Yahoo Tuning Groups Ultimate Backup tuning-math Creating a Temperment /Comma

> From: paulhjelmstad [mailto:paul.hjelmstad@us.ing.com]
> Sent: Friday, July 25, 2003 10:21 AM
> To: tuning-math@yahoogroups.com
> Subject: [tuning-math] Creating a Temperment /Comma
>
>
> I would like to know which temperment/comma connects
> 12, 47, 35, and 23 ets on zoomr.gif. How could I
> calculate these for myself? Thanks! If this is a
> new one, could I name it? What is the 5-limit vector?

paul is referring to the "zoom:10" mouse-over link here

http://sonic-arts.org/dict/eqtemp.htm

... just trying to be helpful ... unfortunately,
i'd be trying to figure out which comma it is
empirically, somehow, but examining the cents errors
of each temperament. Gene and paul erlich know how
to do it easily with algebra.

-monz

🔗Carl Lumma <ekin@lumma.org>

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

🔗Carl Lumma <ekin@lumma.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Carl Lumma <ekin@lumma.org>

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

🔗Carl Lumma <ekin@lumma.org>

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Carl Lumma <ekin@lumma.org>

>As a linear temperament, it isn't very interesting, because its
>complexity is too high. As a means of tempering 12 notes, it becomes
>much more interesting.

Is this because there's a T[n] comma associated with stopping at
12? Or...?

By the way, what wound up being your favorite complexity measure
and why? The last 5-limit list I have from you uses "geometric
complexity"... which is... Oh, that's the really hard-to-understand
Euclidean distance-on-the-lattice-to-the-comma thing, isn't it?

Meanwhile, Dave and Graham, are you guys still using map-based
complexity? Doesn't this run into problems because there are
many different ways to write the map for a given temperament?

Me, I like the idea of number of notes. But then we have to
consider the effects of the comma formed between the ends of
the chain. Has anyone ever looked at error as a function of
chain length for each temperament?

-Carl

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> >As a linear temperament, it isn't very interesting, because its
> >complexity is too high. As a means of tempering 12 notes, it becomes
> >much more interesting.
>
> Is this because there's a T[n] comma associated with stopping at
> 12?

Only if you think a comma would be interesting.

Or...?

There's a historical importance to 12 which is, of course, obvious.
There are just two ways to generate a cyclic group of order 12--one
gives the fourth/fifth generators we usually focus on, the other is
the semitone. Just as with meantone, we can temper 12 notes, or we can
allow ourselves to take something written in 12 notes and extend it to
more using the temperament; it's an idea worth exploring, at any rate.

> By the way, what wound up being your favorite complexity measure
> and why?

I like geometric complexity because it works for all regular temperaments.

The last 5-limit list I have from you uses "geometric
> complexity"... which is... Oh, that's the really hard-to-understand
> Euclidean distance-on-the-lattice-to-the-comma thing, isn't it?

Fraid so.

> Me, I like the idea of number of notes. But then we have to
> consider the effects of the comma formed between the ends of
> the chain. Has anyone ever looked at error as a function of
> chain length for each temperament?

Aren't you conflating temperaments and scales?

🔗Carl Lumma <ekin@lumma.org>

>>>As a linear temperament, it isn't very interesting, because its
>>>complexity is too high. As a means of tempering 12 notes, it
>>>becomes much more interesting.
>>
>> Is this because there's a T[n] comma associated with stopping at
>> 12? Or...?
//
>Just as with meantone, we can temper 12 notes, or we can allow
>ourselves to take something written in 12 notes and extend it to
>more using the temperament;

How does one "take something written in 12 notes and extend it with
the temperament"?

>>By the way, what wound up being your favorite complexity
>>measure and why?
>
>I like geometric complexity because it works for all regular
>temperaments.

Such as planar and higher temperaments. But the 'how many notes'
complexity obviously generalizes to these as well (in the linear
case it's equivalent to Dave/Graham complexity).

>> Me, I like the idea of number of notes. But then we have to
>> consider the effects of the comma formed between the ends of
>> the chain. Has anyone ever looked at error as a function of
>> chain length for each temperament?
>
>Aren't you conflating temperaments and scales?

I understand that temperaments are abstract Things, but they
ultimately must be expressed as scales to be used. If geometric
complexity is so good, the T[n] stuff wouldn't have been such a
surprise (unless, I suppose, consonances formed by way of the
extra "wolf" comma necessarily break consistency -- do they?).

In my view, musically complexity has to boil down to 'how many
notes get me how many consonances'. Say n/i where n is notes
and i is the number of consonant dyads. Dave/Graham get rid of
the denominator by standardizing i to 'a complete n-ad' (Dave
once showed that for linear temperaments, otonal and utonal n-ads
always come in pairs -- I wonder if this is true for planar
temperaments...). I suggest we plot n/i for temperament T for
all n up to some large number, and report the minima. If this
is a bad idea I'd like to know why.

-Carl

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Graham Breed <graham@microtonal.co.uk>

Carl Lumma wrote:

> Such as planar and higher temperaments. But the 'how many notes'
> complexity obviously generalizes to these as well (in the linear
> case it's equivalent to Dave/Graham complexity).

What about equal temperaments?

> In my view, musically complexity has to boil down to 'how many
> notes get me how many consonances'. Say n/i where n is notes
> and i is the number of consonant dyads. Dave/Graham get rid of
> the denominator by standardizing i to 'a complete n-ad' (Dave
> once showed that for linear temperaments, otonal and utonal n-ads
> always come in pairs -- I wonder if this is true for planar
> temperaments...). I suggest we plot n/i for temperament T for
> all n up to some large number, and report the minima. If this
> is a bad idea I'd like to know why.

For a linear temperament, the number of consonances is (roughly) the number of notes minus the complexity. So n-i is a constant. If you then plot n/i, you can replace n with i+c to get (i+c)/i = 1+c/i. As i tends to infinity, this tends to 1, which is as small as it gets. So your measure would depend on how large n is allowed to get.

For an equal temperament, n can only ever take one value, and n/i=1.

For a planar temperament, the number of consonance depends on what generators you choose, and how you construct the scale from them. You can get more otonal than utonal if you like. Taking 5-limit JI as the canonical planar scale, C-E-G gives 1 more otonality than utonality. And C-D-E-G-G#-B has 2 more. But you should get the same value for complexity whether you use otonal or utonal. If you standardize on a parallelogram, like C-E-G-B, than you get equal numbers of each chord.

That's assuming you take the fifth and third as generators. With a tone and semitone, it'd take 5 notes to get 1 triad, and 6 notes to get 1 of each triad (which gets you 2 of 1 kind). I haven't looked into planar temperaments yet, so I don't know if there's an obvious way to get the optimum generators.

I think you still get n/i tends to 1 as n tends to infinity.

Graham

🔗Graham Breed <graham@microtonal.co.uk>

paulhjelmstad wrote:

> I would like to know which temperment/comma connects 12, 47, 35, and > 23 ets on zoomr.gif. How could I calculate these for myself? Thanks!
> If this is a new one, could I name it? What is the 5-limit vector?

Go to

http://microtonal.co.uk/temper/twoet.html

and enter 12 and 47 for the ets, and 5 for the limit. That gives you the mapping

[(1, 0), (2, -5), (3, -8)]

It doesn't show the comma because I don't see the point, and they aren't that easy to calculate for some temperaments. But for the 5-limit case you can work it out from the mapping. It must involve 3**8 and 5**5 with opposite signs, from the octave equivalent mapping (-5, -8). You then multiply (x, 8, -5) by the period part of the mapping to get x + 2*8 - 3*5. For a unison vector, this should be zero, so x = -2*8 + 3*5 = 15-16 = -1. So the comma is (-1 8 -5) or 2**(-1) * 3**8 * 5**(-5) = 6561:6250.

There are different tools runnable online at http://microtonal.co.uk/temper/ and you can get the library for Python, which is free, and play with it as you like. To get commas for a linear temperament as ratios,

>>> map(temper.getRatio, temper.Temperament(12,47,
temper.limit5).getUnisonVectors())
[(6561, 6250)]

Graham

🔗Carl Lumma <ekin@lumma.org>

>> Such as planar and higher temperaments. But the 'how many notes'
>> complexity obviously generalizes to these as well (in the linear
>> case it's equivalent to Dave/Graham complexity).
>
>What about equal temperaments?

For an n-et, one can always think of the generators as 1200 and
1200/n. Then it works the same as for linear temperaments. Or
isn't that what you do?

>> In my view, musically complexity has to boil down to 'how many
>> notes get me how many consonances'. Say n/i where n is notes
>> and i is the number of consonant dyads. Dave/Graham get rid of
>> the denominator by standardizing i to 'a complete n-ad' (Dave
>> once showed that for linear temperaments, otonal and utonal n-ads
>> always come in pairs -- I wonder if this is true for planar
>> temperaments...). I suggest we plot n/i for temperament T for
>> all n up to some large number, and report the minima. If this
>> is a bad idea I'd like to know why.
>
>For a linear temperament, the number of consonances is (roughly)
>the number of notes minus the complexity. So n-i is a constant.
>If you then plot n/i, you can replace n with i+c to get
>(i+c)/i = 1+c/i. As i tends to infinity, this tends to 1, which
>is as small as it gets. So your measure would depend on how
>large n is allowed to get.

Ok, I follow your reasoning but I'm not sure what your conclusion
is. I think we'd limit n to the number of notes in the Fokker
block. At that n, the complexity would be expected to match the
heuristic (for linear temperaments) and geometric complexity (for
everything). But when n gets smaller we need a way to quantify
what happens. It was probably wrong of me to suggest this be some
sort of temperament complexity. And I'm wondering if intervals
achieved through use of the "wolf" break consistency, or whatever
Gene calls it for regular temperaments.

