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?

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- Jul 17th 2006, 08:42 AMwosciMinkowski 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? - Jul 17th 2006, 11:18 AMThePerfectHacker
I cannot understand anything you said?

...Infinite?

...Minkowski metric? there is no such thing?

This is my 17:):)th Post!!! - Jul 17th 2006, 12:42 PMtopsquarkQuote:

Originally Posted by**wosci**

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

or

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

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

-Dan - Jul 17th 2006, 02:25 PMwosci
hello,

lim d(a,b) = Dmax(a,b)

when r-> infinite

d(a,b) is minkowski distance - Jul 17th 2006, 10:08 PMCaptainBlackQuote:

Originally Posted by**wosci**

$\displaystyle

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

$

The problem here is to show that:

$\displaystyle

\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 $\displaystyle z_i=|x_i-y_i|$ then:

$\displaystyle

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

$

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

$\displaystyle

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}$$\displaystyle \le z_j\ \left[1+(N-1) \right]^{1/p}

$

that is:

$\displaystyle

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 $\displaystyle p \to \infty$ are the same and equal to $\displaystyle z_j$, hence:

$\displaystyle

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

$,

which proves the required result.

RonL - Jul 18th 2006, 03:13 AMwosci
Thanks alot :)

- Jul 18th 2006, 04:05 AMCaptainBlackQuote:

Originally Posted by**wosci**

want to fill in yourself.

RonL