Let $x,y > 0$ Consider the equation
$$\ln(\exp(x) + \exp(y)) = x + y$$
Now it is clear (by symmetry) that if we express $y$ as function of $x$ or $x$ as a function of $y$ that is the same function.
So
$$f(x) = y , f(y) = x $$
This implies
$$f(f(x)) = x , f(f(y)) = y$$
So the analytic function $f(x)$ is an involution for $x>0$, yet not a rational function.
In fact we can explictly find $f(x)$.
$$f(x) = x - \ln(\exp(x) - 1)$$
You can see for yourself that $f(f(x)) = x$.
I wondered about continu iterations of this function.
I realized
$$f(x) = \ln\left(\frac{\exp(x)}{\exp(x)-1}\right)$$
Let
$$f(x) = \ln(L(\exp(x)))$$
Then it becomes clear that basicly we are iterating
$$L(x) = \frac{x}{x-1}$$
for $x \neq 0, x \neq 1$.
This is well studied.
But then I started to wonder about complex numbers.
$$\ln(\exp(x) + \exp(y) + \exp(z)) = x + y + z$$
$$\ln(\exp(x) + \exp(y) + \exp(z) + \exp(a) ) = x + y + z + a$$
etc
How do those additions relate to the function $f(x)$ from above ?
Clearly
$$y = f(x) , z = f(x + f(x))$$
works for
$$\ln(\exp(x) + \exp(y) + \exp(z)) = x + y + z$$
however it did use
$$\ln(\exp(x) + \exp(y)) = x + y$$
as a restriction.
This looses a degree of freedom or something.
Notice iterations of $x + f(x)$ correspond to iterations of
$$L_2(x) = \frac{x^2}{x-1}$$
which is more complicated perhaps.
So finally I come to the main question :
$$\ln\left(\sum_n \exp(x_n)\right) = \sum_n x_n$$
What do we know about that equation ?
And in particular :
Let $x_n$ be linear independant and consider
$$A = \ln\left(\sum_{n=1}^{\infty} \exp(x_n)\right) = \sum_{n=1}^{\infty} x_n$$
Does that even make sense ? Does $A$ even converge ?
What can $A$ converge to ? Does this relate to the fixpoints of $\exp(z)$ ?
what do we know about the $x_n$ ?