by using the following procedure. First, we plot it parametrically depending on the parameter $H$, Fig. (\[Figg4\]). Then we fit the resulting curve to various functions. Based on least-square fit, best choice would be a mixture of polynomial and exponential functions, $$\label{fullgofz}
g(z)= a + b\, z + c \frac{\exp(d\, z)+e}{\left[\exp(d\, z)+f\right]^2},$$ where the coefficients read $a=-2.078\pm0.117$, $b=0.091\pm0.007$ and $c=17.332\pm1.553$, $d=0.189\pm0.003$, $e=-0.814\pm0.162$ and $f=2.849\pm0.02$. At small values of $z$, it is clear that the dependence is linear, $$\label{lineargofz2}
g(z) = c + {\cal C} z.$$ Obviously, the intersect $c$ is much smaller than the slope ${\cal C}$. The sign of $g(z)$ can be flipped regarding to the sign of its independent variable $z$. Accordingly, we get $$\label{lineargofz}
g(z)\approx {\cal C} z.$$

To prove this dependence, algebraically, we try to estimate $g(z)$ directly from the division of $F_0$ by $F_1$, which can be approximated by including their first terms only, i.e. $$\label{eq.A1}
g(H)\approx \frac{3(\gamma-2)}{2[1+(1-r)\gamma]}\; H^{1-r},$$ Then, we approximate $z(H)$ to the form, $$\label{eq.A2}
z(H)\approx-\frac{3[1+(1-r)\gamma]}{2(1-r)}\; H^{1-r}.$$ Finally, we now able to derive an approximate estimation for $g(z)$. According to Eq. (\[eq.A1\]) and (\[eq.A2\]), we get $$g(z)\approx \frac{(1-r)(\gamma-2)}{[1+(1-r)\gamma]^2}\; z$$ Amazingly, this expression looks the same as the one we obtained from the numerical 