# Thread: Functions that satisfy a relation.

1. ## Functions that satisfy a relation.

Are there any functions, other than f(x)=Constant, satisfy the relation:
f(kx) = f(x), with the constant k>1?

2. Originally Posted by linshi
Are there any functions, other than f(x)=Constant, satisfy the relation:
f(kx) = f(x), with the constant k>1?
Here's how to do it for k=2. Define f(x) in any way you like, for x in the interval 1≤x<2. Then for every (positive or negative) integer n, define $f(2^nx)=f(x)$ (for 1≤x<2). This function will even be continuous (except at 0), provided that f(x) is continuous in the interval [1,2) and $\lim_{x\to2}f(x)=f(1)$. But it will not be continuous at 0 unless f is constant.

The same construction will work for any k>1, if you define f arbitrarily in the interval [1,k) and then define $f(k^nx)=f(x)$.

3. Originally Posted by Opalg
Here's how to do it for k=2. Define f(x) in any way you like, for x in the interval 1≤x<2. Then for every (positive or negative) integer n, define $f(2^nx)=f(x)$ (for 1≤x<2). This function will even be continuous (except at 0), provided that f(x) is continuous in the interval [1,2) and $\lim_{x\to2}f(x)=f(1)$. But it will not be continuous at 0 unless f is constant.

The same construction will work for any k>1, if you define f arbitrarily in the interval [1,k) and then define $f(k^nx)=f(x)$.
You are defining this function recursively then? I am curious to see if there is an explicit form formula for this (k = 2 as an example) function.

-Dan