# Maclaurin and Taylor Series

> Edexcel International A-Level Further Mathematics · IAL FP2
> Source: https://www.owlsprep.com/study/edexcel-ial-further-math-fp2-maclaurin-and-taylor-series/

This guide covers all Edexcel IAL FP2 content for Maclaurin and Taylor series, including higher order derivatives, series derivations, expansions about arbitrary points, and ODE series solutions, aligned to the 2018 specification.

**Prerequisites:** [Differentiation of standard and composite functions (P3)](https://www.owlsprep.com/study/edexcel-ial-maths-p3-differentiation/); [Second order differential equations (FP2)](https://www.owlsprep.com/study/edexcel-ial-further-math-fp2-second-order-differential-equations/)

## Learning objectives

- Compute third and higher order derivatives for polynomial, trigonometric, exponential and logarithmic functions
- Derive and apply Maclaurin series expansions for standard and simple composite functions up to specified terms
- Derive and apply Taylor series expansions about a general point a up to specified terms
- Use Taylor series methods to find series solutions of first and second order differential equations up to specified terms

## Higher Order Derivatives for Series Expansions

**Higher Order Derivative** — The result of differentiating a function \(f(x)\) \(r\) times with respect to \(x\), where \(r \geq 1\)

*Notation:* f^{(r)}(x)

*Example:* The 3rd derivative of \(\sin x\) is \(-\cos x\)

To construct Maclaurin and Taylor series, you will need to compute derivatives of order 3 and above for simple functions. For composite functions, apply the chain rule repeatedly, and for products, use the product rule as required. You do not need to simplify factorial terms in series coefficients unless explicitly asked to do so.

**Worked example:** Find the first, second, third and fourth derivatives of \(f(x) = \ln(1 + 2x)\) evaluated at \(x=0\).

1. First derivative (chain rule):

   $$f'(x) = \frac{2}{1+2x} \implies f'(0) = 2$$
2. Second derivative:

   $$f''(x) = -\frac{4}{(1+2x)^2} \implies f''(0) = -4$$
3. Third derivative:

   $$f'''(x) = \frac{16}{(1+2x)^3} \implies f'''(0) = 16$$
4. Fourth derivative:

   $$f^{(4)}(x) = -\frac{96}{(1+2x)^4} \implies f^{(4)}(0) = -96$$

> **Exam tip:** Label derivatives clearly with their order when computing them for series questions, as markers award marks for correct derivative values even if you make a substitution error later.

*Calculator:* allowed

## Maclaurin Series Expansions

**Maclaurin Series** — A power series expansion of a function \(f(x)\) centred at \(x=0\), valid for values of \(x\) where the series converges

*Notation:* f(x) = \sum_{r=0}^{\infty} \frac{f^{(r)}(0)}{r!}x^r

*Example:* The Maclaurin series of \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...\)

When asked to derive a Maclaurin series from first principles, you must compute each required derivative explicitly, substitute \(x=0\) into each derivative, then plug the values into the general Maclaurin form. You can use standard expansions from the formula booklet for composite functions unless derivation is explicitly required.

**Worked example:** Derive the Maclaurin series expansion of \(f(x) = \ln(1 + 2x)\) up to and including the term in \(x^4\), and state its validity.

1. Use the derivative values from the previous worked example, plus the base function value:

   $$f(0) = \ln(1) = 0$$
2. Substitute into the general Maclaurin series formula:

   $$f(x) = f(0) + x f'(0) + \frac{x^2}{2!}f''(0) + \frac{x^3}{3!}f'''(0) + \frac{x^4}{4!}f^{(4)}(0)$$
3. Plug in derivative values:

   $$f(x) = 0 + 2x + \frac{x^2}{2}(-4) + \frac{x^3}{6}(16) + \frac{x^4}{24}(-96)$$
4. Simplify terms:

   $$f(x) = 2x - 2x^2 + \frac{8}{3}x^3 - 4x^4$$
5. Validity: Standard \(\ln(1+u)\) is valid for \(|u| < 1\), so \(|2x| < 1 \implies |x| < 0.5

> **Exam tip:** Always state the validity range if asked, using standard ranges from the formula booklet for composite functions.

*Calculator:* allowed

## Taylor Series Expansions About a General Point

**Taylor Series (about point a)** — A power series expansion of a function \(f(x)\) centred at \(x=a\), used to approximate values of \(f(x)\) near \(x=a\)

*Notation:* f(x) = \sum_{r=0}^{\infty} \frac{f^{(r)}(a)}{r!}(x-a)^r

Taylor series follow the same structure as Maclaurin series, except derivatives are evaluated at \(x=a\) instead of \(x=0\), and terms use powers of \((x-a)\) instead of \(x\). They are used when you need an approximation of a function near a point other than 0.

**Worked example:** Find the Taylor series expansion of \(f(x) = \sin x\) in ascending powers of \((x - \pi)\) up to and including the term in \((x - \pi)^3\).

