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Message: 4451 Date: Mon, 27 Aug 2001 00:17:09 Subject: Re: Now I think "the hypothesis" is true :) From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote: > What if we first stick with the case where the chromatic unison > vector is unchanged in size -- so in this case, 2/7-comma meantone. That would work, but it seems to me you are relying on the fact that the 7-et is relatively good. Suppose we take instead 512/375 as our chromatic unison vector, and 10/9 as our commatic unison vector. We then get the [7, 12, 17] system. If we temper out the 10/9 and keep the 512/375 value just, we want to solve the linear system a = 1200 a-2*b+c = 0 9*a-b-3*c = 1200 * log_2(512/375) = 512/375 expressed in cents. Solving this gives us a = 1200 b = 1980.12820 c = 2760.25604 The value for b is 77 cents flat from the 1200*(12/7) mark, well over the allowed amount; we would not have a MOS.
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Message: 4452 Date: Mon, 27 Aug 2001 18:05:46 Subject: Re: 19-consistent goodness of measurement systems From: Carl >I just tested the Maple zeta function routine, and it *does* seem >to work for complex arguments--I had thought otherwise. I'm off to >look at some graphs of et's, and will report anon. Cool! -C.
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Message: 4453 Date: Mon, 27 Aug 2001 02:17:26 Subject: Re: 19-consistent goodness of measurement systems From: Carl > Alas, 311 and 8539 are prime! Gene, does this have anything to do with the zeta function? I've long wondered why many good ETs are prime... 5, 7, 19, 31, 41, 53... notable exceptions being 12, 22, 34, 58, and 72... I know Paul's been asking for your zeta stuff; did I miss it? -Carl
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Message: 4454 Date: Mon, 27 Aug 2001 19:11:05 Subject: Re: Now I think "the hypothesis" is true :) From: Paul Erlich Whoops . . . I meant "at least one step size is larger than the chromatic unison vector" . . .
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Message: 4455 Date: Mon, 27 Aug 2001 03:43:17 Subject: Re: 19-consistent goodness of measurement systems From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "Carl" <carl@l...> wrote: > Gene, does this have anything to do with the zeta function? Only in the sense that the big spikes in the zeta function occur at good et's. The zeta function doesn't have the mojo to do this any better than we can do in a much less abstruse way in any case. > I know Paul's been asking for your zeta stuff; did I miss it? I'd like to find a good program for computing zeta(s+it) and Z(t) (which you might say is zeta(1/2+it) made into a real analytic function of a real variable.) I don't want to go back and write my own again! I think they may be availble for Mathematica, but I'd like to find a Maple version.
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Message: 4456 Date: Mon, 27 Aug 2001 19:10:12 Subject: Re: Now I think "the hypothesis" is true :) From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote: > --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote: > > > What if we first stick with the case where the chromatic unison > > vector is unchanged in size -- so in this case, 2/7-comma meantone. > > That would work, but it seems to me you are relying on the fact that > the 7-et is relatively good. Suppose we take instead 512/375 as our > chromatic unison vector, Whoa -- that's 539 cents! > and 10/9 as our commatic unison vector. We > then get the [7, 12, 17] system. If we temper out the 10/9 and keep > the 512/375 value just, we want to solve the linear system > > a = 1200 > a-2*b+c = 0 > 9*a-b-3*c = 1200 * log_2(512/375) = 512/375 expressed in cents. > > Solving this gives us > > a = 1200 > b = 1980.12820 > c = 2760.25604 > > The value for b is 77 cents flat from the 1200*(12/7) mark, well over > the allowed amount; we would not have a MOS. Carl brought this sort of thing up a while back (actually, he showed that some PBs are not CS, also using very large unison vectors). I replied that there needs to be some notion of "good" PBs. What's the weakest such condition we can come up with? If at least one step size is smaller than the chromatic unison vector . . . will that work?
