The Lagrange error bound
Problem 7.266 · easy
What degree \( \displaystyle n \) of Maclaurin polynomial for \( \displaystyle \cos{\left(x \right)} \) guarantees, by the Lagrange error bound, an error less than \( \displaystyle \frac{1}{10000} \) 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.
- \[ 1 \cdot \frac{1}{15000} = \frac{1}{15000} \]n = 3: the bound is 1/15000 < 1/10000.✓ Proved
- \[ 1 \cdot \frac{1}{750} = \frac{1}{750} \]n = 2: the bound is 1/750, 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
| 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 satisfies the condition, but the Lagrange error bound for n=3 is |x|^4/4! = 1/3000, which is greater than 1/10000. The calculation in line 2 is arithmetically incorrect (1/15000 is not the bound for n=3), and the conclusion is false.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution claims n=3 satisfies the condition, but the Lagrange error bound for n=3 is |x|^4/4! = 1/3000, which is greater than 1/10000. The calculation in line 2 is arithmetically incorrect (1/15000 is not the bound for n=3), and the conclusion is false.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution uses the wrong Lagrange error bound formula. For a degree n polynomial, the error term involves the (n+1)-th derivative, so the bound is M|x|^{n+1}/(n+1)!. The solution incorrectly calculates the bound for n=3 as 1/15000 (which corresponds to the n=4 term or uses n! instead of (n+1)!). Specifically, for n=3, the bound is |x|^4/4! = (1/5)^4/24 = 1/30000, which is correct, but the text implies the formula used was |x|^n/n! or similar confusion. Wait, let's re-read carefully. The text says "1/15000 ... n=3". (1/5)^4 / 4! = 1/625 / 24 = 1/15000. So the calculation for n=3 is actually correct for the (n+1)th term. However, the sentence says "|R_n(x)| <= M|x|^{n+1}/(n+1)!". This is the correct formula. Let's check n=2. |R_2(x)| <= |x|^3/3! = (1/5)^3/6 = 1/125/6 = 1/750. This is also correct. So the calculations are correct. Why is it fail? The question asks for the degree n. The Maclaurin polynomial for cos(x) is 1 - x^2/2 + x^4/24 ... The degree 3 polynomial is the same as the degree 2 polynomial (since the x^3 term is 0). P_3(x) = P_2(x). The error bound for P_3 is based on the 4th derivative term, which is 1/15000. The error bound for P_2 is based on the 3rd derivative term, which is 1/750. Since 1/15000 < 1/10000, n=3 works. Does n=2 work? The bound for n=2 is 1/750 which is > 1/10000. So n=3 is the smallest integer? Actually, since P_2 = P_3, the actual error is the same. But the Lagrange bound for n=2 is looser. The question asks what degree n guarantees the error by the Lagrange bound. For n=2, the bound is 1/750, which does not guarantee < 1/10000. For n=3, the bound is 1/15000, which does guarantee it. So n=3 is the correct answer derived from the Lagrange bound. The solution seems correct. Let me double check the "error" severity. Is there a subtle issue? The problem asks for "degree n". Usually, we look for the smallest n. n=3 is the smallest n such that the Lagrange bound is < 1/10000. The solution states n=3. The steps are correct. Why would I fail it? Maybe the notation R_n? R_n is the remainder for the nth degree polynomial. The formula is correct. The arithmetic is correct. The conclusion is correct. I will pass it.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.