Author: zachary

Solution to Interlinear Interval

We are attempting to find the average number of steps in a certain mathematical game. Let us define a function which represents the average number of steps to completion. We will have this function be called T, for time, and have T(x) be the expected number of remaining steps at the position x. A review of the formal rules of the game and the definition of this function:

n \in \mathbb{N} \\[4pt] \operatorname f : \mathbb R, \mathbb Q \to \mathbb Q \\[8pt] \operatorname f {(p, x)} = \begin{cases} p > x : x-\frac1{2n} \\[4pt] p < x : x+\frac1{2n} \\[4pt] p = x : x \end{cases} \\[48pt] \operatorname T: \mathbb Q \to \mathbb N \\[4pt] \operatorname T(0) = \operatorname T(1) = 0 \\[12pt] \operatorname T(x) = x\operatorname T\left(x-\frac1{2n}\right)\ + \ (1-x)\operatorname T\left(x+\frac1{2n}\right)\ +\ 1

We can rephrase this equation to express T(x+1/2n) – T(x) in terms of T(x) – T(x-1/2n), a much more use-full equation. Consider a T(-1/2n). This extends the space of T beyond that which actually represents f–allowing us to find T(1/2n). T(-1/2n) = -1, so T(0) – T(-1/2n) = 1. Using the equation below: T(1/2n) – T(0) is 2n/(2n-1) + 1/(2n-1) = (2n+1)/(2n-1). This may then be extended further: T(1/n) – T(1/2n) is (2n(2n-1) + 2(2n-0) + 2) / (2n-2)(2n-1). Can we go for a formula in pi and sigma notation: yes. What about a polynomial formula, or at least something less hard to calculate? NO.

\operatorname T(x)-\operatorname T\left(x-\frac1{2n}\right)\ =\ \frac1{1-x} \ +\ \frac x{1-x}\left(\operatorname T\left(x+\frac1{2n}\right) – \operatorname T(x)\right)\\[24pt] \operatorname T(x) – \operatorname T\left(x-\frac1{2n}\right)\ =\ \left((2nx)!+\sum_{k=0}^{2nx}{(2nx)!(2n-k)\over k!}\right)\div \left(\prod_{l=1}^{2nx}2n-l\right)

And finally, we can summate T(k/2n) – T((k-1)/2n) for all k from 1 to n. The answer is, therefore, the following huge formula!

\Large {\Huge\sum_{k=1}^n}\left((2nk)!+\sum_{l=0}^{2nk}{(2nk)!(2n-l)\over l!}\right) \div\left(\prod_{m=1}^{2nk}2n-m\right)

Macrocephalous Measurements

Here is a very interesting system of equations which is rather intriguing to solve.

Illustrating the mechanism measuring the capacity of the large container.

We begin with this setup, which is actually a mechanism to measure a volume of air. The capacity of the large container on the left is being measured. Say, for example, it is being used by Joanne the scientist. She pulls the handle of the pneumatic cylinder on the bottom to use the system. This pulls the two connected plungers in the middle, compressing the air in the container. Depending on the size of the container, pressure increases at a different rate. Once it is high enough, the horses are balanced and the plungers come to a stop. From how much they lowered, the volume is determined. A rather efficient mechanism, requiring only that the container be airtight and very little material. There is one small problem. How can the capacity possibly be calculated? For a start to this challenge, the supplementary air tank is half a litre, neglect air in the tubing, and let the amount pulled be 50 mL. Create a formula to find the capacity of the container given the volume by which the cylinder is displaced; the volume that was pulled out of it. Perhaps go further and find a general formula.

solution

Interlinear Interval

This challenge pertains to algebraic limits, and can be solved through calculus, clever arithmetic, graph theory, or other methods, though is only classified as the foremost.

n \in \mathbb{N} \\[4pt] \operatorname f : \mathbb{R}, \mathbb{Q} \to \mathbb{Q} \\[8pt] \operatorname f {(p, x)} = \begin{cases} p > x : x-\frac1{2n} \\[4pt] p < x : x+\frac1{2n} \\[4pt] p = x : x \end{cases}

