For finding a simple isolated real root $x = \alpha$ of a sufficiently smooth non-linear function $f(x) = 0$ with $f'(\alpha) \ne 0$, the order of convergence of the Newton-Raphson iteration $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ is:

(A) Quadratic ($p = 2$)

(B) Linear ($p = 1$)

(C) Superlinear ($p = 1.618$)

(D) Cubic ($p = 3$)

Answer Key:

A

### 5-Stage Derivation & Analytical Solution: 1. **Stage 1: Taylor Series Expansion of Error:** Let $\epsilon_n = x_n - \alpha$. Expanding $f(x_n)$ and $f'(x_n)$ around $\alpha$: * $f(x_n) = f(\alpha) + \epsilon_n f'(\alpha) + \frac{\epsilon_n^2}{2} f''(\alpha) + \dots = \epsilon_n f'(\alpha) + \frac{\epsilon_n^2}{2} f''(\alpha) + \dots$ * $f'(x_n) = f'(\alpha) + \epsilon_n f''(\alpha) + \dots$ 2. **Stage 2: Iteration Error Relation:** * $\epsilon_{n+1} = \epsilon_n - \frac{\epsilon_n f'(\alpha) + \frac{1}{2}\epsilon_n^2 f''(\alpha)}{f'(\alpha) [1 + \epsilon_n \frac{f''(\alpha)}{f'(\alpha)}]}$. * Binomial expansion yields $\epsilon_{n+1} \approx \frac{f''(\alpha)}{2 f'(\alpha)} \epsilon_n^2$. 3. **Stage 3: Definition of Order of Convergence:** Since $\lim_{n\to\infty} \frac{|\epsilon_{n+1}|}{|\epsilon_n|^2} = \text{constant} > 0$, the order is $p = 2$ (Quadratic). 4. **Stage 4: Distractor Analysis:** Secant method has $p \approx 1.618$ (superlinear). Bisection has $p = 1$ (linear). 5. **Stage 5: Official Reference:** Erwin Kreyszig, *Advanced Engineering Mathematics*, Section 19.2.

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