Question

Consider the iterative formula \( x_{n+1} = \frac{x_n}{2} + \frac{9}{8x_n} \) with \( x_0 = 0.5 \) obtained from the Newton-Raphson method to solve an equation. The successive iterations converge to:

• A. \( 1.1 \)
• B. \( 1.5 \)
• C. \( 1.9 \)
• D. \( \sqrt{2} \)

Hint:

This specific iteration formula is the Newton-Raphson method for finding the square root of \( \frac{9}{4} \). The general form is \( x_{n+1} = \frac{1}{2}(x_n + \frac{a}{x_n}) \).

Answer 1

The Correct Option is B

Concept: Successive iterations of a convergent sequence \( x_{n+1} = g(x_n) \) converge to a fixed point.

• At convergence, we assume \( \lim_{n \to \infty} x_n = L \).

• Therefore, \( \lim_{n \to \infty} x_{n+1} = L \).

• The equation becomes \( L = g(L) \).

Step 1: Set up the fixed-point equation.
Let the limit of the sequence be \( L \). \[ L = \frac{L}{2} + \frac{9}{8L} \]

Step 2: Solve for \( L \).
Subtract \( \frac{L}{2} \) from both sides. \[ L - \frac{L}{2} = \frac{9}{8L} \implies \frac{L}{2} = \frac{9}{8L} \] Cross-multiply to isolate \( L^2 \). \[ 8L^2 = 18 \implies L^2 = \frac{18}{8} = \frac{9}{4} \]

Step 3: Identify the positive root.
Taking the square root: \[ L = \sqrt{2.25} = 1.5 \] Since the initial guess \( x_0 = 0.5 \) is positive, it converges to the positive root. Final Answer: (B)