Tuning Coincidence

Here I go with another potentially ridiculous comparision between
musical set theory (a la Polya) and tuning mathematics.

Consider the importance of 80, 50, 35 as ratios, and in hexachord
set theory.

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 T11 T0
0 2 0 6 0 20 0 6 0 2 0 924 SUM/12 = 80 (Map->Tn)
20 20 20 20 20 20 20 20 20 20 20 20 SUM/12 = 20 (Map->TnI)
2 4 8 X 2 64 2 X 8 4 2 X SUM/12 = 8 (Map->Comp,Tn)
64 0 64 0 64 0 64 0 64 0 64 0 SUM/12 = 32 (Map->Comp,TnI)

There it is. 80 + 20 /2 = 50 (Reduce for mirror image)
Now take (50 + ((32 + 8)/2))=35 (Reduced for complementability)

There are formulas a la Polya for all the above, of course!

Now consider this grid
1 6/5 (6/5)^2 (6/5)^3
1
5/4
(5/4)^-1

With the usual commas (128/125, 648/625 = (128/125 * 81/80)

Also consider (6/5)^2 -> 10/7 with 128/125 comma and 9 -> 7^(-1)

Also consider shifting steps (3,7,11) and (6,2,10) (as a permutation,
just one time each) to obtain this grid with the usual commas
1 3 3^2 3^3
1
5/4
(5/4)^1

As before 3^2->7^(-1) changes the column header to 1 3 7^-1 3/7
With the usual commas, each row ends up by 5. (With 3^2->7(-1) and
also 50/49 and 128/125) Another expression of 7^-1 is 35 then. This
is my shoebox idea - (C[2*2] X C3) where [2*2] is notational, not C2
X C2...

I propose the prevalence of 80, 50 and 35 (81/80, 50/49, 36/35)in
both set theory and tuning theory is not just a coincidence, but a
number-theoretical correspondence. Back to sets-

80 + 20 = 100 / 2 = 50 which also leaves 30 (mirror-image)
50 + 20 = 70 / 2 = 35 leaves 15 (Z-related complements)

30/50 = 3/5 minor third
15/35 = 3/7 minor third (slim)
100/80 = 5/4 major third
70/50 = 7/5 tritone

These are the important ratios for the shoebox idea and also
correspond to 2^(1/2), 2^(1/3) and 2^(1/4) the main divisions of 12
(2 x 2 x 3)

And so forth. I can't reproduce this in 22-tET, for example, which
is greatly simplified in terms of Polya being 2 * prime so it would
be easier to figure out, that is, if the same kind of coincidences
only occurred.

It would help if one could factor 80, 50 and 35 (Like one can with
the 66 pentachords, or at least it's a little easier...)

I tried adding the columns of the Polya grid in different
combinations (even/odd, mod3, mod4, mod6 with unconvincing results...

But I'm not done yet. Of course, I would like to extend all this
to lattices (working on it) 15/14 and 21/20 are prevalent for example.
The shoebox is not the greatest lattice I must admit, but it serves
its purpose.

Thoughts? More to come, stay tuned. (I always wanted to say that)

