Yahoo Tuning Groups Ultimate Backup tuning-math Triangular commas from two superparticular squared commas

I was looking at the relationship between 5/4, 81/64, and 9/7
and found that you get 81/80, 64/63 between 81/64 & 5/4 and 9/7 &
81/64
respectively. Multiplied out you get 36/35 (ratio between 9/7 & 5/4).
36 is a triangular comma and also a square itself.

Taking other superparticular squared commas:

9/8*4/3=3/2
16/15*9/8=6/5
25/24*16/15=10/9
36/35*25/24=15/14
49/48*36/35=21/20
64/63*49/48=28/27
81/80*64/63=36/35
100/99*81/80=45/44

and so forth.

Now triangular numbers can be found from (n+1)(n)/2 where n is the nth
triangular number. So, for 36 you have (9*8)/2. That got me thinking
if there is a way to find all the squared triangular numbers. So
taking n^2-n-2m=0 and using the quadratic formula you get -1+/- sqrt
(1+4(2m)) for m an square. (you get n=(-1+/-17)/2 = (8, -9) when m=36.
Any ideas of how I could find other square triangular numbers, anyone?
It's interesting that for 36/35 you have ((8*9)/2)/(((8*9)-1)/2)).
(No its not...)

But what is kind of interesting is that 36/35=(6/5)/(7/6) and also
(10/9)/(8/7) and ALSO (10/7)/(5/4). Does any other superparticular
ratio have this property? (Also, 36/35 is a squared comma and a
triangular one too.)

Paul Hj

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:
>
> I was looking at the relationship between 5/4, 81/64, and 9/7
> and found that you get 81/80, 64/63 between 81/64 & 5/4 and 9/7 &
> 81/64
> respectively. Multiplied out you get 36/35 (ratio between 9/7 &
5/4).
> 36 is a triangular comma and also a square itself.
>
> Taking other superparticular squared commas:
>
> 9/8*4/3=3/2
> 16/15*9/8=6/5
> 25/24*16/15=10/9
> 36/35*25/24=15/14
> 49/48*36/35=21/20
> 64/63*49/48=28/27
> 81/80*64/63=36/35
> 100/99*81/80=45/44
>
> and so forth.
>
> Now triangular numbers can be found from (n+1)(n)/2 where n is the
nth
> triangular number. So, for 36 you have (9*8)/2. That got me thinking
> if there is a way to find all the squared triangular numbers. So
> taking n^2-n-2m=0 and using the quadratic formula you get -1+/- sqrt
> (1+4(2m)) for m an square. (you get n=(-1+/-17)/2 = (8, -9) when
m=36.
> Any ideas of how I could find other square triangular numbers,
anyone?
> It's interesting that for 36/35 you have ((8*9)/2)/(((8*9)-1)/2)).
> (No its not...)
>
> But what is kind of interesting is that 36/35=(6/5)/(7/6) and also
> (10/9)/(8/7) and ALSO (10/7)/(5/4). Does any other superparticular

I meant ALSO (9/7)/(5/4)

> ratio have this property? (Also, 36/35 is a squared comma and a
> triangular one too.)
>
> Paul Hj
>

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> Any ideas of how I could find other square triangular numbers, anyone?

Those can be found as solutions to a Pell's equation. There are an
infinite number of them, but size grows exponentially. Here is a
relevant links posting:

/tuning-math/message/8380

Other stuff on related topics:

/tuning-math/message/8386

/tuning-math/message/8379

/tuning-math/message/1286

/tuning-math/message/1016

/tuning-math/message/849

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
>
> > Any ideas of how I could find other square triangular numbers,
anyone?
>
> Those can be found as solutions to a Pell's equation. There are an
> infinite number of them, but size grows exponentially. Here is a
> relevant links posting:
>
> /tuning-math/message/8380
>
> Other stuff on related topics:
>
> /tuning-math/message/8386
>
> /tuning-math/message/8379
>
> /tuning-math/message/1286
>
> /tuning-math/message/1016
>
> /tuning-math/message/849

Thanks. Obviously I sublimated this from your postings back then.
Even though I don't remember reading about 81/80*64/63=36/35
>

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

Here it is - (Message 1016)

If t(n) = n(n+1)/2 is the nth triangular number, we may denote these
by T(n) = t(n)/(t(n)-1) = and S(n) = n^2/(n^2-1). We then have

T(n) = S(n)S(n+1),

S(n) = T(2n-1)T(2n),

as we may verify by a simple calculation

Can this be shown algebraically? Or is it just by inspection.
Thanks

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

Which I just did (on a notepad, that is). The question is, how would
someone discover these identities? It couldn't just be trial and error
(like I did, examining different major thirds). I mean, life is short.
Oh by the way the Pell equation is pretty neat. I would like to tie
in the math for 36/35 with the 35 hexachords and 35 pentachords in 12-et
but my brother says that is just numerology. Just a coincidence.

