# GATE1996-2.5

647 views

Newton-Raphson iteration formula for finding $\sqrt[3]{c}$, where $c > 0$ is

1. $x_{n+1}=\frac{2x_n^3 + \sqrt[3]{c}}{3x_n^2}$

2. $x_{n+1}=\frac{2x_n^3 - \sqrt[3]{c}}{3x_n^2}$

3. $x_{n+1}=\frac{2x_n^3 + c}{3x_n^2}$

4. $x_{n+1}=\frac{2x_n^3 - c}{3x_n^2}$

retagged

$x_{n+1} = x_n - \frac{f(x)}{f'(x)} = x_n - \frac{x_n^3 - c}{3x_n^2} = \frac{3x_n^3 - x_n^3 + c}{3x_n^2} = \frac{2x_n^3 + c}{3x_n^2}$

Answer is C. Direct application of the formula. With f(x)=X^{3}-c.

## Related questions

1
650 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 of $f$. The method converges always only if $f$ is a polynomial only if $f(x_o) <0$ none of the above
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
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
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