TITLE: Maximizing the volume of the convex hull of $N$ points in the unit ball
QUESTION [2 upvotes]: Suppose we are given an integer $N\ge4$, and we have to pick $N$ points in a unit ball in $\mathbb R^3$ to maximize the volume of their convex hull. Are those points necessarily on the surface of the ball? Can we generalize the result to $\mathbb R^n$?
For the similar problem in $\mathbb R^2$ the answer seems obvious: we always have to pick the vertices of a regular $N$-gon inscribed in the disk, so they always lie on the disk boundary.

REPLY [4 votes]: Take a point $Q$ in the interior of the convex hull of points $P_1,\cdots,P_N$ and extend the rays $QP_k$ to the points $P_k^\prime$ on the surface of the ball to get a convex hull containing the old yet with volume not less than the old.


TITLE: A relation in a finite ring
QUESTION [0 upvotes]: Let $R$ be a finite ring such that for any $a,b\in R$ there exists $c\in R$ (depending on $a$ and $b$) such that $a^2+b^2=c^2$.
Prove that for any $a,b,c\in R$ there exists  $d\in R$ such that $2abc=d^2$.

REPLY [3 votes]: Use the word "assumption" to refer to the statement: "for any $a, b\in R$
there exists $c\in R$ (depending on $a$ and $b$) such that $a^2+b^2=c^2$". Let the phrase ``desired conclusion'' refer to the statement: "for any $a, b, c\in R$ there exists $d\in R$ such that $2abc=d^2$".
The assumption implies that the set of squares is closed under sum. This implies that if $u$ is a square, then $u+u=2u$ is a square, and
$u+2u=3u$ is a square, etc, until $-u$ is a square (by finiteness).
It follows from this that $(x+1)^2 + (-x^2) + (-1) = 2x$ is a square for any $x$, which is the desired conclusion when $x=abc$.


TITLE: Solving a system of equations in Maple
QUESTION [1 upvotes]: I have a set of equations which I would like to symbolically solve for a defined set of variables using the solve command in Maple. However, being not familiar with the program, it does not give me exactly what I would like. Maybe someone can give me a hing. Here's a simple example to illustrate my issue:
Given the system of equations
\begin{align}
x+b &=2\cdot y + y - a\\
y-a &= b,
\end{align}
using the solve(...,{x,y}) command in Maple gives me the result
\begin{align}
y &=a+b\\
x &= 2\cdot b + 2\cdot a.
\end{align}
What would be sufficient for me is to have one of the variables to be solved for as a function of the second one, i. e.
\begin{align}
y &=a+b\\
x &= 2\cdot y.
\end{align}
How can I do this?

REPLY [1 votes]: If you're looking for a solution where the value of $x$ doesn't involve $a$, you can solve for $a$ and $x$ in terms of $b$ and $y$:

solve(..., [a,x]);

$$[[a = y-b, x = 2 y]]$$
and then rewrite the first equation to give $y$ in terms of $a$ and $b$:

isolate(%[1][1],y);


TITLE: How to prove $(\forall x)(A\to B)\to(\exists x)A\to(\exists x)B$
QUESTION [0 upvotes]: I am stuck on this one, I know I have to use Ax3 which is $$(\forall x)(A\to B)\to(\exists x)A\to(\exists x)B$$ and convert the existential quantifier to universal, but I have problem making it to become just as Ax3
The original was just like the following:
Show that |- $(\forall x)(A\to B)\to(\exists x)A\to(\exists x)B$
Hint: Use the definition of existential quantifier to eliminate ∃

REPLY [0 votes]: I don't think you can use Ax3 to prove that property. I don't know what rules you're allowed to use, but I would do the following: Assume that $\forall x (A \implies B)$ and $\exists x A$. Then eliminate the existential to get some particular $A$. Then apply $\forall x (A \implies B)$ to get a particular $B$. Then use this $B$ as the witness to the final existential.


text:
\section{Algorithms.}\label{sec:algorithms}
\subsection{Graph representations of maps.}

Let $X$ be a topological space and let $N\subset X$. For a family of subsets $\mathcal U\in 2^{X}$ by $\supp{\mathcal U}$ we will denote its geometrical representation $\supp{\mathcal U}=\bigcup_{U\in\mathcal U} U$.

\begin{definition}
We will say that $\mathcal U\in 2^X$ is a cover of $N$ if $\mathcal U$ is a finite set and $N\subset\supp{\mathcal U}$.
\end{definition}

\begin{definition}
 Let $\mathcal U, \mathcal V$ be a finite sets. A multivalued function $f\colon \mathcal U\multimap\mathcal V$ is called a combinatorial map.
\end{definition}

Covers of compact sets as well as combinatorial maps appear naturally in the computer assisted proofs in dynamical systems.
We usually cannot compute the range of a map $f:X\to Y$ over a domain  $U\subset X$. Instead, one can fix covers $\mathcal X, \mathcal Y$ of $X$ and $Y$, respectively, and then compute combinatorial representation of the map in the covers $\mathcal X, \mathcal Y$.
This is usually realized in the following steps.
\begin{itemize}
\item For $U\subset X$ compute a minimal cover $$\mathcal U = \bigcap \left\{ \mathcal V\subset \mathcal X : U\subset \supp{\mathcal V}\right\}$$
\item For $V_i\in\mathcal U$ compute a cover $\mathcal V_i\subset \mathcal Y$ of $f(V_i)$. This step is often realized by means of the interval arithmetic \cite{M}.
In most cases it is very difficult to compute $\mathcal V_i$ as a minimal cover of $f(V_i)$.
\item Define a cover of $f(U)$ by $\bigcup \mathcal V_i$.
\end{itemize}
The above considerations lead to a very natural definition of the combinatorial representation of a map.
\begin{definition}
Let $f\colon X\to Y$ be a map and let us fix covers $\mathcal X,
\mathcal Y$ of $X$ and $Y$, respectively.

We will say that a combinatorial map $\mathcal F\colon\mathcal X\multimap\mathcal Y$ is a combinatorial representation of $f$ if for $x\in X$ and for all $U\in\mathcal X$ such that $x\in U$ holds $f(x)\in \supp{\mathcal F(U)}$.
\end{definition}
For a map $f\colon X\to X$ and a cover $\mathcal X$ of $X$ it is natural to encode a combinatorial representation of $f$ as a directed graph. A finite directed graph is a pair $\mathcal G = \graph{V}{E}$, where
$\mathcal V$ is a finite set whose elements are called vertexes, and $\mathcal E \subset \mathcal V\times \mathcal V$ is a set of selected
(ordered) pairs of vertexes. The elements of $\mathcal E$ are called edges.
\begin{definition}
 We say that a graph $\mathcal G=\graph{V}{E}$ is a representation of a combinatorial map $\mathcal F\colon \mathcal U\multimap\mathcal U$ if
  $\mathcal V=\mathcal U$ and
 \begin{equation*}
  \mathcal E =  \{(U,V)\in\mathcal U\times \mathcal U : V\in \mathcal F(U)\}.
 \end{equation*}
\end{definition}

We will need a notion of the transposed graph and outgoing edges. For a graph $\mathcal G=\graph{V}{E}$ by $\mathcal G^T=\graph{V'}{E'}$ we will denote a graph in which $\mathcal V'=\mathcal V$ and
$$\mathcal E'=\left\{\right(U,W):(W,U)\in\mathcal E\}.$$

For a vertex $V$ in a graph $\mathcal G=\graph{V}{E}$ by $\mathrm{out}(V,\mathcal G)$ we will denote the set of vertexes $$\mathrm{out}(V,\mathcal G) = \{P\in\mathcal V : (V,P)\in\mathcal E\}.$$

config: arxiv


TITLE: Property of absolute convergent serie
QUESTION [1 upvotes]: So I just ran into this inequality for absolutely convergent series:
$$|\sum_{k=1}^\infty a_k | \le \sum_{k=1}^\infty |a_k| $$
I cant seem to find the proof of this inequality, is it available somewhere?

REPLY [3 votes]: The infinite series $\sum_{k=1}^\infty a_k$ is shorthand for $\lim_{k\to\infty} \sum_{i=1}^k a_i$.  For each $k$,
$$
    | a_1 + a_2 + \dots + a_k | \leq |a_1| + |a_2| + \dots + |a_k|
$$
(that is the triangle inequality).  Since both $\sum_{k=1}^\infty a_k$ and $\sum_{k=1}^\infty |a_k|$ converge, we can take the limit of each side of the above inequality and arrive at
$$
    \left|\lim_{k\to\infty}(a_1 + a_2 + \dots + a_k)\right|
\leq \lim_{k\to\infty}\left(|a_1| + |a_2| + \dots + |a_k|\right)
$$
which is what we want.


TITLE: Linear Transformation (Rotation)
QUESTION [1 upvotes]: Let $T: \mathbb R^2 \rightarrow \mathbb R^2$ be the rotation by $π/2$ clockwise about the origin, and let $S: \mathbb R^2 \rightarrow \mathbb R^2$ be the shear along the y-axis given by $S(x, y) = (x, x + y)$. (You may assume that these are linear transformations.)
a) Write down, or compute, the standard matrix representations of T and S.
Solution:
[T] = $\begin{bmatrix} cos(π/2)&sin(π/2) \\ -sin(π/2)&cos(π/2)\end{bmatrix} = \begin{bmatrix} 0&1 \\ -1&0\end{bmatrix}$
$S(1,0) = (1,1) \implies [S(e_1)] = \begin{bmatrix} 1 \\ 1\end{bmatrix}$
$S(0,1) = (0,1) \implies [S(e_1)] = \begin{bmatrix} 0 \\ 1\end{bmatrix}$
$\implies [S] = \begin {bmatrix} 1&0 \\ 1&1 \end{bmatrix}$
b) Use (a) to find the standard matrix representations of (i) S ◦ T (T followed by S), and (ii) T ◦ S (S followed by T).
Solution:
i) S ◦ T: $\begin{bmatrix}1&0 \\ 1&1 \end{bmatrix} \begin{bmatrix} 0&1 \\ -1&0\end{bmatrix} = \begin{bmatrix} 0&1 \\ -1&1\end{bmatrix}$
ii) T ◦ S: $\begin{bmatrix} 0&1 \\ -1&0\end{bmatrix} \begin{bmatrix} 1&0 \\ 1&1\end{bmatrix} =\begin{bmatrix} 1&1 \\ -1&0\end{bmatrix}$
c) Let $C ⊆ \mathbb R^2$ be the unit square whose vertices are $(0, 0), (1, 0), (0, 1)$, and $(1, 1)$.
Draw, on separate axes, the image of C under the transformations (i) S ◦ T and (ii) T ◦ S,
clearly labelling each image. (Please draw large, clear diagrams.)
Can someone please help me with part c? And also check my solution for the first two parts?