>For an equal temperament, n can only ever take one value, and
>n/i=1.

How do you figure the number of consonant intervals equals the
number of notes in the et?

>For a planar temperament, the number of consonance depends on what
>generators you choose, and how you construct the scale from them.
>You can get more otonal than utonal if you like.

Aha.

>Taking 5-limit JI as the canonical planar scale, C-E-G gives 1
>more otonality than utonality. And C-D-E-G-G#-B has 2 more.

Do such blocks necessarily involve unison vectors that are larger
than their smallest 2nd?

>But you should get the same value for complexity whether you use
>otonal or utonal.

Excactly -- which is why they apparently both must be counted
for planar temperaments.

-Carl

🔗Graham Breed <graham@microtonal.co.uk>

Carl Lumma wrote:

> For an n-et, one can always think of the generators as 1200 and
> 1200/n. Then it works the same as for linear temperaments. Or
> isn't that what you do?

In that case, you're turning an et into a linear temperament in an arbitrary way. If you're consistent between ets, it should be the same as taking a fixed proportion of the number of notes to the octave. For comparing equal and linear temperaments it's still arbitrary, and will make ets look more complicated then they are.

>>For a linear temperament, the number of consonances is (roughly)
>>the number of notes minus the complexity. So n-i is a constant.
>>If you then plot n/i, you can replace n with i+c to get
>>(i+c)/i = 1+c/i. As i tends to infinity, this tends to 1, which
>>is as small as it gets. So your measure would depend on how
>>large n is allowed to get.
> > Ok, I follow your reasoning but I'm not sure what your conclusion
> is. I think we'd limit n to the number of notes in the Fokker
> block. At that n, the complexity would be expected to match the
> heuristic (for linear temperaments) and geometric complexity (for
> everything). But when n gets smaller we need a way to quantify
> what happens. It was probably wrong of me to suggest this be some
> sort of temperament complexity. And I'm wondering if intervals
> achieved through use of the "wolf" break consistency, or whatever
> Gene calls it for regular temperaments.

You can conclude whatever you like. What Fokker block? Why bring the heuristic into it? Geometric complexity might work, but I don't understand it.

>>For an equal temperament, n can only ever take one value, and
>>n/i=1.
> > How do you figure the number of consonant intervals equals the
> number of notes in the et?

I thought consonances were chords. Otherwise, why distinguish otonal and utonal? It makes more sense for planar temperaments anyway. A-B-C-G has all the 5-limit consonances, but no triads. This can't happen with linear temperaments.

>>Taking 5-limit JI as the canonical planar scale, C-E-G gives 1
>>more otonality than utonality. And C-D-E-G-G#-B has 2 more.
> > Do such blocks necessarily involve unison vectors that are larger
> than their smallest 2nd?

I don't know. Would it always work out for a well formed periodicity block? I'd prefer the complexity didn't depend on the tuning.

>>But you should get the same value for complexity whether you use
>>otonal or utonal.
> > Excactly -- which is why they apparently both must be counted
> for planar temperaments.

I thought it was why it didn't matter.

Graham

🔗Carl Lumma <ekin@lumma.org>

>> For an n-et, one can always think of the generators as 1200 and
>> 1200/n. Then it works the same as for linear temperaments. Or
>> isn't that what you do?
>
>In that case, you're turning an et into a linear temperament in an
>arbitrary way. If you're consistent between ets, it should be the
>same as taking a fixed proportion of the number of notes to the
>octave. For comparing equal and linear temperaments it's still
>arbitrary, and will make ets look more complicated then they are.

So you suggest taking a fixed proportion of the number of notes to
the octave? What does that mean, and how does one do it?

>>>For a linear temperament, the number of consonances is (roughly)
>>>the number of notes minus the complexity. So n-i is a constant.
>>>If you then plot n/i, you can replace n with i+c to get
>>>(i+c)/i = 1+c/i. As i tends to infinity, this tends to 1, which
>>>is as small as it gets. So your measure would depend on how
>>>large n is allowed to get.
>>
>>Ok, I follow your reasoning but I'm not sure what your conclusion
>>is. I think we'd limit n to the number of notes in the Fokker
>>block. At that n, the complexity would be expected to match the
>>heuristic (for linear temperaments) and geometric complexity (for
>>everything). But when n gets smaller we need a way to quantify
>>what happens. It was probably wrong of me to suggest this be some
>>sort of temperament complexity. And I'm wondering if intervals
>>achieved through use of the "wolf" break consistency, or whatever
>>Gene calls it for regular temperaments.
>
>You can conclude whatever you like. What Fokker block?

Any temperament can be viewed in terms of Fokker blocks.

>Why bring the heuristic into it?

Because it tells you the complexity of the temperament.

>Geometric complexity might work, but I don't understand it.

That makes two of us.

>>>For an equal temperament, n can only ever take one value, and
>>>n/i=1.
>>
>> How do you figure the number of consonant intervals equals the
>> number of notes in the et?
>
>I thought consonances were chords. Otherwise, why distinguish otonal
>and utonal?

I was asking about the o/utonal distinction re. your complexity.
I'm suggesting looking at dyads.

>>>Taking 5-limit JI as the canonical planar scale, C-E-G gives 1
>>>more otonality than utonality. And C-D-E-G-G#-B has 2 more.
>>
>> Do such blocks necessarily involve unison vectors that are larger
>> than their smallest 2nd?
>
>I don't know. Would it always work out for a well formed periodicity
>block? I'd prefer the complexity didn't depend on the tuning.

But tempering adds intervals/notes, which is how you lower complexity.
So the tuning has to matter.

>>>But you should get the same value for complexity whether you use
>>>otonal or utonal.
>>
>> Excactly -- which is why they apparently both must be counted
>> for planar temperaments.

So that's important result number 1 from this little exchange.

-Carl

🔗Graham Breed <graham@microtonal.co.uk>

Carl Lumma wrote:

> So you suggest taking a fixed proportion of the number of notes to
> the octave? What does that mean, and how does one do it?

I don't suggest anything, I only know the problems. If you're counting steps upwards from the 1/1, then a 4:5:6 chord will be the same size as a 2:3 interval. For a consistent et of n notes to the octave, that means the number of steps you need for a complete triad is the nearest integer to 0.585*n. For an inconsistent et it won't be far off. So for comparing one et to another, you get the same result as comparing the number of steps. For comparing equal with linear temperaments, you get 7-equal having the same complexity as meantone, which is underestimating 7-equal.

> Any temperament can be viewed in terms of Fokker blocks.

Usually an infinite number of them!

>>Why bring the heuristic into it?
> > Because it tells you the complexity of the temperament.

It gives you an estimate of something you can calculate exactly. At least, the heuristic I'm thinking of does, but that gives accuracy not complexity.

> I was asking about the o/utonal distinction re. your complexity.
> I'm suggesting looking at dyads.

For a linear temperament, the complexity of a chord is always the same as its most complex interval. So it doesn't matter which you look at. For planar temperaments that needn't be the case, but perhaps it will be for sensibly constructed blocks.

> But tempering adds intervals/notes, which is how you lower complexity.
> So the tuning has to matter.

I noticed some mention of wolves before, so perhaps this is it. It's assumed we're dealing with regular temperaments, where the tempering of each interval is always the same. So even if the tuning happens to give you some alternative approximations you ignore them. Hence the 5:1 in meantone is always four fifths, even with a Pythagorean tuning where a schismic approximation would be closer. If you wanted that, you should have asked for schismic.

In meantone, a fifth is 1 generator step, a major third is 4 generators and a minor third is 3 generators. If that's all you need to calculate the complexity, you don't need the tuning. Most of the complexity measures for linear temperaments work like this. The exception is the "smallest MOS" one, because you need some idea of the size of the generator to know what MOSs are valid. That's a practical problem for a search algorithm, because it means you have to optimize the tuning for any given temperament before you can reject it for being too complex. (Then again, for temperaments defined by a pair of ets, you can get a list of MOS sizes by working backwards on the scale tree, and this will always be sufficient if the seed ets are consistent.)

Because a search for planar temperaments is going to be harder, it would be nice if we could still calculate the complexity independent of tuning. But perhaps planar temperaments generated from equal temperaments can always be assigned well formed Fokker blocks without worrying about the tuning.

Graham

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Graham Breed <graham@microtonal.co.uk>

🔗Carl Lumma <ekin@lumma.org>

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

--- In tuning-math@yahoogroups.com, Graham Breed <graham@m...> wrote:
> paulhjelmstad wrote:
>
> > I would like to know which temperment/comma connects 12, 47, 35,
and
> > 23 ets on zoomr.gif. How could I calculate these for myself?
Thanks!
> > If this is a new one, could I name it? What is the 5-limit vector?
>
> Go to
>
> http://microtonal.co.uk/temper/twoet.html
>
> and enter 12 and 47 for the ets, and 5 for the limit. That gives
you
> the mapping
>
> [(1, 0), (2, -5), (3, -8)]
>
> It doesn't show the comma because I don't see the point, and they
aren't
> that easy to calculate for some temperaments. But for the 5-limit
case
> you can work it out from the mapping. It must involve 3**8 and
5**5
> with opposite signs, from the octave equivalent mapping (-5, -8).
You
> then multiply (x, 8, -5) by the period part of the mapping to get x
+
> 2*8 - 3*5. For a unison vector, this should be zero, so x = -2*8 +
3*5
> = 15-16 = -1. So the comma is (-1 8 -5) or 2**(-1) * 3**8 * 5**(-
5) =
> 6561:6250.
>
> There are different tools runnable online at
> http://microtonal.co.uk/temper/ and you can get the library for
Python,
> which is free, and play with it as you like. To get commas for a
linear
> temperament as ratios,
>
> >>> map(temper.getRatio, temper.Temperament(12,47,
> temper.limit5).getUnisonVectors())
> [(6561, 6250)]
>
>
> Graham
Thanks. This will keep me busy. I've downloaded Python, and I will
work on both websites.

