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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.

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Actually, you should be trying to directly apply the binomial series formula on $\frac1{\sqrt{1-x^2}}$ (i.e., take the exponent to be $-\frac12$ in the binomial theorem). There's a way to properly interpret $\binom{-1/2}{k}$, of course... –  J. M. Feb 2 '12 at 17:48
    
@J.M Thanks. Do you think that there's a solution by develop the approach I chose? –  Jozef Feb 2 '12 at 17:51
    
Might work, but doesn't look to nice. Let $\sum a_nx^n$ be the power series, then $\sum a_nx^n=\int \sqrt{\sum_{n=0}^{\infty}x^{2n}} $. Derivate, square and try to identify the coefficients. You end up with $(\sum (n+1)a_{n+1} x^n)^2 = \sum_{n=0}^{\infty}x^{2n}$. trying to get $a_n$ from here is not that nice.... Also there are many issues, namely you don't know that arcsin has a power series, there could also be issues about how to square a power series.... –  N. S. Feb 2 '12 at 17:58
    
Thanks N.S. @J.M.: How do I get over the problem that my sum is in the denominator ? –  Jozef Feb 2 '12 at 18:23
    
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... –  J. M. Feb 2 '12 at 18:28
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3 Answers

up vote 3 down vote accepted

The generalized binomial thorem states that

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

However, as $\alpha$ is not necesarilly an integer, you need to find the binomial coefficients by terms of the Gamma Function. i.e.

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

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

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

For instance, first note that.

$$ {-\frac{1}{2}\choose {k}} = \frac{\Gamma {\left(\frac{1}{2}\right)}}{\Gamma {(k+1)}\Gamma {\left(\frac{1}{2}-k\right)}}$$

It is very common to find many books showing two thigs:

  1. $\Gamma {\left(\frac{1}{2}\right)} = \sqrt{\pi}$
  2. $\Gamma {\left(\frac{1}{2}-k\right)} = \displaystyle (-1)^k \frac{2^k}{(2k-1)!!} \sqrt{\pi}$

Thus you have

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

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:

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

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On the matter of $\dbinom{-1/2}{k}$, see also this question. –  J. M. Feb 3 '12 at 2:02
    
Thanks a lot @Peter! –  Jozef Feb 3 '12 at 10:35
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Hint: binomial series. http://en.wikipedia.org/wiki/Binomial_series

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Thanks.How do I get over the problem that my sum is in the denominator ? –  Jozef Feb 2 '12 at 19:01
1  
It's not - the exponent is (-1/2), not (1/2). –  marty cohen Feb 3 '12 at 3:56
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This is an expansion about x' = 0, which clearly converges on the disk |x| < 1. But what if an expansion about another nonzero point x' = a is needed? The binomial expansion wouldn't apply, then. And would the radius of convergence be |x - a| < 1?

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