Series expansion of $\left(1 + x^2\right)^{-1/2}$

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...

• Thanks a lot! I got it. Mar 10, 2018 at 22:26
• I think that I have entered this exact sequence at least twice. Mar 10, 2018 at 22:28
• @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. Mar 10, 2018 at 22:29

Let $$u=u(x)=1+x^2$$. For $$k\ge0$$, making use of the Faa di Bruno formula and some properties of the partial Bell polynomials $$B_{n,k}$$, we obtain \begin{align*} \biggl[\frac{1}{(1+x^2)^{1/2}}\biggr]^{(k)}&=\sum_{j=0}^{k}\frac{\textrm{d}^j}{\textrm{d} u^j}\biggl(\frac{1}{u^{1/2}}\biggr) B_{k,j}(2x,2,0,\dotsc,0)\\ &=\sum_{j=0}^{k}\frac{\langle-1/2\rangle_j}{u^{1/2+j}} 2^jB_{k,j}(x,1,0,\dotsc,0)\\ &=\sum_{j=0}^{k}\frac{\langle-1/2\rangle_j}{(1+x^2)^{1/2+j}} 2^j \frac{1}{2^{k-j}}\frac{k!}{j!}\binom{j}{k-j}x^{2j-k}\\ &=\frac{k!}{2^k(1+x^2)^{1/2}} \sum_{j=0}^{k}\biggl\langle-\frac12\biggr\rangle_j\frac{2^{2j}}{j!}\binom{j}{k-j}\frac{x^{2j-k}}{(1+x^2)^{j}}\\ &\to\begin{cases}\displaystyle \frac{(2m)!}{2^{2m}}\biggl\langle-\frac12\biggr\rangle_m\frac{2^{2m}}{m!}\binom{m}{2m-m}, & \text{k=2m is even}\\ 0,& \text{k is odd} \end{cases}\\ &=\begin{cases}\displaystyle (-1)^m\frac{(2m)!(2m-1)!!}{(2m)!!}, & \text{k=2m is even}\\ 0,& \text{k is odd} \end{cases} \end{align*} as $$x\to0$$. Consequently, we acquire $$$$\frac{1}{(1+x^2)^{1/2}}=\sum_{m=0}^\infty(-1)^m\frac{(2m-1)!!}{(2m)!!}x^{2m}$$$$ and $$$$\operatorname{arcsinh}x=\sum_{m=0}^\infty(-1)^m\frac{(2m-1)!!}{(2m)!!}\frac{x^{2m+1}}{2m+1}.$$$$

References

1. Siqintuya Jin, Bai-Ni Guo, and Feng Qi, Partial Bell polynomials, falling and rising factorials, Stirling numbers, and combinatorial identities, Computer Modeling in Engineering & Sciences 132 (2022), no. 3, 781--799; available online at https://doi.org/10.32604/cmes.2022.019941.
2. F. Qi and B.-N. Guo, Explicit formulas for special values of the Bell polynomials of the second kind and for the Euler numbers and polynomials, Mediterr. J. Math. 14 (2017), no. 3, Article 140, 14 pages; available online at https://doi.org/10.1007/s00009-017-0939-1.
3. F. Qi, D.-W. Niu, D. Lim, and Y.-H. Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, J. Math. Anal. Appl. 491 (2020), no. 2, Article 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.