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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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. – Aareyan Manzoor · 11 months, 1 week agoLog in to reply

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 – Keshav Gupta · 11 months ago

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– Aareyan Manzoor · 11 months ago

this is correct!Log in to reply

– Aditya Agarwal · 11 months ago

THIS is not. AS i pointed above. moreover, this polynomial has the property that it has roots which are reciprocal of each other.Log in to reply

– Aareyan Manzoor · 11 months ago

\[f(x)=f(\dfrac{1}{x})\]only constant polys can satisfy this. so \[c+c=(x+x^{-1})c\forall x\in \mathbb{R^+}\] \[c=0\Longrightarrow f(x)=0\]Log in to reply

– Aditya Agarwal · 11 months ago

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.Log in to reply

– Keshav Gupta · 11 months ago

Sorry, but I have substituted \( x \) for \( \frac{1}{x} \) everywhere in the statement. This is valid as \( \frac{1}{x} \) is in domain whenever \( x \) is.Log in to reply

– Aditya Agarwal · 11 months ago

Oh yes, I jad completed most part of it but got stumped at this step. Nice!Log in to reply

@Aareyan Manzoor – Aditya Agarwal · 11 months ago

What do you mean by all the possible derivatives?Log in to reply

– Aareyan Manzoor · 11 months ago

all the possible \[f'(x)\]polynomialdLog in to reply

– Aditya Agarwal · 11 months ago

You mean, suppose \(f(x),g(x)\) satisfy this condition. Then I have to enter \(f'(x),g'(x)\)?Log in to reply

– Aareyan Manzoor · 11 months ago

not if they are both equalLog in to reply

@Aareyan Manzoor @Keshav Gupta @Shrihari B – Aditya Agarwal · 11 months ago

Can some one post a new question?Log in to reply

familiar – Aareyan Manzoor · 11 months, 1 week ago

looksLog in to reply

Find all \( f: \mathbb{Q} \rightarrow \mathbb{Q} \) such that \( f(1) = 2 \) and \( f(xy)=f(x)f(y)-f(x+y)+1 \) – Keshav Gupta · 11 months ago

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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. – Shrihari B · 11 months ago

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– Keshav Gupta · 10 months, 4 weeks ago

Yes that is correct, but you need to prove it for all rationals :PLog in to reply

I suggest u to start this contest after the inequality contest. It is best to have one contest at a time. – Aditya Kumar · 11 months, 1 week ago

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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. – Calvin Lin Staff · 11 months, 1 week ago

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@Calvin Lin Sure Sir – Shrihari B · 11 months, 1 week ago

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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. – Shivam Jadhav · 11 months, 1 week agoLog in to reply

– Shrihari B · 11 months, 1 week ago

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 :)Log in to reply

@Shivam Jadhav (for \(f(x)\) ) Thanks :) – Neelesh Vij · 11 months ago

Can you post the solution for your problem. I am getting the same answer as ofLog in to reply

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 – Shrihari B · 11 months ago

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