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Math Help - Mean value theorem

  1. #1
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    Mean value theorem

    By using the mean value therom prove that |sinx-siny|< or = |x-y| for all x and y.
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  2. #2
    Senior Member TriKri's Avatar
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    Quote Originally Posted by myoplex11 View Post
    By using the mean value therom prove that |sinx-siny|< or = |x-y| for all x and y.
    Notice it won't make any difference if we swap x and y in both equations, so let's say x \geq y. So if it isn't already, just swap x and y.

    Say f_1(x) = |sin(x)-sin(y)| and f_2(x) = |x-y|.
    Since x \geq y, f_2(x) = x-y.

    If sin(x) = sin(y), f_1(x) will be 0 and the relation will be satisfied, since f_2(x) \geq 0.

    If sin(x) > sin(y), f_1'(x) = (sin(x)-sin(y))' = cos(x)
    If sin(x) < sin(y), f_1'(x) = (sin(y)-sin(x))' = -cos(x)

    I none of the cases the derivative is > 1, since x and y is real numbers (I assume).

    Now the mean value theorem states that there is a point on any given section of a smooth curve that has the same derivative as the average derivative of the section. That mean the average derivative a of the section [y, x] of f_1(x) can't be greater than 1, since the derivative at any given point on f_1(x) is \leq 1. Hence, f_1(x) - f_1(y) can't be greater than x - y, since f_1(x) - f_1(y) = a \cdot (x-y), a \leq 1 and x-y is a non-negative number. And f_1(x) - f_1(y) = |sin(x) - sin(y)| - |sin(y) - sin(y)| = (I have to break the line here, LaTeX won't tale any longer expressions)
    = |sin(x) - sin(y)| - 0 = f_1(x). Hence f_1(x) can't be greater than x-y, so |sin(x) - sin(y)| \leq |x-y| for x \geq y. And as I wrote in the begining of this post it won't make any difference if x \leq y.

    I don't know if that last part was completely correct. And if you wan't a more accurate explanation than mine, you'll just have to wait until any of the math gurus on this forum answers your thread.

    -Kristofer
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  3. #3
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    Lightbulb value theorem

    There is x< c <y, if x<y or y<c<x if y<x such that (sinx-siny)/(x-y)= cosc
    or │(sinx-siny)/(x-y)│=│cosc│
    But │cosc│≤ 1
    Then
    │sinx-siny│/│x-y│≤1
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