# 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

Explanation:

```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.

Quiz of this Question

Attention reader! Don’t stop learning now.  Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.

Learn all GATE CS concepts with Free Live Classes on our youtube channel.

My Personal Notes arrow_drop_up