Homework Help: Hypergeometric function problem

1. Jan 1, 2013

matematikuvol

1. The problem statement, all variables and given/known data
Calculate
$$_2F_1(\frac{1}{2},\frac{1}{2},\frac{3}{2};x)$$

2. Relevant equations
$$_2F_1(a,b,c;x)=\sum^{\infty}_{n=0}\frac{(a)_n(b)_n}{n!(c)_n}x^n$$
$$(a)_n=a(a+1)...(a+n-1)$$

3. The attempt at a solution
$$(\frac{1}{2})_n=\frac{1}{2}\frac{3}{2}\frac{5}{2}...\frac{2n-1}{2}$$
$$(\frac{3}{2})_n=\frac{3}{2}\frac{5}{2}\frac{7}{2}...\frac{2n+1}{2}$$
From this relations
$$\frac{(\frac{1}{2})_n}{(\frac{3}{2})_n}=\frac{1}{2n+1}$$
But I don't see how to calculate this to the end...

2. Jan 1, 2013

haruspex

Can you write $$(\frac{1}{2})_n$$ in terms of factorials and powers of 2?

3. Jan 1, 2013

matematikuvol

Yes I could
$$(\frac{1}{2})_n=\frac{(2n-1)!!}{2^n}$$
$$(\frac{3}{2})_n=\frac{(2n+1)!!}{2^{n-1}}$$
So I get
$$_2F_1(\frac{1}{2},\frac{1}{2},\frac{3}{2};x)=\sum^{\infty}_{n=0}\frac{(2n-1)!!}{(2n+1)2^{n+1}n!}x^n=\sum^{\infty}_{n=0}\frac{(2n-1)!!}{2(2n+1)(2n)!!}x^n$$
And I don't know what to do know.

4. Jan 1, 2013

lurflurf

Hint:

$$\arcsin (x) \sim \sum_{k=0}^\infty \dfrac{(2k)!x^{2k+1}}{4^k (k!)^2 (2k+1)} \sim \sum_{k=0}^\infty \dfrac{(2k-1)!!x^{2k+1}}{(2k)!!(2k+1)}$$

I am not sure how to motivate this though. It can be difficult to recognize familiar functions by their power series.

5. Jan 2, 2013

haruspex

Try writing out the general term of the expansion of (1-x)-1/2