问:
Prove stochastic domination via coupling of a one-point perturbation
Let $S_n$ be the usual Simple Symmetric Random Walk on $\mathbb Z$. That is, $X_n \sim \mbox{Ber}_{\pm 1}(\frac 12)$ are iid with $X_0 = 0$ and $S_n = X_0+...+X_n$. (Where $\mbox{Ber}_{\pm 1}(p)$ for $p \in [0,1]$ is $+1$ with probability $p$ and $-1$ otherwise).

This has its own hitting probabilities for various sets and other growth functionals.

Suppose at one $z \in \mathbb Z$, I perturb the probability of the random walk and create a new stochastic process. That is, fix a $p \in [\frac 12,1]$ and define $S'_n$ recursively as follows : $S'_0 = 0$, and

if $S'_{n-1}\neq z$ then $S'_n = S'_{n-1} + \mbox{Ber}_{\pm 1}$.

On the other hand, if $S'_{n-1} = z$ then $S'_n = S'_{n-1} + \mbox{Ber}_{\pm 1}(p)$.

Since $p \geq \frac 12$, the stochastic process has a right drift at the point $z$ since if it reaches $z$ then it is more likely to go right than left.

This is like a one-point perturbation of the stochastic process.

I believe that $S'_n$ dominates $S_n$ stochastically. That is, for every $n$, and every $t \in \mathbb Z$ we have $\mathbb P(S'_n>t) \geq \mathbb P(S_n > t)$. This should be provable from one of the theorems in Shaked and Shanthikumar, which goes through a lot of these "stochastic comparison of random variables" results, so I am not worried about that.

My question is this: can we find an explicit coupling of $S'_n$ and $S_n$, for which $S'_n \geq S_n$ almost surely, thereby implying stochastic dominance of $S'_n$ over $S_n$? Such a coupling non-explicitly exists through Strassen's theorem, but in this simple situation, I would like an explicit example.

While I have seen coupling used to prove Stochastic dominance, I have seen this only in cases where the parameter is uniformly changed (e.g. $\mbox{Bin}(p)$ dominates $\mbox{Bin}(q)$ with $p>q$), but not at one point.

EDIT : Thank you for placing the bounty, @Restless.

答:
I propose something: let $U_1,\cdots,U_n,\cdots$ be independent uniformly distributed on $(0,1)$. For all $n\in\mathbb N^*$, let $X_n=1_{\{U_n<\frac12\}}-1_{\{U_n\ge\frac12\}}$, that is $X_n=1$ if $U_n<\frac12$ and $X_n=-1$ else, so that $X_n\sim\textrm{Ber}_{\pm1}(\frac12)$ and $X_1,\cdots,X_n,\cdots$ are iid. Let $S_0=0$ and for $n\in\mathbb N^*$ let $S_n=X_1+\cdots+X_n$,
$$
X'_n=X_n1_{\{S_{n-1}\neq z\}}+\left(1_{\{U_n<p\}}-1_{\{U_n\ge p\}}\right)1_{\{S_{n-1}=z\}}
$$
and $S'_n=X'_1+\cdots X'_n$. Then for all $n\in\mathbb N^*$, $X_n\le X'_n$, and of course $S_n\le S'_n$.


问:
Suguru puzzles with no given clues
A Suguru puzzle consists of a bunch of polynominoes fit together in a rectangular shape. The goal of the puzzle is to fill all the cells in all the polynominoes with the numbers $1$ to the amount of cells the polynomino has. A pentomino requires you to fill in $1$, $2$, $3$, $4$, and $5$, for example. A number is not allowed to touch the same number, not diagonally either!

Recently I was wondering if it is possible to construct a Suguru puzzle without any given clues (no given numbers) with exactly one solution. I was able to find some examples so I started wondering for which sizes this would work. So my question is:

For which $a$, $b$ is it possible to construct an $a\times b$ Suguru puzzle with no given clues and exactly one solution?

I was able to prove some cases, but not all cases.

I was able to prove that if $a$ and $b$ are both odd, than there exists a possible arrangement with no given clues and exactly one solution unless both $a$ and $b$ are equal to $3$, in which case it is impossible to construct one.

If one side is even and the other one is odd, I was able to proof that there exists an arrangement if the even side is at least $6$ and the odd side at least $3$. If the odd side is equal to $1$ than no arrangement with the other side being even exists.

If the even side is equal to either $2$ or $4$, I am not sure whether it will ever be possible to construct a puzzle with no given clues and exactly one solution. So far I have not been able to find any working examples, so I believe there aren't any at all.

With the help of the $8 \times 8$ configuration which was provided by Richard Tobin, I was also able to prove that an arrangement exists if $a$ and $b$ are both even and at least 8, however, I still haven't figured out what happens if $a$ and $b$ are both even, but not both at least 8.

With the help of the $6 \times 6$ configuration provided by Kris van Bael I was able to prove all $even \times even$ are possible if both evens are at least 6, slightly improving the argument from before.

With the help of Kris van Bael who also provided a working 5x4 I was able to prove all $4 \times odd$ work if the odd is at least 5. It is not hard to prove that $3 \times 4$ is impossible.

This brings all the unsolved cases down to two categories.
$2\times n$ for all $n$ and $4\times n$ for all even $n$

Does anyone know how I could prove all cases, or just some of the cases I have not yet already solved?

ContentLicense: CC BY-SA 4.0

Title: Suguru puzzles with no given clues

Body:
Here is an 8x8 suguru with no clues and one solution (so your conjecture that there are none with both sides even is wrong):

+---+---+---+---+---+---+---+---+
|           |           |       |
|---+---+---+---+---+   +---+---|
|               |   |       |   |
|---+---+   +---+   +---+---+   |
|   |   |   |       |       |   |
|   +   +---+   +   +   +---+   |
|   |       |       |   |       |
|---+   +   +---+---+   +   +---|
|   |       |   |       |   |   |
|   +---+---+   +---+---+---+   |
|   |           |   |           |
|   +---+   +---+   +---+---+   |
|   |   |   |       |       |   |
|---+---+---+---+   +---+   +---|
|               |       |       |
+---+---+---+---+---+---+---+---+


问:
Singleton - Compactness by Subcover Definition
I know that the singleton is compact by the Heine Borel Theorem. However, how can I prove that it is compact using the definition?

I am defining compactness of the set A by if A is compact then every open cover of A have a finite subcover.

On a related note, how can I get better at proving compactness by the definition?

I am considering the usual Rn topology.

Thank you

答:
If $X$ is a singleton $\{x\}$, then any cover of $X$ must contain the open set $\{x\}$ ($\{x\}$ is not open in $\mathbb{R}^n$, but it is relatively open in $X$), so you have a subcover with only one open set, $\{x\}$.

In general, if $X$ is finite, $X=\{x_1,\dots,x_n\}$, then if you are given an open cover, you can pick elements of the cover $A_1,\dots,A_n$ such that $x_i\in A_i$, and then $\{A_1,\dots,A_n\}$ is a finite subcover of $X$.

In general proving compactness by definition is not the most common method, except in some easy cases like finite topological spaces or spaces with the trivial or finite complement topology.

In most cases, you prove compactness using the properties of compact spaces, like:

Product of compact spaces is compact.

Continuous images of compact spaces are compact.

Closed subspaces of compact spaces are compact.

Finite union of compact subspaces is compact.

If your space is T$_2$, arbitrary intersections of compact subspaces are compact.

Or characterizations of compactness, like the Heine Borel theorem.


