How to Substitute an array into an equation and solve it ?

7 vues (au cours des 30 derniers jours)
Kalpha.mc
Kalpha.mc le 15 Nov 2020
Modifié(e) : Ameer Hamza le 16 Nov 2020
So if i had a = [1 2 3] b = [4 5 6] c = [7 8 9]
and the equation is ax^2+bx+c = 0
How would i substitude the array and solve for x ?

Réponses (1)

Ameer Hamza
Ameer Hamza le 15 Nov 2020
Modifié(e) : Ameer Hamza le 15 Nov 2020
For the given value of a, b, and c, the equation have no real solutions. Following shows how to use fsolve() when the equations have solutions
a = [1 2 3];
b = [4 5 6];
c = [-7 -8 -9];
F = @(x) a(:).*x.^2+b(:).*x+c(:);
sol = fsolve(F, rand(3,1))
If you don't have optimization toolbox, then you can use fzero()
a = [1 2 3];
b = [4 5 6];
c = [-7 -8 -9];
sol = zeros(size(a));
for i=1:numel(a)
F = @(x) a(i).*x.^2+b(i).*x+c(i);
sol(i) = fzero(F, rand());
end
  2 commentaires
Ameer Hamza
Ameer Hamza le 15 Nov 2020
No, you are not using subs() correct. Following shows how it needs to be done
syms x a b c
C = a * (x.^2) + (b * x) + c == 0;
x1 = subs(C,{a,b,c},{1,3,5});
solution = double(solve(x1,x));
Ameer Hamza
Ameer Hamza le 16 Nov 2020
Modifié(e) : Ameer Hamza le 16 Nov 2020
If a, b, and c are arrays
syms x a b c
av = [1 2 3];
bv = [4 5 6];
cv = [7 8 9];
C = a * (x.^2) + (b * x) + c == 0;
solution = cell(size(av));
for i = 1:numel(av)
x1 = subs(C,{a,b,c},{av(i),bv(i),cv(i)});
solution{i} = double(solve(x1,x));
end
solution is a cell array, where each cell contain solution for a combination of a, b, and c.

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