Okay so, $x^2$ is square, $x^3$ is cube, $x^4$ is m-cube?

Question

I'm trying to get some intuition, I think it almost makes sense, since you can think of the area of a square is length multiplied by height. The volume of a cube is length multiplied by height, multiplied by depth. I don't understand how to find the volume of the hypercube. What's the correct intuition?

Answer 1

Ok, so what you would call the "volume" of a cube is the "hyper-volume" of a "hypercube". And regular volume of a hypercube is like surface area on a regular cube. To put these concepts into equations:

(by squares, I mean rectangles. by cubes, I mean rectangular prisms. etc.)

$$\begin{align} \mathsf{Hyper\ V} & =2^0\left(\underbrace{l_1[\times(l_2\times[l_3\times(l_4)])]}_{1\times2\times3\times4}\right) \\ \mathsf{Volume} & =2^1\left(\underbrace{l_1\times[l_2\times(l_3+l_4)]}_{1\times2\times(3+4)}+\underbrace{l_2\times[l_3\times(l_4)]}_{2\times3\times4}\right) \\ \mathsf{Area} & =2^2\left(\underbrace{l_1\times(l_2+l_3+l_4)}_{1\times(2+3+4)}+\underbrace{l_2\times(l_3+l_4)}_{2\times(3+4)}+\underbrace{l_3\times(l_4)}_{3\times4}\right) \\ \mathsf{Length/Edge?} & =2^3\left(\underbrace{l_1+l_2+l_3+l_4}_{1+2+3+4}\right) \end{align}$$

See if you can spot the pattern and construct it for regular cubes and squares.

Answer 2

The only reason that "square" and "cube" are used is their source in Euclidean geometry from the corresponding geometrical figures.

Since the Greeks could not visualize higher dimensions (I can not either), no name was given to higher powers.

I just use "fourth power", "fifth power", and, in general, "n-th power".