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

--- In tuning-math@yahoogroups.com, Graham Breed <graham@m...> wrote:
> paulhjelmstad wrote:
>
> > I would like to know which temperment/comma connects 12, 47, 35,
and
> > 23 ets on zoomr.gif. How could I calculate these for myself?
Thanks!
> > If this is a new one, could I name it? What is the 5-limit vector?
>
> Go to
>
> http://microtonal.co.uk/temper/twoet.html
>
> and enter 12 and 47 for the ets, and 5 for the limit. That gives
you
> the mapping
>
> [(1, 0), (2, -5), (3, -8)]
>
> It doesn't show the comma because I don't see the point, and they
aren't
> that easy to calculate for some temperaments. But for the 5-limit
case
> you can work it out from the mapping. It must involve 3**8 and
5**5
> with opposite signs, from the octave equivalent mapping (-5, -8).
You
> then multiply (x, 8, -5) by the period part of the mapping to get x
+
> 2*8 - 3*5. For a unison vector, this should be zero, so x = -2*8 +
3*5
> = 15-16 = -1. So the comma is (-1 8 -5) or 2**(-1) * 3**8 * 5**(-
5) =
> 6561:6250.
>
> There are different tools runnable online at
> http://microtonal.co.uk/temper/ and you can get the library for
Python,
> which is free, and play with it as you like. To get commas for a
linear
> temperament as ratios,
>
> >>> map(temper.getRatio, temper.Temperament(12,47,
> temper.limit5).getUnisonVectors())
> [(6561, 6250)]
>
>
> Graham
This site is cool. I've been playing with it. However, what do I
enter (as an example) in "Temperments from Unison Vectors"?

🔗Carl Lumma <ekin@lumma.org>

>> Any temperament can be viewed in terms of Fokker blocks.
>
>Usually an infinite number of them!

That's why blocks was plural!

>>>Why bring the heuristic into it?
>>
>> Because it tells you the complexity of the temperament.
>
>It gives you an estimate of something you can calculate exactly. At
>least, the heuristic I'm thinking of does, but that gives accuracy
>not complexity.

There's an error heuristic |n-d|/d*log(d), and a complexity heuristic,
log(d).

>> I was asking about the o/utonal distinction re. your complexity.
>> I'm suggesting looking at dyads.
>
>For a linear temperament, the complexity of a chord is always the same
>as its most complex interval. So it doesn't matter which you look at.

But I'm suggesting counting the number of available dyads per number
of notes.

>>But tempering adds intervals/notes, which is how you lower complexity.
>>So the tuning has to matter.
>
>I noticed some mention of wolves before, so perhaps this is it. It's
>assumed we're dealing with regular temperaments, where the tempering
>of each interval is always the same. So even if the tuning happens to
>give you some alternative approximations you ignore them. Hence the
>5:1 in meantone is always four fifths, even with a Pythagorean tuning
>where a schismic approximation would be closer. If you wanted that,
>you should have asked for schismic.

Right. Gene did a search for scales where the wolf was considered to
be another comma. Or something. I'm trying to understand how this
fits into the temperament stuff. His results looked like:

>>Blackwood[10]
>>[0, 5, 0, 8, 0, -14] [[5, 8, 12, 14], [0, 0, -1, 0]]
>>
>>bad 1662.988586 comp 10.25428060 rms 15.81535241
>>graham 5 scale size 10 ratio 2.000000

>In meantone, a fifth is 1 generator step, a major third is 4 generators
>and a minor third is 3 generators. If that's all you need to calculate
>the complexity, you don't need the tuning.

Ok, but it implies some kind of non-JI tuning. Maybe I should have
said "ignoring commas increases intervals/notes".

>Most of the complexity measures for linear temperaments work like
>this. The exception is the "smallest MOS" one,

That doesn't sound like a very good one.

Maybe Gene can prod us to figure out geometric complexity.

/tuning-math/message/5546
/tuning-math/message/5598
/tuning-math/message/5671
/tuning-math/message/5692

-Carl

🔗Graham Breed <graham@microtonal.co.uk>

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "paulhjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> (But I can't find
> > anything for 12, 25, 37, 49 ets etc)
>
> This has comma 262144/253125. Once again the generator is about a
> semitone and five of them give a fourth. This time, however, the
> semitone is a little smaller than 12-et (about 98.3 cents) and four
of
> them give you a major third. Evidently these two belong together and
> should get cute matching names.

Found both of them. They are "diaschizoid" for (12,23,35, and 47) and
"schizoid" for (12,25,37 and 49). I am using Graham's applet to find
more. I used to know how you calculated the generators,(rms) I will
have to review that. Two steps forward and one back...I don't have
the greatest memory in the world. But this newsgroup has been fun.
Thanks everybody.

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Graham Breed <graham@microtonal.co.uk>

Gene Ward Smith wrote:

> What wasn't clear?

You haven't given a general formula, only a lot of specific formulae involving real numbers you don't derive and indices that only work for your particular implementation.

If I had worked out your numbering rule, I've forgotten it now.

Every time you try to explain something, you bring in more jargon terms that I don't understand (I can't speak for anybody else).

The word "metric" in particular is something that's important but you haven't defined.

You keep missing out important steps in explanations, like the need to take the complement at certain points when using wedge products.

Graham

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

Thanks. Where can I get the Library for Python? I have the command
shell but get an error running the above command.

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, Graham Breed <graham@m...> wrote:
> Gene Ward Smith wrote:
>
> > What wasn't clear?
>
> You haven't given a general formula, only a lot of specific
formulae
> involving real numbers you don't derive and indices that only work
for
> your particular implementation.

Not true any more, and I think I posted this. Here is the Maple code
for 7-limit geometric complexity:

gc7 := proc(l)
# l is 7-limit wedgie
sqrt(evalf(1/4*(-ln(5)^4+4*ln(5)^2*ln(7)^2)*l[1]^2
+1/4*(-ln(3)^4+4*ln(3)^2*ln(7)^2)*l[2]^2
+1/4*(-ln(3)^4+4*ln(3)^2*ln(5)^2)*l[3]^2
+1/2*(ln(3)^2*ln(5)^2-2*ln(3)^2*ln(7)^2)*l[1]*l[2]
-1/2*ln(3)^2*ln(5)^2*l[1]*l[3]+
1/2*(ln(3)^4-2*ln(3)^2*ln(5)^2)*l[2]*l[3])) end:

Here is 11-limit linear:

sqrt(evalf((-1/4*ln(5)^2*ln(7)^4-1/4*ln(11)^2*ln(5)^4+ln(7)^2*ln(11)
^2*ln(5)^2)*l[1]^2
+(ln(11)^2*ln(7)^2*ln(3)^2-1/4*ln(3)^4*ln(11)^2-1/4*ln(3)^2*ln(7)^4)*l
[2]^2
+(ln(3)^2*ln(5)^2*ln(11)^2-1/4*ln(3)^2*ln(5)^4-1/4*ln(3)^4*ln(11)^2)*l
[3]^2
+(ln(7)^2*ln(3)^2*ln(5)^2-1/4*ln(3)^2*ln(5)^4-1/4*ln(7)^2*ln(3)^4)*l
[4]^2
+(1/2*ln(3)^2*ln(5)^2*ln(11)^2-ln(11)^2*ln(7)^2*ln(3)^2+1/4*ln(3)^2*ln
(7)^4)*l[1]*l[2]
+(-1/2*ln(3)^2*ln(5)^2*ln(11)^2+1/4*ln(7)^2*ln(3)^2*ln(5)^2)*l[1]*l[3]
-1/4*ln(7)^2*ln(3)^2*ln(5)^2*l[1]*l[4]
+(-ln(3)^2*ln(5)^2*ln(11)^2+1/2*ln(7)^2*ln(3)^2*ln(5)^2-1/4*ln(7)^2*ln
(3)^4+
1/2*ln(3)^4*ln(11)^2)*l[2]*l[3]+(-1/2*ln(7)^2*ln(3)^2*ln(5)^2+1/4*ln
(7)^2*ln(3)^4)*l[2]*l[4]
+(-ln(7)^2*ln(3)^2*ln(5)^2+1/2*ln(3)^2*ln(5)^4+1/4*ln(7)^2*ln(3)^4)*l
[3]*l[4])) end:

11-limit planar:

gpc11 := proc(l)
# l is 11-limit planar wedgie
sqrt(evalf((-1/4*ln(7)^4+ln(7)^2*ln(11)^2)*l[1]^2+(-1/4*ln(5)^4+ln(5)
^2*ln(11)^2)*l[2]^2
+(-1/4*ln(5)^4+ln(5)^2*ln(7)^2)*l[3]^2+(-1/4*ln(3)^4+ln(3)^2*ln(11)^2)
*l[4]^2
+(-1/4*ln(3)^4+ln(3)^2*ln(7)^2)*l[5]^2+(-1/4*ln(3)^4+ln(3)^2*ln(5)^2)
*l[6]^2
-(-1/2*ln(5)^2*ln(7)^2+ln(5)^2*ln(11)^2)*l[1]*l[2]-1/2*ln(5)^2*ln(7)
^2*l[1]*l[3]
+(-1/2*ln(3)^2*ln(7)^2+ln(3)^2*ln(11)^2)*l[1]*l[4]+1/2*ln(3)^2*ln(7)
^2*l[1]*l[5]
+(1/2*ln(3)^2*ln(5)^2-1/2*ln(3)^2*ln(7)^2)*l[1]*l[6]-(-1/2*ln(5)^4+ln
(5)^2*ln(7)^2)*l[2]*l[3]
-(-1/2*ln(3)^2*ln(5)^2+ln(3)^2*ln(11)^2)*l[2]*l[4]+(-ln(3)^2*ln(5)
^2+ln(3)^2*ln(7)^2)*l[2]*l[5]
+1/2*ln(3)^2*ln(5)^2*l[2]*l[6]-(-1/2*ln(3)^2*ln(5)^2+ln(3)^2*ln(7)^2)
*l[3]*l[5]
-1/2*ln(3)^2*ln(5)^2*l[3]*l[6]-(-1/2*ln(3)^4+ln(3)^2*ln(7)^2)*l[4]*l
[5]
+(-1/2*ln(3)^4+ln(3)^2*ln(5)^2)*l[4]*l[6]-(-1/2*ln(3)^4+ln(3)^2*ln(5)
^2)*l[5]*l[6])) end:

It's unfortunately true that these results are not immediate from the
definition of geometric complexity, but must themselves be computed,
so only for 5-limit is this straightforward.

