Difference between revisions of "Polygamma reflection formula"

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==Theorem==
<strong>[[Polygamma reflection relation|Theorem]]:</strong> The following formula holds:
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The following formula holds:
 
$$(-1)^m \psi^{(m)}(1-z)-\psi^{(m)}(z)=\pi \dfrac{\mathrm{d}^m}{\mathrm{d}z^m} \cot(\pi z),$$
 
$$(-1)^m \psi^{(m)}(1-z)-\psi^{(m)}(z)=\pi \dfrac{\mathrm{d}^m}{\mathrm{d}z^m} \cot(\pi z),$$
 
where $\psi^{(m)}$ denotes the [[polygamma]], $\pi$ denotes [[pi]], and $\cot$ denotes the [[cotangent]].
 
where $\psi^{(m)}$ denotes the [[polygamma]], $\pi$ denotes [[pi]], and $\cot$ denotes the [[cotangent]].
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<strong>Proof:</strong> █
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==Proof==
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==References==
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* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Polygamma recurrence relation|next=Polygamma multiplication formula}}: $6.4.7$
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[[Category:Theorem]]
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[[Category:Unproven]]

Latest revision as of 22:46, 17 March 2017

Theorem

The following formula holds: $$(-1)^m \psi^{(m)}(1-z)-\psi^{(m)}(z)=\pi \dfrac{\mathrm{d}^m}{\mathrm{d}z^m} \cot(\pi z),$$ where $\psi^{(m)}$ denotes the polygamma, $\pi$ denotes pi, and $\cot$ denotes the cotangent.

Proof

References