## Tuesday, November 18, 2008

### f(g(x)) = g(f(x)) (a game)

This game is for 2 players.
First, the players each secretly come up with two mathematical functions (two functions per player), f(x) and g(x).
The functions, f(x) and g(x), are defined and real for all real x, and they are continuous.

Next, the players take turns each being the "proposer" and the "solver".

First in a round the proposer reveals his/her functions.

The solver tries to determine (within a given time period) whether
f(g(x)) = g(f(x))
either for (1 of 3 choices) no real x's, a finite number of real x's, or an infinite number of real x's.

If the solver can't do this in a specific finite period of time, then the proposer gets a point. Otherwise the solver gets a point.

The players then switch who is the proposer and who is the solver.

(So, with only the two rounds, this is a low-scoring game.)

Example: f(x) = sin(x). g(x) = e^(x+1).
So, f(g(x)) = sin(e^(x+1)). g(f(x)) = e^(sin(x)+1).

The number of x's where sin(e^(x+1)) = e^(sin(x)+1) is precisely the number of pairs of integers (m,n) such that:
e^(1- pi/2 +2*n*pi) = pi/2 + 2*pi*m (I think).

I don't personally know if this equation is solvable at all for any integers m and n, let alone if there are a finite number of pairs (m,n) or an infinite number of pairs of integers m and n. So, if I was the solver and if the proposer proposed f(x) = sin(x) and g(x) = e^(x+1), then the proposer would get a point that round, and I would get no points that round.