RE: more questions about meantone prime-factoring

Joe Monzo wrote,

>Here is a list of the prime-factor formulae for the ratio
>of the '5th' in several of the most popular meantones:

>1/3-comma MT '5th' = 2^(x/3)*3^(-x/3)*5^(x/3)

>2/7-comma MT '5th' = 2^(x/7)*3^(-x/7)*5^((x*2)/7)

>5/18-comma MT '5th' = 2^(x/9)*3^(-x/9)*5^((x*5)/18)

>7/26-comma MT '5th' = 2^(x/13)*3^(-x/13)*5^((x*7)/26)

>1/4-comma MT '5th' = 5^(x/4)

>2/9-comma MT '5th' = 2^(-x/9)*3^(x/9)*5^((x*2)/9)

>3/14-comma MT '5th' = 2^(x/7)*3^(-x/7)*5^((x*3)/14)

>1/5-comma MT '5th' = 2^(-x/5)*3^(x/5)*5^(x/5)

>1/6-comma MT '5th' = 2^(-x/3)*3^(x/3)*5^(x/6)

>[...]

>I've been trying to find the single overall formula that
>would calculate all of these as variables, but I'm puzzled
>by the way that the sign of the exponents of 2 and 3 are
>switched in some of them, and by the way the denominators
>are doubled or halved in some of them. Any explanations
>and a solution would be greatly appreciated.

First of all, I believe you meant to specify "x=1" for all of the above
formulae, correct? Now, I'm under the impression that you yourself derived
all or most the formulae above, and they're correct. So what you appear to
be asking is, is there a pattern to the exponents of 2, 3, and 5 as a
function of the fraction of a comma by which the comma is tempered? That's
an interesting question, though the reason for asking it may be based on a
misconception. The expressions you wrote above appear to be attempts at some
sort of prime-factorization, but they aren't. The prime-factorization
theorem only applies when all the exponents are integers, and, of course,
the number being factored is an integer as well. For factoring non-integers,
you may be interested in various analogues to the prime-factorization
theorem, where certain irrationals play the role of prime numbers, depending
on the "field" of irrationals you're seeking to factor. But even these
factorizations are only unique when the exponenets are restricted to be
integers.

Anyway, let's get you your general formula. The fifth itself is

2^(-1)*3^(1).

We are dividing this by a fraction (call it f) of a syntonic comma, so we
are multiplying by

(80/81)^f

or

2^(4*f)*3^(-4*f)*5^f

So the answer is

2^(4*f-1)*3^(1-4*f)*5^f

There ya go, Joe!

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