Locus Pocus

While picking random points inside the equilateral triangle with side length 1, you and your colleague notice that some points are “special.” When one of these special points is connected with the triangle’s three vertices, it creates three smaller triangles whose areas form an arithmetic sequence.

After further investigation, you realize these special points all belong to a locus, the total length of which is $L$. What is this length $L$?

As we saw before, given choice of any point $x$ in the triangle, we have three sub-triangles with areas $A_1(x),$ $A_2(x),$ and $A_3(x),$ that are mutually non-overlapping and tile the entirety of the area of the unit equilateral triangle with area $A = \frac{\sqrt{3}}{4}$ square units. If we have chosen our point x in a special way then $A_1$, $A_2$, and $A_3$ are in arithmetic progression and so, without loss of generality we have $A_2(x) = \frac{1}{2} (A_2(x) + A_3(x)),$ or $$2A_2(x) = A_1(x) + A_3(x).$$ Adding $A_2(x)$ to both sides we get $$3A_2(x) = A_1(x)+A_2(x)+A_3(x) = A,$$ so we conclude that $$A_2(x) = \frac{1}{3} A = \frac{\sqrt{3}}{12}.$$

So the locus of special points are those for which at least one of the subtriangles has area $\frac{\sqrt{3}}{12}.$ Without loss of generality, let's assume that the original unit equilateral triangle is located such that its vertices are at $P(-1/2, -\sqrt{3}/6),$ $Q(1/2, -\sqrt{3}/6)$ and $R(0,\sqrt{3}/3).$ Then for instance, for any point $X_t(t,0)$ for $-1/3 \leq t \leq 1/3,$ the area of the triangle $PQX_t$ will be $\sqrt{3}/12.$ So the line segment from $(-1/3,0)$ to $(1/3,0)$ is in the locus of special points and has a length of $2/3.$

However, since the equilateral triangle has trifold rotational symmetry, there will also be a similar sub-locus of length $2/3$ running parallel to the side $PR$ and the side $QR.$ So the locus of special points where the areas of three smaller triangles form an arithmetic sequences has total length of $2$ units.

Random Quadrilaterals

Inspired by Delta Airlines, you and a graphic designer colleague are crafting a new logo for Fiddler Airlines. Your colleague starts with an equilateral triangle with a side length of $1$ unit. Next, you pick a random point inside the triangle, giving you four points: the three corners of the triangle and this fourth point inside. There are several (simple) quadrilaterals that can be drawn whose four vertices are these four points. Your colleague picks one of these quadrilaterals at random, with each having an equal probability. This quadrilateral will be the new logo.

On average, what can you expect the area of the quadrilateral to be?

Since you start with an equilateral triangle with unit side lengths, the total aread of the triangle is $A = \frac{\sqrt{3}}{4}$ square units. For any random chosen point, say $x,$ within this equilateral triangle, there are $\binom{3}{2} = 3$ possible choices of triangles to remove from the larger equilateral triangle to form our simple quadrilateral logo. Let's assume that we enumerate the areas of these three resulting, nonoverlapping triangles as $A_1(x),$ $A_2(x)$ and $A_3(x),$ then by design we have $$A_1(x) + A_2(x) + A_3(x) = A.$$ In order to obtain a quadrilateral, your colleague will uniformly randomly select one of these three triangles to remove from the initial unit equilateral triangle. Let's denote the area of the resulting random quadrilateral, conditioned on your initial choice of the point $x,$ as $Q \mid x,$ then we have \begin{align*} \mathbb{E} \left[ Q \mid x \right] &= \frac{1}{3} ( A - A_1(x) ) + \frac{1}{3} (A - A_2(x)) + \frac{1}{3} (A - A_3(x)) \\ &= A - \frac{A_1(x)+A_2(x)+A_3(x)}{3} \\ &= \frac{2}{3} A = \frac{\sqrt{3}}{6}.\end{align*} Since this equation holds regardless of your choice of $x,$ the expected area of the quadrilateral obtained is $$\mathbb{E}[Q] = \frac{\sqrt{3}}{6}.$$

Dummymandering for Dummies

In the next election cycle you’re back at it, gerrymandering on behalf of the Moon party once more. This time, you can draw a map that results in any set of probabilities $p_1$, $p_2$, $p_3$, $p_3$, $p_5$, $p_6$, and $p_7$ that you’ll win each of the corresponding districts—subject to the constraint that the average of the probabilities is $1/2.$ In other words, $$p_1 + p_2 + p_3 + p_4 + p_5 + p_6 + p_7 = 7/2.$$

The bosses of the Moon party want to go big. More specifically, they want to maximize the party’s chances of winning at least five of the seven districts. However, you are concerned that such a greedy attempt at gerrymandering may backfire, a phenomenon that has become known as dummymandering. When you draw a map that maximizes your chances of winning at least five districts, the probability the Moon party blows it and winds up losing most of the districts is $L_A.$

Your personal preference is to simply maximize the party’s chances of winning at least four of the seven districts—a simple majority of them. When you draw that maximizes these chances, the probability the Moon party loses most of the districts is $L_B.$ How much greater than $L_B$ is $L_A$?

