Why does Landau theory not fail when dealing with a first order phase transition? Here is a problem where I can do the calculation, but I am not understanding the philosophy behind it. It is about Landau theory:
The Landau theory of phase transitions is based on the idea that the free energy of a system can be expanded as a power series of the order parameter. For a second order phase transition, the order parameter develops an expectation value, that evolves continuously from zero, so this power expansion has solid mathematics grounds. For a first order phase transition, however, since there is a jump in the order parameter, the order parameter never gets an infinitesimal value. If so, why is Landau theory still commonly used for the first order phase transition, even if the expansion seems to not be valid at the phase transition?
 A: The key is: Landau theory doesn't assume the order parameter is small. All it assumes is that the free energy is analytic in the order parameter. One then usually expands this free energy up to some order (which is possibly by definition of 'analytic'). It is key to realize that expanding a function in a variable to some order does not mean this variable has to be small! It just means that terms we throw away have to be small, which is a different thing.
Let's take an example. Suppose we have this somewhat unusual looking free energy handed to us, which is indeed analytic:
$$\boxed{F(\phi,T) = \phi^2 - 2 + e^{-\frac{\phi^4}{T}} + \cosh(\phi^3)}$$
For high temperatures, the minimum of the free energy selects $\phi = 0$. Around $T \approx .3$, there is a first order transition to $\phi \neq 0$. The following two graphs give the intuitive picture (the x-axis is the order parameter, the y-axis the free energy):

In Landau theory one usually expands these free energies. For example if we expand it to 8th order, we get
$$F(\phi,T) = \phi^2 - \frac{\phi^4}{T} + \frac{\phi^6}{2} + \frac{\phi^8}{2T^2}$$
To this order, the graph for $T = .25$ looks as follows:

So we see that this already gives a good representation of our free energy in the region $-1 \leq \phi \leq 1$. This is because despite $\phi$ not being small, the terms we have thrown away are.
Note that if one is not interested in quantitative details but rather just wants the intuitive picture, then one can note that $F(\phi,T) = \phi^2 - \frac{\phi^4}{T} + \frac{\phi^6}{2}$ already displays the same qualitative behaviour. Moreover to this order it is easy to solve exactly and one obtains $T_c = \frac{1}{\sqrt{2}} \approx .7$ which is not a great quantitative match to the more exact $T_c = .3$, but the same physics is at play.
