whatever, we can use this method to determine the "right" multiplier for phi by just computing some kind of average of (n2/n1 - d2/d1) * 1/c for all adjacent pairs of intervals. I think that if instead of the tonality diamond we use the ≤ n'th level of the Stern-Brocot tree, then as n -> inf we will indeed get the phi/sqrt(5) thing, because successive ratios in this set will be successive mediants, and I think the majority of them will have a complexity ratio of phi again. It would be interesting to try this and see.

We also note that previous results can be derived from this equation. If we let n1 -> n2 and d1 -> d2, this expression tends toward phi, which is the original approximation I was using. If we instead say that n2 -> n1*phi and d2 -> d1*phi, this instead tends to phi/sqrt(5), which we've been talking about.

Lastly, we can also write this expression as:

phi * (n1*d2 - n2*d1) * 1/(n1*d1) * (-1/c)

which is in terms of three quantities of interest. That -(n1*d2 - n2*d1) is the negative of the determinant (which for noble mediants should be +/- 1), the 1/(n1*d1) is a measure of complexity, and 1/c is the chroma size. I guess if the chroma is positive then the determinant will be negative and so on. But anyway, if we know that our ratios are going to have a determinant of +/- 1, then we can rewrite this as

phi * 1/(n1*d1) * |-1/c|

Anyway, I think this answers a bunch of questions, but sheesh, how does one even get this magical expression? For whatever reason, actually looking at the Taylor series for "a/b" didn't give this, but rather that for "a/c" did, even though it's a much better approximation for "a/b". Well, whatever...

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia I was stuck at the second paragraph for a long time. I had no idea what you were referring to as the "chroma" here. I thought at first you must mean the quotient of the two ratios, i.e. d₁n₂/n₁d₂. But when I tried that, it didn't give sensible numbers for the ratio of the two parts in logarithmic terms. I eventually reverse-engineered it and figured you must mean ln(d₁n₂/n₁d₂). In this case, it matters what log base you use.

If that's correct, then, as a pedagogical note: Why you thought "the natural log of the quotient of the two ratios" or "the difference between the natural logs of the two ratios" could be unpacked from the name "chroma" is beyond me. It would have been good if you had just given a formula for "c".

Yes, it does give much better approximations than simply assuming a ratio of ϕ² between parts, but I don't see how it has any practical value. A vector basis that made use of it would have more dimensions than the feudal prime basis that gives /exact/ results.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan I just explained all of this a few posts up, what I was calling "a" "b" and "c". I explained why it has practical value I'm the paragraph about using it to compute the best approximations.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Yes. 18 posts back, in a different subthread, where you explicitly say the log base doesn't matter. That and the fact that you hadn't previously used the term "chroma" led me to believe that this must be a different use of "c". Never mind.

Re practical use: I assume you're referring to where you write:

"We can use this to determine the best approximations to things. Given some set of important intervals, like a tonality diamond or whatever, we can use this method to determine the "right" multiplier for phi by just computing some kind of average of (n2/n1 - d2/d1) * 1/c for all adjacent pairs of intervals."

I don't see any point in trying to determine the best approximation for the set by computing the average of the approximations for all pair of adjacent elements, because we can more easily determine the best approximation for the set by computing the average of the /exact/ values for all pair of adjacent elements. Or better still, by doing a minimax or miniRMS optimisation of some suitable damage measure that weights the absolute cents errors by some kind of complexity measure.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia I also don't see any point in using pairs of ratios that are not successive convergents of the nobles being approximated. But I do see your point that these should be further limited to (what I prefer to describe as) some combination of low prime limit and low integer limit.

It appears at present, that for the first 32 nobles on the SB-tree we can stay within the 20-integer limit (maybe even the 18-integer limit), and all but two can stay within the 13-prime-limit. Those two (923c and 1530c) have pretty bad damage unless we let them go to the 17-prime limit.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan sorry, you're right, I should have mentioned that in this particular formula the log does need to be base e. I wrote this pretty late last night.

Of course you can just do some kind of minimax optimization and determine the values empirically, and if you do there will likely be various seemingly random values things seem to converge to, such as the 72% or whatever it is. Looking at some of these series gives us some way to see what we are really doing and why these approximations work so well, and seems to give us simple closed-form values for some of the empirically-determined values in terms of phi, which so far have agreed with the empirical estimates to within some reasonable margin of error. I think that's certainly interesting, and what's really interesting to me is that all of this stuff is just another part of the inherent structure of the SB tree.

