Find the volume formula of a simplex proving that $\int_0^1\int_0^{1-x_n}...\int_0^{1-(x_n+...+x_2)}1\,\,\,dx_1...dx_n=\frac 1{n!}$

Question

Definition

If $x_0,...,x_n$ are $(n+1)$ affinely independent point of $\Bbb R^n$ (which means that the vectors $(x_1-x_0),...,(x_n-x_0)$ are linearly independent) then simplex determined by them is the set $$ S:=\Biggl\{x\in\Bbb R^n: x=\alpha_1v_1+...+\alpha_nv_n, \sum_{i=1}^n\alpha_i\le1\,\,\,\text{and}\,\,\,\alpha_i\ge0\,\,\,\text{for all}\,i\Biggl\} $$ where $v_i:=(x_i-x_0)$ for each $i>0$.

So with the previous definition I try to show that the volume of a symplex $S$ is given by the formula $$ v(S)=\Big|\frac{1}{n!}\det\big[(x_1-x_0),...,(x_n-x_0)\big]\Big| $$ for each $n\in\Bbb N$.

First of all we observe that any simplex is the intersection of two parallelopides (see here for details) so that any simplex is rectifiable (indeed any parallelopiped is rectifiable and the intersection of rectifiable sets is rectifiable too) and thus any continuous function is there integrable. Now if $x_0,...,x_n$ are $(n+1)$ point affinely independent we define the transformation $h:\Bbb R^n\rightarrow\Bbb R^n$ through the condition $$ h(x):=A\cdot x+x_0 $$ where $A$ is the matrix whose $j$-th column is the vectors $(x_j-x_0)$ for each $j=1,...,n$. So now we observe that the transformation $h$ carries the simplex $$ E:=\{x\in\Bbb R^n:x_1+...+x_n\le 1\,\,\,\text{and}\,\,\, x_i\ge 0\,\,\,\text{for all}\,i\} $$ onto the simplex $S$ generated by the points $x_0,...,x_n$. So if we prove that $$ v(E):=\frac 1{n!} $$ for all $n\in\Bbb N$ then by the change variable theorem (it is easy to verify that $h$ is a diffeomorphism) $$ v(S)=\int_S 1=\int_E|\det A|=\frac 1{n!}|\det A| $$ for each $n\in\Bbb N$.

So let's start to prove by induction that $$ v(E)=\frac 1{n!} $$ for each $n\in\Bbb N$.

So if $n=1$ then $E=[0,1]$ and so clearly the formula trivially holds. So we suppose that the formula holds for $(n-1)$ and we prove that it holds for $n$.

So we have to prove that $$ \int_E1=\frac 1{n!} $$ and to do this we will use Fubini's formula. So I have to prove that $$ \int_0^1\int_0^{1-x_n}...\int_0^{1-(x_n+...+x_2)}1\,\,\,dx_1...dx_{n-1}dx_n=\frac 1{n!} $$ but unfortunately I don't be able to prove it.

For sake of completeness I point out that it seems that here there is a similar solution to that I gave (see the answer of the professor Blatter) but I don't fully understand it. In particular the solution I linked says that if we define $$ E_\xi:=\{x\in\Bbb R^n:x_1+...+x_{n-1}\le1,\,\,\,\text{and}\,\,\,x_1,...,x_{n-1}\ge 0\,\,\,\text{and}\,\,\,x_n=\xi\} $$ for any $\xi\in[0,1]$ then $E=\bigcup_{\xi\in[0,1]}E_\xi$ and so if we observe that the projection of $E_\xi$ is a $(n-1)$ dimensional simplex then $\int_E 1=\int_0^1(1-x_n)^{n-1}v(E_\xi)\,dx_n=\frac 1 n(1-x_n)^nv(E_\xi)=\frac 1{n!}$ that complete the proof but I don't understand how to prove effectively last equality.

So I ask to prove the last equality and then I ask to prove that $h[E]=S$ too. So could someone help me, please?

Answer 1

With a slight generalization, it becomes easy to prove by induction that for each $a$, we have: $$\int_0^a \int_0^{a-x_n} \cdots \int_0^{a-(x_n+\cdots+x_2)} dx_1 \cdots dx_n = \frac{a^n}{n!}.$$ The base case $n = 1$ is trivial. For the inductive case, we have using the inductive hypothesis that the desired integral is equal to $\int_0^a \frac{(a-x_n)^{n-1}}{(n-1)!} dx_n = \left. -\frac{(a-x_n)^n}{n!}\right|_{x_n=0}^{x_n=a} = \frac{a^n}{n!}$.

(And in fact, via the transformation that you outlined in the question, the general case given above can be proved from the special case $a=1$. This would give another way to proceed in a proof by induction: use the transformation along with the inductive hypothesis to conclude that the function being integrated with respect to $x_n$ is $\frac{(1-x_n)^{n-1}}{(n-1)!}$.)

