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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.

Regards...Cliowa

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# Homework Help: Functional equation f(x)=f(x^2)

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