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Newton's method

Problem 3.772 · medium Video

Use Newton's method to approximate \( \displaystyle \sqrt{5} \), as a root of \( \displaystyle f(x) = x^{2} - 5 \) with \( \displaystyle x_0 = 3 \). Write the iteration \( \displaystyle x_{n+1} = F(x_n) \) and find \( \displaystyle x_1 \) and \( \displaystyle x_2 \) exactly.

\[ \frac{d}{d x} \left(x^{2} - 5\right) = 2 x \]
f′(x).✓ Proved
\[ x - \frac{x^{2} - 5}{2 x} = \frac{x^{2} + 5}{2 x} \]
Newton's formula x − f(x)/f′(x), simplified.✓ Proved
\[ \left. \frac{x^{2} + 5}{2 x} \right|_{\substack{ x=3 }} = \frac{7}{3} \]
x₁ = F(x₀).✓ Proved
\[ \left. \frac{x^{2} + 5}{2 x} \right|_{\substack{ x=\frac{7}{3} }} = \frac{47}{21} \]
x₂ = F(x₁).✓ Proved

Answer \( x_1 = \frac{7}{3},\ x_2 = \frac{47}{21} \approx 2.238095 \)

The graph of f(x) = x² - 5. From x₀ = 3, go up to the curve, follow the tangent down to the x-axis to get x₁ = 7/3, and repeat to get x₂ = 47/21, closing in on a root.

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. 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: fail (error) — The problem explicitly asks to write the iteration formula x_{n+1} = F(x_n), but the solution only provides the simplified expression without stating the recurrence relation. Additionally, the final answer omits the exact value of x_2, providing only the approximation, whereas the problem asks for x_1 and x_2 exactly.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-11-03 — The problem explicitly asks to write the iteration formula x_{n+1} = F(x_n), but the solution only provides the simplified expression without stating the recurrence relation. Additionally, the final answer omits the exact value of x_2, providing only the approximation, whereas the problem asks for x_1 and x_2 exactly.
gpt-oss:20b: pass 2026-11-03
qwen3.6:27b-mlx: fail (error) 2026-11-02 — The solution fails to explicitly write the iteration function x_{n+1} = F(x_n) as requested by the problem statement. It only computes the values.
gpt-oss:20b: pass 2026-11-02
qwen3.6:27b-mlx: pass 2026-11-02 — The solution correctly derives the Newton's method iteration function and computes the first two iterations exactly as requested. The algebraic steps are verified and the final values match the stated answer.
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.