Question

In Newton-Raphson's method write the formula for finding cube root of the number $N$.

Answer 1

Let $x=\sqrt[3]{3N}$

$⟹x^3=N$

$⟹x^3-N=0.$

Newton-Raphson method $\Rightarrow$ $x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$

$f\left(x\right)=x^3-N$

$f^{\prime }\left(x\right)=3x^2$

$\therefore$ $x_{n+1}=x_n-\frac{x_n^3-N}{3x_n^2}$

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

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

$x_{n+1}=\frac{{2x}_n}{3}+\frac{N}{3x_n^2}$

$x_{n+1}=\frac{1}{3}[{2x}_n+\frac{N}{x_n^2}]$