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> Which I just did (on a notepad, that is). The question is, how would
> someone discover these identities?

I discovered them by observing the pattern in question, and them
proving the general result, which turned out to be easy.

>I would like to tie
> in the math for 36/35 with the 35 hexachords and 35 pentachords in 12-et
> but my brother says that is just numerology. Just a coincidence.

Is your brother interested in this stuff?

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
>
> > Which I just did (on a notepad, that is). The question is, how
would
> > someone discover these identities?
>
> I discovered them by observing the pattern in question, and them
> proving the general result, which turned out to be easy.

Yes I guess I rediscovered the one, but I don't think I would have
come up with the reversed one on my own. (Squares based on
Triangulars)
>
> >I would like to tie
> > in the math for 36/35 with the 35 hexachords and 35 pentachords
in 12-et
> > but my brother says that is just numerology. Just a coincidence.
>
> Is your brother interested in this stuff?

My brother is a Neurophysiology PhD Research Fellow. I show him this
stuff but he really doesn't have time for it. He does look at this
newsgroup from time to time when things are slow at work. He helps
me with programming, (I'm not very good at it, I did struggle through
Knuth's first section on the Art of Computer Programming). He's my
little brother by the way. He does research in drug addiction, etc.

If anyone is going to the SMT Conference in Boston, I am going, it
would be fun to see people from this newsgroup there. (I know at
least one person besides me is going). Hopefully they won't kick
me out (I'm not a professor, just a lowly BM /BA). Anyway I paid for
it, so there. But enough about me and my family.

Gene, do you have any plans to write a book on all this? Maybe a
textbook with accompanying workbook? Luckily the search feature is
much stronger now, I can find nearly any post in my memory in just a
few minutes, but it would be nice to have it all in one place.

Not to be a complainer, but nobody has really given me a clear answer
on how to find two commas from the kernel of two temperaments. Maybe
I am looking at things wrong. Not that straightforward perhaps. Paul's
Middle Path paper has some great examples but he said he found them
by inspection. Graham says he finds them by a "dumb" algorithm. You
have wedgies. (Something about triprime commas rings a bell). I think
I'll have to revisit some real early posts on this...I hate to make
you repeat yourself unneccessarily. Unless its for the good of the
group of course! The trouble with the approach I am taking, is
similar to someone trying to learn a foreign language with say,
just a English-German dictionary. It's not that flat. You might be
able to find your way around a little but it doesn't really teach you
all the grammar. So, time to dig out the textbooks. Even Mathworld
really isn't enough, it is actually rather incomplete in certain
areas. Eventually I would like to work my way up the Zeta tunings,
which is how I found out about this group in the first place.

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> Gene, do you have any plans to write a book on all this?

I've thought about it, but people don't seem to like what I have
written. On the other hand, it is nothing like Mazzola, and my commad
of English beats Parizek.

> Not to be a complainer, but nobody has really given me a clear answer
> on how to find two commas from the kernel of two temperaments.

We talked about the 7-limit as I recall, where two 7-limit
temperaments combine to give at most *one* comma. What, exactly, is it
you want to do?

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
>
> > Gene, do you have any plans to write a book on all this?
>
> I've thought about it, but people don't seem to like what I have
> written. On the other hand, it is nothing like Mazzola, and my
commad
> of English beats Parizek.

I'd read it! I guess your web page counts as a kind of book of its
own. Someday I'm going to reread everything on your site, maybe on
a day off.

>
> > Not to be a complainer, but nobody has really given me a clear
answer
> > on how to find two commas from the kernel of two temperaments.
>
> We talked about the 7-limit as I recall, where two 7-limit
> temperaments combine to give at most *one* comma. What, exactly, is
it
> you want to do?

