Gamma Function of One Half/Proof 4
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Theorem
- $\map \Gamma {\dfrac 1 2} = \sqrt \pi$
Proof
Due to the organization of pages at $\mathsf{Pr} \infty \mathsf{fWiki}$, this argument is circular. In particular: This uses Gamma Function of Positive Half-Integer in its proof, which itself uses Gamma Function of One Half, so it's circular. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving this issue. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{CircularStructure}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
\(\ds \map \Gamma {\frac 1 2}\) | \(=\) | \(\ds \frac {0!} {2^0 0!} \sqrt \pi\) | Gamma Function of Positive Half-Integer | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt \pi\) | Factorial of Zero |
$\blacksquare$