Question

After repetition the cube root of $10$ by Newton-Raphson method is _____.

Answer 1

Let $x=\sqrt[3]{310}$

$⟹x^3-10=0$

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

$f\left(x_n\right)=x_n^3-10$

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

(1) $x_1=x_0-\frac{x_0^3-10}{3x_0^2}$ Take $x_0$ closer to $\sqrt[3]{310}$ and $x_0=2$

$x_1=2-\frac{{\left(2\right)}^3-10}{3{\left(2\right)}^2}=2.167$

(2) $x_2=x_1-\frac{x_1^3-10}{3x_1^2}=2.167-0.01249$

$x_2=2.154$

(3) $x_3=x_2-\frac{x_2^3-10}{3x_2^2}=2.154-(-0.00043)$

$=2.1544$

(4) $x_4=x_3-\frac{x_3^3-10}{3x_3^2}=2.1544-(-0.000034)$

$=2.15443$

Since value starts repeating

$\therefore$ $\sqrt[3]{310}=2.1544$