Author Topic: A question about zeroes of the Zeta function  (Read 2440 times)

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Offline Xeda112358

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A question about zeroes of the Zeta function
« on: October 28, 2014, 01:24:11 pm »

The problem:
Is there an ##s\in \mathbb{C}## such that ##\zeta(s)=\zeta(2s)=0##? If so, what is ##\lim_{z\rightarrow s}{\frac{\zeta(s)^{2}}{\zeta(2s)}}##?



Okee dokee, I know this is getting dangerously close to the Riemann Hypothesis, but I swear that wasn't my original intent!


I was playing around with infinite sums and products and I wanted to figure out the sum form of ##\prod_{p}{\frac{1+p^{-s}}{1-p^-s}}##. I started by knowing that ##\prod_{p}{(1-p^{-s})}^{-1} = \zeta(s)##, and from previous work, ##\prod_{p}{\sum_{k=0}^{n-1}{(p^{-s})^{k}}} = \frac{\zeta(s)}{\zeta(ns)}##). From that, I knew ##\prod_{p}{1+p^{-s}} = \frac{\zeta(s)}{\zeta(2s)}##, thus I know the product converges when ##\zeta(2s) \neq 0, \zeta(s) \in \mathbb{C}##. I knew convergence wasn't an issue (for the most part) so I expanded the product some (a reversal of Euler's conversion from ##\zeta(s) = \sum_{k=1}^{\infty}{k^{-s}} = \prod_{p}{(1-p^{-s})}^{-1}##) and obtained:

##\prod_{p}{\frac{1+p^{-s}}{1-p^-s}}= \sum_{k=1}^{\infty}{2^{f(k)}k^{-s}}##,


 where ##f(k)## is the number of prime factors of k. It turns out that this has been known since 1979 and after I had access to the internet, I figured out the typical symbol used for my ##f(k)## in this context is ##\omega(k)##. So that was cool, and I could write that sum and product in terms of the zeta function as ##\frac{\zeta(s)}{\zeta(2s)}\zeta(s) = \frac{\zeta(s)^{2}}{\zeta(2s)}##, so do with that what you will (I tried to use it to find the number of prime factors of a number n, but I didn't get anywhere useful). What I decided to pursue was when this function was zero. As long as ##\zeta(s)=0, \zeta(2s)\neq 0##, we know that it is 0, but I haven't yet figured out if there is ever an instance where ##s\in \mathbb{C}, \zeta(s)=\zeta(2s)=0##. If there is such an s, then the Riemann Hypothesis is false. However, if such an s does not exist, this says nothing about the Riemann Hypothesis :P .
Spoiler For Why the existence of an s would prove RH false:
It is simple: RH says that ##\zeta(s)=0## if and only if ##Re(s)=\frac{1}{2}##. So if ##\zeta(s)=\zeta(2s)=0##, assuming RH is true, then ##Re(s)=Re(2s)=2Re(s)##, which would imply s=0, a contradiction to the Riemann Hypothesis which we just assumed was true!


As well, if there is such an s, I wanted to know if it could be one of those removable discontinuity thingies.


EDIT: fixed a tag and an incorrect restatement of the RIemann Hypothesis :P