TU Wien:Analysis UE (diverse)/Übungen SS19/Beispiel 106

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Revision as of 01:58, 4 December 2008 by Amsmath-bot (talk | contribs) (replaced <amsmath> with <math>)
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Mit Hilfe der Stirling'schen Approximationsformel zeige man, dass <amsmath>\binom{3n}{n} \sim \left(\frac{27}{4}\right)^n \sqrt{\frac{3}{4 \pi n}}</math>


Lösungsvorschlag

Stirling-Formel: <amsmath>n! \sim \sqrt{2 \pi n} \; \left(\frac{n}{\mathrm e}\right)^{n}</math>

oder: <amsmath>n! \sim \sqrt{2 \pi n} \; n^n \; {\mathrm e}^{-n}</math>


<amsmath>\binom{3n}{n} = \frac{(3n)!}{(3n-n)! \; n!} \sim \frac{3^{3n} \; n^{3n} \; e^{-3n} \; \sqrt{6 \pi n}}{2^{2n} \; n^{2n} \; e^{-2n} \; \sqrt{4 \pi n} \; n^n \; e^{-n} \sqrt{2 \pi n}} = \frac{3^{3n} \; \sqrt{3}}{2^{2n} \; \sqrt{4 \pi n}} = \left(\frac{27}{4}\right)^n \sqrt{\frac{3}{4 \pi n}}</math>


q.e.d.

Manül 13:56, 17. Jun 2008 (CEST)