0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
2
$\begingroup$
FullSimplify[Gamma[z] == Gamma[z + n]/Pochhammer[z, n], 
 Assumptions -> Re[z] > 0 && n ∈ PositiveIntegers]

(* True *)

$\endgroup$
1
  • $\begingroup$ Works without assumptions too: FullSimplify[Gamma[z] == Gamma[z + n]/Pochhammer[z, n]] gives True. $\endgroup$
    – Roman
    Mar 6 at 18:22
3
$\begingroup$
Gamma[z] == Gamma[z + n]/Pochhammer[z, n] // FunctionExpand
(*    True    *)
$\endgroup$
5
  • $\begingroup$ 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

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.