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Math Help - Functions that satisfy a relation.

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    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?
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    Quote Originally Posted by linshi View Post
    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).
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    Quote Originally Posted by Opalg View Post
    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
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