Newton's method
Problem 3.514 · medium
Use Newton's method to approximate \( \displaystyle \sqrt{3} \), as a root of \( \displaystyle f(x) = x^{2} - 3 \) 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.
- \[ \frac{d}{d x} \left(x^{2} - 3\right) = 2 x \]f′(x).✓ Proved
- \[ x - \frac{x^{2} - 3}{2 x} = \frac{x^{2} + 3}{2 x} \]Newton's formula x − f(x)/f′(x), simplified.✓ Proved
- \[ \left. \frac{x^{2} + 3}{2 x} \right|_{\substack{ x=1 }} = 2 \]x₁ = F(x₀).✓ Proved
- \[ \left. \frac{x^{2} + 3}{2 x} \right|_{\substack{ x=2 }} = \frac{7}{4} \]x₂ = F(x₁).✓ Proved
Answer \( x_1 = 2,\ x_2 = \frac{7}{4} \approx 1.750000 \)
✓ 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's method iteration function, applies it to the given initial value, and computes the first two approximations exactly as requested.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly derives the Newton's method iteration function, applies it to the given initial value, and computes the first two approximations exactly as requested.gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly derives the Newton iteration formula, applies it to the given initial value, and computes the exact values for x_1 and x_2 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 2026-10-08 with SymPy 1.14.0.