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I'd like to convert expressions such as 

Gamma[(1/10)*(13 + 10*n)]

to 

Gamma[13/10 + n]

I tried the rule

Gamma[r_] -> Gamma[Expand[r]]

but that doesn't seem to work.

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  • 2
    $\begingroup$ You want :> instead of ->. Alternatively just use ExpandAll on the complete expression. $\endgroup$
    – Quantum_Oli
    Aug 23, 2016 at 16:00
  • $\begingroup$ FullSimplify@Gamma[(1/10)*(13 + 10*n)] works as well. $\endgroup$
    – bill s
    Aug 23, 2016 at 20:05
  • $\begingroup$ An alternative is to use FunctionExpand@ Gamma[(1/10)*(13 + 10*n)] $\endgroup$
    – user31159
    Oct 22, 2016 at 22:42
  • $\begingroup$ ExpandAll also works. $\endgroup$
    – Wen Chern
    Oct 23, 2016 at 8:31

2 Answers 2

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There's also ExpandAll:

ExpandAll[Gamma[(1/10)*(13 + 10*n)]]
(*  Gamma[13/10 + n]  *)
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You need to use a delayed version of the rule

Gamma[r_]:>Gamma[Expand[r]]

rather than

Gamma[r_]->Gamma[Expand[r]]

otherwise the Expand is applied to r (returning simply r) before the substitution is attempted.

A slightly briefer alternative is

r_Gamma :> Expand/@r

This matches any expression that has Head Gamma and has the added effect of matching multiple argument versions of Gamma and expanding each of their arguments independently.

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