Question

In Exercises 31–40 find the Maclaurin series for the specified function. Note: These are the same functions as in Exercises 21–30.

$f\left(x\right)=\cos 2x$

Short answer

Ans: The Maclaurin series of the function$f\left(x\right)=\sum _{k=0}^{\infty }\frac{(-4{)}^k}{\left(2k\right)!}{x}^{2k}$

Step-by-step solution

Step 1. Given information:

$f\left(x\right)=\cos 2x$

Step 2. Finding the general form  Maclaurin series:  

Since for any function $f$with derivatives of all orders at the point $x_0=0$, then the Maclurin series is

$f\left(x\right)=f\left(0\right)+f^{\text{'}}\left(0\right)x+\frac{f^{\text{'}\text{'}}\left(0\right)}{2!}{x}^2+\frac{f^{\text{'}\text{'}\text{'}}\left(0\right)}{3!}{x}^3+\frac{f^{\text{'}\text{'}\text{'}}\left(0\right)}{4!}{x}^4+\dots$

Or, we can write the general form Maclurin series of the function $f$ is

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

Step 3. Constructing the table of the Maclaurin series for the function :

So, let us first construct the table of the Maclaurin series for the function$f\left(x\right)=\cos 2x$$\begin{array}{|cccc|}\hline n& f^n\left(x\right)& f^n\left(0\right)& \frac{f^n\left(0\right)}{n!}\\ 0& \cos 2x& 1& 1\\ 1& -2\sin 2x& 0& 0\\ 2& -4\cos 2x& -4& -\frac{4}{2!}\\ 3& 8\sin 2x& 0& 0\\ ·& ·& ·& ·\\ 2k& (-4{)}^k\cos 2x& (-4{)}^k& \frac{(-4{)}^k}{\left(2k\right)!}\\ 2k+1& (-1{)}^{k+1}(2{)}^{2k+1}\sin x& 0& 0\\ ·& ·& ·& ·\\ \hline\end{array}$

Step 4. Finding the Maclaurin series :

Therefore, the Maclaurin series for the function $f\left(x\right)=\cos 2x$is

$1+0·x+\frac{(-4)}{2!}{x}^2+\frac{0}{3!}{x}^3+\frac{16}{4!}{x}^4+\dots$

Or, we can write it as

$f\left(x\right)=\sum _{k=0}^{\infty }\frac{(-4{)}^k}{\left(2k\right)!}{x}^{2k}$