You know what I love about your videos? You make good level math available on yt w/o the fear of views. People switch generally to simple calc1 for views.
I learnt it this way for the first time but I think Heine Borel (equivalent form of Bolzano- Weierstrass) makes this fact intuitive. Cover [0,1] with balls centered at each x in [0,1] with radii 1 (say). Continuity gives a bound for f on each ball. Heine Borel gives a finite subcover and the rest is easy to see. :)
Is this fact just a consequence of the Extreme Value Theorem?... also, since Wienner Process is continuous, that this means that for any compact domain the path will never run to infinity, so it distribution within the compact domain cannot be gaussian (since +/- infinity is never achievable), is this right??
Guru, how about the case of complex numbers? Like some trigonometric functions, ie: sin(x), can be written via Euler's formula, and if x->inf, than e^ix goes some complex infinity.
e^ix, for real x, only takes values on the unit circle. As x approaches positive infinity, e^ix simply traverses the unit circle infinitely many times, so its modulus (absolute value) is always 1. sin(ix), however, equaling i sinh(x), does shoot off to +i times infinity.
Hi, I have a question: Could we also use the monotone sequence theorem because we can find a reordering of x_n such that its monotonically increasing and also bounded and therefore convergent? Im not sure about this but i think its kind of the same argumentation.
One sequence of elements that is reordered is not the same sequence, that's why we use vectored notation (a_n) and not set notation {a_n} same way that reordering a convergent series can make it diverge, consider e.g. Σ (1-1/2+1/3-1/4+1/5..) this converges [ln(2)] since it alternates, monotonic, but Σ(1+1/3+1/5..)-Σ(1/2+1/4+1/6..) = Σ(1+1/3+1/5..)-(1/2)Σ(1+1/2+1/3..) clearly harmonic diverges.
@@tomctutor Thanks, i get your point but isn’t a subsequence of (a_n) defined as (a_m(n)) where m is a bijective function from the Index set into itself? This would imply that m could also be a permutation of the index set wich would be a reordering. So one can choose a reordering of the index set such that the subsequence is monotonically increasing an therefore convergent by the monotone sequence theorem because the sequence is bounded. (Wich is the same result as Bolzano- Weierstraß in IR.) As i said i completely understand what you mean but i think i confused you a little by not mentioning that im talking about subsequences. Anyway, thanks for replying tho.
@@User-gt1lu Good thinking, but strictly maintaining order, i.e. dropping elements of parent sequence, is allowed. As I said do not confuse sequences with sets, a sequence has a strict indexing on N.
He's been making lectures from his fall 2020 class public more gradually. Older comments are from his students, and UA-cam dates videos when they become public.
u cant use the hypothesis that f is continous. becouse u say as true that is continous and then say its not, but u are using the continuity as a true hipothesis, at the same time as the thesis like its not bounded then its not continous.
The assumption is that f is indeed continuous and you show (by way of contradiction) that f must indeed be bounded. If f was not continuous, then the statement vacuously holds since F -> anything is true.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
He explains Analysis stuff in such a natural way that it seems like we're doing High School Algebra
4:45 Gotta love it when you get to use the Bolzano-Weierstrass-(ouch!) theorem in a proof!
delta Dirac impulse in 4:46, δ(t-4:46)
This proof explanation contained several insightful nuggets which helped me to understand it,
I like these analysis videos because I'm not good with it and they help me a lot. Ty Dr Peyam
You know what I love about your videos? You make good level math available on yt w/o the fear of views. People switch generally to simple calc1 for views.
Thanks so much!
Brilliant! Only Dr Peyam can explain these topics so well! 😊
Welcome back Dr Peyam
Thank you so much for making my first year math course so much easier
thank you so much for such an easy explanation , it really helped a lot , thank you so much again
I learnt it this way for the first time but I think Heine Borel (equivalent form of Bolzano- Weierstrass) makes this fact intuitive. Cover [0,1] with balls centered at each x in [0,1] with radii 1 (say). Continuity gives a bound for f on each ball. Heine Borel gives a finite subcover and the rest is easy to see. :)
I like that proof too
Waòoooooo Amazing Dr.PEYAM
2:22 and is still continuous according to the Definition.
Nice video! Brings me back to my analysis days! Just wondering, what courses are you teaching in the Fall?
