Minkowski distance

Printable View

July 17th 2006, 08:42 AM
wosci

Minkowski distance

Hi,
i'm looking for the proof of minkowski metric in infinite that become the largest of the differences of the coordinates of x and y.
can anybody help me please?
July 17th 2006, 11:18 AM
ThePerfectHacker

I cannot understand anything you said?
...Infinite?
...Minkowski metric? there is no such thing?

This is my 17:):)th Post!!!
July 17th 2006, 12:42 PM
topsquark

Quote:

Originally Posted by wosci

Hi,
i'm looking for the proof of minkowski metric in infinite that become the largest of the differences of the coordinates of x and y.
can anybody help me please?

I really don't understand what you are asking for either. However, for the record the Minkowski metric is either
$\begin{bmatrix} 1 & 0 & 0 & 0\\0 & -1 & 0 & 0\\0 & 0 & -1 & 0\\0 & 0 & 0 & -1 \end{bmatrix}$

or

$\begin{bmatrix} -1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{bmatrix}$

depending on your choice of style. (The upper left corner is the $g_{00}$ entry.)

-Dan
July 17th 2006, 02:25 PM
wosci

hello,

lim d(a,b) = Dmax(a,b)
when r-> infinite

d(a,b) is minkowski distance
July 17th 2006, 10:08 PM
CaptainBlack

Quote:

Originally Posted by wosci

hello,

lim d(a,b) = Dmax(a,b)
when r-> infinite

d(a,b) is minkowski distance

The Minkowski distance between two points $x,\ y$ in $\mathbb{R}^N$ of order $p$ is:

$ d_p(x,y)=\left[ \sum_1^N |x_i-y_i|^p \right]^{1/p} $

The problem here is to show that:

$ \lim_{p \to \infty} d_p(x,y)= \max_{i=1,..N} |x_i-y_i| $

To prove this we observe that if we let $z_i=|x_i-y_i|$ then:

$ d_p(x,y)=D(z,p)=\left[ \sum_1^N z_i^p \right]^{1/p} $

and if $z_j=\max_{i=1..N} z_i$

$ z_j \le D(z,p) = z_j \left[1+\sum_{i=1,..N;\ i \ne j} \left(\frac{z_i}{z_j}\right)^p \right]^{1/p}$ $\le z_j\ \left[1+(N-1) \right]^{1/p} $

that is:

$ z_j \le D(z,p) \le z_j\ \left[N \right]^{1/p} $

But the limits of both ends of this chain of inequalities as $p \to \infty$ are the same and equal to $z_j$ , hence:

$ \lim_{p \to \infty} D(z,p)=z_j=\max_{i=1..N} z_i $ ,

which proves the required result.

RonL
July 18th 2006, 03:13 AM
wosci

Thanks alot :)
July 18th 2006, 04:05 AM
CaptainBlack

Quote:

Originally Posted by wosci

Thanks alot :)

That's OK - note I have left some of the minor detail out, which you may
want to fill in yourself.

RonL