Maximum Error in TOP

Hi!

How do I get the maximum Tenney-weighted error in a TOP temperament?
Is it the same as the maximum weighted error of the primes or some
other set of lattice-spanning intervals?

Kalle

--- In tuning-math@yahoogroups.com, "Kalle Aho" <kalleaho@m...> wrote:
> Hi!
>
> How do I get the maximum Tenney-weighted error in a TOP
temperament?
> Is it the same as the maximum weighted error of the primes

Yes.

> or some
> other set of lattice-spanning intervals?

That works too, I believe.

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
> --- In tuning-math@yahoogroups.com, "Kalle Aho" <kalleaho@m...>
wrote:
> > Hi!
> >
> > How do I get the maximum Tenney-weighted error in a TOP
> temperament?
> > Is it the same as the maximum weighted error of the primes
>
> Yes.

Thanks, Paul. Can you explain why this is so?

Kalle

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
> --- In tuning-math@yahoogroups.com, "Kalle Aho" <kalleaho@m...>
wrote:
> > --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
> > wrote:
> > > --- In tuning-math@yahoogroups.com, "Kalle Aho" <kalleaho@m...>
> > wrote:
> > > > Hi!
> > > >
> > > > How do I get the maximum Tenney-weighted error in a TOP
> > > temperament?
> > > > Is it the same as the maximum weighted error of the primes
> > >
> > > Yes.
> >
> > Thanks, Paul. Can you explain why this is so?
> >
> > Kalle

Whoops, I screwed up. Let me try again:

It follows from the prime factorization theorem.

The maximum Tenney-weighted error being T implies that

error(p)/log(p) <= T

so

error(p) <= T*log(p)

for all the primes p. If the factors in a chosen ratio are 2^a 3^b
5^c . . . (each exponent may either be positive or negative), then
the error of the chosen ratio cannot be greater than

T*(|a|*log(2) + |b|*log(3) + |c|*log(5) . . .)

since the errors in the primes that make up the chosen ratio, at
worst, add up without cancellation to the error in the chosen ratio.
The complexity of the ratio, meanwhile, is exactly

(|a|*log(2) + |b|*log(3) + |c|*log(5) . . .)

So the Tenney-weighted error in the ratio cannot be greater than the
second-to-last expression divided by the last expression, i.e., T.

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Kalle Aho" <kalleaho@m...>
> wrote:
> > > --- In tuning-math@yahoogroups.com, "Paul Erlich"
<perlich@a...>
> > > wrote:
> > > > --- In tuning-math@yahoogroups.com, "Kalle Aho"
<kalleaho@m...>
> > > wrote:
> > > > > Hi!
> > > > >
> > > > > How do I get the maximum Tenney-weighted error in a TOP
> > > > temperament?
> > > > > Is it the same as the maximum weighted error of the primes
> > > >
> > > > Yes.
> > >
> > > Thanks, Paul. Can you explain why this is so?
> > >
> > > Kalle
>
> Whoops, I screwed up. Let me try again:
>
> It follows from the prime factorization theorem.
>
> The maximum Tenney-weighted error being T implies that
>
> error(p)/log(p) <= T
>
> so
>
> error(p) <= T*log(p)
>
> for all the primes p. If the factors in a chosen ratio are 2^a
3^b
> 5^c . . . (each exponent may either be positive or negative),
then
> the error of the chosen ratio cannot be greater than
>
> T*(|a|*log(2) + |b|*log(3) + |c|*log(5) . . .)
>
> since the errors in the primes that make up the chosen ratio, at
> worst, add up without cancellation to the error in the chosen
ratio.
> The complexity of the ratio, meanwhile, is exactly
>
> (|a|*log(2) + |b|*log(3) + |c|*log(5) . . .)
>
> So the Tenney-weighted error in the ratio cannot be greater than
the
> second-to-last expression divided by the last expression, i.e., T.

Thanks, I understand it now!

Kalle