Filter löschen
Filter löschen

Indexing polynomials using 'for' loop

4 Ansichten (letzte 30 Tage)
Blazo Arsoski
Blazo Arsoski am 25 Apr. 2015
Bearbeitet: Stephen23 am 26 Apr. 2015
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.

Akzeptierte Antwort

Mohammad Abouali
Mohammad Abouali am 25 Apr. 2015
Bearbeitet: Mohammad Abouali am 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 Kommentare
Mohammad Abouali
Mohammad Abouali am 25 Apr. 2015
you are welcome
Stephen23
Stephen23 am 26 Apr. 2015
Bearbeitet: Stephen23 am 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:

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Polynomials finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by