> If I had worked out your numbering rule, I've forgotten it now.
>
> Every time you try to explain something, you bring in more jargon
terms
> that I don't understand (I can't speak for anybody else).
>
> The word "metric" in particular is something that's important but
you
> haven't defined.

I've posted this before; it's standard math:

http://mathworld.wolfram.com/Metric.html

> You keep missing out important steps in explanations, like the need
to
> take the complement at certain points when using wedge products.

As I've explained before, the way I wrote my Maple code I don't need
to. All that is built into the functions themselves.

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >> >I would like to know which temperment/comma connects 12, 47, 35,
and
> >> >23 ets on zoomr.gif. How could I calculate these for myself?
Thanks!
> >> >If this is a new one, could I name it? What is the 5-limit
vector?
> >>
> >> The comma is 6561/6250, according to Gene's maple, if I did it
right.
> >
> >I get the same. You get a temperament with a generator the size of
a
> >semitone, five of which give a fourth, and eight of which give a
minor
> >sixth, which obviously is compatible with 12-et.
>
> So what I'm missing is... the heuristic says temperaments based on
> this comma will be bad.

which heuristic -- the one for complexity or the one for error?

> And indeed, 23, 47 don't seem to be very
> good. But 12 is good. So what gives?

the heuristic for error applies to the best choice, in some sense. so
something close to 12. but i don't know how you came up with "bad" ~=
"good" inequality here. could you try to be more quantitative?

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Carl Lumma <ekin@lumma.org>

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >> >For a linear temperament, the number of consonances is (roughly)
> >> >the number of notes minus the complexity. So n-i is a constant.
> >> >If you then plot n/i, you can replace n with i+c to get
> >> >(i+c)/i = 1+c/i. As i tends to infinity, this tends to 1, which
> >> >is as small as it gets. So your measure would depend on how
> >> >large n is allowed to get.
> >>
> >> Ok, I follow your reasoning but I'm not sure what your conclusion
> >> is. I think we'd limit n to the number of notes in the Fokker
> >> block.
> >
> >what's "the" fokker block? you mean "a" fokker block?
>
> A, but clearly straightness or something similar needs to be
invoked.
>
> -Carl

not only that, but you're missing a unison vector! a linear
temperament has one fewer defining unison vectors than a periodicity
block in the same space.

🔗Carl Lumma <ekin@lumma.org>

🔗Graham Breed <graham@microtonal.co.uk>

paulhjelmstad wrote:

> Thanks. Where can I get the Library for Python? I have the command > shell but get an error running the above command.

"The library" is the file temper.py, linked as "script" from the source code page. If it's in your Python path, you import it with the command "import temper". If that gives an error, you either need to read the documentation on the Python path, or follow this recipe (which depends on where temper.py is):

>>> import sys
>>> sys.path.append(r'd:/net/microtonal')
>>> import temper

The first "test script" on the source code page gives you examples of how to use the library. The dir and help (depending on the Python version) builtin functions can be useful:

>>> dir(temper)
['BestET', 'EqualTemperament', 'LinearTemperament', 'OddLimit', 'PrimeDiamond', 'PrimeET', 'Temperament', 'TemperamentException', 'TemperamentList', 'TonalityDiamond', 'UserDict', 'UserList', 'WedgableRatio', 'Wedgie', 'WedgieET', '__builtins__', '__doc__', '__file__', '__name__', 'alternativeMappings', 'combinations', 'defaultFigureOfDemerit', 'dotprod', 'euclidianGCD', 'exp', 'factorize', 'factorizeRatio', 'getCombinations', 'getDimensions', 'getDivisors', 'getGenerator', 'getLimitedETs', 'getLinearTemperaments', 'getRatio', 'hcf', 'intervalCompare', 'limit11', 'limit13', 'limit15', 'limit17', 'limit19', 'limit21', 'limit5', 'limit7', 'limit9', 'linearTemperamentsFromIntervals', 'linearTemperamentsFromUnisonVectors', 'log', 'log2', 'makeWedgie', 'new', 'nint', 'normalizeInterval', 'operator', 'primeNumbers', 'primes', 'ratioMatch', 're', 'readVectors', 'reduceWithinTritone', 'simplestOddLimit', 'sqrt', 'string', 'temperOut', 'uniqueWithinTritone', 'unisonVectorsFromIntervals', 'wedgeEquivalent', 'wedgeProduct']
>>> help(temper.BestET)
Help on function BestET in module temper:

BestET(nNotes, consonances, cutoff=1.0)

---

To understand Python itself, see the tutorial at

http://www.python.org/doc/current/tut/tut.html

and you probably downloaded a copy of that with the interpreter.

Graham

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

--- In tuning-math@yahoogroups.com, Graham Breed <graham@m...> wrote:
> paulhjelmstad wrote:
>
> > Thanks. Where can I get the Library for Python? I have the
command
> > shell but get an error running the above command.
>
> "The library" is the file temper.py, linked as "script" from the
source
> code page. If it's in your Python path, you import it with the
command
> "import temper". If that gives an error, you either need to read
the
> documentation on the Python path, or follow this recipe (which
depends
> on where temper.py is):
>
> >>> import sys
> >>> sys.path.append(r'd:/net/microtonal')
> >>> import temper
>
> The first "test script" on the source code page gives you examples
of
> how to use the library. The dir and help (depending on the Python
> version) builtin functions can be useful:
>
> >>> dir(temper)
> ['BestET', 'EqualTemperament', 'LinearTemperament', 'OddLimit',
> 'PrimeDiamond', 'PrimeET', 'Temperament', 'TemperamentException',
> 'TemperamentList', 'TonalityDiamond', 'UserDict', 'UserList',
> 'WedgableRatio', 'Wedgie', 'WedgieET', '__builtins__', '__doc__',
> '__file__', '__name__', 'alternativeMappings', 'combinations',
> 'defaultFigureOfDemerit', 'dotprod', 'euclidianGCD', 'exp', 'factori
ze',
> 'factorizeRatio', 'getCombinations', 'getDimensions', 'getDivisors',

> 'getGenerator', 'getLimitedETs', 'getLinearTemperaments', 'getRatio'
,
> 'hcf', 'intervalCompare', 'limit11', 'limit13', 'limit15', 'limit17'
,
> 'limit19', 'limit21', 'limit5', 'limit7', 'limit9',
> 'linearTemperamentsFromIntervals',
> 'linearTemperamentsFromUnisonVectors', 'log', 'log2', 'makeWedgie',
> 'new', 'nint', 'normalizeInterval', 'operator', 'primeNumbers',
> 'primes', 'ratioMatch', 're', 'readVectors', 'reduceWithinTritone',
> 'simplestOddLimit', 'sqrt', 'string', 'temperOut',
> 'uniqueWithinTritone', 'unisonVectorsFromIntervals', 'wedgeEquivalen
t',
> 'wedgeProduct']
> >>> help(temper.BestET)
> Help on function BestET in module temper:
>
> BestET(nNotes, consonances, cutoff=1.0)
>
>
> ---
>
> To understand Python itself, see the tutorial at
>
> http://www.python.org/doc/current/tut/tut.html
>
> and you probably downloaded a copy of that with the interpreter.
>
>
> Graham

Thanks for all the info. I have a ton of libraries, but no temper.py
I downloaded Python a while ago.

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

--- In tuning-math@yahoogroups.com, "paulhjelmstad"
<paul.hjelmstad@u...> wrote:
> --- In tuning-math@yahoogroups.com, Graham Breed <graham@m...>
wrote:
> > paulhjelmstad wrote:
> >
> > > I would like to know which temperment/comma connects 12, 47,
35,
> and
> > > 23 ets on zoomr.gif. How could I calculate these for myself?
> Thanks!
> > > If this is a new one, could I name it? What is the 5-limit
vector?
> >
> > Go to
> >
> > http://microtonal.co.uk/temper/twoet.html
> >
> > and enter 12 and 47 for the ets, and 5 for the limit. That gives
> you
> > the mapping
> >
> > [(1, 0), (2, -5), (3, -8)]
> >
> > It doesn't show the comma because I don't see the point, and they
> aren't
> > that easy to calculate for some temperaments. But for the 5-
limit
> case
> > you can work it out from the mapping. It must involve 3**8 and
> 5**5
> > with opposite signs, from the octave equivalent mapping (-5, -
8).
> You
> > then multiply (x, 8, -5) by the period part of the mapping to get
x
> +
> > 2*8 - 3*5. For a unison vector, this should be zero, so x = -2*8
+
> 3*5
> > = 15-16 = -1. So the comma is (-1 8 -5) or 2**(-1) * 3**8 * 5**(-
> 5) =
> > 6561:6250.
> >
> > There are different tools runnable online at
> > http://microtonal.co.uk/temper/ and you can get the library for
> Python,
> > which is free, and play with it as you like. To get commas for a
> linear
> > temperament as ratios,
> >
> > >>> map(temper.getRatio, temper.Temperament(12,47,
> > temper.limit5).getUnisonVectors())
> > [(6561, 6250)]
> >
> >
> > Graham
>
> Thanks. Where can I get the Library for Python? I have the command
> shell but get an error running the above command.
Nevermind. I found the library and am running Python scripts, Cool.

