The real 6-dimensional vector space V consists of all homogeneous quadratics \begin{align*} p(x, y, z) \equiv ax^2 + by^2 + cz^2 + 2dyz + 2ezx + 2fxy \end{align*} in \(x, y\) and \(z\), under the usual definitions of addition and multiplication by scalars. Find the dimension of, and write down a basis for,
Positive rational 'weights' \(m_1, \ldots, m_n\) are attached to positive numbers \(a_1, \ldots, a_n\). Use the inequality connecting the arithmetic and geometric means to prove that \begin{align*} \frac{m_1a_1 + \ldots + m_na_n}{m_1 + \ldots + m_n} \geq (a_1^{m_1} a_2^{m_2} \ldots a_n^{m_n})^{1/(m_1 + \ldots + m_n)}. \end{align*} By attaching suitable weights to 1 and \(1 + x/n\), prove that, if \(x\) is positive, \begin{align*} \left(1 + \frac{x}{n+1}\right)^{n+1} \geq \left(1 + \frac{x}{n}\right)^n. \end{align*}
The real polynomial \(f(x)\) has degree 5. Prove that \begin{align*} \int_{-y}^{+y} f(x) dx = \frac{1}{2}y\{f(\lambda y) + f(\mu y) + f(-\lambda y) + f(-\mu y)\} \end{align*} for positive constants \(\lambda\) and \(\mu\) (independent of \(y\) and \(f\)) whose squares are the roots of a certain quadratic equation to be determined.
From the circumcentre \(S\) of a triangle \(ABC\), perpendiculars \(SD\), \(SE\) and \(SF\) are drawn to the sides \(BC\), \(CA\) and \(AB\) respectively, and produced to \(A'\), \(B'\) and \(C'\) so that \(D\), \(E\) and \(F\) are the mid-points of \(SA'\), \(SB'\) and \(SC'\). Prove that the triangles \(ABC\) and \(A'B'C'\) are congruent, that \(AA'\), \(BB'\) and \(CC'\) all have a common mid-point \(M\), and that a rotation about \(M\) moves one triangle to the other.
Suppose \(a > b > 0\). Show that the circle of curvature of the ellipse \begin{align*} x^2/a^2 + y^2/b^2 = 1 \end{align*} at the point \((0, -b)\) is \begin{align*} b^2x^2 + (by + b^2 - a^2)^2 = a^4. \end{align*} The tangent to the ellipse at \((a \cos \theta, b \sin \theta)\) meets the circle at \(P\) and \(Q\), and the lines from \((0, -b)\) to \(P\) and \(Q\) meet the \(x\)-axis at \(X_1\) and \(X_2\). Show that the distance from \(X_1\) to \(X_2\) is equal to the distance between the foci of the ellipse.
A sequence of numbers \(x_0, x_1, \ldots\) is defined by \begin{align*} x_0 &= 0,\\ x_{n+1} &= x_n + \frac{1}{2k}(x^{2k} - x_n^{2k}), \end{align*} where \(-1 \leq x \leq 1\) and \(k\) is a positive integer. Show that \begin{align*} x_0 \leq x_1 \leq \ldots \leq x_n \leq x_{n+1} \leq \ldots \leq |x| \end{align*} and find the limit of the sequence \(x_0, x_1, x_2, \ldots\).
Show that \(e^{x}/x \to \infty\) as \(x \to \infty\). Sketch the graph of the function \begin{align*} f(x) = x \log_e(x) \quad (x > 0). \end{align*} Solve the equation \begin{align*} \int_0^x f(t) dt = 0. \end{align*}
A computer prints out a list of \(M\) integers. Each integer has been chosen independently and at random from the range 1 to \(N\), with equal probability assigned to each of the \(N\) possible values. What is the probability
A machine produces boiled sweets in 100 kg batches, at the rate of 10 tonnes per day. When the machine is working properly, the chance of a given batch being bad is known to be \(\frac{1}{10}\), different batches being independent: however, the machine is prone to develop a fault which increases the chance of bad batches. At the end of a day in which 2 tonnes of bad sweets were produced, the quality control officer reasoned as follows: 'The machine has produced 8 tonnes of good sweets and 2 tonnes of bad, as against the expected tonnages of 9 and 1 respectively. Hence \begin{align*} \chi^2 = (9-8)^2/9 + (2-1)^2/1 = 10/9, \end{align*} which is not significant against \(\chi_1^2\).' Is his argument satisfactory, and is there evidence for the machine being faulty? Does it matter that the same value of \(\chi^2\) is obtained on a day when all the sweets produced are good?
Solution: His argument is erroneous. He essentially is losing information about the size of the sample, we should be using: \begin{align*} && \chi^2 &= \frac{(90-80)^2}{90} + \frac{(20-10)^2}{10} \\ &&&= \frac{100}{9} = 11.11 \end{align*} which is significant at \(p=0.001\), so there is a very strong evidence the machine is faulty. When all the sweets are good the issue is Chi-squared is a two-sided test, so we are still learning that the machine is acting well out of it's distribution which is not good.