Solving a system of recursive functions

[Chef Hoovisan]

I've run across a system of recursive functions (call them f and g). The system looks like this:

f(x) = a f(x-1) + b g(x-1)
g(x) = a g(x-1) + c f(x-1)

I also know that f(0) = 0 and g(0)>0. Finally, I know for other reasons that are too complicated to go into here that the system is somehow related to the trig functions. This is well outside my area of expertise, so I'm hopeful one of you can point me in the right direction so that I can perhaps simplify the system. Thanks in advance.

 

[write2diba]

I've run across a system of recursive functions (call them f and g). The system looks like this:

f(x) = a f(x-1) + b g(x-1)
g(x) = a g(x-1) + c f(x-1)

I also know that f(0) = 0 and g(0)>0. Finally, I know for other reasons that are too complicated to go into here that the system is somehow related to the trig functions. This is well outside my area of expertise, so I'm hopeful one of you can point me in the right direction so that I can perhaps simplify the system. Thanks in advance.

If it's possible can you say what are you trying to do here. Why is f(0) = 0 and g(0)>0? Is it some kind of conditions?
In what way are you trying to simplify the system.Please do explain

 

[Chef Hoovisan]

I'll give you the context, but as I said it's a bit involved. It's a math puzzle to which I know the answer, but do not understand the intuition behind the answer. Here's the puzzle:

A “small” ball with unit mass is resting on a pool table. A “big” ball with mass 16x100ⁿ (n a non-negative integer) is also resting on the pool table. That is, the higher mass ball will have a mass of m∈{ 16, 1600, 160000, …}. A man then strikes the big ball causing it to move toward and then strike the small ball. This collision sends the small off toward the bumper. The direction of all movement is perpendicular to the bumper, so the small ball rebounds from the bumper and returns until it strikes the big ball. Another collision results, sending the small ball back toward the bumper and repeating the process (albeit with different velocities for the balls). The pool table is frictionless and its bumpers are perfectly elastic, so there is no loss of velocity as the balls move along the table or when they rebound after hitting a bumper. Eventually, the collisions slow the big ball enough so that it reverses course and moves away from the small ball rather than toward it, thus ending the process. How many collisions, as a function of n, will there be between the big and small balls? Note that because of the frictionless system, you can ignore any spin in the balls and focus only on conservation of momentum and conservation of energy.

The governing formulas for this situation are shown here: http://en.wikipedia.org/wiki/Elastic_collision . Let m₁ and m₂ be the masses of the two balls and let v₁(n) and v₂(n) be the velocities of the two balls. We then have

v₁(x) = (v₁(x-1)*(m₁-m₂) + 2m₂v₂(x-1)) / (m₁+m₂)

and

v₂(x) = (v₂(x-1)*(m₂-m₁) + 2m₁v₁(x-1)) / (m₂+m₁)

Using a little algebra from there, you get the form I mentioned in the original post. Note that you do need to be careful to reverse the sign of the small ball's velocity to account for its rebound off the wall.

This is one of the most interesting math puzzles I've ever run across (and I've seen many, many of them), so I won't spoil the solution in case you want to work on it. I've seen a few somewhat unsatisfying "solutions," which is what has led me to ponder the recursive nature of the problem and how that might be used to produce a solution that provides good intuition.

 

[da_nang]

Would you say the problem is discrete-time? If so, a Z-transform might help.