Not using Taylor series or Maclaurin series, prove $\frac{1}{\sqrt{1-x^2}}=1+\frac{1\cdot3}{2\cdot4}x^2+\frac{1\cdot3\cdot5}{2\cdot4\cdot6}x^4+\cdots$

I’m trying to find a proof of
$$\frac{1}{\sqrt{1-x^2}} = 1+\frac{1\cdot3}{2\cdot4}x^2+\frac{1\cdot3\cdot5}{2\cdot4\cdot6}x^4+\cdots,$$ which doesn’t need Taylor or Maclaurin series, like the proof of Mercator series and Leibniz series.
I tried to prove it by using calculus, but I couldn’t hit upon a good proof.

• Are you familiar with the generalized binomial theorem? – Crescendo Nov 6 '17 at 17:32
• math.stackexchange.com/questions/746388/… – lab bhattacharjee Nov 6 '17 at 17:47
• I can use it only when coefficient is natural, because I need Taylor or Maclaurin series to get the rational one. – Gymnast Nov 6 '17 at 22:19
• So, labbhattacharjee, I’m sorry it must not at all be what I want to know about ... – Gymnast Nov 6 '17 at 22:21
• The general binomial theorem can be proved without the use of Taylor / Maclaurin series but the proof is not quite well known. See this blog post. – Paramanand Singh Nov 7 '17 at 3:55

$$\frac{(2n-1)!!}{(2n)!!}=\frac{1}{4^n}\binom{2n}{n}=\frac{1}{\pi}\int_{0}^{\pi}\cos^{2n}(\theta)\,d\theta$$ due to $\cos\theta=\frac{e^{i\theta}-e^{-i\theta}}{2}$, the binomial theorem (standard form) and $\int_{-\pi}^{\pi}e^{im\theta}\,d\theta = 2\pi\,\delta(m)$, for any $x$ such that $|x|<1$ we have
$$\sum_{n\geq 0}\frac{(2n-1)!!}{(2n)!!}x^{2n} = \frac{1}{\pi}\int_{0}^{\pi}\sum_{n\geq 0} x^{2n}\cos^{2n}(\theta)\,d\theta = \frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{1-x^2\cos^2\theta}$$ and by enforcing the substitution $\theta=\arctan t$ in the last integral we get $$\sum_{n\geq 0}\frac{(2n-1)!!}{(2n)!!}x^{2n} =\frac{2}{\pi}\int_{0}^{+\infty}\frac{dt}{(1-x^2)+t^2}=\frac{1}{\sqrt{1-x^2}}$$ as wanted.
• Essentially this is using $1/(1-x)=1+x+x^{2}+\dots$ (sum of an infinite GP). It is rather very smart to use this formula and integration to obtain the series for $(1-x^{2})^{-1/2}$. +1 – Paramanand Singh Nov 7 '17 at 3:58