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I just learnt about the Cauchy Schwartz inequality, and I'm so glad I did

@BriTheMathGuy

3 жыл бұрын

Glad to hear it!

@lambdasaturn

3 жыл бұрын

It is also known as the Cauchy-Bunyakovsky-Schwarz inequality

@leif1075

3 жыл бұрын

@@BriTheMathGuy Is there some other restriction on the sequence..are they increasing and you cam solve by trial.and error if you dont knkw the cauchy Schwartz inequality right??And is that sequence supposed to be infinite kr it can be finite? I'm not sure you clarified?

@bigbroiswatchingyou2137

7 ай бұрын

same

Realized I said "n" at the beginning when I should have said "m" and at 1:08-2:07 the sequence should have been "a_i" (not a_n) to match the summation notation. Sorry everyone! I don't know why I couldn't seem to get these indices right 😂

@Nothingtonnobodson

3 жыл бұрын

Hey brian I am a ninth grader and a fan of yours kzread.info/dash/bejne/eodhzqORmJXgos4.html plz see this video it has an interesting radicle equation and even if you don't see the video (which you probably are going to skip ) your videos are great and the integrals are awesome keep it up

@michaeltamajong4659

3 жыл бұрын

I get it! n seems to be more popular than m in math notation🤓🤓

@TheDigiWorld

9 ай бұрын

​@@michaeltamajong4659true

So neat, I’ve never seen that inequality before but it clicked nicely

@BriTheMathGuy

3 жыл бұрын

Great! Thanks for watching!

@aashsyed1277

3 жыл бұрын

@@BriTheMathGuy YOU ARE SO AWESOME!

Man's getting pumped af

@BriTheMathGuy

3 жыл бұрын

😅

Using Cauchy-Schwarz inequality seems a bit problematic, when all you know is that every finite part of the infinite series is less than the entire infinite series. This problem really asks for Hölder's inequality, if you don't want to go through a bunch of arguments, that things should hold in the limit as well.

BTMG - it's insane how good your videos have gotten. I just took a peek at one of your first videos...hand-drawn equations, no face, rough audio. And now you look at today's video and it's a masterpiece. Your equation animations are especially impressive. I have a Fourier series video coming out later today with a surprise ending that I think you'll love (and it's related to one of yours). Lmk what you think!

Please continue doing Putnam problems....🥰

@BriTheMathGuy

3 жыл бұрын

I'll do my best!

@harish6787

3 жыл бұрын

@@BriTheMathGuy ok thanks bro

Please, more infinite sequences, sums and analysis types of problems. They in my opinion hold the most elegant solutions. Thanks for the video.

thanks so much!

@BriTheMathGuy

3 жыл бұрын

You're welcome!

I have to say that I've never come across a similar math problem on a KZread math channel before. This was something a bit different and I really enjoyed it; thanks for sharing!

@BriTheMathGuy

2 жыл бұрын

Wow, thanks! Have a great day!

Great video dude as always. You're amazing

@BriTheMathGuy

3 жыл бұрын

You are!

I love these short, yet informative math videos. My suggestion is you do Fermat's theorem next. Pull out the big guns

@BriTheMathGuy

3 жыл бұрын

Glad you enjoy them! (I'd like to do a video on Fermat's theorem, just might take some time/ton of research)

Dude you looked ripped. Workout video when?

@BriTheMathGuy

3 жыл бұрын

😅

This was so neat OMG

Thanks!

@BriTheMathGuy

2 жыл бұрын

You bet!

I think there is a little intuition in using the Cauchy-Schwarz inequality. I've often seen cases where that inequality comes in handy for no apparent reason. So, might as well try it here.

Amazing!

@BriTheMathGuy

3 жыл бұрын

Thank you! Cheers!

@broroman5979

3 жыл бұрын

*KZread CHANNEL FOR MATHLOVERS* 🛑 *USEFUL FOR MANY OLYMPIADS LIKE AMC , UKET , IOQM etc* --------------- 🛑 *USEFUL FOR JEE MAIN & ADVANCE, ISI, CMI ASPIRANTS* ------------------- 🛑 *TUTORED BY KOWSHIQ KATTAMURI, PURSUING BTECH FROM NIT CALICUT, WITH 5 YEARS OF TEACHING EXPERIENCE* --------------- 🛑 *HE HAS LOT OF ACHIEVEMENTS IN FIELD OF MATHEMATICS* --------------- 🛑 *AUTHOR OF THE ARTICLE "HIDDEN FACTS OF CALCULUS"* ------------- 🛑 *AWARDED WITH ELITE CERTIFICATION FROM NPTEL IN FIELD OF CALCULUS* --------------- 🛑 *KNOWN FOR HIS EXCELLENCE IN CALCULUS AND ALGEBRA * ‼️ *EXCITING PROBLEM OF THE DAY* ‼️ kzread.info/head/PLz_Iv-Gj5IOZ-aGiWnC_SjSA2eGIl3FvI ‼️ *KZread CHANNEL LINK* ‼️ kzread.info/dron/2vtt2gkjDU6YzfsaxuhNSg.html *SUBSCRIBE HIS CHANNEL TO JOIN THE ULTIMATE VOYAGE OF MATHEMATICS* 🔥🔥🔥🔥

Damn the solution is so satisfying.

i like you so much! IT'S SO NICELY done!

