meta user
network profile
| bio | website | |
|---|---|---|
| location | Bogota, Colombia | |
| age | 26 | |
| visits | member for | 1 year, 8 months |
| seen | May 13 at 13:58 | |
| stats | profile views | 69 |
"When you're doing something you're passionate about, stress becomes a feature not a bug."
|
Feb 27 |
comment |
Is this recurrence $O(n^2)$? @BrianM.Scott Thanks. I edit my question. I'm sorry but I'm not a native english speaker so I thought that was the correct translation. One more thing, given that there is not an annihilator for the logarithm is there anything else I can do to prove it without induction? |
|
Feb 27 |
revised |
Is this recurrence $O(n^2)$? added 3 characters in body |
|
Feb 27 |
comment |
Is this recurrence $O(n^2)$? @anon can you provide details about your probe by induction (I'm not so agile with that method for probing recurrences). |
|
Feb 27 |
comment |
Is this recurrence $O(n^2)$? @anon 4. Comes from a nullator table. $<digit>$ gets null with $(E-1)$; $<n>$ with $(E-1)^2$; $<2n+digit>$ with $(E-1)^2$ and so on. (Excuse my mathematical informality but I'm an engineering student not a mathematician). |
|
Feb 27 |
asked | Is this recurrence $O(n^2)$? |
|
Feb 24 |
accepted | Pretty simple question about running time |
|
Feb 24 |
comment |
Pretty simple question about running time Thanks @ArturoMagidin didn't know about Lambert's W function. Actually that's the solution. |
|
Feb 24 |
asked | Pretty simple question about running time |
|
Feb 18 |
comment |
Why this subset of $\mathbb{R}^3$ is not a subspace? So, It was kinda simple :( I'm sorry but didn't knew how to test it. |
|
Feb 18 |
revised |
Why this subset of $\mathbb{R}^3$ is not a subspace? deleted 4 characters in body |
|
Feb 18 |
asked | Why this subset of $\mathbb{R}^3$ is not a subspace? |
|
Feb 16 |
comment |
Orthogonal vectors with given magnitude Ummm now I see, the free parameters are the parameters of the general solution. Thanks again. |
|
Feb 16 |
awarded | Commentator |
|
Feb 16 |
comment |
Orthogonal vectors with given magnitude Thank you very much David. Just one more thing. Why do you say the system has two free parameters? I mean, doesn't has three including also v3? |
|
Feb 16 |
accepted | Orthogonal vectors with given magnitude |
|
Feb 16 |
asked | Orthogonal vectors with given magnitude |
|
Feb 5 |
comment |
If $A^2 = I$ (Identity Matrix) then $A = \pm I$ Thank you @Martin Wanvik, pretty clear explanation. |
|
Feb 5 |
accepted | If $A^2 = I$ (Identity Matrix) then $A = \pm I$ |
|
Feb 5 |
asked | If $A^2 = I$ (Identity Matrix) then $A = \pm I$ |
|
Aug 26 |
accepted | Power a Matrix (Without calculating) |
