Difference between revisions of "Partial derivative of beta function"

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==Theorem==
<strong>[[Partial derivative of beta function|Theorem]]:</strong> The following formula holds:
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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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<strong>Proof:</strong> █
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==Proof==
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==References==
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[[Category:Theorem]]
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[[Category:Unproven]]

Latest revision as of 15:32, 23 June 2016

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

References