f(xo) + hf’(xo) + h2f’’(xo)/2! +-----= 0
Neglecting the second and higher order derivatives, we have
f(xo) + hf’(xo) = 0
Which gives
h = - f(xo)/f’(xo)
A better approximation than xo is therefore given be x1, where
x1 = xo – f(xo)/f’(xo)
Successive approximations are given by x2, x3,----,xn+1, where
xn+1 = xn – f(xn)/f’(xn)
Which is the Newton-Raphson formula.
Example: - Find a real root of the equation x3 -5x + 3 = 0.
Soln: -
Let, f(x) = x3 -5x + 3 = 0
f’(x) = 3x2 - 5
Choosing xo = 1
Step-1:
f(xo) = -1
f(xo) = -2
So, x1 =1 – ½ = 0.5
Step-2:
f(x1) = 0.625
f’(x1) = -4.25
x2 = 0.5 + 0.625/4.25 = 0.647
Step-3:
f(x2) = 0.0356
f’(x2) = -4.139
X3 = 0.647+ 0.0356/4.1393 = 0.6556
Step-4:
f(x3) = 0.0037
f’(x3) = -3.7105
x4 = 0.6556 + 0.0037/3.7105 = 0.6565
The successive approximation 0.6565
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