Find a root of an equation f(x)=x^3+2x^2+x-1 using Muller method
Solution:
Here x^3+2x^2+x-1=0
Let f(x) = x^3+2x^2+x-1
Here
x_0 = 0
x_1 = 1
x_2 = 0.5
1^(st) iteration :
f(x_0)=f(0)=0^3+2*0^2+0-1=-1
f(x_1)=f(1)=1^3+2*1^2+1-1=3
f(x_2)=f(0.5)=0.5^3+2*0.5^2+0.5-1=0.125
h_1=x_1-x_0=1-0=1
h_2=x_2-x_1=0.5-1=-0.5
delta_1=(f(x_1)-f(x_0))/h_1=(3--1)/1=4
delta_2=(f(x_2)-f(x_1))/h_2=(0.125-3)/-0.5=5.75
a=(delta_2-delta_1)/(h_2+h_1)=(5.75-4)/(-0.5+1)=3.5
b=a xx h_2 + d_2=3.5xx-0.5+5.75=4
c=f(x_2)=0.125
x_3=x_2+(-2c)/(b +- sqrt(b^2-4ac))
x_3=x_2+(-2c)/(b +sign(b) sqrt(b^2-4ac))
=0.5+(-2 xx 0.125)/(4 + sqrt(4^2 - 4xx 3.5 xx 0.125))
=0.5+(-0.25)/(4 + sqrt(14.25))
=0.5+(-0.25)/(4 + 3.7749)
=0.4678
Relative percent error
varepsilon_(a^1)=|(x_3-x_2)/x_3| xx 100%=|(0.4678-0.5)/0.4678| xx 100%=6.8729%
Now,
x_0=x_1=1
x_1=x_2=0.5
x_2=x_3=0.4678
2^(nd) iteration :
f(x_0)=f(1)=1^3+2*1^2+1-1=3
f(x_1)=f(0.5)=0.5^3+2*0.5^2+0.5-1=0.125
f(x_2)=f(0.4678)=0.4678^3+2*0.4678^2+0.4678-1=0.008
h_1=x_1-x_0=0.5-1=-0.5
h_2=x_2-x_1=0.4678-0.5=-0.0322
delta_1=(f(x_1)-f(x_0))/h_1=(0.125-3)/-0.5=5.75
delta_2=(f(x_2)-f(x_1))/h_2=(0.008-0.125)/-0.0322=3.6385
a=(delta_2-delta_1)/(h_2+h_1)=(3.6385-5.75)/(-0.0322+-0.5)=3.9678
b=a xx h_2 + d_2=3.9678xx-0.0322+3.6385=3.5109
c=f(x_2)=0.008
x_3=x_2+(-2c)/(b +- sqrt(b^2-4ac))
x_3=x_2+(-2c)/(b +sign(b) sqrt(b^2-4ac))
=0.4678+(-2 xx 0.008)/(3.5109 + sqrt(3.5109^2 - 4xx 3.9678 xx 0.008))
=0.4678+(-0.016)/(3.5109 + sqrt(12.1994))
=0.4678+(-0.016)/(3.5109 + 3.4928)
=0.4656
Relative percent error
varepsilon_(a^2)=|(x_3-x_2)/x_3| xx 100%=|(0.4656-0.4678)/0.4656| xx 100%=0.491%
Now,
x_0=x_1=0.5
x_1=x_2=0.4678
x_2=x_3=0.4656
3^(rd) iteration :
f(x_0)=f(0.5)=0.5^3+2*0.5^2+0.5-1=0.125
f(x_1)=f(0.4678)=0.4678^3+2*0.4678^2+0.4678-1=0.008
f(x_2)=f(0.4656)=0.4656^3+2*0.4656^2+0.4656-1=0
h_1=x_1-x_0=0.4678-0.5=-0.0322
h_2=x_2-x_1=0.4656-0.4678=-0.0023
delta_1=(f(x_1)-f(x_0))/h_1=(0.008-0.125)/-0.0322=3.6385
delta_2=(f(x_2)-f(x_1))/h_2=(0-0.008)/-0.0023=3.5202
a=(delta_2-delta_1)/(h_2+h_1)=(3.5202-3.6385)/(-0.0023+-0.0322)=3.4334
b=a xx h_2 + d_2=3.4334xx-0.0023+3.5202=3.5124
c=f(x_2)=0
x_3=x_2+(-2c)/(b +- sqrt(b^2-4ac))
x_3=x_2+(-2c)/(b +sign(b) sqrt(b^2-4ac))
=0.4656+(-2 xx 0)/(3.5124 + sqrt(3.5124^2 - 4xx 3.4334 xx 0))
=0.4656+(0.0001)/(3.5124 + sqrt(12.3375))
=0.4656+(0.0001)/(3.5124 + 3.5125)
=0.4656
Relative percent error
varepsilon_(a^3)=|(x_3-x_2)/x_3| xx 100%=|(0.4656-0.4656)/0.4656| xx 100%=0.0026%
Approximate root of the equation x^3+2x^2+x-1=0 using Muller method is 0.4656 (After 3 iterations)
| n | x_0 | x_1 | x_2 | f(x_0) | f(x_1) | f(x_2) | a | b | c | x_3 | varepsilon_(a^n |
| 1 | 0 | 1 | 0.5 | -1 | 3 | 0.125 | 3.5 | 4 | 0.125 | 0.4678 | 6.8729 |
| 2 | 1 | 0.5 | 0.4678 | 3 | 0.125 | 0.008 | 3.9678 | 3.5109 | 0.008 | 0.4656 | 0.491 |
| 3 | 0.5 | 0.4678 | 0.4656 | 0.125 | 0.008 | 0 | 3.4334 | 3.5124 | 0 | 0.4656 | 0.0026 |
This material is intended as a summary. Use your textbook for detail explanation.
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