REPLY [2 votes]: Your first two parts are correct.
The point is that $S \circ T$ and $T \circ S$ are linear transformations.
Therefore, let $(x,y) \in C$. Then from what we know about $C$, we have $0 \leq x \leq 1$ and $0 \leq y \leq 1$. Observe that for any linear transformation $L$:
$$
L(x,y) = xL(1,0) + yL(0,1)
$$
Therefore,
$$
L(C) = \{xL(0,1) + yL(1,0) : x,y \in [0,1]\}
$$
From which it is easy to prove that $L(C)$ is the quadrilateral determined by the vertices $L(0,1), L(1,0), L(1,1)$ and $0$.  Calculation of these quantities for $L = S \circ T$ and $L = T \circ S$ gives you the desired conclusions.


TITLE: what is a vector dot del in cylindrical coordinates? $\vec{U}\cdot \nabla$
QUESTION [0 upvotes]: what should be the product of :
$\vec{U}\cdot \nabla$  in a cylindrical coordinate ? for example with a scalar following the product such as $(\vec{U}\cdot \nabla) \Omega$ ? Thank you

REPLY [0 votes]: $\renewcommand\phi\varphi$
The question already has answers here and here. However, those answers only treat the case when one is using an orthonormal basis $\{e_r,e_\phi,e_z\}$. In that case the gradient vector is
$$\nabla f = \partial_rf \,e_r + \frac 1 r \partial_\phi f \,e_\phi +\partial_zf \,e_z$$
So for a vector $u=u_re_r+u_\phi e_\phi+u_ze_z$ you have
$$\begin{align*}
u\cdot \nabla f
&= u_r\partial_rf \, e_r\cdot e_r
  + u_\phi\frac 1 r\partial_\phi f \, e_\phi\cdot e_\phi
  + u_z\partial_zf \, e_z\cdot e_z \\
&= u_r\partial_rf + u_\phi\frac 1 r\partial_\phi f+u_z\partial_zf \\
&= \left(u_r\partial_r + u_\phi\frac 1 r\partial_\phi +u_z\partial_z\right)f
\end{align*}$$
Hence
$$u\cdot \nabla =u_r\partial_r + u_\phi\frac 1 r\partial_\phi +u_z\partial_z.$$

On the other hand, if you are using the holonomic basis $\{v_r,v_\phi,v_z\}$ (that is, $v_r=e_r, v_\phi=re_\phi, v_z=e_z$), then the gradient vector is
$$\nabla f = \partial_rf \, v_r + \frac 1 {r^2}\partial_\phi f \, v_\phi +\partial_zf \, v_z$$
So for a vector $u=u_rv_r+u_\phi v_\phi+u_zv_z$ you have
$$\begin{align*}
u\cdot \nabla f
&= u_r\partial_rf \, v_r\cdot v_r
  + u_\phi\frac 1 {r^2}\partial_\phi f \, v_\phi\cdot v_\phi
  + u_z\partial_zf \, v_z\cdot v_z \\
&= u_r\partial_rf + u_\phi\partial_\phi f+u_z\partial_zf \\
&= \left(u_r\partial_r + u_\phi\partial_\phi +u_z\partial_z\right)f
\end{align*}$$
(the middle factor goes away because $v_\phi\cdot v_\phi=r^2$). Hence
$$u\cdot \nabla =u_r\partial_r + u_\phi\partial_\phi +u_z\partial_z.$$
Again please note that these formulas only hold when the directional derivative operator $u\cdot \nabla$ is applied to a scalar.


TITLE: Evaluate: $\lim_{n\to\infty} \int_a^{\infty}\frac{n^2xe^{-n^2x^2}}{1+x^2}\,dx.$
QUESTION [0 upvotes]: Show that $\displaystyle\lim_{n\to\infty}\displaystyle\int_a^{\infty}\frac{n^2xe^{-n^2x^2}}{1+x^2}\,dx=0$ for $a>0$ but not for $a=0.$

My work: Let the value of the given limit be $L.$ Put $n^2x^2=t$ in the integrand. Then we have:
$$L=\frac12\cdot\lim_{n\to\infty}\int_{n^2a^2}^{\infty}\frac{e^{-t}}{1+t/n^2}\,dt.\tag1$$
Now plugging in $t=n^2a^2+z$ yields:
$$L=\frac12\cdot\lim_{n\to\infty}\int_0^{\infty}\frac{e^{-n^2a^2}\cdot e^{-z}}{1+a^2+z/n^2}\,dz.$$
The last integrand is less than $e^{-z},$ an integrable function in $[0,\infty).$ Also as $n\to\infty,$ the same integrand approaches $0.$ Hence by dominated convergence theorem, we obtain that $L=0.$
The way I have solved the problem, I am not noticing any effect of the sign of $a$ on the value of $L.$ However, if I consider $a=0$ then from $(1)$ using a similar approach I found that $L=\frac12.$ So sign of $a$ does matter. So my main query is what is wrong with my approach? Please give some insights.
Thanks in advance.
EDIT. Another thing that came into my mind just now is the following:
If we substitute $nx=t,$ then:
\begin{align}
L&=\lim_{n\to\infty}\int_{na}^{\infty} \frac{te^{-t^2}}{1+t^2/n^2}\,dt\\\\
&=\lim_{n\to\infty}\int_0^{\infty} \chi_{(na,\infty)}\frac{te^{-t^2}}{1+t^2/n^2}\,dt,
\end{align}
where $\chi_A$ represents the characteristic function of the set $A.$
Here it seems that if $a<0$ then it would affect the value of $L$ and to be honest I'm not quite sure about the dominant integrable function anymore. Please shed some light on the same, particularly for the case $a<0.$

REPLY [2 votes]: Hint
Another way is the squeeze theorem:
$$
\int_a^{\infty}\frac{n^2xe^{-n^2x^2}}{1+x^2}\,dx \le
\int_{a}^{\infty}n^2xe^{-n^2x^2}\,dx=\frac{1}{2}\exp(-n^2a^2).
$$


TITLE: Index notation and total differential
QUESTION [2 upvotes]: I understand that the gradient $\partial_i$ is covariant.
Let f be a function of 3 variables
So I can write the total differential as
$$
df=\partial_1fdx^1+\partial_2fdx^2+\partial_3fdx^3 = \partial_kfdx^k,
$$
summing correctly over the same lower and upper index.
But when I write
$
\displaystyle\partial_i=\frac{\partial}{\partial x^i}
$ (and this is how it is written in all books)
and want to write the total differential, I have to write
$$
df= \frac{\partial f}{\partial x^1}dx^1+\frac{\partial f}{\partial x^2}dx^2+\frac{\partial f}{\partial x^3}dx^3=\frac{\partial f}{\partial x^k}dx^k
$$
and so I am summing over the same upper indices.
My question: Why does an index get raised when it is moved to the denominator?

REPLY [1 votes]: $\partial_i$ is defined as $\dfrac{\partial}{\partial x^i}$. Therefore the index does not get raised when brought to the denominator, rather it is notation for the derivative justified rigorously:
In Cartesian coordinates the partial derivative transforms as a covariant vector
$$\frac{\partial}{\partial \bar{x}^i}=\frac{\partial x^j}{\partial \bar{x}^i}\frac{\partial}{\partial x^j}$$
Covariant vectors are written with lowered indices.


TITLE: Domain of the Stokes operator
QUESTION [0 upvotes]: Let

$\Omega\subseteq\mathbb R^d$ be open ($d\in\mathbb N$)
$\mathcal D:=C_c^\infty(\Omega)^d$ and $$\mathfrak D:=\left\{\phi\in\mathcal D:\nabla\cdot\phi=0\right\}$$
$\mathcal H:=\overline{\mathfrak D}^{\langle\;\cdot\;,\;\cdot\;\rangle_{L^2(\Omega,\:\mathbb R^d)}}$ and $\mathcal V:=H_0^1(\Omega,\mathbb R^d)\cap\mathcal H$

If $$\mathfrak a(u,v):=\sum_{i=1}^d\langle\nabla u_i,\nabla v_i\rangle_{L^2(\Omega,\:\mathbb R^d)}\;\;\;\text{for }u,v\in H_0^1(\Omega,\mathbb R^d)\;,$$ then the Dirichlet-Laplace operator is defined to be $$\Delta_D:H_0^1(\Omega,\mathbb R^d)\to H^{-1}(\Omega,\mathbb R^d)\;,\;\;\;u\mapsto-\mathfrak a(\;\cdot\;,u)\;.$$ If $\tilde\iota:\mathcal V\to H_0^1(\Omega,\mathbb R^d)$ is the inclusion and $\tilde{\operatorname P}:H^{-1}(\Omega,\mathbb R^d)\to\mathcal V'$ is the adjoint of $\tilde\iota$, then $$A_0:\mathcal V\to\mathcal V'\;,\;\;\;u\mapsto-\tilde{\operatorname P}\Delta_D\tilde\iota u$$ is a well-defined bounded, linear and symmetric operator with $$A_0u=\left.(-\Delta_Du)\right|_{\mathcal V}\;\;\;\text{for all }u\in\mathcal V\tag 1$$ and the Stokes operator is defined to be the restriction $A$ of $A_0$ to $$D(A):=\left\{u\in\mathcal V:A_0u\in\mathcal H\right\}\;.\tag 2$$

Now, I've observed the following: If $u\in H_0^1(\Omega)$ is twice weak differentiable (e.g. $u\in H_0^2(\Omega)$), then $u$ has a weak Laplacian $\Delta u\in L_{\text{loc}}^1(\Omega)$ and $$(\Delta_Du)v=\langle v,\Delta u\rangle_{L^2(\Omega,\:\mathbb R^d)}\;\;\;\text{for all }v\in H_0^1(\Omega,\mathbb R^d)\;.\tag 3$$ If $\iota:\mathcal H\to L^2(\Omega,\mathbb R^d)$ is the inclusion and $\operatorname P:L^2(\Omega,\mathbb R^d)\to\mathcal H$ is the orthogonal projection, then $\iota^\ast=\operatorname P$ and hence (using $(3)$) $$(\Delta_Du)v=\langle v,\operatorname P\Delta u\rangle_{L^2(\Omega,\:\mathbb R^d)}\;\;\;\text{for all }v\in\mathcal V\;.\tag 4$$ Thus, if $u\in\mathcal V$, we obtain $$(A_0u)v=\langle v,-\operatorname P\Delta u\rangle_{L^2(\Omega,\:\mathbb R^d)}\;\;\;\text{for all }v\in\mathcal V\;.\tag 5$$ by $(1)$, i.e. $\Delta_Du\in\mathcal H$.

