Given a positive integer $n \geq 3$. Find all $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that for any $n$ positive reals $a_1,...,a_n$, the following condition is always satisfied: $\sum_{i=1}^{n}(a_i-a_{i+1})f(a_i+a_{i+1}) = 0$ where $a_{n+1} = a_1$.
Problem
Source: 2015 Taiwan TST round 1 Day 1 P2
Tags: algebra, function, functional equation, Taiwan TST 2015
14.04.2015 18:19
WLOG n=3 and we can consider it as an old problem.
09.05.2020 21:27
note that if $S_n$ is the set of functions that satisfies $\sum_{i=1}^{n}(a_i-a_{i+1})f(a_i+a_{i+1}) = 0$ then $S_{n} \subseteq S_{n-1}$ so $S_n \subseteq S_3$ so it suffices to consider $n=3$ $$(a-b)f(a+b)+(b-c)f(b+c)+(c-a)f(a+c)=0$$let $P(a,b,c)$ is the previous assertion $P(a,a+d,a+2d) \implies f(2a+d)+f(2a+3d)=2f(2a+2d)$ so $f(x+y)=2f(\frac{x+y}{2})$ so its jensen and so $f(x)=kx \forall x \in \mathbb{Q}$ combining with $(a-b)f(a+b)+(b-c)f(b+c)+(c-a)f(a+c)=0$ $$f(a)(c-b)+f(b)(a-c)+f(c)(b-a)=0$$put $a,b \in \mathbb{Q} \implies f(a)=ka \forall a \in \mathbb{R}^+$ and we win
07.02.2021 13:05
Ali3085 wrote: note that if $S_n$ is the set of functions that satisfies $\sum_{i=1}^{n}(a_i-a_{i+1})f(a_i+a_{i+1}) = 0$ then $S_{n} \subseteq S_{n-1}$ so $S_n \subseteq S_3$ so it suffices to consider $n=3$ $$(a-b)f(a+b)+(b-c)f(b+c)+(c-a)f(a+c)=0$$let $P(a,b,c)$ is the previous assertion $P(a,a+d,a+2d) \implies f(2a+d)+f(2a+3d)=2f(2a+2d)$ so $f(x+y)=2f(\frac{x+y}{2})$ so its jensen and so $f(x)=kx \forall x \in \mathbb{Q}$ combining with $(a-b)f(a+b)+(b-c)f(b+c)+(c-a)f(a+c)=0$ $$f(a)(c-b)+f(b)(a-c)+f(c)(b-a)=0$$put $a,b \in \mathbb{Q} \implies f(a)=ka \forall a \in \mathbb{R}^+$ and we win Watch carefully on your Jenson equation lol. Cuz the true solution is all linear functions $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$