Problems

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Solution:

1. Suppose \(n = 1\), then all polynomials are reflexive (since \(x - a_1\) has the root \(a_1\). Suppose \(n = 2\), then we want \begin{align*} && x^2-a_1x+a_2 &= (x-a_1)(x-a_2) \\ &&&= x^2-(a_1+a_2)x+a_1a_2 \\ \Rightarrow && a_2 &= 0 \\ \end{align*} So all polynomials of the form \(x^2-a_1x\) work and no others. Suppose \(n = 3\) then we want \begin{align*} && x^3-a_1x^2+a_2x-a_3 &= (x-a_1)(x-a_2)(x-a_3) \\ &&&= x^3-(a_1+a_2+a_3)x+(a_1a_2+a_1a_3+a_2a_3)x-a_1a_2a_3 \\ \Rightarrow && a_2+a_3 &= 0 \\ && a_2a_3 &= a_2 \\ \Rightarrow && -a_2^2 &= a_2 \\ \Rightarrow && a_2 &= 0, -1 \\ && -a_1a_2^2 &= -a_2 \\ \Rightarrow && a_2 &= 0, a_2 = 1/a_1 \end{align*} So we need either \(x^3-a_1x\) or \((x+1)^2(x-1) = x^3+x^2-x-1\)
2. Suppose \(n > 3\) then \begin{align*} && \sum a_i^2 &= \left (\sum a_i \right)^2 - 2 \sum_{i < j} a_i a_j \\ && &= a_1^2 - 2a_2 \\ \Rightarrow && 2a_2 &= a_1^2 - \sum a_i^2 \\ &&&= -a_2^2 - a_3^2 - \cdots - a_n^2 \end{align*} So \((a_2+1)^2 = 1-a_3^2 -\cdots -a_n^2\) so if \(a_n > 0\) (or any other \(a_i, i > 2\) for that matter) then we must have \(a_n = \pm 1, a_{3}, \ldots a_{n-1} = 0\), but if \(a_n = \pm 1\) \(x = 0\) is not a root. Therefore we must have \(a_0\) and \(a_i = 0\) for all \(i > 3\)
3. The only reflexive polynomials therefore must be \(x^n - kx^{n-1}\) and \(x^{n+3}+x^{n+2}-x^{n+1}-x^n\)

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Solution:

1. \(\,\)
   

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2. \(\,\) \begin{align*} && y &= \frac{cx}{\sqrt{x^2+p}} \\ && \d y &= \frac{c(x^2+p)-cx^2}{(x^2+p)^{3/2}} \d x \\ &&&= \frac{cp^2}{(x^2+p)^{3/2}} \d x\\ && I &= \int \frac1{(b^2-y^2)\sqrt{c^2-y^2}} \d y \\ &&&= \int \frac{1}{\left ( b^2 - \frac{c^2x^2}{x^2+p} \right) \sqrt{c^2 - \frac{c^2x^2}{x^2+p} }} \d y \\ &&&= \int \frac{(x^2+p)^{3/2}}{((b^2-c^2)x^2+pb^2)\sqrt{c^2p}}\frac{cp}{(x^2+p)^{3/2}} \d x \\ &&&= \int \frac{\sqrt{p}}{((b^2-c^2)x^2+pb^2)} \d x \\ p=1: &&&= \int \frac{1}{(b^2-c^2)x^2+b^2} \d x \end{align*} When \(b = \sqrt{3}, c = \sqrt{2}\) \begin{align*} && I_1 &= \int_1^{\sqrt{2}} \frac{1}{(3 - y^2)\sqrt{2 - y^2}} \d y\\ &&&= \int_{x =1 }^{x=\infty} \frac{1}{3+x^2} \d x \\ &&&= \left [ \frac{1}{\sqrt{3}} \tan^{-1} \frac{x}{\sqrt{3}} \right]_1^\infty \\ &&&= \frac{\pi}{2\sqrt{3}} - \frac{1}{\sqrt{3}} \frac{\pi}{6} \\ &&&= \frac{\pi}{3\sqrt{3}} \end{align*} \begin{align*} && I_2 &= \int_{\frac{1}{\sqrt{2}}}^1 \frac{y}{(3y^2 - 1)\sqrt{2y^2 - 1}} \d y \\ x = \frac1y, \d x = -\frac1{y^2} \d y &&&= \int_{x=\sqrt{2}}^{x=1} \frac{x^2}{(3-x^2)\sqrt{2-x^2}}\cdot \left ( -\frac{1}{x^2} \right ) \d x \\ &&&= \int_1^{\sqrt{2}} \frac{1}{(3-x^2)\sqrt{2-x^2}} \d x \\ &&&= I_1 = \frac{\pi}{3\sqrt{3}} \end{align*}
3. \(\,\) \begin{align*} && I_3 &= \int_{\frac{1}{\sqrt{2}}}^1 \frac{1}{(3y^2 - 1)\sqrt{2y^2 - 1}} \d y \\ x = 1/y, \d x = -1/y^2 \d y &&&= \int_{x=1}^{x=\sqrt{2}} \frac{x}{(3-x^2)\sqrt{2-x^2}} \d x \\ u = x^2, \d u = 2x \d x &&&= \int_{u=1}^{u=2} \frac{\frac12}{(3-u)\sqrt{2-u}} \d u \\ v=2-u, \d v = -\d u &&&= \frac12\int_{v=0}^{v=1} \frac{1}{(1+v)\sqrt{v}} \d v \\ &&&=\left [\tan^{-1}\sqrt{v}\right]_0^1 \\ &&&= \frac{\pi}{4} \end{align*}

