What is wrong expansion point?

A Taylor series about a uses powers of (x - a), not powers of x; only the Maclaurin case (a = 0) uses powers of x.

Pearson Edexcel Edexcel 2021-style practice: Given that y satisfies (dy)/(dx) = x + y with y = 1 at x = 0, find the Taylor series solution for y up to and including the term in x^3.

Differentiate the differential equation repeatedly, evaluate derivatives at x = 0, then assemble the Maclaurin series. At x = 0, y = 1 (given). From (dy)/(dx) = x + y: y'(0) = 0 + 1 = 1 (M1 A1). Differentiate: y'' = 1 + y', so y''(0) = 1 + 1 = 2 (M1 A1). Differentiate again: y''' = y'', so y'''(0) = 2 (A1). Maclaurin series: y = y(0) + y'(0)x + (y''(0))/(2!)x^2 + (y'''(0))/(3!)x^3 + (M1) = 1 + x + (2)/(2)x^2 + (2)/(6)x^3 = 1 + x + x^2 + (1)/(3)x^3 + (A1).

EnglandFurther MathsSyllabus dot point

How do you build power series approximations of functions and use them to solve differential equations?

Maclaurin and Taylor series of standard functions, finding series solutions of differential equations, and using series to approximate functions and limits.

A focused answer to the Edexcel A-Level Further Mathematics Further Pure series content, covering Maclaurin and Taylor series of standard functions, finding series solutions of differential equations, and using power series to approximate functions and evaluate limits.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

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Quick answer

The Maclaurin series expands $f(x)$ about $x = 0$ as $f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots$ , and the Taylor series expands about $x = a$ in powers of $(x - a)$ . Standard expansions include $e^x = \sum \frac{x^n}{n!}$ , $\sin x = x - \frac{x^3}{3!} + \cdots$ , $\cos x = 1 - \frac{x^2}{2!} + \cdots$ and $\ln(1 + x) = x - \frac{x^2}{2} + \cdots$ (valid for $-1 < x \le 1$ ). Series solutions of differential equations come from differentiating the equation repeatedly to get successive derivatives at a point, then assembling the Taylor series.

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What this dot point is asking
Maclaurin and Taylor series
Standard expansions
Series solutions of differential equations
Examples in context
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What this dot point is asking

Edexcel Further Pure wants you to derive Maclaurin and Taylor series of standard functions, quote the standard expansions, find series solutions of differential equations by repeated differentiation, and use power series to approximate function values and evaluate limits. The series-solution-of-a-differential-equation question is a recurring high-tariff item.

Maclaurin and Taylor series

The Maclaurin series is the special case of the Taylor series taken about zero. Both reconstruct a (suitably smooth) function from the values of all its derivatives at a single point: the $n$ th derivative at the centre determines the coefficient of the $n$ th-degree term, divided by $n!$ .

Maclaurin series of a function

Find the first three terms of the Maclaurin series of $f(x) = e^{2x}$ .

Step 1: Evaluate the function and its derivatives at $x = 0$

The Maclaurin formula requires the values $f(0)$ , $f'(0)$ , $f''(0)$ , and so on. Each derivative of $e^{2x}$ brings down a factor of $2$ by the chain rule:

f(x) = e^{2x},\quad f(0) = 1.

f'(x) = 2e^{2x},\quad f'(0) = 2.

f''(x) = 4e^{2x},\quad f''(0) = 4.

Step 2: Assemble the series using the Maclaurin formula

Insert the derivative values into $f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots$ :

e^{2x} = 1 + 2x + \frac{4}{2!}x^2 + \cdots = 1 + 2x + 2x^2 + \cdots.

You can verify this matches the standard expansion $e^u = 1 + u + \frac{u^2}{2!} + \cdots$ with $u = 2x$ .

Final answer: $e^{2x} = 1 + 2x + 2x^2 + \cdots$ (first three terms).

Standard expansions

These expansions are derived once and then quoted, with their intervals of validity listed in the formula booklet. They can be combined, substituted into, differentiated and integrated term by term to obtain new series quickly.

Series solutions of differential equations

To solve a differential equation as a Taylor series about the initial point, differentiate the equation repeatedly. Each differentiation produces a higher derivative in terms of lower ones; evaluating at the initial point (using the given conditions) gives the successive coefficients, which you then assemble into the Maclaurin series.