I think there are several valid ways to approach this. We could say we want to approximate nobles, and then as you say, if we look for convergents we get the phi/sqrt(5) thing. But I also think it's useful to say that we want some binary operator that takes some arbitrary pair of ratios and returns some kind of idealized maximally dissonant interval between them which they "generate." To some extent the structure we are talking about gives you both of those things simultaneously and that's what I've been looking at.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia Well you set yourself a fun puzzle, and you solved it. But for my money, the breakthrough was when you figured out that the magic number for successive convergents was that constant that can be written ϕ/√5. I suggest we give that constant the Greek small letter chi χ as it looks a bit like the X you've been using for the variable, and it's traditional to use lowercase Greek letters for non-(quadratic)-integer values in quadratic fields, and I don't know of any other use of χ in RTT or related math. Then we can write the basis as e.g. 2.3.5.7.13.2ᵡ.3ᵡ.5ᵡ.7ᵡ.11ᵡ.13ᵡ.17ᵡ (yes, a superscript chi exists in unicode, U+1D61).

I think it better to order the basis in that way, rather than interleaving chi powers with 1st powers, because the sub-vector over the first powers of the primes is the vector for the simplest of the two chosen convergents, and the subvector over the chi powers is the vector for the quotient of the two convergents (note: not the vector for the second convergent).

You've called this ratio between the two generating ratios, the "chroma" (I'm assuming you didn't mean for the "chroma" to be strictly the natural log of that ratio). That's not bad, but "chroma" seems to have several connotations in addition to being a ratio between ratios, such as being a comma-sized interval that is not tempered out, and being a "side" of a periodicity block or MOS scale. Neither of which seem relevant here.

Douglas Blumeyer = Cmloegcmluin Xenharmonic Feisbeuk once asked me if I knew of a term that functioned multiplicatively in the way that the term "delta" and its symbol Δ function additively. In other words a term for a small ratio that is the quotient of two other ratios. I did not know of one, but suggested "qoppa" since it sounds a bit like a contraction of "quotient delta". "quotta" might be better in that regard, however "qoppa" happens to be the name of a Greek letter, albeit an archaic one, that looks a bit like a Q. In fact it looks like this: Ϙ.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan Regarding the basis, I got in the habit of writing it like this

|a b c d e> + |f g h i j>*X

So we explicitly write the monzo in terms of a "just part" and a "golden part". This is a little bit nicer, I thought, than either |(a+fX) (b+gX) ... > or something like |a b c d e ; f g h i j> or whatever.

I was just using the term "chroma" loosely here; I wasn't proposing it be the official name of anything. In my own writings, I've used the term "chroma" extensively when talking about MODMOS, as well as for epimorphic scales more generally, as the difference between some pair of steps (or thirds, or etc). But I don't really have much problem if people want to use the term "chroma" in a generalized sense to refer to the quotient of any two ratios or intervals in general, not just two steps in an MOS. In fact I haven't really thought much about it before and have probably done that myself plenty of times without thinking about it, as I did here.

FWIW I had different names for all of this stuff back in the day, although we've gone way further than I ever did. Looking at some of my old notes on this stuff, I apparently used to call feudal primes "phrimes", heh.

Reply

29 w

Mike BattagliaAdmin

BTW, how are you typing all of these ridiculous Unicode characters so easily? I mean damn, superscript phi? Are you just constantly Googling unicode letters and copying and pasting them? I'll probably just keep typing "X" but sure, I don't mind if you want to make it chi for when things are typeset nicely...

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia You wrote: "But I also think it's useful to say that we want some binary operator that takes some arbitrary pair of ratios and returns some kind of idealized maximally dissonant interval between them which they "generate.""

I think we already know that any "kind of idealized maximally dissonant interval" will always be a noble number. Of course there are intervals like the dissonance maximum near 70 cents, and others near the octave and fifth. But those are extremely dependent on properties of the human ear/brain, or on the spectrum of the timbre being used, and therefore can't be considered "idealized".

That's why I don't have any interest in a binary operator that takes an arbitrary pair of ratios. But I wish you luck.

Reply

29 wEdited

Mike BattagliaAdmin

I'm not sure I follow - yes, we're agreeing those are idealized, and now we're talking about approximating them within this other 2.2^X.3.3^X.etc space...

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia "Phrimes" is cute. And [EDIT]

[a b c d e⟩ + [f g h i j⟩×χ

or

[a b c d e⟩ + χ×[f g h i j⟩

is excellent.

