Difference between revisions of "TADM2E 2.19"
From Algorithm Wiki
(Recovering wiki) |
|||
| (3 intermediate revisions by 3 users not shown) | |||
| Line 1: | Line 1: | ||
| − | + | <math> | |
| − | ({1}/{3})^n | + | ({1}/{3})^n < |
| − | 6 | + | 6 < |
| − | \log \log n | + | \log \log n < |
| − | \log n = \ln n | + | \log n = \ln n < |
| − | (\log n)^2 | + | (\log n)^2 < |
| − | n^{\frac{1}{3}} + \log n | + | n^{\frac{1}{3}} + \log n < |
| − | \sqrt{n} | + | \sqrt{n} < |
| − | \frac{n}{\log n} | + | \frac{n}{\log n}< |
| − | n | + | n < |
| − | n \log n | + | n \log n < |
| − | n^2 = n^2 + \log n | + | n^2 = n^2 + \log n < |
| − | n^3 | + | n^3 < |
| − | n - n^3 + 7n^5 | + | n - n^3 + 7n^5 < |
| − | ({3}/{2})^n | + | ({3}/{2})^n < 2^n < |
n! | n! | ||
| − | + | </math> | |
Latest revision as of 01:03, 1 August 2020
$ ({1}/{3})^n < 6 < \log \log n < \log n = \ln n < (\log n)^2 < n^{\frac{1}{3}} + \log n < \sqrt{n} < \frac{n}{\log n}< n < n \log n < n^2 = n^2 + \log n < n^3 < n - n^3 + 7n^5 < ({3}/{2})^n < 2^n < n! $
