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 $$ \prod_{i=1}^n\left(1-\frac z{b_i}\right)^{b_i^2-1} $$


1 Answer 1


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).

  • $\begingroup$ Mmm I would not say this is random since $\frac n2$ on the right is a very special thing. E. g. in the $\tanh$ reformulation it gives zero on the right. Accordingly, your $P(y)$ can be written as$$2yF'(y)-nF(y),$$ where $F$ is the polynomial with $a_i$ as roots. This is what I had in mind in the last sentence about logarithmic derivative. Roots of $F$ are known, and one has to express (one of the) roots of $P$ in terms of them. $\endgroup$ Feb 7, 2017 at 21:57
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    $\begingroup$ Yes I see your point -- "random" is not a good word in the above. On the other hand I think the Galois group computation above is an indication that "simply" expressing the roots of $P$ in terms of the roots of $F$ can't really be done. By the way, in my numerical example four roots are negative (so can't be exp of something) and the other is $2.627581098\ldots$, so there is a unique solution -- it's just not expressible in terms of elementary functions. $\endgroup$ Feb 7, 2017 at 22:05

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