Answer to Question #204175 in Quantitative Methods for Anunay

Question

Find a root of the following equations correct to three decimals using Newton's raphson method: x³ -5x +3=0

Answer 1

$$x^3-5x+3=0$$

$$f(x)=x^3-5x+3$$

$$f'(x)=3x^2-5$$

$$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}$$

$x_0=-2$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} step & x_0 & f(x_0) & x_1 & f(x_1) \\ \hline 1 & -2.0000 & 5 & -2.7143 & -3.4257 \\ \hdashline 2 & -2.7143 & -3.4257 & -2.5140 & -0.3187 \\ \hdashline 3 & -2.5140 & -0.3187 & -2.4912 & -0.0039 \\ \hdashline 4 & -2.4912 & -0.0039 & -2.4909 & 0.0000 \\ \hdashline 5 & -2.4909& -0.0000 & -2.4909 & 0.0000 \\ \hdashline \end{array}$$

$x_{I}=-2.491$

$x_0=1$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} step & x_0 & f(x_0) & x_1 & f(x_1) \\ \hline 1 & 1.0000 & -1.0000 & 0.5000 & 0.6250 \\ \hdashline 2 & 0.5000 & 0.6250 & 0.6471 & 0.0356 \\ \hdashline 3 & 0.6471 & 0.0356 & 0.6566 & 0.0002\\ \hdashline 4 & 0.6566 & 0.0002 & 0.6566 & 0.0002 \\ \hdashline \end{array}$$

$x_{II}=0.657$

$x_0=2$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} step & x_0 & f(x_0) & x_1 & f(x_1) \\ \hline 1 & 2.0000 & 1.0000 & 1.8571 & 0.1195 \\ \hdashline 2 & 1.8571 & 0.1195 & 1.8348 & 0.0028 \\ \hdashline 3 & 1.8348 & 0.0028 & 1.8342 & 0.0000\\ \hdashline 4 & 1.8342 & 0.0000 & 1.8342 & 0.0000 \\ \hdashline \end{array}$$

$x_{III}=1.834$