Let f(n) represent the amount of new money in circulation after n days. So the reference point is day 0, where f(0) = 0.
This is a function by definition because it's impossible for f(x) to simultaneously take on more than one value. Formally speaking, f(a) = f(b) if and only if a = b.
What you mean to be asking is: can we find a closed-form expression for f(n)?
We start with a recursive definition: given f(n-1), we can find f(n).
Start with a simple example. Suppose we have 7 million new money in circulation today. The proportion of new money to old money is 7 million divided by 10 billion. Let this proportion be p. Then of the 50 million chosen for recycling, p*50 million will be new already, and (1-p)*50 million will be old. Recycling new money has no effect, we only need to consider the proportion that is old.
This leads to a recursive definition: (for simplicity, I'll count everything in millions)
f(n+1) = f(n) + (1-(f(n)/10000)) * 50
= f(n) + 50 - (50/10000) * f(n)
= 0.995*f(n) + 50
like you had.
Let's sidetrack just a bit and consider another function.
g(0) = 1
g(n) = 0.5*g(n-1), n > 0
This gives the sequence 1, 0.5, 0.25, 0.125, ...
it is a geometric sequence and has a closed-form expression
g(n) = (1/2)^n
So we can infer that the closed-form expression of f(x) will be exponential.
In particular, there is an upper bound of 10 billion, and this is an example of exponential decay.
I'm not really in the mood to work out the derivation for this, so I'll just write the formula for exponential decay, in the increasing form:
There is also a decreasing form, which would be used to model the amount of old money after n days.
Anyway, this function automatically passes through (0,0) already, and C is going to be the upper limit 10000 since as , we have . To find k, use the fact that f(1) = 50 and solve for k. You should get k = -ln(0.995) = ln(200/199).
The variable t is not restricted to integers, so assuming continuous growth rate, you can find the amount of new money after 2.25 days, if you wanted.
EDIT: It is of interest to note that
Thus we have