Let f ( x ) = 2 x 4 + 6 x 2 + 8 f(x)=2x^4+6x^2+8 f ( x ) = 2 x 4 + 6 x 2 + 8
Use initial guesses x 0 = − 1 , x 1 = 0 , x 2 = 1 x_0=-1, x_1=0, x_2=1 x 0 = − 1 , x 1 = 0 , x 2 = 1
f ( x 0 ) = f ( − 1 ) = 16 f(x_0)=f(-1)=16 f ( x 0 ) = f ( − 1 ) = 16
f ( x 1 ) = f ( 0 ) = 8 f(x_1)=f(0)=8 f ( x 1 ) = f ( 0 ) = 8
f ( x 2 ) = f ( 1 ) = 16 f(x_2)=f(1)=16 f ( x 2 ) = f ( 1 ) = 16
h 0 = x 1 − x 0 = 0 − ( − 1 ) = 1 h_0=x_1-x_0=0-(-1)=1 h 0 = x 1 − x 0 = 0 − ( − 1 ) = 1
h 1 = x 2 − x 1 = 1 − 0 = 1 h_1=x_2-x_1=1-0=1 h 1 = x 2 − x 1 = 1 − 0 = 1
δ 0 = f ( x 1 ) − f ( x 0 ) h 0 = 8 − 16 1 = − 8 \delta_0=\dfrac{f(x_1)-f(x_0)}{h_0}=\dfrac{8-16}{1}=-8 δ 0 = h 0 f ( x 1 ) − f ( x 0 ) = 1 8 − 16 = − 8
δ 1 = f ( x 2 ) − f ( x 1 ) h 1 = 16 − 8 1 = 8 \delta_1=\dfrac{f(x_2)-f(x_1)}{h_1}=\dfrac{16-8}{1}=8 δ 1 = h 1 f ( x 2 ) − f ( x 1 ) = 1 16 − 8 = 8
a = δ 1 − δ 0 h 1 + h 0 = 8 − ( − 8 ) 1 + 1 = 8 a=\dfrac{\delta_1-\delta_0}{h_1+h_0}=\dfrac{8-(-8)}{1+1}=8 a = h 1 + h 0 δ 1 − δ 0 = 1 + 1 8 − ( − 8 ) = 8
b = a × h 1 + δ 1 = 8 × 1 + 8 = 16 b=a\times h_1+\delta_1=8\times1+8=16 b = a × h 1 + δ 1 = 8 × 1 + 8 = 16
c = f ( x 2 ) = f ( 1 ) = 16 c=f(x_2)=f(1)=16 c = f ( x 2 ) = f ( 1 ) = 16
x 3 = x 2 + − 2 c b ± b 2 − 4 a c x_3=x_2+\dfrac{-2c}{b\pm\sqrt{b^2-4ac}} x 3 = x 2 + b ± b 2 − 4 a c − 2 c
x 3 = 1 + − 2 ( 16 ) 16 + 1 6 2 − 4 ( 8 ) ( 16 ) = i x_3=1+\dfrac{-2(16)}{16+\sqrt{16^2-4(8)(16)}}=i x 3 = 1 + 16 + 1 6 2 − 4 ( 8 ) ( 16 ) − 2 ( 16 ) = i
x 0 = − 1 , x 1 = 0 , x 2 = i x_0=-1, x_1=0, x_2=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 ≈ 0.5 + 1.32288 i , x 3 ≈ − 0.5 + 1.32288 i . x_1\approx 0.5+1.32288i, x_3\approx-0.5+1.32288i. x 1 ≈ 0.5 + 1.32288 i , x 3 ≈ − 0.5 + 1.32288 i .
Two other roots can be determined as their conjugate pairs x 2 ≈ 0.5 − 1.32288 i , x_2\approx0.5-1.32288i, x 2 ≈ 0.5 − 1.32288 i , x 4 ≈ − 0.5 − 1.32288 i . x_4\approx-0.5-1.32288i. x 4 ≈ − 0.5 − 1.32288 i .
x 1 ≈ 0.5 + 1.32288 i x_1\approx 0.5+1.32288i x 1 ≈ 0.5 + 1.32288 i
x 2 ≈ 0.5 − 1.32288 i x_2\approx 0.5-1.32288i x 2 ≈ 0.5 − 1.32288 i
x 3 ≈ − 0.5 + 1.32288 i x_3\approx -0.5+1.32288i x 3 ≈ − 0.5 + 1.32288 i
x 4 ≈ − 0.5 − 1.32288 i x_4\approx -0.5-1.32288i x 4 ≈ − 0.5 − 1.32288 i
Comments