91Ó°ÊÓ

Understanding root approximation is crucial for solving equations, especially when using methods like Newton-Raphson. When we approximate the roots of a function, we are looking for a number that the function equals zero. In the exercise, the function is given by \( f(x) = x^2 - a \), and we want to find the roots \( \pm \sqrt{a} \). The Newton-Raphson method helps us get closer to these roots with each iteration. By starting with an initial estimate, such as \( x_1 = 2 \) for \( a = 5 \), we refine this guess through the iteration formula. Each step brings us closer to the accurate root, gradually improving our approximation.

The derivative of a function plays a central role in understanding its behavior, especially when using the Newton-Raphson method. In this scenario, the function is \( f(x) = x^2 - a \). By calculating the derivative, \( f'(x) = 2x \), we gain insight into how the function changes at any point \( x \).
The derivative is pivotal for the Newton-Raphson formula because it helps determine how much we adjust our current approximation in each iteration. It effectively guides the direction and magnitude of our adjustment, thus speeding up the convergence towards actual root values.

The iteration formula from the Newton-Raphson method is a key tool to move our approximate solution forward. For the problem at hand, the iteration formula is derived as:

• Start with the Newton-Raphson formula: \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \)
• Substitute in our specific function and its derivative: \( x_{n+1} = x_n - \frac{x_n^2 - a}{2x_n} \)
• Simplify the formula to get: \( x_{n+1} = \frac{1}{2}\left(x_n + \frac{a}{x_n}\right) \)

Using this formula, each new guess, \( x_{n+1} \), is calculated by averaging the previous guess and the quotient \( \frac{a}{x_n} \). This careful balance helps steer the approximation closer without overshooting.

Convergence evaluation is critical in confirming that our approximation is heading towards the true root and not diverging. With each iteration, we need to check how close our approximation is to the root. Applying the derived formula repeatedly makes the sequence \( x_2, x_3, x_4, \) etc., better approximations of \( \sqrt{a} \).
In practical terms, convergence is assessed by examining the function value at each approximate root. In the exercise, after calculating \( x_4 = 2.23607 \), we evaluate \( f(x_4) = (2.23607)^2 - 5 \approx 0.00002 \). This small result signifies that \( x_4 \) is a very accurate approximation of the true root \( \sqrt{5} \), ensuring that our iterative process is successful.