Problems

Filters
Clear Filters

3 problems found

2025 Paper 3 Q2
D: 1500.0 B: 1500.0
modulus function sequences recurrence relation curve sketching periodic sequences iterated functions fixed points piecewise linear

Let \(f(x) = 7 - 2|x|\). A sequence \(u_0, u_1, u_2, \ldots\) is defined by \(u_0 = a\) and \(u_n = f(u_{n-1})\) for \(n > 0\).

    1. Sketch, on the same axes, the graphs with equations \(y = f(x)\) and \(y = f(f(x))\).
    2. Find all solutions of the equation \(f(f(x)) = x\).
    3. Find the values of \(a\) for which the sequence \(u_0, u_1, u_2, \ldots\) has period 2.
    4. Show that, if \(a = \frac{28}{5}\), then the sequence \(u_2, u_3, u_4, \ldots\) has period 2, but neither \(u_0\) or \(u_1\) is equal to either of \(u_2\) or \(u_3\).
    1. Sketch, on the same axes, the graphs with equations \(y = f(x)\) and \(y = f(f(f(x)))\).
    2. Consider the sequence \(u_0, u_1, u_2, \ldots\) in the cases \(a = 1\) and \(a = -\tfrac79\). Hence find all the solutions of the equation \(f(f(f(x))) = x\).
    3. Find a value of \(a\) such that the sequence \(u_3, u_4, u_5, \ldots\) has period 3, but where none of \(u_0, u_1\) or \(u_2\) is equal to any of \(u_3, u_4\) or \(u_5\).

Solution:

    1. TikZ diagram
    2. If \(a = 1\) then \(u_1 = f(a) = 7-2 = 5\), \(u_2 = f(5) = -3\), \(u_3 = f(-3) = 7-6 = 1\). Therefore it must be the case that \(f(f(f(x))) = x\) for \(x = 1, 5, -3\). Similarly, if \(a = -\tfrac79\) then \(u_1 = f(-\tfrac79) = \tfrac{49}{9}\), \(u_2 = f(\tfrac{49}{9}) = -\tfrac{35}{9}\) and \(u_3 = f(-\tfrac{35}{9}) = -\tfrac79\). Therefore we must also have roots \(x = -\tfrac79, \tfrac{49}{9}, -\tfrac{35}9\). We also have the roots \(x = -7, \tfrac73\) from the first part so we have found all \(8\) roots.
    3. We need \(f(f(f(x))) = 1\) but \(f(f(x)) \neq -3, f(x) \neq 5, x \neq 1\). Suppose \(f(y) = 1 \Rightarrow 7-2|y| = 1 \Rightarrow y = \pm 3\). So \(y = 3\), ie \(f(f(x)) = 3\). Suppose \(f(z) = 3 \Rightarrow 7-2|z| = 3 \Rightarrow z = \pm 2\). Finally we need \(f(x) = \pm 2\), so say \(7-2|x| = 2 \Rightarrow x = \tfrac52\), so we have the sequence \(\tfrac52, 2, 3, 1, 5, -3, 1, \cdots\)as required.

View
2012 Paper 1 Q7
D: 1500.0 B: 1500.0
sequences recurrence relation periodic sequences algebraic manipulation factorisation simultaneous equations linear recurrence

A sequence of numbers \(t_0\), \(t_1\), \(t_2\), \(\ldots\,\) satisfies \[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t_{n+2} = p t_{n+1}+qt_{n} \ \ \ \ \ \ \ \ \ \ (n\ge0), \] where \(p\) and \(q\) are real. Throughout this question, \(x\), \(y\) and \(z\) are non-zero real numbers.

  1. Show that, if \(t_n=x\) for all values of \(n\), then \(p+q=1\) and \(x\) can be any (non-zero) real number.
  2. Show that, if \(t_{2n} = x\) and \(t_{2n+1}=y\) for all values of \(n\), then \(q\pm p=1\). Deduce that either \(x=y\) or \(x=-y\), unless \(p\) and \(q\) take certain values that you should identify.
  3. Show that, if \(t_{3n} = x\), \(t_{3n+1}=y\) and \(t_{3n+2}=z\) for all values of \(n\), then \[ p^3+q^3 +3pq-1=0\,. \] Deduce that either \(p+q=1\) or \((p-q)^2 +(p+1)^2+(q+1)^2=0\). Hence show that either \(x=y=z\) or \(x+y+z=0\).

