Taylor series about a point
Problem 7.404 · medium
Find the Taylor polynomial of degree 3 for \( \displaystyle f(x) = \frac{1}{x} \) centered at \( \displaystyle a = -1 \).
- \[ \left[\begin{matrix}\left. \frac{1}{x} \right|_{\substack{ x=-1 }}\\\left. \frac{d}{d x} \frac{1}{x} \right|_{\substack{ x=-1 }}\\\left. \frac{d^{2}}{d x^{2}} \frac{1}{x} \right|_{\substack{ x=-1 }}\\\left. \frac{d^{3}}{d x^{3}} \frac{1}{x} \right|_{\substack{ x=-1 }}\end{matrix}\right] = \left[\begin{matrix}-1\\-1\\-2\\-6\end{matrix}\right] \]f and its first three derivatives at a.✓ Proved
- T(x) = Σ f⁽ʲ⁾(a)/j! · (x − a)ʲ.
- \[ \lim_{x \to -1^+}\left(\frac{x + \left(x + 1\right)^{3} + \left(x + 1\right)^{2} + 2 + \frac{1}{x}}{\left(x + 1\right)^{3}}\right) = 0 \]T agrees with f to third order at a.✓ Proved
Answer \( T_3(x) = - x - \left(x + 1\right)^{3} - \left(x + 1\right)^{2} - 2 \)
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: fail (misleading) — The limit expression in step 3 uses the wrong signs for the Taylor polynomial; it should be T(x)=−x−(x+1)²−(x+1)³−2, not the expression shown. This would mislead a student about the correct polynomial.qwen3.6:27b-mlx: fail (error) — The stated answer is incorrect; the correct Taylor polynomial is T_3(x) = -1 - (x+1) - (x+1)^2 - (x+1)^3. The provided solution claims T_3(x) = -x - (x+1)^3 - (x+1)^2 - 2, which simplifies to -1 - (x+1)^2 - (x+1)^3, missing the linear term. Furthermore, the limit check in line 3 is flawed because it tests the difference against (x+1)^3, which would be zero even if the linear term were missing (since the error would be O((x+1)^2), and dividing by (x+1)^3 yields infinity, not 0; wait, if the linear term is missing, the error is O((x+1)^2), so the limit of Error/(x+1)^3 is infinity. The limit check actually fails for the stated answer, but the solution claims it equals 0, which is false. The stated answer is wrong and the verification is mathematically invalid.
Every verdict on record (4)
gpt-oss:20b: fail (misleading) 2026-10-07 — The limit expression in step 3 uses the wrong signs for the Taylor polynomial; it should be T(x)=−x−(x+1)²−(x+1)³−2, not the expression shown. This would mislead a student about the correct polynomial.qwen3.6:27b-mlx: fail (error) 2026-10-07 — The stated answer is incorrect; the correct Taylor polynomial is T_3(x) = -1 - (x+1) - (x+1)^2 - (x+1)^3. The provided solution claims T_3(x) = -x - (x+1)^3 - (x+1)^2 - 2, which simplifies to -1 - (x+1)^2 - (x+1)^3, missing the linear term. Furthermore, the limit check in line 3 is flawed because it tests the difference against (x+1)^3, which would be zero even if the linear term were missing (since the error would be O((x+1)^2), and dividing by (x+1)^3 yields infinity, not 0; wait, if the linear term is missing, the error is O((x+1)^2), so the limit of Error/(x+1)^3 is infinity. The limit check actually fails for the stated answer, but the solution claims it equals 0, which is false. The stated answer is wrong and the verification is mathematically invalid.gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: fail (error) 2026-10-07 — The stated answer T_3(x) = -x - (x+1)^3 - (x+1)^2 - 2 is incorrect. The correct Taylor polynomial is -1 - (x+1) - (x+1)^2 - (x+1)^3. The solution fails to convert the derivatives into the correct polynomial terms, and the verification limit in line 3 is mathematically false (the limit does not exist or is not 0 for the given expression), indicating a fundamental error in the derivation.
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.