energy $f(T)$.

To expand the free energy functional (\[eqfree\]) in powers of $\tau = T_c - T = 1 - T$ we observe that in the absence of external fields $q(x)$ is different from $q(1)$ only in a region $[0,x_{\rm max}]$ with $x_{max} = O(\tau)$ [@P7980], so that an expansion in power of $\tau$ must correspond to an expansion of the same order in $x$. Therefore to compute the free energy to order $n$ we insert into eq. (\[eqfree\]) the following expansions: $$\label{eqq1}
 q(1) = \sum_{i=1}^{n-2}\, a_i\,\tau^i$$ and $$\label{eqxq}
 x(q) = \sum_{i=1}^{n-3}\,\sum_{j=0}^{n-3-i} b_{ij}\,q^i\,\tau^j.$$ The coefficients of the expansion of the function $\phi(q,y)$ about $q=q(1)$ and $y=0$ can be obtained by repeated differentiation with respect to $q$ of the equation $$\frac{\partial \phi}{\partial q} = -\frac{1}{2}\left[ 
                    \frac{\partial^{2} \phi}{\partial y^{2}} 
              + x(q)\left( \frac{\partial \phi}{\partial y}\right)^{2}
                                               \right].$$ Differentiating $j$ times with respect to $y$ this equation, mixed derivatives $\phi^{(1,j)}(q,y)$ can be eliminated in favor of derivatives with respect to $y$ only. In the absence of an external magnetic field the last term in equation (\[eqfree\]) reduces to $\phi(0,0)$ greatly simplifying the calculation since at each step we can eliminate all terms containing odd derivatives of $\phi$ with respect to $y$, as for example $(\partial 