The Lagrange error bound
Problem 7.472 · easy
What degree \( \displaystyle n \) of Maclaurin polynomial for \( \displaystyle e^{x} \) guarantees, by the Lagrange error bound, an error less than \( \displaystyle \frac{1}{100000} \) at \( \displaystyle x = \frac{1}{2} \)?
- |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.
- \[ 1 \cdot \frac{1}{215040} = \frac{1}{215040} \]n = 6: the bound is 1/215040 < 1/100000.✓ Proved
- \[ 1 \cdot \frac{1}{15360} = \frac{1}{15360} \]n = 5: the bound is 1/15360, not below 1/100000.✓ Proved
Answer \( n = 6 \)
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 Lagrange error bound formula is incorrect; it should be M|x|^{n+1}/(n+1)!, not M|x|^{n+1}/(n+1)! (the solution text has a typo in the denominator or exponent structure, but more critically, the calculation for n=6 uses 7! = 5040, so 3*(1/2)^7 / 5040 = 3/64/5040 = 3/322560 = 1/107520, which is indeed < 1/100000. However, the text says 1/215040. Let's re-read carefully. The text says |R_n(x)| <= M|x|^{n+1}/(n+1)!. For n=6, bound is 3*(1/2)^7 / 7! = 3/128 / 5040? No, (1/2)^7 = 1/128. 3/128/5040 = 3/645120 = 1/215040. Okay, the calculation is correct. Wait, is the formula written correctly in line 1? "M|x|^{n+1}/(n + 1)!". Yes. Is M=3 valid? Yes. Is the conclusion correct? Yes. Why did I think it was wrong? Let me re-read the prompt's constraints. "Check that... each theorem or test is applied where its hypotheses hold". The hypothesis for Lagrange Error Bound is that f is n+1 times differentiable. e^x is. The bound M must bound |f^{(n+1)}(z)| for z between 0 and x. Here x=1/2. The interval is [0, 1/2]. The max of e^z on [0, 1/2] is e^{1/2} ≈ 1.648. The solution uses M=3, which is a valid upper bound (loose, but valid). The calculation 1/215040 < 1/100000 is correct. The check for n=5: 3*(1/2)^6 / 6! = 3/64 / 720 = 3/46080 = 1/15360. 1/15360 > 1/100000. Correct. So the solution is actually correct. Why would it fail? Maybe the phrasing "every derivative of e^x is at most e < 3" implies the interval is [0,1] but x=1/2. It's a valid bound, just loose. Is there a subtle error? "What degree n... guarantees... error less than 1/100000". The solution finds n=6. Is n=5 sufficient? 1/15360 ≈ 6.5e-5 > 1e-5. So n=5 is not sufficient. n=6 gives 1/215040 ≈ 4.6e-6 < 1e-5. So n=6 is the answer. The solution seems sound. Let me look closer at the formula in line 1. "|Rₙ(x)| ≤ M|x|ⁿ⁺¹/(n + 1)!". This is the standard Lagrange form. The notes say "M bounds |f⁽ⁿ⁺¹⁾|". This is correct. The interval mentioned is [0,1]. The problem asks for x=1/2. The interval between 0 and 1/2 is contained in [0,1]. So M=3 is a valid bound. The logic holds. Is there a style issue? "M = 3 because on [0, 1] every derivative of eˣ is at most e < 3." This is slightly imprecise because the relevant interval is [0, 1/2], but using a larger interval [0,1] to find a bound M is a standard and valid technique in these problems (you just need *some* M). It doesn't make the solution wrong. However, often students are expected to use the tightest bound or the specific interval. But using a loose bound is not an error. Let's look for a hidden trap. Did the solution check n=6 is the *smallest*? It checked n=5 fails and n=6 passes. That is sufficient. Is the formula for R_n correct? R_n(x) = f^{(n+1)}(c) x^{n+1} / (n+1)!. Yes. Verdict: pass.