问:
What condition guarantees that every sequence in a set converges in that set?
I was wondering whether there exists a set, say $A$ such that every sequence $a_n \in A$ converges to some point in $A$. I found out that any set which contains more than one element cannot satisfy such a criteria, since if {${a_1,a_2}$} be such an $A$, then immediately I can construct a sequence $a_n=a_1,a_2,a_1,a_2,...$ which diverges. Hence I came to a conclusion that a singleton set is the only condition that satisfies such a criteria. Are there any more? If yes, what must be the condition?

答: Assuming that you are writing in the context of metric spaces, the only other possibility is when $A=\emptyset$. In this case, your condition is vacuously true.


问:
I am trying to figure out the critical points of a multivariable function $\ f(x,y)=(x-2)(y-2)(x+y-2)$
After multiplying it out I get: $\ f(x,y)=x^2y+xy^2-6xy-2x^2-2y^2+8x+8y-8$

Where I would take the partial derivative

with respect to x and set to zero: $\ y^2+2xy-6y-4x+8=0$,

with respect to y and set to zero: $\ x^2+2xy-6x-4y+8=0$

This is where I am stuck and unable to proceed. Any tips would be helpful.

答:
Just don't multiply out.  The partial derivative with respect to $x$ (note that $(y-2)$ is constant in this case) using the product rule for $(x-2)(x+y-2)$ is
$$f_x(x,y) =(y-2)(x+y-2+x-2).$$
Similarly,
$$f_y(x,y)=(x-2)(x+y-2+y-2).$$
From here it's a piece of cake.


问:
Does Taylor series always converge somewhere to some function
Suppose $f$ is $C^\infty$ on an open set $U$, and $x \in U$. Does $Tf$, the Taylor series centered at $x$, converge in a neighborhood of $x$ to some function(not necessarily the original function $f$)? To be more precise, if $B$ is an open ball centered at $x$ contained in $U$, do we have $Tf$ converges in $B$?

I’m not asking about Taylor series not equal to the original function. It’s a totally different question.

答: Yes, such functions exist. Take a power series centered at $0$ such that its radius of convergence is $0$, such as, say,$$\sum_{n=0}^\infty n^nx^n.\tag1$$Then, according to Borel's lemma, there is a smooth function $f\colon\Bbb R\longrightarrow\Bbb R$ such that$$(\forall n\in\Bbb N):f^{(n)}(0)=n!n^n.$$So, the Taylor series of $f$ centered at $0$ is the power series $(1)$.


问:
a random variable has the density $f\left(x\right)=a+bx^{2}$ . Determine and b so that its mean will be 2/3
A random variable has the density  $f\left(x\right)=a+bx^{2}$ with  $0<x<1$. Determine and b so that its mean will be 2/3.

I'm a little confued trying to get a and b. This is what I already tried:
$$P\left(0<X<1\right)=\int_{0}^{1}ax^{2}dx+\int_{0}^{1}bx^{4}dx$$
$$P\left(0<X<1\right)=a\int_{0}^{1}x^{2}dx+b\int_{0}^{1}x^{4}dx=a\left(\frac{x^{3}}{3}\right)_{0}^{1}+b\left(\frac{x^{5}}{5}\right)_{0}^{1}$$
$$P\left(0<X<1\right)=a\left(\frac{1}{3}\right)+b\left(\frac{1}{5}\right)=\frac{a}{3}+\frac{b}{5}=\frac{5a+3b}{15}$$
$$E\left[X\right]=a\left(\frac{1}{3}\right)+b\left(\frac{1}{5}\right)=\frac{a}{3}+\frac{b}{5}=\frac{5a+3b}{15}$$
$$E\left[X\right]=\frac{2}{3}=\frac{5a+3b}{15}$$
$$10=5a+3b$$
I'm assuming x as a continuous random variable
In this case shouldn't be a=0 b=1?

答:
I think the issue lies in your first line of computation. If I remember correctly we have

$$\mathbb{P}(0<X<1) = \int_{0}^{1} f(x)\, dx.$$

(Notice the missing $x$ in the integrand). Now since we're working with probabilities, this integral must evaluate to 1, which gives you one condition that $a$ and $b$ must satisfiy. Then, you can also compute the expected value

$$\mathbb{E}[X] = \int_{0}^{1} x f(x) \, dx, $$

which should give you another condition that must be satisfied. I didn't compute any further, but I think with these two equations you should be able to determine $a$ and $b$.

Edit: Tito's answer shows you the computation for $\mathbb{E}[X]$, you can proceed similarly for the first integral in my post and that should do the trick.


Title: a random variable has the density $f\left(x\right)=a+bx^{2}$ . Determine and b so that its mean will be 2/3

Body:
A random variable has the density  $f\left(x\right)=a+bx^{2}$ with  $0<x<1$. Determine and b so that its mean will be 2/3.

I'm a little confued trying to get a and b. This is what I already tried:
$$P\left(0<X<1\right)=\int_{0}^{1}ax^{2}dx+\int_{0}^{1}bx^{4}dx$$
$$P\left(0<X<1\right)=a\int_{0}^{1}x^{2}dx+b\int_{0}^{1}x^{4}dx=a\left(\frac{x^{3}}{3}\right)_{0}^{1}+b\left(\frac{x^{5}}{5}\right)_{0}^{1}$$
$$P\left(0<X<1\right)=a\left(\frac{1}{3}\right)+b\left(\frac{1}{5}\right)=\frac{a}{3}+\frac{b}{5}=\frac{5a+3b}{15}$$
$$E\left[X\right]=a\left(\frac{1}{3}\right)+b\left(\frac{1}{5}\right)=\frac{a}{3}+\frac{b}{5}=\frac{5a+3b}{15}$$
$$E\left[X\right]=\frac{2}{3}=\frac{5a+3b}{15}$$
$$10=5a+3b$$
I'm assuming x as a continuous random variable
In this case shouldn't be a=0 b=1?

Body: $$E[X]=\int_0^1 xf(x)dx=\int_0^1(ax+bx^3)dx=a\frac{x^2}{2}+b\frac{x^4}{4}\bigg|_0^1=\frac{2a+b}{4}.$$


问:
Proving that a parameterized curve in 3D is an ellipse and calculating its shape parameters
Consider the following closed curve in 3D, parameterized by $\theta$:
$$
\begin{align}
x &= A\cos(\theta) \\
y &= -2A\sin(\theta) \tag{1}\label{xyz} \\
z &= -C\sin(\theta-\phi)
\end{align}
$$
Two questions:

Is this curve an ellipse? When I graph it, it certainly looks like one. How can I prove that?

Assuming the answer to (1) is affirmative, what are the shape parameters (e.g. semi-major axis, semi-minor axis) of this ellipse? How do I calculate them?

Here's what I've done so far.

It's easy to verify that the curve lies in the plane normal to $(2C\sin\phi, C\cos\phi, -2A)$.

The curve lies on the surfaces
$$
4x^2+y^2=4A^2 \\
(\sin^2\phi)x^2 + (\sin\phi\cos\phi)xy + (\cos^2\phi/4)y^2 - (A^2/C^2)z^2 = 0
$$
so presumably it is their intersection. But I don't know why the intersection of these surfaces must be an ellipse.

