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Example-1
1. Find a root of an equation `f(x)=x^3-x-1` using Muller's method

Solution:
Here `x^3-x-1=0`

Let `f(x) = x^3-x-1`

Here

`x`	0	1	2
`f(x)`	-1	-1	5

`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

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