3. (15 marks ) (Zenos paradox: The dichotomy). One of the classical paradoxes of Zeno runs (more or less) as follows: A pair of dance partners are two units apart and wish to move together, each moving one unit. But for that to happen, they must first each move half a unit. After they have done that, then they must move half of the distance remaining. After that, they must move half the distance yet remaining, and so on. Since there are an infinite number of steps involved, logical difficulties seem to arise and indeed there is puzzlement in the first-year calculus class regarding things like this, although in modern models of analysis this paradox has long since been resolved. Roughly speaking, the applied mathematics view is that after a finite number of steps, the dancers are close enough for all practical purposes! (a) By symmetry, replace one partner with a mirror. Then start the remaining dancer off at so = 0 and the mirror is thus at s' = 1. Assume s, is the position of the dancer after the i-th step. Write a recursive formula for s, with the initial value 80 = 0. (b) Write a simple while loop in MATLAB for your answer in the previous part. Set the initial values s = 0 and i = 0, at each step increment i and compute s based on your formula and let the stopping condition be s > 1. (c) Does this loop terminate? Why? If so, how many iterations does it take? Justify your answer 3. (15 marks ) (Zenos paradox: The dichotomy). One of the classical paradoxes of Zeno runs (more or less) as follows: A pair of dance partners are two units apart and wish to move together, each moving one unit. But for that to happen, they must first each move half a unit. After they have done that, then they must move half of the distance remaining. After that, they must move half the distance yet remaining, and so on. Since there are an infinite number of steps involved, logical difficulties seem to arise and indeed there is puzzlement in the first-year calculus class regarding things like this, although in modern models of analysis this paradox has long since been resolved. Roughly speaking, the applied mathematics view is that after a finite number of steps, the dancers are close enough for all practical purposes! (a) By symmetry, replace one partner with a mirror. Then start the remaining dancer off at so = 0 and the mirror is thus at s' = 1. Assume s, is the position of the dancer after the i-th step. Write a recursive formula for s, with the initial value 80 = 0. (b) Write a simple while loop in MATLAB for your answer in the previous part. Set the initial values s = 0 and i = 0, at each step increment i and compute s based on your formula and let the stopping condition be s > 1. (c) Does this loop terminate? Why? If so, how many iterations does it take? Justify your