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Solution: \begin{align*} && |z-a|^2 &= (z-a)(z-a)^* \\ &&&= (z-a)(z^*-a^*) \\ &&&= zz^*-az^*-a^*z+aa^* \\ &&&= r^2 \end{align*} Therefore the locus of \(P\) is a circle centre \(a\) radius \(r\).

1. \begin{align*} && 0 &= zz^* - az^* - a^*z + aa^* - r^2 \\ &&&= \frac{1}{\omega \omega^{*}} - \frac{a}{\omega^*} - \frac{a^*}{\omega} + aa^*-r^2 \\ \Rightarrow && 0 &= 1-a\omega-a^*\omega^*+(|a|^2-r^2)\omega\omega^* \\ \Rightarrow && 0 &= \omega\omega^* - \left ( \frac{a^*}{|a|^2-r^2}\right)^*\omega - \left ( \frac{a^*}{|a|^2-r^2}\right)\omega^*+\left ( \frac{a^*}{|a|^2-r^2}\right)\left ( \frac{a}{|a|^2-r^2}\right)-\left ( \frac{a^*}{|a|^2-r^2}\right)\left ( \frac{a}{|a|^2-r^2}\right)+ \frac{1}{|a|^2-r^2} \\ &&&= \omega\omega^* - \left ( \frac{a^*}{|a|^2-r^2}\right)^*\omega - \left ( \frac{a^*}{|a|^2-r^2}\right)\omega^*+\frac{|a|^2}{(|a|^2-r^2)^2}-\frac{|a|^2}{(|a|^2-r^2)^2}+ \frac{1}{|a|^2-r^2} \\ &&&=\omega\omega^* - \left ( \frac{a^*}{|a|^2-r^2}\right)^*\omega - \left ( \frac{a^*}{|a|^2-r^2}\right)\omega^*+\frac{|a|^2}{(|a|^2-r^2)^2}- \frac{r^2}{(|a|^2-r^2)^2} \end{align*} Therefore \(\displaystyle \left|\omega-\left ( \frac{a^*}{|a|^2-r^2}\right)\right|^2 = \frac{r^2}{(|a|^2-r^2)^2}\) ie \(\omega\) lies on a circle centre \(\frac{a^*}{|a|^2-r^2}\), radius \(\frac{r}{||a|^2-r^2|}\). If these are the same circle then \(r = \frac{r}{||a|^2-r^2|} \Rightarrow (|a|^2-r^2)^2 = 1\) and \(a = \frac{a^*}{|a|^2-r^2} \Rightarrow a = \pm a^*\), ie \(a\) is purely real or imaginary.
2. This is the same story, except we end up with centre \(\frac{a}{|a|^2-r^2}\), so we do not end up with the same conditions

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Solution:

1. Suppose \(a=b\), ie \begin{align*} && y^2(y^2-a^2) &= x^2(x^2-a^2) \\ \Rightarrow && 0 &= x^4-y^4-a^2(x^2-y^2) \\ &&&= (x^2-y^2)(x^2+y^2-a^2) \end{align*} Therefore we have the lines \(y = \pm x\) and a circle radius \(a\).
   

