⧼exchistory⧽
3 exercise(s) shown, 0 hidden
BBy Bot
Jun 01'24
[math] \newcommand{\smallfrac}[2]{\frac{#1}{#2}} \newcommand{\medfrac}[2]{\frac{#1}{#2}} \newcommand{\textfrac}[2]{\frac{#1}{#2}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\e}{\operatorname{e}} \newcommand{\B}{\operatorname{B}} \newcommand{\Bbar}{\overline{\operatorname{B}}} \newcommand{\pr}{\operatorname{pr}} \newcommand{\dd}{\operatorname{d}\hspace{-1pt}} \newcommand{\E}{\operatorname{E}} \newcommand{\V}{\operatorname{V}} \newcommand{\Cov}{\operatorname{Cov}} \newcommand{\Bigsum}[2]{\mathop{\textstyle\sum}_{#1}^{#2}} \newcommand{\ran}{\operatorname{ran}} \newcommand{\card}{\#} \newcommand{\mathds}{\mathbb}[/math]

Let [math]\Bbar_1(0)[/math] denote the unit ball of [math]\mathbb{R}^d[/math].

  • Use Lemma to compute [math]\lambda^d(\Bbar_1(0))[/math] for [math]d=1,\dots,10[/math].
  • Compute [math]\lim_{d\rightarrow\infty}\lambda^d(\Bbar_1(0))[/math].
  • Show that [math]\lambda^d(\Bbar_1(0))=\medfrac{\pi^{d/2}}{\Gamma(d/2+1)}[/math] holds, where [math]\Gamma\colon(0,\infty)\rightarrow\mathbb{R}[/math] denotes the Gamma function.
BBy Bot
Jun 01'24

Show that in Lemma indeed [math]a_0\in(1,\sqrt{5}-1)[/math] holds.

Hint: Establish first that [math]\exp(x)\leqslant 1+x+x^2/2[/math] holds for [math]x\leqslant0[/math].

BBy Bot
Jun 01'24
[math] \newcommand{\smallfrac}[2]{\frac{#1}{#2}} \newcommand{\medfrac}[2]{\frac{#1}{#2}} \newcommand{\textfrac}[2]{\frac{#1}{#2}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\e}{\operatorname{e}} \newcommand{\B}{\operatorname{B}} \newcommand{\Bbar}{\overline{\operatorname{B}}} \newcommand{\pr}{\operatorname{pr}} \newcommand{\dd}{\operatorname{d}\hspace{-1pt}} \newcommand{\E}{\operatorname{E}} \newcommand{\V}{\operatorname{V}} \newcommand{\Cov}{\operatorname{Cov}} \newcommand{\Bigsum}[2]{\mathop{\textstyle\sum}_{#1}^{#2}} \newcommand{\ran}{\operatorname{ran}} \newcommand{\card}{\#} \renewcommand{\P}{\operatorname{P}} \renewcommand{\L}{\operatorname{L}} \newcommand{\mathds}{\mathbb} [/math]


Let [math]d\geqslant3[/math] and [math]n[/math] be such that [math]2\ln(n)\leqslant d[/math] holds. Show that

[[math]] \P\bigl[\bigl|\bigl\langle{}\medfrac{x^{\scriptscriptstyle(j)}}{\|x^{\scriptscriptstyle(j)}\|},\medfrac{x^{\scriptscriptstyle(k)}}{\|x^{\scriptscriptstyle(k)}\|}\bigr\rangle{}\bigr|\leqslant\medfrac{\sqrt{6\ln n}}{\sqrt{d-1}} \text{ for all }j\not=k\bigr]\geqslant 1-\medfrac{1}{n} [[/math]]

holds, when [math]x^{\scriptscriptstyle(1)},\dots,x^{\scriptscriptstyle(n)}[/math] are drawn uniformly at random from the [math]d[/math]--dimensional unit ball.

Hint: Use that [math]1\leqslant d^2/(d-2\ln n)^2[/math] holds and apply then the Theorem of Total Probability.