# Multiplication theorem (Q98831)

theorem
Language Label Description Also known as
English
Multiplication theorem
theorem

## Statements

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${\displaystyle {\frac {d\xi }{dt}}+\nabla \cdot f(\xi )=0}$
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${\displaystyle \Gamma (z)\;\Gamma \left(z+{\frac {1}{2}}\right)=2^{1-2z}\;{\sqrt {\pi }}\;\Gamma (2z).\,\!}$
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${\displaystyle \Gamma (z)\;\Gamma \left(z+{\frac {1}{k}}\right)\;\Gamma \left(z+{\frac {2}{k}}\right)\cdots \Gamma \left(z+{\frac {k-1}{k}}\right)=(2\pi )^{\frac {k-1}{2}}\;k^{1/2-kz}\;\Gamma (kz)\,\!}$
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${\displaystyle k^{m}\psi ^{(m-1)}(kz)=\sum _{n=0}^{k-1}\psi ^{(m-1)}\left(z+{\frac {n}{k}}\right)}$
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${\displaystyle k\left[\psi (kz)-\log(k)\right]=\sum _{n=0}^{k-1}\psi \left(z+{\frac {n}{k}}\right).}$
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${\displaystyle k^{s}\zeta (s)=\sum _{n=1}^{k}\zeta \left(s,{\frac {n}{k}}\right),}$
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${\displaystyle k^{s}\,\zeta (s,kz)=\sum _{n=0}^{k-1}\zeta \left(s,z+{\frac {n}{k}}\right)}$
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${\displaystyle \zeta (s,kz)=\sum _{n=0}^{\infty }{s+n-1 \choose n}(1-k)^{n}z^{n}\zeta (s+n,z).}$
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${\displaystyle F(s;q)=\sum _{m=1}^{\infty }{\frac {e^{2\pi imq}}{m^{s}}}=\operatorname {Li} _{s}\left(e^{2\pi iq}\right)}$
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${\displaystyle 2^{-s}F(s;q)=F\left(s,{\frac {q}{2}}\right)+F\left(s,{\frac {q+1}{2}}\right).}$
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${\displaystyle k^{-s}F(s;kq)=\sum _{n=0}^{k-1}F\left(s,q+{\frac {n}{k}}\right).}$
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