I've thought of a new approach to the question of defining a canonical

generator for a given linear temperament.

If T is a linear temperament, and r is a rational representative of

the generator (eg, the TM reduced generator), and if TM is the

Tenny-Minkowski reduction function for T, then we define the limit

generator for T as

g = lim n-->infinity (TM(r^n))^(1/n)

in terms of cents, this would be

g = lim n--> infinity cents(TM(r^n))/n

The limit does exist, so this definition defines a unique generator,

without, alas, telling us a good way to compute it. The limit secor in

the 7-limit is about 116.785 cents, which we may compare to the rms

value of 116.573 cents and the minimax value of 116.588 cents. Here

are the ratios of successive terms for the TM reduced sequence for

miracle:

15/14, 16/15, 343/320, 15/14, 16/15, 15/14, 16/15, 15/14, 16/15,

343/320, 15/14, 15/14, 16/15, 15/14, 16/15, 15/14, 343/320, 15/14,

16/15, 15/14, 16/15, 15/14, 343/320, 15/14, 16/15, 15/14, 16/15,

15/14, 343/320, 16/15, 15/14, 16/15, 15/14, 15/14, 2401/2250, 15/14,

15/14, 16/15, 15/14, 15/14, 16/15,343/320, 15/14, 16/15, 15/14, 15/14,

16/15, 343/320, 15/14, 16/15, 15/14, 16/15, 15/14, 15/14, 2401/2250,

15/14, 16/15, 15/14, 15/14, 15/14, 2401/2250, 15/14, 16/15, 15/14,

15/14, 15/14, 2401/2250, 15/14, 16/15, 15/14, 15/14, 16/15,

343/320, 15/14, 16/15, 15/14, 16/15, 15/14, 15/14, 2401/2250, 15/14,

16/15, 15/14, 15/14, 15/14, 2401/2250, 15/14, 16/15, 15/14, 15/14,

15/14, 2401/2250, 15/14, 16/15, 15/14, 15/14, 16/15, 343/320, 15/14...

The limit secor is the limit of geometric averages of this sequence,

(or C(1) summation applied to the logs of the sequence, if you speak

that language.)

I'm uploading a graph of the secors approaching their limit for values

from 300 to 999, and another of the Borel summation method applied to

the first graph.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>

wrote:

> I'm uploading a graph of the secors approaching their limit for values

> from 300 to 999, and another of the Borel summation method applied to

> the first graph.

You'll find them in "photos", where there is, incidentally, lots of

free space.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>

wrote:

> I've thought of a new approach to the question of defining a

canonical

> generator for a given linear temperament.

cool!

> If T is a linear temperament, and r is a rational representative of

> the generator (eg, the TM reduced generator), and if TM is the

> Tenny-Minkowski reduction function for T, then we define the limit

> generator for T as

>

> g = lim n-->infinity (TM(r^n))^(1/n)

>

> in terms of cents, this would be

>

> g = lim n--> infinity cents(TM(r^n))/n

>

> The limit does exist, so this definition defines a unique

generator,

> without, alas, telling us a good way to compute it. The limit secor

in

> the 7-limit is about 116.785 cents, which we may compare to the rms

> value of 116.573 cents and the minimax value of 116.588 cents. Here

> are the ratios of successive terms for the TM reduced sequence for

> miracle:

