

roots of weighted summations of real polynomials - aswanson

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&#60;a&#60;0) is also stable?  Intuitively it seems that would be the case. This is related to a practical idea.
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shunted
Consider 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\.

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aswanson
_Consider 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.

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shunted
I 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
He probably meant that the sequence of partial sums should be stable, as a
whole.

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willchang
Nope. 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.

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aswanson
Those original functions roots are not bounded by 1.

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willchang
[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>)

What's your definition of stable?

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aswanson
I meant the z-plane not s plane but it doesn't matter.

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aswanson
Guys, 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.

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pixcavator
Maybe after all of these answers it's time to fix the question?

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pixcavator
f+(-f)=0.

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aswanson
That 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 a1 _f1+a2_ f2 are bounded as well provided that a1,a2 lie between 0
and 1.

~~~
pixcavator
Suppose 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 a1 _f1+a2_ f2 =
0.

