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

mindsets (@[email protected])

For #PiDay, allow me to share a computer-animated poem/video. It's as splendid as the equation it honors. Hey, you didn't think the only videos I watch for fun are about Marvel movies, did you??? Remember what I had said was the best nonfiction book of 2024. https://www.youtube.com/watch?v=zLzLxVeqdQg

Mindly.Social

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
Numberphile v. Math: the truth about 1+2+3+...=-1/12

YouTube
The Riemann Hypothesis, explained - Cantor’s Paradise

You remember prime numbers, right? Those numbers you can’t divide into other numbers, except when you divide them by themselves or 1? Right. Here is a 3000 year old question: Present an argument or…

Cantor’s Paradise
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
SOME PRODUCTS OVER PRIME NUMBERS [1/2]:
\[\displaystyle\prod_{p\in\mathbb{P}}\dfrac{1}{1-p^{-s}}=\prod_{p\in\mathbb{P}}\left(\sum_{k=0}^\infty\dfrac{1}{p^{ks}}\right)=\sum_{n=1}^\infty\dfrac{1}{n^s}=\zeta(s)\]
\[\displaystyle\prod_{p\in\mathbb{P}}\dfrac{p^s+1}{p^s-1}=\prod_{p\in\mathbb{P}}\left(1+\frac{2}{p^s}+\frac{2}{p^{2s}}+\cdots\right)=\sum_{n=1}^\infty\dfrac{2^{\omega(n)}}{n^s}=\dfrac{\zeta(s)^2}{\zeta(2s)}\]
#PrimeProducts #EulerProduct #RiemannZetaFunction #ZetaFunction #InfiniteProduct