Q. 49 - Calculus

Question

In Exercises 49–56 find the Taylor series for the specified function and the given value of $x_0$. Note: These are the same functions and values as in Exercises 41–48.

$49. \cos \left(x\right),\frac{\pi }{2}$

Step-by-Step Solution

Answer

The Taylor series for the function $f\left(x\right)=\cos \left(x\right)$ at $x=\frac{\mathrm{\pi }}{2}$ is$P_n\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{(2k+1)!}{(x-\frac{\pi }{2})}^{2k+1}$

Step 1. Given data

We have the function $f\left(x\right)=\cos \left(x\right)$

Step 2. Table of the taylor series

Any function $f$ with a derivative of order $n$, the taylor series at $x=\frac{\mathrm{\pi }}{2}$is given by,

$\begin{array}{rl}{P}_n\left(x\right)=& f\left(\frac{\pi }{2}\right)+f^{\mathrm{\prime }}\left(\frac{\pi }{2}\right)(x-\frac{\pi }{2})+\frac{f^{\prime \prime }\left(\frac{\pi }{2}\right)}{2!}{(x-\frac{\pi }{2})}^2+\frac{f^{\prime \prime }\left(\frac{\pi }{2}\right)}{3!}{(x-\frac{\pi }{2})}^3+\frac{f^{\prime \prime \prime \prime }\left(\frac{\pi }{2}\right)}{4!}{(x-\frac{\pi }{2})}^4+\dots \end{array}$

we can write the general of the Taylor series of the function $f$ is,

$P_n\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{f^k\left(x_0\right)}{k!}{(x-x_0)}^n$

So, let us first construct the table of the Taylor series for the function $f\left(x\right)=\cos \left(x\right)$ at $x=\frac{\pi }{2}$

n	$f^n\left(x\right)$	$f^n\left(\frac{\mathrm{\pi }}{2}\right)$	$\frac{f^n\left(\frac{\mathrm{\pi }}{2}\right)}{n!}$
0	$\cos \left(x\right)$	0	0
1	$-\sin \left(x\right)$	-1	-1
2	$-\cos \left(x\right)$	0	0
3	$\sin \left(x\right)$	1	$\frac{1}{3!}$
. . .	. . .	. . .	. . .
2k	$(-1{)}^k\cos x$	0	0
2k+1	$(-1{)}^{k+1}\sin \left(x\right)$	${(-1)}^{k+1}$	$(-1{)}^{k+1}\frac{1}{(2k+1)!}$
. . .	. . .	. . .	. . .

Step 3. Taylor series for the f ( x ) = cos x

The Taylor series for the function $f\left(x\right)=\cos \left(x\right)$ at $x=\frac{\mathrm{\pi }}{2}$ is$\begin{array}{rl}{P}_n\left(x\right)=& 0+-1\cdot (x-\frac{\pi }{2})+\frac{0}{2!}{(x-\frac{\pi }{2})}^2+\frac{1}{3!}{(x-\frac{\pi }{2})}^3+\frac{0}{4!}{(x-\frac{\pi }{2})}^4+\cdots +\frac{0}{\left(2k\right)!}{(x-\frac{\pi }{2})}^{2k}+\frac{(-1{)}^{k+1}}{(2k+1)!}{(x-\frac{\pi }{2})}^{2k+1}+\cdots \end{array}$

Or we can write this as $P_n\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{(2k+1)!}{(x-\frac{\pi }{2})}^{2k+1}$