Question

Find the indicated Maclaurin or Taylor series for the given function about the indicated point, and find the radius of convergence for the series.

$\sin x,x_0=0$

Short answer

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

Step-by-step solution

Given information

The function is$f\left(x\right)=\sin x$

 Find the general of the Maclurin series of the function 

The Maclurin series at $x_0=0$ for any function $f$ with a derivative of order $n$ is given by

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

The function's general Maclurin series is

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

 Make a table of the Maclurin series for the function f(x)=sinx

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

 Find the Maclurin series for the function f(x)=sinx 

The Maclurin series for the function $f\left(x\right)=\sin x$is:

$0+1·x+\frac{0}{2!}{x}^2+\frac{(-1)}{3!}{x}^3+0x^4+\frac{1}{5!}{x}^5+\frac{0}{6!}{x}^6+\dots$

Or, we can write as:

$f\left(x\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{2k+1}$