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GATE2007-28 [closed]

+1 vote

248 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

1.5
$\sqrt{2}$
1.6
1.4

closed with the note: Out of syllabus now

gate2007
numerical-methods
newton-raphson
normal
out-of-syllabus-now

asked Sep 22, 2014 in IS&Software Engineering by Kathleen Veteran (59.7k points)
closed Nov 6, 2016 by jothee | 248 views

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Answer:

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+3 votes

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Newton-Raphson iteration formula for finding $\sqrt[3]{c}$, where $c > 0$ is $x_{n+1}=\frac{2x_n^3 + \sqrt[3]{c}}{3x_n^2}$ $x_{n+1}=\frac{2x_n^3 - \sqrt[3]{c}}{3x_n^2}$ $x_{n+1}=\frac{2x_n^3 + c}{3x_n^2}$ $x_{n+1}=\frac{2x_n^3 - c}{3x_n^2}$

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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

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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

asked Sep 17, 2014 in Numerical Methods by Kathleen Veteran (59.7k points) | 395 views

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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

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Consider the function f(x) = x2 - 2x - 1. Suppose an execution of the Newton-Raphson method to find a zero of f(x) starts with an approximation x0 = 2 of x. What is the value of x2, the approximation of x that algorithm produces after two iterations, rounded to three decimal places? 2.417 2.419 2.423 2.425

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GATE1995-2.15
The iteration formula to find the square root of a positive real number $b$ using the Newton Raphson method is $x_{k+1} = 3(x_k+b)/2x_k$ $x_{k+1} = (x_{k}^2+b)/2x_k$ $x_{k+1} = x_k-2x_k/\left(x^2_k+b\right)$ None of the above

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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 will converge to -1 converge to $\sqrt{2}$ converge to $\sqrt{-2}$ not converge

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