Pustam | पुस्तम | পুস্তম🇳🇵The Riemann Hypothesis: Past, Present and a Letter Through Time
https://arxiv.org/abs/2602.04022
#RiemannHypothesis #Riemann #Conjecture #Hypothesis #ZetaFunction #RiemanZetaFunction #RH #MillenniumPrizeProblems #PrimeNumbers
Khazri bouzidi fethi📢 NEW WORK: "Origin of the constant B ≈ 0.486 in the distribution of prime numbers"
I study the Gaussian decay of a prime-related sum: S(s) ~ e^{-Bs²}. The constant B ≈ 0.486 emerges from three perspectives:
🔹 Arithmetic: B ≈ Σ 1/π(eⁿ) − 1
🔹 Spectral (under RH): B linked to zeros of ζ(s)
🔹 Geometric: the base e is optimal
📄 Full paper: [PDF_LINK]
👤 My other work: [ACADEMIA_LINK]
Open question: Which approach seems most enlightening to you?
#NumberTheory #PrimeNumbers #ZetaFunction #Maths #Mathematics #Research #RiemannHypothesis #Preprint #PrimeDistributionrime_numbers
SuiQing🚀 New preprint out on Cambridge Open Engage:
“A Fully Invertible Global Analytic Model of the Riemann Zeta Function”
I introduce Sui Theory to construct a globally analytic and invertible framework for ζ(s) in Hardy space H²(ℂ⁺).
🔗 Cambridge Open Engage - https://bit.ly/461M67F
A FULLY INVERTIBLE GLOBAL ANALYTIC MODEL OF THE RIEMANN ZETA FUNCTION
Abstract. We introduce Sui Theory and present a globally invertible analytic model for functions in the Hardy space H2(C+) [5, 15]. The construction is based on an atomic system {Φn}, where each atom has the intrinsic properties of admissibility, completeness, stability, and modulator regularity [6, 3]. These properties are built into the atomic definition itself, rather than imposed externally as assumptions. The resulting family {Φn}forms a Riesz basis of H2(C+) [2], yielding a unique and stable expansion f(s) = ∞ cnΦn(s), f ∈H2(C+). n=1 This expansion is globally invertible: coefficients can be recovered from analytic data, and conversely the synthesis of coefficients reconstructs the function uniformly on compact subsets of the domain [12]. The model provides a new framework for analyzing analytic functions with growth and symmetry constraints, and suggests applications to L-functions and other areas of analytic number theory [13, 1, 4].
One of the best things I saw this week: a paper uncovering alien signals in the Riemann Zeta function. April Fools always brings peak creativity. 😅
#Riemann #Zeta #ZetaFunction #RiemanZetaFunction #AprilFool #AprilFools #AprilFoolsDay #Creativity #Math #Maths #NumberTheory #PeakCreativity #Nerd #Nerds #Humor #Humour #Alien #AlienSignals
ƧƿѦςɛ♏ѦਹѤʞ@zvavybir
It does diverge. It has no sum.
However, the uniquely valued #Riemann #ZetaFunction can be analytically continued into the left half-plane where we find zeta(-1)=-1/12 (which 'looks like' 1+2+...). #Cesàro #summation will get you part of the way there also, and, as you say, yields the same result; presumably due to some ultimate cosmic logical rightness :-)
I very strongly recommend BP's superb exposition of this issue
https://www.youtube.com/watch?v=YuIIjLr6vUA
#maths #AnalyticContinuation #Ramanujan
It does diverge. It has no sum.
However, the uniquely valued #Riemann #ZetaFunction can be analytically continued into the left half-plane where we find zeta(-1)=-1/12 (which 'looks like' 1+2+...). #Cesàro #summation will get you part of the way there also, and, as you say, yields the same result; presumably due to some ultimate cosmic logical rightness :-)
I very strongly recommend BP's superb exposition of this issue
https://www.youtube.com/watch?v=YuIIjLr6vUA
#maths #AnalyticContinuation #Ramanujan
Numberphile v. Math: the truth about 1+2+3+...=-1/12
xenonrealityBrand new drop!!! *Bwaaann bwaan bwwaahhh* it's how to solve the #Riemann #zetafunction w an explanation (but not the solve itself) #math
#milleniumpizeproblems
This time just the link. I'm testing to see if that actually shows the views on the metrics
How to, most likely, solve the #riemann #zeta function w an explanation
Interesting integral! #Challenge
\[\displaystyle\int_0^1\dfrac{\ln (x)\ln(1-x)}{x(1-x)}\operatorname{Li}_2(x)\ dx=5\zeta(2)\zeta(3)-8\zeta(5)\]
Where \(\operatorname{Li}_2(x)\) denotes the dilogarithm (or Spence's function), and \(\zeta(x)\) denotes the Riemann zeta function.
#ZetaFunction #Zeta #Dilogarithm #SpenceFunction #Polylogarithm #Integral #DefiniteIntegral #Integration #Integrals #RiemannZetaFunction #Logarithm #Function #LogarithmicFunction
Basel Problem: \(1/1^2 + 1/2^2+ 1/3^2 + 1/4^2+ \cdots = ?\)
Some of the brightest mathematicians, like Newton, Leibniz and (Jacob) Bernoulli, struggled with this simple series.
It was only in \(1734\) that Euler, at the age of \(27\), found that this infinite series converged to \(\pi^2/6\).
\[\displaystyle\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots=\dfrac{\pi^2}{6}\]
#Euler #Leibniz #Newton #Bernoulli #JacobBernoulli #LeonhardEuler #BaselProblem #ZetaFunction #MathHistory
A representation of \(18\) using an analytic continuation of the Dirichlet series and the numbers \(0, 1,2,3,4,5,6\).
\[18=\left|\dfrac{1}{\zeta^2(0)}\left(\dfrac{\mathcal{H}(-6)}{\zeta(-5)}-\dfrac{\mathcal{H}(-2)}{\zeta(-1)}\right)\dfrac{\mathcal{H}(-4)}{\zeta(-3)}\right|\]
where \(\displaystyle\zeta(z)=\sum_{n\geq1}\dfrac{1}{n^z}\) denotes the Riemann zeta function, and \(\displaystyle\mathcal{H}(z)=\sum_{n\geq1}\dfrac{H_n}{n^z}\) denotes the harmonic zeta function.
#ZetaFunction #RiemannZetaFunction #HarmonicZetaFunction #AnalyticContinuation #DirichletSeries #Series #Numbers #Zeta #Harmonic #HarmonicNumbers #Representation #Function #Expression
