Finding the power series of $\arcsin x$

Question

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.

Answer 1

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:

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

Answer 2

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

Answer 3

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?