The distance from the origin is
$$
\begin{align*}
r^2 &= x^2+y^2+z^2 \\
&= A^2\cos^2\theta + 4A^2\sin^2\theta + C^2(-\sin\theta\cos\phi+\cos\theta\sin\phi)^2 \\
&= (A^2+C^2\sin^2\phi)\cos^2\theta + (4A^2+C^2\cos^2\phi)\sin^2\theta - 2C^2\sin\phi\cos\phi\sin\theta\cos\theta \\
&= \frac12 (A^2+C^2\sin^2\phi)(1+\cos 2\theta) + \frac12 (4A^2+C^2\cos^2\phi)(1-\cos 2\theta) - \frac12 C^2 \sin 2\phi \sin 2\theta \\
&= \frac12 (5A^2+C^2) + \frac12 (-3A^2 + C^2(\sin^2\phi-\cos^2\phi))\cos 2\theta - \frac12 C^2 \sin 2\phi \sin 2\theta \\
&= \frac12 (5A^2+C^2) - \frac12 (3A^2+C^2\cos 2\phi)\cos 2\theta - \frac12 C^2 \sin 2\phi \sin 2\theta \\
\end{align*}
$$
so
$$
\frac{d(r^2)}{d\theta} = (3A^2+C^2\cos 2\phi)\sin 2\theta - C^2\sin 2\phi\cos 2\theta.
$$
Setting $d(r^2)/d\theta = 0$ gives
$$
(3A^2+C^2\cos 2\phi)\sin 2\theta - C^2\sin 2\phi\cos 2\theta = 0
$$
or
$$
\tan 2\theta = \frac{\sin 2\phi}{3A^2/C^2+\cos 2\phi}.
$$
This gives the local extrema of the distance function and should be a step towards finding the semi-major and semi-minor axes. In theory, I should use this equation to substitute back into Eq. \eqref{xyz} and be able to get a semi-major axis and semi-minor axis vector. But it looks computationally hard and I decided not to try to proceed further without asking this question first.

Am I heading in the right direction? Is there a way to proceed that is not as computational? (At least for the proof of (1); I expect that (2) is going to get computational at some point no matter what.)

答:
Here is a way which is indeed non-computational.

You surely mean $4x^2+y^2=4$ which is in 3D the equation of an elliptical-based cylinder, belonging to the category of quadric surfaces. It is well known that the intersection of a quadric with a plane is a conic curve. As this curve is bounded
$(|x|\leq A, |y|\leq 2A, |z|\leq C)$, it can be neither an hyperbola nor a parabola ; it remains only the case of an ellipse.


问:
Elementary number theory - prerequisites
Since summer comes with a lot of spare time, I've decided to select a mathematical subject I want to learn as much as possible about over the next three months. That being said, number theory really caught my eye, but I have no prior training in it.

I've decided to conduct my studying effort in a library; I prefer real books to virtual ones, but as I'm not allowed to browse through the books on my own, I have to know beforehand what I'm looking for and this is where I'm kind of lost. I'm not really sure where to start.

Basically I wanna know about the following:

What are the prerequisites?
(I'm currently trained in Linear Algebra, Calculus, Complex Analysis - all on an undergraduate level )

Can you recommend some reading materials?

Thank you.

答:
I recommend Number Theory: Step by Step published in Dec. 2020 by Kuldeep Singh for two reasons. It provides solutions online to EVERY exercise. It uses color.

I'm not the author, but I know of him. He's taught number theory for at least twenty years to first year undergraduates, and his book reflects his experience.

ContentLicense: CC BY-SA 3.0


问:
Evaluate the integral $\int_{\mathbb R} \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx$
I try to calculate the integral
\begin{align}
f(y,z):=\int_{\mathbb R} \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx
\end{align}
on the set $y>0$ and $0<z<1$. I could show by some calculation that in the limit $z\to0$ holds
\begin{align}
f(y,0)=e^{y/2}K_0(y/2),
\end{align}
where $K_0$ denotes the modified Bessel function of the second kind and
\begin{align}
\lim_{z\to1}f(y,z)=f(y,1)=\infty,\\
\lim_{y\to0}f(y,z)=f(0,z)=\infty.
\end{align}
But I can't evaluate the integral in the interior of the defined set. Any help, ideas or hints how to solve this would be appreciated.

Here is a selection of what I've tried:

several substitutions (e.g. $\sinh$,$\dots$),

integration by parts,

tried to evaluate the integral $$\partial_zf(y,z)=\int_{\mathbb R} -yx\cdot \exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)\mathrm dx, $$

tried to evaluate the integrals $$\partial_yf(y,z)=\int_{\mathbb R} -(x^2+zx\sqrt{x^2+1})\cdot \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx $$ and
$$f(y,z)-\partial_yf(y,z)=\int_{\mathbb R} (\sqrt{x^2+1}+zx)\cdot \exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)\mathrm dx. $$

答:
not an answer

I computed numerically the integral in the interval $[0,10]\times[0,1]$
$$F(y,z)=\int_{-\infty }^{\infty } \frac{\exp \left(-y \left(xz\sqrt{x^2+1} +x^2\right)\right)}{\sqrt{x^2+1}} \, dx$$
maybe it can be useful


问:
Using Python, generate 100x100 random matrix whose entries are sampled from the normal distribution.
Generate random vectors using numpy.random. Write code in Python that produces a $100 \times 100$ random matrix whose entries are samples from the normal distribution.

This is what I produced and am unsure if this is correct. I have a question on normal distribution as well. Should my random numbers come from a certain pool? Such as $[0,1]$ or is it possible to get values out of this range?

If anyone knows a better way to post Python code, I would also appreciate that.

答:
Your method is correct. You could do:

```<code>from matplotlib import pyplot as plt
plt.hist(a.reshape(-1, 1))
plt.show()
</code>```

to see a histogram which should reflect a bell curve. a.reshape(-1, 1) puts the matrix into a single vector.

Body:
Your method is correct. You could do:

from matplotlib import pyplot as plt
plt.hist(a.reshape(-1, 1))
plt.show()

to see a histogram which should reflect a bell curve. a.reshape(-1, 1) puts the matrix into a single vector.


答:
Finding the $\phi(n)$ for $25$ was easy, but what it if the $n$ was arbitrarily large?

$$48 \equiv -2( \mod 25)$$

Playing around with $-2(\mod 25)$ so as to get $1$ or $-1(\mod 25)$.

We see that $1024 \equiv -1(\mod 25)$

$$((-2)^{10})^{32} \equiv 1 (\mod 25)$$

$$(-2)^{322} \equiv  4(\mod 25)$$

Title: Evaluating $48^{322} \pmod{25}$

Body: How do I find $48^{322} \pmod{25}$?</p>


问:
Showing $\lim_{\nu\to\infty}\ln(\coth(\frac\nu2))\to2e^{-\nu}$ and $ \lim_{\nu\to0}\ln(\coth(\frac\nu2))\to-\ln(\frac\nu2)$
I am struggling to derive a couple limits I have come across in a paper I am reading.  Both involve the natural log of the hyperbolic cotangent.  The paper seems to be saying the two terms trend the same, or are equivalent, as nu gets large, or small.

$$ \lim_{\nu \rightarrow \infty }\ln\left(\coth\left(\frac{\nu }{2}\right)\right)\rightarrow 2e^{-\nu} $$

$$ \lim_{\nu \rightarrow 0}\ln\left(\coth\left(\frac{\nu }{2}\right)\right)\rightarrow -\ln\left(\frac{\nu }{2}\right) $$

I should mention these are not explicitly stated as limit formulas in the paper.  They are evaluating a boundary term after integrating by parts.  The final expression is obviously convenient for their purposes.

This is the paper for anyone wishing more context.  See top, right of page 2.

"Kramers-Kronig, Bode, and the meaning of zero" (arXiv link) by John Bechhoefer

Thanks

答:
Hint:

For the first limit, rewrite the argument of the log as
$$ \coth(\frac{\nu }{2})=\frac{1+\mathrm e ^{-\nu}}{1-\mathrm e ^{-\nu}},$$
and use that near $u=0$, one has
$$\ln(1+u)=u+o(u),\qquad \ln(1-u)=-u+o(u).$$

For the second limit, add the expansion at order $1$ of $\;\mathrm e^{-\nu}=1-\nu+o(\nu)$.


问:
Defining area between three curves

The area bounded by the circle $x^2 +y^2=8$ , the parabola $x^2 =2y$, and the line $y=x$ is..?

The integration is pretty easy to do but I'm having a difficult time finding the area between three curves in the above figure. What would you define in the following figure as the area between the three curve? / what is the criteria for it?

答:
For each of the potential boundaries you have to specify which side of the boundary contains the area you care about. For the circle  it's natural to think about the inside. For the parabola it's natural to think about the convex region ("inside the U").

There is no natural way to distinguish one side of the line from the other, so you have to specify that explicitly.

You could also say "beneath the parabola" or "outside the circle" if that was what mattered to you.

Algebraically, you would specify one side of a boundary with an inequality.

Title: Defining area between three curves


问:
If $x^5=2$, find $\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$

If $x^5=2$, find
$$\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$$

My attempt-

Since $x^5=2,x^6=2x,x^7=2x^2..$and so on

The equation is equivalent to
$$x^5\left(\frac{1}{2x+x^4}+\frac{1}{2x^2+x^3}+\frac{1}{x^3+x^2}+\frac{1}{x^4+x}\right)$$

Which simplifies to
$$2\left[\frac{3x^4+7x^3+5x^2+8x+12}{2x^4+6x^3+4x^2+6x+12}\right]$$
Now I am struck here.

Any help will be appreciated.

Note: Please don't use root tables and complex numbers too!

答:
I used Sage cell to calculate this.