I thought there might be a nice way to obtain exactly two
commas from the kernel of two 7-limit temperaments. I did dig
up messages 6615 and 8501 which demonstrate how to get 4 commas from
a 7-limit wedgie. The one-comma thing is also valuable, I'll locate
that one too. Thanks.
>

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul_hjelmstad@a...> wrote:
> >
> > > Gene, do you have any plans to write a book on all this?
> >
> > I've thought about it, but people don't seem to like what I have
> > written. On the other hand, it is nothing like Mazzola, and my
> commad
> > of English beats Parizek.
>
> I'd read it! I guess your web page counts as a kind of book of its
> own. Someday I'm going to reread everything on your site, maybe on
> a day off.
>
> >
> > > Not to be a complainer, but nobody has really given me a clear
> answer
> > > on how to find two commas from the kernel of two temperaments.
> >
> > We talked about the 7-limit as I recall, where two 7-limit
> > temperaments combine to give at most *one* comma. What, exactly,
is
> it
> > you want to do?
>
> I thought there might be a nice way to obtain exactly two
> commas from the kernel of two 7-limit temperaments. I did dig
> up messages 6615 and 8501 which demonstrate how to get 4 commas
from
> a 7-limit wedgie. The one-comma thing is also valuable, I'll locate
> that one too. Thanks.

Dear Paul,

Rather than immediately attacking the problem of how to get two 7-
limit commas from the common kernel of two ETs, how about stepping
back to the problem of how to get two 5-limit commas from the kernel
of one ET? It seems to me to make more sense to attack these kinds of
problems in order of increasing difficulty, rather than skipping
around.

-Paul

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
> <gwsmith@s...>
> > wrote:
> > >
> > > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > > <paul_hjelmstad@a...> wrote:
> > >
> > > > Gene, do you have any plans to write a book on all this?
> > >
> > > I've thought about it, but people don't seem to like what I have
> > > written. On the other hand, it is nothing like Mazzola, and my
> > commad
> > > of English beats Parizek.
> >
> > I'd read it! I guess your web page counts as a kind of book of
its
> > own. Someday I'm going to reread everything on your site, maybe on
> > a day off.
> >
> > >
> > > > Not to be a complainer, but nobody has really given me a
clear
> > answer
> > > > on how to find two commas from the kernel of two temperaments.
> > >
> > > We talked about the 7-limit as I recall, where two 7-limit
> > > temperaments combine to give at most *one* comma. What,
exactly,
> is
> > it
> > > you want to do?
> >
> > I thought there might be a nice way to obtain exactly two
> > commas from the kernel of two 7-limit temperaments. I did dig
> > up messages 6615 and 8501 which demonstrate how to get 4 commas
> from
> > a 7-limit wedgie. The one-comma thing is also valuable, I'll
locate
> > that one too. Thanks.
>
> Dear Paul,
>
> Rather than immediately attacking the problem of how to get two 7-
> limit commas from the common kernel of two ETs, how about stepping
> back to the problem of how to get two 5-limit commas from the
kernel
> of one ET? It seems to me to make more sense to attack these kinds
of
> problems in order of increasing difficulty, rather than skipping
> around.
>
> -Paul

You're right, but since I'm a lefty (right-brained) I tend to skip
around a lot.

I guess I thought I had mastered the 5-limit so I was 'moving on up'
to the 7-limit. (Plus I think the 7-limit is really important, given
the prevalence of dominant-7th chords in western music). Yes, how
about two commas from the 5-limit (Like of course, 128/125 and 81/180
from 12-tET...). Gene showed me how to get one comma in the kernel of
four ETs (2 pairs actually). The Pffafian of their wedge products
needs to be zero.

Okay, well lets start with the kernel of one ET in the 5-limit. In
Octave, if I take the kernel of 12 19 28 I don't get anything
sensible. It uses singular value decomposition so it reduces the
commas by sqrt() of a number but I haven't gotten that figured out
yet. So, of course the cross product of (-4, 4, -1) and (7, 0, -3)
gives (12 19 28), but how to go the other way?

Paul Hj

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> Okay, well lets start with the kernel of one ET in the 5-limit. In
> Octave, if I take the kernel of 12 19 28 I don't get anything
> sensible. It uses singular value decomposition so it reduces the
> commas by sqrt() of a number but I haven't gotten that figured out
> yet. So, of course the cross product of (-4, 4, -1) and (7, 0, -3)
> gives (12 19 28), but how to go the other way?

You've got three reduced two-prime commas, namely |-19 12 0>,
|7 0 -3> and |0 28 -19>. These generate the kernel, and any two kernel
elements whose cross-product is <12 19 28| form a kernel basis.

