Question

Give the Maclaurin series for $\sin x$.

Answer 1

Step 1: Maclaurin series explanation

A Maclaurin series is a function that has expansion series that gives the sum of derivatives of that function.

The Maclaurin series of a function $f\left(x\right)$ up to order n may be found using series $[f,x,0,n]$.

It is a special case of the Taylor series when $x=0$.

The general equation of the Maclaurin series is

$\begin{array}{l}f\left(x\right)=f\left(x_0\right)+f\text{'}\left(x_0\right)(x-x_0)+\frac{f”\left(x_0\right)}{2!}(x-x_0{)}^2+\frac{f”\text{'}\left(x_0\right)}{3!}(x-x_0{)}^3+\dots ..\end{array}$

Step 2: Maclaurin series for $\sin x$

Given, $f\left(x\right)=\sin x$

Using $x=0$, the given equation function becomes

$f\left(0\right)=\sin \left(0\right)=0$

Now taking the derivatives of the given function and using x=0, we have

1. $f\prime \left(0\right)=\cos \left(0\right)=1$
2. $f\text{'}\text{'}\left(0\right)=–\sin \left(0\right)=–0=0$
3. $f\text{'}\text{'}\text{'}\left(0\right)=\sin \left(0\right)=0$
4. $f\text{'}\text{'}\text{'}\text{'}\left(0\right)=\sin \left(0\right)=0$

Therefore, we get the series as,

$\begin{array}{l}f\left(x\right)=f\left(0\right)+xf\text{'}\left(0\right)+\frac{x^2}{2!}f\text{'}\text{'}\left(0\right)+\frac{x^3}{3!}f\text{'}\text{'}\text{'}\left(0\right)+\frac{x^4}{4!}f\text{'}\text{'}\text{'}\text{'}\left(0\right)+\dots .\end{array}$

Putting the values in the above series, we get

$\begin{array}{l}\sin x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots \end{array}$

Hence, the Maclaurin series for $\sin x$ is $\begin{array}{l}1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots \end{array}$.