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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 $\sqrt{n + 1} / n!$ ?

R zu
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  • I can't just take the derivative of the volume between the origin and the n-simplex, which is $1^{n}/n!$. – R zu Nov 13 '18 at 00:20
  • For n-simplex, I can calculate on $R^{n}$ instead of $R^{n + 1}$. Can't use change of coordinate formula because the Jacobian is not square. – R zu Nov 13 '18 at 01:13
  • This question and https://math.stackexchange.com/questions/2996301/how-to-integrate-over-the-standard-n-simplex-directly-in-mathbbrn1/2996559#2996559 are duplicates. What is the purpose of this manoeuvre? – Christian Blatter Nov 13 '18 at 17:12
  • I asked this before that. This question asks for proof of the formula. The answers showed the formula is correct. That question ask specifically for using integration with coordinates of $R^{n+1}$ to prove the formula. I linked that question to this question to show the formula is correct before asking the more difficult question. See: https://math.stackexchange.com/questions/2996301/how-to-integrate-over-the-standard-n-simplex-directly-in-mathbbrn1/2996559 – R zu Nov 13 '18 at 17:34

2 Answers2

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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}\cdot\vec{a} -\vec{a}\cdot\vec{e}_{i} - \vec{e}_{1}\cdot\vec{a} + \vec{e}_{1}\cdot\vec{e}_{i}$

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

Length of height vector is:

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

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.

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

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

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

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

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

R zu
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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 $$V_D=\prod_{d\leq D}\frac{h_d}d=\frac1{D!}\sqrt{\frac{D+1}{2^D}}$$ (The factor $2^D$ in the denominator surely could be omitted, if you deal with a simplex with sidelength of $\sqrt2$ units instead.)

--- rk

Dr. Richard Klitzing
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