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2. 1. Since \(x^4 - 4x^2 - y^2(y^2-5)= 0\), we must have \(0 \leq \Delta = 16 + 4y^2(y^2-5) \Rightarrow y^4-5y^2+4 = (y^2-4)(y^2-1) \geq 0\), therefore \(0 \leq y \leq 1\) or \(y \geq 2\) (since we are only considering positive values of \(y\)).
   2. When \((x, y) \approx 0\) the equation is more like \(4x^2 \approx 5y^2\) or \(y \approx \frac{2}{\sqrt{5}}x\) If \(|x|, |y|\) are very large, it is more like \(x^4 \approx y^4\), ie \(y \approx x\)
   3. \(\,\) \begin{align*} && (2y(y^2-5)+y^2(2y))y' &= 2x(x^2-4)+2x^3 \\ \Rightarrow && (4y^3-10y)y' &= 4x^3-8x \end{align*} Therefore the gradient is parallel to the \(x\)-axis when \(x = 0, x = \sqrt{2}\). We need \(x = 0, y \neq 0\), ie \(y = \sqrt{5}\), so \((0, \sqrt{5})\) and \((\sqrt{2}, 0)\) It is parallel to the \(y\)-axis when \(y = 0\) or \(y = \sqrt{\frac52}\), ie \((2, 0)\)
   

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3. \(\,\)
   

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Solution:

1. Let \(A = (0,0,0)\) and then \(B = 22b \mathbf{i}, D = 2d\mathbf{j}, C = 2b\mathbf{i}+2d\mathbf{j}\) and \(V = b \mathbf{i} + d\mathbf{j} + h\mathbf{k}\) We also have \begin{align*} && \tan \alpha &= \frac{h}{d}\\ && \tan \beta &= \frac{d}{b} \\ && \vec{AV} \times \vec{VB} &= \begin{pmatrix} b \\ d \\ h \end{pmatrix} \times \begin{pmatrix} -b \\ d \\ h \end{pmatrix} \\ &&&= \begin{pmatrix} 0 \\ -2bh \\ 2db \end{pmatrix} \\ &&&= 2b \begin{pmatrix} 0 \\ -d \tan \alpha \\ d \end{pmatrix} \\ &&&= k \begin{pmatrix} 0 \\ - \sin \alpha \\ \cos \alpha \end{pmatrix} \end{align*} similarly for the vector perpendicular to the other face it must be \(\begin{pmatrix}-\sin \beta \\ 0 \\ \cos \beta \end{pmatrix}\) Looking at the angle between these perpendicular (to find the angles between the faces we see: \begin{align*} \begin{pmatrix} 0 \\ - \sin \alpha \\ \cos \alpha \end{pmatrix} \cdot \begin{pmatrix}-\sin \beta \\ 0 \\ \cos \beta \end{pmatrix} &= \cos \alpha \cos \beta \end{align*} But this is also \(\pi -\) the angle between the planes, ie \(\cos \theta = \cos \alpha \cos \beta\)
2. \(\,\) \begin{align*} && \cot^2 \phi &= \frac{b^2+d^2}{h^2} \\ && \cot^2 \alpha &= \frac{d^2}{h^2} \\ && \cot^2 \beta &= \frac{b^2}{h^2} \\ \Rightarrow && cot^2 \phi &= \cot^2 \beta+\cot^2 \alpha \end{align*} \begin{align*} && \cos^2 \phi &= \frac{b^2+d^2}{b^2+d^2+h^2} \\ && \cos^2 \alpha &= \frac{d^2}{d^2+h^2} \\ && \cos^2 \beta &= \frac{b^2}{b^2+h^2} \\ && \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1-\cos^2 \theta} &= \frac{\frac{d^2}{d^2+h^2}+\frac{b^2}{b^2+h^2}-2\cdot \frac{d^2}{d^2+h^2} \cdot \frac{b^2}{b^2+h^2}}{1 - \frac{d^2}{d^2+h^2} \cdot\frac{b^2}{b^2+h^2}} \\ &&&= \frac{d^2(b^2+h^2)+b^2(d^2+h^2)-2d^2b^2}{(d^2+h^2)(b^2+h^2)-d^2b^2} \\ &&&= \frac{h^2(b^2+d^2)}{h^2(b^2+d^2+h^2)} \\ &&&= \frac{b^2+d^2}{b^2+d^2+h^2} \\ &&&= \cos^2\phi \end{align*} Also notice that \begin{align*} && \cos^2 \alpha + \cos^2 \beta &\underbrace{\geq}_{AM-GM} 2 \cos \alpha \cos \beta \\ &&&= 2 \cos \theta \\ \Rightarrow && \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1-\cos^2 \theta} &\geq \frac{2 \cos \theta - 2\cos^2 \theta}{1-\cos^2 \theta} \\ &&&= \frac{2\cos \theta}{1+\cos \theta} = \cos \theta \frac{2}{1+\cos \theta} \\ &&&> \cos^2 \theta \\ \Rightarrow && \phi &< \theta \end{align*}