I'm certainly not searching, copying and pasting.Or at least I only had to do that once per codepoint, long ago. The secret weapon is an insanely great utility called WinCompose.

http://wincompose.info/

If you're not using a Windows machine, I think there are similar utilities for Linux and Mac. I don't know of any equivalent for Android or iPhone. But if you are using Windows, as well as installing WinCompose, you'll need my custom sequences.

https://dkeenan.com/WinComposeSequences.txt

Don't bother trying to find the right place to put this file, just choose "Show sequences→Edit" then copy and paste its contents to overwrite the existing examples.

Then scan through it to learn the common prefixes, like ⎄8 for Greek, ⎄6 for superscript, ⎄- for subscript. ⎄ here stands for the right Alt key. You don't hold them down simultaneously, you just type them in sequence. The most often used characters get sequences with double characters like ⎄pp for pi, ⎄22 for superscript 2, ⎄88 for infinity. Some are pictorial, or like overstrikes, such as ⎄v/ for square root sign. Superscript phi is ⎄68f. Superscript chi is ⎄68c.

WinCompose | Easy typing of special characters on Windows™

WINCOMPOSE.INFO

WinCompose | Easy typing of special characters on Windows™

WinCompose | Easy typing of special characters on Windows™

Reply

Remove Preview

29 wEdited

Dave KeenanAuthor

Mike Battaglia Agreed. But then what am /I/ missing? Why do we need, or want, anything other than pairs of successive convergents, for approximating nobles in the 2.2^X.3.3^X.etc space?

Reply

29 w

Dave KeenanAuthor

Edit above, because you were right to have an explicit (non-matrix, non dot product ) multiply sign there, which can be assumed to be mapped over the vector. i.e.

[a b c d e⟩ + [f g h i j⟩×χ

or

[a b c d e⟩ + χ×[f g h i j⟩

Note that I'm using (upright) χ = ϕ/√5 ≈ 0.7236 to represent that specific constant. You should still use (italic) 𝑋 or 𝑥 in typeset material when it's a variable.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan Thanks, I'm on Mac but I'll check that program out and see if I can locate a port of some kind. It looks like there are some "Mac Compose key" programs out there so hopefully that'll be good. I also wish I had something on Android as I am often on mobile when doing stuff.

Regarding why, it's just that I'm trying to get the structure of what's going on. Right now, if you take any two monzos m1 and m2, the approximate noble mediant from m1 -> m2 = m1 + (m2-m1)*X. That's pretty neat. So for 5/4 -> 6/5, we literally get (5/4) * (24/25)^(X). Basically, what we have done is added, for each JI ratio n/d, a new element of the form (n/d)^X. It's really simple and gives great results.

What we've done is now point to some element in the space and say, this is the best representation of n6/5 or n5/4 or whatever, which is equivalent to pointing to some pair of two ratios and saying that's the right pair to generate the noble. However, these other elements still exist. Some of them are approximate phi-weighted mediants that aren't noble, and some of which are approximate noble mediants using other convergents, and so on. And since they're all there I'm trying to figure out what they are. Whatever they are, they all flow naturally from the new "primes" we've added and inherit this complexity function from them and everything.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia AFAIK we haven't yet pointed to pairs of ratios and said those are the right pairs to generate the nobles. But that's certainly what I'm working towards, and I'm very close to making a specific proposal. If you've already chosen a set for the usual 32 nobles, I'd like to see it, with errors.

It seems I'm viewing this in a different way from the way you are viewing it. Of course, this basis supports elements which are not simple rationals and are not the standard approximations of the simple nobles (whatever we decide those to be). Such elements are either (a) useful approximations of simple rationals, or (b) useful approximations of the standard approximations of the nobles, or (c) junk. In cases (a) and (b) what we have are commas that can be made use of by temperaments.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia When you write: "... and inherit this complexity function from them and everything", what complexity function are you referring to?

Reply

29 w

Mike BattagliaAdmin

Dave Keenan there is also option (d), which is transposed versions of other nobles which are not junk. One of the reasons I was so interested in this space is because I think that nothing in it is really junk at all.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia I'm viewing these vectors as representing intervals, not pitches, so I'm automatically including transpositions of pitch. To my way of thinking, an interval is not changed by being transposed. Do you have some meaning for a "transposed interval" that actually changes the size of the interval? I assume you don't mean things like octave extensions of intervals, because I assume you're aware that when you multiply a noble by a rational other than 1, it is no longer noble, because you break its unimodularity. And so it is no longer anti-JI.

