TITLE: Correlation inequality
QUESTION [1 upvotes]: This question is a modified version of my previous post: Maximum correlation inequality.
Let $ X $ be a random variable and suppose $ A $ is a subset of $ \mathbb{R} $. For two functions $ f:\mathbb{R}\mapsto\mathbb{R} $ and $ g:\mathbb{R}\mapsto\mathbb{R} $, let $ \rho(f(X), g(X) | X \in A) $ denote the conditional correlation between $ f(X) $ and $ g(X) $ given $ X \in A $. I am wondering whether
$$
\sup_{f \; \text{monotone}}|\rho(f(X),g(X) | X \in A)| \geq \sup_{f \; \text{monotone}}|\rho(f(X),g(X))|.
$$
This seems intuitively true, since it becomes more difficult to find a good monotone approximation of $ g(x) $ if $ x $ belongs to a larger subset, but I'm not sure how to show it. Any thoughts?

REPLY [1 votes]: First note that if $g$ is monotone, the conjecture is trivially true by picking $f = g$ so that both sides $=1$.
If $g$ is non-monotone though, what happens outside $A$ can easily overwhelm what happens inside $A$.  Here is a non-monotone $g$ that disproves the conjecture:

$X$ is continuous uniform within $S= [1, 100]$
$A = \{1, 2, 3, \dots, 100\}$ (i.e. just the integers of $S$)
$\forall x \in A: g(x) = (x \pmod 2)$
$\forall x \in \mathbb{R} - A: g(x) = x$.

The RHS $=1$ by picking $f(x) = x$.  This $f$ is monotone and achieves $\rho = 1$ because $f=g$ except for set $A$ of measure $0$.
The LHS...  I'm sorry I lack the math skills to formally prove that $\rho = 1$ is unachieveable, but it seems pretty obvious.  In fact, the best "line fit" is roughly horizontal, making the best $f$ and $g$ almost independent.  Thus LHS $\rho \approx 0$.
In both this question and your previous question, the key insight is that what happens in a subset of the support can be overwhelmed by what happens in a larger subset of (or the whole) support.  Without further restrictions in the problem statement, a conditional distribution is its own entity and can be very different from the unconditioned distribution.  So if you're talking about a "global" or "aggregate" characteristic like $\rho$, there is very little you can conclude.
BTW it is easy to modify the above example to use a different continuous $g$.  The proof will be even more vague because it's no longer obvious RHS $=1$, but the intuition is just the same.


TITLE: Norm of $(I+A)^{-1}A$ for self-adjoint semi-positive operator $A$.
QUESTION [2 upvotes]: For a given bounded self-adjoint semi-positive linear operator $A$ on a Hilbert space $X$, i.e. $(Ax,x)\ge 0, \forall x\in X$, I know from here that $(I+A)^{-1}$ is well-defined with $\|(I+A)^{-1}\|\le 1$. Consequently $(I+A)^{-1}A$ is a bounded operator with $\|(I+A)^{-1}A\|\le \|A\|$.
I was wondering whether we can show $\|(I+A)^{-1}A\|\le 1$. I think the statement holds for finite dimensional cases, and I would like to know whether it holds in general.

For finite dimensional case, I think it is OK. By using the eigen-decomposition, I can write $A=UDU^{-1}$ where $U$ is the normalized eigenvectors and $D$ is corresponding (positive or null ) eigenvalues. Then
$(I+A)^{-1}A=U[(I+D)^{-1}D]U^{-1}$, and each entry of $[(I+D)^{-1}D]$ is non-negative and less that $1$, which completes the proof.

REPLY [3 votes]: 1. This easily follows from the continuous functional calculus. Indeed, since $\sigma(A) \subseteq [0, r]$ for some $r \geq 0$ and $f(x) = \frac{x}{1+x} \in C(\sigma(A))$, we have $(I+A)^{-1}A = f(A)$ and
$$ \|f(A)\| = \|f\|_{C(\sigma(A))} \leq f(r) \leq 1. $$
2. A direct approach is also available. We begin by noting that both $(I+A)^{-1}$ and $(I+A)^{-1}A$ are positive semi-definite. These, together with a simple identity $(I+A)^{-1}A = I - (I+A)^{-1}$, yield
$$ 0 \leq \langle x, (I+A)^{-1}A x \rangle = \langle x, x \rangle - \langle x, (I + A)^{-1}x \rangle \leq \|x\|^2. $$
This is enough to conclude that $\|(I+A)^{-1}A\| \leq 1$.


