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Math Help - Approximation of ln

  1. #1
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    Approximation of ln

    Approximation of ln-question.png
    Approximation of ln-answer.png

    My question is about e) x*ln(x). So to find the approximation. It is said:

    x*ln(x) = (1+h)ln(1+h) \approx (1+h)(h-\frac{h^2}{2}) \approx h + \frac{h^2}{2}

    My question is about (h-\frac{h^2}{2}). I thought I studied my approximations manual fairly in depth however I don't recall if anything was said about ln or anything about (h-\frac{h^2}{2})? I am wondering why ln(1+h) seems to be approximately equal to (h-\frac{h^2}{2}) or should I just accept that it is?

    Thanks for any responses...
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  2. #2
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    Re: Approximation of ln

    Hey sepoto.

    What do you have for your Taylor series approximation (up to quadratic term) for ln(1+x)?
    Thanks from sepoto
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  3. #3
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    Re: Approximation of ln

    [deleted]
    Last edited by romsek; December 8th 2013 at 06:14 PM. Reason: chiro is on the case
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  4. #4
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    Re: Approximation of ln

    I am trying to solve the problem without using the Taylor series because I just have not gotten there yet. Is (h-h^2/2) the quadratic approximation of ln(1+h) perhaps? The use of ln is throwing me off I think.


    P.S. I dug this out of the lecture notes. I think this is the identity that I was looking for:

    Approximation of ln-notes.png

    I am trying to get to the Taylor Series but I have so much to do still.
    Last edited by sepoto; December 8th 2013 at 07:08 PM.
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