Answer to Question #287596 in Quantitative Methods for beki

Question

approximate the real root to four decimal places of x3+5x-3=0 (newton raphson method)

Answer 1

Solution:

Here $x^{3}+5 x-3=0$

Let $f(x)=x^{3}+5 x-3$

$\\ \\\frac{d}{d x}\left(x^{3}+5 x-3\right)=3 x^{2}+5 \therefore f(x)=3 x^{2}+5$

Here

Here $f(0)=-3<0\ and\ f(1)=3>0$

$\therefore$ Root lies between 0 and 1

$x_{0}=\frac{0+1}{2}=0.5 \\x_{0}=0.5$

$1^{s t}$ iteration :

$f\left(x_{0}\right)=f(0.5)=0.5^{3}+5 \cdot 0.5-3=-0.375 \\f^{\prime}\left(x_{0}\right)=f(0.5)=3 \cdot 0.5^{2}+5=5.75$

$\begin{aligned} &x_{1}=x_{0}-\frac{f\left(x_{0}\right)}{f\left(x_{0}\right)} \\ &x_{1}=0.5-\frac{-0.375}{5.75} \\ &x_{1}=0.5652 \end{aligned}$

$2^{n d}$ iteration :

$\begin{aligned} &f\left(x_{1}\right)=f(0.5652)=0.5652^{3}+5 \cdot 0.5652-3=0.0067 \\ &f^{\prime}\left(x_{1}\right)=f(0.5652)=3 \cdot 0.5652^{2}+5=5.9584 \end{aligned}$

$\begin{aligned} &x_{2}=x_{1}-\frac{f\left(x_{1}\right)}{f\left(x_{1}\right)} \\ &x_{2}=0.5652-\frac{0.0067}{5.9584} \\ &x_{2}=0.5641 \end{aligned}$

$3^{\text {rd }}$ iteration :

$f\left(x_{2}\right)=f(0.5641)=0.5641^{3}+5 \cdot 0.5641-3=0$

$\begin{aligned} f^{\prime}\left(x_{2}\right)=f(0.5641)=3 \cdot 0.5641^{2}+5=5.9546 \\ x_{3}=x_{2}-\frac{f\left(x_{2}\right)}{f\left(x_{2}\right)} \\ x_{3}=0.5641-\frac{0}{5.9546} \\ x_{3}=0.5641 \\ \text { Approximate root of the equation } x^{3}+5 x-3=0 \\\text { using Newton Raphson mehtod is } 0.5641 \text { (After } 3 \text { iterations) } \end{aligned}$