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

Concept: When an iterative root-finding process converges to a stable value, the value at step \( x_{n+1} \) approaches the value at step \( x_n \). This limiting value \( \alpha \) is called a fixed point of the iteration: \[ \lim_{n \to \infty} x_n = \lim_{n \to \infty} x_{n+1} = \alpha \] Substituting \( \alpha \) into the recurrence relation allows us to reconstruct the original underlying polynomial equation \( f(\alpha) = 0 \).

Step 1: Applying the fixed-point condition to the iterative formula.
The given numerical iteration equation is: \[ x_{n+1} = \frac{1}{2}\left( -\frac{7}{x_n} + x_n \right) \] Let us substitute the fixed-point limit value \( \alpha \) for both \( x_n \) and \( x_{n+1} \): \[ \alpha = \frac{1}{2}\left( -\frac{7}{\alpha} + \alpha \right) \]

Step 2: Solving the algebraic equation for \( \alpha \).
Multiply both sides of the equation by 2 to clear the fraction: \[ 2\alpha = -\frac{7}{\alpha} + \alpha \] Subtract \( \alpha \) from both sides to collect like terms: \[ 2\alpha - \alpha = -\frac{7}{\alpha} \quad \Rightarrow \quad \alpha = -\frac{7}{\alpha} \]

Step 3: Eliminating the denominator variable.
Multiply both sides of the equation by \( \alpha \) (given that \( x_0 \neq 0 \implies \alpha \neq 0 \)): \[ \alpha^2 = -7 \] Rearranging all terms onto one side of the equation: \[ \alpha^2 + 7 = 0 \]

Step 4: Identifying the function \( f(x) \).
Since the fixed-point limit \( \alpha \) represents a root of the equation \( f(x) = 0 \), we replace \( \alpha \) back with the independent variable \( x \): \[ f(x) = x^2 + 7 = 7 + x^2 \] This matches option (C).