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Message: 4457 Date: Mon, 27 Aug 2001 03:55:35 Subject: Re: 19-consistent goodness of measurement systems From: genewardsmith@xxxx.xxx --- In tuning-math@y..., genewardsmith@j... wrote: > I'd like to find a good program for computing zeta(s+it) and Z(t) > (which you might say is zeta(1/2+it) made into a real analytic > function of a real variable.) I don't want to go back and write my > own again! I just tested the Maple zeta function routine, and it *does* seem to work for complex arguments--I had thought otherwise. I'm off to look at some graphs of et's, and will report anon.
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Message: 4458 Date: Mon, 27 Aug 2001 19:16:33 Subject: Re: EDO consistency and accuracy tables (was: A little research...) From: Paul Erlich --- In tuning-math@y..., BobWendell@t... wrote: > Gene said: > > I think microtonalists should use whatever scale suits them; > however > > it would be well if they understood the structure of the system > they > > intend to use. > > Bob answers: > Thanks, Gene, for your complete comments and not just this quote from > it. Regarding the above quote from you, see my response in post #880 > to Paul Erlich and Paul's comments just below that response. > > I'm interested in 72-tET because it is reasonably accurate in its > approximation of just intervals, is strong in consistency, and > reduces the set of pitches available for composition to a finite > number that is conceptually manageable with simple arithmetic you can > do quickly in your head. > > Given time and experience with it, you could learn to think > comprehensively in terms of it and its implications while in the act > of composing. These criteria comprise the essential motivations for > my quest. Sounds great! If you look back a few months in the tuning list archives, you will see a great deal of discussion on 72-tET, and you'll see me advocating it as a certain sort of "standard" . . . I'm sure you'll find some useful information in all that.
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Message: 4459 Date: Mon, 27 Aug 2001 08:46:46 Subject: The Riemann Zeta function and n-et's From: genewardsmith@xxxx.xxx Let z = s + i*t be a complex number with real part s and imaginary part t. If s>1 we can define a function zeta(z) = sum_n n^(-z), where n runs over the positive integers. This function analytically continues over the whole complex plane, giving a meromorphic function with a single simple pole at z=1. The sum is absolutely convergent for s>1, and we may pass in this case to a corresponding infinite product zeta(z)= prod_p (1 - p^(-z))^(-1), where the product is taken over all primes p. This is the key to the zeta function--it encodes information about the primes. If we take logarithms, we get ln(zeta(z))= sum_p -ln(1-p^(-z)) = sum_p ln(1+p^z + p^(2z)+p^3z+...) The real part of this is the log of the absolute value Re(ln(zeta(z)) = ln(|zeta(z)|) = sum_p ln(|1-p^(-z)|). If we fix s and let t vary, then each term of the above sum becomes a periodic function of t with period 2 pi / ln(p). If we look at the first-order approximation, we have ln(1+p^(-z)) ~ p^(-z) = p^(-s) (cos(ln(p) t) - i sin(ln(p) t)), so that the real part of ln(zeta(z)) is approximately sum_p p^(-s) cos(ln(p) t) If we rescale by setting t = 2 pi n /ln(2), we get a sum weighted by factors of 1/p^s of cosine functions with periods of log_2(p). If n is a real number near a good et (e.g. n=12.018) then for the smallest primes, which have the highest weight, we will simultaneously be near a peak value of several of these cosine functions. The functions we are actually summing are also periodic but not exactly cosines; they in fact are an improvement, which take into account the powers of the prime p and so sharpen the peaks near their maximum, particularly for the smaller primes where we are especially concerned to accurately represent prime powers. Since the infinite product converges when s>=1, we are justified in taking values of t=2 pi n /ln(2) which give relatively high maximums for the fixed value of s as representing good scale divisions. As s goes to infinity, this becomes increasingly a matter of finding good values for the fifth, which we can much more easily accomplish via a continued fraction. More interesting are the cases with smaller values of s, as these give more weight to the larger primes and prime powers. It is a little harder to justify taking s into the critical strip between 0 and 1. However, the fact of analytic continuation and the Riemann-Siegel formula helps to make sense of this also, at least up to the critical line s=1/2; this is particularly the case if the Riemann hypothesis is true, in which