Weak Spots of Functional Equations | International Mathematical Olympiad 2011 Shortlist A3

#MathOlympiad #IMO #Algebra

Here is the solution to IMO Shortlist 2011 A3!
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I share Maths problems and Maths topics from well-known contests, exams and also from viewers around the world. Apart from sharing solutions to these problems, I also share my intuitions and first thoughts when I tried to solve these problems.

[littlefermat]

Great video. These are truly good weak spots!
I think that symmetry is the greatest weak spot in this kind of situations.
I recommend that you use the P(x, y) notation when substituting since it is clear and short and it is the most common way in IMO solutions.
I am using it too on my channel functional equations playlist.
Good luck my friend and keep going!

[alperenkoken]

Can you upload another video which explains all tricks we can use in functional equations.

[ignaciobenjamingarridoboba]

I've never tried this kind of problems before and now i figure out is not as complicated as i used to believe, you only need to understand the methods and a lot of practice. Thank you for the video, i hope you keep doing this.

[thebruhtruth]

Amazing you are helping us.. the imo aspirants so much by providing how to look for insights

[qing]

When you have g(x) as linear or constant function, just set x=0 in the original equation and you have f(y)

[melodiesformath]

Hello Letsthinkcritically! My name is Swetha, founder of Melodies for Math. I recently found your channel and subscribed, as I love your content. Your example in this video easily helped me understand how to work with functional equations.

Math is cumulative, so building understanding across concepts is critical. As a group of high school students that explains various math concepts through song, we are so glad to support other channels with similar missions.

I hope you have a great day!

~Swetha from Melodies for Math

[satyamsaurav]

Looking forward for more such illustrations on fe :)

[SONUKUMAR-mbsp]

Let a, b, c be positive real such that a<b<c<2a .find the maximum value of b/a+c/b+a/c .pls solve hmmt 2019

[arifhasnatmubin]

10:41 How is B equal zero ? It is x=1/2 . Please reply how did you get B equal zero ?

[saatvik]

WHICH BOOK SHOULD I PREFER FOR ALGEBRA I AM A BEGGINER IN ALGEBRA

[TVWJ]

10:18 if A*A = A then A = 1 or A=-1. What happens when A = -1? Does this give the same solutions, or are there more solutions?

[deeptnaman]

I think you made the solution more complex, while the solution is pretty straightforward put y = x + h where h is a very small quantity then you will get the derivative relation between f and g

[emilsriram]

Are you allowed to just swap x and y like that in a functional equation?

[hodesdjole]

the functional equations problems posed by various "mathematical problem posing authorities" almost always phrased in a form which seems unbelieveble for a function to satisfy and it ALWAYS(almost always) turns out that the function was constant, linear or maybe even quadratic, it is a simple sleight of hand taking advantage of the fact that the most trivial functions (constant, linear, etc) satisfy unbeliveable constraints, why not show some NON-BORING functions satisfyieng extreme functional conditions

[NoNameAtAll]

how did you come up with the three substitutions to sum up?

[hodesdjole]

suprise suprise, another esoteric complicated looking functional equation has a simple polynomial soulution, as always, why not cover some functional equations with more interesting solutions than g(x)=x and f(x)=x^2+C

[angelishify]

Nekoliko bolj preprosto:
i) g(f(x+y))=f(x)+(2x+y)g(y)
{∂/∂x, ∂/∂y }g(f(x+y))=g'(f(x+y))f'(x+y) ... oba odvoda sta enaka
f'(x)+2g(y) = g(y)+(2x+y)g'(y)
za y=0: f'(x) = - g(0)+2xg'(0) = 2Ax+B in po integraciji: f(x)=Ax^2+Bx+C
za y=0 in x=0 dobimo iz i) g(f(0))=f(0) in za y=-x: f(0)=f(x)+xg(-x) ter g(x)=(f(-x)-f(0))/x = Ax-B

v i) vstavimo y=0: g(f(x))=f(x)+2xg(0); sledi:
A(Ax^2+Bx+C)-B = Ax^2+Bx+C-2xB in (A^2-A)x^2+(AB+B)x+AC-B-C=0 :
A(A-1)=0; B(A+1)=0; C(A-1)=B
1) A=0, B=0, C=0: f(x)=0, g(x)=0 ... trivialna rešitev
2) A=1, B=0, C poljuben: f(x) = x^2+C, g(x) = x
preizkus pa je že narejen v zgornjem videu.

[tayyihcheung]

Your reasoning doesn't seem to be correct.