Q. 51 - 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.

$51. \sin \left(x\right),\mathrm{\pi }$

Step-by-Step Solution

Answer

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

Step 1. Given data

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

Step 2. Table of the taylor series

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

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

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 $f\left(x\right)=\sin \left(x\right)$ at  $x=\mathrm{\pi }$

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

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

The Taylor series for the function $f\left(x\right)=\sin \left(x\right)$ at  $x=\mathrm{\pi }$ is

$\begin{array}{rl}{P}_n\left(x\right)=& 0+-1\cdot (x-\pi )+\frac{0}{2!}(x-\pi {)}^2+\frac{1}{3!}(x-\pi {)}^3+\cdots +\frac{0}{\left(2k\right)!}(x-\pi {)}^{2k}+\frac{(-1{)}^{k+1}}{(2k+1)!}(x-\pi {)}^{2k+1}+\cdots \end{array}$

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