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Math Help - Demonstration

  1. #1
    Super Member dhiab's Avatar
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    Demonstration

    Prove :
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  2. #2
    Senior Member Twig's Avatar
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    Hi

    There might be a better way to do this, but IŽll give one solution.
    Note: x-\frac{x^{2}}{2} is the first two terms of the maclaurin series of  ln(1+x) .

    Anyway, let f(x)=x-\frac{x^{2}}{2}-ln(1+x) , then we want to show that f(x)<0 for all x>0 .

    Solution: f is continous and differentiable for all x>-1 , with f'(x)=\frac{-x^{2}}{1+x} . (I have already simplified the derivative here).

    Note tha we are only interested in  x>0 , so we donŽt have to worry about the denominator of the derivative becoming zero.

    Looking at f'(x) we see that the only stationary point is x=0 . And for all x>0 we have that  f'(x)<0 .

    And because f(0)=0 , and the function is strictly decreasing for all  x>0 , it immediately follows that f(x)<0 .

    Now for the right inequality:

    You could do the same thing for g(x)=x-ln(1+x) , and show that g(x)>0 .
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  3. #3
    MHF Contributor red_dog's Avatar
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    Let 0,\infty)\to\mathbf{R}" alt="f(x)=\ln(1+x)-x, \ f0,\infty)\to\mathbf{R}" />

    f'(x)=-\frac{x}{1+x}<0, \ \forall x>0, so f is strictly decreasing.

    Then, x>0\Rightarrow f(x)<f(0)=0\Rightarrow\ln(1+x)<x

    Try to prove the other inequality in a similar way by using the function

    g(x)=x-\frac{x^2}{2}-\ln(1+x), \ x>0
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