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Solution:

1. \(\mathbf{r} = (a \sin \theta - s) \mathbf{i}+a\cos \theta\mathbf{j}\), so \begin{align*} && \dot{\mathbf{r}} &=(a \dot{\theta} \cos \theta - \dot{s}) \mathbf{i}- a\dot{\theta} \sin \theta \mathbf{j}\\ \\ \text{COM}(\rightarrow): && 0 &= M(-\dot{s}) + m(a \dot{\theta} \cos \theta - \dot{s}) \\ \Rightarrow && \dot{s} &= \frac{ma \dot{\theta} \cos \theta}{m+M} \\ &&&= \left ( 1- \frac{M}{m+M} \right) a\dot{\theta} \cos \theta \\ &&&= (1 - k) a\dot{\theta} \cos \theta \\ \\ \Rightarrow && \dot{\mathbf{r}} &=(a \dot{\theta} \cos \theta - \dot{s}) \mathbf{i}- a\dot{\theta} \sin \theta \mathbf{j} \\ &&&= (a \dot{\theta} \cos \theta - (1 - k) a\dot{\theta} \cos \theta) \mathbf{i}- a\dot{\theta} \sin \theta \mathbf{j} \\ &&&= a\dot{\theta} \left ( k \cos \theta \mathbf{i} - \sin \theta \mathbf{j} \right) \end{align*}
2. \(\,\) \begin{align*} COE: &&\underbrace{0}_{\text{k.e.}}+ \underbrace{mga}_{\text{GPE}} &= \underbrace{\frac12 m \mathbf{\dot{r}}\cdot\mathbf{\dot{r}}}_{\text{k.e. }P} + \underbrace{mg a\cos \theta}_{\text{GPE}} + \underbrace{\frac12 M \dot{s}^2}_{\text{k.e. hemisphere}} \\ \Rightarrow && 2amg(1-\cos \theta) &= a^2m \dot{\theta}^2(k^2 \cos^2 \theta + \sin^2 \theta)+ M(1 - k)^2 a^2\dot{\theta}^2 \cos^2 \theta \\ \Rightarrow && 2mg(1-\cos \theta) &= a \dot{\theta}^2 \left (m\sin^2 \theta + (mk^2 + M(1-k)^2)\cos^2 \theta \right) \\ &&&= a \dot{\theta}^2 \left (m\sin^2 \theta + mk\cos^2 \theta \right) \\ \Rightarrow && 2g(1-\cos \theta) &= a \dot{\theta}^2 \left (\sin^2 \theta + k\cos^2 \theta \right) \\ \end{align*}
3. The equation of motion is \(m \ddot{\mathbf{r}} = \mathbf{R} - mg\mathbf{j}\) and the particle will leave the surface when \(\mathbf{R} = 0\). If we take the component in the directions suggested: \begin{align*} && \ddot{\mathbf{r}} &= a\ddot{\theta}(k \cos \theta \mathbf{i}- \sin \theta \mathbf{j}) + a \dot{\theta}(-k\dot{\theta} \sin \theta \mathbf{i}- \dot{\theta} \cos \theta \mathbf{j}) \\ &&&= ak (\ddot{\theta} \cos \theta - \dot{\theta}^2 \sin \theta) \mathbf{i} -a(\ddot{\theta} \sin \theta + \dot{\theta}^2 \cos \theta) \mathbf{j} \\ \Rightarrow && \mathbf{\ddot{r}} \cdot (\sin \theta \mathbf{i} + k \cos \theta \mathbf{j}) &= ak (\ddot{\theta} \cos \theta - \dot{\theta}^2 \sin \theta) \sin \theta -ak(\ddot{\theta} \sin \theta + \dot{\theta}^2 \cos \theta)\cos \theta \\ &&&= - ak \dot{\theta}^2 \\ && (-g\mathbf{j}) \cdot (\sin \theta \mathbf{i} + k \cos \theta \mathbf{j}) &= -gk \cos \theta \\ \mathbf{R} = 0: && gk \cos \theta &= ak \dot{\theta}^2 \\ \Rightarrow && g \cos \theta &= a \dot{\theta}^2 \end{align*}
4. \(\,\) \begin{align*} && 2g(1-\cos \theta) &= a \dot{\theta}^2(k \cos^2 \theta + \sin^2 \theta) \\ && a \dot{\theta}^2 &= g \cos \alpha \\ \Rightarrow && 2g(1-\cos \alpha) &= g \cos \alpha(k \cos^2 \alpha + (1-\cos^2 \alpha)) \\ \Rightarrow && 0 &= g(k-1)c^3+3gc-2g \\ \Rightarrow && 0 &= (k-1)c^3+3c - 2 \end{align*} When \(c =1, f(c) = k > 0\) when \(c = \frac23, f(c) = k-1 < 0\). Therefore there is a root with \(\cos \alpha > \frac23\)