Again, we will assume that $p_1 \leq p_2 \leq p_3 \leq \cdots \leq p_7.$ We have $X_i \sim B(p_i)$ is a Bernoulli distribution with probability of success $p_i.$ Then the Moons want $\sum_{i=1}^7 X_i \geq 5,$ so we should determine $$f_5(p) = \mathbb{P} \left\{ \sum_{i=1}^7 X_i \geq 5 \right\}.$$ In particular, we could have $$g(p) = \mathbb{P} \left\{ \sum_{i=1}^7 X_i = 7 \right\} = p_1 p_2 \cdots p_7.$$ We have $$h(p)=\mathbb{P} \left\{ \sum_{i=1}^7 X_i = 6 \right\} = (1-p_1) p_2 \cdots p_7 + p_1 (1-p_2) p_3 \cdots p_7 + \cdots + p_1 \cdots p_6 (1-p_7).$$ Finally, we have \begin{align*} k(p) = \mathbb{P} \left\{ \sum_{i=1}^7 X_i = 5 \right\} &= (1-p_1)(1-p_2) p_3 \cdots p_7 + (1-p_1)p_2(1-p_3)p_4 \cdots p_7 + \cdots + (1-p_1)p_2 \cdots p_6 (1-p_7) \\ & \quad + p_1 (1-p_2) (1-p_3) p_4 \cdots p_7 + \cdots + p_1 (1-p_2) p_3 \cdots p_6 (1-p_7) \\ & \quad\quad + \cdots + p_1 p_2 \cdots p_5 (1-p_6)(1-p_7)\end{align*} Putting this altogether we get f_5(p) = g(p) + h(p) + k(p), and for instance, we see that \begin{align*}f_5(p) &= p_1 \Biggl( p_2 \cdots p_7 + (1-p_2) p_3 \cdots p_7 + p_2 (1-p_3) p_4 \cdots p_7 + \cdots + p_2 \cdots p_6 (1-p_7) \Biggr. \\ & \quad\quad\quad\quad \Biggl.+ (1-p_2) (1 - p_3) p_4 \cdots p_7 + \cdots + p_2 \cdots p_5 (1-p_6) (1 - p_7) \Biggr) + \\ & \quad + (1 - p_1) \Biggl( p_2 \cdots p_7 + (1 - p_2) p_3 \cdots p_7 + p_2 (1-p_3) p_4 \cdots p_7 + \cdots + p_2 \cdots p_6 (1 -p_7 ) \Biggr) \\ & \quad\quad\quad + \text{ other terms not involving $p_1$ } \\ &= p_1 \Biggl( (1-p_2)(1-p_3) p_4 \cdots p_7 + (1-p_2) p_3 (1-p_4) p_5 p_6 p_7 + \cdots + p_2 \cdots p_5 (1-p_6) (1-p_7) \Biggr) \\ &\quad + \text{ other terms not involving $p_1$}.\end{align*} Therefore, we see that $\frac{\partial}{\partial p_1} f_5(p) \geq 0$ for all $0 \leq p \leq 1.$ Since the choice of $p_1$ was arbitrary, this holds for all $p_i, i = 1, \dots, 7.$ So it is clear that there will not be some $p^*$ such that $\nabla f_5(p^*) = 0.$ Therefore, we need to look at the boundary $\sum_{i=1}^7 p_i = 7/2,$ along with the other boundaries $0 \leq p_1 \leq p_2 \leq \cdots \leq p_6 \leq p_7 \leq 1.$

In particular, we need to look at potential solutions of the form $$q^{i,j} = \bigl( \overbrace{0, \dots, 0}^{i}, \overbrace{\lambda_{ij}, \dots, \lambda_{ij}}^{7-i-j}, \overbrace{1, \dots, 1}^j \bigr),$$ where $\lambda_{ij} = \frac{7 - 2j}{2(7- i - j)}.$ We see that for each $i, j,$ $$0 \leq q^{i,j}_1 \leq q^{i,j}_2 \leq \cdots \leq q^{i,j}_7 \leq 1,$$ and we have $$\sum_{k=1}^7 q^{i,j}_k = i \cdot 0 + (7-i-j) \cdot \lambda_{ij} + j \cdot 1 = \frac{7}{2}.$$ Plugging $q^{i,j}$ into $f_5(p),$ we see that, since the Moons will automatically lose $i$ districts and win $j$ districts, that $$f_5(q^{i,j}) = \sum_{k=5}^{7-i} \binom{7-i-j}{k-j} (1 - \lambda_{ij})^{7-i-k} \lambda_{ij}^{k-j},$$ which is just the upper tail of a binomial distribution with probability $\lambda_{ij}.$

