Tuning-Math Archive Section 2: 1902

Message: 1902 - Contents - Hide Contents Date: Wed, 31 Oct 2001 00:12:33 Subject: Re: Tribonacci From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:

> --- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote: >> Gene, >>

>> <<When you do that for 12/22,>> >> >> But I wrote N/M, so for a 7-out-of-12-out-of-22 that would be 22/12 > or

>> 0 1 3 5 7 9 11 12 14 16 18 20 22. Then the L-tet (7-tet here) inside >> of that would be 0 3 7 9 12 16 18 22. >> >> The idea is that this should work for any arbitrary A, B, C, ... >> Tribonacci series. >

> This is a scale out of the 7;7+5;12+10 muddle--why do you need to > bring in the Tribonacci constant at all?

Gene, take a deep breath. You're evidently missing something important even though it's being repeated (this has happened before), so you need to slow down and empty your mind of preconceptions, and give your genius room to function. Now take a look at the scales Dan posted (giving cents for all rotations) which were _not_ subsets of ETs at all. Notice how one scale is precisely a subset of the next scale, and each scale has _three_ step sizes. If I'm understanding Dan correctly (and I believe that I am), the relative sizes of the three step sizes are governed by the Tribonacci ratio -- in every iteration. This is similar to how the relative sizes of the two step sizes, in each of the proper MOSs of the Kornerup golden generator, or any other generator which is a noble (CF ends in all 1s) fraction of an octave, is exactly the Golden ratio. If one were to use a different constant from the Tribonacci ratio, and again assuming I'm understanding Dan correctly, one would see that the scale in one iteration would _not_ be precisely a subset of the scale in the next iteration -- just as if, in the sequence of meantone scales with 7, 12, 19, 31, 50, 81 . . . notes, if the ratio of small step to large step were not the Golden Ratio in each scale, then one scale would not be precisely a subset of the next scale in the sequence.

Message: 1903 - Contents - Hide Contents Date: Wed, 31 Oct 2001 00:13:28 Subject: Re: Tribonacci From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> --- In tuning-math@y..., genewardsmith@j... wrote:

>> I don't see that this has anything to do with MOS. If I take Dan at >> his word about the 7-12-22 business, it seems to me you just get >> 1/t = t^2-t-1 (where t is the Tribonacci constant) as a generator.

> Huh? How do you see a single generator operating here? These are > three-step-size scales, while a single generator would produce two- > step-size scales.

Not for a muddle--we could have a 7;7+5;1/t muddle, which would have three step sizes and which *would* be a Tribonacci limit, but it doesn't seem to be what Dan is talking about, since the scales don't correspond.

Message: 1904 - Contents - Hide Contents Date: Wed, 31 Oct 2001 00:16:40 Subject: Re: Tribonacci From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> Gene, take a deep breath. You're evidently missing something > important even though it's being repeated (this has happened before) ...

Sorry to be so frustrating, and thanks for the explanation. I may need things repeated again some time. :)

Message: 1907 - Contents - Hide Contents Date: Thu, 01 Nov 2001 20:23:10 Subject: Re: Scale step iterations From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:

> Similarly, suppose we have steps A of size 1, B of size t+1/t and C > of size t^2, where t^3-t^2-t-1=0.

That doesn't agree with what Dan said. He said that C/A = t, not t^2. I think maybe you just made a typo, Gene?

Message: 1910 - Contents - Hide Contents Date: Thu, 01 Nov 2001 03:27:03 Subject: Re: Tribonacci scales From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote: I was suggesting 0-8-15-19-27-35-42-46 in 46-equal, with step sizes 8748874 and Paul pointed out that this is the Indian Diatonic scale, which in JI is 9/8 10/9 16/15 9/8 9/8 10/9 16/15 in terms of step sizes. Of course, this has the same sizes of intervals as the Western diatonic scale. I got this from what I've been calling a "muddle", and they could be used to produce other 3-interval scales.

Message: 1911 - Contents - Hide Contents Date: Thu, 01 Nov 2001 05:41:47 Subject: Scale step iterations From: genewardsmith@xxxx.xxx Suppose you have a scale with step sizes proportional to 1 and phi, and you multiply by phi. You now have scales with steps of size phi and phi^2, or in other words phi and phi+1, which means a transformation of s small steps and L large steps by s->L, L->s+L; which can be expressed in terms of the matrix [1 1] [0 1] The inverse of this unimodular matrix is [0 1] [1 -1] This tells us to go to smaller step sizes, we send L->s, s->L-s. The characteristic polynomial of the first matrix is x^2-x-1, with root phi, and of the second matrix is x^2+x-1, with root 1/phi. Similarly, suppose we have steps A of size 1, B of size t+1/t and C of size t^2, where t^3-t^2-t-1=0. We can also write this as [1,t^2-1,t], and multiplying by t gives [t,t^2+1,t^2], which is to say [C,B+2A,B+A], so that the transformation matrix is [0 0 1] [2 1 0] [1 1 0] The characteristic polynomial of this is x^3-x^2-x-1, and its inverse is [0 1 -1] [0 -1 2] [1 0 0] with characteristic polynomial x^3+x^2+x-1, with root 1/t. It tells us that to go to smaller step sizes, we send A->B-C, B->2C-B, C->A. It would appear we can do this sort of thing in a lot of different ways.

Message: 1912 - Contents - Hide Contents Date: Thu, 01 Nov 2001 21:38:41 Subject: Re: Wavelet Dreams From: Paul Erlich Why not post this, or just your conclusions, to the tuning list, under the original topic heading?