TITLE: Make this system of 3 equations solvable for $x$, $y$ and $z$
QUESTION [0 upvotes]: I have the system of these three equations:
$$ax = y+z$$
$$by = x+z$$
$$cz = x+y$$
How do I find all $a$, $b$ and $c$ for which the system has real, positive solutions for $x$, $y$ and $z$?
As a comparison, I have a simpler system:
$$ax = y$$
$$by = x$$
For this simpler system, with a substitution I get that $ab = 1$ yields a system that has real positive solutions. I'd like to find something similar for the above system as well. Basically, I'm looking for the relation between $a$, $b$ and $c$, independent from $x$, $y$ and $z$.
This is somewhat similar to this other question, only my equations are different.

REPLY [2 votes]: Multiply the second equation by $c$ to obtain $$bcy=cx+cz=cx+x+y$$ using the third equation, so that $$(bc-1)y=(c+1)x$$
Likewise multiply the first equation by $c$ to obtain equivalently $$(ac-1)x=(c+1)y$$Now multiply this second equation by $c+1$ so that $$(c+1)^2y=(ac-1)(c+1)x=(ac-1)(bc-1)y$$whence $y=0$ or $$c^2+2c+1=abc^2-ac-bc+1$$ so $c=0$ or $$abc-a-b-c=2$$

Note that the condition $c=0$ can be avoided (by symmetric argument) unless $a=b=c=0$

Further note that positive solutions imply $a,b,c\gt 0$ - because e.g. $a=\frac {y+z}x\gt0$.
Also the solution $a=b=c=-1$ has us multiplying by $c+1=0$ which is inadmissible if we want to remain strictly positive.
Suppose we have positive values $a,b,c,x$ - we then find $$y=\frac {ac-1}{c+1}x; z=\frac{ab-1}{b+1}x$$which are positive provided $ac,ab\gt 1$
Writing the condition $abc-a-b-c=2$ in the form $$b(ac-1)=2+a+c$$ the right-hand side is positive, $b$ is positive, so $ac\gt 1$ is enforced by the condition.
So if $a,,b,c\gt 0$ satisfy the condition, there is a unique positive solution for any chosen positive value of $x$.

REPLY [0 votes]: This is an homogeneous system of the type $\pmb{A}\pmb{x}=\pmb{0}$.
Every homogeneous system has at least one solution, known as the zero solution (or trivial solution), which is obtained by assigning the value of zero to each of the variables. If the system has a non-singular matrix ($\det(\pmb{A}) ≠ 0$) then it is also the only solution. If the system has a singular matrix then there is a solution set with an infinite number of solutions.


TITLE: Allowed system energies from quantized Hamiltonian spectra
QUESTION [0 upvotes]: To find the allowed energies for a system, I can find the spectrum of the Hamiltonian $\hat{H}_{\psi}$ given a wavefunction $\psi$ representing the state of the system. 3 cases might happen: either there are boundary conditions that will lead to full quantization of the spectrum, or the spectrum can be partially quantized, partially continuous, or last case, it can be fully continuous, if there are no boundary conditions (the system is free). However, the eigenvalues I find are not necessarily "physical": in the case of the free system in particular, the eigenstates $\psi_k$ might not be normalizable, therefore you can't compute $E_k=\langle\psi_k|\hat{H}|\psi_k\rangle$ (the denominator blows up). So, do I have to insist on the corresponding eigenstate being normalizable to declare that I have found a possible energy for the system? (i.e. it's not enough to find the spectrum of $\hat{H}$)

REPLY [1 votes]: If $\lambda$ is a value in the spectrum of $H$ there always exists a state $\omega$ such that $\lambda =\omega(H)$ and $\Delta_\omega(H)=0$, so every point in the spectrum of $H$ is a possible outcome of an experiment. What can fail to happen is that, in a particular representation, like the Schrödinger representation, the state $\omega$ is a vector state. So either you allow distribution to come into play through the rigged Hilbert space, where Dirac's bras and kets actually live, or you content yourself with approximate eigenvectors.


TITLE: Ramification of the field extension $\mathbb{Q}_p(\zeta_{p^2},p^{1/p})$
QUESTION [0 upvotes]: I am interested in the extension of local fields $\mathbb{Q}_p(\zeta_{p^2},p^{1/p})/\mathbb{Q}_p$. Is it totally ramified?
Here are the partial results that I have:

It is a field extension of degree $p^2(p-1)$. To see that, note that it suffices to show $p^{1/p} \not \in \mathbb{Q}_p(\zeta_{p^2})$. Indeed, otherwise $\mathbb{Q}_p(p^{1/p}) \subset \mathbb{Q}_p(\zeta_{p^2})$, thus $\mathbb{Q}_p(p^{1/p})/\mathbb{Q}_p$ is an abelian extension, hence contains its conjugate $\zeta_p \cdot p^{1/p}$, hence also contains $\zeta_p$, which is of degree $p-1$ a contradiction.

