# Maclaurin series help

1. Sep 11, 2005

### Muzly

I've looked at a number of explanations for the Maclaurin series of cos(x) yet none have given an easily understood answer, i was wondering if anyone has a way of explaining it when it is used as only a part of a function

eg. use that Maclaurin series of cos(x) to obtain the Macalurin series for

h(x)=(cos(3x)-1)/x^2

2. Sep 11, 2005

### lurflurf

The formula for Maclaurin series is
$$f(x)=\sum_{k=0}^\infty \frac{x^k}{k!}f^{(k)}(x)$$
as can be easily derived several ways
-integration by parts
-formal operator manipulation
-matching derivatives at zero
-complex contour integration
This can be applied to cos(x) as
$$\cos(x)=\sum_{k=0}^\infty \frac{x^k}{k!}\cos^{(k)}(0)=\sum_{k=0}^\infty \frac{x^k}{k!}\cos(k\pi/2)=\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}(-1)^k$$
where the index is changed in the final step to indicate odd terms are 0
Often a known Maclaurin series can be used to find a closely related one.
Just apply the known series
h(x)=(cos(3x)-1)/x^2
$$h(x)=\frac{1}{x^2}\left(\sum_{k=0}^\infty \frac{(3x)^{2k}}{(2k)!}(-1)^k-1\right)$$
simplify that into standard form.