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ID: 2586965
User: 24.56.201.36
Article: Gamma function
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\Gamma\left(\tfrac{1}{2}+n\right) &= {(2n)! \over 4^n n!} \sqrt{\pi} = \frac{(2n-1)!!}{2^n} \sqrt{\pi} = \sqrt{\pi} \left[ {n-\frac{1}{2}\choose n} n! \right] \\
\Gamma\left(\tfrac{1}{2}+n\right) &= {(2n)! \over 4^n n!} \sqrt{\pi} = \frac{(2n-1)!!}{2^n} \sqrt{\pi} = \sqrt{\pi} \left[ {n-\frac{1}{2}\choose n} n! \right] \\
\Gamma\left(\tfrac{1}{2}-n\right) &= {(-4)^n n! \over (2n)!} \sqrt{\pi} = \frac{(-2)^n}{(2n-1)!!} \sqrt{\pi} = \frac{\sqrt{\pi}}{{-\frac{1}{2} \choose n} n!}
\Gamma\left(\tfrac{1}{2}-n\right) &= {(-4)^n n! \over (2n)!} \sqrt{\pi} = \frac{(-2)^n}{(2n-1)!!} \sqrt{\pi} = \frac{\sqrt{\pi}}{{-\frac{1}{2} \choose n} n!}
\end{align}</math>
\end{align}</math> Rightmost eqn looks wrong. Try 3 as n for example.


where {{math|''n''!!}} denotes the [[double factorial]] and, when {{math|''n'' {{=}} 0}}, {{math|''n''!! {{=}} 1}}. See [[Particular values of the gamma function]] for calculated values.
where {{math|''n''!!}} denotes the [[double factorial]] and, when {{math|''n'' {{=}} 0}}, {{math|''n''!! {{=}} 1}}. See [[Particular values of the gamma function]] for calculated values.
Reason: ANN scored at 0.880553
Reporter Information
Reporter: Matt (anonymous)
Date: Tuesday, the 7th of July 2020 at 01:15:39 PM
Status: Reviewed - Not included in dataset
Monday, the 14th of March 2016 at 11:41:00 PM #103573
Matt (anonymous)

This equation has three expressions. The rightmost one appears incorrect. Try 3 for n