#mathematics visualization

Happy winter break, friends.

A deeply intuitive, aesthetically pleasing geometrical “proof without words” that the sum of the first n cubes is the square of the nth triangular number, sometimes called Nicomachus’s theorem.

Fractal fungi. 

Where do they lead?

Into a better year, my friends.

An excellent algebraic/geometric proof without words.  Bonus: Some expressions are written in 3D perspective! Fun!

To make sense of this, first observe that the upper shape has volume (a^3)-(b^3). Think of it as taking a cube with side lengths a (volume a^3) and removing (subtracting) from it from a smaller cube having side lengths b.

The shape is cut into three pieces, whose volumes are (a^2)(a-b), ab(a-b), and (b^2)(a-b), respectively. The total volume remains unchanged. Thus we can write 

(a^3)-(b^3)=(a^2)(a-b)+ab(a-b)+(b^2)(a-b). 

Dividing over (a-b) yields the algebraic equation written at the bottom right.

Here are a few pretty variations of the complex sine function. 

A stunning symmetric 7-Venn diagram. Try to figure out how it’s put together (it’s made out of 7 pieces). Hint: follow the curves.

Bruhat-Tits Trees.

Stop snickering. Mathematics is serious. 

jk/also,

Another inspiring “proof without words.” Depicted here is the convergence of a few geometric series. 

This one, I feel, merits a bit of informal explanation. 

The leftmost example can be understood as follows. Numerically, 4 is 2 squared. As such, the quantity 4 can be represented as the area of a 2 x 2 square. Therefore, knowing that the series below the figure converges to 4, we can expect there to be a corresponding geometric representation of the series as a square. Let us begin, then, by visualizing a 2 x 2 grid, whose entire area the (forthcoming) geometric representation of the series must fill.

Now, for the series, it is instructive to consider the first few terms of the sequence to be summed. This sequence of terms will be used to dictate the areas that our chosen shape (squares, in this case) must have. By organizing these smaller squares such that they fully cover our imaginary 2 x 2 grid, we demonstrate the expected convergence.

By this logic, we can “reverse engineer” geometric arrangements--first taking terms of a sequence, then treating them as the area of squares-to-be, which will be distributed throughout the grid, covering any empty spaces. As we go through terms, the total area covered grows closer to 4. Note that, by totaling the areas, we are adding a sequence of numbers--the definition of a series.

The leftmost series begins with n=0. Substituting this into the generating function (3/4)^n gives us our first term, 1. Accordingly, we construct a square whose total area is 1 units squared, a 1 x 1 square. We use it to cover the bottom-left quadrant of our imaginary grid (as depicted in the figure). 

The next term, for n=1, is 3/4. Now, to construct a square of a certain area, we are in essence asking for the square root of that area (which determines the length/width of the square). For the previous term, this is simple. However, constructing a square of length (3^(1/2))/2 is difficult. Why? Because 3/4 does not have a pleasant square root. The numerator is the culprate, seeing as the denominator 4 is a square number. Any ugliness that comes from square-rooting the fraction 3/4 is the result of square-rooting 3. 

For our geometry, this will not do. But because we know the origin of our troubles, we can try to circumvent the issue. The elegant solution in the figure above is to let the number 3/4 represent the area of not one, but several squares. These “sub-squares” areas’ will total 3/4. Thus, to design the squares, we “divide the difficulties” (a useful heuristic of my hero, Kurt Gödel). In this case, we are literally dividing the fraction into three simpler ones, so we have 3(1/4). This generates three smaller squares, each with an area of 1/4 units sq. Taking the square root, we have 1/2 units, the length of our 3 smaller squares. This can be easily constructed with geometry (by comparison to the length of our 1 x 1 square), and thus we have circumvented the “problem.”

Let’s take another look at the above figure. Imagine arranging the smaller squares into “tiers,” by size, in descending order. For convenience, we will start with 0. The 0th tier has just one square. The 1st tier has three squares. The 2nd tier consists of nine squares. And so on.

This confirms the numerical results we obtained from the sequence of terms. We have one 1 x 1 square and three (1/2 x 1/2) squares. 

Let’s try in reverse. Note that the nine squares of tier 2 each appear to be 1/4 the size of the previous tier. Tier 1 squares have area 1/4 units squared, thus tier 2 squares have area 1/16 units squared. As aforementioned, there are nine such squares, totaling 9/16 units squared of area. Tier 2 is therefore 9/16=(3/4)^2 units of area sq. And, by our series, we find the term for n=2 is (3/4)^2=9/16, illustrating the geometrical accuracy of the figure (and it’s method of construction). 

The pattern holds for successive n; there are 3^n squares, each with area (1/4)^n. By arranging successive tiers of squares as above (self-similarly), the empty spaces are filled. We observe through intuition that this pattern can be carried on ad infinitum, filling smaller subsections of space, which together will eventually fill the entire space. 

In accordance with the limit definition of a Riemann sum, the total area of the geometric construction can be made arbitrarily close to 4, provided sufficiently large n.

As for the other two series, note that (2^2)/2=2, and that 2^3=8 to derive the geometric intuitions evidenced by the remaining figures.