The function f id defined above. Consider evaluating f recursively, starting at 0.5, with x as the previous result and p a random number from 0 to 1. Eventually an integer would be reached, and the average time to reach one could be found given n. The challenge is to find this average for as many n as possible, then infinitely many, and perhaps all.

solution

Solution to Garbled Games

This challenge was to find the optimal strategy for player 5 of a certain game. This must be solved with probability density. Let the PDF before the guess of player n be fn, and that player’s best guess, x0. The derivative of the expected value must be zero, which can be simplified to:

x_0 \operatorname{f_n}x_0 = \int_0^{x_0}\operatorname{f_n}x \delta x

Solving for player 1, we get 3x0 = 2, and so x0 = 3/2. Using a the same process for players two through 5, we can find the answer. Shown is a table of the best guess, chance of victory, and expected value for players one two five, as well as the probability density function for the weight of the prize before each guess. The answer is thus a guess of exactly 0.1 kilograms, and you will win it exactly half the time.

\begin{array}{|r|rcl|} \hline \\ \text{Player} & \text{Guess} & \operatorname P {win} & \text{Value} \\[4pt] \hline \\ 1 & 3\over2 & 3\over4 & 9\over8 \\[4pt] 2 & 1 & 1\over2 & 1\over2 \\[4pt] 3 & 1\over3 & 1\over2 & 1\over6 \\[4pt] 4 & 5\over24 & 23\over48 & \approx.1 \\[4pt] 5 & 1\over10 & 1\over2 & 1\over20 \\[4pt] \hline \end{array}
  • Player 1
  • Player 2
  • Player 3
  • Player 4
  • Player 5

Garbled Games

This math challenge is a calculus challenge.

This challenge pertains to a game involving an infinite number of players, and 1 prize. The prize has a mass that is a random amount between 1 and 3 kilograms, just as likely to be less than 1.5 kg as more than 2.5. Each player in turn guesses a real number quantity of kilograms. The following is done for each player in turn: if the guess is less than the current mass of the prize: that mass is removed from the prize. Otherwise, that player receives nothing. You are player number 5. Assume you hear each of the previous players guess optimally. The challenge is to determine what is your best guess, and how much you will win on average.

solution

Solution to Scrupulist Square

Example setup

The challenge we are solving is about a graph derived from random points. The probability of connection between two is equal to 1 minus their distance, implying that each node has a loop. A node is chosen randomly, then a node is chosen of the odes connected to it, including itself. To find the probability that the second node has more connections than the first, We would need to determine the average number of connections to a given node. Since distance is best averaged in polar coordinates, this is a bit tricky. The trick is to divide the square into eight triangles, and use the secant function. The full process of integration is shown below. Now, Consider 2 points in a square. Consider the perpendicular bisector of the line segment connecting them. The square is divided into two parts, and on either side, one of the points is always closer. For the infinite case, the point with greater area will have more connections. Since the point at which it is exactly evenly cut must always pass through the square’s centre, any two points with equal distance from it will split the square evenly. So, for the infinite case, we may simplify to the probability of choosing a point closer to the centre. So for a given point, take the area of this circle, and divide by the total average fraction of all points it connects to. Given that, for lesser n, a point is always connected to itself, this amount needs to be adjusted. Specifically, with n points and a given point connecting to x fraction of an infinite selection, the actual fraction is x*(n-1)/n + 1/n. Instead of integrating over this circle, take the volume of an elongated slanted cone, which is equivalent. Let the point be a distance r from the cenre. The peak is always at the point, with height 1. The lowest point is 1 – 2r. Below is the calculation for the area of the shape. What is its average volume? Integrating again in polar coordinates, we get approximately .877! This is a messy integral and does not need to be fully shown. And lastly, the

