#riemannzetafunction — Public Fediverse posts

🎉🥳 Ah, the Riemann zeta function: the ultimate math party trick that's here to explain why primes are the wallflowers of numbers! Apparently, someone thought it was crucial to complicate Euclid's simple "infinite primes" concept by adding a function that ensures only the nerdiest of minds could understand it. 📈🔢
https://hidden-phenomena.com/articles/rh #RiemannZetaFunction #MathParty #PrimeNumbers #InfinitePrimes #NerdyMath #HackerNews #ngated

What does the Riemann zeta function have to do with the distribution of primes?

https://hidden-phenomena.com/articles/rh

Comments: https://news.ycombinator.com/item?id=48952713

#HackerNews #RiemannZetaFunction #PrimeDistribution #Mathematics #NumberTheory #MathExploration

“If I were to awaken after having slept for a thousand years, my first question would be: Has the Riemann hypothesis been proven?” – David Hilbert

#RiemannHypothesis #DavidHilbert #RiemannZetaFunction #ZetaFunction #UnsolvedProblem #MathHistory #Hilbert #Riemann

On Pi Day 2025, as you might recall, I introduced you (more or less) to 3Blue1Brown. Also known as Grant Sanderson.

If there's any better source of animated math presentations, than Mr. Sanderson, I'm unaware of it.

Key word: "animated." In addition, he is a warm, enthusiastic teacher. And of course he knows his stuff well—how else could he have made the truly fantastic animations?

I haven't watched them all yet. But so far, I especially like 2016's BUT WHAT IS THE RIEMANN ZETA FUNCTION? VISUALIZING ANALYTIC CONTINUATION and 2019's DIFFERENTIAL EQUATIONS, A TOURIST'S GUIDE | DE1.

I may never again be able to ponder some ideas they convey, without seeing those animations in my head. They're that perfect.

So give those two videos a try, if you're comfortable enough with the one I had shared on Pi Day...if now you want a couple which are more challenging. Maybe unforgettable, too.

#3Blue1Brown
#RiemannZetaFunction
#DifferentialEquations
#learning

https://mindly.social/@setsly/114161481212443838

https://www.youtube.com/watch?v=sD0NjbwqlYw

https://www.youtube.com/watch?v=p_di4Zn4wz4

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

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

Tail of the zeta function \(\zeta(s)\):

\[\displaystyle\sum_{n\geq N}\dfrac{1}{n^s}=\dfrac{N^{1-s}}{s-1}+\dfrac{1}{2N^s}-s\int_N^\infty\left(\{t\}-\dfrac{1}{2}\right)t^{-s-1}\ \mathrm{d}t\]

where \(\{t\}\) denotes the fractional part of \(t\).

#ZetaFunction #RiemannZetaFunction #TailOfFunction #Function #Maths #NumberTheory #Analysis

@futurebird
That mathematicians are adding infinite diverging series of positive numbers and coming up with finite negative numbers as the sum, but if I had written that down as an answer in my exams in secondary school, it would have been marked as wrong
https://youtu.be/YuIIjLr6vUA
#YouTube #RiemannZetaFunction #mathematics #maths #math #mathstodon

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