Question

find the roots using newton raphson. Stop until at 15th iteration or when approximate percent relative error is below 0.05%.

3x^4-8x^3-37x^2+2x+40

Answer 1

$$f(x)=3x^4-8x^3-37x^2+2x+40$$

$$f'(x)=12x^3-24x^2-74x+2$$

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

Initial solution $x_0 =-0.5$

$$\begin{matrix} n & x_n & f(x_n) & x_{n+1} \\ 0 & -0.5 & 30.9375 & -1.482143\\ 1 & -1.482143 & -3.719695 & -1.295091 \\ 2 & -1.295091 & 1.168422 & -1.332165 \\ 3 & -1.332165 & 0.034568 & -1.333332 \\ 4 & -1.333332 & 0.000037 & -1.333333 \\ 5 & -1.333333 & 0.000000 & -1.333333 \\ \end{matrix}$$

$$\varepsilon =\big|\dfrac{x_{n+1}-x_n}{x_n}\big|\cdot 100\%$$

$$\varepsilon =\big|\dfrac{-1.482143+0.5}{-0.5}\big|\cdot 100\%=196.4286\%$$

$$\varepsilon =\big|\dfrac{ -1.295091+1.482143}{-1.482143}\big|\cdot 100\%=12.6204\%$$

$$\varepsilon =\big|\dfrac{-1.332165 +1.295091}{ -1.295091}\big|\cdot 100\%=2.8627\%$$

$$\varepsilon =\big|\dfrac{-1.333332+1.332165}{-1.332165}\big|\cdot 100\%=0.0876\%$$

$$\varepsilon =\big|\dfrac{-1.333333+1.333332}{-1.333332}\big|\cdot 100\%=0.000075\%$$

$x=-1.333333$

The root is $-1.333333$