Taylor series about a point
Problem 7.317 · medium
Find the Taylor polynomial of degree 3 for \( \displaystyle f(x) = \sin{\left(x \right)} \) centered at \( \displaystyle a = \frac{\pi}{2} \).
- \[ \left[\begin{matrix}\left. \sin{\left(x \right)} \right|_{\substack{ x=\frac{\pi}{2} }}\\\left. \frac{d}{d x} \sin{\left(x \right)} \right|_{\substack{ x=\frac{\pi}{2} }}\\\left. \frac{d^{2}}{d x^{2}} \sin{\left(x \right)} \right|_{\substack{ x=\frac{\pi}{2} }}\\\left. \frac{d^{3}}{d x^{3}} \sin{\left(x \right)} \right|_{\substack{ x=\frac{\pi}{2} }}\end{matrix}\right] = \left[\begin{matrix}1\\0\\-1\\0\end{matrix}\right] \]f and its first three derivatives at a.✓ Proved
- T(x) = Σ f⁽ʲ⁾(a)/j! · (x − a)ʲ.Reviewed
- \[ \lim_{x \to \frac{\pi}{2}^+}\left(\frac{\frac{\left(x - \frac{\pi}{2}\right)^{2}}{2} + \sin{\left(x \right)} - 1}{\left(x - \frac{\pi}{2}\right)^{3}}\right) = 0 \]T agrees with f to third order at a.✓ Proved
Answer \( T_3(x) = 1 - \frac{\left(x - \frac{\pi}{2}\right)^{2}}{2} \)
✓ 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 constructs the Taylor polynomial. The limit check confirms the approximation order, and the final answer matches the derived polynomial.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly identifies the derivatives and constructs the Taylor polynomial. The limit check confirms the approximation order, and the final answer matches the derived polynomial.gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly identifies the derivatives and constructs the Taylor polynomial. Although the stated answer is technically a degree 2 polynomial (since the degree 3 term is zero), it is the correct Taylor polynomial of degree 3, and the limit check confirms the approximation order.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.