How to prove this beautiful series by using Taylor and Maclaurin series

Question

I have been playing with Taylor and Maclaurin series lately and stumble on this beautiful identity. I don't know to expand the left hand side to yield the right hand side: How to prove: $\dfrac{1}{\sqrt{1-x^2}} =1+\dfrac{1}{2}x^2+\dfrac{1 \cdot3}{2\cdot4}x^4+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}x^6+\dfrac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}x^8...$ I can only expand this as followed: $$\frac{1}{\sqrt{1-x^2}} = 1+\frac{1}{2}x^2+\frac{3}{8}x^4+\frac{5}{16}x^6...,$$

How can you prove this by using Maclaurin series? I need two proofs, one in Maclaurin series and one in binomial theorem. Please don't use the sigma notation too much as I cannot see the pattern.

Answer 1

Very simple: expand $$\frac1{\sqrt{1-u}}=(1-u)^{-\tfrac12}$$ with the binomial series, and substitute $u=x^2$.

Note the general term of the binomial series is $$(-1)^n\frac12\cdot\frac32\dotsm\frac{2n-1}2\,\frac{(-u)^n}{n!}= \frac{1\cdot 3\dotsm(2n-1)}{2^n n!}\,u^n=\frac{(2n-1)!!}{(2n)!!}\,u^{n}.$$ From the final formula, you can deduce instantly the Taylor series for $\arcsin x$.

Answer 2

I have finally been able to derive the series after some manipulations:

The general formula for binomial series is: $(1+x)^k=1+kx+\dfrac{k(k-1)}{2!}x^2+\dfrac{k(k-1)(k-2)}{3!}x^3+\dfrac{k(k-1)(k-2)(k-3)}{4!}x^4...$

$ \dfrac{1}{\sqrt{1-x^2}}=(1-x^2)^{-\frac{1}{2}}=1+(-\dfrac{1}{2})(-x^2)+\dfrac{(-\dfrac{1}{2})(-\dfrac{1}{2}-1)}{2!}(-x^2)^2+\dfrac{(-\dfrac{1}{2})(-\dfrac{1}{2}-1)(-\dfrac{1}{2}-2)}{3!}(-x^2)^3+\dfrac{(-\dfrac{1}{2})(-\dfrac{1}{2}-1)(-\dfrac{1}{2}-2)(-\dfrac{1}{2}-3)}{4!}(-x^2)^4...$

$=1+\dfrac{1}{2}x^2+\dfrac{(-\dfrac{1}{2})(-\dfrac{3}{2})}{2!}(x^4)+\dfrac{(-\dfrac{1}{2})(-\dfrac{3}{2})(-\dfrac{5}{2})}{3!}(-x^6)+\dfrac{(-\dfrac{1}{2})(-\dfrac{3}{2})(-\dfrac{5}{2})(-\dfrac{7}{2})}{4!}(x^8)...$

$=1+\dfrac{1}{2}x^2+\dfrac{1 \cdot3}{2^2\cdot2!}x^4+\dfrac{1\cdot 3\cdot 5}{2^3\cdot 3!}x^6+\dfrac{1\cdot 3\cdot 5\cdot 7}{2^4\cdot 4!}x^8...$

$=1+\dfrac{1}{2}x^2+\dfrac{1 \cdot3}{2\cdot4}x^4+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}x^6+\dfrac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}x^8...$

$$=\sum_{n=0}^{\infty} \dfrac{(2n-1!!)}{2^nn!}x^{2n}$$

Answer 3

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*}