- #1

- 648

- 18

How would you graph a square of any dimension on the cartesian coordinate system?

- Thread starter mesa
- Start date

- #1

- 648

- 18

How would you graph a square of any dimension on the cartesian coordinate system?

- #2

chiro

Science Advisor

- 4,790

- 132

What exactly do you mean? Do you mean instead of representing x you represent x

- #3

- 648

- 18

Any way you like, I just want to see other solutions than mine.

What exactly do you mean? Do you mean instead of representing x you represent x^{2}for a particular vector (where x is the 1st element of the vector)?

- #4

chiro

Science Advisor

- 4,790

- 132

If you represent your system as <x

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 x

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.

- #5

Mentallic

Homework Helper

- 3,798

- 94

Maybe mesa means an actual square (quadrilateral)?

- #6

- 648

- 18

Yes, an actual square.Maybe mesa means an actual square (quadrilateral)?

Sorry about that Chiro, I should have read your post more carefully and been more specific.

- #7

chiro

Science Advisor

- 4,790

- 132

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).

- #8

- 648

- 18

Basically I am looking for (an)other function(s) that will result in a graph of a plain old square on a two dimensional cartesian coordinate system.

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).

If there is more than one solution then pick your favorite, this is new territory for me although it was a great deal of fun working up a solution.

Now my function can graph a square about the origin although I am having difficulty limiting the range/domain outside of it lol. I have also played with trying to make rectangles too off the original function but I changed x to being a dependent variable along with y so transformations and deformations seem to work very differently than what I am used to... Yikes!,

should be fun :)

- #9

- 9,555

- 766

##|x|+|y|=1##Basically I am looking for (an)other function(s) that will result in a graph of a plain old square on a two dimensional cartesian coordinate system.

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

- #10

- 648

- 18

That's a nice solution, and takes care of the domain/range problems I was having.##|x|+|y|=1##

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

Got anything else?

- #11

- 9,555

- 766

You can rotate and translate it to anywhere and resize it by changing the 1:$$That's a nice solution, and takes care of the domain/range problems I was having.

Got anything else?

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

- #12

- 648

- 18

Very interesting, okay from your answers I think I see how I might fix my domain/range issues.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$$

As it stands I have:

[(y^2)^(1/2)-n]^2 = x^2

with n being the number that fixes the size of the square. I used a trig based function that produced a square root of -1 to keep the domain/range in check outside the 'square' but it's just tacked on and even though it sort of works it's kind of crappy so I removed it in search of a proper solution.

- #13

- 45

- 0

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

- #14

- 648

- 18

Isn't there a problem at x=0 or am I missing something?[tex]|x+y|-|x-y| = 1[/tex]

- #15

- 9,555

- 766

[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.Isn't there a problem at x=0 or am I missing something?

- #16

- 45

- 0

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

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 776

- Replies
- 4

- Views
- 2K

- Replies
- 9

- Views
- 4K

- Replies
- 10

- Views
- 2K

- Replies
- 6

- Views
- 1K

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 2K