Answer to Question #287105 in Algebra for samir seikh

Question

. Prove the following statements (the symbols carry their usual meaning):

(a) 2F1(−n, 1, 1; x) = (1 − x)

n

(b) 2F1(−n, 1, 1; −x) = (1 + x)

n

(c) x 2F1(1, 1, 2; x) = log(1 − x)

(d) x 2F1(1, 1, 2; −x) = log(1 + x)

(e) x 2F1

Answer 1

Using hypergeometric series identity:

$_2F_1(a,b;b;z)=(1-z)^{-a}$

(a)

$_2F_1(-n,1;1;x)\implies\>z=x,and\>a=-n$

$\therefore_2F_1(-n,1;1;x)=(1-x)^{-(-n)}$

$=(1-x)^n$

(b)

In $_2F_1(-n,1;1;-x),z=-x\>and \>a=-n$

$\implies\>_2F_1(-n,1;1;-x)=[1-(-x)]^{-(-n)}$

$=(1+x)^n$

(c)

Using hypergeometric series identity;

$_2F_1(1,1;2;-z)=\frac{In(1+z)}{z}$

Then in the statement

$(x)_2F_1(1,1;2;x)$

$x=-z\implies\>z=-x$

$\therefore\>_2F_1(1,1;2;x)=\frac{In(1+-x)}{x}$

$\therefore\>(x)_2F_1(1,1;2;x)=In(1-x)$

(d)

In the statement

$(x)_2F_1(1,1;2;-x)$

$x=z$

$\therefore\>_2F_1(1,1;2;-x)=\frac{In(1+x)}{x}$

$\therefore\>(x)_2F_1(1,1;2;-x)=In(1+x)$

(e)

Assuming the question statement is;

$x_2F_1(\frac{1}{2},\frac{1}{2},\frac{3}{2};x^2)=$ arc Sin $x$

Using hypergeometric series identity

$2F_1(\frac{1}{2},\frac{1}{2};\frac{3}{2};z^2)=\frac{arc\>Sin\>z}{z}$

Then substituting $z=x$ in the identity

$2F_1(\frac{1}{2},\frac{1}{2};\frac{3}{2};x^2)$ $=\frac{arc\>Sin\>x}{x}$

$\implies(x)_2F_1(\frac{1}{2},\frac{1}{2},\frac{3}{2},x^2)=arc\>Sin\>x$