∫Calc Practice

The Lagrange error bound

Problem 7.260 · easy

What degree \( \displaystyle n \) of Maclaurin polynomial for \( \displaystyle e^{x} \) guarantees, by the Lagrange error bound, an error less than \( \displaystyle \frac{1}{10000} \) at \( \displaystyle x = \frac{1}{10} \)?
  1. |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.
  2. \[ 1 \cdot \frac{1}{80000} = \frac{1}{80000} \]
    n = 3: the bound is 1/80000 < 1/10000.✓ Proved
  3. \[ 1 \cdot \frac{1}{2000} = \frac{1}{2000} \]
    n = 2: the bound is 1/2000, not below 1/10000.✓ Proved
Answer \( n = 3 \)

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
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the bound recomputed at n and n − 1

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: inconclusive — reviewer returned a non-object
Every verdict on record (4)
  • qwen3.6:27b-mlx: inconclusive 2026-10-04 — reviewer returned a non-object
  • gpt-oss:20b: pass 2026-10-04
  • qwen3.6:27b-mlx: inconclusive 2026-10-04 — reviewer returned a non-object
  • gpt-oss:20b: pass 2026-10-04

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 2026-10-04 with SymPy 1.14.0.