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Math Help - Mean Value Theorem Proof

  1. #1
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    Mean Value Theorem Proof

    Use the Mean Value Theorem to prove that for all a,b that belongs to the reel, we have :

    |sin(b)-sin(a)| \leq | b-a |
    Last edited by Larrioto; January 26th 2009 at 04:55 PM.
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    Quote Originally Posted by Larrioto View Post
    Use the Mean Value Theorem to prove that for all a,b that belongs to the reel, we have :

    |sin(b)-sin(a)| \leq | b-a |
    well, if you let f(x)=\sin x, then by mean value theorem there exists c between a,b such that f(b)-f(a)=(b-a)f'(c). that means \sin b - \sin a = (b-a)\cos c.

    so |\sin b - \sin a|=|b-a||\cos c| \leq |b-a|, because |\cos c| \leq 1. it wan't very difficult, was it?
    Last edited by mr fantastic; January 27th 2009 at 05:10 AM. Reason: Replaced quote
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    There is a second part to this problem :

    Assume that f :  \Re \mapsto \Re is a differentiable function and that there exists a real number M > 0 such that |f '(x)| \leq M for all x in the reel then show that :

    |f(a) - f(b)|\leq M|b-a|for all x in the reel
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    Quote Originally Posted by Larrioto View Post
    There is a second part to this problem :

    Assume that f :  \Re \mapsto \Re is a differentiable function and that there exists a real number M > 0 such that |f '(x)| \leq M for all x in the reel then show that :

    |f(a) - f(b)|\leq M|b-a|for all x in the reel
    Choose any value c, we know there exists a corresponding interval [a,b] such that |f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\leqslant M and since c was arbitrary this must work for every interval [a,b].


    Another almost identical method would be to assume that there exists an interval [a,b] such that \left|\frac{f(b)-f(a)}{b-a}\right|>M. But since f is differentiable we can find a c\in(a,b) such that |f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\implies |f'(c)|>M which is a contradiction.

    This is called Lipschitz's continuity by the way.

    If you have any questions at all feel free to ask
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    Quote Originally Posted by Mathstud28 View Post
    Choose any value c, we know there exists a corresponding interval [a,b] such that |f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\leqslant M and since c was arbitrary this must work for every interval [a,b].


    Another almost identical method would be to assume that there exists an interval [a,b] such that \left|\frac{f(b)-f(a)}{b-a}\right|>M. But since f is differentiable we can find a c\in(a,b) such that |f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\implies |f'(c)|>M which is a contradiction.

    This is called Lipschitz's continuity by the way.

    If you have any questions at all feel free to ask
    Do you mean that whatever number we choose to be c , it's derivative will be 0 , therefore we can say that if M\geq 0 then it is \geq \left|\frac{f(b)-f(a)}{b-a}\right| ?
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Larrioto View Post
    Do you mean that whatever number we choose to be c , it's derivative will be 0 , therefore we can say that if M\geq 0 then it is \geq \left|\frac{f(b)-f(a)}{b-a}\right| ?
    Im not quite sure what you mean by that...If the first proof doesn't make sense consider the second one. It reads better anyways.
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    ok , i get it , it's just that f(c) can be replaced by f(x), that was confusing me ...
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    Quote Originally Posted by Mathstud28 View Post

    Choose any value c, we know there exists a corresponding interval [a,b] such that |f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right| ...
    this, although an interesting claim, but is not what mean value theorem says: you choose an interval [a,b] and then you find a < c < b that satisfies the condition. what you're doing here is

    reversing things, i.e. you choose c and then you claim that there exists an interval [a,b] containing c with that condition! i'm not sure if this is always possible!
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    Quote Originally Posted by NonCommAlg View Post
    this, although an interesting claim, but is not what mean value theorem says: you choose an interval [a,b] and then you find a < c < b that satisfies the condition. what you're doing here is

    reversing things, i.e. you choose c and then you claim that there exists an interval [a,b] containing c with that condition! i'm not sure if this is always possible!
    Mathworld has an "iff" in it :S...that is why I supplied a second proof that did not require that just in case. Sorry if there has been any confusion.
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    Quote Originally Posted by Mathstud28 View Post

    Mathworld has an "iff" in it :S...that is why I supplied a second proof that did not require that just in case. Sorry if there has been any confusion.
    here's a simple counter-example: let f(x)=x^3 and choose c=0. geometrically it's clear that there are no a \neq b such that f'(0)=\frac{f(b)-f(a)}{b-a}. you can also prove it very easily:

    if there exist such a,b, then: 0=\frac{b^3-a^3}{b-a}=a^2+ab+b^2, which is possible iff a=b=0, because: 0=2a^2 + 2ab+2b^2=(a+b)^2+a^2+b^2.


    regarding Larrioto's question:

    by mean value theorem for any real numbers a,b, there exists c (between a,b) such that f(b)-f(a)=(b-a)f'c). thus: |f(b)-f(a)|=|b-a||f'(c)| \leq M|b-a|.
    Last edited by NonCommAlg; January 26th 2009 at 08:35 PM.
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