I've been playing with q-derivatives over a finite field. It turns out that not all values of q give a derivative that's linear. Over GF(p^n), setting q=x^(p-1) works: Let x^a + x^b = x^c. Then we want D(x^a)+D(x^b) = D(x^a+x^b) = D(x^c). (qx)^a - x^a (qx)^b - x^b (qx)^a + (qx)^b - (x^a + x^b) ------------ + ------------ = ----------------------------- qx - x qx - x qx - x If q = x^(p-1) then we have (x^pa + x^pb) - (x^a + x^b) (x^a + x^b)^p - (x^a + x^b) = --------------------------- = --------------------------- (*) qx - x qx - x x^pc - x^c (qx)^c - x^c = ---------- = ------------ qx - x qx - x At (*) we can only do that because all the inner coefficients on the pth line of Pascal's triangle are divisible by p and therefore congruent to zero. Other values of q work, too, but I haven't looked at them enough to see if there's an intuitive reason for them yet. Once we have a working derivative, we can start doing differential equations. For example, what's the solution to f'(x) = f(x)? Normally it's f(x) = exp(x), but over GF(2^3) mod x^3-x-1 it's f(x) = x^5: Setting q=x, the q-derivative is (xx)^5 - x^5 x^10 - x^5 x^3 - x^5 011 - 111 ------------ = ---------- = --------- = --------- (xx) - x x^2 - x x^2 - x 110 100 x^2 = --- = --- = x^(-2) = x^5 110 x^4 With more variables in our polynomials we can start looking at physics over a finite field: we can solve the wave equation or Schroedinger's equation (pick units so hbar = c = m = 1): instead of complex phases, we get phases mod p. Do the squares still sum to 1? Can we do quantum computation this way? -- Mike Stay staym@clear.net.nz