Taylor series about a point
Problem 7.405 · hard
Find the Taylor polynomial of degree 3 for \( \displaystyle f(x) = e^{x} \) centered at \( \displaystyle a = 2 \).
- \[ \left[\begin{matrix}\left. e^{x} \right|_{\substack{ x=2 }}\\\left. \frac{d}{d x} e^{x} \right|_{\substack{ x=2 }}\\\left. \frac{d^{2}}{d x^{2}} e^{x} \right|_{\substack{ x=2 }}\\\left. \frac{d^{3}}{d x^{3}} e^{x} \right|_{\substack{ x=2 }}\end{matrix}\right] = \left[\begin{matrix}e^{2}\\e^{2}\\e^{2}\\e^{2}\end{matrix}\right] \]f and its first three derivatives at a.✓ Proved
- T(x) = Σ f⁽ʲ⁾(a)/j! · (x − a)ʲ.Reviewed
- \[ \lim_{x \to 2^+}\left(\frac{- \frac{\left(x - 2\right)^{3} e^{2}}{6} - \frac{\left(x - 2\right)^{2} e^{2}}{2} - \left(x - 2\right) e^{2} + e^{x} - e^{2}}{\left(x - 2\right)^{3}}\right) = 0 \]T agrees with f to third order at a.✓ Proved
Answer \( T_3(x) = \frac{\left(x - 2\right)^{3} e^{2}}{6} + \frac{\left(x - 2\right)^{2} e^{2}}{2} + \left(x - 2\right) e^{2} + e^{2} \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: pass — The solution correctly identifies the derivatives and applies the Taylor polynomial formula. The verification step confirms the remainder is of higher order, validating the result.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly identifies the derivatives and applies the Taylor polynomial formula. The verification step confirms the remainder is of higher order, validating the result.gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly identifies the derivatives and applies the Taylor polynomial formula. The final expression matches the stated answer, and the limit check confirms the order of accuracy.
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 2026-10-07 with SymPy 1.14.0.