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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?

Upd: New pictureenter image description here

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 $\Theta$

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  • $\begingroup$ never mind, I just saw the title :p $\endgroup$ May 19 '17 at 22:28
  • $\begingroup$ 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? $\endgroup$ May 19 '17 at 22:38
  • $\begingroup$ 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. $\endgroup$ May 19 '17 at 22:47
  • $\begingroup$ Try to replace $'x'$ by $\tanh^{-1}(x)$ $\endgroup$ May 19 '17 at 23:52
  • $\begingroup$ @BarryCipra, it's only true if the function is continuos. $\endgroup$ May 22 '17 at 8:06
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For functions $f:(0,\infty)\to\mathbb{R}$, the general solution to the functional equation

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

is

$$f(x)=\Theta(\ln x)-\sum_{k=1}^\infty\tanh\left(x\over3^k\right)$$

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).

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