Could someone tell me why Mathematica just returns the input with this expression:


Gamma[2 z]


I expected this result:


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    $\begingroup$ Why should it? The right hand side is much more complicated than the left one. $\endgroup$ Sep 5, 2018 at 21:52
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    $\begingroup$ Yes, it is the $\Gamma$-function. But how on earth should Mathematica tell that you want to have the expression Gamma[2 z] expanded in exactly this way? For example, it could also return Gamma[2 z - 1] (2 z - 1). Or any other identity involving the $\Gamma$-function (there are probably really many identities at least as interesting as the one you gave). What I tried to explain to you: i) Computers cannot read your mind and ii) Mathematica just doesn't work this way. $\endgroup$ Sep 5, 2018 at 22:06
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    $\begingroup$ It's not a command. Don't think of it that way. It's an expression. Mathematica is an expression rewriting language. If you want it to rewrite an expression in a particular way, you must tell it. $\endgroup$
    – John Doty
    Sep 5, 2018 at 22:59
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    $\begingroup$ I'm voting to close this question as off-topic because the issued raised is not really a problem; it is arises from the OP's misunderstanding of the result returned by Mathematica. $\endgroup$
    – m_goldberg
    Sep 5, 2018 at 23:16
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    $\begingroup$ That rhs looks like gamma after she got run over by a reindeer... $\endgroup$ Sep 6, 2018 at 4:18

2 Answers 2


You can use an undocumented internal function to for this purpose:

Simplify`GammaTuplicate[Gamma[2z], 2] //TeXForm

$\frac{2^{2 z-1} \Gamma (z) \Gamma \left(z+\frac{1}{2}\right)}{\sqrt{\pi }}$

Another example:

Simplify`GammaTuplicate[Gamma[3z], 3] //TeXForm

$\frac{3^{3 z-1/2} \Gamma (z) \Gamma \left(z+\frac{1}{3}\right) \Gamma \left(z+\frac{2}{3}\right)}{2 \pi }$


A top level way to look up the identity is through MathematicalFunctionData:

identity = MathematicalFunctionData[Gamma, "MultipliedArgumentFormulas", 
  "IncludedSubexpressions" -> {Gamma[2 _]}][[1]];

Gamma[2 z] == (2^(-1 + 2 z) Gamma[z] Gamma[1/2 + z])/Sqrt[\[Pi]]

An obscure way of getting the identity is to take backward and forward Mellin transforms and to mix in some hackery that prevent simplifications along the way:

Block[{Simplify`SimplifyGamma = # &},
  MellinTransform[MeijerGReduce[InverseMellinTransform[Gamma[2 z], z, s], s], s, z]
 (2^(-1 + 2 z) Gamma[z] Gamma[1/2 + z])/Sqrt[\[Pi]]

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