So, my question is: Shouldn't we be able to define the Stokes operator to be $$\tilde A:D(\tilde A)\to\mathcal H\;,\;\;\;u\mapsto-\operatorname P\Delta u$$ with $D(\tilde A)=H^2(\Omega,\mathbb R^d)\cap\mathcal V$? I've read that this isn't possible, unless $\Omega$ is regular enough, but I don't see why.

REPLY [1 votes]: If $\Omega$ has reentrant corners, then solutions of the Stokes problem will in general have corner singularities which prevent $H^2$ regularity. Therefore, if you define the Stokes operator in the manner you suggest (which you can do, it is a free country), you will not get the "desirable" existence theory.


TITLE: Torsion and curvature of a curve
QUESTION [4 upvotes]: A regular curve $\textbf{$\gamma$}$ in $\mathbb{R}^3$ with curvature $> 0$ is called a generalized helix if its tangent vector makes a fixed angle $\theta$ with a fixed unit vector $\textbf{a}$. Show that the torsion $\tau$ and curvature $\kappa$ of $\textbf{$\gamma$}$ are related by $\tau = ±\kappa \cot \theta$. Show conversely that, if the torsion and curvature of a regular curve are related by $\tau = \lambda \kappa$ where $\lambda$ is a constant, then the curve is a generalized helix.
$$$$
I have done the following:
$\textbf{a}$ is a fixed unit vector: $||\textbf{a}||=1$ and since it is fixed we have that $\textbf{a}'=0$.
$$\textbf{a} \cdot \textbf{t}=||\textbf{a}|| ||\textbf{t}|| \cos \theta = ||\textbf{t}|| \cos \theta \\ \Rightarrow (\textbf{a} \cdot \textbf{t})'=0 \Rightarrow \textbf{a}' \cdot \textbf{t}+\textbf{a} \cdot \textbf{t}'=0 \Rightarrow \textbf{a} \cdot \kappa \textbf{n}=0 \overset{ \kappa >0 }{ \Rightarrow } \textbf{a} \cdot  \textbf{n}=0 \Rightarrow \textbf{a} \bot \textbf{n}$$
$\textbf{t}$, $\textbf{n}$ and $\textbf{b}$ consists an orthonormal basis of $\mathbb{R}^3$.
So, $$\textbf{a}=A \textbf{t}+ B \textbf{n}+ C\textbf{b}$$
We have that $\textbf{a}$ is on the plane spanned by $\textbf{t}$ and $\textbf{b}$.
So, $B=0$.
Therefore, $$\textbf{a}=A \textbf{t}+C\textbf{b}$$
How could we continue?

REPLY [5 votes]: You know $\mathbf{a \cdot t}=\cos{\theta}$, so you also have $A=\cos{\theta}$. Differentiating gives
$$ 0 = \cos{\theta} \, \mathbf{t}' + C' \mathbf{b}+ C \mathbf{b}'. $$
We have $\mathbf{t}'=\kappa \mathbf{n}$, and $\mathbf{b}'=-\tau \mathbf{n}$, so one equation is
$$ 0 = (\kappa\cos{\theta}-\tau C) \mathbf{n} + C' \mathbf{b}, $$
and so
$$ C'=0, \quad \kappa \cos{\theta} = \tau C. $$
Now we have to get to $C$. But this is easy: $\mathbf{a}$ is a unit vector, so $(\mathbf{a} \cdot \mathbf{b})^2+(\mathbf{a} \cdot \mathbf{t})^2 = 1$, and therefore $1=\cos^2{\theta}+C^2$, or $C=\pm \sin{\theta}$, which gives you the first part.

For the second part, let's try and cook up a constant vector $\mathbf{a}$ so that
$$\mathbf{a} \cdot \mathbf{t}=\text{const.}$$
First, differentiating this equation gives $ \kappa \mathbf{a} \cdot \mathbf{n} = 0 $. Assume $\kappa \neq 0$, so the equations are nontrivial, and we again have your equation
$$ \mathbf{a} = A \mathbf{t}+C\mathbf{b}, $$
and we then have, differentiating,
$$ 0 = (\kappa A-\tau C) \mathbf{n} + A' \mathbf{t} + C' \mathbf{b} = 0, $$
which immediately implies that $A'=C'=0$, and $\kappa A -\tau C=0 $. The relationship $\tau=\lambda \kappa$ then implies $A=\lambda C$, and hence
$$ \mathbf{a} = C(\lambda \mathbf{t}+\mathbf{b}). $$
Imposing that $\mathbf{a}$ is a unit vector gives $C^2(1+\lambda^2)=1$. This $\mathbf{a}$ clearly exists and is unique up to sign, and was constructed so that $ (\mathbf{a} \cdot \mathbf{t})'=0 $, so we have that the curve is a generalised helix.


TITLE: Can there be an acceleration when no net force acts on a massless object?
QUESTION [1 upvotes]: If no net force acts on a massless object can there be any acceleration of the object?
My attempt:
$$F=ma\\ \implies a=F/m\\ \implies a=0/0 $$
$\implies a$ can be anything.

REPLY [0 votes]: On a massless object (like an ideal rod) you can specify the acceleration of one end and derive the acceleration of the other end. But you cannot apply forces and derive the acceleration because the inertial properties are zero.
$$ \vec{a}_A = \vec{a}_B + \vec{\alpha} \times \vec{r}_{BA} + \vec{\omega} \times \vec{\omega} \times \vec{r}_{BA} $$


TITLE: On Laplacian spectrum of Graph
QUESTION [0 upvotes]: can somebody refer me the article/source in which it has been proved that $G:=K_1\vee (mK_n)$ is determined by its Laplacian spectrum

REPLY [0 votes]: Rather Laplacian spectrum of join of any two graphs is completely determined by the Laplacian spectra of the constituting graphs.

R. Merris, Laplacian graph eigenvectors, Linear Algebra and its Applications, 278 (1998), 221--236.

See Theorem 2.1, Page number 225.


text:
\begin{document}

\title{Strongly Stable Ideals and Hilbert polynomials}

\author[D.~Alberelli]{Davide Alberelli}
\address{Davide Alberelli \\ University of Osnabr\"uck \\ Institut f\"ur Mathematik \\ Albrechtstr.~28a \\ 49076 Osnabr\"uck \\ Germany.}
\email{\href{mailto:dalberelli@uos.de}{dalberelli@uos.de}}
\urladdr{\url{http://www.mathematik.uni-osnabrueck.de/graduiertenkolleg/personen/alberelli_davide.html}}

\author[P.~Lella]{Paolo Lella}
\address{Paolo Lella\\ Dipartimento di Matematica dell'Universit\`{a} degli Studi di Torino\\
         Via Carlo Alberto 10\\ 10123 Torino\\ Italy.}
\email{\href{mailto:paolo.lella@unito.it}{paolo.lella@unito.it}}
\urladdr{\url{http://www.personalweb.unito.it/paolo.lella/}}

\subjclass[2010]{13P10, 13P99}

\begin{abstract}
The \emph{StronglyStableIdeals} package for \textit{Macaulay2} provides a method to compute all the saturated strongly stable ideals in a given polynomial ring defining schemes with a fixed Hilbert polynomial. A description of the main method and the auxiliary tools is given.
\end{abstract}

\keywords{Strongly stable ideal, Borel-fixed ideal, Hilbert polynomial, Gotzmann number}

\maketitle

\section*{Introduction}

Strongly stable ideals are a key tool in commutative algebra and algebraic geometry. These ideals have nice combinatorial properties that makes them well suited for both theoretical and computational applications. In the case of polynomial rings with coefficients in a field with characteristic zero, the notion of strongly stable ideals coincides with the notion of Borel-fixed ideals. Ideals of the latter type are fixed by the action of the Borel subgroup of triangular matrices of the linear group and they correspond to the possible generic initial ideals by a famous result by Galligo \cite{Galligo}. In the context of Hilbert schemes, Galligo's theorem states that each component and each intersection of components contains at least a point corresponding to a scheme defined by a Borel-fixed ideal. Hence, these ideals represent a promising tool for studying Hilbert schemes and to obtain a complete knowledge of their structure. To this aim, in recent years several authors \cite{LellaRoggero,CioffiRoggero,BCLR,BLR,LR2} developed algorithmic methods based on the use of strongly stable ideals to construct flat families corresponding to open subschemes of a Hilbert scheme. Once we are able to determine such open subschemes, in order to study all the components of a Hilbert scheme, we need to determine all its points corresponding to schemes defined by Borel-fixed ideals, i.e.~all the saturated strongly stable ideals in a polynomial ring with a given Hilbert polynomial. The main feature of the package \emph{StronglyStableIdeals} is a method to compute this set of ideals introduced in \cite{CLMR} and improved in \cite{LellaBorel}. Several other tools are developed and presented in the paper.

\section{Strongly stable ideals}

Let $\KK$ be any field and let $\KK[\boldsymbol{x}]:=\KK[x_0,\ldots,x_n]$ be the polynomial ring in $n+1$ variables with coefficients in $\KK$. We assume $x_0 < x_1 < \cdots < x_n$. We use the multi-index notation to describe monomials, i.e.~for every $\boldsymbol{\alpha} = (\alpha_0,\ldots,\alpha_n) \in \ZZ_{\geq 0}^{n+1}$, $\xx^{\al} := x_0^{\alpha_0}\cdots x_n^{\alpha_n}$ and we denote by $\TT_{n,s}$ the set of monomials of $\KK[\xx]$ of degree $s$. For any monomial $\xx^{\al}$, we denote by $\min \xx^{\al}$ and $\max \xx^{\al}$ the minimal and maximal index of a variable dividing $\xx^{\al}$.

