How to calculate roots for multiple polynomial equations simultaneously i.e. without iterating over them one by one
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Earth Sugandhi
el 30 de Mayo de 2021
Comentada: John D'Errico
el 30 de Mayo de 2021
Actually I have a bunch of quadratic equations(around 1 million equations!!) which I need to solve. I made a matrix of 1 million rows with each row is a vector containing coeff(s) for x^2, x^1 & x^0. i named this matrix M and wrote following code:
answers = zeros(1000000,2);
for i=1:1:length(answers)
answers(i,:) = roots(M(i,:));
end
I was wondering if there's a way we can calculate roots simultaneously for every polynomial equation without iterating over them one by one.
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John D'Errico
el 30 de Mayo de 2021
Editada: John D'Errico
el 30 de Mayo de 2021
In fact, the simple answer is YES. It is trivial. No loop required.
You have a set of QUADRATIC EQUATIONS!!!!!! Surely you learned the quadratic formula for the roots of a quadratic equation?
tic
N = 1e6;
coef = rand(N,3);
D = sqrt(coef(:,2).^2 - 4*coef(:,1).*coef(:,3));
R = [(-coef(:,2) + D)./(2*coef(:,1)), (-coef(:,2) - D)./(2*coef(:,1))];
toc
Elapsed time is 0.115124 seconds.
size(R)
ans =
1000000 2
One million sets of roots, computed in around 1/10 of a second. Since my coefficients are randomly generated, many of those roots will be complex. As a test to verify it worked...
coef(1,:)
ans =
0.562209637790579 0.0918349300613903 0.952524012642822
R(1,:)
ans =
-0.0816732086115523 + 1.29906892840699i -0.0816732086115523 - 1.29906892840699i
roots(coef(1,:))
ans =
-0.0816732086115523 + 1.29906892840699i
-0.0816732086115523 - 1.29906892840699i
Hopefully your quadratics will be better behaved, and have real roots.
2 comentarios
John D'Errico
el 30 de Mayo de 2021
Some years ago, we realized we needed to do the same thing for zillions of cubic polynomials. So we wrote up the cubic formula, fully vectorized. Worked nicely.
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