A student computes the degree-3 Taylor polynomial for sin(x) centered at 0: P₃(x) = x − x³/6. They claim this equals sin(x) for 'small x.' What is the most accurate statement about this claim?
AP₃(x) = sin(x) for all x, because Taylor polynomials converge to the original function
BP₃(x) equals sin(x) exactly at x = 0 and approximates sin(x) for small x, but does not equal it at any other point
CP₃(x) equals sin(x) exactly on some interval around x = 0, but diverges outside it
DP₃(x) is only useful for x > 0, where the approximation is valid
A Taylor polynomial is exact at exactly one point: the center a (here, x = 0). At x = 0, P₃(0) = 0 = sin(0) ✓, and matching derivatives are guaranteed by construction. Everywhere else, P₃(x) approximates sin(x), with error growing as |x| increases. The phrase 'equals sin(x) for small x' is imprecise — it should be 'closely approximates sin(x) for small x.' Convergence to the true function on an interval requires the full infinite Taylor series, not a finite polynomial.
Question 2 Multiple Choice
The Taylor polynomial formula includes f^(k)(a)/k! as the coefficient of (x−a)^k. Why is the k! in the denominator?
ATo keep the polynomial's values bounded as k grows large
BTo cancel the k! factor that appears when differentiating (x−a)^k exactly k times, ensuring the k-th derivative condition is satisfied
CTo normalize the approximation so errors stay proportional to (x−a)
DTo make the formula dimensionally consistent across different functions
When you differentiate the monomial c_k(x−a)^k exactly k times and evaluate at x = a, the power rule produces k! · c_k (and all lower terms vanish, all higher terms contribute zero). Setting k! · c_k = f^(k)(a) gives c_k = f^(k)(a)/k!. The factorial in the denominator is not a normalization convention — it is the exact inverse of the factorial that differentiation produces. This is why the coefficients have this form: they're determined by solving for what makes the k-th derivative match.
Question 3 True / False
The degree-n Taylor polynomial of f centered at a is the unique polynomial of degree at most n that matches f in both value and all derivatives up to order n at x = a.
TTrue
FFalse
Answer: True
This is the defining property of the Taylor polynomial, and 'unique' is the key word. There is exactly one polynomial of degree ≤ n satisfying all n+1 conditions (matching f(a), f'(a), f''(a), ..., f^(n)(a)). This uniqueness is what makes the Taylor polynomial 'the best' degree-n polynomial approximation near a in a precise sense — no other polynomial of the same degree can match as many derivatives at a.
Question 4 True / False
A Taylor polynomial and a Taylor series for the same function f, centered at the same point, represent the same mathematical object.
TTrue
FFalse
Answer: False
A Taylor polynomial is a finite sum — P_n(x) has exactly n+1 terms and provides an approximation with bounded error. A Taylor series is an infinite sum that may converge to f(x) on some interval, but is a fundamentally different object. The polynomial is a partial sum of the series, not the series itself. For example, sin(x) = x − x³/6 + x⁵/120 − ··· is an infinite Taylor series that equals sin(x) everywhere; P₃(x) = x − x³/6 is just the first two non-zero terms and diverges from sin(x) for larger |x|.
Question 5 Short Answer
Why does the Lagrange remainder formula turn Taylor polynomials from a useful approximation into a rigorous engineering tool? What problem does it solve that a plain polynomial cannot?
Think about your answer, then reveal below.
Model answer: A Taylor polynomial tells you what approximation to use, but not how accurate it is at any specific point. The Lagrange remainder R_n(x) = f^(n+1)(c)/(n+1)! · (x−a)^(n+1) gives a concrete bound on the error: if you can bound |f^(n+1)| on the interval, you know the maximum possible error. This lets you certify 'my approximation is within 0.001 of the true value' — a guarantee, not just an intuition. Without it, you have no way to know whether your polynomial is close enough for a given application.
This is how calculators and software compute transcendental functions (sin, cos, exp, ln): they use Taylor polynomials of sufficient degree and use the remainder bound to guarantee that the result is accurate to the last bit of precision. The Lagrange remainder transforms Taylor polynomials from a mathematical curiosity into a constructive tool: 'I need accuracy ε on interval [a−δ, a+δ] — what degree n suffices?' The remainder formula answers this directly.