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  1. MHB Apprentice

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    #1
    In the following equation:
    $$g'(x) = f'\left( x + \frac{c f'(x)}{\sqrt{ 1 + f'(x)^2 }} \right)$$
    find $g(x)$ with respect to $f(x)$ where $c$ is any constant.
    Last edited by I like Serena; November 26th, 2016 at 14:28. Reason: Convert to latex

  2. MHB Journeyman
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    #2
    Quote Originally Posted by wheepep View Post
    In the following equation:



    find g(x) with respect to f(x) where c is any constant.
    The link is broken. Can you type it out?

    -Dan

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    #3
    Quote Originally Posted by wheepep View Post
    In the following equation:
    $$g'(x) = f'\left( x + \frac{c f'(x)}{\sqrt{ 1 + f'(x)^2 }} \right)$$
    find $g(x)$ with respect to $f(x)$ where $c$ is any constant.
    Hi wheepep! Welcome to MHB!

    I'm not sure where we want to go with this, but I can give a couple of observations.

    We can observe that:
    $$|f'(x)| < \sqrt{1+f'(x)^2}$$
    and therefore:
    $$-|c| < \frac{c f'(x)}{\sqrt{ 1 + f'(x)^2 }} <| c|$$


    From a Taylor expansion we get:
    $$g'(x) = f'\left( x + \frac{c f'(x)}{\sqrt{ 1 + f'(x)^2 }} \right) = f'(x)+ \frac{c f'(x)}{\sqrt{ 1 + f'(x)^2 }}f''(x+\theta c)$$
    where $0\le \theta < 1$.


    If we assume that $f''$ is bounded, then we get:
    $$|g'(x) - f'(x)| < |cf''(x+\theta c)| \le |c|M$$
    for some upper bound $M$.
    And therefore:
    $$ -|c|Mx + C < g(x) - f(x) < |c|Mx + C \quad\Rightarrow\quad f(x) - |c|Mx + C < g(x) < f(x) + |c|Mx + C$$
    where $C$ is an integration constant.

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