Multivariable calculus 😁
Sequences and boundedness, reminds me of Bolzano-Weierstrass
If only dr peyam can make a series on analysis 😕
I already did, check out my playlists!
Why couldn't my teacher be like him
Hi will you make videos for math140B? That would be so helpful!
people.tamu.edu/~tabrizianpeyam/Math%20409/math409.html
@@drpeyam thanks!
Is this fact just a consequence of the Extreme Value Theorem?... also, since Wienner Process is continuous, that this means that for any compact domain the path will never run to infinity, so it distribution within the compact domain cannot be gaussian (since +/- infinity is never achievable), is this right??
what if f is only continuous at a single point?
Wait so if the mother sequence goes to infinity that forces the subsequence of that mother sequence to go to infinity?
Yes!
Like if the entire class jumps off a bridge, that definitely means that you and your homies all jumped off the bridge since you're in the class!
Why are the comments so old?
How comments are 8 months ago
Video is uploaded now just 10 min ago 🙄
Wow bro......How did u manage to comment 8 months ago? It is released 6 hrs ago?! Am I hallucinating?
Guru, how about the case of complex numbers? Like some trigonometric functions, ie: sin(x), can be written via Euler's formula, and if x->inf, than e^ix goes some complex infinity.
Really, infinity is somewhere you, I or Mr Euler never reached. Mind you, Euler got quite close and you, my friend, will go around in circles trying.
e^ix, for real x, only takes values on the unit circle. As x approaches positive infinity, e^ix simply traverses the unit circle infinitely many times, so its modulus (absolute value) is always 1. sin(ix), however, equaling i sinh(x), does shoot off to +i times infinity.
Is there any video on simple high school calculus on your channel. I have a exam on Tuesday. 👍
College algebra and Precalculus playlists
Sir where are you from 🙄
Hi, I have a question:
Could we also use the monotone sequence theorem because we can find a reordering of x_n such that its monotonically increasing and also bounded and therefore convergent?
Im not sure about this but i think its kind of the same argumentation.
One sequence of elements that is reordered is not the same sequence, that's why we use vectored notation (a_n) and not set notation {a_n}
same way that reordering a convergent series can make it diverge, consider e.g. Σ (1-1/2+1/3-1/4+1/5..) this converges [ln(2)] since it alternates, monotonic, but
Σ(1+1/3+1/5..)-Σ(1/2+1/4+1/6..) = Σ(1+1/3+1/5..)-(1/2)Σ(1+1/2+1/3..) clearly harmonic diverges.
@@tomctutor Thanks, i get your point but isn’t a subsequence of (a_n) defined as (a_m(n)) where m is a bijective function from the Index set into itself? This would imply that m could also be a permutation of the index set wich would be a reordering. So one can choose a reordering of the index set such that the subsequence is monotonically increasing an therefore convergent by the monotone sequence theorem because the sequence is bounded. (Wich is the same result as Bolzano- Weierstraß in IR.)
As i said i completely understand what you mean but i think i confused you a little by not mentioning that im talking about subsequences. Anyway, thanks for replying tho.
@@User-gt1lu Good thinking, but strictly maintaining order, i.e. dropping elements of parent sequence, is allowed. As I said do not confuse sequences with sets, a sequence has a strict indexing on N.
Excuse me sir, isn't there posibble that f(x0) goes to infinity either? Then there is no contradiction!!??
Is it theorem or fact if you prove it? Or it dose not matter?
It’s a theorem that I prove in the video
Wait.... What??? Comments are older than the video!!!! I mean..... How is it that damn incidence even possible???
He's been making lectures from his fall 2020 class public more gradually. Older comments are from his students, and UA-cam dates videos when they become public.
Magic 🪄
@@iabervon Oh I see,,,, thx 😁
@@drpeyam 🙊🙉🙊🙉🙊
へええ bounded functionちうのがあるの
後半難しくなってごにゃごにゃなったよ😅
Love from India
Nice
u cant use the hypothesis that f is continous. becouse u say as true that is continous and then say its not, but u are using the continuity as a true hipothesis, at the same time as the thesis like its not bounded then its not continous.
I don’t understand. I show that if f is continuous then f is bounded. So I can use the hypothesis that f is continuous
The assumption is that f is indeed continuous and you show (by way of contradiction) that f must indeed be bounded. If f was not continuous, then the statement vacuously holds since F -> anything is true.
👍