Question

Let \( \{x_n\} \) be a real sequence such that \( 7x_{n+1} = x_n^3 + 6 \) for \( n \geq 1 \). Then which of the following statements is/are TRUE?

• A. If \( x_1 = \frac{1}{2} \), then \( \{x_n\} \) converges to 1.
• B. If \( x_1 = \frac{1}{2} \), then \( \{x_n\} \) converges to 2.
• C. If \( x_1 = \frac{3}{2} \), then \( \{x_n\} \) converges to 1.
• D. If \( x_1 = \frac{3}{2} \), then \( \{x_n\} \) converges to -3.

Hint:

To analyze the convergence of a recurrence relation, find its fixed points and check the stability of each fixed point by computing the derivative of the recurrence function.

Answer 1

The Correct Option is A, C

Step 1: Analyzing the recurrence relation.
The recurrence is given by \( 7x_{n+1} = x_n^3 + 6 \), or equivalently: \[ x_{n+1} = \frac{x_n^3 + 6}{7}. \] This is a nonlinear recurrence relation.
Step 2: Finding fixed points.
To find the limit of the sequence, assume that \( \lim_{n \to \infty} x_n = L \). Then, taking the limit on both sides of the recurrence relation, we get: \[ L = \frac{L^3 + 6}{7}. \] Solving \( 7L = L^3 + 6 \), we obtain the cubic equation: \[ L^3 - 7L + 6 = 0. \] Factoring the cubic equation gives: \[ (L - 1)(L^2 + L - 6) = 0. \] The solutions are \( L = 1, -3, 2 \).
Step 3: Analyzing the stability of the fixed points.
- For \( L = 1 \), we compute the derivative of the recurrence function \( f(x) = \frac{x^3 + 6}{7} \). At \( x = 1 \), the derivative is \( f'(1) = \frac{3 \times 1^2}{7} = \frac{3}{7} \), which is less than 1, indicating that \( L = 1 \) is stable. - For \( L = -3 \) and \( L = 2 \), we analyze the stability and find that \( L = 1 \) is the only stable fixed point.
Step 4: Conclusion.
Thus, the correct answers are (A) and (C).