Taylor series about a point
Problem 7.312 · hard
Find the Taylor polynomial of degree 3 for \( \displaystyle f(x) = e^{x} \) centered at \( \displaystyle a = -1 \).
- \[ \left[\begin{matrix}\left. e^{x} \right|_{\substack{ x=-1 }}\\\left. \frac{d}{d x} e^{x} \right|_{\substack{ x=-1 }}\\\left. \frac{d^{2}}{d x^{2}} e^{x} \right|_{\substack{ x=-1 }}\\\left. \frac{d^{3}}{d x^{3}} e^{x} \right|_{\substack{ x=-1 }}\end{matrix}\right] = \left[\begin{matrix}e^{-1}\\e^{-1}\\e^{-1}\\e^{-1}\end{matrix}\right] \]f and its first three derivatives at a.✓ Proved
- T(x) = Σ f⁽ʲ⁾(a)/j! · (x − a)ʲ.
- \[ \lim_{x \to -1^+}\left(\frac{- \frac{\left(x + 1\right)^{3}}{6 e} - \frac{\left(x + 1\right)^{2}}{2 e} - \frac{x + 1}{e} + e^{x} - e^{-1}}{\left(x + 1\right)^{3}}\right) = 0 \]T agrees with f to third order at a.✓ Proved
Answer \( T_3(x) = \frac{\left(x + 1\right)^{3}}{6 e} + \frac{\left(x + 1\right)^{2}}{2 e} + \frac{x + 1}{e} + e^{-1} \)
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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Not checked | — | a sentence; read, not computed |
| 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 | SymPy's series() expansion about a |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution claims to find the Taylor polynomial for e^x, but the limit in step 3 checks a polynomial with negative coefficients against e^x. The correct Taylor polynomial has all positive coefficients (since derivatives of e^x are positive). The limit shown is for -T_3(x) + e^{-1}, not T_3(x).
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution claims to find the Taylor polynomial for e^x, but the limit in step 3 checks a polynomial with negative coefficients against e^x. The correct Taylor polynomial has all positive coefficients (since derivatives of e^x are positive). The limit shown is for -T_3(x) + e^{-1}, not T_3(x).gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly identifies the derivatives and applies the Taylor polynomial formula. The final expression matches the stated answer.gpt-oss:20b: pass 2026-10-05
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/taylor_series_center, checked 2026-10-05 with SymPy 1.14.0.