Answer to Question #142402 in Quantitative Methods for Usman

Question

Use Mu ̈ller’s method to determine the real and complex roots of
f(x)=2x^4 + 6x^2 + 8

Answer 1

Let $f(x)=2x^4+6x^2+8$

Use initial guesses $x_0=-1, x_1=0, x_2=1$

$f(x_0)=f(-1)=16$

$f(x_1)=f(0)=8$

$f(x_2)=f(1)=16$

$h_0=x_1-x_0=0-(-1)=1$

$h_1=x_2-x_1=1-0=1$

$\delta_0=\dfrac{f(x_1)-f(x_0)}{h_0}=\dfrac{8-16}{1}=-8$

$\delta_1=\dfrac{f(x_2)-f(x_1)}{h_1}=\dfrac{16-8}{1}=8$

$a=\dfrac{\delta_1-\delta_0}{h_1+h_0}=\dfrac{8-(-8)}{1+1}=8$

$b=a\times h_1+\delta_1=8\times1+8=16$

$c=f(x_2)=f(1)=16$

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

$$x_3=1+\dfrac{-2(16)}{16+\sqrt{16^2-4(8)(16)}}=i$$

$x_0=-1, x_1=0, x_2=i$

Calling the function Muller with different parameters yields two complex roots of the equation $x_1\approx 0.5+1.32288i, x_3\approx-0.5+1.32288i.$

Two other roots can be determined as their conjugate pairs $x_2\approx0.5-1.32288i,$  $x_4\approx-0.5-1.32288i.$

$$x_1\approx 0.5+1.32288i$$

$$x_2\approx 0.5-1.32288i$$

$$x_3\approx -0.5+1.32288i$$

$$x_4\approx -0.5-1.32288i$$