Taylor and Maclaurin polynomials
Problem 7.137 · hard
Find the Maclaurin polynomial of degree 4 for \( \displaystyle f(x) = \frac{1}{1 - x} \).
- p_n(x) = Σ f⁽ᵏ⁾(0)/k! · xᵏ, for k = 0 to n.Reviewed
- \[ \left. \frac{1}{1 - x} \right|_{\substack{ x=0 }} = 1 \]f⁽0⁾(0).✓ Proved
- \[ \left. \frac{d}{d x} \frac{1}{1 - x} \right|_{\substack{ x=0 }} = 1 \]f⁽1⁾(0).✓ Proved
- \[ \left. \frac{d^{2}}{d x^{2}} \frac{1}{1 - x} \right|_{\substack{ x=0 }} = 2 \]f⁽2⁾(0).✓ Proved
- \[ \left. \frac{d^{3}}{d x^{3}} \frac{1}{1 - x} \right|_{\substack{ x=0 }} = 6 \]f⁽3⁾(0).✓ Proved
- \[ \left. \frac{d^{4}}{d x^{4}} \frac{1}{1 - x} \right|_{\substack{ x=0 }} = 24 \]f⁽4⁾(0).✓ Proved
- \[ 24 x^{4} \cdot 1 \cdot \frac{1}{24} + 6 x^{3} \cdot 1 \cdot \frac{1}{6} + 2 x^{2} \cdot 1 \cdot \frac{1}{2} + x 1 + 1 = x^{4} + x^{3} + x^{2} + x + 1 \]Assemble the polynomial.✓ Proved
Answer \( p_{4}(x) = x^{4} + x^{3} + x^{2} + x + 1 \)
✓ Nihil obstat Lines: 6 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 | 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 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 7 | ✓ 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 applies the definition of the Maclaurin polynomial, computes the derivatives and their values at zero accurately, and assembles the final polynomial correctly.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly applies the definition of the Maclaurin polynomial, computes the derivatives and their values at zero accurately, and assembles the final polynomial correctly.gpt-oss:20b: pass 2026-09-26qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly applies the definition of the Maclaurin polynomial, computes the derivatives and their values at zero accurately, and assembles the final 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 2026-09-26 with SymPy 1.14.0.