A question implying Ganma function

[eljose]

A question implying Ganma function...

Let define the function

Z(x)=\Gamma(x)2^{1-x}\pi^{-x}cos(x\pi/2)

then my question is if [Z(x)]=1=[Z(1-x)] with [z]=a^2+b^^2 then Re(x)=Re(1-x) with []=modulus of the function.

 

[Hurkyl]

Why would it be?

 

[shmoe]

I don't know why you insist on using bizarre non-standard notation. What you call Z(x) is what's usually called \chi(1-x), though it's usually given as \chi(s)=2^s\pi^{s-1}\Gamma(1-s)\sin{s\pi/2} as "x" is almost never used to denote a complex variable. And why not use the usual |s| to denote the modulus of a complex variable? I swear you're trying to avoid widespread conventions on purpose.

With these notation changes, you seem to be asking "if |\chi(s)|=|\chi(1-s)|=1 then must we have the real part of s equal to 1/2?"

The answer is no. \chi(-18)=0 and \chi(-19)= \frac{1856156927625}{8\pi^{20}}=26.45...>1 so we have a real number -19<\alpha<-18 where \chi(\alpha)=1 (\chi is continuous on (-19,-18)). Since \chi(s)\chi(1-s)=1 we necessarily have \chi(1-\alpha)=1 so your conditions are satisfied, yet \alpha\neq 1/2. If you examine the behavior of \chi(s) as s goes to negative infinity along the real axis you can show this sort of thing happens infinitely often.

The converse is true though, if the real part of s=1/2 then |\chi(s)|=|\chi(1-s)|=1. Apply the reflection principle and use \chi(s)\chi(1-s)=1.