The Lagrange error bound
Problem 7.417 · easy
What degree \( \displaystyle n \) of Maclaurin polynomial for \( \displaystyle \cos{\left(x \right)} \) guarantees, by the Lagrange error bound, an error less than \( \displaystyle \frac{1}{100000} \) at \( \displaystyle x = \frac{1}{5} \)?
- |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.
- \[ 1 \cdot \frac{1}{375000} = \frac{1}{375000} \]n = 4: the bound is 1/375000 < 1/100000.✓ Proved
- \[ 1 \cdot \frac{1}{15000} = \frac{1}{15000} \]n = 3: the bound is 1/15000, not below 1/100000.✓ Proved
Answer \( n = 4 \)
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 claims n=3 fails because the bound 1/15000 is not below 1/100000, but for f(x)=cos(x), the n=3 Maclaurin polynomial is identical to the n=4 polynomial (since the x^4 term is 0). Thus, the error for n=3 is actually bounded by the n=4 Lagrange term (1/375000), which is less than 1/100000. Therefore, n=3 is sufficient, and the claim that n=4 is the required degree is incorrect.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution claims n=3 fails because the bound 1/15000 is not below 1/100000, but for f(x)=cos(x), the n=3 Maclaurin polynomial is identical to the n=4 polynomial (since the x^4 term is 0). Thus, the error for n=3 is actually bounded by the n=4 Lagrange term (1/375000), which is less than 1/100000. Therefore, n=3 is sufficient, and the claim that n=4 is the required degree is incorrect.gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution uses the Lagrange error bound formula for a polynomial of degree n, which involves the (n+1)-th derivative. For n=4, the bound is |x|^5/5! = 1/150000, not 1/375000. The value 1/375000 corresponds to the next term (n=5), implying the solver confused the degree n with the order of the derivative or used the wrong factorial. Consequently, the claim that n=4 satisfies the condition is based on an incorrect calculation.
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.