Taylor and Maclaurin polynomials
Problem 7.42 · medium
Find the Maclaurin polynomial of degree 2 for \( \displaystyle f(x) = \ln{\left(x + 1 \right)} \).
- p_n(x) = Σ f⁽ᵏ⁾(0)/k! · xᵏ, for k = 0 to n.Reviewed
- \[ \left. \ln{\left(x + 1 \right)} \right|_{\substack{ x=0 }} = 0 \]f⁽0⁾(0).✓ Proved
- \[ \left. \frac{d}{d x} \ln{\left(x + 1 \right)} \right|_{\substack{ x=0 }} = 1 \]f⁽1⁾(0).✓ Proved
- \[ \left. \frac{d^{2}}{d x^{2}} \ln{\left(x + 1 \right)} \right|_{\substack{ x=0 }} = -1 \]f⁽2⁾(0).✓ Proved
- \[ \left(-1\right) x^{2} \cdot 1 \cdot \frac{1}{2} + x 1 = - \frac{x^{2}}{2} + x \]Assemble the polynomial.✓ Proved
Answer \( p_{2}(x) = - \frac{x^{2}}{2} + x \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ 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.series expands f on its own and matches |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly identifies the formula for the Maclaurin polynomial, computes the necessary derivatives and their values at x=0, and assembles the polynomial correctly.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the formula for the Maclaurin polynomial, computes the necessary derivatives and their values at x=0, and assembles the polynomial correctly.gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the formula for the Maclaurin polynomial, computes the necessary derivatives and their values at zero, and assembles the polynomial correctly.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly applies the definition of the Maclaurin polynomial, computes the necessary derivatives and their values at x=0, and assembles the polynomial correctly.gpt-oss:20b: pass 2026-09-26
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_polynomial, checked 2026-09-28 with SymPy 1.14.0.