Yahoo Tuning Groups Ultimate Backup tuning-math Zeta Function Mysticism (was The Subgroup Naming Conundrum)

On Fri, Nov 4, 2011 at 4:14 AM, Keenan Pepper <keenanpepper@gmail.com> wrote:
>
> Random tantalizing question before I go to sleep:
>
> Is every Riemann zeta function peak somehow associated with a specific infinite val?
>
> Keenan

Yes, I think so. Some of this might be review for you, but I'll just
include it all for completeness' sake. Enjoy the lengthy read.

I'm not sure how this applies to the analytic continuation which uses
gamma and such, but the zeta function can be written as

zeta(s) = Prod_p 1/(1-p^(-s))

This can be written

zeta(s) = Prod_p D(P_p(s)/p^w)

Where
- P_p(s) is the "prime sinusoid," a phasor equal to exp^(-ln(p)*i*Im{s})
- w = Re{s} for whatever line we're looking at
- p^w is hence a weighting factor for the contribution of each prime
- D(x) is a "distortion pedal" which equals 1/(1-x), hence generating
"harmonics" for each prime sinusoid in the final plot

Let's derive some of this stuff, so you can see where I'm coming from.
If s=a+ib, then zeta(s) = Prod_p 1/(1-p^(-(a+ib)))

Let's take a look at that p^(-(a+ib)) in the denominator. I'll call
this the "prime sinusoid." It reduces to p^(-a+ib), which further
reduces to e^[-(a+ib)·ln(p)], which finally reduces to
e^(-ln(p)ib)/(p^a).

The numerator, e^(-ln(p)*ib), is a complex exponential sinusoid. Note
that the frequency increases proportionally to ln(p). You can think of
this sinusoid as existing in a space where the x-axis is "logarithmic
divisions of exp(2pi)," and the y-axis is something like "error." To
make it so we're subdividing 2/1 instead of exp(2pi), we can define a
conversion factor c = (2pi/ln(2)) * b. In fact, let's define P_p(c) =
e^(-ln(p)*ic) = e^(-2*pi*log_2(p)*i*b).

In short, I'm not going to write that awful looking exponential
anymore. I'm going to write P_p(c), and you should know that I mean a
complex exponential sinusoid of frequency log_2(p).

The prime sinusoid can then be written P_p(c)/(p^a). This is REALLY
interesting, because it means that each prime sinusoid ends up being
weighted by 1/p^a. So if you're looking at the critical line, that
means that the prime sinusoid becomes P_p(c)/sqrt(p). This means that
looking at the critical line gives all the primes sqrt(p) weighting.
If we instead wanted to give them 1/p weighting, which is more in
accordance with Tenney/Benedetti Height, we should instead look at the
Re{s} = 1 line, rather than the critical line.

OK, so each prime sinusoid is a weighted function tracing out a
sinusoid in EDO-space. But what do we do with these sinusoids? We want
to cover all integers, not just the primes. But we'd have to add a ton
of sinusoids to do that, right? Well, fear not: the shortcut to doing
this is to literally to add some "distortion" to the signal, so that
we end up getting "harmonic distortion" (powers of primes) and
"intermodulation distortion" (composite ratios of primes) as well.

The "distortion pedal" that the zeta function uses is 1/(1-x) which I
will call D(x): if you take a sine wave of amplitude 1, and run it
through a distortion pedal which applies this equation to each sample
coming in, you'll generate all harmonics at equal volume. If you apply
it to a sine wave at amplitude 1/2, you'll end up getting harmonics
with a 1/N rolloff, and if you apply it to a sine wave at amplitude
1/4, you'll end up getting harmonics with a 1/N^2 rolloff. This is
another way of saying that the Taylor series for 1/(1-x) is 1 + x +
x^2 + x^3 + ... Hence, if you're looking at the critical line, you end
up getting all the powers of primes p^n, each one weighted by
1/sqrt(p^n).

