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I have a problem where I am trying to bound the roots of the sum S of two polynomials to be only in the right (positive) half of the complex plane: $S = P_n + Q_1$. There are some special conditions on the polynomials:

  • $P_n$ is a n-th order polynomial whose roots are all real and located in the right half of the complex plane. Finally, $P_n(0)>0$ for all n.

  • $Q_1$ is linear (a first order polynomial) with a real root that lies in between the roots of $P_n$. The derivative of $Q$ is always positive.

  • $|P_n(0)| > |Q_1(0)|$, $P_n(0) > 0$, and $Q_1(0) < 0$.

As far as I know, these are the only constraints on the polynomials, they are otherwise pretty free to do whatever they like.

I've proven that $S$ has only positive roots in the 4th order case, and I've done some pretty extensive numerical testing on the eigenvalues of the matrix which generates S, and it seems to be the case that indeed all the roots are positive (and possibly complex).

Any help would be greatly appreciated. For some context, this problem basically boils down to proving the stability of a P-matrix with some very nice structure.

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Is $Q_1$ decreasing? Otherwise you could have $P_2 = (x-1)(x-3)$, and $Q_1 = 2(x-2)$, which satisfy your constraints but $S=P_2+Q_1$ has a negative root (since $S(0) <0$, and $S(-1) >0$). –  Zack Wolske Apr 2 '12 at 14:27
    
Then even if it is decreasing, we can just do the same thing with $P$ replaced by $-P$. Maybe you have some kind of growth (derivative) constraints near the roots? –  Zack Wolske Apr 2 '12 at 14:30
    
Ah sorry, one more thing. Q is always INCREASING, and P always has a positive y-intercept (in other words,, P_n(0) > 0 for all n). Original post amended. –  Ed Reznik Apr 2 '12 at 14:41
    
The line and parabola given above also satisfy those conditions; you need something stronger. You can make things like this for larger $n$ as well: for any $k>0$, take $P_{2k} = (x-1)^k(x-3)^k$, $Q_1 = 3^k(x-2)$. Then $S$ always has a root between $-2$ and $0$. –  Zack Wolske Apr 2 '12 at 14:56
    
You are correct again, $|P_n(0)| > |Q_1(0)|$, $P_n(0) > 0$, and $Q_1(0) < 0$. –  Ed Reznik Apr 2 '12 at 16:24
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