Signs of Maclaurin coefficients of $\frac{1}{e^x+q}$

Question

My question is based on question by Robert Israel. There is only one answer by Lucia and also two useful comments (by Lucia too), from which we understand, that if $$\frac{1}{e^x+q}=\sum\limits_{n=0}^{\infty}F_{n}\frac{x^n}{n!}$$ so signs of $F_{n}$ be the same as signs of $$\cos\left((n+1)\arctan\left(\frac{\pi}{\log q}\right)\right)$$ for $q=3$.

Is it obvious from Lucia's answer (not comments)? If not, why is that true? Is it true for any $q$?

Answer 1

To determine the asymptotic behaviour of the MacLaurin coefficients $F_n$ of a function $f(z)$, the answerer implicitly uses an approach due to Darboux (see Olver, Asymptotic and Special Functions (1997) 4.9.2, p. 310). Suppose that $r$ is the distance from the origin of the nearest singularity of $f(z)$, if we can find a ``comparison function'' $g(z)$ with the properties

• $g(z)$ is isomorphic in $0<\left|z\right|<r$

• $f(z)-g(z)$ is continuous in $0<\left|z\right|\le r$

• The coefficients $b_n$ in the Laurent expansion \begin{equation} g(z)=\sum_{-\infty}^\infty b_nz^n \end{equation} have known asymptotic behaviour,

then, for $n\to\infty$, \begin{equation} F_n=b_n+o\left( r^{-n} \right) \end{equation} The function \begin{equation} f(z)=\frac{1}{e^{z}+q} \end{equation} is meromorphic and has poles at \begin{equation} z_n= \ln q+\left( 2n+1 \right)i\pi \end{equation} where $n$ is an integer. The nearest poles are $z_1=\ln q+i\pi $ and $z_{-1}=\bar{z_1}$. Corresponding residues are both $-1/q$. We choose \begin{equation} g(z)=-\frac{1}{q}\left( \frac{1}{z-z_1}+\frac{1}{z-z_{-1}} \right) \end{equation} it satisfies the required conditions with \begin{equation} g(z)=\frac{1}{q}\sum_{n=0}^\infty \left( z_1^{-n-1}+z_{-1}^{-n-1}\right)z^n \end{equation} and thus, by denoting $z_1=\rho e^{i\varphi}$, \begin{equation} b_n=\frac{2}{q}\rho^{-n-1}\cos\left( n+1 \right)\varphi \end{equation} then \begin{equation} F_n\sim\frac{2}{q}\rho^{-n-1}\cos\left( n+1 \right)\varphi \end{equation} Their signs are thus approximatively given by that of $\cos\left(\left( n+1 \right)\arctan\frac{\pi}{\ln q} \right)$

Finally, it is remarked that, when $q=3$, we have $\varphi\simeq \frac{11\pi}{28}$ with a good accuracy. This explains the near 28-periodicity of the coefficients.