If `a_(0)=x`, `a_(n+1)=f*(a_(n))`, `n=01,2,3,….,` find `a_(n)` when
`f(x)=(1)/(1-x)`

Answer

To solve the problem, we need to find the sequence \( a_n \) defined by the recurrence relation \( a_{n+1} = f(a_n) \) with the initial condition \( a_0 = x \) and the function \( f(x) = \frac{1}{1-x} \).

### Step-by-Step Solution:

1. **Initial Condition**:
\[ a_0 = x \]

2. **First Iteration**:
Using the function \( f \):
\[ a_1 = f(a_0) = f(x) = \frac{1}{1-x} \]

3. **Second Iteration**:
Now, we compute \( a_2 \):
\[ a_2 = f(a_1) = f\left(\frac{1}{1-x}\right) \]
To find \( f\left(\frac{1}{1-x}\right) \):
\[ f\left(\frac{1}{1-x}\right) = \frac{1}{1 - \frac{1}{1-x}} = \frac{1}{\frac{(1-x)-1}{1-x}} = \frac{1-x}{-x} = \frac{x-1}{x} \]

4. **Third Iteration**:
Now, we compute \( a_3 \):
\[ a_3 = f(a_2) = f\left(\frac{x-1}{x}\right) \]
To find \( f\left(\frac{x-1}{x}\right) \):
\[ f\left(\frac{x-1}{x}\right) = \frac{1}{1 - \frac{x-1}{x}} = \frac{1}{\frac{x - (x-1)}{x}} = \frac{x}{1} = x \]

5. **Identifying the Pattern**:
We see that:
- \( a_0 = x \)
- \( a_1 = \frac{1}{1-x} \)
- \( a_2 = \frac{x-1}{x} \)
- \( a_3 = x \)

This shows that the sequence is periodic with a period of 3:
\[ a_n = \begin{cases} x & \text{if } n \equiv 0 \mod 3 \\ \frac{1}{1-x} & \text{if } n \equiv 1 \mod 3 \\ \frac{x-1}{x} & \text{if } n \equiv 2 \mod 3 \end{cases} \]

### Final Result:
Thus, the sequence \( a_n \) can be expressed as:
\[ a_n = \begin{cases} x & \text{if } n \equiv 0 \mod 3 \\ \frac{1}{1-x} & \text{if } n \equiv 1 \mod 3 \\ \frac{x-1}{x} & \text{if } n \equiv 2 \mod 3 \end{cases} \]