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 ?


2 Answers 2


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.

  • $\begingroup$ Well actually, after numerical simulation, your numerical values don't seem to work... $\endgroup$ Dec 14, 2018 at 0:17
  • $\begingroup$ @HarmonicSun Yes, I had forgotten to include $\beta$. The values are not as pretty now but you can still easily calculate them. $\endgroup$ Dec 14, 2018 at 0:20
  • $\begingroup$ Ok thanks a lot ! Additionaly, do you think that based on my edit on my initial question, one can derive a prettier closed form for the $a_m$ ? $\endgroup$ Dec 14, 2018 at 0:26
  • $\begingroup$ I might be picky but, in your formula for $a_m$, ${n \choose m-n}$ is not defined for, say, n=0... Or do we take it a 0 conventionaly ? $\endgroup$ Dec 14, 2018 at 0:33
  • $\begingroup$ @HarmonicSun Yes, the Binomial coefficient $\binom{a}{b}$ is understood to be zero when $a < b$ (and $a, b \in \mathbb{Z}$). $\endgroup$ Dec 14, 2018 at 0:39

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}


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  • $\begingroup$ Is it possible to do it via Gegenbauer polynomials? $\endgroup$
    – Gary
    Mar 12, 2023 at 0:41

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