The 4th dimension

[fallow bear]

I fully understand that the 4th dimension is a Tesseract but i bealive that there is a way to show it in different form

[night anvil]

I mean the 4th dimension ain't all that crazy when u learn how to work with mutlivariable functions

[fallow bear]

Could you explain more?

[night anvil]

Like I could find the hypervolume of the n-pyramid bounded by the positively oriented axes in n-space and the n-1 space determined by 1 - x_1 - x_2 - x_3 ... - x_(n-1)

You gotta take a multi-integral for it

So i guess for an example let's calculate the hypervolume of the 4th dimensional hyperpyramid

Bounded by 1-x-y-z

and w is a function of those variables

Then we get $\int_0^1 \int_0^{1-x} \int_0^{1-x-y} \int_0^{1-x-y-z} \dd w \dd z \dd y \dd x$

[potent quartz]

hiidostuff

[night anvil]

And at this point it's just a matter of calculation

You end up getting 1/4!

Or 1/24

And for an n-pyramid of this form it's n-volume will be 1/n!

[fallow bear]

That makes so much more since, Thank you!

[night anvil]

No problem

We can do it for a tesseract as well

It's 4-volume would just be n^4

No need for integrating as things are constant

[fallow bear]

So are you saying the tesseract would just be a shape being repeated 4* bigger and always connected?

[night anvil]

"Bigger" is not really the right way

Its like saying a cube is bigger than a square

Not really able to be compared like that as they live in different spaces

[fallow bear]

So are you saying if the last shape is lets say K and the "new" shape is J could the formula be K*4=J

[night anvil]

Do you mean K would be something like a cube and J would be a tesseract?

[fallow bear]

Yes

Would that be correct to say?

[night anvil]

No because the respective volumes can't scale linearly

It has to be with respect to a new side length

For the tesseract case, every side length is the same

So if we have n as a side length and d as a dimensional measure

Then we know K = n^(d-1)

And K * n = n^d

Which is consistent with an n-square measure

[fallow bear]

What im getting from this is the formula represents all the points of a tesseract. Each point has four coordinates, and each coordinate can take one of two possible values. This creates all the possible combinations of these coordinates, resulting in the total number of points (vertices) of the tesseract?

[pale thistle]

Dimension is basically a measure of how many independent axes there.

[night anvil]

^

And for an axis to be independent to every other, it has to be orthogonal to every other

[fallow bear]

If i am correct there would be 16 combinations of equasions

[pale thistle]

(Sorta. In a more abstract way, no)

[night anvil]

I know but that's like higher level

[fallow bear]

Okay i gotta go thank you for helping me understand

[night anvil]

No problem

[turbid osprey]

This is only true for spatial dimensions

[turbid osprey]

What do you mean when you say THE 4th dimension?

[night anvil]

True but here he was implying the 4th spatial dimension

As implied by trying to visualize it

[turbid osprey]

good catch