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
>
> > Okay, well lets start with the kernel of one ET in the 5-limit. In
> > Octave, if I take the kernel of 12 19 28 I don't get anything
> > sensible. It uses singular value decomposition so it reduces the
> > commas by sqrt() of a number but I haven't gotten that figured out
> > yet. So, of course the cross product of (-4, 4, -1) and (7, 0, -3)
> > gives (12 19 28), but how to go the other way?
>
> You've got three reduced two-prime commas, namely |-19 12 0>,
> |7 0 -3> and |0 28 -19>. These generate the kernel, and any two kernel
> elements whose cross-product is <12 19 28| form a kernel basis.
>
All three generate the kernel? Let's test in Octave. I get -0.33424, -
0.52921, -0.77989. Yes - you obtain 12,19,28. Neat. But I guess the
question Paul E. was posing (if I understand him correctly). Is how
do you find those three commas? (The two-prime commas), knowing only
12, 19, 28 as your input?

Thanx

Paul Hj

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul_hjelmstad@a...> wrote:
> >
> > > Okay, well lets start with the kernel of one ET in the 5-limit.
In
> > > Octave, if I take the kernel of 12 19 28 I don't get anything
> > > sensible. It uses singular value decomposition so it reduces the
> > > commas by sqrt() of a number but I haven't gotten that figured
out
> > > yet. So, of course the cross product of (-4, 4, -1) and (7, 0, -
3)
> > > gives (12 19 28), but how to go the other way?
> >
> > You've got three reduced two-prime commas, namely |-19 12 0>,
> > |7 0 -3> and |0 28 -19>. These generate the kernel, and any two
kernel
> > elements whose cross-product is <12 19 28| form a kernel basis.
> >
> All three generate the kernel? Let's test in Octave. I get -
0.33424, -
> 0.52921, -0.77989. Yes - you obtain 12,19,28. Neat. But I guess the
> question Paul E. was posing (if I understand him correctly). Is how
> do you find those three commas? (The two-prime commas), knowing only
> 12, 19, 28 as your input?
>
> Thanx
>
> Paul Hj
>
Okay I get it - duh. 12,19; 12,28->3,7; 19,28. Ready to move to the 7-
limit. Is the wedge method the best way? At least to find 2-free, 3-
free, 5-free and 7-free commas? A question that pertains to both 5
and 7 limit: Once you find these commas, how do you get the kernel
basis: Let's take (7,0,-3). Does knowing (12,19,28) help you find
(-4,4,-1)? What about the 7-limit? Thanks.

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

Gene wrote:

> > > You've got three reduced two-prime commas, namely |-19 12 0>,
> > > |7 0 -3> and |0 28 -19>. These generate the kernel, and any two
> > > kernel elements whose cross-product is <12 19 28| form a kernel
> > > basis.
> > >

Okay ---

I found (7,0,-3)&(-4,4,-1)
and (-19, 12, 0)& (-4,4,-1)
and (0,28,-19)& the atrocious comma (1,36,-24)

Octave can't do it - I had to do it by hand. Can Python do it?

Paul Hj

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:
>
>
> Gene wrote:
>
> > > > You've got three reduced two-prime commas, namely |-19 12 0>,
> > > > |7 0 -3> and |0 28 -19>. These generate the kernel, and any
two
> > > > kernel elements whose cross-product is <12 19 28| form a
kernel
> > > > basis.
> > > >
>
> Okay ---
>
> I found (7,0,-3)&(-4,4,-1)
> and (-19, 12, 0)& (-4,4,-1)
> and (0,28,-19)& the atrocious comma (1,36,-24)
>
> Octave can't do it - I had to do it by hand. Can Python do it?
>
> Paul Hj
>

For example can Python solve 28c-19b=12 for b and c integral values,
or is that insufficient information?

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul_hjelmstad@a...> wrote:
> >
> > > Okay, well lets start with the kernel of one ET in the 5-limit.
In
> > > Octave, if I take the kernel of 12 19 28 I don't get anything
> > > sensible. It uses singular value decomposition so it reduces the
> > > commas by sqrt() of a number but I haven't gotten that figured
out
> > > yet. So, of course the cross product of (-4, 4, -1) and (7, 0, -
3)
> > > gives (12 19 28), but how to go the other way?
> >
> > You've got three reduced two-prime commas, namely |-19 12 0>,
> > |7 0 -3> and |0 28 -19>. These generate the kernel, and any two
kernel
> > elements whose cross-product is <12 19 28| form a kernel basis.
> >
> All three generate the kernel? Let's test in Octave. I get -
0.33424, -
> 0.52921, -0.77989. Yes - you obtain 12,19,28. Neat. But I guess the
> question Paul E. was posing (if I understand him correctly). Is how
> do you find those three commas? (The two-prime commas), knowing only
> 12, 19, 28 as your input?

