Find all functions $f:Z\rightarrow R$ such that
$f(\frac{x+y}{3})=\frac{f(x)+f(y)}{2}$
Surely $x+y$ is divisible by $3$
Functional equation
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Re: Functional equation
No one interested? OK,here is my solution:
Let \[P(x,y)\Rightarrow f(\frac{x+y}{3})=\frac{f(x)+f(y)}{2}\]
then,
\[P(0,3x)\Rightarrow f(\frac{3x}{3})=f(x)=\frac{f(0)+f(3x)}{2}\]
and
\[P(3x,3x)\Rightarrow f(\frac{6x}{3})=f(2x)=\frac{f(3x)+f(3x)}{2}=f(3x)\]
This implies that
\[f(x)=f(2x)=f(3x)=f(0)\]
also it is easy to check that all constant functions are the solutions.
Let \[P(x,y)\Rightarrow f(\frac{x+y}{3})=\frac{f(x)+f(y)}{2}\]
then,
\[P(0,3x)\Rightarrow f(\frac{3x}{3})=f(x)=\frac{f(0)+f(3x)}{2}\]
and
\[P(3x,3x)\Rightarrow f(\frac{6x}{3})=f(2x)=\frac{f(3x)+f(3x)}{2}=f(3x)\]
This implies that
\[f(x)=f(2x)=f(3x)=f(0)\]
also it is easy to check that all constant functions are the solutions.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
Re: Functional equation
Explain.SANZEED wrote:This implies that
\[f(x)=f(2x)=f(3x)=f(0)\]
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Nur Muhammad Shafiullah | Mahi
Re: Functional equation
Eliminated a step:\[P(x,2x)\Rightarrow f(x)=\frac{f(x)+f(2x)}{2}\Rightarrow f(x)=f(2x)\]
This implies the conclusion.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$