Home / A-Level Further Maths 9231 / Further differentiation / Maclaurin series
Cambridge International AS & A Level · Topic 2.3
A-Level Further Maths 9231 / Topic 2.3 / AS & A Level

Know a function at zero, and you can rebuild it nearby.

A Maclaurin series writes a function as a polynomial whose coefficients come from its derivatives at \(x = 0\). Derive the series for \(e^{x}\), \(\sin x\) and \(\ln(1 + x)\), use successive differentiation for functions such as \(\tan x\) and \(e^{\sin x}\), combine known series, and use the result to approximate values and limits.

Topic 2.3All further differentiation lessons →
The key idea

\(f(x) = f(0) + x f'(0) + \dfrac{x^{2}}{2!} f''(0) + \dfrac{x^{3}}{3!} f'''(0) + \dots\) Every coefficient is a derivative at \(0\) divided by a factorial. When \(f\) is hard to differentiate repeatedly, find a relation such as \(y' = 1 + y^{2}\) and differentiate THAT. When \(f\) is built from standard functions, substitute into and multiply the standard series instead, keeping every term up to the power you need. A series for \(\ln(1 + x)\) only works for \(-1 \lt x \le 1\).

x y −π π −2 −1 1 2 y = x y = x − x³/6 y = x − x³/6 + x⁵/120 y = sin x
Fig. 1: y = sin x (dashed) with its Maclaurin polynomials of degree 1 (y = x), 3 and 5. Near x = 0 all of them hug the curve; each extra term keeps the polynomial close to sin x over a wider interval, and each one eventually runs away from it
Section 01

The Maclaurin formula.

Suppose \(f(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + \dots\) Put \(x = 0\): \(a_{0} = f(0)\). Differentiate and put \(x = 0\): \(a_{1} = f'(0)\). Differentiate again: \(2a_{2} = f''(0)\); again: \(3 \times 2\,a_{3} = f'''(0)\). In general the coefficient of \(x^{r}\) is \(\dfrac{f^{(r)}(0)}{r!}\). So to build the series, list \(f, f', f'', \dots\), evaluate each at \(0\), and divide by the factorial. Fig. 1 shows the first few polynomials for \(\sin x\) closing in on the curve.

\[f(x) = f(0) + x f'(0) + \frac{x^{2}}{2!} f''(0) + \frac{x^{3}}{3!} f'''(0) + \dots + \frac{x^{r}}{r!} f^{(r)}(0) + \dots\]
Standard series (formula list)Valid for
\(e^{x} = 1 + x + \dfrac{x^{2}}{2!} + \dfrac{x^{3}}{3!} + \dots\)all \(x\)
\(\sin x = x - \dfrac{x^{3}}{3!} + \dfrac{x^{5}}{5!} - \dots\)all \(x\)
\(\cos x = 1 - \dfrac{x^{2}}{2!} + \dfrac{x^{4}}{4!} - \dots\)all \(x\)
\(\ln(1 + x) = x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3} - \dots\)\(-1 \lt x \le 1\)
\((1 + x)^{n} = 1 + nx + \dfrac{n(n - 1)}{2!}x^{2} + \dots\)\(|x| \lt 1\) (\(n\) not a positive integer)

You are expected to derive the first few terms of such series, not a general term. Angles are always in radians: the derivative of \(\sin x\) is \(\cos x\) only in radians.

Worked example e to the x from scratch
Use the Maclaurin formula to find the first four terms of the series for \(e^{x}\), and use them to estimate \(e^{0.1}\).
  1. \(f(x) = e^{x}\), and every derivative is also \(e^{x}\). So \(f(0) = f'(0) = f''(0) = f'''(0) = 1\).
  2. \[e^{x} = 1 + x(1) + \frac{x^{2}}{2!}(1) + \frac{x^{3}}{3!}(1) + \dots = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \dots\]
  3. Put \(x = 0.1\): \(1 + 0.1 + 0.005 + 0.000\,1667 = 1.105\,167\) (to 6 d.p.).
  4. A calculator gives \(e^{0.1} = 1.105\,171\): four terms are already correct to 5 decimal places, because \(x = 0.1\) is close to \(0\).
Answer: \(e^{x} = 1 + x + \dfrac{x^{2}}{2} + \dfrac{x^{3}}{6} + \dots\); \(e^{0.1} \approx 1.105\,17\)
Worked example sin x from scratch
Find the Maclaurin series for \(\sin x\) up to and including the term in \(x^{5}\).
  1. The derivatives repeat in a cycle of four: \(f = \sin x\), \(f' = \cos x\), \(f'' = -\sin x\), \(f''' = -\cos x\), \(f^{(4)} = \sin x\), \(f^{(5)} = \cos x\).
  2. At \(x = 0\): \(f(0) = 0\), \(f'(0) = 1\), \(f''(0) = 0\), \(f'''(0) = -1\), \(f^{(4)}(0) = 0\), \(f^{(5)}(0) = 1\).
  3. \[\sin x = 0 + x(1) + 0 + \frac{x^{3}}{3!}(-1) + 0 + \frac{x^{5}}{5!}(1) = x - \frac{x^{3}}{6} + \frac{x^{5}}{120}.\]
  4. Only odd powers appear, as they must for an odd function: \(\sin(-x) = -\sin x\).
Answer: \(\sin x = x - \dfrac{x^{3}}{6} + \dfrac{x^{5}}{120} - \dots\)
Your turn
Find the Maclaurin series for \(\ln(1 + x)\) up to and including the term in \(x^{4}\).
Examiner trap

