Default pure-octaves tuning

What's a good choice for a default pure octaves tuning applicable to
all regular tunings? I think pocpit is a good choice. It is easy
tompute, and to write a program for which covers all the different
cases. And I like the answers it gives; they seem reasonable.

7-limit meantone

pocpit: 696.8843497522732988
convergents: 1,1/2,3/5,4/7,7/12,18/31,187/322,205/353

kees: 696.5784284662087044 (1/4 comma)
convergents: 1,1/2,3/5,4/7,7/12,11/19,18/31,101/174,119/205

not: 697.6458468570556466
convergents: 1,1/2,3/5,4/7,7/12,18/31,25/43,543/934,568/977

least-squares: 696.64796858466541748
convergents: 1,1/2,3/5,4/7,7/12,11/19,18/31,173/298

11-limit miracle

pocpit: 116.6867876772980307
convergents: 1/10,3/31,4/41,7/72,74/761,81/833

kees: 116.6592976228737459
convergents: 1/10,3/31,7/72,213/2191

not: 116.6592976228737459 identical to kees

least-squares: 116.67226429605611606
convergents: 1/10,3/31,4/41,7/72,284/2921,291/2993

On 08/03/07, Gene Ward Smith <genewardsmith@coolgoose.com> wrote:
> What's a good choice for a default pure octaves tuning applicable to
> all regular tunings? I think pocpit is a good choice. It is easy
> tompute, and to write a program for which covers all the different
> cases. And I like the answers it gives; they seem reasonable.

What was wrong with optimizing the standard deviation of the weighted
errors? I don't know what this pocpit is doing, only how you
calculate it. Is it weighted?

My rank 2 standard deviation calculation yields a single parameter. I
think I'm converting it to a generator right...

> 7-limit meantone
>
> pocpit: 696.8843497522732988
> convergents: 1,1/2,3/5,4/7,7/12,18/31,187/322,205/353
>
> kees: 696.5784284662087044 (1/4 comma)
> convergents: 1,1/2,3/5,4/7,7/12,11/19,18/31,101/174,119/205
>
> not: 697.6458468570556466
> convergents: 1,1/2,3/5,4/7,7/12,18/31,25/43,543/934,568/977
>
> least-squares: 696.64796858466541748
> convergents: 1,1/2,3/5,4/7,7/12,11/19,18/31,173/298

optimal Tenney-weighted STD: 696.494 cents

> 11-limit miracle
>
> pocpit: 116.6867876772980307
> convergents: 1/10,3/31,4/41,7/72,74/761,81/833
>
> kees: 116.6592976228737459
> convergents: 1/10,3/31,7/72,213/2191
>
> not: 116.6592976228737459 identical to kees
>
> least-squares: 116.67226429605611606
> convergents: 1/10,3/31,4/41,7/72,284/2921,291/2993

optimal Tenney-weighted STD: 116.633 cents

This one looks different to all of yours. But it's close to the
unstretched TOP-RMS: 116.63272 against 116.63274 cents.

Graham

--- In tuning-math@yahoogroups.com, "Graham Breed" <gbreed@...> wrote:

> What was wrong with optimizing the standard deviation of the weighted
> errors?

Nothing as far as I know. With all this pocpit business I thought I was
following your lead, but it seems I wasn't.

> optimal Tenney-weighted STD: 696.494 cents

For Tenney-weighted least-squares over the 7-limit diamond, I get
696.67589023037823242. This would be one way of interpreting the phase
"standard deviation of the weighted errors".

At 12:27 PM 3/8/2007, you wrote:
>--- In tuning-math@yahoogroups.com, "Graham Breed" <gbreed@...> wrote:
>
>> What was wrong with optimizing the standard deviation of the weighted
>> errors?
>
>Nothing as far as I know. With all this pocpit business I thought I was
>following your lead, but it seems I wasn't.
>
>> optimal Tenney-weighted STD: 696.494 cents
>
>For Tenney-weighted least-squares over the 7-limit diamond, I get
>696.67589023037823242. This would be one way of interpreting the phase
>"standard deviation of the weighted errors".

Graham's just considering primes, not the whole diamond!