@BriTheMathGuy

3 жыл бұрын

Thank you so much!!

Would love to see more Putnam stuff

@BriTheMathGuy

3 жыл бұрын

I'll do my best to make more!

Awsome video...

@BriTheMathGuy

3 жыл бұрын

Glad you thought so!

My way was very ugly, but does use a little more intuition: Let L(m) be the limit for m. Step 1) limit yourself to even numbers for m so that you only need to consider non-negative numbers. Step 2) At least one a_n must be at least 1 (else L(4)

Thank for sharing a great this video ! Love it 😍 so much sir

@BriTheMathGuy

3 жыл бұрын

My pleasure! Thanks for watching!

@MathZoneKH

3 жыл бұрын

I always watch your videos , because your content is my favorite one 😊❤️

@BriTheMathGuy

3 жыл бұрын

@@MathZoneKH Thank you so much!! :)

What if you turned the question around? i.e. is there a sequence meeting this equation for every positive integer m?

Using ramanujan summation u can do it ultimately

Another solution without the C-S inequality. For every k and n we have that Σ(a_m)^(2n) = 2n => (a_k)^(2n) = |a_k| = So, taking the limit as n approaches infinity, we get that |a_k| = 4 = Σ(a_k)^4 = Σ|a_k|^4 = Contradiction.

Not sure if this is correct, but here is my attempt to solve this. Suppose there is such sequence. There may be two possibilities. Possibility one: there's at least one a_n, s.t. |a_n|>1. If there's at least one a_n>1, call it a_{n_0}, then we may choose a sufficiently big m=2k, s.t. (a_{n_0})^(2k)>2k and everything else is non-negative, so the series isn't less then (a_{n_0})^(m) which is already bigger than m. Contradiction, so all a_n should be less then or equal 1. Possibility two: all a_n are s.t |a_n|

@Eulercrosser

3 жыл бұрын

I like how you dealt with the case of at least one |a_n|>1 better than I did.

The sequence exists and is this: a_n = mth root of (m/(2^n))

you just calculate 0=2^2-4=(Sum of a^2)^2-(Sum of a^4)=Sum of squares, like 2*(a1*a2)^2, this means, we have set of numbers (±sqrt(2),0,0,0,...), but it doesn't work for m=1

I sir I am from India can you put videos for jee maths which is one of the toughest exams of the world Pls Pls Pls

Nice one. But how about complex numbers?🤔🤔🤔

@AxelStrem

2 жыл бұрын

I think you might be able to construct such series for complex numbers using the properties of complex roots of 1

I don’t know if I’d say there’s no intuition. Cauchy-Schwarz offers a slick solution, but I think you can show such a sequence cannot exist by considering m = 2k and examining the sizes of the entries in the sequence. It’s not as nice of a solution, but it feels natural to work with the problem this way. The intuition behind where this problem came from is another story.

@tavishu

3 жыл бұрын

Can you expand on that approach?

