The Iron Man hyperspace formula really works (hypercube visualising, Euler’s n-D polyhedron formula)

From Mathologer.

On the menu today are some very nice mathematical miracles clustered around the notion of mathematical higher-dimensional spaces, all tied together by the powers of (x+2). Very mysterious 🙂 Some things to look forward to: The counterparts of Euler’s polyhedron formula in all dimensions, a great mathematical moment in the movie Iron man 2, making proper sense of hupercubes, higher-dimensional shadow play and a pile of pretty proofs.

00:00 Intro
01:17 Chapter 1: Iron man
06:05 Chapter 2: Towel man
11:16 Cauchy’s proof of Euler’s polyhedron formula
17:37 Chapter 3: Beard man
22:16 Tristans proof that (x+2)^n works
26:16 Chapter 4: No man
28:52 Shadows of spinning cubes animation
28:42 Thanks

Here is a link to a zip file with the Mathematica notebooks for creating the cube and hypercube shadows that I discuss at the end of the video in chapter 4.

If you don’t have Mathematica, you can have a look at pdf versions of the programs that are also part of the zip archive or you can use the free CDF player to open the cdf versions of the notebooks.

Something I forgot to mention: There is also another purely algebraic incarnations of this process of growing the cubes. It comes in the form of a recursion formula that connects the different numbers of bits and pieces in consecutive dimensions. That recursion formula is also present at the bottom of the "iron man page". Have a close look 🙂 Also, in the Marvel movies the cube that Tony Stark is holding in the thumbnail of this video is called the Tesseract. Probably worth pointing out that "tesseract" is another name for a 4-d cube. I also built an easter egg into the thumbnail that plays on this fact:

The formulae for n-d tetrahedra and octahedra can be found on this page;

Here is a link to my video on solving the 4d Hyper Rubik’s Cube

Another proof of Moessner for cubes using cubical shells Anthony Harradine and Anita Ponsaing

Here is a really nice video on the 120-cell that I only mentioned in passing.

Noteworthy from the comments:
Today’s video was "triggered" by a comment made by Godfrey Pigott on the last video on Moessner’s miracle in which he pointed out that (x+2)^n captures the vital statistics of the n-dimensional cube.

Z. Michael Gehlke There is an easy way to see this: (x^1 + 2*x^0) describes the parts of a line; all of the cubes are iterated products of lines: n-cube = (1-cube)^n. Therefore, all cubes are described by iterated powers of (x^1 + 2*x^0)^n. (Me: Nice insight. Of course needs some fleshing out to make this work on it’s own, like in the comment by …

HEHEHE I AM A SUPAHSTAR SAGA I came up with an even simpler visual proof. Take a cube of side length x+2. This cube has a volume (x+2)^3. Now, slice the cube six times. Each slicing plane is parallel to a face and 1 unit deeper than the face. Don’t throw away any volume. What you’re left with is an inner cube of side length x (volume x^3), 6 square pieces of volume x^2, 12 edge pieces of volume x, and 8 corner cubes with volume 1 each. Adding up these volumes gives you the original (x+2)^3 volume, so it’s proven. This works in any dimension.

Here is a link to an animation of this idea that I put on Mathologer 2, as a reward to those of you who who are keen enough to actually read these descriptions.

Typo: The numbers of vertices and faces of the dodecahedron got switched.

Today’s music is Floating Branch by Muted.