Algorithm
Example-1
1. Find all roots of polynomial using Bairstow method
`f(x)=x^4-3x^3+3x^2-3x+2` and `r=0.1, s=0.1`
Solution:
`x^4-3x^3+3x^2-3x+2=0`
Let the initial approximation be `r=0.1` and `s=0.1`
Here `a_4=1`, `a_3=-3`, `a_2=3`, `a_1=-3`, `a_0=2`
Iteration=1
`b_4=a_4=1`
`b_3=a_3+rb_4=-3+0.1 * 1=-2.9`
`b_2=a_2+rb_3+sb_4=3 +0.1 * (-2.9) +0.1 * 1=2.81`
`b_1=a_1+rb_2+sb_3=-3 +0.1 * 2.81 +0.1 * (-2.9)=-3.009`
`b_0=a_0+rb_1+sb_2=2 +0.1 * (-3.009) +0.1 * 2.81=1.9801`
`c_4=b_4=1`
`c_3=b_3+rc_4=-2.9+0.1 * 1=-2.8`
`c_2=b_2+rc_3+sc_4=2.81 +0.1 * (-2.8) +0.1 * 1=2.63`
`c_1=b_1+rc_2+sc_3=-3.009 +0.1 * 2.63 +0.1 * (-2.8)=-3.026`
The simultaneous equations for `Deltar` and `Deltas` are
`c_2Deltar+c_3Deltas=-b_1` and `c_1Deltar+c_2Deltas=-b_0`
Substitute values of `c_1,c_2,c_3` and `b_0,b_1`
`2.63Deltar-2.8Deltas=3.009` and `-3.026Deltar+2.63Deltas=-1.9801`
Solving equations using Cramer's rule method
`D=c_2*c_2-c_1*c_3=2.63*2.63-(-3.026)*(-2.8)=-1.5559`
`D_1=b_0*c_3-b_1*c_2=1.9801*(-2.8)-(-3.009)*2.63=2.3694`
`D_2=b_1*c_1-b_0*c_2=(-3.009)*(-3.026)-1.9801*2.63=3.8976`
`Deltar=D_1/D=(2.3694)/(-1.5559)=-1.5228`
and `Deltas=D_2/D=(3.8976)/(-1.5559)=-2.505`
The new `r` and `s` are
`r=r+Deltar=0.1-1.5228=-1.4228`
and `s=s+Deltas=0.1-2.505=-2.405`
The approximate error in r and s
`|epsilon_(a,r)|=|(Deltar)/(r)|xx100%=|(-1.5228)/(-1.4228)|xx100%=107.0282`
`|epsilon_(a,s)|=|(Deltas)/(s)|xx100%=|(-2.505)/(-2.405)|xx100%=104.158`
Iteration=2
`b_4=a_4=1`
`b_3=a_3+rb_4=-3-1.4228 * 1=-4.4228`
`b_2=a_2+rb_3+sb_4=3 -1.4228 * (-4.4228) -2.405 * 1=6.888`
`b_1=a_1+rb_2+sb_3=-3 -1.4228 * 6.888 -2.405 * (-4.4228)=-2.1635`
`b_0=a_0+rb_1+sb_2=2 -1.4228 * (-2.1635) -2.405 * 6.888=-11.4875`
`c_4=b_4=1`
`c_3=b_3+rc_4=-4.4228-1.4228 * 1=-5.8457`
`c_2=b_2+rc_3+sc_4=6.888 -1.4228 * (-5.8457) -2.405 * 1=12.8004`
`c_1=b_1+rc_2+sc_3=-2.1635 -1.4228 * 12.8004 -2.405 * (-5.8457)=-6.3174`
The simultaneous equations for `Deltar` and `Deltas` are
`c_2Deltar+c_3Deltas=-b_1` and `c_1Deltar+c_2Deltas=-b_0`
Substitute values of `c_1,c_2,c_3` and `b_0,b_1`
`12.8004Deltar-5.8457Deltas=2.1635` and `-6.3174Deltar+12.8004Deltas=11.4875`
Solving equations using Cramer's rule method
`D=c_2*c_2-c_1*c_3=12.8004*12.8004-(-6.3174)*(-5.8457)=126.9215`
`D_1=b_0*c_3-b_1*c_2=(-11.4875)*(-5.8457)-(-2.1635)*12.8004=94.8456`
`D_2=b_1*c_1-b_0*c_2=(-2.1635)*(-6.3174)-(-11.4875)*12.8004=160.7128`
`Deltar=D_1/D=(94.8456)/(126.9215)=0.7473`
and `Deltas=D_2/D=(160.7128)/(126.9215)=1.2662`
The new `r` and `s` are
`r=r+Deltar=-1.4228+0.7473=-0.6756`
and `s=s+Deltas=-2.405+1.2662=-1.1388`
The approximate error in r and s
`|epsilon_(a,r)|=|(Deltar)/(r)|xx100%=|(0.7473)/(-0.6756)|xx100%=110.6152`
`|epsilon_(a,s)|=|(Deltas)/(s)|xx100%=|(1.2662)/(-1.1388)|xx100%=111.1915`
Iteration=3
`b_4=a_4=1`
This material is intended as a summary. Use your textbook for detail explanation.
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