Paul Hj

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> Here I go with another potentially ridiculous comparision between
> musical set theory (a la Polya) and tuning mathematics.
>
> Consider the importance of 80, 50, 35 as ratios, and in hexachord
> set theory.
>
> T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 T11 T0
> 0 2 0 6 0 20 0 6 0 2 0 924 SUM/12 = 80 (Map->Tn)
> 20 20 20 20 20 20 20 20 20 20 20 20 SUM/12 = 20 (Map->TnI)
> 2 4 8 X 2 64 2 X 8 4 2 X SUM/12 = 8 (Map->Comp,Tn)
> 64 0 64 0 64 0 64 0 64 0 64 0 SUM/12 = 32 (Map->Comp,TnI)
>
> There it is. 80 + 20 /2 = 50 (Reduce for mirror image)
> Now take (50 + ((32 + 8)/2))=35 (Reduced for complementability)
>
> There are formulas a la Polya for all the above, of course!
>
> Now consider this grid
> 1 6/5 (6/5)^2 (6/5)^3
> 1
> 5/4
> (5/4)^-1
>
> With the usual commas (128/125, 648/625 = (128/125 * 81/80)
>
> Also consider (6/5)^2 -> 10/7 with 128/125 comma and 9 -> 7^(-1)
>
> Also consider shifting steps (3,7,11) and (6,2,10) (as a
permutation,
> just one time each) to obtain this grid with the usual commas
> 1 3 3^2 3^3
> 1
> 5/4
> (5/4)^1
>
> As before 3^2->7^(-1) changes the column header to 1 3 7^-1 3/7
> With the usual commas, each row ends up by 5. (With 3^2->7(-1) and
> also 50/49 and 128/125) Another expression of 7^-1 is 35 then.
This
> is my shoebox idea - (C[2*2] X C3) where [2*2] is notational, not
C2
> X C2...
>
> I propose the prevalence of 80, 50 and 35 (81/80, 50/49, 36/35)in
> both set theory and tuning theory is not just a coincidence, but a
> number-theoretical correspondence. Back to sets-
>
> 80 + 20 = 100 / 2 = 50 which also leaves 30 (mirror-image)
> 50 + 20 = 70 / 2 = 35 leaves 15 (Z-related complements)
>
> 30/50 = 3/5 minor third
> 15/35 = 3/7 minor third (slim)
> 100/80 = 5/4 major third
> 70/50 = 7/5 tritone
>
> These are the important ratios for the shoebox idea and also
> correspond to 2^(1/2), 2^(1/3) and 2^(1/4) the main divisions of 12
> (2 x 2 x 3)
>
> And so forth. I can't reproduce this in 22-tET, for example, which
> is greatly simplified in terms of Polya being 2 * prime so it would
> be easier to figure out, that is, if the same kind of coincidences
> only occurred.
>
> It would help if one could factor 80, 50 and 35 (Like one can with
> the 66 pentachords, or at least it's a little easier...)
>
> I tried adding the columns of the Polya grid in different
> combinations (even/odd, mod3, mod4, mod6 with unconvincing
results...
>
> But I'm not done yet. Of course, I would like to extend all this
> to lattices (working on it) 15/14 and 21/20 are prevalent for
example.
> The shoebox is not the greatest lattice I must admit, but it
serves
> its purpose.
>
> Thoughts? More to come, stay tuned. (I always wanted to say that)
>
> Paul Hj

I've improved on this a little. Take a shoebox (5/4)^3 X (7/5)^2 X
(6/5)^2. (Using 7/5 instead of 10/7) 128/125 is kind of a built in
ratio, (the diesis) Now the other three that appear are: 35/36,
50/49 and 80/81. 80/81 corresponds to 80 hexachords, 50/49 to 50
hexachords (one reduction) 35/36 to 35 hexachords (two reductions).
Now 80/81 (times 125/128) ignores the second dimension and just
takes (6/5)^4 right back to 1 (or 2). So this has no "folds" 50/49
takes two tritones (7/5)^2 to 2, so that has one fold. (Just like 50
hexachords is one reduction...) and 35/36 takes two minor thirds
(6/5)^2 to 7/5. So that represents two folds, (35 hexachords has two
reductions). The numbers is the numerators are 80, 50 and 35.
128/125 or 125/128 kind of stay in the background, and are used in
the first and third dimensions. If you prefer, just use two
dimensions with (6/5)^2->7/5

Another correspondence is 8 X 10, 5 X 10, 5 X 7. The first and
second share a common factor and the second and third do also.
Also, these numbers are as close together as you can get by making
two integers from 2^4 * 5, 5^2 * 2, and 5 * 7 respectively.

Combinatorically, one can show this, remembering that these factors
don't represent literal hexachord arrangements, but possibilities
for arrangements - just like 66 pentachords aren't literally
arranged in 6 * 11 pattern. (Start with 12!/7!5! divide by 12 etc.
I suppose 12*11*10*9*8 could be a pattern but it favors choices from
the front, in order... Oh by the way hit Reply to see the grids
correctly, of course!
>