Take each possible pair of these with the third element zeroed out.
This gives you three vals:

[12 19 0>
[12 0 28>
[0 19 28>

Now take the complement of each and remove any common factors. Voila!

🔗Graham Breed <gbreed@gmail.com>

Paul G Hjelmstad wrote:

> For example can Python solve 28c-19b=12 for b and c integral values, > or is that insufficient information?

The standard distribution has no direct way of doing it. Python as a Turing-complete language is surely capable of implementing an algorithm to do it. Probably there's a library out there. You could search for Diophantine equations, or have a look at PuLP. I'm not sure what it does but it looks very clever:

http://www.jeannot.org/~js/code/index.en.html#PuLP

Graham

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@g...> wrote:
>
> Paul G Hjelmstad wrote:
>
> > For example can Python solve 28c-19b=12 for b and c integral values,
> > or is that insufficient information?
>
> The standard distribution has no direct way of doing it. Python as a
> Turing-complete language is surely capable of implementing an algorithm
> to do it.

This usually the first Diophantine equation presented in elementary
number theory textbooks such as Niven and Zuckerman, and it isn't hard.
The equation ax + by = cz has an infinite number integer solutions if
gcd(a,b) divides c, and I suspect someone has written Python code
already to solve it. Maple, of course, has a built-in routine (called
igcdex) to do it, and it might be possible Paul H. can get Matlab to
do it, using the Maple kernel. Euclidean algorithm or continued
fraction routines, if available, could also be used.

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, Graham Breed <gbreed@g...>
wrote:
> >
> > Paul G Hjelmstad wrote:
> >
> > > For example can Python solve 28c-19b=12 for b and c integral
values,
> > > or is that insufficient information?
> >
> > The standard distribution has no direct way of doing it. Python
as a
> > Turing-complete language is surely capable of implementing an
algorithm
> > to do it.
>
> This usually the first Diophantine equation presented in elementary
> number theory textbooks such as Niven and Zuckerman, and it isn't
hard.
> The equation ax + by = cz has an infinite number integer solutions
if
> gcd(a,b) divides c, and I suspect someone has written Python code
> already to solve it. Maple, of course, has a built-in routine
(called
> igcdex) to do it, and it might be possible Paul H. can get Matlab to
> do it, using the Maple kernel. Euclidean algorithm or continued
> fraction routines, if available, could also be used.
>
Thanks you guys. Now to get back to what I was trying to do -
Finding the kernel basis, knowing just (7,0,-3), (0,28,-19), (-
19,12,0). How would you guys find, for example, (-1,36,-24) as the
kernel basis partner of (0,28,-19)? (Their cross product is (12,19,28)

Thanks

Paul Hj

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> Thanks you guys. Now to get back to what I was trying to do -
> Finding the kernel basis, knowing just (7,0,-3), (0,28,-19), (-
> 19,12,0). How would you guys find, for example, (-1,36,-24) as the
> kernel basis partner of (0,28,-19)? (Their cross product is (12,19,28)

You want to find |a b c> such that the cross product with
|0 28 -19> is <12 19 28|. This is back to Diophantine equations; the
cross product is <19b+28c -19a -28a|. Hence a = -1 is easy, and you
are left with the Diophantine equation 19b+28c = 12.

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
>
> > Thanks you guys. Now to get back to what I was trying to do -
> > Finding the kernel basis, knowing just (7,0,-3), (0,28,-19), (-
> > 19,12,0). How would you guys find, for example, (-1,36,-24) as the
> > kernel basis partner of (0,28,-19)? (Their cross product is
(12,19,28)
>
> You want to find |a b c> such that the cross product with
> |0 28 -19> is <12 19 28|. This is back to Diophantine equations; the
> cross product is <19b+28c -19a -28a|. Hence a = -1 is easy, and you
> are left with the Diophantine equation 19b+28c = 12.
>
Is that how you find the kernel basis? What's special about
128/125&81/80 for example, as a kernel basis? Thanks.

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

What method do you use for the 5-limit?

> > What's special about
> > 128/125&81/80 for example, as a kernel basis? Thanks.
>
> It's TM reduced, meaning it has smaller integers in the integer ratios
> than other kernel bases.
>
Right - and I take it you can't find any simpler commas using these two
commas as factors.

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
>
> > What method do you use for the 5-limit?
>
> I use the same program for all limits, which first finds the two-prime
> commas and then feeds those into first Hermite, then LLL.

You mean (3) two-prime commas in the 5-limit and
(4) three-prime commas in the 7-limit, etc?

