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2024 Paper 2 Q8
D: 1500.0 B: 1500.0

In this question, the following theorem may be used without proof. Let \(u_1, u_2, \ldots\) be a sequence of real numbers. If the sequence is

  • bounded above, so \(u_n \leqslant b\) for all \(n\), where \(b\) is some fixed number
  • and increasing, so \(u_n \leqslant u_{n+1}\) for all \(n\)
then there is a number \(L \leqslant b\) such that \(u_n \to L\) as \(n \to \infty\). For positive real numbers \(x\) and \(y\), define \(\mathrm{a}(x,y) = \frac{1}{2}(x+y)\) and \(\mathrm{g}(x,y) = \sqrt{xy}\). Let \(x_0\) and \(y_0\) be two positive real numbers with \(y_0 < x_0\) and define, for \(n \geqslant 0\) \[ x_{n+1} = \mathrm{a}(x_n, y_n)\,, \] \[ y_{n+1} = \mathrm{g}(x_n, y_n)\,. \]
  1. By considering \((\sqrt{x_n} - \sqrt{y_n})^2\), show that \(y_{n+1} < x_{n+1}\), for \(n \geqslant 0\). Show further that, for \(n \geqslant 0\)
    • \(x_{n+1} < x_n\)
    • \(y_n < y_{n+1}\).
    Deduce that there is a value \(M\) such that \(y_n \to M\) as \(n \to \infty\). Show that \(0 < x_{n+1} - y_{n+1} < \frac{1}{2}(x_n - y_n)\) and hence that \(x_n - y_n \to 0\) as \(n \to \infty\). Explain why \(x_n\) also tends to \(M\) as \(n \to \infty\).
  2. Let \[ \mathrm{I}(p,q) = \int_0^{\infty} \frac{1}{\sqrt{(p^2 + x^2)(q^2 + x^2)}}\,\mathrm{d}x, \] where \(p\) and \(q\) are positive real numbers with \(q < p\). Show, using the substitution \(t = \frac{1}{2}\!\left(x - \dfrac{pq}{x}\right)\) in the integral \[ \int_{-\infty}^{\infty} \frac{1}{\sqrt{\left(\frac{1}{4}(p+q)^2 + t^2\right)(pq + t^2)}}\,\mathrm{d}t, \] that \[ \mathrm{I}(p,q) = \mathrm{I}\!\left(\mathrm{a}(p,q),\, \mathrm{g}(p,q)\right). \] Hence evaluate \(\mathrm{I}(x_0, y_0)\) in terms of \(M\).

1998 Paper 1 Q6
D: 1500.0 B: 1500.0

Let \(a_{1}=\cos x\) with \(0 < x < \pi/2\) and let \(b_{1}=1\). Given that \begin{eqnarray*} a_{n+1}&=&{\textstyle \frac{1}{2}}(a_{n}+b_{n}),\\[2mm] b_{n+1}&=&(a_{n+1}b_{n})^{1/2}, \end{eqnarray*} find \(a_{2}\) and \(b_{2}\) and show that \[a_{3}=\cos\frac{x}{2}\cos^{2}\frac{x}{4} \ \quad\mbox{and}\quad \ b_{3}=\cos\frac{x}{2}\cos\frac{x}{4}.\] Guess general expressions for \(a_{n}\) and \(b_{n}\) (for \(n\ge2\)) as products of cosines and verify that they satisfy the given equations.


Solution: \begin{array}{c|c|c} n & a_n & b_n \\ \hline 1 & \cos x & 1 \\ \hline 2 & \frac12(1 + \cos x) & \sqrt{a_2} \\ &=\frac12(1+2\cos^2 \frac{x}{2}-1)& \sqrt{a_2} \\ &= \cos^2 \frac{x}{2} & \cos \frac{x}{2} \\ \hline 3 & \frac12(\cos^2 \frac{x}{2}+\cos \frac{x}{2}) & \sqrt{a_3\cos \frac{x}{2}} \\ &= \cos \frac{x}{2} \cdot \frac12 (\cos \frac{x}{2}+1) & \sqrt{a_3\cos \frac{x}{2}} \\ &= \cos \frac{x}{2} \cos^2 \frac{x}{4} & \cos \frac{x}{2} \cos \frac{x}{4} \end{array} Claim: \(\displaystyle a_n = \cos \frac{x}{2^{n-1}}\prod_{k=1}^{n-1} \cos \frac{x}{2^k}\), \(\displaystyle b_n = \prod_{k=1}^{n-1} \cos \frac{x}{2^k}\) Claim: \(a_{n+1} = \frac12(a_n + b_n)\) Proof: \begin{align*} && \frac12(a_n + b_n) &= \frac12 \left ( \cos \frac{x}{2^{n-1}}\prod_{k=1}^{n-1} \cos \frac{x}{2^k} + \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \right) \\ &&&= \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \frac12\left (\cos \frac{x}{2^{n-1}} + 1 \right) \\ &&&= \left ( \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \right) \cos^{2} \frac{x}{2^n} \\ &&&= \cos \frac{x}{2^n} \prod_{k=1}^{n} \cos \frac{x}{2^k} \\ &= a_{n+1} \end{align*} Claim: \(b_{n+1} = \sqrt{a_{n+1}b_n}\) Proof: \begin{align*} && \sqrt{a_{n+1}b_n} &= \sqrt{ \cos \frac{x}{2^n} \prod_{k=1}^{n} \cos \frac{x}{2^k} \cdot \prod_{k=1}^{n-1} \cos \frac{x}{2^k} }\\ &&&= \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \sqrt{\cos ^2\frac{x}{2^{n}}} \\ &&&= \prod_{k=1}^{n} \cos \frac{x}{2^k} \\ &&&= b_{n+1} \end{align*}