Tuning-Math Archive Section 2: 1925

Message: 1925 - Contents - Hide Contents Date: Fri, 02 Nov 2001 02:37:04 Subject: Re: Tribonacci scales From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:

> You could start with this--if we replace the steps of a scale with > steps A,B,C by a scale with steps a,b,c, such that A=c, B=a+c and > C=b+c, we plainly will get the old scale as a subscale of the new > scale.

Well, I'm glad I read that right! Now, in the 2-step-size case, consistently putting the new small step _below_ the new large step, or consistently putting the new small step _above_ the new large step, when replacing the old large step, gets you MOS scales at every stage in the construction. In the 3-step-size case, do some or all such consistent ordering rules lead to scales that match the step-size-patterns of Dan's L-out- of-M-out-of-N scales, and/or your muddles?

Message: 1926 - Contents - Hide Contents Date: Fri, 02 Nov 2001 02:41:21 Subject: Re: Tribonacci scales From: Paul Erlich --- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> Hi Paul, > > Sorry. I'm sure you know what I mean, but I must be saying it in too > personal a way--fractional periodicity, scale index, etc. > > By fractional periodicity I just mean the period divided by the GCD of > the scale index (and in a Fibonacci scale I'd call the scale index A, > B, ... and in a Tribonacci scale it'd be A, B, C, ...),

The same A, B, and C that satisfy C/A = t, (A+B)/C = t? Or rather a different A, B, and C (those that "seed" the sequence?)

Message: 1929 - Contents - Hide Contents Date: Fri, 02 Nov 2001 04:16:50 Subject: Re: Scale step iterations From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> I'd think other, similar constructing rules could also make some > sense as ways of obtaining three-step-size scales? For the two-step- > size case, we discussed the Silver Mean case, with ratio sqrt(2)-1 > (IIRC).

That's what I was trying to do with my example. The polynomial (and associated number field) x^3-x^2-x-1 has discriminant -44. We have only two smaller discriminants in absolute value, -31 and -23. They are associated to the smallest and second smallest Pisot numbers (by a theorem of Siegel); x^3-x-1 of disciminant -23 gives us the smallest Pisot number, and x^3-x^2-1 of discriminant -31 the second smallest. These seem like the two most obvious characteristic polynomials to use to look for alternative schemes. They suggest we could for instance try the following replacement schemes: A->b, B->c, C->a+b and A->b, B->c, C->a+c.

Message: 1930 - Contents - Hide Contents Date: Fri, 02 Nov 2001 04:19:01 Subject: Re: Scale step iterations From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:

> That's what I was trying to do with my example. The polynomial (and > associated number field) x^3-x^2-x-1 has discriminant -44. We have > only two smaller discriminants in absolute value, -31 and -23. They > are associated to the smallest and second smallest Pisot numbers (by > a theorem of Siegel); x^3-x-1 of disciminant -23 gives us the > smallest Pisot number, and x^3-x^2-1 of discriminant -31 the second > smallest. These seem like the two most obvious characteristic > polynomials to use to look for alternative schemes. They suggest we > could for instance try the following replacement schemes: A->b, B- >c, > C->a+b and A->b, B->c, C->a+c.

Cool! What does the discriminant tell us?

Message: 1931 - Contents - Hide Contents Date: Fri, 02 Nov 2001 04:55:09 Subject: Re: Scale step iterations From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> Cool! What does the discriminant tell us?

Lots of things, but for our purposes the most significant seems to be that these are the smallest cubic fields in some sense. Just as the Golden Field Q(phi) has the smallest real quadratic field discriminant (at 5) and Q(sqrt(2)) the second smallest (at 8) and so supply the least complicated examples, the cubic fields of discriminants -23, -31 and -44 would seem to be the right place to start. The smallest totally real cubic field has discriminant 49, but this does not give us a Pisot number and that might be important. One polynomial for it has roots 2*cos(2^i pi/7) for i from 1 to 3, and is x^3+x^2-2*x-1.