```<code>var("X")
F.<x> = NumberField(X^5 - 2)
y = x/(x^2 + 1) + x^2/(x^4 + 1) + x^3/(x + 1) + x^4/(x^3 + 1)
print(y)
print(y.minpoly())
</code>```

Output:

```<code>-44/765*x^4 + 122/765*x^3 + 79/765*x^2 + 233/765*x + 1336/765
x^5 - 1336/153*x^4 + 1556/51*x^3 - 8182/153*x^2 + 7210/153*x - 12806/765
</code>```

In the output, the first line is the sum, and the second line is the minimal polynomial of that sum.

None of these is nice enough to be an intended answer. Thus I suspect that the question is slightly wrong.

One possibility is to change $x^5 = 2$ to $x^5 = 1$. Then it is easy to evaluate that the sum is equal to $2$.

Body:
I used Sage cell to calculate this.

var("X")
F.<x> = NumberField(X^5 - 2)
y = x/(x^2 + 1) + x^2/(x^4 + 1) + x^3/(x + 1) + x^4/(x^3 + 1)
print(y)
print(y.minpoly())

Output:

-44/765*x^4 + 122/765*x^3 + 79/765*x^2 + 233/765*x + 1336/765
x^5 - 1336/153*x^4 + 1556/51*x^3 - 8182/153*x^2 + 7210/153*x - 12806/765

In the output, the first line is the sum, and the second line is the minimal polynomial of that sum.

None of these is nice enough to be an intended answer. Thus I suspect that the question is slightly wrong.

One possibility is to change $x^5 = 2$ to $x^5 = 1$. Then it is easy to evaluate that the sum is equal to $2$.


问:
Maximize $\sum{a_i^2}$ given $\sum{a_i}=237$
I recently got stuck with the following problem:

Consider 18 integer numbers in segment $[10; 20]$. Find the maximum possible value of
$$
\sum_{i=1}^{18} {a_i^2}
$$
given that
$$\sum_{i=1}^{18} {a_i}=237$$

I don't really know how to approach this problem. The thing is, solving it like constraint optimization problem I got the solution $a_1=a_2=...=a_n=\frac{237}{18} \approx13.2$

But we need to solve it in integers. And just taking numbers around $13.2$ doesn't make sense to me. Any ideas?

答:
The idea is to suppose all of the $a_i$ are fixed except for two of them, and see what happens if you add $1$ to one and subtract $1$ from the other.  So say for instance $a_1 = x$, $a_2 = y$, and all the other $a_i$ are fixed.  Then adding $1$ to $x$ and subtracting $1$ from $y$ gives the same sum, but what happens to the sum of squares?  $$(x+1)^2 + (y-1)^2 - (x^2 + y^2)$$ is the amount by which the sum of squares increases or decreases.  But this is $$2x + 1 - 2y + 1 = 2(x-y+1).$$  So if $x \ge y$, this means the sum of squares will increase if we add $1$ to $x$ and subtract $1$ from $y$.  Repeated application of this logic means that the sum of squares is actually maximized when $x$ is as large as possible and $y$ is as small as possible.  Then we repeat this logic for the other terms in the sum; it follows that the maximum is attained when we can make as many of the $a_i$ as large as possible.  Since the $a_i$ are restricted to $[10; 20]$ we want the largest $m$ such that $20 m + 10(18 - m) \le 237$, or $m = 5$.  This means $5$ of the $a_i$ are equal to $20$, $12$ of them are $10$, and the last one makes up the remainder, which has value $17$, and the sum of squares is $3489$.

A slightly different way to look at the second part of the solution is to note that since each of the $a_i$ must be at least $10$, we can create an auxiliary sequence $b_1, b_2, \ldots, b_{18}$ with $b_i = a_i - 10$, hence $b_i \in [0; 10]$ and the constraint on them is $$\sum_{i=1}^{18} b_i = 57.$$  Then the same reasoning as before gives five $b_i$ equal to $10$, and one $b_i$ equals $7$.


答:
For the second one, use
$$
\mathop {\lim }\limits_{x \to 0} x\coth x = 1.
$$
For the first one, use
$$
\coth x = \frac{{\cosh x}}{{\sinh x}} = \frac{{1 + e^{ - 2x} }}{{1 - e^{ - 2x} }} = (1 + e^{ - 2x} )(1 + e^{ - 2x}  +\! \mathcal{O}(e^{ - 4x} )) = 1 + 2e^{ - 2x}  +\!\mathcal{ O}(e^{ - 4x} )
$$
as $x\to+\infty$ and
$$
\mathop {\lim }\limits_{x \to 0} \frac{{\log (1 + x)}}{x}.
$$

Title: Showing $\lim_{\nu\to\infty}\ln(\coth(\frac\nu2))\to2e^{-\nu}$ and $ \lim_{\nu\to0}\ln(\coth(\frac\nu2))\to-\ln(\frac\nu2)$

Body:
I am struggling to derive a couple limits I have come across in a paper I am reading.  Both involve the natural log of the hyperbolic cotangent.  The paper seems to be saying the two terms trend the same, or are equivalent, as nu gets large, or small.

$$ \lim_{\nu \rightarrow \infty }\ln\left(\coth\left(\frac{\nu }{2}\right)\right)\rightarrow 2e^{-\nu} $$

$$ \lim_{\nu \rightarrow 0}\ln\left(\coth\left(\frac{\nu }{2}\right)\right)\rightarrow -\ln\left(\frac{\nu }{2}\right) $$

I should mention these are not explicitly stated as limit formulas in the paper.  They are evaluating a boundary term after integrating by parts.  The final expression is obviously convenient for their purposes.

This is the paper for anyone wishing more context.  See top, right of page 2.

"Kramers-Kronig, Bode, and the meaning of zero" (arXiv link) by John Bechhoefer

Thanks


问:
A delta function conjecture: almost any function can be a delta kernel
I have been thinking about delta kernels, and I think I have come up with a surprising result:

If $f\colon\Bbb{R}\to\Bbb{R}$ is such that $\int_{-\infty}^\infty f(x)\,dx=L$ is finite and nonzero, then $g_k(x)=\frac kLf(kx)$ is a delta kernel.

Note that by "delta kernel", I mean a sequence of integrable functions $g_k(x)$ such that for any absolutely integrable function $h\colon\Bbb{R}\to\Bbb{R}$ that is continuous at $0$, $\lim_{k\to\infty}\int_{-\infty}^\infty g_k(x)h(x)\,dx=h(0)$.

I find this an interesting problem to think about, so I thought I'd post it here. I'll post the proof as an answer if no one else does (unless I've missed some extra hypotheses, in which case someone will point it out, I'm sure).

答:
(Note: $\int f$ will be used throughout in place of $\int_{-\infty}^\infty f(x)\,dx$.) Suppose $f:\Bbb{R}\to\Bbb{R}$ is integrable on $\Bbb{R}$, with $\int f=L\ne0$. By the definition of the integral $\int_{-\infty}^\infty$, we know that the one-sided limits $\lim_{y\to\infty}\int_0^y f=L_+$ and $\lim_{y\to-\infty}\int_y^0 f=L_-$ exist and $\int f=L_++L_-=L$. Let $F$ be the antiderivative of $f$, i.e. $F(y)=\int_0^yf(x)\,dx$.

Then $\lim_{x\to-\infty}F(x)=-K_-$ and $\lim_{x\to\infty}F(x)=K_+$. Taking $g_k(x)=\frac kLf(kx)$ we have

$$\int_0^yg_k(x)\,dx=\int_0^{ky}\frac 1Lf(x')\,dx'=\frac 1LF(ky).$$

By the limit, given an $\varepsilon>0$ we can find an $M$ such that $|F(x)-K_+|<|L|\varepsilon$ for all $x\ge M$, and so for any $y>0$ we can choose $k=\lceil M/y\rceil$ so that $ky\ge M$ and hence

$$\left|\int_0^yg_k(x)\,dx-\frac{K_+}L\right|=\frac1{|L|}\left|F(ky)-K_+\right|<\varepsilon,$$

so the function $G_k(y)=\int_0^yg_k(x)\,dx$ is pointwise convergent to $G_\infty(y)=\frac{K_+}L$ for $y>0$. By similar considerations, $G_\infty(y)=-\frac{K_-}L$ for $y<0$ (and $G_\infty(0)=0$, for what it's worth). By this point, it should be clear that $g_k$ is a delta kernel (noting that the difference between $\frac{K_+}L$ and $-\frac{K_-}L$ is $1$).

Note in particular that nowhere did we assume that $f$ was nonnegative, nor even absolutely convergent.


问:
Standard duality argument
In a paper of distributionally robust optimization, there is a step of derivation using the "standard duality argument". It obtains $ \sup _{\mathbb{Q}_{i} \in \mathcal{M}(\Xi)} \inf _{\lambda \geq 0}  \frac{1}{N} \sum_{i=1}^{N} \int_{\Xi} \ell(\xi) \mathbb{Q}_{i}(d \xi) +\lambda\left(\varepsilon-\frac{1}{N} \sum_{i=1}^{N} \int_{\Xi}\left\|\xi-\widehat{\xi}_{i}\right\| \mathbb{Q}_{i}(\mathrm{~d} \xi)\right)$ from
$\sup _{\mathbb{Q}_{i} \in \mathcal{M}(\Xi)} \frac{1}{N} \sum_{i=1}^{N} \int_{\Xi} \ell(\xi) \mathbb{Q}_{i}(\mathrm{~d} \xi)$
s.t. $\quad \frac{1}{N} \sum_{i=1}^{N} \int_{\Xi}\left\|\xi-\widehat{\xi}_{i}\right\| \mathbb{Q}_{i}(\mathrm{~d} \xi) \leq \varepsilon$.

