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

in Numerical Methods by Veteran (52.2k points)
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2 Answers

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Best answer

Answer: C

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}

by Boss (33.8k points)
+2 votes
Answer is C. Direct application of the formula. With f(x)=X^{3}-c.
by Active (1.2k points)

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