The Minkowski distance of order p between two points
$X=(x_{1},x_{2},\ldots ,x_{n}){\text{ and }}Y=(y_{1},y_{2},\ldots ,y_{n})\in {\mathbb {R}}^{n}$
is defined as:
${\displaystyle D\left(X,Y\right)=\left(\sum _{i=1}^{n}|x_{i}-y_{i}|^{p}\right)^{1/p}} \tag{1}$
simplified, let p = 2, and then
${\displaystyle D\left(X,Y\right)=\left(\sum _{i=1}^{n}|x_{i}-y_{i}|^{2}\right)^{1/2}} \tag{2}$
I am trying to construct a connection between equation_2 and this figure that comes from the same wiki page.
Where, the unit circle and x-axis cross at point (1,0) (ignored other crosses)
the Minkowski distance with $p=2$ corresponds to the Euclidean distance, and the point (1,0) is at the unit distance from the centre, so do other point at the unit circle.
Let $X=(x_{1},x_{2})\in {\mathbb {R}}^{2}$, Y = (0,0)
the points at the unit circle could be viewed as a bound, inside which, X was mapped from $R^2$ to $[0,1]^2$ (I am not sure if I use the notation, please feel free to fix it).
Is my understanding right?