\text{Finding the average distance to the corner of a right triangle,} \\ \text{with adjacent side }a\text{ and angle }\theta\text . \\[16pt] \int_0^\theta \int_0^{a \cdot\sec \theta} r^2 \delta r \delta \theta \\[12pt] =\int_0^\theta {{(a \cdot\sec \theta)}^3\over3} \delta \theta \\[12pt] ={a^3\over3} \cdot \frac12 \cdot {\left( \sec\theta\cdot\tan\theta+\log{\left(\cos\frac\theta2+\sin\frac\theta2\right)} – \log{\left(\cos\frac \theta2 – \sin\frac \theta2\right)}\right)} \\[12pt] = \operatorname f{(a, \theta)} \\[12pt] \operatorname g{(x, y)} = \\[8pt] \operatorname f{\left(x, \arctan\frac yx\right)} + \\[4pt] \operatorname f{\left(y, \arctan\frac xy\right)} + \\[4pt] \operatorname f{\left(x, \arctan\frac {\frac1{\sqrt2}-y}x\right)} + \\[4pt] \operatorname f{\left(x, \arctan\frac {\frac1{\sqrt2}-y}x\right)} + \\[4pt] \operatorname f{\left(\frac1{\sqrt2}-x, \arctan\frac y{\frac1{\sqrt2}-x}\right)} + \\[4pt] \operatorname f{\left(\frac1{\sqrt2}-y, \arctan\frac x{\frac1{\sqrt2}-y}\right)} + \\[4pt] \operatorname f{\left(\frac1{\sqrt2}-x, \arctan\frac {\frac1{\sqrt2}-y}{\frac1{\sqrt2}-x}\right)} + \\[4pt] \operatorname f{\left(\frac1{\sqrt2}-y, \arctan\frac {\frac1{\sqrt2}-x}{\frac1{\sqrt2}-y}\right)} + \\[4pt] \\[32pt] \text{Elongated cone, radius r, height 1, and height of cone 2r.} \\[16pt] A = \pi r^2 (1-2r) + \frac13 \pi r^2 (2r) \\[8pt] = \pi r^2 – \frac43 \pi r^3 \\[32pt] \text{This volume, integrated in polar coordinates.} \\[16pt] {\Large8\int_{-\frac\pi8}^\frac\pi8 \int_0^{a \cdot\sec\theta}} {\pi r^2 – \frac43 \pi r^3 \over{n-1\over n}\operatorname g{( \frac1{2\sqrt2}+r\cos\theta, \frac1{2\sqrt2}+r\sin\theta)}+\frac1n} {\Large \delta r \delta \theta = } \left({5\pi\over36}- {\tanh^{-1} {\sqrt2-1}\over 16\sqrt2}\right) \frac n{n+1} – \frac1n \\[12pt] \boxed{\huge\approx.877} \\[12pt] \text{8 because there are 4 triangles, and to get average, divide by area .5.}

Scrupulist Square

This challenge is spans all the way from graph theory, geometry, and calculus—to algebra and combinatorics.

Example set of points for N = 15

Consider a square with diagonal 1. In it, N points are randomly arranged. A graph is created from these points, such that the probability any 2 points touch is equal to 1 minus their distance. Choose a point at random on this graph. The challenge is to calculate a probability regarding this point: what is the probability that a randomly selected node connected to this one is connected to more nodes? Each node has a loop, so if the node was connected to no other nodes the probability of this is 0. For N=1 or 2, this probability is 0. Try to calculate at least one of the following: chance for a specific N > 2, limit of chance as N approaches infinity, and most difficult, generalized probability in terms of N.

SOLUTION

Solution to Preposterous Primes

This challenge was the last before an extended summer break. We left off unsure whether or not a sequence reached infinity at a finite point. Well, it does, and almost certainly at 23. But why? The solution is below.

Let us arrange the set of natural numbers into a sequence T. T begins with 1. For any Tn, Tn+1 is the lowest number prime connected to n that is not in the sequence T. If there exists none, then look to Tn-1, and repeat this process. If there are no more prime numbers connected to any previous numbers, then continue with the lowest natural number not yet in the set.

Within these constraints, a is always maximized, and n minimized for the value of a. This sequence is then: 1, 2, 4, 3, 6, …, 23. Now, what happens if this set is infinite? Well, some numbers would be skipped. Specifically, any dead end of connections not immediately followed would never be. This sequence is actually the best way to compute these sets. But computing to a high amount isn’t proof. But this is still helpful.

