Question

Given \( x_0 \neq 0 \), if the iteration formula \( x_{n+1} = \frac{1}{2}\left( -\frac{7}{x_n} + x_n \right) \), \( n \geq 0 \) is used to find the root of \( f(x) = 0 \), then \( f(x) = \)

• A. \( 9 + x^2 \)
• B. \( 4 + x^2 \)
• C. \( 7 + x^2 \)
• D. \( -7 + x^2 \)

Hint:

This formula matches Newton-Raphson iteration \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \) for finding purely imaginary roots, or a rearranged fixed-point iteration format \( x = g(x) \) designed to isolate roots where \( x^2 = -7 \).

Answer 1

The Correct Option is C

Step 1: Compare the iteration to the standard Newton-Raphson form.
The classic Newton-Raphson iteration for solving \( x^2 = c \) (finding \( \sqrt{c} \)) is \[ x_{n+1} = \frac{1}{2}\left(x_n + \frac{c}{x_n}\right) \] This comes from applying Newton's formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \) to \( f(x) = x^2 - c \).

Step 2: Match the given formula to this pattern.
The given iteration is \[ x_{n+1} = \frac{1}{2}\left(-\frac{7}{x_n} + x_n\right) = \frac{1}{2}\left(x_n + \frac{-7}{x_n}\right) \] Comparing directly with \( \frac{1}{2}\left(x_n + \frac{c}{x_n}\right) \), we read off \( c = -7 \).

Step 3: Recover \( f(x) \).
Since this iteration solves \( x^2 = c \), that is \( f(x) = x^2 - c = 0 \), substituting \( c = -7 \): \[ f(x) = x^2 - (-7) = \boxed{x^2 + 7} \]
which matches option (C).