Find all \(f:\mathbb{R}\mapsto\mathbb{R}\) satisfying \[(x-y)f(x+y)-(x+y)f(x-y)=4xy\left(x^2-y^2\right)\] for all real \(x,y\).
Try this.
Another Tricky FE
Last edited by *Mahi* on Thu Sep 25, 2014 10:58 am, edited 1 time in total.
Reason: Please try posting new problems in separate topics, this way problems don't get lost under a load of replies.
Reason: Please try posting new problems in separate topics, this way problems don't get lost under a load of replies.
- What is the value of the contour integral around Western Europe?
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
Re: Another Tricky FE
Solution:
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
Re: Another Tricky FE
Nice substitution!
My solution: Because \(4xy=(x+y)^2-(x-y)^2\), the equation can be rewritten as \[\dfrac{f(x+y)}{x+y}-(x+y)^2=\dfrac{f(x-y)}{x-y}-(x-y)^2\] so \(f(x)/x - x^2\) is constant. Letting the constant \(c\) and solving leads to \(f(x)=x^3+cx\) for all \(x\).
The similar approach works for the previous Tricky FE I posted.
My solution: Because \(4xy=(x+y)^2-(x-y)^2\), the equation can be rewritten as \[\dfrac{f(x+y)}{x+y}-(x+y)^2=\dfrac{f(x-y)}{x-y}-(x-y)^2\] so \(f(x)/x - x^2\) is constant. Letting the constant \(c\) and solving leads to \(f(x)=x^3+cx\) for all \(x\).
The similar approach works for the previous Tricky FE I posted.
- What is the value of the contour integral around Western Europe?
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
Re: Another Tricky FE
If you're using $\dfrac{f(x)}x -x^2 =$ constant, you have to prove $f(0)=0$ separately as $g(x) = \dfrac{f(x)}x -x^2$ is not defined at $x=0$.
$x=\frac{a+b}2, y=\frac{a-b}2$ can also be handy whenever the terms are sums and products of $x+y \text{ or } x-y$
$x=\frac{a+b}2, y=\frac{a-b}2$ can also be handy whenever the terms are sums and products of $x+y \text{ or } x-y$
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi