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_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)
This material is intended as a summary. Use your textbook for detail explanation.
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