Cauchy Schwarz Proof

[@cipicm]

even after 5 years this video is a masterpiece regarding the clearness of this proof. with no english background in mathematics it was still so easy to understand. thank you a lot.

[@MarkJay]

"What is love?" -Dr. Peyam

[@shaunderoza2321]

The one dislike must from someone who saw the words "discriminant" and "inequality" and got offended.

[@StreuB1]

BEST COMMENT EVER!! LOL No time for snowflakes in mathematics. We've all business here.

[@forklift1712]

probably from people who can't appreciate Dr. Peyam's enthusiasm

[@letladisebesho4503]

This is an excellent, clear proof. Now I finally get it. Thank you 🙏🏽

[@duckymomo7935]

In France, it’s called Cauchy inequality In Germany, it’s Schwarz inequality In Russia, it’s just inequality 😂 LOL

[@drpeyam]

😂😂😂

[@syahimiafiq5914]

Well Russian called it Bunyakovsky inequality

[@jamesclerkmaxwell676]

Syahimi Afiq cauchy was his doctoral advisor. My two cents

[@alexeybekasov4298]

Actually, Bunyakovsky published it 25 years earlier than Schwarz.. ;)

[@joshuafreeman9532]

Nope in France its Cauchy-Schwarz

[@vukstojiljkovic7181]

I love this proof dude! I was confused with the proof that they presented to me on the university, but this one, i love it. Keep making math videos, maybe i will join you in the near future!

[@jamesclerkmaxwell676]

"it's better than watching TV " 😂 I love you Prof. Tough crowd

[@rubentuquerrez]

I'v neve liked the proyections proof, but I love this one. I discover this by myself while I was studying for an exam. This was awesome. Thanks for the video Dr Peyam!

[@datboi_wild1222]

Still dont get it. Im teaching myself quantum physics and im learning linear algebra in order to do that but i cant seem to understand this. Im in 8th grade btw. Dumb it down.

[@nilsdula7693]

You need to let your brain time to develop.

[@thedoublehelix5661]

@@datboi_wild1222 did you figure it out now?

[@datboi_wild1222]

@@thedoublehelix5661 lol, ya

[@tlhomotsemoteme2423]

I am writing in 5 hrs and I needed you to boost my confidence. Thank you.

[@drpeyam]

Good luck!!!

[@afifakimih8823]

I'm a regular viewer of Dr. Peyam...his derivation methods are always unique.!!

[@randywilton3466]

This is an amazing proof! Way easier to understand than the one my real analysis prof gave us. Thank you

[@mokouf3]

The camera man is actually blackpenredpen!

[@StreamsofEdification]

Aggredd

[@jeanlucas2834]

I recognized instantly

[@chessislife3429]

Wow!! You get so many people that have same interest and passion as you, so fortunate

[@RealEverythingComputers]

At 1:40 when you show the summation of the dot product, do you mean i in superscript or subscript because its done in superscript in the video which confuses me with an exponent. Your clarification will be much appreciated. thx

[@drandrewsanchez]

Great work! I love the way you speak!

[@emanuellandeholm5657]

As far as I can find out, Cauchy-Schwarz seems to hold for, not just real inner product spaces, and not just complex inner product spaces, but for all vector spaces! Please correct me if I'm uneducated! :)

[@lamequemaciel6412]

Hey, professor! I was search the proof of Schwarz inequality for the context of Quantum Mechanics, mean that we can to use the complex space. But, I appreciate your proof.

[@abratumwe]

07:08 I didn't understand that part, we also have quadratic curves with positive answers only 🤔😕 What if the polynomial of t is only positive?

[@abratumwe]

You meant negative in y?

[@richardjamieson7548]

In step i of the proof t is defined to be an element of the Real Numbers. In the final step of the proof the the discriminant is said to be negative. That implies t is a complex number. Contradiction? Rik

[@sdmartens22]

Very impressive video as always, i wish you would do some damage on more advanced topics; the world needs you there.