It's Galois group is an explicite group of order $p^{2}(p-1)$, its element its uniquely determined by its action on $\zeta_{p^2}$ and $p^{1/p}$.

Its ramification index is at least $p(p-1)$, indeed, $\mathbb{Q}_p(\zeta_{p^2})/\mathbb{Q}_p)$ is a totally ramified extension of degree $p(p-1)$

But I can not figure out whether it has chance to have an unramified subextension.

REPLY [4 votes]: So you’re taking a $p$-th root of $p$ in $K=\mathbb{Q}_p(\zeta_{p^2})$, a local field that contains all the $p$-th roots of unity. Let $L=K(p^{1/p})$, it has degree $p$ over $K$.
As $K$ is totally ramified over $\mathbb{Q}_p$, $L$ is totally ramified over $\mathbb{Q}_p$ iff $L/K$ is totally ramified. Suppose (for the sake of contradiction) that it is false.
Then $L/K$ is unramified of degree $p$, and thus contained in a cyclotomic extension. Therefore $L/\mathbb{Q}_p$ is contained in a cyclotomic extension, and again we find a contradiction as $\mathbb{Q}_p(p^{1/p})$ is not Galois.


TITLE: How to solve $\int_{S_3^+(0)} \frac{e^w+z}{z+2} dw$
QUESTION [0 upvotes]: In my lecture notes the following integral was computed:
\begin{align*}
\int_{S_3^+(0)} \frac{e^w+z}{z+2} dw.
\end{align*}
There is written: In order to use the Cauchy Integral formula, which is
\begin{align*}
f(z) = \frac{1}{2\pi i} \int_{S_R^+(a)} \frac{f(w)}{w-z} dw,
\end{align*}
we define $f(w) := e^w+w$ and $z := -2$. Clearly $f$ is differentiable in $\mathbb C$ and we obtain
\begin{align*}
\int_{S_3^+(0)} \frac{e^w+z}{z+2} dw = 2\pi i (e^{-2}-2).
\end{align*}
Could please someone explain to me what happened here? I don't understand, why one can set $z = -2$ because then the integrand would not be defined any more? Furthermore I don't understand why the definition of $f$ should fit here.
Thanks in advance.
Edit: If I would change the path $S_3^+(0)$ to $S_1^+(0) = e^{it}$, $t \in [0,2\pi]$, then it turned out by the Deformation Theorem, that the integral is zero. And that is because $S_1^+(0)$ is homotopic to a point in $\mathbb C \setminus \{-2\}$. Could someone explain to me please, why this is the case?

REPLY [1 votes]: An integral over a closed path that does not contain any poles of an entire function (such as $e^w+w$) is zero.  This is because the path is closed, so the starting and ending points where the integral is evaluated are equivalent, and thus we have
$$\int_a^a f(z)dz=F(a)-F(a)=0$$
When the path contains a pole, as with a function like $e^w+w\over w+2$ with a path like $S_3^+(0)\to|w|=3$, then it is possible to supply an infinitesimal cut along the path and provide a different (non-circular) path which cuts out the discontinuity and results in an integral that evaluates to zero.  But this different integral is a transformation of the original, and the residue from the transformation results in the original taking on a non-zero value.
The function $e^w+w\over w+2$ over the path $S_1^+(0)\to |w|=1$ contains no pole, and therefore no transformation is necessary to obtain a path that cuts out the pole, and therefore it evaluates to zero.


TITLE: finding the line segment?
QUESTION [0 upvotes]: let  $a$ and $b$ belong $\mathbb R$. If $P(x,y)$ is the point  in the plane, define $f(P)= ax + by$. let the  line segment $AB$ bisect the line segment $CD$.If $f(A)=5$ , $f(B)=5$ and $f(C)= 10$ , find $f(D).$

I was trying this   question that  AB bisect the  CD that mean  it  mean  they have equal sides and  their  mid point  have  same  coordinates. But my  answer is  f(A)= f(B)=5 and  f(C)= f(D) = 10 and my answer  is  f(D)=10
Is my answer is correct or not, im not sure  i have only used my own logic
if anbody help me i  would be very thankful to him

REPLY [0 votes]: An equation of $AB$ it's $ax+by=5$.
Thus, $$a\cdot\frac{x_C+x_D}{2}+b\cdot\frac{y_C+y_D}{2}=5$$ or
$$10+f(D)=10$$ or
$$f(D)=0$$.


