Find cofficient for $x^n$ for a Generating function

Question

If given a GF $F(x) = \frac{1}{(1-rx)^2}$, how do I find the coefficient for the term $x^n$?

I can tell $F(x) = A(x)^2$, where $A(x)$ is the GF for the sequence $1, r, r^2, r^3, r^4, \ldots$ but I don't know how to find the coefficient at $x^n$.

Thanks.

Answer 1

You need to make the Cauchy product of $A$ with itself :

$$F=A^2=\sum_{n\geq 0}r^nx^n\sum_{n\geq 0}r^nx^n=\sum_{n,m\geq 0}r^{n+m}x^{n+m}=\sum_{N\geq 0}\sum_{n=0}^Nr^Nx^N $$

In the last equality I made the change of variable $(n,m)\mapsto (n,n+m)$ the inverse of this change of variable is given by $(n,N)\mapsto (n,N-n)$ (This is exactly the Cauchy-product of two series here).

Now :

$$F=A^2=\sum_{N\geq 0}\sum_{n=0}^Nr^Nx^N=\sum_{N\geq 0}\left(\sum_{n=0}^N1\right)r^Nx^N=\sum_{N\geq 0}(N+1)r^Nx^N$$

Answer 2

If you use the generalized binomial theorem

$$(a+b)^\alpha=\sum_{k\ge 0}\binom{\alpha}{k}a^{\alpha -k} b^\alpha,\ |a|>|b|$$

then we have

$$F(x)=(1+(-rx))^{-2}=\sum_{k\ge0}\binom{-2}{k}(-rx)^k=\sum_{k\ge0}\binom{k+1}{k}r^kx^k=\sum_{k\ge0}(k+1)r^kx^k$$

so $$[x^k]F(x)=(k+1)r^k$$

P.S.: you dont need take care of convergence because this is just a formal power series.