Two classic slips: forgetting the factorials (writing \(f'''(0)\,x^{3}\) instead of \(\dfrac{f'''(0)}{3!}x^{3}\)), and evaluating at the wrong point. Every derivative is evaluated at \(x = 0\). Also, \(\ln x\) itself has NO Maclaurin series: \(\ln 0\) is undefined, which is why the standard series is for \(\ln(1 + x)\).

Section 02

When the derivatives get messy.

For \(\tan x\) or \(e^{\sin x}\), differentiating four or five times directly creates long product and quotient rules. Two tricks keep it short. First, write \(y = f(x)\) and find a simple relation between \(y\) and its derivatives, then differentiate the RELATION (successive implicit differentiation, from the previous lesson). Second, never simplify more than you must: you only need values at \(x = 0\), so substitute \(x = 0\) as early as possible.

Worked example tan x via y' = 1 + y²
Given \(y = \tan x\), show that \(\dfrac{dy}{dx} = 1 + y^{2}\). Hence find the Maclaurin series for \(\tan x\) up to and including the term in \(x^{3}\).
  1. \(\dfrac{dy}{dx} = \sec^{2} x = 1 + \tan^{2} x = 1 + y^{2}\).
  2. Differentiate the relation (chain rule on \(y^{2}\)): \(y'' = 2y\,y'\). Again (product rule): \(y''' = 2(y')^{2} + 2y\,y''\).
  3. At \(x = 0\): \(y = 0\), so \(y' = 1 + 0 = 1\), \(y'' = 2(0)(1) = 0\), \(y''' = 2(1)^{2} + 0 = 2\).
  4. \[\tan x = 0 + x(1) + \frac{x^{2}}{2!}(0) + \frac{x^{3}}{3!}(2) = x + \frac{x^{3}}{3}.\]
  5. Check: \(\tan 0.1 = 0.100\,335\) and \(0.1 + \tfrac{0.001}{3} = 0.100\,333\).
Answer: \(\tan x = x + \dfrac{x^{3}}{3} + \dots\)
Worked example e to the sin x
Find the Maclaurin series for \(e^{\sin x}\) up to and including the term in \(x^{3}\).
  1. Let \(y = e^{\sin x}\). Then \(y' = y\cos x\).
  2. \(y'' = y'\cos x - y\sin x\) and \(y''' = y''\cos x - y'\sin x - y'\sin x - y\cos x = y''\cos x - 2y'\sin x - y\cos x\).
  3. At \(x = 0\) (\(\cos 0 = 1\), \(\sin 0 = 0\)): \(y = 1\), \(y' = 1\), \(y'' = 1 - 0 = 1\), \(y''' = 1 - 0 - 1 = 0\).
  4. \[e^{\sin x} = 1 + x + \frac{x^{2}}{2!}(1) + \frac{x^{3}}{3!}(0) = 1 + x + \frac{x^{2}}{2}.\]
  5. The \(x^{3}\) coefficient really is zero; say so, rather than leaving the reader to wonder whether you stopped early.
Answer: \(e^{\sin x} = 1 + x + \dfrac{x^{2}}{2} + 0x^{3} + \dots\)
Your turn
Find the Maclaurin series for \(\ln(\cos x)\) up to and including the term in \(x^{4}\).
Examiner trap

Do not substitute \(x = 0\) BEFORE differentiating. \(y'' = 2y\,y'\) is a function you can still differentiate; the number \(0\) is not. Write the general derivative first, differentiate it, and only then plug in \(0\). And keep \(y\) itself in the product rule: from \(y' = y\cos x\), the next derivative has TWO terms.

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In the Maclaurin series of \(f(x)\), what is the coefficient of \(x^{3}\)?

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What is the series for \(e^{-2x}\) up to the term in \(x^{2}\)?

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For which values of \(x\) is the Maclaurin series for \(\ln(1 + x)\) valid?

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For \(y = \tan x\), \(y''' = 2(y')^{2} + 2y\,y''\), and at \(x = 0\) this equals 2. What is the coefficient of \(x^{3}\) in the series for \(\tan x\)?

