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Math Help - MacLaurin series error estimation

  1. #1
    Bar0n janvdl's Avatar
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    MacLaurin series error estimation

    Hey everyone, quick question:

    Approximate ln(0.6) by using the MacLaurin series expansion for ln(1-x), with an error guaranteed to be less than 0.01. Show your error estimate.
    My workings:

    Since we want ln(0.6), but make use of the series expansion for ln(1-x), x = 0.4

    \ln(1-x) =  - \left(x + \frac{x^2}{2} + \frac{x^3}{3} + ... \right)

    \ln(1-x) =  - \sum ^{\infty} _{n=0} (-1) \frac{x^{n+1}}{n+1}<br />

    Let e_n be the error estimate.

    | e_1 | = \frac{0.4^{1 + 1}}{1+1} = 0.08

    | e_2 | = 0.02

    | e_3 | = 0.0064


    So therefore \ln(0.6) = \sum ^{2} _{n=0} (-1) \frac{0.4^{n+1}}{n+1} = -1,101333...
    However, my calculator gives me -0.510826.

    Where did I go wrong?
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  2. #2
    MHF Contributor
    Jester's Avatar
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    Quote Originally Posted by janvdl View Post
    Hey everyone, quick question:



    My workings:



    However, my calculator gives me -0.510826.

    Where did I go wrong?
    Your first series doesn't alternate - your second does. They are not the same.
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Danny View Post
    Your first series doesn't alternate - your second does. They are not the same.
    Sorry, could you please elaborate on this?
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  4. #4
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    Quote Originally Posted by janvdl View Post
    Sorry, could you please elaborate on this?
    Sorry, it doesn't. Why are you writing the series

     <br />
-\sum_{n=0}^{\infty} (-1) \frac{x^{n+1}}{n+1}?<br />

    and further, how are you getting your error bounds?
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