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

🔗Carl Lumma <ekin@lumma.org>

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Carl Lumma <ekin@lumma.org>

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Carl Lumma <ekin@lumma.org>

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Carl Lumma <ekin@lumma.org>

>>>A linear temperament always needs one unison vector less than a
>>>periodicity block. An equal temperament needs the same number,
>>>a planar temperament needs two less, and so on.
>>
>> So this formulation does depend on the space, since the number
>> needed for a block depends on the space!
>
>but the *difference* is always 1, regardless of the space. it's
>called the "codimension".

Ok, but what's happening to this comma? It's the one that doesn't
vanish? Why is it any less "defining" than the others?

I thought it's what *does* vanish that defines things. Wait --
is it that in a PB, everything 'vanishes', even though it doesn't,
and Graham's ambiguous grammar (above) means to say that ets and
pbs have the same number? Then it makes sense.

-Carl

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> >>>A linear temperament always needs one unison vector less than a
> >>>periodicity block. An equal temperament needs the same number,
> >>>a planar temperament needs two less, and so on.
> >>
> >> So this formulation does depend on the space, since the number
> >> needed for a block depends on the space!
> >
> >but the *difference* is always 1, regardless of the space. it's
> >called the "codimension".
>
> Ok, but what's happening to this comma?

what's happening to *which* comma?

> It's the one that doesn't
> vanish?
> Why is it any less "defining" than the others?

let's get straight which comma we're talking about. how about an
example.

> I thought it's what *does* vanish that defines things.

no, for example a 3-d ji pb is defined by 3 independent unison
vectors, none of which vanish.

🔗Carl Lumma <ekin@lumma.org>

>> >>>A linear temperament always needs one unison vector less than a
>> >>>periodicity block. An equal temperament needs the same number,
>> >>>a planar temperament needs two less, and so on.
>> >>
>> >> So this formulation does depend on the space, since the number
>> >> needed for a block depends on the space!
>> >
>> >but the *difference* is always 1, regardless of the space. it's
>> >called the "codimension".
>>
>> Ok, but what's happening to this comma?
>
>what's happening to *which* comma?

The one that the pb needs but the lt doesn't.

>> It's the one that doesn't vanish?
>> Why is it any less "defining" than the others?
>
>let's get straight which comma we're talking about. how about an
>example.

Howabout a 7-limit block defined by 81:80, 25:24, and 36:35?

>> I thought it's what *does* vanish that defines things.
>
>no, for example a 3-d ji pb is defined by 3 independent unison
>vectors, none of which vanish.

But meantone is defined by 81:80, which does vanish.

-Carl

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

--- In tuning-math@yahoogroups.com, "paulhjelmstad"
<paul.hjelmstad@u...> wrote:
> --- In tuning-math@yahoogroups.com, "paulhjelmstad"
> <paul.hjelmstad@u...> wrote:
> > --- In tuning-math@yahoogroups.com, Graham Breed <graham@m...>
> wrote:
> > > paulhjelmstad wrote:
> > >
> > > > I would like to know which temperment/comma connects 12, 47,
> 35,
> > and
> > > > 23 ets on zoomr.gif. How could I calculate these for myself?
> > Thanks!
> > > > If this is a new one, could I name it? What is the 5-limit
> vector?
> > >
> > > Go to
> > >
> > > http://microtonal.co.uk/temper/twoet.html
> > >
> > > and enter 12 and 47 for the ets, and 5 for the limit. That
gives
> > you
> > > the mapping
> > >
> > > [(1, 0), (2, -5), (3, -8)]
> > >
> > > It doesn't show the comma because I don't see the point, and
they
> > aren't
> > > that easy to calculate for some temperaments. But for the 5-
> limit
> > case
> > > you can work it out from the mapping. It must involve 3**8 and
> > 5**5
> > > with opposite signs, from the octave equivalent mapping (-5, -
> 8).
> > You
> > > then multiply (x, 8, -5) by the period part of the mapping to
get
> x
> > +
> > > 2*8 - 3*5. For a unison vector, this should be zero, so x = -
2*8
> +
> > 3*5
> > > = 15-16 = -1. So the comma is (-1 8 -5) or 2**(-1) * 3**8 * 5**
(-
> > 5) =
> > > 6561:6250.
> > >
> > > There are different tools runnable online at
> > > http://microtonal.co.uk/temper/ and you can get the library for
> > Python,
> > > which is free, and play with it as you like. To get commas for
a
> > linear
> > > temperament as ratios,
> > >
> > > >>> map(temper.getRatio, temper.Temperament(12,47,
> > > temper.limit5).getUnisonVectors())
> > > [(6561, 6250)]
> > >
> > >
> > > Graham
> >
> > I'm having fun with Python. I would like to review the connection
between wedgies and commas. Please see below:

Let p = 2^u1 3^u2 5^u3 7^u4 and q = 2^v1 3^v2 5^v3 7^v4, then

p ^ q = [u3*v4-v3*u4,u4*v2-v4*u2,u2*v3-v2*u3,u1*v2-v1*u2,
u1*v3-v1*u3,u1*v4-v1*u4]

Let r be the mapping to primes of an equal temperament given
by r = [u1, u2, u3, u4], and s be given by [v1, v2, v3, v4]. This
means r has u1 notes to the octave, u2 notes in the approximation of
3, and so forth; hence [12, 19, 28, 24] would be the usual 12-equal,
and [31, 49, 72, 87] the usual 31-et. The wedge now is

r ^ s = [u1*v2-v1*u2,u1*v3-v1*u3,u1*v4-v1*u4,u3*v4-v3*u4,
u4*v2-u2*v4,u2*v3-v2*u3]

Whether we've computed in terms of commas or ets, the wedge product
of the linear temperament is exactly the same, up to sign.

If the wedgie is [u1,u2,u3,u4,u5,u6] then we have commas given by

2^u6 3^(-u2) 5^u1
2^u5 3^u3 7^(-u1)
2^u4 5^(-u3) 7^u2
3^u4 5^u5 7^u6

I see that the wedgie was calculated from the two commas, but here
(at the end of Gene's message) we also have 4 commas derived from the
wedgie, how does this work? How are the particular commas above
derived from that particular wedgie. Thanks

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >> >>>A linear temperament always needs one unison vector less than
a
> >> >>>periodicity block. An equal temperament needs the same
number,
> >> >>>a planar temperament needs two less, and so on.
> >> >>
> >> >> So this formulation does depend on the space, since the number
> >> >> needed for a block depends on the space!
> >> >
> >> >but the *difference* is always 1, regardless of the space. it's
> >> >called the "codimension".
> >>
> >> Ok, but what's happening to this comma?
> >
> >what's happening to *which* comma?
>
> The one that the pb needs but the lt doesn't.

well, that comma's presence in defining the system makes the set of
notes finite. without it, the set of notes is infinite, since linear
temperaments contain an infinite number of notes per octave. does
that answer your question?

> >> It's the one that doesn't vanish?
> >> Why is it any less "defining" than the others?
> >
> >let's get straight which comma we're talking about. how about an
> >example.
>
> Howabout a 7-limit block defined by 81:80, 25:24, and 36:35?

ok, so if you temper out two of these, and ignore the third, you get
a linear temperament with an infinite number of notes. is there one
of the three you wanted to use for this purpose?

> >> I thought it's what *does* vanish that defines things.
> >
> >no, for example a 3-d ji pb is defined by 3 independent unison
> >vectors, none of which vanish.
>
> But meantone is defined by 81:80, which does vanish.

yes, but we could just as easily define a "slice" of ji using 81:80
in non-vanishing form -- leading to constructs like monz's and
terpstra's "just interpretations" of meantone (which are of course
deficient since they have 27:20 "wolves" and the like). "defining"
means "delimiting" when it comes to unison vectors -- they section
off a region of the space, but if they vanish, one is effectively
free to move from this region to any another, since one will simply
be using the same set of pitches anyway!

🔗Graham Breed <graham@microtonal.co.uk>

🔗Carl Lumma <ekin@lumma.org>

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> Hey Paul E., you back already?

yup!

> How was/is Washington?

hot. nice.

> I'm hoping you'll get a chance to look at the harmonic
> entropy thread.

i've already replied to everything, so i'm not sure what you're
referring to.

> As far as this temperament stuff goes, I'm not sure what
> I didn't understand, if anything. I'll just voice again
> a lack of introductory materials that explain the basic
> tools needed to do scale building with temperaments.

_the forms of tonality_?

> If a linear temperament has an infinite number of notes,
> what happens when we cast it into a scale? If a 5-limit
> lt has chromatic uv 25:24 and commatic uv 81:80, do we
> have a name for the "wolf" comma formed between the ends
> of the chain?
>
> -Carl

the chromatic uv. that is, if you put the last note on one end of the
chain, it will differ by the chromatic uv compared to where it would
be if you put it on the other end instead. is this what you had in
mind?

