[Modes, Medians and Means: A Unifying Perspective](http://www.johnmyleswhite.com/notebook/2013/03/22/modes-medians-and-means-an-unifying-perspective/) defines the following centers based on the $L\_p$ norms:


$$
\begin{aligned}
\text{mode of x} = \arg \min\_s \sum\_i \lvert x\_i - s \rvert^0 \\
\text{median of x} = \arg \min\_s \sum\_i \lvert x\_i - s \rvert^1 \\
\text{mean of x} = \arg \min\_s \sum\_i \lvert x\_i - s \rvert^2 \\
\end{aligned}
$$


Where:


$$
\begin{aligned}
x = (x\_1, x\_2, \ldots, x\_n) \ &:\ \text{a list of rational numbers} \\
\lvert x \rvert \ &:\ \text{number of elements in the list if $x$ is a list} \\
\end{aligned}
$$


I wanted to extend this to the L4 center:


$$
\begin{aligned}
\arg \min\_s \sum\_i \lvert x\_i - s \rvert^4 = \text{L4 center} \\
\frac{d}{ds} \sum\_i (x\_i - s)^4 = 0 \\
\sum\_i \left[ \frac{d}{ds}(x\_i-s)^4 \right] = 0 \\
\sum\_i \left[ -4(x\_i-s)^3 \right] = 0 \\
\sum\_i (x\_i-s)^3 = 0 \\
\sum\_i \left[ x\_i^3 -3x\_i^2s + 3x\_is^2 - s^3 \right] = 0 \\
\sum\_i x\_i^3 + \sum\_i -3x\_i^2s + \sum\_i 3x\_is^2 + \sum\_i -s^3 = 0 \\
\sum\_i x\_i^3 -3s \sum\_i x\_i^2 + 3s^2 \sum\_i x\_i - s^3 \lvert x \rvert = 0 \\
\end{aligned}
$$


Which is a cubic with the following discriminant:


$$
\begin{aligned}
&+ 162 \lvert x \rvert \sum\_i x\_i \sum\_i x\_i^2 \sum\_i x\_i^3 \\
&- 108 \left(\sum\_i x\_i\right)^3 \sum\_i x\_i^3 \\
&+ 18 \left(\sum\_i x\_i\right)^2 \left(\sum\_i x\_i^2\right)^2 \\
&- 108 \lvert x \rvert \left(\sum\_i x\_i^2\right)^3 \\
&- 27 \left(\lvert x \rvert\right)^2 \left(\sum\_i x\_i^3\right)^2
\end{aligned}
$$


I was wondering if anything could be said about the roots of the L4 center given any list $x$ of rationals? For example, how many real roots, because intuitively I expect only one?



Can anything be said about the roots of the L4 center?
The function $\sum\_i(x\_i-s)^3$ is strictly increasing in $s$, from $-\infty$ to $\infty$. It is also continuous, so there is a unique real root.