Reply

29 w

Mike BattagliaAdmin

Dave Keenan I view them interchangeably, but I just mean the product of various approximate nobles and just ratios (and maybe other approximate nobles). I don't think it really matters from a musical standpoint that, for instance, 3/2 * n5/4 isn't technically noble anymore; it's still a very important interval (or note or however you like to view things) and it still very much is part of what I would call "anti-JI."

Reply

29 wEdited

Mike BattagliaAdmin

Hm, here's an interesting theorem: the elements of this algebra which are approximations of noble numbers all have "golden part" equal to some superparticular ratio. That is, if we look at |a b c d e> + |f g h i j>*phi, it is required that |f g h i j> be either superparticular or the inverse of a superparticular in order for the result to be an approximate noble.

It goes the other way as well. If |f g h i j> is superparticular or inverse superparticular, then the entire thing is either an approx noble, or the product of an approx noble and a just ratio. Sometimes there are multiple |a b c d e> making |f g h i j> noble: for instance, if |f g h i j> is the monzo for 24/25, then we have that this is approx. noble when |a b c d e> is 5/4 or 5/3.

To see this we simply note that the ratio between any two adjacent elements of the SB tree is a superparticular ratio, and that every superparticular ratio appears somewhere in the SB tree.

Reply

29 w

Mike BattagliaAdmin

If the golden part is not a superparticular ratio, then the resulting ratio can simultaneously be viewed either as a phi-weighted mediant of non-adjacent ratios, or a product of multiple nobles and JI ratios. So I guess this property from the feudal numbers is also true here.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia I don't understand why the interval 3/2 * n5/4 (as opposed to its components) is of any musical interest. I would never include it in a target set for optimisation of the tuning of some temperament. I would include 3/2 and n5/4, but if the tuning was such that the errors in the 3/2 and n5/4 added to make a large error in 3/2 * n5/4, I would see this as being of no consequence, since there's nothing audibly special about the interval 3/2 * n5/4.

Reply

29 wEdited

Mike BattagliaAdmin

It's of interest to me because I play it all the time. n5/4, or something in the ballpark, is basically one of the most important intervals (or notes, as I prefer to think of it) that there is. Moving it up a fifth is also pretty important. In general, if some note or interval is important, moving it around by a fifth tends to get you to some other note or interval of interest.

Reply

29 wEdited

Dave KeenanAuthor

Mike Battaglia OK. We're talking about different things. There's seems to be conflation of pitches (notes) and intervals here. Moving it around doesn't change it. If we target the interval n5/4 for optimisation (in a tuning of a regular temperament) then it will be good no matter how you move it around. You don't need to also target 3/2 * n5/4.

Reply

29 wEdited

Mike BattagliaAdmin

Dave Keenan well, my goal with all of this is to build something that we can just put into the existing temperament finder infrastructure and get good results from it. Currently we need to give a list of "primes" and "weights" and it will do a least squares-based search on those primes, and we have theorems showing how such an optimization also optimizes on all intervals. It isn't set up to do some kind of minimax optimization on arbitrary intervals, but my hope with this is that this structure makes everything magically work out because the new "primes" we're adding are very basic and simple and more complex nobles are built up from them in a simple way.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia Sure. That's a laudable goal. But you still don't need to worry about how well it approximates products of nobles and rationals or products of nobles and nobles. Approximating the nobles and rationals well is sufficient.

Reply

29 w

Dave KeenanAuthor

Mike Battaglia I note that the dual-norms-based optimisations only optimise a specific kind of damage over all intervals, namely a damage which is simplicity-weighted absolute error. i.e. they have to allow the absolute error to increase as the complexity increases. But you'd really like the absolute error to decrease as complexity increases, because the more complex a noble is (or a rational) the more easily it loses its anti-JI (or JI) character by mistuning.

But we put up with this aspect of all-interval tunings in exchange for the mathematical convenience.

Reply

29 wEdited

Dave KeenanAuthor

Here's a possible set of pairs of convergents for the 32 nobles, listed with the errors they produce when used in your geometric-mean approximation scheme (using an exponent of 0.7236). Before you look at them, it would be good if you could arrive at your own list independently, so we'd have more confidence where they agree.

viewtopic.php?p=4628#p4628

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

FORUM.SAGITTAL.ORG

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Noble frequency ratios as prime-count vectors in ℚ(√5) - Page 2 - The Sagittal forum

Reply

Remove Preview

29 wEdited

Mike BattagliaAdmin

Thanks Dave, will look at it when I get the chance. I'm sure I will get the same result; still trying to make sense of these other things...

