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

Step 1: Recognise the form of Newton-Raphson.
Newton's method for finding a root of \( f(x) = x^2 - c \) gives the iteration \( x_{n+1} = x_n - \dfrac{x_n^2-c}{2x_n} = \dfrac{x_n}{2} + \dfrac{c}{2x_n} \).

Step 2: Match coefficients with the given formula.
Comparing \( \dfrac{x_n}{2} + \dfrac{9}{8x_n} \) with \( \dfrac{x_n}{2} + \dfrac{c}{2x_n} \) gives \( \dfrac{c}{2} = \dfrac{9}{8} \), so \( c = \dfrac{9}{4} \). The iteration is therefore solving \( x^2 = \dfrac{9}{4} \).

Step 3: Take the root consistent with the starting value.
\[ x = \sqrt{\frac{9}{4}} = 1.5 \] Since \( x_0 = 0.5 \gt 0 \), the sequence converges to the positive root. \[ \boxed{1.5} \]