TITLE: Find the degree $[E:\mathbb{Q}]$
QUESTION [5 upvotes]: Let $p$ a prime number. Find a splitting field $E$ of the polynomial $x^p-2 \in \mathbb{Q}[x]$.
I have done the following:
The solutions of $x^p-2=0$ are :
$$\sqrt[p]{2}, \sqrt[p]{2}\omega, \dots, \sqrt[p]{2}\omega^{p-1}, \text{ where } \omega=e^{\frac{2 \pi i}{p}}$$
Therefore, the splitting field is $E=\mathbb{Q(\sqrt[p]{2}, \omega)}$
Is this correct so far??
How can I find the degree $[E:\mathbb{Q}]$ ??
EDIT:
$[E:\mathbb{Q}]=[\mathbb{Q(\sqrt[p]{2}, \omega)} : \mathbb{Q}]=[\mathbb{Q(\sqrt[p]{2}, \omega)} : \mathbb{Q}(\sqrt[p]{2})][\mathbb{Q(\sqrt[p]{2})} : \mathbb{Q}]=[\mathbb{Q(\sqrt[p]{2}, \omega)} : \mathbb{Q}(\sqrt[p]{2})]p$
Since $Irr(\sqrt[p]{2}, \mathbb{Q})=x^p-2 \Rightarrow [\mathbb{Q(\sqrt[p]{2})} : \mathbb{Q}]=p$.
But how can we find $[\mathbb{Q(\sqrt[p]{2}, \omega)} : \mathbb{Q}(\sqrt[p]{2})]$ ??
EDIT: Proof that the degree is the product of p and p-1:
$$[\mathbb{Q}(\omega, \sqrt[p]{2}) :\mathbb{Q}]=[\mathbb{Q}(\omega, \sqrt[p]{2}):\mathbb{Q}(\sqrt[p]{2})][\mathbb{Q}(\sqrt[p]{2}):\mathbb{Q}]=a \cdot p$$
$$[\mathbb{Q}(\omega, \sqrt[p]{2}):\mathbb{Q}]=[\mathbb{Q}(\omega, \sqrt[p]{2}):\mathbb{Q}(\omega)][\mathbb{Q}(\omega):\mathbb{Q}]=b \cdot (p-1)$$
$$a \cdot p = b \cdot (p-1) \Rightarrow p \mid a \cdot p \overset{(p, p-1)=1}{ \Longrightarrow } p \mid b \Rightarrow p \leq b \tag 1 $$ $$\Rightarrow a \cdot p \leq a \cdot b \Rightarrow b \cdot (p-1) \leq a \cdot b \Rightarrow p-1 \leq a \tag 2 $$
$$\mathbb{Q} \leq \mathbb{Q}(\omega) \leq \mathbb{Q}(\omega, \sqrt[p]{2}) \Rightarrow Irr(\sqrt[p]{2}, \mathbb{Q}(\omega)) \mid Irr(\sqrt[p]{2}, \mathbb{Q}) $$ $$\Rightarrow \deg Irr(\sqrt[p]{2}, \mathbb{Q}(\omega)) \leq \deg Irr(\sqrt[p]{2}, \mathbb{Q}) \Rightarrow b \leq p \tag 3$$
$$\mathbb{Q} \leq \mathbb{Q}(\sqrt[p]{2}) \leq \mathbb{Q}(\omega, \sqrt[p]{2}) \Rightarrow Irr(\omega, \mathbb{Q}(\sqrt[p]{2})) \mid Irr(\omega, \mathbb{Q}) $$ $$ \Rightarrow \deg Irr(\omega, \mathbb{Q}(\sqrt[p]{2})) \leq \deg Irr(\omega, \mathbb{Q}) \Rightarrow a \leq p-1 \tag 4$$
From $(1)$ and $(3)$ we have that $b=p$ and from $(2)$ and $(4)$ we have that $a=p-1$
Therefore, $[\mathbb{Q}(\omega, \sqrt[p]{2}) :\mathbb{Q}]=p(p-1)$.

