Problems

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Solution: If the highest mark is \(k\), then there are \(n-1\) remaining marks to give, and they have to be chosen from the numbers \(1, 2, \ldots, k-1\), ie in \(\binom{k-1}{n-1}\) ways. There are \(n\) numbers to be chosen from \(1, 2, \ldots, m\) in total, therefore \(\displaystyle \mathbb{P}(K=k) = \left.\binom{k-1}{n-1} \right/ \binom{m}{n}\) Since \(K\) can take any of the values \(n, \cdots, m\), we must have \begin{align*} && 1 &= \sum_{k=n}^m \mathbb{P}(K=k) \\ &&&= \sum_{k=n}^m \left.\binom{k-1}{n-1} \right/ \binom{m}{n} \\ \Rightarrow && \binom{m}{n} &= \sum_{k=n}^m \binom{k-1}{n-1} \\ \\ && \mathbb{E}(K) &= \sum_{k=n}^m k \cdot \mathbb{P}(K=k) \\ &&&= \sum_{k=n}^m k \cdot \left.\binom{k-1}{n-1} \right/ \binom{m}{n} \\ &&&= n\binom{m}{n}^{-1} \sum_{k=n}^m \frac{k}{n} \cdot \binom{k-1}{n-1} \\ &&&= n\binom{m}{n}^{-1} \sum_{k=n}^m \binom{k}{n} \\ &&&= n\binom{m}{n}^{-1} \sum_{k=n+1}^{m+1} \binom{k-1}{n+1-1} \\ &&&= n\binom{m}{n}^{-1} \binom{m+1}{n+1} \\ &&&= n \cdot \frac{m+1}{n+1} \end{align*}

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Solution: \begin{align*} && \mathbb{P}(N = k) &= \mathbb{P}(\text{guesses }k-1\text{ correctly then 1 wrong})\\ &&&= \frac12 \cdot \frac{1}{3} \cdots \frac{1}{k-1} \frac{k-1}{k} \\ &&&= \frac{k-1}{k!} \\ &&\mathbb{E}(N) &= \sum_{k=2}^\infty k \cdot \mathbb{P}(N=k) \\ &&&= \sum_{k=2}^{\infty} \frac{k(k-1)}{k!} \\ &&&= \sum_{k=0}^{\infty} \frac{1}{k!} = e \\ && \textrm{Var}(N) &= \mathbb{E}(N^2) - \mathbb{E}(N)^2 \\ && \mathbb{E}(N^2) &= \sum_{k=2}^{\infty} k^2 \mathbb{P}(N=k) \\ &&&= \sum_{k=2}^{\infty} \frac{k^2(k-1)}{k!} \\ &&&= \sum_{k=0}^{\infty} \frac{k+2}{k!} \\ &&&= \sum_{k=0}^{\infty} \frac{1}{k!} + 2 \sum_{k=0}^{\infty} \frac{1}{k!} = 3e \\ \Rightarrow && \textrm{Var}(N) &= 3e-e^2 \end{align*}

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Solution: \begin{align*} && \P(A|H) &= \P(\text{achieve }r\text{ heads immediately}) + \P(\text{don't and then achieve it from having flipped a tail}) \\ &&&= p^{r-1} + (1-p^{r-1}) \cdot \P(A|T) \\ && \P(A|T) &= (1-q^{s-1})\P(A|H) \\ \\ &&\P(A|H) &= p^{r-1}+(1-p^{r-1})(1-q^{s-1})\P(A|H) \\ \Rightarrow && \P(A|H) &= \frac{p^{r-1}}{1-(1-p^{r-1})(1-q^{s-1})} \\ && \P(A|T) &= \frac{(1-q^{s-1})p^{r-1}}{1-(1-p^{r-1})(1-q^{s-1})} \\ && \P(A) &= \frac{(2-q^{s-1})p^{r-1}}{2(1-(1-p^{r-1})(1-q^{s-1}))} \end{align*}

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Solution:

