understood if we consider the reduced determinant at $\varkappa=1$ (typically, $\varkappa_c$ is rather close to 1). In this case $$\det \textbf{T}_{\rm tr} = 4c(x)c(z)c(x+z)-4c^2(x+z)=4c(x+z)\left[c(x)c(z)-c(x+z)\right].
\label{particular-exp}$$ We assume that $c(x)>0$, hence the sign of $\det \textbf{T}_r$ is determined by the factor $c(x)c(z)-c(x+z)$. For the exponential correlation function $c(x)c(z)-c(x+z)$ is exactly zero. If $c(x)$ decays faster than the exponential one, then the sign of $\det \textbf{T}_{\rm tr}$ is positive, and in the opposite case of the more slow decay it is negative. For $\varkappa=0$ $\det \textbf{T}=1$, so for some $0<\varkappa<1$ there is the zero of the determinant for slowly decaying correlation functions. One should note that the separation of the correlation functions to those decaying more slowly than the exponential ones and decaying faster is considered here not for asymptotics of $c(x)$ for $x\rightarrow\infty$ but for the functional dependence of $c(x)$ in the whole $x$ range. If the function $c(x)$ has the exponential asymptotics but decays differently for moderate $x$, then the critical $\varkappa_c$ still could exist (see Fig. \[mix\])

![Dependence of the critical $\varkappa_c$ on $b/a$ for the correlation functions $c(x)=1/\left[(x/a)^3+\exp\left(x/b\right)\right]$ (filed squares) and $c(x)=1/\left[(x/a)^3+(x/b)^5+1\right]$ (empty squares). Solid lines are provided as a guide for an eye. Broken lines indicates $\varkappa_c$ 