>

> 15/14, 16/15, 343/320, 15/14, 16/15, 15/14, 16/15, 15/14, 16/15,

> 343/320, 15/14, 15/14, 16/15, 15/14, 16/15, 15/14, 343/320, 15/14,

> 16/15, 15/14, 16/15, 15/14, 343/320, 15/14, 16/15, 15/14, 16/15,

> 15/14, 343/320, 16/15, 15/14, 16/15, 15/14, 15/14, 2401/2250,

15/14,

> 15/14, 16/15, 15/14, 15/14, 16/15,343/320, 15/14, 16/15, 15/14,

15/14,

> 16/15, 343/320, 15/14, 16/15, 15/14, 16/15, 15/14, 15/14,

2401/2250,

> 15/14, 16/15, 15/14, 15/14, 15/14, 2401/2250, 15/14, 16/15, 15/14,

> 15/14, 15/14, 2401/2250, 15/14, 16/15, 15/14, 15/14, 16/15,

> 343/320, 15/14, 16/15, 15/14, 16/15, 15/14, 15/14, 2401/2250,

15/14,

> 16/15, 15/14, 15/14, 15/14, 2401/2250, 15/14, 16/15, 15/14, 15/14,

> 15/14, 2401/2250, 15/14, 16/15, 15/14, 15/14, 16/15, 343/320,

15/14...

>

> The limit secor is the limit of geometric averages of this

sequence,

> (or C(1) summation applied to the logs of the sequence, if you

speak

> that language.)

so it picks the mode of the distribution? or something?

--- In tuning-math@yahoogroups.com, "wallyesterpaulrus"

<wallyesterpaulrus@y...> wrote:

> so it picks the mode of the distribution? or something?

I'm trying to figure out what it does, but the mode has nothing to do

with it. It averages the generator along a line of generators, and I

see no reason to think that it is independent of generator-pair

choice, so I suppose its canonicity is open to question.

>I'm trying to figure out what it does, but the mode has nothing to do

>with it. It averages the generator along a line of generators, and I

>see no reason to think that it is independent of generator-pair

>choice, so I suppose its canonicity is open to question.

If a "generator-pair" is what I think it is, doesn't this method claim

to give the optimum size(s) *given* a generator-pair? Would we expect

two different gen-pair choices for a temperament to result in the same

tuning if carried out forever at their respective optimum sizes?

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> If a "generator-pair" is what I think it is, doesn't this method claim

> to give the optimum size(s) *given* a generator-pair? Would we expect

> two different gen-pair choices for a temperament to result in the same

> tuning if carried out forever at their respective optimum sizes?

The very question I've been pondering. The answer is no!

This method gives something much nicer than a canonical generator,

namely a canonical homomorphism. For T a regular temperament

(therefore now no longer necessarily linear) and q a rational number,

define TR(q) to be the Tenney-Minkowski reduction function for T, and

TH(q) = lim n--> infinity TR(q^n)^(1/n)

TH turns out to be a canonically defined homomorphism, which can be

thought of as the temperament mapping. To show this, note that if

a and b are rational numbers, then

TH(a)TH(b) = lim TR(a^n)^(1/n) lim TR(b^n)^(1/n) =

lim (TR(a^n)*TR(b^n))^(1/n)

Since the ratio between TH(a^n)TH(b^n) and TH((ab)^n) is bounded, when

we take the nth root they tend towards the same limit, and so TH is a

homomorphic mapping--that is to say TH(1) = 1, TH(a)TH(b) = TH(ab).

This is as slick as goose grease, and whether or not we want to use

this exact tuning is not to me the real question. What puts a smile on

my face is that in this way we can in a theoretical sense define the

temperament as being the mapping. The definition has nothing whatever

to do with octaves or octave equivalence, and in no way depends on the

definition or choice of consonances. I propose to call this the

canonical homomorphism.

>Since the ratio between TH(a^n)TH(b^n) and TH((ab)^n) is bounded, when

>we take the nth root they tend towards the same limit, and so TH is a

>homomorphic mapping--that is to say TH(1) = 1, TH(a)TH(b) = TH(ab).

>

>This is as slick as goose grease, and whether or not we want to use

>this exact tuning is not to me the real question. What puts a smile on

>my face is that in this way we can in a theoretical sense define the

>temperament as being the mapping. The definition has nothing whatever

>to do with octaves or octave equivalence, and in no way depends on the

>definition or choice of consonances. I propose to call this the

>canonical homomorphism.

Wow, sometimes you do get what you've always wanted. Can you show

how this works with an example?

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> Wow, sometimes you do get what you've always wanted. Can you show

> how this works with an example?

First I want to fix what I've written, which is defective, but not, I

think, fatally so.