REPLY [2 votes]: You've done most of the work for yourself, and your proposed splitting field is correct.
We can consider two towers of fields:
$$\mathbb{Q} \subset \mathbb{Q}[\sqrt[p]{2}] \subset \mathbb{Q}[\sqrt[p]{2}, \omega]$$
$$\mathbb{Q} \subset \mathbb{Q}[\omega] \subset \mathbb{Q}[\sqrt[p]{2}, \omega]$$
 where $\omega$ is a $p^{\text{th}}$ root of unity.
In the first tower, the intermediate field is of degree $p$ over $\mathbb{Q}$ (Why?).  In the second tower, the intermediate field is of degree $p-1$ over $\mathbb{Q}$. (again, why?)
Now apply the fact that $[K:F] = [K:E] \cdot [E:F]$ for a tower of fields $F \subset E \subset K$.  Given this, both $p$ and $p-1$ divide the degree of the splitting field.  Further, the degree of the splitting field is at most $p(p-1)$ since, worst case scenario, the minimal polynomial for $\sqrt[p]{2}$ is irreducible over $\mathbb{Q}[\omega]$.
Therefore...


TITLE: Infinite GP in an Infinite AP
QUESTION [0 upvotes]: Let a, b, c, d,....... be a strictly increasing sequence of positive integers in an arithmetic progression. Prove that there is an infinite subsequence of the given sequence whose terms are in a geometric progression.

REPLY [4 votes]: If the arithmetic progression is $\{a,a+m,a+2m,\cdots\}$  then $$\{a, a(1+m),a(1+m)^2,\cdots\}$$ is a subsequence in geometric progression.


text:
\begin{document}
\begin{frontmatter}

\title{Model Predictive Control of a Food Production Unit: A Case Study for Lettuce Production

}

\thanks[footnoteinfo]{This project has received funding from the European Social Fund (ESF)}

\author[First]{Murali Padmanabha}
\author[First]{Lukas Beckenbach}
\author[First]{Stefan Streif}

\address{Technische Universtit\"{a}t Chemnitz,
	Automatic Control and Systems Dynamic Lab, 09107 Chemnitz, Germany}
\address[First]{(e-mail:\{murali.padmanabha, lukas.beckenbach, stefan.streif\}@etit.tu-chemnitz.de)}
\begin{abstract}

Plant factories with artificial light are widely researched for food production in a controlled environment.
For such control tasks, models of the energy and resource exchange in the production unit as well as those of the plant's growth process may be used.
To achieve minimal operation cost, optimal control strategies can be applied to the system, taking into account the availability of resources by control reference specification.
A particular advantage of model predictive control (MPC) is the incorporation of constraints that comply with actuator limitations and general plant growth conditions.
In this work, a model of a production unit is derived including a description of the relation between the actuators' electrical signals and the input values to the model.
Furthermore, a preliminary model based state tracking control is evaluated for production unit containing Lettuce.
It could be observed that the controller is capable to track the reference while satisfying the constraint under changing weather conditions and resource availability.

\end{abstract}

\begin{keyword}
state-space models, predictive control, tracking applications, agriculture, food production
\end{keyword}

\end{frontmatter}

\graphicspath{{./gfx/}}
\section{Introduction}

Conventional controlled environment agriculture based on greenhouses are transforming into highly sophisticated plant factories for continuous food production. These plant factories, also referred to as indoor-vertical-farms and sometimes as urban-farms, utilize infrastructures (warehouses, shipping containers, etc.) with artificial light and precisely controlled climate for production of biomass (e.g. plants, fish, algae) \citep{KOZAI2013}. Regardless of the efficient use of resources (water, land, etc.), sustainability of these farms are eminently criticised due to the high energy consumption for artificial lighting and climate and inefficient byproduct reuse \citep{ALCHALABI201574,GRAAMANS201831}.

A recent study on economically feasible vertical farms by \cite{CONRAD2017} has shown potential for sustainable operation by incorporating multiple production units (plant-unit, fish-unit, etc.) and interconnecting them for byproduct reuse.
Such interconnection imparts additional complexity to the efficient control of production units which are inherently nonlinear due to the underlying biological processes that depend on several states (temperature, humidity, CO$_2$, water etc.) and also the conglomerate of actuators that influences them. Therefore, for efficient operation of such farms, it is necessary to consider resource (electricity, CO$_2$ etc.) change dynamics due to external disturbances together with state and input constraints.

Works of \cite{Henten1994,VANSTRATEN2000221} propose the use of optimal control approaches to manage greenhouse climate (temperature and humidity) for plant production demonstrating economic benefits using mathematical models and weather forecast data.
In particular, model predictive control (MPC) is a widely applied optimal control approach that utilizes a process model, possibly including disturbance specifications, and considers input and state constraints as well as economic factors to track desired references (states and/or inputs) \citep{Kim2002,Ferreau2007,Gu2006}.

