The Lagrange error bound
Problem 7.423 · easy
What degree \( \displaystyle n \) of Maclaurin polynomial for \( \displaystyle \sin{\left(x \right)} \) guarantees, by the Lagrange error bound, an error less than \( \displaystyle \frac{1}{1000000} \) at \( \displaystyle x = 1 \)?
- |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.
- \[ 1 \cdot \frac{1}{3628800} = \frac{1}{3628800} \]n = 9: the bound is 1/3628800 < 1/1000000.✓ Proved
- \[ 1 \cdot \frac{1}{362880} = \frac{1}{362880} \]n = 8: the bound is 1/362880, not below 1/1000000.✓ Proved
Answer \( n = 9 \)
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=8 and n=9 but fails to check n=7. Since the Maclaurin polynomial of degree 7 is identical to that of degree 8 for sin(x), n=7 also satisfies the error bound. The question asks for 'what degree n', implying the smallest such n, which is 7, not 9.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution checks n=8 and n=9 but fails to check n=7. Since the Maclaurin polynomial of degree 7 is identical to that of degree 8 for sin(x), n=7 also satisfies the error bound. The question asks for 'what degree n', implying the smallest such n, which is 7, not 9.qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution incorrectly identifies n=9 as the answer. For n=8, the error bound is 1/9! = 1/362880, which is less than 1/1000000, so n=8 is sufficient. The solution claims n=8 is not sufficient, which is false.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.