How to calculate roots for multiple polynomial equations simultaneously i.e. without iterating over them one by one
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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
2021-5-30
编辑:John D'Errico
2021-5-30
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.
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John D'Errico
2021-5-30
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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