Mathematical inspiration from my bathroom floor

Question

I spend some time each day contemplating the grid pattern of my bathroom vinyl floor covering:

What neat, not too difficult mathematical concepts are embodied in such a pattern? I've thought of a couple. First, the proof that rational numbers are countable:

Second, showing that$$\sum_{n=1}^{\infty}\frac{1}{2^{n}}=1$$

I would love to hear of others.

Answer 1

Here's your basic

$$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$

Answer 2

1. Sums over the grid. The sum of the first $n$ odd numbers is actually $n^2$:

$$1+3+5+\ldots+(2n-3)+(2n-1)=\sum^n_{k=1} (2k-1) = n^2.$$

Moreover, you can express any odd number as a difference of squares: $$2n-1=n^2-(n-1)^2.$$ Or any sum of consecutive odd numbers for that matter: $$\sum^n_{k=m} (2k-1) = \left(\sum^n_{k=1} -\sum^{m-1}_{k=1}\right) (2k-1) = n^2 -(m-1)^2.$$

Or even better: the sum of the first $n$ cubes equals the squared sum of the first $n$ integers:

$$1^3+2^3+3^3+\ldots+(n-1)^3+n^3 =\sum_{k=1}^n k^3 =\left(\sum_{k=1}^n k\right)^2 =\frac{n^2(n+1)^2}{4}.$$

This can also be seen on a grid: from: https://upload.wikimedia.org/wikipedia/commons/2/26/Nicomachus_theorem_3D.svg

Here's more. If you have a sum over a square grid (as in assigning each tile a number), you can compute the same sum over just (a triangular) half of the whole region (plus the diagonal strip):

Let $S=\{1,2,3,\ldots,n\}$. Let $f:S\times S\to$ any field. Then $$\sum_{x=1}^n \ \sum_{y=1}^n f(x,y) = \sum_{x=1}^n \left(f(x,x) +\sum_{y=1}^{x-1} \big(f(x,y)+f(y,x)\big)\right)$$ when the sums are finite.

This is easier to see with integrals: let $g:[0,1]\times[0,1]\to[a,b]\subsetneq\mathbb{R}$, then $$\int^1_0\int^1_0 g(x,y)\ \text{d}y\ \text{d}x=\int^1_0\int^x_0\big(g(x,y)+g(y,x)\big)\ \text{d}y\ \text{d}x.$$

If you place an origin at the centre of your grid, you can also divide the grid in four square smaller grids, or eight triangular grids.

2. Spirals on the grid. You can also draw square spirals on the grid, such as Ulam's.

3. Golden rectangles. Consider drawing squares on the grid following this pattern (starting at the upper $1$ tile):

Applications aside, the numbers $1, 1, 2, 3, 5, 8, 13, \ldots$ are the Fibonacci sequence. Look at the outer rectangle: the more squares you add to it, the closer the ratio of its side lengths gets to the golden ratio $1.618$.

4. Computer screens. The screen you are reading right now is just a huge grid of tiny little dot lights. Computers can't represent continuous stuff, so they approximate continuous with small straight lines made up of square points (pixels). Hopefully, it gives the optical ilusion of continuous smooth curves when seen from afar. This is mathematically relevant as of the use of graphicators.

5. Games/problems that rely heavily on grids: Chess, checkers, sudokus, magic squares, "squaring the square" (tiling an integral square using only other integral squares), crosswords, pixel art, nonograms, Battleship, Dungeons and Dragons, most $2$D Super Mario Bros. videogames, tic-tac-toe, Connect $4$, lines and dots, Tetris, among others.

Answer 3

You can visualize the problem of how many paths there are from one corner of a rectangle to another:

Also, you can visualize Pascal's triangle and some of the patterns within it. For example, here I get the Fibonacci numbers out of it (I will leave it as an exercise to see why this pattern continues):

Answer 4

You can think about the Euler characteristic of a planar region. Draw diagonals for each small square to get a triangulation of the floor. Then check if the Euler characteristic (the alternating sum of vertices, edges and faces) is in fact 1, which it should be since it is homotopic to a point -- unless your tailing is not simply connected!

Answer 5

Pick's Theorem

Given a simple polygon constructed on a grid of equal-distanced points (i.e., points with integer coordinates) such that all the polygon's vertices are grid points, Pick's theorem provides a simple formula for calculating the area $A$ of this polygon in terms of the number i of lattice points in the interior located in the polygon and the number $b$ of lattice points on the boundary placed on the polygon's perimeter:

$$A = i + \frac{b}{2} - 1$$