Simplify $\sum_{s=0: s \text{ even }}^\infty \sum_{m=0: \text{ even }}^\infty b_{s,m}x^s(1-x^2)^{\frac{m}{2}}$

Question

I have the following double sum that I am trying to simplify into a single sum: \begin{align} \sum_{s=0}^\infty \sum_{m=0}^\infty b_{2s,2m}x^{2s}(1-x^2)^{m} \end{align} where $b_{s,m}$' are coefficients that don't have any specific structure.

But I am having difficulty finding the coefficients of the new sum $\sum_{k} a_k x^k$.

Here is what I tried to do. First, it is clear that only even coefficients should be considered.

Second, we can use the Binomial theorem to write \begin{align} (1-x^2)^{m}= \sum_{i=0}^{m} { m \choose i} (-1)^i x^{2i} \end{align} so we can combine everything to \begin{align} \sum_{s=0}^\infty \sum_{m=0}^\infty b_{2s,2m}x^{2s} \sum_{i=0}^{m} { m \choose i} (-1)^i x^{2i} \end{align}

But at this point, I start getting lost.

Second approach I was thinking we can define \begin{align} f(x)=\sum_{s=0}^\infty \sum_{m=0}^\infty b_{2s,2m}x^{2s}(1-x^2)^{m} \end{align} and then find the power series of $f(x)$ around $x=0$. The first coefficients are given by \begin{align} a_0&=\frac{f(0)}{1}=b_{0,0}\\ a_1&=\frac{f'(0)}{1}=0\\ a_2&=\frac{f''(0)}{2}=- \sum_{m=0}^\infty m b_{0,m}+ \sum_{m=0}^\infty b_{2,m} \end{align} but again I am stuck with finding other coefficients.

Answer 1

The sum is $\sum_{s=0}^\infty \sum_{m=0}^\infty b_{2s,2m}x^{2s} \sum_{i=0}^{m} { m \choose i} (-1)^i x^{2i}$

We find the coefficient of $x^{2k}$:

We may have $0 \leq s \leq k$, for a fixed $s$ we require a $x^{2(k-s)}$ term from the $(1-x^2)^m$ factor, in particular we require that $m \geq k-s$. The coefficient of the $x^{2(k-s)}$ term in $(1-x^2)^m$ is ${m \choose (k-s)} (-1)^{k-s}$, so the coefficient of $x^{2k}$ is $\sum_{s=0}^k \sum_{m \geq (k-s)} b_{2s,2m} {m \choose (k-s)} (-1)^{k-s} = c_k$, the final sum is $\sum_{k=0}^{\infty} c_kx^{2k}$