# Root of sum of shifted polynomials

For an arbitrary positive odd integer $k$, I would like to obtain an expression for the root $x_{root} \in \mathbb{R}$ of the following polynomial $$p(x) = \sum_{i=1}^N (x-x_i)^k,$$ where $x_i\in \mathbb{R}$, for all $i \in \{1,2,...,N\}$. For example, when $k=1$, we can trivially see that $x_{root} = \sum_{i=1}^N x_i/N$. I am stuck in finding a general expression for higher $k$'s.

Also, for all $k$ and $x_i$'s under the previously mentioned conditions, $p(x)$ has one real root because it is a sum of non-decreasing functions.

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Maybe there's no neat answer. Try -2, 1, 1 for $k=3$, see whether it looks like there will be any nice answer for that one. – Gerry Myerson Jan 26 '13 at 5:36

Your root is also a root for the general $k$ when $N=1$ or $2$. Since $a^3+b^3=(a+b)(a^2-ab+b^2)$ we have for example for $k=3$, $N=2$ case $$(x-x_1)^3+(x-x_2)^3=(2x-x_1-x_2)((x-x_1)^2-(x-x_1)(x-x_2)+(x-x_2)^2)$$ which has the root $(x_1+x_2)/2$ as well as two more. Possibly the average is the answer for all odd $k$ and all values of $N$. You could try plugging in the average and see what happens for various other $k$ and $N$.
Not for all $N$, surely. Try 1, 2, 6 with average 3; for odd $k$ you get $1+2^k-3^k$, rarely zero. – Gerry Myerson Jan 26 '13 at 5:33