SQA AH style: derive a series-style practice: Find the Maclaurin series for f(x) = e^2x up to and including the term in x^3.

Derivatives: f(x) = e^2x, f'(x) = 2e^2x, f''(x) = 4e^2x, f'''(x) = 8e^2x (1 mark). At x = 0: f(0) = 1, f'(0) = 2, f''(0) = 4, f'''(0) = 8 (1 mark). Maclaurin: f(x) = f(0) + f'(0)x + f''(0)2!x^2 + f'''(0)3!x^3 + (1 mark). = 1 + 2x + 42x^2 + 86x^3 = 1 + 2x + 2x^2 + 43x^3 (1 mark). Markers reward the derivatives, their values at zero, the formula, and the simplified series.

SQA AH style: composite series-style practice: Use the standard expansion of cos x to find the Maclaurin series for cos(3x) up to the term in x^4.

Standard: cos u = 1 - u^22! + u^44! - (1 mark). Replace u = 3x: cos 3x = 1 - (3x)^22 + (3x)^424 - (1 mark). = 1 - 9x^22 + 81x^424 - (1 mark). Simplify: = 1 - 92x^2 + 278x^4 - (1 mark). Markers reward the standard expansion, the substitution, and the simplified terms.

ScotlandMathsSyllabus dot point

How do you build a power series for a function from its derivatives at zero?

Find the Maclaurin series expansion of a function using the standard formula, derive the standard expansions of exponential, logarithmic and trigonometric functions, and use known expansions to build the series of composite or product functions.

A focused answer to the SQA Advanced Higher Mathematics Maclaurin series content, covering the Maclaurin formula, deriving a series from successive derivatives at zero, the standard expansions of the exponential, logarithmic, sine and cosine functions, and combining known expansions for composite or product functions.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-14

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

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What this dot point is asking
The Maclaurin formula
The standard expansions
Building new series
Deriving a less obvious series
Using a series to approximate
Try this

What this dot point is asking

The SQA wants you to represent a function as a power series, a Maclaurin series, built from the function's derivatives evaluated at zero. You should be able to derive a series from scratch and to build new series quickly from the standard expansions you have learned.

The Maclaurin formula

A Maclaurin series is the special case of a Taylor series centred at $x = 0$ . Each coefficient comes from a derivative evaluated at zero, scaled by a factorial.

Deriving the Maclaurin series for

\sin x

from scratch

Derive the Maclaurin series for $\sin x$ .

Step 1: Differentiate repeatedly and spot the cycle

Each differentiation of $\sin x$ cycles through four functions with period four, which means we only need to work out four derivatives before the pattern repeats:

f = \sin x,\quad f' = \cos x,\quad f'' = -\sin x,\quad f''' = -\cos x,\quad f^{(4)} = \sin x,\ \ldots

Step 2: Evaluate each derivative at $x = 0$

The Maclaurin formula needs the value of each derivative at zero. The zero-values of $\sin$ and $-\sin$ contribute nothing; only the cosine terms survive:

f(0) = 0,\quad f'(0) = 1,\quad f''(0) = 0,\quad f'''(0) = -1,\quad f^{(4)}(0) = 0,\ \ldots

Step 3: Substitute into the Maclaurin formula

Every zero coefficient kills its term, leaving only the odd powers. Divide each surviving derivative value by the appropriate factorial:

\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots

Final answer: $\sin x = x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots$ , an alternating series of odd powers.

The standard expansions

Four expansions are worth knowing cold; almost every exam series is built by substituting into one of them.

Building new series

Rather than differentiate from scratch, substitute into or multiply known expansions. Replacing $x$ by $2x$ or $-x^2$ , or multiplying two series and keeping terms up to the required power, is usually far faster than computing high-order derivatives.

Building the series for

xe^x

by multiplying a known expansion

Find the series for $xe^{x}$ up to $x^3$ .

Step 1: Start from the standard exponential series

Rather than differentiate from scratch, use the known expansion for $e^x$ , which saves time. Write enough terms to cover all the powers we need after multiplication:

e^{x} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots

Step 2: Multiply every term by $x$

Multiplying by $x$ shifts each power up by one. The $x^4$ and higher terms will exceed our target order, so we can drop them straight away:

x e^{x} = x + x^2 + \frac{x^3}{2} + \frac{x^4}{6} + \cdots

Step 3: Truncate at $x^3$

We only need terms up to $x^3$ , so discard the $x^4$ term and beyond:

x e^{x} = x + x^2 + \frac{x^3}{2} + \cdots

Final answer: $xe^x = x + x^2 + \dfrac{x^3}{2} + \cdots$ to three terms.

