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Math Help - [SOLVED] Maclaurin series for square root

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Maclaurin series for square root

    Can anyone tell me where I messed up (not getting the correct answer):

    What is the 4th order Maclaurin of: f(x) = \sqrt{9 + x^2}

    1 + u = 9 + x^2 \implies

    \sqrt{1 + u} = 1 + \frac{u}{2} - \frac{u^2}{2} + O(x^3) \implies

    \sqrt{9 + x^2} = 1 + \frac{8 + x^2}{2} - \frac{(8 + x^2)^2}{8} + O(x^6)

    EDIT: The substitution u = \frac{x^2}{9} works better. Then I can just use 3\sqrt{1 + \frac{x^2}{9}}
    Last edited by Spec; February 2nd 2008 at 12:54 PM.
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  2. #2
    Eater of Worlds
    galactus's Avatar
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    I'm getting

    3+\frac{x^{2}}{6}-\frac{x^{4}}{216}+\frac{x^{6}}{3888}-\frac{5x^{8}}{279936}+...............
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  3. #3
    Senior Member Spec's Avatar
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    Yeah I got the correct answer after using a proper substitution.
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