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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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- Thread starter eljose
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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.

- #2

Hurkyl

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Why would it be?

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shmoe

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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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