Question

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

Answer 1

Hint: We recall the iterative formula for root approximations using Newton-Raphson method as $x_{n+1}=x_n-{\displaystyle \frac{f\left(x_n\right)}{f^{{}^{\prime }}\left(x_n\right)}}$. We use the fact that the real cube root of number$N$ is the solution to the equation$x^3-N=0$. We take $f\left(x\right)=x^3-N$ and differentiate with respect to $x$ to find $f^{{}^{\prime }}\left(x\right)$. We put $f\left(x_n\right),f^{{}^{\prime }}\left(x_n\right)$ in the iterative formula to get the required formula. $$$$

Complete step by step answer:
We know that Newton-Raphson is root finding algorithm which produces successively better approximations for roots or zero of real valued function. Like all root finding algorithms it takes initial guess root $x_o$ for a function $f\left(x\right)$ as , the functional value at $x_0$ as $f\left(x_0\right)$ but it also takes first derivative value $f^{{}^{\prime }}\left(x_0\right)$ for the first iteration in the iterative formula
$$x_1=x_o-{\displaystyle \frac{f\left(x_0\right)}{f^{{}^{\prime }}\left(x_0\right)}}$$
 Here $x_1$ is the first approximation. Similarly the ${(n+1)}^{\text{th}}$ approximation can be obtained in the $n^{\text{th}}$ iteration as;
$$x_{n+1}=x_n-{\displaystyle \frac{f\left(x_n\right)}{f^{{}^{\prime }}\left(x_n\right)}}$$
We are asked to find the formula for finding cube root of the number$N$. We know that real cube root of number$N$are the solutions of the equation $x^3=N$ or $x^3-N=0$. Let us assume $f\left(x\right)=x^3-N$. So now zeroes of the function $f\left(x\right)$ are cube root of number $N$.
We use the differential formula ${\displaystyle \frac{d}{dx}}{x}^n=n{x}^{n-1}$ for $n=3$ and differentiate $f\left(x\right)=x^3-N$ with respect to $x$ to have;
$$\begin{array}{rl}& {\displaystyle \frac{d}{dx}}f\left(x\right)={\displaystyle \frac{d}{dx}}(x^3-N)\\ & \Rightarrow f^{{}^{\prime }}\left(x\right)={\displaystyle \frac{d}{dx}}{x}^3-{\displaystyle \frac{d}{dx}}N\\ & \Rightarrow f^{{}^{\prime }}\left(x\right)=3x^2\end{array}$$
So the iteration formula to approximate the root is
$$\begin{array}{rl}& x_{n+1}=x_n-{\displaystyle \frac{f\left(x_n\right)}{f^{{}^{\prime }}\left(x_n\right)}}\\ & \Rightarrow x_{n+1}=x_n-{\displaystyle \frac{x_n^3-N}{3x_n^2}}\\ & \Rightarrow x_{n+1}=x_n-{\displaystyle \frac{x_n^3}{3x_n^2}}+{\displaystyle \frac{N}{3x_n^2}}\\ & \Rightarrow x_{n+1}=x_n-{\displaystyle \frac{x_n}{3}}+{\displaystyle \frac{N}{3x_n^2}}\\ & \therefore x_{n+1}={\displaystyle \frac{2}{3}}{x}_n+{\displaystyle \frac{N}{3x_n^2}}\end{array}$$


Note: We note that we can apply Newton-Raphson method only when the function is differentiable in $(x^{\ast }-x_0,x^{\ast }+x_0)$ where $x^{\ast }$ is the exact root of the function $f\left(x\right)$. We cannot find the complex roots of $N$ or more than one root using Newton-Raphson method. The rate of convergence of Newton-Raphson method is quadratic which means ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|{\displaystyle \frac{x_{n+1}-x^{\ast }}{x_n-x^{\ast }}}\right|=2}$ which is faster than bisection, false position and secant approximation method.