Difference between revisions of "Reciprocal gamma written as an infinite product"
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(Created page with "==Theorem== The following formula holds: $$\dfrac{1}{\Gamma(z)} = ze^{\gamma z} \displaystyle\prod_{k=1}^{\infty} \left( 1 + \dfrac{z}{k}\right)e^{-\frac{z}{k}},$$ where $\Gam...") |
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The following formula holds: | The following formula holds: | ||
$$\dfrac{1}{\Gamma(z)} = ze^{\gamma z} \displaystyle\prod_{k=1}^{\infty} \left( 1 + \dfrac{z}{k}\right)e^{-\frac{z}{k}},$$ | $$\dfrac{1}{\Gamma(z)} = ze^{\gamma z} \displaystyle\prod_{k=1}^{\infty} \left( 1 + \dfrac{z}{k}\right)e^{-\frac{z}{k}},$$ | ||
| − | where | + | where $\dfrac{1}{\Gamma}$ is the [[reciprocal gamma]] function, and $\gamma$ is the [[Euler-Mascheroni constant]]. |
==Proof== | ==Proof== | ||
Revision as of 09:41, 4 June 2016
Theorem
The following formula holds: $$\dfrac{1}{\Gamma(z)} = ze^{\gamma z} \displaystyle\prod_{k=1}^{\infty} \left( 1 + \dfrac{z}{k}\right)e^{-\frac{z}{k}},$$ where $\dfrac{1}{\Gamma}$ is the reciprocal gamma function, and $\gamma$ is the Euler-Mascheroni constant.
Proof
References
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous): §1.1 (3)
