INMO is getting closer and I feel its best if we start contests on various topics. So I have decided to start the first thread on Functional Equations..... The main motto is that everyone should improve their functional equations via this contest.

Rules :

1) I will post the first problem

2) The one who answers my question(with proofs of course) should post the next question and this goes on .....

3) If no one answers a particular question within 24 hours then the problem poser must also post the solution.

4) Someone should post a solution IF AND ONLY IF the solution is complete.

So an easy question to start is as follows :-

1) \( f: \mathbb{R} \longrightarrow \mathbb{R} , \forall x \in \mathbb{R}, x+ f(x) = f( f(x) ) \) . Find all solutions to the equation f(f(x))=0.

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## Comments

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TopNewestSOLUTION TO PROBLEM 2let \(z=f(y),x=2z\). then \[f(2z-z)=f(z)+2z^2+f(2z)-1\] \[f(2z)=1-2z^2\] \[f(x)=1-\dfrac{x^2}{2}\]PROBLEM 3find all polynomials \(f:\mathbb{R^+}\implies\mathbb{R}\) that satisfies \[f(x)+f(\dfrac{1}{x})=(x+\dfrac{1}{x})f(x)\] enter you answer as all the possible derivatives of the polynomial.Log in to reply

Solution to problem 3:

Replace \( x \) with \( \frac{1}{x} \) to get: \( f(x) + f(\frac{1}{x}) = (\frac{1}{x} + x)f(\frac{1}{x}) \)

Comparing with statement of question gives \( f(x) = f(\frac{1}{x}) \)

Now from statement of question, \( 2f(x) = (x + \frac{1}{x})f(x) \)

So, \( f(x) = 0 \) as \( x + \frac{1}{x} \neq 2 \forall x \in \mathbb{R^+} \)

Sorry guys, i dont have a problem that i can think of to post, i give the opportunity to post a problem to @Harsh Shrivastava

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this is correct!

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When we substitute x with 1/x we assume that x equals 1 or -1. (-1 not included in the domain). So this equation holds only for x=1.

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What do you mean by all the possible derivatives? @Aareyan Manzoor

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all the possible \[f'(x)\]polynomiald

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looks familiar

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Can some one post a new question? @Aareyan Manzoor @Keshav Gupta @Shrihari B

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I suggest u to start this contest after the inequality contest. It is best to have one contest at a time.

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Find all \( f: \mathbb{Q} \rightarrow \mathbb{Q} \) such that \( f(1) = 2 \) and \( f(xy)=f(x)f(y)-f(x+y)+1 \)

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I got the solutions for x in N. Please do check that

f(xy)=f(x)f(y)-f(x+y)+1. Put x=y=1. We get f(2)=3.An easy induction yields f(x)=x+1.

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Yes that is correct, but you need to prove it for all rationals :P

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Could you write the equation in Latex so that it is much easier to understand? I've added it at the end of the note. If you agree, please edit it in.

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@Calvin Lin Sure Sir

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Answer to

ProblemI: \[f(x)+x=f(f(x)).....(1)\]Since

LHScontains \(f(x)+x\) therefore degree of \(f(x)\) must be one .Therefore let \(f(x)=ax+b\).

Putting in it \((1)\) we get \[(a+1)x=a^{2}x+ab\]

This implies \(b=0\) , \[a^{2}-a-1=0\]

Therefore \(\boxed{a=\frac{-1\pm\sqrt{5}}{2}}\)

Therefore \(f(x)=ωx\) where \(ω=\frac{-1\pm\sqrt{5}}{2}\).

ProblemII: Find all functions \(f:\)R\(\rightarrow\)Rsuch that \[f(x-f(y))=f(f(y))+xf(y)+f(x)-1\] for all \(x,y\) belonging to Reals.Log in to reply

Your solution does not seem to be correct. Firstly you have assumed f(x) to be a polynomial which is not correct and secondly if we take f(x) to be a polynomial then i don't understand how u can conclude from that equation that b=0. Thirdly u have not found the solutions of f(f(x))=0. So please try again :)

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Can you post the solution for your problem. I am getting the same answer as of @Shivam Jadhav (for \(f(x)\) ) Thanks :)

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We first attempt to prove the injectivity of the function. So let f(x)=f(y). Putting x and y in the equations and equating we get x=y. Now put x=0. We get f(0)=f(f(0)). But since the function is injective the arguments must be equal. So f(0)=0. f(f(0))=f(0)=0. And there cannot be any other solution for f(f(x))=0 as the function is injective

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