Reputation
347
Top tag
Next privilege 500 Rep.
Access review queues
 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. $$gets null with (E-1);$$ 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)