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Re:Executing gradient descent on the earth by pbhjpbhj on Friday April 20, 2018 @09:03AM (#16884385)
Doesn't a sphere require all points to be separated from the centre point by the same distance r, where r>0? A point isn't a sphere.
Re:Executing gradient descent on the earth by gus_massa on Friday April 20, 2018 @11:15AM (#16885434)
Are you taking about the orange peel, the orange juicy part or the whole fruit?

If you consider only the "orange peel", if the center is 0 and r=1, then the 1D version is just two points {-1} and {1}, not only one point.

If you consider only the "whole orange" then the 1D case is the segment [-1, 1], if the center is 0 and r=1

In both case it is equal to the 1D "square" equivalent.

Re:Executing gradient descent on the earth by detaro on Friday April 20, 2018 @09:05AM (#16884393)
Really depends on who you are asking for a definition. E.g. there is the term "point sphere" for the special case of a sphere that is a point.
Re:Executing gradient descent on the earth by pbhjpbhj on Friday April 20, 2018 @09:08AM (#16884426)
Is there any interesting result that comes from casting a point as being any shape (presumably the point-sphere is also a point-dodecahedron, etc.)?
Re:Executing gradient descent on the earth by chestervonwinch on Friday April 20, 2018 @10:43AM (#16885146)
Define a topology on any set you'd like and a sphere is just the set of limit points of any open ball.
Re:Executing gradient descent on the earth by dTal on Friday April 20, 2018 @08:43PM (#16889384)
The 1D case is a line segment, not a point. A point is the 0D case and indeed a degenerate case.