# Homework Help: Power series help

1. Jan 31, 2007

### mkienbau

How do I multiply power series?

1. The problem statement, all variables and given/known data
Find the power series:
$$e^x arctan(x)$$

2. Relevant equations

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}$$

$$arctan(x) = 0 + x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7}$$
3. The attempt at a solution

So do I multiply 1 by 0, x by x and so forth? Or do I go 1 by 0, 1 by x? Or is there another way?

Last edited: Jan 31, 2007
2. Jan 31, 2007

### dimensionless

You have to multiply 1 by the whole acrtan series, x by the whole arctan series, and so on. There might be a way to simplify it though. Wikipedia has this under "power series"

$$f(x)g(x) = \left(\sum_{n=0}^\infty a_n (x-c)^n\right)\left(\sum_{n=0}^\infty b_n (x-c)^n\right)$$

$$= \sum_{i=0}^\infty \sum_{j=0}^\infty a_i b_j (x-c)^{i+j}$$

$$= \sum_{n=0}^\infty \left(\sum_{i=0}^n a_i b_{n-i}\right) (x-c)^n$$

3. Jan 31, 2007

### mkienbau

So I kind of treat it like F.O.I.L.?

4. Jan 31, 2007

### Tom Mattson

Staff Emeritus
Sort of. FOIL is the distrubutive law for (binomial)X(binomial). Here, you've got two infinitely long "polynomials". Obviously, you won't be able to write out all of the terms. :tongue:

5. Jan 31, 2007

### mkienbau

Awesome, I think I got it, I only had to take it out to the $$x^5$$ term.