The Lagrange error bound
Problem 7.478 · easy
What degree \( \displaystyle n \) of Maclaurin polynomial for \( \displaystyle \cos{\left(x \right)} \) guarantees, by the Lagrange error bound, an error less than \( \displaystyle \frac{1}{1000000} \) 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.
- \[ 1 \cdot \frac{1}{10321920} = \frac{1}{10321920} \]n = 7: the bound is 1/10321920 < 1/1000000.✓ Proved
- \[ 1 \cdot \frac{1}{645120} = \frac{1}{645120} \]n = 6: the bound is 1/645120, not below 1/1000000.✓ Proved
Answer \( n = 7 \)
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: passqwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution uses the wrong error bound formula. For a Maclaurin polynomial of degree n, the Lagrange error bound is M|x|^{n+1}/(n+1)!. The solution
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: inconclusive 2026-10-10 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution uses the wrong error bound formula. For a Maclaurin polynomial of degree n, the Lagrange error bound is M|x|^{n+1}/(n+1)!. The solutionqwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution uses the standard Lagrange error bound formula but fails to account for the fact that the Maclaurin series for cos(x) contains only even powers. For n=6, the 7th derivative term is zero, so the error is actually bounded by the n=7 term (or effectively the n=8 term in the standard formula context), meaning n=6 also satisfies the condition. The solution incorrectly concludes n=7 is the minimum.gpt-oss:20b: pass 2026-10-10
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 2026-10-10 with SymPy 1.14.0.