#zetafunction — Public Fediverse posts

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

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

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

📢 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

📢 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

📢 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

📢 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

📢 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

🚀 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

#NumberTheory #Mathematics #ZetaFunction

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

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

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

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

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

Decades-Long Quest for the Irrational: A Breakthrough in Number Theory

In a stunning revelation, mathematicians have expanded upon Roger Apéry's groundbreaking proof of the irrationality of ζ(3), paving the way for a new era in number theory. This innovative approach not...

https://news.lavx.hu/article/decades-long-quest-for-the-irrational-a-breakthrough-in-number-theory

#news #tech #NumberTheory #IrrationalNumbers #ZetaFunction

@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

@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

@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

@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

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

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

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

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

@kat LOL
https://en.wikipedia.org/wiki/Basel_problem
#BaselProblem #ZetaFunction #RiemannZeta #maths #piday

Riemann zeta function \(\zeta(s)\) and \(\displaystyle\sum_{n=1}^\infty n=1+2+3+\cdots=-\dfrac{1}{12}\)

Have you ever heard that the sum of all natural numbers is \(-1/12\)?🤔 Of course not; this doesn't make sense in the usual sum, but using a summation method based on analytic continuation of the Riemann zeta function leads to the following result.

The Riemann zeta function is defined as:
\[\zeta(s)=\displaystyle\sum_{n=1}^\infty\dfrac{1}{n^s}=\dfrac{1}{1^s}+\dfrac{1}{2^s}+\dfrac{1}{3^s}+\cdots\]
for \(s\in\mathbb{C}\) such that \(\Re(s)>1\).
It can be extended to a meromorphic function with only a simple pole at \(s=1\), using analytic continuation and the following functional equation:
\[\zeta(1-s)=2^{1-s}\pi^{-s}\cos\left(\dfrac{\pi s}{2}\right)\Gamma(s)\zeta(s)\]
For \(s=2\), this gives \(\zeta(-1)=\displaystyle\sum_{n=1}^\infty n=-\dfrac{1}{2\pi^2}\zeta(2)=-\dfrac{1}{2\pi^2}\cdot\dfrac{\pi^2}{6}=-\dfrac{1}{12}\), which is a reason for assigning a finite value to the divergent sum/series (zeta function regularization). That is, \(\displaystyle\sum_{n=1}^\infty n=1+2+3+\cdots=-\dfrac{1}{12}\).
#RiemannZetaFunction #ZetaFunction #Riemann #DivergentSum #DivergentSeries #FiniteValue #ZetaRegularization #ZetaFunctionRegularization #NegativeFraction #MeromorphicFunction #AnalyticContinuation

There is this really awesome series on YouTube about the Riemann-Zeta function which is currently running and seems to want to explain all the mathematic objects related to it properly it would be awesome if you could show the video series some love while it's running. https://youtu.be/4bzSFNCiKrk
#mathematics #mathstodon #primenumbers #riemann #zetafunction

PRODUCTS OVER PRIME NUMBERS [2/2]:
\[\displaystyle\prod_{p\in\mathbb{P}}\left(1+\dfrac{1}{p(p+1)}\right)=\prod_{p\in\mathbb{P}}\dfrac{1-p^{-3}}{1-p^{-2}}=\dfrac{\zeta(2)}{\zeta(3)}=\dfrac{\pi^2}{6\zeta(3)}\]
\[\displaystyle\prod_{p\in\mathbb{P}}\left(1+\dfrac{1}{p(p-1)}\right)=\prod_{p\in\mathbb{P}}\dfrac{1-p^{-6}}{(1-p^{-2})(1-p^{-3})}=\dfrac{\zeta(2)\zeta(3)}{\zeta(6)}=\dfrac{315}{2\pi^4}\zeta(3)\]
#PrimeProducts #RiemannZetaFunction #EulerProduct #ZetaFunction #InfiniteProduct #NumberTheory

Me looking at a formula of :ramanujan: Ramanujan: Yeah, just do partial fractions of the infinite sum, telescope two of the terms, and look at the remaining value of the #ZetaFunction.

Me counting stitches on my #knitting for DPNs: 48 divided by 3 is 18, right?