It is confusing for me. Since if I try to use the strong duality, the optimal value of dual problem is solved from the minimax problem $\inf_{\lambda\ge 0}\sup_{\mathbb{Q} \in \mathcal{M}(\Xi) } \dots$. I am wondering how $\sup_{\mathbb{Q} \in \mathcal{M}(\Xi) } \inf_{\lambda\ge 0}$ is derived in their paper.

Any hint would be appreciated.

答: They do not use strong duality in that step. The two expressions are equal because the infimum in the first expression is $-\infty$ when the constraint in the second expression is not satisfied, so the supremum operator ensures that the constraint is satisfied also in the first expression.

Title: Standard duality argument


问:
Number theory problem book
Can anyone suggest me some good book that has problems on classical elementary number theory with solutions?</p>

答: I recommend Number Theory: Step by Step published in Dec. 2020 by Kuldeep Singh for two reasons. It provides solutions online to EVERY exercise. It uses color.

Title: Number theory problem book


问:
Prove or disprove $\bar{A}*\overline{(B \cup C)} = \overline{(A*(B \cup C))}$
I am trying to learn Discrete mathematics by myself, from a well known book by Kenneth H. Rosen.

I have tried to solve this problem, but, after evaluating the left side, I get stuck how to proceed with right hand side,
Here is how, I solved $\text{LHS}$ as follows:
 (Note : symbol $*$ is for cartesian product of sets)

$\text{let (a,x) }\epsilon \text{ } \bar{A}*\overline{(B\cup C)}$

$\Rightarrow \{(a,x)\text{ }|\text{ }x\text{ }\epsilon\ \overline{(B \cup C)} \}$

$\Rightarrow \{(a,x)\text{ }|\text{ }x\text{ }\epsilon\ {(\bar{B} \cap \bar{C})} \}$

$\Rightarrow \{(a,x)\text{ }|\text{ }x\text{ } \epsilon \bar{B} \wedge x\text{ }\epsilon\bar{C} \}$

I don't know if it's correct or not.

I solved $\text{LHS}$ using above approach, but I am not able to think of a starting point for solving the $\text{RHS}$, any suggestion or hint will help a lot.

P.S: I have just started learning discrete mathematics, so, please don't judge on asking for a naive question.

答:
BIG HINT

Let the universal set be $\{0,1\}$ with $A=B=C=\{0\}$.

Then LHS=$\{(1,1)\}$. What is the RHS?


答:
These results are called asymptotics, so I will use some notation that is normal for that kind of analysis. For the first one, we know that $\coth x \to 1$ when $x>>1$ so rewrite the log expression as

$$\log\left(\frac{e^{\frac{\nu}{2}}+e^{-\frac{\nu}{2}}}{e^{\frac{\nu}{2}}-e^{-\frac{\nu}{2}}}\right) = \log\left(\frac{e^{\frac{\nu}{2}}-e^{-\frac{\nu}{2}}+2e^{-\frac{\nu}{2}}}{e^{\frac{\nu}{2}}-e^{-\frac{\nu}{2}}}\right) = \log\left(1+\frac{2e^{-\frac{\nu}{2}}}{e^{\frac{\nu}{2}}-e^{-\frac{\nu}{2}}}\right)$$

Since $\nu$ is very large, in the denominator we have $e^{\frac{\nu}{2}}-e^{-\frac{\nu}{2}} \approx e^{\frac{\nu}{2}}$. Continuing on we get

$$\log\left(1+\frac{2e^{-\frac{\nu}{2}}}{e^{\frac{\nu}{2}}-e^{-\frac{\nu}{2}}}\right) \approx \log\left(1+2e^{-\nu}\right) = 2e^{-\nu}+O\left(e^{-2\nu}\right)$$

by Taylor series. The $O$ means the largest extra term that differentiates the true values of the two expressions is only as big as $e^{-2\nu}$ times a constant.

For the second one, use small angle approximation

$$\log \coth\left(\frac{\nu}{2}\right) = -\log \tanh\left(\frac{\nu}{2}\right) \approx -\log \frac{\nu}{2}$$

since $\tanh x \approx x$ for $x<<1$


问:
How to solve this particular differential equation?
for this question I'm stuck on solving the following equation. Can anyone please help me out? Here's what I have so far

$$\dfrac{du}{dt} = \dfrac{1-t-u}{t+u}$$

$du = (\frac{1}{t+u} -1) dt $

Body:
$$\dfrac{du}{dt} = \dfrac{1-t-u}{t+u}$$
$$u'=\dfrac {1}{t+u}-1$$
$$u'+1=\dfrac {1}{t+u}$$
$$(u+t)'=\dfrac {1}{t+u}$$
That's easy to integrate now.
$$\int (u+t)\, d(u+t)=\int \, dt$$
$$(u+t)^2=2t+C$$


问:
Inverse | Modulo | Power
Describe the inverse of $5$ modulo $18$ as a positive power of $5\pmod{18}$.

I've got that the inverse of $5$ is $11$, but is this question asking to find a $t$ such that
$$ 11=5^t\pmod{18}?$$

答:
$$5^2=25\equiv 7\equiv-11 \pmod {18}, \quad 5^3\equiv 5\cdot7\equiv-1 \pmod {18}$$

$$\implies 11=(-11)(-1)\equiv 5^2\cdot5^3\equiv5^5 \pmod {18}$$

Now, as $5^3\equiv-1 \pmod {18}, 5^6=(5^3)^2\equiv(-1)^2\equiv1$

$\implies 5^{6u}=(5^6)^u\equiv1^u\equiv1\pmod{18}$ for any integer $u$

So, the general solution will be $t=6u+5$ as $5^{5+6u}\equiv1\cdot5^5\equiv1\cdot11 \pmod {18}$

If we need $5+6u>0, u\ge 0$

Title: Inverse | Modulo | Power


问:
Converting vector field from cartesian to cylindrical and finding the curl
The question is this:

(a) Find the curl of the vector field $ = y \hat{x}  + \hat{} + \hat{}$ in Cartesian coordinates. (b) Rewrite  in cylindrical coordinates. (c) Find ∇× explicitly in cylindrical coordinates.

I've worked out (a) to be $x \hat{x} - y\hat{y} + (2x-2y)\hat{z}$ but I feel like I keep messing up on converting the vector field to cylindrical coordinates.

I know $ \rho = \sqrt{x^2 + y^2}$, $\theta = arctan(y/x)$, and $z = z$
and  have done $\rho = \sqrt{y^4+x^4}$, $\theta = arctan(x^2/y^2)$ and $z = xy = r^2cos\theta sin\theta$

I feel like i'm missing something to use this though because of the partial derivatives in the $\nabla$xV formula.

Any guidance would be appreciated.

答:
$\vec{V}= (y, x, xy)$

Please note the $\hat{z}$ component of the curl is zero and not $(2x-2y)$.

So, $ \ \nabla \times \vec{V} = (x,-y,0)$.

In cylindrical coordinates,

$\left
\{\begin{array}
{l} x = \rho \cos\varphi, y = \rho \sin\varphi, z = z \\ \hat{x} = \cos\varphi \hat{\rho} - \sin\varphi \hat{\varphi} \\ \hat{y} = \sin\varphi \hat{\rho} + \cos\varphi \hat{\varphi} \\ \hat{z} = \hat{z}
\end{array}
\right.$

Refer wiki

So the vector field can be re-written in cylindrical coordinates as

$\vec{V} = \rho \sin \varphi (\cos\varphi \hat{\rho} - \sin\varphi \hat{\varphi}) + \rho \cos \varphi (\sin\varphi \hat{\rho} + \cos\varphi \hat{\varphi}) + \rho^2 \sin \varphi \cos\varphi \hat{z}$

Rearrange this in $\hat{\rho}, \hat{\varphi}, \hat{z} \ $ components and that is your vector field in cylindrical coordinates.

Again refer to the same link that gives you formula to find curl of the vector field in cylindrical coordinates as the question asks you to explicitly find curl in cylindrical coordinates which means you cannot convert the curl found in cartesian coordinates to cylindrical using the above conversion I showed.


问:
Notation for extension of a linear map
Let be $V=V_1\oplus V_2$ a vector space.
Let be $\phi_1$ a linear map on $V_1$.
I would like to define a linear map $\phi_1'$ on $V$ such that for each $v\in V$ we have
$$\tag{1}
\phi_1'(v)=\begin{cases}
\phi_1(v)&\text{if $v\in V_1$},\\
v&\text{if $v\in V_2$},\\
\cdots &\text{otherwise, linearly extended}
\end{cases}
$$

My question is: does the following intuitive notation exist, or is there any other notation about it?
$$\tag{2}
\phi_1'=\phi_1\oplus\operatorname{id}_{V_2}
$$

My goal is to write for a map $\phi$ on $V$ with invariant subspace $V_1$ and $V_2$
$$\tag{3}
\phi=\phi|_{V_1}\oplus\operatorname{id}_{V_2}
+\operatorname{id}_{V_1}\oplus\phi|_{V_2}
$$

EDIT

This would also be something, but not as concise as the other one
$$
\phi=(\iota_{1V}\circ\phi|_{V_1}\circ\pi_1 +
  \iota_{2V}\circ\operatorname{id}_{V_2}\circ\pi_2 )
\circ (\iota_{1V}\circ\operatorname{id}_{V_1}\circ\pi_1 +
  \iota_{2V}\circ\phi|_{V_2}\circ\pi_2 )
$$

答:
This notation is perfectly fine and actually quite common. In fact, it can introduced in a more general context: given vector spaces $V_1,V_2$ and linear maps $f_1\colon V_1\to W$ and $f_2\colon V_2\to W$ one can (uniquely!) define
$$f_1\oplus f_2\colon V_1\oplus V_2\to W,\,(v_1,v_2)\mapsto f_1(v_1)+f_2(v_2)$$
