### 3 Differentiation of Maclaurin series

We have already noted that, by the binomial series,

$\phantom{\rule{2em}{0ex}}\frac{1}{1-x}=1+x+{x}^{2}+{x}^{3}+\cdots \phantom{\rule{2em}{0ex}}\left|x\right|<1$

Thus, with $x$ replaced by $-x$

$\phantom{\rule{2em}{0ex}}\frac{1}{1+x}=1-x+{x}^{2}-{x}^{3}+\cdots \phantom{\rule{2em}{0ex}}\left|x\right|<1$

We have previously obtained the Maclaurin expansion of  $ln\left(1+x\right)$ :

$\phantom{\rule{2em}{0ex}}ln\left(1+x\right)=x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}-\frac{{x}^{4}}{4}+\cdots \phantom{\rule{2em}{0ex}}-1

Now, we differentiate both sides with respect to $x$ :

$\phantom{\rule{2em}{0ex}}\frac{1}{1+x}=1-x+{x}^{2}-{x}^{3}+\cdots$

This result matches that found from the binomial series and demonstrates that the Maclaurin expansion of a function $f\left(x\right)$ may be differentiated term by term to give a series which will be the Maclaurin expansion of $\frac{df}{dx}.$

As we noted in Section 16.4 the derived series will have the same radius of convergence as the original series.

Find the Maclaurin expansion of ${\left(1-x\right)}^{-3}$ and state its radius of convergence.

First write down the expansion of ${\left(1-x\right)}^{-1}$ :

$\frac{1}{1-x}=1+x+{x}^{2}+\cdots \phantom{\rule{2em}{0ex}}\left|x\right|<1$

Now, by differentiation, obtain the expansion of $\frac{1}{{\left(1-x\right)}^{2}}$ :

$\frac{1}{{\left(1-x\right)}^{2}}=\frac{d}{dx}\left(1+x+{x}^{2}+\cdots \phantom{\rule{0.3em}{0ex}}\right)=1+2x+3{x}^{2}+4{x}^{3}$

Differentiate again to obtain the expansion of ${\left(1-x\right)}^{-3}$ :

$\frac{1}{{\left(1-x\right)}^{3}}=\frac{1}{2}\frac{d}{dx}\left(\frac{1}{{\left(1-x\right)}^{2}}\right)=\frac{1}{2}\left[2+6x+12{x}^{2}+20{x}^{3}+\cdots \phantom{\rule{0.3em}{0ex}}\right]=1+3x+6{x}^{2}+10{x}^{3}+\cdots$ Finally state its radius of convergence:

The final series: $1+3x+6{x}^{2}+10{x}^{3}+\cdots$ has radius of convergence $R=1$ since the original series has this radius of convergence. This can also be found directly using the formula $R=\underset{n\to \infty }{lim}\left|\frac{{b}_{n}}{{b}_{n+1}}\right|$ and using the fact that the coefficient of the ${n}^{th}$ term is ${b}_{n}=\frac{1}{2}n\left(n+1\right)$ .