Home›Calculus 2›The Lagrange error bound›Problem 7.265

The Lagrange error bound

Problem 7.265 · easy

What degree \( \displaystyle n \) of Maclaurin polynomial for \( \displaystyle \sin{\left(x \right)} \) guarantees, by the Lagrange error bound, an error less than \( \displaystyle \frac{1}{1000} \) at \( \displaystyle x = \frac{1}{10} \)?

|Rₙ(x)| ≤ M|x|ⁿ⁺¹/(n + 1)!, where M bounds |f⁽ⁿ⁺¹⁾|; here M = 1 because |sin| and |cos| are at most 1.
\[ 1 \cdot \frac{1}{6000} = \frac{1}{6000} \]
n = 2: the bound is 1/6000 < 1/1000.✓ Proved
\[ 1 \cdot \frac{1}{200} = \frac{1}{200} \]
n = 1: the bound is 1/200, not below 1/1000.✓ Proved

Answer \( n = 2 \)

Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.

The full receipt

Line	Status	Checked by	Detail
1	Not checked	—	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	the bound recomputed at n and n − 1

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The solution fails to account for the fact that the Maclaurin polynomial of degree 2 for sin(x) is identical to that of degree 1 (since the x^2 term is 0). Therefore, the error bound for n=2 is the same as for n=1, which is 1/200, not 1/6000. The bound 1/6000 corresponds to n=3.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to account for the fact that the Maclaurin polynomial of degree 2 for sin(x) is identical to that of degree 1 (since the x^2 term is 0). Therefore, the error bound for n=2 is the same as for n=1, which is 1/200, not 1/6000. The bound 1/6000 corresponds to n=3.
gpt-oss:20b: pass 2026-10-04
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution incorrectly calculates the Lagrange error bound for n=1. For n=1, the bound is M|x|^2/2! = 1*(1/10)^2/2 = 1/200, which is indeed > 1/1000. However, for n=2, the bound is M|x|^3/3! = 1*(1/10)^3/6 = 1/6000, which is < 1/1000. The calculation is actually correct, but the reasoning is flawed because the Maclaurin polynomial of degree 2 for sin(x) is identical to the degree 1 polynomial (since the x^2 term is 0). The question asks for the degree n. If n=1, the error bound using the n=1 formula is 1/200. If we use n=2, the bound is 1/6000. Since the polynomial of degree 2 is the same as degree 1, one might argue n=1 is sufficient if we consider the actual error, but the Lagrange bound for n=1 is not sufficient. The Lagrange bound for n=2 is sufficient. So n=2 is the correct answer based on the bound. Wait, let me re-read. The solution says n=2 gives 1/6000 < 1/1000. This is correct. It says n=1 gives 1/200 > 1/1000. This is correct. So n=2 is the smallest n such that the bound is less than 1/1000. Is there a smaller n? n=0: bound is 1/10 = 0.1 > 0.001. So n=2 is the first one. The solution is actually correct. Why did I think it was wrong? Let me check the prompt again. 'Review it.' The solution claims n=2. The calculations support n=2. Is there a nuance? For sin(x), the term for n=2 is 0. So P_2(x) = P_1(x). The error for P_1(x) is bounded by 1/200. The error for P_2(x) is bounded by 1/6000. The question asks for the degree n of the Maclaurin polynomial. If we choose n=1, the bound is 1/200, which does not guarantee error < 1/1000. If we choose n=2, the bound is 1/6000, which does guarantee error < 1/1000. Even though P_1 = P_2, the 'degree n' refers to the order of the polynomial used in the error bound formula. The Lagrange error bound for a polynomial of degree n uses the (n+1)-th derivative. So for n=1, we use the 2nd derivative. For n=2, we use the 3rd derivative. The bound for n=2 is tighter. So n=2 is the correct answer. The solution is correct. I will mark it as pass.
gpt-oss:20b: pass 2026-10-04

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/lagrange_error, checked 2026-10-04 with SymPy 1.14.0.