1. We can obtain a \(6\) from \(4+2+0, 4+1+1, 3+3+0, 3+2+1, 2+2+2\). So \(x^6\) from \(4,2,0\) can happen in \(6\) ways and gets us a coefficient of \(1 \cdot 3 \cdot 5\). \(x^6\) from \(4,1,1\) can happen in \(3\) ways and gets us a coefficient of \(5 \cdot (-2) \cdot (-2)\). \(x^6\) from \(3,3,0\) can happen in \(3\) ways and gets us a coefficient of \((-4) \cdot (-4) \cdot 1\). \(x^6\) from \(3,2,1\) can happen in \(6\) ways and gets us a coefficient of \((-4) \cdot 3 \cdot (-2)\). \(x^6\) from \(2,2,2\) can happen in \(1\) ways and gets us a coefficient of \(3 \cdot 3 \cdot 3\). This leaves us with a total coefficient of: \(6 \cdot 15 + 3 \cdot 20 + 3 \cdot 16 + 6 \cdot 24 + 1 \cdot 27 = 369\)
2. \begin{align*} (1+x)^{-2} &= 1 + (-2)x+\frac{(-2)\cdot(-3)}{2!} x^2 + \frac{(-2)(-3)(-4)}{3!}x^3 + \cdots \\ &= 1 -2x+3x^2-4x^3+5x^4+\cdots \\ \end{align*} The coefficient of \(x^6\) in the expansion of \((1+x)^{-6}\) will be \(\frac{(-6)(-7)(-8)(-9)(-10)(-11)}{6!} = \frac{11!}{6!5!} = 462\). The coefficient of \(x^6\) in the expansion of \((1 -2x+3x^2-4x^3+5x^4+\cdots)^3\) will be the same as the coefficient of \(x^6\) in the expansion of \((1 -2x+3x^2-4x^3+5x^4-6x^5+7x^6)^3\), ie it will be \(462\)

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Solution: Consider the linear \(3\times 3\) system in \(yz, zx, xy\), then \begin{align*} \left(\begin{array}{ccc|c} 2 & 1 & -5 & 2 \\ 1 & -1 & 2 & 1 \\ 1 & -2 & 6 & 3 \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 1 & -2 & 6 & 3 \\ 2 & 1 & -5 & 2 \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 0 & -1 & 4 & 2 \\ 0 & 3 & -9 & 0 \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 0 & -1 & 4 & 2 \\ 0 & 0 & 3 & 6 \\ \end{array}\right) \\ \end{align*} Therefore \(yz = 2, zx = 6, xy = 3 \Rightarrow (xyz)^2 = 36 \Rightarrow xyz = \pm 6\). If \(xyz = 6, x = 3, y = 1, z = 2\), if \(xyz = -6, x = -3, y = -1, z = -2\)

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Solution: \begin{align*} && \cos 4u &= 2\cos^2 2u - 1 \\ &&&= 2 (2\cos^2 u - 1)^2 - 1 \\ &&&= 2(4\cos^4u - 4\cos^2 u + 1) - 1\\ &&&= 8\cos^4u - 8\cos^2 u + 1 \end{align*} \begin{align*} && I &= \int_{-1}^1 \frac{1}{\sqrt{1+x}+\sqrt{1-x}+2} \d x \\ x = \cos t, \d x = - \sin t \d t: &&&= \int_{t = \pi}^{t=0} \frac{1}{\sqrt{1+\cos t} + \sqrt{1-\cos t} + 2} (- \sin t ) \d t \\ &&&= \int_0^\pi \frac{\sin t}{\sqrt{2 \cos^2 \frac{t}{2}}+\sqrt{2 \sin^2 \frac{t}{2}}+2} \d t \\ &&&= \int_0^\pi \frac{\sin t}{\sqrt{2}(\cos \frac{t}{2} + \sin \frac{t}{2}) + 2} \d t \\ &&&= \int_0^\pi \frac{\sin t}{\sqrt{2}(\sqrt{2} \cos (\frac{t}{2}-\frac{\pi}{4})) + 2} \d t \\ &&&= \int_0^\pi \frac{\sin t}{2(1+\cos (\frac{t}{2}-\frac{\pi}{4}))} \d t \\ &&&= \int_0^\pi \frac{\sin t}{4\cos^2(\frac{t}{4}-\frac{\pi}{8})} \d t \\ \\ u = \tfrac{t}{4} -\tfrac{\pi}{8}, \d u = \tfrac14 \d t:&&&=\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin (4u+\frac{\pi}{2})}{4 \cos^2 u} 4 \d u \\ &&&= \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\cos4u}{\cos^2 u} \d u \\ &&&= \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 4 (2 \cos^2 u-1)-4 + \sec^2 u \d u \\ &&&= \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 4 \cos 2u-4 + \sec^2 u \d u \\ &&&= \left [2\sin 2u - 4u + \tan u \right]_{-\pi/8}^{\pi/8} \\ &&&= 4 \sin \frac{\pi}{4} - \pi+ 2\tan \frac{\pi}{8} \\ &&&= \frac{4}{\sqrt{2}} - \pi + 2\sqrt{2}-2 \\ &&&= 4\sqrt{2} - \pi - 2 \end{align*}

