3
$\begingroup$

I want to simplify the following

Gamma[1 + d - k]

given the constraint that 1 + d - k is a positive integer. I have tried using

FullSimplify[Gamma[1 + d - k], {1 + d - k ∈ Integers, 1 + d - k > 0}]

However, this simply returns Gamma[1 + d - k]. I am expecting (d - k)! to be returned. Is there another way to simplify this expression with Mathematica?

$\endgroup$
  • 1
    $\begingroup$ You may have to use Gamma[1 + d - k] /. Gamma[x_] :> Factorial[x - 1]. $\endgroup$ – flinty Jul 27 at 15:25
  • 2
    $\begingroup$ The easiest way is Gamma[1 + d - k] /. Gamma[x_] :> (x - 1)! Note that this relationship holds for all x, i.e., Gamma[x] == (x - 1)! // FullSimplify evaluates to True $\endgroup$ – Bob Hanlon Jul 27 at 15:28
4
$\begingroup$

You are almost there

$Version
(* 12.1.0 for Mac OS X x86 (64-bit) (April 13, 2020) *)

FullSimplify[Gamma[1+ d - k],Element[d,Integers] && Element[k,Integers] && 1+d-k > 0]
(* (d-k)! *)
| improve this answer | |
$\endgroup$

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.