exercise:E6f17799c6: Difference between revisions
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Let <math>\Bbar_1(0)</math> denote the unit ball of <math>\mathbb{R}^d</math>. | Let <math>\Bbar_1(0)</math> denote the unit ball of <math>\mathbb{R}^d</math>. | ||
<ul style="list-style-type:lower-roman"><li> Use [[#LEM-9-2 |Lemma]] to compute <math>\lambda^d(\Bbar_1(0))</math> for <math>d=1,\dots,10</math>. | <ul style="list-style-type:lower-roman"><li> Use [[guide:D9c33cd067#LEM-9-2 |Lemma]] to compute <math>\lambda^d(\Bbar_1(0))</math> for <math>d=1,\dots,10</math>. | ||
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Latest revision as of 00:18, 2 June 2024
[math]
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\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.