Find a root of an equation `f(x)=x^3+2x^2+x-1` using Muller's method

Solution:
Here `x^3+2x^2+x-1=0`

Let `f(x) = x^3+2x^2+x-1`

Here

`x`	0	1
`f(x)`	-1	3

`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

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