Jtdylktuy

The standard $n$-simplex contains all points $vec{x} in mathbb{R}^{n + 1}$ such that $0 le x_i le 1$ and $vec{x} cdot vec{1} = 1$

The standard 2-simplex is an equilateral triangle with side length $sqrt{2}$ and vertices at (1, 0, 0), (0, 1, 0), and (0, 0, 1). The area is $sqrt{3}/2$.

The standard 1-simplex is a line with vertices at (1, 0) and (0, 1). The length is $sqrt{2}$

What is the area of the standard $n$-simplex?

Is it $sqrt{n + 1} / n!$ ?

From this SE-quest you have the height $h_D$ of the unit-sided regular simplex as
$$h_D=sqrt{frac{D+1}{2D}}$$
Further you obviously have the dimensional recursion on the volume $V_D$
$$V_D=frac1D V_{D-1} h_D$$
With the obvious recursion start of $V_1=1$ you thus get
$$V_D=prod_{dleq D}frac{h_d}d=frac1{D!}sqrt{frac{D+1}{2^D}}$$
(The factor $2^D$ in the denominator surely could be omitted, if you deal with a simplex with sidelength of $sqrt2$ units instead.)

--- rk

I found an answer by integrating over the $n$-simplex moved to $mathbb{R}^{n}$.

The proof would be much simpler if it is possible to directly integrate over the $n$-simplex, which is in $mathbb{R}^{n+1}$.

Average of all vertex except first one:

$vec{a} = frac{1}{n} sum_{i = 2}^{n + 1} vec{e}_i$

Height $vec{h}$ is between one vertex and the average of the remaining vertex.

$vec{h} = vec{a} - vec{e}_{1}$

Height vector is perpendicular to all vectors in the base:

$vec{h} cdot (vec{a} - vec{e}_i ) = (vec{a} - vec{e}_{1}) cdot (vec{a} - vec{e}_i)$

$ = vec{a}cdotvec{a}
-vec{a}cdotvec{e}_{i}
- vec{e}_{1}cdotvec{a}
+ vec{e}_{1}cdotvec{e}_{i}$

$ = frac{n + 1}{n} -frac{1}{n} - 0 + 1 = 0$

Length of height vector is:

$l_{n} = sqrt{vec{h}cdotvec{h}} = sqrt{vec{a}cdot vec{a} + vec{e}_{1}cdot vec{e}_{1}} = sqrt{left(frac{1}{n^{2}}sum_{i=2}^{n+1} vec{e}_{i}cdotvec{e}_{i} right ) + 1}=sqrt{frac{1}{n} + 1}=sqrt{frac{n + 1}{n}}$

Let $A_{n}(s)$ be area of the standard $n$-simplex scaled by $s$ along each of all axes.

$A_{1} = sqrt{2}s$

Integrate along the height vector for the scaled standard $(n + 1)$-simplex.

$A_{n+1}(s) = int_{0}^{sl_{n+1}} A_{n}left (frac{sx}{sl_{n+1}} right)dx = sl_{n+1}int_{0}^{1} A_{n}(su)du$

$A_{2}(s) = sl_{2}int_{0}^{1} sqrt{2}sudu = l_{2} sqrt{2} frac{1}{2}s^{2}= frac{sqrt{3}}{2}s^{2}$

$A_{3}(s) = sl_{3}int_{0}^{1} sl_{2}int_{0}^{1} sqrt{2}sudu$

$$A_{n}(s) = s^{n} left( prod_{1}^{n}l_{n} right) int_{0}^{1} dots int_{0}^{1} udu$$

$$A_{n}(s) = frac{ s^{n} sqrt{n+1}}{n!}$$