Recurrence Relations: Quadratic recurrence: fixed points & monotonicity — TMJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
The diagram below shows the graph of \(y = x^2 - x - 2\). The two roots of the equation \(x^2 - x - 2 = 0\) are \(-1\) and \(2\).
A sequence of real numbers \(x_1, x_2, x_3, \ldots\) satisfies the recurrence relation \(x_{n+1} = x_n^{\,2} - 2\) for \(n \geq 1\).
- It is given that as \(n \to \infty\), \(x_n \to L\). Show that \(L\) can take values \(-1\) or \(2\).
- Hence or otherwise, determine the behaviour of the sequence when \(x_1 = -1\).
- By considering \(x_{n+1} - x_n\), prove that \(x_{n+1} < x_n\) if \(-1 < x_n < 2\).
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(a) Let \(x_n \to L\) as \(n \to \infty\). Then \(x_{n+1} \to L\). \[\begin{aligned} L &= L^2 - 2 \\ L^2 - L - 2 &= 0 \\ (L + 1)(L - 2) &= 0 \\ L &= -1 \text{ or } L = 2 \quad \text{(shown)}. \end{aligned}\]
(b) Method 1: Since \(x_1 = -1 = L\), the sequence remains constant at \(-1\).
Method 2: Using GC, the sequence remains constant at \(-1\).
(c) \[\begin{aligned} x_{n+1} - x_n &= x_n^{\,2} - 2 - x_n \\ &= x_n^{\,2} - x_n - 2. \end{aligned}\] If \(-1 < x_n < 2\), then \(x_n^{\,2} - x_n - 2 < 0\) from the graph.
Hence \(x_{n+1} - x_n < 0 \Rightarrow x_{n+1} < x_n\).