The Lagrange error bound
Problem 7.347 · 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 \( \displaystyle x = \frac{1}{2} \)?
- |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.
- \[ 1 \cdot \frac{1}{46080} = \frac{1}{46080} \]n = 5: the bound is 1/46080 < 1/10000.✓ Proved
- \[ 1 \cdot \frac{1}{3840} = \frac{1}{3840} \]n = 4: the bound is 1/3840, not below 1/10000.✓ Proved
Answer \( n = 5 \)
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 fails to verify that n=4 does not satisfy the condition (1/3840 > 1/10000), so it does not prove that n=5 is the minimal degree. Additionally, the Lagrange error bound formula is stated with |x|^{n+1}, but for sin(x), the (n+1)-th derivative bound M=1 is correct, yet one must ensure the specific term used corresponds to the actual remainder term for the chosen n; for n=5, the next non-zero term is degree 7, so the bound using n=5 directly (degree 6 term is 0) is loose but valid, but the logic for minimality is incomplete.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution fails to verify that n=4 does not satisfy the condition (1/3840 > 1/10000), so it does not prove that n=5 is the minimal degree. Additionally, the Lagrange error bound formula is stated with |x|^{n+1}, but for sin(x), the (n+1)-th derivative bound M=1 is correct, yet one must ensure the specific term used corresponds to the actual remainder term for the chosen n; for n=5, the next non-zero term is degree 7, so the bound using n=5 directly (degree 6 term is 0) is loose but valid, but the logic for minimality is incomplete.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution uses the wrong Lagrange error bound formula, omitting the evaluation point x = 1/2. The correct bound is M * |x|^(n+1) / (n+1)!, not M / (n+1)!. Consequently, the calculated bounds are incorrect (e.g., for n=3, the actual bound is 1/384 which is > 1/10000, but the solution's logic is flawed from the start). While n=5 is indeed sufficient, the derivation provided is mathematically invalid.gpt-oss:20b: pass 2026-10-06
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 2026-10-06 with SymPy 1.14.0.