Newton's method
Problem 3.462 · medium
Use Newton's method on \( \displaystyle f(x) = x^{3} + 2 x + 3 \) with \( \displaystyle x_0 = -2 \). Write the iteration \( \displaystyle x_{n+1} = F(x_n) \) and find \( \displaystyle x_1 \) and \( \displaystyle x_2 \) exactly.
- \[ \frac{d}{d x} \left(x^{3} + 2 x + 3\right) = 3 x^{2} + 2 \]f′(x).✓ Proved
- \[ x - \frac{x^{3} + 2 x + 3}{3 x^{2} + 2} = \frac{2 x^{3} - 3}{3 x^{2} + 2} \]Newton's formula x − f(x)/f′(x), simplified.✓ Proved
- \[ \left. \frac{2 x^{3} - 3}{3 x^{2} + 2} \right|_{\substack{ x=-2 }} = - \frac{19}{14} \]x₁ = F(x₀).✓ Proved
- \[ \left. \frac{2 x^{3} - 3}{3 x^{2} + 2} \right|_{\substack{ x=- \frac{19}{14} }} = - \frac{439}{413} \]x₂ = F(x₁).✓ Proved
Answer \( x_1 = - \frac{19}{14},\ x_2 = - \frac{439}{413} \approx -1.062954 \)
Every line of this solution was proved by the computer algebra system SymPy. 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 | ✓ 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 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | two Newton steps in 30-digit floating point |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly derives the Newton iteration function, simplifies it algebraically, and computes the first two iterations exactly as requested. The steps are logically sound and the arithmetic is verified.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly derives the Newton iteration function, simplifies it algebraically, and computes the first two iterations exactly as requested. The steps are logically sound and the arithmetic is verified.gpt-oss:20b: fail (error) 2026-10-06 — The computed value for x₂ is incorrect; the algebraic simplification of Newton’s iteration at x₁ = –19/14 does not yield –439/413. The correct value is –3073/2889 (≈ –1.0631).qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly derives the Newton's method iteration function, simplifies the algebra, and computes the first two iterations exactly as requested.
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/newtons_method, checked 2026-10-06 with SymPy 1.14.0.