[@stevenbi7495]

Watching this before Linear Algebra Final!!! Absolutely love this proof!!

[@ayushdudhani]

But sometimes after taking square roots inequality sign changes doesn't it??

[@drpeyam]

No, things are positive here

[@ayushdudhani]

@@drpeyam not getting can u explain it in detail

[@bluemonk9480]

I don't understand why the proof suddenly decides to look towards the discriminant. It makes sense to use it since it guarantees that b^2 - 4ac

[@josephinell314]

Such an elegant proof! Thank you for sharing.

[@Loathepotion]

This was good-you kinda didn't address the part where you wrote out a piece (cut off for some reason)but good stuff.

[@arjunnarasimhan8027]

What was your thought process behind Step 1? How did you get (u+tv)(u+tv)?

[@pheasant139]

Very elegant proof. Easy to understand. Thank you!

[@loganreina2290]

We were JUST proving this for homework in my honors calc class for our homework! Neat!

[@TruongNguyen-pl9cd]

why can u use the disciminant ? and why it have to be less then 0

[@drpeyam]

That’s the magical part about the proof. And it’s negative because otherwise there would be 2 roots, so at least somewhere the function would be negative

[@TruongNguyen-pl9cd]

@@drpeyam okay but how can I explain in my proof ? We just can use the discriminant ? Same like we use root and so on ?

[@drpeyam]

If you want do a contradiction. You have a quadratic function in t, and if the discriminant is positive then you would have 2 roots, which gives a contradiction etc

[@NabeelMustafaSMM]

please make video on Minkowski inequality ..I really missed that

[@RealEverythingComputers]

Great video! That was a really darn good proof!

[@fellipetoffolo4226]

very nice, but wouldn't the complex values of t contradict something? Or its okay because (u+tv)(u+tv) is the norm of something, thus will be real anyway?

[@drpeyam]

This proof is for real spaces only

[@fellipetoffolo4226]

ok, thank you Dr

[@sandnstars8]

Thanks for explaining the notation! It helped a lot!

[@drandrewsanchez]

I have a question. By listing the three inequalities regarding the descriminant, aren't you fixing the vectors in R2. Would this proof still be valid for vectors in R3 or RN?

[@TheRedfire21]

yes since you are only working on the real plane; modulus of u and v are real numbers, the dot product of u.v are real numbers, and t is an scalar.

[@drpeyam]

SanCHEneering It's valid in all of R^n :) The polynomial is quadratic, but the vectors u and v are in R^n

[@leonardromano1491]

Do they need to explicitly be in R^n? Wouldn't it be sufficient to be on a real-scalar-product-space?

[@drpeyam]

That's correct, it works on any real inner product space!

[@avdrago7170]

A way that I thought to prove this inequality when looking at the thumbnail is that with the the angle between two vectors x, cos(x)=|u * v|/(||u||||v||). So, |u*v|= cos(x)||u|| ||v||. Now, |cos(x)| is always less than or equal to 1, so when you multiply it to a separate quantity, you are scaling that quantity such that it is less than or equal to its original value. So, in the end, |u*v| is less than or equal to ||u|| ||v||

[@drpeyam]

Your method only works in R^n, unfortunately

[@avdrago7170]

Dr. Peyam's Show ? How does this matter

[@aliyss]

​@@avdrago7170 --> I think ​ Dr. Peyam's Show meant R^2.

[@minyhailabera3616]

the proof was short and understandable way.Thanks a lot.

[@starfire8679]

Really great video!! ❤And your great personality makes everything so interesting and fun 👌

[@lLl-fl7rv]

Thanks a lot man!! You saved my life one day before my final exam xD

[@railgun.__.7055]

Luckily we do not have to prove it in high school. (It was a question in AL Pure Maths)

[@LegendOfMurray]

great proof, Dr. Peyam!

[@TheRedfire21]

hi peynam could you proof this by definition of the dot product? or does it restrict your proof? ex: |u.v| = (by definition) ||u|| ||v|| cos@ ≤ (since cos@ is between 0 and 1) ||u|| ||v||

[@drpeyam]

Sebastian Cor Your proof is correct, but only works in R^2; this proof is valid for more general dot products!