So now we have harmonic distortion for each prime sinusoid, which
generates sinusoids for powers of those primes as well and adds them
together. Thus, each distorted prime sinusoid is the sum of all the
prime sinusoid and the sinusoids of its powers - each one tracing out
a "goodness" curve in EDO-space.

So what do we do? Just add all these distorted prime sinusoids
together? Then we'll end up with a curve that adds all of these
sinusoids, so that when all the peaks are aligned, we get a peak, and
when all the troughs are aligned, we get a low. Sounds good, right -
but how will we get 15 this way? That's not a power of any prime, so
the summed output of sinusoids will also not be a power of any prime.

Actually, the problem is that we approached the problem of distortion
basically like the Allman Brothers did: add distortion over each note
separately, then play them all at the same time. You don't get any
"crunch!" The "crunch" that's missing is literally intermodulation
distortion, which is also what we're missing here: we need to multiply
a sinusoid of frequency log(5) with one of frequency log(3) to get one
of frequency log(5)+log(3) = log(15). We quite literally need to
generate "sum" and "difference" tones to create a sinusoid of
frequency 15.

So rather than adding all of these distorted prime sinusoids together,
we're going to MULTIPLY them all together, ring modulator style. You
convolve all of their individual spectra together, and blam: your
signal now consists of a sinusoid for every integer, each weighted at
1/sqrt(n) if you're looking at the critical line, all summed together.
When the peaks align, you get a zeta peak, when they don't align, you
get a trough. Or, in other words,

zeta(s) = Prod_p 1/(1-p^(-s)) = Sum_n 1/n^s

Bonus question: what if you want to sum over all the rationals instead
of all the integers? That'd be even more relevant to music theory. I
suspect the way to go about doing this is to replace P_p(c) =
e^(-2*pi*log_2(p)*i*b) with P_p(c) = cos(-2*pi*log_2(p)*i*b), so that
you get negative frequencies as well, meaning that the convolution
when you multiply them all together will give you some "undertones" as
well.

I'm not 100% sure how to generalize this to the analytic continuation
that Riemann used, which throws gamma in there somewhere. If anyone
sees how to do it, let me know.

-Mike

🔗Mike Battaglia <battaglia01@gmail.com>

On Fri, Nov 4, 2011 at 6:38 AM, Mike Battaglia <battaglia01@gmail.com> wrote:
> On Fri, Nov 4, 2011 at 4:14 AM, Keenan Pepper <keenanpepper@gmail.com> wrote:
>>
>> Random tantalizing question before I go to sleep:
>>
>> Is every Riemann zeta function peak somehow associated with a specific infinite val?
>>
>> Keenan
>
> Yes, I think so. Some of this might be review for you, but I'll just
> include it all for completeness' sake. Enjoy the lengthy read.

Funny, and after all that, I didn't answer your question. 6:38 AM, bleh.

So to answer your question - yes, there is a val associated with each
zeta peak, as well as with each point on the zeta function. The peaks
show up when a ton of peaks in the distorted prime sinusoids align.
These peaks will also line up with when the peaks in the original
prime sinusoids align.

So for any point in the zeta function, to find out what each prime is
mapped to - well, it's rather simple: you can find out what 2 is
mapped to because that's what the x-coordinate itself reflects. And
then you can find out what 3 is mapped to by reconfiguring the axes to
reflect ED3/1's instead of EDOs, and looking at the x coordinate
there. And then you can do it for 5, etc. In fact, you can just say
that the val v(p) for the peak of x-coordinate "q" that you want will
correspond to v(p) = q*log_2(p).

So now you have a val with a bunch of fractional entries. What do you
do? Your two options are:

1) Round them off, hence generating the w-limit patent val for your
peak. This corresponds to finding the "peak" for the prime sinusoid
you want which is nearest the coordinate you're looking at, and using
that. The nth peak can be viewed as n-EDp.
2) Work out TE error for the w-limit, and find the rounded-off
equation for v(p) which minimizes it. I'm not sure what this
corresponds to.

