Answer to Question #139992 in Quantitative Methods for xerin

Question

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

2x^5+x^4-2x-1=0

Answer 1

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

$$=(2x+1)(x^4-1)=(2x+1)(x^2-1)(x^2+1)=$$

$$=(2x+1)(x+1)(x-1)(x^2+1)$$

$x1=-1, x2=-\dfrac{1}{2}, x3=1$

We know that there is a solution for the equation $2x^5+x^4-2x-1=0$ in $[0.5,1.5].$

$$x=\big(\dfrac{-x^4+2x+1}{2}\big)^{1/5}$$

$$x_{i+1}=\big(\dfrac{-x_i^4+2x_i+1}{2}\big)^{1/5}$$

$$\begin{matrix} i & x_i \\ 0 & 0.5\\ 1 & 0.993670 \\ 2 & 1.001239\\ 3 & 0.999751\\ 4 & 1.000050\\ 5 & 0.999990\\ 6 & 1.000002\\ 7 &1.000000 \\ 8 & 1.000000\\ \end{matrix}$$

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

$$\varepsilon =\big|\dfrac{0.993670-0.5}{0.5}\big|\cdot 100\%=98.734\%$$

$$\varepsilon =\big|\dfrac{1.001239-0.993670}{0.993670}\big|\cdot 100\%=0.7617\%$$

$$\varepsilon =\big|\dfrac{0.999751-1.001239}{1.001239}\big|\cdot 100\%=0.1486\%$$

$$\varepsilon =\big|\dfrac{1.000050-0.999751}{0.999751}\big|\cdot 100\%=0.0299\%$$

$$\varepsilon =\big|\dfrac{0.999990-1.000050}{1.000050}\big|\cdot 100\%=0.0060\%$$

$$\varepsilon =\big|\dfrac{0.999998-0.999990}{0.999990}\big|\cdot 100\%=0.0016\%$$

$$\varepsilon =\big|\dfrac{1.000002-0.999998}{0.999998}\big|\cdot 100\%=0.0012\%$$

$$\varepsilon =\big|\dfrac{1.000000-1.000002}{1.000002}\big|\cdot 100\%=0.0002\%$$

If $\varepsilon\leq0.05\%,$ then $x=0.999998$

The root is $1.00000$