case ln(zeta(z)) is analytic in the strip up to the critical line. We have a functional equation relating values of s to values of 1/2-s, so we don't want to push it past the critical line in any case. When we continue past s=1, we may follow a line of steepest ascent up to a high value of absolute value for zeta, and particularly when we get a good et which does well for a number of primes we might expect to reach a high maximum. We may adjust the zeta function along the critical line by setting Z(t) = zeta(1/2 + i t) pi^(-i t) Gamma(1/4 + i t/2)/|Gamma(1/4 + i t/2)| this makes Z into a real function of the real variable t, whose absolute value is the same as that of zeta(1/2+i t). Here Gamma is the Gamma function, where Gamma(z) = (z-1)! for complex z. We have an approximate formula for Z, Z(t) ~ 2 sum_{n <= L} cos(t ln(n) - theta(t))/sqrt(n) where L = sqrt(t / 2 pi) and theta has an asymptotic expression theta(t) ~~ (t/2) ln(t / 2 pi) - t/2 - pi/8 + 1/ (48 t) + 7/(5760 t^3) + ... which is extremely accurate for our purposes. We see that if theta(t) is a multiple of pi, we have again a weighted sum of cosines. Hence we are justified at looking at maximums of Z(t) particularly at "Gram points", which are points where theta(t) is an integer multiple of pi and so cos(theta(t))=+-1, sin(theta(t))=0. If we divide theta by pi and set r = t / 2 pi, we get f(r) = theta(t / 2 pi)/pi ~ r ln(r) - r - 1/8 and Gram points will be integer values of f(r). We can then use Newton's method to find the Gram point near a value of r where f(r) is close to an integer G. In this case, if we ignore the terms in inverse powers of r in the asymptotic formula we simply need iterate r' = (G + r + 1/8) / ln(r) If n is the number of steps in an n-et, then r = n/ln(2). For example, if n=12 then r=17.32... and f(r) = 31.927..., so we pick the Gram point where G=32. Applying Newton's method, we get an adjusted value r=17.337... which corresponds to n=12.017..., where octaves are flat by 1.764... cents. We may call such a tuning a Gram tuning, and it is interesting to consider what the flatness or sharpness of the Gram tuning for a given et is telling us, and whether they make practical sense. I found an amazing applet on the web, which will graph Z(t) for anyone who wants to investigate this. (Don't you just love Java?) Unfortunately, it isn't set up for microtonalists, but if you multiply your et n by the magic number 2 pi / ln(2), you get a value of t which corresponds to that et and around which it is interesting to consider the graph. Maybe if we ask nicely he will put up a version for music theorists; he might be interested by the interest! The url is: Riemann Hypothesis in a Nutshell *
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Message: 4461 Date: Mon, 27 Aug 2001 21:27:10 Subject: Re: Now I think "the hypothesis" is true :) From: Carl >/.../ What's the weakest such condition we can come up with? If >at least one step size is smaller than the chromatic unison >vector . . . will that work? Well, in this non-CS example from way-back, |4 -1| |0 2| = 8 1/1 135/128 9/8 5/4 45/32 3/2 27/16 15/8 The smallest 2nd is 135:128, and the chromatic UV is 32:25. >Whoops . . . I meant "at least one step size is larger than the >chromatic unison vector" . . . The largest 2nd is 9:8, so this condition is not met, though still not 100% on why the 81:80 can't be chromatic (isn't this just a matter of choice?). IIRC, we agreed way back that all steps being larger than all vectors would give CS, so this is indeed weaker. That's non-CS for you. Not exactly clear on what property Gene has found. -Carl
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Message: 4462 Date: Mon, 27 Aug 2001 21:37:51 Subject: Re: Now I think "the hypothesis" is true :) From: Paul Erlich --- In tuning-math@y..., "Carl" <carl@l...> wrote: > >/.../ What's the weakest such condition we can come up with? If > >at least one step size is smaller than the chromatic unison > >vector . . . will that work? > > Well, in this non-CS example from way-back, > > |4 -1| > |0 2| = 8 > 1/1 135/128 9/8 5/4 45/32 3/2 27/16 15/8 > > The smallest 2nd is 135:128, and the chromatic UV is 32:25. > > >Whoops . . . I meant "at least one step size is larger than the > >chromatic unison vector" . . . > > The largest 2nd is 9:8, so this condition is not met, Cool . . . so my proposal looks good so far. > though > still not 100% on why the 81:80 can't be chromatic (isn't this > just a matter of choice?). You haven't tempered any unison vectors out yet, so at this point, there's really no difference between commatic and chromatic. If you called the 81:80 chromatic and 25:16 commatic, you'd have to temper out the 25:16 first before looking at the step sizes.