I went to all the trouble of giving a music-theoretic interpretation
for Euler's product formula to illuminate that the answer "just find
the patent val" actually has a zeta-related interpretation to it. But
that's the simple answer.

But hey, now that I have this music-theoretic interpretation all
worked out, can someone hook me up with a music-theoretic view of the
Riemann hypothesis? :)

-Mike

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> On Fri, Nov 4, 2011 at 6:38 AM, Mike Battaglia <battaglia01@...> wrote:
> > On Fri, Nov 4, 2011 at 4:14 AM, Keenan Pepper <keenanpepper@...> wrote:
> >>
> >> Random tantalizing question before I go to sleep:
> >>
> >> Is every Riemann zeta function peak somehow associated with a specific infinite val?
> >>
> >> Keenan
> >
> > Yes, I think so. Some of this might be review for you, but I'll just
> > include it all for completeness' sake. Enjoy the lengthy read.
>
> Funny, and after all that, I didn't answer your question. 6:38 AM, bleh.
>
> So to answer your question - yes, there is a val associated with each
> zeta peak, as well as with each point on the zeta function. The peaks
> show up when a ton of peaks in the distorted prime sinusoids align.
> These peaks will also line up with when the peaks in the original
> prime sinusoids align.

So, a more specific and interesting question that's closer to what I was actually thinking is this:

Is there a *one-to-one* correspondence between zeta peaks and a certain kind of infinite val that is somehow *optimal*?

The really vague picture I have in mind is this: Patent vals are optimal for mapping primes themselves. So the only reason we'd ever want to choose a different val is to improve the mapping of ratios between primes. We have a preference for vals that tune lots of primes the same direction (all flat or all sharp), because this improves the tuning of the ratios between primes. But if we're tuning most of the primes sharp, for example, that's equivalent to using a slightly smaller step for our rank-1 temperament (and tuning most of the primes to the nearest step). So this could be interpreted as a kind of "force" that pushes you toward smaller or larger rank-1 steps.

Perhaps the zeta peaks are related to stable "equilibrium points", where this force becomes zero and rank-1 temperaments from either side are pushed toward it. Zeta zeros could be related to unstable equilibria, and define the "basins of attraction".

Keenan

🔗Paul <phjelmstad@msn.com>

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@> wrote:
> >
> > On Fri, Nov 4, 2011 at 6:38 AM, Mike Battaglia <battaglia01@> wrote:
> > > On Fri, Nov 4, 2011 at 4:14 AM, Keenan Pepper <keenanpepper@> wrote:
> > >>
> > >> Random tantalizing question before I go to sleep:
> > >>
> > >> Is every Riemann zeta function peak somehow associated with a specific infinite val?
> > >>
> > >> Keenan
> > >
> > > Yes, I think so. Some of this might be review for you, but I'll just
> > > include it all for completeness' sake. Enjoy the lengthy read.
> >
> > Funny, and after all that, I didn't answer your question. 6:38 AM, bleh.
> >
> > So to answer your question - yes, there is a val associated with each
> > zeta peak, as well as with each point on the zeta function. The peaks
> > show up when a ton of peaks in the distorted prime sinusoids align.
> > These peaks will also line up with when the peaks in the original
> > prime sinusoids align.
>
> So, a more specific and interesting question that's closer to what I was actually thinking is this:
>
> Is there a *one-to-one* correspondence between zeta peaks and a certain kind of infinite val that is somehow *optimal*?
>
> The really vague picture I have in mind is this: Patent vals are optimal for mapping primes themselves. So the only reason we'd ever want to choose a different val is to improve the mapping of ratios between primes. We have a preference for vals that tune lots of primes the same direction (all flat or all sharp), because this improves the tuning of the ratios between primes. But if we're tuning most of the primes sharp, for example, that's equivalent to using a slightly smaller step for our rank-1 temperament (and tuning most of the primes to the nearest step). So this could be interpreted as a kind of "force" that pushes you toward smaller or larger rank-1 steps.
>
> Perhaps the zeta peaks are related to stable "equilibrium points", where this force becomes zero and rank-1 temperaments from either side are pushed toward it. Zeta zeros could be related to unstable equilibria, and define the "basins of attraction".
>
> Keenan
>

So, does the critical line (where the zeta zeroes are) define
"badness" or "goodness"?

