Finding the power series of $\arcsin x$

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.

-
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

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:

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

-
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
You really really don't need the $\Gamma$ function to define $\binom{\alpha}k$, as long as $k$ is a positive integer. – Thomas Andrews Sep 6 '14 at 12:35
@ThomasAndrews True. ${}{}{}{}$ – Pedro Tamaroff Sep 6 '14 at 15:35

Hint: binomial series. http://en.wikipedia.org/wiki/Binomial_series

-
Thanks.How do I get over the problem that my sum is in the denominator ? – Jozef Feb 2 '12 at 19:01
It's not - the exponent is (-1/2), not (1/2). – marty cohen Feb 3 '12 at 3:56