From evaluating products of Wigner 6-j symbols I end up with expressions containing fractions of Gamma functions. For example,

$\sqrt{\frac{\Gamma (2 N)}{\Gamma (2 N+3)}},$

where $N$ is a positive integer. This expression can be simplified to

$\frac{1}{2 \sqrt{N (N+1) (2 N+1)}}.$

However, when executing

FullSimplify[Sqrt[Gamma[2*N]/Gamma[3 + 2*N]],{N >= 1, N \[Element] Integers}]

the unaltered expression containing the Gamma functions is returned. Funny enough, running FullSimplify on the inverse, that is,

FullSimplify[(Sqrt[Gamma[2*N]/Gamma[3 + 2*N]])^(-1),{N >= 1, N \[Element] Integers}]

does return

2*Sqrt[NN*(1 + NN)*(1 + 2*NN)]

However, as my expressions are more complicated than the example given here, simplifying the reciprocal is not an option for me. How can I make mathematica simplify this fractions of factorials?

P.S. I tried the rule suggested in this answer, but had no success.

  • 2
    $\begingroup$ Try: FullSimplify[ Sqrt[Gamma[2*n]/Gamma[3 + 2*n]] // FunctionExpand, {n >= 1, n ∈ Integers}] $\endgroup$ – Mariusz Iwaniuk Jan 25 '18 at 13:35
  • 2
    $\begingroup$ dont use capital N for your own symbol by the way. $\endgroup$ – george2079 Jan 25 '18 at 17:08

Use FunctionExpand:

FunctionExpand[Sqrt@(Gamma[2 N]/Gamma[3 + 2 N])]

(* 1/2 Sqrt[1/(N (1 + N) (1 + 2 N))] *)


Simplify[Simplify[Sqrt[Gamma[2 n]]/Sqrt[Gamma[3 + 2 n]],
{n >= 1, n ∈ Integers}] // FunctionExpand, {n >= 1, n ∈ Integers}]

$$\frac{1}{2 \sqrt{n \left(2 n^2+3 n+1\right)}}$$

| improve this answer | |
  • $\begingroup$ Thanks, that'a a nice function to remember. Do you known how to get the same result when the square root is in the nominator and denominator, e.g., FunctionExpand[Sqrt[Gamma[2*NN]]/Sqrt[Gamma[3 + 2*NN]]] does not simplify. $\endgroup$ – Paul Jan 25 '18 at 14:56
  • $\begingroup$ Found it here $\endgroup$ – Paul Jan 25 '18 at 15:05

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