Home›Calculus 2›The Lagrange error bound›Problem 7.346

The Lagrange error bound

Problem 7.346 · 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}{10000} $ at $ \displaystyle x = 1 $?

|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.
\[ 1 \cdot \frac{1}{40320} = \frac{1}{40320} \]
n = 7: the bound is 1/40320 < 1/10000.✓ Proved
\[ 1 \cdot \frac{1}{5040} = \frac{1}{5040} \]
n = 6: the bound is 1/5040, not below 1/10000.✓ Proved

Answer $ n = 7 $

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: pass
qwen3.6:27b-mlx: fail (error) — The solution uses the wrong error bound formula. For a Maclaurin polynomial of degree n, the Lagrange error bound involves the (n+1)-th derivative, so the term is M|x|^(n+1)/(n+1)!. However, for sin(x), the (n+1)-th derivative is either sin or cos. If n=7, the next non-zero term is n=8 (since the 8th derivative is -sin(x) and the 9th is -cos(x), wait. The Maclaurin series for sin(x) is x - x^3/3! + x^5/5! - x^7/7! + ... The polynomial of degree 7 is P_7(x) = x - x^3/6 + x^5/120 - x^7/5040. The next term in the series is x^9/9!. The Lagrange error bound for P_n(x) is |R_n(x)| <= M |x|^(n+1) / (n+1)!. For n=7, this is M |x|^8 / 8!. Since the 8th derivative of sin(x) is -sin(x), M=1. So the bound is 1/8! = 1/40320. This is indeed < 1/10000. For n=6, P_6(x) is the same as P_7(x) because the 7th degree term is non-zero? No, P_6(x) = x - x^3/6 + x^5/120. The error bound is M |x|^7 / 7! = 1/5040. This is > 1/10000. So n=7 is correct. Wait, let me re-read the solution's logic. Line 1 states the formula correctly. Line 2 checks n=7. Line 3 checks n=6. The logic holds. Why did I think it was wrong? Ah, the question asks for the degree n. The solution concludes n=7. The check for n=6 shows the bound is too large. The check for n=7 shows the bound is small enough. Is there a smaller n? n=5: bound is 1/6! = 1/720 > 1/10000. n=4: bound is 1/5! = 1/120. n=3: 1/24. n=2: 1/6. n=1: 1. n=0: 1. So n=7 is the smallest integer. The solution is actually correct. Let me double check the 'error' vs 'misleading'. The solution says 'n=7: the bound is 1/40320'. This corresponds to the term x^8/8!. But the actual next term in the Taylor series for sin(x) after x^7 is x^9/9!. The Lagrange bound is an upper bound, not necessarily the next term. The Lagrange bound for n=7 uses the 8th derivative. The 8th derivative of sin(x) is -sin(x). At x=1, |-sin(1)| < 1. So M=1 is a valid bound. The bound is 1/8!. This is correct. The solution is sound.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-06
qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution uses the wrong error bound formula. For a Maclaurin polynomial of degree n, the Lagrange error bound involves the (n+1)-th derivative, so the term is M|x|^(n+1)/(n+1)!. However, for sin(x), the (n+1)-th derivative is either sin or cos. If n=7, the next non-zero term is n=8 (since the 8th derivative is -sin(x) and the 9th is -cos(x), wait. The Maclaurin series for sin(x) is x - x^3/3! + x^5/5! - x^7/7! + ... The polynomial of degree 7 is P_7(x) = x - x^3/6 + x^5/120 - x^7/5040. The next term in the series is x^9/9!. The Lagrange error bound for P_n(x) is |R_n(x)| <= M |x|^(n+1) / (n+1)!. For n=7, this is M |x|^8 / 8!. Since the 8th derivative of sin(x) is -sin(x), M=1. So the bound is 1/8! = 1/40320. This is indeed < 1/10000. For n=6, P_6(x) is the same as P_7(x) because the 7th degree term is non-zero? No, P_6(x) = x - x^3/6 + x^5/120. The error bound is M |x|^7 / 7! = 1/5040. This is > 1/10000. So n=7 is correct. Wait, let me re-read the solution's logic. Line 1 states the formula correctly. Line 2 checks n=7. Line 3 checks n=6. The logic holds. Why did I think it was wrong? Ah, the question asks for the degree n. The solution concludes n=7. The check for n=6 shows the bound is too large. The check for n=7 shows the bound is small enough. Is there a smaller n? n=5: bound is 1/6! = 1/720 > 1/10000. n=4: bound is 1/5! = 1/120. n=3: 1/24. n=2: 1/6. n=1: 1. n=0: 1. So n=7 is the smallest integer. The solution is actually correct. Let me double check the 'error' vs 'misleading'. The solution says 'n=7: the bound is 1/40320'. This corresponds to the term x^8/8!. But the actual next term in the Taylor series for sin(x) after x^7 is x^9/9!. The Lagrange bound is an upper bound, not necessarily the next term. The Lagrange bound for n=7 uses the 8th derivative. The 8th derivative of sin(x) is -sin(x). At x=1, |-sin(1)| < 1. So M=1 is a valid bound. The bound is 1/8!. This is correct. The solution is sound.
qwen3.6:27b-mlx: inconclusive 2026-10-06 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution incorrectly applies the Lagrange error bound formula, using $n+1$ in the factorial and power instead of $n+1$ for the derivative order b
gpt-oss:20b: pass 2026-10-06

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 2026-10-06 with SymPy 1.14.0.