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

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

    A periodic function is where if you have any K>0, then for all $\displaystyle x \in \Re $, $\displaystyle f(x + K) = f(x) $. And we need to show that a set $\displaystyle 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.

    $\displaystyle | f(x+K) - f(x) | = 0 < \epsilon $ for all $\displaystyle \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; Feb 9th 2008 at 10:34 AM.
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  2. #2
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    Do you know the theorem that says: If $\displaystyle f:\mathbb{R} \to \mathbb{R}$ is continuous then on any closed interval $\displaystyle [a,b]$ f attains a maximum and a minimum in that interval?
    If so, then what about the interval $\displaystyle [0,K]$?
    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 $\displaystyle f:\mathbb{R} \to \mathbb{R}$ is continuous then on any closed interval $\displaystyle [a,b]$ f attains a maximum and a minimum in that interval?
    If so, then what about the interval $\displaystyle [0,K]$?
    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 $\displaystyle f(0+K) = f(0) $? and $\displaystyle f(K+K) = f(K)$ ? Don't know where to go with this. *embarrassed*
    Last edited by Joel24; Feb 9th 2008 at 12:12 PM.
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  4. #4
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    Quote Originally Posted by Joel24 View Post
    Over that interval, does $\displaystyle f(0) = 0 $? and $\displaystyle f(K) = K$ ? So the maximum is K and minimum is 0?
    No absolutely not. Apply the theorem to f. Then there are points $\displaystyle
    p \in [0,K]\,\& \,q \in [0,K]$ such that $\displaystyle \forall x \in [0,K] \Rightarrow \quad f(q) \leqslant f(x) \leqslant f(p)$.

    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 $\displaystyle
    p \in [0,K]\,\& \,q \in [0,K]$ such that $\displaystyle \forall x \in [0,K] \Rightarrow \quad f(q) \leqslant f(x) \leqslant f(p)$.

    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, $\displaystyle f(x+K) = f(x) $

    We need to show that for all real x, $\displaystyle 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]

    $\displaystyle f(K+99) = f(99) $ which could be greater than $\displaystyle 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 $\displaystyle \forall x\left[ {f(x + nK) = f(x)} \right]$.
    Lets say that $\displaystyle t > K$ then using floor function, $\displaystyle j = \left\lfloor {\frac{t}{K}} \right\rfloor \,\& \,r = t - Kj$.
    Is it clear that $\displaystyle r \in [0,K]\,\& \,t = jK + r$.
    Can you finish?
    Last edited by Plato; Feb 9th 2008 at 02:07 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 $\displaystyle \forall x\left[ {f(x + nK) = f(x)} \right]$.
    Lets say that $\displaystyle t > K$ then using floor function, $\displaystyle j = \left\lfloor {\frac{t}{K}} \right\rfloor \,\& \,r = \frac{t}{K} - j$.
    Is it clear that $\displaystyle r \in [0,K]\,\& \,t = jK + r$.
    Can you finish?
    Yes, I think so. Thank you very much.

    What does $\displaystyle 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 $\displaystyle j = \left\lfloor {\frac{t}{K}} \right\rfloor $ mean?
    The floor function is often called the greatest integer function.
    $\displaystyle \left\lfloor {\frac{{99.1}}{{11}}} \right\rfloor = 9$.
    Please, see my edit of the reply above: $\displaystyle r = t - Kj$.
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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 $\displaystyle f(x) = \tan x$ whenever it is defined an $\displaystyle f(x)= 0$ elsewhere then $\displaystyle f$ 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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