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2002 Paper 2 Q5
D: 1600.0 B: 1495.1
proof by induction recurrence relation sequences quadratic logistic map fixed points periodic orbits inequality

The numbers \(x_n\), where \(n=0\), \(1\), \(2\), \(\ldots\) , satisfy \[ x_{n+1} = kx_n(1-x_n) \;. \]

  1. Prove that, if \(0 < k < 4\) and \(0 < x_0 < 1\), then \(0 < x_n < 1\) for all \(n\,\).
  2. Given that \(x_0=x_1=x_2 = \cdots =a\,\), with \(a\ne0\) and \(a\ne1\), find \(k\) in terms of \(a\,\).
  3. Given instead that \(x_0=x_2=x_4 = \cdots = a\,\), with \(a\ne0\) and \(a\ne1\), show that \(ab^3 -b^2 +(1-a)=0\), where \(b=k(1-a)\,\). Given, in addition, that \(x_1 \ne a\), find the possible values of \(k\) in terms of \(a\,\).

Solution:

  1. Consider \(f(x) = x(1-x) = x - x^2 = \tfrac14 - (x - \tfrac12)^2\) which is clearly in \((0,\tfrac14)\) when \(x \in (0,1)\), therefore if \(0 < k < 4\) then \(f(x) \in (0, 1)\) and so by induction \(x_n \in (0,1)\).
  2. Suppose \(a = g(a)\) then \(a = ka(1-a) \Rightarrow 1 = k(1-a) \Rightarrow k = \frac{1}{1-a}\) (since \(a \neq 0, 1\))
  3. If \(g(g(a)) = a\) then \begin{align*} && a &= kg(a)(1-g(a)) \\ &&&= k^2a(1-a)(1-ka(1-a)) \\ &&&= -k^3a^2(1-a)^2 + k^2a(1-a) \\ \Rightarrow && 1 &= -k^3a(1-a)^2 + k^2(1-a) \\ \Rightarrow && 1-a &= -k^3a(1-a)^3+k^2(1-a)^2 \\ \Rightarrow && 1-a &= -ab^3+b^2 \\ \Rightarrow && 0 &= ab^3-b^2+(1-a) \end{align*} Note that \begin{align*} && 0 &= ab^3-b^2+(1-a) \\ &&&= (b-1)(ab^2-(1-a)b - (1-a)) \end{align*} and since \(b \neq 1\) (otherwise \(x_2 =0\) which is a contradiction) we must have \(b = \frac{1-a \pm \sqrt{(1-a)^2+4a(1-a)}}{2a} = \frac{1-a\pm \sqrt{1+2a-3a^2}}{2a}\) and so \(k = \frac{b}{1-a} = \frac{1-a \pm \sqrt{1+2a-3a^2}}{2a(1-a)}\)

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