Johannes Kloos
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 Apr12 awarded Nice Answer Mar31 comment Push Down Automata What have you tried so far? Also, do you know how to construct a PDA for $a^nb^n$? If you have one for $a^nb^n$ and one for $a^nc^n$, what can you do with that? Mar5 awarded Yearling Feb22 comment When is matrix multiplication commutative? Here's an example why this condition is not neccesary: Take any non-diagonalizable matrix $A$. It will always commute with the unit matrix $1$, but $A$ and $1$ are clearly not simultaneously diagonalizable. Feb12 comment When is matrix multiplication commutative? It's not neccesary, AFAICT. I can't think of a counter-example right know, though. Jan31 awarded Revival Jan22 comment $f(x) \in R[X]$ irreducible $\Rightarrow (f(x))$ ideal? Just to make sure: Do you mean by $(f)$ the ideal generated by $f$? Jan11 comment Powers of $2\times 2$ matrices, such that $A^n = I$ Do you know about rotation matrices? Dec19 awarded Caucus Sep30 awarded Explainer Sep25 awarded Nice Answer Sep13 comment What is a polynomially bounded function? Your argument would be correct if, say, $2^x-1$ was actually a polynomial. But by definition, a polynomial is of the form $a_n x^n + a^{n-1} x^{n-1} + \cdots + a_0$ for some parameters $a_0, \ldots, a_n$, and $2^x-1$ is not of this form. Sep13 revised What is a polynomially bounded function? Not particularly computer science-related, texified Sep2 reviewed Approve spectral-graph-theory tag wiki excerpt Aug2 reviewed Approve Finding the volume of $(x-4)^2+y^2 \leqslant 4$ Aug2 reviewed Reject Finitely generated ideal in Boolean ring; how do we motivate the generator? Jul6 comment Help with function proof Indeed, my bad. Ignore my previous comment. Jul6 comment Help with function proof @CameronWilliams: If you prefer, write $\mathbb{N}_0$. My natural numbers start at 0. Jul6 comment Help with function proof To everybody trying to show a proof: Please make sure that what you want to prove is actually correct. In particular, consider $A = B = \mathbb{N}$, $f(x) = 0$, $Y = 2\mathbb{N}$ (the set of even numbers). Jul6 comment Who named “Quotient groups”? Skimming the paper, Hölder uses the term "Faktorgruppe" (factor group), and refers to an expression "factor of composition" that he attributes to Jordan.