--- In

tuning-math@yahoogroups.com, Graham Breed <gbreed@g...> wrote:

>

> I'm making progress on proving that all pairs of a set of equal

> temperaments can find all R2 temperaments with certain properties.

Here's another thought.

Suppose we start with an R2 wedgie, and divide through each

coefficient by the corresponding product of logs base 2,

1/log2(q)*log2(r) for the q-r coefficient. Then we get a weighted

wedgie W = <<w1 w2 .., wn||, and C = max(|wi|) is a complexity measure.

If v is a val belonging to the temperament, let w be the corresponding

weighted val, dividing through by 1/log2(q) for the prime q term. Then

the wedge product W^w = zero trival. Therefore, if u = w/N, where the

val is an N-et val, we also have W^u = zero trival.

If j = <1 1 1 ... 1| is the JI val, we have that the maximum of the

absolute values of the coefficients of W^j, call it E, is an error

measurement for W. We also have that the maximum of the absolute

values of j-u, call it e, is an error measurement for u, and hence for

the original val w.

Now

E = ||W^j|| = ||W^(j-u)|| <= 3 ||W|| ||j-u|| = 3Ce

Here here "3" appears because there are three added products in the

computation of the trival.