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Message: 7352 Date: Wed, 20 Aug 2003 23:04:48 Subject: Re: Cartesian product of two ET scales From: Paul Erlich excellent, carlos. not to mention the fact that balzano sneaks the 2:1 ratio in as the octave, contradicting his own purported independence from ratios. --- In tuning-math@xxxxxxxxxxx.xxxx Carlos <garciasuarez@y...> wrote:

> I understand and thanks for the observation. After the discussion,

is clear

> for me that this is a mere coincidence. For example ET31 is prime

and has not

> such a decomposition, yet the scale is a useful one. > > It would seem that Balazano found some aspects wich are specific of

12Et and

> too quickly he assumed, without a really good reason, that this had

to be

> generalized. This is now clear in the introductory statement to his

article,

> in which he seeks to generalized diatonic scales without regard to

any

> ratios. What for me is a mistake, based on several aspects,

including the

> history of the diatonic scales. > > Thanks > > Carlos > > > > > > > On Wednesday 20 August 2003 13:40, Graham Breed wrote:

> > hstraub64 wrote:

> > >Maybe one point worth to notice is that ET 12 is itself a direct > > >product of two cyclic groups, Z3 and Z4. Applications might be

found

> > >starting from there.

> > > > That's what Balzano did. > > > > > > > > > > > > To unsubscribe from this group, send an email to: > > tuning-math-unsubscribe@xxxxxxxxxxx.xxx > > > > > > > > Your use of Yahoo! Groups is subject to

Yahoo! Terms of Service * [with cont.] (Wayb.)

Message: 7354 Date: Wed, 20 Aug 2003 12:40:17 Subject: Re: Cartesian product of two ET scales From: Graham Breed hstraub64 wrote:

>Maybe one point worth to notice is that ET 12 is itself a direct >product of two cyclic groups, Z3 and Z4. Applications might be found >starting from there. > >

That's what Balzano did.

Message: 7355 Date: Thu, 21 Aug 2003 20:10:32 Subject: Re: Classificiation of musical scales From: Graham Breed Carlos wrote:

>Motivated by a recent dicussion in the group I have tried to provide a >comprehensive and clear classification of musical scales. > >I have collected the ideas in the attached short file. Writting all this in >the text of the email seemed a bit too much, besides I added a small figure >to clarify the ideas. The file is in PDF. > >

It's easier to get at, and takes up less bandwidth, if you quote it in the e-mail. I also saw something about Yahoo removing attachments from the archives.

>Comments would be most wellcome. > >

It may work as well as any other branching classification. I prefer to think of scales as having a number of orthogonal properties -- so the equivalence interval may be an octave, the scale may approximate some subset of just intonation, and so on. You're counting the dimensions according to the number of odd primes in the JI being apporoximated, are you? Like Lindley and Turner-Smith. Graham

Message: 7358 Date: Thu, 21 Aug 2003 21:13:23 Subject: Re: Classificiation of musical scales From: Graham Breed Carlos wrote:

>I understand the orthogonality in the sense that, just thinking about just >intervals as generators, each inteval is an indepent dimension and then a >particular interval in the interval space is a projection in each axis. > >Is this what you are meaning? > >

I mean "orthogonal" in the sense of the parameters being independent. So the Bohlen Pierce is a nonoctave just intonation scale. But there's nothing about it being just intonation that tells you it isn't likely to have the octave as an equivalence. However, your classification first puts it in a "nonoctave" category that means it has no connection with other just intonation scales.

>To be more precise, I guess one would have to define a scalar product type of >operation. I guess if you define the vectors in the usual sense >(first_JI_interval, second_JI_interval, ...) an so on and use the regular >product among vectors, but of course we are taking about vectorial space in >which the the components of the vector are number mod_N. I am not sure if >this arrays will have all the properties of a proper vector space. > >Is this the case? Is there a writting about this somewhere? > >

We have been talking about scalar products recently, although that's a different question. What we would end up with is a lattice (in the algebraic sense of the word) but I don't know much about them.