[@leonardromano1491]

Actually the angle is defined through the Cauchy-Schwarz-Inequality so you can't define it the other way around. ( |uv|≤||u|| ||v|| -> -1≤uv/(||u|| ||v||) ≤ 1 and -1≤cos(phi)≤1 so there exists an open set U where for all phi in U cos(phi)=uv/(||u|| ||v||) ) The Proof only works for real Dotproducts because complex ones also conjugate the left vector.

[@tmogoreanu]

Can you please make a video with Newton-Leibniz formula proof?

[@drpeyam]

Anatolie Mogoreanu The proof of the product rule? I think you've read my mind, because it's coming on Friday :)

[@tmogoreanu]

Actually I meant the definite integral calculation formula F(b)-F(a)

[@drpeyam]

Ooooh, the FTC!!! It's on my to-do list :P

[@tmogoreanu]

Great! Thanks in advance)

[@sunbreezy3935]

Thank you very much. This is an excellent clear proof. Is there any chance that you could also do a proof of the Minkowski inequality.

[@fardinzaman904]

Thank you very much . Absolutely clear and easy . You saved me

[@mathunt1130]

You can reduce the proof to a line using geometric algebra.

[@drpeyam]

How?

[@mathunt1130]

@@drpeyam Simply take two vectors U and V and examine the product (UV)(VU) and expand using the geometric product. In a few simple steps (on one line) you get |U|^2|V|^2=(U.V)^2+|UxV|^2>= (U.V)^{2}

[@joudy5767]

Can’t thank you enough!! Excellent proof :)

[@Zhinoi]

This is a really nice and neat proof

[@danieledwin8927]

If you want to prove it for two real numbers are you allowed to say the numbers are just vectors with one component? If not make a video with the proof for the case for two numbers, preferrably from the axioms

[@drpeyam]

For real numbers the statement is trivial, it becomes |ab|

[@sweetychaudhary277]

thanks 😊 your video help me a lot to understand this proof ☺️

[@kenmeyer100]

Wow, someone who actually pronounces both names correctly

[@favoriteflavor46]

camera man just watches like others in that room.

[@DavidPumpernickel]

dank, this works for non-euclidean metrics, too

[@syahimiafiq5914]

I've show my cat this proof Now she still a cat

[@gabrielnegrini8889]

amazing explanation

[@alimghazzawi3700]

At first glims i thought of course it’s smaller or equal because the dot product value equals the multiplication of the brooms times cos theta and cos theta its biggest value 1 and smallest value is-1 but i dont know of course there is a reason why this isn’t a valid proof i don’t understand high level algebra i am an engineer student.

[@michaelsohnen6526]

why not start a proof from the fact that "A dot B = ||A|| ||B|| cos(theta)" ?

[@michaelsohnen6526]

we know -1

[@drpeyam]

No, this only works in R2; my proof works in Rn and more!

[@michaelsohnen6526]

Dr. Peyam's Show thanks!

[@luizr.8316]

Divertido e fantástico! Parabéns pelo trabalho!

[@MathematicsMadeSimple1]

Good presentation!

[@aryammlg6833]

Great proof, but it would be even better if u had gone over the equality case and shown that it only happens when one is a scalar multiple of the other 👍💕💕

[@ndriqa]

Thank you!!! Great channel ❤️

[@boydmwansa8750]

Very nice proof indeed so easy to understand

[@varungoyal4732]

thanks alot Dr Peyam. this video helped me alot.

[@vijay85321]

I got a little but happy to see the proof!

[@pibeeulotro9603]

Beautiful proof, thanks.

[@stevekaszycki8629]

neat proof! and it comes with at least two bad puns.

[@mokoepa]

This is beautiful... Perfection...

