Dave KeenanAuthor
We need a UFD to make the whole monzo thing work. Right?
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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.
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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.
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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.
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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.
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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.
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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′.
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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.
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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.
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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 🙂
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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.
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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 . . .
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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.
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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.
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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.
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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.
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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.
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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.
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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
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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?
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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. 🙂
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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).
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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?
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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.
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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?
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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'?
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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?
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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?
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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.
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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?
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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".
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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′?
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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.
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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.
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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.
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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.
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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".
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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?
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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.
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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.
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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.
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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
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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.
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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.
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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
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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.
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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.]
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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?
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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.
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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.
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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.
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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).
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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?
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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.
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31 w
Mike BattagliaAdmin
There is no need to be rude, but thanks, I guess.
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31 w
Dave KeenanAuthor
Mike Battaglia I apologise.
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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".
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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.
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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.
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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.
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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).
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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
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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?
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31 w
Paul ErlichAdmin
(Also worth noting is that 62 = 31×2 shows up pretty strongly)
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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
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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 ...
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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.
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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.
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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.
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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.
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30 w
Steve Martin
Paul Erlich ah yes I see
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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)
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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
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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
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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
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30 w
Paul ErlichAdmin
Dave Keenangreat book!!!!!!!!
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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.
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30 wEdited
Paul ErlichAdmin
Dave Keenanit's deep in storage! 🤣
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30 w
Steve Martin
Yes, thank you!
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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
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30 w
Steve Martin
Thanks again, Paul, I'd not used that one before
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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
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30 w