Since we are already at it, and without the need to rigorously reprove anything, we can also define $$f_4(p) = \mathbb{P} \left\{ \sum_{i=1}^7 X_i \geq 4 \right\}.$$ We can reassert that when maximizing for $f_4$ that there will be no feasible solutions p^* with \nabla f_4(p^*) = 0, so we will also look for $f_4(q^{i,j})$ as previously define, where we additionally have $$f_4(q^{i,j}) = \sum_{k=4}^{7-i} \binom{7-i-j}{k-j} (1 - \lambda_{ik})^{7-i-k} \lambda_{ij}^{k-j}.$$

We obviously cannot have $j \gt 3,$ since there is only $7/2$ total probability to go around. Similarly, we see that $f_5(q^{5,j}) = 0,$ for each $j,$ since you have already allowed the Suns to win 3 districts and hence can only win a maximum of 4. Additionally, whenever So for $i, j \in \{0,1,2, 3\},$ we can calculate the values of $f_5$ and $f_4$ using some handy-dandy calculators, whether Excel or numpy or whatever you fancy, to obtain the following table:

$i$	$j$	$\lambda_{ij}$	$f_5(q^{i,j})$	$f_4(q^{i,j})$
$0$	$0$	$1/2$	$0.2265625$	$0.5000000$
$1$	$0$	$7/12$	$0.2082593$	$0.5097931$
$2$	$0$	$7/10$	$0.1680700$	$0.5282200$
$3$	$0$	$7/8$	$0.0000000$	$0.5861864$
$0$	$1$	$5/12$	$0.2030319$	$0.4902069$
$1$	$1$	$1/2$	$0.1875000$	$0.5000000$
$2$	$1$	$5/8$	$0.1525878$	$0.5187988$
$3$	$1$	$5/8$	$0.0000000$	$0.5787037$
$0$	$2$	$3/10$	$0.1630800$	$0.4717800$
$1$	$2$	$3/8$	$0.1680700$	$0.4812012$
$2$	$2$	$1/2$	$0.1516113$	$0.5000000$
$3$	$2$	$3/4$	$0.1516113$	$0.5625000$
$0$	$3$	$1/8$	$0.1630800$	$0.4138184$
$1$	$3$	$1/6$	$0.0740741$	$0.4212963$
$2$	$3$	$1/4$	$0.0625000$	$0.4375000$
$3$	$3$	$1/2$	$0.0000000$	$0.5000000$

We further note that the probability of losing the election is $L(p) = 1 - f_4(p).$ Therefore, we see that if the Moons want to greedily maximize the probability of winning at least 5 districts, then they need to pick $$p^A = (1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2),$$ which will give probability of losing the election of $$L_A = 1 - f_4(p^A) = \frac{1}{2}.$$ However, we see that $p^B = (0, 0, 0, 7/8, 7/8, 7/8, 7/8)$ is the maximizer of $f_4,$ which means that $$L_B = 1 - f_4(p^B) = 1- \frac{2401}{4096} = \frac{1695}{4096} = 0.4138184\dots.$$ Therefore, we see that the Moons' greedy decision to maximize the probability of winning 5 or more districts cost them $$L_A - L_B = \frac{353}{4096} = 8.61816\dots \%$$ in winning percentage.

Non-representative Democracy

Fiddler Nation has an upcoming election between its two major political parties: the Suns and the Moons. You are drawing a new electoral map, and you have been tasked with gerrymandering on behalf of your party, the Moons. Fortunately, you know exactly how every citizen will vote in the election.

Fiddler Nation has seven districts, each with an equal population. For simplicity, assume that the Moons win any district in which they receive at least half of the vote.

What is the smallest total fraction of Fiddler Nation’s population needed to vote Moon such that it’s possible for the Moons to win a majority of the seven districts?

