[seqfan] Re: generating-function-ology

[Ed Jeffery]

Peter,

If I'm not mistaken again:

If A(x) = f(x) / g(x) is a rational function, then f(x) and g(x) are both
polynomials, so A(x) * g(x) / f(x) = 1, and your inverse is just the
expansion of b(x) = g(x) / f(x) to series. For example, for the inverse of
the Fibonacci series, you just get the inverse of the generating function
(up to an offset) for the Fibonacci sequence:

(1)     1 / (1 + x + 2*x*2 +3*x^3 + 5*x^4 + ...) = 1 - x - x^2.

Multiplying (1) through by 1 + x + 2*x^2 + ..., you can see (although this
is no proof) that all terms in x^k, for k > 0, on the right vanish except
for the coefficient of x^0 which is equal to 1.

Finally, if A(x) is a series and not a polynomial, then its inverse could
be another series or it could be a rational function.

LEJ

> if A(x) is the generating function for the sequence a(i), is there an
> equation for the generating function of b(i) = 1/a(i) ?

> thanks,
> Peter Lawrence.