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Maclaurin series help

  1. Sep 11, 2005 #1
    I've looked at a number of explanations for the Maclaurin series of cos(x) yet none have given an easily understood answer, i was wondering if anyone has a way of explaining it when it is used as only a part of a function

    eg. use that Maclaurin series of cos(x) to obtain the Macalurin series for

    h(x)=(cos(3x)-1)/x^2
     
  2. jcsd
  3. Sep 11, 2005 #2

    lurflurf

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    The formula for Maclaurin series is
    [tex]f(x)=\sum_{k=0}^\infty \frac{x^k}{k!}f^{(k)}(x)[/tex]
    as can be easily derived several ways
    -integration by parts
    -formal operator manipulation
    -matching derivatives at zero
    -complex contour integration
    This can be applied to cos(x) as
    [tex]\cos(x)=\sum_{k=0}^\infty \frac{x^k}{k!}\cos^{(k)}(0)=\sum_{k=0}^\infty \frac{x^k}{k!}\cos(k\pi/2)=\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}(-1)^k[/tex]
    where the index is changed in the final step to indicate odd terms are 0
    Often a known Maclaurin series can be used to find a closely related one.
    Just apply the known series
    h(x)=(cos(3x)-1)/x^2
    [tex]h(x)=\frac{1}{x^2}\left(\sum_{k=0}^\infty \frac{(3x)^{2k}}{(2k)!}(-1)^k-1\right)[/tex]
    simplify that into standard form.
     
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