The Lagrange error bound
Problem 7.422 · easy
What degree \( \displaystyle n \) of Maclaurin polynomial for \( \displaystyle \sin{\left(x \right)} \) guarantees, by the Lagrange error bound, an error less than \( \displaystyle \frac{1}{10000} \) 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.
- \[ 1 \cdot \frac{1}{240000} = \frac{1}{240000} \]n = 3: the bound is 1/240000 < 1/10000.✓ Proved
- \[ 1 \cdot \frac{1}{6000} = \frac{1}{6000} \]n = 2: the bound is 1/6000, not below 1/10000.✓ Proved
Answer \( n = 3 \)
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: fail (error) — The solution checks n=2 and n=3 but fails to check n=1. Although n=3 is a valid answer, the problem asks for 'what degree', implying the minimal such degree. Since the bound for n=1 is 1/200 < 1/10000, n=1 is the minimal degree satisfying the condition, making n=3 incorrect as the primary answer.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution checks n=2 and n=3 but fails to check n=1. Although n=3 is a valid answer, the problem asks for 'what degree', implying the minimal such degree. Since the bound for n=1 is 1/200 < 1/10000, n=1 is the minimal degree satisfying the condition, making n=3 incorrect as the primary answer.qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly applies the Lagrange error bound formula with M=1 and verifies that n=3 satisfies the condition while n=2 does not.gpt-oss:20b: pass 2026-10-08
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 2026-10-08 with SymPy 1.14.0.