Maths › Further 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.
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:
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! − ….
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:
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.
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.