Limiting behavior of $x$ given that $ 1=\sum^N_{i=1}\frac{\alpha_i}{x+M\alpha_i} $ as $N,M\rightarrow\infty$?

Question

Consider the following implicit equation for $x$. $$ 1=\sum^N_{i=1}\frac{\alpha_i}{x+M\alpha_i} $$ I am trying to see what a limiting behavior of $x$ is, as $N$ and $M$ explode to infinity. I require that $M/N = \mathcal{O}(1)$, as $N\rightarrow\infty$ and $M\rightarrow\infty$. I also assume that $\alpha_i = \mathcal{O}(1)$.

My gut feeling is that $x=\mathcal{O}(N)$, but I am not sure how to arrive at this result in a rigorous fashion.

Answer 1

I will make additional assumptions:

Assumption. $N \geq M$ and $\alpha_i > 0$ for each $i$.

Now define $f(x)$ by the sum in the right-hand side:

$$ f(x) = \sum_{i=1}^{N} \frac{\alpha_i}{x + M\alpha_i}. $$

Clearly, $f(x)$ is a strictly decreasing function for $x \geq 0$. Also, let $S = \sup_i \alpha_i$ and note that $S < \infty$ by OP's assumption. Then

$$ f(SN) = \sum_{i=1}^{N} \frac{\alpha_i}{SN + M\alpha_i} \leq \sum_{i=1}^{N} \frac{\alpha_i}{SN} \leq 1, $$

whereas

$$ f(0) = \sum_{i=1}^{N} \frac{\alpha_i}{M\alpha_i} = \frac{N}{M} \geq 1. $$

Since $f$ is strictly decreasing on $[0, \infty)$, there exists a unique non-negative solution of $f(x) = 1$, and that unique solution lies in the interval $[0, SN]$.