Brilliant Functional Equations Contest!

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.

Easy Math Editor

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Comments

SOLUTION TO PROBLEM 2 let $z=f(y),x=2z$ . then $f(2z-z)=f(z)+2z^2+f(2z)-1$ $f(2z)=1-2z^2$ $f(x)=1-\dfrac{x^2}{2}$ PROBLEM 3 find 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 - 5 years, 5 months ago

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

Keshav Gupta - 5 years, 5 months ago

this is correct!

Aareyan Manzoor - 5 years, 5 months ago

@Aareyan Manzoor – THIS is not. AS i pointed above. moreover, this polynomial has the property that it has roots which are reciprocal of each other.