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 ${{x}_{n+1}}={{x}_{n}}-\dfrac{f\left( {{x}_{n}} \right)}{{{f}^{'}}\left( {{x}_{n}} \right)}$. 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 ${{f}^{'}}\left( x \right)$. We put $f\left( {{x}_{n}} \right),{{f}^{'}}\left( {{x}_{n}} \right)$ 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 ${{f}^{'}}\left( {{x}_{0}} \right)$ for the first iteration in the iterative formula
\[{{x}_{1}}={{x}_{o}}-\dfrac{f\left( {{x}_{0}} \right)}{{{f}^{'}}\left( {{x}_{0}} \right)}\]
 Here ${{x}_{1}}$ is the first approximation. Similarly the ${{\left( n+1 \right)}^{\text{th}}}$ approximation can be obtained in the ${{n}^{\text{th}}}$ iteration as;
\[{{x}_{n+1}}={{x}_{n}}-\dfrac{f\left( {{x}_{n}} \right)}{{{f}^{'}}\left( {{x}_{n}} \right)}\]
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 $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ for $n=3$ and differentiate $f\left( x \right)={{x}^{3}}-N$ with respect to $x$ to have;
\[\begin{align}
  & \dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( {{x}^{3}}-N \right) \\
 & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{d}{dx}{{x}^{3}}-\dfrac{d}{dx}N \\
 & \Rightarrow {{f}^{'}}\left( x \right)=3{{x}^{2}} \\
\end{align}\]
So the iteration formula to approximate the root is
\[\begin{align}
  & {{x}_{n+1}}={{x}_{n}}-\dfrac{f\left( {{x}_{n}} \right)}{{{f}^{'}}\left( {{x}_{n}} \right)} \\
 & \Rightarrow {{x}_{n+1}}={{x}_{n}}-\dfrac{x_{n}^{3}-N}{3x_{n}^{2}} \\
 & \Rightarrow {{x}_{n+1}}={{x}_{n}}-\dfrac{x_{n}^{3}}{3x_{n}^{2}}+\dfrac{N}{3x_{n}^{2}} \\
 & \Rightarrow {{x}_{n+1}}={{x}_{n}}-\dfrac{{{x}_{n}}}{3}+\dfrac{N}{3x_{n}^{2}} \\
 & \therefore {{x}_{n+1}}=\dfrac{2}{3}{{x}_{n}}+\dfrac{N}{3x_{n}^{2}} \\
\end{align}\]


Note: We note that we can apply Newton-Raphson method only when the function is differentiable in $\left( {{x}^{*}}-{{x}_{0}},{{x}^{*}}+{{x}_{0}} \right)$ where ${{x}^{*}}$ 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 $\displaystyle \lim_{n\to \infty }\left| \dfrac{{{x}_{n+1}}-{{x}^{*}}}{{{x}_{n}}-{{x}^{*}}} \right|=2$ which is faster than bisection, false position and secant approximation method.