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Finding a Maclaurin Series

  1. May 22, 2005 #1
    I need some peer review of this quesion. Let [tex]f(x)=x^2e^{-2x^2}[/tex]

    (a) find the maclaurin series of f

    (b) find the 100th degree maclaurin polynomial of f.


    for part a i have:[tex]\sum_{n=0}^{\infty}\frac{(-2)^nx^{2n+2}}{n!}[/tex]

    and for b:[tex]\sum_{n=0}^{99}\frac{(-2)^nx^{2n+2}}{n!}[/tex]
     
    Last edited: May 22, 2005
  2. jcsd
  3. May 23, 2005 #2
    does anyone have a review for me?
     
  4. May 23, 2005 #3

    Galileo

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    It looks good. I'm not sure what you mean by the 100th degree MacLaurin polynomial, but if you mean the expansion up to the 100th power of x you have too many terms, since it goes up to the 2(99)+2=200th power of x.
     
  5. May 23, 2005 #4

    dextercioby

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    Are u sure...?

    What is [tex]f^{\mbox{iv}}(x) [/tex]...?And then,of course,[tex]f^{\mbox{iv}}(x) [/tex]...?

    Daniel.
     
  6. May 26, 2005 #5

    This is the 100th degree maclaurin polynomial let [tex]100=2n+2[/tex] so that [tex]n=49[/tex] now we have this:[tex]\sum_{n=0}^{49}\frac{(-2)^nx^{2n+2}}{n!}[/tex]

    my oginal post of part (b) was incorrect. now this is correct
     
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