Let's let $p_i \in [0,1]$ be the proportion of voters in district $i$ that vote for the Moons. We see that the Moons will win district $i$ if and only if $p_i \geq 0.5.$ So without loss of generality, we assume that $p_1 \leq p_2 \leq \dots \leq p_6 \leq p_7,$ so if the Moons are only going to win 4 districts, then they would win districts 4 through 7, that is we would need $p_i \geq 0.5, i = 4, \dots, 7.$ Since each district has the same population, say $1/7$ of the total, we have that the total number of votes for the Moons would be $\frac{1}{7} \sum_{i=1}^7 p_i.$ For the classic problem, we want \begin{align*} \min \,\, & \frac{1}{7} \sum_{i=1}^7 p_i \\ \text{suct that} \,\, & 0 \leq p_1 \leq p_2 \leq \dots \leq p_6 \leq p_7 \leq 1 \\ & p_i \geq \frac{1}{2}, \, i = 4, 5, 6, 7\end{align*} We see that the minimizer is $$p^*_1 = p_2^* = p_3^* = 0, p^*_4 = p^*_5 = p^*_6 = p_7^* = 1/2,$$ so the smallest total proportion of the entire voting population to vote for the Moons that could result in a Moon victory is $$p^* = \frac{1}{7} \left( 3 \cdot 0 + 4 \cdot \frac{1}{2} \right) = \frac{2}{7}$$ of the population, which is indeed a pitifully low portion of the populace.

Surf's Up 2: WaveMania

The surfer rides the crest of the combined wave, $y = 2\sin(x−t) + \sin(x+2t),$ for a very long time. What is her average speed, in meters per second?

As we saw in the Classic answer, the surfer's horizontal speed with as a function of time is given by $$\dot{x}(t) = \frac{ 2(1-\cos (3t))}{5 + 4 \cos(3t)},$$ while her horizontal location $$x(t) = \pi \left\lceil \frac{x}{\pi} \right\rceil + \tan^{-1} \left( \frac{\cos (2t) + 2\cos t}{\sin (2t) - 2 \sin t} \right).$$

There are obviously two ways of measuring the average speed: (a) integrate the periodic function $\dot{x}(t)$ over some large time period from $0$ to $T \gg 1;$ or (b) rely on the relationship between distance and average speed to take $\bar{x} = (x(T) - x(0)) / T$ for some large $T \gg 1.$ For kicks, let's do both!

Since $\dot{x}$ is periodic with period $2\pi/3,$ we get that \begin{align*}\bar{x} &= \frac{1}{T} \int_0^T \dot{x}(t)\, dt\\ &=\frac{1}{T} \left( \int_0^{\frac{2\pi}{3} \lfloor 3T/2\pi \rfloor} \dot{x}(t) \,dt + \int_{\frac{2\pi}{3} \lfloor 3T/2\pi \rfloor}^T \dot{x}(t) \,dt \right).\end{align*} Firstly, we see that \begin{align*}I = \int_0^{2\pi/3} \dot{x}(t) \,dt &= 2 \int_0^{\pi/3} \frac{2(1-\cos (3t))}{5 + 4 \cos (3t)} \,dt \\ &= 2\int_0^\infty \frac{ 2 \left( 1 - \frac{1 - s^2}{1+ s^2} \right)}{ 5 + 4 \frac{1-s^2}{1+s^2} } \, \frac{ 2 ds }{ 3 ( 1 + s^2 ) } \\ &= \frac{16}{3} \int_0^\infty \frac{s^2}{(9+s^2)(1+s^2)} \,ds,\end{align*} where we have taken advantage of the powerful and mysterious tangent half-angle substitution $s = \tan \frac{3t}{2},$ where $\cos 3t = \frac{1 - s^2}{1+ s^2}$ and $dt = \frac{2ds}{3(1+s^2)}.$ Using the method of partial fractions we see that $$\frac{16}{3} \frac{s^2}{(9+s^2)(1+s^2)} = \frac{6}{9+s^2} - \frac{2}{3(1+s^2)},$$ and since $\int \frac{du}{a^2+u^2} = \frac{1}{a} \tan^{-1} (u/a) + C,$ so we have \begin{align*}I &= \frac{16}{3} \int_0^\infty \frac{ s^2 }{ (9+s^2) (1+s^2) } \, ds = \int_0^\infty \frac{6}{9+s^2} \,ds - \frac{2}{3} \int_0^\infty \frac{ds}{1+s^2} \\ &= 6 \left[ \frac{1}{3} \tan^{-1} \frac{s}{3} \right]_0^\infty - \frac{2}{3} \left[ \tan^{-1} s \right]_0^\infty \\ &= 2 \left( \frac{\pi}{2} - 0 \right) - \frac{2}{3} \left( \frac{\pi}{2} - 0 \right) \\ &= \frac{2\pi}{3}.\end{align*} We have $$\int_0^{\frac{2\pi}{3} \lfloor 3T/2\pi \rfloor} \dot{x}(t) \,dt = \left\lfloor \frac{3T}{2\pi} \right\rfloor \int_0^{2\pi/3} \dot{x}(t)\,dt = \left\lfloor \frac{3T}{2\pi} \right\rfloor I = \frac{2\pi}{3} \left\lfloor \frac{3T}{2\pi} \right\rfloor.$$ Since we already saw that $0 \leq \dot{x}(t) \leq 4,$ we have \begin{align*}\bar{x} &=\frac{1}{T} \left( \int_0^{\frac{2\pi}{3} \lfloor 3T/2\pi \rfloor} \dot{x}(t) \,dt + \int_{\frac{2\pi}{3} \lfloor 3T/2\pi \rfloor}^T \dot{x}(t) \,dt \right) \\ & \geq \frac{1}{T} \left( \frac{2\pi}{3} \left\lfloor \frac{3T}{2\pi} \right\rfloor + 0 \right) = \frac{2\pi}{3T} \left\lfloor \frac{3T}{2\pi} \right\rfloor =: m(T).\end{align*} Additionally, since $0 \leq T - \left\lfloor \frac{3T}{2\pi} \right\rfloor \frac{2\pi}{3} \leq \frac{2\pi}{3},$ we have $$\bar{x} \leq \frac{1}{T} \left( \frac{2\pi}{3} \left\lfloor \frac{3T}{2\pi} \right\rfloor + 4 \left( T - \left\lfloor \frac{3T}{2\pi} \right\rfloor \frac{2\pi}{3} \right) \right) \leq \frac{2\pi}{3T} \left\lfloor \frac{3T}{2\pi} \right\rfloor + \frac{8\pi}{3T} =: M(T).$$ Since we further have that for any $z \gt 0$ that $1 - \frac{1}{z} \leq \frac{\lfloor z \rfloor}{z} \leq 1,$ we have that $\lim_{T \to \infty} m(T) = \lim_{T \to \infty} M(T) = 1,$ so by the Squeeze Theorem, we have that the surfer's average speed is $\hat{x} = 1$ m/s, as $T \to \infty.$