Is there a wedge method in the 5-limit like the one you use in the 7-
limit (Message 6615). Or do you just take, for example (7,0,-3),(0,28,-
19),(-19,12,0) and do first Hermite and then LLL? Sorry to repeat
questions so much. One thing I am trying to do is work this out
geometrically: Two lines (commas) intersect in the 5-limit, giving a
temperament, two planes (commas) intersect in the 7-limit, giving a
rank-two temperament, etc.

Thanks

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> > I use the same program for all limits, which first finds the two-prime
> > commas and then feeds those into first Hermite, then LLL.
>
> You mean (3) two-prime commas in the 5-limit and
> (4) three-prime commas in the 7-limit, etc?

No, we are talking about vals. Hence, there are three two-prime commas
in the 5-limit (usually), six in the 7-limit, and so forth.

> Is there a wedge method in the 5-limit like the one you use in the 7-
> limit (Message 6615).

The vals here are like the bivals used for rank-two temperaments.

> One thing I am trying to do is work this out
> geometrically: Two lines (commas) intersect in the 5-limit, giving a
> temperament, two planes (commas) intersect in the 7-limit, giving a
> rank-two temperament, etc.

It's not an intersection, but a union. The subspace generated by the
commas defines the temperament. A line in 3-space passing through the
origin is a point in the projective plane, so 5-limit temperaments can
be considered points in the plane. A plane in 4-space passing through
the origin is a line in projective space, and thus in the 7-limit rank
two temperaments can be seen as lines in projective space.

Equivalently, we can take the bival point of view, and define the
temperament by the subspace generated by two vals. This gives us a
picture where the 5-limit rank-two temperaments are lines, not points,
in the projective plane. These are the two dual views from Paul's
diagrams. In the 7-limit, we get a dual picture, still with lines as
representing temperaments, but with the points now being projective
vals, not projective commas.

Finally, in the 7-limit we can view the rank-two temperaments as
*points*, not lines, on a hypersurface in 5-dimensional projective space.

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
>
> > > I use the same program for all limits, which first finds the
two-prime
> > > commas and then feeds those into first Hermite, then LLL.
> >
> > You mean (3) two-prime commas in the 5-limit and
> > (4) three-prime commas in the 7-limit, etc?
>
> No, we are talking about vals. Hence, there are three two-prime
commas
> in the 5-limit (usually), six in the 7-limit, and so forth.

Yes, C(3,2) and C(4,2).. C(5,2). Why do you call them vals? Aren't
they monzos? What defines a val?
>
> > Is there a wedge method in the 5-limit like the one you use in
the 7-
> > limit (Message 6615).
>
> The vals here are like the bivals used for rank-two temperaments.

So, if the cross product generates (12,19,28) that is considered a
bival? Obviously its a lot simpler in the 5-limit.

>
> > One thing I am trying to do is work this out
> > geometrically: Two lines (commas) intersect in the 5-limit,
giving a
> > temperament, two planes (commas) intersect in the 7-limit, giving
a
> > rank-two temperament, etc.
>
> It's not an intersection, but a union. The subspace generated by the
> commas defines the temperament. A line in 3-space passing through
the
> origin is a point in the projective plane, so 5-limit temperaments
can
> be considered points in the plane. A plane in 4-space passing
through
> the origin is a line in projective space, and thus in the 7-limit
rank
> two temperaments can be seen as lines in projective space.

(Warning: You might find this question irritating: In Paul's zoom
diagrams he assumes octave-equivalence, so we are dealing with
one less dimension, visually. I was talking about actual
intersections of comma-lines in his diagrams. In your more abstract
description:

A line in 3-space (passing thru the origin) is -> a) one comma b) two
commas c) something else - more abstract

A plane in 4-space (passing thru the origin) is -> a) one comma b)
two commas c) something else

(I told you it would be irritating!).
>
> Equivalently, we can take the bival point of view, and define the
> temperament by the subspace generated by two vals. This gives us a
> picture where the 5-limit rank-two temperaments are lines, not
points,
> in the projective plane. These are the two dual views from Paul's
> diagrams. In the 7-limit, we get a dual picture, still with lines as
> representing temperaments, but with the points now being projective
> vals, not projective commas.
>
> Finally, in the 7-limit we can view the rank-two temperaments as
> *points*, not lines, on a hypersurface in 5-dimensional projective
space.

Did you mean 4-dimensional projective space? How did you get up to 5?
The dual views thing is pretty interesting. I think I will be able
to comprehend this once I am clear on comma-space. (First view).

Thanx

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> Yes, C(3,2) and C(4,2).. C(5,2). Why do you call them vals? Aren't
> they monzos? What defines a val?

