Patent val page: Request to relative newbies

I've updated the patent val page to include examples of the way I've come
to think of patent vals.

http://xenharmonic.wikispaces.com/Patent+val

This is the kind of thing I would have liked to have seen back when I was
starting to work through these things. I'd like people to read it --
especially if you're currently struggling with vals and regular mapping --
to see if the examples help.

It may be that some of this information should be spread onto the "vals and
tuning space" page, here --
http://xenharmonic.wikispaces.com/Vals+and+Tuning+Space -- but there's
already a lot of content there. Maybe there should be a "vals for
non-mathematicians" section, though. I'm interested in people's thoughts.

Regards,
Jake

Nice work. My only point of contention is

"A val defines a regular temperament, which is the deliberate
introduction of an error into one or more primes."

I think you mean that a val defines a rank 1 temperament. I don't see
why it would have to be regular, and there are other temperaments that
require more than one val to define. In general, it takes n vals to
define a rank-n temperament.

-Mike

On Fri, Aug 19, 2011 at 10:17 PM, Jake Freivald <jdfreivald@...> wrote:
>
> I've updated the patent val page to include examples of the way I've come to think of patent vals.
>
> http://xenharmonic.wikispaces.com/Patent+val
>
> This is the kind of thing I would have liked to have seen back when I was starting to work through these things. I'd like people to read it -- especially if you're currently struggling with vals and regular mapping -- to see if the examples help.
>
> It may be that some of this information should be spread onto the "vals and tuning space" page, here -- http://xenharmonic.wikispaces.com/Vals+and+Tuning+Space -- but there's already a lot of content there. Maybe there should be a "vals for non-mathematicians" section, though. I'm interested in people's thoughts.
>
> Regards,
> Jake

Jake>"http://xenharmonic.wikispaces.com/Patent+val"

Look good, though still a bit of a long read.

Perhaps an oversimplification of a patent val, as I understand it, is
A) A scale which takes 31 steps to get to exactly 2/1 is 31EDO IE x^31 = 2/1 where x is a step in 31EDO, that's how you get the first 2 in 31EDO's patent val.
B) Now how many steps gets you closest to 3/1 in that scale (3 being the second prime)?  That step number is the second number in your val.
C) Now how many steps gets you closest to 5/1 in that scale (3 being the third prime)?  That step number is the third number in your val.
D) Keep doing this till a higher prime, such as 9,11, or 13...to make a more complete val.

   The advantage of this is...when you know how far off the primes are...you know how far off simple musical ratios like 3/2 (5th) and 5/3 (6th) are likely to be and can quickly choose EDOs according to which ratios they represent well.

----------------------------------------
   In addition, even though 2/1 maps with no error in an EDO...3/1, 5/1...will have error.  That error is the comma.  Hence even though 2 maps perfectly to an EDO, 3 doesn't, so there is always a comma where the EDO's representation of 3/2 meets 2/1. 

  That's where I get a bit confused...as I'd think the error of the prime 3 vs. its representation in 12EDO to the 12th power would represent the error of its intersection with 2/1...the error also known as 80/81.  But I'm still not sure...

On Fri, Aug 19, 2011 at 10:44 PM, Michael <djtrancendance@...> wrote:
>
> Jake>"http://xenharmonic.wikispaces.com/Patent+val"
> Look good, though still a bit of a long read.
>
> Perhaps an oversimplification of a patent val, as I understand it, is
> A) A scale which takes 31 steps to get to exactly 2/1 is 31EDO IE x^31 = 2/1 where x is a step in 31EDO, that's how you get the first 2 in 31EDO's patent val.
> B) Now how many steps gets you closest to 3/1 in that scale (3 being the second prime)?  That step number is the second number in your val.
> C) Now how many steps gets you closest to 5/1 in that scale (3 being the third prime)?  That step number is the third number in your val.
> D) Keep doing this till a higher prime, such as 9,11, or 13...to make a more complete val.

Jake is surely a better teacher than I am. I've been trying to teach
you this for years now :)

The only thing is that vals exist that aren't the patent vals. For
example, the patent val for 17-equal in the 5-limit is <17 27 39|,
which maps the major third to the 353 cent "neutral third" sized
interval. However, there's also the <17 27 40| val, which maps the
major third to the 424 cent "supermajor third" sized interval. There's
also the <17 12938129831 -15| val, which is a bit absurd, but still
exists, mathematically.

At any rate, it turns out that the 17-equal val that has the lowest
error over all 5-limit dyads is actually <17 27 40|, despite that <17
27 39| is the patent val. Of course, you might also decide that
17-equal doesn't support the 5-limit at all, and treat it as a
2.3.7.11.13 temperament. Then the val you'd use is <17 27 48 59 63|,
where the columns reflect 2, 3, 7, 11, and 13 respectively, skipping
5. A val that reflects a subgroup in this way is sometimes called an
"sval."

>    The advantage of this is...when you know how far off the primes are...you know how far off simple musical ratios like 3/2 (5th) and 5/3 (6th) are likely to be and can quickly choose EDOs according to which ratios they represent well.

It's also just a way of seeing where each prime lays out in JI for
every equal temperament.

>    In addition, even though 2/1 maps with no error in an EDO...3/1, 5/1...will have error.  That error is the comma.  Hence even though 2 maps perfectly to an EDO, 3 doesn't, so there is always a comma where the EDO's representation of 3/2 meets 2/1.

Vals don't say anything of tuning. You can always stretch an EDO if
you want so that 3/1 or 5/1 is pure and 2/1 is not. They're more
abstract and represent where, in an EDO, you're saying the tempered
primes are. Once you map the primes you can see what commas vanish, or
even mix two EDOs together to get a rank 2 temperament.

>   That's where I get a bit confused...as I'd think the error of the prime 3 vs. its representation in 12EDO to the 12th power would represent the error of its intersection with 2/1...the error also known as 80/81.  But I'm still not sure...

No, the difference between 12 3/2's and a few octaves is the
Pythagorean comma, or 531441/524288. 81/80 is the difference between 4
3/2's and one 5/1, or the syntonic comma, and it defines meantone
temperament. Both vanish in 12-equal. Here's proof:

81/80 is 3^4/(2^4*5). Another way to look at this is 2^-4 * 3^4 *
5^-1, or |-4 4 -1> for short. This is called a "monzo" and is the
other important part of all of this.

So the patent val for 12 is <12 19 28|. Let's apply the homomorphism:
<12 19 28|-4 4 -1> = (12)*(-4) + (19)*(4) + (28)*(-1) = 0. So it looks
like 81/80 maps to 0 steps of 12-equal. Now let's see what happens to
3/2: <12 19 28|-1 1 0> = (12)*(-1) + (19)*(1) + (28)*(0) = 7, and
indeed 3/2 is 7 steps out of 12.

-Mike

--- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> The only thing is that vals exist that aren't the patent vals. For
> example, the patent val for 17-equal in the 5-limit is <17 27 39|,
> which maps the major third to the 353 cent "neutral third" sized
> interval. However, there's also the <17 27 40| val, which maps the
> major third to the 424 cent "supermajor third" sized interval. There's
> also the <17 12938129831 -15| val, which is a bit absurd, but still
> exists, mathematically.

Vals which are not any good as equal divisions are not thereby absurd. For instance, <0 1 4 10| doesn't make much sense as a val for "0edo", but it does make sense as the mapping of generators of a fifth in meantone. If you want to consider it in terms of tempering things out, it tempers out 81/80, 126/125 and 225/224 like septimal meantone, but also 2. 2 is a "comma" of sorts; technically, a kernel element at any rate.

On Fri, Aug 19, 2011 at 11:34 PM, genewardsmith
<genewardsmith@...t> wrote:
>
> --- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> > The only thing is that vals exist that aren't the patent vals. For
> > example, the patent val for 17-equal in the 5-limit is <17 27 39|,
> > which maps the major third to the 353 cent "neutral third" sized
> > interval. However, there's also the <17 27 40| val, which maps the
> > major third to the 424 cent "supermajor third" sized interval. There's
> > also the <17 12938129831 -15| val, which is a bit absurd, but still
> > exists, mathematically.
>
> Vals which are not any good as equal divisions are not thereby absurd. For instance, <0 1 4 10| doesn't make much sense as a val for "0edo", but it does make sense as the mapping of generators of a fifth in meantone. If you want to consider it in terms of tempering things out, it tempers out 81/80, 126/125 and 225/224 like septimal meantone, but also 2. 2 is a "comma" of sorts; technically, a kernel element at any rate.

Oh, right. That's a good point. I just didn't want to scare the hell
out of these people quite yet :)

-Mike

Mike,

> I think you mean that a val defines a rank 1 temperament. I don't see
> why it would have to be regular, and there are other temperaments that
> require more than one val to define. In general, it takes n vals to
> define a rank-n temperament.

It's stuff like this that drives me crazy, because it means that there are
still terminological issues that I don't get. It's probably because I don't
find the lattice stuff all that intuitive, and that's where dimensions and
ranks come in.

It's okay. I'll figure out all the rank-2+ stuff and figure out what people
mean by "the 12&31 temperament" or what have you, and then I'll try to write
them up, too. And have you shoot them down until I get it right, I hope. :)

Thanks,
Jake

I'll be following this thread closely, and reading and rereading your
writing and updates to the patent val page. So far I already see much of
what helped me earlier this week, both on that page and in other responses
to this thread. Until I get a chance to dive deeper, here are a couple
thoughts related to accessibility:

1. My newbiness didn't understand until the other day that when y'all say
"primes" you're often referring directly to the prime ~harmonics~ (not just
the prime numbers). Your email the other day went to the trouble to spell
out "3/1", "5/1", etc. (right after referring to "the primes"), and I had an
"oh doyy!" moment. It's a simple notion, makes complete sense that you'd say
it that way, and it's been very common in the lingo I've read for months ...
but if everyone is using the shorthand without defining it, a newb won't
necessarily make that connection (until he realizes and feels so dumb).
There are probably other reasons I wasn't seeing that, but anyway.

2. Similarly related to what Mike and Gene have been saying about ranks,
different kinds of vals, non-patent vals, etc... I'm starting to get that
stuff, maybe, but I need to grasp onto something more concrete first.
Assuming your purpose in the intro/beginning of an article is to be more
accessible, it's often essential to say something like "this gets more
complicated in practice, but ignore that for the moment" or "exceptions to
this will be covered later". As the son of an EE, and as someone who's spent
the last 2+ years of professional life turning myriad engineers' expert
gobbledygook into readable documentation, I have an all-too-keen sense of
this sort of thing :-)

3. A thought on organization. A "procedural" or "tutorial" approach has been
very helpful in getting me over the humps ("this is your starting point,
then you count the steps to 3/1 to obtain this value, then this goes here,
OK, now you've got a val! Let's look at it...") -- see your email from
earlier this week. Right now, the article examples do that, but they sorta
flip-flop between 12edo and 31edo. I'd recommend running the *whole process*
for 12edo, as an intro to the val concept, lay down the patent val with a
few numbers (e.g. 2^(19/12) = 2.99661...). Then set them aside. At this
point the newb is still thinking "OK, looks good, whatever you say...". Then
run it again for 31edo, generate the new val and corresponding numbers, and
then compare them to each other. This gives an opportunity to show how vals
convey useful information -- i.e. THIS number here shows that 31edo has
better approximations of just intervals in THIS limit than 12edo, etc. --
then use these already-fleshed-out examples to build on the concepts and get
into the complications.

Hopefully the above reveals some of my misunderstandings, and I'll hear
about it...

Jason

On Fri, Aug 19, 2011 at 21:17, Jake Freivald <jdfreivald@...> wrote:

> **
>
>
> I've updated the patent val page to include examples of the way I've come
> to think of patent vals.
>
> http://xenharmonic.wikispaces.com/Patent+val
>
> This is the kind of thing I would have liked to have seen back when I was
> starting to work through these things. I'd like people to read it --
> especially if you're currently struggling with vals and regular mapping --
> to see if the examples help.
>
> It may be that some of this information should be spread onto the "vals and
> tuning space" page, here --
> http://xenharmonic.wikispaces.com/Vals+and+Tuning+Space -- but there's
> already a lot of content there. Maybe there should be a "vals for
> non-mathematicians" section, though. I'm interested in people's thoughts.
>
> Regards,
> Jake
>

On Fri, Aug 19, 2011 at 11:46 PM, Jake Freivald <jdfreivald@...> wrote:
>
> Mike,
>
> > I think you mean that a val defines a rank 1 temperament. I don't see
> > why it would have to be regular, and there are other temperaments that
> > require more than one val to define. In general, it takes n vals to
> > define a rank-n temperament.
>
> It's stuff like this that drives me crazy, because it means that there are still terminological issues that I don't get. It's probably because I don't find the lattice stuff all that intuitive, and that's where dimensions and ranks come in.

