British Math Olympiad | 2009 Round 2 Question 3

Рет қаралды 36,742

3 жыл бұрын

We solve a nice functional equation problem from the British math olympiad.
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Пікірлер: 108

@schumensch 3 жыл бұрын

Awesome content! I'm a student from germany who is actively participating in Math-Olympiade type contests here and this is my favourite channel showcasing these types of problems.

@friedrichotto5675 3 жыл бұрын

Welcher Jahrgang bist du? Schon mal in die Bundesrunde geschafft?

@AnkhArcRod 3 жыл бұрын

I love your content. No abuses, no trivialities, just beautiful math on board. Finally, I found a channel that does not hold back punches. Will you be doing expositionary content as well?

@goodplacetostop2973 3 жыл бұрын

13:00 Wherever you are I hope you’re having a great day/night. Love you all. Other than that, I was thinking about an AMA. Maybe I’ll do that some day for anyone caring about me lol

@pbj4184 3 жыл бұрын

Count me in!

@djvalentedochp 3 жыл бұрын

what's an AMA?

@goodplacetostop2973 3 жыл бұрын

DJ VALENTE DO CHP AMA = Ask Me Anything. I have people asking questions about me from time to time, so yeah

@djvalentedochp 3 жыл бұрын

@@goodplacetostop2973 It would be great to watch it 😂

@aakashchakrabarty7714 3 жыл бұрын

Ya sure just mention us the time

@elgourmetdotcom 3 жыл бұрын

Beautiful problem Michael thanks a lot! I appreciate that you always include exercises and theory of different levels of difficulty for the broad mathematical fans audice. Thanks!

@sahilbaori9052 3 жыл бұрын

Since the equation is homogenous, we can say that it'll work for any "a" without actually plugging and checking it.

@henrikolsson3626 3 жыл бұрын

What does homogenous mean? I might know it as I don't learn math in English, is it that I can swap x and y around and it won't affect the equation?

@quantumcaffeine 3 жыл бұрын

@@henrikolsson3626 "Homogeneous" means (broadly) "all the same". In this case, it means that the function f always appears to the same power: there are only linear terms in f. Given that, and the fact that f is a linear function, we know that changing a will simply scale each of the terms by the same amount and the equation will still be valid. Hope that helps!

@sahilbaori9052 3 жыл бұрын

@Adam Romanov Okay, first off, f(x)=ax is not the solution of your functional equation. And if you are setting a=0 anyway, then f(x) = a(sinx) will also satisfy :/ Second, I only made the homogeneous argument once I saw that the solution of the equation is a polynomial.

@CauchyIntegralFormula 3 жыл бұрын

@Adam Romanov We can make the argument that if f(x) is a solution, then g(x) = cf(x) is also a solution for any constant c. This is really easy to see: take the equation and multiply through by c, and distribute as appropriate. The fact that this distribution works cleanly is what is meant by homogeneous

@CauchyIntegralFormula 3 жыл бұрын

@@sahilbaori9052 But the point stands that f(x) = ax containing all possible solutions, and the equation being homogeneous, don't necessarily give you that a can be any real number. You could also find that a can only be 0, and you could also find that there are no solutions whatsoever. Those would both be consistent with the homogeneity. The homogeneity only tells you that if any non-zero real number works, then they all work, but A=>B is not equivalent to just B.

@himanshusalunkhe88 3 жыл бұрын

I dont really comment on youtube videos, but I just couldnt resist saying that you are way underrated. Your content is gold and i really wish you had way more subscribers and i hope you are happy.

@joshuajodrian7364 3 жыл бұрын

great content as always

@germandeu 3 жыл бұрын

Good teaching makes this kind of questions look so easy.

