Composition of functions to the nth time

Question

I am trying to create an equation that shows me how long it will take with an annual withdrawal taken from compounded interest in a present value. I am stuck because I am wanting to take the composition of a function n times until my remaining value is zero or less than zero: For example:

If your future value is determined by the equation:

$$ F(p)=1.05p $$

In dollars.

You then take a fixed amount each year, $w=100$ after interest has been compounded. I then am stuck by having to take the composition of a function of itself until the $F(p) \leq 0$

Long Version:

$$ F(1000)=1050; 1050-w= 950 $$

$$ F(950)=997.5; 997.5-w=897.5 $$

$$ F(897.5) = 942.38; 942.38-w= 842.38 $$

So on and so forth, that leads me to my wall that I hit. I feel to express this correctly i have to use $F(F(p)-w)-w$ n times. Can someone please help me with this thought process?

How can I express this mathematical dilemma?

Answer 1

You can define recurrence relation on the balance. $B_0$ is the beginning balance money and $$B_n = f(B_{n-1}) - w$$ is the balance after $n$ withdrawals.

When $f$ is linear like multiplication as in your example, there is easy way to solve for explicit formula for $B_n$. When $f$ is not a simple function, I do not think there is a way. Suppose that $f(x) = ax$ with $a = 1.05$ in your example. We will try to find some $v$ such that $$B_n - v = a(B_{n-1} - v)$$ Evidently, we must have $$(a-1)v = w$$ so $$v = \frac{w}{a-1}.$$

Now the sequence $C_n = B_n - v$ can be easily solved explicitly $C_n = a^n C_0$ which allows you to deduce $$B_n = C_n + v = a^n (B_0 - v) + v.$$ Then solving for $n$ is a matter of solving equation.

Answer 2

One way to work out a problem like this is to suppose that at the very beginning, you take the original balance of the account, $p$ and split it into $n$ smaller accounts: the first account will sit for a year and then you withdraw the entire amount; the second account will sit for two years and then you withdraw the entire amount; and so forth. To make this match the original problem, we want the amount in the $k$th account to be just enough so that in $k$ years it grows to exactly the amount, $w$, that you need to withdraw at the end of the $k$th year.

At the interest rate of $5\%$ that is given in the question, in order to withdraw $w$ after one year, the first account needs to start with a balance of $\frac{w}{1.05}$. In order to withdraw $w$ after two years, the second account needs to start with $\frac{w}{1.05^2}$. In general the $k$th account, which needs to grow to a balance of $w$ after $k$ years, must start with a balance of $\frac{w}{1.05^k}$. So the original balance has to be split among $n$ accounts like this:

$$ p = \frac{w}{1.05} + \frac{w}{1.05^2} + \cdots + \frac{w}{1.05^{n-1}} + b $$

where $0 < b \leq \frac{w}{1.05^{n}}$, that is, $b$ is the amount in the last account, which lets you withdraw an amount up to $w$ after $n$ years, but not more (because if it were more, after withdrawing $w$ you would still have money left over for year $n+1$).

You may recognize the sum in the equation above as a geometric series with the initial term $\frac{w}{1.05}$ and ratio $\frac{1}{1.05}$. You can express this in a closed form, then find the maximum value of $n$ for which the sum is less than the initial value $p$.