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Message: 4463 Date: Mon, 27 Aug 2001 21:34:48 Subject: Re: Now I think "the hypothesis" is true :) From: Carl [I wrote...] >That's non-CS for you. Not exactly clear on what property >Gene has found. Sorry for the 3rd-person, Gene! [Gene wrote...] >The value for b is 77 cents flat from the 1200*(12/7) mark, well >over the allowed amount; we would not have a MOS. So it's MOS, then! -Carl
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Message: 4464 Date: Mon, 27 Aug 2001 21:55:27 Subject: Re: Now I think "the hypothesis" is true :) From: Carl >> The largest 2nd is 9:8, so this condition is not met, > > Cool . . . so my proposal looks good so far. Well, to me the key thing about the Hypothesis is that L-s is the chromatic vector. So if this proposal gives MOS, then the hypothesis is true. But it seems there are scales with more than two sizes of 2nd that meet the proposal condition... >>though still not 100% on why the 81:80 can't be chromatic (isn't >>this just a matter of choice?). > >You haven't tempered any unison vectors out yet, so at this point, >there's really no difference between commatic and chromatic. If you >called the 81:80 chromatic and 25:16 commatic, you'd have to temper >out the 25:16 first before looking at the step sizes. Oh, right. It blows my mind that this could work, or that anybody could think of it. -Carl
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Message: 4465 Date: Mon, 27 Aug 2001 22:39:57 Subject: Re: Now I think "the hypothesis" is true :) From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote: > Carl brought this sort of thing up a while back (actually, he showed > that some PBs are not CS, also using very large unison vectors). I > replied that there needs to be some notion of "good" PBs. What's the > weakest such condition we can come up with? If at least one step size > is smaller than the chromatic unison vector . . . will that work? I don't know what the weakest condition is, and I haven't thought about the one you propose above. I'm not sure what the point of it is- -a step of the 7-et is larger than 25/24, after all. However, the point of what I was sketching out before was that it is certainly possible to come up with conditions which make sense and suffice to produce a MOS. What's CS?
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Message: 4466 Date: Mon, 27 Aug 2001 22:55:23 Subject: Re: Now I think "the hypothesis" is true :) From: Carl --- In tuning-math@y..., genewardsmith@j... wrote: > What's CS? The property that every interval in a scale appears in only one interval class. For example, 3:2 appears only as a 5th in the diatonic scale... but in 12-tET, the tritone appears as both a 4th and a 5th, so the diatonic scale in 12-tET is non-CS. CS is our acronym for Constant Structures. ...You don't want to know. :) Actually, terms like MOS and CS come to us from Erv Wilson, who admits he has trouble naming the things he thinks about, and suggests we call them whatever we like. But, names seem to stick. -Carl
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Message: 4468 Date: Tue, 28 Aug 2001 16:52:16 Subject: Re: Now I think "the hypothesis" is true :) From: genewardsmith@xxxx.xxx --- In tuning-math@y..., BobWendell@t... wrote: > In Just Intonation, these pitches form intervals with C that are not > equal. Your definition said every interval appears in only one class-- however, these are two different intervals. Naturally, they will be represented as different in many systems, and will be the same only in those systems that has their ratio in the kernel.
