# Power series representation of a function

1. Oct 6, 2009

### quasar_4

1. The problem statement, all variables and given/known data

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

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 $$f(x) = \int_{0}^{x} f(t) dt = \sum_{n=0}^{\infty} \frac{a_n x^{n+1}}{n+1}$$. 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 $$sin(t^3) \approx \sum_{n=0}^{\infty} \frac{(-1)^n (x^3)^{2n+1}}{(2n+1)!}$$. Now do I just change this to the sum
$$\sum_{n=0}^{\infty} \frac{(-1)^n (x^3)^{2n+3}}{(2n+3)! (n+1)}$$? It's not at all clear to me that this is what I'm supposed to do...

2. Oct 6, 2009

### Staff: Mentor

I think you might confuse yourself by having two different definitions for f.
$$f(x) = \int_{0}^{x} sin(t^3) dt = \int_{0}^{x}\sum_{n=0}^{\infty} \frac{(-1)^n (t^3)^{2n+1}}{(2n+1)!} dt$$
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.