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Math Help - Applying the Extreme Value Theorem

  1. #1
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    Continuity - minimum and maximum

    Hi. I'd really appreciate any help you can give me on this.

    If  f: \Re \to \Re is a continuous, periodic function, then prove  f has a maximum and minimum.

    ================================================== =======

    A periodic function is where if you have any K>0, then for all  x \in \Re ,  f(x + K) = f(x) . And we need to show that a set S = \{ f(x) | x \in \Re \} has a maximum and minimum.

    I'm thinking that we need to show it is bounded first of all. But I have no idea how I could do that. I'm just jotting down some thoughts, so I don't know whether they are relevant.

     | f(x+K) - f(x) | = 0 < \epsilon for all  \epsilon > 0

    This question in the textbook is under the section of the "The extreme value theorem" so I think that must be relavent.
    Please help.
    Regards,
    Joel.
    Last edited by Joel24; February 9th 2008 at 11:34 AM.
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  2. #2
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    Do you know the theorem that says: If f:\mathbb{R} \to \mathbb{R} is continuous then on any closed interval [a,b] f attains a maximum and a minimum in that interval?
    If so, then what about the interval [0,K]?
    If not, then you do need to prove that theorem.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Do you know the theorem that says: If f:\mathbb{R} \to \mathbb{R} is continuous then on any closed interval [a,b] f attains a maximum and a minimum in that interval?
    If so, then what about the interval [0,K]?
    If not, then you do need to prove that theorem.
    Thank you, Plato.

    I think I mentioned that theorem in my post. If I'm not mistaken that is the extreme value theorem (at least that's what is written in a textbook - not sure if that is the universal name for it).

    Over the interval [0, K], I'm not sure. Over that interval, does  f(0+K) = f(0) ? and  f(K+K) = f(K) ? Don't know where to go with this. *embarrassed*
    Last edited by Joel24; February 9th 2008 at 01:12 PM.
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  4. #4
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    Quote Originally Posted by Joel24 View Post
    Over that interval, does  f(0) = 0 ? and  f(K) = K ? So the maximum is K and minimum is 0?
    No absolutely not. Apply the theorem to f. Then there are points <br />
p \in [0,K]\,\& \,q \in [0,K] such that \forall x \in [0,K] \Rightarrow \quad f(q) \leqslant f(x) \leqslant f(p).

    Now you need to use the property of a periodic function to prove that these are the maximum and minimum for all reals.
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  5. #5
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    Quote Originally Posted by Plato View Post
    No absolutely not. Apply the theorem to f. Then there are points <br />
p \in [0,K]\,\& \,q \in [0,K] such that \forall x \in [0,K] \Rightarrow \quad f(q) \leqslant f(x) \leqslant f(p).

    Now you need to use the property of a periodic function to prove that these are the maximum and minimum for all reals.
    Yes, I realised and edited my mistake just before your post. My concern is that these max and minimums don't hold. (I know they do, but I don't understand why)

    A function is periodic if there exists a K>0 such that for all real x,  f(x+K) = f(x)

    We need to show that for all real x,  f(x+K) \leq f(p)

    But we only considered the period [0,K]. There has to be a number greater than K in the set of real numbers. What if k=98. Then consider the interval [-7,K+99]

     f(K+99) = f(99) which could be greater than  f(p) . :S

    Apologies if I'm going off track. I'm just trying to understand this.
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  6. #6
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    K is the period of f. If n is an integer then \forall x\left[ {f(x + nK) = f(x)} \right].
    Lets say that t > K then using floor function, j = \left\lfloor {\frac{t}{K}} \right\rfloor \,\& \,r = t - Kj.
    Is it clear that r \in [0,K]\,\& \,t = jK + r.
    Can you finish?
    Last edited by Plato; February 9th 2008 at 03:07 PM.
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  7. #7
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    Quote Originally Posted by Plato View Post
    K is the period of f. If n is an integer then \forall x\left[ {f(x + nK) = f(x)} \right].
    Lets say that t > K then using floor function, j = \left\lfloor {\frac{t}{K}} \right\rfloor \,\& \,r = \frac{t}{K} - j.
    Is it clear that r \in [0,K]\,\& \,t = jK + r.
    Can you finish?
    Yes, I think so. Thank you very much.

    What does j = \left\lfloor {\frac{t}{K}} \right\rfloor mean?
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  8. #8
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    Quote Originally Posted by Joel24 View Post
    What does j = \left\lfloor {\frac{t}{K}} \right\rfloor mean?
    The floor function is often called the greatest integer function.
    \left\lfloor {\frac{{99.1}}{{11}}} \right\rfloor  = 9.
    Please, see my edit of the reply above: r = t - Kj.
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  9. #9
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    Nowhere in your initial post do you mention the function is periodic on a closed interval. Consider f(x) = \tan x whenever it is defined an f(x)= 0 elsewhere then f does not attain a maximum or minimum eventhough the function is periodic.
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  10. #10
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    Quote Originally Posted by ThePerfectHacker View Post
    Nowhere in your initial post do you mention the function is periodic on a closed interval.
    But the post made it clear that the function is continuous on the reals.
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