 Following \cite{GreenGIN}, we call \emph{increasing} and \emph{decreasing elementary moves} the elements of the field of fractions $\KK(x_0,\ldots,x_n)$
\begin{equation*}
\up{i} := \frac{x_{i+1}}{x_i},\ i < n\qquad\text{and}\qquad \down{j} := \frac{x_{j-1}}{x_j},\ j > 0.
\end{equation*}
We say that an elementary move $\mathrm{e}_i^{\text{\tiny +/$-$}}$ is \emph{admissible} for a monomial $\xx^{\al}$ if $\mathrm{e}_i^{\text{\tiny +/$-$}}(\xx^{\al})$ is contained in $\KK[\xx]$, i.e.~if $\alpha_i > 0$.

\begin{definition} An ideal $I \subset \KK[\xx]$ is called \emph{strongly stable} if
\begin{enumerate}
\item $I$ is a monomial ideal;
\item for every $\xx^{\al} \in I$ and for every admissible increasing move $\up{i}$, the monomial $\up{i}(\xx^{\al})$ in contained in $I$.
\end{enumerate}
\end{definition}

We recall that a strongly stable ideal is a Borel-fixed ideal. We now summarize some properties holding in general for Borel-fixed ideals and useful in this context.

\begin{proposition}[{\cite[Section 2]{GreenGIN}}]\label{prop:BorelProperties} Let $I \subset \KK[\xx]$ be a strongly stable ideal.
\begin{enumerate}[(i)]
\item\label{it:BorelProperties_i} The regularity of $I$ is equal to the maximal degree of a generator (straightforward using the Eliahou-Kervaire resolution \cite{EliahouKervaire}).
\item\label{it:BorelProperties_ii} Let $\mathfrak{m}$ be the irrelevant ideal of $\KK[\xx]$. Then, $(I:\mathfrak{m}) = (I : x_0)$, so that the ideal $I$ is saturated if no generator involves the variable $x_0$.
\item\label{it:BorelProperties_iii} The variable $x_0$ is a regular element for $I$, i.e. the multiplication by $x_0$ induces the short exact sequence
\[
0\ \longrightarrow\ \faktor{\KK[\xx]}{I}(t-1) \ \xrightarrow{\cdot x_0}\ \faktor{\KK[\xx]}{I}(t)\ \longrightarrow \faktor{\KK[\xx]}{(x_0,I)}(t)\ \longrightarrow\ 0.
\]
\end{enumerate}
\end{proposition}

\section{Hilbert polynomials}
The Hilbert polynomial $p(t)$ of a homogeneous ideal $I \subset \KK[\xx]$ is the numerical polynomial such that for all $s$ sufficiently large
\[
\dim_\KK \left(\faktor{\KK[\xx]}{I}\right)_s = \dim_\KK \left(\faktor{\KK[\xx]_s}{I_s}\right) = \binom{n+s}{n} - \dim_{\KK} I_s = p(s).
\]
Obviously, not every numerical polynomial is a Hilbert polynomial of some homogenous ideal. Those being Hilbert polynomials have been completely described by Gotzmann \cite{Gotzmann}.

\begin{gotzmannDec}\label{th:gotzmannDec}
A numerical polynomial $p(t) \in \QQ[t]$ is a Hilbert polynomial if, and only if, it can be decomposed as
\begin{equation}\label{eq:gotzmannDec}
p(t) = \binom{n+a_1}{a_1} + \binom{n+a_2-1}{a_2} + \ldots + \binom{n+a_r-(r-1)}{a_r},\qquad a_1 \geq a_2 \geq \cdots\geq a_r \geq 0.
\end{equation}
\end{gotzmannDec}

This decomposition is strictly related to the Macaulay's decomposition
\[
p(t) = \sum_{k=0}^d \left[\binom{t+k}{k+1} - \binom{t+k-m_k}{k+1}\right]
\]
where $d = \deg p(t)$. For all $n\geq d+1$ the saturated lexicographic ideal in $\KK[x_0,\ldots,x_n]$ with Hilbert polynomial $p(t)$ is
\[
(x_n,\ldots,x_{d+2},x_{d+1}^{b_d+1},x_{d+1}^{b_d} x_d^{b_{d-1}+1},\ldots,x_{d+1}^{b_d} x_d^{b_{d-1}} \cdots x_1^{b_0}),
\]
where
\[
b_d = \# \{ a_j\ \vert\ a_j = d \} = m_d \quad\text{and}\quad b_k = \# \{ a_j\ \vert\ a_j = k \} = m_k - m_{k+i},\ 0\leq i < d.
\]

The description of the lexicographic ideal in terms of the Gotzmann's decomposition suggests the following theorem.

\begin{gotzmannReg}
The regularity of a saturated ideal $I \subset \KK[\xx]$ with Hilbert polynomial $p(t)$ is at most $r$, where $r$ is the number of terms in the decomposition \eqref{eq:gotzmannDec} and it is called \emph{Gotzmann number} of $p(t)$.
\end{gotzmannReg}

\begin{example}
The package \emph{StronglyStableIdeals} provides the method \texttt{isHilbertPolynomial} to determine if a numerical polynomial is a Hilbert polynomial.

\smallskip

\begin{Verbatim}
Macaulay2, version 1.6
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition,
               ReesAlgebra, TangentCone

i1 : loadPackage ("StronglyStableIdeals");

i2 : R = QQ[t]; -- the ring of numerical polynomials

i3 : p = 4*t;  q = 5*t-6;

i5 : isHilbertPolynomial p

o5 = true

i6 : isHilbertPolynomial q

o6 = false
\end{Verbatim}
Gotzmann's and Macaulay's decompositions of a Hilbert polynomial can be computed with the methods \texttt{gotzmannDecomposition} and \texttt{macaulayDecomposition}. These methods returns the list of the terms of the decomposition. The summand $\binom{t+e}{c}$ is constructed with the command \texttt{projectiveHilbertPolynomial(c,c-e)}.

\smallskip

\begin{Verbatim}
i7 : gotzmannDecomposition p

o7 = {P , - P  + P , - 2*P  + P , - 3*P  + P , P , P }
       1     0    1       0    1       0    1   0   0

o7 : List

i8 : macaulayDecomposition p

o8 = {- P  + P , 7*P  - P , - P  + P , - 10*P  + 5*P  - P }
         0    1     0    1     1    2        0      1    2

o8 : List
\end{Verbatim}
Finally, the saturated lexicographic ideal $L$ with Hilbert polynomial $p(t)$ in the polynomial ring $\KK[\xx]$ can be computed with the method \texttt{lexIdeal} and its regularity is equal to the Gotzmann number of $p(t)$.

\smallskip

\begin{Verbatim}
i9 : L = lexIdeal(p,QQ[x,y,z,w])

             4 2   5
o9 = ideal (y z , y , x)

o9 : Ideal of QQ[x, y, z, w]

i10 : regularity L == gotzmannNumber p

o10 = true
\end{Verbatim}
\end{example}

\section{The main algorithm}
In this section, we outline the strategy of the main algorithm. This algorithm was firstly described in \cite{CLMR} and then optimized in \cite{LellaBorel}. The same problem has been previously discussed in \cite{Reeves} and an alternative algorithm was later presented in \cite{MN}.

We need to relate the properties of a strongly stable ideal with its Hilbert polynomial. If $I$ is a strongly stable ideal, for each $s$ the monomial basis of the homogeneous piece $I_s$ of the ideal is a subset of $\TT_{n,s}$ closed by increasing elementary moves. We call \emph{Borel sets} such subsets of $\TT_{n,s}$ (see Figure \ref{fig:BorelSet} for an example). Proposition \ref{prop:BorelProperties}\emph{(\ref{it:BorelProperties_i})} implies that for every saturated strongly stable ideal of $\KK[\xx]$ with Hilbert polynomial $p(t)$ and regularity at most $s$, the monomial basis of the homogeneous piece of degree $s$ of the ideal is a Borel set with $p(s)$ elements, i.e.~we can consider the following map
\begin{equation}\label{eq:injective}
\left\{\begin{array}{c} \text{saturated strongly stable ideals in }\KK[\xx] \text{ with}\\   \text{Hilbert polynomial } p(t) \text{ and regularity } \leq s \end{array}\right\} \hookrightarrow \left\{ \begin{array}{c} \text{Borel sets of }\TT_{n,s}\\ \text{with } p(s) \text{ elements}\end{array}\right\}.
\end{equation}
Moreover, Gotzmann's Regularity Theorem states that in order to consider all the saturated strongly stable ideals with a given Hilbert polynomial, it suffices to consider \eqref{eq:injective} for $s$ equal to the Gotzmann number.
Obviously, not every Borel set in $\TT_{n,s}$ with $p(s)$ elements corresponds to an ideal with Hilbert polynomial $p(t)$. To identify the image of the previous map, we recall a definition and a proposition by Mall.

\begin{figure}
\begin{center}
\begin{tikzpicture}[>=latex,scale=0.65]

\node (300) at (0,0) [inner sep = 1pt] {\scriptsize $x_2^3$};
\node (210) at (2,0) [inner sep = 1pt] {\scriptsize $x_2^2 x_1$};
\node (120) at (4,0) [inner sep = 1pt] {\scriptsize $x_2 x_1^2$};
\node (030) at (6,0) [inner sep = 1pt] {\scriptsize $x_1^3$};

\draw [<-,very thin] (300) -- (210);
\draw [<-,very thin] (210) -- (120);
\draw [<-,very thin] (120) -- (030);

\node (201) at (2,-1.2) [inner sep = 1pt] {\scriptsize $x_2^2 x_0$};
\node (111) at (4,-1.2) [inner sep = 1pt] {\scriptsize $x_2 x_1 x_0$};
\node (021) at (6,-1.2) [inner sep = 1pt] {\scriptsize $x_1^2 x_0$};

\draw [->,very thin] (201) -- (210);
\draw [->,very thin] (111) -- (120);
\draw [->,very thin] (021) -- (030);

\draw [<-,very thin] (201) -- (111);
\draw [<-,very thin] (111) -- (021);

\node (102) at (4,-2.4) [inner sep = 1pt] {\scriptsize $x_2 x_0^2$};
\node (012) at (6,-2.4) [inner sep = 1pt] {\scriptsize $x_1 x_0^2$};

\draw [->,very thin] (102) -- (111);
\draw [->,very thin] (012) -- (021);

\draw [->,very thin] (012) -- (102);

\node (003) at (6,-3.6) [inner sep = 1pt] {\scriptsize $x_0^3$};

\draw [->,very thin] (003) -- (012);