[@abstractalien12345]

Thank you so much! Greetings from Chicago :)

[@drpeyam]

The city of Lou Malnati’s 😋😋😋

[@PervezAli112]

sir its awesome proof

[@MrJapogm]

This channel is also very cool

[@jennifera.3208]

very good explanation :)

[@malicksoumare370]

amazing! The other proofs are just incomprehensible

[@mihlalimjacu6984]

thank you sir this is so much clear☺

[@joshuafreeman9532]

absolute genius

[@OmarOmar-cj9rk]

هل تتكلم اللغة العربية؟

[@joudy5767]

نعم

[@OmarOmar-cj9rk]

لقد مضى على هذا التعليق عام، وها أنت ذا يا joudy تستمتع بمشاهدة هذا المقطع الممتع. وفقك الله وسهل لك طريقا تلتمس فيه علما نافعا.

[@伊藤暉人-s6w]

数学が世界共通言語って感じがして最高。普段コーシーシュワルツの不等式、と書いていますが英字で書きたいなと思い英語で検索しました。高校生です

[@jonsnow9246]

More videos on Maths!

[@vecter]

That's a great proof and all but what in god's name motivated that initial formula to begin with? This is what makes people think math is so inaccessible. Once you come up with that formula, everything else is just algebra, but no insight is given into the intuition or meaning behind the original key formula.

[@drpeyam]

It's actually extremely useful in Analysis and Partial Differential Equations when you want to estimate the product of two things! Basically, think of it as follows: For a function f, let ||f|| = the square root of the integral of (f(x))^2, which you can think of as the 'energy' of f. Similarly we can define the energy of g, and finally define f . g = integral of f(x) g(x), which you can think of as the 'interaction' between f and g. Now suppose that f and g have finite energies, that is ||f|| < oo and ||g|| < oo, then the C-S inequality says that |f.g|

[@vecter]

Thanks for that lengthy reply! I wasn't talking about the CS inequality itself which is immensely useful, I'm talking about the formula you introduce at 3:00. That is a black magic equation from which the proof magically falls out. What is the motivation or intuition behind the starting point of this proof?

[@drpeyam]

Oh, I see! I agree, I did kinda pull it out of my hat, which makes the proof magical :P But I guess the intuition is that you want some inequality involving u and v, and that formula provides us with a good starting point!

[@vecter]

Surely there must be some geometric intuition behind it also? I'll think more about it, but the intuition you just mentioned isn't really much of an answer tbh. I feel like the key ideas behind proofs always have some intuition behind them, which is what originally drove the person who dicovered that proof.

[@0x8055]

it dare saying it comes from phisics as dr peyam stated. im my engeneering classes as good as 15yrs ago i remember thinking something like "thats obvious, u always lose some energy on the interaction". so as phisics is a good client for math its good to have vectors abeying cs inequality.

[@inthebackwiththerabbish]

8:12 baby don't hurt me

[@azizketata3241]

i like your energy .

[@witthawasphanthawimol5534]

Thank you very much😊

[@hazza6915]

Thinking about the square root step. x^2=y^2 doesn’t necessarily imply x=y

[@andreaLA222]

Thanks for the video!

[@NabeelMustafaSMM]

excellent way

[@martinsaidi1303]

neatly done, heyy!!

[@ronoguern]

You are great! Thanks!

[@ketara1234od]

THANK YOU !

[@souverain1er]

Very nice

[@chelangatedwin6112]

👍👍👍 your the best

[@jgoep2310]

You're great!

[@joshuafreeman9532]

mind frickin blown

[@fredericchopin255]

9:17 no supreme t-shirt mmmm

[@fountainovaphilosopher8112]

Hello!

[@rasikajayathilaka3516]

He was very good!

[@wreza97]

WooooooW 🙏❤️

[@couldia_chan6310]

GOOD!我靠，这证明可以，挺好的。

[@teshh_958]

its a great show im happy😅😅

[@koungmeng]

wow blackpenredpen is in your vdo nice

[@algonte]

Bravo

[@anavictoriaarmas3184]

OMG, I love you

[@arthurlbn]

Amazing nice