Problems

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Solution: Consider the angle from the origin, then \(P = (\cos \theta, \sin \theta)\) where \(\theta \sim U(0, 2\pi)\), and \(X = \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta}\) \begin{align*} \mathbb{E}[X] &= \int_0^{2\pi} \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta} \frac1{2\pi} \d \theta \\ &= \frac1{2\pi}\int_0^{2\pi} \sqrt{2 - 2\cos \theta} \d \theta \\ &= \frac{1}{2\pi}\int_0^{2\pi} \sqrt{4\sin^2 \frac{\theta}{2}} \d \theta \\ &= \frac{1}{\pi}\int_0^{2\pi} \left |\sin \frac{\theta}{2} \right| \d \theta \\ &= \frac{1}{\pi} \left [ -2\cos \frac{\theta}{2} \right]_0^{2\pi} \\ &= \frac1{\pi} \l 2 + 2\r \\ &= \frac{4}{\pi} \end{align*} \begin{align*} \mathbb{E}(X^2) &= \frac1{2\pi}\int_0^{2\pi} (\cos \theta - 1)^2 + \sin^2 \theta \d \theta \\ &= \frac1{2\pi}\int_0^{2\pi} 2 - 2 \cos \theta \d \theta \\ &= \frac{4\pi}{2\pi} \\ &= 2 \\ \end{align*} \(\Rightarrow\) \(\mathrm{Var}(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2 = 2 - \frac{16}{\pi^2} = \frac{2\pi^2 - 16}{\pi^2}\).

Where the line makes a length longer than \(\frac{4}{\pi}\) it will make an angle at the origin of \(2\sin^{-1} \frac{2}{\pi}\). Therefore the probability of being larger than this is \(\frac{2\pi - 2 \times 2\sin^{-1} \frac{2}{\pi}}{2 \pi} = 1 - \frac{2}{\pi} \sin^{-1} \frac{2}{\pi} \approx 0.560\)

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Solution: \begin{align*} \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\frac{\cos^{2}\theta}{1-\sin\theta\sin2\alpha}\,\mathrm{d}\theta &= \int_{u = \frac12 \pi}^{u = -\frac12 \pi} \frac{\cos^2 (-u)}{1-\sin(-u) \sin 2 \alpha} -\d u \tag{\(u = -\theta\)} \\ &= \int_{\frac12 \pi}^{-\frac12 \pi} \frac{\cos^2 u}{1+\sin u \sin 2 \alpha} -\d u \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 u}{1+\sin u \sin 2 \alpha} \d u \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1+\sin \theta \sin 2 \alpha} \d \theta \\ \end{align*} Since \(\displaystyle \frac{1}{(1-a^2u^2)} = \frac12 \l \frac{1}{1+au} + \frac1{1-au} \r\) \begin{align*} \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1+\sin \theta \sin 2 \alpha} \d \theta &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{1-\sin^2 \theta}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{(1-\sin ^2\theta \sin^2 2 \alpha) \frac{1}{\sin^2 2\alpha} + 1 - \cosec^2 2\alpha}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{1 - \sin^2 \theta \sin^2 2 \alpha} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{1 - \sin^2 \theta (1-\cos^2 2 \alpha)} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{\cos^2 \theta +\sin^2 \theta \cos^2 2 \alpha} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sec^2 \theta}{1 +\tan^2 \theta \cos^2 2 \alpha} \d \theta \\ \end{align*} Finally, using the substitution \(u =|\cos 2 \alpha | \tan \theta, \d u = |\cos 2 \alpha |\sec^2 \theta \d \theta\) \begin{align*} \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sec^2 \theta}{1 +\tan^2 \theta \cos^2 2 \alpha} \d \theta &= |\sec 2\alpha|\int_{u = -\infty}^{u = \infty} \frac{1}{1 + u^2} \d u \\ &= |\sec 2 \alpha|\pi \end{align*} and so \begin{align*} I &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha |\sec 2 \alpha|\pi \\ &= \frac{\pi}{\sin^2 2\alpha} \l 1-\cos 2\alpha \r \\ &= \frac{\pi}{4\sin^2 \alpha\cos^2 \alpha} \l 2 \sin^2 \alpha \r \\ &= \frac{\pi}{2 \cos^2 \alpha} = \frac{\pi}{2} \sec^2 \alpha \end{align*} When \(\alpha\) small enough that the modulus doesn't flip the sign. When if \(\frac{1}{4}\pi<\alpha<\frac{1}{2}\pi\) we have: \begin{align*} I &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha |\sec 2 \alpha|\pi \\ &= \frac{\pi}{\sin^2 2\alpha} \l 1+\cos 2\alpha \r \\ &= \frac{\pi}{4\sin^2 \alpha\cos^2 \alpha} \l 2 \cos^2 \alpha \r \\ &= \frac{\pi}{2 \sin^2 \alpha} = \frac{\pi}{2} \cosec^2 \alpha \end{align*}