A particular focus of the current investigation is to set up a suitable predictive control scheme for a plant biomass production unit under the following considerations:
\begin{enumerate*} [1)]
\item resource availability from other production units (CO$_2$, H$_2$O, etc.),
\item changes in weather,
\item actuator limitations and operation costs,
\item relevant state constraints for plant survival, and
\item dynamics not included in the plant model but necessary for plant growth (day-night pattern)

\end {enumerate*}.

In this work, a detailed mathematical model describing various dynamics of a production unit is presented along with the hardware limitations and effects of the disturbance on the system's states.
A dynamical model of the plant growth presented in literature is considered for the subject produced in the production unit.
References for states representing the optimal growth conditions, disturbances (resource availability and weather) and input references for light intensity are specified.

Given these references, control operation of a tracking MPC is evaluated and the influence of the disturbance is investigated.
Although disturbances may pose a distinct difficulty in predictive controlling, in general, a robustness analysis for the controller is bypassed by employing specific disturbance functions over the prediction horizon.

These functions are available through short-term weather forecasts and render the process model time-variant.

The following Section \ref{sec:modeling} consolidates the model equations of the system under study.
Then, in Section \ref{sec:control}, the online optimization of the predictive controller is introduced for climate tracking.
Section \ref{sec:result} evaluates the performance of the predictive controller while discussing certain issues related with the reachability of the reference.
A conclusion and outlook is given in Section \ref{sec:conclusion}.

\input{modelling}

\input{control}
\input{results_discussion}

\section{conclusion}\label{sec:conclusion}

This work reviews the application of a nonlinear predictive controller on a growth chamber for climate tracking control.

Specifically, a disturbance affected prediction model of the container-plant-environment interaction was considered, in which particular trajectories have been substituted for the disturbance.
Additionally, references for temperature, humidity and CO$_2$ concentration and light intensity have been specified, representing best growth conditions.

The controller is applied on the nonlinear system, relaxing the binary input constraint to ease the computation.
It has been observed that the (heuristic) optimal plant growth environment could be tracked within the limitation of the actuators, while using a short horizon in combination with large sensor and actuator sampling.

As for the nonlinearity of the model, further investigations could consider utilizing adaptive control methods in finding efficient control actions that require less computational resources.

In a future work, the MPC based controller developed in this work shall be implemented to run on a resource constrained embedded PC.
\bibliography{bib/MPC,bib/ADP_RL,bib/ClassicOptControl,bib/NonlinearControl,bib/ConstructiveMath,bib/MPCappl_extra,bib/PlantModel}

\end{document}

config: arxiv


TITLE: Pull back from $dx \wedge dy \wedge dz$ to $dx \wedge dy$
QUESTION [0 upvotes]: What is the pull-back of $dx \wedge dy \wedge dz$ under $\varphi : \mathbb{R}^2 \rightarrow \mathbb{R}^3 : (a,b) \rightarrow (0,a,b)$? I would guess that
$$
\varphi^{\ast}(dx \wedge dy \wedge dz) = \pm da \wedge db,
$$
where I am not sure about the sign, but why is this the case?

REPLY [2 votes]: If I let $(a,b)$ be standard coordinates on $\mathbb{R}^2$ and $(x,y,z)$ be standard coordinates on $\mathbb{R}^3$, then the transformation is given by the equations
$$ x = 0 \qquad \qquad y = a \qquad \qquad z = b $$
and thus the pullback is
$$ \varphi^*(\mathrm{d} x \wedge \mathrm{d} y \wedge \mathrm{d} z) = \mathrm{d} 0 \wedge \mathrm{d} a \wedge \mathrm{d} b = 0 $$

Of course, there's a simpler shortcut here; the pullback of a 3-form is a 3-form, and there are no nonzero 3-forms on a 2-dimensional manifold.