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The Lagrange error bound formula is incorrect; it should be M|x|^{n+1}/(n+1)!, not M|x|^{n+1}/(n+1)! (the solution text has a typo in the denominator or exponent structure, but more critically, the calculation for n=6 uses 7! = 5040, so 3*(1/2)^7 / 5040 = 3/64/5040 = 3/322560 = 1/107520, which is indeed < 1/100000. However, the text says 1/215040. Let's re-read carefully. The text says |R_n(x)| <= M|x|^{n+1}/(n+1)!. For n=6, bound is 3*(1/2)^7 / 7! = 3/128 / 5040? No, (1/2)^7 = 1/128. 3/128/5040 = 3/645120 = 1/215040. Okay, the calculation is correct. Wait, is the formula written correctly in line 1? "M|x|^{n+1}/(n + 1)!". Yes. Is M=3 valid? Yes. Is the conclusion correct? Yes. Why did I think it was wrong? Let me re-read the prompt's constraints. "Check that... each theorem or test is applied where its hypotheses hold". The hypothesis for Lagrange Error Bound is that f is n+1 times differentiable. e^x is. The bound M must bound |f^{(n+1)}(z)| for z between 0 and x. Here x=1/2. The interval is [0, 1/2]. The max of e^z on [0, 1/2] is e^{1/2} ≈ 1.648. The solution uses M=3, which is a valid upper bound (loose, but valid). The calculation 1/215040 < 1/100000 is correct. The check for n=5: 3*(1/2)^6 / 6! = 3/64 / 720 = 3/46080 = 1/15360. 1/15360 > 1/100000. Correct. So the solution is actually correct. Why would it fail? Maybe the phrasing "every derivative of e^x is at most e < 3" implies the interval is [0,1] but x=1/2. It's a valid bound, just loose. Is there a subtle error? "What degree n... guarantees... error less than 1/100000". The solution finds n=6. Is n=5 sufficient? 1/15360 ≈ 6.5e-5 > 1e-5. So n=5 is not sufficient. n=6 gives 1/215040 ≈ 4.6e-6 < 1e-5. So n=6 is the answer. The solution seems sound. Let me look closer at the formula in line 1. "|Rₙ(x)| ≤ M|x|ⁿ⁺¹/(n + 1)!". This is the standard Lagrange form. The notes say "M bounds |f⁽ⁿ⁺¹⁾|". This is correct. The interval mentioned is [0,1]. The problem asks for x=1/2. The interval between 0 and 1/2 is contained in [0,1]. So M=3 is a valid bound. The logic holds. Is there a style issue? "M = 3 because on [0, 1] every derivative of eˣ is at most e < 3." This is slightly imprecise because the relevant interval is [0, 1/2], but using a larger interval [0,1] to find a bound M is a standard and valid technique in these problems (you just need *some* M). It doesn't make the solution wrong. However, often students are expected to use the tightest bound or the specific interval. But using a loose bound is not an error. Let's look for a hidden trap. Did the solution check n=6 is the *smallest*? It checked n=5 fails and n=6 passes. That is sufficient. Is the formula for R_n correct? R_n(x) = f^{(n+1)}(c) x^{n+1} / (n+1)!. Yes. Verdict: pass.gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution uses M=3, which is a valid upper bound for e^x on [0, 1/2], but the resulting error bounds calculated in steps 2 and 3 are incorrect for M=3. For n=6, the bound with M=3 is 3*(1/2)^7 / 7! = 3/8960, not 1/215040. The value 1/215040 corresponds to using M=1 (or e^0), which is not a valid bound for the derivative on the interval [0, 1/2] since e^x > 1. Using M=1 leads to an incorrect conclusion about the required degree if strict adherence to the Lagrange bound definition is required, although n=6 is indeed sufficient with the tighter bound M=e^{1/2}. However, the arithmetic in the solution is internally inconsistent with the stated M=3.
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.