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 start with the given recursive relation:

1. **Initialization**: We have \( a_0 = x \).
2. **Recursive Definition**: The next term is defined as \( a_{n+1} = f(a_n) \) where \( f(x) = \frac{1}{1-x} \).

Now, we will compute the first few terms to identify a pattern.

### Step 1: Calculate \( a_1 \)
Using the recursive definition:
\[ a_1 = f(a_0) = f(x) = \frac{1}{1-x} \]

### Step 2: Calculate \( a_2 \)
Now, we find \( a_2 \):
\[ a_2 = f(a_1) = f\left(\frac{1}{1-x}\right) \]
Substituting into the function \( f \):
\[ a_2 = \frac{1}{1 - \frac{1}{1-x}} = \frac{1}{\frac{(1-x)-1}{1-x}} = \frac{1-x}{-x} = \frac{x-1}{x} \]

### Step 3: Calculate \( a_3 \)
Next, we compute \( a_3 \):
\[ a_3 = f(a_2) = f\left(\frac{x-1}{x}\right) \]
Substituting into the function \( f \):
\[ a_3 = \frac{1}{1 - \frac{x-1}{x}} = \frac{1}{\frac{x - (x-1)}{x}} = \frac{x}{1} = x \]

### Step 4: Identify the Pattern
From our calculations, we have:
- \( a_0 = x \)
- \( a_1 = \frac{1}{1-x} \)
- \( a_2 = \frac{x-1}{x} \)
- \( a_3 = x \)

It appears that \( a_3 = a_0 \). This suggests that the sequence is periodic with a period of 3.

### Step 5: Generalize \( a_n \)
Based on the periodicity observed:
- If \( n \equiv 0 \mod 3 \), then \( a_n = x \)
- If \( n \equiv 1 \mod 3 \), then \( a_n = \frac{1}{1-x} \)
- If \( n \equiv 2 \mod 3 \), then \( a_n = \frac{x-1}{x} \)

Thus, we can express \( a_n \) as:
\[ a_n = \begin{cases} x & \text{if } n \equiv 0 \mod 3 \\ \frac{1}{1-x} & \text{if } n \equiv 1 \mod 3 \\ \frac{x-1}{x} & \text{if } n \equiv 2 \mod 3 \end{cases} \]