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Solution:

We can approximate the integral by \(N\) trapeziums, each with height \(x_{i+1}-x_{i} = \frac{x_N-x_0}{N} = \frac{h}{N}\). The will have area \(\frac{(f(x_i)+f(x_{i+1}))h}{2}\) and summing all these areas we will get: \[\frac12 h \l f(x_0) + f(x_1) + f(x_1)+f(x_2) + \cdots + f(x_{N-1})+f(x_N) \r = \frac12 h \l f(x_0) +2 f(x_1) + + \cdots +2f(x_{N-1})+f(x_N) \r\] But this is approximately the integral \(\displaystyle \int_{x_0}^{x_N} f(x) \d x\) \begin{align*} && \int_1^n \ln x \d x &= [x \ln x]_1^n - \int_1^n x \cdot \frac{1}{x} \d x \\ &&&= n \ln n - n+1 \\ &&&\approx \frac12 \l \ln 1 + 2\sum_{k=2}^{n-1} \ln k + \ln n \r \\ &&&= \ln (n!) - \frac12 \ln n \\ \Rightarrow && \ln (n!) &\approx n \ln n + \frac12 \ln n - n + 1 \\ \Rightarrow && n! &\approx \exp(n \ln n + \frac12 \ln n - n + 1) \\ &&&=n^{n+\frac12}e^{1-n} \end{align*} Since \(\ln x\) is a concave function, we should expect all the trapeziums to all lie under the curve, therefore this is always an underestimate for the integral, ie \(n! < g(n)\) \begin{align*} && \int_1^n \ln x \d x &= n \ln n - n+1 \\ &&&\approx \frac12 k^{-1} \l \ln 1 + 2\sum_{r=1}^{k(n-1)-1} \ln \l 1+\frac{r}{k} \r + \ln n \r \\ &&&=\frac{1}{2k} \l 2\sum_{r=1}^{k(n-1)-1} \l \ln(k+r) - \ln k)\r + \ln n\r \\ &&&=\frac1{k} \l \ln ((k+k(n-1)-1)!) - \ln(k!) - (k(n-1)-1) \ln k+\frac12\ln n \r \\ &&&=\frac1{k} \l \ln ((kn-1)!) - \ln(k!) - (k(n-1)-1) \ln k+\frac12 \ln n \r \\ &&&=\frac1{k} \l \ln ((kn)! ) -\ln k -\ln n - \ln(k!) - (k(n-1)-1) \ln k+\frac12\ln n \r \\ &&&= \frac1{k} \l \ln ((kn)! ) - \ln(k!) - (k(n-1)) \ln k - \frac12 \ln n\r \\ \Rightarrow && \ln ((kn)!) &\approx kn \ln n - kn + k + \ln(k!) + (k(n-1)) \ln k + \frac12 \ln n\\ \Rightarrow && (kn)! &\approx n^{kn+\frac12}e^{-k(n-1)}k!k^{k(n-1)} \\ &&&= n^{kn+\frac12} k! \l \frac{e}{k} \r^{k(1-n)} \end{align*} I would expect this approximation to be a better approximation for \((kn)!\) since it is created using a finer mesh.

