1. ## 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.

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

This is my 17th Post!!!

3. 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.
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

4. hello,

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

d(a,b) is minkowski distance

5. 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

6. Thanks alot

7. 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