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

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

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

`d/(dx)(2x^3-2x-5)=6x^2-2`

`d/(dx)(2x^3-2x-5)`

`=d/(dx)(2x^3)-d/(dx)(2x)-d/(dx)(5)`

`=6x^2-2-0`

`=6x^2-2`

`d/(dx)(6x^2-2)=12x`

`d/(dx)(6x^2-2)`

`=d/(dx)(6x^2)-d/(dx)(2)`

`=12x-0`

`=12x`

`:. f'(x) = 6x^2-2`

`:. f ''(x) = 12x`

Here

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

Here `f(1) = -5 < 0 and f(2) = 7 > 0`

`:.` Root lies between `1` and `2`

`x_0 = (1 + 2)/2 = 1.5`

`x_0 = 1.5`


`1^(st)` iteration :

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

`f'(x_0)=f'(1.5)=6*1.5^2-2=11.5`

`f''(x_0)=f'(1.5)=12*1.5=18`

`x_1=x_0-(2*f(x_0)*f'(x_0))/(2*f'(x_0)^2-f(x_0)*f''(x_0))`

`x_1=1.5-(2xx-1.25xx11.5)/(2xx(11.5)^2-(11.5)xx(18))`

`x_1=1.6002`


`2^(nd)` iteration :

`f(x_1)=f(1.6002)=2*1.6002^3-2*1.6002-5=-0.0057`

`f'(x_1)=f'(1.6002)=6*1.6002^2-2=13.3633`

`f''(x_1)=f'(1.6002)=12*1.6002=19.2021`

`x_2=x_1-(2*f(x_1)*f'(x_1))/(2*f'(x_1)^2-f(x_1)*f''(x_1))`

`x_2=1.6002-(2xx-0.0057xx13.3633)/(2xx(13.3633)^2-(13.3633)xx(19.2021))`

`x_2=1.6006`


`3^(rd)` iteration :

`f(x_2)=f(1.6006)=2*1.6006^3-2*1.6006-5=0`

`f'(x_2)=f'(1.6006)=6*1.6006^2-2=13.3715`

`f''(x_2)=f'(1.6006)=12*1.6006=19.2072`

`x_3=x_2-(2*f(x_2)*f'(x_2))/(2*f'(x_2)^2-f(x_2)*f''(x_2))`

`x_3=1.6006-(2xx0xx13.3715)/(2xx(13.3715)^2-(13.3715)xx(19.2072))`

`x_3=1.6006`


Approximate root of the equation `2x^3-2x-5=0` using Halleys method is `1.6006` (After 3 iterations)

`n`	`x_0`	`f(x_0)`	`f'(x_0)`	`f''(x_0)`	`x_1`	Update
1	1.5	-1.25	11.5	18	1.6002	`x_0 = x_1`
2	1.6002	-0.0057	13.3633	19.2021	1.6006	`x_0 = x_1`
3	1.6006	0	13.3715	19.2072	1.6006	`x_0 = x_1`

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