This is an Opt In Archive . We would like to hear from you if you want your posts included. For the contact address see About this archive. All posts are copyright (c).
- Contents - Hide Contents - Home - Section 10 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950
850 - 875 -
Message: 875 - Contents - Hide Contents Date: Mon, 27 Aug 2001 00:17:09 Subject: Re: Now I think "the hypothesis" is true :) From: genewardsmith@j... --- 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.
Message: 876 - Contents - Hide Contents 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
Message: 877 - Contents - Hide Contents Date: Mon, 27 Aug 2001 03:43:17 Subject: Re: 19-consistent goodness of measurement systems From: genewardsmith@j... --- 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.
Message: 878 - Contents - Hide Contents Date: Mon, 27 Aug 2001 03:55:35 Subject: Re: 19-consistent goodness of measurement systems From: genewardsmith@j... --- 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.
Message: 879 - Contents - Hide Contents
Date: Mon, 27 Aug 2001 08:46:46
Subject: The Riemann Zeta function and n-et's
From: genewardsmith@j...
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 * [with cont.] (Wayb.)
Message: 881 - Contents - Hide Contents Date: Mon, 27 Aug 2001 18:04:53 Subject: Re: 19-consistent goodness of measurement systems From: Carl>> >ene, 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.That's by definition, right? (that is, doesn't actually explain anything). -Carl
Message: 883 - Contents - Hide Contents 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.
Message: 884 - Contents - Hide Contents 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" . . .
Message: 885 - Contents - Hide Contents 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?
Message: 886 - Contents - Hide Contents 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.
Message: 888 - Contents - Hide Contents 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
Message: 889 - Contents - Hide Contents 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.
Message: 890 - Contents - Hide Contents 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
Message: 891 - Contents - Hide Contents Date: Mon, 27 Aug 2001 21:55:27 Subject: Re: Now I think "the hypothesis" is true :) From: Carl>> >he 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
Message: 892 - Contents - Hide Contents Date: Mon, 27 Aug 2001 22:39:57 Subject: Re: Now I think "the hypothesis" is true :) From: genewardsmith@j... --- 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?
Message: 893 - Contents - Hide Contents 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
Message: 894 - Contents - Hide Contents Date: Tue, 28 Aug 2001 01:22:15 Subject: Further examples of Zeta function tunings From: genewardsmith@j... This came up on the other list. A "Gram tuning" is a tuning derived from the Gram point associated to the large value of |Z(t)| assoicated to an et--in practice, this seems to mean the nearest Gram point. A "Z tuning" is the tuning derived from the actual value where |Z(t)| achieves its large local maximum near the n-et referred to before. Here are some further examples: 15 Gram point 45 Gram tuning = 15.052, 4.14 cents flat Z tuning = 15.053, 4.26 cents flat 19 Gram point 63 Gram tuning = 18.954, 2.93 cents sharp Z tuning = 18.948, 3.29 cents sharp 22 Gram point 78 Gram tuning = 22.025, 1.35 cents flat Z tuning = 22.025, 1.37 cents flat 72 Gram point 378 Gram tuning = 71.954, .763 cents sharp Z tuning = 71.951, .823 cents sharp It would be interesting to compare this tunings to those arrived at by other methods.
Message: 895 - Contents - Hide Contents Date: Tue, 28 Aug 2001 06:14:04 Subject: Re: Now I think "the hypothesis" is true :) From: genewardsmith@j... --- In tuning-math@y..., "Carl" <carl@l...> wrote:> --- 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.It seems to me that in a 12-et, a tritone would always be 6 steps. Can you clarify?
Message: 896 - Contents - Hide Contents Date: Tue, 28 Aug 2001 10:37 +0 Subject: More microtemperaments From: graham@m... I've altered my temperament finding program to accept only temperaments with a worst error of less than 2.8 cents. I think this is the cutoff for a microtemperament. Results are at <3 4 5 7 8 9 10 12 15 16 18 19 22 23 25 26 27 2... * [with cont.] (Wayb.)> <4 5 6 9 10 12 15 16 18 19 22 26 27 29 31 35 36... * [with cont.] (Wayb.)> <5 12 19 22 26 27 29 31 41 46 50 53 58 60 68 70... * [with cont.] (Wayb.)> <22 26 29 31 41 46 58 72 80 87 89 94 111 113 11... * [with cont.] (Wayb.)> <26 29 41 46 58 72 80 87 94 111 113 121 130 149... * [with cont.] (Wayb.)> <29 41 58 72 80 87 94 111 121 130 149 159 183 1... * [with cont.] (Wayb.)> Some of them don't have as many as 10 results. Graham
Message: 897 - Contents - Hide Contents Date: Tue, 28 Aug 2001 10:20:02 Subject: Re: More microtemperaments From: genewardsmith@j... --- In tuning-math@y..., graham@m... wrote:> I've altered my temperament finding program to accept only temperaments > with a worst error of less than 2.8 cents. I think this is the cutoff for > a microtemperament.Is there somewhere where the meaning of this results is documented? It's hard to tell what to think of an error less than 2.8 cents when one doesn't know what the error is a departure from, for instance.
Message: 898 - Contents - Hide Contents Date: Tue, 28 Aug 2001 11:29 +0 Subject: Re: More microtemperaments From: graham@m... In-Reply-To: <9mfr8i+di6b@e...> In article <9mfr8i+di6b@e...>, genewardsmith@j... () wrote:> Is there somewhere where the meaning of this results is documented? > It's hard to tell what to think of an error less than 2.8 cents when > one doesn't know what the error is a departure from, for instance.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. Graham
0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950
850 - 875 -