\node at (0,-4.2) [] {};
\draw [rounded corners,thick] (-1,0.6) -- (5,0.6) -- (5,-1.8) -- (1,-1.8) -- (1,-0.6) -- (-1,-0.6) -- cycle;
\end{tikzpicture}
\qquad\quad
\begin{tikzpicture}[>=latex,scale=0.65]
\node (400) at (-2,1.2) [inner sep = 1pt] {\scriptsize $x_2^4$};
\node (310) at (0,1.2) [inner sep = 1pt] {\scriptsize $x_2^3 x_1$};
\node (220) at (2,1.2) [inner sep = 1pt] {\scriptsize $x_2^2 x_1^2$};
\node (130) at (4,1.2) [inner sep = 1pt] {\scriptsize $x_2 x_1^3$};
\node (040) at (6,1.2) [inner sep = 1pt] {\scriptsize $x_1^4$};

\draw [<-,very thin] (400) -- (310);
\draw [<-,very thin] (310) -- (220);
\draw [<-,very thin] (220) -- (130);
\draw [<-,very thin] (130) -- (040);

\node (300) at (0,0) [inner sep = 1pt] {\scriptsize $x_2^3 x_0$};
\node (210) at (2,0) [inner sep = 1pt] {\scriptsize $x_2^2 x_1 x_0$};
\node (120) at (4,0) [inner sep = 1pt] {\scriptsize $x_2 x_1^2 x_0$};
\node (030) at (6,0) [inner sep = 1pt] {\scriptsize $x_1^3 x_0$};

\draw [->,very thin] (300) -- (310);
\draw [->,very thin] (210) -- (220);
\draw [->,very thin] (120) -- (130);
\draw [->,very thin] (030) -- (040);

\draw [<-,very thin] (300) -- (210);
\draw [<-,very thin] (210) -- (120);
\draw [<-,very thin] (120) -- (030);

\node (201) at (2,-1.2) [inner sep = 1pt] {\scriptsize $x_2^2 x_0^2$};
\node (111) at (4,-1.2) [inner sep = 1pt] {\scriptsize $x_2 x_1 x_0^2$};
\node (021) at (6,-1.2) [inner sep = 1pt] {\scriptsize $x_1^2 x_0^2$};

\draw [->,very thin] (201) -- (210);
\draw [->,very thin] (111) -- (120);
\draw [->,very thin] (021) -- (030);

\draw [<-,very thin] (201) -- (111);
\draw [<-,very thin] (111) -- (021);

\node (102) at (4,-2.4) [inner sep = 1pt] {\scriptsize $x_2 x_0^3$};
\node (012) at (6,-2.4) [inner sep = 1pt] {\scriptsize $x_1 x_0^3$};

\draw [->,very thin] (102) -- (111);
\draw [->,very thin] (012) -- (021);

\draw [->,very thin] (012) -- (102);

\node (003) at (6,-3.6) [inner sep = 1pt] {\scriptsize $x_0^4$};

\draw [->,very thin] (003) -- (012);

\draw [rounded corners, thick] (-3,1.8) -- (7,1.8) -- (7,0.6) -- (5,0.6) -- (5,-1.8) -- (1,-1.8) -- (1,-0.6) -- (-1,-0.6) -- (-1,0.6) -- (-3,0.6) -- cycle;
\end{tikzpicture}
\end{center}
\caption{\label{fig:BorelSet} The Borel sets defined in $\TT_{2,3}$ and $\TT_{2,4}$ by the ideal $(x_2^2,x_2x_1,x_1^4) \subset \KK[x_0,x_1,x_2]$.}
\end{figure}

\begin{definition}[{\cite[Definition 2.7]{Mall1}}] Let $B \subset \TT_{n,s}$ be a Borel set. The set $B^{(i)} := \{ \xx^{\al} \in B\ \vert\ \min \xx^{\al} = i \}$ is called \emph{$i$-growth class} of $B$. The sequence $\text{gv}(B) := (\vert B^{(0)}\vert,\ldots,\vert B^{(n)}\vert)$ is called \emph{growth vector} of $B$.
\end{definition}

\begin{proposition}[{\cite[Proposition 3.2]{Mall1}}]
Let $I\subset \KK[\xx]$ be a strongly stable ideal generated by the monomials of a Borel set $B \subset \TT_{n,s}$ and let $p(t)$ be its Hilbert polynomial. Then,
\begin{equation}\label{eq:hilbPolyGV}
p(t) = \binom{n+t}{n} - \sum_{k=0}^n \left\vert B^{(k)}\right\vert \binom{k+t-s}{k},\qquad \forall\ t \geq s.
\end{equation}
\end{proposition}

Now, the point is to determine the growth vector of a Borel set $B \subset \TT_{n,s}$ starting from the Hilbert polynomial. The first difference polynomial is
\[
\begin{split}
\Delta p(t) = p(t) - p(t-1) & {} = \binom{n+t}{n} - \binom{n+t-1}{n} - \sum_{k=0}^n \left\vert B^{(k)}\right\vert \left[\binom{k+t-s}{k} - \binom{k+t-s-1}{k}\right] = {} \\
& {} = \binom{n+t-1}{n-1} - \sum_{k=1}^n \left\vert B^{(k)}\right\vert \binom{k+t-s-1}{k-1}
\end{split}
\]
so that repeating this computation and evaluating these polynomials for $t = s$, we have
\[
 \sum_{k=i}^n \left\vert B^{(k)}\right\vert = \binom{n+s-i}{n-i} - \Delta^{i} p(s),\qquad \forall\ i = 0,\ldots,n,
\]
where $\Delta^0 p (t) = p(t)$ and $\Delta^{i} p(t) = \Delta^{i-1} p(t) - \Delta^{i-1} p(t-1)$ (notice that $\Delta^i p(t) = 0$ for $i > \deg p(t)$). These conditions allow to deduce that $\vert B^{(n)}\vert = 1$ and for $i < n$
\[
 \left\vert B^{(i)} \right\vert = \sum_{k=i}^n \left\vert B^{(k)}\right\vert - \sum_{k=i+1}^n \left\vert B^{(k)}\right\vert = \binom{n+s-i-1}{n-i} - \Delta^i p(s) + \Delta^{i+1} p(s).
\]
Let us call the sequence of such integers \emph{growth vector} of $p(t)$ in degree $s$ and let us denote it by
$\text{gv}_s\big(p(t)\big)$.

\begin{proposition}[{cf.~\cite[Theorem 3.3]{LellaBorel}}]\label{prop:main}
Let $p(t)$ be a Hilbert polynomial. There is a bijective map
\begin{equation}\label{eq:identification}
\begin{array}{ccccc}
\medskip
\left\{\begin{array}{c} \text{saturated strongly stable ideals} \\ \text{in }\KK[\xx] \text{ with Hilbert polynomial} \\ p(t) \text{ and regularity } \leq s \end{array}\right\} && \xleftrightarrow{1:1} && \left\{ \begin{array}{c} \text{Borel sets of }\TT_{n,s}\\ \text{with } p(s) \text{ elements and}\\ \text{growth vector } \textnormal{gv}_{s}\big(p(t)\big)\end{array}\right\}\\
\smallskip
 I && \longrightarrow && \text{monomial basis of } I_s \\
 \text{saturation of } (B) && \longleftarrow && B
\end{array}
\end{equation}
\end{proposition}

In order to determine the Borel sets of Proposition \ref{prop:main}, we use a recursive algorithm based on Proposition \ref{prop:BorelProperties}\emph{(\ref{it:BorelProperties_iii})}. Indeed, if $I \subset \KK[x_0,\ldots,x_n]$ is a strongly stable ideal with Hilbert polynomial $p(t)$ and $B$ is the associated Borel set in $\TT_{n,s}$, then the subset $B' = \{\xx^{\al} \in B\ \vert\ \min \xx^{\al} > 0\} \subset B$ is a Borel set in $\TT_{n-1,s}$ corresponding to the strongly stable ideal $I' = (x_0,I) \cap \KK[x_1,\ldots,x_n] \subset \KK[x_1,\ldots,x_n]$ with Hilbert polynomial $\Delta p(t)$.

\begin{example} We want to find the strongly stable ideals of the polynomial ring $\KK[x_0,x_1,x_2]$ with regularity at most 5 defining schemes with Hilbert polynomial $p(t) = t+6$. The Gotzmann number of $p(t)$ is $6$ and its growth vector in degree $5$ is $\text{gv}_5(t+6) = (5,4,1)$. In order to accomplish our task, we start considering the strongly stable ideals of $\KK[x_1,x_2]$ with Hilbert polynomial $\Delta p(t) = 1$ and regularity at most $5$ corresponding to Borel sets with growth vector $\text{gv}_{5}(\Delta p(t)) = (4,1)$. There is a unique Borel set $B' = \left\{x_2^5,x_2^4 x_1, x_2^3 x_1^2, x_2^2 x_1^3, x_2 x_1^4\right\}$. Since $x_1^5$ is not contained in $B'$, any Borel set $B \subset \TT_{2,5}$ with growth vector $(5,4,1)$ could not contain the monomials that can be obtained from $x_1^5$ by applying decreasing elementary moves, i.e.~$x_1^4 x_0$, $x_1^3 x_0^2$, $x_1^2 x_0^3$, $x_1 x_0^4$ and $x_0^5$. In Figure \ref{fig:example2}, there are the Borel sets and the corresponding strongly stable ideals with Hilbert polynomial $t+6$.
\end{example}

\begin{figure}
\begin{center}
\begin{tikzpicture}[>=latex,scale=0.6]
\node (500) at (-4,2.4) [inner sep = 1pt] {\tiny $x_2^5$};
\node (410) at (-2,2.4) [inner sep = 1pt] {\tiny $x_2^4 x_1$};
\node (320) at (0,2.4) [inner sep = 1pt] {\tiny $x_2^3 x_1^2$};
\node (230) at (2,2.4) [inner sep = 1pt] {\tiny $x_2^2 x_1^3$};
\node (140) at (4,2.4) [inner sep = 1pt] {\tiny $x_2 x_1^4$};
\node (050) at (6,2.4) [inner sep = 1pt] {\tiny $x_1^5$};

\draw [<-,very thin] (500) -- (410);
\draw [<-,very thin] (410) -- (320);
\draw [<-,very thin] (320) -- (230);
\draw [<-,very thin] (230) -- (140);
\draw [<-,very thin] (140) -- (050);

