Results 1 to 4 of 4

Thread: Trouble with Integration

  1. #1
    Member grgrsanjay's Avatar
    Joined
    May 2010
    From
    chennai,tamil nadu
    Posts
    143
    Thanks
    1

    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
    $\displaystyle \int_{-1}^{1}f[f(x)] dx$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46

    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 $\displaystyle \int_{-1}^{1}f[f(x)] dx$
    Necessary condition: Taking into account that $\displaystyle g(x)=3$ is continuous in $\displaystyle \mathbb{R}$ and satisfies $\displaystyle g(x)=g(2x)\;\wedge \;g(1)=3$ for all $\displaystyle x\in \mathbb{R}$ then, if the solution does not depend on $\displaystyle f$ we necessarily have $\displaystyle \int_{-1}^{1}f[f(x)] \;dx=\int_{-1}^{1}3 \;dx=6$ .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1

    Re: Trouble with Integration

    We can show that f is necessarily constant.

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

    Now let $\displaystyle n \to \infty$ and apply the continuity of f.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46

    Re: Trouble with Integration

    Another way: the solutions of the functional equation $\displaystyle f(2x)=f(x)$ are $\displaystyle f(x)=\phi(\log x)$ where $\displaystyle \phi$ is any periodic function with period $\displaystyle \log 2$ . According to the hypothesis, necessarily $\displaystyle f(x)=3$ .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trouble with Integration by Parts
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Apr 8th 2011, 09:36 AM
  2. Integration trouble!
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Apr 6th 2010, 05:38 AM
  3. Some trouble with trig integration.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 2nd 2009, 06:41 PM
  4. trouble with the bounds of integration
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Oct 21st 2008, 03:23 PM
  5. Trouble with integration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jun 10th 2008, 06:40 PM

Search Tags


/mathhelpforum @mathhelpforum