# Functions that satisfy a relation.

• Jun 10th 2008, 04:19 AM
linshi
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?
• Jun 10th 2008, 09:23 AM
Opalg
Quote:

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 $\displaystyle f(2^nx)=f(x)$ (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 $\displaystyle \lim_{x\to2}f(x)=f(1)$. 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 $\displaystyle f(k^nx)=f(x)$.
• Jun 10th 2008, 12:53 PM
topsquark
Quote:

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 $\displaystyle f(2^nx)=f(x)$ (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 $\displaystyle \lim_{x\to2}f(x)=f(1)$. 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 $\displaystyle f(k^nx)=f(x)$.

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