I love how happy he gets when he teaches!! I hope all teachers were like this haha
@Newtonissac6
6 жыл бұрын
Man, you are the biggest life saver ever. Just now, I was doing all these inequalities in Hilbert spaces and I have been stuck on this one particular problem for a while now. I am gonna go grab some popcorn. Let the video begin...
5 жыл бұрын
Man I love you, the way you make these problemas that easy is just incredible!
@mridultyagi2995
2 жыл бұрын
I was solving sammer Bansal book for jee advance . And there was a problem mixed with calculus and this inequality and your video helped me . thanks
@bosepukur
6 жыл бұрын
your videos are wonderful , need more of such excellent work
@TVSuchty
3 жыл бұрын
This is a cool easy and short proof. Very elegant.
@user-xb1vc9no6x
2 жыл бұрын
I really like the way you talk. I can feel your passion about math even though I am only watching the video!
@drpeyam
2 жыл бұрын
Thanks so much!!!
@joelvirolainen590
6 жыл бұрын
Rewinding rules! I love these videos!
@aminahanjum3592
7 ай бұрын
studying for lin alg final was very down bc my prep didnt feel good enough but seeing your video made me happy thanks for teaching with enthusiasm
@narayandhakal3526
Жыл бұрын
I'm also left handed ✋️ 😅 Thank you for making this video!
@FranqiePR
3 жыл бұрын
Thanks Dr. Peyam!
@elchaflan100
6 жыл бұрын
manage to keep a smile while explaining, i dont know if i should take him serious
@JoaoGabriel-eb9nd
Жыл бұрын
No, no one smiles doing math, he’s a reptilian no cap
@stydras3380
6 жыл бұрын
that indeed was quick
@lucifer.333
8 ай бұрын
please continue making such videos🤙👍
@mkhan28.
2 жыл бұрын
Thank you so much. ❤️
@EL-eo8dh
6 жыл бұрын
Very cool!
@sachi5427
2 жыл бұрын
the best proof of cauchy schwarz i've seen!
@drpeyam
2 жыл бұрын
Thank you!!
@Zoonofski
3 жыл бұрын
You legend!
@AndDiracisHisProphet
6 жыл бұрын
Man, I remember when I was a little Erstsemester I thought "What is this stupid thing good for!? " Was I naive at that time :/
@justwest
6 жыл бұрын
AndDiracisHisProphet Do you say "Erstsemester" in english? Is that actually a word, like Kindergarten?
@AndDiracisHisProphet
6 жыл бұрын
Nicht, dass ich wüsste
@lin1450
5 жыл бұрын
@@AndDiracisHisProphet xDDD
@flooreijkelboom1693
2 жыл бұрын
Love this proof
@taubone9257
6 жыл бұрын
Neat! Link to the "swag" proof?
@drpeyam
6 жыл бұрын
Proof of the Cauchy-Schwarz inequality kzitem.info/news/bejne/tIZ5v3eMmmd1oII
@ap6872
Жыл бұрын
Nice sessions Sir I am from India 🇮🇳...
@demudu
Жыл бұрын
Very cool
@yuvalpaz3752
6 жыл бұрын
I don't remember this proof, but this inequality has so many ways to prove it you can't expect from one to remember them all. Well, in Euclidean space this come straight out of the definition of dot product(as |cos(x)|
@YorangeJuice
2 жыл бұрын
Nice!
@Noporeh2
5 жыл бұрын
Interesting, i'm used to prove the triangle inequality using CS but you're doing it the other way around. So i'm curious, how do you prove the triangle inequality ?
@FranciscoTornay
4 жыл бұрын
You don't need the general triangle inequality for this The proof relies only on the special case of a right triangle, which is obvious from Pythagoras' theorem
@jabunapg1387
2 жыл бұрын
From the definition of ||x||
@cameronspalding9792
4 жыл бұрын
If I was proving this: I would consider which is a quadratic function in t which is non negative hence has a non positive determinant
@drpeyam
4 жыл бұрын
Check this out: Cauchy Schwarz Proof kzitem.info/news/bejne/tIZ5v3eMmmd1oII
@Rolljack
6 жыл бұрын
THANK S a lot
@cancermelon2155
6 жыл бұрын
Cool! I also like the proof made by Serge Lang (maybe it's the standard proof idk) If v = 0 easy. If v ≠ 0, let x = (v • v) , y = - (u • v). Then, since: (u x + v y) • (u x + v y) ≥ 0 dividing by: (v • v) we get: (u • u)(v • v) - (u • v)^2 ≥ 0 QED
@drpeyam
6 жыл бұрын
Yep, that’s also a nice proof!
