-1 -0.5  0  0.5 -10  0  10 g(z) z Fig. 4: The parametric dependence of g ( z ) on z is given by open symbols. Eq. (50) is depicted as solid line. The dotted line represents the linear fit, Eq. (51). (4). 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, g ( z ) = a + b z + c exp( d z ) + e [exp( d z ) + f ] $_{2}$, (50) where the coefficients read a = − 2 . 078 ± 0 . 117, b = 0 . 091 ± 0 . 007 and c = 17 . 332 ± 1 . 553, d = 0 . 189 ± 0 . 003, e = − 0 . 814 ± 0 . 162 and f = 2 . 849 ± 0 . 02. At small values of z , it is clear that the dependence is linear, g ( z ) = c + C z. (51) Obviously, the intersect c is much smaller than the slope C . The sign of g ( z ) can be flipped regarding to the sign of its independent variable z . Accordingly, we get g ( z ) ≈ C z. (52) 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. g ( H ) ≈ 3( γ − 2) 2[1 + (1 r ) γ ] H $^{1}$− $^{r}$, (53) Then, we approximate z ( H ) to the form, z ( H ) ≈ − 3[1 + (1 − r ) γ ] 2(1 r ) H $^{1}$− $^{r}$. (54) 18