@seitanarchist

3 жыл бұрын

​@@tavishu Sure! (Warning: this is very long for a youtube comment XD ) Consider m = 2k for k >=1. The sum condition in the problem now becomes that the sum from n = 1 to infinity of (a_n)^(2k) is equal to 2k. A key observation is that (a_n)^(2k) is nonnegative for all positive integers n and k. But how big can the terms get? Let's focus on the case k = 1 for now, so that m = 2. If the terms (a_n)^2 sum to 2, then the terms of the sequence can't be very large at all. Certainly, we must have (a_n)^2 1) |a_n| 2) |a_n| >= 1 for exactly two values of n. In this case, if a third term is anything other than zero, the terms (a_n)^2 sum to something greater than 2. If either of the nonzero terms is actually greater than one, then again the sum is greater than 2. So we must have |a_n| = 1 for these two values of n, and a_n = 0 for all the rest. But then for any other value of k, the terms (a_n)^(2k) still sum to 2, contradicting the sum condition. 3) |a_n| >= 1 for exactly one value of n. Let's give it a specific name, N, to work with it. This case has two sub-cases. (3a) |a_N| = 1. In this case, a modification of case (1) works: the sum of all the fourth power terms must again be less than the sum of the squared terms, because (a_N)^2 = (a_N)^4 = 1, but (a_n)^4 (3b) |a_N| > 1. This case is a bit trickier, because if k > 1, then (a_N)^2 But this relation has to be true for *every* k >=1. It cannot be: the expression (a_N)^(2k) eventually grows much faster than 2k as k increases. One only has to ask: when is (a_N)^(2k) > 2k, or equivalently, when is |a_N|^k > sqrt(2k)? Applying a logarithm to both sides and doing a little manipulation, we obtain log |a_N| > log(sqrt(2k)) / k = log(2k) / (2k). The function log(2k)/(2k) has limit zero as k increases, so it will eventually be smaller than the fixed positive number log |a_N|. This tells us that (a_N)^(2k) > 2k for some value of k, which contradicts the sum condition. Having (finally) exhausted all cases, we seem to have ruled out the existence of such a sequence. I am sure this solution could be pared down a bit, but it's what I came up with before watching Bri's solution. It takes a bit of effort to write it down, especially without knowing your background. There is nothing slick about this, but sometimes the clunkier method can be a bit more intuitive. Most of the facts I used are rather elementary, although the bit about exponential growth and logarithms at the end involves some slightly heavier machinery. I certainly wouldn't want to spend time writing this out on the Putnam! The Cauchy-Scwharz solution, by contrast, is elegant and is just one of many examples of how powerful that inequality can be.

If you let a be a positive real number greater than zero then m = ∞ should work. Elsewise great video Bri, love the shorter more concise vids.

@pinguino55h40

3 жыл бұрын

I don't think this works because infinity is not a number (in fact infinity does not exist) so you can't just substitute infinity in an expression. You can take limits in infinity but you can't manipulate infinity as if it was a concrete number because it isn't. Furthermore, you can't say one side of the equality equals the other just because both diverge since there are different "magnitudes" of infinity. Some infinities are "bigger than others. More than magnitudes I like to think about them as the pattern that the infinity follows. For example, in this case when m grows the left side will grow in a different way than the right side. The right side is a linear function while the left side is not, it is not difficult to see that the left side will grow much faster than the right side when m tends to infinity if every term of the sequence is greater than one, so this equation will not hold when m goes to infinity if every term of the sequence is greater than one. Likewise, if every term of the sequence is less than one, when m tends to infinity the left side will go to 0 while the right side will blow up to infinity in a linear way. It is a matter of analyzing the pattern in which a function changes when you let a variable tend to infinity. To maintain equality you need to ensure that the pattern will hold for concrete values of m, not "infinity" which is not a concrete value.

@magicmulder

3 жыл бұрын

The question was if there is a series (a_n) that works for every m = 1, 2, 3, ... Not if there is one series and one m.

Correct me if I'm wrong, but isn't this step function a solution? a_n = 1 , if 1

@m4riel

3 жыл бұрын

I mean, it really seems that there are no smooth/continuous/analytic functions that satisfy it, but the problem doesn't seem to specify which kind of sequences are valid.

Alright, alright... WAIT WHAT IS THAT IT?!

I enjoy math problem solving, but I realize that I do not find it easy solving open ended problems such as IMO, Putnam, IMC. I really want to be better at discovering tricks and methods, using the math knowledge I have.

Do you intentionally post right after Michael Penn releases a video?

@BriTheMathGuy

3 жыл бұрын

Nah, lately I've just been posting at the same time each Monday morning. Thanks for watching!

@faiselbutt2944

3 жыл бұрын

@@BriTheMathGuy Thanks for clarifying!

there is a different solution if we limit ourselves to the situation m = 2l assuming a solution exists we immediately can deduce a solution to the problem sum(b_n^l)=2l with b_n = a_n^2. this implies that all b_n>=0. the right-hand side of this problem is monotonously increasing with l this implies that there has to be at least 1 b_n with b_n >1 because else the left-hand side would be decreasing. But for all x >1 there exist l such that x^l >2l this implies b_n

set a_1=m^(1/m) and a_i=0 for i>1

@magicmulder

3 жыл бұрын

The same (a_n) series has to work for *all* m. That's the point.

This one was really interesting😐😯

@BriTheMathGuy

3 жыл бұрын

Glad you thought so! Thanks for watching!

I just put in 0 for m and since anything to the power 0 is 1, the sequence becomes 1+1+1+1+1+1........ = 0 Which is obviously false Maybe not as rigorous but I think it still works

I wonder if he said: "I'll see you in that one" in his first video. He did not.

@BriTheMathGuy

3 жыл бұрын

I wish I had!

Hey can you please prove why all non happy number always form a loop .

@BriTheMathGuy

3 жыл бұрын

I'm not familiar with this concept!

