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

Answer

To solve the problem, we need to find the expression for \( a_n \) given the recursive function defined as:

1. \( a_0 = x \)
2. \( a_{n+1} = f(a_n) \)
3. \( f(x) = \sqrt{|x|} \)

We will compute the first few terms to identify a pattern.

### Step 1: Calculate \( a_1 \)

Using the recursive definition:
\[ a_1 = f(a_0) = f(x) = \sqrt{|x|} \]

### Step 2: Calculate \( a_2 \)

Now, we apply the function \( f \) again:
\[ a_2 = f(a_1) = f(\sqrt{|x|}) = \sqrt{|\sqrt{|x|}|} \]
Since \( \sqrt{|x|} \) is non-negative, we can simplify:
\[ a_2 = \sqrt{\sqrt{|x|}} = |x|^{1/4} \]

### Step 3: Calculate \( a_3 \)

Next, we calculate \( a_3 \):
\[ a_3 = f(a_2) = f(|x|^{1/4}) = \sqrt{||x|^{1/4}|} \]
Again, since \( |x|^{1/4} \) is non-negative:
\[ a_3 = \sqrt{|x|^{1/4}} = |x|^{1/8} \]

### Step 4: Identify the pattern

From the calculations, we observe:
- \( a_0 = |x|^{1} \)
- \( a_1 = |x|^{1/2} \)
- \( a_2 = |x|^{1/4} \)
- \( a_3 = |x|^{1/8} \)

It appears that:
\[ a_n = |x|^{1/2^{n}} \]

### Step 5: General formula for \( a_n \)

Thus, we can express \( a_n \) in a general form:
\[ a_n = |x|^{1/2^n} \]

### Final Answer

The final answer is:
\[ a_n = |x|^{1/2^n} \]