Alternatively, let's try to directly bound $(x(T) - x(0))/ T$ for large $T \gg 1.$ First, let's look at the behavior of $\dot{x}$ on $[0, 2\pi/3].$ We see that there are two solutions to $\dot{x} = 1,$ namely $t_1 = \frac{2\pi}{9}$ and $t_2 = \frac{4\pi}{9}.$ The tangent lines at $t_1$ and $t_2$ are given by $\dot{x}(t_1) (t-t_1) + x(t_1) = t + \pi/3$ and $\dot{x}(t_2) (t-t_2) + x(t_2) = t + 2\pi/3,$ respectively. Since we can differentiate again to obtain $$\ddot{x}(t) = \frac{54 \sin (3t)}{(5+4\cos(3t))^2},$$ we see that $x(t)$ is concave up from $[0, \pi/3]$ and concave down from $[\pi/3, 2\pi/3].$ Since the concavity of a function determines whether tangent lines lie above or below the curves, we see that we have $t+\pi/3 \leq x(t) \leq t + 2\pi/3,$ at least for $[0, 2\pi/3];$ however, due to the periodicity of $x$, we can extend this result to show that for all $t \geq 0$ we have $$ t + \frac{\pi}{3} \leq x(t) \leq t + \frac{2\pi}{3}.$$ Therefore, we have $$1 - \frac{\pi}{6T} = \frac{ T + \frac{\pi}{3} - \frac{\pi}{2} }{T} \leq \bar{x} = \frac{x(T) - x(0)}{T} \leq \frac{T + \frac{2\pi}{3} - \frac{\pi}{2}}{T} = 1 + \frac{\pi}{6T}.$$ Again, since both the upper and lower bounds converge to $1$, we see that as $T \to \infty,$ the surfers average speed will converge to $1$ m/s.

Surf's Up

A surfer is riding along a crest of a sinusoidal wave given by the function $y = 2·sin(x−t),$ where $y$ is the height above sea level in meters, $x$ is horizontal position in meters, and $t$ is time in seconds. Based on this, the surfer knows that her speed—and the speed of the wave—is $1$ meter per second.

But hold on—it appears that another wave is approaching from the opposite direction. This wave has half the amplitude but twice the speed: $y = sin(x+2t).$

The two waves, the forward and the backward, ”interfere,” meaning they combine. In other words, the height of the water at position $x$ and time $t$ is given by $y = 2·sin(x−t) + sin(x+2t).$ Nevertheless, the surfer stays atop a crest of the combined wave. As the wave moves along, what are the surfer’s slowest and fastest speeds, in meters per second?

