Find all functions $f:\mathbb{Q}\rightarrow\mathbb{R} \setminus \{ 0 \}$ such that \[(f(x))^2f(2y)+(f(y))^2f(2x)=2f(x)f(y)f(x+y)\] for all $x,y\in\mathbb{Q}$
Problem
Source: 2015 Taiwan TST Round 1 Quiz 1 Problem 2
Tags: function, Taiwan, Function equations, algebra, Taiwan TST 2015
12.07.2015 16:49
wanwan4343 wrote: Find all functions $f:\mathbb{Q}\rightarrow\mathbb{R} \setminus \{ 0 \}$ such that \[(f(x))^2f(2y)+(f(y))^2f(2x)=2f(x)f(y)f(x+y)\] for all $x,y\in\mathbb{Q}$ Let $P(x,y)$ be the assertion $f(x)^2f(2y)+(f(y)^2f(2x)=2f(x)f(y)f(x+y)$ $f(x)$ solution implies $cf(x)$ solution and so WLOG $f(0)=1$ (since $0\notin f(\mathbb Q)$) $P(x,0)$ $\implies$ $f(2x)=f(x)^2$ and $P(x,y)$ becomes $f(x+y)=f(x)f(y)$ Hence the solution : $\boxed{f(x)=b e^{ax}}$ $\forall x\in\mathbb Q$, which indeed is a solution, whatever is $a\in\mathbb R$, $b\in\mathbb R\setminus\{0\}$
25.02.2018 17:41
What about non continuous solutions?
25.02.2018 17:48
WolfusA wrote: What about non continuous solutions? What is your definition of continuous / non continuous function when domain is $\mathbb Q$ ???????
25.02.2018 22:29
$f:\mathbb Q\to \mathbb R$ is continuous on set $\mathbb Q $ iff for every sequence $(x_n)$ where $\forall n \quad x_n\in \mathbb Q$ and $\lim_{n\to \infty}x_n=y$ holds $\lim_{x_n\to y} f(x_n)=f(y)$
26.02.2018 06:53
WolfusA wrote: What about non continuous solutions? There's no need to consider it, isn't it? Because the domain is rational number, then we can directly apply Cauchy Functional equation as pco did.
26.02.2018 08:56
WolfusA wrote: $f:\mathbb Q\to \mathbb R$ is continuous on set $\mathbb Q $ iff for every sequence $(x_n)$ where $\forall n \quad x_n\in \mathbb Q$ and $\lim_{n\to \infty}x_n=y$ holds $\lim_{x_n\to y} f(x_n)=f(y)$ Huhhh ! And what about $y\notin\mathbb Q$ so that $f(y)$ is undefined ?
26.02.2018 21:06
I see your point. So what is the criterium then for using Cauchy's functional equation, because I thought it's that function is continuous.
27.02.2018 06:15
Given that the domain is $\mathbb{Q}$, you just apply Cauchy's functional equation the normal way on $\log{f(x)}+\log{f(y)}=\log{f(x+y)}$.
01.03.2018 21:19
This solutions also works. $\forall x\in\mathbb Q$ $f(x)=bs^{ax}$ whatever is $a\in\mathbb R$, $s,b\in\mathbb R\setminus\{0\}$,
01.03.2018 21:26
InCtrl wrote: Given that the domain is $\mathbb{Q}$, you just apply Cauchy's functional equation the normal way on $\log{f(x)}+\log{f(y)}=\log{f(x+y)}$. You can't do that, because $f:\mathbb{Q}\rightarrow\mathbb{R} \setminus \{ 0 \}$, and you can't take a logarithm of negative number.
01.03.2019 17:07
WolfusA wrote: InCtrl wrote: Given that the domain is $\mathbb{Q}$, you just apply Cauchy's functional equation the normal way on $\log{f(x)}+\log{f(y)}=\log{f(x+y)}$. You can't do that, because $f:\mathbb{Q}\rightarrow\mathbb{R} \setminus \{ 0 \}$, and you can't take a logarithm of negative number. Actually, f(2x)=f(x)^2, so nothing would be negative.