GATE CSE 1996 | Question: 2.5

Question

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

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

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

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

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

Answer 1

Answer: C

Answer 2

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