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Is there a closed form (or approximation) for a hypergeometric function of form:

$_2F_1(1,b+c;c;\frac{1}{2}) \quad \text{where} \; b,c \in \mathbb{N}$ ?

I researched all identities in http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/03/04/ but nothing seems to work for me.

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1 Answer 1

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Case $1$: $c\neq1$

Then $_2F_1\left(1,b+c;c;\dfrac{1}{2}\right)$

$=~_2F_1\left(b+c,1;c;\dfrac{1}{2}\right)$

$=\dfrac{1}{B(1,c-1)}\int_0^1\dfrac{(1-x)^{c-2}}{\left(1-\dfrac{x}{2}\right)^{b+c}}dx$

$=\dfrac{2^{b+c}(c-1)!}{0!(c-2)!}\int_0^1\dfrac{(1-x)^{c-2}}{(2-x)^{b+c}}dx$

$=2^{b+c}(c-1)\int_0^1\dfrac{(2-x-1)^{c-2}}{(2-x)^{b+c}}dx$

$=2^{b+c}(c-1)\int_0^1\sum\limits_{n=0}^{c-2}\dfrac{C_n^{c-2}(-1)^n(2-x)^{c-n-2}}{(2-x)^{b+c}}dx$

$=2^{b+c}(c-1)\int_0^1\sum\limits_{n=0}^{c-2}\dfrac{(-1)^n(c-2)!(2-x)^{-b-n-2}}{n!(c-n-2)!}dx$

$=2^{b+c}(c-1)\left[-\sum\limits_{n=0}^{c-2}\dfrac{(-1)^n(c-2)!(2-x)^{-b-n-1}}{n!(c-n-2)!(-b-n-1)}\right]_0^1$

$=\left[\sum\limits_{n=0}^{c-2}\dfrac{(-1)^n(c-1)!2^{b+c}}{n!(c-n-2)!(b+n+1)(2-x)^{b+n+1}}\right]_0^1$

$=\sum\limits_{n=0}^{c-2}\dfrac{(-1)^n(c-1)!2^{b+c}}{n!(c-n-2)!(b+n+1)}-\sum\limits_{n=0}^{c-2}\dfrac{(-1)^n(c-1)!2^{c-n-1}}{n!(c-n-2)!(b+n+1)}$

Case $2$: $c=1$

Then $_2F_1\left(1,b+1;1;\dfrac{1}{2}\right)$

$=~_2F_1\left(b+1,1;1;\dfrac{1}{2}\right)$

$=\left(1-\dfrac{1}{2}\right)^{-b-1}$

$=\dfrac{1}{2^{b+1}}$

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