Suppose a large number of particles are bouncing back and forth between x = 0 an

Question

Suppose a large number of particles are bouncing back and forth between x = 0 and

x = 1, except that at each endpoint some escape. Let r be the fraction reflected

each time; then (1?r) is the fraction escaping. Suppose the particles start at x = 0

heading toward x = 1; eventually all particles will escape. Write an infinite series

for the fraction which escape at x = 1 and similarly for the fraction which escape at

x = 0. Sum both the series. What is the largest fraction of the particles which can

escape at x = 0? (Remember that r must be between 0 and 1.)

Explanation / Answer

call a(n) the number of particles that remains after the nth bounce, with a(0) being the original number leaving at x=0, then a(1)=ra(0) after the bounce at x=1, then a(2)=ra(1) after the bounce at x=0, etc.
So a(n)=r a(n-1), n>0.
Also, a(2n-1) (n>0) is the number after a bounce at x=1, and a(2n) is after a bounce at x=0.

But we want what escapes: call b(n) the number that escapes after the nth bounce, n>0.
We have b(1) = (1-r) a(0), and b(n) = (1-r) a(n-1).
Then b(n+1) = (1-r) a(n)=(1-r) r a(n-1) = r b(n),
so that b(n) = r^(n-1) b(1) = (1-r) r^(n-1) a(0).
Again, b(2n-1) (n>0) is what escapes at x=1, and b(2n) (n>0) is at x=0.

Now, the total number that escapes at x=0 is:
A = Sum[n:1->infinity, b(2n)]
= Sum[n:1->infinity, (1-r) a(0) r^(2n-1)]
= r (1-r) a(0) Sum[n:1->infinity, r^(2n-2)]
= r (1-r) a(0) Sum[n:0->infinity, r^(2n)]
= r (1-r) a(0) [1/(1 - r^2)]
= r a(0) / (1 + r)

The total number that escapes at x=1 is:
B = Sum[n:1->infinity, b(2n-1)]
= Sum[n:1->infinity, (1-r) a(0) r^(2n-2)]
= (1-r) a(0) Sum[n:1->infinity, r^(2n-2)]
= (1-r) a(0) Sum[n:0->infinity, r^(2n)]
= (1-r) a(0) [1/(1 - r^2)]
= a(0) / (1 + r)

I'm not sure about the last question. If it's in total, then I guess it's r/(1+r). If it's at any individual moment, then it should be during b(2), and the fraction is r(1-r).  

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