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. Note by Shrihari B
3 years, 9 months ago

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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.

- 3 years, 9 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

- 3 years, 9 months ago

this is correct!

- 3 years, 9 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.

- 3 years, 9 months ago

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

- 3 years, 9 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.

- 3 years, 9 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.

- 3 years, 9 months ago

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

- 3 years, 9 months ago

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

- 3 years, 9 months ago

all the possible $f'(x)$polynomiald

- 3 years, 9 months ago

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

- 3 years, 9 months ago

not if they are both equal

- 3 years, 9 months ago

looks familiar

- 3 years, 9 months ago

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

- 3 years, 9 months ago

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

- 3 years, 9 months ago

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

- 3 years, 9 months ago

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.

- 3 years, 9 months ago

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

- 3 years, 9 months ago

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.

Staff - 3 years, 9 months ago

@Calvin Lin Sure Sir

- 3 years, 9 months ago

Answer to ProblemI: $f(x)+x=f(f(x)).....(1)$

Since LHS contains $f(x)+x$ 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$\rightarrow$R such that $f(x-f(y))=f(f(y))+xf(y)+f(x)-1$ for all $x,y$ belonging to Reals.

- 3 years, 9 months 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 :)

- 3 years, 9 months ago

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

- 3 years, 9 months ago

The solution is straight forward. Here it goes :

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

- 3 years, 9 months ago