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Taylor series
A Taylor polynomial about x = a uses derivatives evaluated at a and powers of x − a. Use Taylor's theorem or a stated remainder bound to justify accuracy. Do not assume that an infinite Taylor series equals the original function without a convergence argument.
Builds on Maclaurin series and The product, quotient and chain rules.
IN THIS TOPIC
- Quote the Taylor polynomial of f about x = a in powers of (x − a), and assemble it from a derivative table.
- Expand about a point close to the value being approximated, and say why the choice reduces the error.
- State the remainder condition that justifies an accuracy claim, and distinguish a polynomial from an infinite series.
COMMON MISCONCEPTION
Series expansions only work near zero; away from the origin a function cannot be turned into a polynomial.
A Taylor polynomial can be centred anywhere the derivatives exist. About x = a it runs in powers of (x − a), with the coefficient of (x − a)r the rth derivative at a divided by r factorial, and a = 0 is the Maclaurin special case rather than the general one. Expanding about a point close to the value wanted keeps |x − a| small, and the remainder carries it to the power n + 1.
Expanding about a general point
A Maclaurin polynomial matches the derivatives of f at 0. A Taylor polynomial of degree n about x = a matches them at a instead, using powers of x − a.
The coefficient of (x − a)r is the rth derivative at a divided by r factorial. Setting a = 0 recovers the Maclaurin polynomial as a special case. Check the current formula booklet for your awarding organisation; whether or not the general form is supplied, you must know when it applies and how to derive it.
Taylor's theorem gives a remainder such as f(n+1)(ξ)(x − a)n+1/(n + 1)! under its usual hypotheses. A smaller |x − a| can reduce the error when the derivative bound is comparable. An infinite Taylor series represents f only where it converges to f; smoothness alone is not sufficient.
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 the expansion point
Expand about a point where the derivatives are known exactly and which lies close to the value being approximated. 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 more terms for the same accuracy, because |x − a| appears in the remainder to the power n + 1.
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 |x − a| = 0.2 and the third derivative is bounded by 1, so the remainder is at most 0.2³/3! = 0.0013.
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. Expanding about the nearest perfect square keeps |x − a| small, which is what keeps the remainder small.
ASSESSMENT FOCUS
- Set out a derivative table at the expansion point before assembling anything, and show the values it produces.
- Leave the expansion in powers of (x − a) as asked; multiplying the brackets out returns an ordinary polynomial in x.
- The factorials divide the derivative values, so the quadratic term carries a 2! and the cubic term a 3!.
- Justify an accuracy claim from the remainder, not from the function being smooth.
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, and a = 0 gives the Maclaurin case.
Expand about a point where the derivatives are exact and which is close to the target value; the remainder carries (x − a) to the power n + 1.
An infinite Taylor series represents f only where it converges to f; smoothness alone is not sufficient.
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.
USED LATER IN
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- Quote the Taylor polynomial of f about x = a in powers of (x − a), and assemble it from a derivative table.
- Expand about a point close to the value being approximated, and say why the choice reduces the error.
- State the remainder condition that justifies an accuracy claim, and distinguish a polynomial from an infinite series.
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