🔗Carl Lumma <ekin@lumma.org>

>> How was/is Washington?
>
>hot. nice.

'dyou get a recording of the gig?

>> I'm hoping you'll get a chance to look at the harmonic
>> entropy thread.
>
>i've already replied to everything, so i'm not sure what
>you're referring to.

Mm, maybe nothing then.

>> As far as this temperament stuff goes, I'm not sure what
>> I didn't understand, if anything. I'll just voice again
>> a lack of introductory materials that explain the basic
>> tools needed to do scale building with temperaments.
>
>_the forms of tonality_?

Is fantastic. But I don't think it covers many of the
topics on this list. It isn't online, and doesn't have any
automation.

>>A linear temperament always needs one unison vector less
>>than periodicity block. An equal temperament needs the same
>>number, a planar temperament needs two less, and so on.
//
>does that answer your question?

First off, how many does an et need:

(a) Same as pb.
(b) Same as lt.

??????????????

I think (a), which makes my blurb from a few posts back
correct.

>> thought it's what *does* vanish that defines things.
>
>no, for example a 3-d ji pb is defined by 3 independent unison
>vectors, none of which vanish.

But a pb isn't a temperament. Temperaments are defined by
the commas that vanish, no?

>> If a linear temperament has an infinite number of notes,
>> what happens when we cast it into a scale? If a 5-limit
>> lt has chromatic uv 25:24 and commatic uv 81:80, do we
>> have a name for the "wolf" comma formed between the ends
>> of the chain?
>
>the chromatic uv. that is, if you put the last note on one
>end of the chain, it will differ by the chromatic uv compared
>to where it would be if you put it on the other end instead.
>is this what you had in mind?

I was speaking of the (generator + chromatic uv) interval,
which I will call the wolf.

1. Does the wolf have any significance apart from the
chromatic uv? Gene's T[n] stuff seems to show it does.

2. How is the difference between the pitches of a new note
at either end of the chain the same regardless of the length
of the chain? If I chain three 700-cent intervals, this
"chromatic uv" as you define it above is 100 cents. If
chain two of them it is 300 cents!

-Carl

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >> How was/is Washington?
> >
> >hot. nice.
>
> 'dyou get a recording of the gig?

yup, dave recorded it on minidisc . . . we did the tunes on our demo
plus 4 more, which we should be laying down in the studio soon . . .

>
> >>A linear temperament always needs one unison vector less
> >>than periodicity block. An equal temperament needs the same
> >>number, a planar temperament needs two less, and so on.
> //
> >does that answer your question?
>
> First off, how many does an et need:
>
> (a) Same as pb.

yes, that's what graham wrote in what you quoted above.

> (b) Same as lt.

no, a lt needs one less, as graham and i told you.

> >> thought it's what *does* vanish that defines things.
> >
> >no, for example a 3-d ji pb is defined by 3 independent unison
> >vectors, none of which vanish.
>
> But a pb isn't a temperament. Temperaments are defined by
> the commas that vanish, no?

yes.

> >> If a linear temperament has an infinite number of notes,
> >> what happens when we cast it into a scale? If a 5-limit
> >> lt has chromatic uv 25:24 and commatic uv 81:80, do we
> >> have a name for the "wolf" comma formed between the ends
> >> of the chain?
> >
> >the chromatic uv. that is, if you put the last note on one
> >end of the chain, it will differ by the chromatic uv compared
> >to where it would be if you put it on the other end instead.
> >is this what you had in mind?
>
> I was speaking of the (generator + chromatic uv) interval,
> which I will call the wolf.

ok!

> 1. Does the wolf have any significance apart from the
> chromatic uv? Gene's T[n] stuff seems to show it does.

how so?

> 2. How is the difference between the pitches of a new note
> at either end of the chain the same regardless of the length
> of the chain?

of course it isn't! if you'd chosen a different chromatic unison
vector, you'd most likely end up with a different number of notes per
octave (the fokker determinant tells you that), and thus a different
length for each chain.

> If I chain three 700-cent intervals, this
> "chromatic uv" as you define it above is 100 cents.

how do you get that? that seems way too small. and since the scale is
not even an MOS (or what we used to call MOS), you're not logically
justified to call it a chromatic uv at all. i wasn't defining it
above, i was explaining what happens when you use one. and, now that
i look, you made an incorrect statement in your question, which i
should have corrected:

"If a 5-limit lt has chromatic uv 25:24 and commatic uv 81:80,"

that's not an lt at all -- it's an MOS scale (in fact, it's the
meantone diatonic scale).

> If
> chain two of them it is 300 cents!

yes, that looks right, you have C-G-D vs. G-D-A, and A-C is 300
cents. 32:27 is what you might call this unison vector. this 3-tone
scale is nice and useful, but not entirely common.

🔗Graham Breed <graham@microtonal.co.uk>

Gene Ward Smith wrote:

> Not true any more, and I think I posted this. Here is the Maple code > for 7-limit geometric complexity:

Sorry, yes, you have shown the derivation of real numbers. But still no general algorithm or full explanation of the indexing rules.

> It's unfortunately true that these results are not immediate from the > definition of geometric complexity, but must themselves be computed, > so only for 5-limit is this straightforward.

Yes, I chased this down, and found a definition here:

/tuning-math/message/5546

Given that it is so straightforward, why do you exclude it from your other listings?

So, I've got as far as converting that into Python:

import math, temper

def complexity5(u, primes=None):
primes = primes or temper.primes[:u.maxBasis()]
return math.log(2) * math.sqrt(
primes[0]**2 * u[1,]**2 +
primes[0]**2 * u[1,] * u[2,] +
primes[1]**2 * u[2,]**2)

>>If I had worked out your numbering rule, I've forgotten it now.
>>
>>Every time you try to explain something, you bring in more jargon > > terms > >>that I don't understand (I can't speak for anybody else).
>>
>>The word "metric" in particular is something that's important but > > you > >>haven't defined.
> > > I've posted this before; it's standard math:
> > http://mathworld.wolfram.com/Metric.html

*That's* standard math, yes, but it doesn't say anything about applying a metric to an exterior algebra. And in the geometric complexity definiton:

/tuning-math/message/4533

you give what is, as far as I can work out, a function of a single rational number as the "metric". Yet the definition you pointed to says a metric is a function of two variables! It looks more like a norm than a metric to me, but I still don't know how to apply a norm to an exterior algebra.

>>You keep missing out important steps in explanations, like the need > > to > >>take the complement at certain points when using wedge products.
> > > As I've explained before, the way I wrote my Maple code I don't need > to. All that is built into the functions themselves.

Do we have to go through this again? I thought I finally pinned you down before. Here:

/tuning-math/message/5673

you say "...I store the wedgies as lists, and reverse the ordering when I compute from commas, etc...". Doesn't that mean that reversing the ordering is your implementation of the complement?

Besides which, whatever your Maple code does, your explanations only defined a wedge product. Not a complement operation or a dual space or however you claim to get around using a complement. So a step is missing.

Except for here:

/tuning-math/message/5583

where you give different formulae for the wedge products of intervals and vals. And say "These two types of wedge product can be
indentified with each other, by `Poincare duality'. It does not matter
whether the wedgie comes from commas or vals, therefore." Not only do you use an undefined jargon term that you can't expect us to know, but you use it to dismiss something that clearly does matter, because you (or I, at least) don't get the right results without it.

Or maybe you meant that the geometric complexity of a wedgie is equal to that of its complement?

And yes, Poincarï¿½ Duality is in Mathworld, but it's completely impenetrable.

Graham

🔗Carl Lumma <ekin@lumma.org>

>yup, dave recorded it on minidisc . . . we did the tunes on
>our demo plus 4 more, which we should be laying down in the
>studio soon . . .

Suh-weet.

>> >>A linear temperament always needs one unison vector less
>> >>than periodicity block. An equal temperament needs the same
>> >>number, a planar temperament needs two less, and so on.
>> //
>> >does that answer your question?
>>
>> First off, how many does an et need:
>>
>> (a) Same as pb.
>
>yes, that's what graham wrote in what you quoted above.

His gramar was ambiguous, as I said before!

>> (b) Same as lt.
>
>no, a lt needs one less, as graham and i told you.

You never mentioned ets that I saw.

>> >> thought it's what *does* vanish that defines things.
>> >
>> >no, for example a 3-d ji pb is defined by 3 independent unison
>> >vectors, none of which vanish.
>>
>> But a pb isn't a temperament. Temperaments are defined by
>> the commas that vanish, no?
>
>yes.

Thank god.

>> >> If a linear temperament has an infinite number of notes,
>> >> what happens when we cast it into a scale? If a 5-limit
>> >> lt has chromatic uv 25:24 and commatic uv 81:80, do we
>> >> have a name for the "wolf" comma formed between the ends
>> >> of the chain?
>> >
>> >the chromatic uv. that is, if you put the last note on one
>> >end of the chain, it will differ by the chromatic uv compared
>> >to where it would be if you put it on the other end instead.
>> >is this what you had in mind?
>>
>> I was speaking of the (generator + chromatic uv) interval,
>> which I will call the wolf.
>
>ok!
>
>> 1. Does the wolf have any significance apart from the
>> chromatic uv? Gene's T[n] stuff seems to show it does.
>
>how so?

Did you catch that thread. Gene's probably the better one to
explain it. Or maybe search for "T[n]".

He basically looked at not-necessarily-MOS n for popular
temperaments T, and found that the wolf was sometimes itself
consonant, which added to the utility of the tuning. Or
something. I've been struggling to understand how this fits
into the temperament terminology. For 5-limit lts, does it
represent a 3rd comma? Does the use of it to form consonant
intervals break regularity/consistency? Etc.

