Question

How to find the Maclaurin series of $f\left(x\right)=\cos \left(x\right)$

Answer 1

Step 1. Define the Maclaurin series

A Maclaurin series is a function that has expansion series that gives the sum of derivatives of function. The Maclaurin series of function $f\left(x\right)$ up to order $n$ may be found using Taylor series when $x_0=0$

General equation of Maclaurin series is

$$\begin{gathered} f\left(x\right)=f\left(x_0\right)+f^{\text{'}}\left(x_0\right)\left(x-x_0\right)+\frac{f^{\text{'}\text{'}}\left(x_0\right)}{2!}·{\left(x-x_0\right)}^2+.....+\frac{f^n\left(x_0\right)}{n!}·{\left(x-x_0\right)}^n \\ =\sum _{n=0}^{\infty }\frac{f^n\left(x_0\right)}{n!}·{\left(x-x_0\right)}^n \end{gathered}$$

At $x_0=0$

$f\left(x\right)=\sum _{n=0}^{\infty }\frac{f^n\left(0\right)}{n!}·{\left(x\right)}^n$

Step 2. Find the derivative of the function

$\begin{array}{rcl}f\left(x\right)& =& \cos \left(x\right)\Rightarrow f\left(0\right)=1\\ f\text{'}\left(x\right)& =& -\sin \left(x\right)\Rightarrow f\text{'}\left(0\right)=0\\ f\text{'}\text{'}\left(x\right)& =& -\cos \left(x\right)\Rightarrow f\text{'}\text{'}\left(0\right)=-1\\ f\text{'}\text{'}\text{'}\left(x\right)& =& \sin x\Rightarrow f\text{'}\text{'}\text{'}\left(0\right)=0\\ f^{iv}\left(x\right)& =& \cos \left(x\right)\Rightarrow f^{iv}\left(0\right)=1\\ f^v\left(x\right)& =& -\sin \left(x\right)\Rightarrow f^v\left(0\right)=0\\ f^{vi}\left(x\right)& =& -\cos \left(x\right)\Rightarrow f^{vi}\left(0\right)=-1\end{array}$

Step 3. Find the Maclaurin series of the function

$f\left(x\right)=f\left(0\right)+x·f\left(0\right)+\frac{x^{2}f\text{'}\text{'}\left(0\right)}{2!}+....+\frac{x^n{f}^n\left(0\right)}{n!}$

Putting the above series we get

$$\begin{gathered} f\left(x\right)=\cos \left(x\right) \\ =\sum _{n=0}^{\infty }\frac{{(-1)}^n·{\left(x\right)}^{2n}}{\left(2n\right)!} \\ =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+.... \end{gathered}$$

Hence, the Maclaurin series of $f\left(x\right)=\cos \left(x\right)$ is $1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+....$