Problems

Filters
Clear Filters

4 problems found

2021 Paper 3 Q8
D: 1500.0 B: 1500.0
recurrence relation sequences and series proof by induction convergence inequality telescoping product

A sequence \(x_1, x_2, \ldots\) of real numbers is defined by \(x_{n+1} = x_n^2 - 2\) for \(n \geqslant 1\) and \(x_1 = a\).

  1. Show that if \(a > 2\) then \(x_n \geqslant 2 + 4^{n-1}(a-2)\).
  2. Show also that \(x_n \to \infty\) as \(n \to \infty\) if and only if \(|a| > 2\).
  3. When \(a > 2\), a second sequence \(y_1, y_2, \ldots\) is defined by \[ y_n = \frac{Ax_1 x_2 \cdots x_n}{x_{n+1}}, \] where \(A\) is a positive constant and \(n \geqslant 1\). Prove that, for a certain value of \(a\), with \(a > 2\), which you should find in terms of \(A\), \[ y_n = \frac{\sqrt{x_{n+1}^2 - 4}}{x_{n+1}} \] for all \(n \geqslant 1\). Determine whether, for this value of \(a\), the second sequence converges.

Solution:

  1. Claim \(x_n \geqslant 2 + 4^{n-1}(a-2)\) Proof: (By induction) Base case: Note that when \(n = 1\), \(x_1 = a = 2 + 1 \cdot(a - 2)\). Inductive step, suppose true for some \(n\), then \begin{align*} && x_{n+1} &= x_n^2 - 2 \\ &&&\geq (2+4^{n-1}(a-2))^2 - 2 \\ &&&= 4 + 4^{2n-2}(a-2)^2 + 4^n(a-2) - 2 \\ &&&= 2 + 4^{n}(a-2) + 4^{2n-2}(a-2)^2 \\ &&&\geq 2 + 4^{n+1-1}(a-2) \end{align*} as required,
  2. (\(\Leftarrow\)) Suppose \(a > 2\) then \(x_n \geq 2+4^{n-1}(a-2) \to \infty\) as required. Suppose \(a < -2\) then \(x_2 > 4 -2 = 2\) so the sequence starting from \(x_2\) clearly diverges for the same reason. (\(\Rightarrow\)) suppose \(|x_n| \leq 2\) then \(x_{n+1} = x_n^2 - 2 \leq 2\) so the sequence is bounded and cannot tend to \(\infty\).
  3. Suppose \(y_n = \frac{Ax_1x_2 \cdots x_n}{x_{n+1}}\) and notice that \(x_{n+1}^2 - 4 = (x_n^2 -2)^2 - 4 = x_n^4 - 4x_n^2 = x_n^2(x_n^2-4)\). In particular, \(\frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}} = \frac{x_n\sqrt{x_n^2-4}}{x_{n+1}} = \frac{x_n x_{n-1} \cdots x_1 \sqrt{x_1^2-4}}{x_{n+1}}\) Therefore if \(A = \sqrt{a^2-4}\) \(y_{n+1} = \frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}}\). Notice that \begin{align*} && y_n &= \frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}} \\ &&&= \sqrt{1 - \frac{4}{x_{n+1}^2}} \to 1 \end{align*}

View
2019 Paper 2 Q4
D: 1500.0 B: 1500.0
trigonometry telescoping product infinite product double angle formula differentiation tan cot product notation series limits

You are not required to consider issues of convergence in this question. For any sequence of numbers \(a_1, a_2, \ldots, a_m, \ldots, a_n\), the notation \(\prod_{i=m}^{n} a_i\) denotes the product \(a_m a_{m+1} \cdots a_n\).

  1. Use the identity \(2 \cos x \sin x = \sin(2x)\) to evaluate the product \(\cos(\frac{\pi}{9}) \cos(\frac{2\pi}{9}) \cos(\frac{4\pi}{9})\).
  2. Simplify the expression $$\prod_{k=0}^{n} \cos\left(\frac{x}{2^k}\right) \quad (0 < x < \frac{1}{2}\pi).$$ Using differentiation, or otherwise, show that, for \(0 < x < \frac{1}{2}\pi\), $$\sum_{k=0}^{n} \frac{1}{2^k} \tan\left(\frac{x}{2^k}\right) = \frac{1}{2^n} \cot\left(\frac{x}{2^n}\right) - 2 \cot(2x).$$
  3. Using the results \(\lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1\) and \(\lim_{\theta\to 0} \frac{\tan \theta}{\theta} = 1\), show that $$\prod_{k=1}^{\infty} \cos\left(\frac{x}{2^k}\right) = \frac{\sin x}{x}$$ and evaluate $$\sum_{j=2}^{\infty} \frac{1}{2^{j-2}} \tan\left(\frac{\pi}{2^j}\right).$$

