# Antiderive complex function f(z) and express as power series

1. Aug 20, 2013

### Jd303

Let F(z) be the anti-derivative of the function f(z) = cos(z^3) with F(0) = 0. Express F(z) as a power series around z=0, giving both the first 3 non-zero terms and the general (nth) term.

Hey guys really struggling with this integration and how to then express this as a power series. Any help at all would be greatly appreciated because I am getting nowhere!

2. Aug 20, 2013

### Goa'uld

Start with the Taylor series of cosine. Is that enough to get you going?

3. Aug 21, 2013

### haruspex

You don't need to integrate anything. Just write out the general Taylor series for this unknown function F in the vicinity of 0.

4. Aug 21, 2013

### HallsofIvy

Staff Emeritus
How is he going to find the anti-derivative without integrating?

Jd303, Goa'uld's suggestion is best- write out the Taylor series for cos(z), replace "z" by z3, then integrate "term by term".

5. Aug 21, 2013

### Jd303

Thanks guys I think I got it:
Let u = z^3
find Taylor series for f(u)
sub back in z^3
integrate term by term to find F(z)

6. Aug 21, 2013

### haruspex

You mean cos(u), right?
Another way: F(z) = F(0) + z F'(0) + z2F"(0)/2! + ...
You are given F(0), and you can substitute for F' etc. using f.