A "regular temperament" is, for example, the opposite of a
well-temperament or a circulating temperament. Werckmeister is also
defined by the <12 19 28| val, but just happens to be an irregular
tuning for it - but these are both "rank 1" temperaments, because
every note in it is generated by a single generator (1 step out of
12). Meantone, on the other hand, is rank 2, because every note in it
is generated by a combination of a generator and a period, or "two
generators" if you're really hip. You could potentially have an
irregular tuning for meantone where the fifths are different sizes,
but it would still be a rank 2 temperament.

If you start with 5-limit JI, you have 3 dimensions - one for 2, one
for 3, and one for 5. So 5-limit JI is a rank-3 "temperament" (to use
the term liberally). If you temper out 81/80, the entire "5" axis is
now reachable by sequences of 4 steps along the "3" axis and -4 steps
along the "2" axis, and so for all intents and purposes it no longer
exists as an independent degree of freedom. The 5 axis is now
redundant, and there are only two truly free dimensions in the system
- one representing 2 and one representing 3, with the axis for 5 being
"encoded" into the other two axes. This is the meantone lattice.

Not so coincidentally, meantone is a temperament that has a generator
of 3/1 and a period of 2/1, which are the axes that we're left with
above. Hmm......

-Mike

> Vals which are not any good as equal divisions are not thereby absurd.
> For instance, <0 1 4 10| doesn't make much sense as a val for "0edo",
> but it does make sense as the mapping of generators of a fifth in
> meantone.

Gene, if I understand the above correctly, you mean that the meantone
generator is a fifth, four of them reach a third (C-G-D-A-E), and 10 of them
reach a dominant seventh (....B-Gb-Db-Ab-Eb-Bb).

This looks like a notational convention that means something slightly
different from the way the patent val is used in mapping. Considering the
700-cent fifth as the generator, 4 * 700 cents = 2800 cents, which is close
to 3/1; meanwhile, 10 * 700 cents = 7000 cents, but 7/1 is 3368.83. So,
instead of the number of steps to a given prime, this val contains terms
that tell you how many generators it takes to get you to a given pitch
class. Yes?

Is a val generally notation that identifies an iteration counter, which can
be applied to any situation? In other words, is this a tuning thing like a
monzo, or is this a general mathematical notation?

> If you want to consider it in terms of tempering things out, it tempers
out
> 81/80, 126/125 and 225/224 like septimal meantone,

Because:

81/80 = (2^-4)*(3^4)*(5^-1) = 0 steps + 4*1 step + -1*4 steps = 0+4-4 = 0.

126/125 = (2^1)*(3^2)*(5^-3)*(7^1) = 0 steps + 2*1 steps + -3*4 steps + 1*10
steps = 0+2-12+10 = 0 steps.

225/224 = (2^-5)*(3^2)*(5^2)*(7^-1) = 0 + 2*1 + 2*4 + -1*10 = 2+8-10 = 0
steps.

Yes?

That's pretty cool. The fact that it maps a generator without even knowing
what the precise generator is kinda blows my mind. This val, then, doesn't
tell you what error you're introducing into the primes, because it will work
with any meantone fifth, but I think it must tell you the relative error
among the primes. Right? Which would mean that this val also defines an
abstract temperament?

Mike said,

> Oh, right. That's a good point. I just didn't want to scare the hell
> out of these people quite yet :)

Aahh! Numbers!!!!1!

I don't mind being "scared". I mind my eyes glazing over, which they're not
yet. :)

Thanks,
Jake

On Sat, Aug 20, 2011 at 1:19 AM, Jake Freivald <jdfreivald@...> wrote:
>
>  > Vals which are not any good as equal divisions are not thereby absurd.
> > For instance, <0 1 4 10| doesn't make much sense as a val for "0edo",
> > but it does make sense as the mapping of generators of a fifth in
> > meantone.
>
> Gene, if I understand the above correctly, you mean that the meantone generator is a fifth, four of them reach a third (C-G-D-A-E), and 10 of them reach a dominant seventh (....B-Gb-Db-Ab-Eb-Bb).
>
> This looks like a notational convention that means something slightly different from the way the patent val is used in mapping. Considering the 700-cent fifth as the generator, 4 * 700 cents = 2800 cents, which is close to 3/1; meanwhile, 10 * 700 cents = 7000 cents, but 7/1 is 3368.83. So, instead of the number of steps to a given prime, this val contains terms that tell you how many generators it takes to get you to a given pitch class. Yes?

It's both. The rabbit hole with vals runs deep.

> Is a val generally notation that identifies an iteration counter, which can be applied to any situation? In other words, is this a tuning thing like a monzo, or is this a general mathematical notation?

I don't know what you mean by "iteration counter," but both vals and
monzos are general mathematical notations that have meaning
irrespective of any particular tuning...

> > If you want to consider it in terms of tempering things out, it tempers out
> > 81/80, 126/125 and 225/224 like septimal meantone,
>
> Because:
>
> 81/80 = (2^-4)*(3^4)*(5^-1) = 0 steps + 4*1 step + -1*4 steps = 0+4-4 = 0.
>
> 126/125 = (2^1)*(3^2)*(5^-3)*(7^1) = 0 steps + 2*1 steps + -3*4 steps + 1*10 steps = 0+2-12+10 = 0 steps.
>
> 225/224 = (2^-5)*(3^2)*(5^2)*(7^-1) = 0 + 2*1 + 2*4 + -1*10 = 2+8-10 = 0 steps.
>
> Yes?

Right.

> That's pretty cool. The fact that it maps a generator without even knowing what the precise generator is kinda blows my mind.

