It is well known that the 2nd moment (physics: moment of inertia) is minimized when centered around the mean (physics: center of mass). However, I recently realized that higher moments are NOT minimized at the same point. Specifically let me ask the 1-dimensional, discrete version of the question:
Given point masses $\{m_i\}$ all lying along the $x$-axis, with positions $\{x_i\}$. What value of $x$ would minimize $\sum m_i |x-x_i|^k$, for values of $k>2$? (If we normalize $\sum m_i=1$, then an equivalent interpretation is a real-valued random variable $X$ taking value $x_i$ with probability $m_i$.)
Here are some partial solutions:
(A) For $k=2$, the optimal $x$ is the (weighted) mean, aka center of mass, $x^* = \frac{\sum m_i x_i}{\sum m_i}$.
(B) For $k \rightarrow \infty$, the optimal $x$ is the point that minimizes the maximum individual "deviation" $d_i = |x-x_i|$, so the optimum is halfway between the leftmost and rightmost points, i.e. $x^* = (\min_i x_i + \max_i x_i)/2$
(C) For any $k\ge2$, if there are just two point masses, the optimal $x$ is the value satisfying: $m_1 d_1^{k-1} = m_2 d_2^{k-1}$, where $d_i = |x-x_i|$. (Easy proof by differentiating $m_1 d_1^k + m_2 d_2^k$.)
Any idea how to solve the general problem?
(Updated Note: For odd $k$, a different version of the problem would be to minimize $\sum m_i (x - x_i)^k$ -- but that sum has no minimum because $\sum m_i (x - x_i)^k \rightarrow -\infty$ as $x \rightarrow -\infty$.)
