Answer to Question #139399 in Quantitative Methods for elle

Question

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

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

Answer 1

$$2x^5+x^4-2x-1=0$$ $$x^4(2x+1)-(2x+1)=0$$ $$(2x+1)(x^4-1)=0$$ $$(2x+1)(x^2+1)(x^2-1)=0$$ $$(2x+1)(x^2+1)(x+1)(x-1)=0$$

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

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

$$f'(x)=10x^4+4x^3-2$$

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

Initial solution $x_0=0$

$$f(0)=-1, f'(0)=-2, x_1=0-{-1 \over -2}=-0.5$$ $$\begin{matrix} n & x_n & f(x_n) & f'(x_n) & x_{n+1} & f(x_{n+1}) \\ 1 & -0.5 & 0 & -1.875 \end{matrix}$$

$x=-0.5$

Initial solution $x_0=2$

$$f(2)=75, f'(0)=190, x_1=2-{75 \over190}=1.60526$$ $$\begin{matrix} n & x_n & f(x_n) & x_{n+1} & f(x_{n+1}) \\ 1 & 1.60526 & 23.74820 & 1.31189 & 7.10986 \\ 2 & 1.31189 & 7.10986 & 1.11790 & 1.81772 \\ 3 & 1.11790 & 1.81772 & 1.02326 & 0.29343 \\ 4 & 1.02326 & 0.29343 & 1.00111 & 0.01332 \\ 5 & 1.00111 & 0.01332 & 1.00000 & 0.00003\\ 6 & 1.00000 & 0 & 1.00000 \end{matrix}$$

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

$$\varepsilon =\big|\dfrac{1.31189-1.60526}{1.60526}\big|\cdot 100\%=18.26\%$$

$$\varepsilon =\big|\dfrac{1.11790-1.31189}{1.31189}\big|\cdot 100\%=14.79\%$$

$$\varepsilon =\big|\dfrac{1.02326-1.11790}{1.11790}\big|\cdot 100\%=8.47\%$$

$$\varepsilon =\big|\dfrac{1.00111-1.02326}{1.02326}\big|\cdot 100\%=2.16\%$$

$$\varepsilon =\big|\dfrac{1.00000-1.00111}{1.00111}\big|\cdot 100\%=0.11\%$$

$$\varepsilon =\big|\dfrac{1.00000-1.00000}{1.00000}\big|\cdot 100\%=0\%$$

$x=1$