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2017 Paper 2 Q2
D: 1600.0 B: 1516.0
sequences recurrence relation periodicity necessary and sufficient conditions algebraic manipulation proof

The sequence of numbers \(x_0\), \(x_1\), \(x_2\), \(\ldots\) satisfies \[ x_{n+1} = \frac{ax_n-1}{x_n+b} \,. \] (You may assume that \(a\), \(b\) and \(x_0\) are such that \(x_n+b\ne0\,\).) Find an expression for \(x_{n+2}\) in terms of \(a\), \(b\) and \(x_n\).

  1. Show that \(a+b=0\) is a necessary condition for the sequence to be periodic with period 2. Note: The sequence is said to be periodic with period \(k\) if \(x_{n+k} = x_n\) for all \(n\), and there is no integer \(m\) with \(0 < m < k\) such that \(x_{n+m} = x_n\) for all \(n\).
  2. Find necessary and sufficient conditions for the sequence to have period 4.

Solution: \begin{align*} x_{n+2} &= \frac{ax_{n+1}-1}{x_{n+1}+b} \\ &= \frac{a \frac{ax_n - 1}{x_n+b}-1}{\frac{ax_n - 1}{x_n+b}+b} \\ &= \frac{a(ax_n-1)-(x_n+b)}{ax_n-1+b(x_n+b)} \\ &= \frac{(a^2-1)x_n-(a+b)}{(a+b)x_n+b^2-1} \end{align*}

  1. If \(x_{n+2} = x_n\) then \begin{align*} && x_n &= \frac{(a^2-1)x_n-(a+b)}{(a+b)x_n+b^2-1} \\ \Rightarrow && 0 &=(a+b)x_n^2+(b^2-a^2)x_n+(a+b) \\ &&&= (a+b)(x_n^2+(a-b)x_n + 1) \end{align*} If \(x_{n+1} = x_n\) then \(x_n^2+(a-b)x_n + 1\) and since our sequence has period \(2\) rather than \(1\) it must be the case this is non-zero. Therefore \(a+b =0\).
  2. \begin{align*} x_{n+4} &= \frac{(a^2-1)x_{n+2}-(a+b)}{(a+b)x_{n+2}+b^2-1} \\ &= \frac{(a^2-1)\frac{(a^2-1)x_{n}-(a+b)}{(a+b)x_{n}+b^2-1} -(a+b)}{(a+b)\frac{(a^2-1)x_{n}-(a+b)}{(a+b)x_{n}+b^2-1} +b^2-1} \\ &= \frac{((a^2-1)^2-(a+b)^2)x_n -(a^2+b^2-2)(a+b)}{(a^2+b^2-2)(a+b)x_n + (b^2-1)^2-(a+b)^2} \end{align*} If \(x_{n+4} = x_n\) then \begin{align*} x_n &=\frac{((a^2-1)^2-(a+b)^2)x_n -(a^2+b^2-2)(a+b)}{(a^2+b^2-2)(a+b)x_n + (b^2-1)^2-(a+b)^2} \\ 0 &= (a^2+b^2-2)(a+b)x_n^2 + \l (b^2-1)^2-(a^2-1)^2 \r x_n+(a^2+b^2-2)(a+b) \\ &= (a^2+b^2-2)(a+b)x_n^2+(b^2-a^2)(a^2+b^2-2)x_n + (a^2+b^2-2)(a+b) \\ &= (a^2+b^2-2)(a+b)(x_n^2+(b-a)x_n + 1) \end{align*} Since we do not want \(x_n\) to be periodic with period \(1\) we must have the quadratic in \(x_n\) \(\neq 0\). If \(a+b = 0\) then \(x_n\) is periodic with period \(2\) since \(x_{n+2} = \frac{(a^2-1)x_n}{((-a)^2-1)} = x_n\). Therefore it is necessary that \(a^2+b^2-2 = 0\). If \(a^2+b^2-2= 0\) then \begin{align*} x_{n+4} &= \frac{((a^2-1)^2-(a+b)^2)x_n}{(b^2-1)^2-(a+b)^2} \\ &=\frac{((a^2-1)^2-(a+b)^2)x_n}{((2-a^2)-1)^2-(a+b)^2} \\ &=\frac{((a^2-1)^2-(a+b)^2)x_n}{((1-a^2)^2-(a+b)^2} \\ &= x_n \end{align*} Therefore it is sufficient too. So our conditions are \(a+b \neq 0, \, \, x_n^2+(a-b)x_n + 1 \neq 0\) and \(a^2+b^2-2 = 0\)

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2007 Paper 2 Q5
D: 1600.0 B: 1488.1
trigonometry function iteration tan addition formula composite functions periodicity inverse trigonometric functions substitution

In this question, \(\f^2(x)\) denotes \(\f(\f(x))\), \(\f^3(x)\) denotes \(\f( \f (\f(x)))\,\), and so on.

  1. The function \(\f\) is defined, for \(x\ne \pm 1/ \sqrt3\,\), by $$ \f(x) = \ds \frac{x+\sqrt3} {1-\sqrt3\, x }\,. $$ Find by direct calculation \(\f^2(x) \) and \(\f^3(x)\), and determine \(\f^{2007}(x)\,\).
  2. Show that \(\f^n(x) = \tan(\theta + \frac 13 n\pi)\), where \(x=\tan\theta\) and \(n\) is any positive integer.
  3. The function \(\g(t)\) is defined, for \(\vert t\vert\le1\) by \(\g(t) = \frac {\sqrt3}2 t + \frac 12 \sqrt {1-t^2}\,\). Find an expression for \(\g^n(t)\) for any positive integer \(n\).

