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Math Help - Mean Value Theorem Revisited!

  1. #1
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    Mean Value Theorem Revisited!

    Let f(x)=\tan^{-1} x, and 0 < a < b. By MVT there exists a < c < b such that f(b)-f(a)=(b-a)f'(c). Prove that \sqrt{ab} < c < \sqrt{\frac{a^2 + b^2}{2}}. (obviously c is unique.)


    Source: Alex Lupas.

    Note: This is just a simple example. The importance of the problem is that it looks at a less discussed side of mean value theorem, i.e. the location of c.
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  2. #2
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    Well, simple f'=1/(1+x^2) so f'(c)=1/(1+c^2)=(arctg(b)-arctg(a))/(b-a)
    c^2=(b-a)/(arctg(b)-arctg(a))-1
    c=sqrt(((b-a)/(arctg(b)-arctg(a))-1)
    I need to show that (b-a)/(arctg(b)-arctg(a))-1>ab, and I know that the maximum is attained when it equals pi/2, not sure how to continue from here, any hints?
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    Let f(x)=\tan^{-1} x, and 0 < a < b. By MVT there exists a < c < b such that f(b)-f(a)=(b-a)f'(c). Prove that \sqrt{ab} < c < \sqrt{\frac{a^2 + b^2}{2}}. (obviously c is unique.)


    Source: Alex Lupas.

    Note: This is just a simple example. The importance of the problem is that it looks at a less discussed side of mean value theorem, i.e. the location of
    c.
    Claim: c^2 > ab

    Clearly \frac{b-a}{1+ab} = \tan\left(\frac{b-a}{1+c^2}\right)

    Since b > a > 0 and \tan x > x, \forall x > 0, we get \frac{b-a}{1+ab} = \tan\left(\frac{b-a}{1+c^2}\right) > \frac{b-a}{1+c^2} \Rightarrow c^2 > ab

    Claim: c^2 < \frac{a^2 + b^2}{2}

    This part of the proof is not so neat

    By a straightforward manipulation, c^2 = \frac{b - a}{\tan^{-1} b - \tan^{-1} a} - 1 = \frac{\tan x - \tan y}{x - y} - 1, where \tan x = b, \tan y = a.

    Hence we have to prove that c^2 < \frac{\tan^2 x + \tan^2 y}{2}

    The \tan^{-1} function has range from -\frac{\pi}{2} to \frac{\pi}{2}, we will assume x and y to be in that range.

    b > a > 0 \Rightarrow x > y > 0 which leads to:

    \frac{\tan x - \tan y}{x - y} = \frac1{\cos^2 x \cos^2 y} \left(\cos x \cos y\frac{\sin (x - y)}{(x - y)} \right) <  \frac1{\cos^2 x \cos^2 y} \cos x \cos y < \frac1{\cos^2 x \cos^2 y}\frac{\cos^2 x + \cos^2 y}{2}  = \frac{\sec^2 x + \sec^2 y}{2} = \frac{\tan^2 x + \tan^2 y}{2} + 1

    Thus c^2 = \frac{\tan x - \tan y}{x - y} - 1 < \frac{\tan^2 x + \tan^2 y}{2}

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