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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
1 year, 11 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.

Aareyan Manzoor - 1 year, 11 months ago

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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 - 1 year, 10 months ago

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this is correct!

Aareyan Manzoor - 1 year, 10 months ago

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@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.

Aditya Agarwal - 1 year, 10 months ago

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

Aareyan Manzoor - 1 year, 10 months ago

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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.

Aditya Agarwal - 1 year, 10 months ago

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@Aditya Agarwal 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.

Keshav Gupta - 1 year, 10 months ago

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

Aditya Agarwal - 1 year, 10 months ago

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

Aditya Agarwal - 1 year, 10 months ago

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

Aareyan Manzoor - 1 year, 10 months ago

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

Aditya Agarwal - 1 year, 10 months ago

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@Aditya Agarwal not if they are both equal

Aareyan Manzoor - 1 year, 10 months ago

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

Aditya Agarwal - 1 year, 10 months ago

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looks familiar

Aareyan Manzoor - 1 year, 11 months ago

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

Keshav Gupta - 1 year, 10 months ago

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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.

Shrihari B - 1 year, 10 months ago

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

Keshav Gupta - 1 year, 10 months ago

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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.

Aditya Kumar - 1 year, 11 months 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 - 1 year, 11 months ago

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@Calvin Lin Sure Sir

Shrihari B - 1 year, 11 months ago

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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.

Shivam Jadhav - 1 year, 11 months ago

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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 :)

Shrihari B - 1 year, 11 months ago

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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 :)

Neelesh Vij - 1 year, 10 months ago

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@Neelesh Vij 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

Shrihari B - 1 year, 10 months ago

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