6
$\begingroup$

According to this site this expression involving Gamma is valid:

$$\prod_{k=0}^{n-1}\Gamma\left(\dfrac{k+z}{n}\right) = n^{\frac{1}{2}-z}(2\pi)^{\frac{n-1}{2}}\Gamma(z)$$

However it seems that Mathematica cannot simplify it (at least not version 10.0). Is thre a way to enforce it?

$\endgroup$
3
$\begingroup$

With version 12.1.0, for any positive integer value of $\,n\,$ you can use FunctionExpand to verify with

ex[n_] := Product[Gamma[(k - 1 + z)/n], {k, n}] ==
   Gamma[z](2Pi)^(n/2 - 1/2)n^(1/2 - z) // FunctionExpand

For example, ex[3] returns True. However, using ex[n] does not evaluate to a truth value.

$\endgroup$
2
$\begingroup$

Workaround:

$$\Gamma \left(\frac{k}{n}+\frac{z}{n}\right)=\frac{\Gamma \left(\frac{k}{n}+\frac{z}{n}+1\right)}{\frac{k}{n}+\frac{z}{n}}$$

$Version
(* 12.1.1 for Microsoft Windows (64-bit) (June 9, 2020) *)

Product[Gamma[k/n + z/n + 1]/(k/n + z/n), {k, 0, n - 1}] // FunctionExpand

(* n^(1/2 - z) (2 \[Pi])^(-(1/2) + n/2) Gamma[z] *)
$\endgroup$
3
  • 1
    $\begingroup$ Even Product[Gamma[k/n + z/n], {k, 0, n - 1}] works in version 12.0. $\endgroup$
    – user64494
    Jul 13 '20 at 16:26
  • 1
    $\begingroup$ +1 Working from the original form: Product[Gamma[(k + z)/n], {k, 0, n - 1}] /. Gamma[x_] :> Gamma[x + 1]/x // FunctionExpand // Simplify[#, n > 0] & $\endgroup$
    – Bob Hanlon
    Jul 13 '20 at 16:26
  • $\begingroup$ @user64494.If: Product[Gamma[(k + z)/n], {k, 0, n - 1}] dosen't work, but: Product[Gamma[(k + z)/n] // ExpandAll, {k, 0, n - 1}] works fine. $\endgroup$ Jul 13 '20 at 16:30

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.