Reply

29 w

Mike BattagliaAdmin

Dave Keenan I thought about this tonight and I think the way it should work is as follows:

1. Choose some basis set of "good" intervals which are deemed to "exist"

2. Take noble mediants of every pair of neighboring ratios in the set

I think starting with nobles and choosing convergents is backwards. First we should see what intervals we even have as generators, and then generate whatever nobles we can from them. Sometimes a noble can be generated in more than one way and I think that's alright for different basis sets.

I also think it's alright if the approximations we get are different for the "same noble" with different basis sets. For instance, if 11/9 doesn't "exist" to us, then we don't really need to worry about avoiding it as some kind of landmine in between 5/4 and 6/5, in the way that we would if we're playing it all over the place in 11-limit chords and training the listener's ear to follow that interval as a harmonic, consonant entity. Thus, it wouldn't matter in this situation if the tuning were a few cents closer to 11/9.

So I can imagine different answers to your question based on the music and the ratios we are trying to represent (and avoid, for nobles).

I have some ideas about systematic ways to do this and will write them up when I get a second...

Reply

29 w

Mike BattagliaAdmin

Dave Keenan FWIW, I played around with a bunch of different series approximations and here is what I found:

As two possiblities, we can choose, as our basis set, either the set of all ratios such that max(n,d) <= N, or the set of all ratios such that sqrt(n*d) <= N. The choice will not matter all that much. We will also only look at ratios between 1/1 and 2/1, although it turns out not matter all that much what we choose.

For almost any value of N, and for pretty much any of these choices, we will typically get that the best possible X is somewhere in that 70-75% range. In general, it seems X is slightly higher for lower values of N and then quickly settles in. It is really robust for a broad range of parameters and series approximations.

Probably the max(n,d) version is the best. So my current suggestion is this:

1. Pick whatever N-odd-limit you want, and then use all adjacent pairs as convergents, and see what nobles you get. These are, a priori, the only convergents we care about.

2. Just use something in the ballpark of that X = phi/sqrt(5). There is probably no point in messing with it for each different N; it seems to be good enough for basically every situation.

3. As you keep increasing N, new, better approximations will pop up for the same noble.

Note that the old approximations will also still "exist" in the space alongside the new ones as they pop up. That is, ApproxNoble(5/4, 6/5) still exists even when we get to N=11, where it is more properly viewed as ApproxNoble(6/5, 11/9). I don't think this is that much of a problem. The first one can be viewed as an approximation of the max-dissonance point between 5/4 and 6/5 which doesn't view 11/9 as "existing", and the second one does. And so on with ApproxNoble(11/9, 17/14). I can make this rigorous with some more endless series approximations but that's the idea and I think it works well.

This viewpoint gives pretty much everything we could ever want, I think, and I'm mostly happy with it.

Reply

29 w

Mike BattagliaAdmin

Some other thoughts, mostly to myself for later:

Noble numbers minimize the two quantities

lim inf d->∞ |r - n/d|*d^2

lim inf d->∞ |log(r) - log(n/d)|*nd

where n is chosen for each d so that n/d is closest to r. This quantity is the reciprocal of what is called the "Markov constant." It tells you that as you get better and better rational approximations, the weighted error asymptotically approaches this value. In other words, we are talking about the relative error of rational approximations "at infinity." Noble numbers maximize this relative error.

It is often true that some convergent early on in the sequence can perform better than the asymptotic performance of all convergents at infinity. In general, we can ask: what is the best rational approximation, and how good is it? So we can instead look at these related quantities, again with the best "n" for each "d":

min_d |r - n/d|*d^2

min_d |log(r) - log(n/d)|*nd

so we have replaced the "lim inf" with a "min" on all rationals. This evaluates each r as the weighted error of its best possible rational approximation on all rationals. This is closer to how models like HE work, as well as what our ears care about; it is perhaps a better measure of the "irrationality" of a number than the aforementioned quantity.

This metric will rank different nobles differently and I am curious to see how it does it. Also, it turns out that it isn't necessarily the noble numbers which maximize this quantity. For instance, the noble between 5/4 and 6/5 ranks as better approximable than 1+sqrt(2). I haven't found anything scoring higher than phi itself.

Reply

29 wEdited

Mike BattagliaAdmin

I asked about this at MSE: https://math.stackexchange.com/.../meas ... w-good-the...

One epiphany that makes this somewhat easy to see is that, for instance, there are noble numbers arbitrarily close to 3/2, such as all of those of the form [1;1,1,N,1,1,1,1,1,1,...] for some huge N.

