2F1(1/2,1;3/2;z^2)=log((1+z)/(1-z))/(2z)

From specialfunctionswiki
Revision as of 21:18, 26 June 2016 by Tom (talk | contribs) (Created page with "==Theorem== The following formula holds: $${}_2F_1 \left( \dfrac{1}{2}, 1 ; \dfrac{3}{2}; z^2 \right)= \dfrac{1}{2z} \log \left( \dfrac{1+z}{1-z} \right),$$ where ${}_2F_1$ de...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

Theorem

The following formula holds: $${}_2F_1 \left( \dfrac{1}{2}, 1 ; \dfrac{3}{2}; z^2 \right)= \dfrac{1}{2z} \log \left( \dfrac{1+z}{1-z} \right),$$ where ${}_2F_1$ denotes the hypergeometric 2F1 and $\log$ denotes the logarithm.

Proof

References

Retrieved from "http://specialfunctionswiki.org/index.php?title=2F1(1/2,1;3/2;z%5E2)%3Dlog((1%2Bz)/(1-z))/(2z)&oldid=6809"