What is the Maclaurin formula?

The Maclaurin series expands f(x) as a power series about x = 0:

What is combining series?

Build new series by multiplying two known ones (keeping terms up to the required power), substituting (for example x → 2x or x → x^2), or differentiating/integrating term by term. This is usually faster than repeated differentiation.

SEAB Original-style practice: Find the Maclaurin series of f(x) = e^2x up to and including the term in x^3.

f(x) = e^2x, f'(x) = 2e^2x, f''(x) = 4e^2x, f'''(x) = 8e^2x. At x = 0: f(0) = 1, f'(0) = 2, f''(0) = 4, f'''(0) = 8. Maclaurin: f(x) = 1 + 2x + 42!x^2 + 83!x^3 + = 1 + 2x + 2x^2 + 43x^3 + Markers reward the derivatives evaluated at 0, the Maclaurin formula with factorials, and the correct first four terms.

SingaporeMathsSyllabus dot point

How does the Maclaurin series approximate a function as a power series, and how do we use the standard expansions?

Derive and use the Maclaurin series of a function, apply the standard series for common functions, and use series to obtain approximations

A focused answer to the H2 Mathematics outcome on the Maclaurin series. The general formula, deriving a series from repeated differentiation, the standard expansions, combining them, and approximating function values.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

SEAB wants you to derive the Maclaurin series of a function by repeated differentiation, use the standard series for common functions, combine them by multiplication, substitution or differentiation, and use a truncated series to approximate function values.

The answer

The Maclaurin formula

The Maclaurin series expands $\mathrm{f}(x)$ as a power series about $x = 0$ :

\mathrm{f}(x) = \mathrm{f}(0) + \mathrm{f}'(0)x + \frac{\mathrm{f}''(0)}{2!}x^2 + \frac{\mathrm{f}'''(0)}{3!}x^3 + \cdots

Each coefficient uses a higher derivative evaluated at $0$ , divided by the factorial of the power.

Deriving a series

Differentiate $\mathrm{f}$ repeatedly, evaluate each derivative at $x = 0$ , and substitute into the formula. Look for a pattern to write the general term where possible.

The standard series

The expansions you should know (with their validity):

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \ (\text{all } x)

\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots, \quad \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \ (\text{all } x)

\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \ (-1 < x \leq 1), \quad (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots \ (|x| < 1)

Combining series

Build new series by multiplying two known ones (keeping terms up to the required power), substituting (for example $x \to 2x$ or $x \to x^2$ ), or differentiating/integrating term by term. This is usually faster than repeated differentiation.

Approximation

For small $x$ , the first few terms give an accurate value. The smaller $x$ is, the fewer terms are needed, because successive terms shrink rapidly.

Worked example

Find the Maclaurin series of $\mathrm{f}(x) = \ln(1 + 2x)$ up to the term in $x^3$ , and use it to estimate $\ln 1.2$ .

Step 1: Use the standard series with a substitution

In $\ln(1 + u) = u - \dfrac{u^2}{2} + \dfrac{u^3}{3} - \cdots$ put $u = 2x$ :

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

Step 2: Simplify each term

= 2x - \frac{4x^2}{2} + \frac{8x^3}{3} - \cdots = 2x - 2x^2 + \frac{8}{3}x^3 - \cdots

Step 3: Choose x for the estimate

$\ln 1.2 = \ln(1 + 0.2) = \ln(1 + 2x)$ with $2x = 0.2$ , so $x = 0.1$ .

Step 4: Substitute

\ln 1.2 \approx 2(0.1) - 2(0.1)^2 + \frac{8}{3}(0.1)^3 = 0.2 - 0.02 + 0.00267 = 0.18267

The true value is $0.18232$ , so three terms give good accuracy.

Examples in context

Example 1. Small-angle approximation. Truncating $\sin x \approx x$ and $\cos x \approx 1 - \dfrac{x^2}{2}$ for small $x$ underlies the pendulum and optics approximations in physics, all coming directly from the Maclaurin series.

Example 2. Linearising a model. Approximating $e^{kt} \approx 1 + kt$ for small $kt$ converts exponential growth into a linear estimate over short times, the basis of "for small changes" reasoning in economics and science.

Try this

Q1. Write the Maclaurin series of $\cos x$ up to the term in $x^4$ . [2 marks]

Cue. $1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \cdots$ .

Q2. Find the series of $e^{-x}$ up to $x^2$ using the standard series. [2 marks]

Cue. Substitute $-x$ into $e^x$ : $1 - x + \dfrac{x^2}{2} - \cdots$ .

Q3. State the validity range of the expansion of $\ln(1 + x)$ . [1 mark]

Cue. $-1 < x \leq 1$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksFind the Maclaurin series of

\mathrm{f}(x) = e^{2x}

up to and including the term in

x^3

Show worked answer →

$\mathrm{f}(x) = e^{2x}$ , $\mathrm{f}'(x) = 2e^{2x}$ , $\mathrm{f}''(x) = 4e^{2x}$ , $\mathrm{f}'''(x) = 8e^{2x}$ .

At $x = 0$ : $\mathrm{f}(0) = 1$ , $\mathrm{f}'(0) = 2$ , $\mathrm{f}''(0) = 4$ , $\mathrm{f}'''(0) = 8$ .

Maclaurin: $\mathrm{f}(x) = 1 + 2x + \dfrac{4}{2!}x^2 + \dfrac{8}{3!}x^3 + \cdots = 1 + 2x + 2x^2 + \dfrac{4}{3}x^3 + \cdots$

Markers reward the derivatives evaluated at $0$ , the Maclaurin formula with factorials, and the correct first four terms.

Original4 marksUsing the standard series for

\sin x

and

e^x

, find the Maclaurin series of

e^x \sin x

up to the term in

x^3

Show worked answer →

$e^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots$ and $\sin x = x - \dfrac{x^3}{6} + \cdots$ .

Multiply, keeping terms up to $x^3$ :
$\left(1 + x + \dfrac{x^2}{2} + \cdots\right)\left(x - \dfrac{x^3}{6} + \cdots\right) = x + x^2 + \dfrac{x^3}{2} - \dfrac{x^3}{6} + \cdots$

$= x + x^2 + \left(\dfrac{1}{2} - \dfrac{1}{6}\right)x^3 + \cdots = x + x^2 + \dfrac{x^3}{3} + \cdots$

Markers reward using the standard series, multiplying and collecting terms up to $x^3$ , and the correct coefficients.