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I'm trying to find the power series of $\arcsin x$.

This is what I did so far: (\arcsin x)'=\frac{1}{\sqrt{1-x^2}}, so $\arcsin x=\int \sqrt{\sum_{n=0}^{\infty}x^{2n}}$.

(for $|x|<1$)

Any hints for what should I do further?

Thanks a lot.

  • 0
    I don't know. I wouldn't go through that anyway, since I know that $\frac1{\sqrt{1-x^2}}$ definitely has a Maclaurin series...2012-02-02

2 Answers 2

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The generalized binomial thorem states that

$(1+x)^{\alpha} = \sum_{k=0}^{\infty} {{\alpha\choose {k}} x^k}$

The definition still holds

$ {\alpha\choose {k}} =\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}$

As people are suggesting, you need to see that

$\sin^{-1} x = \int \frac{dx}{\sqrt{1-x^2}}$

thus you can integrate the binomial series of

$(1-x^2)^{-\frac{1}{2}} = \sum_{k=0}^{\infty} (-1)^k {{-\frac{1}{2}\choose {k}} x^{2k}}$

Using the definition shows that

$ {-\frac{1}{2}\choose {k}} = {( - 1)^k}\frac{{(2k - 1)!!}}{{k!{2^k}}} = {( - 1)^k}\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}$

Putting this in the sum produces:

${\left( {1 - {x^2}} \right)^{-\frac{1}{2}}} = \sum\limits_{k = 0}^\infty {\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}{x^{2k}}} $

Finally this yields:

${\sin ^{ - 1}}x = \sum\limits_{k = 0}^\infty {\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}\frac{{{x^{2k + 1}}}}{{2k + 1}}} $

Remember that by definition

$0!! = 1 $ $(-1)!! = 1$

If you're not comfortable with that simply put:

${\sin ^{ - 1}}x = x + \sum\limits_{k = 1}^\infty {\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}\frac{{{x^{2k + 1}}}}{{2k + 1}}} $

  • 0
    How do you show this: $ {-\frac{1}{2}\choose {k}} = {( - 1)^k}\frac{{(2k - 1)!!}}{{k!{2^k}}} = {( - 1)^k}\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}$2018-01-01
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Hint: binomial series. http://en.wikipedia.org/wiki/Binomial_series

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    It's not - the exponent is (-1/2), not (1/2).2012-02-03