Series solution of a differential equation

Find the Maclaurin series up to $x^2$ for the solution of $\frac{dy}{dx} = y^2 + x$ with $y = 1$ at $x = 0$ .

Step 1: Read off the initial values from the given conditions

The initial condition gives $y(0) = 1$ directly. To find $y'(0)$ , substitute $x = 0$ and $y = 1$ into the differential equation:

y'(0) = y(0)^2 + 0 = 1^2 + 0 = 1.

Step 2: Differentiate the equation to find the next derivative

To find $y''$ , differentiate both sides of $y' = y^2 + x$ with respect to $x$ . Using the chain rule on $y^2$ :

y'' = 2y\, y' + 1.

Evaluate at $x = 0$ using the values already found:

y''(0) = 2(1)(1) + 1 = 3.

Step 3: Assemble the Maclaurin series

Insert $y(0) = 1$ , $y'(0) = 1$ , and $y''(0) = 3$ into the Maclaurin formula:

y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \cdots = 1 + x + \frac{3}{2}x^2 + \cdots.

Final answer: $y = 1 + x + \dfrac{3}{2}x^2 + \cdots$ .

Examples in context

Taylor series connect to much of the course. The standard expansions for $e^x$ , $\sin x$ and $\cos x$ make Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ transparent, linking series to complex numbers, and the same idea gives the series for $\sinh$ and $\cosh$ in hyperbolic functions. Series provide an alternative to the numerical methods of root finding and integration, and the truncation error of those methods is analysed through the remainder term of a Taylor series. Series solutions of differential equations complement the exact integrating-factor and auxiliary-equation methods when no closed-form solution exists. Power series also evaluate awkward limits by cancelling leading terms.

Try this

Q1. Write down the first three terms of the Maclaurin series for $\cos x$ . [2 marks]

Cue. $1 - \frac{x^2}{2!} + \frac{x^4}{4!}$ .

Q2. Use the series for $e^x$ to estimate $e^{0.1}$ to three terms. [2 marks]

Cue. $1 + 0.1 + \frac{0.01}{2} = 1.105$ .

Q3. Find the Maclaurin series for $\ln(1 + x)$ up to the term in $x^3$ . [3 marks]

Cue. $x - \frac{x^2}{2} + \frac{x^3}{3}$ .

Exam-style practice questions

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

Edexcel 20186 marksFind the Maclaurin series for

f(x) = \ln(1 + 2x)

up to and including the term in

x^3

, and state the interval of validity.

Show worked answer →

Differentiate repeatedly and evaluate at $x = 0$ , or adapt the standard $\ln(1 + x)$ series.

Using the standard expansion $\ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots$ with $u = 2x$ (M1):

$\ln(1 + 2x) = 2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} - \cdots$ (M1 A1).

$= 2x - 2x^2 + \frac{8x^3}{3} - \cdots$ (A1 A1).

Validity: the standard series converges for $-1 < u \le 1$ , so $-1 < 2x \le 1$ , i.e. $-\frac{1}{2} < x \le \frac{1}{2}$ (A1).

Edexcel 20217 marksGiven that

y

satisfies

\frac{dy}{dx} = x + y

with

y = 1

x = 0

, find the Taylor series solution for

y

up to and including the term in

x^3

Show worked answer →

Differentiate the differential equation repeatedly, evaluate derivatives at $x = 0$ , then assemble the Maclaurin series.

At $x = 0$ , $y = 1$ (given). From $\frac{dy}{dx} = x + y$ : $y'(0) = 0 + 1 = 1$ (M1 A1).

Differentiate: $y'' = 1 + y'$ , so $y''(0) = 1 + 1 = 2$ (M1 A1). Differentiate again: $y''' = y''$ , so $y'''(0) = 2$ (A1).

Maclaurin series: $y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \cdots$ (M1) $= 1 + x + \frac{2}{2}x^2 + \frac{2}{6}x^3 = 1 + x + x^2 + \frac{1}{3}x^3 + \cdots$ (A1).

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Sources & how we know this

Pearson Edexcel A-Level Further Mathematics (9FM0) specification — Pearson Edexcel (2017)