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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?
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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?
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Nov
24
revised Why aren't all homomorphic abelian groups isomorphic as well?
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