Problems

Solution: Consider \(E = \frac12 m \dot{x}^2 - \frac{Gm}{x}\), notice that \begin{align*} && \dot{E} &= m \dot{x} \ddot{x} + \frac{Gm}{x^2} \dot{x} \\ &&&= \dot{x} \underbrace{\left (m\ddot{x} + \frac{Gm}{x^2} \right)}_{=0 \text{ by N2}} \end{align*} Therefore \(E\) is conserved. Therefore if \(x \to \infty\) \(\frac12 m V^2 - \frac{Gm}{h} = \frac12 m u^2 - 0 \geq 0\) so \(V^2 \geqslant 2G/h\) Since \(E \approx 0\) we want to solve \begin{align*} && \dot{x} &= -\sqrt{\frac{2G}{x}} \\ \Rightarrow && -\int_h^0 \sqrt{x} \d x &= \int_0^T \sqrt{2G} \d t \\ \Rightarrow && \frac{2h^{3/2}}{3} &= \sqrt{2G}T \\ \Rightarrow && T &= \frac{\sqrt{2}h^{3/2}}{3\sqrt{G}} = \frac13 \sqrt{\frac{2h^3}{G}} \end{align*}

Solution:

1. We cannot say this, since we do not know how many people were flying or walking each year.
2. This is difficult to say without knowing the variance. We might expect this to have quite a skewed distribution (one big air crash causes lots of deaths infrequently) so it's impossible to know, although it is substantially lower.
3. If we have 1080 deaths annually, we should expect ~3 deaths per day. While a day with \(7\) deaths might seem unlikely, over the course of a year it is very likely to occur. (Perhaps the weather was bad). It is also probably a case of selective reporting, we are seeing this data point because it's notable and being reported rather than because it is significant).
4. This is certainly the most alarming, a ~25% increase is very unlikely without something else going on. (We'd expect it to be ~Po(1080) approximalely N(1080, 1080) but then this is many standard deviations away). However we also know that other factors could drive this (more walking, more people, change in reporting etc)

Solution: If you arrive between 12 to and 12 past, it will take 60 minutes + how many minutes you wait at the station. If you arrive between 12 past and 12 to, it will take 75 minutes plus waiting at the station. Let's say arrival time \(X \sim U(0,60)\) minutes past the hour, then travel time is. Let's say there are two random variables, \(X_{fast} \sim U(0,24)\) \(X_{slow} \sim U(0, 36)\). Then if you wait for a fast train your expected wait time is \(72\), if you wait for a slow time, your expected wait time is \(75 + 18 = 93\). There is a \(\frac{24}{60} = \frac{4}{10}\) chance of being in the first case, and \(\frac{6}{10}\) chance of the second, ie: \(\frac{4}{10} \cdot 72 + \frac{6}{10} \cdot 93 = \frac{846}{10} = 84.6\) expected wait time. Suppose the time the trains so the expected fraction of time waiting for the fast train is \(t\) and the slow train is \(1-t\). Then the expected time is: \begin{align*} t \l 30t + 60 \r + (1-t) \l 30(1-t) + 75 \r &= 60t^2 -75t + 105 \\ &= 60 \l t^2 - \frac{5}{4}t \r + 105 \\ &= 60 \l t - \frac{5}{8} \r^2 - ? + 105 \\ \end{align*} Threfore we should choose \(x\) such that \(t = \frac58\), which is \(~37.5\) minutes after the slower train, \(25.5\) minutes past.

Solution:

1. \begin{align*} \lim_{n \to \infty} \frac{n}{n+1} &= \lim_{n \to \infty} \left (1 - \frac{1}{n+1} \right ) \\ &\underbrace{=}_{\text{sum of limits}} \lim_{n \to \infty} 1 - \lim_{n \to \infty} \frac{1}{n+1}\\ &= 1 \end{align*}
2. \begin{align*} \lim_{n \to \infty} \frac{5n+1}{n^2-3n+4} &= \lim_{n \to \infty} \frac{5/n + 1/n^2}{1-3/n+ 4/n^2} \\ &\underbrace{=}_{\text{ratio of limits}} \frac{\displaystyle \lim_{n \to \infty}(5/n + 1/n^2)}{\displaystyle \lim_{n \to \infty}(1-3/n+ 4/n^2)} \\ &= \frac{0}{1} = 0 \end{align*}
3. \begin{align*} && \lvert \frac{\sin n}{n} \rvert &\leq \frac{1}{n} \quad \quad (n \geq 1) \\ \Rightarrow && \lim_{n \to \infty} \lvert \frac{\sin n}{n} \rvert &\leq \lim_{n \to \infty}\frac{1}{n} \\ &&&= 0\\ \Rightarrow && \lim_{n \to \infty} \frac{\sin n}{n} &= 0 \end{align*}
4. First note that \(\displaystyle \lim_{x \to 0} \frac{\sin x}{x} \to 1\), then \(\frac1n\) is a sequence converging to zero, therefore \(\frac{\sin 1/n}{1/n}\) also must tend to \(1\).
5. Note that \(\lim_{x \to \infty} \tan^{-1} x = \frac{\pi}{2}\) and since \(n\) is a sequence tending to infinity we must have \(\lim_{n \to \infty} \tan^{-1} n = \frac{\pi}{2}\)
6. \begin{align*} \lim_{n \to \infty} \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+2}-\sqrt{n}} &= \lim_{n \to \infty} \dfrac{\frac{1}{\sqrt{n+1}+\sqrt{n}}}{\frac{2}{\sqrt{n+2}+\sqrt{n}}} \\ &= \frac12 \lim_{n \to \infty} \dfrac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\\ &= \frac12 \lim_{n \to \infty} \dfrac{\sqrt{1+2/n}+\sqrt{1}}{\sqrt{1+1/n}+\sqrt{1}}\\ &= \frac12 \end{align*}

