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Maclaurin series

Every function on the booklet's series page is a polynomial in the making. Match each derivative at zero and the series writes itself, term by factorial term.

Year FMEDEXCEL 9FM0 CP2 4.5, 4.6

Builds on The product, quotient and chain rules and The general binomial expansion.

IN THIS TOPIC

  • Derive a Maclaurin series from repeated differentiation at zero.
  • Quote the standard series and state where ln(1 + x) and arctan x are valid.
  • Adapt standard series to composites, products and quotients.

COMMON MISCONCEPTION

A curved function like sin x cannot be approximated by polynomials to any accuracy you like.

Matching derivatives at zero

Suppose f(x) = a₀ + a₁x + a₂x² + a₃x³ + … and keep differentiating, setting x = 0 each time. Each substitution isolates one coefficient, so a₀ = f(0), a₁ = f'(0), and in general r! ar = f(r)(0). That forces the Maclaurin series:

f(x)=f(0)+f′(0)x+f′′(0)2!x2+…\text{f}(x) = \text{f}(0) + \text{f}'(0)x + \frac{\text{f}''(0)}{2!}x^{2} + …IN THE FORMULAE BOOKLET

That general form heads the booklet's Maclaurin page, so read it off rather than reconstructing it under pressure. For ex every derivative at 0 is 1, giving 1 + x + x²/2! + x³/3! + …. For sin x the derivatives cycle through 0, 1, 0, −1, so only odd powers survive and the signs alternate, leaving x − x³/3! + x⁵/5! − ….

sin x with its Maclaurin approximations: the line y = x holds near zero, the cubic x − x³/6 holds much furtherπy = xsin xx − x³/6
FIG. 1sin x against its first two Maclaurin approximations: the line y = x hugs the curve near zero and x − x³/6 clings on much further.

WORKED EXAMPLE

How good is three terms?

Estimate sin 0.5 using x − x³/3! + x⁵/5!.

0.5 − 0.125/6 + 0.03125/120 = 0.5 − 0.02083… + 0.00026…

That is 0.479427 to 6 decimal places, and the true value is 0.479426.

Three terms of a polynomial pinned the sine of half a radian to five decimal places.

One distinction is worth keeping straight. Cutting the series after a few terms gives a polynomial approximation, and any function with enough derivatives at zero has one. The infinite series is a stronger claim: its sum returns f(x) only where the function is analytic near the point. Each of the booklet's functions is, and the range printed beside its series is where that promise holds.

Standard series and their reach

The same booklet page gives you ex, sin x, cos x, ln(1 + x) and arctan x, and (1 + x)n is printed separately under Binomial series. ex, sin x and cos x converge for every x; the other two come with printed intervals, and the interval is part of the result. The logarithm is the delicate one:

ln(1+x)=x-x22+x33-x44+…(-1<x≤1)\text{ln}(1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + … (-1 < x ≤ 1)IN THE FORMULAE BOOKLET

For |x| > 1 the terms grow instead of shrinking and the series diverges, so it gives no information about the function there. The excluded endpoint is subtler: at x = −1 the terms shrink like 1/n, yet their sum −1 − 1/2 − 1/3 − … still runs off without limit, which is why the window is open at that end. Composites come from substitution. Replace x by 2x throughout to get ln(1 + 2x), now valid for −1/2 < x ≤ 1/2, because the substituted quantity has to stay inside the original window.

Where the ln series works: the window (−1, 1], and the halved window that 2x drags it to-2-1012-2-1012ln(1 + x) valid for −1 < x ≤ 1ln(1 + 2x) valid for −½ < x ≤ ½open end excluded, closed end included
FIG. 2The validity window for ln(1 + x): open at −1, closed at 1. Substituting 2x for x squeezes the window to half the width.

arctan x carries a window of its own. The booklet prints its series, x − x³/3 + x⁵/5 − …, for −1 ≤ x ≤ 1, closed at both ends. The function itself is defined for every real x, but beyond that window the terms grow and the series diverges, so quote the interval alongside the series just as you would for the logarithm.

WORKED EXAMPLE

A composite series

Find the series for e2x up to the x³ term.

Substitute 2x into 1 + x + x²/2! + x³/3!.

e2x = 1 + 2x + 4x²/2 + 8x³/6 = 1 + 2x + 2x² + (4/3)x³ + …

At x = 0.1 the four terms give 1.22133, and the true e0.2 is 1.22140. Already three decimal places from a cubic.

INDEPENDENT PRACTICE

A logarithm, term by term

Write down the series for ln(1 + x) up to x⁴ and use it to estimate ln 1.5. Comment on the accuracy.

Show the working

With x = 0.5: 0.5 − 0.125 + 0.041667 − 0.015625 = 0.401.

The true value is 0.405 to 3 decimal places, so four terms give barely two.

The logarithm converges slowly compared with ex or sin x. Near the edge of its window, each extra term adds much less accuracy.

When nothing standard fits

Two moves cover almost everything else. Multiply series together and collect powers, which handles things like ex sin x = (1 + x + x²/2 + …)(x − x³/6 + …) = x + x² + x³/3 + …. Or go back to first principles and differentiate repeatedly, simplifying each derivative before substituting x = 0, since an unsimplified third derivative is where the arithmetic usually collapses.

WORKED EXAMPLE

Repeated differentiation on an awkward function

Find the Maclaurin series for ln(1 + sin x) up to the x³ term.

f(0) = ln 1 = 0.

f'(x) = cos x/(1 + sin x), so f'(0) = 1.

Simplify before going further: f''(x) = −1/(1 + sin x), giving f''(0) = −1.

f'''(x) = cos x/(1 + sin x)², so f'''(0) = 1.

Assembling: ln(1 + sin x) = x − x²/2 + x³/6 − …

That second derivative collapsed from a quotient-rule mess to a single fraction, and the third derivative was then a one-line chain rule. Simplify at every stage, not at the end.

ASSESSMENT FOCUS

  • 'Derive' means differentiate. Show f(0), f'(0) and f''(0) explicitly before assembling anything.
  • Quote each printed interval of validity with its series: ex, sin x and cos x hold for every x, ln(1 + x) needs −1 < x ≤ 1, and arctan x needs −1 ≤ x ≤ 1.
  • After substituting into ln's or arctan's series, transform the validity window with the same substitution.
  • Keep factorials unevaluated until the last line. 8/3! is easier to check than 1.333….

CHECK YOURSELF

Using the series for ex, write down the series for e−x up to the x³ term, and hence the series for ½(ex − e−x).

Show a hint

Replace x by −x, then subtract term by term: even powers cancel.

Show the answer

e−x = 1 − x + x²/2 − x³/6 + …. Subtracting and halving removes the even powers, leaving ½(ex − e−x) = x + x³/6 + …, odd powers only, all signs positive.

Maclaurin: the coefficient of xr is the rth derivative at zero over r factorial.

Substitute into standard series for composites, and carry ln's and arctan's windows through the substitution.

Simplify each derivative before evaluating it at zero.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

PDF · 53 KB · print edition · white background

Question workbook

PDF · 53 KB · screen edition · navy

Answer book

PDF · 57 KB · fully worked · navy
7 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the maclaurin series questions page.

CHECK YOUR PROGRESS

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  • Derive a Maclaurin series from repeated differentiation at zero.
  • Quote the standard series and state where ln(1 + x) and arctan x are valid.
  • Adapt standard series to composites, products and quotients.

Open the full revision checklist to see every objective in the course in one place.