Cauchy Schwarz Proof

[@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

[@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.

[@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

[@MarkJay]

"What is love?" -Dr. Peyam

[@letladisebesho4503]

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

[@jamesclerkmaxwell676]

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

[@mokouf3]

The camera man is actually blackpenredpen!

[@StreamsofEdification]

Aggredd

[@jeanlucas2834]

I recognized instantly

[@afifakimih8823]

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

[@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!!!

[@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!

[@randywilton3466]

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

[@chessislife3429]

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

[@stevenbi7495]

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

[@josephinell314]

Such an elegant proof! Thank you for sharing.

[@drandrewsanchez]

Great work! I love the way you speak!

[@pheasant139]

Very elegant proof. Easy to understand. Thank you!

[@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.

[@loganreina2290]

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

[@minyhailabera3616]

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

[@sandnstars8]

Thanks for explaining the notation! It helped a lot!

[@RealEverythingComputers]

Great video! That was a really darn good proof!

[@joudy5767]

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

[@fardinzaman904]

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

[@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

[@lLl-fl7rv]

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

[@sdmartens22]

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

[@syahimiafiq5914]

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

[@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! :)

[@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

[@mokoepa]

This is beautiful... Perfection...

[@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.

[@LegendOfMurray]

great proof, Dr. Peyam!

[@sweetychaudhary277]

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

[@gabrielnegrini8889]

amazing explanation

[@Zhinoi]

This is a really nice and neat proof

[@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.

[@arjunnarasimhan8027]

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

[@luizr.8316]

Divertido e fantástico! Parabéns pelo trabalho!

[@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.

[@starfire8679]

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

[@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?

[@varungoyal4732]

thanks alot Dr Peyam. this video helped me alot.

[@boydmwansa8750]

Very nice proof indeed so easy to understand

[@pibeeulotro9603]

Beautiful proof, thanks.

[@vijay85321]

I got a little but happy to see the proof!

[@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!

[@ndriqa]

Thank you!!! Great channel ❤️

[@DavidPumpernickel]

dank, this works for non-euclidean metrics, too

[@mihlalimjacu6984]

thank you sir this is so much clear☺

[@joshuafreeman9532]

absolute genius

[@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

[@jennifera.3208]

very good explanation :)

[@railgun.__.7055]

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

[@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)

[@abstractalien12345]

Thank you so much! Greetings from Chicago :)

[@drpeyam]

The city of Lou Malnati’s 😋😋😋

[@witthawasphanthawimol5534]

Thank you very much😊

[@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

[@andreaLA222]

Thanks for the video!

[@MrJapogm]

This channel is also very cool

[@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.

[@NabeelMustafaSMM]

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

[@ronoguern]

You are great! Thanks!

[@stevekaszycki8629]

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

[@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

[@MathematicsMadeSimple1]

Good presentation!

[@ketara1234od]

THANK YOU !

[@inthebackwiththerabbish]

8:12 baby don't hurt 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}

[@azizketata3241]

i like your energy .

[@kenmeyer100]

Wow, someone who actually pronounces both names correctly

[@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

[@PervezAli112]

sir its awesome proof

[@malicksoumare370]

amazing! The other proofs are just incomprehensible

[@martinsaidi1303]

neatly done, heyy!!

[@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 👍💕💕

[@NabeelMustafaSMM]

excellent way

[@souverain1er]

Very nice

[@rasikajayathilaka3516]

He was very good!

[@favoriteflavor46]

camera man just watches like others in that room.

[@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|

[@chelangatedwin6112]

👍👍👍 your the best

[@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!

[@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.

[@jgoep2310]

You're great!

[@joshuafreeman9532]

mind frickin blown

[@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

[@fredericchopin255]

9:17 no supreme t-shirt mmmm

[@algonte]

Bravo

[@jonsnow9246]

More videos on Maths!

[@teshh_958]

its a great show im happy😅😅

[@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.

[@fountainovaphilosopher8112]

Hello!

[@wreza97]

WooooooW 🙏❤️

[@couldia_chan6310]

GOOD!我靠，这证明可以，挺好的。

[@arthurlbn]

Amazing nice

[@prozpectyoung0558]

1 St

[@BlackPillHurts]

Wow super clever wowoww

[@koungmeng]

wow blackpenredpen is in your vdo nice

[@KarlMarg]

I love you

[@drpeyam]

❤️