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.
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.
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!
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
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.
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
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 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}
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?
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||
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.
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?
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 👍💕💕
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
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
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
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.
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.
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.
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
"What is love?"
-Dr. Peyam
This is an excellent, clear proof. Now I finally get it. Thank you 🙏🏽
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
"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 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 am writing in 5 hrs and I needed you to boost my confidence. Thank you.
Good luck!!!
I'm a regular viewer of Dr. Peyam...his derivation methods are always unique.!!
The camera man is actually blackpenredpen!
Aggredd
I recognized instantly
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
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
Great work! I love the way you speak!
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.
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! :)
Very elegant proof. Easy to understand. Thank you!
Watching this before Linear Algebra Final!!! Absolutely love this proof!!
Such an elegant proof! Thank you for sharing.
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?
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
Very impressive video as always, i wish you would do some damage on more advanced topics; the world needs you there.
We were JUST proving this for homework in my honors calc class for our homework! Neat!
Great video! That was a really darn good proof!
Really great video!! ❤And your great personality makes everything so interesting and fun 👌
Thanks for explaining the notation! It helped a lot!
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
Thanks a lot man!! You saved my life one day before my final exam xD
the proof was short and understandable way.Thanks a lot.
Can’t thank you enough!! Excellent proof :)
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.
Thank you very much . Absolutely clear and easy . You saved me
This was good-you kinda didn't address the part where you wrote out a piece (cut off for some reason)but good stuff.
camera man just watches like others in that room.
What was your thought process behind Step 1? How did you get (u+tv)(u+tv)?
thanks 😊 your video help me a lot to understand this proof ☺️
great proof, Dr. Peyam!
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}
please make video on Minkowski inequality ..I really missed that
Luckily we do not have to prove it in high school.
(It was a question in AL Pure Maths)
This is a really nice and neat proof
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 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.
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.
Wow, someone who actually pronounces both names correctly
I've show my cat this proof
Now she still a cat
dank, this works for non-euclidean metrics, too
Thank you!!! Great channel ❤️
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)
Thank you so much! Greetings from Chicago :)
The city of Lou Malnati’s 😋😋😋
thanks alot Dr Peyam.
this video helped me alot.
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
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
neat proof! and it comes with at least two bad puns.
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
👍💕💕
Beautiful proof, thanks.
I got a little but happy to see the proof!
thank you sir this is so much clear☺
Very nice proof indeed so easy to understand
8:12 baby don't hurt me
Good presentation!
Divertido e fantástico! Parabéns pelo trabalho!
very good explanation :)
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
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|
amazing explanation
This is beautiful... Perfection...
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
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!
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.
This channel is also very cool
More videos on Maths!
amazing! The other proofs are just incomprehensible
sir its awesome proof
Thank you very much😊
absolute genius
Thanks for the video!
neatly done, heyy!!
Thinking about the square root step. x^2=y^2 doesn’t necessarily imply x=y
9:17 no supreme t-shirt mmmm
You are great! Thanks!
i like your energy .
excellent way
👍👍👍 your the best
Hello!
هل تتكلم اللغة العربية؟
نعم
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.
wow blackpenredpen is in your vdo nice
You're great!
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.
THANK YOU !
mind frickin blown
Very nice
its a great show im happy😅😅
WooooooW 🙏❤️
GOOD!我靠，这证明可以，挺好的。
He was very good!
عندكم الاساتدة بحال الحماااااق
1
St
Wow super clever wowoww