this is fantastic! you managed to explain in so clearly without making an 1h long video. Thank you very much :) Math makes so much more sense when you finally understand what you are doing
ok but what does an infinit radius means for the taylor expension of sin / cosine ? i heard it means that the approximation of the function is correcte on the radius thus sin is always a good approximation
Depends a bit on your definition of periodic, but the general answer is no. If we aren't talking about analytic functions the answer is trivially no, so let's assume we are taking a function analytic on some domain and extending it. In the real variable case we can always continue a real analytic function to a complex analytic function in some neighborhood, but the neighborhood could have infinitesimal width (examples are known, and to make it periodic use a bump function and tesselate). In the complex case things get more interesting, but the answer is still no. In that case the question comes down to *can* we take a function analytic on some disk and extend it. Turns out the answer is sometimes no, for a myriad of reasons (holonomy, natural domains, etc) so it depends on how you wish to get your periodic function. We unfortunately can't just tesselate copies using bump functions, because holomorphicity is a very stringent requirement. Of interest here are doubly periodic functions on the plane, which actually cannot be analytic everywhere (but are meromorphic), so we cannot hope to duplicate the real variable case in two directions. Now, why do I mention all this complex analysis? I do so since the radius of convergence of a series is the nearest distance to the nearest complex singularity, so unless your real, periodic function has an analytic continuation to the whole plane you cannot hope for infinite radius of convergence.
If I might add something, the Laguerre Polya class is quite interesting here. It's the unfirom limits on compact subsets of polynomials having only real zeros, and encapsulates pretty much every example of a "naturally periodic" function you can think of. It turns out these things have Genus 1 Hadamard Products and do have analytic continuations to the plane, so there is a sense in which the answer to your question is yes. As such, things like bump functions prevent us from stating your claim is true in general, but it is true for "sufficiently nice" periodic functions.
This actually clarifies this whole concept. But I have a question, how would this workd for cosx centred around pi/4? I've tried this but can't seem to get an expression for the taylor series as it alternates in the pattern: +--++--++. I eventually ended up saying since we're going to take the absolute value of it in the limit that those patterns wouldn't matter, but I don't like that justification. What do you think?
Given that you have a second order numerator with respect to x and a second order denominator with respect to n, doesn't that mean that the ratio test doesn't strictly converge to 0 for all values of x? If you let x approach infinity as well, the limit still evaluates to 1/4, which means that there are no values of x for which the sequence doesn't converge, but without having studied the field in depth, I'd assume that's a contingency that's supposed to be considered.
It is another problem, double limit is really tricky, and there is commutativity of the order in which you approach the limits in specific conditions. Here, the limit for x is infinite, and the limit for n is 0, meaning that approaching the other limit aftewards won't even change the result. Anyway, there is no reason to do so here, the demonstration proves that it does converge to 0 for any real value of x, and that's actually already good enough ^_^. Btw, the d'Alembert ratio test for the *power series* (series of term an*x^n, in that specific case, the Taylor series of sine and cosine) takes the ration an+1/an rather than the whole thing like he did an+1*x^(n+1)/an*x^n. Basically, the version of d'Alembert ration test for power series specifically doesn't even take the x parameter into account, and you end up with the limit of 1/((2n+1)(2n+2)) rather than x^2/((2n+1)(2n+2)). You can try to solve it by yourself, you will find the same radius. You are right to be concerned, because it is not really obvious. But I believe he wanted to keep it clear by not introducing power series. Just remember that the demonstration for d'Alembert ratio test for power series use the d'Alembert ratio test for series in general. I suggest you to look for the demonstration if you are interested, it is not too difficult to understand, just be sure to have basic knowledge of power series before stepping in. Tl;dr : there is no point of having the limit as x approaches infinity, d'Alembert ration test only use the limit as n approaches infinity.
Although the answer is correct, we don't have a general term for a_n here, as every time, one of a_n or a_(n+1) would be zero. So, how will the ratio form of radius of convergence work here!
![Radius of Convergence - The Natural Logarithm [ ln(x+1), considering the boundary cases! ]](http://i.ytimg.com/vi/9WIiM2nXiUw/mqdefault.jpg)
![Seungmin "그렇게, 천천히, 우리(As we are)" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/kAzmhLHePqU/mqdefault.jpg)
If only you uploaded this a week ago. My test would've gone much better.
oh yeah yeah :c
this is fantastic! you managed to explain in so clearly without making an 1h long video. Thank you very much :) Math makes so much more sense when you finally understand what you are doing
Oh God this takes me back to my high school days. Great video as always
U did a great job explaining
Thanks! =)
Nice explanation with fun, keep it up bro....
When are you gonna do something on differential geometry/ manifolds ?
ok but what does an infinit radius means for the taylor expension of sin / cosine ? i heard it means that the approximation of the function is correcte on the radius thus sin is always a good approximation
Does all periodic functions have infinite radius of convergence?
