/sci/ - Science & Math

[Anonymous No.4678777]

Hey /sci/fags, I have a question... what would an object from a dimension higher than ours look like?

[Anonymous No.4678788]

Like this

[EK !EKFnQG0UBo No.4678800]

>>4678788
wow, thats cool

[Anonymous No.4678804]

>>4678777

[Anonymous No.4678825]

Pick related, it's Odin from the higher dimension of Asgard.

[Anonymous No.4678832]

>>4678800
>>4678788
>>4678777

Hmm, I GET your points.

[Anonymous No.4678842]

>>4678832

1/10 not a clever sentence to match a GET statement. I gave you the 1 point because you found the dubs and trips.

[Anonymous No.4678866]

>>4678825
No, it's Thor from Age of Mythology. See me after class.

[Anonymous No.4678888]

>>4678777
>>4678788
A good way to understand this is to build the 3D pattern of this hypercube (also called tesseract) by analogy with the 2D pattern of a cube.

What you do to get the 2D pattern of a cube is:
-take a square
-add 1 square on each side of it
-add another square at the end (that does make the 6 faces of a cube)

Then you fold your 2D pattern to get all the opposed edges (lines here) to touch.

Now in 3D:
-take a cube
-add 1 cube on each face of it (7 cubes)
-add another square at the end (8 cubes = 8 "faces")

Now you "fold" your pattern to get all the opposed edges (the are now squares) to touch.

[Anonymous No.4678912]

You would see it as a 3 dimensional object. Imagine that we would live in only 2 dimensions, on a piece of paper. If someone would put a nail trough the paper, the only thing that we would see would be the part on the same plane as our paper. We would see it as 2d, but in reality it would be just an infinitely thin slice.

[Anonymous No.4678918]

>>4678888
Forgot to draw the "opposed edges" on pic

There will be 6 of these on the 3D pattern

[Anonymous No.4678926]

>>4678888
Woah! Hold on a second here! I know you got quads and shit, but how do you know you have to add only ONE cube once you add one cube at each side?

[Anonymous No.4678965]

>>4678926
I'm not him, but I thing it's for the sake of the analogy

[Anonymous No.4679038]

>>4678926
>because it works
You can actually count the number of faces for a hypercube because it is easy to enumerate all its vertices (there 8 "faces" = 8 cubes)
(0,0,0,0)
(1,0,0,0)
(1,1,0,0)
... (there should never be more than one unit between two vertices)

You have <span class="math"> 4^2 = 16 [/spoiler] vertices in the hypercube : from (0,0,0,0) to (1,1,1,1).
There are 8 vertices on each cubes.
In 3D, 1 vertices of a face (=square) is shared by 3 faces because 3D (no more dimensions to stick another square there)
There are 4 vertices on a square so 4/3 vertices for each square on the cube. if N is the number of faces, N*4/3 = exact number of vertices = 8 (on a cube), therefore, N = 3*2 = 6 faces on a cube.

So in 4D, 1 vertices of a face (=cube) is shared by 4 faces.
N is the number of "faces" N*8/4 = 16, therefore N = 16/2 = 8 faces.

We need 8 cubes.

[Anonymous No.4679076]

>>4679038
>between two vertices
I meant between two following vertices

[Anonymous No.4679100]

>>4679038
>(there 8 "faces" = 8 cubes)
And you can scrap that part too.

Also, the general formula for n the number of dimensions and N the number of (hyper)faces of the hypercube is <div class="math"> N = \frac {n2^n}{2^{n-1}} = 2n </div>