Do I have the right idea about the zeta function?

On Tue, Jun 28, 2011 at 3:18 AM, Mike Battaglia <battaglia01@gmail.com> wrote:
>>
>> Then there's zeta tuning, which seems to be the right thing, but
>> I don't know how to express it as a weighting function so I
>> feel like I don't really understand it.
>
> I believe that Gene has said it weights the primes roughly equal to
> sqrt(n), or something like that, but also that it weights full
> integers and not just primes. So that's in line with Tenney height.
> Perhaps adapting it to something Graham's covered in composite.pdf
> would do the trick.

Actually, looking back at Keenan's post here:

/tuning-math/message/19140

Keenan wrote:
> If you want to include only a finite set of primes, then you don't even use your
> zeta function calculator at all. You just plot
>
> Prod_("primes" p to include) 1/(1-p^(-s))
>
> If you only include one prime you always get a periodic function, like this:
> https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B9CMyeCjAMQGZ\GI4Y2QyODQtNjA0Yi00MDUxLTk3MGYtNDk4MDY5OTMxOTY4&hl=en
> (Note that all these PDFs are vector plots, so you can zoom in to see more
> detail)

I think those p^(-s)'s are going to look like (complex) sinusoids
along the critical line, and the 1/(1-x) operator will basically turn
them into sawtooth waves that are phase shifted by 90 degrees, which
is what the waveform in Keenan's graph is (it looks like
UUUUUUUUUUUUU). Note very carefully how each peak has a "field of
attraction" around it, which in this case is this sort of tapered Vos
curve type shape you see here (the phase-shifted sawtooth wave):

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B9CMyeCjAMQGZGI4Y2QyODQtNjA0Yi00MDUxLTk3MGYtNDk4MDY5OTMxOTY4&hl=en

Then, when you add in another prime, you multiply it pointwise by the
first one, which generates this graph

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B9CMyeCjAMQGMzg0MmQ5MjgtN2E2OC00M2FkLTg4ZmYtZTg4ZGQzZjY4NWNm&hl=en

As you keep multiplying by more and more primes, you'll keep placing
selective pressure on the peaks in the graph that end up being in the
fields of attraction of each primes. The zeta function does this all
at once. Do I have the right idea here?

-Mike

On Tue, Jun 28, 2011 at 3:42 AM, Mike Battaglia <battaglia01@gmail.com> wrote:
>
> As you keep multiplying by more and more primes, you'll keep placing
> selective pressure on the peaks in the graph that end up being in the
> fields of attraction of each primes. The zeta function does this all
> at once. Do I have the right idea here?

Actually, and furthermore, if all of this is correct (which it looks
like it is), then the sqrt(n) weighting of the primes comes because
we're looking at the critical line of the zeta function, which is
Re=-0.5, right? Because then p^(-(-0.5+it)) = p^(0.5-it)=
p^(0.5)*p^(-it) = sqrt(p)*p^(-it), and that sqrt(p) term appears. So
if we wanted to weight primes by 1/p instead, rather than 1/sqrt(p),
would we look at the Re = -1 line instead of the Re = -0.5 line?

And if we really wanted to improve it further, rather than doing Prod
1/(1-p^(-s)), we'd replace the 1/(1-p^(-s)) functions with something
that more resembles the known psychoacoustic perception of tuning
error, such as a Gaussian-shaped field of attraction around each peak
to resemble the fields of attraction in harmonic entropy, for which
something like 1/(2-p^(-s))-1 might conceivably do the trick. Or we
could work out whatever shape that TE error uses to make it conform
with that.

