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Integrating infinite sums

  1. Nov 27, 2007 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data
    Using the macluarin's expansion for sinx show that [itex]\int sinx dx=-cosx+c[/itex]


    2. Relevant equations

    [tex]sinx=\sum_{n=0} ^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}[/tex]

    3. The attempt at a solution
    Well I can easily write out some of the series and just show that it is equal to -cosx

    but if I integrate the representation for the infinite series i get

    [tex]\int sinx dx= \sum_{n=0} ^\infty \frac{(-1)^nx^{2n+2}}{(2n+1)!}[/tex]
    shouldn't -cosx be:
    [tex]\int sinx dx= \sum_{n=0} ^\infty \frac{(-1)^{n+1}x^{2n+2}}{(2n+1)!}[/tex]

    and also I am supposed to get [itex]x^{2n}[/itex] not what I got
     
  2. jcsd
  3. Nov 27, 2007 #2

    Avodyne

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    Your denominators should be (2n+2)! in the last two lines.

    With this correction, your last series is cos(x)-1; write out the first few terms to see.
     
  4. Nov 27, 2007 #3

    rock.freak667

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    ah yes I made a typo...but even if I change it
    how do I manipulate

    [tex]\sum_{n=0} ^\infty \frac{(-1)^nx^{2n+2}}{(2n+2)!}[/tex] (what I integrated and got)

    Into

    [tex]-\sum_{n=0} ^\infty \frac{(-1)^nx^{2n}}{(2n)!}[/tex] (What I am supposed to get)
     
  5. Nov 28, 2007 #4

    HallsofIvy

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    Let j= n+1 and see what you get with j as the index.
     
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