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Jul
31
revised How following matrices equation is solved?
math type
Jul
31
comment How following matrices equation is solved?
What do the $[]$ mean?
Jul
2
comment Show that $A^k$ has eigenvalues $\lambda^k$ and eigenvectors $v$.
I find your proof more simple, more direct, more general and more obvious, thus much better.
Jun
24
awarded  Popular Question
Jun
21
awarded  Tenacious
Jun
19
revised Why are these curves' points called rational?
added 2 characters in body
Jun
19
comment Why are these curves' points called rational?
Thanks for the answer and explanation
Jun
19
accepted Why are these curves' points called rational?
Jun
19
comment Why are these curves' points called rational?
Ok, thank you very much for the answer and the example. Which by the way... is it right? $i^4+i^4=2=\sqrt 2^4=4$? Should the curve have a $2$ coefficient multiplying the $x$ and $y$ terms?
Jun
19
asked Why are these curves' points called rational?
Jun
17
comment Proposed proof of topological result
Other than that, I find it perfect. Try to separate paragraphs though, it's a bit hard to read.
Jun
17
comment Proposed proof of topological result
What's $(B,d)$ and $(A,d)$?
Jun
17
reviewed Approve Proposed proof of topological result
Jun
17
answered How to factor $ax^{2}+bxy+cy^{2},\,a\neq 0$?
Jun
17
revised $ (x x^T)^{-1}$, efficient matrix inversion for matrix composed as product of a vector with itself?
edited a wrong order of factor in the first denominator of the second formula
Jun
12
answered what will be the parameterization of cone
Jun
12
comment Solutions for $|x^2-5x+2|=4$
@BetterWorld That means they should also come as a solution when you consider the negative sign of the modulus, try it. You should discard those solutions when you get the from the positive values, and when consider the negative, $x=1$ should come out again.
Jun
12
comment Solutions for $|x^2-5x+2|=4$
@BetterWorld That just means they're not solutions, because they fail to satisfy the hypothesis that $x^2-x-1$ is positive, on which the solution depends.
Jun
12
reviewed Approve Limit of $\large{\frac{1}{2^x}}$ as $x \to \infty$
Jun
12
comment Solutions for $|x^2-5x+2|=4$
You're very welcome! :) No need for sirs.