Pretty straightforward problem, I have a summation and no idea how to find its zeroes.
I need to find the zeroes in terms of $x$ for any array of positive constants $a_0, a_1, ..., a_n$.
Summation:
$$\sum_{i=0}^{n}\frac{-x^2+3a_ix-2(a_i)^2}{(a_i)^2 \lvert x-a_i\rvert}$$
Note that the upper polynomial is factorable as $(x-a_i)(-x+2a_i)$, if that helps.
On
I will be using $k$ instead of $i$ just for personal preference.
$$\sum_{k=0}^n\frac{(x-a_k)(-x+2a_k)}{(a_k)^2|x-a_k|}=\sum_{k=0}^n\frac{\text{sgn}(x-a_k)(-x+2a_k)}{a_k^2}$$
where $\text{sgn}(\mu)$ is the sign (+/-) of $\mu$.
For any negative $x$ value, we have
$$\text{sgn}(x-a_k)=-1$$
Also see that $\sum_{k=0}^n\frac{\text{sgn}(x-a_k)(-x+2a_k)}{a_k^2}$ has only one root by the fundamental theorem of algebra.
So, if we assume the root occurs for $x<\min(a_k)$, we will have
$$\sum_{k=0}^n\frac{\text{sgn}(x-a_k)(-x+2a_k)}{a_k^2}=\sum_{k=0}^n\frac{2a_k-x}{a_k^2}$$
Sadly, it is clear that for any negative number $x$, the summation will be larger than $0$.
Assume the root occurs at $x>2\max(a_k)$ so that we may have $\text{sgn}(x-a_k)=1$
$$\sum_{k=0}^n\frac{\text{sgn}(x-a_k)(-x+2a_k)}{a_k^2}=\sum_{k=0}^n\frac{2a_k-x}{a_k^2}$$
This time we see that for any $x>2\max(a_k)$, the result will be a negative value, so that will not be a root.
I cannot manage to conclude anything beyond this, so the best answer I can give you is
$$\min(a_k)<x<\max(a_k)$$
For convenience, sort the $a_i$ in increasing order. The $a_i$ divide the real line into $n+1$ intervals, on each of which your summand is linear, since $(x - a_i)/|x - a_i| = \pm 1$. It's easy to find the zeros of a linear function.