Solution: \begin{align*} && y & =x\frac{\mathrm{d}y}{\mathrm{d}x}-\cosh\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right) \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{\d y}{\d x} + x\frac{\d ^2 y}{\d x^2} - \sinh \left ( \frac{\d y}{\d x} \right) \frac{\d^2 y}{\d x^2} \\ \Rightarrow && 0 &= \frac{\d^2 y}{\d x^2} \left ( x - \sinh \left ( \frac{\d y}{\d x}\right)\right) \end{align*} Therefore \(\frac{\d^2y}{\d x^2} = 0\) or \( x - \sinh \left ( \frac{\d y}{\d x}\right) = 0\) as required. \begin{align*} && \frac{\d ^2 y}{\d x^2} &= 0 \\ \Rightarrow && y &= ax + b \\ \\ && 0 &= x - \sinh \left ( \frac{\d y}{\d x}\right) \\ \Rightarrow && \frac{\d y}{\d x} &= \sinh^{-1} (x) \\ \Rightarrow && y &= x \sinh^{-1} x - \sqrt{x^2+1} + C \end{align*} Since it is necessary the solution satisfies one of those equations, we just need to check if either of these types of solutions work for our differential equation, ie \begin{align*} && ax + b &\stackrel{?}{=} ax - \cosh(a) \\ \Rightarrow && b &= -\cosh(a) \\ \Rightarrow && y &= ax -\cosh(a) \\ \\ && x \sinh^{-1} x - \sqrt{x^2+1} + C &\stackrel{?}{=} x\sinh^{-1} x - \cosh ( \sinh^{-1} x) \\ &&&= \sinh^{-1} x -\sqrt{x^2+1} \\ \Rightarrow && C &= 0 \end{align*} Therefore the general solutions are, \(y = ax - \cosh(a)\) and \(y = x \sinh^{-1} x - \sqrt{x^2+1}\)

Solution: Claim: If \(f(x) = c\) then \(\lozenge f(x) = 0\) Proof: Consider \(g(x) = x\) then \begin{align*} (1) && \lozenge(f(x)g(x)) &= g(x) \lozenge f(x) + f(x) \lozenge g(x) \\ \Rightarrow && \lozenge(c x) &= x \lozenge f(x) + c \lozenge x \\ (4) && \lozenge(c x) &= c \lozenge x \\ \Rightarrow && 0 &= x \lozenge f(x) \\ \Rightarrow && \lozenge f(x) &= 0 \end{align*} \begin{align*} (1) && \lozenge(x^2) &= x \lozenge x + x \lozenge x \\ (3) &&&= 2 x \cdot 1 \\ &&&= 2x \\ \\ (1) && \lozenge (x^3) &= x^2 \lozenge x + x \lozenge (x^2) \\ &&&= x^2 \cdot \underbrace{1}_{(3)} + x \cdot\underbrace{ 2x}_{\text{previous part}} \\ &&&= 3x^2 \end{align*} Claim: \(\lozenge h(x) = \frac{\d }{\d x} ( h(x))\) for any polynomial \(h\). Proof: (By (strong) induction on the degree of \(h\)). Base case: True, we proved this in the first part of the question. Inductive step: Assume true for all polynomials of degree less than or equal to \(k\). Then consider \(n = k+1\). We can write \(h(x) = ax^{k+1} + h_k(x)\) where \(h_k(x)\) is a polynomial of degree less than or equal to \(k\). Then notice: \begin{align*} && \lozenge (h(x)) &= \lozenge (ax^{k+1} + h_k(x)) \\ (2) &&&= \lozenge (ax^{k+1})+ \lozenge (h_k(x)) \\ &&&=\underbrace{a\lozenge (x^{k+1})}_{(4)}+ \underbrace{\frac{\d}{\d x} (h_k(x))}_{\text{inductive hypothesis}}\\ &&&= a \underbrace{\left (x \lozenge x^k + x^k \lozenge x \right)}_{(1)} + \frac{\d}{\d x} (h_k(x)) \\ &&&= a \left ( x \cdot \underbrace{k x^{k-1}}_{\text{inductive hyp.}} + x^k \cdot \underbrace{1}_{(3)} \right) + \frac{\d}{\d x} (h_k(x)) \\ &&&= (k+1)a x^k + \frac{\d}{\d x} (h_k(x)) \\ &&&= \frac{\d }{\d x} \left ( ax^{k+1} + h_k(x) \right) \\ &&&= \frac{\d }{\d x} (h(x)) \end{align*} Therefore since our statement is true for \(n=0\) and if it is true for \(n=k\) it is true for \(n=k+1\) by the principle of mathematical induction it is true for all \(n \geq 0\)