Timeline for What's a nice argument that shows the volume of the unit ball in $\mathbb R^n$ approaches 0?

Current License: CC BY-SA 2.5

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Dec 20, 2022 at 14:34	comment	added	user7427029		@B.Bischof: \begin{align} 2^n \left(2 \sqrt{\frac{2}{n}}\right)^{n/2} &= 2^n 2^{n/2} \sqrt{\frac{2}{n}}^{n/2} \\ &= 2^{3n/2} \sqrt{\frac{2}{n}}^{n/2} \\ &= (2^3)^{n/2} \sqrt{\frac{2}{n}}^{n/2} \\ &= 8^{n/2} \sqrt{\frac{2}{n}}^{n/2} \\ &= \left(8 \sqrt{\frac{2}{n}}\right)^{n/2} \\ &= \left(\sqrt{64} \sqrt{\frac{2}{n}}\right)^{n/2} \\ \end{align}
Aug 2, 2016 at 9:47	comment	added	David Feldman		Alternatively one could observe that at most 4 coordinates have magnitude at least $\sqrt{1/5}$. Then one gets the volume estimate $$\left(\begin{array}{c} n\\4\end{array}\right) \left(\sqrt{4/5}\right)^{n-4}$$ so now you don't even have two exponentials to fight it out.
Jun 9, 2010 at 22:43	comment	added	Greg Kuperberg		BTW, the software now makes it look like it was my argument. But it wasn't; it was contributed by fedya.
Jun 9, 2010 at 20:17	history	edited	Randomblue	CC BY-SA 2.5	making it look even nicer; added 4 characters in body
Dec 10, 2009 at 17:39	history	edited	Greg Kuperberg	CC BY-SA 2.5	added 16 characters in body
Dec 10, 2009 at 17:34	history	edited	Greg Kuperberg	CC BY-SA 2.5	added 159 characters in body
Dec 9, 2009 at 17:01	history	edited	Reid Barton	CC BY-SA 2.5	added 24 characters in body
Dec 9, 2009 at 4:57	comment	added	B. Bischof		Unfortunately, I do not see the last part of your argument, perhaps I am being dense. Could you explain how you conclude the volume is bounded by (128/n)^(n/4)?
Dec 9, 2009 at 2:33	comment	added	Greg Kuperberg		A very nice argument!
Dec 9, 2009 at 1:14	history	edited	Reid Barton	CC BY-SA 2.5	fixed signs
Dec 9, 2009 at 0:13	history	answered	fedja	CC BY-SA 2.5