Difference between revisions of "Partial derivative of beta function"
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<strong>[[Partial derivative of beta function|Theorem]]:</strong> The following formula holds: | <strong>[[Partial derivative of beta function|Theorem]]:</strong> The following formula holds: | ||
| − | $$\dfrac{\partial}{\partial x} B(x,y)=B(x,y) \left( \dfrac{\Gamma'(x)}{\Gamma(x)} - \dfrac{\Gamma'(x+y)}{\Gamma(x+y)} \right) = B(x,y)(\psi(x) - \psi(x+y),$$ | + | $$\dfrac{\partial}{\partial x} B(x,y)=B(x,y) \left( \dfrac{\Gamma'(x)}{\Gamma(x)} - \dfrac{\Gamma'(x+y)}{\Gamma(x+y)} \right) = B(x,y)(\psi(x) - \psi(x+y)),$$ |
where $B$ denotes the [[Beta function]], $\Gamma$ denotes the [[gamma function]], and $\psi$ denotes the [[digamma function]]. | where $B$ denotes the [[Beta function]], $\Gamma$ denotes the [[gamma function]], and $\psi$ denotes the [[digamma function]]. | ||
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Revision as of 21:13, 27 April 2015
Theorem: The following formula holds: $$\dfrac{\partial}{\partial x} B(x,y)=B(x,y) \left( \dfrac{\Gamma'(x)}{\Gamma(x)} - \dfrac{\Gamma'(x+y)}{\Gamma(x+y)} \right) = B(x,y)(\psi(x) - \psi(x+y)),$$ where $B$ denotes the Beta function, $\Gamma$ denotes the gamma function, and $\psi$ denotes the digamma function.
Proof: █
