Yahoo Tuning Groups Ultimate Backup tuning Another tuning-math question -- multivals in 3D

2012/3/23 Petr Pařízek <petrparizek2000@...>
>
> Hi tuners.
>
> For 2D multivals, I have absolutely no problem in finding that, for
> example,
> 5-equal and 7-equal make a wedgie of <<1 4 4||. But now, suppose we have
> three 7-limit vals like this:
> [< 11 17 25 30 ]
> < 8 13 19 23 ]
> < 7 11 16 20 ]>
> If they were two of them, I would know what I should multiply with what.
> But
> for three, I have no idea. Can someone help?
>
> Thanks in advance.

You can always just work it out by hand if you want. For instance,
here's what you really want

<11 17 25 30| ^ <8 13 19 23| ^ <7 11 16 20|

This is short for

(11a + 17b + 25c + 30d) ^ (8a + 13b + 19c + 23d) ^ (7a + 11b + 16c + 20d)

Now just multiply the polynomials out. Exterior algebra is
associative, so you can do the first one first, like this:

[(11a + 17b + 25c + 30d) ^ (8a + 13b + 19c + 23d)] ^ (7a + 11b + 16c + 20d)

So let's just do the first polynomial multiplication first; we'll
replace that by the multivector "u" above:

u ^ (7a + 11b + 16c + 20d)
let u = (11a + 17b + 25c + 30d) ^ (8a + 13b + 19c + 23d)

OK, so let's simplify u. What do we do? Well, first, let's expand the
whole thing out:

(11a + 17b + 25c + 30d) ^ (8a + 13b + 19c + 23d)
[11a^8a + 11a^13b + 11a^19c + 11a^23d] + [17b^8a + 17b^13b + 17b^19c +
17b^23d] + [25c^8a + 25c^13b + 25c^19c + 25c^23d] + [30d^8a + 30d^13b
+ 30d^19c + 30d^23d]

Yikes, what a mess! Fortunately, this simplifies down a lot. Just
remember the rule that for any vector v, v^v = 0. So we can eliminate
all terms above that have an a^a or a b^b or a c^c or a d^d in them:

[11a^13b + 11a^19c + 11a^23d] + [17b^8a + 17b^19c + 17b^23d] + [25c^8a
+ 25c^13b + 25c^23d] + [30d^8a + 30d^13b + 30d^19c]

OK, so only 12 terms left - not bad, but still not ideal. The second
thing to remember is the wedge product is bilinear, so ka^lb =
(kl)(a^b), where k and l are scalars. So let's reduce all these
numbers

[143a^b + 209a^c + 253a^d] + [136b^a + 323b^c + 391b^d] + [200c^a +
325c^b + 575c^d] + [240d^a + 390d^b + 570d^c]

OK, last step: the rule to remember when doing wedge products out by
hand is that b^a = -a^b. So terms like 136b^a are really just -136a^b.
So let's finish the job

[143a^b + 209a^c + 253a^d] + [-136a^b + 323b^c + 391b^d] + [-200a^c -
325b^c + 575c^d] + [-240a^d - 390b^d - 570c^d]

OK, now adding it all up, we finally get

7a^b + 9a^c + 13a^d - 2b^c + b^d + 5c^d

or, written as a multival,

<<7 9 13 -2 1 5||.

So that's just the first part. Now your original multiplication has
been reduced to

<<7 9 13 -2 1 5|| ^ <7 11 16 20|

And so you just have to do it again. Write out all the terms and work
it out. This is what I get if I do: <<<1 -7 -4 1|||

Anyway, it's a lot faster to program this though, I assure you :)

-Mike

🔗Petr Pařízek <petrparizek2000@...>

🔗Mike Battaglia <battaglia01@...>

2012/3/23 Petr Pařízek <petrparizek2000@...>
>
> I wrote:
>
> > thanks a lot for such a detailed explanation. I'll probably read it
> > several
> > times before I can fully appreciate it.
>
> Another way of finding relationships between multivals and step sizes
> would
> be to find a general solution of the following three equations:
> u(1)*x + u(2)*y + u(3)*z = A
> v(1)*x + v(2)*y + v(3)*z = B
> w(1)*x + w(2)*y + w(3)*z = C
> If I knew the general values of "x", "y", "z", then my question would be
> 100% answered.

What's A, B, and C, and what are u, v, and w?

It's easy to solve in general. Your equations can be rewritten in
matrix form as follows:

[u(1) u(2) u(3)] [x]
[v(1) v(2) v(3)] [y]
[w(1) w(2) w(3)] [z]

And then that you want that to multiply and equal [A B C]' on the
right. So just find the inverse of the matrix on the left, and then
multiply the inverse times [A B C]' on the right.

-Mike

🔗Petr Pařízek <petrparizek2000@...>

Mike wrote:

> What's A, B, and C,

The logarithmic sizes of the basic intervals (for example, 1200 cents, ~1901.955 cents, ~2786.314 cents).

> and what are u, v, and w?

These are triples of step sizes which add up to one of the basic intervals. For example, in the case of the previously mentioned temperament where 11x+8y+7z add up to 1200 cents, the "u"s are equal to 11, 8, 7, respectively.

> It's easy to solve in general. Your equations can be rewritten in
> matrix form as follows:
>
> [u(1) u(2) u(3)] [x]
> [v(1) v(2) v(3)] [y]
> [w(1) w(2) w(3)] [z]
>
> And then that you want that to multiply and equal [A B C]' on the
> right. So just find the inverse of the matrix on the left, and then
> multiply the inverse times [A B C]' on the right.

Okay, thanks. Once I get over this, the rest will be "plain sailing", I think.

Petr

🔗genewardsmith <genewardsmith@...>

--- In tuning@yahoogroups.com, Petr PaÅÃzek <petrparizek2000@...> wrote:
>
> Hi Mike,
>
> thanks a lot for such a detailed explanation. I'll probably read it several
> times before I can fully appreciate it.

In case you have Maple or can read Maple code as pseudocode, here's how to compute it in Maple:

cho := proc(m, n) combinat[choose](m, n) end:

zerlist := proc(n)
# list of n 0s
local i, u;
u := NULL;
for i from 1 to n do
u := u,0 od;
[u] end:

relpar := proc(u, v)
# relative parity of two permutations
local t;
t := table('antisymmetric');
t[op(u)] := 1;
t[op(v)];
end:

pari := proc(u)
# parity of permutation u
local v;
v := sort(u);
relpar(u, v) end:

mvec := proc(l)
# multivector wedge product of vector list l
local c, i, j, k, q, r, t, u, v, w;
u := combinat[permute](nops(l));
c := cho(nops(l[1]), nops(l));
w := zerlist(nops(c));
for i from 1 to nops(c) do
t := c[i];
r := 0;
for j from 1 to nops(u) do
v := u[j];
q := pari(v);
for k from 1 to nops(v) do
q := q * l[v[k], t[k]] od;
r := r+q od;
w[i] := w[i]+r od;
w end: