# expansion of hypergeometric series [closed]

I am completely new to mathematica. I am not sure how to calculate the expression 2F1. Is it possible to a closed form solution in terms of z of the hypergeometric series 2F1(-a, N/2-a, N/2; z) and 2F1(1-a, N/2-a, 1+N/2; z) when N>2a and N is any positive natural number.

• Try Hypergeometric2F1 – wuyudi Sep 29 at 12:40

Clear["Global*"]

nmax = 7;

Table[{n, Assuming[n/2 > a,
Hypergeometric2F1[-a, n/2 - a, n/2, z] // FunctionExpand //
FullSimplify]}, {n, 1, nmax}] //
Prepend[#, {"n", Hypergeometric2F1[-a, n/2 - a, n/2, z]}] & //
Grid[#, Frame -> All] & // TraditionalForm


Table[{n, Assuming[n/2 > a,
Hypergeometric2F1[1 - a, n/2 - a, 1 + n/2, z] // FunctionExpand //
FullSimplify]}, {n, 1, nmax}] //
Prepend[#, {"n", Hypergeometric2F1[1 - a, n/2 - a, 1 + n/2, z]}] & //
Grid[#, Frame -> All] & // TraditionalForm


Consequently, even for specific values of n these hypergeometric functions do not generally reduce to simpler functions.

For general $$n$$ and $$a$$ no, but for many values it can simplify. For instance:

a = 3/2;

Table[Hypergeometric2F1[-a, n/2 - a, n/2, z] // FunctionExpand // FullSimplify, {n, 1, 5}]

{
3 z + 1,
(8 (z - 1) EllipticK[z] + 2 (z + 7) EllipticE[z])/(3 π),
1,
(4 (z - 1) (z + 3) EllipticK[z] + 4 ((7 - 2 z) z + 3) EllipticE[z])/(15 π z),
(Sqrt[z] ((8 - 3 z) z + 3) + 3 (z - 1)^3 ArcTanh[Sqrt[z]])/(16 z^(3/2))
}
`

Capital $$N$$ is a builtin function in Mathematica, so don't use it as a variable.