Indexing polynomials using 'for' loop

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I need help about an exercise that I've been given from my professor. Six polynomials are given:
  • p1 = [0 2 0 -2 6 -1]; % the first polynomial is p1(x)=2x^4-2x^2+6x-1
  • p2 = [1 3 0 -2 0 0]; % the second polynomial is p2(x)=x^5+3x^4-5x^2
  • 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];
Using 'polyval(p(i),2/3)' function I should create new vector 'A' that contains all the values of the polynomials from 'p1' to 'p6'(i=1:6) for x=2/3. Then find the minimum and maximum value of the vector. I want to know how can I loop through the polynomials to find their value and put that value in the new vector. Thank you in advance.

Risposta accettata

Mohammad Abouali
Mohammad Abouali il 25 Apr 2015
Modificato: Mohammad Abouali il 25 Apr 2015
%%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.
  3 Commenti
Mohammad Abouali
Mohammad Abouali il 25 Apr 2015
you are welcome
Stephen23
Stephen23 il 26 Apr 2015
Modificato: Stephen23 il 26 Apr 2015
@Blazo Arsoski: and please note the advice that solution one is "not recommended" dues to relying on eval. Read more to know why:

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