Maths › Further Pure 1 › Taylor series
Taylor series
Maclaurin series anchored at zero; Taylor series anchor anywhere. Move the anchor to where the action is and the same factorial machinery approximates any well-behaved function near any point.
Builds on Maclaurin series and The product, quotient and chain rules.
IN THIS TOPIC
- Quote the Taylor series of f about x = a in powers of (x − a).
- Compute the derivative values at the anchor and assemble the series.
- Choose the anchor closest to the point being approximated.
WHAT YOU PROBABLY THINK
Series expansions only work near zero; away from the origin a function cannot be turned into a polynomial.
Moving the anchor
A Maclaurin series matches every derivative of f at 0. The Taylor series plays the same game at any anchor x = a:
The coefficient of (x − a)r is the rth derivative at a over r!. Setting a = 0 recovers Maclaurin exactly, so nothing about the opening claim survives: the origin was never special, only convenient.
WORKED EXAMPLE
sin x in powers of (x − π)
Expand sin x in ascending powers of (x − π), up to the cubic term.
Derivatives at π: sin π = 0, cos π = −1, −sin π = 0, −cos π = 1.
So sin x = −(x − π) + (x − π)³/3! + … = −(x − π) + (x − π)³/6 + …
Check at x = π + 0.3: the series gives −0.3 + 0.0045 = −0.2955, and sin(π + 0.3) = −0.2955 to four decimal places.
Choosing where to stand
The series is most accurate near its anchor, so anchor where the derivatives are known exactly and the target point is close. To approximate cos of an angle near 60°, expand about π/3, where cos and sin take exact values; expanding about 0 would need many more terms for the same accuracy.
WORKED EXAMPLE
cos x about π/3
Expand cos x about x = π/3, up to the term in (x − π/3)².
Values at π/3: cos = 1/2, derivative −sin = −√3/2, second derivative −cos = −1/2.
cos x = 1/2 − (√3/2)(x − π/3) − (1/4)(x − π/3)² + …
At x = π/3 + 0.2 the three terms give 0.3170, against the true 0.3180: two decimal places from a quadratic, because the anchor sat close.
TRY IT UNSEEN
A square root without a calculator
Expand √x about x = 4 up to the (x − 4)² term, and use it to estimate √4.4.
Show the working
Derivatives at 4: √4 = 2, then 1/(2√x) = 1/4, then −1/(4x^(3/2)) = −1/32.
√x = 2 + (x − 4)/4 − (x − 4)²/64 + …
At x = 4.4: 2 + 0.1 − 0.0025 = 2.0975, against the true 2.0976. The anchor at the nearest perfect square did the heavy lifting.
THE EXAM BIT
- Set out a derivative table at the anchor before assembling anything; most marks live there.
- Write the expansion in powers of (x − a) as asked; expanding brackets back into powers of x undoes the point.
- The factorials divide the derivative values; forgetting 2! on the quadratic term is the standard slip.
- State the anchor explicitly: 'about x = a' is part of a complete answer.
CHECK YOURSELF
Write down the first three terms of the Taylor series of ex about x = 2.
Show a hint
Every derivative of the exponential at 2 is e².
Show the answer
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Taylor about a: coefficient of (x − a)r is the rth derivative at a over r factorial.
Anchor where derivatives are exact and the target is near; Maclaurin is the a = 0 special case.
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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- Quote the Taylor series of f about x = a in powers of (x − a).
- Compute the derivative values at the anchor and assemble the series.
- Choose the anchor closest to the point being approximated.
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.