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//Example 22.3
//Bisection Method
//Page no. 764
clc;clear;close;
deff('y=f(x)','y=x^3-x^2+x-1')
printf('N01\tN02\tN11\tN12\tN21\tNet31\tO31\tN41\tN42\n------------------------------------------------------------------------\n')
N01=[0,1,0.5,0.75,0.875,0.938,0.969,0.984,0.992,0.996,0.998,0.999,1,1]
N02(1)=2
for i=2:13
N02(i)=1;
end
for i=1:13
net31(i)=f(N01(i+1))*f(N01(i))
if net31(i)>0 then
O31(i)=1;
else
O31(i)=0;
end
N41(i)=(1-O31(i))*(N01(i))+O31(i)*N01(i+1)
N42(i)=(1-O31(i))*N01(i+1)+O31(i)*N02(i)
if i==2 then
printf('%.3f\t%.3f\t%.3f\t%.3f\t%.3f\t%.3f\t%.3f\t%.3f\t%.3f\n',0,N02(i),f(N01(i)),N01(i+1),f(N01(i+1)),net31(i),O31(i),N41(i),N42(i))
else
printf('%.3f\t%.3f\t%.3f\t%.3f\t%.3f\t%.3f\t%.3f\t%.3f\t%.3f\n',N01(i),N02(i),f(N01(i)),N01(i+1),f(N01(i+1)),net31(i),O31(i),N41(i),N42(i))
end
end
printf('\n\nTherefore the solution is %.3f',N42(13))
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