I have the following function $f(x)$ given a pair of positive real parameter $s$ and $t$:
\begin{aligned} f(x) &= 1, \text{if $-max(s,t)<x<0$};\\ f(0) &= 1.5; \\ f(x) &= f(x-s)+f(x-t), \text{if $x > 0$}. \end{aligned}
I have explored some properties of this function (with some not-so-strict proof that I'll omit here):
- $f(x)$ grows exponentially
- $f(x) = 1 + \sum_{p=0}^{\infty}\sum_{q=0}^{\infty}\frac{(p+q)!}{p!q!}u(x - (ps+qt))$, where $u(x)$ is the unit step function with $u(0) = 0.5$
- (obtained using Laplace transform) $f(x) = \sum_{r}\frac{e^{rx}}{r(se^{-sr}+te^{-tr})}$, where $r$ are all the complex roots of equation $e^{-sr}+e^{-tr} = 1$. The summation order is using $r$ that's closer to the origin first.
In the third bullet point, the equation $e^{-sr}+e^{-tr} = 1$ always has a single positive real root, which I call $r = R$. I want to show that its corresponding term is the "dominant term" of the function. That is, if we define $$ F(x) = \frac{e^{Rx}}{R(se^{-sR}+te^{-tR})} $$ I want to show the limit $$ \lim_{x\to\infty} \frac{f(x)}{F(x)} = 1 $$ which is equivalent to showing that $$ \lim_{x\to\infty} \sum_{r\neq R}\frac{e^{(r-R)x}}{r(se^{-sr}+te^{-tr})} = 0 $$
I actually discovered that this isn't always true. In particular, whenever $s/t\in\mathbb{Q}$, this limit doesn't exist, as it oscillates forever. What I haven't fully understand is the $s/t\notin\mathbb{Q}$ case, for which I suspect that the limit does exist.
I noticed that when $s/t\notin\mathbb{Q}$, $R$ has the single largest real components among all roots, so each individual term $e^{(r-R)x}$ has a limit of 0. I wish I could exchange the limit and summation operators to prove that the sum also has a limit of 0, but I can't find a theorem to back it up. Tannery's theorem doesn't work, as the sum isn't even absolute convergent (hence there is no hope for me to find the bound that appears in the theorem).
So my question is: is there any tools I can use here to exchange the limit operator, or is there a different ways to find the limit using $f$'s other properties? Or maybe the limit actually doesn't exist, and if so, is there a way to show it?
I ended up finding a tricky solution for the specific problem.
First let's introduce a smooth version of the unit jump function $$ u_a(x) = \begin{cases} 0 &\quad (x < 0)\\ x/a &\quad (0\le x \le a)\\ 1 &\quad (x > a)\\ \end{cases} $$ And use it to construct a smooth version of $f(x)$: $$ f_a(x) = 1 + \sum_{p=0}^{\infty}\sum_{q=0}^{\infty}\frac{(p+q)!}{p!q!}u_a(x-(ps+qt)) = \sum_r\frac{(1-e^{-ar})e^{rx}}{ar^2(se^{-sr}+ts^{-tr})} $$ which luckily converges absolutely (because this is a $\sum 1/n^2$-like series, which is because roots $r$ has almost periodic layout, which is intuitive to see but tedious to prove), and we can find bounds to apply Tannery's theorem. The limit is $$ \lim_{x\to\infty} \frac{f_a(x)}{F(x)} = \frac{1-e^{-aR}}{aR} $$
At the same time, we have $$ u_a(x)\le u(x) \le u_a(x+a) $$ which means $$ f_a(x)\le f(x) \le f_a(x+a) $$ and then their limit $$ \frac{1-e^{-aR}}{aR}=\lim_{x\to\infty} \frac{f_a(x)}{F(x)}\le \liminf_{x\to\infty} \frac{f(x)}{F(x)}\le \limsup_{x\to\infty} \frac{f(x)}{F(x)} \le \lim_{x\to\infty} \frac{f_a(x+a)}{F(x)}=\frac{e^{aR}-1}{aR} $$ By choosing arbitrarily small $a$, we squeeze the limit $$ 1\le \liminf_{x\to\infty} \frac{f(x)}{F(x)}\le \limsup_{x\to\infty} \frac{f(x)}{F(x)}\le1 $$
So I guess the takeaway from this is that, if the series you want to exchange limit and sum doesn't work well, try looking at its neighbors.