Missing details on the divergence of the series for |z| > R, when 0 1/|z| for infinitely many n. So |a_n z^n| > 1 for infinitely many n. Finally recall that a requirement for the series to converge is that the terms of the series convergence to 0. So we obtain divergence.
This is probably a dumb question, but when you apply the root test and say "this means |a_n|^(1/n) > 1/|z| for infinitely many n", aren't you already assuming it diverges, in order for the inequality not to be the other way around? Feels circular, but I'm almost certainly missing something.
@@Alex_Deam Not dumb at all. I can see why you would be suspicious, as the formula for R is only presented after theorem, yet I have seemingly used it in the proof of the theorem. I think the best way to recoincile it will be to rewrite the theorem in two parts as follows: . (Part 1) Given a power series sum_n a_n z^n, define R by the formula in 14:52. Then, (Case 1) 0
@@strikeemblem2886 Yeah I think writing in reverse order makes more sense, but also I think I just lacked an understanding of the fundamentals of why the root test etc even works to begin with. It all makes sense now having read up on it.
5:48 I’m not sure why there can’t be any conditional convergence beyond that R. Of course there is no absolute convergence, but you said it diverged beyond that R, and I don’t think you proved why.
@@deltalima6703 I don’t care too much, I have a doctorate and I’m in industry so I don’t need to hear the proofs because I have the intuition already. My comment here (and pretty much every criticism I give to him and other math UA-camrs) is just to bring it to the attention of everyone else
I did learn a few things, being more thorough to find all possible solutions for example. Mostly I just learned from other channels to get some background to make sense of what he is saying. I picked up geometric algebra to make sense of clifford algebra, for example. Overall, its a great topic, but I will look for a better teacher. I unsubbed from here too.
I think it's because the modulus of z^n only depends on the modulus of z, so whether z^n converges or not only depends on whether z is inside or outside the circle.
Thank you a lot for your work, its amazing have the oportunity to learn advanced math without moving home. I will finish engineer this year and next year i want to make a master in math, this help me a lot. I will colaborate with the channel as soon as posible, your work is amazing and you are giving opportunities for many people arround the world. The future will need people that understand math at a deep level(as an quasi-engineer i think that is very basic how math is teaching to us, is more like learn rules to solve problems without understand whats really happening)
15:10 this limit might not exist if a_n diverges, but we could take the the reciprocal of the limsup of nth root of a_n, i.e. the largest of its subsequences' limits.
I've been catching up with your 60 video real analysis playlist and it ended with integrability and didn't cover taylor series, power series, or absolute convergence. (It did do sequences and series of functions plus pointwise vs uniform convergence and their consequences for continuity and differentiability). Am I missing any videos that didn't get included in the playlist or did you cover the entire course material there and there aren't videos for those topics? Thanks!
You could also say that, in order to define the sort of useful mathematical object that a power series is, you need a constant term, and your notation is being idiomatic in including it in the sum rather than outside. The notation is really trying to make sense for the infinity part and be convenient for writing all of the coefficients, not be an expression you could always evaluate without referring to the power series definition.
0^0 is not undefined. It is 1. z^0 = 1 for all z in C. In fact even in rings with zero divisors (so not every element is invertible) we always have a^0=e for all elements a, where e is the identity
Why can't there be convergent values outside of the circle? It won't converge for all values with that R, but it hasn't been proven that it will converge for none.
When you say that something is a particular branch of a function in this video, you really mean that it agrees at one point with that branch and doesn't have any branch cuts that intersect the region inside the radius of convergence. So it's not saying where the branch cut for the logarithm should go, just that it isn't forced to intersect a unit circle around 1.
Echoing the existing, non spam, comment, it seems like it hasn't been disproved that there might be some special Z bigger than R, that does converge, but not absolutely
Missing details on the divergence of the series for |z| > R, when 0 1/|z| for infinitely many n. So |a_n z^n| > 1 for infinitely many n. Finally recall that a requirement for the series to converge is that the terms of the series convergence to 0. So we obtain divergence.
This is probably a dumb question, but when you apply the root test and say "this means |a_n|^(1/n) > 1/|z| for infinitely many n", aren't you already assuming it diverges, in order for the inequality not to be the other way around? Feels circular, but I'm almost certainly missing something.
@@Alex_Deam Not dumb at all. I can see why you would be suspicious, as the formula for R is only presented after theorem, yet I have seemingly used it in the proof of the theorem. I think the best way to recoincile it will be to rewrite the theorem in two parts as follows:
.