1. Compute derivatives evaluated at \(a = \pi\):

   $$f(\pi) = \sin \pi = 0, f'(\pi) = \cos \pi = -1, f''(\pi) = -\sin \pi = 0, f'''(\pi) = -\cos \pi = 1$$
2. Substitute into the general Taylor series formula:

   $$f(x) = f(\pi) + (x-\pi)f'(\pi) + \frac{(x-\pi)^2}{2!}f''(\pi) + \frac{(x-\pi)^3}{3!}f'''(\pi)$$
3. Plug in values:

   $$f(x) = 0 + (x-\pi)(-1) + \frac{(x-\pi)^2}{2}(0) + \frac{(x-\pi)^3}{6}(1)$$
4. Simplify:

   $$f(x) = -(x-\pi) + \frac{1}{6}(x-\pi)^3$$

> **Exam tip:** You do not need to expand \((x-a)\) terms unless explicitly asked; leaving them in factored form is acceptable for full marks.

*Calculator:* allowed

## Series Solutions of Differential Equations

To find a series solution of a differential equation, repeatedly differentiate the given ODE, use initial conditions to evaluate each derivative at the given point (almost always \(x=0\) for FP2 questions), then substitute the values into the Maclaurin series form for \(y(x)\).

**Worked example:** Find the series solution of \(\frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0\) with initial conditions \(y=1\) and \(\frac{dy}{dx}=0\) at \(x=0\), up to and including the term in \(x^4\).

1. State initial values: \(y(0)=1\), \(y'(0)=0\). Rearrange ODE to isolate highest derivative:

   $$y'' = -x y' - y \implies y''(0) = -0 -1 = -1$$
2. Differentiate ODE once to find third derivative:

   $$y''' = -y' - x y'' - y' = -x y'' - 2y' \implies y'''(0) = -0 - 0 = 0$$
3. Differentiate again to find fourth derivative:

   $$y^{(4)} = -y'' - x y''' - 2y'' = -x y''' - 3y'' \implies y^{(4)}(0) = -0 - 3(-1) = 3$$
4. Substitute into Maclaurin series for \(y(x)\):

   $$y = 1 + 0x + \frac{x^2}{2!}(-1) + \frac{x^3}{3!}(0) + \frac{x^4}{4!}(3)$$
5. Simplify:

   $$y = 1 - \frac{1}{2}x^2 + \frac{1}{8}x^4$$

> **Exam tip:** Show all steps of differentiating the ODE, as these are awarded method marks even if you make an arithmetic error when evaluating derivatives.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using powers of \(x\) instead of \((x-a)\) for Taylor series expansions about \(a \neq 0\)
  - Why it fails: Taylor series are centred at \(a\), so terms must be powers of the difference between \(x\) and \(a\)
  - Correct: Always use \((x-a)\) as the base for all terms in a Taylor series expansion about \(a\)
- **Wrong:** Quoting a standard expansion from the formula booklet when the question asks for derivation from first principles
  - Why it fails: Derivation questions require explicit demonstration of derivative calculation, not just recall of formulae
  - Correct: Compute all required derivatives explicitly, evaluate them at the centre point, and substitute into the general series formula for derivation questions
- **Wrong:** Forgetting to divide each series term by the factorial of the derivative order
  - Why it fails: The general series formula includes a denominator of \(r!\) for the rth derivative term, so omitting it leads to incorrect coefficients
  - Correct: Write the factorial denominator explicitly for each term before substituting derivative values
- **Wrong:** Making differentiation errors when deriving higher order derivatives for ODE series solutions
  - Why it fails: Differentiation errors propagate to all higher order derivatives, leading to incorrect coefficients for all higher terms
  - Correct: Check each differentiation step carefully, and verify that initial condition substitution gives consistent derivative values
- **Wrong:** Conflating Maclaurin/Taylor series with the FP2 method of differences topic
  - Why it fails: These are separate topics with distinct techniques, so applying the wrong method gains no marks
  - Correct: Use derivative calculations for Maclaurin/Taylor series questions, and telescoping sum methods for method of differences questions

## Cheatsheet

| Expansion Type | General Formula | Use Case |
| --- | --- | --- |
| Maclaurin Series | \(f(x) = \sum_{r=0}^n \frac{f^{(r)}(0)}{r!}x^r + ...\) | Expansions centred at \(x=0\), standard function expansions |
| Taylor Series (about \(a\)) | \(f(x) = \sum_{r=0}^n \frac{f^{(r)}(a)}{r!}(x-a)^r + ...\) | Expansions centred at any point \(x=a\), approximations near \(a\) |
| ODE Series Solution | Compute \(y^{(r)}(a)\) via repeated ODE differentiation, substitute into series form | Power series solutions of ODEs with given initial conditions |
| Standard Expansions | Provided in formula booklet: \(e^x, \sin x, \cos x, \ln(1+x), \arctan x\) | Shortcut for composite function expansions when derivation is not required |

## What's next

Now that you have mastered Maclaurin and Taylor series for Edexcel IAL FP2, you can apply these skills to solve a wide range of exam questions, including function value approximations, composite function expansions, and series solutions for complex differential equations. This topic is often combined with other FP2 content such as polar coordinates and further calculus, so cross-topic practice is critical for exam readiness. Make sure to complete past paper questions on this topic to familiarise yourself with common phrasing and mark scheme requirements, and review the official FP2 formula booklet to confirm which expansions are provided for exam use.

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/edexcel-ial-further-math-fp2-maclaurin-and-taylor-series/