Maclaurin series of a functionAQA A-Level Further Maths: Revision notes

Section 1

Maclaurin's theorem

If $f(x)$ can be differentiated as often as needed at $x=0$ , its Maclaurin series is $f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\dots+\frac{f^{(r)}(0)}{r!}x^r+\dots$ The coefficient of $x^r$ is $\frac{f^{(r)}(0)}{r!}$ . So $f^{(r)}(0)=r!\times(\text{coefficient of }x^r)$ , which lets you work backwards from a known series to a derivative at $0$ . The function and every derivative you use must exist at $x=0$ : $\ln x$ has no Maclaurin series, but $\ln(1+x)$ does.

Key termsMaclaurin seriesderivative at zero

Common mistake

Using $f^{(r)}(0)$ as the coefficient. The coefficient of $x^r$ is $\frac{f^{(r)}(0)}{r!}$ .

Section 2

Finding a series by repeated differentiation

Method: differentiate, simplify, then evaluate at $0$ each time, in a table of $f(0),f'(0),f''(0),\dots$ . Example, $f(x)=\tan x$ : $f'=\sec^2x$ , $f''=2\sec^2x\tan x$ , $f'''=4\sec^2x\tan^2x+2\sec^4x$ . So $f(0)=0$ , $f'(0)=1$ , $f''(0)=0$ , $f'''(0)=2$ and $\tan x=x+\frac{x^3}{3}+\dots$ Example, $f(x)=\ln\cos x$ : $f'=-\tan x$ , $f''=-\sec^2x$ , giving $f(0)=0$ , $f'(0)=0$ , $f''(0)=-1$ , $f^{(4)}(0)=-2$ and $\ln\cos x=-\frac{x^2}{2}-\frac{x^4}{12}-\dots$ Substituting a small $x$ then gives approximations, e.g. $\tan0.2\approx0.2+\frac{0.008}{3}=0.2027$ .

Key termsrepeated differentiation

Exam tip

Simplify each derivative before differentiating again, and evaluate at $0$ as soon as you have it.

Exam tip

If $f$ is an even function only even powers appear, and if $f$ is odd only odd powers: use this to check your series, e.g. $\tan x$ is odd.

Section 3

The general term

When derivatives follow a pattern you can write the general term. If $f^{(r)}(0)$ is known for every $r$ , the term in $x^r$ is $\frac{f^{(r)}(0)}{r!}x^r$ .

$f(x)=\frac{1}{1-2x}=(1-2x)^{-1}$ : $f^{(r)}(0)=2^rr!$ , so the term is $2^rx^r$ .
$f(x)=\mathrm{e}^{2x}$ : $f^{(r)}(0)=2^r$ , so the term is $\frac{2^rx^r}{r!}$ .
$f(x)=\sin2x$ : odd powers only, $\frac{(-1)^r(2x)^{2r+1}}{(2r+1)!}$ for $r=0,1,2,\dots$
$f(x)=\ln(1+x)$ : $f^{(r)}(0)=(-1)^{r+1}(r-1)!$ , so the term is $\frac{(-1)^{r+1}x^r}{r}$ . Use $(-1)^r$ or $(-1)^{r+1}$ to produce alternating signs, and check your formula by substituting the first few values of $r$ .

Key termsgeneral term

Common mistake

Writing $\frac{(-1)^rx^r}{r}$ for $\ln(1+x)$ : the first term must be $+x$ , so the sign is $(-1)^{r+1}$ .

Section 4

Using a differential equation to find the series

For some functions it is easier to find a relation between the derivatives and then differentiate it repeatedly. For $y=\mathrm{e}^{\sin x}$ : $y'=\cos x\,y,\qquad y''=(\cos^2x-\sin x)y,\qquad y'''=(-2\cos x\sin x-\cos x)y+(\cos^2x-\sin x)y'.$ At $x=0$ : $y=1$ , $y'=1$ , $y''=1$ , $y'''=0$ , so $y=1+x+\frac{x^2}{2}+0\cdot x^3+\dots$ You can check by substituting the series for $\sin x$ into $\mathrm{e}^u$ : the $x^3$ terms $-\frac16+\frac16$ cancel. Agreement between two methods is a strong check.

Key termsimplicit relation

Exam tip

Keep $y$ and $y'$ in your expressions and use their values at $0$ at the end, rather than expanding them.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Maclaurin series of a function

Let $f(x)=\ln(\cos x)$ for $-\frac{\pi}{2}<x<\frac{\pi}{2}$ .
Given that $f'''(0)=0$ and $f^{(4)}(0)=-2$ , find the first two non-zero terms of the Maclaurin series of $f(x)$ .2 marks
Let $f(x)=\dfrac{1}{1-2x}$ .
Deduce an expression for $f^{(r)}(0)$ in terms of $r$ .2 marks
Let $f(x)=\tan x$ .
Show that $f'''(0)=2$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).