A question implying Ganma function

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A question implying Ganma function....

Let define the function

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

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.
 

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Hurkyl
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Why would it be?
 
  • #3
shmoe
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I don't know why you insist on using bizarre non-standard notation. What you call [tex]Z(x)[/tex] is what's usually called [tex]\chi(1-x)[/tex], though it's usually given as [tex]\chi(s)=2^s\pi^{s-1}\Gamma(1-s)\sin{s\pi/2}[/tex] as "x" is almost never used to denote a complex variable. And why not use the usual [tex]|s|[/tex] 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 [tex]|\chi(s)|=|\chi(1-s)|=1[/tex] then must we have the real part of s equal to 1/2?"

The answer is no. [tex]\chi(-18)=0[/tex] and [itex]\chi(-19)= \frac{1856156927625}{8\pi^{20}}=26.45...>1[/itex] so we have a real number [tex]-19<\alpha<-18[/tex] where [tex]\chi(\alpha)=1[/tex] ([tex]\chi[/tex] is continuous on (-19,-18)). Since [tex]\chi(s)\chi(1-s)=1[/tex] we necessarily have [tex]\chi(1-\alpha)=1[/tex] so your conditions are satisfied, yet [tex]\alpha\neq 1/2[/tex]. If you examine the behavior of [tex]\chi(s)[/tex] 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 [tex]|\chi(s)|=|\chi(1-s)|=1[/tex]. Apply the reflection principle and use [tex]\chi(s)\chi(1-s)=1[/tex].
 

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