Message: 1932 - Contents - Hide Contents Date: Fri, 02 Nov 2001 09:18:03 Subject: How to Tribonacci From: genewardsmith@xxxx.xxx The basis I got for the Tribonacci process seemed a little mysterious, but the mystery is solved. The transformation matrix is actually the transpose of the Tribonacci one, meaning the usual operation is on vals, and I was looking at a transposed one on [1,t^2-t,t]. The transformation matrix one gets from multiplication by t sends C->a, B->a+c, C->b+c, and is [0 0 1] [1 0 1] [0 1 1] If we take the diatonic scale Dan started with, we have intervals of 9/8, 10/9 and 16/15, with a matrix [ 4 -1 -1] [ 1 -2 1] [-3 2 0] The inverse of this is [2 2 3] [3 3 5] [4 5 7] If we set our "a" value so that a*(2+2*(t^2-t)+3*t) = 1, and then use the next two rows to calculate the corresponding approximations of 3 and 5, we get log_2(3) ~ a*(3 + 3*(t^2-t)+5*t) 2.1 cents sharp log_2(5) ~ a*(4 + 5*(t^2-t)+7*t) -3.5 cents flat If we now transform the above val matrix by the Tribonacci transformation, we get [2 2 3] [0 0 1] [2 3 7] [3 3 5] [1 0 1] = [3 5 11] [4 5 7] [0 1 1] [5 7 16] If we do the same calculation with this new val matrix we get the same approximation to 3 and 5 as before; instead of a Golden Meantone we are getting a Tribonacci 3 *and* 5. The sequence in question is 2 2 3 7 12 22 41 75 ... and the fifth is indeed the same fifth I started out by yowling about, but we need the Tribonacci Third also to define this thing.

Message: 1934 - Contents - Hide Contents Date: Fri, 02 Nov 2001 19:15:54 Subject: Re: How to Tribonacci From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> I'd like to be wrong though, so here's a couple for you to try: > > 2, 1, 4, ... > 1, 4, 2, ...

I need to get corresponding Tribonacci sequences for the 3 and the 5, as well as the 2, so I extend your first example to 2 1 4 7 12 23 ... 3 2 6 11 19 36 ... 4 3 9 16 28 53 ... If t is the Tribonacci constant, this gives me an octave multiplier of d = 2+(t^2-t)+4t = 2+3t+t^2, a 3 of (3 + 2(t^2-t)+6t)/d, 17 cents flat, and a 5 of (4+3(t^2-t)+9t)/d, 14 cents flat. Do you think you could do the other one, or is this not clear?

Message: 1935 - Contents - Hide Contents Date: Fri, 02 Nov 2001 20:36:48 Subject: Re: How to Tribonacci From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> Gene, > > If you lattice out the results you gave in 2D they create a 2, 3, 2, > ... scale. What I'm looking for are results that would lattice out to > either a 2, 1, 4, ... or a 1, 4, 2, ... scale.

If you start with a block with steps of size 27/25, 9/8, and 10/9, which corresponds to the first three steps I gave, and approximate using the approximate 3 and 5 I gave, you get 27/25:9/8:10/9 approximated by 1:(t^2-t):t. The 27/25 comes in at about 110 cents, 9/8 at about 170 cents, and 10/9 at about 202.5 cents (so they are effectively reversed.) The JI block is 1-10/9-6/5-4/3-3/2-5/3-9/5-(2), which has 2 27/25, 1 9/8, and 4 10/9. Since the 10/9 is largest, this is actually a 2, 1, 4 scale, which is what you wanted.

Message: 1936 - Contents - Hide Contents Date: Sat, 03 Nov 2001 02:12:09 Subject: A non-Tribonacci example From: genewardsmith@xxxx.xxx Let's see what can be done with the following recurrence: 6 1 2 7 3 9 10 12 19 22 31 41 53 72 94 ... 9 2 3 11 5 14 16 19 30 35 49 65 84 114 149 ... 14 2 5 16 7 21 23 28 44 51 72 95 123 167 218 ... The characteristic polynomial is x^3-x-1, and in terms of the power basis [1, x, x^2] the matrix corresponding to x is [0 0 1] m = [1 0 1] [0 1 0] if tm denotes the transpose of m, we want a transpose basis, in which multiplication by x acts like tm, not m. This means we want to find a matrix n such that n^(-1) m n = tm; and since the first basis element in both cases may as well be 1, we want the first row of n to be [1 0 0]. This determines n, and solving for it gives us [1 0 0] n = [0 0 1] [0 1 0] This means the transpose basis, corresponding to [1,t^2-t,t] in the case of Tribonacci, will be [1,x^2,x]. From the above we now have that the approximation to log_2(3) will be given by (9+28x^2+3*x)/(6+x^2+2*x) = (356-7*x+20x^2)/241 and to log_2(5) by (14+2*x^2+5*x)/(6+x^2+2*x) = (565+18*x-17*x^2)/241 This works out to a fifth flat by -0.75 cents and a third flat by -2.85 cents.