>> 2. How is the difference between the pitches of a new note
>> at either end of the chain the same regardless of the length
>> of the chain?
>
>of course it isn't! if you'd chosen a different chromatic unison
>vector, you'd most likely end up with a different number of
>notes per octave (the fokker determinant tells you that), and
>thus a different length for each chain.

So the chromatic uv tells you the length! So maybe the T[n]
stuff was just about the chromatic uv!

>> If I chain three 700-cent intervals, this
>> "chromatic uv" as you define it above is 100 cents.
>
>how do you get that? that seems way too small.

(F)-C-G-D-A-(E); F-E=100cents

>since the scale is not even an MOS (or what we used to call MOS),
>you're not logically justified to call it a chromatic uv at all.

Ok, this is a bombshell. How can we flesh this out a bit?

>i wasn't defining it above, i was explaining what happens when
>you use one. and, now that i look, you made an incorrect statement
>in your question, which i should have corrected:
>
>"If a 5-limit lt has chromatic uv 25:24 and commatic uv 81:80,"
>
>that's not an lt at all -- it's an MOS scale (in fact, it's the
>meantone diatonic scale).

Is this because lts don't have chromatic uvs?

>> If chain two of them it is 300 cents!
>
>yes, that looks right, you have C-G-D vs. G-D-A, and A-C is 300
>cents. 32:27 is what you might call this unison vector. this
>3-tone scale is nice and useful, but not entirely common.

It is MOS, IIRC.

-Carl

🔗Paul Erlich <perlich@aya.yale.edu>

> >> 1. Does the wolf have any significance apart from the
> >> chromatic uv? Gene's T[n] stuff seems to show it does.
> >
> >how so?
>
> Did you catch that thread.

yes.

> found that the wolf was sometimes itself
> consonant, which added to the utility of the tuning.

ok.

> Or
> something. I've been struggling to understand how this fits
> into the temperament terminology. For 5-limit lts, does it
> represent a 3rd comma?

3rd? 5-limit lts are defined by 1 comma.

> Does the use of it to form consonant
> intervals break regularity/consistency? Etc.

consistency is only defined for ets. it doesn't break regularity,
though it may break the universality of the mapping, if the wolf is
another form of a consonant interval already mapped, and you want to
use it as a consonance.

> >> 2. How is the difference between the pitches of a new note
> >> at either end of the chain the same regardless of the length
> >> of the chain?
> >
> >of course it isn't! if you'd chosen a different chromatic unison
> >vector, you'd most likely end up with a different number of
> >notes per octave (the fokker determinant tells you that), and
> >thus a different length for each chain.
>
> So the chromatic uv tells you the length! So maybe the T[n]
> stuff was just about the chromatic uv!

yes (in many cases it is), apparently you're the one who missed a lot
of that thread. in fact, gene didn't even understand what a chromatic
unison vector was until he posted his T[n] idea and i explained the
relationship to him.

>
> >> If I chain three 700-cent intervals, this
> >> "chromatic uv" as you define it above is 100 cents.
> >
> >how do you get that? that seems way too small.
>
> (F)-C-G-D-A-(E); F-E=100cents

that's not three 700-cent intervals; it's a pentatonic scale, so four
700-cent intervals. this is just the gentle introduction to fokker
periodicity blocks, part 1, revisited.

>
> >since the scale is not even an MOS (or what we used to call MOS),
> >you're not logically justified to call it a chromatic uv at all.
>
> Ok, this is a bombshell. How can we flesh this out a bit?

well, your scale above *is* an MOS, so we're fine. the point is that
the chromatic uv would have to be capable of producing a ji variant
of the scale when it, along with any commatic uvs defining the
temperament, are used to define a fokker periodicity block.

> >i wasn't defining it above, i was explaining what happens when
> >you use one. and, now that i look, you made an incorrect statement
> >in your question, which i should have corrected:
> >
> >"If a 5-limit lt has chromatic uv 25:24 and commatic uv 81:80,"
> >
> >that's not an lt at all -- it's an MOS scale (in fact, it's the
> >meantone diatonic scale).
>
> Is this because lts don't have chromatic uvs?

not until you start choosing finite scales from the lt. before that,
you've got an infinite number of *potential* chromatic unison
vectors, but none that are operative as yet.

> >> If chain two of them it is 300 cents!
> >
> >yes, that looks right, you have C-G-D vs. G-D-A, and A-C is 300
> >cents. 32:27 is what you might call this unison vector. this
> >3-tone scale is nice and useful, but not entirely common.
>
> It is MOS, IIRC.

yes, and as always, for any generic interval that comes in two
different sizes, the difference between the sizes is the chromatic
uv.

🔗Carl Lumma <ekin@lumma.org>

>> Or something. I've been struggling to understand how this fits
>> into the temperament terminology. For 5-limit lts, does it
>> represent a 3rd comma?
>
>3rd? 5-limit lts are defined by 1 comma.

Eep! 2nd, that is!

>> Does the use of it to form consonant
>> intervals break regularity/consistency? Etc.
>
>consistency is only defined for ets.

What Gene said about regular temperaments sounded
equivalent to me.

>it may break the universality of the mapping, if the wolf is
>another form of a consonant interval already mapped, and you
>want to use it as a consonance.

Yeah, that's what I'm thinkin'. Let's take kleismic[8] as an
example. It's non-MOS but a good scale. It's proper, too.

>> So the chromatic uv tells you the length! So maybe the T[n]
>> stuff was just about the chromatic uv!
>
>yes (in many cases it is), apparently you're the one who missed
>a lot of that thread. in fact, gene didn't even understand what
>a chromatic unison vector was until he posted his T[n] idea and
>i explained the relationship to him.

In fact I think this is coming back.

>>>since the scale is not even an MOS (or what we used to call MOS),
>>>you're not logically justified to call it a chromatic uv at all.
>>
>> Ok, this is a bombshell. How can we flesh this out a bit?
>
>well, your scale above *is* an MOS, so we're fine. the point is that
>the chromatic uv would have to be capable of producing a ji variant
>of the scale when it, along with any commatic uvs defining the
>temperament, are used to define a fokker periodicity block.

And what happens in the case of kleismic[8]?

>> >i wasn't defining it above, i was explaining what happens when
>> >you use one. and, now that i look, you made an incorrect statement
>> >in your question, which i should have corrected:
>> >
>> >"If a 5-limit lt has chromatic uv 25:24 and commatic uv 81:80,"
>> >
>> >that's not an lt at all -- it's an MOS scale (in fact, it's the
>> >meantone diatonic scale).
>>
>> Is this because lts don't have chromatic uvs?
>
>not until you start choosing finite scales from the lt. before that,
>you've got an infinite number of *potential* chromatic unison
>vectors, but none that are operative as yet.

Got it. Thanks!

-Carl

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> >consistency is only defined for ets.
>
> What Gene said about regular temperaments sounded
> equivalent to me.

a definition of consistency for regular temperaments? doesn't seem
possible, as they're infinite, so you can always find better and
better approximations to anything.

> >it may break the universality of the mapping, if the wolf is
> >another form of a consonant interval already mapped, and you
> >want to use it as a consonance.
>
> Yeah, that's what I'm thinkin'. Let's take kleismic[8] as an
> example. It's non-MOS but a good scale. It's proper, too.

ok, and what does the wolf do here? i'm in a rush so can't figure it
out right now . . .

> And what happens in the case of kleismic[8]?

i'll have to answer this from the office tomorrow.

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, Graham Breed <graham@m...> wrote:
> Yes, I chased this down, and found a definition here:
>
> /tuning-math/message/5546
>
> Given that it is so straightforward, why do you exclude it from
your
> other listings?

It follows immediately from the definition.

> So, I've got as far as converting that into Python:
>
> import math, temper
>
> def complexity5(u, primes=None):
> primes = primes or temper.primes[:u.maxBasis()]
> return math.log(2) * math.sqrt(
> primes[0]**2 * u[1,]**2 +
> primes[0]**2 * u[1,] * u[2,] +
> primes[1]**2 * u[2,]**2)
>
>
> >>If I had worked out your numbering rule, I've forgotten it now.
> >>
> >>Every time you try to explain something, you bring in more jargon
> >
> > terms
> >
> >>that I don't understand (I can't speak for anybody else).
> >>
> >>The word "metric" in particular is something that's important but
> >
> > you
> >
> >>haven't defined.
> >
> >
> > I've posted this before; it's standard math:
> >
> > http://mathworld.wolfram.com/Metric.html
>
> *That's* standard math, yes, but it doesn't say anything about
applying
> a metric to an exterior algebra. And in the geometric complexity
definiton:
>
> /tuning-math/message/4533
>
> you give what is, as far as I can work out, a function of a single
> rational number as the "metric".

A single rational number, in terms of 5-limit note-classes, is a two-
dimensional vector. Does that explain it for you?

🔗Graham Breed <graham@microtonal.co.uk>

Me:
>>Given that it is so straightforward, why do you exclude it from > your >>other listings?

Gene:
> It follows immediately from the definition.

So it follows from a definition that nobody understands! (Well does anybody understand it? Speak up!) That isn't any use at all.

Me:
>>/tuning-math/message/4533
>>
>>you give what is, as far as I can work out, a function of a single >>rational number as the "metric". Gene:
> A single rational number, in terms of 5-limit note-classes, is a two-
> dimensional vector. Does that explain it for you?