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@...> wrote:
>
> At 12:27 PM 3/8/2007, you wrote:
> >--- In tuning-math@yahoogroups.com, "Graham Breed" <gbreed@> wrote:
> >
> >> What was wrong with optimizing the standard deviation of the
weighted
> >> errors?
> >
> >Nothing as far as I know. With all this pocpit business I thought
I was
> >following your lead, but it seems I wasn't.
> >
> >> optimal Tenney-weighted STD: 696.494 cents
> >
> >For Tenney-weighted least-squares over the 7-limit diamond, I get
> >696.67589023037823242. This would be one way of interpreting the
phase
> >"standard deviation of the weighted errors".
>
> Graham's just considering primes, not the whole diamond!

That would give 696.95213773673908187.

On 09/03/07, Gene Ward Smith <genewardsmith@coolgoose.com> wrote:
> --- In tuning-math@yahoogroups.com, Carl Lumma <ekin@...> wrote:
> >
> > At 12:27 PM 3/8/2007, you wrote:
> > >--- In tuning-math@yahoogroups.com, "Graham Breed" <gbreed@> wrote:
> > >
> > >> What was wrong with optimizing the standard deviation of the
> weighted
> > >> errors?
> > >
> > >Nothing as far as I know. With all this pocpit business I thought
> I was
> > >following your lead, but it seems I wasn't.
> > >
> > >> optimal Tenney-weighted STD: 696.494 cents
> > >
> > >For Tenney-weighted least-squares over the 7-limit diamond, I get
> > >696.67589023037823242. This would be one way of interpreting the
> phase
> > >"standard deviation of the weighted errors".

For an odd-limit diamond, you should be able to Kees-weight the
intervals within the diamond. In that case, the mean error is zero,
so the RMS and standard deviation are identical.

> > Graham's just considering primes, not the whole diamond!
>
> That would give 696.95213773673908187.

Standard deviation or RMS?

At my optimal STD point for 7-limit meantone, the primes are,

1200.0
1896.4936084
2785.9744336
3364.93608399

cents. The Tenney-weighted primes are

1.0
0.997128537498
0.999878233285
0.998845347732

Their sum is

3.995852118514839

The sum of squares is

3.991713830375573

The standard deviaton of weighted primes is the same as the standard
deviation of errors, so

sqrt(3.991713830375573/4 - 3.995852118514839*3.995852118514839/16)

= 0.0011502299434728799

There may be floating point imprecision -- it diasgrees with the
standard calculation in the last 6 digits.

For your optimal tuning, the primes are:

1200.0
1896.95213774
2787.80855095
3369.52137737

cents. Tenney weighted:

1.0
0.997369620666
1.00053649274
1.00020644311

Their sum is 3.9981125565174049 and their sum squared
3.9962323623734703. This gives a standard deviation of

sqrt(3.9962323623734703/4 - 3.9981125565174049*3.9981125565174049/16)

= 0.0012608259133705998

so my standard deviation's smaller than yours and there's enough
working to see if I made a mistake calculating it.

Graham

--- In tuning-math@yahoogroups.com, "Graham Breed" <gbreed@...> wrote:

> At my optimal STD point for 7-limit meantone, the primes are,
>
> 1200.0
> 1896.4936084
> 2785.9744336
> 3364.93608399

I don't see why you call this a worked example. If I take the minimal
square deviation of error(p)/log2(p), not from 1, but from the mean of
error(3)/log2(3), error(5)/log2(5) and error(7)/log2(7) I still don't
check your figures. I got

1200.0
1895.9345297751362659
2783.7381191005450638
3359.3452977513626594

from that.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:

> I don't see why you call this a worked example. If I take the minimal
> square deviation of error(p)/log2(p), not from 1, but from the mean of
> error(3)/log2(3), error(5)/log2(5) and error(7)/log2(7) I still don't
> check your figures.

Sorry, this should read log2("3")/log2(3), log2("5")/log2(5), log2
("7")/log2(7).

