Assumption Xn= k f(Xn)=l and F'(Xn)= m
Newton-Raphson Method
This method is generally use to improve the result obtained by one of the previous method. Let x0 be an approximate roof of f(x)=0 and let x1=x0+h be the correct roof so that f(x1)=0. Expanding f(x0+h) by taylors series we obtain
F(x) = f(x0) +h f (x0)+ ........ f (x0)+…=0
Neglecting the second and higher order derivatives we have f(x0) + h f (x0)=0 which gives
h= - \[\frac{f(x_0)}{f'(x_0)}\]
A better approximation than x0 is therefore given by x1, where
X1= x0+ h = x0 – \[\frac{f(x_0)}{f'(x_0)}\]
Successive approximation are given by x2, x3,… xn+1= xn+1, where
Xn+1 = xn - \[\frac{f(x_n)}{f'(x_n)}\]
Which is Newton Raphson formula .
Xn+1 = Xn -\[\frac{f(x_n)}{f'(x_n)}\]
So that the Newton Raphson process has a second order or quadratic convergence.
Geometrically , in Newton-Raphson method a tangent to curve is drawn at point [x0, f(x0)] and the point of intersection of this tangent and x axis is taken as x1 which is the next value of the iterate of course x1 is closer to root than x0. It can be use to solving both algebraic and transcendental equations and it can also be used when the roots are complex .
The equation x3-x2+4x-4=0 is to be solved using the Newton Raphson method . if x=2 is taken as the initial approximation of the solution then the next approximation using this method will be ?
x0 = 2
F(x) = x3-x2+4x- 4
F(x)=3x2-2x+4
F(x0)f(2)= 8
F(x0)= f(2) = 12
X1 = x0-\[\frac{f(x_0)}{f'(x_0)}\] = 2 - \[\frac{8}{12}\]= \[\frac{4}{3}\]
