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We are given the recursive sequence:

\[
x_n = \frac{1}{2} x_{n-1} + \frac{A}{2x_{n-1}}, \quad n \geq 1
\]

where \( A \) is a positive number, and the task is to show that this sequence converges to \( \sqrt{A} \) when \( x_0 > 0 \). Additionally, we need to consider what happens if \( x_0 < 0 \).

### 1. Analyze the recursive relation:

The sequence can be seen as a form of the **Babylonian method** (also called the **Newton-Raphson method**) for approximating the square root of a number. The recurrence relation is constructed to bring the values of \( x_n \) closer to \( \sqrt{A} \).

- If \( x_n \) converges to some value \( L \), then in the limit as \( n \to \infty \), we should have \( x_n = x_{n-1} = L \).
  
Substituting \( x_n = L \) into the recurrence relation:

\[
L = \frac{1}{2}L + \frac{A}{2L}
\]

Multiplying both sides by 2 to eliminate the fraction:

\[
2L = L + \frac{A}{L}
\]

Simplifying:

\[
L = \frac{A}{L}
\]

Multiplying both sides by \( L \):

\[
L^2 = A
\]

Thus, \( L = \sqrt{A} \) (since \( A > 0 \) and we are assuming \( x_0 > 0 \), we take the positive square root).

### 2. Show that the sequence converges:

The recursive relation tends to pull each term of the sequence toward \( \sqrt{A} \). We can observe that if \( x_0 \) is positive, subsequent terms are also positive, and the sequence remains bounded and monotonic.

- For \( x_n \to \sqrt{A} \), the difference \( |x_n - \sqrt{A}| \) decreases with each step, meaning the sequence converges to \( \sqrt{A} \).

### 3. What happens if \( x_0 < 0 \)?

If \( x_0 < 0 \), we observe that the sequence will behave differently. Let's rewrite the recursion:

\[
x_n = \frac{1}{2}x_{n-1} + \frac{A}{2x_{n-1}}
\]

For negative values of \( x_{n-1} \), the term \( \frac{A}{2x_{n-1}} \) will be negative, potentially causing the sequence to oscillate or diverge, because the Babylonian method is typically designed for positive initial guesses to approximate square roots.

In this case, depending on the magnitude of \( x_0 \), the sequence could either:
- Diverge if the oscillations grow too large, or
- Approach \( -\sqrt{A} \) (a negative square root), which is not typically the solution expected for this type of method, since we usually look for positive roots.

### Conclusion:

- If \( x_0 > 0 \), the sequence converges to \( \sqrt{A} \).
- If \( x_0 < 0 \), the sequence may diverge or oscillate.

Would you like more details or have any questions?

Here are five related questions:
1. How does the Babylonian method work for finding square roots?
2. Can the sequence ever converge to \( -\sqrt{A} \)?
3. How does the initial value \( x_0 \) affect the convergence speed?
4. Is the sequence guaranteed to converge for all positive \( A \)?
5. What are other numerical methods for finding square roots?

**Tip**: Recursive sequences are powerful in numerical analysis, especially for iterative approximations of solutions like square roots.

Show that if A is any positive number, then the sequence defined by x_n = 1/2 * x_(n-1) + A/(2 * x_(n-1)), for n ≥ 1, converges to √A whenever x_0 > 0. What happens if x_0 < 0?

Explore the convergence of the recursive sequence x_n = 1/2 * x_(n-1) + A/(2 * x_(n-1)) to √A for positive A. This analysis involves understanding recursive relations and the Babylonian method. Learn what happens when x_0 is negative.

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