Difficulty: Intermediate to Advanced | Prerequisites: Derivatives, integrals, geometric series, ratio test
Taylor series and polynomials are the bridge between calculus and approximation. They let you replace complicated functions with polynomials that behave identically near a chosen point, and they give you a way to bound exactly how much error that replacement introduces. This topic typically carries 20+ points on the MATH 126 final (Problems 7 and 8 on this exam), so it is worth investing serious revision time.
A Taylor polynomial approximates a function near a base point using derivatives. The error is bounded by the Lagrange remainder. A Taylor series is the infinite version, valid on its interval of convergence. You can build new series from known ones (eˣ, sin x, cos x, 1/(1 − x)) by substitution, differentiation, or integration.
Taylor polynomial (Tₙ(x))
The degree-n polynomial that matches a function f and its first n derivatives at a base point b. Formula: Tₙ(x) = Σₖ₌₀ⁿ f⁽ᵏ⁾(b)/k! · (x − b)ᵏ. Think of it as the best polynomial approximation of f near the point b.
Taylor series
The infinite version of the Taylor polynomial: Σₖ₌₀^∞ f⁽ᵏ⁾(b)/k! · (x − b)ᵏ. When it converges, it equals f(x) exactly.
Base point (b) / centre
The point around which the Taylor expansion is built. The approximation is most accurate near b and gets worse as you move away. When b = 0, the series is also called a Maclaurin series.
Interval of convergence
The set of x-values for which the Taylor series converges. Found using the ratio test or root test. Always symmetric about the base point (possibly the entire real line).
Lagrange remainder / error bound
The error from using Tₙ(x) instead of f(x) is |Rₙ(x)| ≤ M|x − b|ⁿ⁺¹/(n + 1)!, where M is an upper bound on |f⁽ⁿ⁺¹⁾(c)| for c between b and x. In simple terms, you find the worst-case value of the next derivative and plug in.
Known basic series
The four series you should have memorised: 1/(1 − x) = Σ xᵏ, eˣ = Σ xᵏ/k!, sin x = Σ (−1)ᵏx²ᵏ⁺¹/(2k + 1)!, cos x = Σ (−1)ᵏx²ᵏ/(2k)!. All centred at 0.
Compute f(b), f'(b), f''(b), ... up to the required degree.
Assemble: Tₙ(x) = f(b) + f'(b)(x − b) + f''(b)(x − b)²/2! + ... + f⁽ⁿ⁾(b)(x − b)ⁿ/n!
Worked example: f(x) = ln(2x − 1), base point b = 1, degree 2. f(1) = ln(1) = 0. f'(x) = 2/(2x − 1), so f'(1) = 2. f''(x) = −4/(2x − 1)², so f''(1) = −4. T₂(x) = 0 + 2(x − 1) + (−4)(x − 1)²/2 = 2(x − 1) − 2(x − 1)².
Identify which x-value to plug in. Often the problem gives a nearby value of the original function to approximate.
Substitute into Tₙ(x).
Worked example: Approximate ln(1.1) using T₂(x) = 2(x − 1) − 2(x − 1)² for f(x) = ln(2x − 1). We need ln(1.1), so set 2x − 1 = 1.1, giving x = 1.05. T₂(1.05) = 2(0.05) − 2(0.05)² = 0.1 − 0.005 = 0.095.
Identify n (the degree of the polynomial used) and compute the (n + 1)th derivative.
Find M, an upper bound on |f⁽ⁿ⁺¹⁾(c)| for c between b and x. Choose the value of c that maximises the derivative.
Compute |Rₙ(x)| ≤ M|x − b|ⁿ⁺¹/(n + 1)!
Worked example: For f(x) = ln(2x − 1), n = 2, b = 1, x = 1.05. f'''(x) = 16/(2x − 1)³. For c between 1 and 1.05, |f'''(c)| is maximised at c = 1 (since the denominator is smallest there). M = 16/1³ = 16. Error ≤ 16 · |0.05|³ / 3! = 16 · 0.000125 / 6 = 0.002/6 = 1/3000.
Start with a known series (eˣ, sin x, cos x, 1/(1 − x)).
Substitute: replace x with the expression in the new function.
Integrate or differentiate term by term as needed.
Worked example: f(x) = ∫₀ˣ e^(2t²) dt. Start with eᵘ = Σₖ₌₀^∞ uᵏ/k!. Replace u with 2t²: e^(2t²) = Σₖ₌₀^∞ (2t²)ᵏ/k! = Σₖ₌₀^∞ 2ᵏt²ᵏ/k!. Integrate term by term from 0 to x: f(x) = Σₖ₌₀^∞ 2ᵏx²ᵏ⁺¹/((2k + 1) · k!). The interval of convergence is (−∞, ∞) since eᵘ converges everywhere.
The coefficient of (x − b)ⁿ in the Taylor series equals f⁽ⁿ⁾(b)/n!.
So f⁽ⁿ⁾(b) = n! times the coefficient of (x − b)ⁿ.
If the series has only certain powers of x (e.g. only odd powers), then derivatives at orders corresponding to missing powers are zero.
