Prove that the following equation has only one solution in the interval $[-\text{min}(a_i), +\infty]$:
$f(x) = \left(\sum_{i=1}^n \frac{1}{a_i + x}\right)\times \left(\sum_{i=1}^n \frac{a_i b_i}{(a_i + x)^2}\right) - \left(\sum_{i=1}^n \frac{a_i}{a_i + x}\right) \times \left(\sum_{i=1}^n \frac{b_i}{(a_i + x)^2}\right) =0$
$x\in \mathbb{R}$
$n>1$.
$a_i$ and $b_i,\,\, i=1,..., n$, are all positive real numbers.
$a_1> a_2> ... > a_n$
$a_i \neq a_j\quad\forall i, j = 1,..,n, i\neq j$.
The context of the problem is as follows:
- $a_i$ are the squared singular values of a random matrix.
- $b_i$ are the diagonal elements of a sample covariance matrix.
- I tried solving the equation for different parameter values using Newton's method. It always converges when initialized at zero.
- The plot of the function in the interval of interest always shows increasing function from the start of the interval, zero crossing, a maximum, then the function continuously decreases. The function approaches zero as $x$ increases following the peak value.
We note/can prove the following:
- $f(x)$ has $n$ discontinuities at $-a_i$.
- $f(x)$ is continuous in the interval of interest $[-\text{min}(a_i), +\infty]$ = $[-a_n, +\infty]$.
- $\lim_{x \to +\infty} = 0$.
- $\lim_{x \to -a_n} = -\infty$.
- $f(x)$ is infinitely differentiable in the interval of interest.
- $f(x)$ can be put in the form: $f(x) = \sum_{\{i,j\}} \frac{a_i - a_j}{(a_i + x)(a_j + x)} \left(\frac{b_i}{a_i + x} - \frac{b_j}{a_j + x} \right)$ for all permutations of $\{i,j\}\subset\{1,...,n\}, i< j$.
I tried the following steps to prove that $f(x)$ has a single root in the interval $[-\text{min}(a_i), +\infty]$:
Assume $x_0\in [-\text{min}(a_i), +\infty]$ is a root.
Tried to show that $x_0$ is not a solution of $f^{'}(x)$, but the derivative expression is far more complex to figure that out.
Any suggestion for an alternative way to obtain the required proof?