The Newton-Raphson method starts with an initial guess $x_0$ for the root and iteratively improves this guess using the formula:

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

Here:

• $f(x)$ is the function for which we want to find the root.
• $f'(x)$ is the derivative of $f(x)$ .
• $x_n$ is the current approximation of the root.
• $x_{n+1}$ is the next approximation.

1. Choose an Initial Guess $x_0$ : Start with a reasonable estimate for the root. The choice of $x_0$ can affect whether the method converges to the correct root.
2. Calculate the Next Approximation: Use the Newton-Raphson formula to find $x_1$ :

$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$

3. Iterate: Repeat the process to get subsequent approximations $x_2, x_3, \dots$ until the difference between successive approximations is smaller than a desired tolerance level (e.g., $|x_{n+1} - x_n| < \epsilon$ ).
4. Check for Convergence: Ensure that the method is converging to a root. If the values of $x_n$ are not approaching a stable value, reconsider the initial guess or check the function's behaviour.

Example : Let's use the Newton-Raphson method to find the root of the equation:

$f(x) = x^3 - 2x - 5 = 0$

Step-by-Step Solution:

5. Function and Derivative:

$f(x) = x^3 - 2x - 5$

$f'(x) = 3x^2 - 2$

6. Initial Guess $x_0$ : Suppose we start with $x_0 = 2$.

7. First Iteration:

$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{2^3 - 2 \times 2 - 5}{3 \times 2^2 - 2} = 2 - \frac{8 - 4 - 5}{12 - 2} = 2 - \frac{-1}{10} = 2.1$

8. Second Iteration:

$x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2.1 - \frac{(2.1)^3 - 2(2.1) - 5}{3(2.1)^2 - 2}$

First, calculate $f(2.1)$ and f$'(2.1)$ :

$f(2.1) = (2.1)^3 - 2(2.1) - 5 = 9.261 - 4.2 - 5 = 0.061$

$f'(2.1) = 3(2.1)^2 - 2 = 3 \times 4.41 - 2 = 13.23 - 2 = 11.23$

So,

$x_2 = 2.1 - \frac{0.061}{11.23} \approx 2.0946$

9. Further Iterations: Continue iterating until the difference between $x_{n+1}$ and $x_n$ is very small. In this case, the root converges to approximately $x \approx 2.0946$ .

The Newton-Raphson method is a powerful tool for finding roots of equations, but it requires a good initial guess and careful handling of functions with certain characteristics (like multiple roots or small derivatives). By iterating with the formula $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ you can quickly converge to an accurate solution.