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Power series representation of a function

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Find a power series representation for f(x) using termwise integration, where [tex]f(x) = \int_{0}^{x} sin(t^3) dt [/tex].


    2. Relevant equations



    3. The attempt at a solution

    I've never done this before, but apparently, if I have a power series representation for sin(t^3), I can use the fact that [tex]f(x) = \int_{0}^{x} f(t) dt = \sum_{n=0}^{\infty} \frac{a_n x^{n+1}}{n+1} [/tex]. Is that right?

    So my approach was to use the power series representation of sine, replacing sin x with sin t^3 - this gave me [tex]sin(t^3) \approx \sum_{n=0}^{\infty} \frac{(-1)^n (x^3)^{2n+1}}{(2n+1)!} [/tex]. Now do I just change this to the sum
    [tex]\sum_{n=0}^{\infty} \frac{(-1)^n (x^3)^{2n+3}}{(2n+3)! (n+1)} [/tex]? It's not at all clear to me that this is what I'm supposed to do...
     
  2. jcsd
  3. Oct 6, 2009 #2

    Mark44

    Staff: Mentor

    I think you might confuse yourself by having two different definitions for f.
    [tex]f(x) = \int_{0}^{x} sin(t^3) dt = \int_{0}^{x}\sum_{n=0}^{\infty} \frac{(-1)^n (t^3)^{2n+1}}{(2n+1)!} dt[/tex]
    Now you're set up to integrate term-by-term to get an antiderivative, which you can evaluate at x and at 0.

    I think you made a mistake with your exponent. (t^3)^(2n+1) = t^(6n + 3). When you integrate that, you get something different from what you show.
     
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