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Taylor series

Edexcel IAL Further Mathematics · WFM02 Further Pure Mathematics 2

A Taylor series expands a function about a point a other than 0, in powers of (x - a). It is the tool for functions with no Maclaurin series, such as ln x, and for accurate values near x = a.

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The Taylor series

f(x) = f(a) + f'(a)(x - a) + f''(a)(x - a)2/2! + f'''(a)(x - a)3/3! + ...

An equivalent form, in the formula booklet: f(x + a) = f(a) + f'(a)x + f''(a)x2/2! + ....

The Maclaurin series is the case a = 0.

Method

1. Find the derivatives you need, as functions of x.

2. Evaluate them all at x = a.

3. Quote the series and substitute.

Leave the answer in powers of (x - a), and remember the factorials: the June 2025 examiners' report notes some students using series with the factorial denominators missing.

Choosing a

The series is most accurate near x = a, so expand about a point close to where you want a value and where f(a) is easy: for √4.2 expand about 4, for sin 3 expand about π.

Example: sin x about π

f(π) = 0, f'(π) = cos π = -1, f''(π) = -sin π = 0, f'''(π) = -cos π = 1.

So sin x = -(x - π) + (x - π)3/3! - ....

Worked example

Find the Taylor series for ln x about x = 1, up to the term in (x - 1)3.

  1. f(1) = 0; f'(x) = 1/x, f'(1) = 1; f''(x) = -1/x2, f''(1) = -1; f'''(x) = 2/x3, f'''(1) = 2.
  2. ln x = 0 + (x - 1) - (x - 1)2/2! + 2(x - 1)3/3!.

Answer: ln x = (x - 1) - (x - 1)2/2 + (x - 1)3/3 - ...

Key idea: f(x) = Σ fr(a)(x - a)r/r!. Evaluate the derivatives at a and keep the answer in powers of (x - a).

Check yourself

  1. Find the coefficient of (x - 1)2 in the Taylor series for ln x about x = 1. Give your answer as a fraction in its simplest form.
    Show answer

    -1/2. f''(1) = -1, so the coefficient is -1/2! = -1/2.

  2. Find the coefficient of (x - 2)2 in the Taylor series for ex about x = 2, to 3 decimal places.
    Show answer

    3.695. Every derivative of ex is ex, so f''(2) = e2 and the coefficient is e2/2 = 3.695.

  3. What is the first non-zero term of the Taylor series for cos x about x = π/2?
    • -(x - π/2)
    • (x - π/2)
    • 1
    • -(x - π/2)2/2
    Show answer

    -(x - π/2). cos(π/2) = 0 and f'(π/2) = -sin(π/2) = -1.

  4. Use the Taylor series for √x about x = 4, up to the term in (x - 4)2, to estimate √4.2, to 4 decimal places.
    Show answer

    2.0494. f(4) = 2, f'(4) = 1/4, f''(4) = -1/32. √4.2 ≈ 2 + 0.2/4 - (0.2)2/64 = 2.0494.

  5. Find the coefficient of (x - π/4) in the Taylor series for tan x about x = π/4.
    Show answer

    2. f'(x) = sec2 x, and sec2(π/4) = 1/(1/√2)2 = 2.

  6. Why is sin 3 best estimated with a Taylor series about π rather than the Maclaurin series?
    • The Maclaurin series of sin x is not valid at x = 3
    • sin π is undefined
    • Taylor series always have fewer terms
    • 3 is close to π, so few terms give an accurate value
    Show answer

    3 is close to π, so few terms give an accurate value. Taylor series are most accurate close to the point of expansion; 3 - π ≈ -0.14 is small.

  7. Find the coefficient of (x - 2)3 in the Taylor series for 1/x about x = 2. Give your answer as a fraction in its simplest form.
    Show answer

    -1/16. f'''(x) = -6/x4, so f'''(2) = -6/16 = -3/8. The coefficient is (-3/8)/3! = -1/16.

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Last updated 8 November 2026. Spotted a mistake? Tell us.