@piyushpathak1186

3 жыл бұрын

@@BriTheMathGuy a happy number is a number which eventually reaches 1 when replaced by the sum of the square of each digit. For instance, 13 is a happy number because 1^2 + 3^2 = 10 and then 1^2 + 0^2 = 1, and . On the other hand, 4 is not a happy number because 0^2+4^2 = 16 , 1^2 + 6^2 = 37 ...eventually reaches 2 ^2 + 0^2 = 4 so the process continues in an infinite cycle without ever reaching 1. Wikipedia definition Since there are Infinite number How they know all non happy number always form a loop without ever reaching to 1 There must be a way to prove it. Anyone in the comment section know how to prove it please share . Thanks in advance 😊😊😊

Come on

Aww i thought it was possible!! I would take a common series like.. 1+1/2+1/4...........= 2 And square out each element 1^2 + (1/√2)^2 + (1/2)^2 +......... = 2 Then 1,1/√2,1/2... Is my sequence Where am i wrong here?

@physira7551

3 жыл бұрын

Oh i see every positive number m I understand

So a high school Putnam test taker should be able to execute a proof by contradiction and know the Cauchy-Schwartz Inequality?

@zmaj12321

3 жыл бұрын

Yes. exactly. Proof by contradiction is an essential proof type, and the Cauchy-Schwartz inequality is a particularly helpful one when trying to create proofs.

You didn’t answer the question, you inverted it in your head. You said finding one sequence that does not work disproves it. In reality, where the question is “is there such a sequence”, finding one sequence that DOES work would PROVE it.

Very unfortunate use of 'sub n' in the summations indexed by 'i' - or did I just miss the whole point?

@BriTheMathGuy

3 жыл бұрын

Yeah you're right! I messed up - sorry about that.

@Rodhern

3 жыл бұрын

@@BriTheMathGuy No problem at all. It is that old property of mistakes, when you first create them they will patiently hide on the blackboard in plain sight, then when others look at them they jump off the board with commotion, the sneaky buggers. The thing that threw me off was the conviction in your tone of voice saying "a sub n" - did you record the sound before or after the graphics? [edit: spelling].

@BriTheMathGuy

3 жыл бұрын

@@Rodhern I record video/sound before adding in the text/animation, things just slip through the cracks sometimes 😐

I'm scared what is this

@BriTheMathGuy

3 жыл бұрын

You can do it!

Helpful for JEE Advanced

@BriTheMathGuy

3 жыл бұрын

Glad you think so!

OK but what If you not demand its working for every m but only that there exists an m? Then you couldnt use cauchy schwarz because you only now that the sum is equal to ONE m and not for every

@magicmulder

3 жыл бұрын

The question of existence of one solution is trivial, just use m = 2 and a_n = 1/n.

@kurax9115

3 жыл бұрын

@@magicmulder (1/1)^2+(1/2)^2+(1/3)^2+... = Sum k=1 to infty 1/k^2 == pi^2/6 Not 2 An example would be for m=1 with a_n=(1/2)^n

a_1 = mth root of m, every other number is 0...

@HA7DN

3 жыл бұрын

Oh you mean the SAME sequence has to work for every m? Damn I can't read...

Can u pick up an insanely tricky jee adv question for ur next vdo

@BriTheMathGuy

3 жыл бұрын

I'll look into it!

@shubhajyotidebnath5651

3 жыл бұрын

@@BriTheMathGuy Thanks keep up ur hardwork

0:13 ok then guess I clicked on the wrong video

I think you use manim, do you?

@BriTheMathGuy

3 жыл бұрын

I’m not sure what that is 🤷

@YT-uo7fc

3 жыл бұрын

@@BriTheMathGuy It is math animations software initially developed by Grant Sanderson which software you use ? I suggest you to check out manim.

what just happened??? :-)

@BriTheMathGuy

2 жыл бұрын

🤷‍♂️

والله انك رجال

@BriTheMathGuy

3 жыл бұрын

I don't know what this says but have a great day!

@user-ps1dm4fc4l

3 жыл бұрын

@@BriTheMathGuy والله انك رجال = God you are men =you are creative 😂😂😂😂👍

“Is there an infinite sequence” didn’t you just find one sequence that didn’t work? Why does that prove that there are no infinite sequences whatsoever that would satisfy this?

@BriTheMathGuy

2 жыл бұрын

The assumption was that such a sequence existed (we didn't know exactly what it was, we just assumed it and used its properties). After we saw that this (arbitrary) sequence led to a contradiction, we can end the problem. :)

@johnhippisley9106

2 жыл бұрын

@@BriTheMathGuy Ahhh I see. Thanks!

Can you take jee advanced question

@BriTheMathGuy

3 жыл бұрын

I'll do my best!

First

@BriTheMathGuy

3 жыл бұрын

Thanks for watching!