Difference between revisions of "2F1(1/2,1;3/2;-z^2)=arctan(z)/z"

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(Created page with "==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 2F...")
 
 
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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=}}: 15.1.5
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* {{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