If $h(x)$ is the generating function for $a_r$, what is the generating function of $$\frac{h(x)}{(1-x)^2}$$
Let $h(x)$ be written as
$$h(x) = \sum_{r} a_r x^r $$
Consider more simply
$$\frac{h(x)}{1-x} = \frac{1}{1-x} h(x) =\sum_{r} x^r \sum_{r} a_r x^r$$
I tried to expand this and see what I could get
$$(1+x+x^2+x^3+\dots+x^r+\dots)(a_0+a_1x+a_2x^2+a_3x^3+\dots+a_rx^r+\dots)$$
there are two ways to simplify the product, either
$$a_0(1+x+x^2+\dots)+a_1(x+x^2+x^3+\dots)+ a_2(x^2+x^3+x^4+\dots)+\dots= \sum_ra_r\sum_{k\ge r}x^k$$
or
$$a_0 + (a_0+a_1)x+(a_0+a_1+a_2)x^2+(a_0+a_1+a_2+a_3)x^3 = \sum_r \left(\sum_{k\le r}a_k \right)x^r$$
obviously this is only for one factor of $\frac{1}{1-x}$ but I assume If I can get help for this I can extend it to two factors.
I'm not sure what form the answer is expected to be in? Because I could say the generating function is
$$\frac{h(x)}{1-x} = h\left(\sum_{k \ge r}x^k\right)$$
but I'm not sure that makes any sense. I was expecting to say something like
$$\frac{h(x)}{1-x} \mapsto h(x^2)$$
(the $x^2$ is not intentional, just some idea of what I believe the answer could look like)
Any help for the case of $\frac{1}{1-x}$ would be great and then I could extend it to $\frac{1}{(1-x)^2}$
Best Answer
First, note that
$$\frac{d}{dx} \frac{1}{1-x}=\frac{1}{(1-x)^2}.$$
Thus, taking the derivative of the power series we have:
$$\frac{1}{(1-x)^2}=\sum_{k=0}^\infty (k+1)x^k.$$
Let
$$h(x)=\sum_{n=0}^\infty a_nx^n.$$
Then taking the convolution of these two series, we have:
$$\frac{h(x)}{(1-x)^2}=\sum_{n=0}^\infty \sum_{k=0}^n (k+1)a_{(n-k)}x^n.$$