TITLE: Is there a standard notation for the collection $\{R\cap S:R\in\pi,S\in\sigma\}$ where $\pi$ and $\sigma$ are collections of subsets of a set $X$?
QUESTION [2 upvotes]: Let $X$ be a set, let $\wp(X)$ be the collection of all subsets of $X$, and let $\pi$ and $\sigma$ be any subcollections of $\wp(X)$.
Consider the collection
\begin{equation*}
\{R\cap S:R\in\pi,S\in\sigma\},
\end{equation*}
is there a standard notation for this collection?
My interest in this collection comes from the following.
Let $\wp^2(X)$ denote the collection of all subsets of $\wp(X)$,
and let $\lesssim$ denote the "reversed refinement order" on $\wp^2(X)$, i.e., $\pi\lesssim\sigma$ if and only if for each $R$ in $\pi$ there is a $S$ in $\sigma$ such that $R\subseteq S$,
for any $\pi,\sigma\in\wp^2(X)$.
Then $\lesssim$ is a preorder on $\wp^2(X)$.
Let $\sim$ be the relation on $\wp^2(X)$ defined for any $\pi$ and $\sigma$ in $\wp^2(X)$ by $\pi\sim\sigma$ if and only if $\pi\lesssim\sigma$ and $\sigma\lesssim\pi$.
Then $\sim$ is an equivalence relation on $\wp^2(X)$,
and a partial order $\leq$ on the collection $\wp^2(X)/\sim$ of equivalence classes $[\pi]$ ($\pi\in\wp^2(X)$) can be defined by
\begin{equation*}
[\pi]\leq[\sigma]\quad\text{if and only if}\quad\pi\lesssim\sigma.
\end{equation*}
If $\upsilon$ is the collection $\{R\cap S:R\in\pi,S\in\sigma\}$,
then the equivalence class $[\upsilon]$ would be the greatest lower bound of $[\pi]$ and $[\sigma]$ under the partial order $\leq$.

REPLY [4 votes]: $$
\def\leq{\leqslant}
\def\geq{\geqslant}
\def\cP#1{{\cal P}(#1)}
\def\PpS{{\cal P}^\uparrow(S)}
$$
I don't know of a specific notation for this case, but as your operation is an instance of a more general construction, you might be interested to know about this more general setting.
If $S$ is a semigroup, the set $\cP{S}$ of subsets of $S$ is also
a semigroup, for the product defined, for every $X, Y \in \cP{S}$, by
$$
 XY = \{ xy \mid x \in X, y \in Y \}
$$
This semigroup is called the power semigroup (or global semigroup)  of $S$. Thus your construction is just the power semigroup of the semigroup $S = (\cP{X}, \cap)$.
There is a similar notion for ordered semigroups. An ordered semigroup is a semigroup $S$ endowed with a partial order $\leq$ such that $x \leq y$ implies $zx \leq zy$ and $xz \leq yz$, that is, a stable preorder.
Let $(S,\leq)$ be an ordered semigroup. One can define a preorder relation $\leq$ on $\cP{S}$ by setting $X \leq Y$ if and only if, for
each $y\in Y$, there exists $x \in X$ such that $x \leq y$. It is easy to see that this defines a stable preorder on $\cP{S}$. Denote by $\sim$ the equivalence relation defined by $X \sim Y$ if and only if $X \leq Y$ and $Y \leq X$. Then $\sim$ is a congruence on $\cP{S}$ and the preorder $\leq$ induces a stable partial order on the semigroup $\cP{S}/{\sim}$.
But there is a more direct way to obtain the resulting ordered semigroup. Let $\PpS$ be the set of upper sets of $(S, \leq)$. Define the product of two elements $U$ and $V$ of $\PpS$ as the upper set generated by $UV$ and take inclusion as the order on $\PpS$. I let you verify that you obtain the same ordered semigroup as above.
In the second part of your post, you apply this construction to the ordered semigroup $(S, \leq) = (\cP{X}, \cap, \subseteq)$, except that you reverse the order at the very end.


