Maclaurin series for arcsin^2 x.

[SQUlSHY]

How do you prove that

?

I tried starting with the series of arcsin x (which is pretty standard), then squaring it, but what I got is a convolution that I can't quite simplify.

A nudge in the right direction would be helpful. smile

As an aside, I really wish Gaia's devs would, if not add LaTeX support, at least fix the BBCode parser. But we all know how that went. >.<

[Lord Tourettes]

What! This would be easier with integrals haha

[SQUlSHY]

Lord Tourettes

What! This would be easier with integrals haha

I've tried differentiating/integrating it, too, in the hope of finding an easy series, but had no luck whatsoever; there is still a lingering product of two terms with distinct series. sad

Perhaps I should discuss the motivation of this problem; yes, indeed, it came from an integral that popped into my head on Christmas Eve.

The integral was

This integral is doable without said series expansion. Sketch of a solution: Use a partial fraction decomposition to break it into two equal integrals, and each of these integrals can be done with a bit of Maclaurin series manipulation. :3

However, an alternate approach is to use the Maclaurin series of log(1 - y), where y = x(1 - x); what you get is a linear combination of Beta integrals that conveniently sums up to a multiple of the given series evaluated at x = 1/2.

Thing is, while also I got it using this formula, I saw this said formula on some paper that I can't find anymore (I saved it to my PC years ago, then my hard disk crashed), so I can't find the proof.

[Layra-chan]

You can get the series for (asin(x))^2 by [differentiating and then multiplying by sqrt(1-x^2)] twice. That gives you 2 on the left and some series on the right which you can extract a recurrence relation from by matching coefficients. No series multiplication needed.