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 1.433Day: August 4, 2019
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=102.3m