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authorGregor Kleen <gkleen@yggdrasil.li>2016-04-19 15:21:02 +0200
committerGregor Kleen <gkleen@yggdrasil.li>2016-04-19 15:21:02 +0200
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1\begin{align*}
2 L &= \lim_{n \to \infty} \frac{n \cdot 2^n}{n^n} \\
3 &= \lim_{n \to \infty} \frac{2^{\log_2(n) + n}}{2^{\log_2(n) \cdot n}} \\
4 &= \lim_{n \to \infty} 2^{n + \log_2(n) - \log_2(n) \cdot n} \\
5 n + \log(n) - \log(n) \cdot n &\leq 2n - \log(n) \cdot n \\
6 &= (2 - \log(n)) \cdot n \xrightarrow{n \to \infty} - \infty \\
7 L &= 0
8\end{align*}
9
10$n \cdot 2^n$ und $n^n$ sind also sogar in der selben Komplexitätsklasse.