The generator for <0 1 4 10| is 3/1... think about it a sec. :)

> This val, then, doesn't tell you what error you're introducing into the primes, because it will work with any meantone fifth, but I think it must tell you the relative error among the primes. Right? Which would mean that this val also defines an abstract temperament?

It doesn't tell you anything about the error among the primes or the
tuning you have to use; everything you're dealing with is -entirely-
in the realm of abstract temperament. It doesn't even tell you that
the columns stand for 2, 3, 5, and 7; we're just assuming that after
the fact. You are most directly dealing with purely mathematical
things that apply to higher dimensional spaces and to groups. All of
this stuff applies to any group at all. I've been wondering for a
while if it might have some use in image compression, actually, which
also has to do with vertical and horizontal frequencies and hence also
has to do with primes (although don't ask me what a
porcupine-compressed bitmap would look like just yet).

For example, let's take the random 2x3 matrix

[7 11 17]
[12 19 28]

Then let's put it in reduced row echelon form:

[1 0 -4]
[0 1 4]

Great, elementary matrix operations, no tuning or math involved at
all. But now, once we take a tuning perspective on what just happened
above, the first matrix consists of the 7-equal patent val on top of
the 12-equal patent val, and the rref version of it also happens to be
the period/generator mapping for meantone, which is also the
temperament common to 7-equal and 12-equal, and hence the 7&12
temperament.

-Mike

--- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> So, instead of the number of steps to a given prime, this val contains terms that tell you how many generators it takes to get you to a given pitch class. Yes?

Not necessarily octave pitch classes; period pitch classes, where the period is some fraction of an octave.

> It's both. The rabbit hole with vals runs deep.

To tell you how to get to a given prime, you need another val to go with, of course. Rank two takes two vals, and <0 1 4 10| is just one. If you use 3/2 as a generator, you pair it with <1 1 0 -3|, if 3 is the generator, <1 0 -4 -13|. The first choice adds 5 as a "comma" to meantone, the second 3.

All of
> this stuff applies to any group at all.

Actually, just to finitely generated free groups.

> [7 11 17]
> [12 19 28]
>
> Then let's put it in reduced row echelon form:
>
> [1 0 -4]
> [0 1 4]
>
> Great, elementary matrix operations, no tuning or math involved at
> all. But now, once we take a tuning perspective on what just happened
> above, the first matrix consists of the 7-equal patent val on top of
> the 12-equal patent val, and the rref version of it also happens to be
> the period/generator mapping for meantone, which is also the
> temperament common to 7-equal and 12-equal, and hence the 7&12
> temperament.

OK, except that <7 11 17| is 7c, and what you get is helmholtz, 7c&12, not meantone, 7&12.

--- In tuning@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:

> All of
> > this stuff applies to any group at all.
>
> Actually, just to finitely generated free groups.

Since I was being pedantic, I suppose I should have said "finitely generated free abelian groups."

On Sat, Aug 20, 2011 at 3:27 AM, genewardsmith
<genewardsmith@...> wrote:
>
> --- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> > So, instead of the number of steps to a given prime, this val contains terms that tell you how many generators it takes to get you to a given pitch class. Yes?
>
> Not necessarily octave pitch classes; period pitch classes, where the period is some fraction of an octave.

Jake wrote that, not me...

> All of
> > this stuff applies to any group at all.
>
> Actually, just to finitely generated free groups.

Who's to say we couldn't apply it to infinitely generated groups? We
could replace vals with some function v and monzos with some function
m, so long as a bijection exists between the two.

I'm not sure I understand the full ramifications of a non-free abelian
group, but I've had my eye set on developing a "fuzzy group" structure
for all of this for a while now; something where three fuzzy 10/9's or
two fuzzy 27/25's gets you to a fuzzy 4/3 would be nice... So it
sounds like my goal is to do the exact opposite of what you laid out
above.

> > [7 11 17]
> > [12 19 28]
> >
> > Then let's put it in reduced row echelon form:
> >
> > [1 0 -4]
> > [0 1 4]
> >
> > Great, elementary matrix operations, no tuning or math involved at
> > all. But now, once we take a tuning perspective on what just happened
> > above, the first matrix consists of the 7-equal patent val on top of
> > the 12-equal patent val, and the rref version of it also happens to be
> > the period/generator mapping for meantone, which is also the
> > temperament common to 7-equal and 12-equal, and hence the 7&12
> > temperament.
>
> OK, except that <7 11 17| is 7c, and what you get is helmholtz, 7c&12, not meantone, 7&12.

I meant <7 11 16|, good catch. The rref form of the one I posted is actually

[1 0 15]
[0 1 -8]

Which is schismatic temperament, as you mentioned, except I guess the
hip thing to do these days is to call it helmholtz temperament.

-Mike

Jake Freivald <jdfreivald@...> wrote:

> Is a val generally notation that identifies an iteration
> counter, which can be applied to any situation? In other
> words, is this a tuning thing like a monzo, or is this a
> general mathematical notation?

One concept is to map primes to generator steps. For equal
temperaments, a generator is a scale step. I'm not sure if
this is precisely what Gene calls a "val".

There are other mappings that can be useful musically. For
example, <-1 -1 . . . -1] is the inversion mapping. I
think Gene was using them before they were called
neo-Riemannian.

Graham

MikeB>"No, the difference between 12 3/2's and a few octaves is the Pythagorean comma, or 531441/524288. 81/80 is the difference between 4 3/2's and one 5/1, or the syntonic comma, and it defines meantone temperament. Both vanish in 12-equal."

   Here's the thing...now I get how/why commas vanish.  But given a val...how do I know which commas matter?
    Given I don't know this...I could easily try to find intersections between all sorts of different ratios from the different primes IE "where 3/1 meets 5/1...where 5/1 meets 7/1....where 3/1 meets 7/1).  It seems people on here give tons of examples of commas...with no hint as to why they chose those commas over other possible commas.  How do you know which commas are important IE given a val, and no other knowledge or hints on the side, how do I find the important commas?

On Sat, Aug 20, 2011 at 5:38 AM, Michael <djtrancendance@...> wrote:
>
> MikeB>"No, the difference between 12 3/2's and a few octaves is the Pythagorean comma, or 531441/524288. 81/80 is the difference between 4 3/2's and one 5/1, or the syntonic comma, and it defines meantone temperament. Both vanish in 12-equal."
>
>    Here's the thing...now I get how/why commas vanish.  But given a val...how do I know which commas matter?