@jaredarell Жыл бұрын

In love with this problem

@youssefamen6872 3 жыл бұрын

Great work. May you pls take a look at PUTNAM 2001 A6

@utsav8981 3 жыл бұрын

Sir Great Explanation😊👍👍❤

@jeffreyhellrung 3 жыл бұрын

An alternative way to finish this solution is, from yf(x^2)+xf(y^2)=(x+y)f(xy), take y=x^3 and use f(x^6)=x^2f(x^4), and you pretty straightforwardly deduce f(x^2)=xf(x). Use this to further simplify the functional equation to xy(f(x)+f(y))=(x+y)f(xy), then take y=1.

@myxail0 3 жыл бұрын

nice

@TechToppers 3 жыл бұрын

Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m) = 20 and s(33m) = 120, what is the value of s(3m)? Can someone please help me with this? It came in India. PRMO 2019 25th August Paper

@Yochillybro 3 ай бұрын

3 years late but here you go,s(33m) can be written as S(11*3m) which is equal to s(3m*10+3m*1) since multiplying by 10 just adds a zero at the end, the sum of the digits does not change. hences(33m) is equal to s(6m), now since we know that s(33m)=120, s(6m)=120 so, s(3m)=s(6m)/2=60!!, see you at ioqm 2025

@cycklist 3 жыл бұрын

It's mathS in Britain mate! MathS!

@sahilbaori9052 3 жыл бұрын

Hello fellow bri ish

@birdboat5647 3 жыл бұрын

mafs

@kuldeepparashar7266 3 жыл бұрын

Thanks sir

@SF-wf6gm Жыл бұрын

As I read the problem it just reminded me the polynomial decomposition of the sum of two cubes. So if working with f(x)=x, would have worked for f(x)=ax as well. But of course that wouldn't be the proof for a generic solution. Great content!

@AlexSasha_ 3 жыл бұрын

Thank you, it is very interesting

@008Nikita 3 жыл бұрын

Русский

@buxeessingh2571 3 жыл бұрын

There are problems in advanced calculus books (e.g., Rudin's _Principles of Mathematical Analysis_) with functional equations for logarithms and exponentials. All of the hints you see here apply to those problems as well.

@WhiterockFTP 3 жыл бұрын

so this is equivalent to saying f is linear. I wonder if we can generalize this to 2 or even n dimensions?

@lucascruz3977 3 жыл бұрын

What is a good book for Coordinate Geometry with tough exercises?

@malawigw 3 жыл бұрын

100k subs let's go!

@bekhaddaderrar2111 3 жыл бұрын

It's awesome

@SalvoLoki 3 жыл бұрын

I solved it too! 1) x=y=0 - > f(0)=0 2) y=0 - > f(x^3)=x*f(x^2) 3) y=x - > f(x)=-f(-x) 4) y=1 and y=-1 give me two identities, adding it: f(x^3)=x*f(x^2)+x*f(1)-f(x) 5) from 2) and 4), f(x)=x*f(1) So, f(x)=ax with a=f(1) (it's an independent value). Very nice!

@martinschulz6832 3 жыл бұрын

12:36 "I've got a little itch back there, would you mind?”

@q45dedos 3 жыл бұрын

How do we know that don't exist non linear functions that satisfy the condition?

@lexhariepisco2119 3 жыл бұрын

Sir can you do some Philippine Mathematical Olympiad problems?

@xl000 3 жыл бұрын

Can you give use your routine to have a body like that ?

@ishdx9374 3 жыл бұрын

True

@Dana.king11 3 жыл бұрын

in 4:23 it is true only if y=0 so why we change whole equation

@Dana.king11 3 жыл бұрын

thank you bro ❤❤

@leif1075 3 жыл бұрын

At 9 :37 you could also let x equal 1 and it works out..

@Someone-cr8cj 3 жыл бұрын

by which criteria is a competition math question worthy of being featured in a video?

@malawigw 3 жыл бұрын

It has to have a good place to stop

@schweinmachtbree1013 3 жыл бұрын

@@malawigw god damnit

@AnkhArcRod 3 жыл бұрын

The criteria of fun math that most KZfaqrs don't like to touch in the fear that it does not reach masses.

