11b - Riemann Zeta Function Has Only One Pole

In this video we show that the Riemann Zeta function has only one pole in the domain σ＞0.

This would be a difficult proof, but thanks to a clever idea by user @leoli1 at math stackexchange the proof isn't difficult at all.

Slides:

Blog:

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The following pieces of music are freely licensed for use in this video.

[sil]

You would have to prove that ln2/ln3 is irrational. Instead you might just write a*ln3=b*ln2 and taking exponential e^x on both sides to get 3^a=2^b, which is a simple equation in integers and fundamental theorem of arithmetic implies it has no solution (except for a=b=0 of course). It is essentially equivalent to proving the irrationality.

[larenmunday]

Wow! You are a brilliant teacher! This makes some tricky things seem so simple because of your logical and clear method of explanation. Thank you 👍

[datamoon]

The whole "could be" discussion of the poles is unclear to me when you equate the two expressions (ln 2 and ln 3). Why couldn't there be some poles contributed by the ln 2 expression and other poles coming from the ln 3 expression? And presumably, there could be similar ln x expressions that could all contribute non-overlapping poles as well? I don't get what allows you to equate them. Once they're equal, it all follows quite nicely, but I'm missing the logic in the step, and your meaning seems to be hiding behind the words "could be". Can you clarify?

Excellent and amazing stuff, btw. It's Sunday night and I have basically spent the entire weekend watching this whole series straight through from the start. This point is the only one so far that hasn't become clear to me by the end of the video. You are an excellent teacher and obviously know what you are talking about, even if I sometimes need to hear what you mean rather than what you say ;) Thank you so much for providing this.

[benheideveld]

Actually ζ(s) converges continuously
to +∞ for s->1 along any complex curve. I would say it is a very well behaved function, except that its analytical continuation becomes nonsensically zero for all even negative whole numbers. That is odd:-)