Difference between revisions of "2F1(1/2,1;3/2;-z^2)=arctan(z)/z"
From specialfunctionswiki
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==References== | ==References== | ||
| − | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=2F1(1/2,1;3/2;z^2)=log((1+z)/(1-z))/(2z)|next=2F1(1/2,1/2;3/2;z^2)=arcsin(z)/z}}: 15.1.5 | + | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=2F1(1/2,1;3/2;z^2)=log((1+z)/(1-z))/(2z)|next=2F1(1/2,1/2;3/2;z^2)=arcsin(z)/z}}: $15.1.5$ |
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] | ||
Latest revision as of 23:16, 12 July 2016
Theorem
The following formula holds: $${}_2F_1 \left( \dfrac{1}{2}, 1; \dfrac{3}{2} ; -z^2 \right)=\dfrac{\arctan(z)}{z},$$ where ${}_2F_1$ denotes the hypergeometric 2F1 and $\arctan$ denotes the inverse tangent.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $15.1.5$