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Solution: \begin{align*} && 0 &= z^4+z^3+z^2+z+1 \\ \Rightarrow && 0 &= z^2+z+1+z^{-1}+z^{-2} \tag{\(z \neq 0\)} \\ &&&= \left ( z+z^{-1} \right)^2-2 + z+z^{-1} + 1 \\ &&&= u^2+u-1 \\ \Rightarrow && u &= \frac{-1 \pm \sqrt{5}}{2} \\ \Rightarrow && z+z^{-1} &= \frac{-1 \pm \sqrt{5}}{2} \\ \Rightarrow && 0 &= z^2-\left ( \frac{-1 \pm \sqrt{5}}{2}\right)z+1 \\ \Rightarrow && z &= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\left ( \frac{-1 \pm \sqrt{5}}{2}\right)^2-4}}{2} \\ &&&= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{1+5\mp2\sqrt{5}-16}{4}}}{2} \\ &&&= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{-10\mp2\sqrt{5}-16}{4}}}{2} \\ &&&= \frac{-1\pm\sqrt{5}}{4} \pm i\frac{\sqrt{10\pm 2\sqrt{5}}}{4} \end{align*} Since \(z^4+z^3+z^2+z+1 = 0\) we can multiply both sides by \(z-1\) to obtain \(z^5-1 = 0\). Therefore if \(z = r(\cos \theta + i \sin \theta)\) we see that \(z^5 = 1 \Rightarrow r^5 (\cos 5 \theta + i \sin 5 \theta) = 1 \Rightarrow r = 1, 5 \theta = 2n \pi\) ie \(z = \cos \frac{2n\pi}{5} + i\sin \frac{2n \pi}{5}\). We are looking for a solution in the first quadrant, therefore \(\cos \frac{2\pi}{5} = \frac{-1 + \sqrt{5}}4\) and \(\sin \frac{2\pi}{5} = \frac{\sqrt{10+2\sqrt{5}}}{4}\)

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Solution: \(12\) has factors \(1,2,3,4,6,12\) of which \(4\) are neither \(1\) nor \(12\). \(16\) has factors \(1,2,4,8,16\) of which \(3\) are neither \(1\) nor \(16\). If \(N = p_1^{m_1} \cdots p_r^{m_r}\) then \(N\) has \((m_1+1)\cdots(m_r+1)\) factors since we can have between \(0 \leq k \leq m_i\) of the \(i\)th prime factor, whcih is \(m_i+1\) possibilities. We then need to subtract two for the proper factors, ie \((m_1+1)\cdots(m_r+1) - 2\).

1. We need \((m_1 + 1) \cdots (m_r + 1) - 2 = 12\) ie \((m_1+1) \cdots (m_r +1) = 14\). Since \(14 = 2 \times 7\) we should have two prime factors, ie of the form \(p_1^6p_2^1\). The smallest number of this form is \(2^6 \cdot 3 = 192\)
2. We wish to make \((m_1+1)\cdots(m_r+1) \geq 14\) with \(2^{m_1} \cdot 3^{m_2} \cdots p_r^{m_r}\) as small as possible and \(m_1 \geq m_2 \geq \cdots \geq m_r\). With \(1\) prime, the best we can do is \(2^{13}\). With \(2\) primes the best we can do is \(2^p \cdot 3^q\) with \((p+1)(q+1) \geq 14\). \begin{array}{c|c} q & p & N \\ \hline 0 & 13 & 2^{13} \\ 1 & 6 & 2^6 \cdot 3^1 \\ 2 & 5 & 2^5 \cdot 3^2 \\ 3 & 3.& 2^3 \cdot 3^3 \end{array} So the best here is \(2^6 \cdot 3\) With \(3\) primes, the best we can do is \(2^p 3^q 5^r\) with \(p \geq q \geq r\) and \((p+1)(q+1)(r+1) \geq 14\) \begin{array}{c|c|c|c} r & q & p & N \\ \hline 1 & 1 & 3 & 2^3 \cdot 3 \cdot 5 \\ 1 & 2 & 2& 2^2 \cdot 3^2 \cdot 5 \\ 2 & 2& 2 & 2^2 \cdot 3^2 \cdot 5^2 \end{array} With four primes we can achieve it immediately with \(2\cdot3\cdot5\cdot 7\) and more primes clearly wont help, so our best options is \(120\)