Do you mean how do you know which commas some val tempers out? Well,
if it's tempered out, then <v|m> should equal 0. So for <12 19 28|,
which is 12-equal's 5-limit patent val, then <12 19 28|a b c> should
equal 0. Therefore, 12a + 19b + 28c should equal 0. Any combination of
a, b, and c that ends up turning into 0 is tempered out by 12-equal.
Some solutions to this problem are 81/80, 128/125, 648/625, and
2048/2025.

-Mike

Mike wrote:

> [1 0 -4]
> [0 1 4]
>
> Great, elementary matrix operations,

- #1. Is there some sort of "tutorial for less mathematically skilled people" on the web which could enlighten me on how do I do those "elementary matrix operations"?

- #2. If you take something like 19&27 (semisixths), for example, how do you find the mapping then? Isn't it necessary to somehow take into account the fact that neither 3/1 nor 5/1 can be expressed with 0 periods?

- #3. Aha, maybe what you're saying sheds some light on why so many people prefer to write 2D mappings rotated by 90� compared to the way I do it.
I personally regard the period/generator mapping of each prime as one "meaningful" piece of information and therefore at first I liked Graham's older scripts more than the newer ones, even though the older ones weren't able to test for partial limits.
This concept can also be applied to EDO mappings, not only to period/generator mappings. For example, if we think of meantone as 7&5, then the 5-limit EDO mapping is:
[(7, 5)
(11, 8)
(16, 12)]
Which suggests a 12-tone MOS adding up to an octave (7+5) where 2/1 is approximated by 7 steps of A + 5 steps of B, 3/1 is approximated by 11 steps of A + 8 steps of B, 5/1 is approximated by 16 steps of A + 12 steps of B. It contains no information on how large the two steps in question are or what the ratio of their sizes should be. If A has 0 cents, the minor second vanishes (i.e. there will be a difference between C and D but there'll be none between C and Db). If B has 0 cents, the chroma vanishes (there will be a difference between C and Db but not between C and C#).

Petr

>"Do you mean how do you know which commas some val tempers out? Well,

if it's tempered out, then <v|m> should equal 0. So for <12 19 28|,

which is 12-equal's 5-limit patent val, then <12 19 28|a b c> should

equal 0. Therefore, 12a + 19b + 28c should equal 0. "

Ah...so that's the simplest algebraic form that solves it. Makes sense...

--- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> > Actually, just to finitely generated free groups.
>
> Who's to say we couldn't apply it to infinitely generated groups? We
> could replace vals with some function v and monzos with some function
> m, so long as a bijection exists between the two.

Countably infinitely generated groups is what I first used on tuning-math, but decided that sticking to a fixed p-limit was better.

The multiplicative group of the positive rationals is free abelian of infinite rank; it has generators given by the primes. The set of homomorphisms from that to the integers has the structure of a group (you can add to homomorphisms and get another) but it isn't free: it's a kind of group called a Baer–Specker group. If you take the p-adic valuations for each prime, you get a basis for a free subgroup. In bra-ket form, the positive rationals are |e2 e3 e5 ... >, where all but fintely many of the prime exponents ep are zero. The dual Baer-Specker group is <v2 v3 v5 ...| where any sequence of integers can be used. The free abelian subgroup is <v2 v3 v5 ...| were all but finitely many of the vp are zero, that makes it noncanonically isomorphic to the positive rationals group. So which way do you do it is one question to start with.

> Which is schismatic temperament, as you mentioned, except I guess the
> hip thing to do these days is to call it helmholtz temperament.

One of of the many things Paul wanted in the terminology department.

On Sat, Aug 20, 2011 at 6:18 AM, Petr Parízek <petrparizek2000@...> wrote:
>
> Mike wrote:
>
> > [1 0 -4]
> > [0 1 4]
> >
> > Great, elementary matrix operations,
>
> - #1. Is there some sort of "tutorial for less mathematically skilled
> people" on the web which could enlighten me on how do I do those "elementary
> matrix operations"?

I dunno, this is just matrix algebra - I'm kind of learning as I go
along. It would be useful if someone with more math knowledge than I
did made a "linear algebra for music theorists" page on the xenwiki
sometime.

> - #2. If you take something like 19&27 (semisixths), for example, how do you
> find the mapping then? Isn't it necessary to somehow take into account the
> fact that neither 3/1 nor 5/1 can be expressed with 0 periods?

Yeah, reduced row echelon form doesn't work too well with that. I
think what you would want in that case is the "normal val list," which
you can find here -
http://xenharmonic.wikispaces.com/Abstract+regular+temperament

-Mike

On Sat, Aug 20, 2011 at 11:13 AM, genewardsmith
<genewardsmith@...t> wrote:
>
> Countably infinitely generated groups is what I first used on tuning-math, but decided that sticking to a fixed p-limit was better.
>
> The multiplicative group of the positive rationals is free abelian of infinite rank; it has generators given by the primes. The set of homomorphisms from that to the integers has the structure of a group (you can add to homomorphisms and get another) but it isn't free: it's a kind of group called a Baer–Specker group.

So a val would denote an element in a Baer-Specker group then. What
does this mean? If it's not free, that means that there's a torsion
element in it? What would that imply for vals...?

> If you take the p-adic valuations for each prime, you get a basis for a free subgroup.

I thought that the p-adic valuation for each prime would be 1... what
does this mean? Do you mean for each prime up to some limit?

> In bra-ket form, the positive rationals are |e2 e3 e5 ... >, where all but fintely many of the prime exponents ep are zero. The dual Baer-Specker group is <v2 v3 v5 ...| where any sequence of integers can be used. The free abelian subgroup is <v2 v3 v5 ...| were all but finitely many of the vp are zero, that makes it noncanonically isomorphic to the positive rationals group. So which way do you do it is one question to start with.

I don't understand this until I know what you mean by p-adic
valuations giving you the basis for a free subgroup above... I was
envisioning a setup where you defined a rank-n temperament by giving n
elements in what you called the Baer-Specker group. Otherwise, you'd
need to define an infinite - 1 amount of commas to define a rank-2
temperament, which I'm not sure how to formalize.