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Solution:

1. Since the impulse is along the line of centres, the velocities are as show in the diagram. Additionally, vertical velocity is unchanged, so: \(v \sin (\theta + \alpha) = u \sin \alpha\) \begin{align*} \text{COM}(\rightarrow): && u \cos\alpha &= v \cos(\alpha + \theta) + w \\ \Rightarrow && w &= u \cos \alpha - v \cos (\alpha + \theta) \end{align*}
2. Since the approach speed (horizontally) is \(u \cos \alpha\) the speed of separation is \(e u \cos \alpha\), in particular \(w - v \cos(\theta + \alpha) = e u \cos \alpha\) or \(w = v \cos (\theta + \alpha) + e u \cos \alpha\). \begin{align*} && w &= w \\ && v \cos (\theta + \alpha) + e u \cos \alpha &= u \cos \alpha - v \cos (\alpha + \theta) \\ \Rightarrow && \frac{u \sin \alpha}{\sin (\alpha + \theta)} \cos (\theta + \alpha) + e u \cos \alpha &= u \cos \alpha - \frac{u \sin \alpha}{\sin (\alpha + \theta)} \cos (\alpha + \theta) \\ \Rightarrow && \sin \alpha \cos(\theta + \alpha) + e \sin (\alpha+\theta)\cos \alpha &= \sin(\alpha+\theta) \cos \alpha - \cos(\alpha+\theta)\sin \alpha \\ &&&= \sin ((\alpha+\theta)-\alpha) \\ &&&= \sin \theta \end{align*} as required. \begin{align*} && \sin \theta &= \cos(\theta+ \alpha)\sin \alpha + e \sin (\theta + \alpha) \cos \alpha \\ &&&= \cos \theta \cos \alpha \sin \alpha - \sin \theta \sin^2 \alpha + e \sin \theta \cos ^2 \alpha + e \cos \theta \sin \alpha \cos \alpha \\ \Rightarrow && \tan \theta \sec^2 \alpha &= \tan \alpha - \tan \theta \tan^2 \alpha + e \tan \theta + e \tan \alpha \\ \Rightarrow && \tan \theta (1 + \tan^2 \alpha+\tan^2 \alpha-e) &= \tan \alpha + e \tan \alpha \\ \Rightarrow && \tan \theta &= \frac{(1+e)\tan \alpha}{1-e + 2\tan^2 \alpha} \end{align*} We seek to maximise \(y = \frac{x}{c+2x^2}\), \begin{align*} && \frac{\d y}{\d x} &= \frac{c+2x^2-4x^2}{(c+2x^2)^2} \\ &&&= \frac{c-2x^2}{(c+2x^2)^2} \end{align*} Therefore the maximum will occur at \(x = \sqrt{c/2}\), ie \(\tan \alpha = \sqrt{(1-e)/2}\) and theta will be \(\displaystyle \frac{(1+e)\sqrt{(1-e)/2}}{2(1-e)} =\frac{1}{2\sqrt{2}} \frac{1+e}{\sqrt{1-e}}\)