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Solution: \begin{align*} && g(0) &= f(0)+f(-0) = 2f(0) = 2 \\ && g'(x) &= f'(x) - f'(-x) \\ && g'(0) &= f'(0) - f'(-0) = 0 \\ && g''(x) &= f''(x) +f''(-x) \\ \Rightarrow && g''(x) + g(x) &= f''(x) +f''(-x) + f(x) + f(-x) \\ &&&= f''(x)+ f(-x) +f''(-x) + f(x) \\ &&&= x + 3 \cos 2x + (-x + 3 \cos (-2x) ) \\ &&&= 6 \cos 2x \\ \end{align*} Considering the homogeneous part, we should expected a solution of the form \(g(x) = A \sin x + B \cos x\). Seeking an integrating factor of the form \(g(x) = C \cos 2x\) we see that \(-4C \cos 2x + C \cos 2x = 6 \cos 2x \Rightarrow -3C = 6 \Rightarrow C = -2\). Therefore the general solution is \begin{align*} && g(x) &= A\sin x + B \cos x - 2\cos 2x \\ && g(0) &= B - 2 = 2\\ && g'(0) &= A = 0 \\ \Rightarrow && g(x) &= 4\cos x - 2\cos 2x \\ \end{align*} \begin{align*} && h(0) &= f(0) - f(-0) = 0 \\ && h'(x) &= f'(x) + f'(-x) \\ && h'(0) &= f'(0) + f'(-0) = -2 \\ && h''(x) &= f''(x) - f''(-x) \\ \Rightarrow && h''(x) - h(x) &= f''(x) - f''(-x) -( f(x) - f(-x)) \\ &&&= f''(x) +f(-x)- (f''(-x) + f(x)) \\ &&&= x + 3\cos 2x - (-x + 3 \cos(-2x)) \\ &&&= 2x \end{align*} Considering the homogeneous part, we should expect a solution of the form \(Ae^x + Be^{-x}\). For a specific integral, we can take \(-2x\), ie \begin{align*} && h(x) &= Ae^x + Be^{-x} - 2x \\ && h(0) &= A+B =0 \\ && h'(0) &= A-B-2 =-2 \\ \Rightarrow && A &=B = 0 \\ \Rightarrow && h(x) &= -2x \end{align*} Therefore \(f(x) = \frac12(f(x) + f(-x)) + \frac12(f(x) -f(-x)) = 2\cos x - \cos 2x -x\)

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Solution:

By symmetry the centre of mass will lie on the main axis. Taking the plane faces as \(x = 0\) we have the following centers of mass: \begin{align*} && \text{COM} && \text{Mass} \\ \text{Hemisphere} && -\frac{3a}{8} && \frac{2\pi a^3}{3} \\ \text{Cone} && \frac{\lambda a}{4} && \frac{\lambda \pi a^3}{3} \\ \text{Toy} && \bar{x} && \frac{(\lambda + 2)\pi a^3}{3} \\ \end{align*} Therefore, \begin{align*} && \frac{(\lambda + 2)\pi a^3}{3} \cdot \bar{x} &= -\frac{3a}{8} \cdot \frac{2\pi a^3}{3} + \frac{\lambda a}{4} \cdot \frac{\lambda \pi a^3}{3} \\ \Rightarrow && (\lambda + 2) \bar{x} &= \frac{(\lambda^2 -3)a}{4} \end{align*} Therefore the centre of mass will be inside the hemisphere (and it will always move to an upright position) iff \(\bar{x} < 0 \Leftrightarrow \lambda < \sqrt{3}\). For the toy to topple from this position, \(\bar{x}\) must be longer than it would need to be to form a right-angled triangle with the vertical at the plane face. The angle at this point will be \(\theta\), so we need: \(\bar{x} > a\tan \theta = a \frac{a}{\lambda a} = \frac{a}{\lambda}\) \begin{align*} && \bar{x} &> \frac{a}{\lambda} \\ \Leftrightarrow && \frac{(3-\lambda^2)a}{4(\lambda + 2)} &> \frac{a}{\lambda} \\ \Leftrightarrow && {(3-\lambda^2)\lambda} &> {4(\lambda + 2)} \\ \Leftrightarrow && -\lambda^3-\lambda -8 &> 0 \\ \end{align*} Contradiction! Therefore it can never topple when laid on its side.

