There seems to be a generalization of the notion of the "asymptotic series" to a notion of "trans-series". One "simple" example of this seems to arise in trying to solve the Euler's equation by power-series, $$\frac {d\phi (z)} {dz } + A\phi(z) = \frac{A}{z} $$
A power series solution to this is, $$\phi_0(z) = \sum_{n=0}^{\infty} \frac{a_n } {z^{n+1}} $$ where $a_n = A^{-n}n!$. This series is an asymptotic series with zero-radius of convergence because the coefficients grow factorially in $n$.
- Now what I don't understand is as to how from all this it follows that one can construct a ``family of formal solutions to this ODE based on $\phi_0(z)$" as, $\phi(z) = \phi_0(z) + Ce^{-{Az}}$
The above is apparently an example of a "trans-series".
- I don't undertand as to how this has two small parameters $z$ and $e^{-\frac{A}{z} }$?
Here the strength of the non-perturbative effect $(A)$ and the divergence of the asymptotic series are related.
- But also in what sense is the $Ce^{-\frac{A}{z}}$ ``non-analytic at $z=\infty$"? That all the possible Taylor series coefficients go to $0$ on taking the $z=\infty$ limit? Is this a signature of being a trans-series?
The equation in question, $(2.17)$ in the book, is $$ \phi'(z)+A\phi(z)=\frac Az\tag{1} $$ Note that $(1)$ can be written in a way to generate an asymptotic series solution: $$ \phi(z)=\frac1z-\frac{\phi'(z)}{A}\tag{2} $$ Iterating this as a recursion, we get an asymptotic approximation to a solution, $(2.18)$ in the book: $$ \phi_0(z)=\frac1z+\frac1{Az^2}+\frac2{A^2z^3}+\frac6{A^3z^4}+\cdots\tag{3} $$ Given any two solutions to $(1)$, their difference must satisfy the homogeneous equation $$ \phi_h'(z)+A\phi_h(z)=0\tag{4} $$ The solution to $(4)$ is $$ \phi_h(z)=Ce^{-Az}\tag{5} $$ Thus, as mentioned in $(2.19)$ of the book, the general solution would be $$ \phi_0(z)+Ce^{-Az}\tag{6} $$ I think one point that the author is trying to make is that the asymptotic series for all of these solutions is given in $(3)$ since $\phi_h(z)$ decays much faster than any power of $\frac1z$. Unfortunately, there seem to be some confusing typos in point 2. following $(2.19)$ related to an earlier comment, $(2.4)$ in the book.