Question

The real sequence generated by the iterative scheme
\[ x_n = \frac{x_{n-1}}{2} + \frac{1}{x_{n-1}}, \quad n \geq 1 \]

• A. converges to \( \sqrt{2} \), for all \( x_0 \in \mathbb{R} \setminus \{ 0 \} \)
• B. converges to \( \sqrt{2} \), whenever \( x_0>\frac{\sqrt{2}}{3} \)
• C. converges to \( \sqrt{2} \), whenever \( x_0 \in (-1, 1) \setminus \{ 0 \} \)
• D. diverges for any \( x_0 \neq 0 \)

Hint:

For iterative sequences, find the limit by assuming the sequence converges to a value and solving the corresponding equation.

Answer 1

The Correct Option is B

The given iterative scheme is: \[ x_n = \frac{x_{n-1}}{2} + \frac{1}{x_{n-1}} \] This is a recurrence relation for the sequence \( x_n \). To find the limit of this sequence, assume it converges to \( L \). As \( n \to \infty \), we have: \[ L = \frac{L}{2} + \frac{1}{L} \] Multiplying both sides by \( L \), we get: \[ L^2 = \frac{L^2}{2} + 1 \] Simplifying: \[ L^2 - \frac{L^2}{2} = 1 \] \[ \frac{L^2}{2} = 1 \] \[ L^2 = 2 \] Thus, \( L = \sqrt{2} \) or \( L = -\sqrt{2} \). Since the initial value \( x_0>\frac{\sqrt{2}}{3} \), the sequence will converge to \( \sqrt{2} \), as the sequence is positive and decreasing. Therefore, the sequence converges to \( \sqrt{2} \) for \( x_0>\frac{\sqrt{2}}{3} \).