Can I assume that $0^0 = 0$ when dealing with the center of order $0$ in statistics?

Question

I came up to the concept of centers in statistics. Honestly, I never heard about this before few days ago.

Given a collection of numbers $\{x_i\}$, $i \in \{1, \ldots, N\}$, the center of order $r$ is defined as the number $c$ which minimizes the following function:

$$f_r(c) = \sum_{i=1}^N \left|x_i - c\right|^r.$$

In particular, it comes out that:

1. The center of order $0$ is the mode of the collection $\{x_i\}$.
2. The center of order $1$ is the median of the collection $\{x_i\}$.
3. The center of order $2$ is the mean of the collection $\{x_i\}$.

---

For fun, I was able to prove point 3 (very easy) since $f$ is smooth. Point 2 is somehow hard, but I think I can solve it. Instead, for point 1, I'm having some troubles.

Consider the function $f_0(c)$. If $c \neq x_i$, then $x_i - c \neq 0$ and then $$\left|x_i - c\right|^0 = 1.$$

Therefore, if I choose $c$ such that $c \neq x_i ~\forall i$, then I get that:

$$f_0(c) = N.$$

Now, $\color{red}{\text{I assume in this context that}~ 0^0 = 0}$. That is, if $c = x_j$ for some $j$, I get:

$$\left|x_i - c\right|^0 = 0.$$

If the value of $x_j$ is repeated exactly $m$ times, then I get:

$$f_0(x_j) = N - m.$$

Therefore if $x_j$ correspond to the mode of the collection, then $m$ is maximum, and hence $f_0(x_j)$ attains its minimum.

Is this "proof" correct?

---

A numerical "evidence"

I consider the collection $\{1, 2, 2, 2, 3, 4, 5, 6\}$. In this case, the mean is $3.125$, the median is $2.5$ (between $2$ and $3$) and the mode is $2$.

Hereafter, I plot the function $f_r(c)$ for $r=2$ (mean) and $r=1$ (median). Also I plot the function for $r=0.1$, $r=0.01$, $r=0.001$ and $r=0.0001$.

Answer 1

The idea is nice, but I think the mode needs to be more carefully defined.

Of course, $0^0$ is undefined. Moreover, if you are taking the limit of $x^0$ as $x \rightarrow 0,$ that limit is 1. Consequently, in many software packages 0^0 returns $1,$ on the assumption that the base $0$ is likely to be an 'underflow' (roughly, a number too small to handle within the the range of rational values the computer can represent), and not a 'genuine' $0.$

In R statistical software for example:

0^0
## 1

n = 1:6;  x = 1/(n^(n^n));  b=x^0  
cbind(n, x, b)   # 6 x 3 matrix
     n             x b
[1,] 1  1.000000e+00 1
[2,] 2  6.250000e-02 1
[3,] 3  1.311373e-13 1
[4,] 4 7.458341e-155 1
[5,] 5  0.000000e+00 1  # x is underflow
[6,] 6  0.000000e+00 1  # x is underflow

Unfortunately, this doesn't work out so well in the (presumably rarer) case, in which it's the exponent that tends to $0.$ Thus programmers do have to be careful some of the time.

n = 1:6;  x = 1/(n^(n^n));  c=0^x  
cbind(n, x, c)
     n             x c
[1,] 1  1.000000e+00 0
[2,] 2  6.250000e-02 0
[3,] 3  1.311373e-13 0
[4,] 4 7.458341e-155 0
[5,] 5  0.000000e+00 1
[6,] 6  0.000000e+00 1

Answer 2

I disagree

$$f(x)=0^0 = a^x$$

using maclaurin series

$$a^x = 1 + x\ln a + \frac{x^2}{2!} (\ln a)^2 +. . . + \frac{x^n}{n!} (\ln a)^n$$

thus when $a = 0 , x= 0$

$$0^0 = 1 - \frac{0}{0} +\frac{0}{0} +. . . +(-1)^n+\frac{1}{0} \quad \text{because in} \quad 0 = -\frac{1}{0} \quad \text{i.e} \quad -\infty$$

hence $0^0 = 1 - \frac{0}{0} = \frac{0}{0}$

. : we can't assume that $0^0 = 0$

even the limit as $k \to 0$ , if the function is $k^k$ is not absolutely true