We wanted to figure out the value of the fathomless fraction. We can start by saying the value of this fraction equals x1. x1 = 1 + 1 / x2, and so on. We can define this a recursive sequence, where xn = n + 1 / xn+1. But this doesn’t help if we don’t know any of the later terms. It approaches infinity, of course. xn>n, but xn<n+1. Maybe we could figure out the value of a term based on the previous one? xn-1 = n-1 + 1 / xn. Subtract n-1 from both sides. xn-1 -n +1 = 1 / xn. Take the reciprocal of both sides to find that xn = 1 / xn-1-n+1. We can make an alternate version of the fathomless fraction, call it fathomless equivalent fraction, or FEF for short. We soon realize that both lead to the exact same fraction. But x1 = x-1+1, and so xn=x-n+1, and we can make another fraction. But it returns to x1! It’s like with √x+1, it cannot be simplified. The answer is simply ≈1.433.
x_1=1+\frac1{2+\frac1{3+\frac1{4+…}}}\\[16pt] x_n=n+\frac1{x_{n+1}}\\[16pt] x_n=\frac1{x_{n-1}-n+1}\\[16pt] x_1=\frac1{\frac1{1+x_{-1}}}\\[16pt] x_n=x_{-n}+n\\[16pt] x_1=1+\frac1{2-2+\frac1{x_1-1}}=x_1\\[16pt] x_1 \approx 1.433Tag: solution
Solution to Baffling Bungee
So, we’re trying to find Bob’s final height after his bungee jump. The spring constant of C5H8, normal rubber, is ≈88. The equation for the force is 1/2 kx2, where k is the spring constant and x is the portion of the natural length. When Bob begins his jump, the rubber is inactive. It’s length is zero. Thus the force of it’s pull is ≈88*02. But gravity pulls down with a force of ≈9.807 m/s2. After this we cannot continue. The simplest way is to code. Simulate the bungee jump in many microscopic steps. Calculate the pull of the elastic at each point. This will show where the lowest point of the jump is. The stable point is easily calculated. 44x2≈9.807. x2≈9.807 / 44 ≈ 0.2229. x ≈ 0.4721. The bottom of the jump is a whopping 101.7 metres down, making the answer ≈102.3 metres above the water.
PE=\frac{kx^2}2\\[16pt] 44x^2=9.807\\ x=0.4721\\[16pt] h_f=\Delta x+\approx0.5279m\\ \quad h_f=h_i-\approx0.4721m\because h_i=\Delta x+1m\\ \Delta x=101.7m\\ \Rightarrow h_i=101.7m+1m\\ \therefore h_f=101.7m+0.5279m=102.3mSolution to Recursive Renewals
We want to find the optimal strategy for renewing a book at the Leibniz Library. First, let’s look at the maximum number of days you can keep a book before renewing it. 8 at first, then 7, 7, 6, and then 4. And that’s it, you cannot renew the book again. The fine equation will start above 0 when you renew the book a fifth time. But the hold equation will max out at 100% after only 10 days. The best is clearly 16 days. But it’s very likely it will be put on hold and you’ll be stuck with 14, or worse, 8. After 2 days, the hold equation ≈22%, and the loss would be 1 day if you don’t know. But if you don’t renew, you will probably gain a day. So don’t renew then. But on day 3, you can lose 2 days, and the hold equation now ≈24%. It’s less than 33%, so the day gained would still be worth it. On day 4, the hold equation is now at ≈29%, more than 25%, so you renew. We start again on day 5, with ≈40%, it’s less than 50%, so we don’t renew. But on day 6, we have ≈51%, far more than 33%. Although we lose the possibility of an extra day, it’s quite unlikely, and not nearly enough to make it worth not renewing. So we ought to renew. Not until day 8 can we gain another day. By then the hold equation is already ≈79%. It doesn’t matter if it has been renewed yet, if it hasn’t yet we do. By the time we can gain another day, the hold equation will have reached 100%. So our answer is to renew on days 4, 6, and 8.
f(x)=d^2+5d-7cos(r)+r^3-100\\[10pt] h(x)=\frac{t^2+3sin(t)+15}{100}Solution to Spinning Stick
We wanted to find the velocity of the ball when it rolls off of the spinning stick. The velocity of a point stick will be 2π metres per minute times it’s distance from the origin, in whatever direction it’s traveling in. This is 90° clockwise of the direction of the stick. This vector makes a circle with radius 2π metres, and it completes a rotation at the same speed. Knowing this, we find the centripetal force the stick experiences to be 4π2 metres per minute squared multiplied by the distance, another 90° clockwise of the stick. This makes centripetal force towards the origin, as you probably already know. If to spin in a circle the ball must accelerate, then it wont. The stick does not hold it in, and so it will quickly fly away. Since acceleration is proportional to distance, velocity increases acceleration. If you try solving Captivating Circles 5, you will see that such relationships relate to e. The rotation doesn’t interfere at all, though you’d think it would. When you have they are the same size and the proportions of change equal, their sum increases exponentially. Every unit of time when one would double the other if the other wasn’t returning the favour, they both multiply by e. They multiply at the same rate even if they aren’t the same size. The difference will decrease by a factor of e in the same duration of time. If one increases the other faster by a factor of n, the other will approach being larger by a factor of √n. The distance from the correct ratio will also decrease by a factor of e in the same duration of time as before. Using this knowledge, we can calculate the velocity, using the duration of time shown before as the unit for time, and take it’s derivative to get distance. Solve this equation to equal 2, and we find the time when the ball leaves the stick. Put time back into the equation for velocity, and we have our answer. 12.3134 metres per minute.
