∫Calc Practice

Taylor series about a point

Problem 7.405 · hard

Find the Taylor polynomial of degree 3 for \( \displaystyle f(x) = e^{x} \) centered at \( \displaystyle a = 2 \).
  1. \[ \left[\begin{matrix}\left. e^{x} \right|_{\substack{ x=2 }}\\\left. \frac{d}{d x} e^{x} \right|_{\substack{ x=2 }}\\\left. \frac{d^{2}}{d x^{2}} e^{x} \right|_{\substack{ x=2 }}\\\left. \frac{d^{3}}{d x^{3}} e^{x} \right|_{\substack{ x=2 }}\end{matrix}\right] = \left[\begin{matrix}e^{2}\\e^{2}\\e^{2}\\e^{2}\end{matrix}\right] \]
    f and its first three derivatives at a.✓ Proved
  2. T(x) = Σ f⁽ʲ⁾(a)/j! · (x − a)ʲ.
    Reviewed
  3. \[ \lim_{x \to 2^+}\left(\frac{- \frac{\left(x - 2\right)^{3} e^{2}}{6} - \frac{\left(x - 2\right)^{2} e^{2}}{2} - \left(x - 2\right) e^{2} + e^{x} - e^{2}}{\left(x - 2\right)^{3}}\right) = 0 \]
    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
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxa sentence; read, not computed
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0SymPy's series() expansion about a

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-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 2026-10-07
  • qwen3.6:27b-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 2026-10-07
  • qwen3.6:27b-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.