I want to (asymptotic function) approximate the series $f(x) = \sum_{n=0}^\infty 2^{-n} \tanh(3^n x)$ around $x = 0$.
I found the equation for $f(x)$:
$$f(x) = \dfrac{1}{2}\ f(3x) + \tanh(x)$$
for $f(x)=\dfrac{1}{2}f(3x) \Rightarrow f(x) = A\ x^{\log_32}$
let $A = A(x) \Rightarrow A(x) = \Theta(\ln(x))$ where $\Theta(x) $ any function with period $\ln3$
An alternative justification for the $\log_3 2$ exponent is as follows: suppose we want to approximate $f$ by some $x^a$, where $a>0$ so that $\lim_{x\to 0} x^{-a}f(x)=c$ with $0<c<\infty$. The functional equation shows $$c=\frac {3^a}2c+\begin{cases} 0 & \text{if }a<1 \\ 1 &\text{if } a=1 \\ \infty & \text{if }a>1\end{cases}$$ Which forces $a<1$ and $a=\log_3 2$.
Given this value of $a$, instead of looking at $\lim x^{-a}f(x)$ as $x\to 0$ we look at $F(x)=3^{-ax}f(3^x)=2^{-x}f(3^x)$ as $x\to -\infty$. Putting $g(x)=2^{-x}\tanh(3^x)$ shows $F(x)=\sum_{n\ge 0}g(x+n)$. Because $g(x)\le(\frac 32)^x$, $\sum_{n\le 1} g(x+n)\le 2(\frac 32)^x$ and thus $\bar F(x)=\sum_{n=-\infty}^{\infty}g(x+n)$ tends to $F$ as $x\to -\infty$. But $\bar F$ has period $1$ so all information about it can be gleaned from its behavior on $[0,1]$.
Since we also have $g(x)\le 2^{-x}$, for any $x$ in the unit interval, $g(x+n)\le 2^{1-n}$ and $g(x-n)\le (\frac 32)^{1-n}$ so that $$\left|\bar F(x)-\sum_{-N}^N g(x+n)\right|\le 2^{1-N}+3\left(\frac 23\right)^N $$ Therefore, the series converges uniformly, and does so at a geometric rate. This means one can obtain a very good approximation for $\bar F$ and this in turn approximates $F$ in the limit.