I thought the point was to find a comma basis for a val, sich as
<12 19 28 34|.

> (Warning: You might find this question irritating: In Paul's zoom
> diagrams he assumes octave-equivalence, so we are dealing with
> one less dimension, visually. I was talking about actual
> intersections of comma-lines in his diagrams. In your more abstract
> description:
>
> A line in 3-space (passing thru the origin) is -> a) one comma b) two
> commas c) something else - more abstract

It isn't anything about commas until you relate it to commas. However,
the set of all lines passing through the origin in 3-space is model of
the projective plane, and equated with it.

> A plane in 4-space (passing thru the origin) is -> a) one comma b)
> two commas c) something else

A line in projective space.

> > Finally, in the 7-limit we can view the rank-two temperaments as
> > *points*, not lines, on a hypersurface in 5-dimensional projective
> space.
>
> Did you mean 4-dimensional projective space? How did you get up to 5?

No, there are six numbers to a wedgie, but it is a projective object,
and so really is a point in 5D projective space. Wedgies are
constrained to lie on a hypersurface in this space, and that is a 4D
object (algebraic variety.)

🔗Paul Erlich <perlich@aya.yale.edu>

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
>
> > Yes, C(3,2) and C(4,2).. C(5,2). Why do you call them vals?
Aren't
> > they monzos? What defines a val?
>
> I thought the point was to find a comma basis for a val, sich as
> <12 19 28 34|.

Yes. So, can you find 6 two-prime commas in the 7-limit from
a single val like <12 19 28 34| or do you need a wedgie? Is the
process similar to what you use to find 4 3-prime commas from a
wedgie? Thanks.

>
> > (Warning: You might find this question irritating: In Paul's zoom
> > diagrams he assumes octave-equivalence, so we are dealing with
> > one less dimension, visually. I was talking about actual
> > intersections of comma-lines in his diagrams. In your more
abstract
> > description:
> >
> > A line in 3-space (passing thru the origin) is -> a) one comma
b) two
> > commas c) something else - more abstract
>
> It isn't anything about commas until you relate it to commas.
However,
> the set of all lines passing through the origin in 3-space is
model of
> the projective plane, and equated with it.
>
> > A plane in 4-space (passing thru the origin) is -> a) one comma
b)
> > two commas c) something else
>
> A line in projective space.

Still don't get it. I need to study this further.

> > > Finally, in the 7-limit we can view the rank-two temperaments
as
> > > *points*, not lines, on a hypersurface in 5-dimensional
projective
> > space.
> >
> > Did you mean 4-dimensional projective space? How did you get up
to 5?
>
> No, there are six numbers to a wedgie, but it is a projective
object,
> and so really is a point in 5D projective space. Wedgies are
> constrained to lie on a hypersurface in this space, and that is a
4D
> object (algebraic variety.)
>
I see.

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
>
> > (Warning: You might find this question irritating: In Paul's zoom
> > diagrams he assumes octave-equivalence,
>
> Not in the recent ones, I don't.
>
> > so we are dealing with
> > one less dimension, visually.
>
> It's still one less dimension, even without the octave-equivalence.

Okay. That's where I get stuck. What do you think would be the best
way to understand the 7-limit? I was hoping on a model where commas
were planes and temperament pairs were lines. Looks like I will need to
scrap that view? Is the 5-limit the only place you get temperaments
determining commas and commas determining temperaments in a manner
that is easy to visualize?

Paul Hj

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> Okay. That's where I get stuck. What do you think would be the best
> way to understand the 7-limit? I was hoping on a model where commas
> were planes and temperament pairs were lines. Looks like I will need to
> scrap that view?

No need to; that can be done in projective 3-space. A 7-limit val
defines a line through the origin in 4-space, which can be taken as a
point in projective 3-space. Two vals define a plane through the
origin, corresponding to a projective line. In that sense of
"temperament pair", they therefore give a line. Three vals determine a
subspace, which equates to a projective plane, and which can be
equated to a projective monzo, or "promo".

🔗Paul Erlich <perlich@aya.yale.edu>

Sure you can -- you have six choices of which two primes out of four
to use! So in this case, the 6 two-prime commas would be

2^19/3^12
2^7/5^3
2^17/7^6
3^29/5^19
3^34/7^19
5^17/7^14

> or do you need a wedgie?

The val *is* the wedgie in this case . . .

> Is the
> process similar to what you use to find 4 3-prime commas from a
> wedgie? Thanks.

Yes.

