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Functional equation f(x)=f(x^2)

  1. Jan 14, 2006 #1
    Hello everybody
    I'm given a continuous function f (from the real numbers to the real numbers) which I know obeys the following functional equation:
    [tex] f(x)=f(x^{2})[/tex]
    How can I proof that this function is constant?
    I started out like this: Looking at a number x in [tex][0,1[[/tex] I said to myself that:
    [tex]f(x)=f(x^{2})=f(x^{4})=f(x^{16})=...\rightarrow f(0).[/tex].
    As this relation holds for all x in [0,1[ (looking at [tex]x^{2^{n}} (n \rightarrow \infty[/tex]), it should be true that f(x)=f(-x)=f(0) (the minus because of symmetry, the square). I can also conclude that f(0)=f(1), for f is continuous. But how do I proceed? I would like to show that this function is constant for all x in the real numbers?
    Any help will be appreciated.
    Last edited: Jan 14, 2006
  2. jcsd
  3. Jan 14, 2006 #2


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    If x is a positive real, you also have [tex]f(x)=f(\sqrt{x})[/tex]. Use this for x>1.
  4. Jan 14, 2006 #3
    Oh my gosh, I should have gotten that one on my own.
    Now it seems clear. Thanks alot, shmoe.
    Best regards
  5. Jan 15, 2006 #4


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    If [tex]f:\mathbb{R}\rightarrow\mathbb{R}[/tex] is a continuous function such that [tex] f(x)=f(x^{2}),\forall x\in\mathbb{R}[/tex], then

    i. f is an even function since [tex] f(x)=f(x^{2})\Rightarrow f(-x)=f((-x)^{2})=f(x^{2})=f(x)[/tex]

    ii. [tex] f(x)=f(x^{2^{n}}),\forall n\in\mathbb{Z}[/tex] which can be proved inductively, or simply reasoned much as you have above:

    for positive integer n,

    [tex]f(x)=f(x^{2})=f(x^{4})=f(x^{16})=\cdots =f(x^{2^{n}})[/tex]

    and for negative integer n=-k,

    [tex]f(x)=f(\sqrt{x})=f(\sqrt{\sqrt{x}})=f(\sqrt{\sqrt{\sqrt{x}}})=\cdots =f(x^{2^{-k}})[/tex]
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