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Maclaurin power series

  1. Nov 27, 2006 #1


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    I was hoping someone could check my work:

    Find the maclaurin power series for the function:

    a. f(x)=1/(4x^2+1)

    b. f(x)= [tex]\int[/tex] e^-x^2 dx

    For a I got [tex](-1)^n*2nx^n[/tex]. For b I dont know where to start.
    Last edited: Nov 27, 2006
  2. jcsd
  3. Nov 27, 2006 #2


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    For b I can get the approx to be


    So i get (x^-2n)/n!
  4. Nov 27, 2006 #3
    1 (a) [tex] f(x) = \frac{1}{1+4x^{2}} [/tex]

    This is equaled to [tex] \sum_{n=0}^{\infty} (-4)^{n}x^{2n} [/tex]

    (b) [tex] e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!} [/tex]

    Thus [tex] e^{-x^{2}} = \sum_{n=0}^{\infty} (-1)^{n}\frac{x^{2n}}{n!} [/tex] and

    [tex] \int \sum_{n=0}^{\infty} (-1)^{n}\frac{x^{2n}}{n!} = \sum_{n=0}^{\infty} (-1)^{n}\frac{x^{2n+1}}{(2n+1)n!} [/tex]
  5. Nov 27, 2006 #4


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    Thanks for the help. I just got b and was coming back to post my answer.
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