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Ask a Question. We get positive traces for the coefficients up to order $n = 25$. Unlike in the unbarred case, we cannot identify the dominant (most probable) path corresponding to the sign change in the trace. For $N=4$, we see only a slight sign change in the trace but no trace at all for $N=6$.

![Lambda stability for barred Gauss-Hermite Monte Carlo paths, with $N=6$. By its definition, the radius of absolute convergence of $\mathcal{L}(\lambda)$ is $4\sqrt{\pi}$ for $d=1$, where the number of bars $N$ is 2. The path is plotted over the interval $[-10,10]$ for $t=0,1,2,\dots$.[]{data-label=»fig:LambdaStability»}](./LambdaStabilityBarred.eps){width=»10cm»}

We have not investigated Gauss-Hermite paths for arbitrary $N$, but we expect the path to show up for any order $n$ for $\lambda$ close enough to zero in the sense that $\lambda_{nm}$ is negligibly small. However, we find a strong dependence on $N$, and it is plausible that no trace will exist for $N$ large enough.

To conclude, we have shown that for each $\varepsilon \in (0,1)$, the spectrum of $\mathcal{L}(\lambda)$ is continuous with respect to $\lambda$, and the set $\lambda_n^{(\alpha)}$ of absolute convergence of the spectrum is $1/n^{1-\alpha}$-attractive. The other statements have not been proved but are believed to be true.

For $\lambda \in \mathbb{R}_+$ the eigenvalues of the trace lattice Laplacian are all real and enumerated in an increasing sequence. The convergence of the sequence $\lambda_n^{(\alpha)}$ to $0$ can be characterized using the resolvent of \$

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