Maths › Further Pure 1 › Taylor series
Taylor series
Maclaurin series are anchored at zero. Taylor series anchor anywhere. Move the anchor to the point you care about and the same factorial formula approximates a smooth function near whichever point you choose.
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), and assemble it from a derivative table.
- Choose the anchor closest to the point being approximated.
COMMON MISCONCEPTION
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 booklet's page is headed Maclaurin's and Taylor's Series, but every expansion printed on it is about 0. The (x − a) form is not there. Learn it. The coefficient of (x − a)r is the rth derivative at a divided by r factorial. Setting a = 0 recovers Maclaurin exactly. 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
Accuracy is a local matter. Anchor where the derivatives are known exactly and the target point is close by. To approximate the cosine 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, and the arithmetic would be worse.
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 0.3168, against the true 0.3180. Two decimal places from a quadratic, because the anchor sat close.
INDEPENDENT PRACTICE
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/(4x3/2) = −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.
ASSESSMENT FOCUS
- Set out a derivative table at the anchor before assembling anything. Most of the marks are awarded for that table.
- Leave the expansion in powers of (x − a) as asked. Multiplying the brackets back out undoes the whole point.
- The factorials divide the derivative values, and forgetting 2! on the quadratic term is the standard slip.
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
ex = e² + e²(x − 2) + (e²/2)(x − 2)² + … Every derivative of the exponential is the same function, so e² scales every term.
Taylor about a: the coefficient of (x − a)r is the rth derivative at a over r factorial.
Anchor where the 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.
Or read them with their worked answers on the taylor series questions page.
CHECK YOUR PROGRESS
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- Quote the Taylor series of f about x = a in powers of (x − a), and assemble it from a derivative table.
- Choose the anchor closest to the point being approximated.
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