-Mike

--- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> So a val would denote an element in a Baer-Specker group then. What
> does this mean? If it's not free, that means that there's a torsion
> element in it? What would that imply for vals...?

Being torsion-free for non-finitely generated abelian groups is not the same as being free, it's the same as being flat, but I don't see how that helps. Aside from the Baer-Specker group, an example of a flat but not free abelian group is the group of the rational numbers under addition.

> > If you take the p-adic valuations for each prime, you get a basis for a free subgroup.
>
> I thought that the p-adic valuation for each prime would be 1... what
> does this mean? Do you mean for each prime up to some limit?

No, I mean you get an infinite list of p-adic valuations, one for each prime, and that is your basis.

> I don't understand this until I know what you mean by p-adic
> valuations giving you the basis for a free subgroup above... I was
> envisioning a setup where you defined a rank-n temperament by giving n
> elements in what you called the Baer-Specker group.

Certainly, but then I ask why not just stick to a finite limit group which contains them?

Otherwise, you'd
> need to define an infinite - 1 amount of commas to define a rank-2
> temperament, which I'm not sure how to formalize.

It's not formalizing it which is the problem, but dealing with the resultant mess.

On Sun, Aug 21, 2011 at 8:44 PM, genewardsmith
<genewardsmith@...> wrote:
>
> --- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> > So a val would denote an element in a Baer-Specker group then. What
> > does this mean? If it's not free, that means that there's a torsion
> > element in it? What would that imply for vals...?
>
> Being torsion-free for non-finitely generated abelian groups is not the same as being free, it's the same as being flat, but I don't see how that helps. Aside from the Baer-Specker group, an example of a flat but not free abelian group is the group of the rational numbers under addition.

I guess I don't understand what a "free" group is then. Perhaps I'm
misusing the word "torsion." A free abelian group, in my
understanding, is one in which every element can be represented in
only one way relative to some basis. If we're in a group that doesn't
have that property, then every element can be represented in more than
one way relative to some basis. This means that the identity element
can be represented in more than one way as well, which I assume means
there'd have to be more than one identity element... (?) If I
understand this then I'll better be able to express my question about
torsion above.

> > I thought that the p-adic valuation for each prime would be 1... what
> > does this mean? Do you mean for each prime up to some limit?
>
> No, I mean you get an infinite list of p-adic valuations, one for each prime, and that is your basis.

OK, I see. So you mean our basis would be |1 0 0 0 0 0 0 ...>, |0 1 0
0 0 0 0 0 ...>, etc.

> > I don't understand this until I know what you mean by p-adic
> > valuations giving you the basis for a free subgroup above... I was
> > envisioning a setup where you defined a rank-n temperament by giving n
> > elements in what you called the Baer-Specker group.
>
> Certainly, but then I ask why not just stick to a finite limit group which contains them?

Because limits can be messy and cause problems, especially when you
start looking at subgroups of limits. What's the best val for 13-equal
in the 5-limit? How about the 7-limit? How about the 11-limit? It
changes. How about the infinite limit?

I also just feel like it would be cool to work out :)

> Otherwise, you'd
> > need to define an infinite - 1 amount of commas to define a rank-2
> > temperament, which I'm not sure how to formalize.
>
> It's not formalizing it which is the problem, but dealing with the resultant mess.

The comma "81/80" by itself already has a resultant mess. We've been
dealing with the resultant mess by saying "this is the 7-limit 81/80
planar temperament" or "this is the 5-limit 81/80 linear temperament"
or "this is the 7-limit 81/80 and 49/48 linear temperament."

I simply wonder if there's another way to deal with it that doesn't
involve limits and subgroups, so that we don't have a million
different temperaments for the same essential thing. We'd be at the
point where we're finding what the best mappings are across all primes
for some MOS, and there'd be no point to ever go beyond that. Sure,
there'd still be utility in limits, but this would be fun too.

-Mike

On Sun, Aug 21, 2011 at 8:44 PM, genewardsmith
<genewardsmith@...> wrote:
>
> > I don't understand this until I know what you mean by p-adic
> > valuations giving you the basis for a free subgroup above... I was
> > envisioning a setup where you defined a rank-n temperament by giving n
> > elements in what you called the Baer-Specker group.
>
> Certainly, but then I ask why not just stick to a finite limit group which contains them?

Actually, ignore everything else I said. Here's the best answer to
this yet - temperament searches. Go to Graham's temperament finder and
look at how the list changes as you go from the 5-limit to the
31-limit. The best temperaments depend on what limit or subgroup you
use.

It would be nice to know what the best temperaments in the world are
without regard to limit, of any rank, in general. That alone is why I
want to figure this out for an infinite rank group.

-Mike

I've modified the patent val page some more. I'd especially like
knowledgeable people to look at the top part to make sure I didn't screw up
the language. My purpose here is to get the language into a form that a
newbie and non-mathematician can understand, without inaccuracy and without
making it useless to mathematicians.

Also, I said:
> > That's pretty cool. The fact that it maps a generator without even
> > knowing what the precise generator is kinda blows my mind.

Mike B replied:
> The generator for <0 1 4 10| is 3/1... think about it a sec. :)

I got that part already: 1 step of the generator gets me to 3/1, so 3/1 =
the generator. But it doesn't tell me the *precise* generator that I need to
use; I can do the math with a 700-cent 12-EDO fifth, or a 696.8-cent 31-EDO
fifth, or any meantone fifth, and it should work.

Really, though, I say "any meantone fifth" as a guess, because Gene
mentioned meantone. I'm not sure how to tell when this breaks. Suppose I use
a slightly sharp 3/2, maybe 709.091 from 22 EDO or 703.03030 from 99 EDO.
That yields 436 or 412 cents, respectively -- still major / supermajor
thirds, but not really 5/4 thirds.

I guess this is like any mapping process: There's a judgment call involved.
When you map a 696.8-cent fifth and get a 7/4 of 968 cents, that works; when
you map a 709.091-cent fifth and get a 7/4 of 1091 cents, it doesn't. You've
introduced too much error into the primes for it to work right. Just like a
12 EDO val of <12 19 21| would be valid as a val, but a lousy map in the
5-limit.