@renesperb
Жыл бұрын
A very simple proof is as follows : consider the quadratic function Sum(u+t*v)^2 (1 to n)> = 0. Then the discriminant condition is just the Cauchy-Schwarz inequality . The same way it follows in a Hilbert-space if you replace the sum by the general scalar product.
@drpeyam
Жыл бұрын
I did that one already
@danieledwin8927
3 жыл бұрын
At 2:02 Shouldn’t it be the projection of u onto v rather than u hat? Sorry if I’m watching some dodgy version of the correct thing or if I made a mistake
@drpeyam
3 жыл бұрын
u hat is better, especially if it’s clear what you’re projecting on, the proj notation is lengthy and confusing
@EmissaryOfSmeagol
6 жыл бұрын
Would saying that |u*v|
@drpeyam
6 жыл бұрын
That only works for vectors in R^2, my proof is more general
@PuzzleQodec
6 жыл бұрын
Makes me wonder if the concept of an angle could be generalized to higher dimensions, or to vector spaces in general. Would it make sense to 'define' an angle theta between u and v as the inverse cosine of |u∙v| / ||u|| ||v||?
@ryliur
3 жыл бұрын
At 0:50, when v is represented as a zero vector, does that not also apply for u?
@drpeyam
3 жыл бұрын
u and v are independent of each other, so if v = 0, u can still be anything we want
@ryliur
3 жыл бұрын
@@drpeyam Sorry I may have worded my question a little awkwardly. If we set u = 0 instead of v, would we also be "done" with the proof? I was just slightly caught off guard by the emphasis of setting *only* v to be a zero vector.
@drpeyam
3 жыл бұрын
I see! Yeah it doesn’t matter, you could set u = 0, but then in your second case you would divide by u and not by v. It’s basically symmetric
@ryliur
3 жыл бұрын
@@drpeyam Thanks for the explanation!
@AndDiracisHisProphet
6 жыл бұрын
Hey, Peyam. I have to apologize. I made STEVE call you, to (ab)use you as a voice-over app. I am sincerely sorry that he stole your precious time. :/ . . . . Naaah I'm just kidding, that was hilarious :D Next time teach him how to properly pronounce "Weierstraß" :D
@drpeyam
6 жыл бұрын
Hahaha, it’s ok, it was really funny and a great idea!
@myonlynick
6 жыл бұрын
2:00 u-hat how come to be the formula as shown? i Know that u-hat= ( u / ||u|| )
@drpeyam
6 жыл бұрын
That’s a different u^, my u^ is the projection of u on v, check out my video on orthogonal projections
@ElizaberthUndEugen
5 жыл бұрын
||û|| = cos(u, v) ||u||, ok? then we have ||u|| ||v|| cos(u, v) --------------------------- v ||v|| ||v|| ||v|| cancles once so we have ||u|| cos(u, v) --------------------------- v ||v|| Now v/||v|| is a unit vector in direction of v and ||u|| cos(u, v) is the length of the projection of u onto v (in other words ||û||, as stated above). Thus v ----- ||u|| cos(u, v) ||v|| is û.
@MiroslavMakaveli
6 жыл бұрын
yes lets do some limits
@MiroslavMakaveli
6 жыл бұрын
they do computer science and have done math
@123bubbleup
4 жыл бұрын
You are adorable
@drpeyam
4 жыл бұрын
🥰🥰🥰
@ricbisa6092
2 жыл бұрын
13 people didn't understand the proof.
@drpeyam
2 жыл бұрын
😂😂😂
@aaryan505
3 жыл бұрын
R U Indian?
@zacharieetienne5784
6 жыл бұрын
Slightly flawed proof. What if u is in the span of v? Trivial case though
@drpeyam
6 жыл бұрын
If u is in the span of v, then u = c v and so v = (1/c) u, so v is in the span of u (and c is nonzero since u is nonzero)
@zacharieetienne5784
6 жыл бұрын
+Dr. Peyam's Show Well yes, but that doesn't prove much. You've just gotta do the same thing you did with v = 0, except substitute u = cv. Both sides should be equal and thus less than or equal to.
@drpeyam
6 жыл бұрын
Sorry, misread your question! It’s actually a not a flaw in the proof, but a feature of the proof! If u is in the span of v, then not only is the formula valid, but moreover u^ = u, so we get equality! And more than that, this is the *only* way we get equality in the C-S inequality! Neat, huh?
Пікірлер: 68