Section 03

Combining series and using them.

Once you know the standard series, many others come without differentiating at all. Replace \(x\) by \(2x\), \(-x\) or \(x^{2}\) in a standard series; multiply two series, keeping every product up to the power you need; add or subtract logs. Then use the polynomial: put in a small \(x\) to approximate a value, or replace each function in a limit by its series and cancel the powers of \(x\).

Worked example A product of two series
Find the Maclaurin series for \(e^{x}\cos x\) up to and including the term in \(x^{3}\).
  1. \(e^{x} = 1 + x + \dfrac{x^{2}}{2} + \dfrac{x^{3}}{6} + \dots\) and \(\cos x = 1 - \dfrac{x^{2}}{2} + \dots\) (the next term of \(\cos x\) is in \(x^{4}\), too high).
  2. \[e^{x}\cos x = \left(1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6}\right)\left(1 - \frac{x^{2}}{2}\right)\]
  3. Constant: \(1\). \(x\): \(x\). \(x^{2}\): \(\dfrac{x^{2}}{2} - \dfrac{x^{2}}{2} = 0\). \(x^{3}\): \(\dfrac{x^{3}}{6} + x\left(-\dfrac{x^{2}}{2}\right) = \dfrac{x^{3}}{6} - \dfrac{x^{3}}{2} = -\dfrac{x^{3}}{3}\).
  4. Check by the formula: \(f = e^{x}\cos x\) gives \(f'(0) = 1\), \(f''(0) = 0\), \(f'''(0) = -2\), and \(\dfrac{-2}{3!} = -\dfrac{1}{3}\). The two methods agree.
Answer: \(e^{x}\cos x = 1 + x - \dfrac{x^{3}}{3} + \dots\)
Worked example A fast series for ln 2
(a) Show that \(\ln\!\left(\dfrac{1 + x}{1 - x}\right) = 2x + \dfrac{2x^{3}}{3} + \dfrac{2x^{5}}{5} + \dots\) for \(|x| \lt 1\). (b) Hence find an approximation to \(\ln 2\), to 4 decimal places.
  1. (a) \(\ln\!\left(\dfrac{1 + x}{1 - x}\right) = \ln(1 + x) - \ln(1 - x)\). Replace \(x\) by \(-x\) in the standard series:
  2. \[\ln(1 - x) = -x - \frac{x^{2}}{2} - \frac{x^{3}}{3} - \frac{x^{4}}{4} - \frac{x^{5}}{5} - \dots\]
  3. Subtracting, the even powers cancel and the odd powers double: \(2x + \dfrac{2x^{3}}{3} + \dfrac{2x^{5}}{5} + \dots\) Both series need \(-1 \lt x \lt 1\).
  4. (b) \(\dfrac{1 + x}{1 - x} = 2\) gives \(x = \tfrac{1}{3}\), inside the range of validity.
  5. \[\ln 2 \approx 2\left(\frac{1}{3} + \frac{1}{81} + \frac{1}{1215}\right) = \frac{842}{1215} = 0.6930\ \text{(4 d.p.)}\]
  6. The true value is \(0.6931\). Compare \(\ln(1 + x)\) at \(x = 1\): four terms give \(1 - \tfrac{1}{2} + \tfrac{1}{3} - \tfrac{1}{4} = 0.583\); that series converges very slowly at the edge of its range.
Answer: (a) shown   (b) \(\ln 2 \approx 0.6930\)
Your turn
Use the series for \(\cos x\) to find \(\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x^{2}}\).
Examiner trap

When multiplying series, do not drop cross terms: in \(e^{x}\cos x\) the \(x^{3}\) term gets a contribution from \(x \times \left(-\dfrac{x^{2}}{2}\right)\) as well as from \(\dfrac{x^{3}}{6} \times 1\). And never use a series outside its range: \(\ln(1 + x)\) at \(x = 2\) gives \(2 - 2 + \tfrac{8}{3} - 4 + \dots\), which never settles on \(\ln 3\).

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Find the Maclaurin series for \(e^{2x}\cos x\) up to and including the term in \(x^{2}\).

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Use Maclaurin series to find \(\displaystyle\lim_{x \to 0} \frac{x - \sin x}{x^{3}}\).

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Find the first two non-zero terms of the Maclaurin series for \(\ln\!\left(\dfrac{1 + x}{1 - x}\right)\).

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Find the Maclaurin series for \(\sinh^{-1} x\) up to and including the term in \(x^{3}\).

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Six original Cambridge-style questions
\(\cos x\) from scratch, \(\ln(1 + \sin x)\) by successive differentiation, \(\ln(1 + 2x)\) with its range of validity, a product of two standard series, a limit, and a series for \(\ln(1 + e^{x})\) from a derivative relation.