closed by
1,053 views
1 votes
1 votes
closed with the note: Out of syllabus now

Consider the series $x_{n+1} = \frac{x_n}{2}+\frac{9}{8x_n},x_0 = 0.5$ obtained from the Newton-Raphson method. The series converges to

  1. 1.5
  2. $\sqrt{2}$
  3. 1.6
  4. 1.4
closed by
Answer:

Related questions

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}$
3.1k
views
2 answers
6 votes
Kathleen asked Sep 23, 2014
3,056 views
The Newton-Raphson method is to be used to find the root of the equation $f(x)=0$ where $x_o$ is the initial approximation and $f'$ is the derivative ... convergesalwaysonly if $f$ is a polynomialonly if $f(x_o) <0$none of the above
1.5k
views
0 answers
0 votes
Kathleen asked Sep 17, 2014
1,492 views
A piecewise linear function $f(x)$ is plotted using thick solid lines in the figure below (the plot is drawn to scale).If we use the Newton-Raphson method to find ... respectively1.3, 1.3, and 0.6 respectively1.3, 0.6, and 1.3 respectively
5.4k
views
1 answers
1 votes
Kathleen asked Sep 11, 2014
5,372 views
The Newton-Raphson iteration $x_{n+1} = \frac{1}{2}\left(x_n+\frac{R}{x_n}\right)$ can be used to compute thesquare of R reciprocal of R square root of R logarithm of R