Dear Professor, you can always write a comment containing timestamps + corrections on your videos and then pin it so everyone can see it. I can't imagine how exhausting it must be to redo the whole thing! As always, thank you very much for doing these lectures.
The reparametrisation for Dirichlet series means taking the sequence b_n = a_n/n^(s_0). This way its Dirichet series is b_1/1^s + b_2/2^s + ... = a_1/1^(s_0 + s) + a_2/2^(s_0 + s) + ... with Re(s)>0. Saying that a_1 + a_2 + a_3 + ... in case s_0 = 0 is convergent is the analogous of saying b_1 + b_2 + b_3 + ... is convergent in the general case, which is our hypothesis.
could you write what the mistake was in the description (or at least a timestamp to where the differences are) so that people who watched the original don't need to rewatch the whole video again? also, I think it would be fine in future videos just to include corrections in the video description (maybe with a note in the video title), rather than spending all the extra time re-recording a new video. thanks for all the lectures!
9:14 Unless one was introduced to quantifiers at a tender age of 12 (by Zbigniew Semadeni, known ia for Extremally Disconnected Space and once responsible for mathematical education, from his radio talks on how mathematics should be taught) and nudged to use this notation all the way since 15 y.o. In fact, I have an inner anxiety of not understanding the formulation until I’ve written it in quantifier form.
I'm sure I'm missing something very obvious, but the series 1-1+1-1+.. is not convergent. It is bounded sure, but not convergent. So how can we use the result that a1+a2+a3+... is convergent implies a1/1^s + a2/2^s + ... is convergent, to show that 1/1^s - 1/2^s + 1/3^s -... is convergent for Re(s) > 0 ?
He claimed the theorem at 28:30 held for any s_0 and in the proof took s_0 = 0 for simplicity. Since we already know 1/1^s - 1/2^s + 1/3^s -... is convergent for s > 0 and s real, the result follows for any s in C with Re(s) > 0.
Yes, you're correct. It converges uniformly on any open of the form Re>r, where r > 1. The reason of working only "locally" is that Zeta has a pole at 1, so it diverges by getting closer and closer, but if you pick any r > 1, then it's bounded on Re(s)>r (do not Liouville's theorem be source of confusion, it requires holomorphy on the whole complex plane, which is not our case). Boundedness is sufficient to show 1/1^s + 1/2^s + 1/3^s + ... is uniformly convergent. There is a very neat way to show this by using convergence in the space of continuous functions.
Divergence is dual to convergence. Integration is dual to differentiation. Syntropy (convergence) is dual to increasing entropy (divergence) -- the 4th law of thermodynamics! There is a dual process to that of increasing entropy in physics.
Dear Professor, you can always write a comment containing timestamps + corrections on your videos and then pin it so everyone can see it. I can't imagine how exhausting it must be to redo the whole thing! As always, thank you very much for doing these lectures.
Typos (do not affect the conclusion)
33:05 should be s instead of 1/s
33:18 should be Re(s) instead of 1/Re(s)
Yes, thanks. I seem to have reached to point where fixing errors introduces an equal number of new errors.
@@richarde.borcherds7998 conservation of minor errors
@@paul_tee "99 little bugs in the code, 99 little bugs! Take one down, pass it around... 137 little bugs in the code!"
The reparametrisation for Dirichlet series means taking the sequence b_n = a_n/n^(s_0). This way its Dirichet series is b_1/1^s + b_2/2^s + ... = a_1/1^(s_0 + s) + a_2/2^(s_0 + s) + ... with Re(s)>0.
Saying that a_1 + a_2 + a_3 + ... in case s_0 = 0 is convergent is the analogous of saying b_1 + b_2 + b_3 + ... is convergent in the general case, which is our hypothesis.
could you write what the mistake was in the description (or at least a timestamp to where the differences are) so that people who watched the original don't need to rewatch the whole video again?
also, I think it would be fine in future videos just to include corrections in the video description (maybe with a note in the video title), rather than spending all the extra time re-recording a new video.
thanks for all the lectures!
The corrected error was in the definitions of uniform convergence and point-wise convergence. He originally left out a quantifier
9:14 Unless one was introduced to quantifiers at a tender age of 12 (by Zbigniew Semadeni, known ia for Extremally Disconnected Space and once responsible for mathematical education, from his radio talks on how mathematics should be taught) and nudged to use this notation all the way since 15 y.o. In fact, I have an inner anxiety of not understanding the formulation until I’ve written it in quantifier form.
Great Lecture!!
Love this video! Thank you
Dear Professor, Can you do online course on measure theory.
I'm sure I'm missing something very obvious, but the series 1-1+1-1+.. is not convergent. It is bounded sure, but not convergent. So how can we use the result that a1+a2+a3+... is convergent implies a1/1^s + a2/2^s + ... is convergent, to show that 1/1^s - 1/2^s + 1/3^s -... is convergent for Re(s) > 0 ?
I also do not understand this, and it is quite frustrating.
He claimed the theorem at 28:30 held for any s_0 and in the proof took s_0 = 0 for simplicity. Since we already know 1/1^s - 1/2^s + 1/3^s -... is convergent for s > 0 and s real, the result follows for any s in C with Re(s) > 0.
Wait, so the zeta function does not converge uniformly in general for Re(s) > 1 but it does locally...? Or am I getting it wrong?
Yes, you're correct. It converges uniformly on any open of the form Re>r, where r > 1. The reason of working only "locally" is that Zeta has a pole at 1, so it diverges by getting closer and closer, but if you pick any r > 1, then it's bounded on Re(s)>r (do not Liouville's theorem be source of confusion, it requires holomorphy on the whole complex plane, which is not our case). Boundedness is sufficient to show 1/1^s + 1/2^s + 1/3^s + ... is uniformly convergent. There is a very neat way to show this by using convergence in the space of continuous functions.
Can most of these technicalities be found in Tom Apostol's book on Dirichlet series and modular functions?
You assume that the sum of the a_n converges, but isn't our sequence a_n = (-1)^(n+1)?
or does our reparameterization change the sequence so that its sum does converge?
Yes, the reparameterization so that s_0 = 0 changes the sequence. I should have made this clearer.
yee
Goldilocks convergence!
Divergence is dual to convergence.
Integration is dual to differentiation.
Syntropy (convergence) is dual to increasing entropy (divergence) -- the 4th law of thermodynamics!
There is a dual process to that of increasing entropy in physics.