in $\sqrt{\epsilon} \equiv \sqrt{4-D}$ predict [@br85] $$U^* = 0.59684\dots .
\label{eq:15b}$$ This follows from the expansion $$\begin{aligned}
\frac{\langle m^4 \rangle}{\langle m^2
\rangle^2} &=& \frac{3}{4}\frac{\Gamma^2(3/4)}{\Gamma^2(5/4)}
\Bigg[ 1 - x_0 \sqrt{6} \left(
\frac{\Gamma(9/4)}{\Gamma(7/4)} +
\frac{\Gamma(5/4)}{\Gamma(3/4)}
-2\frac{\Gamma(7/4)}{\Gamma(5/4)} \right) \nonumber \\
&+& 6 x_0^2 \left(
\frac{\Gamma(9/4)\Gamma(5/4)}{\Gamma(7/4)\Gamma(3/4)}
+3 \frac{\Gamma^2(7/4)}{\Gamma^2(5/4)} - 4 \right) \Bigg]
\label{eq:15c} \\
&=& \frac{12}{R^2}\left[ 1 - x_0 \sqrt{6} \left(
\frac{2}{3}R - \frac{6}{R} \right) + 6 x_0^2 \left(
\frac{5}{48} R^2 + \frac{27}{R^2} -4 \right) \right],
\nonumber\end{aligned}$$ where $x_0 = -1.7650848012 \frac{5}{2\sqrt{33}}
\sqrt{\epsilon} = -0.76815456 \, \sqrt{\epsilon}$, and in the last line we have abbreviated $R \equiv
\Gamma(1/4)/\Gamma(3/4) = 2.95867512\dots$. Notice also that (\[eq:15\]) can be rewritten as $$U_L = \frac{2}{3} - \frac{\sigma^2_{\chi}}{3\chi^2},
\label{eq:15a}$$ where $\sigma^2_{\chi} \equiv V^2 \beta^2 \left( \langle m^4 \rangle
 - \langle m^2 \rangle ^2 \right)$ is the variance of the susceptibility. In the $T \rightarrow \infty$ limit we get $\sigma^2_{\chi} \rightarrow \frac{2}{3} \chi^2$ and at criticality, inserting the field theory prediction (\[eq:15b\]), this implies $\sigma^2_{\chi} \approx 0.21 \chi^2$.

Figure 2 shows the crossings of the $U_L(\beta)$ curves on a large scale, which gives a first estimate of ($\beta_c,U^*$). To extract more precise values of $U^*$ and $\beta_c$ from our data we used that the locations of the crossing point $\beta^{\times} 