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[Gene Ward Smith, Yahoo tuning-math message 12906 (Sun Oct 16, 2005 9:27 pm)]

Given a projection matrix defining a tuning for a rank-two temperament, there will be a left eigenspace of dimension two corresponding to the eigenvalue 1. If the projection map gives pure octaves, one of the eigenvectors will be the octave, that is, |1 0 0 ... 0>. Ignoring 2, the rest of the eigenspace is defined as multiples of a single vector. Choosing a multiple such that the coefficients are integers, and therefore the result is a monzo, is always possible if the projection matrix is defined over the rationals; ie, has rational coefficients. In this case we can choose a monzo so that the corresponding interval q satisfies 1<q<2. There will be two of these; we can make the choice unique by requiring that 1<q<sqrt(2), but I prefer choosing the one with the least Tenney height. We may call this, and the corresponding monzo, the eigenmonzo for the temperament tuning. Given a temperament, the eigenmonzo determines the tuning.

For least squares tunings, the projection matrix defined over the rationals and the corresponding eigenmonzo is readily obtained. The minimax tuning also gives a rational projection matrix, and if the two tunings are distinct, then all the other tunings can be defined in terms of t*(least sqares) + (1-t)*(minimax), where "least squares" and "minimax" denote matrices. The minimax matrix, for someone who has a simplex algorithm over the rationals implemented, is easy to find. Those, like me, who only have a floating point version have alternative methods. One involves the fact that the set of possible eigenmonzos for a minimax tuning is finite.

For a minimax tuning with respect to a consonance set C, there will be two elements u and v of C such that the errors of u and v are equal, and maximal among all errors for C. Hence, of u/v and v/u reduced to the octave, the one with minimal Tenney height will be the eigenmonzo. This gives us a finite set of possibilities for any C, and we can find the minimax tuning simply by checking all the possible eigenmonzos. For the 7-limit, for instance, possible eigenmonzos for minimax tunings are the 21 values 49/48, 25/24, 21/20, 15/14, 35/32, 7/6, 25/21, 49/40, 5/4, 9/7, 21/16, 7/5, 35/24, 3/2, 49/30, 5/3, 7/4, 25/14, 9/5, 15/8, and 35/18.

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[Gene Ward Smith, Yahoo tuning-math message 12907 (Sun Oct 16, 2005 10:23 pm)]

Subject: Some 7-limit minimax eigenmonzos

Decimal 3/2
Dominant seventh 7/5
Diminished 21/16
Blackwood 25/21
Augie 7/5
Pajara 35/24
Mavila 7/4
Negri 3/2
Kemun 3/2
Augmented 7/6
Godzilla 5/4
Meantone 5/4
Injera 35/24
Doublewide 3/2
Porcupine 5/4
Unidec 7/5
Superpyth 7/5
Mothra 5/4
Muggles 7/6
Beatles 5/3
Flattone 7/5
Magic 3/2
Myna 3/2
Sensi 7/4
Orwell 7/5
Miracle 5/3
Valentine 7/6
Mothra 5/4
Garibaldi 7/6
Superkleismic 7/5
Squares 5/4
Semififths 5/4
Diaschismic 7/4
Octacot 5/4
Tritonic 7/6
Rodan 7/5
Shrutar 7/4
Catakleismic 7/4
Hemiwuerschmidt 3/2
Hemikleismic 7/6
Hemithirds 7/6
Wizard 7/5
Waage 35/24
Slender 5/3
Amity 7/5
Hemififths 5/4
Pontiac 7/5
Tertiaseptal 7/5
Enneadecal 7/4
Ennealimmal 5/4
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[Gene Ward Smith, Yahoo tuning-math message 12912 (Mon Oct 17, 2005 2:10 pm)]

Subject: Re: Eigenmonzos

--- In tuning-math@yahoogroups.com, Carl Lumma wrote:

> You lost me on this one. You might say what you're trying to
> do here.

Some ideas to keep in mind from this are that rank two temperament tunings can often be defined by giving a single interval which the tuning tunes exactly. This interval is easily found for least squares and minimax tunings, and for minimax tunings, the intervals are simple. The method works when the tuning uses fractional monzos, that is, monzos with integer coefficients. Of course, this allows one to get arbirarily close to any given tuning.

Hence, for instance, instead of 1/4-comma meaantone we could speak of 5/4-eigenmonzo meantone. Similarly, we might want to consider 5/3-eigenmonzo miracle, or 9/5-eigenmonzo miracle, etc.

We have, for meantone,

1/3-comma = 5/3-eigenmonzo
2/7-comma = 25/24-eigenmonzo
1/4-comma = 5/4-eigenmonzo
1/5-comma = 16/15-eigenmonzo
1/6-comma = 45/32-eigenmonzo
1/7-comma = 135/128-eigenmonzo
1/11-comma = 10935/8192-eigenmonzo
0-comma = 3/2-eigenmonzo
7/26-comma = 78125/73728-eigenmonzo
. . . . . . . . .
[Gene Ward Smith, Yahoo tuning-math message 12915 (Mon Oct 17, 2005 3:19 pm)]

Subject: Re: Eigenmonzos

It might be better yet to point out how these are related in the 5-limit meantone case. For q-comma meantone, the corresponding projection matrix is

[|1 0 0>, |4q 1-4q q>, |16q-4 4-16q 4q>]

The non-octave 1 left eigenvectors are multiples of |0 1-4q q>. If we pick a value for q, let's say q=5/17, then we can susbstitute and get a rational monzo: |0 -3/17 5/17>. Clearing denominators gives |0 -3 5>, or 3125/27. Reducing to the octave gives v=3125/1728. Since v has a lower Tenney height than 2/v, v is the eigenmonzo.

. . . . . . . . .
[Paul Erlich, Yahoo tuning-math message 12921 (Mon Oct 17, 2005 4:05 pm)]

Subject: Re: Eigenmonzos

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" wrote:

> There will be two of these; we can make the choice
> unique by requiring that 1 > with the least Tenney height.

Maybe this is a nitpick, but I think this is an unfortunate mixing of octave-equivalent and non-octave-equivalent paradigms, so to speak. In the former, I believe Kees expressibility (aka "odd-limit") is relevant, and Tenney HD isn't. After all, we're talking about interval classes here, right?

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