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Math Help - 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
    MHF Contributor

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    Re: Intermediate-value theorem

    Define r(\theta) to be the radiation at angle \theta from whatever starting point you want. Define f(\theta) to be r(\theta)- r(\theta+ \pi). Then 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 r(\theta)= r(\theta+ \pi) for all \theta or there exist [tex]\theta[tex] such that f(\theta) is positive and f(\theta+ \pi) is negative. Since r(\theta) is continuous, so is f(\theta) and there must be some \theta such that 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 g(\theta) = f(\theta) - f(\theta + \pi) ? It follows from a fairly fundamental theorem about functions that g is continuous on [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 f(0) = f(\pi) and we are done.

    Now, for the other two cases we must be more clever. We must first realise that f(0) = f(2\pi) and then apply the intermediate value theorem to 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; June 1st 2013 at 09:49 AM. Reason: Duplicate answer
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  4. #4
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    Re: Intermediate-value theorem

    thanks both of you
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