I have the following function: $$f(x) = arcsinh(x)$$ which i want to express as a series of powers. To do that I thought: $$\frac{df}{dx} = \frac{1}{\sqrt{1 - x^2}}$$ so $$f = \int\frac{1}{\sqrt{1 - x^2}}dx= \int\left(1 - x^2\right)^{-1/2}dx$$ and now i will expand that using the binomial expansion $(1 + x)^a$. I tried several ways to make that expansion but I couls not get a compact expression. After some search I saw that: $$\left(1 - x\right)^{-1/2} = \sum_{k=0}^{\infty}\frac{(-1)^k(2k)!}{2^{2k}(k!)^2}x^k$$ But I dont understand how this expression was obtained.Any other ideas for expressing my function as a series of powers or a demonsstration as to how this expression is obtined?


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

\begin{align} \binom{-1/ 2}{n} &= \frac{\prod_{k=0}^{n-1}(-\tfrac12-k)}{n!} = \frac{\prod_{k=0}^{n-1}\big[\tfrac12(-1)(2k+1)\big]}{n!}\\ &= \frac{(-1)^{n}}{2^nn!} \prod_{k=0}^{n-1}(2k+1)\\ &= \frac{(-1)^{n}}{2^nn!} \frac{\prod_{k=0}^{n-1}\big[(2k+1)(2k+2)\big]}{\prod_{k=0}^{n-1}\big[2(k+1)\big]}\\ &= \frac{(-1)^{n}}{2^nn!} \frac{\prod_{k=0}^{2n-1}(k+1)}{2^n\prod_{k=0}^{n-1}(k+1)} = \frac{(-1)^{n}}{2^nn!} \frac{(2n)!}{2^n n!}\\ &= \frac{(-1)^n (2n)!}{2^{2n} (n!)^2} % \end{align}

then (fixing your sign mistake), where the power series about $0$ for $\arcsin(x)$ is defined,

$$\arcsin(x) = \int \frac{dx}{(1-x^2)^{1/ 2}} = \int\sum_{n=0}^{\infty} \binom{-1/ 2}{n} (-x^2)^{n} = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} \binom{-1/ 2}{n}x^{2n+1}$$ and you can do the rest...

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  • $\begingroup$ Thanks a lot! I got it. $\endgroup$ – Desperados Mar 10 '18 at 22:26
  • $\begingroup$ I think that I have entered this exact sequence at least twice. $\endgroup$ – marty cohen Mar 10 '18 at 22:28
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    $\begingroup$ @martycohen As have I, but always as a lemma to an answer. For the life of me I've never managed to actually memorize it and always end up rederiving it. $\endgroup$ – adfriedman Mar 10 '18 at 22:29

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