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In the Newton-Raphson method, an initial guess of $x_0= 2 $ is made and the sequence $x_0,x_1,x_2\:\dots$ is obtained for the function 

$$0.75x^3-2x^2-2x+4=0$$

Consider the statements

  1. $x_3\:=\:0$
  2. The method converges to a solution in a finite number of iterations.

Which of the following is TRUE?

  1. Only I
  2. Only II
  3. Both I and II
  4. Neither I nor II
asked in Numerical Methods by Veteran (101k points) | 362 views

1 Answer

+7 votes
Best answer

$x_{n+1}  = x_n - \frac{f(x_n)}{f'(x_n)}$ for Newton-Raphson method (See the link below)

$f(x) = 0.75 x^3 -2x^2-2x + 4$

$ \implies f(2) = -2$

$f'(x) = 2.25x^2 - 4x -2$

$\implies f'(2) = 9 - 8 - 2 = -1$

So, $x_1 = 2 - \frac{-2}{-1} =2-2 = 0$

$x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0 - \frac{4}{-2} = 2$

Since $x_2 = x_0$, we will get $x_3 = x_1 = 0$. 

So, $x_3 = 0$, and the method never converges. A choice. 

https://www.math.ubc.ca/~anstee/math104/104newtonmethod.pdf

answered by Veteran (358k points)
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Sir, What is mean by convergence of a function. Please suggest some books / online resources for learning such terminology and concepts
0
Say $f(1) = 1.9, f(2) = 1.95, f(3) = 1.97, \ldots f(100) = 1.999, f(10000) = 1.999999, f(1000000) = 2$, here we say $f$ converges to the value 2. These things you have to think a lot as it is mathematics. But the above question is from Numerical Methods -- now not in GATE syllabus.


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