menu
search
Log In
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

Subjects

Network Sites

 GO Mechanical
 GO Electrical
 GO Electronics
 GO Civil
 CSE Doubts

GATE2007-28 [closed]

1 vote
409 views

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 with the note: Out of syllabus now
in IS&Software Engineering
closed by 409 views
← Prev. Next →
Next Qn. in Sub. →
Answer:

Related questions

4 votes
2 answers
1
770 views
GATE1996-2.5
Newton-Raphson iteration formula for finding $\sqrt[3]{c}$, where $c > 0$ is $x_{n+1}=\frac{2x_n^3 + \sqrt[3]{c}}{3x_n^2}$ $x_{n+1}=\frac{2x_n^3 - \sqrt[3]{c}}{3x_n^2}$ $x_{n+1}=\frac{2x_n^3 + c}{3x_n^2}$ $x_{n+1}=\frac{2x_n^3 - c}{3x_n^2}$
Newton-Raphson iteration formula for finding $\sqrt[3]{c}$, where $c > 0$ is $x_{n+1}=\frac{2x_n^3 + \sqrt[3]{c}}{3x_n^2}$ $x_{n+1}=\frac{2x_n^3 - \sqrt[3]{c}}{3x_n^2}$ $x_{n+1}=\frac{2x_n^3 + c}{3x_n^2}$ $x_{n+1}=\frac{2x_n^3 - c}{3x_n^2}$
asked Oct 9, 2014 in Numerical Methods Kathleen 770 views
5 votes
2 answers
2
959 views
GATE1999-1.23
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 of $f$. The method converges always only if $f$ is a polynomial only if $f(x_o) <0$ none of the above
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 of $f$. The method converges always only if $f$ is a polynomial only if $f(x_o) <0$ none of the above
asked Sep 23, 2014 in Numerical Methods Kathleen 959 views
0 votes
0 answers
3
669 views
GATE2003-42
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 the roots of \(f(x)=0\) using \(x_0, x_1,\) and \(x_2\) respectively as initial guesses, the roots obtained ... 0.6 respectively 0.6, 0.6, and 1.3 respectively 1.3, 1.3, and 0.6 respectively 1.3, 0.6, and 1.3 respectively
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 the roots of \(f(x)=0\) using \(x_0, x_1,\) and \(x_2\) respectively as initial guesses, the roots obtained would be ... and 0.6 respectively 0.6, 0.6, and 1.3 respectively 1.3, 1.3, and 0.6 respectively 1.3, 0.6, and 1.3 respectively
asked Sep 17, 2014 in Numerical Methods Kathleen 669 views
1 vote
1 answer
4
1.5k views
GATE2008-22
The Newton-Raphson iteration $x_{n+1} = \frac{1}{2}\left(x_n+\frac{R}{x_n}\right)$ can be used to compute the square of R reciprocal of R square root of R logarithm of R
The Newton-Raphson iteration $x_{n+1} = \frac{1}{2}\left(x_n+\frac{R}{x_n}\right)$ can be used to compute the square of R reciprocal of R square root of R logarithm of R
asked Sep 12, 2014 in Numerical Methods Kathleen 1.5k views
Dark theme
Muffin by Hélio Chun
Thanks to Q2A
...