How many positive roots can $\sum_{i}\frac{a_i}{x+b_i}$ have where $b_i$'s are all positive? [closed]

Question

What is the maximum number of positive roots $\sum_{i}^N\frac{a_i}{x+b_i}$ can have where $b_i$'s are all positive? (everything here is a real number. To provide context, I encountered this problem while doing theoretical neuroscience research where I am modeling a biological neuronal network as an artificial neural network.)

$x$ is our variable, $a_i$ is a constant (can be either positive or negative), and $b_i$ is always a positive constant. $a_i$ and $b_i$ have unique values at each $i$.

In other words, how many $x>0$ can satisfy $\sum_{i}^N\frac{a_i}{x+b_i}=0$?

If $a_i$ happens to be all positive or negative, I see that there are no roots at positive $x$. For example, the following shows $y=\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}$. You can see that there are poles at -1, -2, and -3 (which are $-b_i$'s), and the roots exist between the poles. Since $b_i$'s are all positive, the roots between the poles need to be all negative.

However, if $a_i$'s are a mix of positives and negatives (and $b_i$'s are still all positive), there can be root(s) outside the poles, making it possible to have a root when $x>0$. For example, $y=\frac{1}{x+1}+\frac{1}{x+2}-\frac{3}{x+3}$ is $0$ at $x>0$ as shown below:

If I zoom in to the positive $x$ part, we see the following:

It overshoots below 0, and then asymptotically approaches 0.

So far, no matter how large my $N$ is, a randomly generated function $\sum_{i}^N\frac{a_i}{x+b_i}$ seemed to have only one root at positive $x$, if there was any, when I swept through $x$ on my computer. However, I still believe the number of positive roots should be dependent on $N$. Any thoughts?

Answer 1

If $N$ is the number of summands, then this is a rational function of degree $N$, and zero at infinity. Therefore, the total number of roots (counting multiplicity) in the plane is $N-1$. All of them can be positive. Indeed, let $P$ and $Q$ be polynomials with all roots positive and distinct, $\deg P=N-1,\deg Q=N$. Then $P/Q$ is of the form as in your, by partial fraction decomposition and has $N-1$ positive roots.

Remark: if all $b_j>0$ and all $a_j>0$, each interval $(b_j,b_{j+1}), 1\leq j\leq N-1,$ contains exactly one zero.