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How can I expand $\frac{1}{\sqrt{1-x^2}}$ by using the binomial series? I know how to expand $\frac{1}{\sqrt{1-x}}$, but I have no idea how to expand $\frac{1}{\sqrt{1-x^2}}$. Simply differentiate this makes expanding way to complicated.

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Simply substitute $x^2$ for $x$ in the expansion of$(1-x)^{-1/2}$

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  • $\begingroup$ so the expansion of that is $1+\sum_{n=1}^{\infty}\frac{1*3*5*....*(2n-1)}{2^nn!}(1-x)^{-\frac{2n+1}{2}}$. is $1+\sum_{n=1}^{\infty}\frac{1*3*5*....*(2n-1)}{2^nn!}(1-x^2)^{-\frac{2n+1}{2}}$ a right answer? $\endgroup$ – V150 Jun 9 '14 at 9:21
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HINT:

$$\frac1{\sqrt{1-x^2}}=(1-x^2)^{-\frac12}$$

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If $$f(x) = \sum_{k=0}^\infty a_k x^k,$$

then $$f(x^2) = \sum_{k=0}^\infty a_k x^{2k}.$$

This means that if you define $$b_k=\begin{cases}a_{l}& k=2l\\0&\text{otherwise}\end{cases},$$ you have $$f(x^2) = \sum_{k=0}^\infty b_k x^k$$

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