MacLaurin series of $f(x) = \frac{1}{\sqrt{1 - x^2}}$ I'd like to show that the MacLaurin series of the function in question is $ \sum_{n=0}^{\infty}\binom{-1/2}{n} x^{2n}$, where $ \binom{-1/2}{n} $ is the generalized binomial coefficient.
I know and I can prove, that in general the MacLaurin series of $(1+x)^\alpha = \sum_{n=0}^{\infty}\binom{\alpha}{n} x^{n} $ (for $|x| < 1$), thus
$$ f(x) = \frac{1}{\sqrt{1 - x^2}} = (1-x)^{-1/2}(1+x)^{-1/2} = \left(\sum_{n=0}^{\infty}{(-1)}^n\binom{-1/2}{n} x^{n}\right)  \left(\sum_{n=0}^{\infty}\binom{-1/2}{n} x^{n} \right)$$
Or equivalently from the Cauchy product of two power-series:
$$ \sum_{n=0}^{\infty} x^{n} \sum_{k=0}^{n}{(-1)^k \binom{-1/2}{k} \binom{-1/2}{n-k}} $$
It's easy to see, that when $n$ is odd, then
$$ \sum_{k=0}^{n}{(-1)^n \binom{-1/2}{k} \binom{-1/2}{n-k}} =0 $$
because the terms cancel each other in the sum.
On the other hand I cannot see that $\sum_{k=0}^{n}{(-1)^n \binom{-1/2}{k} \binom{-1/2}{n-k}} = (-1)^n\binom{-1/2}{n/2}$ for $n$ even. I looked into Vandermonde's identity and its different forms, but I couldn't find anything that I could directly apply. I also wrote out the terms of the $\binom{-1/2}{k} \binom{-1/2}{n-k} $ product, but I couldn't find a way to simplify it. All I managed to do, is to check numerically that the identity holds for the first couple $n=2, 4, 6, 8, \dots$
Could anyone point me to the right direction?
 A: You already know that $$\frac{1}{\sqrt{1+y}}= \sum_{n=0}^\infty \binom{-1/2}{n} y^n$$ Now set $y=-x^2$ to get $$\frac{1}{\sqrt{1-x^2}}= \sum_{n=0}^\infty \binom{-1/2}{n}(-1)^n x^{2n}$$
A: Perhaps your question is why we have
$$(1+x)^{\alpha} \cdot (1-x)^{\alpha} = (1-x^2)^{\alpha}$$ with binomial series, that is
$$(\sum_{n\ge 0} \binom{\alpha}{n} x^n) \cdot (\sum_{n\ge 0 }(-1)^n\binom{\alpha}{n} x^n) = \sum_{n\ge 0} (-1)^n \binom{\alpha}{n} x^{2n}$$
So we need to show that for every $n\ge 0$ integer we have
\begin{eqnarray}\sum_{p+q = 2n}(-1)^q \binom{\alpha}{p}\binom{\alpha}{q} &=& (-1)^{n} \binom{\alpha}{n}\\ 
\sum_{p+q = 2n+1}(-1)^q \binom{\alpha}{p}\binom{\alpha}{q} &=& 0
\end{eqnarray}
For every $2n$ or $2n+1$ we get an indentity in $\alpha$ that is polynomial. Now, it is easy to check the identity for every $\alpha = N$ natural, since it follows from the equality $(1+x)^N \cdot (1-x)^N = (1-x)^{2N}$. We conclude that the equality for $\alpha$ is valid in general, so we have an identity.
It is an interesting question.
A: Letting $n=2m$ even we use generating functions in order to show the binomial identity
\begin{align*}
\color{blue}{\sum_{k=0}^{2m}(-1)^k\binom{-\frac{1}{2}}{k}\binom{-\frac{1}{2}}{2m-k}=\binom{-\frac{1}{2}}{m}\qquad\qquad m\geq 0}\tag{1}
\end{align*}
It is convenient to use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write for instance
\begin{align*}
[z^k](1+z)^{\alpha}=\binom{\alpha}{k}\tag{2}
\end{align*}

We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^{2m}}&\color{blue}{(-1)^k\binom{-\frac{1}{2}}{k}\binom{-\frac{1}{2}}{2m-k}}\\
&=\sum_{k=0}^\infty [z^k](1+z)^{-\frac{1}{2}}[u^{2m-k}](1-u)^{-\frac{1}{2}}\tag{3}\\
&=[u^{2m}](1-u)^{-\frac{1}{2}}\sum_{k=0}^\infty [z^k](1+z)^{-\frac{1}{2}}u^k\tag{4}\\
&=[u^{2m}](1-u)^{-\frac{1}{2}}(1+u)^{-\frac{1}{2}}\tag{5}\\
&=[u^{2m}]\left(1-u^2\right)^{-\frac{1}{2}}\\
&\,\,\color{blue}{=\binom{-\frac{1}{2}}{m}}\tag{6}
\end{align*}
and the claim (1) follows.

Comment:

*

*In (3) we apply the coefficient of operator twice. We also set the upper limit of the sum to $\infty$ which does not change the sum, since the coeffcient of $u^{2m-k}$ is zero if $k>2m$.


*In (4) we factor out terms independent from $k$. We also apply the rule $[u^{p-q}]A(u)=[u^p]u^qA(u)$.


*In (5) we use the substitution rule of the coefficient of operator.
\begin{align*}
A(u)=\sum_{k=0}^\infty a_k u^k=\sum_{k=0}^\infty [z^k]A(z)u^k
\end{align*}


*In (6) we select the coefficient of $u^{2m}$.
Note: Here we closely follow the proof of Theorem 2.1.1 in Integral Representation and the Computation of Combinatorial Sums by G. P. Egorychev.
