\phi / \partial y)^{2}$ in the previous equation, since all these vanish if evaluated at $y=0$ being $\phi(q,y)$ and even function of $y$.

Collecting all terms with the same power of $\tau$ the free energy functional (\[eqfree\]) is written as $$f = \sum_{i=0}^{n}\, c_i[\{a\},\{b\}]\, \tau^i.$$ This expression must be stationary with respect to variations of $a$’s and $b$’s for any $\tau$. Imposing stationarity of each $c_i$ we can find the value of the parameters $a$ and $b$. For example to order $\tau^6$ we have: $$\begin{aligned}
\label{eq:qx6}
 q(x) = \left(
               \frac{1}{2} + \frac{3}{2}\,\tau + 2\,\tau^3 
               - 9\, \tau^4 + \frac{336}{5}\,\tau^5
        \right)\, x
       &+& \left(
               -\frac{1}{8} + \frac{25}{8}\,\tau + 3\,\tau^2
               + 38\,\tau^3
         \right)\, x^3
\nonumber\\
       &+& \left(
               -1 - 9\,\tau - 30\,\tau^2
         \right)\, x^4
       + \left(
               \frac{351}{320} + \frac{9189}{320}\,\tau
         \right)\, x^5
        - \frac{27}{5}\, x^6\end{aligned}$$ and $$\label{eq:xmax6}
 x_{\rm max} = 2\,\tau - 4\,\tau^2 + 12\,\tau^3 - 69\,\tau^4
               +\frac{2493}{5}\,\tau^5 - \frac{20544}{5}\,\tau^6.$$ By using this procedure we have obtained the free energy up to order $30$, $q(x)$ to order $13$ and $q(1)$ to order $14$ because despite the fact that the free energy is evaluated to order $n$, the variational relations allow to determine $x(q)$ only to order $[(n-3)/2]$ and $q(1)$ only to order $[(n-1)/2]$.

