Timeline for Solving recurrence equation $T(n)=T(n^{2/3})+17$
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Mar 20, 2019 at 22:02 | vote | accept | John D | ||
| Mar 20, 2019 at 5:16 | answer | added | ryan | timeline score: 1 | |
| Mar 19, 2019 at 21:53 | answer | added | ryan | timeline score: 1 | |
| Mar 16, 2019 at 13:58 | comment | added | gnasher729 | I’d recommend using a spreadsheet to find Say T(1000); that should give you some idea. | |
| Mar 16, 2019 at 12:24 | comment | added | Yuval Filmus | The answer will be proportional to the number of times that you have to raise a number to the power $2/3$ until it gets below an arbitrary constant. I'm sure you can work that out on your own. | |
| Mar 16, 2019 at 9:55 | history | edited | David Richerby | CC BY-SA 4.0 |
Formatting
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| Mar 16, 2019 at 9:43 | history | edited | John D | CC BY-SA 4.0 |
added 1 character in body
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| Mar 16, 2019 at 6:52 | history | edited | John L. | CC BY-SA 4.0 |
Fixed typos.
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| Mar 16, 2019 at 6:45 | answer | added | John L. | timeline score: 3 | |
| Mar 15, 2019 at 16:38 | history | edited | dkaeae | CC BY-SA 4.0 |
added 23 characters in body
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| Mar 15, 2019 at 16:32 | comment | added | dkaeae | Have you tried applying any of the methods listed here? | |
| Mar 15, 2019 at 15:45 | review | First posts | |||
| Mar 15, 2019 at 16:32 | |||||
| Mar 15, 2019 at 15:40 | history | undeleted | John D | ||
| Mar 15, 2019 at 15:39 | history | deleted | John D | via Vote | |
| Mar 15, 2019 at 15:38 | history | asked | John D | CC BY-SA 4.0 |