Real Analysis and I haven't gotten along very well the past 13 weeks but this has brought some great intuition to some things that were quite difficult to understand. Thank you.
Today I studied sequence of functions the entire day. It was quite interesting. Then I watched your video, which was a very nice way to summarise the things I've studied throughout the day. Also your graphs helped me better understand pointwise and uniform convergence.Thank you very much!
You are welcome! The middle point has the most extremal distance from both sides. And you don't need to calculate the sup-norm exactly. But, yes, it could be equal to 2 :)
If its not real analysis then you don't need, calculus isn't heavy proof like that, maybe in calc 2 or even early, you may encounter unconditionally convergence , which means that the absolute value doest converge so rearranging the n terms of an series will change the answer, and if it's absolutely convergence, means that its absolutely value is indeed converges which implies that rearranging its nterms of the series will not change the answer, hopefully you understand, if you didn't understand something tell me
The lim_n ||f_n - g ||_oo means that when n=1 I have a list of (possible infinite) values (f(x0) -g(x0), f(x1) -g(x1), ..., f(x2) -g(x_2), ....) where I take the supremum? When a I have n=1, n=2 I have a list of list of values and so on? And in the limit lim_n ||f_n - g ||_oo means I take the supremum over all list of values? I have this doubt because you said that "we started with a sequence of function but we ended with an ordinary sequence of numbers" so I struggle a little with that.
@@brightsideofmaths Perfect. So if I called e_n = |f(x) - fn(x)| the "error" I have a sequence of errors (so an infinite countable list of values). And when we take the limit over that, lim_n || f_n - f||_oo = lim_n sup{e_n} and said that lim_n || f_n - f||_oo = lim_n sup{e_n} = 0 we are saying that the limit of the supremum of e_n is zero. So at this point we have an infinite list of values where each one is the supremum of each e_n, and in the limit that list of supremums is zero? That's strong! If a translated this to a computer (and we treated the infinite sequence as a finite sequence, just as an approximation. Like the vandermonde matrix) it's like take two for loops and see that the plot of differences goes to zero? Well, maybe 3 for loops if we do that for every point x in the domain of fn (I)?
f_n(x) =s_n^x where s is a sequence that converges to e. Intuitively this seems like it should converge to e^x but it doesn't uniformly converge because the supremum is positive infinity
I suppose you can Define your way out of this by saying that a series of functions is uniformly convergent over a non-compact set if its uniformly convergent over all compact subsets
This is the best explanation of uniform convergence I have ever seen on RUclips.
Dude you are doing god's work. You are saving students from these tyrannical professor. Keep up the great work man I hope you get more donation.
Thank you very much :)
These videos are uniformly most helpful!
Glad you like them!
You win, this was easily the best explanation!
Wow, thanks!
I like these videos. I can only cost less time to consider and try to understand a new knowldege by taking your class.
Thank you very much :) And thanks for your support!
Real Analysis and I haven't gotten along very well the past 13 weeks but this has brought some great intuition to some things that were quite difficult to understand. Thank you.
Today I studied sequence of functions the entire day. It was quite interesting. Then I watched your video, which was a very nice way to summarise the things I've studied throughout the day. Also your graphs helped me better understand pointwise and uniform convergence.Thank you very much!
Hello! Many thanks for your videos! At 5:59, why did you simply take the mid-point? Is it correct to say that the sup-norm = 2?
You are welcome! The middle point has the most extremal distance from both sides. And you don't need to calculate the sup-norm exactly. But, yes, it could be equal to 2 :)
Thank you. I finally understand uniform convergence. I will share your channel to my friends who need to understand math deeply.
Wow, even earlier than expected :D
Thanks a lot!
This is exactly what I need! Excellent video and thank you!
Hast meinen Tag gerettet. Danke dir :)
I will follow a calculus course coming semester. This video helped a lot to prepare for it. Excellent teaching skills
If its not real analysis then you don't need, calculus isn't heavy proof like that, maybe in calc 2 or even early, you may encounter unconditionally convergence , which means that the absolute value doest converge so rearranging the n terms of an series will change the answer, and if it's absolutely convergence, means that its absolutely value is indeed converges which implies that rearranging its nterms of the series will not change the answer, hopefully you understand, if you didn't understand something tell me
@@soccerbels7947 I guess you commented on the wrong comment :)
@@sharonnuri i dont understand sorry
@@sharonnuri did i missundestand? Sorry if i did, im not native English
@@soccerbels7947 No problem
Good vid
Excellent explanation, thank you!
Great presentation! Very intuitive.. what is the program you use to write with your tablet? Very neat
Thanks! This is the nice and free program Xournal.
@@brightsideofmaths thank you very much!
awesome explanation
Glad it was helpful!
Thanks
Thank you very much :)
The lim_n ||f_n - g ||_oo means that when n=1 I have a list of (possible infinite) values (f(x0) -g(x0), f(x1) -g(x1), ..., f(x2) -g(x_2), ....) where I take the supremum? When a I have n=1, n=2 I have a list of list of values and so on? And in the limit lim_n ||f_n - g ||_oo means I take the supremum over all list of values? I have this doubt because you said that "we started with a sequence of function but we ended with an ordinary sequence of numbers" so I struggle a little with that.
For a sequence of functions you have infinitely many values and infinitely many ordinary sequences of numbers :)
@@brightsideofmaths Perfect. So if I called e_n = |f(x) - fn(x)| the "error" I have a sequence of errors (so an infinite countable list of values). And when we take the limit over that, lim_n || f_n - f||_oo = lim_n sup{e_n} and said that lim_n || f_n - f||_oo = lim_n sup{e_n} = 0 we are saying that the limit of the supremum of e_n is zero. So at this point we have an infinite list of values where each one is the supremum of each e_n, and in the limit that list of supremums is zero? That's strong! If a translated this to a computer (and we treated the infinite sequence as a finite sequence, just as an approximation. Like the vandermonde matrix) it's like take two for loops and see that the plot of differences goes to zero? Well, maybe 3 for loops if we do that for every point x in the domain of fn (I)?
It's an epsilon sausage. Lol how I've remembered this
Yeah :D
It could be wurst!
The supremum norm seems not to be defined for non bounded functions. Why isn't that a problem?
f_n(x) =s_n^x where s is a sequence that converges to e. Intuitively this seems like it should converge to e^x but it doesn't uniformly converge because the supremum is positive infinity
I suppose you can Define your way out of this by saying that a series of functions is uniformly convergent over a non-compact set if its uniformly convergent over all compact subsets
@@chair547 You can say this but this is a different notion of convergence.
@@brightsideofmaths so the sequence i defined isn't uniformly convergent?
@@chair547 You didn't give any domain for the functions, so I cannot answer the question :)
Why for the last example at the jump the value is above?
Look at the functions on the left. They all have this point in common. Therefore, it stays in the limit.
@@brightsideofmaths thanks for the explanation. I thought it was just an arbitrary definition
Shouldn’t 6:04 be 0.5 instead?
No, we go from -1 to 1 :)
@@brightsideofmaths yea sorry my bad
Sir next All video upload plz
All videos?