The first step to higher dimensional thinking!

**Keywords**: Pythagoras's theorem in higher dimensions, Polytopes, n-cube, Iteration in higher dimensions, Recurrence relation, Induction

Welcome to this blog post. In this post, we will look at the fundamentals of geometry.

**Fundamentals of Geometry.**

This is our starting point to visualise higher, non-integer or even negative dimensions.

You gotta learn to walk before you run.

First, let's discuss Pythagoras's Theorem: If you have a right triangle as shown below, then:

or

Now let's imagine a three-dimensional version of this formula (a cuboid is shown below):

or

This is because if there was a diagonal as shown:

then

and

Therefore:

So we have learnt that in an n-D case, Pythagoras's theorem gets extended to:

Also, another interesting thing to see is a table comparing numbers of points, edges, faces and solids in higher dimensions:

For future reference, let's call an n-D shape an n-D polytope (polygon is a 2-D polytope, polyhedron is a 3-D polytope).

Points Edges Faces Solids Hyper-Solids

0-D 1 0 0 0 0

1-D 2 1 0 0 0

2-D 4 4 1 0 0

3-D 8 12 6 1 0

4-D 16 32 24 8 1

5-D 32 80 80 40 10

6-D 64 192 240 160 60

7-D 128 448 672 560 280

8-D 256 1024 1792 1792 1120

9-D 512 2304 4608 5376 4032

10-D 1024 5120 11520 15360 13440

11-D 2048 11264 28160 42240 42240

...

This seems so weird. Our intuition tells us that there must be more "small" things (points etc.) than "big" things (solids, faces etc.) but the results show that there is an "equality point" where the number of nth dimensional polytopes is equal to the number of (n+1)th dimensional polytopes. For the number of points and edges, that equality point is 2-D (a square has both 4 points and 4 edges). For the number of edges and faces, the equality point is 5-D. For faces and solids, the equality point is 8-D. Wait a minute! How do I know the number of faces in a 9-D cube when I cannot even see it? The answer is iteration!

Let's take points, for example. I know that when transitioning between an n-cube to an (n+1)-cube, the number of points doubles. So, the relevant recurrence relation is Points(n+1) = Points(n)*2. Therefore, Points(n) = Points(0)*(2^n). Points(0) =1, so Points(n) = **2^n**.

For edges, going from an n-cube to an (n+1)cube, I know that every edge is duplicated and every point forms a new edge with its duplicated copy (see the title GIF to see what I mean).

Therefore, the relevant recurrence relation is Edges(n) = Edges(n-1)*2+Points(n-1) =

Edges(n-1)*2+2^(n-1)

Therefore, Edges(n) = Edges(0) + (2^(n-1)) + (2^(n-2)*2) + (2^(n-3)*4) ... (2^0 *2^(n-1)) =

For faces, going from an n-cube to an (n+1)cube, I know that every face is duplicated and every edge forms a new face with its duplicated copy (see the title GIF to see what I mean).

Therefore, the relevant recurrence relation is Faces(n) = Faces(n-1)*2+Edges(n-1) =

Faces(n-1)*2+(n-1)*2^(n-2)

Therefore, Faces(n) = Faces(0) + ((n-1)*(2^(n-2))) + ((n-2)*2^(n-3)*2) + ((n-3)*2^(n-4)*4) ... (1*2^1 *2^(n-1)) =

For solids, going from an n-cube to an (n+1)cube, I know that every solid is duplicated and every face forms a new solid with its duplicated copy (see the title GIF to see what I mean).

Therefore, Solids(n) = Solids(n-1)*2+Faces(n-1) = Solids(n-1)*2+n*(n-1)*2^(n-4)

Therefore, by induction, Solids(n) =

**For any m-cube, going from an n-cube to an (n+1)cube, I know that every m-cube is duplicated and every (m-1)-cube forms a new m-cube with its duplicated copy (see the title GIF to see what I mean).**

**Therefore, C(m,n) = C(m,n-1)*2+C(m-1,n-1)**

**Therefore, by induction, C(m,n) = n!/(m!*(n-m)!) * 2^(n-m)**

Or, in a cleaner view:

Great! This is great! Now we can form an equation for the nth "equality point" as we were discussing before. Let's say that for an n-cube, the number of m-cubes and number of (m+1)-cubes are equal. Then C(m,n)=C(m+1,n).

Therefore:

So for equal numbers of Solids and Hyper-Solids in the n-cube, n must be 11!

These might seem like very specific puzzles ðŸ§© but these show the exact things we need for the future parts of this topic: recursive reasoning, stating physical properties as formulas whilst still respecting the physical and not abstract nature of these higher dimensional structures and just playing around with maths!

Also, I have recently learnt of a new way to demonstrate polytopes: SchlÃ¤fli symbols! The symbols for a point and a line are ( ) and { } respectively. A polygon with n sides has a symbol {n} (triangle: {3}, square: {4} etc.). The next part gets complicated. For a polyhedron with m n-sided polygon faces around a point, the symbol is {n, m}. This sentence is very anfractuous and warrants an example. A cube, for instance, has 3 squares (with obviously ðŸ™„ 4 sides) around each point. Therefore the symbol for a cube is {4,3}. For higher dimensions, if the (n-1)th polytope has a symbol of {a,b,c,d...,j} and the corresponding nth polytope has k of these (n-1)th polytopes around each peak, (3-D peak: point; 4-D peak: line; 5-D peak: face etc.) the symbol would be {a,b,c,d...,j,k}. For n-cubes, the SchlÃ¤fli symbol would be {4,3,3,3...,3} with (n-2) 3s in there.

Part 2 will be about cross-raumartig-zeitartig* (#letsmakenewwords) dimensions!

Hope you like it!ðŸ˜ƒ

*This word is an extension of the existing words "raumartig" and "zeitartig" which are derived from the German "raum" (meaning space) and "zeit" (meaning time).

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