Newton's method
Problem 3.413 · medium
Use Newton's method on \( \displaystyle f(x) = x^{3} + x - 2 \) with \( \displaystyle x_0 = 1 \). Write the iteration \( \displaystyle x_{n+1} = F(x_n) \) and find \( \displaystyle x_1 \) and \( \displaystyle x_2 \) exactly.
- \[ \frac{d}{d x} \left(x^{3} + x - 2\right) = 3 x^{2} + 1 \]f′(x).✓ Proved
- \[ x - \frac{x^{3} + x - 2}{3 x^{2} + 1} = \frac{2 x^{3} + 2}{3 x^{2} + 1} \]Newton's formula x − f(x)/f′(x), simplified.✓ Proved
- \[ \left. \frac{2 x^{3} + 2}{3 x^{2} + 1} \right|_{\substack{ x=1 }} = 1 \]x₁ = F(x₀).✓ Proved
- \[ \left. \frac{2 x^{3} + 2}{3 x^{2} + 1} \right|_{\substack{ x=1 }} = 1 \]x₂ = F(x₁).✓ Proved
Answer \( x_1 = 1,\ x_2 = 1 \approx 1.000000 \)
✓ Nihil obstat 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. 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 | ✓ 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 iterations. Since x=1 is a root of f(x)=0, the method correctly converges immediately to x_1=1 and x_2=1.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly derives the Newton iteration function, simplifies it algebraically, and computes the iterations. Since x=1 is a root of f(x)=0, the method correctly converges immediately to x_1=1 and x_2=1.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly derives the Newton iteration formula and accurately computes the iterations. Since x=1 is a root of f(x), the method converges immediately, making x_1 = 1 and x_2 = 1 correct.gpt-oss:20b: pass 2026-10-04
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 2026-10-04 with SymPy 1.14.0.