Timeline for Solving recurrence relation $T(2n) \leq T(n) + T(n^a)$
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 28, 2014 at 7:33 | comment | added | Yuval Filmus | @QiangLi Thanks for spotting the error. The lower bound does hold, however, once the argument is suitably modified. | |
| Dec 28, 2014 at 7:32 | history | edited | Yuval Filmus | CC BY-SA 3.0 |
added 41 characters in body
|
| Dec 28, 2014 at 6:14 | comment | added | Qiang Li | It seems that $\log^k(2n) = (\log 2 + \log n)^k$. The Lower bound doesn't hold. | |
| Dec 8, 2014 at 21:22 | history | answered | Yuval Filmus | CC BY-SA 3.0 |