Is there a way to find all analytic functions $f:\mathbb{R} \to \mathbb{R}$ which are a solution of the functional equation (where $f^{-1}$ denotes inverse of the function) $$f^{-1}(-x)=-f(x)$$ with the additional condition $$\left. \frac{df}{dx} \right|_{x=0}=1?$$
Asked
Active
Viewed 77 times
2
-
1By $f^{-1}$ you mean $\frac{1}{f}$ ? – Tuvasbien Apr 28 '20 at 20:44
-
No, I mean the inverse of $f$. – Fizikus Apr 28 '20 at 20:45
-
3If you call $g=-f$ you obtain the functional equation $g(g(x))=x$, $g'(0)=-1$ – Apr 28 '20 at 20:47
-
1@Gae.S.: Indeed. So $g$ is an involution with $g'(0) = -1$. – Fizikus Apr 28 '20 at 20:49
-
1Conversely, if $g$ is an involution with $g'(0)=-1$ then $-g$ satisfies the conditions in the question. This answer characterizes continuous involutions on $\mathbb R$. By applying this characterization to a power series around $0$, maybe something intelligent could be said about the function. – Jack M Apr 28 '20 at 23:52