Euler's Insight - A Function That Generates Values of the Riemann Zeta Function

In this video I show a particular identity which can be used to solve the Basel problem, as well as find exact forms for the Riemann zeta function evaluated at any positive even number.

[prwf]

What a great proof of the Basel problem! Very clever and easy to understand.

[RSLT]

Absolutely fascinating! Looking forward to watching more videos like this in the future!

[cameronspalding]

The sum for 1/(1-z) is given by 1+z+z^2+z^3+… .
Note that when n is a positive integer then 1/(1-z) - (1+z+z^2+…+z^(n-1))=z^n/(1-z), this error is bounded for |z|<=1 (except when z=1). The fact that this error is bounded means that when we integrate the sum it is convergent for
|z|<=1 and z<>1, thus
-log(1-z)= z+z^2/2+z^3/3+z^4/4+…

[glauberk]

I just found your channel, and your content is amazing! This demo is incredible and new to me. However, I noticed you considered 0<x<2π but used x=0 to evaluate zeta(2). The end result is correct, but it might be more rigorous to choose x=π and x=π/2. Keep up the great work!

[cndr]

Thank you for this presentation, you're a great pedagogue and i'm very grateful to you. While i'm not a mathetician at all, at 12:55, i'd suggest you to choose x=Pi, instead of x= - inf. That way we could remember that semi-harmonics series equals log(2). It seems to me a good idea but without certitude that it's right 😅 in anycase, thank you again.

[stevenismart]

Does anyone know of any explanations for that convergence of that powerseries? It would be interesting to see how that works

[quite_unknown_]

Are you sure all your steps in the beginning are justified? By that I mean, are you allowed to switch the summation and derivative without proving uniform convergence?

[williammartin]

What is the app you used to graph things at the beginning of the video

[cameronspalding]

Im(-log(1-e^ix))=sum_n=1 to infinity of sin(nx)/n for all real x (except when cos(x)=1 iff x is an integer multiple of 2*pi).

[satyam-isical]

Looks like he's teaching in a spaceship 😮

[WielkiKaleson]

Man, you are being somewhat sloppy (in a physicist-like way), but sounds like you're really cheating :-). I'd say that the first sum is OK (not 100% obvious, but you we have to start somewhere and leave some proofs to be found elsewhere). So, once we believe the first sum is OK, there is standard trick with adding in front lim(a -> 0^+) and multiplying by a taming factor exp(-an). The other way out is to just silently ignore a problem that has a well-known solution.