Which means, getting back to using sharp fifths, I should be able to adjust
mappings if things get out of whack. Instead of <0 1 4 10|, I could use <0 1
9 20|. Using the 22-EDO fifth, that would get me a major third of 381.819
and a harmonic 7th of 982 cents. It would also mean that there is a
substantially "higher complexity for 5 and 7" in that tuning, if I
understand this correctly.

> the first matrix consists of the 7-equal patent val on top of
> the 12-equal patent val, and the rref version of it also happens
> to be the period/generator mapping for meantone, which is also
> the temperament common to 7-equal and 12-equal, and hence the
> 7&12 temperament.

I've wondered what people meant by things like "7&12 temperament", but I was
chewing on other things. Thanks for the example -- it came at an opportune
time for my brain to absorb it.

Regards,
Jake

--- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> I guess I don't understand what a "free" group is then. Perhaps I'm
> misusing the word "torsion." A free abelian group, in my
> understanding, is one in which every element can be represented in
> only one way relative to some basis.

It's a group in which every element is a unique *finite sum* of basis elements. This leads to the universal property of free abelian groups, which says that if you map the basis into any abelian group using any function, it extends uniquely to a group homomorphism. This means, for instance, that if you map the free group of positive rational numbers by mapping the primes to anything you like in some abelian group, for instance the real numbers thought of as cents, then that extends to a homomorphism from the positive rationals to the reals.

> Because limits can be messy and cause problems, especially when you
> start looking at subgroups of limits. What's the best val for 13-equal
> in the 5-limit? How about the 7-limit? How about the 11-limit? It
> changes. How about the infinite limit?

The zeta tuning gives you an infinite limit val, if you can find a use for it, and patent-type rounding another.

--- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> It would be nice to know what the best temperaments in the world are
> without regard to limit, of any rank, in general. That alone is why I
> want to figure this out for an infinite rank group.

It's not a well-defined question, but you could always combine zeta mappings for zeta integral/peak/gap edos.

--- In tuning@yahoogroups.com, Jake Freivald <jdfreivald@...> wrote:
>
> I've modified the patent val page some more. I'd especially like
> knowledgeable people to look at the top part to make sure I didn't screw up
> the language.

I don't like this as a definition: "A p-limit val contains the number of steps it takes to get to each prime number up to p, in prime number order". You need to make it clear you are talking about vals for some equal division, not in general.

> I don't like this as a definition: "A p-limit val contains the
> number of steps it takes to get to each prime number up to p, in
> prime number order". You need to make it clear you are talking
> about vals for some equal division, not in general.

Thanks, Gene, this is exactly the kind of thing I'm nervous about. :)

When you recently mentioned the way you use a val as a
definition/description of meantone, it still involved "the number of steps
it takes to get to each prime number", but that was the number of generator
steps instead of EDO steps. That's why I didn't mention what kind of steps
we're talking about here. If vals always (at least from a tuning
perspective) define the number of steps -- of whatever kind -- to get to
primes, then maybe I could clarify that and still have this introduction be
okay.

Rather than clutter up the tuning list, I'll take this part of the
conversation to the discussion page, starting with a tweaked definition.

Thanks,
Jake

On Mon, Aug 22, 2011 at 12:04 PM, genewardsmith
<genewardsmith@...> wrote:
>
> --- In tuning@yahoogroups.com, Jake Freivald <jdfreivald@...> wrote:
> >
> > I've modified the patent val page some more. I'd especially like
> > knowledgeable people to look at the top part to make sure I didn't screw up
> > the language.
>
> I don't like this as a definition: "A p-limit val contains the number of steps it takes to get to each prime number up to p, in prime number order". You need to make it clear you are talking about vals for some equal division, not in general.

Why should it have to be equal? The 12-equal patent val describes
werckmeister just as well as 12-equal, doesn't it?

-Mike

On Mon, Aug 22, 2011 at 11:50 AM, genewardsmith
<genewardsmith@...t> wrote:
>
> --- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> It's a group in which every element is a unique *finite sum* of basis elements. This leads to the universal property of free abelian groups, which says that if you map the basis into any abelian group using any function, it extends uniquely to a group homomorphism.

OK, I see.

> This means, for instance, that if you map the free group of positive rational numbers by mapping the primes to anything you like in some abelian group, for instance the real numbers thought of as cents, then that extends to a homomorphism from the positive rationals to the reals.

Right.

> The zeta tuning gives you an infinite limit val, if you can find a use for it, and patent-type rounding another.

I've been going nuts with the zeta function for a while now. I think I
understand the low-level gist of it now. But how are you using it to
derive vals?

Also, have you considered looking at the real = -1 line of the zeta
function rather than the real = -0.5 line? Consider how that would
weight the integers differently.

-Mike

Mike Battaglia <battaglia01@...> wrote:

> I guess I don't understand what a "free" group is then.
> Perhaps I'm misusing the word "torsion." A free abelian
> group, in my understanding, is one in which every element
> can be represented in only one way relative to some
> basis. If we're in a group that doesn't have that
> property, then every element can be represented in more
> than one way relative to some basis. This means that the
> identity element can be represented in more than one way
> as well, which I assume means there'd have to be more
> than one identity element... (?) If I understand this
> then I'll better be able to express my question about
> torsion above.

I've been reading about group theory again. Free abelian
groups are on a different page to free groups. I don't
even know if free abelian groups are free groups that are
abelian, or an independent concept. A free abelian group
is simple though: it's made up of infinite cyclic groups.
An infinite cyclic group is something that's called cyclic
but doesn't cycle.

Equal temperaments in an octave equivalent space are cyclic
groups. The order is the number of notes to the octave.
So 12-equal is the cyclic group of order 12. An infinite
cyclic group is a cyclic group of order infinity. Equal
temperaments in octave specific space are infinite cyclic
groups. Each independent prime in just intonation behaves
like a non-octave equal temperament.

Of course, not all groups we might look at are free abelian
because I already said that octave-equivalent equal
temperaments aren't. Finite cyclic groups are torsion
groups. In that case torsion means that you can add an
element to itself and eventually end up where you started.