I think not necessarily
The tangent function doesn't
Depends a bit on your definition of periodic, but the general answer is no. If we aren't talking about analytic functions the answer is trivially no, so let's assume we are taking a function analytic on some domain and extending it. In the real variable case we can always continue a real analytic function to a complex analytic function in some neighborhood, but the neighborhood could have infinitesimal width (examples are known, and to make it periodic use a bump function and tesselate). In the complex case things get more interesting, but the answer is still no. In that case the question comes down to *can* we take a function analytic on some disk and extend it. Turns out the answer is sometimes no, for a myriad of reasons (holonomy, natural domains, etc) so it depends on how you wish to get your periodic function. We unfortunately can't just tesselate copies using bump functions, because holomorphicity is a very stringent requirement. Of interest here are doubly periodic functions on the plane, which actually cannot be analytic everywhere (but are meromorphic), so we cannot hope to duplicate the real variable case in two directions. Now, why do I mention all this complex analysis? I do so since the radius of convergence of a series is the nearest distance to the nearest complex singularity, so unless your real, periodic function has an analytic continuation to the whole plane you cannot hope for infinite radius of convergence.
If I might add something, the Laguerre Polya class is quite interesting here. It's the unfirom limits on compact subsets of polynomials having only real zeros, and encapsulates pretty much every example of a "naturally periodic" function you can think of. It turns out these things have Genus 1 Hadamard Products and do have analytic continuations to the plane, so there is a sense in which the answer to your question is yes. As such, things like bump functions prevent us from stating your claim is true in general, but it is true for "sufficiently nice" periodic functions.
schmeck. I dont know what it means just thought it would fit in this comment section.
Ahhhhh ! You cannot use the ratio test for a series that has an infinite amount of zeroes !
it makes sense
This actually clarifies this whole concept.
But I have a question, how would this workd for cosx centred around pi/4? I've tried this but can't seem to get an expression for the taylor series as it alternates in the pattern: +--++--++.
I eventually ended up saying since we're going to take the absolute value of it in the limit that those patterns wouldn't matter, but I don't like that justification.
What do you think?
Had to solve a very similar problem in my january's analysis exam
When's the q/a coming out papa? Nice vid btw
@@PapaFlammy69 eks dee
Ty!!
Given that you have a second order numerator with respect to x and a second order denominator with respect to n, doesn't that mean that the ratio test doesn't strictly converge to 0 for all values of x? If you let x approach infinity as well, the limit still evaluates to 1/4, which means that there are no values of x for which the sequence doesn't converge, but without having studied the field in depth, I'd assume that's a contingency that's supposed to be considered.
It is another problem, double limit is really tricky, and there is commutativity of the order in which you approach the limits in specific conditions. Here, the limit for x is infinite, and the limit for n is 0, meaning that approaching the other limit aftewards won't even change the result. Anyway, there is no reason to do so here, the demonstration proves that it does converge to 0 for any real value of x, and that's actually already good enough ^_^. Btw, the d'Alembert ratio test for the *power series* (series of term an*x^n, in that specific case, the Taylor series of sine and cosine) takes the ration an+1/an rather than the whole thing like he did an+1*x^(n+1)/an*x^n. Basically, the version of d'Alembert ration test for power series specifically doesn't even take the x parameter into account, and you end up with the limit of 1/((2n+1)(2n+2)) rather than x^2/((2n+1)(2n+2)). You can try to solve it by yourself, you will find the same radius.
You are right to be concerned, because it is not really obvious. But I believe he wanted to keep it clear by not introducing power series. Just remember that the demonstration for d'Alembert ratio test for power series use the d'Alembert ratio test for series in general. I suggest you to look for the demonstration if you are interested, it is not too difficult to understand, just be sure to have basic knowledge of power series before stepping in.
Tl;dr : there is no point of having the limit as x approaches infinity, d'Alembert ration test only use the limit as n approaches infinity.
Although the answer is correct, we don't have a general term for a_n here, as every time, one of a_n or a_(n+1) would be zero. So, how will the ratio form of radius of convergence work here!
Yes same like with sinhx coshx
I have my intro to real ANALysis 2 exam this Thursday and this helped a lot❤️
Will you solve problems from your subreddit? 🤔
@@PapaFlammy69 Gotta prepare myself for some ridiculous problems then, lol.
In memory of hentai haven
@@PapaFlammy69 I knew you were a man of culture😌
Isn’t it analytic everywhere, (-∞, ∞)
The equals sign should be level with the main fractional bar 😠😠😠 all math is now RUINED
pretty juicy
First
The smallest prime
@@hyunwoopark9241 IN ORDER TO BE NUMBER ONE BE ODD :- coolleo