Cauchy Schwarz Proof
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- Опубліковано 4 жов 2024
- This is one of my favorite math proofs! Usually the Cauchy-Schwarz inequality is proven using projections, but this proof is completely elementary. It is taken from Pugh's Real Mathematical Analysis-book. But beware it doesn't work for complex inner product spaces. Bon appétit! :)
Note: I forgot to mention: If a = 0, then v = 0, but then the statement is trivial.
The one dislike must from someone who saw the words "discriminant" and "inequality" and got offended.
BEST COMMENT EVER!! LOL No time for snowflakes in mathematics. We've all business here.
probably from people who can't appreciate Dr. Peyam's enthusiasm
The camera man is actually blackpenredpen!
Aggredd
"What is love?"
-Dr. Peyam
This is an excellent, clear proof. Now I finally get it. Thank you 🙏🏽
In France, it’s called Cauchy inequality
In Germany, it’s Schwarz inequality
In Russia, it’s just inequality
😂 LOL
😂😂😂
Well Russian called it Bunyakovsky inequality
Syahimi Afiq cauchy was his doctoral advisor. My two cents
Actually, Bunyakovsky published it 25 years earlier than Schwarz.. ;)
Nope in France its Cauchy-Schwarz
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.
"it's better than watching TV " 😂 I love you Prof. Tough crowd
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!
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.
You need to let your brain time to develop.
@@datboi_wild1222 did you figure it out now?
@@thedoublehelix5661 lol, ya
I am writing in 5 hrs and I needed you to boost my confidence. Thank you.
Good luck!!!
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!
I'm a regular viewer of Dr. Peyam...his derivation methods are always unique.!!
This is an amazing proof! Way easier to understand than the one my real analysis prof gave us. Thank you
Wow!! You get so many people that have same interest and passion as you, so fortunate
Such an elegant proof! Thank you for sharing.
Great work! I love the way you speak!
Very elegant proof. Easy to understand. Thank you!
Watching this before Linear Algebra Final!!! Absolutely love this proof!!
Great video! That was a really darn good proof!
Thanks for explaining the notation! It helped a lot!
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.
the proof was short and understandable way.Thanks a lot.
We were JUST proving this for homework in my honors calc class for our homework! Neat!
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
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! :)
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
Thank you very much . Absolutely clear and easy . You saved me
Really great video!! ❤And your great personality makes everything so interesting and fun 👌
thanks 😊 your video help me a lot to understand this proof ☺️
This was good-you kinda didn't address the part where you wrote out a piece (cut off for some reason)but good stuff.
Divertido e fantástico! Parabéns pelo trabalho!
Very impressive video as always, i wish you would do some damage on more advanced topics; the world needs you there.
Can’t thank you enough!! Excellent proof :)
This is a really nice and neat proof
thanks alot Dr Peyam.
this video helped me alot.
amazing! The other proofs are just incomprehensible
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
Thanks a lot man!! You saved my life one day before my final exam xD
Very nice proof indeed so easy to understand
Luckily we do not have to prove it in high school.
(It was a question in AL Pure Maths)
amazing explanation
great proof, Dr. Peyam!
camera man just watches like others in that room.
Thank you!!! Great channel ❤️
I got a little but happy to see the proof!
Good presentation!
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||
Your method only works in R^n, unfortunately
Dr. Peyam's Show ? How does this matter
@@avdrago7170 --> I think Dr. Peyam's Show meant R^2.
This is beautiful... Perfection...
I've show my cat this proof
Now she still a cat
Thank you so much! Greetings from Chicago :)
The city of Lou Malnati’s 😋😋😋
very good explanation :)
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.
please make video on Minkowski inequality ..I really missed that
Beautiful proof, thanks.
You can reduce the proof to a line using geometric algebra.
How?
@@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}
dank, this works for non-euclidean metrics, too
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
👍💕💕
What was your thought process behind Step 1? How did you get (u+tv)(u+tv)?
Wow, someone who actually pronounces both names correctly
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?
This proof is for real spaces only
ok, thank you Dr
neat proof! and it comes with at least two bad puns.
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
For real numbers the statement is trivial, it becomes |ab|
Thank you very much😊
thank you sir this is so much clear☺
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?
You meant negative in y?
But sometimes after taking square roots inequality sign changes doesn't it??
No, things are positive here
@@drpeyam not getting can u explain it in detail
THANK YOU !
Can you please make a video with Newton-Leibniz formula proof?
Anatolie Mogoreanu The proof of the product rule? I think you've read my mind, because it's coming on Friday :)
Actually I meant the definite integral calculation formula F(b)-F(a)
Ooooh, the FTC!!! It's on my to-do list :P
Great! Thanks in advance)
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
neatly done, heyy!!
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||
Sebastian Cor Your proof is correct, but only works in R^2; this proof is valid for more general dot products!
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.
This channel is also very cool
Thanks for the video!
i like your energy .
You are great! Thanks!
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.
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?
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.
SanCHEneering It's valid in all of R^n :) The polynomial is quadratic, but the vectors u and v are in R^n
Do they need to explicitly be in R^n? Wouldn't it be sufficient to be on a real-scalar-product-space?
That's correct, it works on any real inner product space!
Hi, I love your videos!
A Ramanujan equation proof would be awesome if you would make a video about it.
Thinking about the square root step. x^2=y^2 doesn’t necessarily imply x=y
Very nice
absolute genius
👍👍👍 your the best
why not start a proof from the fact that "A dot B = ||A|| ||B|| cos(theta)" ?
we know -1
No, this only works in R2; my proof works in Rn and more!
Dr. Peyam's Show thanks!
why can u use the disciminant ? and why it have to be less then 0
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
@@drpeyam okay but how can I explain in my proof ? We just can use the discriminant ? Same like we use root and so on ?
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
sir its awesome proof
excellent way
More videos on Maths!
mind frickin blown
8:12 baby don't hurt me
You're great!
WooooooW 🙏❤️
Bravo
He was very good!
its a great show im happy😅😅
GOOD!我靠，这证明可以，挺好的。
Amazing nice
Hello!
Wow super clever wowoww
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.
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|
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?
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!
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.
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.
9:17 no supreme t-shirt mmmm
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