Here, the sum may or not be internal. This is the universal property of the categorical coproduct of vector spaces. The coproduct happens to coincide with the product in the finite case, which is (either by definition of by convention) nothing else than the direct sum. This map does exactly the same as the one you defined by hand.

Also, given two linear maps $f,g\colon V\to W$ there is a natural notion of their sum $f+g$ defined pointwise. This is an interesting obersevation as it gives $\operatorname{Hom}(V,W)$ a very natural (abelian) group structure (with identity the trivial map).

Thus, your map $\phi$ is perfectly valid and there is not need for an inconcise version using composition.

Title: Notation for extension of a linear map


问:
Meaning cone, ray, fan for polytopes
I'm trying to understand some mathematical operations and definitions for my project. Could you explain the meanings? $P_b =\{x∈\mathbb{R}^d : Ax≤b\}$
is a polytope. Let's have a 10x3 matrix $Ax≤b$

$$
\begin{bmatrix}
0 & 1 & 1 \\
1 & 0 &1 \\
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1 \\
0 &  0& 1\\
0 &  -1& 0\\
0 &  -1& 1\\
-1 &  0& 0\\
-1 &  0& 1
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}
\leq
\begin{bmatrix}
4\\
4\\
3\\
3\\
0\\
2\\
0\\
1\\
0\\
1
\end{bmatrix}
$$

Let us denote the set of such right hand side vectors $b$ by
$U(A)=\{b∈\mathbb{R}^m : P_b \neq ∅\}.$

Switching from the H-representation to its V- representation, the cone U (A) is generated by 9 rays and 3 lines in $Z^{10}$.
The normal cone of a face F of a polytope P in $R^d$ is the set
$N(F;P)=\{v∈\mathbb{R}^dd : v^⊤x=h(P,v)~for~all~ x∈F\}.$
The dimension of the normal cone of a $k$-dimensional face is $(d − k)$. The normal fan $N(P)$ of $P$, which is the collection of the normal cones of all faces of $P$ , is a complete fan in $\mathbb{R}^d$.

I have difficulty in understanding of the terms.
So, what is the exact meaning of $U(A)$ here? What are the $b_{1,2,...}$ letters and how obtained? How is the 9 rays and 3 lines expression written? Can you explain also meanings of normal fan,cone and complete fan?

Article can be found from here. I want to grasp all of the article also I have to, but at least up to there is enough.

@Edit:
I think it's a kinda Millenium Problem. If not, isn't there any helper mathematician?

答:
This answer is rather a story, not a pointed answer.
The intuition will be supported by sage code and examples. Sage is free, installed by one click on linux boxes.

The definitions are already given, so there is nothing to do in this regard,
i will only try to digest them, based on the example. And i will  add a 2D-example.

Since $P(A,b)$ looks like a house, i'll call it so.

(1) First of all, we consider the polytope, (solid) polyhedral set from the OP given by
$$
\underbrace{
\begin{bmatrix}
0 & 1 & 1 \\
1 & 0 &1 \\
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1 \\
0 &  0& 1\\
0 &  -1& 0\\
0 &  -1& 1\\
-1 &  0& 0\\
-1 &  0& 1
\end{bmatrix}}_{=:A}
\underbrace{
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}}_{=:v}
\leq
\underbrace{
\begin{bmatrix}
4\\
4\\
3\\
3\\
0\\
2\\
0\\
1\\
0\\
1
\end{bmatrix}}_{=:b}\ .
$$
This is explicitly the set of all $v=(x,y,z)\in\Bbb R^3$ satisfying all the following conditions:
$$
\left\{
\begin{aligned}
y+z  & \le 4\ ,\\
x+z  & \le 4\ ,\\
x    & \le 3\ ,\\
y    & \le 3\ ,\\
-z   & \le 0\ ,\\
 z   & \le 2\ ,\\
-y   & \le 0\ ,\\
-y+z & \le 1\ ,\\
-x   & \le 0\ ,\\
-x+z & \le 1\ .
\end{aligned}
\right.
$$
(It is convenient to swicht tacitly between $v=(x,y,z)$, simpler when writing humanly the inequations,
and $x=(x_1,x_2,x_3)$, the sage preference.)
Note that each equation is an equation determining a half-space,
delimited by a (hyper)plane in the 3D-space.
(In 2D, the space is a plane, a hyperplane is an affine set of codimension one, so it is a line, et caetera.)
Each such half-space is a convex set,
i.e. together with two points in it, the segment joining them is also in it.

The intersection of convex sets is a convex set.
So we expect that the intersection of the half-planes determined by the above inequalities,
i.e. the polytope
$$
P(A,b):=\{\ v\in\Bbb R^d=\Bbb R^3\ :\
\underbrace{Av}_{\in\Bbb R^m=\Bbb R^{10}}
\ge
\underbrace{0 }_{\in\Bbb R^m=\Bbb R^{10}} \ \}
$$
is also a convex set.

Among the above inequalities some are "very simple", for instance from $x\le 3$ and $-x\le 0$
we obtain $x\in[0,3]$. Similar restrictions appear for $y,z$, explicitly
$y\in[0,3]$ and $z\in [0,2]$, so the whole "solid polyhedral set" $P(A,b)$, the whole house is at any rate
contained in the "box" $[0,3]\times[0,3]\times[0,2]$.
The ground of the house is the (solid) square $[0,3]\times[0,3]\times[0]$, and it is a $2$--dimensional (or $1=3-2$ codimensional) face.
The roof has an upper platform, the intersection with the house with the (hyper)plane $z=2$,
and it is the (solid) square $[1, 2]\times[1,2]\times[2]$.
We have side walls, which are also faces of dimension $2$ (i.e. codimension $1$), for instance $[0,3]\times[0]\times[0,1]$.
And on the roof there are also four dimension $2$ faces that let the water flow downwards.

Think about $d=3$ in the example as the dimension of the space containing the polyhedral set.
And $m=10$ is the number of conditions (inequalities) used as plane delimiters for some half-spaces.
The intersection of these half-spaces is our set of interest, it is a convex set, a (solid) polyhedral set. It may be that such a set is either bounded or unbounded. Our set in the example is bounded. So it is determined by its vertices (=extremal points). (In the sense that the convex hull of the set of these vertices is $P(A,b)$. In less mathematical terms, somehow intuitively...) One can put a mass (a real quantity $\ge 0$ in this physical world i accept today) in each such extremal point. Then this physical system has a mass center. All such possible mass centers are the given polyhedral set.

(1-2D) It may be good (at least for drawing) to also have a similar 2D-example.
I will use the following set of inequalities, also allowing only $\le$ (so no $\ge$) while writing them:
$$
\left\{
\begin{aligned}
-x   & \le 0\ ,\\
x    & \le 3\ ,\qquad\text{ so }x\in[0,3]\\
-y   & \le 0\ ,\\
y    & \le 2\ ,\qquad\text{ so }y\in[0,2]\\
-x+y & \le 1\ ,\\
 x+y & \le 4\ .
\end{aligned}
\right.
$$
I hope that this is simpler to spot. From the first four inequalities
we obtain $(x,y)\in[0,3]\times[0,2]$. And the other two conditions
are "biting" from two corners. Is the whole system of inequalities "feasible",
i.e. can we find some point satisfying all these conditions?
Yes, $(0,0)$ is one such point. Some of the conditions are satisfied with the equality.
But we can even do "better", the point $(1,1)$ satisfies them all with strict inequalities.
So it is an "interior point", and we have indeed a $d=2$--dimensional "polyhedron",
well a "solid polygon". The corresponding $A,b$ data can be extracted from:
$$
\underbrace{
\begin{bmatrix}
-1 & 0 \\
1 & 0 \\
0 & -1\\
0 & 1\\
-1 &  1\\
 1 &  1
\end{bmatrix}}_{=:A}
\underbrace{
\begin{bmatrix}
x\\
y
\end{bmatrix}}_{=:v}
\leq
\underbrace{
\begin{bmatrix}
0\\
3\\
0\\
2\\
1\\
4
\end{bmatrix}}_{=:b}\ .
$$

(2-2D)
It may be a good idea to get computer support. For (1-2D) first.
In sage, the above set can be declared as follows. The sage convention wants from us to rewrite the above,
using $\ge$, and moving all terms on one side, so we pass from $Av\le b$ to $0\le b-Av$ and finally to $b-Av\ge 0$:
$$
\underbrace{
\begin{bmatrix}
0 & 1 & 0 \\
3 & -1 & 0 \\
0 & 0 & 1\\
2 & 0 & -1\\
1 & 1 & -1\\
4 & -1 &  -1
\end{bmatrix}}_{=[b, -A}
\begin{bmatrix}
1\\
x\\
y
\end{bmatrix}
\ge
\begin{bmatrix}
0\\
0\\
0\\
0\\
0\\
0
\end{bmatrix}\ .
$$
Now we extract the coefficients and ask for...

