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I have to determine the measue of $$\Delta_n=\lbrace (x_1,...,x_n) \in \mathbb{R}^n : \sum_{k=1}^{n} x_k \leq 1, x_k \geq 0, k=1,...,n \rbrace$$ The hint that I have is express $\Delta_n$ in terms of $\Delta_{n-1}$.

My attempt was to calculate the case $n = 2$ and $n = 3$, where I got $\frac{1}{2}$ and $\frac{1}{6}$ respectively. From there I could write the measure of $\Delta_n$ as $$\int_{\Delta_{n-1}}dx_{n-1} \cdots dx_1 \int_{x_n=0}^{1- \sum_{k=1}^{n-1}}dx_n$$ and from there integrate step by step. The problem is that I think it is too long a process and I would like to know if there is a faster way to find the measure of $\Delta_n$.

Note: I was able to write the previous expression since in the cases $n = 2$ I wrote: $$\int_{x=0}^{1}\int_{y=0}^{1-x}dydx$$ and for the case $n=3$ : $$\int_{x=0}^{1}\int_{y=0}^{1-x}\int_{z=0}^{1-x-y}dzdydx$$

GHR01
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1 Answers1

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This way, let $f_n(x)$ be (for $x\geqslant 0$ and $n>0$) the measure of $$\left\{(x_1,\ldots,x_n)\in\mathbb{R}_+^n : \sum_{k=1}^n x_k\leqslant x\right\}.$$ Then $f_1(x)=x$ and $f_n(x)=\int_0^x f_{n-1}(x-x_n)\,dx_n$ for $n>1$.

By induction, $f_n(x)=x^n/n!$, and the answer is $f_n(1)=1/n!$.

metamorphy
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  • I do not see why it is true that $f_n(x)= \int_{0}^{x} f_{n-1}(x-x_n)dx_n$. Can you explain in more detail, please? – GHR01 May 05 '21 at 05:28