Maclaurin Series — H2 Maths help

A Maclaurin series expresses a function as an infinite power series centred at $x = 0$ . It lets you approximate complicated functions with polynomials — invaluable for limits, small-angle work, and integration in H2 Maths.

What you need to know

The Maclaurin formula is $f(x) = f(0) + f'(0)x + \dfrac{f''(0)}{2!}x^2 + \dfrac{f'''(0)}{3!}x^3 + \cdots$ ; each coefficient is a derivative evaluated at $x = 0$ .
Standard series (must know with validity):
- $e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots$ , valid for all $x \in \mathbb{R}$
- $\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots$ , valid for all $x \in \mathbb{R}$
- $\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots$ , valid for all $x \in \mathbb{R}$
- $\ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots$ , valid for $-1 < x \leq 1$
- $(1+x)^n = 1 + nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3 + \cdots$ , valid for $|x| < 1$ (any real $n$ ; terminates when $n$ is a non-negative integer)
To derive a series, either differentiate $f$ repeatedly and apply the formula, or substitute/combine/multiply existing standard series — whichever is less laborious.
Small-angle approximations (for $x$ in radians, $x \approx 0$ ): $\sin x \approx x$ , $\cos x \approx 1 - \dfrac{x^2}{2}$ , $\tan x \approx x$ .
The validity range matters when the series is used for approximations or infinite sums; always state it for $\ln(1+x)$ and $(1+x)^n$ .

Worked example 1 — deriving a series by repeated differentiation

Find the Maclaurin series for $f(x) = e^x \cos x$ up to and including the term in $x^3$ .

Compute successive derivatives and evaluate at $x = 0$ :

f(x) = e^x \cos x \implies f(0) = 1

f'(x) = e^x \cos x - e^x \sin x = e^x(\cos x - \sin x) \implies f'(0) = 1

f''(x) = e^x(\cos x - \sin x) + e^x(-\sin x - \cos x) = -2e^x \sin x \implies f''(0) = 0

f'''(x) = -2e^x \sin x - 2e^x \cos x = -2e^x(\sin x + \cos x) \implies f'''(0) = -2

Substituting into the Maclaurin formula:

e^x \cos x = 1 + x + \frac{0}{2!}x^2 + \frac{-2}{3!}x^3 + \cdots = 1 + x - \frac{x^3}{3} + \cdots

Worked example 2 — combining standard series

Find the Maclaurin series for $\ln(1 + \sin x)$ up to and including the term in $x^3$ .

Use the standard series $\sin x = x - \dfrac{x^3}{6} + \cdots$ and $\ln(1+u) = u - \dfrac{u^2}{2} + \dfrac{u^3}{3} - \cdots$ , substituting $u = \sin x$ . Keep only terms up to $x^3$ :

u = x - \frac{x^3}{6} + \cdots

u^2 = \left(x - \frac{x^3}{6} + \cdots\right)^2 = x^2 - \frac{x^4}{3} + \cdots \approx x^2 \quad (\text{to } x^3)

u^3 = x^3 + \cdots

Substituting:

\ln(1 + \sin x) = \left(x - \frac{x^3}{6}\right) - \frac{x^2}{2} + \frac{x^3}{3} - \cdots = x - \frac{x^2}{2} + \frac{x^3}{6} + \cdots

Worked example 3 — small-angle approximation

Given that $x$ is small, show that $\dfrac{1 - \cos 2x}{\sin^2 x} \approx 2$ .

Using $\cos 2x \approx 1 - \dfrac{(2x)^2}{2} = 1 - 2x^2$ and $\sin x \approx x$ :

\frac{1 - \cos 2x}{\sin^2 x} \approx \frac{1 - (1 - 2x^2)}{x^2} = \frac{2x^2}{x^2} = 2

Common mistakes

Wrong factorials: $\cos x$ starts $1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots$ , not $1 - \dfrac{x^2}{2} + \dfrac{x^4}{24}$ rewritten carelessly — the factorials are $2! = 2$ and $4! = 24$ , so those forms are the same, but forgetting a factorial entirely (writing $x^3/3$ instead of $x^3/3! = x^3/6$ for $\sin x$ ) is a common slip.
Sign errors in $\ln(1+x)$ : the series alternates $+x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots$ ; substituting a negative argument (e.g. $\ln(1-x)$ ) flips every sign.
Forgetting the validity range: $(1+x)^n$ and $\ln(1+x)$ are only valid for $|x| < 1$ and $-1 < x \leq 1$ respectively — always state this if the question asks for the range of validity.
Truncating too early when multiplying series: when combining two series, carry enough terms in each factor before multiplying, or you will miss contributions to the final degree you need.