Find a root of an equation `f(x)=2x^3-2x-5` using Ridder'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_1=1` and `x_2=2`

`f_1=-5 < 0` and `f_2= 7 > 0`


`1^(st)` iteration :

`x_3=0.5(x_1+x_2)=0.5(1+2)=1.5`

`f_3=2*1.5^3-2*1.5-5=-1.25`

`s=sqrt(f_3^2-f_1*f_2)=sqrt(-1.25^2-(-5)*7)=6.0467`

`x_4=x_3+-(x_3-x_1)*f_3/s`

Here, `f_1 < f_2`, So we use minus sign

`x_4=1.5-(1.5-1)*-1.25/6.0467`

`x_4=1.6034`

`f_4=2*1.6034^3-2*1.6034-5=0.037`

As the root lies in the interval `(x_3,x_4)`, we let

`x_1=x_3=1.5` and `f_1=f_3=-1.25`

`x_2=x_4=1.6034` and `f_2=f_4=0.037`

which are the starting point for the next iteration

`2^(nd)` iteration :

`x_3=0.5(x_1+x_2)=0.5(1.5+1.6034)=1.5517`

`f_3=2*1.5517^3-2*1.5517-5=-0.6314`

`s=sqrt(f_3^2-f_1*f_2)=sqrt(-0.6314^2-(-1.25)*0.037)=0.667`

`x_4=x_3+-(x_3-x_1)*f_3/s`

Here, `f_1 < f_2`, So we use minus sign

`x_4=1.5517-(1.5517-1.5)*-0.6314/0.667`

`x_4=1.6006`

`f_4=2*1.6006^3-2*1.6006-5=0`

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

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