Let \( p_1<p_2 \) be the two fixed points of the function \( g(x) = e^x - 2 \), where \( x \in {R} \). For \( x_0 \in {R} \), let the sequence \( (x_n)_{n \geq 1} \) be generated by the fixed-point iteration \[ x_n = g(x_{n-1}), \quad n \geq 1. \] Which one of the following is/are correct?
Let \( p_1<p_2 \) be the two fixed points of the function \( g(x) = e^x - 2 \), where \( x \in {R} \). For \( x_0 \in {R} \), let the sequence \( (x_n)_{n \geq 1} \) be generated by the fixed-point iteration \[ x_n = g(x_{n-1}), \quad n \geq 1. \] Which one of the following is/are correct?
Show Hint
- \( (x_n)_{n \geq 0} \) converges to \( p_1 \) for any \( x_0 \in (p_1, p_2) \)
- \( (x_n)_{n \geq 0} \) converges to \( p_2 \) for any \( x_0 \in (p_1, p_2) \)
- \( (x_n)_{n \geq 0} \) converges to \( p_2 \) for any \( x_0>p_2 \)
- \( (x_n)_{n \geq 0} \) converges to \( p_1 \) for any \( x_0<p_1 \)
The Correct Option is A
Solution and Explanation
Step 1: Fixed points of \( g(x) \). The fixed points \( p_1 \) and \( p_2 \) satisfy \( g(p) = p \), which corresponds to solving \( e^p - 2 = p \).
Step 2: Behavior of the iteration. The convergence behavior depends on the derivative of \( g(x) \) at the fixed points: \[ g'(x) = e^x. \] - At \( p_1 \), \( |g'(p_1)|<1 \), implying that \( p_1 \) is an attracting fixed point. - At \( p_2 \), \( |g'(p_2)|>1 \), implying that \( p_2 \) is a repelling fixed point.
Step 3: Convergence analysis. - For \( x_0 \in (p_1, p_2) \), the sequence \( (x_n) \) converges to \( p_1 \) because \( p_1 \) is the attracting fixed point. - For \( x_0<p_1 \), the sequence also converges to \( p_1 \) due to the monotonic behavior of \( g(x) \) in this region. - For \( x_0>p_2 \), the sequence does not converge to \( p_2 \), as \( p_2 \) is repelling.
Step 4: Conclusion. The correct answers are \( {(1), (4)} \).
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