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  • 0 posts edited
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  • 33 votes cast
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)