Are the symmetries of any truncation of a asymptotic series preserved when summing to all orders? I am a researcher in Physics. Recently I have to deal with some asymptotic series.
I have a naive question. Usually for a convergent series, if any truncation of the series respects some symmetry, then the sum of the series respects the same symmetry. I wonder whether asymptotic series has the same property. Thank you in advance for your answers!
 A: An asymptotic expansion does not determine a function uniquely: infinitely many functions can possess the same asymptotic expansion. There are theorems (by Watson, Nevanlinna, etc.) that give conditions for uniqueness (certain type of error bounds must hold in large enough sectors of the complex plane). The expansions are valid in sectors and can change accross certain complex rays (Stokes lines). Thus, even basic properties like evenness can be difficult to study because the expansion may be valid for $z$ but not for $-z$. Consider for example the asymptotic expansion of the logarithm of the gamma function:
$$
\log \left( {\frac{{\Gamma (z)}}{{z^{z - 1/2} e^{ - z} \sqrt {2\pi } }}} \right) \sim \sum\limits_{k = 1}^\infty  {\frac{{B_{2k} }}{{2k(2k - 1)z^{2k - 1} }}} 
$$
(with $B_{2k}$ being the Bernoulli numbers). The series is an odd one, but the function on the left is not odd. It turns out that the rays $\arg z =\pm \frac{\pi}{2}$ are Stokes lines, and the correct expansion to use for $\Re (z)<0$ ($\Im (z)\neq 0$) is
$$
\log \left( {\frac{{\Gamma (z)}}{{z^{z - 1/2} e^{ - z} \sqrt {2\pi } }}} \right) \sim \log (1 - e^{ \pm 2\pi iz} ) + \sum\limits_{k = 1}^\infty  {\frac{{B_{2k} }}{{2k(2k - 1)z^{2k - 1} }}} ,
$$
where the upper or lower sign is chosen according as $\Im (z)>0$ or $\Im(z)<0$. You can show this via the reflection formula for the gamma function. Another example is the modified Bessel function $I_0(z)$:
$$
I_0 (z) \sim \frac{{e^z }}{{\sqrt {2\pi z} }}\sum\limits_{k = 0}^\infty  {( - 1)^k \frac{{a_k }}{{z^k }}}  \pm i\frac{{e^{ - z} }}{{\sqrt {2\pi z} }}\sum\limits_{k = 0}^\infty  {\frac{{a_k }}{{z^k }}} 
$$
for large $z$ in the sectors $0 <  \pm \arg z < \pi$ (the $a_k$'s are certain rational  umbers). The rays $\arg z=0$ (positive real line) and $\arg z =\pm \pi$ are Stokes lines. To obtain the asymptotics on a Stokes line, one usually takes the average of the expansions on the two sides of it. Thus, for $z>0$, we have
$$I_0 (z) \sim \frac{{e^z }}{{\sqrt {2\pi z} }}\sum\limits_{k = 0}^\infty  {( - 1)^k \frac{{a_k }}{{z^k }}}.$$
Hence, in this case, the expansion enherites the realness of the function. These were just a few thoughts that came into my mind. This is a vast and exciting topic.