\frac d{dx}\frac{e^t-e^{-t}}{2\sqrt{2\pi}}=\frac{e^t+e^{-t}}{2\sqrt{2\pi}}\\[16pt] t=log(2\sqrt{2\pi}+\sqrt{8π-1})\approx2.295\\[16pt] \overrightarrow v\approx\frac{e^{2.295}-e^{-2.295}}{2\sqrt{2\pi}}\approx12.313Solution to Ridiculous Radical
We wanted to find a way to factor out the 1 from √x+1 to find it’s derivative. It is impossible to factor out the 1. The answer is thus simply √x+1 / 2x+2.
\frac d{dx}\sqrt{x+1}=\frac{\sqrt{x+1}}{2x+2}Solution to Machiavellian Matrix
We wanted to solve the Machiavellian Matrix. We can began by dividing the top row by xyz to simplify it to ( 1 1 1 1 ). Then we subtract row 1 from row 3 to fin that x = -1/3. This makes x2 = 1/9. Then we can get a system of equations for y and z, that y + z = 4/3, and y2 + z2 = 8/9. If y + z = 4/3 then z = 4/3 – y, and we can substitute to get the quadratic equation. Solve to get (y – 2/3) * (y – 2/3) = 0, so y = 2/3. Our answer is thus that x = -1/3, y = 2/3, and z = 2/3.
\left(\begin{matrix} 1 & 1 & 1 & 1 \\ x & y & z & 1 \\ 4 & 1 & 1 & 0 \end{matrix}\right) \overrightarrow{R_3-R_1→R_1} \left(\begin{matrix} 3 & 0 & 0 & -1 \\ x & y & z & 1 \\ 4 & 1 & 1 & 0 \end{matrix}\right) \\[10pt] y^2+\left(\frac43-y\right)^2-\frac89=0Solution to Peculiar Pets
We wanted to find what portion of the population owns a dog. First, we know that 60% of the population own at least one pet. Since all pets are randomly distributed, people with 1 pet must have a three fifths chance of owning a second pet, and so on. This makes the average number of pets per person equal to 3/5 + (3/5)2 + (3/5)3 + … . The general formula for a sequence like this is shown below. In this case, we get 1.5. Now, we make an equation. Let the ratio of dogs to cats to fish to people be x:y:z:1. We know that 3x = 2y, x+y+z = 1.5, and 6y / (1-y) = 5z / (1-z). We can easily substitute y to get x+1.5x+z = 1.5 and 9x / (1-1.5x) = 5z / (1-z). Substituting z is impossible. We must use a matrix. In this case it is long and difficult and we will not show it. The positive solution is x = 1/3, z = 2/3, and so y = 1/2. Since the ratio of dogs to people is 1:3, 1/(3+1) of the population must own dogs to make 1/(3+1-1) = 1/3. And so our answer is that 1/4 of the population owns at least 1 dog.
\sum_{k=1}^\infin\left(\frac ab \right)^k=\frac a{b-a}Solution to Fascinating Fruits
We wanted to find the probability that the row started by a pear ends with one. Since no conditions limit the number of fruits, there are infinite solutions. Our answer is thus one third.
Solution to Puzzling Probabilities
We wanted to find the probability, x, that Bob wins a game of Stupid Spin at Scam Casino. The probability Bob gets a point is a, otherwise Scam Casino gets one. The first to b points wins. First, the maximum spins in a round is 2b – 1, where 1 player wins by just 1 point. If we always play this many spins, you will always win at b points. We can simplify the problem to whomever has more points after 2b – 1 spins. Bob wins if the Casino has 0 to b-1 points. This gives us the first part of the equation. The probability of any given score for the Casino is equal to the number of ways that score can happen times the probability of it happening. The number of ways is easily done with the choose function. The probability of that is a to the power of the number of points scored by Bob, part three, multiplied by 1-a to the power of the number of points scored by the casino, the final part. We combine these 4 parts to get our answer, the equation below.
\sum_{k=0}^{b-1} {2b-1 \choose k}a^{2b-k-1}(1-a)^{k}Solution to Crazy Capture
We wanted to find the minimum number of police officers to capture a target. Remember that all units and the target are points with constant speed, perfect logic, and instant reaction time. Consider the lines perpendicular to those through the target and officers. Clearly, the target cannot cross any of these lines, the officers would always block it off. Furthermore, the target cannot delay all the officers in this way at once. The inactive ones can close in. It follows that a closed shape can catch the target. The smallest 2D shape is a triangle. We realize that we can generalize this. In higher dimensions, we use perpendicular planes and such. But the smallest figure is always the simplex! The simplex has n+1 vertices. Therefore, our answer is that 3 constables are needed to catch the target in 2 dimensions.