MacLaurin series of $f(x) = \frac{1}{\sqrt{1 - x^2}}$

Question

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?

Answer 1

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 $$\frac{1}{\sqrt{1-x^2}}= \sum_{n=0}^\infty \binom{-1/2}{n}(-1)^n x^{2n}$$

Answer 2

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.

Answer 3

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.

Answer 4

At the site https://math.stackexchange.com/a/4657751, it was established that \begin{equation*} \biggl[\frac{1}{\sqrt{1-x^2}\,}\biggr]^{(k)} =\frac1{\sqrt{1-x^2}\,}\frac{k!}{2^k} \frac1{x^k}\sum_{j=0}^{k}\frac{2^{j}(2j-1)!!}{j!}\binom{j}{k-j}\frac{x^{2j}}{(1-x^2)^{j}}, \quad k\ge0. \end{equation*} Therefore, we derive \begin{align*} \lim_{x\to0}\biggl[\frac{1}{\sqrt{1-x^2}\,}\biggr]^{(k)} &=\frac{k!}{2^k} \lim_{x\to0}\sum_{j=0}^{k}\frac{2^{j}(2j-1)!!}{j!}\binom{j}{k-j}\frac{x^{2j-k}}{(1-x^2)^{j}}\\ &=\begin{cases} 0, & k=2m+1\\ (2m)! \dfrac{(2m-1)!!}{(2m)!!}, & k=2m \end{cases} \end{align*} for $k,m\ge0$. Consequently, we acquire \begin{equation*} \frac{1}{\sqrt{1-x^2}\,} =\sum_{m=0}^\infty \dfrac{(2m-1)!!}{(2m)!!}x^{2m}, \quad |x|<1. \end{equation*} See also the web site https://math.stackexchange.com/a/4659020.