No, that makes even less sense. You seem to be suggesting that the two coefficients of an octave-equivalent 5-limit vector constitute the parameters for a metric. From where I'm sitting, that's gibberish. It means those two coefficients are supposed to be points in some abstract space. And to be symmetric, the metric would have to treat the two coefficients as interchangeable, so you certainly can't multiply one by log(3) and the other by log(5) like you do in geometric complexity. As for the g(x,x)=0 rule, that means the "distance" of 15:8 is zero, the same as a unison! And even if you could somehow (using a process you haven't explained) get geometric complexity from such a measure, it would only be defined in the 5-limit. That's a pervers way of getting a very simple result.

I'm guessing that "note-classes" are something to do with octave equivalence -- there's another piece of jargon you haven't defined.

Graham

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, Graham Breed <graham@m...> wrote:
> Me:
> >>Given that it is so straightforward, why do you exclude it from
> > your
> >>other listings?
>
> Gene:
> > It follows immediately from the definition.
>
> So it follows from a definition that nobody understands! (Well
does
> anybody understand it? Speak up!) That isn't any use at all.

Up until now I had thought you understood geometric complexity. You
remarked once that it might be better to use logs base two for the
definition, which is true and insightful.

> Me:
> >>/tuning-math/message/4533
> >>
> >>you give what is, as far as I can work out, a function of a
single
> >>rational number as the "metric".
>
> Gene:
> > A single rational number, in terms of 5-limit note-classes, is a
two-
> > dimensional vector. Does that explain it for you?
>
> No, that makes even less sense. You seem to be suggesting that the
two
> coefficients of an octave-equivalent 5-limit vector constitute the
> parameters for a metric. From where I'm sitting, that's
gibberish. It
> means those two coefficients are supposed to be points in some
abstract
> space.

Correct.

And to be symmetric, the metric would have to treat the two
> coefficients as interchangeable, so you certainly can't multiply
one by
> log(3) and the other by log(5) like you do in geometric complexity.

Not correct. All that is required is that the matrix B(i,j) that you
are summing over (as sum_{i,j} B(i,j) x_i x_j) be a symmetric matrix
with real and positive eigenvalues.

Here's the usual grab bag from Weisstein:

http://mathworld.wolfram.com/SymmetricBilinearForm.html

http://mathworld.wolfram.com/QuadraticForm.html

http://mathworld.wolfram.com/PositiveDefiniteQuadraticForm.html

http://mathworld.wolfram.com/HermitianInnerProduct.html

As
> for the g(x,x)=0 rule, that means the "distance" of 15:8 is zero,
the
> same as a unison!

Eh? We are looking at g([0,0], [1, 1]) to find the distance of 15/8
to 1. If we find the distance of 15/8 to itself, we get zero, which
is correct.

And even if you could somehow (using a process you
> haven't explained) get geometric complexity from such a measure, it
> would only be defined in the 5-limit. That's a pervers way of
getting a
> very simple result.

No, because in other cases we use the metric induced on wedge
products by the metric we started out from.

> I'm guessing that "note-classes" are something to do with octave
> equivalence -- there's another piece of jargon you haven't defined.

I thought "note-classes" was standard music theory jargon for the
equivalence classes defined by octave equivalence.

🔗Carl Lumma <ekin@lumma.org>

🔗Graham Breed <graham@microtonal.co.uk>

Gene Ward Smith wrote:

> Up until now I had thought you understood geometric complexity. You > remarked once that it might be better to use logs base two for the > definition, which is true and insightful. No, I said I didn't understand it. I could tell from the examples that it involves logs of primes, and that any logarithms should be to base 2 in this context.

In much the same way, I can see that the units depend on the codimension? of the wedgie. So an interval has complexity measured in octaves, but a linear temperament (or pair of commas) is in octaves^2. Would it make sense to take root so that everything's in octaves?

> Not correct. All that is required is that the matrix B(i,j) that you > are summing over (as sum_{i,j} B(i,j) x_i x_j) be a symmetric matrix > with real and positive eigenvalues.

I've found that matrix! It's

(2g1 g1 g1 ...)
( g1 2g2 g2 ...)
( g1 g2 2g3 ...)
(... ... ... ...)

where gi is the square of the logarithm of the (i-1)th prime number. The octave specific equivalent will have a zero eigenvalue. Is that a problem?

Here's the general function for intervals. I still don't know how to do more complex wedgies.

def intervalComplexity(u, primes=None):
"""
Geometric complexity for any interval u,
expressed as a wedgie
"""
maxBasis = u.maxBasis()
primes = primes or temper.primes[:maxBasis]
result = 0.0
for i in range(maxBasis):
for j in range(i, maxBasis):
result += primes[i]**2 * u[i+1,]*u[j+1,]
# back two levels of indentation
return math.log(2) * math.sqrt(result)

> Here's the usual grab bag from Weisstein:
> > http://mathworld.wolfram.com/SymmetricBilinearForm.html

I'm not sure the point of that.

> http://mathworld.wolfram.com/QuadraticForm.html

Yes, that looks familiar, and you did mention it in the original message.

> http://mathworld.wolfram.com/PositiveDefiniteQuadraticForm.html

And that's a special case of the above.

> http://mathworld.wolfram.com/HermitianInnerProduct.html

Indeed, I thought it looked more like an inner product than a metric. We can ignore the complex bit, can't we?

> Eh? We are looking at g([0,0], [1, 1]) to find the distance of 15/8 > to 1. If we find the distance of 15/8 to itself, we get zero, which > is correct.

Hey, that's the first time you've mentioned that the metric is from the origin to the point! That's one of the steps that's missing from the original definition.

>>I'm guessing that "note-classes" are something to do with octave >>equivalence -- there's another piece of jargon you haven't defined.
> > I thought "note-classes" was standard music theory jargon for the > equivalence classes defined by octave equivalence.

It's always best to define things that aren't common currency here, in case somebody doesn't know or there's an ambiguity with some other context. I've heard of "pitch classes" which are octave equivalent, but also ignore fine tuning and so work like equal temperaments. I think that makes them "finite cyclic groups". Agmon doesn't seem to call his integer pairs anything.

Graham

🔗Gene Ward Smith <gwsmith@svpal.org>

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

> In much the same way, I can see that the units depend on the
> codimension? of the wedgie. So an interval has complexity measured
in
> octaves, but a linear temperament (or pair of commas) is in
octaves^2.
> Would it make sense to take root so that everything's in octaves?

You could if you wished, I suppose.

>
> > Not correct. All that is required is that the matrix B(i,j) that
you
> > are summing over (as sum_{i,j} B(i,j) x_i x_j) be a symmetric
matrix
> > with real and positive eigenvalues.
>
> I've found that matrix! It's
>
> (2g1 g1 g1 ...)
> ( g1 2g2 g2 ...)
> ( g1 g2 2g3 ...)
> (... ... ... ...)
>
> where gi is the square of the logarithm of the (i-1)th prime
number.

I think I used half of that matrix.

> The octave specific equivalent will have a zero eigenvalue. Is
that a
> problem?

I don't think so--we are ignoring 2. That should make you happy. :)

> Here's the general function for intervals. I still don't know how
to do
> more complex wedgies.

You can either convert to an orthonormal basis, or use my worked-out
formulas.

> > Here's the usual grab bag from Weisstein:
> >
> > http://mathworld.wolfram.com/SymmetricBilinearForm.html
>
> I'm not sure the point of that.

This gives the inner product you asked about below.

> > http://mathworld.wolfram.com/QuadraticForm.html
>
> Yes, that looks familiar, and you did mention it in the original
message.
>
> > http://mathworld.wolfram.com/PositiveDefiniteQuadraticForm.html
>
> And that's a special case of the above.
>
> > http://mathworld.wolfram.com/HermitianInnerProduct.html
>
> Indeed, I thought it looked more like an inner product than a
metric.
> We can ignore the complex bit, can't we?

Since everything in sight is real, yes. It makes things that much
easier.

> > Eh? We are looking at g([0,0], [1, 1]) to find the distance of
15/8
> > to 1. If we find the distance of 15/8 to itself, we get zero,
which
> > is correct.
>
> Hey, that's the first time you've mentioned that the metric is from
the
> origin to the point! That's one of the steps that's missing from
the
> original definition.

That's the norm, or size of the interval, which is where I started
from (the L(p/q) function.) You can formulate things in terms of
Euclidean metrics, Euclidean norms, inner products, bilinear forms,
or quadradic forms, and pass from one to another. It's all closely
related.

Creating a Temperment /Comma

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

🔗monz@attglobal.net

🔗Carl Lumma <ekin@lumma.org>

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

🔗Carl Lumma <ekin@lumma.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Carl Lumma <ekin@lumma.org>

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🔗Carl Lumma <ekin@lumma.org>

🔗paulhjelmstad <paul.hjelmstad@us.ing.com>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Carl Lumma <ekin@lumma.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Carl Lumma <ekin@lumma.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

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🔗Graham Breed <graham@microtonal.co.uk>

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🔗Graham Breed <graham@microtonal.co.uk>

🔗Carl Lumma <ekin@lumma.org>

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🔗Paul Erlich <perlich@aya.yale.edu>

🔗Carl Lumma <ekin@lumma.org>

🔗Carl Lumma <ekin@lumma.org>

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Carl Lumma <ekin@lumma.org>

🔗Carl Lumma <ekin@lumma.org>

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Carl Lumma <ekin@lumma.org>

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🔗Paul Erlich <perlich@aya.yale.edu>

🔗Graham Breed <graham@microtonal.co.uk>

🔗Carl Lumma <ekin@lumma.org>

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Carl Lumma <ekin@lumma.org>

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Graham Breed <graham@microtonal.co.uk>

🔗Carl Lumma <ekin@lumma.org>

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Carl Lumma <ekin@lumma.org>

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Graham Breed <graham@microtonal.co.uk>

🔗Gene Ward Smith <gwsmith@svpal.org>