Home›Calculus 1›Newton's method›Problem 3.770

Newton's method

Problem 3.770 · medium Video

Use Newton's method on \( \displaystyle f(x) = x^{3} - 2 x - 1 \) 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.

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

Answer \( x_1 = -1,\ x_2 = -1 \approx -1.000000 \)

The graph of f(x) = x³ - 2x - 1. From x₀ = -1, go up to the curve, follow the tangent down to the x-axis to get x₁ = -1, and repeat to get x₂ = -1, 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 function and accurately computes x1 and x2. Since x0 = -1 is a root of f(x), the method correctly stabilizes at -1.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-11-03 — The solution correctly derives the Newton iteration function and accurately computes x1 and x2. Since x0 = -1 is a root of f(x), the method correctly stabilizes at -1.
gpt-oss:20b: pass 2026-11-03
qwen3.6:27b-mlx: pass 2026-11-02 — The solution correctly derives the Newton iteration formula and accurately computes the fixed point x = -1. Since x_0 = -1 is a root, x_1 and x_2 are correctly identified as -1.
gpt-oss:20b: pass 2026-11-02
qwen3.6:27b-mlx: pass 2026-11-02 — The solution correctly derives the Newton iteration function and accurately computes the iterations. Since x_0 = -1 is a root of the polynomial, the method correctly converges immediately to x_1 = -1 and x_2 = -1.
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.