Answer to Question #142401 in Quantitative Methods for Usman

Question

Use Mu ̈ller’s method to determine the real and complex roots of
f (x) = x^3− x^2 + 2x − 2

Answer 1

Let $f(x)=x^3-x^2+2x-2$

Use initial guesses $x_0=-2, x_1=0, x_2=2$

$f(x_0)=f(-2)=-18$

$f(x_1)=f(0)=-2$

$f(x_2)=f(2)=6$

$h_0=x_1-x_0=0-(-2)=2$

$h_1=x_2-x_1=2-0=2$

$\delta_0=\dfrac{f(x_1)-f(x_0)}{h_0}=\dfrac{-2-(-18)}{2}=8$ $\delta_1=\dfrac{f(x_2)-f(x_1)}{h_1}=\dfrac{6-(-2)}{2}=4$

$a=\dfrac{\delta_1-\delta_0}{h_1+h_0}=\dfrac{4-8}{2+2}=-1$

$b=a\times h_1+\delta_1=-1\times2+4=2$

$c=f(x_2)=f(2)=6$

$$x_3=x_2+\dfrac{-2c}{b\pm\sqrt{b^2-4ac}}$$

$$x_3=2+\dfrac{-2(6)}{2+\sqrt{2^2-4(-1)(6)}}=0.35424869$$

Relative percent error

$$\varepsilon_{a^1}=|\dfrac{x_3-x_2}{x_3}|\cdot100\%$$

$$\begin{matrix} n & 1 & 2 & 3 \\ x_0 & -2 & 0 & 2 \\ x_1 & 0 & 2 & 0.354249 \\ x_2 & 2 & 0.354249 & 0.828618 \\ f(x_0) & -18 & -2 & 6 & \\ f(x_1) & -2 & 6 & -1.3725396 \\ f(x_2) & 6 & -1.3725396 & -0.460436 \\ a & -1 & 1.354249 & 2.182867 \\ b & 2 & 2.250984 & 2.958255 \\ c & 6 & -1.3725396 & -0.460436 \\ x_3 & 0.354249 & 0.828618 & 0.969597 \\ \varepsilon_{a^n} & 464.575131 & 57.248251 & 14.539952 \end{matrix}$$

$$\begin{matrix} n & 4 & 5 & 6 \\ x_0 & 0.354249 & 0.828618 & 0.969597 \\ x_1 & 0.828618 & 0.969597 & 1.001174 \\ x_2 & 0.969597 & 1.001174 & 0.999998 \\ f(x_0) & -1.372539 & -0.460436 & -0.089389 \\ f(x_1) & -0.460436 & -0.089389 & 0.003525 \\ f(x_2) & -0.089389 & 0.003525 & -0.000006 \\ a & 1.152467 & 1.799389 & 1.970769\\ b & 2.794410 & 2.999252 & 3.000028 \\ c & -0.089389 & 0.003525 & -0.000006 \\ x_3 & 1.001174 & 0.999998 & 1 \\ \varepsilon_{a^n} & 3.154012 & 0.117612 & 0.000204 \end{matrix}$$

Approximate root of the equation $x^3-x^2+2x-2=0$ using Muller method is $1$ (After 6 iterations)