Message: 1939 - Contents - Hide Contents Date: Sat, 03 Nov 2001 06:58:15 Subject: Blocks for my example From: genewardsmith@xxxx.xxx Here are 9 and 10 note Fokker blocks to go with my example: 1-16/15-75/64-5/4-4/3-3/2-8/5-128/75-15/8-(2), with steps of 16/15-1125/1024-16/15-16/15-9/8-16/15-16/15-1125/1024-16/15 We have 6 16/15, 2 1125/1024 and 1 9/8. The 10-note block goes 1-16/15-256/225-5/4-4/3-45/32-3/2-8/5-225/128-15/8-(2), with steps of 16/15-16/15-1125/1024-16/15-135/128-16/15-16/15-1125/1024-16/15-16/15 This is 7 16/15, 2 1125/1024 and 1 135/128.

Message: 1942 - Contents - Hide Contents Date: Sun, 04 Nov 2001 07:05:58 Subject: Re: How to Tribonacci From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> If you take a Tribonacci like 1, 1, 4, ..., it's something that > shouldn't be based on 3s and 5s. Or rather there's nothing saying it > should (and from the looks of it I think it would probably be 7s and > 11s). Anyway, using your method, if I'm understanding correctly, > deciding the primes (or 2D planes) is arbitrary. Is this correct or am > I mistaken?

You have three generators, including the 2, and so the most obvious thing is to approximate 2,3 and 5. What would make more sense? I don't see why you would want to do 2,7 and 11 instead, as you seem to be suggesting.

> What would be ideal would be something akin to the 1D generator > achieved by way of adjacent fractions in a Fibonacci series that is > not tied into an arbitrary function--well calling that generator some > rational approximation is arbitrary, but that comes after the fact and > not before it.

We start off with certain scale steps, and nothing prevents us from using these as generators instead--it's a change of basis, but no more. You could, for instance, have approximate values for 9/8, 16/15 and 1125/1024 as your three generators, just as you could use 9/8 and 4/3 instead of 3/2 and 2 for the Golden case.

> Also, like the 1D Fibonacci case, I would think that successive scales > in a given series should create the same 2D "generators". Take the > prototypical 2, 2, 3, ... example--what's the Fokker block for 2, 3, > 7, ...? And how's that and the ones that follow related to the 3s and > 5s of the 2, 2, 3, ...?

This is determined by your transpose basis, which tells you how your intervals break up into smaller ones.

Message: 1943 - Contents - Hide Contents Date: Sun, 04 Nov 2001 23:55:08 Subject: Block recurrences From: genewardsmith@xxxx.xxx I was thinking of Dan's question, and something occurred to me which is interesting in its own right, as well as for this business, namely that the blocks themselves have recurrence relationships. If we start from the JI diatonic scale, with steps of size 16/15, 10/9 and 9/8, we can write the steps as a matrix and invert it: [ 4 -1 -1]^(-1) [2 2 3] [ 1 -2 1] = [3 3 5] [-3 2 0] [4 5 7] Applying the Tribonacci recurrence to the columns of the inverted matrix gives: 2 2 3 7 12 22 ... 3 3 5 11 19 35 ... 4 5 7 16 28 51 ... The starting matrix is unimodular, and the property is preserved by the transformation, which can be viewed as multiplication by a unimodular matrix. Hence each of the 3x3 matrices we get is unimodular, and each therefore defines a block. We have as inverse matricies ones which represent <16/15,10/9,9/8>, <25/24/135/128,16/15>, <81/80,128/125,25/24> ... and so forth. The rule to go from one to the next is <r1,r2,r3>--><r2/r1,r3/r1,r1>, which is the Tribonacci transpose operation. We get in this way 5- limit blocks with 7,12,22,41 ... notes to the octave, which approximate to the Tribonacci recurrence scales we get by starting from r1=1/d, r2=(t^2-t)/d, r3=t/d where d=2+2(t^2-t)+3t and applying the same rule. This can be regarded as a generalization of what we would get by inverting the Pythagorean pentatonic intervals of 256/243 and 9/8, obtaining [2 5] [3 8] and extending this to 2 5 7 12 19 31 50 ... 3 8 11 19 30 49 79 ... The Pythagorean scales are turned into meantone versions by using the meantone 3 of (8 phi + 3)/(5 phi + 2) = (19 - phi)/11. One can also attempt a generalization of the basis change from two vals to octave plus generator, which might be from three vals to 2,3 and generator, but this no longer is canonical, and the above seems more interesting.

