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author | Gregor Kleen <gkleen@yggdrasil.li> | 2016-04-19 15:21:02 +0200 |
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committer | Gregor Kleen <gkleen@yggdrasil.li> | 2016-04-19 15:21:02 +0200 |
commit | a61b998270e8f955c603de96775e072a555f735d (patch) | |
tree | c89c8cd2ecab33eecc4fefb4a8089a8df208d17a /ss2016/algodat/01/H1-2.md | |
parent | 6e70a46771d3b28edc80a88de22b98df47147dc1 (diff) | |
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AlgoDat 01
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diff --git a/ss2016/algodat/01/H1-2.md b/ss2016/algodat/01/H1-2.md new file mode 100644 index 0000000..7a50a2a --- /dev/null +++ b/ss2016/algodat/01/H1-2.md | |||
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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. | ||