The Lagrange error bound
Problem 7.421 · easy
What degree \( \displaystyle n \) of Maclaurin polynomial for \( \displaystyle \sin{\left(x \right)} \) guarantees, by the Lagrange error bound, an error less than \( \displaystyle \frac{1}{1000} \) 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.
- \[ 1 \cdot \frac{1}{5040} = \frac{1}{5040} \]n = 6: the bound is 1/5040 < 1/1000.✓ Proved
- \[ 1 \cdot \frac{1}{720} = \frac{1}{720} \]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
| 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 claims n=6 is the answer, but the Maclaurin polynomial of degree 6 for sin(x) is identical to the degree 5 polynomial (since the 6th der
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: inconclusive 2026-10-08 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution claims n=6 is the answer, but the Maclaurin polynomial of degree 6 for sin(x) is identical to the degree 5 polynomial (since the 6th derqwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution claims n=5 fails because 1/720 is not below 1/1000, but for n=5, the error bound is actually 1/6! = 1/720, which is indeed > 1/1000, so n=5 does fail. However, for n=6, the bound is 1/7! = 1/5040, which is < 1/1000. The logic is actually correct for these specific numbers. Wait, let me re-read carefully. The problem asks for the degree n. For sin(x), the Maclaurin polynomial of degree n includes terms up to x^n. If n=5, the polynomial is x - x^3/6 + x^5/120. The next term is x^7/5040 (since the x^6 term is 0). The Lagrange error bound uses the (n+1)-th derivative. For n=5, the bound is M|x|^6/6!. M=1. So bound is 1/720. 1/720 > 1/1000. So n=5 is not enough. For n=6, the polynomial is the same as n=5 (since the 6th degree term is 0). The bound is M|x|^7/7! = 1/5040. 1/5040 < 1/1000. So n=6 works. The solution says n=6 works and n=5 fails. This is correct. Why did I think it was wrong? Ah, usually for sin(x), because the even terms are zero, n=5 and n=6 give the same polynomial. The question asks for the degree n. If n=5, the error bound formula gives 1/720. If n=6, the error bound formula gives 1/5040. Both are valid applications of the theorem. The smallest n satisfying the condition is 6? No, wait. If n=5, the actual error is approx 1/5040, but the *bound* is 1/720. The question asks what n *guarantees* an error less than 1/1000 *by the Lagrange error bound*. So we need the bound to be < 1/1000. For n=5, bound is 1/720 > 1/1000. So n=5 does not guarantee it via the bound. For n=6, bound is 1/5040 < 1/1000. So n=6 guarantees it. The answer n=6 is correct. The solution is correct. I will mark it pass.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.