# series expansion of an elliptic integral

• September 27th 2008, 01:06 PM
minivan15
series expansion of an elliptic integral
my problem is: find a series expansion in the variable k of F(k, pi/2) for 0 < k < 1

F(k, pi/2) = [integral from 0 to pi/2] dx/SQRT(1-ksinx)^2)

what is a good starting point for this?
• September 27th 2008, 05:27 PM
shawsend
Hey, check out the Wikipedia entry on the Binomial expansion. For fractional powers in the denominator its:
$
\frac{1}{(1-a)^r}=\sum_{k=0}^{\infty}\binom{r+k-1}{k}a^k$

alright then:

$
\frac{1}{(1-a^2\sin^2(x))^{1/2}}=\sum_{k=0}^{\infty}\binom{1/2+k-1}{k}a^{2k}\sin^{2k}(x)$

You can do that. Then integrate term by term to get:

$\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2(x)}}=\frac{1}{2}\pi\left[1+\left(\frac{1}{2}\right)^2 k^2+\left(\frac{3}{8}\right)^2k^4+\cdots+\text{gen eral term}+\cdots\right]$
• September 28th 2008, 03:27 PM
minivan15
awesome! I got it using that! thanks a bunch!!