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Message: 4469 Date: Tue, 28 Aug 2001 17:47:25 Subject: Re: Now I think "the hypothesis" is true :) From: Carl Lumma >>>What's CS? >> >>The property that every interval in a scale appears in only >>one interval class. For example, 3:2 appears only as a 5th >>in the diatonic scale... but in 12-tET, the tritone appears >>as both a 4th and a 5th, so the diatonic scale in 12-tET is >>non-CS. > >It seems to me that in a 12-et, a tritone would always be 6 steps. >Can you clarify? Scale steps. What recent threads have called "steps" are actually 2nds. Then, there's 3rds (major and minor), etc. Here's what Rothenberg calls a "interval matrix" for the diatonic scale, 1sts 2nds 3rds 4ths 5ths 6ths 7ths 8ths ionian 0 2 4 5 7 9 11 12 dorian 0 2 3 5 7 9 10 12 phrygian 0 1 3 5 7 8 10 12 lydian 0 2 4 [6] 7 9 11 12 mixolydian 0 2 4 5 7 9 10 12 aeolian 0 2 3 5 7 8 10 12 locrian 0 1 3 5 [6] 8 10 12 The tritone is what R. calls an "ambiguous interval". CS is equivalent to no ambiguous intervals. In meantone, the augmented 4th is smaller than the diminished 5th, so the meantone diatonic is CS. In Pythagorean tuning, the aug. 4th is larger than the dim. 5th, so it is also CS. But Rothenberg would call this scale "improper", since its scale steps overlap in interval space. When certain other conditions are met, R. claims proper scales make possible a compositional style in which melodies may be transposed across the modes of a scale without loosing their identity. -Carl
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Message: 4470 Date: Tue, 28 Aug 2001 17:53:55 Subject: Re: Now I think "the hypothesis" is true :) From: Carl Lumma >>In Just Intonation, these pitches form intervals with C that are >>not equal. > >Your definition said every interval appears in only one class-- >however, these are two different intervals. Naturally, they will be >represented as different in many systems, and will be the same only >in those systems that has their ratio in the kernel. Just for the record, that was Bob, not me. "Interval class" is just a bad way to say "scale step". "Every interval" is just a bad way to say "every acoustic interval". Does that help? See also my previous message in this thread. Gene, was it ever decided if a kernel is equivalent to a set of unison vectors, as we use them? -Carl
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Message: 4471 Date: Tue, 28 Aug 2001 18:38:08 Subject: Re: Now I think "the hypothesis" is true :) From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote: > --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote: > > > Carl brought this sort of thing up a while back (actually, he > showed > > that some PBs are not CS, also using very large unison vectors). I > > replied that there needs to be some notion of "good" PBs. What's > the > > weakest such condition we can come up with? If at least one step > size > > is smaller than the chromatic unison vector . . . will that work? > > I don't know what the weakest condition is, and I haven't thought > about the one you propose above. I'm not sure what the point of it is- > -a step of the 7-et is larger than 25/24, after all. You must have missed the message where I corrected this -- I meant "at least one step size is larger than the chromatic unison vector". > > However, the point of what I was sketching out before was that it is > certainly possible to come up with conditions which make sense and > suffice to produce a MOS. > > What's CS? Every specific interval size is always subtended by the same number of steps. Seems to be synonymous with "good" PBs in the untempered JI case.
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Message: 4472 Date: Tue, 28 Aug 2001 18:44:10 Subject: Re: Now I think "the hypothesis" is true :) From: Paul Erlich --- In tuning-math@y..., BobWendell@t... wrote: > In 12-tET, six steps can be used "enharmonically" to represent either > an augmented fourth or a diminished fifth. If we think in the key of > C, this implies that the sixth step above C can be used as either an > F# or a Gb. > > In Just Intonation, these pitches form intervals with C that are not > equal. In 12-tET, the irrational approximation of both intervals (sq > root of 2) lies between F# and Gb. On the other hand, in Just > Intonation F# is 45/32 of the frequency of C(1.40625*Fc)and Gb is > 36/25 (1.44)of C. In Just Intonation there are actually several possible ratios for F#, as well as for Gb. So for these purposes, a well-defined tuning such as Pythagorean or meantone would have been better.