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Solution: Let \({\displaystyle I_{m,n}=\int\cos^{m}x\sin nx\,\mathrm{d}x,}\) Then \begin{align*} && I_{m,n} &= \int\cos^{m}x\sin nx\,\mathrm{d}x \\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] - \frac{m}{n} \int \sin^{m-1} x \cos x \cos n x \d x \\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] + \frac{m}{n} \int \sin^{m-1} x (\cos (n-1)x -\sin x \sin nx) \d x\\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] + \frac{m}{n} \int \sin^{m-1} x \cos (n-1)x \d x-\frac{m}{n} I_{m,n} \\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] + \frac{m}{n} I_{m-1,n-1} -\frac{m}{n} I_{m,n} \\ \Rightarrow && nI_{m,n} &= -\cos^m x \cos n x + mI_{m-1,n-1} -mI_{m,n}\\ \Rightarrow && (m+n)I_{m,n} &= -\cos^m x \cos n x + mI_{m-1,n-1} \end{align*}

1. Note that \(I_{2m,0} = 0\) (the integrand is 0) and \(I_{0, 2m} = 0\) (symmetry for our limits). \(\displaystyle \left [-\cos^m x \cos n x \right]_0^\pi = \l - (-1)^m (-1)^n \r - \l -1 \r = 1 - (-1)^{m+n} = 0\) since \(m+n\) is even. Therefore all reductions are \(I_{m,n} = \frac{I_{m-1,n-1}}{m+n}\) terminating at \(0\), so all values are zero
2. \begin{align*} \int_{0}^{\frac{\pi}{2}}\sin^{2}x\sin3x\,\mathrm{d}x &= \int_{0}^{\frac{\pi}{2}}(1-\cos^2x)\sin3x\,\mathrm{d}x \\ &= I_{0,3} - I_{2,3} \\ &= \frac13 - \frac15 \l \left [-\cos^2 x \cos 3 x \right]_0^{\pi/2} + 2 \cdot I_{1,2} \r \\ &= \frac13 - \frac15 \l 1 + \frac23 \l \left [-\cos x \cos 2 x \right]_0^{\pi/2} + 1\cdot I_{0,1} \r \r \\ &= \frac13 - \frac15 -\frac2{15} - \frac2{15} \\ &= \frac{5}{15} - \frac{3}{15} - \frac{4}{15} \\ &= -\frac2{15} \end{align*}

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Solution: The initial volume of water in \(A\) is: \begin{align*} \pi \int_0^4 x^2 \, \d y &= \pi \int_0^4 y \d y \\ &= \pi [ \frac{y^2}{2}]_0^4 \\ &= 8\pi \end{align*} We assume that no water is in the tube as it is `thin'. Therefore we must have: \begin{align*} && 8\pi &= \pi \int_0^{h-1} x^2 \d y +\pi \int_0^{h} x^2 \d y \\ &&&= \pi \int_0^{h-1} y \d y +\pi \int_0^{h} \l \sinh \frac{x}{2}\r^2 \d y \\ &&&= \pi \left [\frac{y^2}{2} \right]_0^{h-1} + \pi \int_0^h \frac{-1+\cosh y}{2}\d y \\ &&&= \pi \frac{(h-1)^2}{2} + \pi \left [ -\frac{y}{2} +\frac{\sinh y}{2}\right]_0^h \\ &&&= \pi \frac{(h-1)^2}{2} -\pi \frac{h}{2} + \pi \frac{\sinh h}{2} \\ \Rightarrow && 0 &= h^2-2h+1-h+\sinh h -16 \\ &&&= h^2 -3h+\sinh h - 15 \\ \Rightarrow && 15 &= h^2 -3h+\sinh h \end{align*}

