%%Solution 1: (not recommended)
p1 = [0 2 0 -2 6 -1];
p2 = [1 3 0 -2 0 0];
p3 = [0 0 3 1 0 -10];
p4 = [0 0 0 -4 16 -5];
p5 = [1 -1 0 1 0 -1];
p6 = [3 -12 0 0 0 7];
polyValue=zeros(6,1);
for i=1:6
polyValue(i)=eval(sprintf('polyval(p%d,2/3)',i));
end
fprintf('Solution 1:(not recommended)\nmin: %f, max: %f.\n',min(polyValue),max(polyValue));
%%Solution 2:
p = [0 2 0 -2 6 -1; ...
1 3 0 -2 0 0; ...
0 0 3 1 0 -10; ...
0 0 0 -4 16 -5; ...
1 -1 0 1 0 -1; ...
3 -12 0 0 0 7];
polyValue=zeros(6,1);
for i=1:6
polyValue(i)=polyval(p(i,:),2/3);
end
fprintf('Solution 2:\nmin: %f, max: %f.\n',min(polyValue),max(polyValue));
Once you run it you get this results:
Solution 1:(not recommended)
min: -8.666667, max: 5.024691.
Solution 2:
min: -8.666667, max: 5.024691.
As you can see the difference between the solutions is slightly how the polynomial coefficients are stored.
