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Thread: Intermediate-value theorem

  1. #1
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    Intermediate-value theorem

    Intermediate-value theorem-rv2-y-_1d-5qf7cms-em.jpg

    can someone help me on this question.
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  2. #2
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    Re: Intermediate-value theorem

    Define $\displaystyle r(\theta)$ to be the radiation at angle $\displaystyle \theta$ from whatever starting point you want. Define $\displaystyle f(\theta)$ to be $\displaystyle r(\theta)- r(\theta+ \pi)$. Then $\displaystyle f(\theta+\pi)= r(\theta+ \pi)- r(\theta+ 2\pi)= r(\theta+ \pi)- r(\theta)= -f(\theta)$. So either f is identically 0, in which case $\displaystyle r(\theta)= r(\theta+ \pi)$ for all $\displaystyle \theta$ or there exist [tex]\theta[tex] such that $\displaystyle f(\theta)$ is positive and $\displaystyle f(\theta+ \pi)$ is negative. Since $\displaystyle r(\theta)$ is continuous, so is $\displaystyle f(\theta)$ and there must be some $\displaystyle \theta$ such that $\displaystyle f(\theta)= 0$
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  3. #3
    Junior Member nimon's Avatar
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    Re: Intermediate-value theorem

    What if we considered the function $\displaystyle g(\theta) = f(\theta) - f(\theta + \pi) $? It follows from a fairly fundamental theorem about functions that $\displaystyle g$ is continuous on $\displaystyle [0,\pi]$. We must then be in one of three cases:
    1. g(0) = 0
    2. g(0) < 0
    3. g(0) > 0


    In the first case, we have $\displaystyle f(0) = f(\pi)$ and we are done.

    Now, for the other two cases we must be more clever. We must first realise that $\displaystyle f(0) = f(2\pi)$ and then apply the intermediate value theorem to $\displaystyle g$. Do you think you can finish this off?

    Edit: I was writing my answer at the same time as another user. I'll leave it here anyway, it might be useful.
    Last edited by nimon; Jun 1st 2013 at 09:49 AM. Reason: Duplicate answer
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    Re: Intermediate-value theorem

    thanks both of you
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