Solveeit Logo
Login

Question

Question: In Newton-Raphson's method write the formula for finding the cube root of the number \(N\). \[\]...

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

Explanation

Solution

We recall the iterative formula for root approximations using Newton-Raphson method as xn+1=xn−f(xn)f′(xn){{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 numberNN is the solution to the equationx3−N=0{{x}^{3}}-N=0. We take f(x)=x3−Nf\left( x \right)={{x}^{3}}-N and differentiate with respect to xx to find f′(x){{f}^{'}}\left( x \right). We put f(xn),f′(xn)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 xo{{x}_{o}} for a function f(x)f\left( x \right) as , the functional value at x0{{x}_{0}} as f(x0)f\left( {{x}_{0}} \right) but it also takes first derivative value f′(x0){{f}^{'}}\left( {{x}_{0}} \right) for the first iteration in the iterative formula
x1=xo−f(x0)f′(x0){{x}_{1}}={{x}_{o}}-\dfrac{f\left( {{x}_{0}} \right)}{{{f}^{'}}\left( {{x}_{0}} \right)}
Here x1{{x}_{1}} is the first approximation. Similarly the (n+1)th{{\left( n+1 \right)}^{\text{th}}} approximation can be obtained in the nth{{n}^{\text{th}}} iteration as;
xn+1=xn−f(xn)f′(xn){{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 numberNN. We know that real cube root of numberNNare the solutions of the equation x3=N{{x}^{3}}=N or x3−N=0{{x}^{3}}-N=0. Let us assume f(x)=x3−Nf\left( x \right)={{x}^{3}}-N. So now zeroes of the function f(x)f\left( x \right) are cube root of number NN.
We use the differential formula ddxxn=nxn−1\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} for n=3n=3 and differentiate f(x)=x3−Nf\left( x \right)={{x}^{3}}-N with respect to xx to have;

& \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{aligned}$$ So the iteration formula to approximate the root is $$\begin{aligned} & {{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{aligned}$$ **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.