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I am a professor of Mathematics at the University of Pisa, Italy.


Dec
17
comment Vector fields whose divergence are proper maps
yes, now I see the point
Dec
16
comment Polynomials of low degree that clone polynomials of higher degree
I think so.$\phantom{}$
Dec
16
comment Polynomials of low degree that clone polynomials of higher degree
Here $\deg_k$ is the degree w.r.to $x_k$. For any (non-zero) polynomial $q$, one has $\deg_k(q)=0$ if and only if $q$ does not depend on $x_k$. Then of course $\delta_kq=0$ . So e.g. $p_1\in\mathbb{R}\mathbb[x_2,x_3,\dots,x_{16}]$.
Dec
16
comment Polynomials of low degree that clone polynomials of higher degree
Yes, in other words the sum is over all $\epsilon=\sum_{j\in R}e_j$, and $R$ varies among the $2^4$ subsets of $S$
Dec
16
revised Polynomials of low degree that clone polynomials of higher degree
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Dec
16
revised Polynomials of low degree that clone polynomials of higher degree
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Dec
16
comment Polynomials of low degree that clone polynomials of higher degree
Here $e_k$ is the k-th element of the standard basis, so $x+e_k=(x_1,x_2,\dots,x_k+1,\dots,x_{16})$
Dec
16
revised Polynomials of low degree that clone polynomials of higher degree
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Dec
16
comment Polynomials of low degree that clone polynomials of higher degree
Yes, I used the same notation as in your post, $x:=(x_1,\dots,x_{16})$
Dec
16
comment Polynomials of low degree that clone polynomials of higher degree
(I only now realize there is another answer; sorry)
Dec
16
answered Polynomials of low degree that clone polynomials of higher degree
Dec
16
comment Polynomials of low degree that clone polynomials of higher degree
is the coefficient ring $\mathbb{Z}/2\mathbb{Z}$?
Dec
15
comment Why do we teach calculus students the derivative as a limit?
We always learn by students. Not really in the topic, but I need to tell this recent one. A student showed me a couple of exercises on limits he had done (a bit worried). "Is it 1?" "Correct!" I said. "And this one? I got -2..." "Correct too, very good" said I. And he... "But isn't this in contradiction with the principle of uniqueness of the limit?".
Dec
15
comment Metrics mappings which are metrics
And of course, if $\phi:Z\to Z$ is any bijection and $d$ is a metric on $Z$, $(x,y)\mapsto d(\phi(x), \phi(y))$ is a metric too.
Dec
15
comment What is the number of equitriangulations of the n-cube?
Call a nice triangulation of $I^n$ special if, for any face f (of any dimension) of $I^n$, it induces a triangulation on f by simplices all having a common vertex. I think a question with some hope to be answered is, counting the number of these "special triangulations" of $I^n$ (see comments below David Eppstein's answer).
Dec
15
comment What is the number of equitriangulations of the n-cube?
So I guess there is a lower bound of the form $$A(n)\ge 2^nB(n-1)-2^{n-1}C(n-2)$$ where A(n) B(n) C(n) count the number of (nice, pulled) triangulations, respectively on the n-cube, on a bouquet of n+1 n-cubes, on a complex of (n+1)(n+2) n-cubes, as described above.
Dec
15
comment What is the number of equitriangulations of the n-cube?
Also, we can fix an "axial" triangulation (= all simplices have two opposite vertices V, V' in common, that is a diagonal of the cube as a common edge) fixing a n-2 dimensional triangulation on the union of all those n-2 dimensional faces of the cube, that do not have neither V nor V' as vertex. These are n(n-1) (just one half of the total number 2n(n-1) of n-2 dimensional faces of $I^n$).
Dec
15
comment What is the number of equitriangulations of the n-cube?
So in dimension n we can fix a pulled partition from a vertex V (= by simplices all having V as a vertex) exactly by defining a triangulation of the bouquet of n facets of the cube around the vertex opposite to V.
Dec
15
comment What is the number of equitriangulations of the n-cube?
@Włodzimierz, yes, I mean triangulations (though the OP is also interested in partitions). I wrote a comment below.
Dec
15
comment What is the number of equitriangulations of the n-cube?
In other words, for any vertex V of the cube, following TMA's construction, we have 8 triangulations made by simplices that share that vertex V. So there are 64 such "conic" triangulations, but this way we count more times these triangulations that admit more common vertices, and these are indeed possible exactly for 2 opposite vertices of a diagonal. If we fix a diagonal of the cube there is exactly one nice triangulation all of whose simplices share that diagonal. So I see 60 conic nice triangulations of a cube.