Question

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

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

The correct option is C $x_{n+1}=\frac{2x_n^3+c}{3x_n^2}$

Let $x=\sqrt[3]{3c}$, then $x^3-c=0$
$\Rightarrow f\left(x\right)=x^3-c=0$ and $f^{\prime }\left(x\right)=3x^2$
Newton-Raphson method is given by
$x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }(x_n}$ $=x_n-\frac{(x_n^3-c)}{3x_n^2}$
$=\frac{3x_n^3-x_n^3+c}{3x_n^2}$ $=\frac{2x_n^3+c}{3x_n^2}$