TITLE: What are the smallest amount of numbers required to generate all the even numbers?
QUESTION [0 upvotes]: Viewpoint 1
To generate the even numbers $<n$ the smallest amount of numbers we need about $O (\sqrt{n})$ (by summing only $2$ elements)
Hence, consider the series:
$$ ( \sum_{r=1}^n x^{b_r})^2 = \sum_{i=1}^n \sum_{j=1}^n x^{b_i + b_j}= \sum_{r=2}^n c_{r} x^{2r}  $$
where $b_i$ is an arbitrary natural number.
For example for $n= 57$
$$ (x^2 + x^4 + x^8 + x^{12} + x^{16} + x^{18}  + x^{36}+ x^{38})^2= \sum_{r=1}^{28} c_r x^{2r} $$
and then taking the square root and the limit:
$$ \lim_{x \to 1^-} ( \sum_{r=1}^n x^{b_r}) = \sqrt{\sum_{r=2}^\infty c_{r} x^{2r}} = \text{number of terms}$$
Proceeding in our example:
$$ \lim_{x \to {1^-}}(x^2 + x^4 + x^8 + x^{12} + x^{16} + x^{18}  + x^{36}+ x^{38})^2= 8 \approx \sqrt{57} $$
Hence, to generate all the even numbers:
$$ \lim_{n \to \infty }\lim_{x \to 1^-} ( \sum_{r=1}^n x^{b_r})^2  \sim O(\sqrt{n})$$
Viewpoint 2
However I do not think we can say that is the case to generate all the even numbers.
We need:
Let the series which generates all even numbers be
$$ (\lim_{n \to \infty} \sum_{r=1}^n  x^{a_r})^2= \sum_{i=1}^n \sum_{j=1}^n x^{a_i + a_j} = \sum_{r=2}^\infty c_{r} x^{2r} $$
where $a_i$ is an arbitrary natural number.
Assuming Goldbach's conjecture is true we can use the following we can use the following example:
$$ (\lim_{n \to \infty} \sum_{r=1}^n  x^{p_r})^2 = \sum_{r=2}^\infty c_{r} x^{2r} $$
Where $p_r$ is the $r$'th prime
$$ \lim_{n \to \infty} (\sum_{r=1}^n  x^{p_r})^2 = \sum_{r=2}^\infty c_{r} x^{2r} $$
And the number of terms would approach required to generate the even powered series:
$$ \implies  \lim_{n \to \infty} (\sum_{r=1}^n  x^{a_r}) = \sqrt{\sum_{r=2}^\infty c_r x^{2r} }= \text{number of terms}\sim \alpha(x)  $$
$x \to 1^-$
Again proceeding with our example:
$$ \lim_{n \to \infty} (\sum_{r=1}^n  x^{p_r})\sim \frac{1}{(x-1)\ln(1-x)} $$
where $x \to 1^-$
Where $ \alpha(x) $ would  be the amount to generate all the even numbers. Hence, the number of elements we need are:
$$ \implies \lim_{n \to \infty} \sum_{r=1}^n a_r x^r \sim \alpha(x)  $$
where $x \to 1^-$
Question
Which viewpoint is correct? and why?
P.S: This is related to Reformulation of Goldbach's Conjecture as optimization problem correct? however I did not have enough space in the comments to dispute my misconception/concept.
Point of Dispute (Edit)
How does one extend the finite case $< n $ to $n \to \infty $
Viewpoint 2 seems to suggest they should have no connection.

REPLY [2 votes]: Given $n$, choose $K$ even such that $K^2>2n$, and use the numbers $$2,4,6,8,\dots,K-4,K-2,K,2K,3K,4K,\dots,(K/2)K$$ That uses about $K$ numbers, where $K$ is about $\sqrt2\sqrt n$.
One can do better than that, but that's enough to show you that $c\sqrt n$ can be achieved for some constant $c$.
Much, much more information is available by following my comment and searching for $$\rm additive\ basis\ of\ order\ 2$$


TITLE: Ratio between subsegments of space diagonal of a cube
QUESTION [3 upvotes]: Let $ABCDEFGH$ be a cube where $ABCD$ is the ground face and $E$ is about $A$, $F$ above $B$, and so on. Consider the space diagonal $AG$, and call $P$ the orthogonal projection of $B$ on $AG$.
The question is to show that $2|AP| = |PG|$. One can easily calculate that the space diagonal has length $\sqrt{3}$, and by the congruence of $ABG$ and $ABP$ we find that $|AP| = \sqrt{3}/3$. Hence $|PG| = 2\sqrt{3}/3$, and we are done. This, however, requires some calculations.
Can we show $2|AP| = |PG|$, without calculating any of the lengths?
So, by some similarity or symmetry argument. One observation is that $P$ is in the plane $BDE$, so by symmetry there is a point $P'$ in the plane $CFH$, and in this setup it is sufficient to prove that $|AP| = |PP'|$.

REPLY [1 votes]: Construction: Join $BG$.
Let the lenght of side $AB$ be $a$, so $AG=a\sqrt{3}$. By Pythagoras' theorem, $BG=a\sqrt{2}$.
Clearly, $BG$ is perpendicular to $AB$ as the whole plane $BCGF$ is perpendicular to $AB$.
As a result, we get a right angled $ΔABG$ where $BP⊥AG$.
Clearly, $ΔAPB \sim ΔABG$, so$$\frac{AP}{AB}=\frac{AB}{AG} \Longrightarrow \frac{AP}{a}=\frac{a}{a\sqrt{3}} \Longrightarrow AP=\frac{a}{\sqrt{3}}.$$
Similarly, $ΔBPG \sim ΔABG$, so$$\frac{PG}{BG}=\frac{BG}{AG} \Longrightarrow \frac{PG}{a\sqrt{2}}=\frac{a\sqrt{2}}{a\sqrt{3}} \Longrightarrow PG=\frac{2a}{\sqrt{3}}.$$
Now, $$\frac{AP}{PG}=\frac{1}{2},$$
therefore $$2|AP|=|PG|.$$
Quod erat demonstrandum…