2,132 views
7 votes
7 votes

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

1 Answer

Best answer
9 votes
9 votes

$x_{n+1}  = x_n - \frac{f(x_n)}{f'(x_n)}$ 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

selected by
Answer:

Related questions

3 votes
3 votes
2 answers
1
Kathleen asked Sep 29, 2014
11,956 views
The Newton-Raphson method is used to find the root of the equation $X^2-2=0$. If the iterations are started from -1, the iterations willconverge to -1converge to $\sqrt{2...
3 votes
3 votes
1 answer
2
gatecse asked Sep 21, 2014
6,153 views
Newton-Raphson method is used to compute a root of the equation $x^2 - 13 = 0$ with 3.5 as the initial value. The approximation after one iteration is3.5753.6763.6673.607...
4 votes
4 votes
2 answers
3
0 votes
0 votes
3 answers
4
Kathleen asked Oct 8, 2014
2,485 views
The iteration formula to find the square root of a positive real number $b$ using the Newton Raphson method is$x_{k+1} = 3(x_k+b)/2x_k$$x_{k+1} = (x_{k}^2+b)/2x_k$$x_{k+1...