MAST10006 · Calculus 2
Taylor Series
A Taylor polynomial is the best polynomial that agrees with f at a chosen centre x = a: it shares the same value, slope, curvature and all higher derivatives up to order n there. Push the degree to infinity and — for the standard functions — the polynomial becomes the function, so every transcendental function in the course is secretly a power series. A Maclaurin series is just a Taylor series centred at 0, which keeps the algebra clean. This chapter has three jobs: know the six standard Maclaurin series cold (ex, sin, cos, 1/(1−x), log(1+x), (1+x)α); build new series by substituting, multiplying, differentiating or integrating a known one rather than computing derivatives from scratch; and bound the error of a truncated polynomial with the Lagrange remainder. The off-by-one to watch is that the remainder uses the (n+1)-th derivative and the power |x − a|n+1.
What this chapter covers
- 014.1 The Taylor / Maclaurin polynomial and the AHA overlay
- 02The standard Maclaurin catalogue — six series to memorise
- 03Building new series by substitute / multiply / differentiate / integrate
- 044.3 The Lagrange remainder and the error bound
- 05The bounding recipe (maximise over the whole interval)
Worked example: bound the error of a Taylor approximation
- +1The polynomial: P2(x) = 1 + x + x²/2, so P2(0.2) = 1 + 0.2 + 0.02 = 1.22.
- +1Write the remainder: R2(x) = f′′′(c)/3! · x³ = ec/6 · x³ for some c between 0 and 0.2 (here a = 0, n = 2, so the derivative is the third).
- +1Bound the derivative over the whole interval: since c < 0.2 and e0.2 < 2, take M = 2.
- +1Plug in: |R2(0.2)| ≤ M/3! · |0.2|³ = (2/6)(0.008) ≈ 2.7×10−3. So e0.2 = 1.22 ± 0.0027, correct to two decimals.
Key terms
- Taylor polynomial
- The degree-n polynomial Pn(x) = ∑k=0n f(k)(a)/k! · (x−a)k that matches f and its first n derivatives at the centre a. The k! in each denominator is exactly what makes the derivatives come out right when you differentiate the polynomial back.
- Maclaurin series
- A Taylor series centred at a = 0, so x − a = x. Almost every standard series in the course is a Maclaurin series; the six catalogue entries are the alphabet from which everything else is built.
- Radius of convergence
- The interval of x-values on which a power series actually equals its function. Substitution and multiplication keep the radius; term-by-term differentiation and integration keep the radius too (the endpoints may change). Stay inside it or the series is meaningless.
- Lagrange remainder
- The exact leftover Rn(x) = f(n+1)(c)/(n+1)! · (x−a)n+1 for some c between a and x, with the clean error bound |Rn| ≤ M/(n+1)! · |x−a|n+1 where M maximises the (n+1)-th derivative. The remainder index is n+1, not n.
- Build-by-substitution
- The workflow of producing a new series by substituting, multiplying, differentiating or integrating a catalogue series — e.g. e−x² from eu, or arctan x by integrating 1/(1+x²) — instead of computing derivatives from scratch.
Taylor Series FAQ
Do I have to derive each Maclaurin series from scratch in the exam?
Almost never. The six catalogue series (ex, sin, cos, 1/(1−x), log(1+x), (1+x)α) are expected to be quoted, not re-derived. The exam then asks you to produce a series for a function not in the catalogue, which you do by substituting, multiplying, differentiating or integrating one of the six — far faster and far less error-prone than tabulating derivatives.
What is the difference between a Taylor and a Maclaurin series?
A Maclaurin series is just a Taylor series centred at a = 0. The Taylor series of f about a general centre a uses powers of (x − a); setting a = 0 makes x − a = x and keeps the algebra clean, which is why nearly every standard series you meet is a Maclaurin series.
What is the most common off-by-one error?
The remainder index. If your polynomial's highest term is xn, the error term carries xn+1 and the (n+1)-th derivative — not the n-th. The matching factorial is (n+1)!. Getting either index wrong is the single off-by-one that costs the most marks on remainder questions.
How do I get the series for arctan x without computing 20 derivatives?
Integrate a geometric series. Start from 1/(1−u) = ∑un, substitute u = −x² to get 1/(1+x²) = ∑(−1)nx2n, then integrate term by term (since d/dx arctan x = 1/(1+x²)) to obtain arctan x = x − x³/3 + x&sup5;/5 − …. The constant of integration is 0 because arctan 0 = 0.
Exam move
Memorise the six catalogue series so you can write them instantly, paying attention to where each starts: the sin series begins at x (odd powers, 2n+1), the cos series at 1 (even powers, 2n), and the log(1+x) series at n = 1 with a leading +x. Then drill the four build moves — substitute, multiply, differentiate, integrate — until producing a new series is mechanical; this is what the exam actually rewards. For remainder questions, follow a fixed recipe: fix the degree and centre, write the (n+1)-th derivative, maximise it over the whole interval to get M, then substitute into |Rn| ≤ M/(n+1)! · |x−a|n+1. Carry an A4-line reminder that the remainder index is n+1 and the factorial matches the power — that one fact rescues most lost marks here.
