Comment by srean

Comment by srean 17 hours ago

1 reply

I can't edit my comment anymore so let me elaborate a bit here.

What is this sneaky connection between squared Euclidean and Cartesian coordinates that I mentioned ? Why are they such a compatible pair ?

The answer is the Pythagorean theorem.

The squared Euclidean distances decomposes nicely along orthogonal (perpendicular) directions.

    d^2 = x^2 + y^2.
The Cartesian coordinates decomposes a point along orthogonal (perpendicular) axes as well, which we know is special for squared Euclidean distances.

The other metrics considered in the blog post decompose as, for lack of a better name, Fermat's last theorem decomposition.

    d^n = x^n + y^n
Now if we were to use a coordinate system that decomposes points like that, that would be interesting to explore. I don't know of coordinate systems that do that.

This much is true, forget about integral triples (lattice points) for integral n > 2.

lupire 9 hours ago

I don't understand. The Cartesian coordinate system works fine in any norm. See the OP article.