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul_hjelmstad@a...> wrote:
> >
> > > (Warning: You might find this question irritating: In Paul's
zoom
> > > diagrams he assumes octave-equivalence,
> >
> > Not in the recent ones, I don't.
> >
> > > so we are dealing with
> > > one less dimension, visually.
> >
> > It's still one less dimension, even without the octave-
equivalence.
>
> Okay. That's where I get stuck. What do you think would be the best
> way to understand the 7-limit? I was hoping on a model where commas
> were planes and temperament pairs were lines.

Temperament pairs? You mean possible tunings of 2D temperaments?

> Looks like I will need to
> scrap that view?

Why? If you accept my substitution above, and add that ETs are points
(which I think you want too), there's no need to scrap anything.

> Is the 5-limit the only place you get temperaments
> determining commas and commas determining temperaments in a manner
> that is easy to visualize?

In the 7-limit, you need 3 dimensions, but with VRML (maybe Robert
Walker can help), you can rotate a 3D diagram around and, as long as
the planes you're drawing are quite transparent (or you're willing to
concentrate on the lines and points only), you should be able to
visualize the whole picture fairly well.

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

I meant like 12&19 which form a line in 3-D space. ETs themselves are
still points. I know how to extract two 7-limit temperaments using
the adjutant of the matrix of two 7-limit commas (plus 1 0 0 0 and a
chromatic unison vector ahead of them). So two planes determine a
line. It's trickier going the other way - but with wedgies Gene shows
us how to get 4 3-prime commas. (Message 6615). Sorry a little too
much information...

> > Looks like I will need to
> > scrap that view?
>
> Why? If you accept my substitution above, and add that ETs are
points
> (which I think you want too), there's no need to scrap anything.

I'm glad to know that. I like a little concreteness among all the
abstract theory!
>
> > Is the 5-limit the only place you get temperaments
> > determining commas and commas determining temperaments in a manner
> > that is easy to visualize?
>
> In the 7-limit, you need 3 dimensions, but with VRML (maybe Robert
> Walker can help), you can rotate a 3D diagram around and, as long
as
> the planes you're drawing are quite transparent (or you're willing
to
> concentrate on the lines and points only), you should be able to
> visualize the whole picture fairly well.
>
I'll look into that.

Paul Hj

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul_hjelmstad@a...> wrote:
> > >
> > > --- In tuning-math@yahoogroups.com, "Paul Erlich"
<perlich@a...>
> > wrote:
> > > >
> > > > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > > > <paul_hjelmstad@a...> wrote:
> > > >
> > > > > (Warning: You might find this question irritating: In
Paul's
> > zoom
> > > > > diagrams he assumes octave-equivalence,
> > > >
> > > > Not in the recent ones, I don't.
> > > >
> > > > > so we are dealing with
> > > > > one less dimension, visually.
> > > >
> > > > It's still one less dimension, even without the octave-
> > equivalence.
> > >
> > > Okay. That's where I get stuck. What do you think would be the
> best
> > > way to understand the 7-limit? I was hoping on a model where
> commas
> > > were planes and temperament pairs were lines.
> >
> > Temperament pairs? You mean possible tunings of 2D temperaments?
>
> I meant like 12&19 which form a line in 3-D space.

Right, that line represents all possible tunings of meantone, a 2D
temperament. "Temperament pairs" might suggest something like
meantone&miracle, but I don't think that's what you meant.

> ETs themselves are
> still points. I know how to extract two 7-limit temperaments using
> the adjutant of the matrix of two 7-limit commas (plus 1 0 0 0 and
a
> chromatic unison vector ahead of them). So two planes determine a
> line. It's trickier going the other way - but with wedgies Gene
shows
> us how to get 4 3-prime commas. (Message 6615). Sorry a little too
> much information...
>
>
> > > Looks like I will need to
> > > scrap that view?
> >
> > Why? If you accept my substitution above, and add that ETs are
> points
> > (which I think you want too), there's no need to scrap anything.
>
> I'm glad to know that. I like a little concreteness among all the
> abstract theory!

Me too . . .

> > > Is the 5-limit the only place you get temperaments
> > > determining commas and commas determining temperaments in a
manner
> > > that is easy to visualize?
> >
> > In the 7-limit, you need 3 dimensions, but with VRML (maybe
Robert
> > Walker can help), you can rotate a 3D diagram around and, as long
> as
> > the planes you're drawing are quite transparent (or you're
willing
> to
> > concentrate on the lines and points only), you should be able to
> > visualize the whole picture fairly well.
> >
> I'll look into that.
>
> Paul Hj
>