Functional Equation

For discussing Olympiad Level Algebra (and Inequality) problems
User avatar
Nadim Ul Abrar
Posts:244
Joined:Sat May 07, 2011 12:36 pm
Location:B.A.R.D , kotbari , Comilla
Functional Equation

Unread post by Nadim Ul Abrar » Mon Sep 24, 2012 5:30 pm

Find all $f:R \to R$ so that
$f(x^3+y^3)=xf(x^2)+yf(y^2) $ for all $x,y \in R$
$\frac{1}{0}$

User avatar
Phlembac Adib Hasan
Posts:1016
Joined:Tue Nov 22, 2011 7:49 pm
Location:127.0.0.1
Contact:

Re: Functional Equation

Unread post by Phlembac Adib Hasan » Mon Sep 24, 2012 10:05 pm

Let $P(x,y)$ be the assertion.The following things can be proved easily:
$f(0)=0;f(x^3)=xf(x^2); f(x+y)=f(x)+f(y)\forall (x,y)\in \mathbb R^2$
$f((x+y)^3)=(x+y)f((x+y)^2)=(x+y)(f(x^2)+2f(xy)+f(y^2))$
$f((x+y)^3)=f(x^3)+f(y^3)+3f(xy(x+y))$
Comparing these two we find that \[xf(y)+yf(x)+2(x+y)f(xy)=3f(xy(x+y))\]\[\Longrightarrow f(x^2)=\frac {xf(1)+(2x-1)f(x)}{2}\]So $f(x^6)=\displaystyle \frac {x^3f(1)+(2x^3-1)xf(x^2)}{2}$
$f(x^6)=x^2f(x^4)=x^2\left (\displaystyle \frac {x^2f(1)+(2x^2-1)f(x^2)}{2}\right )$
Comparing these two, we get\[(x-1)f(x^2)=(x-1)x^2f(1)\]Let's assume $x\neq 1$.So $f(x^2)=x^2f(1)$.The last formula also works for $x=1$.So $f(x^3)=xf(x^2)=x^3f(1)\; \; \forall x\in \mathbb R$.
So the only function satisfying $P(x,y)$ is $f(x)=cx\; \; \forall x\in \mathbb R$ where $c$ is a fixed real.

Post Reply