4 ms·
roots of weighted summations of real polynomials
Does anyone know off hand if the summation of a set of real polynomials with roots bounded in the complex plane (stable), with each polynomial weighted by a real value a (1<a<0) is also stable? Intuitively it seems that would be the case. This is related to a practical idea.
- pixcavator 19y agof+(-f)=0.
- aswanson 19y agoThat was not the question. The question was if the roots of f1 and f2 (both of order n) are bounded by 1 in the complex plane, can it be proven that the roots of a1f1+a2f2 are bounded as well provided that a1,a2 lie between 0 and 1.
- pixcavator 19y agoSuppose such f is given that the roots of f (of order n) are bounded by 1 in the complex plane. Now let f1=f, f2=-f, a1=1/2, a2=1/2. Then a1f1+a2f2 = 0.
- shunted 19y agoConsider the set of polynomials {z^n} where n ranges from 0 to infinity using even integers. All the zeros of the polynomials in the set are zero and so stable. Consider the power series expansion of .5 Cos(z). It is a weighted sum of the polynomials of our set and the zeros of .5 Cos(z) are clearly not bounded. I use .5 Cos(z) so that the coefficient of each term of the power series expansion is a value that satisfies 0 < a < 1.
- aswanson 19y agoConsider the set of polynomials {z^n} where n ranges from 0 to infinity using even integers. All the zeros of the polynomials in the set are zero and so stable But consider the case where n=0,2. Then you have z^0+z^2, with a zero at +/- j. So that statement isn't totally true.
- shunted 19y agoI provided an example where the answer to the question posed is false. There are examples where it is true. Since a counterexample exists then the conjecture is considered false. A conjecture can be false even though there exist instances where it is true.
- pixcavator 19y agoHe probably meant that the sequence of partial sums should be stable, as a whole.
- willchang 19y agoNope. Easy counterexample: f1 = x^2 + 2x (roots: 0, -2) f2 = -x - 1 (root: -1) (f1 + f2) / 2 has one root which is 0.618 (the golden ratio) Now you have to tell us what your idea was.
- aswanson 19y agoThose original functions roots are not bounded by 1.
- pixcavator 19y agoReplace each x with 2x.
- willchang 19y ago[A real polynomial P is said to be stable if all its roots lie in the left half-plane.](http://mathworld.wolfram.com/StablePolynomial.html http://mathworld.wolfram.com/StablePolynomial.html) What's your definition of stable?
- aswanson 19y agoI meant the z-plane not s plane but it doesn't matter.
- pixcavator 19y agoMaybe after all of these answers it's time to fix the question?
- aswanson 19y agoGuys, you are totally right. I found a simple, 1st order counterexample. This was more an exercise in my wanting to believe it to be true than math. Sorry to waste your time.