Worked example: Find f⁽²⁰²³⁾(0) for f(x) = ∫₀ˣ e^(2t²) dt = Σₖ₌₀^∞ 2ᵏx²ᵏ⁺¹/((2k + 1) · k!). We need the coefficient of x²⁰²³, so 2k + 1 = 2023, giving k = 1011. The coefficient is 2¹⁰¹¹/((2023) · 1011!). Then f⁽²⁰²³⁾(0) = 2023! · 2¹⁰¹¹/(2023 · 1011!) = 2022! · 2¹⁰¹¹/1011!.
Taylor polynomial of degree n at base b:
Tₙ(x) = Σₖ₌₀ⁿ f⁽ᵏ⁾(b)/k! · (x − b)ᵏ
Lagrange error bound:
|Rₙ(x)| ≤ M · |x − b|ⁿ⁺¹ / (n + 1)! where M = max |f⁽ⁿ⁺¹⁾(c)| for c between b and x
Four basic Maclaurin series:
1/(1 − x) = Σ xᵏ, |x| < 1
eˣ = Σ xᵏ/k!, all x
sin x = Σ (−1)ᵏx²ᵏ⁺¹/(2k + 1)!, all x
cos x = Σ (−1)ᵏx²ᵏ/(2k)!, all x
Extracting derivatives from series:
f⁽ⁿ⁾(b) = n! · [coefficient of (x − b)ⁿ in the Taylor series]
Taylor approximations are how calculators and computers evaluate functions like sin, cos, eˣ, and ln. Your calculator does not "know" what sin(0.3) is; it evaluates a polynomial approximation and bounds the error to ensure the displayed digits are correct. In physics, small-angle approximations (sin θ ≈ θ) are first-order Taylor polynomials.
Students often confuse the base point b with the value of x they are approximating at. The base point is where you know the function exactly; x is the point where you want the approximation.
For the error bound, M must be an upper bound on the (n + 1)th derivative on the entire interval from b to x, not just at b.
The interval of convergence of a derived series (obtained by substitution or integration) is determined by the substitution, not just copied from the original.
When finding f⁽ⁿ⁾(0) from a series, students forget to multiply the coefficient by n!. The coefficient of xⁿ is f⁽ⁿ⁾(0)/n!, not f⁽ⁿ⁾(0).
⚠️ Problems 7 and 8 on this exam are worth 28 points combined (28% of the exam). Taylor series is the single highest-value topic.
⚠️ The "find f⁽ⁿ⁾(0)" trick (reading derivatives off series coefficients) is a classic exam question. Practise it until it is automatic.
⚠️ For error bounds, the exam says "as sharp as possible." This means choosing M carefully, not just picking a huge overestimate.
Fill in the blank: The Maclaurin series for eˣ is Σ ______. (xᵏ/k!)
True or false: The Taylor series for sin x centred at 0 contains only odd powers of x. (True.)
Fill in the blank: f⁽ⁿ⁾(0) = ______ times the coefficient of xⁿ in the Maclaurin series. (n!)
True or false: The interval of convergence of eˣ is (−1, 1). (False. It is (−∞, ∞).)
Q: Find the second Taylor polynomial T₂(x) for f(x) = ln(2x − 1) based at b = 1.
A: f(1) = 0, f'(1) = 2, f''(1) = −4. T₂(x) = 2(x − 1) − 2(x − 1)².
Q: Use T₂(x) to approximate ln(1.1).
A: Set 2x − 1 = 1.1, so x = 1.05. T₂(1.05) = 2(0.05) − 2(0.0025) = 0.1 − 0.005 = 0.095.
Q: Find an upper bound on the error of this approximation.
A: f'''(x) = 16/(2x − 1)³. On [1, 1.05], max is at x = 1: M = 16. Error ≤ 16(0.05)³/6 = 1/3000 ≈ 0.000333.
Q: Find the Taylor series for f(x) = ∫₀ˣ e^(2t²) dt based at b = 0, in Σ-notation.
A: f(x) = Σₖ₌₀^∞ 2ᵏx²ᵏ⁺¹ / ((2k + 1) · k!).
Q: Find the open interval of convergence for this series.
A: (−∞, ∞), since eᵘ converges for all u.
Q: Find f⁽²⁰²³⁾(0) for f(x) = ∫₀ˣ e^(2t²) dt.
A: The x²⁰²³ term has k = 1011. Coefficient = 2¹⁰¹¹/(2023 · 1011!). So f⁽²⁰²³⁾(0) = 2023! · 2¹⁰¹¹/(2023 · 1011!) = 2022! · 2¹⁰¹¹/1011!.
Taylor series connect to the convergence tests from earlier in MATH 126 (ratio test, root test). The error bound is an application of the mean value theorem from Calculus I. In more advanced courses (MATH 324, physics, engineering), Taylor expansions are used to linearise nonlinear equations, analyse stability of equilibria, and derive numerical methods.
Taylor polynomial, Taylor series, Maclaurin series, interval of convergence, radius of convergence, Lagrange remainder, error bound, Taylor's theorem, power series, ratio test, known series, substitution into series, term-by-term integration, extracting derivatives, MATH 126, Calculus III, UW Math 126