Solution:

  1. Suppose \(t_n = x\) for all \(n\), then we must have \begin{align*} && x &= p x + q x \\ \Leftrightarrow && 1 &= p+q \end{align*} and this clearly works for any value of \(x\).
  2. Suppose \(t_{2n} = x, t_{2n+1} = y\) for all \(n\), then \begin{align*} && x &= py + q x \\ && y &= px + q y \\ \Rightarrow && 0 &= py + (q-1) x \\ && 0 &= px + (q-1) y \\ \Rightarrow && p &= (q-1) \frac{x}{y} = (q-1) \frac{y}{x} \\ \Rightarrow && \frac{y}{x} = \pm 1 & \text{ or } q = 1, p = 0 \\ \Rightarrow && y = \pm x & \text{ or } (p,q) = (0,1) \\ \end{align*}
  3. Suppose \(t_{3n} = x\), \(t_{3n+1}=y\) and \(t_{3n+2}=z\) , so \begin{align*} && x &= pz + qy \\ && y & = px + qz \\ && z &= py + qx \\ \\ && z &= p(px+qz) + q(pz + qy) \\ &&&= p^2x + 2pqz + q^2 y \\ &&&= p^2(pz+qy) + 2pqz + q^2(px+qz) \\ &&&= p^3 z + p^2qy + 2pqz + q^2p x + q^3 z \\ &&&= (p^3+q^3+2pq)z + pq(py+qx) \\ &&&= (p^3 + q^3 + 2pq)z + pq z \\ &&&= (p^3 + q^3 + 3pq)z \\ \Rightarrow && 0 &= p^3 + q^3 + 3pq- 1 \\ &&&= (p+q-1)(p^2+q^2+1+p+q-pq) \\ &&&= \tfrac12(p+q-1)((p-q)^2+(p+1)^2+(q+1)^2) \end{align*} Therefore \(p+q = 1\) or \((p-q)^2+(p+1)^2+(q+1)^2 = 0 \Rightarrow p = q = -1\). If \(p+q = 1\), then \(z = py + (1-p)x\) and \(x = p(py+(1-p)x) + (1-p)y \Rightarrow (1-p+p^2)x = (1-p+p^2)y \Rightarrow x = y \Rightarrow x= y = z\). If \(p = q = -1\) then adding all the equations we get \(x + y + z = -2(x+y+z) \Rightarrow x + y + z = 0\)
Note that what is actually going on here is that solutions must be of the form \(t_n = \lambda^n\) so the only way to be constant is for \(\lambda = 1\) to be a root, the only way for it to be \(2\)-periodic is for \(\lambda = -1\) to be a root, and the only way for it to be \(3\)-periodic is for \(\lambda = 1, \omega, \omega^2\) to be the roots (although we see this via the classic \(x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x + \omega y + \omega^2 z)(x+\omega^2 y +\omega z)\) which is because of the real constraint in the question.

View
2008 Paper 2 Q1
D: 1600.0 B: 1500.7
sequences recurrence relation complex numbers fixed points periodic sequences simultaneous equations quadratic equations

A sequence of points \((x_1,y_1)\), \((x_2,y_2)\), \(\ldots\) in the cartesian plane is generated by first choosing \((x_1,y_1)\) then applying the rule, for \(n=1\), \(2\), \(\ldots\), \[ (x_{n+1}, y_{n+1}) = (x_n^2-y_n^2 +a, \; 2x_ny_n+b+2)\,, \] where \(a\) and \(b\) are given real constants.

  1. In the case \(a=1\) and \(b=-1\), find the values of \((x_1,y_1)\) for which the sequence is constant.
  2. Given that \((x_1,y_1) = (-1,1)\), find the values of \(a\) and \(b\) for which the sequence has period 2.

View