Source: Math 126 Spring 2023 Final Exam, University of Washington
Tags: Taylor polynomial, Taylor series, Maclaurin series, error bound, Taylor remainder, interval of convergence, nth derivative from series, MATH 126, Calculus III
Difficulty: Intermediate to Advanced | Prerequisites: Derivatives, integration, geometric series, ratio test
Taylor series let you replace complicated functions with polynomials that approximate them near a chosen base point. This is how calculators evaluate sin, cos, eˣ, and ln internally. The error bound tells you how good the approximation is. Extracting the nth derivative from a series representation is a classic exam technique that avoids computing derivatives directly. If you are comfortable with basic series convergence tests, you are ready.
Build Taylor polynomials from derivatives at the base point. Use known series (eˣ, sin x, cos x, 1/(1−x)) as building blocks for more complex functions. The Taylor remainder theorem gives error bounds. The interval of convergence tells you where the series is valid.
Taylor polynomial (Tₙ(x))
A polynomial of degree n that approximates f(x) near a base point b. Built from the values of f, f′, f″, ... , f⁽ⁿ⁾ at b. Think of it as the best polynomial fit to the function at that point.
Taylor series
The infinite series you get by letting n go to infinity in the Taylor polynomial. When it converges, it equals the original function.
Maclaurin series
A Taylor series with base point b = 0. The standard series for eˣ, sin x, cos x, and 1/(1−x) are all Maclaurin series.
Taylor remainder (error bound)
The difference |f(x) − Tₙ(x)|, bounded by M|x − b|ⁿ⁺¹ / (n+1)!, where M is the maximum of |f⁽ⁿ⁺¹⁾| on the interval between b and x. In simple terms, it tells you the worst-case error of your polynomial approximation.
Interval of convergence
The set of x values where a Taylor series converges. For eˣ, sin x, and cos x, the interval is (−∞, ∞). For 1/(1−x), it is (−1, 1). Substitutions and compositions do not change the radius of convergence when the substitution is linear.
The nth Taylor polynomial of f at b is: Tₙ(x) = f(b) + f′(b)(x−b) + f″(b)(x−b)²/2! + ... + f⁽ⁿ⁾(b)(x−b)ⁿ/n!
Example: f(x) = ln(2x−1), base point b = 1.
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.
To approximate a value like ln(1.1), rewrite it in terms of f at the base point.
ln(1.1) = ln(2(1.05)−1) = f(1.05).
T₂(1.05) = 2(0.05) − 2(0.05)² = 0.1 − 0.005 = 0.095.
|f(x) − Tₙ(x)| ≤ M|x−b|ⁿ⁺¹ / (n+1)!, where M = max of |f⁽ⁿ⁺¹⁾| on the interval.
Example: bounding the error of T₂(1.05) for f(x) = ln(2x−1).
f‴(x) = 16/(2x−1)³. On [1, 1.05], f‴ is maximised at x = 1: M = 16.
Error ≤ (16)(0.05)³/3! = 16(0.000125)/6 = 1/3000 ≈ 0.000333.
Sharper bound using [0.95, 1.05]: M = 16/(0.9)³, giving error ≤ 0.000457.
Start from a known series and apply substitutions, multiplication, or integration.
Standard series to memorise:
eˣ = Σ xᵏ/k!, sin x = Σ (−1)ᵏx²ᵏ⁺¹/(2k+1)!, cos x = Σ (−1)ᵏx²ᵏ/(2k)!, 1/(1−x) = Σ xᵏ.
Example: find the Taylor series for f(x) = ∫₀ˣ e^{2t²} dt at b = 0.
Start with eᵘ = Σ uᵏ/k!. Replace u with 2t²: e^{2t²} = Σ 2ᵏt²ᵏ/k!.
Integrate term by term: ∫₀ˣ e^{2t²} dt = Σ 2ᵏx²ᵏ⁺¹/((2k+1)k!).
eˣ converges for all x. Replacing x with 2t² and integrating does not change this.
So f(x) = ∫₀ˣ e^{2t²} dt has interval of convergence (−∞, ∞).
If the Taylor series of f at 0 has the term cₙxⁿ, then f⁽ⁿ⁾(0) = n! · cₙ.
Example: f(x) = Σ 2ᵏx²ᵏ⁺¹/((2k+1)k!). Find f⁽²⁰²³⁾(0).
The x²⁰²³ term has 2k+1 = 2023, so k = 1011.
Coefficient: 2¹⁰¹¹/((2023)(1011!)).
f⁽²⁰²³⁾(0) = 2023! · 2¹⁰¹¹/(2023 · 1011!) = 2022! · 2¹⁰¹¹/1011!.
T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(b)}{k!}(x - b)^k|f(x) - T_n(x)| \leq \frac{M|x - b|^{n+1}}{(n+1)!}f^{(n)}(0) = n! \cdot c_n \quad \text{where } c_n \text{ is the coefficient of } x^n \text{ in the Maclaurin series}Students often confuse the base point b with the value they are approximating. In the exam example, the base point is b = 1 and the approximation is at x = 1.05.
When building series from known series, students sometimes forget to adjust the factorial or index after substitution. Track every piece carefully.
The error bound requires finding the maximum of |f⁽ⁿ⁺¹⁾| on the relevant interval, not just evaluating it at the base point.
For f⁽ⁿ⁾(0) problems, students frequently forget to multiply the coefficient by n!. The coefficient cₙ in the series is f⁽ⁿ⁾(0)/n!, so f⁽ⁿ⁾(0) = n! · cₙ.
⚠️ Taylor polynomial construction (find Tₙ, use it to approximate, bound the error) is a three-part question worth 14 points on this exam.
⚠️ Series-building from known series (substitution + integration) and finding f⁽ⁿ⁾(0) from the result is a classic final-exam problem.
⚠️ The interval of convergence for eˣ-based series is always (−∞, ∞). State this explicitly to earn full marks.
Fill in the blank: The second Taylor polynomial of f at b = 1 uses f(1), f′(1), and ______. (f″(1))
True or False: The Maclaurin series for eˣ converges only on (−1, 1). (False, it converges on (−∞, ∞).)
Fill in the blank: If the coefficient of x⁵ in a Maclaurin series is 3, then f⁽⁵⁾(0) = ______. (5! · 3 = 360)
True or False: To bound the error of T₂(x), you need the third derivative of f. (True.)
Q: Find T₂(x) for f(x) = ln(2x−1) 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 to approximate ln(1.1).
A: ln(1.1) = f(1.05). T₂(1.05) = 2(0.05) − 2(0.0025) = 0.095.
Q: Bound the error of this approximation.
A: f‴(x) = 16/(2x−1)³. On [1, 1.05], max is 16. Error ≤ 16(0.05)³/6 = 1/3000.
Q: Find the Taylor series for ∫₀ˣ e^{2t²} dt.
A: Σ_{k=0}^∞ 2ᵏx²ᵏ⁺¹/((2k+1)k!). Interval of convergence: (−∞, ∞).
Q: Find f⁽²⁰²³⁾(0) for this function.
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 numerical methods (how computers evaluate transcendental functions), to differential equations (series solutions), and to Fourier series in more advanced courses. The error bound is a special case of the mean value theorem applied repeatedly.
Taylor polynomial, Taylor series, Maclaurin series, Taylor remainder theorem, error bound, Lagrange remainder, interval of convergence, radius of convergence, known series substitution, nth derivative from series, MATH 126, Calculus III, UW Math 126