Solution:

  1. \(\,\) \begin{align*} && f(x) &= \frac{x+\sqrt3}{1-\sqrt3x} \\ \Rightarrow && f(f(x)) &= \frac{f(x)+\sqrt3}{1-\sqrt3f(x)} \\ &&&= \frac{\frac{x+\sqrt3}{1-\sqrt3x}+\sqrt3}{1-\sqrt3 \frac{x+\sqrt3}{1-\sqrt3x}} \\ &&&= \frac{x+\sqrt{3}+\sqrt3(1-\sqrt3x)}{1-\sqrt3x-\sqrt3(x+\sqrt3)} \\ &&&= \frac{-2x+2\sqrt3}{-2-2\sqrt3x} \\ &&&= \frac{x-\sqrt3}{1+\sqrt3 x} \\ \\ && f^3(x) &= f^2(f(x)) \\ &&&= \frac{f(x)-\sqrt3}{1+\sqrt3 f(x)} \\ &&&=\frac{\frac{x+\sqrt3}{1-\sqrt3x}-\sqrt3}{1+\sqrt3 \frac{x+\sqrt3}{1-\sqrt3x}} \\ &&&= \frac{(x+\sqrt3)-\sqrt3(1-\sqrt3 x)}{(1-\sqrt3x)+\sqrt3 (x+\sqrt3)} \\ &&&= \frac{-2x}{-2} = x \\ \\ && f^{2007}(x) &= x \end{align*}
  2. If \(x = \tan \theta\) then \(f(x) = \frac{\tan \theta + \tan \frac{\pi}{3}}{1 - \tan \frac{\pi}{3} \tan \theta} = \tan (\theta + \frac{\pi}{3})\) and hence \(f^n(x) = \tan (\theta + \frac{n \pi}{3})\)
  3. Note that if \(t = \sin \theta\) then \(g(t) = \cos \frac{\pi}{6} t\sin \theta + \frac12 \cos \theta = \sin(\theta + \frac{\pi}6)\) therefore \(g^n(t) = \sin(\sin^{-1}(t) + \frac{n\pi}{6})\)

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1994 Paper 2 Q3
D: 1600.0 B: 1500.0
functional equation trigonometric cosine symmetry periodicity proof even function

The function \(\mathrm{f}\) satisfies \(\mathrm{f}(0)=1\) and \[ \mathrm{f}(x-y)=\mathrm{f}(x)\mathrm{f}(y)-\mathrm{f}(a-x)\mathrm{f}(a+y) \] for some fixed number \(a\) and all \(x\) and \(y\). Without making any further assumptions about the nature of the function show that \(\mathrm{f}(a)=0\). Show that, for all \(t\),

  1. \(\mathrm{f}(t)=\mathrm{f}(-t)\),
  2. \(\mathrm{f}(2a)=-1\),
  3. \(\mathrm{f}(2a-t)=-\mathrm{f}(t)\),
  4. \(\mathrm{f}(4a+t)=\mathrm{f}(t)\).
Give an example of a non-constant function satisfying the conditions of the first paragraph with \(a=\pi/2\). Give an example of an non-constant function satisfying the conditions of the first paragraph with \(a=-2\).

Solution: Let \(P(x,y)\) be the statement that the functional equation holds, then: \begin{align*} P(0,0): && f(0) &= f(0)f(0)-f(a)f(a) \\ \Rightarrow && 1 &= 1 - f(a)^2 \\ \Rightarrow && f(a)^2 &= 0 \\ \Rightarrow && f(a) &= 0 \end{align*}

  1. \begin{align*} P(0,t): && f(-t) &= f(0)f(t) - f(a)f(a-t) \\ \Rightarrow && f(-t) &= f(t) - 0 \\ \Rightarrow && f(t) &= f(-t) \end{align*}
  2. \begin{align*} P(a,a): && f(0) &= f(a)f(a)-f(0)f(2a) \\ \Rightarrow && 1 &= 0 - f(2a) \\ \Rightarrow && f(2a) &= -1 \end{align*}
  3. \begin{align*} P(2a,t): && f(2a-t) &= f(2a)f(t) - f(-a)f(a+t) \\ \Rightarrow && f(2a-t) &= -f(t)-f(a)f(a+t) \\ &&&= -f(t)-0 \\ \Rightarrow && f(2a-t) &= -f(t) \end{align*}
  4. \begin{align*} && f(4a+t) &= f(2a-(-2a-t)) \\ &&&=-f(2a+t) \\ &&&=-f(2a-(-t)) \\ &&&=f(-t) \\ &&&=f(t) \end{align*}
Let \(f(x) = \cos x\) then \(f(\frac{\pi}{2}-x) = \sin x\) and \(f(\frac{\pi}{2}+y) = -\sin y\) so the equation becomes \(\cos(x-y) = \cos x \cos y + \sin x \sin y\) which is the normal cosine addition formula. Similarly, consider \(f(x) = \cos \frac{\pi}{4} x\).

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