Let $\beta>0$ be given. For each $n\geq 2$, let $\Delta_n=\det M_n$ denote the determinant of the following matrix: \begin{align} M_n = \begin{pmatrix} 2+\epsilon^2 & -1 & 0 & 0 & 0 & -1 \\ -1 & 2+\epsilon^2 & -1 & 0 & 0 & \ddots \\ 0 & -1 & 2+\epsilon^2 & -1 & 0 & \ddots \\ 0 & 0 & -1 & 2+\epsilon^2 & -1 & \ddots \\ 0 & 0 & 0 & -1 & 2+\epsilon^2 & \ddots \\ -1 & \ddots & \ddots & \ddots & \ddots & \ddots \\ \end{pmatrix}, \qquad \epsilon = \frac{\beta}{n} \end{align} How can one evaluate the limit \begin{align} \Delta = \lim_{n\to\infty}\Delta_n ? \end{align} This problem arose in the context of evaluating a certain path integral in a physics problem. I know how to determine the eigenvalues of each $M_n$, but computing the determinant by taking the product of eigenvalues leads to a product I can't evaluate. I also tried deriving a recursion relation for the $\Delta_n$ and showing that in the limit $n\to\infty$, the recursion relation can be regarded as a differential equation whose solution subject to certain initial data determines $\Delta$, but that failed as well.

Any insights would be appreciated.


When I compute the eigenvalues $\lambda_k$ of each $M_n$, I obtain \begin{align} \lambda_k = 4\sin^2\left(\frac{\pi k}{n}\right)+\left(\frac{\beta}{n}\right)^2, \qquad k=0,1,\dots, n-1 \end{align} from which it follows that \begin{align} \Delta_n = \prod_{k=0}^{n-1}\left[4\sin^2\left(\frac{\pi k}{n}\right)+\left(\frac{\beta}{n}\right)^2\right] \end{align} which I don't have the foggiest idea of how to evaluate. I do, however, have a conjecture for the answer which comes from the fact that this limit of determinants came from a path integral which can be computed in other ways. My conjecture (which I have checked numerically using mathematica to some extent) is \begin{align} \Delta = 4\sinh^2\frac{\beta}{2} \end{align}

  • $\begingroup$ Don't you get something like the Euler product formula for $\sinh$ with the product of eigenvalues? $\endgroup$ – Raskolnikov Oct 9 '13 at 20:47
  • $\begingroup$ @Raskolnikov Hmm I don't think so; see the addendum. $\endgroup$ – joshphysics Oct 9 '13 at 21:24
  • $\begingroup$ Actually, it does look alot like that formula. It would just require a careful limiting procedure. $\endgroup$ – Raskolnikov Oct 9 '13 at 22:21
  • $\begingroup$ If you take the logarithm of your product formula for $\Delta_n$, it looks a lot like a Riemann sum for an integral. Can you use that to evaluate the limit? $\endgroup$ – Greg Martin Oct 10 '13 at 0:56
  • $\begingroup$ @GregMartin I dunno, I'll look into that. Thanks. $\endgroup$ – joshphysics Oct 10 '13 at 1:14

We answer this question using the information in the addendum. We proceed in three steps. (1) We compute the eigenvalues of each $M_n$ and notice that the problem reduces to evaluating a limit of a sequence of products. (2) We show that the product formula proven in another math.SE question can be used to evaluate each term in this sequence of products. (3) We take the $n\to\infty$ limit of the resulting expression.

Step 1.

For each $n\geq 2$, let $U_n = ((U_n)_{ij}) $ denote the $n\times n$ matrix defined by \begin{align} (U_n)_{ij} = \left\{\begin{array}{cc} 1, & j=i+1 \\ 1, & (i,j) = (n,1) \\ 0, & \mathrm{otherwise} \end{array}\right. \end{align} then we have \begin{align} M_n = (2+\epsilon^2)I_n-U_n-U_n^\dagger \end{align} where $I_n$ is the $n\times n$ identity, and $\dagger$ denotes complex conjugate + transpose (which in this case is of course just the transpose since the matrices are real). Let's suppress the subscript $n$'s for notational simplicity for a moment.

It is straightforward to show that each $U_n$ has the property $(U_n)^n = I$; it's $n^\mathrm{th}$ matrix power is the identity. This tells us that each eigenvalue $\lambda$ satisfies $\lambda^n = \lambda$, namely its eigenvalues are the $n^\mathrm{th}$ roots of unity $e^{2\pi ik/n}$ with $k=0,1,\dots, n-1$ with corresponding eigenvectors $v_k$. It is also not hard to see that $U_n^\dagger$ shares the same eigenvectors $v_k$ with the complex conjugated corresponding eigenvalues $e^{-2\pi ik/n}$. It follows that the eigenvalues of $M_n$ are \begin{align} \lambda_k = 2+\epsilon^2 - e^{2\pi ik/n} - e^{-2\pi ik/n} = 4\sin^2\left(\frac{\pi k}{n}\right)+\epsilon^2 \end{align} The determinant of $M_n$ is the product of its eigenvalues, so we obtain \begin{align} \Delta_n = \prod_{k=0}^{n-1}\left[4\sin^2\left(\frac{\pi k}{n}\right)+\left(\frac{\beta}{n}\right)^2\right] \end{align} as claimed in the addendum. So problem reduces to evaluating the limit of a sequence of finite products; \begin{align} \Delta = \lim_{n\to \infty}\left(\prod_{k=0}^{n-1}\left[4\sin^2\left(\frac{\pi k}{n}\right)+\left(\frac{\beta}{n}\right)^2\right]\right). \end{align}

Step 2.

The following product identity holds: \begin{align} \prod_{k=0}^{n-1}\left[\sinh^2y+\sin^2\left(x+\frac{k\pi}{n}\right)\right]=2^{1-2n}(\cosh(2ny) -\cos(2nx)) \end{align} see, for example, the following math.SE question (which, incidentally, was inspired by the question as hand):

How would one discover this finite product identity?

If we set $x=0$, $y=\sinh^{-1}(\beta/(2n))$ and multiply this product by $4^n$, then we obtain the following identity as a special case: \begin{align} \prod_{k=0}^{n-1}\left[4\sin^2\left(\frac{\pi k}{n}\right)+\left(\frac{\beta}{n}\right)^2\right] = 2\left\{\cosh\left[2n\sinh^{-1}\left(\frac{\beta}{2n}\right)\right]-1\right\} \end{align} The desired expression is thus the $n\to\infty$ limit of the expression on the right hand side.

Step 3.

For each $\beta >0$, the large $n$ limit allows us to approximate $\sinh^{-1}(\beta/(2n)) \sim \beta/(2n)$ so the expression on the right hand side goes to $2(\cosh(\beta)-1)$ as $n\to\infty$. Now simply recall the half-angle formula $\sinh(\theta/2) = ((\cosh\theta-1)/2)^{1/2}$ which gives the conjectured result; \begin{align} \Delta = 4\sinh^2\left(\frac{\beta}{2}\right). \end{align}


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