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Newton's method

Problem 3.774 · medium Video

Use Newton's method on \( \displaystyle f(x) = x^{3} - 3 x + 1 \) 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.

\[ \frac{d}{d x} \left(x^{3} - 3 x + 1\right) = 3 x^{2} - 3 \]
f′(x).✓ Proved
\[ x - \frac{x^{3} - 3 x + 1}{3 x^{2} - 3} = \frac{2 x^{3} - 1}{3 x^{2} - 3} \]
Newton's formula x − f(x)/f′(x), simplified.✓ Proved
\[ \left. \frac{2 x^{3} - 1}{3 \left(x^{2} - 1\right)} \right|_{\substack{ x=2 }} = \frac{5}{3} \]
x₁ = F(x₀).✓ Proved
\[ \left. \frac{2 x^{3} - 1}{3 \left(x^{2} - 1\right)} \right|_{\substack{ x=\frac{5}{3} }} = \frac{223}{144} \]
x₂ = F(x₁).✓ Proved

Answer \( x_1 = \frac{5}{3},\ x_2 = \frac{223}{144} \approx 1.548611 \)

The graph of f(x) = x³ - 3x + 1. From x₀ = 2, go up to the curve, follow the tangent down to the x-axis to get x₁ = 5/3, and repeat to get x₂ = 223/144, closing in on a root.

✓ 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. How we verify

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: pass
qwen3.6:27b-mlx: pass — The solution correctly derives the Newton iteration formula, simplifies it algebraically, and computes the first two iterations exactly as requested. No sentences require review as all steps are verified equations.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-11-03 — The solution correctly derives the Newton iteration formula, simplifies it algebraically, and computes the first two iterations exactly as requested. No sentences require review as all steps are verified equations.
gpt-oss:20b: pass 2026-11-03
qwen3.6:27b-mlx: pass 2026-11-02
gpt-oss:20b: pass 2026-11-02
qwen3.6:27b-mlx: pass 2026-11-02 — The solution correctly derives the Newton iteration function and computes the exact values for x_1 and x_2. The algebraic simplification and substitutions are correct.
gpt-oss:20b: pass 2026-11-02

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 2026-11-03 with SymPy 1.14.0.