Let's assume that at time $t = 0$, that the surfer is perch above the crest of the wave at $x(0)=\frac{\pi}{2},$ with $y(0)=3.$ Since at this point we have $\frac{\partial y}{\partial x}=0,$ for any $t \gt 0,$ we can take the derivative of $y$ with respect to $x$ and solve for $x$ in terms of time $t.$ That is, for any t, we have \begin{align*}\frac{\partial y}{\partial x} &= 2 \cos ( x - t ) + \cos (x + 2t) \\ &= 2 \cos x \cos t + 2 \sin x \sin t + \cos x \cos (2t) - \sin x \sin (2t) \\ &= \left( \cos (2t) + 2 \cos t \right) \cos x - \left( \sin (2t) - 2 \sin t \right) \sin x.\end{align*} So by setting $\frac{\partial y}{\partial x} = 0,$ we get $$\tan x = \frac{ \cos (2t) + 2\cos t }{ \sin (2t) - 2\sin t }.$$ One way to solve this equation is to set $$x(t) = \tan^{-1} \left( \frac{ \cos (2t) + 2\cos t }{ \sin (2t) - 2\sin t } \right);$$ however, this will have a bunch of nasty jump discontinuities that likely would be fairly gnarly for the surfer to deal with. Instead, let's resolve this by ensuring that $x(t)$ is a nondecreasing, continuous function of $t,$ which would lead us to $$x(t) = \pi \left\lceil \frac{t}{\pi} \right\rceil + \tan^{-1} \left( \frac{\cos (2t) + 2\cos t }{ \sin (2t) - 2\sin t } \right).$$

Since we want to compute the surfer's horizontal speed at various times, we can ignore the almost everywhere constant initial term, $\pi \lceil t/\pi \rceil,$ and calculate \begin{align*}\dot{x}(t) = \frac{d}{dt} x(t) &= \frac{1}{1 + \left( \frac{\cos (2t) + 2 \cos t }{\sin (2t) - 2 \sin t} \right)^2} \frac{d}{dt} \left( \frac{ \cos (2t) + 2\cos t}{\sin (2t) - 2\sin t} \right).\end{align*} We first see that \begin{align*}\frac{d}{dt} \left( \frac{\cos (2t) + 2\cos t}{\sin (2t) - 2\sin t} \right) &= \frac{ (\sin (2t) - 2\sin t) (-2\sin (2t) - 2\sin t) - (\cos (2t) + 2\cos t) (2\cos (2t) - 2 \cos t)}{ (\sin(2t) - 2 \sin t)^2} \\ &= \frac{ -2 \sin^2 (2t) + 2 \sin t \sin (2t) + 4 \sin^2 t - 2 \cos^2 (2t) - 2 \cos t \cos (2t) + 4 \cos^2 (2t)}{ (\sin (2t) - 2 \sin t)^2} \\ &= \frac{ -2 - 2 \left( \cos t \cos (2t) - \sin t \sin (2t) \right) + 4 }{ (\sin (2t) - 2 \sin t)^2} \\ &= \frac{ 2( 1 - \cos (3t))}{ (\sin (2t) - 2 \sin t)^2}. \end{align*} Secondly, we see that \begin{align*}\frac{1}{1 + \left( \frac{ \cos (2t) + 2 \cos t }{ \sin (2t) - 2 \sin t } \right)^2} &= \frac{ (\sin (2t) - 2 \sin t)^2}{ (\sin (2t) - 2 \sin t)^2 + (\cos (2t) - 2 \cos t)^2 } \\ &= \frac{ (\sin (2t) - 2 \sin t)^2}{ \sin^2 (2t) - 4 \sin t \sin (2t) + 4 \sin^2 t + \cos^2 (2t) + 4 \cos t \cos (2t) + 4 \cos^2 t} \\ &= \frac{( \sin (2t) - 2 \sin t)^2}{ 1 + 4 \left( \cos t \cos (2t) - \sin t \sin (2t) \right) + 4 } \\ &= \frac{ (\sin (2t) - 2 \sin t)^2}{ 5 + 4 \cos (3t)}.\end{align*} Therefore, putting everything together we get $$\dot{x}(t) = \frac{ (\sin (2t) - 2 \sin t)^2 }{ 5 + 4 \cos (3t)} \frac{ 2 ( 1 - \cos (3t) )}{ (\sin (2t) - 2 \sin t)^2} = \frac{ 2(1 - \cos (3t))}{ 5 + 4 \cos (3t) }.$$

