Find a root of an equation `f(x)=x^3+2x^2+x-1` using Ridder's method
Solution:
Here `x^3+2x^2+x-1=0`
Let `f(x) = x^3+2x^2+x-1`
Here
`x_1=0` and `x_2=1`
`f_1=-1 < 0` and `f_2= 3 > 0`
`1^(st)` iteration :
`x_3=0.5(x_1+x_2)=0.5(0+1)=0.5`
`f_3=0.5^3+2*0.5^2+0.5-1=0.125`
`s=sqrt(f_3^2-f_1*f_2)=sqrt(0.125^2-(-1)*3)=1.7366`
`x_4=x_3+-(x_3-x_1)*f_3/s`
Here, `f_1 < f_2`, So we use minus sign
`x_4=0.5-(0.5-0)*0.125/1.7366`
`x_4=0.464`
`f_4=0.464^3+2*0.464^2+0.464-1=-0.0055`
As the root lies in the interval `(x_3,x_4)`, we let
`x_1=x_3=0.5` and `f_1=f_3=0.125`
`x_2=x_4=0.464` and `f_2=f_4=-0.0055`
which are the starting point for the next iteration
`2^(nd)` iteration :
`x_3=0.5(x_1+x_2)=0.5(0.5+0.464)=0.482`
`f_3=0.482^3+2*0.482^2+0.482-1=0.0586`
`s=sqrt(f_3^2-f_1*f_2)=sqrt(0.0586^2-0.125*(-0.0055))=0.0642`
`x_4=x_3+-(x_3-x_1)*f_3/s`
Here, `f_1 > f_2`, So we use plus sign
`x_4=0.482+(0.482-0.5)*0.0586/0.0642`
`x_4=0.4656`
`f_4=0.4656^3+2*0.4656^2+0.4656-1=0`
Approximate root of the equation `x^3+2x^2+x-1=0` using Ridder's method is `0.4656` (After 2 iterations)
This material is intended as a summary. Use your textbook for detail explanation.
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