1. Algorithm & Example-1 `f(x)=x^3-x-1`
Example-1
1. Find a root of an equation `f(x)=x^3-x-1` using Muller method
Solution:
Here `x^3-x-1=0`
Let `f(x) = x^3-x-1`
Here
`x_0 = 1`
`x_1 = 2`
`x_2 = 1.5`
`1^(st)` iteration :
`f(x_0)=f(1)=1^(3)-1-1=-1`
`f(x_1)=f(2)=(2)^(3)-2-1=5`
`f(x_2)=f(1.5)=(1.5)^(3)-1.5-1=0.875`
`h_1=x_1-x_0=2-1=1`
`h_2=x_2-x_1=1.5-2=-0.5`
`delta_1=(f(x_1)-f(x_0))/h_1=(5--1)/1=6`
`delta_2=(f(x_2)-f(x_1))/h_2=(0.875-5)/-0.5=8.25`
`a=(delta_2-delta_1)/(h_2+h_1)=(8.25-6)/(-0.5+1)=4.5`
`b=a xx h_2 + d_2=4.5xx-0.5+8.25=6`
`c=f(x_2)=0.875`
`x_3=x_2+(-2c)/(b +- sqrt(b^2-4ac))`
`x_3=x_2+(-2c)/(b +sign(b) sqrt(b^2-4ac))`
`=1.5+(-2 xx 0.875)/(6 + sqrt(6^2 - 4xx 4.5 xx 0.875))`
`=1.5+(-1.75)/(6 + sqrt(20.25))`
`=1.5+(-1.75)/(6 + 4.5)`
`=1.33333`
Relative percent error
`varepsilon_(a^1)=|(x_3-x_2)/x_3| xx 100%=|(1.33333-1.5)/1.33333| xx 100%=12.5%`
Now,
`x_0=x_1=2`
`x_1=x_2=1.5`
`x_2=x_3=1.33333`
`2^(nd)` iteration :
`f(x_0)=f(2)=(2)^(3)-2-1=5`
`f(x_1)=f(1.5)=(1.5)^(3)-1.5-1=0.875`
`f(x_2)=f(1.33333)=(1.33333)^(3)-1.33333-1=0.03704`
`h_1=x_1-x_0=1.5-2=-0.5`
`h_2=x_2-x_1=1.33333-1.5=-0.16667`
`delta_1=(f(x_1)-f(x_0))/h_1=(0.875-5)/-0.5=8.25`
`delta_2=(f(x_2)-f(x_1))/h_2=(0.03704-0.875)/-0.16667=5.02778`
`a=(delta_2-delta_1)/(h_2+h_1)=(5.02778-8.25)/(-0.16667+-0.5)=4.83333`
`b=a xx h_2 + d_2=4.83333xx-0.16667+5.02778=4.22222`
`c=f(x_2)=0.03704`
`x_3=x_2+(-2c)/(b +- sqrt(b^2-4ac))`
`x_3=x_2+(-2c)/(b +sign(b) sqrt(b^2-4ac))`
`=1.33333+(-2 xx 0.03704)/(4.22222 + sqrt(4.22222^2 - 4xx 4.83333 xx 0.03704))`
`=1.33333+(-0.07407)/(4.22222 + sqrt(17.11111))`
`=1.33333+(-0.07407)/(4.22222 + 4.13656)`
`=1.32447`
Relative percent error
`varepsilon_(a^2)=|(x_3-x_2)/x_3| xx 100%=|(1.32447-1.33333)/1.32447| xx 100%=0.66908%`
Now,
`x_0=x_1=1.5`
`x_1=x_2=1.33333`
`x_2=x_3=1.32447`
`3^(rd)` iteration :
`f(x_0)=f(1.5)=(1.5)^(3)-1.5-1=0.875`
`f(x_1)=f(1.33333)=(1.33333)^(3)-1.33333-1=0.03704`
`f(x_2)=f(1.32447)=(1.32447)^(3)-1.32447-1=-0.00105`
`h_1=x_1-x_0=1.33333-1.5=-0.16667`
`h_2=x_2-x_1=1.32447-1.33333=-0.00886`
`delta_1=(f(x_1)-f(x_0))/h_1=(0.03704-0.875)/-0.16667=5.02778`
`delta_2=(f(x_2)-f(x_1))/h_2=(-0.00105-0.03704)/-0.00886=4.29796`
`a=(delta_2-delta_1)/(h_2+h_1)=(4.29796-5.02778)/(-0.00886+-0.16667)=4.1578`
`b=a xx h_2 + d_2=4.1578xx-0.00886+4.29796=4.26112`
`c=f(x_2)=-0.00105`
`x_3=x_2+(-2c)/(b +- sqrt(b^2-4ac))`
`x_3=x_2+(-2c)/(b +sign(b) sqrt(b^2-4ac))`
`=1.32447+(-2 xx -0.00105)/(4.26112 + sqrt(4.26112^2 - 4xx 4.1578 xx -0.00105))`
`=1.32447+(0.0021)/(4.26112 + sqrt(18.17461))`
`=1.32447+(0.0021)/(4.26112 + 4.26317)`
`=1.32472`
Relative percent error
`varepsilon_(a^3)=|(x_3-x_2)/x_3| xx 100%=|(1.32472-1.32447)/1.32472| xx 100%=0.01861%`
Approximate root of the equation `x^3-x-1=0` using Muller method is `1.32472`
| `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 | 1 | 2 | 1.5 | -1 | 5 | 0.875 | 4.5 | 6 | 0.875 | 1.33333 | 12.5 | | 2 | 2 | 1.5 | 1.33333 | 5 | 0.875 | 0.03704 | 4.83333 | 4.22222 | 0.03704 | 1.32447 | 0.66908 | | 3 | 1.5 | 1.33333 | 1.32447 | 0.875 | 0.03704 | -0.00105 | 4.1578 | 4.26112 | -0.00105 | 1.32472 | 0.01861 |
This material is intended as a summary. Use your textbook for detail explanation.
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