Find a root of an equation `f(x)=2x^3-2x-5` using Muller's method

Solution:
Here `2x^3-2x-5=0`

Let `f(x) = 2x^3-2x-5`

Here

`x`	0	1	2
`f(x)`	-5	-5	7

`x_0 = 1`

`x_1 = 2`

`x_2 = 1.5`


`1^(st)` iteration :

`f(x_0)=f(1)=2*1^(3)-2*1-5=-5`

`f(x_1)=f(2)=2(2)^(3)-2(2)-5=7`

`f(x_2)=f(1.5)=2(1.5)^(3)-2(1.5)-5=-1.25`

`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=(7--5)/1=12`

`delta_2=(f(x_2)-f(x_1))/h_2=(-1.25-7)/-0.5=16.5`

`a=(delta_2-delta_1)/(h_2+h_1)=(16.5-12)/(-0.5+1)=9`

`b=a xx h_2 + d_2=9xx-0.5+16.5=12`

`c=f(x_2)=-1.25`

`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 -1.25)/(12 + sqrt(12^2 - 4xx 9 xx -1.25))`

`=1.5+(2.5)/(12 + sqrt(189))`

`=1.5+(2.5)/(12 + 13.74773)`

`=1.5971`

Relative percent error
`varepsilon_(a^1)=|(x_3-x_2)/x_3| xx 100%=|(1.5971-1.5)/1.5971| xx 100%=6.07953%`

Now,
`x_0=x_1=2`

`x_1=x_2=1.5`

`x_2=x_3=1.5971`


`2^(nd)` iteration :

`f(x_0)=f(2)=2(2)^(3)-2(2)-5=7`

`f(x_1)=f(1.5)=2(1.5)^(3)-2(1.5)-5=-1.25`

`f(x_2)=f(1.5971)=2(1.5971)^(3)-2(1.5971)-5=-0.04672`

`h_1=x_1-x_0=1.5-2=-0.5`

`h_2=x_2-x_1=1.5971-1.5=0.0971`

`delta_1=(f(x_1)-f(x_0))/h_1=(-1.25-7)/-0.5=16.5`

`delta_2=(f(x_2)-f(x_1))/h_2=(-0.04672--1.25)/0.0971=12.39272`

`a=(delta_2-delta_1)/(h_2+h_1)=(12.39272-16.5)/(0.0971+-0.5)=10.19419`

`b=a xx h_2 + d_2=10.19419xx0.0971+12.39272=13.38253`

`c=f(x_2)=-0.04672`

`x_3=x_2+(-2c)/(b +- sqrt(b^2-4ac))`

`x_3=x_2+(-2c)/(b +sign(b) sqrt(b^2-4ac))`

`=1.5971+(-2 xx -0.04672)/(13.38253 + sqrt(13.38253^2 - 4xx 10.19419 xx -0.04672))`

`=1.5971+(0.09343)/(13.38253 + sqrt(180.99718))`

`=1.5971+(0.09343)/(13.38253 + 13.45352)`

`=1.60058`

Relative percent error
`varepsilon_(a^2)=|(x_3-x_2)/x_3| xx 100%=|(1.60058-1.5971)/1.60058| xx 100%=0.21753%`

Now,
`x_0=x_1=1.5`

`x_1=x_2=1.5971`

`x_2=x_3=1.60058`


`3^(rd)` iteration :

`f(x_0)=f(1.5)=2(1.5)^(3)-2(1.5)-5=-1.25`

`f(x_1)=f(1.5971)=2(1.5971)^(3)-2(1.5971)-5=-0.04672`

`f(x_2)=f(1.60058)=2(1.60058)^(3)-2(1.60058)-5=-0.00028`

`h_1=x_1-x_0=1.5971-1.5=0.0971`

`h_2=x_2-x_1=1.60058-1.5971=0.00348`

`delta_1=(f(x_1)-f(x_0))/h_1=(-0.04672--1.25)/0.0971=12.39272`

`delta_2=(f(x_2)-f(x_1))/h_2=(-0.00028--0.04672)/0.00348=13.33768`

`a=(delta_2-delta_1)/(h_2+h_1)=(13.33768-12.39272)/(0.00348+0.0971)=9.39535`

`b=a xx h_2 + d_2=9.39535xx0.00348+13.33768=13.37039`

`c=f(x_2)=-0.00028`

`x_3=x_2+(-2c)/(b +- sqrt(b^2-4ac))`

`x_3=x_2+(-2c)/(b +sign(b) sqrt(b^2-4ac))`

`=1.60058+(-2 xx -0.00028)/(13.37039 + sqrt(13.37039^2 - 4xx 9.39535 xx -0.00028))`

`=1.60058+(0.00056)/(13.37039 + sqrt(178.7779))`

`=1.60058+(0.00056)/(13.37039 + 13.37079)`

`=1.6006`

Relative percent error
`varepsilon_(a^3)=|(x_3-x_2)/x_3| xx 100%=|(1.6006-1.60058)/1.6006| xx 100%=0.00131%`


Approximate root of the equation `2x^3-2x-5=0` using Muller method is `1.6006`

`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	-5	7	-1.25	9	12	-1.25	1.5971	6.07953
2	2	1.5	1.5971	7	-1.25	-0.04672	10.19419	13.38253	-0.04672	1.60058	0.21753
3	1.5	1.5971	1.60058	-1.25	-0.04672	-0.00028	9.39535	13.37039	-0.00028	1.6006	0.00131

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