Topic: Engineering Mathematics
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: C. $x_{n+1}=\frac{2x_{n}^{3}+C}{3x_{n}^{2}}$
Explanation: Let $x = \sqrt[3]{C}$, which implies $x³ = C$, or $f(x) = x³ - C = 0$.
The derivative is $f'(x) = 3x^2$.
The Newton-Raphson formula is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$.
$x_{n+1} = x_n - \frac{x_n^3 - C}{3x_n^2} = \frac{3x_n^3 - x_n^3 + C}{3x_n^2} = \frac{2x_n^3 + C}{3x_n^2}$.