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2014 Paper 1 Q6
D: 1500.0 B: 1474.3
trigonometry recurrence relation proof by induction double angle formula substitution sequences sin² quadratic recurrence

  1. The sequence of numbers \(u_0, u_1, \ldots \) is given by \(u_0=u\) and, for \(n\ge 0\), \begin{equation} u_{n+1} =4u_n(1- u_n)\,. \tag{\(*\)} \end{equation} In the case \(u= \sin^2\theta\) for some given angle \(\theta\), write down and simplify expressions for \(u_1\) and \(u_2\) in terms of \(\theta\). Conjecture an expression for \(u_n\) and prove your conjecture.
  2. The sequence of numbers \(v_0, v_1, \ldots\) is given by $v_0= v \text{ and, for }n\ge 0$, \[ v_{n+1} = -pv_n^2 +qv_n +r\,, \] where \(p\), \(q\) and \(r\) are given numbers, with \(p\ne0\). Show that a substitution of the form \(v_n =\alpha u_n +\beta\), where \(\alpha\) and \(\beta\) are suitably chosen, results in the sequence \((*)\) provided that \[ 4pr = 8 +2q -q^2 \,. \] Hence obtain the sequence satisfying \(v_0=1\) and, for \(n\ge0\), \(v_{n+1} = -v_n^2 +2 v_n +2 \,\).

Solution:

  1. Suppose \(u_0 = u = \sin^2 \theta\) then \begin{align*} && u_1 &= 4 u_0 (1-u_0) \\ &&&= 4 \sin^2 \theta ( 1- \sin^2 \theta) \\ &&&= 4 \sin^2 \theta \cos^2 \theta \\ &&&= (2 \sin \theta \cos \theta)^2 \\ &&&= (\sin 2 \theta)^2 = \sin^2 2 \theta \\ \\ && u_2 & = 4u_1 (1-u_1) \\ &&&= 4 \sin^2 2\theta \cos^2 2 \theta \\ &&&= \sin^2 4 \theta \end{align*} Claim: \(u_n = \sin^2 2^n \theta\). Proof: (By Induction) Base case is clear, suppose it's true for \(n=k\), then \begin{align*} && u_{k+1} &= 4u_k(1-u_k) \\ &&&= 4 \sin^2 2^k \theta(1-\sin^2 2^k \theta) \\ &&&= (2 \sin 2^k \theta \cos 2^k \theta)^2 \\ &&&= (\sin 2^{k+1} \theta)^2 \\ &&&= \sin^2 2^{k+1} \theta \end{align*} Therefore since it is true for \(n = 1\) and if it's true for \(n = k\) it is true for \(n=k+1\) it must be true for all \(k\).
  2. Suppose \(v_n = \alpha u_n + \beta\) then \begin{align*} && (\alpha u_{n+1}+\beta) &= -p(\alpha u_n + \beta)^2 + q(\alpha u_n + \beta) + r \\ &&&= -p\alpha^2u_n^2+\alpha(q-2p\beta) u_n -p \beta^2 +q \beta+r \\ \Rightarrow && u_{n+1} &= u_n(q-2p\beta -p \alpha u_n) -(p\beta^2-(q-1)\beta-r) \end{align*} So if \(\alpha = \frac{4}{p}\) and \(q-2p\beta = 4\) ie \(\beta = \frac{q-4}{2p}\) then we also need the constant term to vanish, ie \begin{align*} 0 &&&= p\beta^2-(q-1)\beta+r \\ &&&= p \left (\frac{q-4}{2p} \right)^2 - (q-1) \frac{q-4}{2p} - r \\ \Rightarrow && 0 &= p(q-4)^2 -(q-1)(q-4)2p - 4p^2r \\ \Rightarrow && 0 &= (q-4)^2-2(q-1)(q-4)-4pr \\ &&&= q^2-8q+16-2q^2+10q-8-4pr \\ \Rightarrow && 4pr &= -q^2+2q+8 \end{align*} Suppose \(v_{n+1} = -v_n^2 + 2v_n +2\) then since \(4\cdot 1 \cdot 2 = 8\) and \(8 + 4 -4 = 8\) we can apply our method. \(v_n = 4u_n + \frac{-2}{2} = 4u_n -1 = 4\sin^2 (2^{n-1} \pi)-1\)

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2012 Paper 2 Q8
D: 1600.0 B: 1485.7
sequences recurrence relation inequality strictly increasing sequence algebraic manipulation second order recurrence quadratic recurrence

The positive numbers \(\alpha\), \(\beta\) and \(q\) satisfy \(\beta-\alpha >q\). Show that \[ \frac{\alpha^2+\beta^2 -q^2}{\alpha\beta}-2> 0\,. \] The sequence \(u_0\), \(u_1\), \(\ldots\) is defined by \(u_0=\alpha\), \(u_1=\beta\) and \[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u_{n+1} = \frac {u_{n}^2 -q^2}{u_{n-1}} \ \ \ \ \ \ \ \ \ \ \ (n\ge1), \] where \(\alpha\), \(\beta\) and \(q\) are given positive numbers (and \(\alpha\) and \(\beta\) are such that no term in the sequence is zero). Prove that \(u_n(u_n+u_{n+2}) = u_{n+1}(u_{n-1}+u_{n+1})\,\). Prove also that \[ u_{n+1} -pu_n + u_{n-1}=0 \] for some number \(p\) which you should express in terms of \(\alpha\), \(\beta\) and \(q\). Hence, or otherwise, show that if \(\beta> \alpha+q\), the sequence is strictly increasing (that is, \(u_{n+1}-u_n > 0\) for all \(n\)). Comment on the case \(\beta =\alpha +q\).

