Ytukyg

Let $T_n={x_ige0:x_1+cdots+x_nle1}$. I know $T_n$ is tetrahedron.

My question: How can I compute the volume of $T_n$ for every $n$?

First suppose that we want to find the volume of $T_n(a)$, then by change of variables $X=aU$, you can see that we have $V(T_n(a))=color{red}{a^n}V(T_n(1))$.

Since $x_1+cdots+x_nle1$ if and only if $x_nle1$ and $x_1+cdots+x_{n-1}le 1-x_n$, we have
$$begin{align}V(T_n(1))&=int_{x_nle 1}left(int_{x_1+cdots+x_{n-1}le1-x_n}dx_1cdots dx_{n-1}right)dx_n\
&=V(T_{n-1}(1))int_{x_nle1}color{red}{(1-x_n)^{n-1}}dx_n=frac1 nV(T_{n-1}(1))end{align}$$
The numbers $V(T_n(1))$ satisfy in the above recursion formula, so
$$V(T_n(1))=frac1{n!}.$$

One another way is to consider the following integral

$$I=int_{T_n(a)}e^{-(x_1+cdots+x_n)}dx_1cdots dx_n$$
Since $V(T_n(a))=a^nV(T_n(1))$,
$$I=int_0^infty e^{-a}dV(T_n(a))=V(T_n(1))int_0^infty n a^{n-1}e^{-a}da=n!V(T_n(1))$$
But also we have
$$I=left(int_0^infty e^{-x}dxright)^n=1$$
Hence,
$$V(T_n(1))=frac1{n!}.$$

What is the probability that a random sequence of $n$ numbers in $[0,1]$ is sorted (lowest to highest)? that is, let $y_1,y_2,dots,y_n$ be independent uniform random variables in $[0,1]$. Then the probability $P(y_1leq y_2leq dots leq y_n)=frac{1}{n!}$.

Let $A=(a_{ij})$ be the matrix such that $a_{ij}=1$ if $igeq j$ and $a_{ij}=0$ if $i<j$. Then $det A = 1$ since $A$ is lower-triangular and $a_{ii}=1$ for all $i$. On the other hand, if you take the two sets:

$$S={(x_1,dots,x_n)^T: 0leq x_ileq 1: sum x_i leq 1}$$

and $$T={(x_1,dots,x_n)^T: 0leq x_1leq x_2dots leq x_nleq 1}$$

Then $T$ is the image of $S$ under the transformation $A$, that is, $AS=T$. So $$frac{1}{n!}=mathrm{vol}(T) = det Acdotmathrm{vol}(S) = mathrm{vol}(S)$$

Hint: The general rule is that the $n$-volume of a simplex is $frac 1n$ times the $ (n-1)$ volume of the base times the height, which you can prove by integration. The height is $1$, so you have a recurrence.

Try using induction. $T_{1}$ is just in the interval $[0,1]$, which has volume (length) $1$. $T_{2}$ is a right triangle with volume (area) $1/2$. Now imagine the case for $n=3$. When we slice the tetrahedron $T_{3}$ at some height $zin[0,1]$, we get a cross section that looks like $T_{2}$. But as $z$ gets bigger, the cross section gets smaller. Try to convince yourself that in general we have
$$
v(T_{n})=int_{0}^{1}(1-x)^{n-1}v(T_{n-1}),mathrm{d}x=frac{1}{n}v(T_{n-1})
$$
Then, using the fact that $v(T_{1})=1$, we have $v(T_{n})=1/n!$. Note that we scale by $(1-x)^{n-1}$ instead of by $1-x$ because it is the linear dimensions of the $T_{n-1}$ slice that scale by $1-x$, which translates to the $mathbb{R}^{n-1}$ volume of the slice scaling by $(1-x)^{n-1}$.