Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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

1 Answer 1

up vote 2 down vote accepted

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 $N$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.