infinitely differentiable but not analytical for any $ x $, in particular the Taylor expansion of the function $ q(x) $ around any $ 0<x<x_{max} $ does not converge but is asymptotic.

This singular behaviour is not connected neither with the replica limit nor with the Parisi Ansatz, actually it originates from the singularities in the complex plain of the initial condition of the Parisi equation: $ f(1,y)=\ln 2\cosh \beta y $. This is clearly seen for the replica-symmetric solution$$q=\int ^{+\infty }_{-\infty }\frac{dz}{\sqrt{2\pi }}e^{\frac{z^{2}}{2}}\tanh ^{2}(\beta \sqrt{q}z)$$ In this case it is easy to prove that the expansion of $ (1-T^{2}) $ in powers of $ p=\beta ^{2}q $ is asymptotic because it corresponds to substitute $ \tanh ^{2}z $ in the integrand with its Taylor expansion which is not convergent on the whole real axes. Then one can prove that the expansion of $ q $ in powers of $ \tau =1-T $ is asymptotic recalling that standard manipulation (e.g. multiplication, division, inversion...) on an asymptotic expansion in power series do not change its character. A detailed treatment of the RSB solution is much more complex, but the origin of the asymptotic character is likely to be the same. Indeed 