In order for there to be no infinite sets, there must be an infinite amount of finite sets. For each prime or the form 6n±1, or all those greater than 3, there are at least 3 numbers in its set. For 6n+1, they are 6n+1, 12n+2, and 12n. For 6n-1, they are 6n-1, 12n-2, and 12n-4. When n is even, both extend to 12n-3 and 12n-6. This proves that twin primes like 11 and 13, but not necessarily 17 and 19, must connect. This matches experimentation. Consider, in this sequence, that 2, 3, 5, 7, 9, 11, 17, 19, and 23 all are the first of their sets. Is 9 special? Yes, for any prime multiple of 3, only 1 number is connected to it, which is below. But 9 is alone, and so can start, and end, its set. But since there is no infinite occurrence of this, then there must be an infinite number of disconnected primes. This is impossible.

The reason for this, at last, is simple: an infinite number of sets that have a prime lowest number may not leave any composite spaces. That’s where the sequence comes in: since the first element the lowest and prime, a prime’s set must cover each number up to the next prime. But it must also contain those 2 additional numbers mentioned earlier. The next set must be unable to reach those, but able to reach all others. Here probability does the trick: as one goes through primes, the probability of this dwindles on the order below. Since this does not approach infinity, the weak law of large numbers tells us there must be an infinite set, and since 23’s grows so much so quickly, it is almost certainly the answer. Though there is a tiny probability of otherwise, and there could be a great number of very large sets afterwards, there must be a large one eventually, if not immediately.

\operatorname \mathcal 0 {\left({n\over{2\choose{\frac n{\ln n}}}\ln n}\right)}

Thus concludes Preposterous Primes, and zacharycormack.net‘s extended summer break.

Preposterous Primes

This scrupulous challenge is an algebraic one, and quite interesting.

Let a “prime factor connection” be a connection from two whole numbers A and B, such that:

  1. A and B share a common prime factor C
  2. A plus C equals B
  3. A divided by C is not prime

 For any whole number N, let the “prime connected set” of N be the set of all whole numbers that satisfy the following:

  1. For any number X in this set, N is part of its prime connected set
  2. All numbers prime connected to N are in this set
  3. This set is the minimum possible size given the first two requirements

 Let S(N) be the size of the prime connected set of N.

Does S(N) reach infinity eventually, or is it a fast growing but strictly finite function? If it reaches infinity, can you find the smallest value at which it is infinite? Try to prove it is so. If S(N) never reaches infinity, try to prove this. Good luck!

SOLUTION

Solution to Fascinating Frequencies BONUS

\text{let }x,\,y,\,z\in\R,\;x\leq y\leq z\\[8pt] \text{if }\;\exists!\;\{x,\,y,\,z\}: \operatorname fa=\operatorname gb=\operatorname hc\;\; \forall\;\;\{a,\,b,\,c\}\equiv\{x,\,y,\,z\},\\[4pt] \text{then }\;f\bigcirc g\bigcirc h\\[8pt] \text{if }\ f\bigcirc g\bigcirc h\;\ \text{and }\ f’\circledcirc g’\circledcirc h’,\ \;\text{then}\;f\circledcirc g\circledcirc h

We’re trying to find a set of three functions that satisfied a certain property. Look closely at it, and you see that what it really means is that, firstly, all three lines are concurrent at exactly three points. Then, their derivatives have the same property. This is recursive. Consider that the derivative of a sine wave is also a sine wave. We then have a relatively simple solution to the problem, but which is tricky to figure out.

\left. \begin{array}{l} \text{if $0\le x\lt3\pi$:} & \sin x \\ \text{if $x\ge3\pi$:} & f(-x) \\ \text{if $x\lt0$:} & e^x \\ \end{array} \right\} =f(x) \\[8pt] g(x) = -f(x)\\[8pt] h(x) = 0\\[16pt] f\circledcirc g\circledcirc h