\node (400) at (-2,1.2) [inner sep = 1pt] {\tiny $x_2^4 x_0$};
\node (310) at (0,1.2) [inner sep = 1pt] {\tiny $x_2^3 x_1 x_0$};
\node (220) at (2,1.2) [inner sep = 1pt] {\tiny $x_2^2 x_1^2 x_0$};
\node (130) at (4,1.2) [inner sep = 1pt] {\tiny $x_2 x_1^3 x_0$};
\node (040) at (6,1.2) [inner sep = 1pt] {\tiny $x_1^4 x_0$};

\draw [<-,very thin] (410) -- (400);
\draw [<-,very thin] (320) -- (310);
\draw [<-,very thin] (230) -- (220);
\draw [<-,very thin] (140) -- (130);
\draw [<-,very thin] (050) -- (040);

\draw [<-,very thin] (400) -- (310);
\draw [<-,very thin] (310) -- (220);
\draw [<-,very thin] (220) -- (130);
\draw [<-,very thin] (130) -- (040);

\node (300) at (0,0) [inner sep = 1pt] {\tiny $x_2^3 x_0^2$};
\node (210) at (2,0) [inner sep = 1pt] {\tiny $x_2^2 x_1 x_0^2$};
\node (120) at (4,0) [inner sep = 1pt] {\tiny $x_2 x_1^2 x_0^2$};
\node (030) at (6,0) [inner sep = 1pt] {\tiny $x_1^3 x_0^2$};

\draw [->,very thin] (300) -- (310);
\draw [->,very thin] (210) -- (220);
\draw [->,very thin] (120) -- (130);
\draw [->,very thin] (030) -- (040);

\draw [<-,very thin] (300) -- (210);
\draw [<-,very thin] (210) -- (120);
\draw [<-,very thin] (120) -- (030);

\node (201) at (2,-1.2) [inner sep = 1pt] {\tiny $x_2^2 x_0^3$};
\node (111) at (4,-1.2) [inner sep = 1pt] {\tiny $x_2 x_1 x_0^3$};
\node (021) at (6,-1.2) [inner sep = 1pt] {\tiny $x_1^2 x_0^3$};

\draw [->,very thin] (201) -- (210);
\draw [->,very thin] (111) -- (120);
\draw [->,very thin] (021) -- (030);

\draw [<-,very thin] (201) -- (111);
\draw [<-,very thin] (111) -- (021);

\node (102) at (4,-2.4) [inner sep = 1pt] {\tiny $x_2 x_0^4$};
\node (012) at (6,-2.4) [inner sep = 1pt] {\tiny $x_1 x_0^4$};

\draw [->,very thin] (102) -- (111);
\draw [->,very thin] (012) -- (021);

\draw [->,very thin] (012) -- (102);

\node (003) at (6,-3.6) [inner sep = 1pt] {\tiny $x_0^5$};

\draw [->,very thin] (003) -- (012);

\draw [rounded corners, thick] (-5,3) -- (5,3) -- (5,1.8) -- (3,1.8) -- (3,-0.6) -- (-1,-0.6) -- (-1,0.6) -- (-3,0.6) -- (-3,1.8) -- (-5,1.8) -- cycle;

\draw [black, dashed] (-3,2.9) -- (-3,1.8);
\draw [black, dashed] (-3,1.8) -- (3,1.8);
\end{tikzpicture}
\qquad
\begin{tikzpicture}[>=latex,scale=0.6]
\node (500) at (-4,2.4) [inner sep = 1pt] {\tiny $x_2^5$};
\node (410) at (-2,2.4) [inner sep = 1pt] {\tiny $x_2^4 x_1$};
\node (320) at (0,2.4) [inner sep = 1pt] {\tiny $x_2^3 x_1^2$};
\node (230) at (2,2.4) [inner sep = 1pt] {\tiny $x_2^2 x_1^3$};
\node (140) at (4,2.4) [inner sep = 1pt] {\tiny $x_2 x_1^4$};
\node (050) at (6,2.4) [inner sep = 1pt] {\tiny $x_1^5$};

\draw [<-,very thin] (500) -- (410);
\draw [<-,very thin] (410) -- (320);
\draw [<-,very thin] (320) -- (230);
\draw [<-,very thin] (230) -- (140);
\draw [<-,very thin] (140) -- (050);

\node (400) at (-2,1.2) [inner sep = 1pt] {\tiny $x_2^4 x_0$};
\node (310) at (0,1.2) [inner sep = 1pt] {\tiny $x_2^3 x_1 x_0$};
\node (220) at (2,1.2) [inner sep = 1pt] {\tiny $x_2^2 x_1^2 x_0$};
\node (130) at (4,1.2) [inner sep = 1pt] {\tiny $x_2 x_1^3 x_0$};
\node (040) at (6,1.2) [inner sep = 1pt] {\tiny $x_1^4 x_0$};

\draw [<-,very thin] (410) -- (400);
\draw [<-,very thin] (320) -- (310);
\draw [<-,very thin] (230) -- (220);
\draw [<-,very thin] (140) -- (130);
\draw [<-,very thin] (050) -- (040);

\draw [<-,very thin] (400) -- (310);
\draw [<-,very thin] (310) -- (220);
\draw [<-,very thin] (220) -- (130);
\draw [<-,very thin] (130) -- (040);

\node (300) at (0,0) [inner sep = 1pt] {\tiny $x_2^3 x_0^2$};
\node (210) at (2,0) [inner sep = 1pt] {\tiny $x_2^2 x_1 x_0^2$};
\node (120) at (4,0) [inner sep = 1pt] {\tiny $x_2 x_1^2 x_0^2$};
\node (030) at (6,0) [inner sep = 1pt] {\tiny $x_1^3 x_0^2$};

\draw [->,very thin] (300) -- (310);
\draw [->,very thin] (210) -- (220);
\draw [->,very thin] (120) -- (130);
\draw [->,very thin] (030) -- (040);

\draw [<-,very thin] (300) -- (210);
\draw [<-,very thin] (210) -- (120);
\draw [<-,very thin] (120) -- (030);

\node (201) at (2,-1.2) [inner sep = 1pt] {\tiny $x_2^2 x_0^3$};
\node (111) at (4,-1.2) [inner sep = 1pt] {\tiny $x_2 x_1 x_0^3$};
\node (021) at (6,-1.2) [inner sep = 1pt] {\tiny $x_1^2 x_0^3$};

\draw [->,very thin] (201) -- (210);
\draw [->,very thin] (111) -- (120);
\draw [->,very thin] (021) -- (030);

\draw [<-,very thin] (201) -- (111);
\draw [<-,very thin] (111) -- (021);

\node (102) at (4,-2.4) [inner sep = 1pt] {\tiny $x_2 x_0^4$};
\node (012) at (6,-2.4) [inner sep = 1pt] {\tiny $x_1 x_0^4$};

\draw [->,very thin] (102) -- (111);
\draw [->,very thin] (012) -- (021);

\draw [->,very thin] (012) -- (102);

\node (003) at (6,-3.6) [inner sep = 1pt] {\tiny $x_0^5$};

\draw [->,very thin] (003) -- (012);

\draw [rounded corners, thick] (-5,3) -- (5,3) -- (5,0.6) -- (1,0.6) -- (1,-0.6) -- (-1,-0.6) -- (-1,0.6) -- (-3,0.6) -- (-3,1.8) -- (-5,1.8) -- cycle;

\draw [black, dashed] (-3,2.9) -- (-3,1.8);
\draw [black, dashed] (-3,1.8) -- (5,1.8);
\end{tikzpicture}
\end{center}
\caption{\label{fig:example2} On the left the Borel set corresponding to the ideal $(x_2^3,x_2^2 x_1,x_2 x_1^4)$ and on the right the one corresponding to $(x_2^3,x_2^2 x_1^2,x_2 x_1^3)$. The dashed lines separate the growth classes of the Borel sets.}
\end{figure}

This package provides the method \texttt{stronglyStableIdeals} to compute the set of strongly stable ideals of a given polynomial ring with fixed Hilbert polynomial and bounded regularity.

\medskip

\begin{Verbatim}
i11 : stronglyStableIdeals(4*t,QQ[x,y,z,w])

                  5   4 2                     2   4    5                2     2   4                2   3
o11 = {ideal (x, y , y z ), ideal (x*z, x*y, x , y z, y ), ideal (x*y, x , x*z , y ), ideal (x*y, x , y )}

o11 : List

i12 : stronglyStableIdeals(4*t,QQ[x,y,z,w],MaxRegularity=>4)

                    2     2   4                2   3
o12 = {ideal (x*y, x , x*z , y ), ideal (x*y, x , y )}

o12 : List
\end{Verbatim}

\section{Segment ideals}

The transitive closure of the order relation
\begin{equation}\label{eq:BorelOrder}
 \xx^{\al} >_B \xx^{\boldsymbol{\beta}} \quad \Longleftrightarrow\quad \xx^{\boldsymbol{\beta}} = \down{i}(\xx^{\al})
\end{equation}
induces a partial order on the set of monomials of any degree called \emph{Borel order}. Every graded term ordering turns out to be a refinement of this partial order. Since a Borel set $B$ is closed with respect to the Borel order, i.e.~$\xx^{\al} >_B \xx^{\boldsymbol{\beta}},\ \xx^{\boldsymbol{\beta}} \in B \Rightarrow \xx^{\al} \in B$, it is natural to ask whether there exists a term ordering $\prec$ preserving the same property. For instance, for the lexicographic ideal, the graded lexicographic order separates, in each degree, the monomials contained in the ideal from those not contained in it. In \cite{CLMR} several notions of segment ideals are introduced.

\begin{definition}[{\cite[Definitions 3.1, 3.7]{CLMR}}]
A Borel set $B \subset \TT_{n,s}$ is called \emph{segment} if there exists a term ordering $\prec$ such that $\xx^{\al} \succ \xx^{\boldsymbol{\beta}}$, for all $\xx^{\al} \in B$ and $\xx^{\boldsymbol{\beta}} \in \TT_{n,s}\setminus B$.