Solution:

  1. \begin{align*}\cos(\frac{\pi}{9}) \cos(\frac{2\pi}{9}) \cos(\frac{4\pi}{9}) &= \frac{\sin(\frac{2\pi}{9}) \cos(\frac{2\pi}{9}) \cos(\frac{4\pi}{9})}{2 \sin \frac{\pi}{9}} \\ &= \frac{\sin(\frac{4\pi}{9})\cos(\frac{4\pi}{9})}{4 \sin \frac{\pi}{9}} \\ &= \frac{\sin(\frac{8\pi}{9})}{8 \sin \frac{\pi}{9}} \\ &= \frac{1}{8} \end{align*}
  2. Let \(\displaystyle P_n = \prod_{k=0}^{n} \cos\left(\frac{x}{2^k}\right)\). Claim: \(P_n = \frac{\sin 2x}{2^{n+1} \sin \l \frac{x}{2^n} \r}\). Proof: This is true for \(n = 0\), assume true for \(n-1\) \begin{align*} \sin\l \frac{x}{2^{n}} \r P_n &= P_{n-1} \cos\l \frac{x}{2^{n}} \r \sin\l \frac{x}{2^{n}} \r \\ &= P_{n-1} \frac{1}{2} \sin\l \frac{x}{2^{n-1}} \r \\ &= \frac{\sin 2x}{2^{n} \sin \l \frac{x}{2^{n-1}}\r} \frac{1}{2} \sin\l \frac{x}{2^{n}} \r \\ &= \frac{\sin 2x}{2^{n+1}} \end{align*} Hence \(P_n = \frac{\sin 2x}{2^{n+1} \sin \l \frac{x}{2^n}\r}\) Taking logs, we determine that: \begin{align*} && \sum_{k=0}^n \ln \cos \l \frac{x}{2^k} \r &= \ln \sin 2x - \ln \sin \l \frac{x}{2^n} \r - (n+1) \ln 2 \\ \Rightarrow && \sum_{k=0}^n \frac{1}{2^k} \tan \l \frac{x}{2^k} \r &= -2 \cot 2x + \frac{1}{2^n} \cot \l \frac{x}{2^n} \r - 0 \\ \end{align*} as required.
  3. As \(n \to \infty\) \(\frac{x}{2^n} \to 0\), so \(\frac{\sin \frac{x}{2^n}}{\frac{x}{2^n}} = \frac{2^n \sin \frac{x}{2^n}}{x} \to 1\) \begin{align*}\prod_{k=1}^{\infty} \cos\left(\frac{x}{2^k}\right) &= \lim_{n \to \infty} \frac{\sin x}{2^n \sin \l \frac{x}{2^n} \r} \\ &= \lim_{n \to \infty} \frac{\sin x}{x \frac{2^n \sin \l \frac{x}{2^n} \r}{x} } \\ &= \lim_{n \to \infty} \frac{\sin x}{x} \\ \end{align*} \begin{align*} \sum_{j=2}^{\infty} \frac{1}{2^{j-2}} \tan\left(\frac{\pi}{2^j}\right) &= \sum_{j=0}^{\infty} \frac{1}{2^{j}} \tan\left(\frac{1}{2^j}\frac{\pi}{4}\right) \\ &= \lim_{n \to \infty} \l -2 \cot \frac{\pi}{2} + \frac{1}{2^n} \cot \l \frac{\pi}{4 \cdot 2^n} \r\r \\ &= \frac{4}{\pi} \lim_{n \to \infty} \l \frac{1}{2^n} \frac{\pi}{4} \cot \l \frac{\pi}{4 \cdot 2^n} \r\r \\ &\to \frac{\pi}{4} \end{align*}

View
2005 Paper 1 Q7
D: 1500.0 B: 1516.0
telescoping product product notation trigonometry simplification cotangent partial fractions sine rule compound angle formula

The notation \(\displaystyle \prod^n_{r=1} \f (r)\) denotes the product $\f (1) \times \f (2) \times \f(3) \times \cdots \times \f(n)$. %For example, \(\displaystyle \prod_{r=1}^4 r = 24\). %Simplify \(\displaystyle \prod^n_{r=1} \frac{\g (r) }{ \g (r-1) }\). %You may assume that \(\g (r) \neq 0\) for any integer \(0 \le r \le n \). Simplify the following products as far as possible:

  1. \(\displaystyle \prod^n_{r=1} \l \frac{r+ 1 }{ r } \r\,\);
  2. \(\displaystyle \prod^n_{r=2} \l \frac{r^2 -1}{r^2 } \r\,\);
  3. $\displaystyle \prod^n_{r=1} \l {\cos \frac{2\pi }{ n} + \sin \frac{2\pi}{ n} \cot \frac{\l 2r-1 \r \pi }{ n} }\r\,$, where \(n\) is even.

