Question about graphing a shape on the cartesian coordinate system

Hey mesa.

What exactly do you mean? Do you mean instead of representing x you represent x² for a particular vector (where x is the 1st element of the vector)?

 

Well you can do it in a few different ways.

If you represent your system as <x²,y,z> then your co-ordinate system will look like a normal Cartesian system with the negative region for x < 0 removed entirely.

However if you want to represent <x,y,z>, then you will get a contracted transformed version of the x-axis, but if you are going from the x² to x, then you have the branch problem since you lose information of the sign by squaring the term.

What happens is that for x as opposed to its square, the rate of change along the axis is not uniform or linear, but non-linear: in the linear case if you double the magnitude on the x-axis you double the x value but in a non-linear if you double the magnitude the value is not simple double: it is less or greater than.

 

Maybe mesa means an actual square (quadrilateral)?

 

Ohh I see what you're saying now.

It depends on the number of dimensions and the particular attribute you are trying to visualize.

Typically what happens in high-dimensional spaces, is that a specific kind of projection is used to create a visual of that attribute in a two or three dimensional space.

The short answer is that it depends on the attribute you are trying to visualize, but the kinds of things would most likely be similar if not equivalent to the ones used in a normal cartesian geometry. (It's hard to give specifics without specifics).

 

##|x|+|y|=1##

##max \{ |x|,|y|\} = 1##

 

You can rotate and translate it to anywhere and resize it by changing the 1:$$

|(x-a)\cos(\alpha)-(y-b)\sin(\alpha)|+|(x-a)\sin(\alpha)+(y-b)\cos(\alpha)| = 1$$

 

[tex]|x+y|-|x-y| = 1[/tex]

 

That's just a typo, he means ##|x+y|+|x-y|=1##. Same as my first one but squared up with the axes.

 

You're right, thanks.