Ok, so far so good for our surfer. We see that since $5 + 4 \cos (3t) \geq 1$ for all $t \leq 0,$ that the denominator of her speed is always well defined. Furthermore, we see that if we define $$f(s) = \frac{ 2(1-s) }{5 + 4s}$$ then we have $$f'(s) = \frac{-18}{(5+4s)^2} \leq 0$$ for all $s \in \mathbb{R},$ so $f$ is a monotone decreasing function of $s.$ Therefore, in particular, since $-1 \leq \cos (3t) \leq 1$ for all $t \geq 0,$ we see that if $-1 \leq s = \cos (3t) \leq 1,$ then we have $$0 =\frac{ 2 ( 1 - 1 )}{ 5 + 4 \cdot 1} = \dot{x} \left( \frac{1}{3} \cos^{-1} 1 \right) \leq \dot{x} (t) \leq \dot{x} \left( \frac{1}{3} \cos^{-1} (-1) \right) = \frac{ 2(1 - (-1))}{ 5 + 4 \cdot -1} = 4.$$ So the surfer's slowest and fastest speeds in meters per second are 0 and 4, respectively.

Just keep swimming, normally ...

Another pond that’s also long and straight contains many, many koi. Initially, they’re all placed at the center of the pond, after which they all swim off at random velocities chosen from a normal distribution with a mean of 0 and a standard deviation of 1 mile per hour. A fish with a positive velocity initially swims toward one end of the pond, while a fish with a negative velocity initially swims toward the other end. The fish then swim back and forth, over and over again.

As before, you place a sensor in the water that measures the speed of each passing fish. All speeds are recorded as positive; direction no longer matters. You collect many such measurements over the course of several days, after which you read out the data and compute the average speed the sensor detected. Toward what value will this average speed converge?

As we saw in the Classic problem, with the added caveat that the sensor will pick up absolute speed, rather than signed velocities, if we have $v_i \sim \mathcal{N}(0,1),$ $i = 1, \dots, n,$ then we have $$\hat{v}_n = \frac{\sum_{i=1}^n v_i^2}{\sum_{i=1}^n |v_i|}.$$ Since we have many fish, let's define break the numerator and denominator up and try to use the Central Limit Theorem to understand what will happen in this case.

Let's define $A_n = \sum_{i=1}^n v_i^2$ and $B_n = \sum_{i=1}^n |v_i|.$ Let's first establish the prerequisites for the CLT. Firstly we see that for each $i = 1, \dots, n,$ we have $$\mathbb{E} |v_i| = 2 \int_0^\infty v \frac{e^{-v^2/2}}{\sqrt{2\pi}} \,dv = \sqrt{\frac{2}{\pi}} \left[ - e^{-v^2/2} \right]_{v=0}^{v=\infty} = \sqrt{\frac{2}{\pi}} \lt \infty,$$ and since each $v_i \sim \mathcal{N}(0,1)$ we have $$\mathbb{E} v_i^2 = 1.$$ Further we have $$var(|v_i|) = \mathbb{E} \left( |v_i| - \sqrt{\frac{2}{\pi}} \right)^2 = \mathbb{E} v_i^2 - 2\sqrt{\frac{2}{\pi}} \mathbb{E} |v_i| + \left( \sqrt{\frac{2}{\pi}} \right)^2 = 1 - \frac{2}{\pi} \lt \infty$$ and $$var(v_i^2) = \mathbb{E} \left( v_i^2 - 1 \right)^2 = \mathbb{E} v_i^4 - 2 \mathbb{E} v_i^2 + 1 = 3 - 2 + 1 = 2 \lt \infty.$$ Therefore, we see from the CLT, that $A_n \to \mathcal{N} (n, 2n)$ and $B_n \to \mathcal{N} ( n\sqrt{\frac{2}{\pi}}, n(1- \frac{2}{\pi}) ),$ uniformly in distribution. In particular, since we have each $v_i$ is independent we have $$\mathbb{E} \left[ v_i^2 |v_j| \right] = \mathbb{E} \left[ v_i^2 \right] \mathbb{E} |v_j| = \sqrt{\frac{2}{\pi}}$$ if $i \ne j$ and \begin{align*}\mathbb{E} |v_i|^3 &= 2 \int_0^\infty v^3 \frac{e^{-v^2/2}}{\sqrt{2\pi}} \,dv\\ &= \sqrt{\frac{2}{\pi}} \left[ \left( -v^2e^{-v^2/2} \right)_{v=0}^{v=\infty} + 2\int_0^\infty ve^{-v^2/2} \,dv \right]\\ &= 2 \sqrt{\frac{2}{\pi}},\end{align*} so we further see that \begin{align*}corr(A_n, B_n) &= \mathbb{E} \left[ \left( \frac{A_n - n}{\sqrt{2n}} \right) \left( \frac{B_n - n \sqrt{\frac{2}{\pi}}}{\sqrt{n(1 - \frac{2}{\pi})}} \right) \right]\\ &= \frac{\sqrt{\pi}}{n\sqrt{2(\pi-2)}} \mathbb{E} \left[ \sum_{i=1}^n \sum_{j=1}^n v_i^2 |v_j| - n \sqrt{\frac{2}{\pi}} \sum_{i=1}^n v_i^2 - n \sum_{j=1}^n |v_j| + n^2 \sqrt{\frac{2}{\pi}} \right] \\ &= \frac{\sqrt{\pi}}{n \sqrt{2(\pi-2)}} \left( n \left(2 \sqrt{\frac{2}{\pi}} \right) + (n^2 - n) \sqrt{\frac{2}{\pi}} - n^2 \sqrt{\frac{2}{\pi}} \right) \\ &= \frac{\sqrt{\pi}}{ n \sqrt{2(\pi -2)}} \left( n \sqrt{\frac{2}{\pi}} \right) = \frac{1}{\sqrt{\pi - 2}}.\end{align*} So in particular, if we define $Z_A, Z_B \sim \mathcal{N}(0,1)$ with $\mathbb{E} Z_AZ_B = \frac{1}{\sqrt{\pi-2}},$ then we can define $A_n = n + \sqrt{2n} Z_A$ and $B_n = n\sqrt{\frac{2}{\pi}} + \sqrt{n (1 - \frac{2}{\pi})} Z_B.$