Message: 1913 - Contents - Hide Contents Date: Thu, 01 Nov 2001 05:49:33 Subject: An example From: genewardsmith@xxxx.xxx Suppose r^3-r-1=0, and we have steps of size A, B, C in sizes proportional to 1:r:r^2. Then we can preserve this by sending A->C-A, B->A, C->B and get a larger scale.

Message: 1914 - Contents - Hide Contents Date: Thu, 01 Nov 2001 22:04:18 Subject: Re: Tribonacci scales From: Paul Erlich --- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> I'm still not sure... Having historical precedents like the syntonic > diatonic to use for comparison sake helps, and indeed the 2, 2, 3, ... > checks out favorably... but what are some other traditional/historical > three-stepsize scales besides the 2, 2, 3, ... that I could check > against? > > thanks, > > --Dan Stearns

Why not just create some others from other series -- e.g., 1, 1, 1, . . . 1, 1, 2, . . . 1, 2, 2, . . . 2, 3, 3, . . . and just examine and enjoy them for their own sake? Not sure what you want "historical precedent" for . . . especially when this process clearly ignores any frequency ratios more complex than 2/1, while the syntonic diatonic clearly doesn't!

Message: 1915 - Contents - Hide Contents Date: Thu, 01 Nov 2001 09:13:01 Subject: Golden basis From: genewardsmith@xxxx.xxx Someone (Paul?) mentioned that Kornerup wanted a sort of basis for everything based on phi--or at least, that's what I got out of it, after it passed through my weird filter. One sort of basis idea is that the algebraic integers Z[phi] are dense, so that you can approximate log2(3) etc as closely as you like; for instance we can approximate 3 by 2^(21-12 phi) and 5 by 2^(-98 + 62 phi). The ring of integers is preserved under multiplication by phi, which is a unit, so a shrinking and expanding process keeps us in the same sort of basis. I don't know if this is what Kornerup meant, of course.

Message: 1916 - Contents - Hide Contents Date: Thu, 01 Nov 2001 22:47:56 Subject: Re: Tribonacci scales From: Paul Erlich --- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> Hi Paul, > > You've misunderstood me. What I'm interested in checking against, just > for the sake of my own curiosity, is the ordering of the scale steps > (a vexing problem to generalize if you don't have something like a 1D > generator doing it for you), and not the properties relative JI (or > whatnot) of the scales themselves.

I think I understood that, but as Gene inadvertantly pointed out by bringing up the Indian diatonic, and which you can immediately see by comparing it with the syntonic diatonic, is that the ordering of scale steps can be contradictory from one historical example to another . . . so I don't see what kind of "checking" you're expecting to acheive here. Rather, it seems that your L-out-of-M-out-of-N method for determining the ordering, and perhaps variations where you rotate these ETs relative to one another by continuous amounts before doing the "out-of" operation, forms a paradigm unto itself, that perhaps a powerful mathematician (Gene?) can find some governing principles behind.

> > I had to check out dozens of different Tribonacci combinations just in > the process of making sure that things worked and were generalized (in > doing this I found that fractional periodicity doesn't work quite the > way I might've expected it would).

Can you expound on what you mean bt "fractional peridocity"? Also, can you verify that the generating rules, and other relationships, that Gene put forth in his latest post here are correct?

Message: 1920 - Contents - Hide Contents Date: Fri, 02 Nov 2001 01:54:50 Subject: Re: Scale step iterations From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> --- In tuning-math@y..., genewardsmith@j... wrote:

>> Similarly, suppose we have steps A of size 1, B of size t+1/t and C >> of size t^2, where t^3-t^2-t-1=0.

> That doesn't agree with what Dan said. He said that C/A = t, not t^2. > I think maybe you just made a typo, Gene?

Do I ever post anything without a mistake in it somewhere, I wonder? Dan set C/A = (A+C)/B = t, so that C = t A. If we set A to 1, C is t, and (1+t)/B = t, so B = (1+t)/t = 1+1/t = t^2-t. Then t * [1, 1+1/t, t] = [t, t+1, t^2] = [C, A+C, B+C], leading to a matrix of [0 0 1] [1 0 1] [0 1 1] with characteristic polynomial x^3-x^2-x-1 and inverse matrix [-1 1 0] [-1 0 1] [ 1 0 0] so that to go to smaller steps we transform by A->B-A, B->C-A, and C->A.

Message: 1921 - Contents - Hide Contents Date: Fri, 02 Nov 2001 01:55:31 Subject: Re: Tribonacci scales From: Paul Erlich --- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> <<Can you expound on what you mean bt "fractional peridocity"?>> > > I mean dividing the period by the GCD of A, B and C--but this does > work the way you'd expect it would (I was thinking of the 1D to 2D > conversions). > Still mystified.

Message: 1922 - Contents - Hide Contents Date: Fri, 02 Nov 2001 02:00:51 Subject: Re: Tribonacci scales From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> I'll have to take a better look at it.

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. Is this what you've been doing?

Message: 1923 - Contents - Hide Contents Date: Fri, 02 Nov 2001 02:09:45 Subject: Re: Scale step iterations From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:

> Dan set C/A = (A+C)/B = t, so that C = t A. If we set A to 1, C is t,

Ah . . . much better.

> and (1+t)/B = t, so B = (1+t)/t = 1+1/t = t^2-t. Then > t * [1, 1+1/t, t] = [t, t+1, t^2] = [C, A+C, B+C], leading to a > matrix of > > [0 0 1] > [1 0 1] > [0 1 1]

So the constructing rule is small(old) = Large(new) medium(old) = small(new) + Large(new) Large(old) = medium(new) + Large(new) Right? 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).