Describe the Newton-Raphson method for finding roots of equations. Provide an example.

The Newton-Raphson method is an iterative numerical method used to find the roots of equations. It is based on the idea of approximating the root of a function by using the tangent line at a given point.

The method starts with an initial guess for the root, denoted as x0. Then, it iteratively improves this guess by using the formula:

xn+1 = xn - f(xn)/f'(xn)

where xn+1 is the new approximation for the root, xn is the previous approximation, f(xn) is the value of the function at xn, and f'(xn) is the derivative of the function at xn.

This process is repeated until the desired level of accuracy is achieved or until a maximum number of iterations is reached.

Now, let's consider an example to illustrate the Newton-Raphson method. Suppose we want to find the root of the equation f(x) = x^3 - 2x - 5.

Step 1: Choose an initial guess, let's say x0 = 2.

Step 2: Calculate the value of the function and its derivative at x0:
f(x0) = (2)^3 - 2(2) - 5 = 1
f'(x0) = 3(2)^2 - 2 = 8

Step 3: Use the Newton-Raphson formula to update the approximation:
x1 = x0 - f(x0)/f'(x0) = 2 - 1/8 = 1.875

Step 4: Repeat steps 2 and 3 until the desired level of accuracy is achieved. Let's continue the iterations:

For x1:
f(x1) = (1.875)^3 - 2(1.875) - 5 = -0.5781
f'(x1) = 3(1.875)^2 - 2 = 5.1094
x2 = x1 - f(x1)/f'(x1) = 1.875 - (-0.5781)/5.1094 = 1.8393

For x2:
f(x2) = (1.8393)^3 - 2(1.8393) - 5 = -0.0067
f'(x2) = 3(1.8393)^2 - 2 = 4.9987
x3 = x2 - f(x2)/f'(x2) = 1.8393 - (-0.0067)/4.9987 = 1.8393

The iterations can be continued until the desired level of accuracy is achieved. In this example, the root of the equation f(x) = x^3 - 2x - 5 is approximately x = 1.8393.

It is important to note that the Newton-Raphson method may not always converge or may converge to a different root if multiple roots exist. Therefore, it is necessary to carefully choose the initial guess and verify the convergence of the method.