Series expansion for irrational functions such as $\dfrac{1}{\sqrt{1+x+x^2}}$

Question

So far I have been able to expand most rational and transcendental functions into the Maclaurin series. But irrational functions such as this one $\dfrac{1}{\sqrt{1+x+x^2}}$ confuse me. I try to use the binomial series formula but the general formula only gives me $2$ terms. What I have tried to do is to set up a dummy variable $X=x+x^2$ so that my function has the form $(1+X)^\frac{1}{2}$ and expand them, then reinsert the variable into the expression.

Is this method correct or not? Is there a more general binomial series formula to deal with non-factorable irrational expression?

Answer 1

Hint:

As $1-x^3=(1-x)(1+x+x^2),$

$$(1+x+x^2)^{-1/2}=(1-x)^{1/2}(1-x^3)^{-1/2}$$

Use https://en.m.wikipedia.org/wiki/Binomial_series assuming $|x|<1$

Answer 2

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

References

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