Timeline for Lower bound in recurrence relation
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 14 at 12:58 | vote | accept | Xd00fg | ||
| Aug 14 at 12:48 | comment | added | Aleksei Kulikov | @Forest if you've spent any time thinking about my answer (instead of arrogantly saying that I can't do something which was not specified in your post) you would see that this would only change the base case of induction, and it still works. | |
| Aug 14 at 12:45 | history | edited | Aleksei Kulikov | CC BY-SA 4.0 |
deleted 166 characters in body
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| Aug 13 at 23:42 | comment | added | Xd00fg | See my updated question. | |
| Aug 13 at 23:38 | comment | added | Xd00fg | To avoid $v_0$, you can consider $n_i\leq 2^i n_{i-1}+2v_{i-1}$ instead of $ n_i=2^i n_{i-1}+v_{i-2}+v_{i-1}$. | |
| Aug 13 at 23:31 | comment | added | Aleksei Kulikov | Then tell me please, what is $n_2$? | |
| Aug 13 at 23:20 | comment | added | Xd00fg | You can't choose $v_0$ as fraction, in fact each $v$ is always positive integer. And you can't consider $v_0$ in calculation, you're bound to start from $v_1.$ | |
| Aug 13 at 22:31 | history | answered | Aleksei Kulikov | CC BY-SA 4.0 |