The Lagrange error bound
Problem 7.350 · easy
What degree \( \displaystyle n \) of Maclaurin polynomial for \( \displaystyle e^{x} \) guarantees, by the Lagrange error bound, an error less than \( \displaystyle \frac{1}{1000} \) at \( \displaystyle x = \frac{1}{5} \)?
- |Rₙ(x)| ≤ M|x|ⁿ⁺¹/(n + 1)!, where M bounds |f⁽ⁿ⁺¹⁾|; here M = 3 because on [0, 1] every derivative of eˣ is at most e < 3.
- \[ 1 \cdot \frac{1}{5000} = \frac{1}{5000} \]n = 3: the bound is 1/5000 < 1/1000.✓ Proved
- \[ 1 \cdot \frac{1}{250} = \frac{1}{250} \]n = 2: the bound is 1/250, not below 1/1000.✓ 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: inconclusive — reviewer returned a non-objectqwen3.6:27b-mlx: fail (error) — The Lagrange error bound formula is incorrect; it should be M|x|^{n+1}/(n+1)!, but the solution uses M|x|^{n+1}/(n+1)! in the text yet calculates values corresponding to M|x|^{n+1}/(n+1)!? Let's check: For n=3, bound is 3*(1/5)^4 / 24 = 3/625/24 = 3/15000 = 1/5000. The solution says 1/5000. For n=2, bound is 3*(1/5)^3 / 6 = 3/125/6 = 3/750 = 1/250. The solution says 1/250. The calculations are actually correct for the formula stated. However, the problem asks for the degree n that guarantees error < 1/1000. n=3 gives 1/5000 < 1/1000. n=2 gives 1/250 > 1/1000. So n=3 is the answer. Wait, is n=3 the *minimal* degree? Yes. Is the setup correct? M=3 is a valid bound for e^x on [0, 1/5] since e^(1/5) < 3. The formula is correct. The calculations are correct. The conclusion is correct. Why would it fail? Ah, look at line 1: "M bounds |f^{(n+1)}|". For n=3, we need bound on 4th derivative. e^x is increasing, max on [0, 1/5] is e^{1/5} ~ 1.22. Using M=3 is a valid upper bound, just loose. The logic holds. Is there a subtle error? "degree n of Maclaurin polynomial". The remainder term R_n(x) corresponds to the polynomial of degree n. The formula |R_n(x)| <= M |x|^{n+1} / (n+1)! is correct. The calculations: n=3 -> 3*(1/5)^4/24 = 3/15000 = 1/5000. Correct. n=2 -> 3*(1/5)^3/6 = 3/750 = 1/250. Correct. 1/5000 < 1/1000. 1/250 > 1/1000. So n=3 is the smallest integer. The answer is correct. The sentences are correct. Wait, did I miss something? "M = 3 because on [0, 1] every derivative...". The interval is [0, 1/5], not [0, 1]. But since [0, 1/5] is a subset of [0, 1], and e^x < 3 on [0, 1], it is also < 3 on [0, 1/5]. This is technically true, though slightly imprecise to cite [0, 1] when the point is 1/5. It's not an error, just loose. Is it misleading? No. Is it an error? No. Let me re-read carefully. "M = 3 because on [0, 1] every derivative of e^x is at most e < 3". This is a true statement. It justifies M=3. The rest follows. I see no error. Pass.
Every verdict on record (4)
gpt-oss:20b: inconclusive 2026-10-06 — reviewer returned a non-objectqwen3.6:27b-mlx: fail (error) 2026-10-06 — The Lagrange error bound formula is incorrect; it should be M|x|^{n+1}/(n+1)!, but the solution uses M|x|^{n+1}/(n+1)! in the text yet calculates values corresponding to M|x|^{n+1}/(n+1)!? Let's check: For n=3, bound is 3*(1/5)^4 / 24 = 3/625/24 = 3/15000 = 1/5000. The solution says 1/5000. For n=2, bound is 3*(1/5)^3 / 6 = 3/125/6 = 3/750 = 1/250. The solution says 1/250. The calculations are actually correct for the formula stated. However, the problem asks for the degree n that guarantees error < 1/1000. n=3 gives 1/5000 < 1/1000. n=2 gives 1/250 > 1/1000. So n=3 is the answer. Wait, is n=3 the *minimal* degree? Yes. Is the setup correct? M=3 is a valid bound for e^x on [0, 1/5] since e^(1/5) < 3. The formula is correct. The calculations are correct. The conclusion is correct. Why would it fail? Ah, look at line 1: "M bounds |f^{(n+1)}|". For n=3, we need bound on 4th derivative. e^x is increasing, max on [0, 1/5] is e^{1/5} ~ 1.22. Using M=3 is a valid upper bound, just loose. The logic holds. Is there a subtle error? "degree n of Maclaurin polynomial". The remainder term R_n(x) corresponds to the polynomial of degree n. The formula |R_n(x)| <= M |x|^{n+1} / (n+1)! is correct. The calculations: n=3 -> 3*(1/5)^4/24 = 3/15000 = 1/5000. Correct. n=2 -> 3*(1/5)^3/6 = 3/750 = 1/250. Correct. 1/5000 < 1/1000. 1/250 > 1/1000. So n=3 is the smallest integer. The answer is correct. The sentences are correct. Wait, did I miss something? "M = 3 because on [0, 1] every derivative...". The interval is [0, 1/5], not [0, 1]. But since [0, 1/5] is a subset of [0, 1], and e^x < 3 on [0, 1], it is also < 3 on [0, 1/5]. This is technically true, though slightly imprecise to cite [0, 1] when the point is 1/5. It's not an error, just loose. Is it misleading? No. Is it an error? No. Let me re-read carefully. "M = 3 because on [0, 1] every derivative of e^x is at most e < 3". This is a true statement. It justifies M=3. The rest follows. I see no error. Pass.qwen3.6:27b-mlx: inconclusive 2026-10-06 — reviewer returned a non-objectgpt-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.