The Lagrange error bound
Problem 7.476 · easy
What degree \( \displaystyle n \) of Maclaurin polynomial for \( \displaystyle e^{x} \) guarantees, by the Lagrange error bound, an error less than \( \displaystyle \frac{1}{1000000} \) at \( \displaystyle x = \frac{1}{2} \)?
- |Rₙ(x)| ≤ M|x|ⁿ⁺¹/(n + 1)!, where M bounds |f⁽ⁿ⁺¹⁾|; here M = 3 because on [0, 1] every derivative of eˣ is at most e < 3.Reviewed
- \[ 1 \cdot \frac{1}{3440640} = \frac{1}{3440640} \]n = 7: the bound is 1/3440640 < 1/1000000.✓ Proved
- \[ 1 \cdot \frac{1}{215040} = \frac{1}{215040} \]n = 6: the bound is 1/215040, not below 1/1000000.✓ Proved
Answer \( n = 7 \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
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 a valid upper bound M=3 for the derivative on the interval [0, 1/2]. The arithmetic checks confirm that n=7 is the smallest integer satisfying the error condition.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly applies the Lagrange error bound formula with a valid upper bound M=3 for the derivative on the interval [0, 1/2]. The arithmetic checks confirm that n=7 is the smallest integer satisfying the error condition.gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly applies the Lagrange error bound formula with a valid upper bound M=3 for the derivative on the interval [0, 1/2]. The calculations for n=6 and n=7 correctly identify that n=7 is the smallest integer satisfying the error condition.
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 2026-10-10 with SymPy 1.14.0.