0.0008.      \label{eq:17}$$ The theoretical prediction (\[eq:15b\]) based on the $\sqrt{\epsilon}$-expansion is about $4\%$ smaller than this value. The deviation is somewhat less than for the XY model where it is about $6\%$ [@r13; @r14].

To determine the derivative ${dU_L}/{d\beta}$, we first used a finite difference approximation at our estimate of $\beta_c$. But we observed that the results depended very sensitively on the interval of the linear approximation, and on the kind of finite difference derivative used, i.e., backward, forward, or symmetric derivative. We have therefore chosen another method [@r26] to calculate the slope of $U_L(\beta)$ that is less sensitive to systematic errors. We took the thermodynamic derivative of $U_L$ with respect to $\beta$, which can be written as $$\begin{aligned}
{{dU_L} \over {d\beta }} &=&
{1 \over {3\langle m^2\rangle ^2}} \left\{
{\left\langle {m^4} \right\rangle\langle E\rangle
-2{{\left\langle {m^4} \right\rangle\langle m^2E\rangle } \over
{\left\langle {m^2} \right\rangle} }
+\langle m^4E \rangle
}
\right\}\hfill\cr  &=&(1-U_L)\left\{ {\langle E\rangle -2{{\langle m^2E\rangle
}  \over {\left\langle {m^2} \right\rangle}}+{{\langle m^4E\rangle } \over
{\left\langle {m^4} \right\rangle}}} \right\}. \label{eq:18a}\end{aligned}$$ The expectation values for $\langle m^kE \rangle(\beta)$ at any temperature $\beta$ can be calculated from $$\langle m^kE\rangle (\beta )={{\sum\limits_E {E\langle \langle
m^k\rangle \rangle (E)P_{\beta_0}(E)e^{-(\beta -\beta _0)E}}}  \over
{\sum\limits_E {P_{\beta_0}(E)e^{-(\beta -\beta _0)E}}}},$$ generalizing eq. 