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Solution:

\begin{align*} \uparrow: && v &= u + at \\ \Rightarrow && T &= \frac{a \omega \sin \theta}{g} \\ && s &= ut + \frac12 gt^2 \\ \Rightarrow && s &= a\omega \sin \theta \cdot \frac{a \omega \sin \theta}{g} - \frac12 g \left ( \frac{a \omega \sin \theta}{g} \right) ^2 \\ &&&= \frac1{2g} a^2 \omega^2 \sin^2 \theta \\ s < \text{distance to top}: && \frac1{2g} a^2 \omega^2 \sin^2 \theta &< a(1- \cos \theta) \\ \Rightarrow && \omega^2 &< \frac{2g}{a} \frac{1-\cos \theta}{\sin^2 \theta} \\ &&&= \frac{2g}{a} \frac{2 \sin^2 \tfrac12 \theta}{4 \sin^2 \tfrac12 \theta \cos^2 \tfrac12 \theta} \\ &&&= \frac{g}{a} \sec^2 \tfrac12 \theta \\ &&&\leq \frac{g}{a} \tag{since it holds for all \(\theta\) it holds for min \(\theta\)}\\ \Rightarrow && \omega_0 &= \sqrt{\frac{g}{a}} \\ \\ && \text{max height} &= \frac1{2g} a^2 \omega^2 \sin^2 \theta - a(1-\cos \theta) \\ &&&= \frac1{2g} a^2 \omega^2 (1-\cos^2 \theta) - a(1-\cos \theta) \\ &&&= \frac{a}{2} \left (- \frac{\omega^2}{\omega_0^2} \cos^2 \theta + 2 \cos \theta + \frac{\omega^2}{\omega_0^2}-2 \right) \\ &&&= \frac{a}{2} \left (-\left (\frac{\omega_0}{\omega}- \frac{\omega}{\omega_0} \cos \theta \right)^2 + \frac{\omega^2}{\omega_0^2}-2+\frac{\omega_0^2}{\omega^2} \right) \\ &&&= \frac{a}{2} \left ( \frac{\omega}{\omega_0} - \frac{\omega_0}{\omega} \right)^2 - \frac{a}{2} \left (\frac{\omega_0}{\omega}- \frac{\omega}{\omega_0} \cos \theta \right)^2 \end{align*} If \(\omega > \omega_0\) we can find a \(\theta\) such that the second bracket is \(0\), hence the maximium height is as desired.

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Solution:

1. \(\,\) \begin{align*} && F_Y(\lambda) &= \mathbb{P}(Y < \lambda) \\ &&&= \prod_i \mathbb{P}(X_i < \lambda) \\ &&&= \lambda^n \\ \Rightarrow && f_Y(\lambda) &= \begin{cases} n \lambda^{n-1} & \text{if } 0 \leq \lambda \leq 1 \\ 0 & \text{otherwise} \end{cases} \\ \\ && \E[Y] &= \int_0^1 \lambda f_Y(\lambda) \d \lambda \\ &&&= \int_0^1 n \lambda^n \d \lambda \\ &&&= \frac{n}{n+1} \\ && \E[Y^2] &= \int_0^1 \lambda^2 f_Y(\lambda) \d \lambda \\ &&&= \int_0^1 n \lambda^{n+1} \d \lambda \\ &&&= \frac{n}{n+2} \\ \Rightarrow && \var[Y] &= \E[Y^2]-(\E[Y])^2 \\ &&&= \frac{n}{n+2} - \frac{n^2}{(n+1)^2} \\ &&&= \frac{(n+1)^2n-n^2(n+2)}{(n+2)(n+1)^2} \\ &&&= \frac{n[(n^2+2n+1)-(n^2+2n)]}{(n+2)(n+1)^2} \\ &&&= \frac{n}{(n+2)(n+1)^2} \end{align*}
2. Using the same reasoning, we can see that \begin{align*} && 1-F_Z(t) &= \mathbb{P}(\text{all lights still on after t}) \\ &&&= \prod_i e^{-t/\lambda} \\ &&&= e^{-nt/\lambda} \\ \\ \Rightarrow && F_Z(t) &= 1-e^{-nt/\lambda} \end{align*} Therefore \(Z \sim Exp(\frac{n}{\lambda})\) and the time to first failure is \(\lambda/n\)