Let functions $f, g: \mathbb{R}^+ \to \mathbb{R}^+$ satisfy the following: \[ f(g(x)y + f(x)) = (y+2015)f(x) \]for every $x,y \in \mathbb{R}^+$. (a) Prove that $g(x) = \frac{f(x)}{2015}$ for every $x \in \mathbb{R}^+. $ (b) State an example of function that satisfy the equation above and $f(x), g(x) \ge 1$ for every $x \in \mathbb{R}^+$.
Problem
Source: INAMO 2015 Shortlist A6 ; INAMO 2015 Problem 04
Tags: algebra, functional equation, function
30.12.2018 13:09
GorgonMathDota wrote: Let functions $f, g: \mathbb{R}^+ \to \mathbb{R}^+$ satisfy the following: \[ f(g(x)y + f(x)) = (y+2015)f(x) \]for every $x,y \in \mathbb{R}^+$. (a) Prove that $g(x) = \frac{f(x)}{2015}$ for every $x \in \mathbb{R}^+. $ Let $P(x,y)$ be the assertion $f(g(x)y+f(x))=(y+2015)f(x)$ $f(x)$ constant is not a solution So let $x,z>0$. Since non constant, $\exists y$ such that $f(x)\ne f(y)$ WLOG $f(x)>f(y)$ $P(x,z)$ $\implies$ $f(g(x)z+f(x))=(z+2015)f(x)$ $P(y,\frac{g(x)z+f(x)-f(y)}{g(y)})$ $\implies$ $f(g(x)z+f(x))=(\frac{g(x)z+f(x)-f(y)}{g(y)}+2015)f(y)$ Subtracting, we get $z(f(x)g(y)-g(x)f(y))=(f(x)-f(y))(f(y)-2015g(y))$ $\forall z>0$ This implies (since available whatever is $z$) $f(x)g(y)=g(x)f(y)$ and $f(y)=2015g(y)$ And so $f(x)=2015g(x)$ and $f(y)=2015 g(y)$ Q.E.D. GorgonMathDota wrote: (b) State an example of function that satisfy the equation above and $f(x), g(x) \ge 1$ for every $x \in \mathbb{R}^+$. Choose for example $f(x)=\max(2015,2015x)$ and $g(x)=\max(1,x)$
30.12.2018 13:24
pco wrote: GorgonMathDota wrote: Let functions $f, g: \mathbb{R}^+ \to \mathbb{R}^+$ satisfy the following: \[ f(g(x)y + f(x)) = (y+2015)f(x) \]for every $x,y \in \mathbb{R}^+$. (a) Prove that $g(x) = \frac{f(x)}{2015}$ for every $x \in \mathbb{R}^+. $ Let $P(x,y)$ be the assertion $f(g(x)y+f(x))=(y+2015)f(x)$ $f(x)$ constant is not a solution So let $x,z>0$. Since non constant, $\exists y$ such that $f(x)\ne f(y)$ WLOG $f(x)>f(y)$ $P(x,z)$ $\implies$ $f(g(x)z+f(x))=(z+2015)f(x)$ $P(y,\frac{g(x)z+f(x)-f(y)}{g(y)})$ $\implies$ $f(g(x)z+f(x))=(\frac{g(x)z+f(x)-f(y)}{g(y)}+2015)f(y)$ Subtracting, we get $z(f(x)g(y)-g(x)f(y))=(f(x)-f(y))(f(y)-2015g(y))$ $\forall z>0$ This implies (since available whatever is $z$) $f(x)g(y)=g(x)f(y)$ and $f(y)=2015g(y)$ And so $f(x)=2015g(x)$ and $f(y)=2015 g(y)$ Q.E.D. GorgonMathDota wrote: (b) State an example of function that satisfy the equation above and $f(x), g(x) \ge 1$ for every $x \in \mathbb{R}^+$. Choose for example $f(x)=\max(2015,2015x)$ and $g(x)=\max(1,x)$ Thank you very much!