It can easy be shown by induction that the determinant of the $(N-1)\times (N-1)$ matrix $$\begin{pmatrix} 2 & -1 & & \\ -1 & 2 & \ddots & \\ & \ddots & \ddots & -1\\ & & -1 & 2 \end{pmatrix}$$ equals $N$. (Up to a factor) this matrix corresponds to the representing matrix of the discrete Laplacian on an interval with Dirichlet boundary conditions, on an equidistant partition.

Now it the partition is not equidistant, say it is $0 = \tau_0 < \tau_1 < \dots < \tau_N$ with increments $\Delta_j := \tau_j - \tau_{j-1}$, then (up to a factor) we end up with the matrix $$\begin{pmatrix} \Delta_1 + \Delta_2 & -\Delta_2 & & \\ -\Delta_2 & \Delta_2 + \Delta_3 & \ddots & \\ & \ddots & \ddots & -\Delta_{N-1}\\ & & -\Delta_{N-1} & \Delta_{N-1}+\Delta_N \end{pmatrix}.$$ Is the determinant of this matrix still explicitly computable? If not, can we at least compute some (suitably normalized) limit of the determinant as the mesh of the partition goes to zero?

\edit: user35593 answered in the comments that the determinant is equal to $$ \prod_{j=1}^N {\Delta_j}\sum_{i=1}^N\frac{1}{\Delta_i},$$ which is indeed easy to check by induction. Thank you very much!

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    $\begingroup$ You can prove by induction that the determinant is $\prod_{i=1}^N \Delta_i \sum_{i=1}^N \frac{1}{\Delta_i}$ $\endgroup$ – user35593 Jan 19 '16 at 9:15
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    $\begingroup$ By user35593's comment (which I didn't check but looks reasonable to me) you have lower and upper bounds on $ |det|$ in terms of what you expect, as soon as your partition is "approximately equidistant" $0<C_1\leq \frac{\Delta_i}{\Delta_j}\leq C_2$. However the scalings are different in the OP's 2 matrices: in order to stay consistent the first matrix should be multiplied by the uniform mesh $\Delta$, otherwise it doesn't make sense. Also, these matrices are the discretization of the positive operator "negative laplacian", not of the laplacian. $\endgroup$ – leo monsaingeon Jan 19 '16 at 9:39

You already have the answer and proof, so let me just use this post to advertise some basic graph theory :). Take a cycle graph on $N$ vertices, and weigh each edge by $\Delta_i$, $i=1,2,\dots N$. Then your matrix is a cofactor of the Laplacian matrix of this graph. By Kirchhoff's theorem, this is the (weighted) count of the spanning trees of your graph. An $N$-cycle has $N$ spanning trees, and their weights are $\frac{\prod_{i=1}^N \Delta_i}{\Delta_j}$, when you omit the $j$th edge. Therefore the cofactor is equal to $\prod_{i=1}^N \Delta_i \left(\sum_{j=1}^N\frac{1}{\Delta_j}\right)$.

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