Functional equation $f(x) = f(3x) + \tanh(x)$

The equation itself: $$f(x) = f(3x) + \tanh(x)$$ So firstly I'm solving homogeneous equation: $$f(x)=f(3x)$$ so is just periodic function $\Theta(\ln x)$ with period $\ln 3$. So: $$F(x) = \Theta(\ln x) + \hat{f}(x)$$where $\hat{f}(x)$ is the particular solution of equation. Any tips how to find some?

It seems like its really bad around 0, but going smooth on big values. And my current guess that its behavior around $x=0$ is strongly connected to period of $\Theta$

• never mind, I just saw the title :p May 19 '17 at 22:28
• Around zero, I would do a telescoping infinite sum like this: $f(\frac{x}{3^{n+1}})=f(\frac{x}{3^{n}})+\tanh(\frac{x}{3^{n+1}})$ No? May 19 '17 at 22:38
• For a suitable domain, you can sum on the telescoping series to get: $\sum\limits_{n=0}^{\infty} f(\frac{x}{3^{n+1}})-f(\frac{x}{3^{n}}) = \sum\limits_{n=0}^{\infty} \tanh(\frac{x}{3^{n+1}}) = f(0)-f(x)$ Of course this is true assuming $f$ is continuous which is garanteed I guess because of the equation involving $\tanh$ which is in turn continuous. This might give a hint on how the particular solution might be constructed. May 19 '17 at 22:47
• Try to replace $'x'$ by $\tanh^{-1}(x)$ May 19 '17 at 23:52
• @BarryCipra, it's only true if the function is continuos. May 22 '17 at 8:06

For functions $f:(0,\infty)\to\mathbb{R}$, the general solution to the functional equation
$$f(x)=f(3x)+\tanh(x)$$
$$f(x)=\Theta(\ln x)-\sum_{k=1}^\infty\tanh\left(x\over3^k\right)$$
where $\Theta:\mathbb{R}\to\mathbb{R}$ is any function satisfying $\Theta(x+\ln3)=\Theta(x)$.
Note that the infinite series is absolutely convergent for any $x$, since $\tanh(x/3^k)\approx x/3^k$ for large enough $k$. It defines a continuous (indeed, smooth) function on all of $\mathbb{R}$. If you want the function $f(x)$ to have a limit as $x\to0^+$, then you need for $\Theta$ to be constant; otherwise $f$ will approach all the values $\Theta$ takes on (as in the OP's figure).