Message: 1945 - Contents - Hide Contents Date: Sun, 04 Nov 2001 08:07:54 Subject: Re: How to Tribonacci From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> <<What would make more sense?>> > Again, what would make more sense would be something akin to the 1D > generator achieved by way of adjacent fractions in a Fibonacci series > that is not tied into an arbitrary function.

I'll take a shot at it, but meanwhile down below I show why I think this is already very much like the Golden Meantone.

> <<I don't see why you would want to do 2,7 and 11 instead, as you seem > to be suggesting.>> > You could see the 1, 1, 4, ... as a PB based on 2, 7, and 11 with UVs > of 353/343 and 121/112. Why? Say we think of this scale as having a > two stepsize cardinality at 1, 5, ..., the 1D generator would have > nothing (or nothing "most obvious") to do with a 3 or a 5. A 1D chain > of 7s would work if you've got to approximate it by a most likely > rational.

If you like that better, I suppose you could do it that way.

> <<This is determined by your transpose basis, which tells you how your > intervals break up into smaller ones.>> > > Okay, what exactly does the near 3/2 and near 5/4 of the 2, 2, 3, ... > have to do with the 2, 3, 7, ... (please use examples, I'm slow)?

The near 3/2 and near 5/4 has to do with more than just 2,3,7... The recurrence is really one on a 3D vector, just as what you call "the 1D case" (which actually has two dimensions) involves a recurrence on a 2D vector. If you wanted to work with some other vectors, they could have the same 2,3,7... even so. In other words, the Golden Meantone really is 7 12 19 31 50 ... 11 19 30 49 79 ... We don't pick the closest approximation to log_2(3), even when a better one becomes available, we stick with the one we get by taking the ratio of the above. We can write this as a sequence of mediants: med(11/7, 19/12) = 30/19, med(19/12, 30/19) = 49/31 ... The corresponding mediants (which might be generalized mediants) are easily found for other recurrences: med(3/2,3/2,5/3) = 11/7 med(4/2,5/2,7/3) = 16/7; then med(3/2,5/3,11/7) = 19/12 and med(5/2,7/2,16/7) = 28/12 and so forth. This is your 223 Tribonacci example. Just as we find the golden generators using [1, phi] we use [1,t^2-t,t] to get a weighted mediant, which tells us the limit of the sequence. The two situations seem quite adequately analogous to me.

Message: 1947 - Contents - Hide Contents Date: Mon, 05 Nov 2001 20:57:27 Subject: Re: Blocks for my example From: genewardsmith@xxxx.xxx

> 3 ~ 2^2 m s^(-2)

This should be 3 ~ 2^2 m^(-1) s^(-1) In matrix form, the map is [1 0 0] [2 -1 -1] [2 1 0] [3 0 -2] [4 -2 1] This sends the 11-limit to the <2,m,s> system.

Message: 1948 - Contents - Hide Contents Date: Mon, 05 Nov 2001 03:08:55 Subject: Re: A non-Tribonacci example From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:

> Let's see what can be done with the following recurrence: > > 6 1 2 7 3 9 10 12 19 22 31 41 53 72 94 ... > 9 2 3 11 5 14 16 19 30 35 49 65 84 114 149 ... > 14 2 5 16 7 21 23 28 44 51 72 95 123 167 218 ... > > The characteristic polynomial is x^3-x-1, and in terms of the power > basis [1, x, x^2] the matrix corresponding to x is > > [0 0 1] > m = [1 0 1] > [0 1 0] > > if tm denotes the transpose of m, we want a transpose basis, in which > multiplication by x acts like tm, not m. This means we want to find a > matrix n such that n^(-1) m n = tm; and since the first basis element > in both cases may as well be 1, we want the first row of n to be > [1 0 0]. This determines n, and solving for it gives us > > [1 0 0] > n = [0 0 1] > [0 1 0] > > This means the transpose basis, corresponding to [1,t^2-t,t] in the > case of Tribonacci, will be [1,x^2,x]. From the above we now have > that the approximation to log_2(3) will be given by > > (9+28x^2+3*x)/(6+x^2+2*x) = (356-7*x+20x^2)/241 > > and to log_2(5) by > > (14+2*x^2+5*x)/(6+x^2+2*x) = (565+18*x-17*x^2)/241 > > This works out to a fifth flat by -0.75 cents and a third flat by > -2.85 cents.

Kornerup eat your heart out! (Not really jumping into this right now, just observing from a distance . . .)