I used Mathematica to prove that

$$\Gamma (z+1) = z \Gamma (z)$$

is true by showing

ForAll[z, Re[z] > 0, Gamma[z + 1] == z Gamma[z]] // Resolve
( True )

Now, I want to show that

$$\Gamma (z) = \frac {\Gamma (z+n)}{(z)_n}$$

is true for all Re(z)>0, n in N, by showing a similar ForAll but I do not know how to add the assumption that both Re(z)>0, n are in N.

Note that $(z)_n$ stands for Pochhammer[z,n]

My question is how to create a similar ForAll which resolves to true for the second identity.

  • $\begingroup$ I don't think it's true: Gamma[z] == Gamma[z + n + 1]/Pochhammer[z, n] // FunctionExpand gives Gamma[z] == (n + z) Gamma[z]. $\endgroup$
    – Roman
    Mar 3 at 15:08
  • $\begingroup$ I corrected that in the question, my mistake. $\endgroup$ Mar 3 at 15:11
FullSimplify[Gamma[z] == Gamma[z + n]/Pochhammer[z, n], 
 Assumptions -> Re[z] > 0 && n ∈ PositiveIntegers]

(* True *)

  • $\begingroup$ Works without assumptions too: FullSimplify[Gamma[z] == Gamma[z + n]/Pochhammer[z, n]] gives True. $\endgroup$
    – Roman
    Mar 6 at 18:22
Gamma[z] == Gamma[z + n]/Pochhammer[z, n] // FunctionExpand
(*    True    *)
  • $\begingroup$ But only for n in N, surely? How can this restriction be added, do you know? $\endgroup$ Mar 3 at 15:15
  • $\begingroup$ No, these functions all have analytic continuations. $\endgroup$
    – Roman
    Mar 3 at 15:16
  • $\begingroup$ Gamma IS an analytic continuation, but does Pochhammer[z,n] has one too? $\endgroup$ Mar 3 at 15:20
  • $\begingroup$ It all seems to work, this was quite a learning experience. - Yeah, analytic continuation defined in terms of Gamma. :-) - dlmf.nist.gov/5.2#iii $\endgroup$ Mar 3 at 15:23
  • 1
    $\begingroup$ The analytic continuation of Pochhammer[z, n] is literally defined through your equation as the ratio of two $\Gamma$ functions. $\endgroup$
    – Roman
    Mar 3 at 15:33

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