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Math Help - differentiation prove..

  1. #1
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    differentiation prove..

    there is a function which is differentiable continuously on [a,b] .
    suppose
    |f'(x)|<1
    for all  x\epsilon [a,b]
    prove that there exists
    0<=k<1
    <br />
 x_1\epsilon [a,b]<br />
    <br />
 x_2\epsilon [a,b]<br />
    that this equation is true
    <br />
|f(x_1)-f(x_2)|<=K|x_1-x_2|<br />
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  2. #2
    Member Abu-Khalil's Avatar
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    If you take f(x)=0.9x and K<0.9 the equation isn't true :S
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  3. #3
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    i got a solution to this..but i cant understand it.
    they make mean value theorem on f'(x) [x1,x2] so x2>=x1
    <br />
\frac{f'(x_2)-f'(x_1)}{x_2-x_1}=f'(c)<br />
    <br />
|f'(x_2)-f'(x_1)|=|f'(c)||x_2-x_1|<br />
    then they say that by the second law of weirsstas there is a minimum and maximum
    so there is l,L on [a,b] so
    f'(L)>=f'(x)>=f'(l) for every x in [a,b]

    i cant understand what is the second law of weirshtass
    that gives two extreme points??

    there is another things but i got stuck on this
    ??
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  4. #4
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by transgalactic View Post
    i got a solution to this..but i cant understand it.
    they make mean value theorem on f'(x) [x1,x2] so x2>=x1
    <br />
\frac{f'(x_2)-f'(x_1)}{x_2-x_1}=f'(c)<br />
    <br />
|f'(x_2)-f'(x_1)|=|f'(c)||x_2-x_1|<br />
    then they say that by the second law of weirsstas there is a minimum and maximum
    so there is l,L on [a,b] so
    f'(L)>=f'(x)>=f'(l) for every x in [a,b]

    i cant understand what is the second law of weirshtass
    that gives two extreme points??

    there is another things but i got stuck on this
    ??
    I correct yours.
    By the means of the mean value theorem, we have
    f(s)-f(t)=f^{\prime}(\xi)(s-t) for all s\leq tand for some \xi\in[s,t]\subset[a,b].
    Taking absolute value of both sides, we have
    |f(s)-f(t)|=|f^{\prime}(\xi)||s-t|\leq\mu|s-t|,
    where
    \mu:=\max\{|f^{\prime}(\zeta)|:\zeta\in[a,b]\}.
    Note that \mu\in(0,1).
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  5. #5
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    what is the meaning of "I correct yours."

    ??
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  6. #6
    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by transgalactic View Post
    what is the meaning of "I correct yours."

    ??
    I corrected your answer, because you used the terms f^{\prime} in the numerator, may be it was just a typo.
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  7. #7
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    it wasnt a mistake the function is continuesly differentiable
    so i can do this thing with f'(x)
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  8. #8
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    Quote Originally Posted by transgalactic View Post
    there is a function which is differentiable continuously on [a,b] .
    suppose
    |f'(x)|<1
    for all  x\epsilon [a,b]
    prove that there exists
    0<=k<1
    <br />
x_1\epsilon [a,b]<br />
    <br />
x_2\epsilon [a,b]<br />
    that this equation is true
    <br />
|f(x_1)-f(x_2)|<=K|x_1-x_2|<br />
    This question may have been posted and answered at other websites that provide maths help.
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  9. #9
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    it have been answered there
    i am trying to solve it.
    Last edited by mr fantastic; March 3rd 2009 at 11:04 PM. Reason: Removed link
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  10. #10
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by transgalactic View Post
    it have been answered there
    i am trying to solve it.
    Please read my first post (#4) carefully
    Last edited by mr fantastic; March 3rd 2009 at 11:05 PM. Reason: Removed link from quote
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