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GATE | GATE-CS-2007 | Question 28

  • Difficulty Level : Medium
  • Last Updated : 28 Jun, 2021

Consider the series Xn+1 = Xn/2 + 9/(8 Xn), X0 = 0.5 obtained from the Newton-Raphson method. The series converges to
(A) 1.5
(B) sqrt(2)
(C) 1.6
(D) 1.4

Answer: (A)


As per Newton Rapson's Method, 

Xn+1  = Xn − f(Xn)/f′(Xn)

Here above equation is given in the below form

Xn+1 = Xn/2 + 9/(8 Xn)

Let us try to convert in Newton Rapson's form by putting Xn as
first part.
Xn+1  = Xn - Xn/2 + 9/(8 Xn)
                 = Xn - (4*Xn2 - 9)/(8*Xn) 

So    f(X)  =  (4*Xn2 - 9)
 and  f'(X) =  8*Xn 

So clearly f(X) = 4X2 − 9. We know its roots are ±3/2 = ±1.5, but if we start from X0 = 0.5, according to equation, we cannot get negative value at any time, so answer is 1.5 i.e. option (A) is correct.

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