As n goes to infinity, from the center of a regular simplex, the limit of angles between vertices?

Calculus Level 3

In regular n-dimensional regular simplices, all subobjects of the same class are congruent.

For a given dimensionality $n$ , the number of m-dimensional subobjects is $\left(\begin{array}{c}n+1\\m+1\\\end{array}\right)$ .

Per a question from Jon Haussmann, I corrected the illustration that had been above.

Designating the angle at the center of the simplex between any two vertices of the $n$ dimensional regular simplex as $\text{angle}_n$ , what is $\lim\limits_{n\to\infty}\text{angle}_n$ ? The angle is the same regardless of which vertices are selected.

The angle measure should be in radians and given to 6 decimal places.

The answer is 1.570796.

1 solution

A Former Brilliant Member
Dec 20, 2018

Using a simple symmetry argument, with the center of the $n$ -dimensional regular simplex at the origin, one vertex at $(1,0,\ldots)$ and all the other vertices (however many that there are) are equally distributed on the other side of the origin, $\text{angle\_n}$ is $\arccos(-\frac1{n+1})$ . $\lim\limits_{n\to\infty}\frac1{n+1}$ is $0$ . $arccos(0)$ is $\frac\pi2$ .

As indicated by Jon Haussman, the previous version of the illustration above was incorrect. I believe that I have fixed those errors. The problem answer was and is correct.

Isn't the angle $120^\circ$ for the equilateral triangle?

Jon Haussmann - 2 years, 4 months ago

As n goes to infinity, from the center of a regular simplex, the limit of angles between vertices?

The answer is 1.570796.

1 solution

2 pending reports

Vote up reports you agree with