Consider the following situation. Suppose we deposit $1000$ in a bank at 10 percent interest. We ask the question: if we leave this money untouched for $n$ years, how much money will we have in our account at the end of this period? For simplicity, we assume that the 10 percent interest is added to our account once each year at the end of the year.

This is one of the simplest examples of an

The equation $A_n=1.1A_{n-1}$ is an example of a (first-order) difference equation . In such an equation, we use information from the previous year (or other fixed time interval) to determine the current information, then we use this information to determine next year's amount, and so forth. We solve this difference equation by the process to iteration. The iterative process involved is multiplication by $1.1$. That is, if we define the function $f(x)=1.1x$, then our savings balances are determined by repeatedly applying this function:
$$ A_1= f(A_0)$$
$$A_2=f(A_1)$$
$$A_3=f(A_2)$$
and so forth. Note that we may also write
$$ A_2=f(f(A_0))=f \circ f(A_0)$$
$$A_3=f(f(f(A_0))) = f \circ f \circ (A_0)$$
to clearly indicate that we compose $f$ with itself repeatedly to obtain the successive balances.
Since $f(x)=1.1x$, we have $$ f(f(x))=(1.1)^2 x$$
$$f(f(f(x)))=(1.1)^3 x,$$
and, in general, the $n$th iteration of the function yields
$$f\underbrace{ \circ \cdots \circ }_{n \: times }f =(1.1)^n x$$
So to find $A_n$, we merely compute $(1.1)^n$ and multiply by $A_0$. For example, using a calculater or computer, you may easily check that $A_{10}= \$ 2593,74$ and $A_{50}=\$117.390,85$.

This is one of the simplest examples of an

*iterative process or dynamical system*. Let's denote the amount we have in the bank at the end of the $n$th year by $A_n$. Our problem is to determine $A_n$ for some given number of years $n$. We know that $A_0$, our initial deposit, is $1000$. After $1$ year we add percent to this amount to get our new balance. That is, $$ A_1=A_0+0.1A_0=1.1A_0$$ In our specific case, $A_1=\$ 1100$. At the end of the second year, we perform the same operation $$ A_2=A_1+0.1A_1=1.1A_1$$ so that $A_2=\$1210$. Continuing, $$A_3=1.1 A_2$$ $$A_4=1.1A_3$$ $$ \vdots$$ $$ A_n=1.1 A_{n-1}$$ Thus, we can recursively determine the amount $A_n$ once we know the previous year's belong.The equation $A_n=1.1A_{n-1}$ is an example of a (first-order)