A Possible Proof Of The Riemann Hypothesis
Creators: name: KAPLAN, Yunus-Emre
Publisher: name: Zenodo
Year: 2025
Subjects: Riemann Hypothesis
Asymptotic Analysis
Prime Numbers
Imbricated Quadratic Equations
Integrals
Zeta Function
Eta Function
Taylor Expansion

description: This study is my approach to the Riemann Hypothesis, which potentially provides elements for a proof of it, or even an actual proof.

I start by breaking down the equality Eta(s)=0 (which is implied by the nullness of the continued Zeta(s)) by expressing |Eta(s)|^2=0 as Re(Eta(s))^2 + Im(Eta(s))^2 = 0, expressed with cosine and sine functions, then I use trigonometric identities to further break down the sum into a sum of squares purely relying on Re(s) and a double sum relying on both Re(s) and Im(s).

The emergence of the sum of squares sum(1/k^(2*Re(s)),k=1..n) is what shows us we're on the right path, for we can compare it with (n^(1-2*Re(s))-1)/(1-2*Re(s)) if Re(s)!=1/2 or ln(n) if Re(s)=1/2, and since we want to prove Re(s)=1/2, that's a very good sign to see 1-2*Re(s) appear.

Using the bounded nature of the cosine function, I reformulate the equality as a problem of quadratic equations, to get rid of the dependence on Im(s) as well as the oscillations it induces,

I then consider Re(s) as a map a[n] of complex numbers tending to a real number (n being the number of terms in Zeta(s) or Eta(s)), rather than a fixed value (since we deal with infinity),

It has already been proved that for all the nontrivial zeros, their real part is in ]0,1[,

So, I then use integrals and asymptotic analysis to observe three cases of asymptotic behaviours:



when a[n] converges to a real number in ]0,1/2[

when a[n] converges to a real number in ]1/2,1[

when a[n] converges to 1/2


Then I conclude that the only logically consistent case is when a[n] converges to 1/2.

Actually, I even find that the real part has to be 1/2 + o(1/ln(n)²), and that it is actually the imaginary part that has a slower (so more prominent) rate of convergence i*k*Pi(b+(ln(n)+4-gamma)/ln(n)²) where b=Im(s) once n has tended to infinity.

NOTES:

- I have yet to specify in my paper (as I did in the paragraph right before this one) that the real part is 1/2 + o(1/ln(n)²), not necessarily a plain 1/2, this is the complete necessary condition on the real part, and as n tends to infinity we are sure it is worth 1/2. I plan to do it as soon as possible.

- I noticed have yet to change some things again, in the part where I sum up the expression of the sums based on the limit of a[n], I forgot to precise that p->infinity for the 1/2 case, for the sum of squares to be = gamma+ln(n)(1+sum of correction terms up to the p-th order) + o(1) .Because ln(n)(1+sum of correction terms up to the p-th order) is equal to (n^(-2*epsilon[n]) - 1)/(-2*epsilon[n])) only as p->infinity.I plan to do it as soon as possible as well.

- In the upcoming version, I may also start the paper by directly writing s[n]=a[n]+i*b[n], n being the order of truncation of the Zeta series, instead of starting to consider Re(s) as a map a[n] only at the 3rd part.

In the current version, I aimed to make every statement as accurate as possible.

In the previous versions:

- I had completed my reasoning regarding some conditions I rely on, namely the condition ensuring that the difference between the sums and the corresponding integrals be a real in [0,1], and specifying that it is the condition Re(s) in ]0,1[ that allow us to replace r[n] by its asymptotic equivalent -1/n and (1-r[n]) by 1 in the expression of the limit obtained as a conclusion of the part with quadratic equations

- I had also made changes on the way I handle Taylor expansions, and deleted redundant parts.

I keep making corrections (and thus new versions) as I spot errors, ambiguous expressions and incomplete statements.

I'm not far from the final version (I hope).

Feedback would be appreciated at: emrekaplan1978@gmail.com

Despite the quite consequent number of downloads, I've received no feedback so far, positive or negative, despite my email address being public.So a review would really be most welcomed.