# Why Mathematica does nothing with the expression Gamma[2 z]? [closed]

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

Gamma[2z]


Gamma[2 z]

?

I expected this result:

$Γ(2z)=\frac{2^{2z-1}Γ(z)Γ(z+1/2)}{\sqrt\pi}$

• Why should it? The right hand side is much more complicated than the left one. Sep 5, 2018 at 21:52
• 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. Sep 5, 2018 at 22:06
• 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. Sep 5, 2018 at 22:59
• 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. Sep 5, 2018 at 23:16
• That rhs looks like gamma after she got run over by a reindeer... Sep 6, 2018 at 4:18

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

SimplifyGammaTuplicate[Gamma[2z], 2] //TeXForm


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

Another example:

SimplifyGammaTuplicate[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]];

Activate[identity[z]]

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[{SimplifySimplifyGamma = # &},
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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