In studying relations between elementary functions, I decided to try to better understand them via their Maclaurin series. I began to study inverse relations.
It's clear that $e^x - 1$ has inverse $\log(1+x)$. I began looking at their odd and even portions.
Looking at $\frac{e^x - e^{-x}}{2}$, which is of course $\sinh(x)$, I thought that $\sinh^{-1}(x)$ might result from looking at the odd part of $\log(1+x)$. But, to my surprise, $\frac{\log(1+x) - \log(1-x)}{2}$ is equal to $\tanh^{-1}(x)$.
Now, I had no reason to believe that I'd be able to derive the inverse of $\sinh(x)$ in this manner, but it seemed reasonable to try. What I did not expect is that I'd derive the inverse of $\frac{\sinh(x)}{\cosh(x)}$ in this manner. There seems to be a deeper reason here.
My general questions are:
- If $f$ and $f^{-1}$ are known, are there any known conditions under which the even/odd portions of each will also be inverses? How about their imaginary portions? (By which I mean going from, say, $\sinh(x)$ to $\sin(x)$ via $\frac{\sinh(ix)}{i} = \sin(x)$)
- Is there any known reason this process yields $\tanh^{-1}$ -- a ratio of the even and odd? This was surprising.
I'm still keen to explore, but I thought to raise this question as it seems to be an elementary topic and there may be many known results already.
Thank you.