TransferOperator

The transfer operator of a complex-valued function $g(x)$ is defined as:

$$ T(g)(z) = \sum_{w:g(w)=z} \frac{1}{g'(w)} $$

where the sum is taken over all the inverse branches $w$ of $g$.

To calculate the transfer operator of $g(x)=\tanh(\ln(1+x^2))$, we first need to find the inverse branches of $g$. Since $\tanh z$ is an even function and $\ln(1+x^2)$ is an odd function, it follows that $g(x)$ is an even function of $x$. Therefore, we only need to consider the inverse branches for $x\geq0$.

Let $w = \ln(1+x^2)$, then $x^2 = e^w - 1$. Solving for $x$, we get:

$$x = \pm \sqrt{e^w-1}$$

Since we only consider $x\geq0$, we take the positive branch, which gives:

$$x = \sqrt{e^w-1}$$

To find the derivative of $g$ at $w$, we use the chain rule:

$$g'(x) = \frac{d}{dx}\tanh(\ln(1+x^2)) = \operatorname{sech}^2(\ln(1+x^2))\cdot\frac{2x}{1+x^2}$$

Therefore, the transfer operator of $g(x)$ is:

\begin{align*} T(g)(z) &= \sum_{w:g(w)=z} \frac{1}{g'(w)}
&= \frac{1}{g'(\ln(1+z^2))}
&= \frac{1}{\operatorname{sech}^2(\ln(1+z^2))\cdot\frac{2z}{1+z^2}}
&= \frac{(1+z^2)\operatorname{cosh}^2(\ln(1+z^2))}{2z\operatorname{sinh}^2(\ln(1+z^2))} \end{align*}

Therefore, the transfer operator of $g(x)=\tanh(\ln(1+x^2))$ is $T(g)(z) = \frac{(1+z^2)\operatorname{cosh}^2(\ln(1+z^2))}{2z\operatorname{sinh}^2(\ln(1+z^2))}$, where $z$ is a complex number.