MathsFurther algebra and series › Maclaurin series

Maclaurin series

Any well-behaved function is a polynomial in the making: match every derivative at zero and the series writes itself, term by factorial term.

Year FMEDEXCEL 9FM0 CP2

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 series for ex, sin x, cos x and ln(1 + x), with validity for the last.
  • Adapt standard series to composites such as e2x or ln(1 + 2x).

WHAT YOU PROBABLY THINK

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: 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 + …

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, with alternating signs: 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; the true value is 0.479426.

Three terms of a polynomial pinned the sine of half a radian to five decimal places, which retires the opening claim.

Standard series and their reach

The exam booklet lists ex, sin x, cos x and ln(1 + x). The first three converge for every x; the logarithm is the delicate one:

ln(1 + x) = x - x22 + x33 - x44 + … (-1 < x ≤ 1)

Outside that window the terms grow instead of shrinking and the series says nothing. Composites come from substitution: replace x by 2x throughout to get ln(1 + 2x), valid now for −1/2 < x ≤ 1/2, since the substituted quantity must 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.

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; the true e^0.2 is 1.22140. Already three decimal places from a cubic.

TRY IT UNSEEN

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 buys much less accuracy.

THE EXAM BIT

  • Derive means differentiate: show f(0), f'(0), f''(0) explicitly before assembling the series.
  • Quote validity for ln(1 + x) whenever it appears; the other standard series need no caveat.
  • After substituting into ln'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

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

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

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.

CHECK YOUR PROGRESS

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  • Derive a Maclaurin series from repeated differentiation at zero.
  • Quote the series for ex, sin x, cos x and ln(1 + x), with validity for the last.
  • Adapt standard series to composites such as e2x or ln(1 + 2x).

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

No animated video for this topic yet; these notes stand alone.