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I heard that to expand $\frac{1}{\sqrt{1-x^2}}$, I have to expand $\frac{1}{\sqrt{1-x}}$ by binomial series and then just replace $x$ to $x^2$. Using binomial series, I found that $\frac{1}{\sqrt{1-x}}$ becomes $1+\sum_{n=1}^{\infty}\frac{1*3*5*....*(2n-1)}{2^nn!}(1-x)^{-\frac{2n+1}{2}}$. So is $ 1+\sum_{n=1}^{\infty}\frac{1*3*5*....*(2n-1)}{2^nn!}(1-x^2)^{-\frac{2n+1}{2}}$ would be an right binomial expansion of $\frac{1}{\sqrt{1-x^2}}$?

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  • $\begingroup$ Are you sure about the powers of $1-x$ ? $\endgroup$ – Yves Daoust Jun 9 '14 at 9:50
  • $\begingroup$ Forgot to multiply $x^n$. I will anend that. $\endgroup$ – V150 Jun 9 '14 at 9:55
  • $\begingroup$ i think it's more like $1+\sum_{n=1}^{\infty}\frac{1*3*5*....*(2n-1)}{2^nn!}x^n$. So, the answer will be $1+\sum_{n=1}^{\infty}\frac{1*3*5*....*(2n-1)}{2^nn!}x^{2n}$. Am I getting it right? $\endgroup$ – V150 Jun 9 '14 at 10:01
  • $\begingroup$ Yep, but don't forget the half exponent and the signs. $\endgroup$ – Yves Daoust Jun 9 '14 at 10:06
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rewrite the function as $(1-x^2)^{-\frac{1}{2}} = \sum_{k=0}^{\infty} \binom{-\frac{1}{2}}{k}(-x)^{2k}$. Now expand $\binom{-\frac{1}{2}}{k}$, you'll get something qutie similar to you expression in the second line. This will be your solution.

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