recategorized by
2,630 views
0 votes
0 votes

The iteration formula to find the square root of a positive real number $b$ using the Newton Raphson method is

  1. $x_{k+1} = 3(x_k+b)/2x_k$

  2. $x_{k+1} = (x_{k}^2+b)/2x_k$

  3. $x_{k+1} = x_k-2x_k/\left(x^2_k+b\right)$

  4. None of the above

recategorized by

3 Answers

Best answer
3 votes
3 votes
Answer: B

$x_{k+1} = x_k - \frac{f(x_k)}{f'(x)} = x_k - \frac{(x_k^2 - b)}{2x_k} = \frac{2x^2_k - x^2_k + b}{2x_k} = \frac{x^2_k + b}{2x_k}$
0 votes
0 votes
Answer is D. F(x)=x^2-b. Applying the general formula we can find the answer.

Related questions

12.5k
views
2 answers
3 votes
Kathleen asked Sep 29, 2014
12,485 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 $\sqrt{2}$converge to $\sqrt{-2}$not converge
734
views
0 answers
0 votes
makhdoom ghaya asked Nov 9, 2016
734 views
Which of the following statements is true in respect of the convergence of the Newton-Rephson procedure?It converges always under all circumstances.It does ... a root where the second differential coefficient vanishes.None of the above.
1.9k
views
2 answers
4 votes
Kathleen asked Oct 9, 2014
1,911 views
Newton-Raphson iteration formula for finding $\sqrt[3]{c}$, where $c > 0$ is$x_{n+1}=\frac{2x_n^3 + \sqrt[3]{c}}{3x_n^2}$x_{n+1}=\frac{2x_n^3 - \sqrt[3]{c}}{3x_n^2}$x_{n+1}=\frac{2x_n^3 + c}{3x_n^2}$x_{n+1}=\frac{2x_n^3 - c}{3x_n^2}$
2.3k
views
1 answers
7 votes
go_editor asked Sep 28, 2014
2,262 views
In the Newton-Raphson method, an initial guess of $x_0= 2 $ is made and the sequence $x_0,x_1,x_2\:\dots$ ... of iterations.Which of the following is TRUE?Only IOnly IIBoth I and IINeither I nor II