Maclaurin Series

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The Maclaurin Series is a tool used by mathematicians to approximate complex functions such as \(\sin(x)\) and \(e^x\) using simple polynomials, found by taking an infinite sum of the function’s derivatives evaluated at \(0\). This A-Level Further Maths guide covers the Maclaurin Series formula, how to apply it, common mistakes to avoid, a teacher’s guide for sequencing the topic, a full video tutorial, a free printable worksheet, and four exam-style questions with handwritten worked solutions. Download the free PDF worksheet or watch the video tutorial to teach or revise the topic.

What is Maclaurin Series?

The Maclaurin Series allows you to approximate the value of a complex function using the sum of its derivatives evaluated at \(0\). It is a specific case of the broader Taylor Series, centred at the origin.

At A-Level, the Maclaurin Series is used not only to approximate complex functions, but also to solve difficult differential equations and to evaluate complex limits.

Maclaurin Series formula

The standard formula is:

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

Where:

• \(f^{(r)}(0)\) is the \(r\)-th derivative of the function evaluated at \(0\)
• \(r!\) is the factorial of \(r\)
• \(x^r\) is the \(r\)-th power of \(x\)

Teacher’s Guide

In order to use the Maclaurin series effectively, students must understand where it comes from and why it applies. If they derive the sum of the function themselves and see why the approximate answer appears, it becomes much easier to visualise how the series works and why the coefficients take the form they do.

After understanding the initial proof, students find the Maclaurin Series far more accessible, and they can move more quickly onto example questions.

They can then apply the Maclaurin series to more complex functions where they need other differentiation rules such as the chain rule and the product rule, for example:

\frac{d}{dx}\left(e^{x}\sin x\right)

Maclaurin Series – Step by step

Follow these steps to use the Maclaurin Series to approximate a function. This example finds an approximation of \(\ln\left(\frac{5}{2}\right)\) using the series expansion of \(\ln\left(\frac{1}{1-2x}\right)\) up to the \(x^3\) term.

Step 1 – Differentiate the function

Use the laws of logarithms to rearrange \(\ln\left(\frac{1}{1-2x}\right)\) as \(\ln\left((1-2x)^{-1}\right)\) and then as \(-\ln(1-2x)\). Differentiate using the chain rule to get \(\frac{2}{1-2x}\), which can be written as \(2(1-2x)^{-1}\).

Step 2 – Differentiate the function again

Use the chain rule again to find the second derivative: \(4(1-2x)^{-2}\).

Step 3 – Differentiate the function a third time

Differentiate once more to find the third derivative: \(16(1-2x)^{-3}\).

Step 4 – Apply the Maclaurin Series

Evaluate each derivative at \(0\) to find the coefficients, then substitute into the formula

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

At \(x = 0\): \(f(0) = 0\), \(f'(0) = 2\), \(f”(0) = 4\) and \(f”'(0) = 16\). The expansion up to the \(x^3\) term is therefore

2x + 2x^{2} + \frac{8}{3}x^{3}

Step 5 – Find the value of \(x\) for the approximation

Set \(\frac{5}{2}\) equal to \(\frac{1}{1-2x}\) to find the value of \(x\) to substitute into the series. This gives \(x = \frac{3}{10}\). Substituting into the expansion produces \(0.852\) to 3 d.p., which is close to the actual value \(\ln\left(\frac{5}{2}\right) \approx 0.916\).

Common mistakes with Maclaurin Series

• Forgetting the factorial – After substituting the derivative at \(0\) into the formula, some students forget to divide by \(r!\), not just by \(r\).
• Expanding to the wrong order – Many students do not expand to enough terms, while others go further than the question asks.
• Using the wrong standard series – When using the formula-book expansions for \(\sin(x)\) and \(\cos(x)\), students sometimes mix the two up.
• Not substituting correctly – For a function such as \(e^{2x}\), students may write \(1 + 2x + \frac{x^{2}}{2!} + \cdots\) instead of \(1 + 2x + \frac{(2x)^{2}}{2!} + \cdots\).
• Losing negatives – With a function such as \(\sin(-x)\), the negative sign is often missed when substituting into the series.
• Multiplying series incorrectly – When asked to multiply two expansions, students often miss terms or fail to combine like powers of \(x\).

Video: Maclaurin Series

Watch the following video, which shows how to derive the Maclaurin series formula and then use it to approximate difficult functions.

Want to practise alongside the video? Download the lesson pack and pause after each example to try the next question before the worked solution.

What the video covers

• Derivation of the Maclaurin Series – Explains the theory behind expressing a function as a power series by finding successive derivatives and evaluating them at \(x = 0\) to reveal the factorial pattern in the coefficients.
• Applying the Maclaurin Series – Worked example for the expansion of \(e^{\frac{1}{3}x}\) up to the \(x^{4}\) term.
• Approximating values – Demonstrates how to use the derived series for \(e^{\frac{1}{3}x}\) to approximate \(\sqrt[3]{e^{2}} = e^{\frac{2}{3}}\).

The written step-by-step guide above follows the same overall approach as the video. Use the video for live modelling in class, or the text sections for independent revision.

Free Maclaurin series worksheet (PDF)

Download the free exam-question worksheet to practise or set for homework. The same four exam-style questions featured on this page are included, with space for working and full worked solutions.

Exam Style Questions

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Frequently asked questions

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