>For example in the 1/4 comma tunning, you have tempered fith such that 4 >firfhts equates a pure third. That would be one generator and one dimension. >Then you need another generator to complete you octave set. That would be >wolf fifth. > >

Ah, well, in group theory terms an octave equivalent meantone only has one generator -- the fifth. The wolf is simply one of the intervals that gets generated (by 11 fifths). So meantone would be one dimensional (or two dimensional if you include the octave as a generator).

>In this case we of all the product that you could generate in the interval 2d >space, for the actual scale you would take only points in the temperd fifht >axis and then only one point outside that axis that would be needed to >complete the cycle. > >Is this Ok with you? > >

You don't need the point outside the axis. Only the tempered fifth is a generator. Graham

Message: 7359 Date: Thu, 21 Aug 2003 20:21:14 Subject: Re: Classificiation of musical scales From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Carlos <garciasuarez@y...> wrote:

> Motivated by a recent dicussion in the group I have tried to

provide a

> comprehensive and clear classification of musical scales. > > I have collected the ideas in the attached short file. Writting all

this in

> the text of the email seemed a bit too much, besides I added a

small figure

> to clarify the ideas. The file is in PDF.

Yahoo has quit supporting file attachments; you might put this up in the files area so we all can see it.

Message: 7360 Date: Thu, 21 Aug 2003 20:49:31 Subject: Re: Classificiation of musical scales From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx Carlos <garciasuarez@y...> wrote:

> Motivated by a recent dicussion in the group I have tried to

provide a

> comprehensive and clear classification of musical scales. > > I have collected the ideas in the attached short file. Writting all

this in

> the text of the email seemed a bit too much, besides I added a

small figure

> to clarify the ideas. The file is in PDF. > > Comments would be most wellcome. > > Thanks > > Carlos

it says attachment not stored. you need to put it in the files folder or somewhere like that.

Message: 7362 Date: Fri, 22 Aug 2003 05:28:21 Subject: Re: Classificiation of musical scales From: Gene Ward Smith

> Thanks for discussion.

You've still not put your pdf file up under files, so many of us are locked out of the discussion. This is not the way to do things.

Message: 7366 Date: Sat, 30 Aug 2003 08:18:18 Subject: Linear recurrences for generators From: Gene Ward Smith If the characteristic polynomial of a linear recurrence has a single root with the maximum absolute value, and if this root is positive, we can use the corresponding linear recurrences much in the manner golden meantone. If the coefficients of the polynomial are confined to {-1,0,1} this is particularly nice. I find there are exactly 800 polynomials between degree two and eight which are irreducible and have the above property. I've uploaded a file to the files section listing the maximal root and the polynomial, with url Yahoo groups: /tuning-math/files/Gene/polly * [with cont.] We find for example that using a fourth as generator, and requiring our meantone forth to be between the 31 and 50 et fourth, gives us x^8-x^6-x^3-x^2-x+1 We get a meantone fifth for the same range from x^8-x^7-x^6+x^5-x^3-x^2+1 We can find a 2/secor between the 103 and 113 et values from x^8-x^7-x^6-x^5-x^3+x^2-x+1

Message: 7367 Date: Sat, 30 Aug 2003 23:21:11 Subject: Magic recurrence From: Gene Ward Smith Here's a recurrence example. If we take x^7-x^4-x-1 as a characteristic polynomial, we have 1.24564 as the large root, which can be used as a generator for magic. Corresponding linear recurrences are of the form C[n] = C[n-3] + C[n-6] + C[n-7] An example would be 3, 8, 9, 10, 15, 16, 22, 26, 33, 41, 51, 64, 79, 99, 123, 153, 191, 167, 248, 289, 324, 467, 570, 639, 882, 1107, 1252, 1673, 2144, 2461... If anyone can figure out an actual use for this stuff I would be interested.

