mebassett
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 Nov 26 awarded Benefactor Nov 26 accepted ideal calculation and relations Nov 25 comment Finding the smallest decision tree of a Boolean function I think learning an optimal decision tree is an NP hard problem, but the standard algorithms (id3. C4.5) will get you pretty close Nov 25 comment Nonzero polynomials of degree $n$ in $D[x]$ have at most $n$ distinct roots in $D$, assuming $D$ is only an integral ring. you are right, that example alone is a counterexample to my assertion. I obviously haven't though about it deeply enough. Nov 24 comment does the $\zeta_K$ function of a function field determine the genus of that function field? yes, of course. I don't know I didn't say that despite taking the time to write it out! Nov 24 comment Basis in R4 with only two vectors in the set linear independence means they must not lie on the same line. most vectors in $\mathbb{R}^4$ do not lie on the line formed by your two vectors, so picking randomly is a pretty good method. Nov 24 awarded Promoter Nov 24 asked does the $\zeta_K$ function of a function field determine the genus of that function field? Nov 24 comment Nonzero polynomials of degree $n$ in $D[x]$ have at most $n$ distinct roots in $D$, assuming $D$ is only an integral ring. it might be true, depending on which $x$ and $a$, but in general, no. But that is OK, so long as the coefficients of $f$ are on the same side of the indeterminate $x$. Then you define the matrix multiplication so its consistent with evaluation of the polynomial, and everything follows. The point is that in a noncom ring, you are OK so long as you do everything all on the right side. or all on the left side. just don't go swapping the two around. Nov 24 comment Nonzero polynomials of degree $n$ in $D[x]$ have at most $n$ distinct roots in $D$, assuming $D$ is only an integral ring. I am not assuming $D$ is commutative. But I might have swapped the order of multiplication around because I am sloppy. The result still holds whether or not $D$ is commutative. Nov 24 awarded Yearling Nov 24 answered Nonzero polynomials of degree $n$ in $D[x]$ have at most $n$ distinct roots in $D$, assuming $D$ is only an integral ring. Nov 24 comment Why is abelianness such a precious property? not a typo, actually, but me not remembering properly all my facts about finite abelian groups. :) Nov 24 revised Why is abelianness such a precious property? added 6 characters in body Nov 24 answered I think I'm making an error in my implementation of least squares but I'm not sure (python). Nov 24 revised Why is abelianness such a precious property? added 870 characters in body Nov 24 revised Why is abelianness such a precious property? added 870 characters in body Nov 24 answered Why is abelianness such a precious property? Nov 24 revised Why aren't all homomorphic abelian groups isomorphic as well? deleted 58 characters in body Nov 24 revised Why aren't all homomorphic abelian groups isomorphic as well? edited body