Yahoo Tuning Groups Ultimate Backup tuning-math Is the TE tuning basis-invariant?

No the TE error doesn't depend on the basis

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Sent from my Nokia phone

------Original message------
From: Mike Battaglia <battaglia01@gmail.com>
To: <tuning-math@yahoogroups.com>
Date: Tuesday, January 24, 2012 4:58:39 AM GMT-0500
Subject: [tuning-math] Is the TE tuning basis-invariant?

I spent a lot of time working with TE error tonight, when I noticed
that the TE error of a temperament depends on what basis you pick. Is
the same thing then true for the TE tuning of a temperament as well?
It would seem to be, and furthermore this would appear to be the case
for any norm you apply on the lattice except for L1 - is this correct?

-Mike

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Yahoo! Groups Links

🔗genewardsmith <genewardsmith@sbcglobal.net>

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> I spent a lot of time working with TE error tonight, when I noticed
> that the TE error of a temperament depends on what basis you pick. Is
> the same thing then true for the TE tuning of a temperament as well?

TE and TOP are prime-based, and the whole thing doesn't work so well if you use a non-prime-power basis in some subgroup.

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
>
>
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@> wrote:
> >
> > I spent a lot of time working with TE error tonight, when I noticed
> > that the TE error of a temperament depends on what basis you pick. Is
> > the same thing then true for the TE tuning of a temperament as well?
>
> TE and TOP are prime-based, and the whole thing doesn't work so well if you use a non-prime-power basis in some subgroup.

What do you mean by "doesn't work so well"? Is it just more difficult to compute, or is it basis-dependent even if you compute it correctly?

It seems to me like the problem you run into would be that, in the 2.5/3.7/3 subgroup for example, you have 5/3 with some Tenney height and 7/3 with some other Tenney height, but their ratio (7/3)/(5/3) = 7/5 has a Tenney height that's not simply the sum of the other two (it's smaller). This affects both TE and TOP.

The way I would explain this is that the eigenvalues of the metric tensor are not parallel to the generator axes, as they are when all the generators are primes. I hope that makes sense to someone.

Keenan

🔗genewardsmith <genewardsmith@sbcglobal.net>

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:

> It seems to me like the problem you run into would be that, in the 2.5/3.7/3 subgroup for example, you have 5/3 with some Tenney height and 7/3 with some other Tenney height, but their ratio (7/3)/(5/3) = 7/5 has a Tenney height that's not simply the sum of the other two (it's smaller). This affects both TE and TOP.

Exactly; Tenney height wants primes or prime powers as subgroup generators.

> The way I would explain this is that the eigenvalues of the metric tensor are not parallel to the generator axes, as they are when all the generators are primes. I hope that makes sense to someone.

It would make even more sense without the reference to tensor fields.

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
> > The way I would explain this is that the eigenvalues of the metric tensor are not parallel to the generator axes, as they are when all the generators are primes. I hope that makes sense to someone.
>
> It would make even more sense without the reference to tensor fields.

The metric tensor in this case is not a field, but a constant tensor. It's the same as W^-2 in http://xenharmonic.wikispaces.com/Tenney-Euclidean+metrics .

Also, I obviously meant to say "eigenvectors", not "eigenvalues".

Keenan

🔗genewardsmith <genewardsmith@sbcglobal.net>

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:

> > It would make even more sense without the reference to tensor fields.
>
> The metric tensor in this case is not a field, but a constant tensor.

I think "metric tensor" by definition is a tensor field, but I probably shouldn't have said anything since this has bugger-all to do with music. It would be nice if people just said "metric form" but they usually make more of mouthful of it with stuff like "positive definite symmetric bilinear form". Bleh.

🔗Keenan Pepper <keenanpepper@gmail.com>

🔗Mike Battaglia <battaglia01@gmail.com>

How do you weight 5/3? Is it 1/log(5*3), or 1/log(sqrt(5^2+3^2))?

-Mike

On Jan 24, 2012, at 12:11 PM, Keenan Pepper <keenanpepper@gmail.com> wrote

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...>
wrote:
>
>
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@> wrote:
> >
> > I spent a lot of time working with TE error tonight, when I noticed
> > that the TE error of a temperament depends on what basis you pick. Is
> > the same thing then true for the TE tuning of a temperament as well?
>
> TE and TOP are prime-based, and the whole thing doesn't work so well if
you use a non-prime-power basis in some subgroup.

What do you mean by "doesn't work so well"? Is it just more difficult to
compute, or is it basis-dependent even if you compute it correctly?

It seems to me like the problem you run into would be that, in the
2.5/3.7/3 subgroup for example, you have 5/3 with some Tenney height and
7/3 with some other Tenney height, but their ratio (7/3)/(5/3) = 7/5 has a
Tenney height that's not simply the sum of the other two (it's smaller).
This affects both TE and TOP.

The way I would explain this is that the eigenvalues of the metric tensor
are not parallel to the generator axes, as they are when all the generators
are primes. I hope that makes sense to someone.

Keenan

🔗Mike Battaglia <battaglia01@gmail.com>

On Tue, Jan 24, 2012 at 8:27 AM, gbreed@gmail.com <gbreed@gmail.com> wrote:
>
> No the TE error doesn't depend on the basis

But the weighted RMS error of the same tuning will change depending on
what you set the basis to, right? For example, consider a tuning where
3/5 is sharp and 5/1 is flat, so that 15/1 is perfectly just. The
weighted RMS error of 2/1, 3/1, 15/1 will be lower than the weighted
RMS error of 2/1, 3/1, 5/1, despite both being bases for the same
temperament.

