Question

find the roots using simple fixed point iteration. 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

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

$$=(3x^4-3x^3) -(5x^3-5x^2)-(42x^2-42x)-(40x+40)=$$

$$=(x-1)(3x^3-5x^2-42x-40)=$$

$$=(x-1)((3x^3+6x^2)-(11x^2+22x)-(20x+40))=$$

$$=(x+2)(x-1)(3x^2-11x-20)=$$

$$=(x+2)(x-1)((3x^2-15x+(4x-20))=$$

$$=(x+2)(x-1)(x-5)(3x+4)$$

$x1=-2, x2=-\dfrac{4}{3}\approx-1.333, x3=1,x4=5$

We know that there is a solution for the equation $3x^4-8x^3-37x^2+2x+40=0$ in $[0,2].$

$$x=\big(\dfrac{3x^4-8x^3+2x+40}{37}\big)^{1/2}$$

$$x_{i+1}=\big(\dfrac{3x_i^4-8x_i^3+2x_i+40}{37}\big)^{1/2}$$

$$\begin{matrix} i & x_i \\ 0 & 0\\ 1 & 1.039750 \\ 2 & 0.994489\\ 3 & 1.000742\\ 4 & 0.999900\\ 5 & 1.000014\\ 6 & 0.999998\\ 7 & 1.000000 \\ 8 & 1.000000\\ \end{matrix}$$

$$\varepsilon =\big|\dfrac{n_{i+1}-n_i}{n_i}\big|\cdot 100\%$$

$$\varepsilon =\big|\dfrac{1.039750-0}{0}\big|\cdot 100\%=undefined$$

$$\varepsilon =\big|\dfrac{0.994489-1.039750}{1.039750}\big|\cdot 100\%=4.5512\%$$

$$\varepsilon =\big|\dfrac{1.000742-0.994489}{0.994489}\big|\cdot 100\%=0.6248\%$$

$$\varepsilon =\big|\dfrac{0.999900-1.000742}{1.000742}\big|\cdot 100\%=0.0841\%$$

$$\varepsilon =\big|\dfrac{1.000014-0.999900}{0.999900}\big|\cdot 100\%=0.0114\%$$

$$\varepsilon =\big|\dfrac{0.999998-1.000014}{1.000014}\big|\cdot 100\%=0.0016\%$$

$$\varepsilon =\big|\dfrac{1.000000-0.999998}{0.999998}\big|\cdot 100\%=0.0002\%$$

$x=1.000014$

The root is $1.000014$