Let's define use the properties of the natural logarithm to see if we can discern the distribution of $\hat{v}_n = A_n / B_n.$ We see that \begin{align*}\ln \hat{v} = \ln \left(\frac{A_n}{B_n}\right) &= \ln \left( \frac{ n + \sqrt{2n} Z_A }{ n\sqrt{\frac{2}{\pi}} + \sqrt{ n(1-\frac{2}{\pi})} Z_B } \right) \\ &= \ln \left( \frac{n}{n \sqrt{\frac{2}{\pi}}} \frac{1 + \sqrt{\frac{2}{n}} Z_A}{1 + \sqrt{ \frac{\pi -2}{2n} } Z_B } \right)\\ &= \ln \sqrt{\frac{\pi}{2}} + \ln \left( 1 + \sqrt{\frac{2}{n}} Z_A \right) - \ln \left( 1 + \sqrt{\frac{ \pi - 2 }{ 2n} } Z_B \right) \\ & \approx \ln \sqrt{\frac{\pi}{2}} + \sqrt{\frac{2}{n}} Z_A - \sqrt{\frac{\pi-2}{2n}} Z_B,\end{align*} where we take advantage of the Taylor approximation $\ln (1+t) = t + O(t^2).$ Let's define $\tilde{Z} = \sqrt{\frac{2}{n}} Z_A - \sqrt{\frac{\pi-2}{2n}} Z_B \sim \mathbb{N}(0,\nu^2),$ where we can calculate \begin{align*}\nu^2 &= \left( \sqrt{\frac{2}{n}} \right)^2 + \left( \frac{\pi - 2}{2n} \right)^2 - 2 \left( \sqrt{\frac{2}{n}} \right) \left( \sqrt{\frac{\pi-2}{2n}} \right) \mathbb{E} Z_AZ_B \\&= \frac{2}{n} + \frac{\pi-2}{2n} - 2 \sqrt{ \frac{2}{n} \frac{\pi - 2}{2n} \frac{1}{\pi - 2} } = \frac{\pi-2}{2n}.\end{align*} Therefore, we see that $\hat{v}$ is approximately log-normally distributed as $\hat{v} \sim \text{Log}\mathcal{N} (\mu, \sigma^2)$ with parameters $\mu = \ln \sqrt{\frac{\pi}{2}}$ and $\sigma^2 = \frac{\pi-2}{2n}.$ From here we see that if there are many fish with normally distribution signed velocities, then the average speed recorded by our sensor will converge to $$\lim_{n\to \infty} \mathbb{E} \hat{v}_n = \lim_{n\to \infty} \exp \left( \ln \sqrt{\frac{\pi}{2}} + \frac{1}{2} \left( \frac{\pi-2}{2n} \right) \right) = \lim_{n\to\infty} \sqrt{\frac{\pi}{2}} \exp \left( \frac{\pi - 2}{4n} \right) = \sqrt{\frac{\pi}{2}}.$$