🔗Keenan Pepper <keenanpepper@gmail.com>

🔗phjelmstad@msn.com

🔗Paul <phjelmstad@msn.com>

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul" <phjelmstad@> wrote:
> > So, does the critical line (where the zeta zeroes are) define
> > "badness" or "goodness"?
>
> I'm not sure what the question is. I was thinking of the critical line, but you could use any real part you wanted, I guess.
>
> Keenan
>

Here's the part on 'goodness' from xenharmonic wiki. I'm confused because i thought things got worse on the critical line, not better:

Into the critical strip
So long as s is greater than or equal to one, the absolute value of the zeta function can be seen as an error measurement. However, the rationale for that view of things departs when s is less than one, particularly in the critical strip, when s lies between zero and one. As s approaches the value s=1/2 of the critical line, the information content, so to speak, of the zeta function concerning higher primes increases and it behaves increasingly like a badness measure (or more correctly, since we have inverted it, like a goodness measure.) The quasi-symmetric functional equation of the zeta function tells us that past the critical line the information content starts to decrease again, with 1-s and s having the same information content. Hence it is the zeta function between s=1/2 and s=1, and especially the zeta function along the critical line s=1/2, which is of the most interest.

🔗Keenan Pepper <keenanpepper@gmail.com>

🔗phjelmstad@msn.com

🔗Keenan Pepper <keenanpepper@gmail.com>

🔗PAUL HJELMSTAD <phjelmstad@msn.com>

Thanks for the explanation of renormalizing zeta tuning on my latest question. I am a little unclear, (I will read all of Mike's post later) on what an infinite valactually is, but short of that, what do you mean, this "force" becomes zero at zeta peaks, I guess I don't understand what the force is, where it pushes rank-1 temperaments to the peaks, and why it would create smaller or larger rank-1 steps, based on not all primes being tuned sharp or flat (I do grasp patent val etc),
The trouble with this stuff is that understanding this and that but not all of it doesn't cut it, but I am getting closer.
To: tuning-math@yahoogroups.com
From: keenanpepper@gmail.com
Date: Fri, 4 Nov 2011 16:07:29 +0000
Subject: [tuning-math] Re: Zeta Function Mysticism (was The Subgroup Naming Conundrum)

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> On Fri, Nov 4, 2011 at 6:38 AM, Mike Battaglia <battaglia01@...> wrote:

> > On Fri, Nov 4, 2011 at 4:14 AM, Keenan Pepper <keenanpepper@...> wrote:

> >>

> >> Random tantalizing question before I go to sleep:

> >>

> >> Is every Riemann zeta function peak somehow associated with a specific infinite val?

> >>

> >> Keenan

> >

> > Yes, I think so. Some of this might be review for you, but I'll just

> > include it all for completeness' sake. Enjoy the lengthy read.

> Funny, and after all that, I didn't answer your question. 6:38 AM, bleh.

> So to answer your question - yes, there is a val associated with each

> zeta peak, as well as with each point on the zeta function. The peaks

> show up when a ton of peaks in the distorted prime sinusoids align.

> These peaks will also line up with when the peaks in the original

> prime sinusoids align.

So, a more specific and interesting question that's closer to what I was actually thinking is this:

Is there a *one-to-one* correspondence between zeta peaks and a certain kind of infinite val that is somehow *optimal*?

Keenan