Math people:

I am looking for a proof of a conjecture I made. I need to give two definitions. For distinct real numbers $x_1, x_2, \ldots, x_k$, define $\sigma(x_1, x_2, \ldots, x_k) =1$ if $(x_1, x_2, \ldots, x_k)$ is an even permutation of an increasing sequence, and $\sigma(x_1, x_2, \ldots, x_k) =-1$ if $(x_1, x_2, \ldots, x_k)$ is an odd permutation of an increasing sequence. For example, $\sigma(2, 1, 10, 8) = 1$ because $(2,1,10,8)$ is an even permutation of $(1,2,8,10)$, and $\sigma(2, 1, 8, 10) = -1$ because $(2,1,8,10)$ is an odd permutation of $(1,2,8,10)$. For real $B$, $n\geq 1$ and distinct real numbers $\mu_1, \mu_2, \ldots, \mu_n, \gamma_1, \gamma_2, \ldots, \gamma_n$, let $M(B;\mu_1,\mu_2,\ldots,\mu_n;\gamma_1,\gamma_2,\ldots,\gamma_n)$ be the $n$-by-$n$ matrix defined by

$$ M(B;\mu_1,\mu_2,\ldots,\mu_n;\gamma_1,\gamma_2,\ldots,\gamma_n)_{i,j}=\frac{\exp(-B\gamma_j)}{\mu_i+\gamma_j}+\frac{\exp(B\gamma_j)}{\mu_i-\gamma_j}. $$

My conjecture is the following: if $n \geq 1$, $B \geq 0$, and $\mu_1, \mu_2, \ldots, \mu_n, \gamma_1, \gamma_2, \ldots, \gamma_n$ are distinct *positive* numbers with
$0<\mu_1 < \mu_2 < \cdots < \mu_n$ and $0<\gamma_1 < \gamma_2 < \cdots < \gamma_n$ , then

$$\operatorname{sgn}(\operatorname{det}(M(B;\mu_1,\mu_2,\ldots,\mu_n;\gamma_1,\gamma_2,\ldots,\gamma_n))) = (-1)^{\frac{n(n+1)}{2}} \sigma(\mu_1, \mu_2, \ldots, \mu_n, \gamma_1, \gamma_2, \ldots, \gamma_n). $$

Of course $\operatorname{sgn}(x)$ is the sign of $x$, which is $1$, $-1$, or $0$. I have proven this is true for $n=1$ and $n=2$. For $n$ between $3$ and $20$, I have run thousands of experiments in Matlab using randomly generated $\mu$'s and $\gamma$'s. In a set of one thousand experiments, the conjectured equation will typically hold every single time, or might fail once or twice, with the determinant (with the wrong sign) being extremely small, so perhaps roundoff error is the culprit.

UPDATE: let $d(B)$ be the determinant of the matrix, where the other parameters should be clear from context. $d(B)$ is an analytic function of $B$. It suffices to show that $\frac{\partial^m d}{\partial B^m}$ has the desired sign at $B=0$ for all $m \geq 0$. Unfortunately, the determinant of $\frac{\partial M}{\partial B}$ is not the same thing as $\frac{\partial^m d}{\partial B^m}$ (if it were, properties of Cauchy matrices would yield the desired conclusion). Since the conjecture is true for $n=1$, that means that the displayed formula for $M_{i,j}$ above, and all its derivatives with respect to $B$, have the same sign as $\mu_i - \gamma_j$ at $B=0$, and $M_{i,j}$ has that sign for all positive $B$. I proved the conjecture for $n=2$, by computing the determinant of $M$ and its derivatives with respect to $B$ at $B=0$, and looking at the six possible orderings of $\mu_1, \mu_2, \gamma_1$, and $\gamma_2$ given the restrictions $\mu_1 < \mu_2$ and $\gamma_1 < \gamma_2$. I had some help from Maple multiplying out, simplifying and factoring algebraic expressions. I am trying to prove the general case by induction on $n$, expanding the determinant along the last row or column, but the determinants of the $n-1$-by-$n-1$ minors don't seem to necessarily have the ``right'' signs.

Thanks to some comments provided below, unless I am confused, the conjecture can be proven for $B=0$ and large positive $B$, for any $n$, using properties of Cauchy matrices.