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Math Help - Maclaurin series estimate

  1. #1
    Senior Member mollymcf2009's Avatar
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    Maclaurin series estimate

    How many term of the MAclaurin series for ln(1+x) do you need to use to estimate ln(1.4) to within 0.001?

    I have tried this 10 different ways and I keep getting bigger values.

    The MacLaurin series is:

    ln(1+x) = \sum^{\infty}_{n=0} (-1)^n \frac{x^{n+1}}{n+1}
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Note this is an alternating series so you can use |S-Sn| < a(n+1)

    replace x with .4

    you get (-1)^(n+1) *.4^(n+1)/(n+1)

    |S-Sn| < .4^(n+2)/(n+2) < .001

    Use a table to find the smallest n to satisfy this inequality
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  3. #3
    MHF Contributor chisigma's Avatar
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    For the alternating series, if we stop the interation after the summing of the term a_{n}, the [obvious] error criterion is...

    |e_{n}|<|a_{n+1}| (1)

    Computing the 'infinite sum'...

    \ln (1+x)= - \sum_{n=1}^{\infty}(-1)^{n}\cdot \frac{x^{n}}{n} (2)

    ... the condition |a_{n+1}|<.001 is met for n= 5, i.e. five term of (2) are enough and that gives...

    \ln(1.4)\approx .336 (3)

    The 'exact' value is...

    \ln(1.4) =.336472236621\dots (4)

    Kind regards

    \chi \sigma
    Last edited by chisigma; April 20th 2009 at 10:00 AM.
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  4. #4
    MHF Contributor chisigma's Avatar
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    The series expansion...

    \ln (1+z) = -\sum_{n=1}^{\infty} (-1)^{n} \frac{z^{n}}{n} (1)

    ... converges in genearal for complex values of z if |z|<1. It is interesting the computation of (1) for z=\frac{i}{\sqrt{3}}. Summing 'only' 44 complex terms of (1) we obtain...

    \ln (1+ \frac{i}{\sqrt{3}}) = \ln \frac{2}{\sqrt{3}} + i\cdot \frac {\pi}{6} = .143841036226\dots + i\cdot .523598775598\dots

    ... where both real and imaginary part have an error less than 10^{-12} ...

    Kind regards

    \chi \sigma
    Last edited by chisigma; April 20th 2009 at 10:01 AM.
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  5. #5
    MHF Contributor Calculus26's Avatar
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    n= 4 will work
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