Answer 2

The answer you linked gives the idea. For instance, consider a triangle in $\mathbb{R}^2$ with an edge of length $1$ along the $x$-axis and the third vertex at height $1$. How to find the area of this triangle to be $0.5(1)(1)=\frac{1}{2!}$? Cut it into horizontal segments. The segment at the base has length $1$, the one at the top has length $0$, and the lengths in between decrease linearly. So we could write $w(y)=1-y$ for the width of the segment at height $y$. Then the area of the whole triangle is $$\int_0^1 w(y)\,dy = \int_0^1 (1-y)\,dy = \frac{1}{2}.$$ A similar strategy will give the volume of the $n$-simplex. All you need is the area of an $(n-1)$-simplex at the base and to note that the this area will decrease proportional to (height)$^{n-1}$.

Answer 3

First of all I summarise to follow what I above defined to point out some things.

Definition

If $x_0,\ldots,x_n$ are $(n+1)$ affinely independent points of $\Bbb R^n$ then the simplex determined by them is the set $$ S:=\Biggl\{x\in\Bbb R^n: x=x_0+\alpha_1v_1+...+\alpha_nv_n,\,\sum_{i=1}^n\alpha_i\le1\,\,\,\text{and}\,\,\,\alpha_i\ge 0\,\,\,\text{for all}\,i\Biggl\} $$ where $v_i:=(x_i-x_0)$ for each $i>0$.

So with this definition it is clear that a simplex can be always individuated by one point $x_0$, that we call origin, and by vectors of a base $v_1,...,v_n$, that we call directions. Finally we call the points $(x_0+v_1),...,(x_0+v_n)$ vertex.

So let's start solve the question.

Theorem

If $S$ is a simplex generated by the affinely independent points $x_0,...,x_n$ of $\Bbb R^n$ then $$ v(S)=\Big|\frac{1}{n!}\det\big[v_1,...,v_n\big]\Big| $$ where $v_i:=(x_i-x_0)$ for each $i=1,...,n$.

Proof. So if $x_0,...,x_n$ are $(n+1)$ affinely independent points then we define the tranformation $h:\Bbb R^n\rightarrow\Bbb R^n$ through the condition $$ h(x):=A\cdot x+x_0 $$ where $A$ is the matrix whose $j$-th column is for each $j=1,...,n$ the vector $v_j$ as above defined. First of all we observe that the transformation $h$ is a diffeomorphism because the vectors $v_1,...,v_n$ are linearly independet so that the matrix $A$ is not singualr. Then we observe that the trasformation $h$ map the simplex $E$ with origin the intersection $O$ of the axes and with directions the vector of canonical base $\mathcal E:=\{e_1,...,e_n\}$ onto the simplex $S$ because $h$ map the vertex of $E$ in the vertex of $S$. Now we suppose that $$ v(E)=\frac 1{n!} $$ so that we observe that in this particular case from the change of variables theorem it would follow that $$ v(S)=\int_S 1=\int_E |\det A|=|\det A|\int_E 1=|\det A|v(E)=\frac 1{n!}|\det A| $$ and so the theorem would hold. So let's prove the theorem using induction. So clearly if $n=1$ then $S=[a,b]$ for some $a,b\in\Bbb R$ and so the formula trivially holds. So we suppose that the formula holds for $(n-1)$ and we prove that it holds of $n$. So we know that it is sufficient to prove that $$ \int_E 1=\int_0^1\int_0^{1-x_n}...\int_0^{1-(x_n+...+x_2)}1\,\,\,dx_1...dx_{n-1}dx_n=\frac 1{n!} $$ and exactly we can do this knowing that the theorem holds for any siplex of $\Bbb R^{n-1}$ by the inductive hypotesis. So fixing $x_n$ let be $\xi:=(1-x_n)$ and we define $$ T:=\big\{x\in\Bbb R^{n-1}:x_1+...+x_{n-1}\le\xi\,\,\,\text{and}\,\,\,x_i\ge 0\,\,\,\text{for all}\,\,\,i=1,...,(n-1)\big\} $$ that is $T$ is the simplex of $\Bbb R^{n-1}$ with origin the intersection $O$ of axes and with directions those vectors that are proportional to the vectors of canonical base and of length $\xi$. So by the inductive hypotesis we know that $$ v(T)=\Biggl|\frac 1{(n-1)!}\det[\xi e_1,...,\xi e_n]\Biggl|=\frac 1{(n-1)!}(1-x_n)^{n-1} $$ and moreover by definition of volume $$ v(T)=\int_T 1=\int_0^{1-x_n}...\int_0^{1-(x_2+...+x_{n-1}+x_n)}1dx_1...dx_{n-1} $$ so that we conclude that $$ \int_0^1\int_0^{1-x_n}...\int_0^{1-(x_n+...+x_2)}1\,\,\,dx_1...dx_{n-1}dx_n= \\ \int_0^1\Biggl[\int_0^{1-x_n}...\int_0^{1-(x_n+...+x_2)}1\,\,\,dx_1...dx_{n-1}\Biggl]dx_n= \\ \int_0^1\frac 1{(n-1)!}(1-x_n)^{n-1}dx_n=-\frac 1{n!}(1-x_n)^n\Biggl|_0^1=\frac 1{n!} $$ and so the theorem holds.