Maclaurin seriesEdexcel International A Level Further Maths: Revision notes

The third and higher order derivatives of $f(x)$ are written $f'''(x)$, $f^{(4)}(x)$, $\dots$, $f^{(n)}(x)$, or $\frac{d^3y}{dx^3}$ and so on. They are found by differentiating repeatedly, using the product, quotient and chain rules.

• $f(x)=\mathrm{e}^{3x}$: every differentiation multiplies by $3$, so $f^{(n)}(x)=3^n\mathrm{e}^{3x}$.
• $f(x)=\cos2x$: $f'=-2\sin2x$, $f''=-4\cos2x$, $f'''=8\sin2x$, $f^{(4)}=16\cos2x$.
• $f(x)=\ln(1+x)$: $f'=(1+x)^{-1}$, $f''=-(1+x)^{-2}$, $f'''=2(1+x)^{-3}$, $f^{(4)}=-6(1+x)^{-4}$. For a function such as $y=\tan x$ it is quicker to use a relationship. Since $y'=1+y^2$, differentiating gives $y''=2yy'$, then $y'''=2y'^2+2yy''$, and so on, with every derivative expressed in terms of lower ones.

A function with derivatives of all orders at $0$ can be written as a power series. Suppose $f(x)=a_0+a_1x+a_2x^2+a_3x^3+\dots$. Putting $x=0$ gives $a_0=f(0)$. Differentiating and putting $x=0$ gives $a_1=f'(0)$; differentiating again gives $2a_2=f''(0)$; and in general $n!\,a_n=f^{(n)}(0)$. Hence the Maclaurin series $f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f'''(0)+\dots+\frac{x^n}{n!}f^{(n)}(0)+\dots$ Using the first few terms gives a polynomial approximation that is accurate for small $|x|$. Taking more terms improves the approximation. Example: for $f(x)=\mathrm{e}^{3x}$, $f^{(n)}(0)=3^n$, so $\mathrm{e}^{3x}=1+3x+\frac92x^2+\frac92x^3+\dots$

Deriving each series from the formula:

• $\mathrm{e}^x$: all derivatives are $\mathrm{e}^x$, which is $1$ at $0$, so $\mathrm{e}^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$ (all $x$).
• $\sin x$: derivatives at $0$ run $0,1,0,-1,0,1,\dots$, so $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots$ (all $x$).
• $\cos x$: derivatives at $0$ run $1,0,-1,0,1,\dots$, so $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots$ (all $x$).
• $\ln(1+x)$: $f(0)=0$ and $f^{(n)}(0)=(-1)^{n-1}(n-1)!$, so $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots$ (valid for $-1<x\le1$). Notice that $\sin x$ is odd (only odd powers) and $\cos x$ is even (only even powers).

Series for other functions are found either directly from the formula, or by substituting into a known series, or by combining series.

• Substitution: replace $x$ in $\mathrm{e}^x$ by $2x$ to get $\mathrm{e}^{2x}=1+2x+2x^2+\frac43x^3+\dots$. Replace $x$ by $-x$ in $\ln(1+x)$ to get $\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\dots$
• Multiplication: $\mathrm{e}^x\sin x=\left(1+x+\frac{x^2}{2}\right)\left(x-\frac{x^3}{6}\right)=x+x^2+\frac{x^3}{3}+\dots$ (collect terms up to the required power only).
• Differentiating or integrating a series term by term: $\frac{d}{dx}\ln(\cos x)=-\tan x$, so integrating the series for $\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+\dots$ gives $\ln(\cos x)=-\frac{x^2}{2}-\frac{x^4}{12}-\frac{x^6}{45}-\dots$
• Trig identities: $\sin^2x=\frac{1-\cos2x}{2}=x^2-\frac{x^4}{3}+\dots$

Approximations. Substitute a small value of $x$ into the first few terms. For $\ln(1+x)$ with $x=\frac13$ via $\ln\frac{1+x}{1-x}=2x+\frac23x^3+\dots$ you can estimate $\ln2\approx\frac23+\frac{2}{81}=0.691$. Limits. Replace each function by its series and cancel the leading power. For example $\lim_{x\to0}\frac{\mathrm{e}^x-1-x}{x^2}=\lim\left(\frac12+\frac x6+\dots\right)=\frac12$. Series from relationships. If $y$ satisfies a relationship between its derivatives (such as $y'=1+y^2$ for $y=\tan x$), use it to evaluate $y^{(n)}(0)$ in turn and build the series. State the highest power you are working to, and give approximations to the accuracy asked for; the series is only a good approximation for small $|x|$ (and for $\ln(1+x)$ only when $-1<x\le1$).