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Math Help - Proving a point in an interval

  1. #1
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    Proving a point in an interval

    Let f be continuous on the interval [0,1] to Reals and such that f(0) = f(1). Prove that there exists a point c in [0,1/2] such that f(c) = f(c + (1/2)).
    [Hint: Consider g(x) = f(x) - f(x + (1/2))]

    I'm not really sure how to go about proving this. Any help would be much appreciated.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by jkru View Post
    Let f be continuous on the interval [0,1] to Reals and such that f(0) = f(1). Prove that there exists a point c in [0,1/2] such that f(c) = f(c + (1/2)).
    [Hint: Consider g(x) = f(x) - f(x + (1/2))]
    What can be said about g ?
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  3. #3
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    what is g?

    g(x) is going to be the difference between a point x and (x + 1/2) when applied to the function f(x).
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Yes...

    f is defined on [0,1] so g(x)=f(x)-f(x+1/2) is defined on [0,1/2] because both x and x+1/2 have to lie in [0,1].

    We are told that f(0)=f(1), can you use this to get some information on g(0) and g(1/2) ?
    Last edited by flyingsquirrel; November 24th 2008 at 12:52 PM. Reason: typo
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  5. #5
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    So, we can say that g(0) = g(1/2)?
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by jkru View Post
    So, we can say that g(0) = g(1/2)?
    There was a typo in my post, I've corrected it, sorry.

    As g(x)=f(x){\color{red}-}f(x+1/2), one has g(0)=f(0)-f(1/2) and g(1/2)=f(1/2)-f(1)=f(1/2)-f(0) hence g(0)=-g(1/2). In other words, if g(0) is positive then g(1/2) is negative and if g(0) is negative then g(1/2) is positive.
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  7. #7
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    Ohh, that makes sense. But, how does this help us prove a point c in [0,1/2]? Are we supposed to say that g(0) and g(1/2) are opposite in sign that there exists a point such that g(c) = 0?
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  8. #8
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by jkru View Post
    Are we supposed to say that g(0) and g(1/2) are opposite in sign that there exists a point such that g(c) = 0?
    Yes ! Can you justify this ?
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  9. #9
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    I think there's a theorem in my book that I'm allowed to use. I think it's something along the lines of:

    With a continuous function the two points are connected, so the function itself must cross the axis, thus resulting in an intermediate value.

    Is that correct? Not sure if I explained it right.
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  10. #10
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by jkru View Post
    With a continuous function the two points are connected, so the function itself must cross the axis, thus resulting in an intermediate value.

    Is that correct?
    Yes it is . This theorem is called the intermediate value theorem.
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  11. #11
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    Thanks for your help!
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