Solution:

  1. \(\,\) \begin{align*} \prod^n_{r=1} \left ( \frac{r+ 1 }{ r } \right) &= \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{n-1}{n-2} \cdot \frac{n}{n-1} \cdot \frac{n+1}{n} \\ &= \frac{n+1}{1} = n+1 \end{align*}
  2. \(\,\) \begin{align*} \prod^n_{r=2} \left ( \frac{r^2 -1}{r^2 } \right) &= \prod^n_{r=2} \left ( \frac{(r -1)(r+1)}{r^2 } \right) \\ &= \left ( \frac{1}{2} \cdot \frac{3}{2} \right) \cdot \left ( \frac{2}{3} \cdot \frac{4}{3} \right) \cdots \left ( \frac{r-1}{r} \cdot \frac{r+1}{r}\right) \cdots \frac{n-1}{n} \cdot \frac{n+1}{n} \\ &= \frac{1}{n} \cdot \frac{n+1}{2} \\ &= \frac{n+1}{2n} \end{align*}
  3. When \(n\) is odd, the product is undefined, since we have a \(\cot \pi\) lurking in there. \begin{align*} \prod^n_{r=1} \left ( {\cos \frac{2\pi }{ n} + \sin \frac{2\pi}{ n} \cot \frac{ (2r-1 ) \pi }{ n} } \right) &= \prod^n_{r=1} \left ( {\cos \frac{2\pi }{ n} + \sin \frac{2\pi}{ n} \frac{\cos \frac{ (2r-1 ) \pi }{ n}}{\sin\frac{ (2r-1 ) \pi }{ n}} } \right) \\ &= \prod^n_{r=1} \frac{1}{\sin\frac{ (2r-1 ) \pi }{ n}} \left ( {\cos \frac{2\pi }{ n} \sin\frac{ (2r-1 ) \pi }{ n} + \sin \frac{2\pi}{ n} \cos \frac{ (2r-1 ) \pi }{ n} } \right) \\ &= \prod^n_{r=1} \frac{1}{\sin\frac{ (2r-1 ) \pi }{ n}} \sin \left ( \frac{2\pi}{n} + \frac{(2r-1)\pi}{n} \right) \\ &= \prod^n_{r=1} \frac{1}{\sin\frac{ (2r-1 ) \pi }{ n}} \sin \left ( \frac{(2r+1)\pi}{n} \right) \\ &= \frac{\sin \frac{3\pi}{n}}{\sin \frac{\pi}{n}} \cdot \frac{\sin \frac{5\pi}{n}}{\sin \frac{3\pi}{n}} \cdots \frac{\sin \frac{(2n+1)\pi}{n}}{\sin \frac{(2n-1)\pi}{n}} \\ &= \frac{\sin \frac{(2n+1)\pi}{n}}{\sin \frac{\pi}{n}} \\ &= 1 \end{align*}

View
2002 Paper 1 Q10
D: 1516.0 B: 1470.2
momentum and collisions elastic collision recurrence relation bicycle pump particle bouncing telescoping product relative velocity sequences

A bicycle pump consists of a cylinder and a piston. The piston is pushed in with steady speed~\(u\). A particle of air moves to and fro between the piston and the end of the cylinder, colliding perfectly elastically with the piston and the end of the cylinder, and always moving parallel with the axis of the cylinder. Initially, the particle is moving towards the piston at speed \(v\). Show that the speed, \(v_n\), of the particle just after the \(n\)th collision with the piston is given by \(v_n=v+2nu\). Let \(d_n\) be the distance between the piston and the end of the cylinder at the \(n\)th collision, and let \(t_n\) be the time between the \(n\)th and \((n+1)\)th collisions. Express \(d_n - d_{n+1}\) in terms of \(u\) and \(t_n\), and show that \[ d_{n+1} = \frac{v+(2n-1)u}{v+(2n+1)u} \, d_n \;. \] Express \(d_n\) in terms of \(d_1\), \(u\), \(v\) and \(n\). In the case \(v=u\), show that \(ut_n = \displaystyle \frac {d_1} {n(n+1)}\). %%%%%Verify that \(\sum\limits_1^\infty t_n = d/u\).

View