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Math Help - Trouble with Integration

  1. #1
    Member grgrsanjay's Avatar
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    Trouble with Integration

    f:R ---> R is a continuous function and f(x) = f(2x) is true for all real numbers
    If f(1)=3 then find the value of
    \int_{-1}^{1}f[f(x)] dx
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Trouble with Integration

    Quote Originally Posted by grgrsanjay View Post
    f:R ---> R is a continuous function and f(x) = f(2x) is true for all real numbers If f(1)=3 then find the value of \int_{-1}^{1}f[f(x)] dx
    Necessary condition: Taking into account that g(x)=3 is continuous in \mathbb{R} and satisfies g(x)=g(2x)\;\wedge \;g(1)=3 for all x\in \mathbb{R} then, if the solution does not depend on f we necessarily have \int_{-1}^{1}f[f(x)] \;dx=\int_{-1}^{1}3 \;dx=6 .
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    Re: Trouble with Integration

    We can show that f is necessarily constant.

    Since f(x) = f(2x) for all x,
    f(x/2) = f(x) .....(just substitute x/2 for x)
    f(x/(2^2)) = f(x) ..... similarly
    ...
    f(x/(2^n)) = f(x)

    Now let n \to \infty and apply the continuity of f.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Trouble with Integration

    Another way: the solutions of the functional equation f(2x)=f(x) are f(x)=\phi(\log x) where \phi is any periodic function with period \log 2 . According to the hypothesis, necessarily f(x)=3 .
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