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Message: 4473 Date: Tue, 28 Aug 2001 19:18:26 Subject: Re: More microtemperaments From: genewardsmith@xxxx.xxx --- In tuning-math@y..., graham@m... wrote: > The departure is from a JI odd limit. That is, all odd numbers up to the > one you pick are involved in ratios, and then you octave reduce. The > "mapping by steps" is your homomorphism. I think the best way to answer my questions may be to do some detective work and then ask about anything which remains unclear. I'm going therefore to take a look at the first 5-limit example, and add commentary. My understanding is that we want 3, 5 and 5/3 within 2.8 cents. If a is how sharp our "3" is, and b is how sharp the "5" is, then we want a and b in the hexagonal region determined by |a| <= 2.8, |b| <= 2.8, |a-b| <= 2.8 We need further conditions to determine the tuning, so let's look at what Graham does. "3 4 5 6 7 8 9 10 12 15 16 18 19 22 23 24 25 26 27 28" I don't know what these are. "5/19, 317.0 cent generator basis: (1.0, 0.26416041678685936)" If we set r = 0.26416041678685936, then 5/19 is a convergent for r. It's not clear why it is singled out; convergents for r are 1/3, 1/4, 4/15, 5/19, 9/34,14/53, ... "mapping by period and generator: [(1, 0), (0, 6), (1, 5)]" This seems to explain where r came from: if we send (a, b) to a + b*r, then (1,0) goes to log_2(2) = 1, (0,6) goes to 6*r which turns out to be log_2(3), and (1,5) goes to 1+5*r which is the approximation of log_2(5) we get when both octaves and fifths are exact and [-6,-5,6] is in the kernel. Hence, r = 3^(1/6). Is Graham's basic condition that all primes up to the last will be exactly represented? "mapping by steps: [(15, 4), (24, 6), (35, 9)]" It seems as if this may have something to do with the convergents to r. We have the 4-et [4, 6, 9] from the convergent 1/4 and the 15-et [15, 24, 35] from the convergent 4/15. We may then proceed to the others: [19, 30, 44] = [ 4, 6, 9] + [15, 24, 35] [34, 54, 79] = [15, 24, 35] + [19, 30, 44] [53, 84, 123] = [19, 30, 44] + [34, 54, 79] after which a slew of semiconvergents come in. Graham says this is "my homomorphism", but I'm getting a whole collection. "unison vectors: [[-6, -5, 6]]" 2^(-6)*3^(-5)*6^5 = 15625/15552 is the unison vector for anything using the matrix M = [0 1] [6 5] to approximate log_2(3) and log_2(5) using 1 and r' as a basis, so that [1, r']M = [6r', 1+5r'] and so 6 r' approximates log_2(3) and 1 + 5 r' approximates log_2(5)--the column vector V = [ 1 ] [ 6 r'] [1+5r'] has unison vectors generated by [-6, -5, 6]. "highest interval width: 6" How did we get to intervals and scales? "complexity measure: 6 (7 for smallest MOS)" How is this defined? "highest error: 0.001126 (1.351 cents)" 5/3 is off by this amount. "unique" What is unique? This system is so close to the 53-et that it would seem to make sense to adjust the fifth, the octave or both and make it exactly the 53-et.
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Message: 4474 Date: Tue, 28 Aug 2001 19:44:34 Subject: Re: Now I think "the hypothesis" is true :) From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "Carl Lumma" <carl@l...> wrote: > Gene, was it ever decided if a kernel is equivalent to a set > of unison vectors, as we use them? Here are some questions: (1) Is the unison a unison vector? (2) If q is a unison vector, is q^2 a unison vector? (3) Are products of unison vectors unique--that is, if we have unison vectors {v1, ... vn} and v1^e1 * ... * vn^en = q, are the exponents ei determined?
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