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Next: List of Maclaurin Expansions

## 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

cos ( x ) = d d x sin ( x ) = d d x ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! x 2 k + 1 = d d x ( x − x 3 3 ! + x 5 5 ! − … ) = 1 − x 2 2 ! + x 4 4 ! − … = ∑ k = 0 ∞ ( − 1 ) k x 2 k ( 2 k ) ! {\displaystyle {\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:

lim k → ∞ | ( − 1 ) k + 1 ( 2 ( k + 1 ) ) ! x 2 ( k + 1 ) / ( − 1 ) k ( 2 k ) ! x 2 k | = lim k → ∞ ( 2 k ) ! ( 2 k + 2 ) ! | x | 2 = lim k → ∞ 1 ( 2 k + 1 ) ( 2 k + 2 ) | x | 2 = 0 {\displaystyle {\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 *e*^{x}. 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 |
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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
- we already know the radius of convergence of sin(x),

the radius of convergence of cos(x) will be the same as sin(x). However, we haven't introduced that theorem in this module. You may want to ask your instructor if you are expected to know this theorem.

## Summary

The Maclaurin Expansion of cos(x) |
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The Maclaurin series expansion for cos(x) is given by
cos ( x ) = ∑ k = 0 ∞ ( − 1 ) k x 2 k ( 2 k ) ! = 1 − x 2 2 ! + x 4 4 ! − x 6 6 ! … {\displaystyle \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 } This formula is valid for all real values of |

Previous: Maclaurin Expansion of sin(x)

Next: List of Maclaurin Expansions

Can you explain what k means. Also for example when x = 1 what is k.

Why if we try to approximate the cosine function it works so much better compared to the sine function?

PS: It is really important. I’m doing a research for school and this would help a lot, email me if possible.

Many thanks,

Rafael