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:
The reader will see that the Maclaurin expansion is the Taylor expansion obtained if ${x}_{0}$ is chosen to be zero.
Task!
Obtain the Taylor series expansion of $\frac{1}{1x}$ about $x=2$ . (That is, find a power series in powers of $\left(x2\right)$ .)
First, obtain the first three derivatives and the ${n}^{\text{th}}$ derivative of $f\left(x\right)=\frac{1}{1x}$ :
${f}^{\prime}\left(x\right)=\frac{1}{{\left(1x\right)}^{2}},\phantom{\rule{2em}{0ex}}{f}^{\u2033}\left(x\right)=\frac{2}{{\left(1x\right)}^{3}},\phantom{\rule{2em}{0ex}}{f}^{\u2034}\left(x\right)=\frac{6}{{\left(1x\right)}^{4}},\phantom{\rule{2em}{0ex}}\cdots \phantom{\rule{1em}{0ex}}{f}^{\left(n\right)}\left(x\right)=\frac{n!}{{\left(1x\right)}^{n+1}}$
Now evaluate these derivatives at ${x}_{0}=2$ :
${f}^{\prime}\left(2\right)=1,\phantom{\rule{2em}{0ex}}{f}^{\u2033}\left(2\right)=2,\phantom{\rule{2em}{0ex}}{f}^{\u2034}\left(2\right)=6,\phantom{\rule{2em}{0ex}}{f}^{\left(n\right)}\left(2\right)={\left(1\right)}^{n+1}n!$
Hence, write down the Taylor expansion of $f\left(x\right)=\frac{1}{1x}$ about $x=2$ :
$\frac{1}{1x}=1+\left(x2\right){\left(x2\right)}^{2}+{\left(x2\right)}^{3}+\cdots +{\left(1\right)}^{n+1}{\left(x2\right)}^{n}+\cdots $
Exercises
 Show that the series obtained in the last Task is convergent if $\leftx2\right<1.$
 Sketch the linear, quadratic and cubic approximations to $\frac{1}{1x}$ obtained from the series in the last task and compare to $\frac{1}{1x}$ .

In the following diagrams some of the terms from the Taylor series are plotted to compare with
$\frac{1}{\left(1x\right)}$
.