Taylor series expansion of $\frac{1}{\sqrt{1-\beta x(x+1)}}$

Question

I am trying to find the taylor series expansion about $0$ (maclaurin series) of $$x \rightarrow \frac{1}{\sqrt{1-\beta x(x+1)}} \text{ with } \beta \in \mathbb{R}^{+*}$$

I've tried using the taylor series expansion of $$\frac{1}{\sqrt{1-X}} = \sum_{n=0}^{\infty}4^{-n}{2n \choose n}X^n \text{ }\text{ }\text{ }\text{ with } \text{ } X=\beta x(x+1)$$

But I can't turn it into a power series because of the $(x+1)^n$...

I've also tried to derive $$\frac{1}{n!}\cdot\frac{\text{d}^n}{\text{d}x^n}\left(\frac{1}{\sqrt{1-\beta x(x+1)}}\right)_{x=0}$$ But no results so far...

Edit : with a more powerful method, I found that, if we call $\left(a_n\right)_{n\in\mathbb{N}}$ the coefficients of the taylor series expansion $\left(\frac{1}{\sqrt{1-\beta x(x+1)}}=\sum_{n=0}^{\infty}a_nx^n\right)$, the sequence $\left(a_n\right)_{n\in\mathbb{N}}$ is then defined by : $$\forall n\geq3, \text{} na_n=\beta\left(n-\frac{1}{2}\right)a_{n-1}+\beta\left(n-1\right)a_{n-2}$$ $$\text{with }\text{ }a_1 = \frac{\beta}{2} \text{ , } a_2 = \frac{3}{8}\beta^2+\frac{1}{2}\beta$$

It's definitely a step forward, but I don't know how to proceed from there. Is there a way to handle sequences that are defined by such a way ?

Answer 1

I would set $y = \beta x(x+1)$. With that: \begin{align} \frac1{\sqrt{1 - y}} &= \sum_{n=0}^{\infty} 4^{-n} \binom{2n}{n} y^n\\ &= \sum_{n=0}^{\infty} 4^{-n} \binom{2n}{n} \beta^n x^n (x+1)^n\\ &= \sum_{n=0}^{\infty}\sum_{k=0}^{n} 4^{-n} \binom{2n}{n} \binom{n}{k} \beta^n x^n x^k \\ &= \sum_{m=0}^{\infty} a_m x^m \end{align} where $m = n + k$, so $a_m = \sum_{n=0}^{m} 4^{-n}\binom{2n}{n}\binom{n}{m-n} \beta^n$.

$a_m$ is simply an number which you can calculate doing some sums and products. For $m = 0, \dots, 3$ the $a_m$ are $1,\frac{\beta }{2},\frac{1}{8} \beta (3 \beta +4),\frac{1}{16} \beta ^2 (5 \beta +12)$.

Mathematica "simplifies" $a_m$ to $$ a_m = 4^{-m} \binom{2 m}{m} \beta ^m \, _2F_1\left(\frac{1}{2}-\frac{m}{2},-\frac{m}{2};\frac{1}{2}-m;-\frac{4}{\beta }\right) $$ where $_2F_1$ denotes the Hypergeometric function.

Answer 2

By the Faa di Bruno formula and some properties of the partial Bell polynomials, we have \begin{align} \frac{\operatorname{d}^n}{\operatorname{d}x^n}\biggl[\frac1{\sqrt{1-\beta x(x+1)}\,}\biggr] &=\sum_{k=0}^n\biggl\langle-\frac12\biggr\rangle_k\frac1{[1-\beta x(x+1)]^{1/2+k}} B_{n,k}(-\beta(2x+1), -2\beta,0,\dotsc,0)\\ &=\sum_{k=0}^n\biggl\langle-\frac12\biggr\rangle_k\frac1{[1-\beta x(x+1)]^{1/2+k}} (-2\beta)^kB_{n,k}\biggl(x+\frac12, 1,0,\dotsc,0\biggr)\\ &\to\sum_{k=0}^n\biggl\langle-\frac12\biggr\rangle_k (-2\beta)^kB_{n,k}\biggl(\frac12, 1,0,\dotsc,0\biggr),\quad x\to0\\ &=\sum_{k=0}^n(-1)^k\frac{(2k-1)!!}{2^k} (-2\beta)^k \frac{1}{2^{n-k}}\frac{n!}{k!}\binom{k}{n-k}\frac1{2^{2k-n}}\\ &=n!\sum_{k=0}^n\frac{(2k-1)!!}{(2k)!!} \binom{k}{n-k} \beta^k, \end{align} where we used \begin{equation}\label{Bell-x-1-0-eq} B_{n,k}(x,1,0,\dotsc,0) =\frac{1}{2^{n-k}}\frac{n!}{k!}\binom{k}{n-k}x^{2k-n}. \end{equation} Consequently, we conclude \begin{equation} \frac1{\sqrt{1-\beta x(x+1)}\,} =\sum_{n=0}^\infty\Biggl[\sum_{k=0}^n\frac{(2k-1)!!}{(2k)!!} \binom{k}{n-k} \beta^k\Biggr]x^n. \end{equation}

References

1. 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, Art. 140, 14 pages; available online at https://doi.org/10.1007/s00009-017-0939-1.
2. F. Qi and M.-M. Zheng, Explicit expressions for a family of the Bell polynomials and applications, Appl. Math. Comput. 258 (2015), 597--607; available online at https://doi.org/10.1016/j.amc.2015.02.027.
3. Feng Qi, Da-Wei Niu, Dongkyu Lim, and Yong-Hong Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, Journal of Mathematical Analysis and Applications 491 (2020), no. 2, Paper No. 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.