A student wants to approximate e^0.1 using the Taylor series centered at 0, with error below 0.0001. She knows |f⁽ⁿ⁺¹⁾(t)| ≤ 2 for all t ∈ [0, 0.1]. Using the Lagrange remainder, what must she find?
AThe smallest n such that 2/(n+1)! · (0.1)ⁿ⁺¹ < 0.0001
BThe smallest n such that 2ⁿ⁺¹/(n+1)! < 0.0001, ignoring the (x−a) factor
CShe should use n = 4 because the fifth derivative of eˣ equals eˣ ≈ 1
DShe must use n = ∞ because only the full Taylor series is exact
The Lagrange remainder bound is |Rₙ(x)| ≤ M/(n+1)! · |x−a|ⁿ⁺¹, where M bounds |f⁽ⁿ⁺¹⁾| on the interval and x−a = 0.1. She needs to find n so this bound is below 0.0001. Option B forgets to include the crucial (0.1)ⁿ⁺¹ factor, which shrinks rapidly and makes the bound tighter. Option D is the key misconception: the remainder theorem tells you exactly when a finite polynomial gives sufficient accuracy — the whole point is to avoid needing infinitely many terms.
Question 2 Multiple Choice
For f(x) = sin x and its Taylor series centered at 0, the series converges to sin x for all x. What does Taylor's theorem with remainder tell you about Rₙ(x) as n → ∞?
ARₙ(x) = 0 for some sufficiently large finite n
BRₙ(x) → 0 as n → ∞, which is exactly what it means for the series to converge to sin x
CThe (n+1)-th derivative of sin x must vanish, forcing Rₙ to zero
DRₙ(x) converges to a constant correction term that the series approximates away
Taylor's theorem with remainder makes this precise: the Taylor series Σ f⁽ᵏ⁾(a)/k! · (x−a)ᵏ converges to f(x) if and only if Rₙ(x) → 0 as n → ∞. For sin x, the (n+1)-th derivative is bounded by 1 in absolute value, so |Rₙ(x)| ≤ |x|ⁿ⁺¹/(n+1)! → 0 for every x (factorials dominate any fixed power). Option A is wrong: the series is infinite — no finite truncation is exact. This equivalence between series convergence and remainder decay is the main theoretical payoff of Taylor's theorem.
Question 3 True / False
The Lagrange remainder Rₙ(x) looks exactly like the (n+1)-th term of the Taylor polynomial, except that the derivative is evaluated at an unknown intermediate point c between a and x rather than at a.
TTrue
FFalse
Answer: True
This is precisely right. The n-th Taylor polynomial has terms f⁽ᵏ⁾(a)/k! · (x−a)ᵏ. The Lagrange remainder is f⁽ⁿ⁺¹⁾(c)/(n+1)! · (x−a)ⁿ⁺¹ for some c ∈ (a, x). Structurally it is the next Taylor term, but with the derivative evaluated at the intermediate point c instead of a. You cannot find c explicitly, but you can bound f⁽ⁿ⁺¹⁾ over the interval to get a concrete error bound without knowing c.
Question 4 True / False
If a smooth function f has a Taylor series that converges for most x, then the series necessarily converges to f(x).
TTrue
FFalse
Answer: False
This is a subtle but important falsehood. A function can be smooth (have derivatives of all orders everywhere) and have a convergent Taylor series that converges to the wrong value — or converges to f(x) only at the center. The canonical example is f(x) = e^{−1/x²} (defined as 0 at x = 0): all its derivatives at 0 are 0, so its Taylor series is identically 0, converging everywhere — but not to f(x) for x ≠ 0. Taylor's theorem with remainder identifies the actual condition: the series converges to f(x) at x if and only if Rₙ(x) → 0 as n → ∞.
Question 5 Short Answer
Why isn't it sufficient to know that a function has derivatives of all orders and a convergent Taylor series to conclude the series equals the function? What condition does Taylor's theorem with remainder actually require?
Think about your answer, then reveal below.
Model answer: The condition is that the remainder Rₙ(x) → 0 as n → ∞. A convergent Taylor series is just a convergent series of numbers — it is not guaranteed to converge to f(x) unless the remainder vanishes in the limit. Taylor's theorem with remainder identifies precisely this gap: f(x) = Pₙ(x) + Rₙ(x), and the series converges to f(x) exactly when the error Rₙ(x) disappears as we take more terms.
This distinction between 'the Taylor series converges' and 'the Taylor series converges to f' is what distinguishes real analysis from naive calculus. In calculus courses, students often assume these are the same. The Lagrange remainder provides the tool to verify the stronger condition: if you can show |Rₙ(x)| ≤ M/(n+1)! · |x−a|ⁿ⁺¹ → 0, you have genuinely proven the series represents the function — not just that an infinite sum exists.