# Science:Infinite Series Module/Units/Unit 3/3.2 Taylor Series/3.2.05 The Maclaurin Series Expansion for cos

## Example

Find the Maclaurin series expansion for cos(x) at x = 0, and determine its radius of convergence.

## Complete Solution

### Step 1: Find the Maclaurin Series

{\begin{aligned}\cos(x)&={\frac {d}{dx}}\sin(x)\\&={\frac {d}{dx}}\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}x^{2k+1}\\&={\frac {d}{dx}}{\Big (}x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\ldots {\Big )}\\&=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\ldots \\&=\sum _{k=0}^{\infty }{\frac {(-1)^{k}x^{2k}}{(2k)!}}\end{aligned}} ### Step 2: Find the Radius of Convergence

The ratio test gives us:

{\begin{aligned}\lim _{k\rightarrow \infty }{\Big |}{\frac {(-1)^{k+1}}{(2(k+1))!}}x^{2(k+1)}{\Bigg /}{\frac {(-1)^{k}}{(2k)!}}x^{2k}{\Big |}&=\lim _{k\rightarrow \infty }{\frac {(2k)!}{(2k+2)!}}|x|^{2}\\&=\lim _{k\rightarrow \infty }{\frac {1}{(2k+1)(2k+2)}}|x|^{2}\\&=0\end{aligned}} Because this limit is zero for all real values of x, the radius of convergence of the expansion is the set of all real numbers.

## Explanation of Each Step

### Step 1

To find the series expansion, we could use the same process here that we used for sin(x) and ex. But there is an easier method. We can differentiate our known expansion for the sine function.

If you would like to see a derivation of the Maclaurin series expansion for cosine, the following video provides this derivation.

Cosine Taylor Series at 0
Derivation of the Maclaurin series expansion for cosine.
This video can be found on the Kahn Academy website, and carries a Creative Commons copyright (CC BY-NC-SA 3.0).

### Step 2

This step was nothing more than substitution of our formula into the formula for the ratio test.

## Possible Challenges

### When Can We Differentiate a Power Series?

For the purposes of this module, we will always assume that we can. There is however a theorem on differentiating and integrating power series, which you are not expected to know, that tells us that a power series can only be differentiated if it has a radius of convergence that is greater than zero.

A page in this module on this theorem is coming. In the mean time, this page on Wikipedia may help.

### Did We Have to Test for Convergence?

The short answer is: no. The theorem mentioned above tells us that, because

• we derived the series for cos(x) from the series for sin(x) through differentiation, and
$\cos(x)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}x^{2k}}{(2k)!}}=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}\ldots$ 