I wrote:
I'd like to have an algebraic rep for S like the formula for the q-derivative, but I have no idea how to go about it.
Sorry--this question has an obvious answer if q is a constant (S x^n = x^(n+1)/[n]), but in this case q is a function. Better stated, the operator I'm trying to find a one-sided inverse for is f(x^(p^j)) - f(x) Df = -----------------, x^(p^j) - x i.e., the deformation is a Frobenius map. Then x^(np^j-1) - 1 Dx^n = -------------- x^(n-1) = [n] x^(n-1). x^(p^j-1) - 1 Thanks, Mike ----- Original Message Follows -----
I've been playing with the q-derivative operator Df = (f(qx)-f(x))/(qx-x) over a finite field. It's obviously singular, since adding a constant to f makes no difference to the derivative.
Given a matrix rep for D, I can use Gauss-Jordan elimination to find a matrix rep for an integral operator S such that DSf = f iff there exists some F such that DF = f. I'd like to have an algebraic rep for S like the formula for the q-derivative, but I have no idea how to go about it.
Can anyone help? Thanks,
-- Mike Stay staym@clear.net.nz http://www.cs.auckland.ac.nz/~msta039
-- Mike Stay staym@clear.net.nz http://www.cs.auckland.ac.nz/~msta039