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Solution:

1. Let \(X\) be the number of people arriving on a given day, and \(Y\) be the number taking sand, then \begin{align*} && \mathbb{P}(Y = k) &= \sum_{x=k}^{\infty} \mathbb{P}(x \text{ arrive and }k\text{ of them take sand}) \\ &&&= \sum_{x=k}^{\infty} \mathbb{P}(X=x)\mathbb{P}(k \text{ out of }x\text{ of them take sand})\\ &&&= \sum_{x=k}^{\infty} e^{-\lambda} \frac{\lambda^x}{x!}\binom{x}{k}p^k(1-p)^{x-k}\\ &&&= e^{-\lambda} \left ( \frac{p}{1-p} \right)^k \sum_{x=k}^{\infty} \frac{((1-p)\lambda)^x}{k!(x-k)!} \\ &&&= e^{-\lambda} \left ( \frac{p}{1-p} \right)^k \frac{((1-p)\lambda)^k}{k!} \sum_{x=0}^{\infty} \frac{((1-p)\lambda)^x}{x!} \\ &&&= e^{-\lambda} \left ( \frac{p}{1-p} \right)^k \frac{((1-p)\lambda)^k}{k!}e^{(1-p)\lambda)} \\ &&&= e^{-p\lambda} \frac{(p\lambda)^k}{k!} \end{align*} which is precisely a Poisson with parameter \(p\lambda\). Alternatively, \(Y = B_1 + B_2 + \cdots + B_X\) where \(B_i \sim Bernoulli(p)\) so \(G_Y(t) = G_X(G_B(t)) = G_X(1-p+pt) = e^{-\lambda(1-(1-p+pt))} = e^{-p\lambda(1-t)}\) so \(Y \sim Po(\lambda)\) Alternatively, alternatively, let \(Z\) be the number of people not taking sand, so \begin{align*} && \mathbb{P}(Y = y, Z= z) &= \mathbb{P}(X=y+z) \cdot \binom{y+z}{y} p^y(1-p)^z \\ &&&= e^{-\lambda} \frac{\lambda^{y+z}}{(y+z)!} \frac{(y+z)!}{y!z!} p^y(1-p)^z \\ &&&=\left ( e^{-p\lambda} \frac{(p\lambda)^y}{y!} \right) \cdot \left ( e^{-(1-p)\lambda} \frac{((1-p)\lambda)^z}{z!}\right) \end{align*} So clearly \(Y\) and \(Z\) are both (independent!) Poisson with parameters \(p\lambda \) and \((1-p)\lambda\)
2. The amount taken is \(Sk + S(1-k)k + \cdots +Sk(1-k)^{Y-1} = Sk\cdot \frac{1-(1-k)^Y}{k} = S(1-(1-k)^Y)\) so \begin{align*} \E[\text{taken sand}] &= \E \left [ S(1-(1-k)^Y)\right] \\ &= S-S\E\left [(1-k)^Y \right] \\ &= S - SG_Y(1-k)\\ &=S - Se^{-p\lambda(1-(1-k))} \tag{pgf for Poisson} \\ &= S\left (1-e^{-kp\lambda} \right) \end{align*}
3. The fraction of grains the assistant takes home is: \((1-k)^Yk\), which has expected value \(ke^{-kp\lambda}\). This the the probability he takes home the golden grain. When \(k = 0\) the probability is \(0\) which makes sense (no-one takes home any sand, including the merchant's assistant). As \(k \to 1\) we get \(e^{-p\lambda}\) which is the probability that no-one gets any sand other than him. \begin{align*} && \frac{\d }{\d k} \left ( ke^{-kp\lambda} \right) &= e^{-kp\lambda} - (p\lambda)ke^{-kp\lambda} \\ &&&= e^{-kp\lambda}(1 - (p\lambda)k) \end{align*} Therefore maximised at \(k = \frac{1}{p\lambda}\). (Clearly this is a maximum just by sketching the function)

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Solution: For every element in \(S\) we can choose whether or not it appears in a subset of \(S\), therefore there are \(2^n\) choices so \(2^n\) distinct subsets.

1. \(\mathbb{P}(1 \in A_1) = \frac12\) (since \(1\) is in exactly half the subsets)
2. \(\,\) \begin{align*} && \mathbb{P}(A_1 \cap A_2 = \emptyset) &= \mathbb{P}(i \not \in (A_1 \cap A_2) \forall i) \\ &&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1 \cap A_2) \right) \\ &&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1)\mathbb{P}(i \in \cap A_2) \right) \\ &&&= \prod_{i=1}^n \left ( 1-\frac12 \cdot \frac12\right) \\ &&&= \left (\frac34 \right)^n \end{align*}
3. \(\,\) \begin{align*} && \mathbb{P}(A_1 \cap A_2 \cap A_3 = \emptyset) &= \mathbb{P}(i \not \in (A_1 \cap A_2 \cap A_3) \forall i) \\ &&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1 \cap A_2 \cap A_3) \right) \\ &&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1)\mathbb{P}(i \in \cap A_2))\mathbb{P}(i \in \cap A_3) \right) \\ &&&= \prod_{i=1}^n \left ( 1-\frac12 \cdot \frac12 \cdot \frac12\right) \\ &&&= \left (\frac78 \right)^n \end{align*} Similarly, \(\displaystyle \mathbb{P}(A_1 \cap A_2 \cap \cdots \cap A_m = \emptyset) = \left ( \frac{2^m-1}{2^m} \right)^n\)
4. \(\,\) \begin{align*} && \mathbb{P}(A_1 \subseteq A_2) &= \mathbb{P}(A_1 \cap A_2^c = \emptyset) \\ &&&= \left (\frac34 \right)^n \\ \\ && \mathbb{P}(A_1 \subseteq A_2 \subseteq A_3) &= \prod_{i=1}^n \mathbb{P}(\text{once }i\text{ appears it keeps appearing}) \\ &&&= \prod_{i=1}^n \frac{\#\{(0,0,0), (0,0,1), (0,1,1), (1,1,1) \}}{2^3} \\ &&&= \prod_{i=1}^n \frac{4}{8} \\ &&&= \frac{1}{2^n} \\ \\ && \mathbb{P}(A_1 \subseteq A_2 \subseteq \cdots \subseteq A_m) &= \prod_{i=1}^n \frac{m+1}{2^m} \\ &&&= \left ( \frac{m+1}{2^m} \right)^n \end{align*}

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Solution:

\begin{align*} && a^2x &= x(b-x)^2 \\ \Rightarrow && 0 &= x((b-x)^2-a^2) \\ &&&= x(b-a-x)(b+a-x)\\ && y &= x(b-x)^2 \\ \Rightarrow && y' &= (b-x)^2-2x(b-x) \\ P(b-a,a^2(b-a)): &&y' &= (b-(b-a))^2-2(b-a)(b-(b-a)) \\ &&&= a^2-2a(b-a) = a(3a-2b) \\ \Rightarrow && y &= a(3a-2b)(x-(b-a)) + a^2(b-a) \\ &&&= a(3a-2b)x + (b-a)(a^2-3a^2+2ba) \\ &&&= a(3a-2b)x + (b-a)2a(b-a) \\ &&&= a(3a-2b)x + 2a(b-a)^2 \\ \end{align*} Therefore the tangent at \(P\) is \(a(3a-2b)x + 2a(b-a)^2\) The area between the curve and \(OP\) is \begin{align*} &&S &= \int_0^{b-a} \left (x(b-x)^2-a^2x \right) \d x\\ &&&= \left [\frac{x^2}{2}b^2 - \frac{2x^3}{3}b +\frac{x^4}{4} - \frac{a^2x^2}{2}\right]_0^{b-a} \\ &&&= (b-a)^2 \tfrac12 (b^2-a^2) - \tfrac23(b-a)^3b + \tfrac14(b-a)^4 \\ &&&= \tfrac1{12}(b-a)^3(6(b+a)-8b+3(b-a)) \\ &&&= \tfrac1{12}(b-a)^3(b+3a) \end{align*} The area \([OPR] = T= \tfrac12 \cdot (b-a) \cdot 2a(b-a)^2 = a(b-a)^3\) Clearly \(S > \frac4{12}(b-a)^3a = \frac13T\)