Taylor & Maclaurin polynomials
The Taylor polynomial of degree n about a is Σ_{k=0}^{n} f^(k)(a)(x−a)^k/k!, matching f in value and the first n derivatives at a. With a = 0 it is called Maclaurin. Four standard expansions are worth memorising — eˣ, sin x, cos x and 1/(1−x) — because most exam series are built from them by substitution.
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Method: how to approach it
The order below is what examiners expect to see, and each step carries its own marks.
- Compute derivatives at the centreEvaluate f(a), f′(a), f″(a), … and look for the pattern before writing the general term.
- Assemble with the factorialsThe kth term is f^(k)(a)(x−a)^k/k!. Dropping the k! is the classic error.
- Reuse standard seriesFor e^(−x²), substitute −x² into the eˣ series rather than differentiating repeatedly.
- Bound the error with the remainderLagrange form: |Rₙ| ≤ M|x−a|^(n+1)/(n+1)! where M bounds the (n+1)th derivative.
Worked example
Find the Maclaurin polynomial of degree 3 for f(x) = eˣ and estimate e^0.1.
- Every derivative of eˣ is eˣ, so f^(k)(0) = 1 for all k.
- P₃(x) = 1 + x + x²/2! + x³/3! = 1 + x + x²/2 + x³/6.
- At x = 0.1: 1 + 0.1 + 0.005 + 0.000167.
- Sum: 1.105167.
Answer. P₃(x) = 1 + x + x²/2 + x³/6, giving e^0.1 ≈ 1.10517 — correct to five decimal places.
Where marks get dropped
These are the specific errors that cost credit on taylor & maclaurin polynomials questions — QED's rubric penalises each of them separately.
- Omitting the k! denominators, which makes every term after the linear one wrong.
- Expanding about 0 when the question specifies a different centre. The powers must be (x − a)^k.
- Using a series outside its interval of convergence. 1/(1−x) requires |x| < 1, and substituting x = 2 gives nonsense.
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Taylor & Maclaurin polynomials — frequently asked questions
Which standard series should I know?
eˣ = Σxᵏ/k!; sin x = x − x³/3! + x⁵/5! − …; cos x = 1 − x²/2! + x⁴/4! − …; 1/(1−x) = Σxᵏ for |x| < 1. Everything else is built from these.
How accurate is a Taylor polynomial?
The Lagrange remainder gives |Rₙ| ≤ M|x−a|^(n+1)/(n+1)!, so accuracy improves rapidly with degree and degrades with distance from the centre.
Why centre at a point other than 0?
To approximate near that point. Expanding ln x about 1 works; about 0 it is impossible, since ln is undefined there.
The rest of Calculus
Limits, derivatives, integrals and their applications. Each subtopic below has its own method, worked example and mark-losing traps.
- 1Limits & continuity
- 2Product, quotient & chain rules
- 3Tangent lines & linear approximation
- 4Extrema & optimisation
- 5Indefinite & definite integrals
- 6Substitution & integration by parts
- 7Series & convergence basics
- 8Implicit differentiation & related rates
- 9Mean value & intermediate value theorems
- 10Improper integrals & convergence
- 11Taylor & Maclaurin polynomials
- 12L’Hôpital’s rule & indeterminate forms
- 13Curve sketching: asymptotes, concavity & inflection
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