-Mike

🔗gbreed@gmail.com

The RMS error will depend on the tuning. The TE error doesn't.
Changing the prime intervals means changing the space the vals are embedded in. The basis for a temperament is a set of vals. The TE error is the same regardless which basis from the same space defines the same temperament

----------
Sent from my Nokia phone

------Original message------
From: Mike Battaglia <battaglia01@gmail.com>
To: <tuning-math@yahoogroups.com>
Date: Tuesday, January 24, 2012 5:01:06 PM GMT-0500
Subject: Re: [tuning-math] Is the TE tuning basis-invariant?

On Tue, Jan 24, 2012 at 8:27 AM, gbreed@gmail.com <gbreed@gmail.com> wrote:
>
> No the TE error doesn't depend on the basis

-Mike

------------------------------------

Yahoo! Groups Links

🔗Mike Battaglia <battaglia01@gmail.com>

Is TE error not defined as the weighted RMS error of the tuning that
minimizes weighted RMS error?

-Mike

On Jan 24, 2012, at 5:50 PM, "gbreed@gmail.com" <gbreed@gmail.com> wrote:

The RMS error will depend on the tuning. The TE error doesn't.
Changing the prime intervals means changing the space the vals are embedded
in. The basis for a temperament is a set of vals. The TE error is the same
regardless which basis from the same space defines the same temperament

----------
Sent from my Nokia phone

On Tue, Jan 24, 2012 at 8:27 AM, gbreed@gmail.com <gbreed@gmail.com> wrote:
>
> No the TE error doesn't depend on the basis

-Mike

------------------------------------

Yahoo! Groups Links

🔗genewardsmith <genewardsmith@sbcglobal.net>

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@> wrote:
> > TE and TOP are prime-based, and the whole thing doesn't work so well if you use a non-prime-power basis in some subgroup.
>
> How did you calculate the POTE tunings for all the non-prime-power subgroups in the "Chromatic pairs" article?

Badly--just relative to the subgroup.

🔗gbreed@gmail.com

The rms error will depend on the intervals you average it over. Those intervals needn't form a basis. If they're primes it helps to mix in the standard deviation. Primes are special in that they have minimal complexity. A different basis should lead to an appropriate metric

----------
Sent from my Nokia phone

On Tue, Jan 24, 2012 at 8:27 AM, gbreed@gmail.com <gbreed@gmail.com> wrote:
>
> No the TE error doesn't depend on the basis

-Mike

------------------------------------

Yahoo! Groups Links

🔗gbreed@gmail.com

TE error can be defined different ways. The simplest one mathematically is that it's the angle between the JI line and the temperament line

----------
Sent from my Nokia phone

------Original message------
From: Mike Battaglia <battaglia01@gmail.com>
To: "tuning-math@yahoogroups.com" <tuning-math@yahoogroups.com>
Date: Tuesday, January 24, 2012 5:55:53 PM GMT-0500
Subject: Re: [tuning-math] Is the TE tuning basis-invariant?

Is TE error not defined as the weighted RMS error of the tuning that
minimizes weighted RMS error?

-Mike

On Jan 24, 2012, at 5:50 PM, "gbreed@gmail.com" <gbreed@gmail.com> wrote:

The RMS error will depend on the tuning. The TE error doesn't.
Changing the prime intervals means changing the space the vals are embedded
in. The basis for a temperament is a set of vals. The TE error is the same
regardless which basis from the same space defines the same temperament

----------
Sent from my Nokia phone

On Tue, Jan 24, 2012 at 8:27 AM, gbreed@gmail.com <gbreed@gmail.com> wrote:
>
> No the TE error doesn't depend on the basis

-Mike

------------------------------------

Yahoo! Groups Links

🔗Mike Battaglia <battaglia01@gmail.com>

On Wed, Jan 25, 2012 at 2:40 AM, gbreed@gmail.com <gbreed@gmail.com> wrote:
>
> The rms error will depend on the intervals you average it over. Those intervals needn't form a basis. If they're primes it helps to mix in the standard deviation. Primes are special in that they have minimal complexity. A different basis should lead to an appropriate metric

OK, I'll clarify my question. TOP minimizes the max Tenney-weighted
error over the entire limit. Does TE minimize the RMS Tenney-weighted
error over the entire limit as well? Or just of the basis, and then we
assume the effect filters out in a reasonable way?

-Mike

🔗gbreed@gmail.com

TE optimizes the rms error of the prime limit subset of a tenney limit as the limit tends to infinity. It doesn't depend on the weighting. It does depend on how you define the limit so it isn't as simple as saying it optimizes the infinite prime limit.
I wrote a pdf trying to explain it all

----------
Sent from my Nokia phone

------Original message------
From: Mike Battaglia <battaglia01@gmail.com>
To: <tuning-math@yahoogroups.com>
Date: Thursday, January 26, 2012 4:19:03 AM GMT-0500
Subject: Re: Re: [tuning-math] Is the TE tuning basis-invariant?

-Mike

------------------------------------

Yahoo! Groups Links

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, "gbreed@..." <gbreed@...> wrote:
>
> TE optimizes the rms error of the prime limit subset of a tenney limit as the limit tends to infinity. It doesn't depend on the weighting. It does depend on how you define the limit so it isn't as simple as saying it optimizes the infinite prime limit.
> I wrote a pdf trying to explain it all

I believe Graham is referring to http://x31eq.com/composite.pdf ("RMS-Based Error and Complexity Measures Involving Composite Intervals"), specifically Theorem 3.

Keenan