(Part 1) Given a power series sum_n a_n z^n, define R by the formula in 14:52. Then, (Case 1) 0
@@strikeemblem2886 Yeah I think writing in reverse order makes more sense, but also I think I just lacked an understanding of the fundamentals of why the root test etc even works to begin with. It all makes sense now having read up on it.
@@Alex_Deam i'm glad its fine now! 🙂
5:48 I’m not sure why there can’t be any conditional convergence beyond that R. Of course there is no absolute convergence, but you said it diverged beyond that R, and I don’t think you proved why.
He never responds. I already unsubbed from the main channel. Time to ditch this one too I suppose.
@@deltalima6703 I don’t care too much, I have a doctorate and I’m in industry so I don’t need to hear the proofs because I have the intuition already. My comment here (and pretty much every criticism I give to him and other math UA-camrs) is just to bring it to the attention of everyone else
I did learn a few things, being more thorough to find all possible solutions for example. Mostly I just learned from other channels to get some background to make sense of what he is saying. I picked up geometric algebra to make sense of clifford algebra, for example.
Overall, its a great topic, but I will look for a better teacher. I unsubbed from here too.
@@noahtaul thank u for that!
I think it's because the modulus of z^n only depends on the modulus of z, so whether z^n converges or not only depends on whether z is inside or outside the circle.
Thank you a lot for your work, its amazing have the oportunity to learn advanced math without moving home. I will finish engineer this year and next year i want to make a master in math, this help me a lot. I will colaborate with the channel as soon as posible, your work is amazing and you are giving opportunities for many people arround the world. The future will need people that understand math at a deep level(as an quasi-engineer i think that is very basic how math is teaching to us, is more like learn rules to solve problems without understand whats really happening)
15:10 this limit might not exist if a_n diverges, but we could take the the reciprocal of the limsup of nth root of a_n, i.e. the largest of its subsequences' limits.
The result you state that fn converges uniformly on |z|
I've been catching up with your 60 video real analysis playlist and it ended with integrability and didn't cover taylor series, power series, or absolute convergence. (It did do sequences and series of functions plus pointwise vs uniform convergence and their consequences for continuity and differentiability). Am I missing any videos that didn't get included in the playlist or did you cover the entire course material there and there aren't videos for those topics? Thanks!
I think I ended up doing lectures in-class for those topics.
16:40 If you substitute z0 you reach an undefined 0^0 . Are you assuming that this equals 1?
That is a common convention for power series. It makes sense, because the exponent is kept constant at 0, while z approaches 0. This limit is 1.
@@kristianwichmann9996 Ah thank you that makes more sense
You could also say that, in order to define the sort of useful mathematical object that a power series is, you need a constant term, and your notation is being idiomatic in including it in the sum rather than outside. The notation is really trying to make sense for the infinity part and be convenient for writing all of the coefficients, not be an expression you could always evaluate without referring to the power series definition.
0^0 is not undefined. It is 1. z^0 = 1 for all z in C. In fact even in rings with zero divisors (so not every element is invertible) we always have a^0=e for all elements a, where e is the identity
Why can't there be convergent values outside of the circle? It won't converge for all values with that R, but it hasn't been proven that it will converge for none.
almost finish the series, many proofs have missing details and need a lot of extra work.
Sir what are the reference books? From which books this material has been covered?
When you say that something is a particular branch of a function in this video, you really mean that it agrees at one point with that branch and doesn't have any branch cuts that intersect the region inside the radius of convergence. So it's not saying where the branch cut for the logarithm should go, just that it isn't forced to intersect a unit circle around 1.
Good explanation professor keep going
Why does fn converge uniformly to f in the power series?
You state that but what’s the proof?
Weirstrass M-Test and the fact that if a Power Series converges at z_0, it converges absolutely on |z|
Never mind it's not true, it needs not converge uniformly on |z|
@@anshumanagrawal346 so what’s the answer to the question?
@@Happy_Abe What's your question
fn converges uniformly to f if replacing z by r. it requires a bit of triangle inequality for the complex case.
Anyway, this series is terrible
Echoing the existing, non spam, comment, it seems like it hasn't been disproved that there might be some special Z bigger than R, that does converge, but not absolutely
No, if a Power series converges at z_0, it converges absolutely for all z satisfying
|z-a|
Can't wait for the Fourier Series!!
Books?
This is very subpar. Many results stated are incomplete or even straight up incorrect.