Division of a Particle

Question

At time 0, a particle resides at the point 0 on the real line. Within 1 second, it divides into 2 particles that fly in opposite directions and stop at distance 1 from the original particle. Within the next second, each of these particles again divides into 2 particles flying in opposite directions and stopping at distance 1 from the point of division, and so on. Whenever particles meet they annihilate (leaving nothing behind).

How many particles will there be at time $2^1$$^1$ + 1?

Here I observed that at time $2^n$ there would be two particles and at time time $2^n$ + 1. There would be 4 particles. But don't know how to prove it. I was trying with mathematical Induction but couldn't do it. Please help.

Hint: Look at Pascal's triangle. See if you can spot a connection. — Arthur, Aug 26 '19 at 14:57
I recommend drawing it (particles as points) out on a graph, with time (n) on the vertical axis, since there are only 12 values of n. — Joe, Aug 26 '19 at 14:57
@Joe There are thousands of values of $n$. We're going to $2^{11}$ here. — Arthur, Aug 26 '19 at 15:01
Oh, I see. I read too quickly. I thought it was n=11, not $n=2^{11}+1$ — Joe, Aug 26 '19 at 15:04
if you have 2 particles at any stage, how come you have 4 in the next step, considering the annihilation? — trula, Aug 26 '19 at 15:09
@trula, if the two are far enough apart, they can both split without any particles colliding. — Joe, Aug 26 '19 at 15:14
Ironically, if you draw the first 10 rows, as I suggested based on reading the problem wrong, you may SEE the connection to Pascal’s triangle that @Arthur mentioned. — Joe, Aug 26 '19 at 15:15

score 3 · Accepted Answer · answered Aug 26 '19 at 15:51

Using induction, prove that at $n=2^k$, there are two particles at positions $x=-2^k$ and $x=2^k$.

Base case: at $n=2^1$, there are two particles at positions $x=-2$ and $x=2$.

Inductive step: if there are two particles at positions $x=-2^k$ and $x=2^k$ when $n=2^k$, then:

After another $2^k$ seconds those two particles will have each split multiple times, but in systems that would not have interacted until the last split, because for one system every particle had $x<0$, but for the other every particle had $x>0$. At the $(2^k)^{th}$ second, each system would have resulted in two particles (each), but two of the particles collide at $x=0$ and annihilate. Therefore, there are two particles at time $n=2^k + 2^k = 2^{k+1}$, located at $x=-2^{k+1}$ and $x=2^{k+1}$.

Division of a Particle

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