Gene Ward Smith wrote:
> --- In tuning-math@yahoogroups.com, "Graham Breed" <gbreed@...> wrote:
> >>At my optimal STD point for 7-limit meantone, the primes are,
>>
>>1200.0
>>1896.4936084
>>2785.9744336
>>3364.93608399
> > I don't see why you call this a worked example. If I take the minimal > square deviation of error(p)/log2(p), not from 1, but from the mean of
> log2("3")/log2(3), log2("5")/log2(5) and log2("7")/log2(7) I still don't > check your figures. I got

I called my original post a worked example because it was an example with working. Of course, as you snipped the working out, that won't be apparent. If you showed me your working, it might help to show where you're going wrong.

From your text though, even after following your correction, I can see that you missed log2("2")/log2(2) and your STD would be lower than mine in that case. So perhaps that's it.

> 1200.0
> 1895.9345297751362659
> 2783.7381191005450638
> 3359.3452977513626594
> > from that.

The weighted primes here are [1.0, 0.99683458804886971, 0.99907562642660341, 0.9971857825304784]. Their mean is 0.99827399925148796. Their mean squared is 0.99655269703279803. That gives a standard deviation of

sqrt(0.99655269703279803-0.99827399925148796*0.99827399925148796)

which gives 0.0013112784747261357, which is higher than the (imprecise) figure I gave of 0.0011502299434728799.

Graham

I spotted an error in one of the equations in my prime errors and complexities paper. None of the results are affected but maybe you were wondering what I was doing. I also added some comments to make the STD optimization clearer. It's in the files section here (still having trouble with my website).

Graham

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:

> I called my original post a worked example because it was an
example
> with working. Of course, as you snipped the working out, that
won't be
> apparent.

I didn't snip the working out; there was no working.

If you showed me your working, it might help to show where
> you're going wrong.

Assuming I *am* going wrong, something not yet established.

Lets call the fifth we are seeking x. Then we have

e3 = (x+1)/log2(3)
e5 = 4x/log2(5)
e7 = (10x-3)/log2(7)

The average of these is m = (e3+e5+e7)/3.

Now form the expression (e3-m)^2+(e5-m)^2+(e7-m)^2, differentiate,
and solve for x. The result is x = 0.57994544147928022162, which is
to say, a fifth of 695.9345 cents.

> From your text though, even after following your correction, I can
see
> that you missed log2("2")/log2(2) and your STD would be lower than
mine
> in that case. So perhaps that's it.

I didn't "miss" it. Why in the world would anyone include it?
log2("2")/log2(2) = 1, and it's not going to make a difference unless
you insist on tossing it into the average also. If that is the case,
don't blame anyone if they didn't guess that!

If I do it that way, I now get x = 0.58011169254875471610, which is
to say, 696.13403105850565932.

Gene Ward Smith wrote:
> --- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:
> >>I called my original post a worked example because it was an > example >>with working. Of course, as you snipped the working out, that > won't be >>apparent. > > I didn't snip the working out; there was no working.

Whatever you want to call the stuff you snipped out, please go back and check your answers against it so we can find where the mistake is.

> If you showed me your working, it might help to show where >>you're going wrong.
> > Assuming I *am* going wrong, something not yet established.

You certainly *were* going wrong, and given your record the chances are you're still going wrong.

> Lets call the fifth we are seeking x. Then we have > > e3 = (x+1)/log2(3)
> e5 = 4x/log2(5)
> e7 = (10x-3)/log2(7)
> > The average of these is m = (e3+e5+e7)/3.

Now, did you check this average to see if it's the same as my average?

> Now form the expression (e3-m)^2+(e5-m)^2+(e7-m)^2, differentiate, > and solve for x. The result is x = 0.57994544147928022162, which is > to say, a fifth of 695.9345 cents.

Right, which is completely meaningless because you left the octaves out. So you're going wrong exactly where I said you were going wrong.

>> From your text though, even after following your correction, I can > see >>that you missed log2("2")/log2(2) and your STD would be lower than > mine >>in that case. So perhaps that's it.
> > I didn't "miss" it. Why in the world would anyone include it? > log2("2")/log2(2) = 1, and it's not going to make a difference unless > you insist on tossing it into the average also. If that is the case, > don't blame anyone if they didn't guess that!

