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.

  • $\begingroup$ Are you familiar with the generalized binomial theorem? $\endgroup$ – Crescendo Nov 6 '17 at 17:32
  • $\begingroup$ math.stackexchange.com/questions/746388/… $\endgroup$ – lab bhattacharjee Nov 6 '17 at 17:47
  • $\begingroup$ I can use it only when coefficient is natural, because I need Taylor or Maclaurin series to get the rational one. $\endgroup$ – Gymnast Nov 6 '17 at 22:19
  • $\begingroup$ So, labbhattacharjee, I’m sorry it must not at all be what I want to know about ... $\endgroup$ – Gymnast Nov 6 '17 at 22:21
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ 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 $\endgroup$ – Paramanand Singh Nov 7 '17 at 3:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.