Question #142402
Use Mu ̈ller’s method to determine the real and complex roots of
f(x)=2x^4 + 6x^2 + 8
1
Expert's answer
2020-11-05T13:57:32-0500

Let f(x)=2x4+6x2+8f(x)=2x^4+6x^2+8

Use initial guesses x0=1,x1=0,x2=1x_0=-1, x_1=0, x_2=1

f(x0)=f(1)=16f(x_0)=f(-1)=16

f(x1)=f(0)=8f(x_1)=f(0)=8

f(x2)=f(1)=16f(x_2)=f(1)=16

h0=x1x0=0(1)=1h_0=x_1-x_0=0-(-1)=1

h1=x2x1=10=1h_1=x_2-x_1=1-0=1

δ0=f(x1)f(x0)h0=8161=8\delta_0=\dfrac{f(x_1)-f(x_0)}{h_0}=\dfrac{8-16}{1}=-8

δ1=f(x2)f(x1)h1=1681=8\delta_1=\dfrac{f(x_2)-f(x_1)}{h_1}=\dfrac{16-8}{1}=8

a=δ1δ0h1+h0=8(8)1+1=8a=\dfrac{\delta_1-\delta_0}{h_1+h_0}=\dfrac{8-(-8)}{1+1}=8

b=a×h1+δ1=8×1+8=16b=a\times h_1+\delta_1=8\times1+8=16

c=f(x2)=f(1)=16c=f(x_2)=f(1)=16

x3=x2+2cb±b24acx_3=x_2+\dfrac{-2c}{b\pm\sqrt{b^2-4ac}}

x3=1+2(16)16+1624(8)(16)=ix_3=1+\dfrac{-2(16)}{16+\sqrt{16^2-4(8)(16)}}=i


x0=1,x1=0,x2=ix_0=-1, x_1=0, x_2=i

Calling the function Muller with different parameters yields two complex roots of the equation x10.5+1.32288i,x30.5+1.32288i.x_1\approx 0.5+1.32288i, x_3\approx-0.5+1.32288i.

Two other roots can be determined as their conjugate pairs x20.51.32288i,x_2\approx0.5-1.32288i,  x40.51.32288i.x_4\approx-0.5-1.32288i.


x10.5+1.32288ix_1\approx 0.5+1.32288i

x20.51.32288ix_2\approx 0.5-1.32288i


x30.5+1.32288ix_3\approx -0.5+1.32288i

x40.51.32288ix_4\approx -0.5-1.32288i


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