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Math, Madison, food, the Orioles, books, my kids.Tue, 17 Oct 2017 22:29:15 +0000hourly1http://wordpress.com/Comment on Game report: Cubs 5, Brewers 0 by berseliusx
https://quomodocumque.wordpress.com/2017/09/24/game-report-cubs-5-brewers-0/#comment-29349
Tue, 17 Oct 2017 22:29:15 +0000http://quomodocumque.wordpress.com/?p=5981#comment-29349To be fair, most Cubs fans would probably boo Sosa if he showed up at Wrigley Field. Though for far more frustrating reasons.
]]>Comment on Maryland flag, my Maryland flag by Andrew Gelman
https://quomodocumque.wordpress.com/2017/08/02/maryland-my-maryland/#comment-29304
Fri, 13 Oct 2017 00:53:20 +0000http://quomodocumque.wordpress.com/?p=5960#comment-29304We had to memorize the names of the counties too! Garrett Allegheny Washington Frederick Carroll . . .
]]>Comment on Trace test by JSE
https://quomodocumque.wordpress.com/2017/10/05/trace-test/#comment-29237
Fri, 06 Oct 2017 22:23:11 +0000http://quomodocumque.wordpress.com/?p=5983#comment-29237Exactly! It’s about all partial sums, not just pairwise sums. It’s a test for whether a subset of the roots could be the set of roots of an irreducible algebraic factor.
]]>Comment on Trace test by Jason Starr
https://quomodocumque.wordpress.com/2017/10/05/trace-test/#comment-29236
Fri, 06 Oct 2017 22:21:30 +0000http://quomodocumque.wordpress.com/?p=5983#comment-29236Now I believe that I understand. The example in your post is for an irreducible quadratic factor. The analogous result holds for a proper irreducible factor of any degree, but you have to consider a partial sum of roots with more than two summands. Let $F(x,y)$ be an (inhomogeneous) polynomial of total degree $n$ and such that the coefficient of $x^n$ is nonzero. Via Puiseux series, we can factor $F(x,y)$ as $(x-r_1(y))…(x-r_n(y))$. If $F(x,y)$ has a nontrivial factorization in $k[x,y]$, then there exists a nonempty proper subset of these roots, say $r_1(y),…,r_m(y)$ after reordering, whose partial sum $r_1(y)+…+r_m(y)$, as a Puiseux series, is actually an (affine) linear polynomial in $y$. Thus, if every nonempty, proper partial sum is not (affine) linear in $y$, then $F(x,y)$ is irreducible. This is straightforward to prove using elementary algebra via the observation I made in my previous comment: for both $F(x,y)$ and every irreducible factor of $F(x,y)$, since the $x$-degree equals the total degree, the “next-to-leading” coefficient in $x$ is an (affine) linear polynomial in $y$. Note, this sufficient criterion for irreducibility is not necessary: the polynomial $(x-y)^4-y^3$ is irreducible, but there are pair sums that are affine linear in $y$.
]]>Comment on Trace test by Jason Starr
https://quomodocumque.wordpress.com/2017/10/05/trace-test/#comment-29234
Fri, 06 Oct 2017 16:23:59 +0000http://quomodocumque.wordpress.com/?p=5983#comment-29234Typo correction: Those first two pair sums should have constant term $-1$.
]]>Comment on Trace test by Jason Starr
https://quomodocumque.wordpress.com/2017/10/05/trace-test/#comment-29233
Fri, 06 Oct 2017 16:18:29 +0000http://quomodocumque.wordpress.com/?p=5983#comment-29233@JSE. Maybe I misunderstood your post. I thought that we are to sum up roots in pairs. For the polynomial $x^3+x^2-y^2$, the “pair sums” of roots are, to order $y^3$, $y+s_1 + y^2$, $-y+s_2+y^2$, and $s_1+s_2$. If we sum up all three roots, we get the coefficient of $x^2$, which is $1$. In general, if we assume that the polynomial has total degree $n$ and that the coefficient of $x^n$ is nonzero (so equal to $1$ after scaling), then certainly the coefficient of $x^{n-1}$ has degree zero or one in $y$, or else the degree is strictly larger than $n$. Maybe that is the point about root sums being linear.
]]>Comment on Trace test by JSE
https://quomodocumque.wordpress.com/2017/10/05/trace-test/#comment-29232
Fri, 06 Oct 2017 14:12:44 +0000http://quomodocumque.wordpress.com/?p=5983#comment-29232Wait now I’m confused again about whether I was confused. The first factor x^3 + x^2 – y^2, considered as a polynomial in x, has three roots, whose sum is -1, right? I don’t know if I get which two roots you mean.
]]>Comment on Trace test by JSE
https://quomodocumque.wordpress.com/2017/10/05/trace-test/#comment-29231
Fri, 06 Oct 2017 14:08:52 +0000http://quomodocumque.wordpress.com/?p=5983#comment-29231Crap, I messed this up somehow, I started writing the post thinking about homogeneous forms and then changed it. Give me a sec and I’ll work this out.
]]>Comment on Trace test by Jason Starr
https://quomodocumque.wordpress.com/2017/10/05/trace-test/#comment-29230
Fri, 06 Oct 2017 13:33:53 +0000http://quomodocumque.wordpress.com/?p=5983#comment-29230I think that I misunderstood something. Begin with F(x,y) = (x^3+x^2-y^2)x. The extra x factor is just there to insure that there is a nontrivial factorization, which I believe was part of the setup. There are power series roots r_1 = y + s_1 and r_2 = -y + s_2. These are roots of the first factor for F(x,y). The sum of these roots is not linear in y.
]]>Comment on Messing around with word2vec by How to Develop Word Embeddings in Python with Gensim | A bunch of data
https://quomodocumque.wordpress.com/2016/01/15/messing-around-with-word2vec/#comment-29229
Thu, 05 Oct 2017 21:18:36 +0000http://quomodocumque.wordpress.com/?p=5194#comment-29229[…] Messing Around With Word2vec, 2016 […]
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