Home›Calculus 2›The Lagrange error bound›Problem 7.349

The Lagrange error bound

Problem 7.349 · easy

What degree \( \displaystyle n \) of Maclaurin polynomial for \( \displaystyle \cos{\left(x \right)} \) guarantees, by the Lagrange error bound, an error less than \( \displaystyle \frac{1}{10000} \) at \( \displaystyle x = \frac{1}{2} \)?

|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.
Reviewed
\[ 1 \cdot \frac{1}{46080} = \frac{1}{46080} \]
n = 5: the bound is 1/46080 < 1/10000.✓ Proved
\[ 1 \cdot \frac{1}{3840} = \frac{1}{3840} \]
n = 4: the bound is 1/3840, not below 1/10000.✓ Proved

Answer \( n = 5 \)

Lines: 2 proved, 1 reviewed. 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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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: pass — The solution correctly applies the Lagrange error bound formula with M=1. The calculations for n=4 and n=5 are correct, and the conclusion that n=5 is the smallest integer satisfying the condition is sound.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-06
qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly applies the Lagrange error bound formula with M=1. The calculations for n=4 and n=5 are correct, and the conclusion that n=5 is the smallest integer satisfying the condition is sound.
qwen3.6:27b-mlx: fail (error) 2026-10-06 — The Lagrange error bound formula is incorrect; it should be M|x|^(n+1)/(n+1)!, not M|x|^(n+1)/(n+1)! (the exponent in the numerator is correct, but the denominator is missing the factorial symbol in the text description, though the calculations imply it. However, the main error is that for n=4, the bound is 1/3840 which is > 1/10000, so n=4 fails. For n=5, the bound is 1/46080 which is < 1/10000, so n=5 works. Wait, let's re-read carefully. The formula stated is `M|x|ⁿ⁺¹/(n + 1)!`. This is correct. The calculations: n=5 -> 1/2^6 / 6! = 1/64 / 720 = 1/46080. Correct. n=4 -> 1/2^5 / 5! = 1/32 / 120 = 1/3840. Correct. 1/3840 ≈ 0.00026 > 0.0001. 1/46080 ≈ 0.00002 < 0.0001. So n=5 is indeed the answer. Why did I think it was wrong? Let me check the question again. 'What degree n ... guarantees ... error less than 1/10000'. The solution says n=5. The steps show n=4 fails and n=5 passes. This seems correct. Is there a subtle issue? The problem asks for the degree n. The Maclaurin polynomial of degree n for cos(x) has terms up to x^n. If n is odd, the term is 0, so P_n(x) = P_{n-1}(x). The error bound for P_n is the same as for P_{n-1} if the n-th derivative is 0? No, the Lagrange remainder R_n(x) uses the (n+1)-th derivative. For n=4, R_4 uses f^(5). f^(5)(x) = -sin(x). Max is 1. Bound is 1/3840. For n=5, R_5 uses f^(6). f^(6)(x) = -cos(x). Max is 1. Bound is 1/46080. Since 1/46080 < 1/10000, n=5 works. Since 1/3840 > 1/10000, n=4 does not work. So n=5 is the smallest integer. The solution is correct. Wait, is the formula in line 1 written correctly? `M|x|ⁿ⁺¹/(n + 1)!`. Yes. Is the calculation correct? Yes. Is the conclusion correct? Yes. Why would it fail? Maybe the prompt implies I should find an error. Let me look closer. `1/46080 = 1/46080`. This is a tautology. It's not a derivation. But the instructions say 'equation, unchecked ... do not re-check the algebra'. So I accept the numbers. The logic holds. Is there a 'misleading' aspect? The solution jumps from n=4 and n=5 without explaining why we check these specific n. But it's a standard search. Is it possible that n=3 works? R_3 uses f^(4)=cos. Bound 1/2^4/4! = 1/16/24 = 1/384. Too big. n=2? R_2 uses f^(3)=-sin. Bound 1/2^3/3! = 1/8/6 = 1/48. Too big. So n=5 is indeed the first one. The solution is sound. I will mark it pass.
gpt-oss:20b: pass 2026-10-06

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 2026-10-06 with SymPy 1.14.0.