Hard Riddle #1

@CalbernandHowbe said in #19:

@TakeThePawnOrLose no this percentage will never reach 100%, the limit as this series goes to infinity is 50%. (n-0.5)/2n as n goes to infinity (which is the series you wrote out just now)

But you reach X because as the rubber band expands then the robot’s position from X also expands, so that the % of journey completed remains constant through expansion
The series isn't n-0.5/2n, it is 0.5n/n+1. That also goes to 0.5 though so you are correct.

I don't know why you guys are using series while we have a continuous evolution over time.
Using the method I explained in #9, you can see that at time t, the robot has covered a distance equal to t*ln(t).

VSauce! Kevin here!...

@polylogarithmique said in #22:

I don't know why you guys are using series while we have a continuous evolution over time.
Using the method I explained in #9, you can see that at time t, the robot has covered a distance equal to t*ln(t).
Hmmm, I don't fully understand how you got to 1/t, so I won't say you did it wrong, but I also can't say you did it correctly.

@TakeThePawnOrLose said in #21:

The series isn't n-0.5/2n, it is 0.5n/n+1. That also goes to 0.5 though so you are correct.
The two are equivalent (it’s a limit not a sum)

Polylog means that the rubber band doesn’t suddenly increase by 1 m every second, it is continuous, ie 1mm every millisecond

@TakeThePawnOrLose :
I have been away from differential calculus for a long time, but here is my best shot. The answer is yes, the robot will reach the end, and is capable of encircling the rubber band infinitely many times. This might be a bit long but I want to be fully rigorous, because that is the heart of mathematics.
Begin by stretching the rubber band into a straight line by cutting it at point x. Thus we have a line starting at the origin and going right along the x-axis with starting and ending point X. Let d denote the distance that the robot has covered at time t, and let L denote the length of the line at time t. The question then becomes, what is the lim(t -> inf) (d/L)?
Clearly L = 1 + t. Also, at any point a along the line, we are moving forward at a speed of a/1+t, because the line drags everything along with it, at a speed proportional to the point on the line. By this logic, the rate of the robot at time t is .5 + d/1+t. Also, since distance = (rate)(time), d = (.5 + d/(1+t))*(t).
This is explicit in d so after solving for d we get d = (t^2 + t)/2. Thus the limit is the limit of d/L = t/2 which approaches infinity, so the robot can encircle as many times as it desires. However, an interesting note, the amount of seconds required to traverse the band each time goes up exponentially, I believe at a speed proportional to the triangular numbers. I won't go into detail but in particular, if N(X) is the number of seconds required to traverse the band the xth time, and T(x) is the xth triangular number, then N(x)/T(x) = c, for all x.
Hopefully this makes sense and if I messed up somewhere, please tell me.

@TakeThePawnOrLose said in #24:

Hmmm, I don't fully understand how you got to 1/t, so I won't say you did it wrong, but I also can't say you did it correctly.
As I said earlier, rescaling everything by a factor t you can view the robot as having speed 1/t on a circle of constant circumference 1 [because at time t the circle has circumference t]. You then integrate the speed from 1 to t, and re-multiply by t.

You can also think of it in terms of angular speed. Angular speed of the robot: speed/radius=2pi/t.
The lenght of the arc covered by the robot at time t is the radius * the integral from 1 to t of (angular speed)=tln(t).

@ChessMathNerd not sure how you obtain L=1+t and a+1/t for the speed.
Also triangular number are quadratic, not exponential.

Edit: ok for L=1+t since you start at t=0.

@TakeThePawnOrLose
I thought this forum was meant to be fun and not for intellectuals, sorry, my bad!