@backyard282 3 жыл бұрын

In the end you didn't have to plug anything to prove that a can be anything because notice that for any linear function passing through origin, i.e. f(x) = ax, we have f(1)=a.

@joshuadavey6773 3 жыл бұрын

I like the Dune t shirt

@aks7451 3 жыл бұрын

let x=y to see that f(x³)=xf(x²). this means f(x^n)=x^{n/3}*f(x^{2n/3}). When n=1, then f(x)=x^{1/3}*f(x^{2/3}). Continuing this with n=2/3 gives f(x)=x^{1/3+2/9}*f(x^{4/9}). Continuing this with n=4/9 gives f(x)=x^{1/3+2/9+4/27}*f(x^{8/27}). We can repeat the process and see that f(x)=x*f(1).

@marcelgunadi2429 3 жыл бұрын

But can you do Euler’s identity?

@TechToppers 3 жыл бұрын

What's your idea bro??

@xiaoyang 3 жыл бұрын

wow

@hnnagarathna7286 3 жыл бұрын

Trinomial (three termed object ) lol 😂

@demr4921 Жыл бұрын

If I find f(x²)=0 can I then say f(1)=0 because f(1²)=f(x²)=0?

@pitreason 3 жыл бұрын

So I found another solution to this problem: I got the equation f(x^3)=xf(x^2) like it was in the video, than you can see that this equation defines function f iteratively: f(x^3)=x^(1+2/3+4/9+8/27+...+(2/3)^N) * f(x^(2*(2/3)^N)) You can prove that by induction easily. Than what you have in exponent of the first x on the rhs is geometric series 1+2/3+4/9+8/27+...+(2/3)^N and it approaches 1/(1-2/3)=3 when N goes to inf And what you got is f(x^3)=x^3*f(1). And the rest of the solution is similar to what is shown in the video One thing with that solution that cannot be fully right is we made an assumption that f is continuous in 1 to make f(x^(2*(2/3)^N)) approach f(1)

@CauchyIntegralFormula 3 жыл бұрын

Yeah, you've made an assumption of continuity here that we don't necessarily know to be true. You actually have to be very careful about continuity arguments in functional equations, because there might be pathological non-continuous solutions. For example, the very simple "for all reals x and y, f(x+y)=f(x)+f(y)" might seem to only be solved by f(x) = ax for a real. But that's only true for continuous f; a Hamel basis of the reals over the rationals gives rise to many other solutions, because there are actually many linear functions of R when we relax continuity

@pitreason 3 жыл бұрын

I guess it can be proven somehow that this is continuous. Maybe I'll do it some other time;)

@CauchyIntegralFormula 3 жыл бұрын

The fastest way to prove that this is continuous is to solve the problem a different way, and then note that all solutions are continuous

@swenji9113 3 жыл бұрын

@@pitreason I don't believe you'll get anything faster than resolving the whole problem. Continuity has absolutely no reason to hold here, nor in any algebraic functional equation. Actually it's quite often that functional equations have very few continuous solutions but many monstruously complicated non-continuous ones. Take f(x+y) = f(x)+f(y) for example. This might look simple but it's not even possible to determine if it has non-continuous solutions or not ...

@pitreason 3 жыл бұрын

Swenji look guys I’ve just suggested another way to solve this, I don’t insist that this one is better than showed in the video

@mathssolverpoint6059 3 жыл бұрын

I have solved this question in 2 minute in my head without copy and pen by comparing polynomial degrees rules

@AaronRotenberg 3 жыл бұрын

Did you do it in a way that proves that there cannot _also_ be non-polynomial solutions? For comparison, assuming the axiom of choice, Cauchy's functional equation f(x + y) = f(x) + f(y) has non-continuous solutions over the reals.

@Icenri 3 жыл бұрын

It's 'a' cubed at the end.

