Solving the Analysis Question: Ball Movement in Room

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The discussion focuses on the mathematical analysis of a ball's movement within a defined region bounded by a convex, decreasing function f:(0,\infty)→ℝ, where f(x)>0 and lim_{x→∞}f(x)=0. The ball reflects off the walls according to the law of reflection, with its angle of incidence remaining constant upon hitting the bottom wall and reversing slightly upon hitting the top wall. The analysis concludes that the cumulative effect of these reflections, influenced by the slope of the wall and the angle of incidence, will eventually cause the ball to exit the room. The relationship between the ball's trajectory and the wall's slope can be approximated using integrals related to the derivative of the function.

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amcavoy
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This is a problem I found on the internet. In fact, I'm not even in analysis but I'm interested in how it's done.

Consider a smooth function [itex]f:(0,\infty)\rightarrow\mathbb{R}[/itex] such that [itex]f(x)>0[/itex] and [itex]\lim_{x\to\infty}f(x)=0[/itex]. f(x) is convex and decreasing. Consider now a region between the positive x-axis and f(x). Think of it as a room and the positive real axis and f(x) as the walls. Now take a ball and throw it inside the room. When it touches the wall, it reflects by the usual law. Prove that after some time the ball will come back and exit the room.
 

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Each time the ball hits the bottom, it's angle doesn't change, but each time it hits the top, it's angle goes backwards a little, and the idea is that the sum of all the backwards bumps will eventually turn the thing around. The influence of each bump would be a function of the angle that the ball is traveling at when it hits the top wall and the slope of the top wall where the bump occurs. Actually, the angle that the ball is traveling at will depend only on the angle it was traveling at after the previous bump with the top, which depends on the slope of the wall at that point plus the angle of the ball during that collision. But again that angle only depends on the previous collision, and you'll find that the only things you'll probably have to take into account are the initial angle and the slope at each point of collision. But you may as well set the initial angle to 0. So you'll end up with some sum whose terms are defined in part recursively, and in part based on the slope of the wall. You'll probably find that this sum can be approximated with an integral that is related to the derivative of the graph, and the integral when evaluated will give you a number that would suggest that the total amount that the ball is pushed back is large enough to send it backwards.
 
think of a laser beam instead of a ball an the optical law that angle of incidence is angle of reflection. The unit normal drawn from the graph towards the x-axis will always point towards the "left" since the second derivative of the graph is always positive.
 

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