Octave equivalent regular temperaments are, in general,
mixed groups. The period has torsion of order of the
number of periods to an octave. Other intervals, like
fifths in meantone, don't have torsion. Equal temperaments
are a special case of torsion groups where the period is
equal to a scale step. Where the period is equal to the
octave, the octave-equivalent temperament is torsion-free
and has a rank one lower than the octave-specific
equivalent.

Graham

Mike Battaglia <battaglia01@...> wrote:
> Actually, ignore everything else I said. Here's the best
> answer to this yet - temperament searches. Go to Graham's
> temperament finder and look at how the list changes as
> you go from the 5-limit to the 31-limit. The best
> temperaments depend on what limit or subgroup you use.

Error and complexity measures depend on a prime limit.
That doesn't mean just intonation has to be defined with a
prime limit. I'm happy with just intonation being a free
abelian group and prime limits being subgroups of it.

You can, in fact, calculate error and complexity of an
epimorphism of infinite-rank just intonation. You define
the weightings of primes you want to ignore to be zero.

There's a fair amount you can do with infinite-rank just
intonation. You can define the mappings for equal
temperaments according to some rule like patent vals. You
can define higher rank temperaments as compositions (or
unions) of equal temperaments. You can specify the
tunings. You can map ratios into those temperaments,
generating mappings for new primes as you need them. You
can determine if a given interval maps to a unison vector
for the temperament, but you can't write out a basis for
the unison vectors.

There's something to be said for this interpretation in
musical terms. If you specify 5-limit meantone and tune it
up, you might still play an augmented sixth. That's going
to sound like an approximate 7:4. It doesn't matter if you
set out to approximate the 7-limit, or optimized with the
7-limit in mind, that interval can still come up and it's
still going to sound like the 7-limit.

Graham

On Tue, Aug 23, 2011 at 12:55 PM, Graham Breed <gbreed@...> wrote:
>
> Mike Battaglia <battaglia01@...> wrote:
> > Actually, ignore everything else I said. Here's the best
> > answer to this yet - temperament searches. Go to Graham's
> > temperament finder and look at how the list changes as
> > you go from the 5-limit to the 31-limit. The best
> > temperaments depend on what limit or subgroup you use.
>
> Error and complexity measures depend on a prime limit.
> That doesn't mean just intonation has to be defined with a
> prime limit. I'm happy with just intonation being a free
> abelian group and prime limits being subgroups of it.
>
> You can, in fact, calculate error and complexity of an
> epimorphism of infinite-rank just intonation. You define
> the weightings of primes you want to ignore to be zero.

Why do they have to depend on a prime limit in general? There should
be some way to generalize it to infinite-rank, right? I've been
working a lot with the zeta function lately to try and figure that
out.

> There's a fair amount you can do with infinite-rank just
> intonation. You can define the mappings for equal
> temperaments according to some rule like patent vals.

I was thinking it might be possible to work out a more complex rule
that gives you optimal TE error as well.

> You can define higher rank temperaments as compositions (or
> unions) of equal temperaments. You can specify the
> tunings. You can map ratios into those temperaments,
> generating mappings for new primes as you need them. You
> can determine if a given interval maps to a unison vector
> for the temperament, but you can't write out a basis for
> the unison vectors.

If we're in countably infinite-rank JI, then the basis for the unison
vectors should end up being an enumerable set, right? In my romantic
moments I like to imagine every temperament could have a function f(n)
which spits out some sort of convergent and infinite-limit series of
unison vectors. Then you could define a temperament either by giving
f(n), or by giving a set of vals, each of which would have to have
their coefficients represented by a similar sort of function function.
Then the name of the game is to find out what these functions would
look like, what class of functions they'd be, and which ones represent
things that the auditory system wants to hear. Easier said than done,
but I think it's possible and would be a fun project to do. I'm just
trying to get my head wrapped around the w-limit space first, though.

> There's something to be said for this interpretation in
> musical terms. If you specify 5-limit meantone and tune it
> up, you might still play an augmented sixth. That's going
> to sound like an approximate 7:4. It doesn't matter if you
> set out to approximate the 7-limit, or optimized with the
> 7-limit in mind, that interval can still come up and it's
> still going to sound like the 7-limit.

Yes! Exactly.

-Mike

Mike Battaglia <battaglia01@...> wrote:

> Why do they have to depend on a prime limit in general?
> There should be some way to generalize it to
> infinite-rank, right? I've been working a lot with the
> zeta function lately to try and figure that out.

Yes. If you have a function with a clearly defined value
as the prime limit approaches infinity, you can use that.

> > There's a fair amount you can do with infinite-rank just
> > intonation. You can define the mappings for equal
> > temperaments according to some rule like patent vals.
>
> I was thinking it might be possible to work out a more
> complex rule that gives you optimal TE error as well.

Maybe tuning-math would be a better place to discuss it.
And not in a thread with "newbies" in the subject line.

> If we're in countably infinite-rank JI, then the basis
> for the unison vectors should end up being an enumerable
> set, right? In my romantic moments I like to imagine
> every temperament could have a function f(n) which spits
> out some sort of convergent and infinite-limit series of
> unison vectors. Then you could define a temperament
> either by giving f(n), or by giving a set of vals, each
> of which would have to have their coefficients
> represented by a similar sort of function function. Then
> the name of the game is to find out what these functions
> would look like, what class of functions they'd be, and
> which ones represent things that the auditory system
> wants to hear. Easier said than done, but I think it's
> possible and would be a fun project to do. I'm just
> trying to get my head wrapped around the w-limit space
> first, though.

You can return the simplest unison vector where any given
prime is the highest with a non-zero coefficient.
"Simplest" would depend on how you measure complexity.

Graham

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

> I've been reading about group theory again. Free abelian
> groups are on a different page to free groups. I don't
> even know if free abelian groups are free groups that are
> abelian, or an independent concept.

It's different. They are only the same for rank one, which is free both as a group and an abelian group.

A free abelian group
> is simple though: it's made up of infinite cyclic groups.

Sadly, not true if it isn't finitely generated; that's what the Baer-Specker business (ZxZxZx... for infinite Z's, or in other words sequences of integers which you add) was all about.

> An infinite cyclic group is something that's called cyclic
> but doesn't cycle.

An amazingly horrible bit of jargon, but standard. You can also call it Z, meaning the integers but you're only looking at addition.