Message: 7368 Date: Sun, 31 Aug 2003 10:28:17 Subject: Re: Magic recurrence From: Graham Breed I've found another meta-magic. To get equal beating for 8:9:10, 9-8 = 10-9 10+8 = 2*9 5 + 4 = 3**2 4x + 4 = x**10 This has a solution of 1.24548... or 0.3167...octaves (380.04 cents). The sequence is X[n] = 4(X[n-10] + X[n-9]) It converges, but very slowly. Here are some chunks -- there may be gaps 256, 256, 256, 256, 256, 256, 256, 512, 1216, 1856, 2048, 2048, 2048, 2048, 2048, 2048, 3072, 6912, 12288, 15616 16384, 16384, 16384, 16384, 16384, 20480, 39936, 76800, 111616, 128000 I think both magic and meantone work for JI-like equivalence because each prime is a positive number of generators. So the RHS terms are always positive. Graham

Message: 7369 Date: Sun, 31 Aug 2003 03:59:33 Subject: Another recurrence table From: Gene Ward Smith Here is another table to go with the previous one: Yahoo groups: /tuning-math/files/Gene/pil * [with cont.] I now only go up to degree four, but allow coefficients of absolute value <= 2. The Wilson fifth appears; I note it is next to a superpythagorean fifth with polynomial -1-2*x+x^2-x^3+x^4. Both have associated linear recurrences; you are invited to examine the table and see what else you find: 1.220744 -1-x+x^4 1.272020 -1-x^2+x^4 1.290649 -1-x+x^2-x^3+x^4 1.308571 -2+x-x^3+x^4 1.324718 -1-x+x^3 1.353210 -2+x-x^2+x^3 1.380278 -1-x^3+x^4 1.386471 1-2*x-x^2+x^4 1.395337 -1-2*x+x^4 1.431683 -1-x+2*x^2-2*x^3+x^4 1.446857 -2-x+x^2-x^3+x^4 1.465571 -1-x^2+x^3 1.484028 -2+x+x^2-2*x^3+x^4 1.490216 1-x-2*x^2+x^4 1.494530 -2-2*x+x^4 1.507982 -1-2*x+x^2-x^3+x^4 1.512876 -1+x-x^2-x^3+x^4 1.521380 -2-x+x^3 1.543689 -2+2*x-2*x^2+x^3 1.553774 -1-2*x^2+x^4 1.558980 1-2*x-x^3+x^4 1.566383 -2-x-x^2+x^4 1.569175 2-2*x-2*x^2+x^4 1.618034 -1-x+x^2 1.652892 -2-2*x^2+x^4 1.669798 -2+x-x^2-x^3+x^4 1.683772 1-2*x-2*x^2+x^4 1.690284 -1+2*x-2*x^2-x^3+x^4 1.695621 -2-x^2+x^3 1.710644 -1-x-2*x^2+x^4 1.712985 -1-2*x+2*x^2-2*x^3+x^4 1.716673 -2+2*x-2*x^3+x^4 1.722084 1-x-x^2-x^3+x^4 1.725340 -2-x-x^3+x^4 1.754878 -1+x-2*x^2+x^3 1.769292 -2-2*x+x^3 1.785371 -2-x-2*x^2+x^4 1.790044 -2+x^2-2*x^3+x^4 1.794310 -1-2*x-x^3+x^4 1.801938 1-2*x-x^2+x^3 1.821049 2-2*x-x^2-x^3+x^4 1.824620 -2+2*x-2*x^2-x^3+x^4 1.830415 -2-2*x+2*x^2-2*x^3+x^4 1.839287 -1-x-x^2+x^3 1.853560 -2-x^2-x^3+x^4 1.866760 -1+x-2*x^3+x^4 1.873709 -2-2*x-x^3+x^4 1.883204 1-2*x+x^2-2*x^3+x^4 1.897179 -1-x+x^2-2*x^3+x^4 1.899321 -2-2*x-2*x^2+x^4 1.905166 1-2*x^2-x^3+x^4 1.914390 -1+x-2*x^2-x^3+x^4 1.921290 1-2*x-x^2-x^3+x^4 1.927562 -1-x-x^2-x^3+x^4 2.065995 -1-2*x-x^2-x^3+x^4 2.075483 -1-2*x^2-x^3+x^4 2.081019 1-x-2*x^2-x^3+x^4 2.089219 -1-2*x+x^2-2*x^3+x^4 2.106919 -1-2*x^3+x^4 2.117689 1-x-2*x^3+x^4 2.123408 -2-2*x-x^2-x^3+x^4 2.132242 -1+2*x-x^2-2*x^3+x^4 2.139219 -2-2*x^2-x^3+x^4 2.147899 -1-2*x-x^2+x^3 2.157509 2-2*x-2*x^2-x^3+x^4 2.162858 -2-2*x+x^2-2*x^3+x^4 2.190328 -2-2*x^3+x^4 2.205569 -1-2*x^2+x^3 2.217157 1-2*x-2*x^2-x^3+x^4 2.222694 2-2*x-2*x^3+x^4 2.226862 -2+2*x-x^2-2*x^3+x^4 2.246980 1-x-2*x^2+x^3 2.256663 -2-x-2*x^2-x^3+x^4 2.269531 -2-2*x-x^2+x^3 2.277452 -1-x-2*x^3+x^4 2.296630 1-2*x-2*x^3+x^4 2.316512 -1-2*x-2*x^2-x^3+x^4 2.332190 2+2*x-2*x^2-2*x^3+x^4 2.339057 -2-x-2*x^3+x^4 2.348594 1-x^2-2*x^3+x^4 2.359304 -2-2*x^2+x^3 2.389870 -2+x-x^2-2*x^3+x^4 2.414214 -1-2*x+x^2 2.440110 2-x-x^2-2*x^3+x^4 2.463379 -2-2*x-2*x^3+x^4 2.470979 -1-x^2-2*x^3+x^4 2.481194 2-2*x-2*x^2+x^3 2.496698 1-x-x^2-2*x^3+x^4 2.538616 -1+2*x-2*x^2-2*x^3+x^4 2.546818 -1-x-2*x^2+x^3 2.568115 1+x-2*x^2-2*x^3+x^4 2.573272 2-2*x-x^2-2*x^3+x^4 2.589308 -2+2*x-2*x^2-2*x^3+x^4 2.592053 -1-x-x^2-2*x^3+x^4 2.633439 -2-x-x^2-2*x^3+x^4 2.647632 2-2*x^2-2*x^3+x^4 2.658967 -2-x-2*x^2+x^3 2.662978 -1+x-2*x^2-2*x^3+x^4 2.691740 1-2*x^2-2*x^3+x^4 2.696797 -1-2*x-x^2-2*x^3+x^4 2.704028 -2+x-2*x^2-2*x^3+x^4 2.732051 -2-2*x+x^2 2.760627 2-x-2*x^2-2*x^3+x^4 2.797136 1-x-2*x^2-2*x^3+x^4 2.803989 -2-2*x^2-2*x^3+x^4 2.831177 -1-2*x-2*x^2+x^3 2.858732 2-2*x-2*x^2-2*x^3+x^4 2.863131 -1-x-2*x^2-2*x^3+x^4 2.890054 1-2*x-2*x^2-2*x^3+x^4 2.919640 -2-2*x-2*x^2+x^3 2.947712 -1-2*x-2*x^2-2*x^3+x^4 2.974449 -2-2*x-2*x^2-2*x^3+x^4

Message: 7370 Date: Sun, 31 Aug 2003 22:57:15 Subject: x^4 - x^2 - 1 From: Gene Ward Smith The subject line is a polynomial whose big root is a sharp 11/8, and a PV number. It works for the 11-limit linear temperament whose wedgie is [12, 5, -9, 1, -20, -48, -40, -35, -15, 34] and whose "icon", or prime mapping, is given by [[1, -4, 0, 7, 3], [0, 12, 5, -9, 1]] and whose TM reduced basis is <126/125, 176/175, 1344/1331> I assume the point is to use recurrence values to construct a scale, which opens the question of what an intelligent way would be for doing that.

