Hypergeometric value

Question

Is their a closed form for the following

$${}_2F_1 \left(a,b;c;\frac{1}{2} \right)$$

I would use the following

$${}_2F_1 \left(a,b;c;x \right)= \frac{\Gamma(c)}{\Gamma(c-b)\Gamma(b)} \int^1_0 t^{b-1}(1-t)^{c-b-1} (1-xt)^{-a} \, dt $$

But it wasn't a success !

Edit: Corrected integral representation (swapped arguments in $\Gamma$ fraction)

Since, they didn't have a closed form, they invented the cool notation. Maybe you have to look for some hypergeometric identities. You can find a closed form anytime for special $a,b,c$. — Torsten Hĕrculĕ Cärlemän, Jul 26 '13 at 22:29
@TorstenHĕrculĕCärlemän there are closed form especially for $ {}_2 F_1(a,b;c;1)$ using the beta function . — Zaid Alyafeai, Jul 26 '13 at 22:30
I was talking about this function. I am not completely sure about the non-existence. — Torsten Hĕrculĕ Cärlemän, Jul 26 '13 at 22:40
The general case seems uncovered, but there are identities such as these for certain dependencies between $a,b,c$ and some transformation to main argument $-1$. — ccorn, Jul 26 '13 at 22:48
@ZaidAlyafeai: The identities follow from Euler's transformation and a theorem by Kummer — ccorn, Jul 26 '13 at 23:12
Thanks @ccorn , I was sure that we can use the Euler's transformation. But I've gotta look at the Kummer theorem !. I think it can be solved by the integral representation . — Zaid Alyafeai, Jul 26 '13 at 23:22
@ZaidAlyafeai: I have corrected the integral representation (pending peer review). With that, the proof should work along the lines you imagine. — ccorn, Jul 27 '13 at 00:42
Thanks @ccorn , I would expand $(1+t)^{-a}$ using the power expansion and see if I get the result ! — Zaid Alyafeai, Jul 27 '13 at 01:03

score 1 · Answer 1 · answered Oct 30 '14 at 16:55

I would write this as a comment but I don't have the privilege yet. This is the best documentation on special functions, first book in these series gives a through Hypergeometric functions.

Higher Transcendental Functions ,H. Bateman and A. Erdelyi,

Hope this helps.

score 0 · Answer 2 · edited Dec 11 '14 at 13:26

There are closed forms for the 2F1 but their exact form depend on the values and the relations between the parameters.

The trivial case is if $a$ or $b$ is a negative integer: the function becomes a finite sum of rational expressions in the parameters. if you choose $a=-4$ and if you name $d$ the value you have fixed at $1/2$, you got

$ 1 - \frac{4 b d}{c} +\frac{6 b (b+1) d^2}{c (c+1)} - \frac{4 b (b+1) (b+2) d^3}{c (c+1) (c+2)} - \frac{(-b-3) b (b+1) (b+2) d^4}{c (c+1) (c+2) (c+3)}$

and setting $d = 1/2$:

$ 1 -\frac{2 b}{c} + \frac{3 (b+1) b}{2 c (c+1)} - \frac{(b+1) (b+2) b}{2 c (c+1) (c+2)} - \frac{(-b-3) (b+1) (b+2) b}{16 c (c+1) (c+2) (c+3)} $

if $a$ or $b$ is equal to $c$, they cancel each other, giving a simpler $_1F_0$.

$_2F_1( a, b ; b ; d) = _1F_0(a; d) = (1-d)^{-a} $
for small rationals, there are more interesting closed forms involving elementary functions and elliptic functions and their inverses. Among classical cases

$_1F_0(1/2 ; d) = \frac{1}{\sqrt{1- d}} $

$_2F_1(1/2, 1/2 ; 3/2; d) = \frac{1}{\sqrt d} \sin^{-1} (\sqrt d )$

$_2F_1(1/2, 1 ; 3/2; d) = \frac{1}{\sqrt d} \tanh^{-1} (\sqrt d )$

score 0 · Answer 3 · answered Mar 25 '17 at 20:05

I have found a few solutions for $z=1/2$ in these references:

K. Oldham, J. Myland, & J. Spanier, An Atlas of Functions, Ch. 60, Springer.

A. Erdelyi, Higher Transcendental Functions, Vol. 1 (and particularly, Sec. 2.8), Krieger Publishing.

NIST Handbook of Mathematical Functions.

Bear in mind, however, that these are for specific values of $a, b, \text{and } c$. The are no general solutions for $z=1/2$. I hope you find something there that meets your needs

Hypergeometric value

3 Answers3