Maclaurin seriesEdexcel A-Level Further Maths: Revision notes

The Maclaurin series of $\mathrm{f}(x)$ expresses it as an infinite polynomial in $x$: $\mathrm{f}(x)=\mathrm{f}(0)+x\mathrm{f}'(0)+\frac{x^2}{2!}\mathrm{f}''(0)+\cdots+\frac{x^r}{r!}\mathrm{f}^{(r)}(0)+\cdots$ Differentiate repeatedly, evaluate at $x=0$ and substitute. Example: for $\mathrm{f}(x)=\mathrm{e}^{2x}$, $\mathrm{f}^{(r)}(x)=2^r\mathrm{e}^{2x}$, so $\mathrm{f}^{(r)}(0)=2^r$ and the general term is $\frac{2^rx^r}{r!}$; the series begins $1+2x+2x^2+\frac43x^3+\cdots$. For $\mathrm{f}(x)=\tan x$: $\mathrm{f}(0)=0$, $\mathrm{f}'(0)=1$, $\mathrm{f}''(0)=0$, $\mathrm{f}'''(0)=2$, so $\tan x=x+\frac{x^3}{3}+\cdots$.

These are given in the formulae booklet; you must recognise and use them.

• $\mathrm{e}^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^r}{r!}+\cdots$, valid for all $x$.
• $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots+\frac{(-1)^rx^{2r+1}}{(2r+1)!}+\cdots$, valid for all $x$.
• $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots+\frac{(-1)^rx^{2r}}{(2r)!}+\cdots$, valid for all $x$.
• $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots+\frac{(-1)^{r+1}x^r}{r}+\cdots$, valid for $-1<x\leq1$.
• $(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\cdots$, valid for $|x|<1$ (for non-integer or negative $n$).

Build new series from the standard ones instead of differentiating. Substitution. Replace $x$ by an expression: $\ln(1+3x)=3x-\frac{9x^2}{2}+9x^3-\cdots$. The range of validity changes with it: $-1<3x\leq1$, so $-\frac13<x\leq\frac13$. Multiplication. Multiply the series and collect powers up to the term needed: $\mathrm{e}^x\sin x=\left(1+x+\frac{x^2}2+\cdots\right)\left(x-\frac{x^3}6\right)=x+x^2+\frac{x^3}{3}+\cdots$. Series inside a series. For $\ln(1+\sin x)$, put $y=\sin x=x-\frac{x^3}6$ into $y-\frac{y^2}2+\frac{y^3}3$ and keep terms up to $x^3$: $x-\frac{x^2}{2}+\frac{x^3}{6}$. Adding or subtracting. $\ln\frac{1+x}{1-x}=\ln(1+x)-\ln(1-x)=2x+\frac23x^3+\cdots$, valid for $-1<x<1$.

A truncated series is a polynomial approximation, accurate near $x=0$.

• Estimates: $\mathrm{e}^{0.1}=\mathrm{e}^{2(0.05)}\approx1+0.1+0.005+0.000167=1.10517$ using $\mathrm{e}^{2x}$ with $x=0.05$.
• Integration: integrate term by term: $\int_0^{0.5}\mathrm{e}^x\sin x\,\mathrm{d}x\approx\left[\frac{x^2}2+\frac{x^3}3+\frac{x^4}{12}\right]_0^{0.5}=0.172$.
• Logarithms: with $x=\frac13$, $\ln\frac{1+x}{1-x}=\ln2\approx2x+\frac23x^3=0.691$, so the series gives a usable estimate even though only two terms are kept. Use values of $x$ that lie inside the range of validity, and keep $x$ small for good accuracy.