Message: 7371 Date: Sun, 31 Aug 2003 09:57:51 Subject: Re: Magic recurrence From: Graham Breed Gene Ward Smith wrote:

>If anyone can figure out an actual use for this stuff I would be >interested. > >

The original motivation was for difference tones to match, but the patterns would be so complex for what you're looking at I don't think that's relevant. Another approach would be to keep to three non-zero coefficients (two on the RHS) that are +/- powers of two. That means first order difference tones between octave equivalents will match. One case would be quartics with RHS coefficients from {-8, -4, -2, -1, 1, 2, 4, 8}. If they're of the form X[n] = A*X[n-alpha] + B*X[n-beta] where 0 < alpha < beta <= 4, the choices of alpha and beta are 4C2 or 6. For each of A and B thre are 8 choices. So the total number of series will be 6*8**2 = 384. So it would be possible to check all of them, but I haven't. Some won't work. Expand it to quintics, and allow +/-16, there'll be 1000 different options. I've come up with a meta-slendric. It keeps the equivalences from JI 4 + 3 = 7 8-7 = 7-6 7-6 = 4-3 8-7 = 4-3 The equation is x**4 = 4-2x, and the root is 1.4390... or 0.1936... octaves. The trouble is, the sequence X[n] = 4X[n-4] - 2X[n-3] doesn'tmake much sense, and I don't think it converges. I don't really know what to do about this. The problem is the -2, so it goes up and down. If, above, you made A and B positive it'd make more sense, and that seems to be what you're doing with your examples. But it would also lose some useful scales like this one. You could re-write it x**4 = 2x+4, and take a negative root, but I don't think that'd solve the problem. Graham

