3
$\begingroup$

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.

$\endgroup$
  • 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
5
$\begingroup$

Use FunctionExpand:

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

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

Edited:

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

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.