Let m and n be integers, and suppose we are seeking a MOS of m scale

steps within an n-et. We may first reduce to the case where m and n

are relatively prime, by setting d=gcd(m,n) and M=m/d, N=n/d. The

scale will then repeat an interval of repetition N d times.

Then we have the following theorem:

(Theorem EZ) Let 0<a<M, 0<b<N be integers such that

|a/b Â– M/N| < 1/(bM)

Then b/N is a generator for a MOS with M scale steps in an N-et.

Proof: If we multiply the above inequality on both sides by b/M, we

obtain

|a/M Â– b/N| < 1/M^2

which entails a/M is a semiconvergent to b/N, and hence we have a MOS.

Corollary: If a/b is the penultimate convergent to M/N (i.e., the

closest rational approximation with denominator less than N) then b/N

is the generator for a MOS.

Proof: The condition implies |a/b Â– M/N| = 1/(bN), and since N>M this

implies |a/b Â– M/N| < 1/(bM).

If b is small compared to M, the above becomes both sufficient *and*

necessary. If b<=M, then 1/(bM)<=1/b^2 and a/b must be a

semiconvergent to M/N. If b<=2M, then 1/(bM)<=1/(2b^2) and a/b is a

convergent to M/N.

Can you use the same argument to tighten up my "proof"?

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> Can you use the same argument to tighten up my "proof"?

It's quite possible EZ could help with that; I'll think about it.

--- In tuning-math@y..., genewardsmith@j... wrote:

> Let m and n be integers, and suppose we are seeking a MOS of m scale

> steps within an n-et.

I hope you realise that MOS exist which are not in any ET?

> We may first reduce to the case where m and n

> are relatively prime, by setting d=gcd(m,n) and M=m/d, N=n/d. The

> scale will then repeat an interval of repetition N d times.

>

> Then we have the following theorem:

Very nice. Now can you give us the _algorithm_ to find all (generator,

period) pairs (say in ET steps) for a MOS of m steps within an n-ET?

On second thoughts, much more useful would be an algorithm to give us

the generator and period (in cents) for a tempering of any

octave-equivalent periodicity block where all but one unison vector is

tempered out. Or even a proof that such is possible.

Regards,

-- Dave Keenan

--- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote:

> I hope you realise that MOS exist which are not in any ET?

Yes, but the theorem won't apply to them. :)

> Very nice. Now can you give us the _algorithm_ to find all

(generator,

> period) pairs (say in ET steps) for a MOS of m steps within an n-ET?

The Euclidean algorithm will always give one such pair by the

corollary to EZ. It is also easy to find semiconvergents for M/N and

test them. In general we only need to look at the range a <= M/2,

b <= N/2, which makes the above quite a bit more powerful.

Of course, this doesn't quite answer your question, but it's a start.