The original problem that gave rise to this question is to solve
$$
\sum\_{i=1}^n\frac1{1+e^{-(x-c\_i)}}=\frac n2
$$
(it comes from the need to determine difficulty of a polytomous item under the graded model in Item Response Theory).


For $n=2$ the (well, a) solution is $\frac{c\_1+c\_2}2$, so it should behave as certain generalized mean of the $c\_i$. The question is whether there might be some explicit expression for it.


I just tried some obvious transformations, but without success. For example, the equation is equivalent to
$$
\sum\_{i=1}^n\tanh\frac{c\_i-x}2=0.
$$
Also, one may switch to polynomials: denoting $e^{-x}$ by $y$ and $e^{-c\_i}$ by $a\_i$ it becomes
$$
\sum\_{i=1}^n\frac1{1+y/a\_i}=\frac n2
$$
(solution being now the geometric mean of the $a\_i$ for $n=2$). Modulo some transformations this can be reformulated as finding solutions of $t\frac{d\log p}{dt}=\frac n2$ in terms of roots of a given polynomial $p(t)$ of degree $n$.


...decided to add yet another reformulation: one has to find an extremum (a root of the derivative) for
$$
\prod\_{i=1}^n\left(1-\frac z{b\_i}\right)^{b\_i^2-1}
$$



Does this kind of mean value have an explicit form?
Using your $y$ and $a\_i$ coordinates, one is seeking a solution to
$$\sum\_{i=1}^n\frac{1}{1+y/a\_i}=\frac{n}{2}.$$ Multiplying all this out this becomes some equation of the form $P(y)/Q(y)=0$ with $P(y)$ and $Q(y)$ polynomials in $y$ with coefficients involving the $a\_i$.


So one seeks solutions to $P(y)=0$. Now if we let $\sigma\_j$ be the $j$th elementary symmetric function in the $a\_i$ (so $\sigma\_1=a\_1+a\_2+\cdots+a\_n$ and $\sigma\_n=a\_1a\_2\cdots a\_n$ then it seems to me that (perhaps up to a constant -- these things are only defined up to constants) we have
$$P(y)=ny^n+(n-2)\sigma\_1 y^{n-1}+(n-4)\sigma\_2 y^{n-2}+\cdots-n\sigma\_n.$$
I didn't check this for all $n$ but I did check it for $n=5$.


This is kind-of bad, because it shows that $y$ is a root of a "random" polynomial. For example if one sets $n=5$ and $a\_1,a\_2,a\_3,a\_4,a\_5=1,2,3,4,5$ then we get a degree 5 polynomial $5y^5 + 45y^4 + 85y^3 - 225y^2 - 822y - 600$
with rational coefficients whose Galois group over the rationals is $S\_5$. 


This suggests to me that there will be no easy formula in general (unless you allow your formula to be powerful enough to spit out roots of quintics).


