B Does this help solve the Riemann Hypothesis?

[MevsEinstein]

TL;DR Summary

I created an equation with the real part of the input of the zeta function on the RHS and a complex expression on the LHS

Hello PF!

If $\Re (s)$ is the real part of $s$ and $\Im (s)$ is the imaginary part, then t is very easy to prove that $$\zeta (s) = \zeta ( \Re (s) ) \zeta ( \Im (s)i) - \displaystyle\sum_{n=1}^\infty \frac{1}{n^{\Re (s)}} [\displaystyle\sum_{k \in S, \mathbb{Z} \S = n} \frac{1}{k^{\Im(s)i}}]$$ Since the second term simplifies to $\zeta ( \Re (s) ) \zeta ( \Im (s)i) - \zeta (s)$. ($\mathbb{Z} \S = n$ is actually $\mathbb{Z}$ \$S$ = $n$). Now, solving for $\Re (s)$ gets us $$\zeta^{-1} ({\frac{\zeta (s) + \displaystyle\sum_{n=1}^\infty \frac{1}{n^{\Re (s)}}[ \displaystyle\sum_{k \in S, \mathbb{Z} \S = n} \frac{1}{k^{\Im (s)i}}]}{\zeta (\Im (s)i)}}) = \Re (s)$$ Setting $\zeta (s)$ to zero gives us the equation we were looking for. Now I am thinking, could this help solve the Riemann hypothesis? Or will this just spit out $\Re (s)$? I mean, the inverse zeta function is very complicated (page 43 of https://arxiv.org/pdf/2106.06915.pdf ), so it's hard to tell.

 

[fresh_42]

The short answer is a simple no.

And there is no such thing as an inverse zeta-function. Let us assume for a second that there is $\zeta^{-1}.$ Then $\zeta(s)=0$ means $s=\zeta^{-1}(\zeta(s))=\zeta^{-1}(0)$. However, there are infinitely many zeros of the zeta-function. How could that be just a single value $\zeta^{-1}(0)?$ The pre-image $\zeta^{-1}(\{0\})$ is an entire, infinite set. No way to make this a function.

You have to be careful with the zeta-function. E.g., $\zeta(-1)=-\dfrac{1}{12}$ but $1+2+3+\ldots \neq -\dfrac{1}{12}.$ I am not sure what you are doing here so I cannot comment your equations. If you want to even understand the Riemann hypothesis then you should study
https://www.physicsforums.com/insig...thesis-and-ramanujans-sum/#toggle-id-1-closed

 

[nuuskur]

To be fair, your first sentence beginning with

I created an equation..

is already alarming and strongly advises to ignore what follows.

 

[martinbn]

Now, solving for $\Re (s)$ gets us $$\zeta^{-1} ({\frac{\zeta (s) + \displaystyle\sum_{n=1}^\infty \frac{1}{n^{\Re (s)}}[ \displaystyle\sum_{k \in S, \mathbb{Z} \S = n} \frac{1}{k^{\Im (s)i}}]}{\zeta (\Im (s)i)}}) = \Re (s)$$

Ignoring all other possible problems, how is this solving for $\Re (s)$? It appears on both sides!

 

[MevsEinstein]

how is this solving for $\Re (s)$? It appears on both sides!

Oops, sorry about that.