1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    rbj

    User Avatar

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook