The Lagrange error bound
Problem 7.425 · 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 \( \displaystyle x = \frac{1}{10} \)?
- |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.
- \[ 1 \cdot \frac{1}{240000} = \frac{1}{240000} \]n = 3: the bound is 1/240000 < 1/10000.✓ Proved
- \[ 1 \cdot \frac{1}{6000} = \frac{1}{6000} \]n = 2: the bound is 1/6000, 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 incorrectly applies the Lagrange error bound formula, using n+1 in the exponent and factorial instead of k (the degree of the first non-zero omitted term). For cos(x), the n=3 polynomial is identical to the n=2 polynomial, so the error is determined by the 4th derivative term, not the 3rd. The calculation for n=3 yields 1/240000, which is correct for the 4th degree term, but the label 'n=3' implies the error bound for a 3rd degree polynomial, which is technically the same as n=2 in this context but confusingly presented. More critically, the formula stated in line 1 is |R_n(x)| <= M|x|^{n+1}/(n+1)!, which is the standard form. However, for cos(x), f'''(0)=0, so the n=3 Maclaurin polynomial is P_3(x) = 1 - x^2/2. The remainder R_3(x) is bounded by M|x|^4/4! because the 3rd degree term is zero? No, the Lagrange remainder for P_n is based on the (n+1)-th derivative. For n=3, it uses the 4th derivative. The bound is |x|^4/4! = (1/10)^4/24 = 1/240000. This is < 1/10000. So n=3 works. Does n=2 work? P_2(x) = 1 - x^2/2. The remainder R_2(x) uses the 3rd derivative. f'''(c) = sin(c). Bound is |x|^3/3! = 1/6000. 1/6000 > 1/10000. So n=2 does not guarantee the error. The solution claims n=3 is the answer. Is n=3 the *minimal* degree? P_1(x) = 1. R_1 uses 2nd derivative. Bound |x|^2/2 = 1/200 > 1/10000. P_0(x) = 1. R_0 uses 1st derivative. Bound |x| = 1/10 > 1/10000. So n=3 is indeed the first integer n where the standard Lagrange bound guarantees the error. Wait, P_2 and P_3 are the same polynomial. Usually, we look for the smallest n. Since P_2 = P_3, if n=2 fails the bound check (using the n=2 remainder formula), but n=3 passes (using the n=3 remainder formula), is n=3 the correct answer? Yes, because the question asks for the degree n of the Maclaurin polynomial. The polynomial of degree 3 is the same as degree 2, but the error bound formula for n=3 is tighter (or rather, utilizes the fact that the 3rd derivative term is zero in the expansion, effectively jumping to the 4th derivative in the remainder estimate if one considers the next non-zero term, but strictly speaking, Lagrange remainder for n=3 uses f^{(4)}. The solution's logic is sound: it checks n=3, finds bound < threshold, checks n=2, finds bound > threshold. Thus n=3 is the smallest n satisfying the condition via the Lagrange bound. The only potential issue is if the question implies the polynomial must have a non-zero x^n term, but it doesn't. The solution is actually correct. Let me re-read carefully. Line 1 states the formula. Line 2 checks n=3. Line 3 checks n=2. Conclusion n=3. This is correct. Why did I think it was wrong? Ah, sometimes students confuse the index of the remainder. For cos(x), the Taylor series is 1 - x^2/2 + x^4/24... P_2(x) = 1 - x^2/2. P_3(x) = 1 - x^2/2. The error for P_2 is bounded by |x|^3/3!. The error for P_3 is bounded by |x|^4/4!. Since |x|^4/4! < |x|^3/3!, the bound for n=3 is much smaller. The question asks for the degree n that *guarantees* the error. n=3 guarantees it. n=2 does not. So n=3 is the correct answer. The solution is correct.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution incorrectly applies the Lagrange error bound formula, using n+1 in the exponent and factorial instead of k (the degree of the first non-zero omitted term). For cos(x), the n=3 polynomial is identical to the n=2 polynomial, so the error is determined by the 4th derivative term, not the 3rd. The calculation for n=3 yields 1/240000, which is correct for the 4th degree term, but the label 'n=3' implies the error bound for a 3rd degree polynomial, which is technically the same as n=2 in this context but confusingly presented. More critically, the formula stated in line 1 is |R_n(x)| <= M|x|^{n+1}/(n+1)!, which is the standard form. However, for cos(x), f'''(0)=0, so the n=3 Maclaurin polynomial is P_3(x) = 1 - x^2/2. The remainder R_3(x) is bounded by M|x|^4/4! because the 3rd degree term is zero? No, the Lagrange remainder for P_n is based on the (n+1)-th derivative. For n=3, it uses the 4th derivative. The bound is |x|^4/4! = (1/10)^4/24 = 1/240000. This is < 1/10000. So n=3 works. Does n=2 work? P_2(x) = 1 - x^2/2. The remainder R_2(x) uses the 3rd derivative. f'''(c) = sin(c). Bound is |x|^3/3! = 1/6000. 1/6000 > 1/10000. So n=2 does not guarantee the error. The solution claims n=3 is the answer. Is n=3 the *minimal* degree? P_1(x) = 1. R_1 uses 2nd derivative. Bound |x|^2/2 = 1/200 > 1/10000. P_0(x) = 1. R_0 uses 1st derivative. Bound |x| = 1/10 > 1/10000. So n=3 is indeed the first integer n where the standard Lagrange bound guarantees the error. Wait, P_2 and P_3 are the same polynomial. Usually, we look for the smallest n. Since P_2 = P_3, if n=2 fails the bound check (using the n=2 remainder formula), but n=3 passes (using the n=3 remainder formula), is n=3 the correct answer? Yes, because the question asks for the degree n of the Maclaurin polynomial. The polynomial of degree 3 is the same as degree 2, but the error bound formula for n=3 is tighter (or rather, utilizes the fact that the 3rd derivative term is zero in the expansion, effectively jumping to the 4th derivative in the remainder estimate if one considers the next non-zero term, but strictly speaking, Lagrange remainder for n=3 uses f^{(4)}. The solution's logic is sound: it checks n=3, finds bound < threshold, checks n=2, finds bound > threshold. Thus n=3 is the smallest n satisfying the condition via the Lagrange bound. The only potential issue is if the question implies the polynomial must have a non-zero x^n term, but it doesn't. The solution is actually correct. Let me re-read carefully. Line 1 states the formula. Line 2 checks n=3. Line 3 checks n=2. Conclusion n=3. This is correct. Why did I think it was wrong? Ah, sometimes students confuse the index of the remainder. For cos(x), the Taylor series is 1 - x^2/2 + x^4/24... P_2(x) = 1 - x^2/2. P_3(x) = 1 - x^2/2. The error for P_2 is bounded by |x|^3/3!. The error for P_3 is bounded by |x|^4/4!. Since |x|^4/4! < |x|^3/3!, the bound for n=3 is much smaller. The question asks for the degree n that *guarantees* the error. n=3 guarantees it. n=2 does not. So n=3 is the correct answer. The solution is correct.qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution incorrectly applies the Lagrange error bound formula, using n! instead of (n+1)! for the denominator. For n=2, the correct bound is |x|^3/3! = 1/6000, which is less than 1/10000, so n=2 suffices. The stated answer n=3 is incorrect because a lower degree already satisfies the condition.gpt-oss:20b: pass 2026-10-08
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.