Deriving a less obvious series

Some functions need a little setup before the standard list helps. The series for $\ln(1 + x)$ , for instance, is most easily found by integrating the geometric series for $\dfrac{1}{1 + x}$ term by term, and the series for $\tan^{-1} x$ comes from integrating the series for $\dfrac{1}{1 + x^2}$ . When no shortcut is available, fall back on the definition and differentiate repeatedly, watching for a repeating pattern in the derivatives that lets you write a general term.

Deriving the series for

\ln(1+x)

from its derivatives

Find the Maclaurin series for $\ln(1 + x)$ from its derivatives.

Step 1: Differentiate repeatedly

Differentiate $f = \ln(1 + x)$ three times. Notice that each differentiation of $\dfrac{1}{(1+x)^n}$ brings down an extra factor and increases the power:

f = \ln(1 + x),\quad f' = \frac{1}{1 + x},\quad f'' = -\frac{1}{(1 + x)^2},\quad f''' = \frac{2}{(1 + x)^3}.

Step 2: Evaluate each derivative at $x = 0$

Substitute $x = 0$ into each expression. At zero, $(1+x)^n = 1$ , so only the numerators survive:

f(0) = 0,\quad f'(0) = 1,\quad f''(0) = -1,\quad f'''(0) = 2.

Step 3: Assemble the series using the Maclaurin formula

Divide each derivative value by the corresponding factorial and collect into a series:

\ln(1 + x) = 0 + x - \frac{1}{2!}x^2 + \frac{2}{3!}x^3 - \cdots = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots

Final answer: $\ln(1 + x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots$ , valid for $-1 < x \le 1$ .

Using a series to approximate

Truncating a Maclaurin series gives a polynomial approximation to the function near $x = 0$ ; the more terms you keep, the better the fit close to the origin. This is how a calculator estimates $\sin$ , $\cos$ and $e^x$ . The approximation is most accurate for small $x$ and degrades as $x$ moves away from zero, which is why the standard expansions carry conditions such as $-1 < x \le 1$ for $\ln(1 + x)$ .

Try this

Q1. Write the Maclaurin series for $e^{x}$ up to $x^3$ . [2 marks]

Cue. $1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6}$ .

Q2. Use the $\sin x$ series to find the series for $\sin(2x)$ up to $x^3$ . [2 marks]

Cue. $\sin 2x = 2x - \dfrac{(2x)^3}{6} = 2x - \dfrac{4x^3}{3}$ .

Exam-style practice questions

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

AH style: derive a series4 marksFind the Maclaurin series for

f(x) = e^{2x}

up to and including the term in

x^3

Show worked answer →

Derivatives: $f(x) = e^{2x}$ , $f'(x) = 2e^{2x}$ , $f''(x) = 4e^{2x}$ , $f'''(x) = 8e^{2x}$ (1 mark).

At $x = 0$ : $f(0) = 1$ , $f'(0) = 2$ , $f''(0) = 4$ , $f'''(0) = 8$ (1 mark).

Maclaurin: $f(x) = f(0) + f'(0)x + \dfrac{f''(0)}{2!}x^2 + \dfrac{f'''(0)}{3!}x^3 + \cdots$ (1 mark).

$= 1 + 2x + \dfrac{4}{2}x^2 + \dfrac{8}{6}x^3 = 1 + 2x + 2x^2 + \dfrac{4}{3}x^3$ (1 mark). Markers reward the derivatives, their values at zero, the formula, and the simplified series.

AH style: composite series4 marksUse the standard expansion of

\cos x

to find the Maclaurin series for

\cos(3x)

up to the term in

x^4

Show worked answer →

Standard: $\cos u = 1 - \dfrac{u^2}{2!} + \dfrac{u^4}{4!} - \cdots$ (1 mark).

Replace $u = 3x$ : $\cos 3x = 1 - \dfrac{(3x)^2}{2} + \dfrac{(3x)^4}{24} - \cdots$ (1 mark).

$= 1 - \dfrac{9x^2}{2} + \dfrac{81x^4}{24} - \cdots$ (1 mark).

Simplify: $= 1 - \dfrac{9}{2}x^2 + \dfrac{27}{8}x^4 - \cdots$ (1 mark). Markers reward the standard expansion, the substitution, and the simplified terms.

Related dot points

Sources & how we know this

SQA Advanced Higher Mathematics Course Specification — SQA (2019)