Recurrence Relations: Logistic map: convergence for different \(u_1\) — EJC 2025 H2 Math Prelim Paper 2
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Question
A sequence \(u_1, u_2, u_3, \ldots\) is defined by \[ u_{n+1} = 2u_n(1 - u_n) \] for all positive integers \(n\).
- Find the possible limits of the sequence, if the sequence is convergent.
- For each of the following, evaluate \(u_2\) and \(u_3\), and state whether the sequence is convergent:
- \(u_1 = 0\),
- \(u_1 = 0.01\),
- \(u_1 = -0.01\).
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(a) Let \(L\) be the limit of the sequence \(u_{n+1} = 2u_n(1 - u_n)\). Setting \(u_{n+1} = u_n = L\): \[\begin{aligned} L &= 2L(1 - L)\\ 2L^2 - L &= 0\\ L(2L - 1) &= 0 \end{aligned}\]
\[ L = 0 \quad \text{or} \quad L = \tfrac{1}{2} \](b) Apply \(u_{n+1} = 2u_n(1 - u_n)\) for each starting value.
Case 1: \(u_1 = 0\). \[ u_2 = 2(0)(1 - 0) = 0, \qquad u_3 = 0. \] The sequence is constant at \(0\) and is convergent (to \(0\)).
Case 2: \(u_1 = 0.01\). \[\begin{aligned} u_2 &= 2(0.01)(1 - 0.01) = 0.0198 \quad (\text{exact})\\ u_3 &= 2(0.0198)(1 - 0.0198) = 0.0388 \quad (\text{3 s.f.}) \end{aligned}\] Terms approach the limit \(L = 0\), so the sequence is convergent.
Case 3: \(u_1 = -0.01\). \[\begin{aligned} u_2 &= 2(-0.01)(1 - (-0.01)) = -0.0202 \quad (\text{exact})\\ u_3 &= 2(-0.0202)(1 - (-0.0202)) = -0.0412 \quad (\text{3 s.f.}) \end{aligned}\] Terms move away from \(0\) (magnitudes grow), so the sequence is not convergent.