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Solution: We start by noticing that \(\displaystyle \tan^{-1} t = \int_0^t \frac{1}{1+x^2} \d x\). Consider the geometric series \(1-x^2+(-x^2)^2+ \cdots + (-x^2)^n = \frac{1-(-x^2)^{n+1}}{1+x^2}\). Therefore, \((1+x^2)(1-x^2+(-x^2)^2+ \cdots + (-x^2)^n) = 1-(-x^2)^{n+1}\) or \(1 = (1+x^2)(1-x^2+x^4-\cdots+(-1)^nx^{2n}) +(-1)^{n+1}x^{2n+2}\) \begin{align*} \tan^{-1} t &= \int_0^t \frac{1}{1+x^2} \d x \\ &= \int_0^t \frac{(1+x^2)(1-x^2+x^4-\cdots+(-1)^nx^{2n}) +(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\ &= \int_0^t (1-x^2+x^4-\cdots+(-1)^nx^{2n})\d x + \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\ &= t - \frac{t^3}{3}+\frac{t^5}{5}-\cdots + (-1)^n \frac{t^{2n+1}}{2n+1}+\int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\ &= \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1} + \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\ \end{align*} Therefore we can say (for \(0 \leq t \leq 1\)) \begin{align*} \left | \tan^{-1} t - \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1} \right | &= \left | \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \right | \\ &\leq \left | \int_0^t x^{2n+2} \d x \right | \\ &= \frac{t^{2n+3}}{2n+3} \\ \\ \left | \tan^{-1} t - \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1} \right | &= \left | \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \right | \\ &\geq \left | \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{1+1} \d x \right | \\ &= \frac{t^{2n+3}}{2(2n+3)} \\ \end{align*} Since \(\tan^{-1} 1 = \frac{\pi}{4}\) we must have that: \begin{align*} \lim_{n \to \infty} \left | \frac{\pi}{4} - \sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)} \right | \to 0 \Rightarrow \lim_{n \to \infty} 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)} \to \pi \end{align*} However, \begin{align*} && \left | 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)} - \pi \right | &\geq 4 \frac{1}{2(2n+3)} \\ && &= \frac{2}{2n+3} \\ \\ && \frac{2}{2n+3} \geq 10^{-2} \\ \Leftrightarrow && 200 \geq 2n+3 \\ \Leftrightarrow && 197 \geq 2n \\ \Leftrightarrow && 98.5 \geq n \\ \end{align*} Therefore we need more than \(98\) terms to get two decimal places of accuracy. Not great!

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Solution: \(\displaystyle \int_x^{x+1} nP_{n-1}(x) \, dx = P_n(x+1) - P_n(x)\) Claim: \(P_{n}(x+1)-P_{n}(x)=nx^{n-1},\) for \(n \geq 1\) Proof: (By induction). (Base case, \(n=1\)). \(P_1(x) = x - \frac12\), \(P_1(x+1) - P_1(x) = 1 x^{0}\) as required. Assume the equation is true for \(n = k\). So \(P_k(x+1) - P_k(x) = kx^{k-1}\) now consider \begin{align*} P_{k+1}(x+1) - P_{k+1}(x) &= \int_0^{x+1} (k+1) P_k(t) \d t + P_{k+1}(0)- \int_0^{x} (k+1) P_k(t) \d t - P_{k+1}(0) \\ &= \int_0^x (k+1)(P_k(t+1)-P_k(t)) \d t + \int_0^1 (k+1)P_k(t) \d t \\ &= (k+1)x^{k} + 0 \end{align*} So by induction we are done. \begin{align*} n\sum_{m=0}^{k}m^{n-1} &= \sum_{m=0}^{k}n \cdot m^{n-1} \\ &= \sum_{m=0}^{k}\l P_n(m+1)-P_n(m) \r \\ &= P_n(k+1) - P_n(0) \end{align*} We need to find \(P_4\) \begin{align*} P_0(x) &= 1 \\ P_1(x) &= x - \frac12 \\ P_2(x) &= x^2 -x - \int_0^1 \l x^2 - x \r \d x \\ &= x^2 - x + \frac16 \\ P_3(x) &= x^3 -\frac{3}{2}x^2 + \frac12x - \int_0^1 \l x^3 -\frac{3}{2}x^2 + \frac12x \r \d x \\ &= x^3 -\frac{3}{2}x^2 + \frac12x \\ P_4(x) &= x^4 - 2x^3 + x^2 + c \end{align*} Therefore the sum we are interested in is \(\frac14 \l P_4(1001) - P_4(0) \r = \frac14 (1001)^2 (1001-1)^2 = (1001 \cdot 500)^2 = (500500)^2\)

