∫Calc Practice

The Lagrange error bound

Problem 7.555 · easy

What degree \( \displaystyle n \) of Maclaurin polynomial for \( \displaystyle \cos{\left(x \right)} \) guarantees, by the Lagrange error bound, an error less than \( \displaystyle \frac{1}{1000} \) at \( \displaystyle x = 1 \)?
  1. |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.
  2. \[ 1 \cdot \frac{1}{5040} = \frac{1}{5040} \]
    n = 6: the bound is 1/5040 < 1/1000.✓ Proved
  3. \[ 1 \cdot \frac{1}{720} = \frac{1}{720} \]
    n = 5: the bound is 1/720, not below 1/1000.✓ Proved
Answer \( n = 6 \)

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
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the bound recomputed at n and n − 1

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The Lagrange error bound formula is incorrect; it should use (n+1) in the exponent and factorial, but the solution implies checking n=6 against 1/504
Every verdict on record (4)
  • gpt-oss:20b: pass 2026-10-13
  • qwen3.6:27b-mlx: inconclusive 2026-10-13 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The Lagrange error bound formula is incorrect; it should use (n+1) in the exponent and factorial, but the solution implies checking n=6 against 1/504
  • qwen3.6:27b-mlx: inconclusive 2026-10-13 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution uses the standard Lagrange error bound formula, which requires checking the (n+1)-th derivative. For n=5, the bound involves the 6th de
  • gpt-oss:20b: pass 2026-10-13

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 2026-10-13 with SymPy 1.14.0.