### 4 The Taylor series

The Taylor series is a generalisation of the Maclaurin series being a power series developed in powers of $\left(x-{x}_{0}\right)$ rather than in powers of $x$ . Thus

##### Key Point 15

Taylor Series

If the function $f\left(x\right)$ can be differentiated as often as required at $x={x}_{0}$ then:

$f\left(x\right)=f\left({x}_{0}\right)+\left(x-{x}_{0}\right){f}^{\prime }\left({x}_{0}\right)+\frac{{\left(x-{x}_{0}\right)}^{2}}{2!}{f}^{″}\left({x}_{0}\right)+\cdots$
This is called the Taylor series of $f\left(x\right)$ about the point ${x}_{0}$ .

The reader will see that the Maclaurin expansion is the Taylor expansion obtained if ${x}_{0}$ is chosen to be zero.

Obtain the Taylor series expansion of $\frac{1}{1-x}$ about $x=2$ . (That is, find a power series in powers of $\left(x-2\right)$ .)

First, obtain the first three derivatives and the ${n}^{\text{th}}$ derivative of $f\left(x\right)=\frac{1}{1-x}$ :

${f}^{\prime }\left(x\right)=\frac{1}{{\left(1-x\right)}^{2}},\phantom{\rule{2em}{0ex}}{f}^{″}\left(x\right)=\frac{2}{{\left(1-x\right)}^{3}},\phantom{\rule{2em}{0ex}}{f}^{‴}\left(x\right)=\frac{6}{{\left(1-x\right)}^{4}},\phantom{\rule{2em}{0ex}}\cdots \phantom{\rule{1em}{0ex}}{f}^{\left(n\right)}\left(x\right)=\frac{n!}{{\left(1-x\right)}^{n+1}}$

Now evaluate these derivatives at ${x}_{0}=2$ :

${f}^{\prime }\left(2\right)=1,\phantom{\rule{2em}{0ex}}{f}^{″}\left(2\right)=-2,\phantom{\rule{2em}{0ex}}{f}^{‴}\left(2\right)=6,\phantom{\rule{2em}{0ex}}{f}^{\left(n\right)}\left(2\right)={\left(-1\right)}^{n+1}n!$

Hence, write down the Taylor expansion of $f\left(x\right)=\frac{1}{1-x}$ about $x=2$ :

$\frac{1}{1-x}=-1+\left(x-2\right)-{\left(x-2\right)}^{2}+{\left(x-2\right)}^{3}+\cdots +{\left(-1\right)}^{n+1}{\left(x-2\right)}^{n}+\cdots$

##### Exercises
1. Show that the series obtained in the last Task is convergent if $\left|x-2\right|<1.$
2. Sketch the linear, quadratic and cubic approximations to $\frac{1}{1-x}$ obtained from the series in the last task and compare to $\frac{1}{1-x}$ .
1. In the following diagrams some of the terms from the Taylor series are plotted to compare with $\frac{1}{\left(1-x\right)}$ .