Asymptotics of the unique root of a polynomial equation defined as a sum of rational expressions

Question

Let $\lambda_1\ge \ldots \ge \lambda_n \gt 0$. Define a function $F:\mathbb R_+ \to \mathbb R_+$ by

$$ F(t) = t^2\sum_{i=1}^n\frac{\lambda_i^2}{(\lambda_i + t)^2}. $$

It is clear that $F$ is strictly increasing with supremum $F(\infty)=\sum_{i=1}^n\lambda_i^2=:M$. Let $0 \le s \le M$. It is clear that the equation $F(t) = s$ must have a unique solution $t(s)$ in $\mathbb R_+$.

For example, if $\lambda_i =1$, then $F(t) \equiv t^2M/(1+t)^2$. So, for every $0 \le s \le M$, we deduce that $t(s) = \sqrt{\epsilon/(1-\epsilon)}$, where $\epsilon := \sqrt{s/M}$.

Now, for varying $n$, let $n_1=n_1(n) \in [n]$ and $b=b(n) \in [0,1]$ be given, and set $\lambda_i = 1_{i \le n_1} + b1_{i \ge n_1+1}$ for all $i \in [n]$. For concreteness, you may take $n_1(n) \equiv 1$ and $b(n) \equiv 1/n$. Note that in this case, $M=n_1 + n_2 b$, where $n_2 := n - n_1$.

Question. Given $s=\epsilon^2 M$ with $\epsilon \in [0,1]$ fixed, is there an an asymptotically valid (in the limit $n \to \infty$) expression for $t_n(s)$ ?

I'd also be very interested in a systematic way of answering the above question for different choices of the sequences $n_1=n_1(n)$ and $b=b(n)$.

Answer 1

If the $\lambda_i$ take only two different values namely $\lambda_\max > \lambda_\min$, the equation $F(t)=s$ can be written $$At^2(\lambda_\min+t)^2 + Bt^2(\lambda_\max+t)^2 = s (\lambda_\max+t)^2 (\lambda_\min+t)^2,$$ where $A$ and $B$ are positive constants with sum $M$. When $s \in~[0,M[$, this yields an algebraic equation with degree $4$, which is solvable by radicals. View https://en.wikipedia.org/wiki/Quartic_equation

Answer 2

If we know the dependency of $\lambda_i$ to $i$, it could be interesting to look at the zero of functions $$f(t) = t^2\sum_{i=1}^n\frac{\lambda_i^2}{(t+\lambda_i )^2}-s\quad \quad \text{and}\quad\quad g(t) = t^2\int_{1}^n\frac{\lambda_i^2}{(t+\lambda_i )^2}\,di-s$$ Using for example $\lambda_i=\frac 1i$ $$f(t)=\psi ^{(1)}\left(\frac{1}{t}+1\right)-\psi ^{(1)}\left(\frac{1}{t}+1+n\right)-\epsilon ^2 H_n^{(2)}\tag 1$$ $$g(t)=\frac{(n-1) t^2}{(t+1) (n t+1)}-\epsilon ^2 H_n^{(2)}\tag 2$$

Use $(2)$ to generate the estimate $t_0$ and perform one single iteration of Newton method. This would give $$t_1=t_0-\frac{\epsilon ^2 H_n^{(2)}+\psi ^{(1)}\left(\frac{1}{t_0}+1+n\right)-\psi ^{(1)}\left(\frac{1}{t_0}+1\right)}{\psi ^{(2)}\left(\frac{1}{t_0}+1\right)-\psi ^{(2)}\left(\frac{1}{t_0}+1+n\right)}\, t_0^2$$