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> On Tue, Jun 28, 2011 at 3:18 AM, Mike Battaglia <battaglia01@...> wrote:
> >>
> >> Then there's zeta tuning, which seems to be the right thing, but
> >> I don't know how to express it as a weighting function so I
> >> feel like I don't really understand it.
> >
> > I believe that Gene has said it weights the primes roughly equal to
> > sqrt(n), or something like that, but also that it weights full
> > integers and not just primes. So that's in line with Tenney height.
> > Perhaps adapting it to something Graham's covered in composite.pdf
> > would do the trick.
>
> Actually, looking back at Keenan's post here:
>
> /tuning-math/message/19140
>
> Keenan wrote:
> > If you want to include only a finite set of primes, then you don't even use your
> > zeta function calculator at all. You just plot
> >
> > Prod_("primes" p to include) 1/(1-p^(-s))
> >
> > If you only include one prime you always get a periodic function, like this:
> > https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B9CMyeCjAMQGZ\GI4Y2QyODQtNjA0Yi00MDUxLTk3MGYtNDk4MDY5OTMxOTY4&hl=en
> > (Note that all these PDFs are vector plots, so you can zoom in to see more
> > detail)
>
> I think those p^(-s)'s are going to look like (complex) sinusoids
> along the critical line, and the 1/(1-x) operator will basically turn
> them into sawtooth waves that are phase shifted by 90 degrees, which
> is what the waveform in Keenan's graph is (it looks like
> UUUUUUUUUUUUU). Note very carefully how each peak has a "field of
> attraction" around it, which in this case is this sort of tapered Vos
> curve type shape you see here (the phase-shifted sawtooth wave):
>
> https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B9CMyeCjAMQGZGI4Y2QyODQtNjA0Yi00MDUxLTk3MGYtNDk4MDY5OTMxOTY4&hl=en
>
> Then, when you add in another prime, you multiply it pointwise by the
> first one, which generates this graph
>
> https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B9CMyeCjAMQGMzg0MmQ5MjgtN2E2OC00M2FkLTg4ZmYtZTg4ZGQzZjY4NWNm&hl=en
>
> As you keep multiplying by more and more primes, you'll keep placing
> selective pressure on the peaks in the graph that end up being in the
> fields of attraction of each primes. The zeta function does this all
> at once. Do I have the right idea here?
>
> -Mike

What's a "VOS?" - thanks

The Vos curve is exp(abs(x)).

Sent from my iPhone

On Jun 28, 2011, at 11:48 AM, "Paul" <phjelmstad@msn.com> wrote:

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> On Tue, Jun 28, 2011 at 3:18 AM, Mike Battaglia <battaglia01@...> wrote:
> >>
> >> Then there's zeta tuning, which seems to be the right thing, but
> >> I don't know how to express it as a weighting function so I
> >> feel like I don't really understand it.
> >
> > I believe that Gene has said it weights the primes roughly equal to
> > sqrt(n), or something like that, but also that it weights full
> > integers and not just primes. So that's in line with Tenney height.
> > Perhaps adapting it to something Graham's covered in composite.pdf
> > would do the trick.
>
> Actually, looking back at Keenan's post here:
>
> /tuning-math/message/19140
>
> Keenan wrote:
> > If you want to include only a finite set of primes, then you don't even
use your
> > zeta function calculator at all. You just plot
> >
> > Prod_("primes" p to include) 1/(1-p^(-s))
> >
> > If you only include one prime you always get a periodic function, like
this:
> >
https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B9CMyeCjAMQGZ
\GI4Y2QyODQtNjA0Yi00MDUxLTk3MGYtNDk4MDY5OTMxOTY4&hl=en
> > (Note that all these PDFs are vector plots, so you can zoom in to see
more
> > detail)
>
> I think those p^(-s)'s are going to look like (complex) sinusoids
> along the critical line, and the 1/(1-x) operator will basically turn
> them into sawtooth waves that are phase shifted by 90 degrees, which
> is what the waveform in Keenan's graph is (it looks like
> UUUUUUUUUUUUU). Note very carefully how each peak has a "field of
> attraction" around it, which in this case is this sort of tapered Vos
> curve type shape you see here (the phase-shifted sawtooth wave):
>
>
https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B9CMyeCjAMQGZGI4Y2QyODQtNjA0Yi00MDUxLTk3MGYtNDk4MDY5OTMxOTY4&hl=en
>
> Then, when you add in another prime, you multiply it pointwise by the
> first one, which generates this graph
>
>
https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B9CMyeCjAMQGMzg0MmQ5MjgtN2E2OC00M2FkLTg4ZmYtZTg4ZGQzZjY4NWNm&hl=en
>
> As you keep multiplying by more and more primes, you'll keep placing
> selective pressure on the peaks in the graph that end up being in the
> fields of attraction of each primes. The zeta function does this all
> at once. Do I have the right idea here?
>
> -Mike

What's a "VOS?" - thanks