User:Tetracube/Triangular numbers and n-simplices

This is a little study on the properties of triangular numbers and their relationship with the n-simplices.

Inverse triangular numbers
The usual definition of a triangular number is:


 * $$T_n = \frac{n(n+1)}{2}$$

This, of course, implies that the inverses of the triangular numbers are:


 * $$T_n^{-1} = \frac{1}{T_n} = \frac{2}{n(n+1)}$$

This fraction can be rewritten as:


 * $$T_n^{-1} = \frac{2}{n} - \frac{2}{n+1}$$

Consider now the sum of consecutive inverse triangular numbers:


 * $$\sum_{i=n}^k T_i^{-1} = \sum_{i=n}^k \left(\frac{2}{n} - \frac{2}{n+1}\right)$$

where $$n < k$$. This sum may be expanded thus:


 * $$\sum_{i=n}^k T_i^{-1} = \frac{2}{n} - \frac{2}{n+1} + \frac{2}{n+1} - \frac{2}{n+2} + \cdots + \frac{2}{k-1} - \frac{2}{k} + \frac{2}{k} - \frac{2}{k+1}$$

Notice that the second and third terms of the sum cancel each other out, and so do every subsequent pair of terms until the second last term. This sum is therefore a telescoping series: all except the first and last terms sum to 0. So:


 * $$\sum_{i=n}^k T_i^{-1} = \frac{2}{n} - \frac{2}{k+1}$$

Ratios of square and triangular numbers
Let $$S_n = n^2$$ be the $$n$$'th square number. Consider the following ratio of a square number with a triangular number:


 * $$\frac{S_n}{T_n} = \frac{n^2}{\left(\frac{n(n+1)}{2}\right)} = n^2\left(\frac{2}{n(n+1)}\right) = \frac{2n}{n+1}$$

Notice that $$2n = (2n+2)-2 = 2(n+1) - 2$$. So:


 * $$\frac{2n}{n+1} = \frac{2(n+1)}{n+1} - \frac{2}{n+1} = 2 - \frac{2}{n+1}$$

That is to say:


 * $$\frac{S_n}{T_n} = 2 - \frac{2}{n+1}$$

In other words, the ratio of the $$n$$'th square number and the $$n$$'th triangular number falls short of 2 by exactly $$\frac{2}{n+1}$$.

Now consider the ratio of the $$(n+1)$$'th square number with the $$n$$'th triangular number:


 * $$\frac{S_{n+1}}{T_n} = (n+1)^2\left(\frac{2}{n(n+1)}\right) = \frac{2(n+1)}{n} = \frac{2n+2}{n} = 2 + \frac{2}{n}$$

In other words, the ratio of the $$(n+1)$$'th square number with the $$n$$'th triangular number exceeds 2 by exactly $$\frac{2}{n}$$.

Now consider the following sum of two ratios:


 * $$\frac{S_n}{T_n} + \frac{S_{k+1}}{T_k} = 2 - \frac{2}{n+1} + 2 + \frac{2}{k} = 4 - \left(\frac{2}{n+1} - \frac{2}{k}\right)$$

But from the previous section:


 * $$\frac{2}{n+1} - \frac{2}{k} = \sum_{i=n+1}^{k-1} T_i^{-1}$$

Therefore:


 * $$\frac{S_n}{T_n} + \sum_{i=n+1}^{k-1} T_i^{-1} + \frac{S_{k+1}}{T_k} = 4$$

for all $$1\le n < k$$. This sum is essentially the sum of inverse triangular numbers from $$n$$ to $$k$$, with the first term multiplied by $$n^2$$ and the last term multiplied by $$(k+1)^2$$.

Relation to n-simplices
Given a fixed integer $$n$$, consider the following matrix product:


 * $$A_n =

\left[\begin{array}{ccccc} -1 & -1 & -1 & \cdots & -1 \\ 1 & -1 & -1 & \cdots & -1 \\ 0 &  2 & -1 & \cdots & -1 \\ 0 &  0 &  3 & \cdots & -1 \\ & \vdots && \ddots & \vdots \\ 0 &  0 &  0 & \cdots & n \end{array}\right] \left[\begin{array}{ccccc} 1 & 0 & 0 & \cdots & 0 \\ 0 & \frac{1}{\sqrt{3}} & 0 & \cdots & 0 \\ 0 & 0 & \frac{1}{\sqrt{6}} & \cdots & 0 \\ & \vdots & & \ddots & 0 \\ 0 & 0 & 0 & \cdots & \frac{1}{\sqrt{T_n}} \end{array}\right] $$

The sum of each column in the matrix on the right is 0, so the vector sum of all the rows of the resulting matrix $$A_n$$ (each row being considered as an n-dimensional vector) is the zero vector.

Furthermore, the difference between any two rows of $$A_n$$ is of the form:



\pm\left[\begin{array}{cccccccccc} 0 & \cdots & 0 & \frac{j}{\sqrt{T_j}} & \frac{1}{\sqrt{T_{j+1}}} & \cdots & \frac{1}{\sqrt{T_{k-1}}} & \frac{k+1}{\sqrt{T_k}} & 0 & \cdots \end{array}\right] $$

The magnitude of this difference vector is therefore:


 * $$\sqrt{

\frac{j^2}{T_j} + \frac{1}{T_{j+1}} + \cdots + \frac{1}{T_{k-1}} + \frac{(k+1)^2}{T_k} } = \sqrt{ \frac{S_j}{T_j} + \sum_{i=j+1}^{k-1} T_i^{-1} + \frac{S_{k+1}}{T_k} }$$

From the result of the previous section, the sum under the square root is simply 4. Therefore, the magnitude of the difference between any two row vectors is constantly 2.

Treating the rows of $$A_n$$ as a set of vectors in n-dimensional space, the fact that their sum is zero indicates that their convex hull is an origin-centered n-dimensional polytope. Furthermore, since the magnitude of the difference between any two of them is 2, they are $$(n+1)$$ equidistant points. The only n-dimensional polytope that is the convex hull of $$(n+1)$$ equidistant points is the regular n-simplex.

Therefore, the rows of $$A_n$$ are the coordinates of a regular n-simplex centered on the origin, having edge length 2.