Newton's method
Problem 3.506 · medium
Use Newton's method to approximate \( \displaystyle \sqrt{13} \), as a root of \( \displaystyle f(x) = x^{2} - 13 \) 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.
- \[ \frac{d}{d x} \left(x^{2} - 13\right) = 2 x \]f′(x).✓ Proved
- \[ x - \frac{x^{2} - 13}{2 x} = \frac{x^{2} + 13}{2 x} \]Newton's formula x − f(x)/f′(x), simplified.✓ Proved
- \[ \left. \frac{x^{2} + 13}{2 x} \right|_{\substack{ x=3 }} = \frac{11}{3} \]x₁ = F(x₀).✓ Proved
- \[ \left. \frac{x^{2} + 13}{2 x} \right|_{\substack{ x=\frac{11}{3} }} = \frac{119}{33} \]x₂ = F(x₁).✓ Proved
Answer \( x_1 = \frac{11}{3},\ x_2 = \frac{119}{33} \approx 3.606061 \)
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.
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: fail (error) — The solution fails to explicitly write the iteration formula x_{n+1} = F(x_n) as requested by the problem statement. It only provides the simplified algebraic expression for F(x) in line 2, but does not state the recurrence relation itself.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution fails to explicitly write the iteration formula x_{n+1} = F(x_n) as requested by the problem statement. It only provides the simplified algebraic expression for F(x) in line 2, but does not state the recurrence relation itself.gpt-oss:20b: pass 2026-10-08gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: fail (error) 2026-10-08 — The calculated value for x_2 is incorrect; substituting x_1 = 11/3 into the iteration formula yields 248/119, not 119/33. Additionally, the solution fails to explicitly state the iteration function F(x) as requested by the problem.
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.