Let $I \subset \KK[\xx]$ be a saturated strongly stable ideal.
\begin{enumerate}[(i)]
\item $I$ is called \emph{hilb-segment} if the Borel set $I \cap \TT_{n,r}$ is a segment, where $r$ is the Gotzmann number of the Hilbert polynomial of $I$.
\item $I$ is called \emph{reg-segment} if the Borel set $I \cap \TT_{n,m}$ is a segment, where $m$ is the regularity of $I$.
\item $I$ is called \emph{gen-segment} if there exists a term ordering $\prec$ such that $\xx^\alpha \succ \xx^{\boldsymbol{\beta}}$ for each minimal generator $\xx^{\al}$ of degree $s$ of $I$ and for all $\xx^{\boldsymbol{\beta}} \in \TT_{n,s} \setminus I_s$.
\end{enumerate}
\end{definition}

These notions are very important in the construction of flat families based on properties of Gr\"obner bases and in general for the study of the Hilbert scheme. The package provides three methods for determining whether a strongly stable ideal may be some type of segment (and, in case, gives the term ordering). This methods use tools of the package \emph{gfanInterface} so that the term ordering is given as the corresponding weight vector of the variables.

\medskip

\begin{Verbatim}
i13 : SSI = stronglyStableIdeals(7,3,MaxRegularity=>5)

               2     2   5           2   4     3             2   2     3   4
o13 = {ideal (x , x x , x ), ideal (x , x , x x ), ideal (x x , x x , x , x )}
               2   2 1   1           2   1   2 1           2 1   2 1   2   1

o13 : List

i14 : isHilbSegment(SSI#0)

o14 = (true, {7, 3, 1})

o14 : Sequence

i15 : isRegSegment(SSI#1)

o15 = (false, )

o15 : Sequence

i16 : isGenSegment(SSI#1)

o16 = (true, {4, 3, 1})

o16 : Sequence
\end{Verbatim}

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }

\providecommand{\MRhref}[2]{
  \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
}
\providecommand{\href}[2]{#2}


TITLE: Other any other versions of identity theorem.
QUESTION [3 upvotes]: The classical identity theorem states that:

If $f(x)$ a real analytic function on a domain $D \subset \mathbb{R}$.
  Suppose $f(x)=0$ on  some $M \subset D$ such that $M$ has an
  accumulation point in $D$. Then, $f(x)=0$  on $D$.

My question:
I  am wondering whether the identity theorem can be shown to hold under different assumptions on the function $f(x)$?
Here are two more specific questions:

Can we relax the analyticity assumption to something weaker? For example, infinitely differentiable on $D$.
Can the condition of analyticity be changed to an entirely new condition that is not necessarily weaker?

Here we are interested only in real valued function.  I was thinking that for now, we can stay away from complex analysis, but of course, we don't have to.
Any reference, if there are any, would be greatly appreciated.
Thanks.

REPLY [2 votes]: No, the function $f(x)= \exp(-1/x^2)\sin (1/x), x\ne 0,$ $f(0)=0,$ is a counterexample. We have $f\in C^\infty(\mathbb R),$ and $f$ is real analytic on $\mathbb R\setminus \{0\}.$ Because $f(1/(n\pi)) = 0$ for $n=1,2,\dots,$ the zero set of $f$ has an acumulation point at $0.$
Quasi-analytic functions are a class of functions that will do what you want here. See https://en.wikipedia.org/wiki/Quasi-analytic_function


TITLE: Can a voltmeter itself disturb the electrochemical potential of battery?
QUESTION [0 upvotes]: I'm trying to understand the effect on a system when a voltage of a system is measured by connecting a voltmeter.
For example let's say we are trying to measure the voltage of a Li-ion battery. The junction is as follows:
Lead(Nickel alloy) | Current collector(Copper) | Anode active material(Lithium-intercalated graphite) | Electrolyte | Cathode active material(Lithium transition metal oxide) | Current collector(Aluminium) | Lead(Nickel alloy)
Lead, current collector and active material would be in thermodynamic equilibrium(Fermi level(a.k.a electrochemical potential) alignment) at each side and the electrolyte works as a border(electrically insulated).
To measure the voltage of the battery, we must connect the voltmeter to both leads of the battery. Each probe will be connected to each side of the battery (anode and cathode lead.)
Assume this voltmeter is a high-end one, there is almost zero current flowing through the voltmeter (infinite input impedance.)
Although it is, I'm concerned that this behavior itself could disturb the voltage of the battery because the Fermi level of each side of the battery will change.
In the situation of anode side when the probe of voltmeter is connected, this makes another metal-metal junction:
Voltmeter | Probe(Stainless steel(?)) | Lead(Nickel alloy) | Current collector(Copper) | Anode active material(Lithium-intercalated graphite) |...
And this would also happen in cathode side when the other probe of the voltmeter is connected to the cathode lead.
So the voltage displayed in the voltmeter would be somewhat distorted because this is the result after some electrons from the voltmeter flow into or out of the lead of the battery since voltmeter measures Fermi level difference between two probes.
For summary, my questions are:

If composition of probes are different from that of a system we want to measure voltage, would the result be somewhat distorted because of Fermi level alignment? If it is, can we calculate how much the Fermi level changes?

Along with question 1, if the two probes of voltmeter are the same material and have dimension, does this effect cancel out so we can measure exact voltage of the system?

Along with question 2, if we have to use additional wire material of which is different from the original one to measure voltage, can we calculate the actual voltage of the system by compensating some constant value?

In the same spirit, does the material of current collector also affect the Fermi level of active material which we actually want to measure?

I've never seen or heard this kind of effect discussed. Is it because the contribution of the effect mentioned above is negligible or am I just wrong?

REPLY [0 votes]: although you broke the rules by putting four questions into your post, I think all of them can be answered in one response, as follows:
Any dissimilar metal junction is going to develop a very small voltage (of order ~ millivolts) across it. In a context of measuring voltages of order ~ ten volts across a stack of cells, those millivolt signals will be utterly swamped out.
Furthermore, a sensitive voltmeter is always zeroed and calibrated before use against a precision voltage source, so as to null out contact voltages before taking measurements.


TITLE: Is angular momentum conserved only for the center point of the centripetal force?
QUESTION [1 upvotes]: From my book,
Consider a centripetal force is applying on a object:
$$\vec{F}=f(r)\hat{r}$$
The torque for a particular point is,
$$\vec{N}=\vec{r}\times\vec{F}=\vec{r}\times{f(r)\hat{r}}$$
$$\vec{N}=r\hat{r}\times{f(r)\hat{r}}$$
$$\vec{N}=rf(r)(\hat{r}\times\hat{r})=\vec{0}$$
So the angular momentum is conserved for any centripetal force.
I tried to figure out this using gravitational force that is applied to a satellite in orbit from earth.
I firstly selected the origin as the center of the earth, in this view point force is constant and it is anti-parallel to the position vector of the satellite. So the torque of the satellite always $\vec{N}=\vec{0}$, the angular momentum of the satellite is conserved.
Then I selected another origin that is random. For this origin, the position vector is anti-parallel to the gravitational force vector on the satellite for one situation. For all other situations the position vector and the force vector are not anti-parallel so the torque of the satellite with respect to the non-center of centripetal force origin is $\vec{N}\ne\vec{0}$.
So, Is the angular momentum conserved only when we choose the origin as the origin of the centripetal force?

REPLY [2 votes]: So, Is the angular momentum conserved only when we choose the origin as the origin of the centripetal force?

Linear momentum is conserved when there are no external forces.
Angular momentum is conserved when there are no external torques.
By selecting an axis that that is not coincident with the earth, you've created a situation where the earth's force creates a torque on the satellite (when considered from this new axis).  So yes, in this situation we would not expect the angular momentum of the satellite to be conserved.
When we set up problems, we will often consider angular momentum about a particular axis because it simplifies the problem.  Both calculations are valid, but we have to know more information and do more work when we pick an axis that is not coincident with the earth's center of mass in this case.

REPLY [1 votes]: In the way you've framed the problem, yes.
However, if you had two bodies, of mass $m_{1,2}$ and positions $\mathbf r_{1,2}$, defining the relative position $\mathbf r=\mathbf r_2-\mathbf r_1$ with a force
$$\mathbf F=f(r)\hat r$$
acting between them, angular momentum would be conserved w.r.t. any point.


TITLE: path connected subspace $A$ of $X$, $ i:A\to X$ inclusion. Why is the induced map of $i$ on homology injective?
QUESTION [3 upvotes]: X is a topological space and $A\subset X$ is a path-connected subspace of X and $i:A\to X$ is the inclusion. I want to know, why the induced map of i on singular homology of dimension zero over $\mathbb{Z}$, namely $H_0(i)=(i)_0:H_0(A)\to H_0(X)\; [v]\mapsto [i(v)]$, is injective.
My first thoughts are: if i consider the pair $(X,A)$, i have a long exact sequence
$$\ldots\to H_{n+1}(X,A)\overset{\delta_n+1}{\to}H_n(A)\overset{(i)_n}{\to}H_n(X)\overset{(j)_n}{\to}H_n(X,A)\overset{\delta_n}{\to}H_{n-1}(A)\to\ldots$$ with $\delta_n$is the boundary map and $j:(X,\emptyset)\to (X,A)$ is the inclusion map on pairs and $(j)_*$ is the induced map of j. I consider the section of the sequence:
$$\ldots H_1(X)\overset{(j)_1}{\to} H_{1}(X,A)\overset{\delta_1}{\to}H_0(A)\overset{(i)_0}{\to}H_0(X)\overset{(j)_0}{\to}H_0(X,A)\overset{\delta_0}{\to}0$$.
Why $(i)_0$ is injective? I only know: $(j)_0$ is surjective, because $Im(j)_0=H_0(X,A)=ker(\delta)_0$ But i need to know, why $ Im(\delta_1)=ker(i)_0$ has to be zero, but i don't know why. Could you help me?  Furthermore i know: $H_0(A)\cong\mathbb{Z}$ and $H_0(X)\cong \mathbb{Z}^{\text{|path-connected-components of X|}}$. But i don't know how to continue...Regards

REPLY [0 votes]: The long exact sequence is not really helpful here. What we need to do is look at chain groups. Any two 0-chains in $A$ of the form $1f$ where $f$ is a constant map are homologous and map to a generator when we pass to homology. 0 chains of the form $1(i\cdot f)$ generate the direct summand in $H_0(X)$ corresponding to the path component containing $A$. Therefore a generator of $H_0(A)$ is mapped to a nonzero element in $H_0(X)$, so since the groups are free abelian the map is injective.


TITLE: Borel measurability is a local property
QUESTION [12 upvotes]: I am looking at Exercise 5.2 (page 44) in "Real Analysis for Graduate Students" by Richard Bass.

