Home›Calculus 2›The Lagrange error bound›Problem 7.479

The Lagrange error bound

Problem 7.479 · easy

What degree \( \displaystyle n \) of Maclaurin polynomial for \( \displaystyle \sin{\left(x \right)} \) guarantees, by the Lagrange error bound, an error less than \( \displaystyle \frac{1}{100000} \) 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: pass
qwen3.6:27b-mlx: fail (error) — The solution incorrectly identifies n=4 as the answer. Since the 4th derivative of sin(x) is -sin(x), the 4th degree Maclaurin polynomial is identical to the 3rd degree polynomial (the x^4 term is 0). The error bound for n=4 is actually determined by the 5th derivative, yielding a much smaller error (1/1500000), which satisfies the condition. However, n=3 also satisfies the condition (error bound 1/15000 is > 1/100000? No, 1/15000 = 0.000066... which is < 0.00001? No. 1/15000 = 6.6e-5 > 1e-4 is false. 1/15000 > 1/100000. So n=3 fails. Wait. 1/15000 = 0.0000666. 1/100000 = 0.00001. 0.0000666 > 0.00001. So n=3 fails. n=4 bound is 1/1500000 = 6.6e-7 < 1e-4. So n=4 works. But is it the *guaranteed* degree? The question asks for the degree n. Usually, we look for the smallest n. For sin(x), P_3(x) = P_4(x). The error for P_3 is bounded by M|x|^4/4! = 1/15000. This is not < 1/100000. The error for P_4 is bounded by M|x|^5/5! = 1/1500000. This is < 1/100000. So n=4 is indeed the smallest integer n such that the Lagrange bound for degree n is less than the tolerance. The solution's logic is slightly garbled but the conclusion n=4 is correct. Let me re-read carefully. Line 1: Formula is correct. M=1 is correct. Line 2: Claims n=4 bound is 1/375000. Let's check: |x|^5 / 5! = (1/5)^5 / 120 = (1/3125) / 120 = 1/375000. Correct. And 1/375000 < 1/100000. Correct. Line 3: Claims n=3 bound is 1/15000. Let's check: |x|^4 / 4! = (1/5)^4 / 24 = (1/625) / 24 = 1/15000. Correct. And 1/15000 is NOT < 1/100000. Correct. So the solution correctly identifies that n=3 fails and n=4 passes. The answer n=4 is correct. Why did I think it was wrong? Wait, is there a nuance? "Degree n of Maclaurin polynomial". P_3(x) = x - x^3/6. P_4(x) = x - x^3/6 + 0x^4. They are the same polynomial. Does the question imply distinct polynomials? Usually, n refers to the order of the remainder term used in the bound. If we use n=3, the remainder is R_3. If we use n=4, the remainder is R_4. Since the coefficient of x^4 is 0, R_3(x) = f(x) - P_3(x) and R_4(x) = f(x) - P_4(x) are the same error. However, the *bound* used for R_3 is M|x|^4/4! and for R_4 is M|x|^5/5!. The actual error is the same, but the *guarantee* provided by the Lagrange bound for n=3 is loose (1/15000), while for n=4 it is tight (1/375000). The question asks what degree n *guarantees* the error. Using n=3, the bound does not guarantee it. Using n=4, the bound does guarantee it. So n=4 is the correct answer to the question as phrased. Is there any error? Line 1 says "M bounds |f^(n+1)|". For n=4, we need bound on f^(5). f^(5)(x) = cos(x). |cos(x)| <= 1. Correct. For n=3, we need bound on f^(4). f^(4)(x) = -sin(x). |sin(x)| <= 1. Correct. The solution seems perfectly sound. Let me double check the arithmetic. 1/375000 = 2.66e-6. 1/100000 = 1e-5. 2.66e-6 < 1e-5. True. 1/15000 = 6.66e-5. 6.66e-5 < 1e-5. False. So n=3 fails, n=4 passes. Is it possible that n=2 works? Bound for n=2: |x|^3/3! = (1/125)/6 = 1/750 = 0.00133. Not < 1e-5. So n=4 is the smallest n. The solution is correct. Wait, look at Line 1 again. "|R_n(x)| <= M|x|^{n+1}/(n+1)!". This is the standard Lagrange error bound. Is there a subtle issue with "degree n"? If P_3 and P_4 are the same, is the degree 3 or 4? The polynomial P_4 has degree 3 technically (since the x^4 term is 0), but it is called the 4th degree Maclaurin polynomial in the context of Taylor series truncation. The question asks for "degree n ... polynomial ... guarantees ... by the Lagrange error bound". This phrasing ties the guarantee to the index n in the remainder formula R_n. So n=4 is the correct index. I will mark it as pass.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-10
qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution incorrectly identifies n=4 as the answer. Since the 4th derivative of sin(x) is -sin(x), the 4th degree Maclaurin polynomial is identical to the 3rd degree polynomial (the x^4 term is 0). The error bound for n=4 is actually determined by the 5th derivative, yielding a much smaller error (1/1500000), which satisfies the condition. However, n=3 also satisfies the condition (error bound 1/15000 is > 1/100000? No, 1/15000 = 0.000066... which is < 0.00001? No. 1/15000 = 6.6e-5 > 1e-4 is false. 1/15000 > 1/100000. So n=3 fails. Wait. 1/15000 = 0.0000666. 1/100000 = 0.00001. 0.0000666 > 0.00001. So n=3 fails. n=4 bound is 1/1500000 = 6.6e-7 < 1e-4. So n=4 works. But is it the *guaranteed* degree? The question asks for the degree n. Usually, we look for the smallest n. For sin(x), P_3(x) = P_4(x). The error for P_3 is bounded by M|x|^4/4! = 1/15000. This is not < 1/100000. The error for P_4 is bounded by M|x|^5/5! = 1/1500000. This is < 1/100000. So n=4 is indeed the smallest integer n such that the Lagrange bound for degree n is less than the tolerance. The solution's logic is slightly garbled but the conclusion n=4 is correct. Let me re-read carefully. Line 1: Formula is correct. M=1 is correct. Line 2: Claims n=4 bound is 1/375000. Let's check: |x|^5 / 5! = (1/5)^5 / 120 = (1/3125) / 120 = 1/375000. Correct. And 1/375000 < 1/100000. Correct. Line 3: Claims n=3 bound is 1/15000. Let's check: |x|^4 / 4! = (1/5)^4 / 24 = (1/625) / 24 = 1/15000. Correct. And 1/15000 is NOT < 1/100000. Correct. So the solution correctly identifies that n=3 fails and n=4 passes. The answer n=4 is correct. Why did I think it was wrong? Wait, is there a nuance? "Degree n of Maclaurin polynomial". P_3(x) = x - x^3/6. P_4(x) = x - x^3/6 + 0x^4. They are the same polynomial. Does the question imply distinct polynomials? Usually, n refers to the order of the remainder term used in the bound. If we use n=3, the remainder is R_3. If we use n=4, the remainder is R_4. Since the coefficient of x^4 is 0, R_3(x) = f(x) - P_3(x) and R_4(x) = f(x) - P_4(x) are the same error. However, the *bound* used for R_3 is M|x|^4/4! and for R_4 is M|x|^5/5!. The actual error is the same, but the *guarantee* provided by the Lagrange bound for n=3 is loose (1/15000), while for n=4 it is tight (1/375000). The question asks what degree n *guarantees* the error. Using n=3, the bound does not guarantee it. Using n=4, the bound does guarantee it. So n=4 is the correct answer to the question as phrased. Is there any error? Line 1 says "M bounds |f^(n+1)|". For n=4, we need bound on f^(5). f^(5)(x) = cos(x). |cos(x)| <= 1. Correct. For n=3, we need bound on f^(4). f^(4)(x) = -sin(x). |sin(x)| <= 1. Correct. The solution seems perfectly sound. Let me double check the arithmetic. 1/375000 = 2.66e-6. 1/100000 = 1e-5. 2.66e-6 < 1e-5. True. 1/15000 = 6.66e-5. 6.66e-5 < 1e-5. False. So n=3 fails, n=4 passes. Is it possible that n=2 works? Bound for n=2: |x|^3/3! = (1/125)/6 = 1/750 = 0.00133. Not < 1e-5. So n=4 is the smallest n. The solution is correct. Wait, look at Line 1 again. "|R_n(x)| <= M|x|^{n+1}/(n+1)!". This is the standard Lagrange error bound. Is there a subtle issue with "degree n"? If P_3 and P_4 are the same, is the degree 3 or 4? The polynomial P_4 has degree 3 technically (since the x^4 term is 0), but it is called the 4th degree Maclaurin polynomial in the context of Taylor series truncation. The question asks for "degree n ... polynomial ... guarantees ... by the Lagrange error bound". This phrasing ties the guarantee to the index n in the remainder formula R_n. So n=4 is the correct index. I will mark it as pass.
qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly applies the Lagrange error bound formula with M=1 and verifies that n=4 satisfies the inequality while n=3 does not.
gpt-oss:20b: pass 2026-10-10

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.