Zeros of Riemann Zeta Function- Euler Product and Functional Equation

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xfootiecrazeesarax
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#1
Report Thread starter 4 years ago
#1
1. The problem statement, all variables and given/known data

Question

Use the functional equation to show that for :

a) k \in Z^+ that  \zeta (-2k)=0
b) Use the functional equation and the euler product to show that these are the only zeros of \zeta(s) for Re(s)<0 . And conclude that the other zeros are all located in the critical strip: 0\leq Re(s) \leq 1 . Show that these are symmetric about s=1/2

2. Relevant equations

Euler product:  \zeta(s)=\Pi^{p}\frac{1}{1-p^{-s}} defined for Re(s)>1

Functional equation: \zeta(s)=\chi(s)\zeta(1-s)

where \chi(s)=2^s\pi^{s-1}\sin(\frac{\pi s}{2}) \Gamma(1-s)

Also have Z(s)=\pi^{\frac{-s}{2}} \Gamma(\frac{s}{2}) \zeta(s) which we know has simple poles at s=0, 1

So from this we can see that the  \Gamma (s) that gave poles for Z(s) gives arise to the zeros of \zeta(s) at s=-2k so that's the trivial zeros done.

3. The attempt at a solution

From the Euler product define for Re(s) > 1 we can see that  \zeta (s) does not vanish for  Re(s) >1 .

I think to make the rest of the conclusions about the critical strip and being symmetrically distributed about Re(s)=1/2 I need to use the functional equation.

But I'm not sure what to do... I want to look where it is positive and negative I guess. But with sin and \Gamma which are positive and negative at different ranges of s I'm not really sure what to do.. any hint greatly appreciated.
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DJMayes
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#2
Report 4 years ago
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Hi,

I'm not sure what bits of this you've done and haven't done. Your attempt at a solution looks like it's on the right lines, but it's a little confusingly ordered.

First thing you should do (as you have done) is show that  \zeta(s) \neq 0 for  \Re(s)>1 . Then, from the functional equation:

\zeta(s) =2^s\pi^{s-1}\sin(\frac{\pi s}{2}) \Gamma(1-s)\zeta(1-s)

I believe you can deduce that for  \Re(s)<0 ,  \zeta(s)=0 \iff \sin(\frac{\pi s}{2})= 0 .

I think the symmetry of zeroes arises from plugging  s=\frac{1}{2}+\delta into the functional equation and proving that the  \sin and  \Gamma functions are non-zero within a suitable range.
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