Comment by bubblyworld

Comment by bubblyworld 18 hours ago

It's an infinite family of metrics - you provide an n (a positive integer) and get back a metric.

So if you pick n=1 you get d(x, y) = |x| + |y|, which is the taxicab metric. You can apply this metric to a Euclidean space of whatever dimension you like, just substituting the appropriate definition of |x| and |y|. For 1-dimensional space you would use |x| = abs(x[0]), for 2-dimensional space you would use |x| = sqrt(x[0]**2 + x[1]**2), etcetera. Hope that helps.

FabHK 18 hours ago

And this works for any positive n (need not be an integer, and I think the author has an example with 1.5). Normally the letter p is used instead.

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brewmarche 17 hours ago

I don’t think it’s a metric for 0 < p < 1 though.

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bubblyworld 13 hours ago

Sorry, on rereading this I've actually written a bunch of nonsense. It's true that each choice of n gives you back a metric, but the |x|, |y|, |z| actually refer to the components of your vectors. So in 1-dim space you just have |x|, in 2-dim space you have |x| and |y|, 3-dim space has |x|, |y| and |z| and so forth.

So in 3-dimensional space, for n=3, the "length" of a vector u=(u_x, u_y, u_z) is: d(u) = (|u_x|^3 + |u_y|^3 + |u_z|^3)^(1/3)

In 2-dimensional space, for n=4 and u=(u_x, u_y) you get the following: d(u) = (|u_x|^4 + |u_y|^4)^(1/4)

If you want to use the norm as a metric to work out the "distance" between two vectors u and v then you just compute d(u-v) where u-v is ordinary vector subtraction.

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