Let $f:(0, 1)\to \mathbb{R}$ be a function such that for every $x\in (0, 1)$,
  there exist $r>0$ and a Borel measurable function $g$, both depending
  on $x$, such that $f$ and $g$ agree on $(x-r, x+r)\cap (0, 1)$. Prove
  that $f$ is Borel measurable.

Attempt. For each $x\in (0, 1)$, let's denote $r_{x}$ and $g_{x}$ to be the quantities given in the problem statement. By choosing $r_x$ sufficiently small, we may assume that $(x-r_{x}, x+r_{x})\subseteq (0, 1)$. Then for $a\in\mathbb{R}$, we have
$$
f^{-1}((a, \infty))=\{y: f(y)>a\}=\bigcup_{x\in (0, 1)} \{y: g_{x}(y)>a\}\cap (x-r_{x}, x+r_{x})
$$
Since $g_{x}$ is Borel measurable, we know that $\{y: g_{x}(y)>a\}\cap (x-r_{x}, x+r_{x})$ is a Borel set. But the union above is uncountable; so how do we show that $f^{-1}((a, \infty))$ is a Borel set?
Attempt 2. Since $\mathbb{Q}$ is countable and dense in $(0, 1)$, we can try looking at the points $x\in\mathbb{Q}\cap (0, 1)$. But then the intervals $(x-r_{x}, x+r_{x})$ need not cover $(0, 1)$. The standard example is as follows. Suppose $\{q_1, q_2, …\}$ is enumeration of rationals in $(0, 1)$, and let $\varepsilon>0$. If we let $r_{q_{j}}$ to be $\varepsilon 2^{-j}$, then the measure of $\bigcup_{j=1}^{\infty} (q_j-r_{q_{j}}, q_{j}+r_{q_{j}})$ is at most $2\varepsilon$.
Attempt 3. If each $g_{x}$ is continuous, we would be done. In that case, $$\{y: g_{x}(y)>a\}\cap (x-r_{x}, x+r_{x})$$ would be an open set; so as a union of open sets, $f^{-1}((a, \infty))$ would be an open set, and hence Borel.

REPLY [3 votes]: By hypothesis, for every $x\in (0, 1)$ we find Borel measurable function $g_{x}$ and $r_x>0$ such that $f=g_{x}$ on the open interval $(x-r_x, x+r_x)\subset (0, 1)$. This gives an open cover of $(0, 1)$:
$$
(0, 1) = \bigcup_{x\in (0, 1)} (x-r_{x}, x+r_x)
$$
Since $(0, 1)$ is a Lindelöf space, this open cover has a countable subcover, i.e. we can find $x_1, x_2, x_3, …$ in $(0, 1)$ such that
$$
(0, 1)=\bigcup_{n=1}^{\infty} (x_{n}-r_{x_{n}}, x+r_{x_{n}})
$$
Now for any $a>0$, we have that
$$
f^{-1}((a, \infty))=\{y: f(y)>a\}=\bigcup_{n=1}^{\infty} \{y: g_{x_n}(y)>a\}\cap (x_n-r_{x_{n}}, x_n+r_{x_{n}})
$$
Each $\{y: g_{x_n}(y)>a\}\cap (x_n-r_{x_{n}}, x_n+r_{x_{n}})$ is a Borel measurable set. As a countable union of measurable sets, $f^{-1}((a, \infty))$ is measurable. Thus, $f$ is Borel measurable.


TITLE: Combinatorics problem involving a regular expression
QUESTION [4 upvotes]: If formal definition of a regular expression is required, please comment and I'll add it.
I was given the assignment of determining how many words of length $n$ can be produced from a regular expression (or equivalently - how many words of length $n$ are there in the regular language that corresponds to the regular expression).
So let $\sum=\{0,1,2\}$ be our alphabet.
The regular expression is: $(01^*|10^*)(0^+|1^+)(0|1|2)^*$, and I need to determine how many words of length $n$ can be produced from it.
I wrote my answer but later found out it was wrong.
What really bugs me is that I can't seem to find what's wrong with it, and I would really appreciate some help and guidance.
So I first had to use some identities, because I had no idea what to make of it...
$(01^*|10^*)(0^+|1^+)(0|1|2)^*=(01^*|10^*)(00^*|11^*)(0|1|2)^*=(01^*|10^*)(0|1)(0|1|2)^*=(01^*|10^*)(0|1)\sum^*=(01^*|10^*)(0\sum^*|1\sum^*)=(01^*0\sum^*|01^*1\sum^*|10^*0\sum^*|10^*1\sum^*)$
And now it is easier to see that every word of length $n$ must have one of the following patterns:
$0\hspace{1 mm}1^k\hspace{1 mm}0\hspace{1 mm}\sum^m$
$0\hspace{1 mm}1^k\hspace{1 mm}1\hspace{1 mm}\sum^m$
$1\hspace{1 mm}0^k\hspace{1 mm}1\hspace{1 mm}\sum^m$
$1\hspace{1 mm}0^k\hspace{1 mm}0\hspace{1 mm}\sum^m$
Where $0\leq k,m\leq n-2$ and $k+m=n-2$.
So the total number of words of length $n$ is: $4\cdot(\sum_{m=0}^{n-2}3^m)$
Where's the flaw in this answer, and how else should I have approached this?

REPLY [2 votes]: Thanks to @Callus for helping me figure out the flaw in my solution.
After some adjustments, I figured out how to solve it.
My revised answer does not involve inclusion-exclusion (but if anyone is kind enough to write his or her solution using inclusion-exclusion I'll much appreciate it [I will also tag his or her answer as 'answer', of course]).
Here it is:
$(01^*|10^*)(0^+|1^+)(0|1|2)^*=(01^*|10^*)(00^*|11^*)\sum^*=$
$(01^*|10^*)(0|1)\sum^*=(01^*0|01^*1|10^*0|10^*1)\sum^*=$
$(01^*0|011^*|100^*|10^*1)\sum^*=(01^*0|01|10|10^*1)\sum^*=(00|01|10|11)\sum^*$
The last transition is justified by the fact that the case where $01^k0\sum^*$ for $k>0$ is already included in the pattern $01|\sum^*$, the same goes for the case where $10^k1\sum^*$ for $k>0$, which already included in the pattern $10|\sum^*$.
It is now easy to see that there are $4\cdot 3^{n-2}$ words of length $n$.


TITLE: "inverse" of an integral
QUESTION [1 upvotes]: I am reading an economics paper Melitz and Ottaviano (2008), but I find it hard to understand some math in the model.
The preference is given by :

, which implies a demand function:

Then the paper says the "inverse demand function can then be inverted to yield the linear market demand system for these varieties":

But what operations do they take in getting the function $q_i$?
I am very confused on dealing with the $Q_c$ variable, which is defined to be :

Thanks.

REPLY [1 votes]: Actually, just take an integral on both sides of equation 2 w.r.t. $i$, and we get an expression of $Q_c$. Then plug the $Q_c$ back into equation 2.


TITLE: Understanding $\lambda$ in the definition of Poisson distributions
QUESTION [3 upvotes]: I am trying to understand the meaning of $\lambda$ in Poisson distributions. I know that it is the average rate of occurrences of the event, but I have not been able to fully understand what that means.
In "A First Course in Probability" by Sheldon Ross, the author says that a Poisson distribution may be used as an approximation for a binomial distribution with parameters $(n,p)$ when $n$ is large and $p$ is small enough so that $np$ is of moderate size.

What does $np$ being of moderate size mean? What is considered as of moderate size?
Why does $np$ have to be of moderate size? Why can a Poisson distribution not be used as an approximation for a binomial distribution if $np$ is too big or too small?

Also, in other books, I read that a Poisson distribution is the limiting case of a binomial distribution when $\lambda=np$ is constant as $n\to\infty$.
Under what conditions, is $\lambda$ constant as $n\to\infty$?
I am new to probability. If someone can provide the intuition behind Poisson distributions (specifically about $\lambda$), I would greatly appreciate it.

REPLY [3 votes]: This is a fantastic question, as it gets at the heart of a common misconception. The answer to the question "Why does $np$ have to be of moderate size?" is that it does not! The only thing that is needed for Poisson approximation is that $p$ is small.
To make this precise, we need a way to measure the quality of a probabilistic approximation. If $X$ and $Y$ are random variables, we define the total variation distance between them to be
$$
d_{\text{TV}}(X, Y)=\sup_{A} |P(X\in A)-P(Y\in A)|
$$
If this distance is small, say it is at most $\epsilon$, then you can use the distribution of $Y$ to calculation the probability of an event involving $X$, and have an absolute error of at most $\epsilon$.

Theorem Let $X\sim \text{Binomial}(n,p)$, let $\lambda=np$, and let $Y\sim \text{Poisson}(\lambda)$. Then
$$
d_\text{TV}(X,Y)\le \min\{p,np^2\}\tag1
$$

Therefore, as long as $p$ is small, the quality of the Poisson approximation is good.
This same idea is true in greater generality. For a Binomial distribution, there are $n$ independent and equiprobable events; we can drop the requirement of equiprobability. Let $E_1,\dots,E_n$ be independent events, with $P(E_i)=p_i$ for $i\in \{1,\dots,n\}$. If we let $X$ be the number of events which occur, and let $Z$ be a Poisson random variable with the same mean as $X$, meaning that $Z\sim \text{Poisson}(\lambda)$ with $\lambda=\sum_{i=1}^n p_i$, then
$$
d_\text{TV}(X, Z)\le \min\left\{\frac1{\lambda}\sum_{i=1}^n p_i^2, \,\,\sum_{i=1}^n p_i^2\right\}\tag2
$$
In the case where $p_1=\dots=p_n=p$, we recover $(1)$. For a proof of $(2)$, see Approximate Computations of Expectations, by Charles Stein. Equation $(2)$ appears on page 89.

Some comments on the semantic parts of your questions.

"Moderate" is not a precise term; it just means "neither larger nor small," but large and small are subjective.

When people say that $\lambda=np$ is constant as $n\to\infty$, it just means that $n$ is getting larger, and for each particular instance of $n$, the corresponding value of $p$ is equal to $\lambda/n$. Saying

The Binomial$(n,p)$ distribution as $n\to\infty$ with $\lambda=np$ constant.

means the exact same thing as

The Binomial$(n,\tfrac{\lambda}n)$ distribution as $n\to\infty$.

So, Binomial$(10,0.5)$ is pretty close to a Poisson$(5)$ distribution, Binomial$(50,0.1)$ is even closer, Binomial$(500,0.01)$ is extremely close, etc.