Message: 7372 Date: Sun, 31 Aug 2003 22:14:54 Subject: Graham-style recurrence polynomials, degree four From: Gene Ward Smith Here they are for degree four. I took out everything where the large root was a power of two, but I might have well have removed all the ones where it was an integer, since they look pretty useless. 1.220744 x^4-x-1 1.324718 x^4-x^2-x 1.353210 x^4-x-2 1.380278 x^4-x^3-1 1.395337 x^4-2*x-1 1.465571 x^4-x^3-x 1.494530 x^4-2*x-2 1.521380 x^4-x^2-2*x 1.533751 x^4-x-4 1.543689 x^4-x^3-2 1.618034 x^4-2*x^2-x 1.618034 x^4-x^3-x^2 1.642935 x^4-2*x-4 1.663252 x^4-4*x-1 1.695621 x^4-x^3-2*x 1.727754 x^4-4*x-2 1.748403 x^4-x^3-4 1.767868 x^4-x-8 1.769292 x^4-2*x^2-2*x 1.796322 x^4-x^2-4*x 1.835087 x^4-4*x-4 1.849421 x^4-2*x-8 2.040042 x^4-8*x-1 2.077216 x^4-8*x-2 2.106919 x^4-2*x^3-1 2.114908 x^4-4*x^2-x 2.144700 x^4-8*x-4 2.166313 x^4-x^2-8*x 2.190328 x^4-2*x^3-2 2.205569 x^4-2*x^3-x 2.214320 x^4-4*x^2-2*x 2.259801 x^4-8*x-8 2.320233 x^4-2*x^3-4 2.330746 x^4-2*x^2-8*x 2.359304 x^4-2*x^3-2*x 2.382976 x^4-4*x^2-4*x 2.394859 x^4-x^3-8*x 2.414214 x^4-2*x^3-x^2 2.507450 x^4-2*x^3-8 2.561553 x^4-x^3-4*x^2 2.594313 x^4-2*x^3-4*x 2.649436 x^4-4*x^2-8*x 2.732051 x^4-2*x^3-2*x^2 2.888969 x^4-8*x^2-x 2.931142 x^4-2*x^3-8*x 2.945995 x^4-8*x^2-2*x 3.000000 x^4-3*x^3 3.051374 x^4-8*x^2-4*x 3.236068 x^4-2x^3-4*x^2 3.236068 x^4-8*x^2-8*x 3.372281 x^4-x^3-8*x^2 3.414214 x^4-4*x^3+2*x^2 3.709275 x^4-4*x^3+4*x 3.732051 x^4-4*x^3+x^2 3.861009 x^4-4*x^3+8 3.866198 x^4-4*x^3+2*x 3.934317 x^4-4*x^3+4 3.935432 x^4-4*x^3+x 3.967988 x^4-4*x^3+2 3.984188 x^4-4*x^3+1 4.015445 x^4-4*x^3-1 4.030545 x^4-4*x^3-2 4.059780 x^4-4*x^3-4 4.060647 x^4-4*x^3-x 4.114825 x^4-4*x^3-8 4.117942 x^4-4*x^3-2*x 4.224170 x^4-4*x^3-4*x 4.236068 x^4-4*x^3-x^2 4.411139 x^4-4*x^3-8*x 4.449490 x^4-4*x^3-2*x^2 4.828427 x^4-4*x^3-4*x^2 5.000000 x^4-5*x^3 5.464102 x^4-4*x^3-8*x^2 6.000000 x^4-6*x^3 6.828427 x^4-8*x^3+8*x^2 7.000000 x^4-7*x^3 7.464102 x^4-8*x^3+4*x^2 7.741657 x^4-8*x^3+2*x^2 7.870865 x^4-8*x^3+8*x 7.872983 x^4-8*x^3+x^2 7.936496 x^4-8*x^3+4*x 7.968502 x^4-8*x^3+2*x 7.984283 x^4-8*x^3+8 7.984314 x^4-8*x^3+x 7.992164 x^4-8*x^3+4 7.996088 x^4-8*x^3+2 7.998045 x^4-8*x^3+1 8.001952 x^4-8*x^3-1 8.003901 x^4-8*x^3-2 8.007790 x^4-8*x^3-4 8.015534 x^4-8*x^3-8 8.015564 x^4-8*x^3-x 8.031009 x^4-8*x^3-2*x 8.061549 x^4-8*x^3-4*x 8.121294 x^4-8*x^3-8*x 8.123106 x^4-8*x^3-x^2 8.242641 x^4-8*x^3-2*x^2 8.472136 x^4-8*x^3-4*x^2 8.898979 x^4-8*x^3-8*x^2 9.000000 x^4-9*x^3 10.000000 x^4-10*x^3 12.000000 x^4-12*x^3

Message: 7373 Date: Sun, 31 Aug 2003 09:28:44 Subject: Re: Magic recurrence From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:

> The original motivation was for difference tones to match, but the > patterns would be so complex for what you're looking at I don't think > that's relevant. Another approach would be to keep to three non-zero > coefficients (two on the RHS) that are +/- powers of two. That means > first order difference tones between octave equivalents will match.

Why do we want the difference tones to match?

> I've come up with a meta-slendric.

I've got a nice recurrence for tritonic, the 7-limit temperament with [5, -11, -12, -29, -33, 3] as a wedgie. What's nice about it is that the big root is a PV number, meaning all of its conjugates are less than one in absolute value, so that the ratio converges quickly. The polynomial is x^4-2*x^3-2*x^2-x+1, so the recurrence goes c[n] = 2 c[n-1] + 2 c[n-2] + c[n-3] - c[n-4] It doesn't have your RHS property, but I could search for that.

Message: 7374 Date: Sun, 31 Aug 2003 09:43:11 Subject: Re: Magic recurrence From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:

> Why do we want the difference tones to match?

I think I just figured this out. Cool!

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