A curve is given parametrically in plane polar coordinates by \((r, \theta) = (e^t, 2\pi t)\) \((0 \leq t < \infty)\). Sketch the section of the curve for \(n \leq t \leq n + 1\), where \(n\) is an integer. Calculate the length of this section, and the area enclosed by it and the line \(\theta = 0\), \(r^n \leq r \leq r^{n+1}\).
A second order linear differential equation for \(y\) is given by \[\frac{d^2y}{dx^2} + P(x)\frac{dy}{dx} + Q(x)y = 0.\] Make the substitution \(y = uv\), where \(u\) and \(v\) are functions of \(x\), and obtain a differential equation for \(u\) in terms of \(P\), \(Q\) and \(v\). What first order differential equation must \(v\) satisfy in order to eliminate the term in \(du/dx\)? Hence, or otherwise, solve \[\frac{d^2y}{dx^2} - \frac{1}{x}\frac{dy}{dx} + \left(1+\frac{3}{4x^2}\right) y = 0\] for \(x > 0\) when \(y\) satisfies the conditions \[y = (\frac{1}{2}\pi)^{\frac{1}{2}} \text{ at } x = \frac{1}{2}\pi,\] \[y=0 \text{ at } x = \pi.\]
Let \(\displaystyle L(x) = \int_1^x \frac{ds}{s}\) for \(x > 0\).
In two dimensions, show that the relation \[\mathbf{l.m} = l_1m_1+l_2m_2\] is equivalent to \[\mathbf{l.m} = lm\cos\theta,\] where \((l_1, l_2)\) are the Cartesian components of \(\mathbf{l}\), \(l\) is the length of \(\mathbf{l}\), and \(\theta\) is the angle between \(\mathbf{l}\) and \(\mathbf{m}\). Using vectors, prove the triangle inequality, that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.
\(A\), \(B\), \(C\) and \(D\) are complex numbers. Describe the set of points in the complex plane that satisfy the equation \[Az\bar{z}+Bz+C\bar{z}+D = 0.\] You should distinguish carefully between the cases that arise, and you may find it helpful to consider first the case \(A = 0\).
(i) Stones are thrown at random into \(n\) tin cans. Let \(P(m)\) be the probability that all the tin cans contain at least one stone after \(m\) throws. Show that \[n\left(1-\frac{1}{n}\right)^m \geq 1-P(m) \geq \left(1-\frac{1}{n}\right)^m.\] (ii) Two points \(x\) and \(y\) are chosen at random in the interval \(0 \leq t \leq 1\). What is the probability that \(|x - y| \geq 1/5\)?
A coin which has the probability \(p\) of falling heads is tossed repeatedly until exactly \(k\) heads have been obtained. Show that the probability that this requires \(n\) tosses is \[\binom{n-1}{k-1}p^k(1-p)^{n-k} \quad (n = k, k+1, \ldots).\] Show that this probability is the coefficient of \(z^n\) in the expansion of \[\left(\frac{pz}{1-(1-p)z}\right)^k.\] By differentiating this series, or otherwise, deduce the mean of \(n\).
Do you think that the following deductions are correct? Explain your reasons simply but clearly. (i) The average age at death of generals is considerably higher than that for the whole population. This shows that generals take care not to expose themselves to danger. (ii) I have tossed this coin twice and it came down heads each time. Therefore it is probably an unfair coin. (iii) I have tossed this coin 1000 times and it came down heads 276 times. Therefore it is probably an unfair coin.
Copies of a daily newspaper, which appears six times a week, are examined for misprints over a long period. It is discovered that the probability of there being one or more misprints in a given issue is \(\frac{1}{3}\). What is the most likely number of misprints in a week? What assumptions have you made? Find an expression for the probability that there will be fewer than the most likely number in a week. [\(\log_e (3/2)\) is approximately 0.4055.]
Solution: Assuming misprints are independent and occur at a constant average rate, then they are distributed as a Poisson random variable. The sum of Poisson random variables is Poisson, so the distribution of number of misprints in a week is \(Pois(2)\). The mode of a \(Pois(\lambda)\) is \(\lfloor \lambda \rfloor\), so the most likely is \(2\). Let \(X \sim Pois(2)\), then \begin{align*} \mathbb{P}(X < 2) &= \mathbb{P}(X = 0) + \mathbb{P}(X = 1) \\ &= e^{-2}\left ( \frac{2^0}{0!} + \frac{2^1}{1!}\right) \\ &= \frac{3}{e^2} \end{align*}