Solution: \begin{align*} && \beta - \alpha &> q \\ \Rightarrow &&(\beta - \alpha)^2 &> q^2 \\ \Rightarrow && \beta^2 +\alpha^2 - 2\beta \alpha &> q^2 \\ \Rightarrow && \alpha^2+\beta^2-q^2 -2 \beta \alpha &> 0 \\ \Rightarrow && \frac{\alpha^2+\beta^2-q^2}{\alpha\beta} - 2 &> 0 \end{align*} \begin{align*} && u_n(u_n+u_{n+2}) &= u_n \cdot \left (u_n + \frac {u_{n+1}^2 -q^2}{u_{n}}\right) \\ &&&= u_n^2 + u_{n+1}^2-q^2 \\ &&&= u_n^2 + u_{n+1}^2 - (u_n^2-u_{n-1}u_{n+1}) \\ &&&= u_{n+1}^2 + u_{n+1}u_{n-1} \\ &&&= u_{n+1}(u_{n-1}+u_{n+1}) \\ \\ && u_{n+1}-pu_n+u_{n-1} &= -pu_n+\frac{u_{n}(u_{n-2}+u_n)}{u_{n-1}} \\ &&&= \frac{u_n(u_{n}-pu_{n-1}+u_{n-2})}{u_{n-1}} \end{align*} Therefore if \(u_2 -pu_1 + u_0 = 0\) it is always zero, ie if \begin{align*} && u_2 &= p\beta - \alpha \\ && u_2 &= \frac{\beta^2-q^2}{\alpha} \\ \Rightarrow && \frac{\beta^2-q^2}{\alpha} &= p\beta - \alpha \\ \Rightarrow && p &= \frac{\alpha^2+\beta^2-q^2}{\alpha\beta} \end{align*} If \(\beta > \alpha + q\) we must have that \(p > 2\), and so \(u_{n+1}-u_n = (p-1)u_n - u_{n-1} > u_n-u_{n-1} > 0\), therefore the sequence is strictly increasing. If \(\beta = \alpha + q\) the sequence follows \(u_{n+1} - 2u_n + u_{n-1} =0\) and so \(u_{n+1}-u_n = u_n - u_{n-1}\) for all \(n\) (which is still increasing - it's an arithmetic progression with common difference \(\beta - \alpha\)).

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2004 Paper 1 Q8
D: 1500.0 B: 1547.8
sequences recurrence relation convergence proof by induction inequality strictly increasing quadratic recurrence fixed points

A sequence \(t_0\), \(t_1\), \(t_2\), \(...\) is said to be strictly increasing if \(t_{n+1} > t_n\) for all \(n\ge{0}\,\).

  1. The terms of the sequence \(x_0\,\), \(x_1\,\), \(x_2\,\), \(\ldots\) satisfy $$ \ds x_{n+1}=\frac{x_n^2 +6}{5} $$ for \(n\ge{0}\,\). Prove that if \(x_0 > 3\) then the sequence is strictly increasing.
  2. The terms of the sequence \(y_0\,\), \(y_1\,\), \(y_2\,\), \(\ldots\) satisfy $$ \ds y_{n+1}= 5-\frac 6 {y_n} $$ for \(n\ge{0}\,\). Prove that if \(2 < y_0 < 3\) then the sequence is strictly increasing but that \(y_n<3\) for all \(n\,\).

Solution:

  1. Suppose \(x_n> 3\) then \begin{align*} && x_{n+1} &= \frac{x_n^2+9-3}{5} \\ &&& \geq \frac{2\sqrt{x_n^2 \cdot 9} - 3}{5} \\ &&&= \frac{6x_n -3}{5} = x_n + \frac{x_n-3}{5} \\ &&&> x_n > 3 \end{align*} Therefore if \(x_i > 3 \Rightarrow x_{i+1} > x_i\) and \(x_{i+1} > 3\) so by induction \(x_n\) strictly increasing for all \(n\).
  2. Suppose \(2 < y_n < 3\) then \begin{align*} && y_{n+1} &= 5 - \frac6{y_n} \\ &&&< 5 - \frac63 = 3 \\ \\ && y_{n+1} &= 5 - \frac4{y_n} - \frac{2}{y_n} \\ \\ &&&= y_n + 5 - \frac{2}{y_n} - \left ( y_n + \frac4{y_n} \right) \\ &&&\geq y_n + 5 - \frac{2}{y_n} - 2\sqrt{y_n \frac{4}{y_n}} \\ &&&= y_n + 1 - \frac{2}{y_n} \\ &&&> y_n \end{align*} Therefore if \(y_n \in (2,3)\) we have \(y_{n+1} \in ( y_n, 3)\) and so \(y_n\) is strictly increasing and bounded.

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