# 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