The Lagrange error bound
Problem 7.424 · easy
What degree \( \displaystyle n \) of Maclaurin polynomial for \( \displaystyle \sin{\left(x \right)} \) guarantees, by the Lagrange error bound, an error less than \( \displaystyle \frac{1}{100000} \) 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.Reviewed
- \[ 1 \cdot \frac{1}{240000} = \frac{1}{240000} \]n = 3: the bound is 1/240000 < 1/100000.✓ Proved
- \[ 1 \cdot \frac{1}{6000} = \frac{1}{6000} \]n = 2: the bound is 1/6000, not below 1/100000.✓ Proved
Answer \( n = 3 \)
Lines: 2 proved, 1 reviewed. 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: pass — The solution correctly applies the Lagrange error bound formula with M=1 and x=1/10. It accurately calculates that n=3 yields an error bound less than 1/100000, while n=2 does not, leading to the correct conclusion.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly applies the Lagrange error bound formula with M=1 and x=1/10. It accurately calculates that n=3 yields an error bound less than 1/100000, while n=2 does not, leading to the correct conclusion.qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution claims n=3 is the answer, but the Maclaurin polynomial of degree 3 for sin(x) is identical to the polynomial of degree 2 (since the x^3 term is 0). The error bound for n=2 is 1/6000, which is not less than 1/100000. The bound for n=3 is 1/240000, which is sufficient, but since P_3(x) = P_2(x), the actual error is determined by the next non-zero term (n=4), or one must strictly use the Lagrange remainder for n=3. However, the standard interpretation is that if P_n satisfies the bound, n is the answer. But wait: P_3(x) = x - x^3/6. The remainder R_3(x) uses f^{(4)}. |f^{(4)}| <= 1. Bound is |x|^4/24 = (1/10)^4/24 = 1/240000. This is < 1/100000. So n=3 works. Does n=2 work? R_2(x) uses f^{(3)}. |f^{(3)}| <= 1. Bound is |x|^3/6 = 1/6000. This is > 1/100000. So n=2 does not guarantee the error. The answer n=3 is correct. Why did I think it was fail? Let me re-read carefully. The question asks for the degree n. P_3 is degree 3. The bound holds. The answer is correct. Wait, is there a nuance? For sin(x), the odd terms are non-zero. P_1(x) = x. R_1 bound: x^2/2 = 1/200. P_2(x) = x. R_2 bound: x^3/6 = 1/6000. P_3(x) = x - x^3/6. R_3 bound: x^4/24 = 1/240000. Since 1/240000 < 1/100000, n=3 is sufficient. Is n=2 sufficient? No. So n=3 is the smallest integer. The solution is correct. Why did I flag it? Ah, I see. The solution says 'n=3: the bound is 1/240000'. This is correct. 'n=2: the bound is 1/6000'. This is correct. The conclusion is n=3. This 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.