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//Eg-4.3
//pg-145
clear
clc
A=[-3 0 1];
[r,c] = size(A);
n = r;
xl=1;
xu=3;
fx=poly(A,'x','c')
// using formula |xnew^n-alpha|<=xhigh-xlow/2^n
xhigh=3;
xlow=1;
t = 1+ (6*2.303/log(2));
disp(t)
printf('Analytically 2^(1-n)<10^(-6), solving that implies n > %f\n so n = %d',t,t+1)
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