@CM63_France 3 жыл бұрын

Hi, This was before hair cutting, so I understand my request about the camera is not taken into account 😄

@Anonymous-ov1vh 3 жыл бұрын

It was a quite *_EASY_* question. I have done by differentiating partially with respect to x and then replacing x with 0 and y with x in the final equation. We get f(x) = x*f'(0). Where we can't find value of f'(0) bcz for any value of f'(0) all equations are being satisfied. Thereby f(x) = k*x (bcz f'(0) is a constant , so i have written k in place of f'(0) where k belongs to R). ~ ♥️ from INDIA

@jeffreyhellrung 3 жыл бұрын

I think the issue with your solution is you have to first justify that f is differentiable.

@Anonymous-ov1vh 3 жыл бұрын

@@jeffreyhellrung I have assumed that the function is differentiable and when i have solved the function comes out to be a linear function, therefore my assumption is correct. If I would have assumed that the function is non differnetiable then there will be no way to solve it! I'am gratefull that u have took interest in my solution. ❤

@jeffreyhellrung 3 жыл бұрын

@@Anonymous-ov1vh In a sense, you are correct. However, fundamentally, if you add additional assumptions to the original problem, you end up solving a different problem :/ For example, if you make a different assumption, say that the function is differentiable *and* f(1) = 1, then now you would conclude that the solution is f(x) = x, but this doesn't end up capturing the totality of all solutions to the original problem!

@Anonymous-ov1vh 3 жыл бұрын

@@jeffreyhellrung We can't assume that f(1) is a particular number but we can assume that f(1) is finite . We have to assume a broader possibility, and should prove the counter statement contradictory. (•‿•) By the way you are from which country ?

@jeffreyhellrung 3 жыл бұрын

@@Anonymous-ov1vh USA; you?

@switchoxford2787 3 жыл бұрын

Hello I'm Japanese collage student. I have a favorite KZfaqr, but it's a Japanese KZfaqr called Yobinori Takumi. I do a video called (Integral of the Week) once a week, so please take a look.

@clutcheu2248 3 жыл бұрын

im a 14 year old kid who isn't good at math why am i here

@chessematics 3 жыл бұрын

4:10 try not to cover the board like this, it hampers concentration and disturbs the view

@ancientwisdom7993 3 жыл бұрын

wow ! really ? I apologize on behalf of entire mathematical community that your precious concentration was hampered. Maybe Michael just become transparent so your view is not blocked when he is working on lower part of the board. SMH

@chessematics 3 жыл бұрын

@@ancientwisdom7993 see how blackpenredpen writes on the lower part of board. He never blocks vision completely

@myxail0 3 жыл бұрын

idk, Prof articulates his answers very well and if he would write on the board in such a way it would just take longer for him, and in my opinion it doesnt really affect the quality of the video. Blackpenredpen is just too proficient with his writing skills lol

@Packerfan130 Жыл бұрын

your habit of stating the answer before showing it really tunes me out You said you want to plug f(x) = ax to see what a can be and before you even do that, you say that a can be anything so there's no point in watching you plug f(x) = ax in because we already know. How about just plug in f(x) = ax and solve for a and find out what a can be after you solve it

@onderozenc4470 3 жыл бұрын

Who wants to win a pack of chocolate ?

@alonmaayan2938 3 жыл бұрын

Am I the only one finding this video kinda slow and aimed for kids like in 4:45 , I prefer the harder ones

@malawigw 3 жыл бұрын

Kids needs to learn math too

@TechToppers 3 жыл бұрын

Directly BMO??😅

@TechToppers 3 жыл бұрын

@PontusLarsson1 3 жыл бұрын

For the digit sum (base 10), we have the following two properties: s(10*a) = s(a), for all integers a s(a + b)

@TechToppers 3 жыл бұрын

From where did you get to know these properties??

@PontusLarsson1 3 жыл бұрын

@@TechToppers the first property is trivial as we are working in base 10. Multiplying any number by 10 adds a zero-digit at the end of the number. E.g. s(2) = 2 = 2 + 0 = s(20) = s(10*2). The second property is not immediate, but for simplicity consider integers 1