For a given natural number $k > 1$, find all functions $f:\mathbb{R} \to \mathbb{R}$ such that for all $x, y \in \mathbb{R}$, $f[x^k + f(y)] = y +[f(x)]^k$.
Problem
Source: China TST 2001, problem 3
Tags: function, algebra, polynomial, functional equation, algebra unsolved
23.05.2005 02:28
f(f(x))=x+a^k x^k +y + 2a^k=f(f(x^k+y+a^k)=f(f(x^k+f(f(y)))=f(f(y)+f(x)^k)=f(f(x))^k + y = [x+a^k]^k + y so we get that a=0. hence f(f(x))=x. (*) also f(x^k)=f(x)^k (**). f(x) is one to one and onto. one can easily see that f(x+y)=f(x)+f(y) (***) so f(x)=x for all x in Q.
23.05.2005 14:05
The additivity, along with $f(x^k)=f^k(x)$ for some $k > 1$, implies at once that $f(x)=x$ for all $x \in \mathbb{R}$. Indeed, for $n \in \mathbb{Z}$, we have $f(x+n)=f(x)+n$, so \[ \sum_j \binom{k}{j}f^j(x)n^{k-j} = (f(x)+n)^k = f^k(x+n) = f\big( (x+n)^k \big) = \sum_j \binom{k}{j}f(x^j)n^{k-j}. \] Fixing $x$, we may view this as a polynomial identity with respect to $n$. Comparing the coefficients of $n^{k-1}$, we see that $f(x^2) = f(x)^2 \geq 0$, so $f(x) \geq 0$ for $x \geq 0$ which, along with the additivity, implies that $f(x)$ increases monotonously. Hence, $f(x) \equiv x$.
23.05.2005 14:08
A slightly more difficult problem is this: find all functions $f: \mathbb{R}^+ \to \mathbb{R}$ satisfying this condition.
24.05.2005 01:10
Satisfying which condition? (Edit: never mind. The functional equation is the same, but for functions of positive reals rather than all real numbers.)
24.05.2005 02:11
I think he means the same condition above
24.05.2005 14:06
Yes, the problem I suggested is this: for a fixed integer $k>1$, find all functions $f: \mathbb{R}^+ \to \mathbb{R}$ satisfying $f(x^k+f(y)) = y + f(x)^k$ for all $x,y \in \mathbb{R}^+$ for which $x^k + f(y) > 0$. (So that we may not plug in $y=0$ or $x=0$ to make the problem trivial ).
24.05.2005 15:18
Is this generalisable to say, n variables.
02.05.2006 10:01
.For k odd you miss f(x)=-x
25.04.2015 19:08
vess wrote: The additivity, along with $f(x^k)=f^k(x)$ for some $k > 1$, implies at once that $f(x)=x$ for all $x \in \mathbb{R}$. Could you prove that?