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Solution: Suppose we divide a sphere of radius \(r\) up into slices of thickness \(\delta x\). Then the force acting on \(P\) will be: \begin{align*} F &= \sum_{\text{slices}} 2\pi mk\rho\delta x\left(1-\frac{b}{(a^{2}+b^{2})^{\frac{1}{2}}}\right) \\ &= \sum_{i=-r/\delta x}^{r/\delta x} 2\pi mk\frac{M}{\frac43 \pi r^3}\delta x\left(1-\frac{i \delta x}{((1-(i\delta x)^2)+(i \delta x)^{2})^{\frac{1}{2}}}\right) \\ &\to \int_{-r}^r \frac{1}{2} \frac{mkM}{r^3}(1-t) \d t \\ &=\frac{mkM}{r^2} \end{align*} We can see that the particle will have a force attracting it towards the centre, with magnitude \(\frac{kmM}{r^2}\), therefore and since \(\frac{\d v}{\d t} = \frac{\d v}{\d r} \frac{\d r}{\d t}\) we must have: \(v \frac{\d v}{\d r}m = - \frac{kmM}{r^2}\) and dividing by \(m\) we get exactly the result we seek. \begin{align*} && v \frac{\d v}{\d r} &= \frac{-kM}{r^2} \\ \Rightarrow && \frac{v^2}{2}+C &= \frac{kM}{r} \\ r = R, v =0: && C &= \frac{kM}{R} \\ \Rightarrow && v^2&= 2kM\left ( \frac1r - \frac1R\right ) \\ \Rightarrow && \frac{\d r}{\d t} &= -\sqrt{2kM\left ( \frac1r - \frac1R\right )} \\ \Rightarrow && -\sqrt{2kM}T &= \int_{r=R}^{r=R\cos^2 \alpha} \frac{1}{\sqrt{\frac1r-\frac1R}} \d r \\ r = R\cos^2 \theta: && -\sqrt{2kM}T &= \int_{\theta = 0}^{\theta = \alpha} \frac{\sqrt{R}}{\sqrt{\sec^2 \theta - 1}} \cdot R \cdot 2 \cdot (-\cos \theta) \cdot \sin \theta \d \theta \\ \Rightarrow && T &= \sqrt{\frac{R^3}{2kM}} \int_0^\alpha \frac{2 \cos \theta \sin \theta}{\sqrt{\sec^2 \theta - 1}} \d \theta \\ &&&= \sqrt{\frac{R^3}{2kM}} \int_0^\alpha \frac{2 \cos \theta \sin \theta}{\tan \theta} \d \theta \\ &&&= \sqrt{\frac{R^3}{2kM}} \int_0^\alpha 2\cos^2 \theta \d \theta \\ &&&= \sqrt{\frac{R^3}{2kM}} \int_0^\alpha 1 + \cos 2 \theta\d \theta \\ &&&= \sqrt{\frac{R^3}{2kM}} \left [1 + \frac12 \sin 2 \theta \right]_0^\alpha \\ &&&= \sqrt{\frac{R^3}{2kM}} \left (\alpha + \frac12 \sin 2 \alpha \right) \\ \end{align*}