It appears that a denominator may be a polynomial whose roots are the reciprocal-products 1/(r(i)*s(j)), where r(i) ranges through all the roots of q(x) and s(j) ranges through all the roots of v(x). Clark Kimberling
--This polynomial A(x) is expressible in terms of the original polynomials q(x) and v(x) by the use of "resultant" -- you can evaluate using determinants without computing any roots, just do rational operations. Specifically, the polynomial B(x) whose roots are the products r(i)*s(j) should be B(x) = Resultant_y [q(y), y^degree(v) * v(x/y)] and then A(x) is the "reverse" of B(x) to reciprocate its roots, A(x) = B(1/x) * x^degree(B). The "numerator" polynomial Kimberling also wanted could then be obtained by simply solving for it by given the known values of the first degree # of outputs, i.e. series coefficients, of the generating function. This is just a polynomial multiplication. -- Warren D. Smith http://RangeVoting.org