# Zeros of linear partial fractions

I am interested in somehow characterizing the zeros of the following function:

$P(z) = \sum_{i=1}^n \frac{\alpha_i}{z+\lambda_i}$,

with $\lambda_1\geq \lambda_2 \geq \ldots \geq \lambda_n=0$ and $\sum_{i=1}^n \alpha_i = 0$.

Furthermore, the values $\alpha_i$ are in general non-integer and may even be complex. The poles $\lambda_i$, however, are all real.

I initially had hoped to use something like the Gauss-Lucas theorem for this, but the conditions on $\alpha_i$ seem to prevent this.

While it would be great to make some strong statements about the zeros, I would be more then satisfied to find conditions guaranteeing that the zeros lie in the left-half of the complex plane.

Any thoughts/suggestions would be great. Thanks!

• Either you meant "the poles $-\lambda_i$" or the sign in the denominator is wrong. – joriki Mar 2 '11 at 8:29
• yes, you are correct, the poles are in fact $-\lambda_i$. – 1yen Mar 2 '11 at 9:17
• we discourage simultaneous cross posts to another forum of the same question. Your question is also asked at mathoverflow. Please follow Jeff's suggestion in the linked Meta thread in the future. – Willie Wong Mar 3 '11 at 9:23
• I'll come back to this later; for now I'll note that the stuff here might help a fair bit with answering your question. – J. M. isn't a mathematician May 1 '11 at 14:54

For given $\lambda_i$, you have $n$ non-zero, linearly independent functions and you form a linear combination of them with arbitrary complex coefficients $\alpha_i$ -- that means you can pick $n-1$ points in general position and make them zeros of $P$ by a suitable choice of the coefficients (e.g. by picking one of the functions and mimicking its function values at the $n-1$ chosen points by a suitable linear combintaion of the other $n-1$ functions). Conversely, since you can bring all the fractions to a common denominator and the numerator will then be of degree $n-1$, the function has exactly $n-1$ zeros in the general case. So generally you have $n-1$ zeros that can be almost anywhere you like (except for the poles, and sets of measure zero where the function values are linearly dependent), and the only thing that's left to characterize is what sort of special cases you can get.
(I implicitly assumed that the $\lambda_i$ are different and the $\alpha_i$ are non-zero -- if either of these assumptions isn't fulfilled, you can reduce to a case with lower $n$ where they are.)
• Thanks It turns out $\alpha_i$ are not exactly arbitrary. Perhaps I can give you some more information. The $\lambda_i$'s are actually the eigenvalues of the combinatorial graph Laplacian ($L(\mathcal{G})$). The Laplacian has a modal decomposition $L(\mathcal{G})=U\Lambda(\mathcal{G})U^T$, and the coefficients $\alpha_i = U_{ji}U_{ki}$ (the product of the ith column in the j and k rows, for fixed j,k). So, in fact, some $\alpha_i$ can be zero, and their structure is somehow connected to the poles. Simulations seem to suggest the zeros are in LHP. Just not sure if (how!) I can prove it. – 1yen Mar 2 '11 at 9:23
• @1yen: I just realized I ignored your restriction $\sum\alpha_i=0$ (sorry about that), but that alone only reduces the number of zeros you can choose freely by one, it doesn't restrict them to the LHP. Of course some properties of the Laplacian might do that -- perhaps it would help if you explain why you're looking at this particular problem with these particular $\alpha_i$. – joriki Mar 2 '11 at 9:57
• In control systems, a popular multi-agent distributed system model has dynamics $\dot{x}=-L(\mathcal{G})x$ Consider now the system $\dot{x}=-L(\mathcal{G})x+bu, y=c^Tx$; $b$ and $c$ are unit-coordinate vectors. We assume that a single agent can inject a control, and we can observe another agent. Now we have an input-output system, and there is a corresponding transfer-function associated with it. This TF can be expressed as the partial fraction written above. The location of the zeros of this function have implications for controlling the network; i.e., find a control of the form $u=Ky$. – 1yen Mar 2 '11 at 10:21