The Markov constant will evaluate these numbers as being maximally inapproximable, attaining the maximum value of sqrt(5). Basically, the Markov constant only cares about how large the numbers get within the "tail" of the continued fraction, and since the tail is an infinite stream of 1's, which is the minimum possible, it is maximally inapproximable.

The metric that I'm talking about would care very much about these huge numbers at the *beginning* of the continued fraction, which are points where the number has an extremely good simple rational approximation. This is also related to how models like HE work, and is possibly a good launching point for a number-theoretic view of HE.

If we use this metric as some kind of "irrationality measure" that is perhaps more sophisticated than the Markov constant, then this metric will give us some kind of interesting measure of the complexity of a noble number, which I think will be related to the point on the SB tree where it first starts zig-zagging, e.g. should be related to the complexity of the numbers "n_/_" in that notation you are using. If we mix in nobles and silver aristocratic numbers, for instance, I'm not quite sure how things will be ranked, but some of the silver ones will be ranked higher than some of the nobles and so on.

We also have that for HE, with the Laplace distribution and 1/(nd) weighting (rather than 1/sqrt(nd)), as s->0 the maxima will be... well, maxima of this metric, whatever they are, which again will be some interesting thing that is looking at the coefficients in the beginning of the continued fraction rather than the tail.

Measuring how good the best possible rational approximation of a real number is

MATH.STACKEXCHANGE.COM

Measuring how good the best possible rational approximation of a real number is

Measuring how good the best possible rational approximation of a real number is

Reply

29 w

Mike BattagliaAdmin

Dave Keenan may think this is interesting so I will tag him, although this is only tangentially related to our other discussion about looking for convergents to nobles.

Reply

29 w

רועי סיני

Mike Battaglia Very interesting. If you are still interested in this now, it seems like both numbers ending in infinite 1s and numbers ending in infinite 2s (and probably numbers ending in infinite repetition of any positive integer) can be local minima of this function. For example, the number that gets the lowest value between 1 and 4/3 is 2 - 1/√2 = [1; 3, 2, 2, 2, ...] = 1.29289...

However, numbers that switch between 1 and 2, e.g. (1+√3)/2 = [1; 2, 1, 2, ...] can always be improved (or worsened, depending on your perspective) in any neighborhood you choose.

Reply

2 d

[End of facebook thread (as of 1-May-2023).]

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

Wow, I'm very flattered. I will now post important stuff here as well as in facebook for the convenience of everyone.

Also, there are recent edits I made to the post with the factoring algorithm, which are not in the version you copied:

About the divisibility of the original number by feudal factors of split ordinary primes, I wrote "both of them cannot work because if they do then the number would divide the ordinary prime which is their product and we have already factored out all of the ordinary primes, so if this prime divides the norm multiple times then one of the feudal primes has to divide the norm the same number of times, and therefore if one of them succeeded the first time you can skip checking the other one and continue dividing by it, i.e. multiplying by the other one and dividing by the product, the same number of times that the corresponding ordinary prime divides the norm"

I also added this: "After you divided by all the primes, you get a unit, which is a power of the golden ratio. If one of its parts is negative the exponent is negative, otherwise it's not. Then you simply can multiply or divide repeatedly by ϕ until you get 1 (or alternatively if b, the golden part of the unit, has absolute value > 1, there is a closed formula for the magnitude of the exponent – round(log(|b|*√5, ϕ)))."

Also, there are recent edits I made to the post with the factoring algorithm, which are not in the version you copied:

About the divisibility of the original number by feudal factors of split ordinary primes, I wrote "both of them cannot work because if they do then the number would divide the ordinary prime which is their product and we have already factored out all of the ordinary primes, so if this prime divides the norm multiple times then one of the feudal primes has to divide the norm the same number of times, and therefore if one of them succeeded the first time you can skip checking the other one and continue dividing by it, i.e. multiplying by the other one and dividing by the product, the same number of times that the corresponding ordinary prime divides the norm"

I also added this: "After you divided by all the primes, you get a unit, which is a power of the golden ratio. If one of its parts is negative the exponent is negative, otherwise it's not. Then you simply can multiply or divide repeatedly by ϕ until you get 1 (or alternatively if b, the golden part of the unit, has absolute value > 1, there is a closed formula for the magnitude of the exponent – round(log(|b|*√5, ϕ)))."

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

I started a new thread on Facebook with a post describing my algorithm(s) for factoring feudal numbers into primes: https://www.facebook.com/groups/xenharm ... 837568088/