Knowing the identity theorem Analytic Continuation seems basically trivial. All other sources I have found make it seem much more complicated than "find the powerseries. You're done". Why do we have the extra steps of going through many regions of convergence? Is this to find a path connected domain?
Thank you very much! I always try to make topic transparent. The Taylor series only exist in a disc. So you might need many discs to cover the whole domain.
@@brightsideofmaths I'm not sure I understand that. Do you have a video explaining why the Taylor series only exists on a disk? If the disks overlap, isn't that the same series by the identity theorem? And I should be able to convert the Taylor series at a point to another power series at an arbitrary point. Doesn't that remove the necessity of a disk?
@@brightsideofmaths right. So if I have two overlapping disks for which the coefficients coincidence in the overlap, which is an open set, then they must coindide everywhere in the domain by the identity theorem. Or did I understand that incorrectly?
I have a question about the example: It is a requirement to f(x) to be a smooth function? As example, think in this function f(x)=1/4*(1-x/2+|1-x/2|)^2 Does the theorem still holds?
Great Video. Following my comments on your last video. So I like to repeat my question. How can I say this in exact math terms: Assume zeta (s)=zeta(1-s). Therefore because zeta (s)-zeta(1-s) =0 the Sum 1/n^s - Sum 1/n^(1-s) or sum(( 1/n^s) -1/n^(1-s)) converge to zero in the critical strip. I want to use the identity theorem. The challenge is power series 1/n^s is a divergence in the critical strip. Please advise
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I'm loving this series so far, thank you so much for your work !!
Glad you enjoy it!
Knowing the identity theorem Analytic Continuation seems basically trivial.
All other sources I have found make it seem much more complicated than "find the powerseries. You're done".
Why do we have the extra steps of going through many regions of convergence? Is this to find a path connected domain?
Thank you very much! I always try to make topic transparent. The Taylor series only exist in a disc. So you might need many discs to cover the whole domain.
@@brightsideofmaths I'm not sure I understand that. Do you have a video explaining why the Taylor series only exists on a disk?
If the disks overlap, isn't that the same series by the identity theorem?
And I should be able to convert the Taylor series at a point to another power series at an arbitrary point. Doesn't that remove the necessity of a disk?
Same series means that the coefficients coincide. @@narfwhals7843
@@brightsideofmaths right. So if I have two overlapping disks for which the coefficients coincidence in the overlap, which is an open set, then they must coindide everywhere in the domain by the identity theorem. Or did I understand that incorrectly?
Yes! But don't forget the expansion point for the series.@@narfwhals7843
So, is the result in this video the same thing as analytic continuation, or am I mixing things up?
It is :)
Very well described!
Glad you think so! :)
Great video. Thank you very much. Any video regarding the uniqueness of analytic continuation?
Yes, this video here :)
I have a question about the example: It is a requirement to f(x) to be a smooth function?
As example, think in this function
f(x)=1/4*(1-x/2+|1-x/2|)^2
Does the theorem still holds?
Yes, we need smooth functions here.
can you explain a bit more why if D is open set, then accumulation point will be in D?
What are other possibilities for the accumulation point? And what does "open" mean? :)
Great Video. Following my comments on your last video. So I like to repeat my question. How can I say this in exact math terms: Assume zeta (s)=zeta(1-s). Therefore because zeta (s)-zeta(1-s) =0 the Sum 1/n^s - Sum 1/n^(1-s) or sum(( 1/n^s) -1/n^(1-s)) converge to zero in the critical strip. I want to use the identity theorem. The challenge is power series 1/n^s is a divergence in the critical strip. Please advise
Thank you very much! I guess my comment section is not the right place to discuss such questions.
@@brightsideofmaths .Sure I Just emailed you.