1. ## Can this be made into a function

There are 10 billion dollars circulating around. 50 Million (0.5% of total amount) is taken in everyday and replaced by new money. When the money is taken in everyday the 50 million is unbiased as to whether it is newly released or old money and is replaced regardless.

My understanding is you need to know the partial sum of the previous term (n-1) in order to determine n.

For example:
Total amount on Day 1: 50 million added
Day 2: 99.5% of day 1's "new money" retained, 50 million added
Day 3: 99.5% of day 2's "new money" retained, 50 million added etc. etc.

This seems like it works out to
inf
(SIGMA) 50 000 000 + Sn-1
n = 1

where Sn-1 is the partial sum of the nth-1 term.

Is there no way to turn this into a function, for example if i wanted to know how much new money was in circulation on the nth day and did not know the n-1 day's total "new" money.

P.S. I was able to determine a function in Excel, for example C1's cell would read: =0.995*B2+50000000, is the computer able to determine/display the nth amount?

2. Originally Posted by someguy456345
There are 10 billion dollars circulating around. 50 Million (0.5% of total amount) is taken in everyday and replaced by new money. When the money is taken in everyday the 50 million is unbiased as to whether it is newly released or old money and is replaced regardless.

My understanding is you need to know the partial sum of the previous term (n-1) in order to determine n.

For example:
Total amount on Day 1: 50 million added
Day 2: 99.5% of day 1's "new money" retained, 50 million added
Day 3: 99.5% of day 2's "new money" retained, 50 million added etc. etc.

This seems like it works out to
inf
(SIGMA) 50 000 000 + Sn-1
n = 1

where Sn-1 is the partial sum of the nth-1 term.

Is there no way to turn this into a function, for example if i wanted to know how much new money was in circulation on the nth day and did not know the n-1 day's total "new" money.

P.S. I was able to determine a function in Excel, for example C1's cell would read: =0.995*B2+50000000, is the computer able to determine/display the nth amount?
Here's how I would think about it. (I repeat your work at the beginning with different presentation.)

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

Let's sidetrack just a bit and consider another function.

Let

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:

$\displaystyle f(t) = C \left( 1 - e^{-kt} \right), k > 0$

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 $\displaystyle t \to \infty$, we have $\displaystyle \left(1 - e^{-kt}\right) \to 1$. 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

$\displaystyle e^{-(-ln(0.995))t}$

$\displaystyle =e^{ln(0.995)t}$

$\displaystyle =\left(e^{ln(0.995)}\right)^t$

$\displaystyle =0.995^t$

Thus we have

$\displaystyle f(t)=10000(1-0.995^t)$