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Homework Help: Maclaurin series of a function

  1. Nov 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the maclaurin series of:

    [tex] f(x) = \int_{0}^{x}(e^{-t^2}-1) dt [/tex]

    3. The attempt at a solution

    I know [tex] e^t = \sum_{n=0}^{∞} \frac{t^n}{n!} [/tex]

    Simple substitution gives me:

    [tex] e^{-t^2} = \sum_{n=0}^{∞}\frac{(-t^2)^n}{n!} [/tex]

    Which I rewrote as

    [tex] e^{-t^2} = \sum_{n=0}^∞\frac{(-1)^n(t^{2n})}{n!}[/tex]

    Since I notice that when n = 0, the term simplifies to 1,

    [tex] e^{-t^2} - 1 = \sum_{n=1}^∞ \frac{(-1)^n(t^{2n})}{n!}[/tex]

    Going back to the original expression and substituting the infinite series

    [tex] \int_0^x\sum_{n=1}^∞ \frac{(-1)^n(t^{2n})}{n!} dt [/tex]

    Now, I'm integrating with respect to t, so after using the power rule and applying the upper limit (lower limit simplifies to 0)

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

    And this is what I would think is the maclaurin representation of the function, but it's wrong and I have no idea why. Would anyone please point to where I'm going wrong with this?
    Last edited: Nov 15, 2012
  2. jcsd
  3. Nov 15, 2012 #2


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    How did your lower limit n=0 appear in the appear in the answer? It was n=1 in the step before.
  4. Nov 15, 2012 #3
    Sorry about that, typo from copying and pasting.
  5. Nov 15, 2012 #4


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    How do you know it's wrong?

    If you have the correct answer, please give it.

    You can change the sum to start at n=0, if that's the problem.
  6. Nov 15, 2012 #5


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    Well, then I'm having trouble finding anything wrong with it. But does the answer you are checking against expect you to shift the lower limit to n=0?
  7. Nov 15, 2012 #6
    Your replies made me look at the answer list a little harder and I just noticed why my answer isn't matching up.

    The answer listed is
    [tex] \sum_{n=0}^∞\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)(n+1)!} [/tex]

    They shifted the lower bound to start at 0 and compensated by replacing n by (n+1).

    Pretty frustrating considering I've been hitting my head against the wall thinking I was doing something wrong all this time...

    Thanks a lot for the help to both of you. Just learned about power series about a week ago so still not too used to working with them.
  8. Nov 15, 2012 #7


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    there isn't anything wrong with it. this

    [tex] f(x) = \int_{0}^{x}(e^{-t^2}-1) dt = \sum_{n=1}^\infty \frac{(-1)^n x^{2n+1}}{n!(2n+1)} [/tex]

    is in fact correct.
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