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How do I find the zeros of:

1/Gamma[1 - I x]

The zeros should be $ -i\, n$ with $ n>0 $

I tried with Solve[1/Gamma[1 - I x]==0,x], but I get the message:

Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.

So I tried with Reduce[1/Gamma[1 - I x]==0,x], but I get False as a result

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  • $\begingroup$ Use the identity Gamma[z] Gamma[1 - z] == Pi/Sin[Pi z] with z=I x... $\endgroup$ – José Antonio Díaz Navas Apr 14 '18 at 20:16
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f[x_] = 1/Gamma[1 - I x];

Plot[f[y*I], {y, -5, 5}, PlotRange -> All]

enter image description here

FindRoot[f[y*I] == 0, {y, #}] & /@ Range[-5.25, -1.25]

(* {{y -> -5.}, {y -> -4.}, {y -> -3.}, {y -> -2.}, {y -> -1.}} *)

Assuming[{Element[n, Integers], n > 0},
 Limit[f[-m*I], m -> n]]

(* 0 *)
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  • $\begingroup$ haven`t you started from the solution? I mean, in your reasoning, why you know already the argument of your plotted function? I am a bit puzzled, as I thought that the OP want to know the solution for $1/\Gamma(1-iz)=0$ by using MMA functions... $\endgroup$ – José Antonio Díaz Navas Apr 15 '18 at 12:58
  • $\begingroup$ @JoséAntonioDíazNavas - Function is a given. x has to be pure imaginary to give a real argument to Gamma. Plot of the function suggests that the roots are near the negative integers. FindRoot verified that starting the search near a negative integer would converge to the integer. However, Gamma is undetermined (ComplexInfinity) at negative integers, so the roots only exist in the Limit. The last Mathematica statement verifies root for all negative integers. $\endgroup$ – Bob Hanlon Apr 15 '18 at 14:45
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Here is my proposal. The product of two gamma functions $\Gamma(z)$ and $\Gamma(w)$, with arguments satisfying the condition that $z+w$ is an integer, as in our case with $z=ix$ and $w=1-ix$, then:

$$\Gamma(z) \Gamma(1 - z) =\frac{\pi}{\sin(\pi z)}$$

So:

Reduce[Gamma[I x] Sin[I \[Pi] x]/\[Pi] == 0, x]

$$c_1\in \mathbb{Z}\land \left(\left(c_1\leq -1\land \left(x=2 i c_1\lor x=2 i c_1+i\right)\right)\lor \left(i x\notin \mathbb{Z}\land \left(x=2 i c_1\lor x=\frac{2 i \pi c_1+i \pi }{\pi }\right)\right)\right)$$

which includes your expected result.

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