The functions $\tanh x$ and $\arctan x$ have a similar graph. Is there a formula to transform $\tanh x$ to $\arctan x$?
[Math] Relationship between $\tanh x$ and $\arctan x$
trigonometry
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Best Answer
The two functions resemble each other on the real line, but they're not the only sigmoidal functions around. There are functions as simple as $\dfrac{x}{\sqrt{1+x^2}}$ and as complicated as $\dfrac2{\sqrt \pi}\int_0^x e^{-t^2}\mathrm dt$ that also have the s-shape.
Also, the resemblance ends when we look at their behavior for complex arguments:
The former possesses branch cuts, while the latter exhibits periodicity on the imaginary axis.
On the other hand, a look at the Maclaurin series for $\tan\tanh\,z$ and $\mathrm{artanh}\arctan\,z$ gives a clue as to why there is some resemblance between $\arctan$ and $\tanh$:
$$\begin{align*} \tan\tanh\,z&=z-\frac{z^5}{15}+\frac{z^7}{45}-\frac{z^9}{2835}-\frac{13 z^{11}}{4725}+\cdots\\ \mathrm{artanh}\arctan\,z&=z+\frac{z^5}{15}-\frac{z^7}{45}+\frac{64 z^9}{2835}-\frac{71 z^{11}}{4725}+\cdots \end{align*}$$