You did miss it. If you didn't miss it, why wasn't it in your calculation? You include it because it makes a difference. You don't have to guess anything -- I told you exactly what to do. Why do have to keep whining every time somebody catches you making a mistake?

> If I do it that way, I now get x = 0.58011169254875471610, which is > to say, 696.13403105850565932.

I get an STD of 0.0012194306927808627 for that. Which is higher than my value. So perhaps you could check it.

Graham

At 07:44 PM 3/14/2007, you wrote:
>I spotted an error in one of the equations in my prime errors and
>complexities paper. None of the results are affected but maybe you were
>wondering what I was doing. I also added some comments to make the STD
>optimization clearer. It's in the files section here (still having
>trouble with my website).
>
>
> Graham

Speaking of that, I tried to access your green pages the other
night (the classic page on neutral thirds) and couldn't. It's
down now as well. Do you think you can get your pages back up?

-Carl

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:

> > Lets call the fifth we are seeking x. Then we have
> >
> > e3 = (x+1)/log2(3)
> > e5 = 4x/log2(5)
> > e7 = (10x-3)/log2(7)
> >
> > The average of these is m = (e3+e5+e7)/3.
>
> Now, did you check this average to see if it's the same as my
average?

Given that it involves an unknown x, it can't be.

> Right, which is completely meaningless because you left the octaves
out.

None of this makes much sense on the face of it, so how was I to know
I was supposed to include the octave? You need to spell things out
very, very clearly when you are doing something which seems
unmotivated, since people will tend to rationalize it to what they
think it "must mean".

> have to guess anything -- I told you exactly what to do. Why do
have to
> keep whining every time somebody catches you making a mistake?

The mistake was yours, and now you are whining. It was yours because
you failed to be clear about what your rather bizarre definition
meant. Do you think we can read minds? Can you even explain why this
proceedure makes sense?

> > If I do it that way, I now get x = 0.58011169254875471610, which
is
> > to say, 696.13403105850565932.
>
> I get an STD of 0.0012194306927808627 for that. Which is higher
than my
> value. So perhaps you could check it.

Perhaps you could tell us how you calculate your value.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:

> > Right, which is completely meaningless because you left the octaves
> out.
>
> None of this makes much sense on the face of it, so how was I to know
> I was supposed to include the octave?

I've been thinking about why you might want to do this stuff, and what
I came up with is that you want the errors to be close to each other,
because they cancel when you take ratios, and you like ratios of primes.
Hence, you might try to cluster them around their mean, and not around
the mean which also includes reciprocals of primes, which is 1.
However, if this is the motivation it makes more sense *not* to include
octaves. Of course, you can include an arbitary amount of octave as a
free parameter also.

Gene Ward Smith wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith" > <genewardsmith@...> wrote:
> > >>>Right, which is completely meaningless because you left the octaves >>
>>out. >>
>>None of this makes much sense on the face of it, so how was I to know >>I was supposed to include the octave? > > I've been thinking about why you might want to do this stuff, and what > I came up with is that you want the errors to be close to each other, > because they cancel when you take ratios, and you like ratios of primes.
> Hence, you might try to cluster them around their mean, and not around > the mean which also includes reciprocals of primes, which is 1. > However, if this is the motivation it makes more sense *not* to include > octaves. Of course, you can include an arbitary amount of octave as a > free parameter also.

Originally I wanted a simple approximation to the optimally stretched RMS error. That happens to be the standard deviation if you use Tenney weighting. It doesn't work without the octaves considered.

The post hoc logic is that it favors smaller intervals, which I think are the most important, and will be the most common with multi-note chords. It does that by considering ratios of prime powers (not only primes).

So what's special about octaves? Why shouldn't 3:2, 4:3, 8:5, 5:4, and so on be considered on an equal footing to other ratios of prime powers? You get an unreasonably small error for 19-equal if you only consider 5:3 and friends.

You could weight the octave differently, yes. That would work like any other weighting (and probably not as easily as I suggest in the PDF -- I need to fix that). But as a standard weighting, what's wrong with Tenney?

Graham