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

`f_1=-1 < 0` and `f_2= 5 > 0`


`1^(st)` iteration :

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

`f_3=1.5^3-1.5-1=0.875`

`s=sqrt(f_3^2-f_1*f_2)=sqrt(0.875^2-(-1)*5)=2.4012`

`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)*0.875/2.4012`

`x_4=1.3178`

`f_4=1.3178^3-1.3178-1=-0.0293`

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

`x_1=x_3=1.5` and `f_1=f_3=0.875`

`x_2=x_4=1.3178` and `f_2=f_4=-0.0293`

which are the starting point for the next iteration

`2^(nd)` iteration :

`x_3=0.5(x_1+x_2)=0.5(1.5+1.3178)=1.4089`

`f_3=1.4089^3-1.4089-1=0.3878`

`s=sqrt(f_3^2-f_1*f_2)=sqrt(0.3878^2-0.875*(-0.0293))=0.4195`

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

Here, `f_1 > f_2`, So we use plus sign

`x_4=1.4089+(1.4089-1.5)*0.3878/0.4195`

`x_4=1.3247`

`f_4=1.3247^3-1.3247-1=0`

Approximate root of the equation `x^3-x-1=0` using Ridder's method is `1.3247` (After 2 iterations)

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