The function x3 2x2 4x 8 has a double root at x 2 use th

The function x^3 - 2x^2 - 4x + 8 has a double root at x = 2, use the standard Newton-Raphson (Equation 6.6 in the book) to solve for the root at x = 2. Compare and discuss the rate of convergence using an initial guess of x_0 = 1.2

Solution

We know that formula for Newton Raphson method,

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

given that f(x_{n) =} x_n³-2x_n²-4x_n+8

df(x_n)/dx = f\'(x_n) = 3x_n²-4x_n-4

x_n+1 = x_{n -} (x_n³-2x_n²-4x_n+8)/(3x_n²-4x_n-4)

x_n+1 = (3x³_n-4x²_n-4x_n-x_n³+2x_n²+4x_n-8)/ (3x_n²-4x_n-4)

x_n+1 = (2x³_n-2x²_n-8)/(3x_n²-4x_n-4) ----------- equation (1)

now given that x₀ = 1.2

simply put value of x₀ in above euuation 1.

we will calculate x₁ then by puttiing the value of x₁ in equation 1 we will calculate x₂ and so on.

As we will calculate further our answer will approach toward 1.9999. which is the equal to the root 2.

the convergence rate of Newton-Raphson method is -

alpha = scaling factor / No. of Iterations * |f(r)|

where r is the estimated root and in case f (r)= 0, the formula fails.