A wedgie titbit

Suppose we have an et val. We may define the error vector, as a point
in val space, as the difference v/v[1] - JIP, so that the coefficients
are the errors (in octave units) of the primes in the pure-octave
tuning (hence, the first coefficient is zero.)

Suppose now we have two et vals v1 and v2, we may call the respective
error vectors e1 and e2. Now define

j1 = e1 ^ JIP, j2 = e2 ^ JIP.

Then v1[1] * v2[1] * (j1 - j2) is approximately v1^v2, and is exactly
v1^v2 for the OE part; moreover the OE part is v1[1]*v2[1]*(e2-e1).

This means the OE part of the wedgie, assuming that v1 and v2 don't
have torsion, is simply found by the differences of the errors of the
respective vals. Consequently, high errors in the vals for a given
prime tend to translate to high complexity for that prime for the
temperament obtained by wedging them, and this is especially true if
the errors have opposite signs.

Consider the denominators of the convergents for log2(3/2), which go
2, 5, 12, 41, 53, 306, 665, ... By taking the corresponding pairs of
vals, we obtain temperaments with complexity 1 for 3/2:

2&5: 16/15 <0 1 -1|
5&12: 81/80 <0 1 4|
12&41: 32805/32768 <0 1 -8|
41&53: 32805/32768
53&306: |69 -45 1> <0 1 45|
306&665: |-554 351 -1> <0 1 351|

Each of the commas has 5 to only the power +-1, and correspondingly
has a complexity for 3 of 1.

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

> Then v1[1] * v2[1] * (j1 - j2) is approximately v1^v2, and is exactly
> v1^v2 for the OE part; moreover the OE part is v1[1]*v2[1]*(e2-e1).

I should probably explain why this is. We have

e1 = v1/v1[1] - JIP
e2 = v2/v2[1] - JIP

Hence v1^v2 = v1[1]*(JIP+e1) ^ v2[1] * (JIP+e2) =
v1[1]*v1[2] * (JIP+e1)^(JIP+e2) =
v1[1]*v2[1]*(e1^JIP - e2^JIP + e1^e2)

Since e1 and e2 are small, with zeros for the 2 coefficient, e1^e2
will be small, with zeros for the OE coefficients. Hence

v1[1]*v2[1]*(e1-e2)^JIP

will differ from v1^v2 by a small amount, with the OE part of it the
same as the OE part of v1^v2. Moreover, since the 2 coefficient of JIP
is 1 and the 2 coefficient of e1-e2 is zero, the OE part of
(e1-e2)^JIP is just what you get from e2-e1.

If we define the prime complexity P of a R2 temperament as the maximum
absolute value of the OE part of the wedige, then the prime complexity
is less than or equal to the Graham complexity. If we consider only et
vals with errors less than or equal to e, and up to size n, then we
have 2 n^2 e <= P for wedgies obtained by taking these vals in pairs.
If we measure e in cents, that becomes 2 n^2 e <= 1200 P. These
formulas remain true if we use Graham complexity instead.