```<code>b_A = matrix(QQ, 6, 3, [[0,  1,  0],
                        [3, -1,  0],
                        [0,  0,  1],
                        [2,  0, -1],
                        [1,  1, -1],
                        [4, -1, -1],])
P = Polyhedron(ieqs=b_A)
</code>```

Now we can interactively ask for:

```<code>sage: P.inequalities()
(An inequality (-1, -1) x + 4 >= 0,
 An inequality (-1, 0) x + 3 >= 0,
 An inequality (0, -1) x + 2 >= 0,
 An inequality (1, 0) x + 0 >= 0,
 An inequality (0, 1) x + 0 >= 0,
 An inequality (1, -1) x + 1 >= 0)
sage: P.vertices()
(A vertex at (3, 0),
 A vertex at (2, 2),
 A vertex at (3, 1),
 A vertex at (1, 2),
 A vertex at (0, 1),
 A vertex at (0, 0))
sage: P.show()
</code>```

Here, sage uses $x$ instead of $(x_1,x_2)$.
Yes, the picture shown is

and we have the claimed vertices.
And the show can go on with the polyhedral set in the OP:

We can ask for the faces in each dimension,
and for the corresponding normal cones. But first we will introduce $h(P, u)$ in (4).

(2) The sage code and dialog for the set $P(A,b)$ from the OP is:

```<code>A = matrix( QQ, 10, 3,
        [[0, 1, 1], [1, 0, 1], [1, 0, 0], [0, 1, 0], [0, 0, -1],
        [0, 0, 1], [0,-1, 0], [0,-1, 1], [-1,0, 0], [-1,0,  1]])

b = matrix( QQ, 10, 1,
        [4, 4, 3, 3, 0, 2, 0, 1, 0, 1])

b_A = block_matrix([[b, -A]])
P = Polyhedron(ieqs=b_A)
</code>```

This initializes P in sage.
(And use it also below.)
Then we can interactively ask for:

```<code>sage: P.inequalities()
(An inequality (-1, 0, -1) x + 4 >= 0,
 An inequality (-1, 0, 0) x + 3 >= 0,
 An inequality (0, -1, -1) x + 4 >= 0,
 An inequality (0, -1, 0) x + 3 >= 0,
 An inequality (0, 0, -1) x + 2 >= 0,
 An inequality (1, 0, 0) x + 0 >= 0,
 An inequality (0, 0, 1) x + 0 >= 0,
 An inequality (0, 1, -1) x + 1 >= 0,
 An inequality (0, 1, 0) x + 0 >= 0,
 An inequality (1, 0, -1) x + 1 >= 0)
sage: P.vertices()
(A vertex at (0, 3, 1),
 A vertex at (0, 0, 1),
 A vertex at (2, 2, 2),
 A vertex at (3, 3, 1),
 A vertex at (3, 3, 0),
 A vertex at (2, 1, 2),
 A vertex at (1, 1, 2),
 A vertex at (3, 0, 0),
 A vertex at (3, 0, 1),
 A vertex at (1, 2, 2),
 A vertex at (0, 0, 0),
 A vertex at (0, 3, 0))
sage: P.show()
Launched html viewer for Graphics3d Object
</code>```

Then i get the image in the browser. It is a 3D-image, and we can rotate it.

(3) After these preliminaries, let us understand the set $$U(A)\ .$$

Using the above two examples. In the definition of the (solid) polyhedral set
$P(A,b)$ we use $A$, $b$ as constants, they are previously fixed.
And we have a non-trivial, non-empty set $P(A,b)$ in some cases, the input data
$(A,b)$ is called feasible then, and $\emptyset$ else.

Now we fix $A$, and let $b$ run among all $b\in\Bbb R^m$. Think at the corresponding
sets $P(A,b)$ as "deformations" of the given special situation of some $b_0$.
I will try to illustrate this for the 2D-case.

Imagine we change the two $x$-conditions, and the two $y$-conditions,
and the two remaining conditions "slightly".
$$
\left\{
\begin{aligned}
-x   & \le 0\ ,\\
x    & \le 3\ ,\qquad\text{ so }x\in[0,3]\\
-y   & \le 0\ ,\\
y    & \le 2\ ,\qquad\text{ so }y\in[0,2]\\
-x+y & \le 1\ ,\\
 x+y & \le 4\ .
\end{aligned}
\right.
\qquad\text{ into }\qquad
\left\{
\begin{aligned}
-x   & \le -0.1\ ,\\
x    & \le 3.2\ ,\qquad\text{ so }x\in[0.1\ ,\ 3.2]\\
-y   & \le -0.3\ ,\\
y    & \le 1.89\ ,\qquad\text{ so }y\in[0.3\ ,\ 1.89]\\
-x+y & \le 1.1\ ,\\
 x+y & \le 4.1\ .
\end{aligned}
\right.
$$
Then the corresponding picture is:

What happened? Every delimiting line was "slighlitly" moved parallely to itself
corresponding to the deformations of the corresponding conditions.
Some half-spaces are now "bigger" (w.r.t. the inclusion), other are "smaller".
So chosing other $b$-values means moving the delimiting lines (in 2D, planes in 3D) in a parallel way, and then $P(A,d)$ is the new intersection of half-planes (in 2D, half-spaces in 3D) given by the new fronteers.

Now we are playing the following games starting with the original $P(A,b)$,
$b=(0,3,0,2,1,4)\in\Bbb R^m=\Bbb R^6$, which implements the $m=6$ delimiting fronteers.
We move the $\color{red}0$ from
$b_0=(\color{red}{0},3,0,2,1,4)$ towards $3$, getting closer and closer to $3$.
This moves the delimiting rectangular box
$[\color{red}{0},3]\times[0,2]$ till it becomes smaller and smaller, of the shape
$[\color{red}{3-\epsilon},3]\times[0,2]$, $\epsilon>0$.
Yes, some conditions become superfluous now, but still, we have some thin (solid) trapezium
left as $P(A,b)$. If we move this first component $b_1$ of $b=(b_1,b_2,b_3,b_4,b_5,b_6)\in\Bbb R^6$ beyond $3$, then there is nothing left, $P(A,b)=\emptyset$.

So far we have moved only one boundary condition. We can move two of them, but let us
move all of them, we try to get closer to the situation where the set $P(A,b)$ becomes
smaller and smaller, becomes a point (or something living in less dimensions) and "vanishes".
A way of thinking of a six-dimensional move that does this is as follows.
Let us focus on some fixed point say $(1,1)$. Now we move all boundaries as we would move the lens of a camera (parallely),

lens camera movement

so that we have one point when all boundaries touch $w'=(1,2)$, say.
Which is the value for this $b$-parameter. Well we plug in this value $w'=(x,y)=(1,2)$ into
the function
$v\to Av$, and get $w'\to Aw'=b'=(-1,1,-2,2,1,3)$.

Note that when we plug in the $b$-value as it was in the 2D-example,
$b=(0,3,0,2,1,4)$, we can always slightly change each component, still having a feasible
$P(A,b)$. But for the above $b'$ we have to be careful. There is a strict movement
in each component, either in direction "bigger" or in direction "smaller", but not both.
We can do the same with $w''=(1,1)$ and $w^\dagger=(0,1)$, getting
$b''=Aw''=(-1,1,-1,1,0,2)$ and $b^\dagger=Aw^\dagger=(0,0,-1,1,1,1)$.

And observe that parallely to the fact that $w', w'', w^\dagger$ are on a line, also
the points $b',b'',b^\dagger$ are on a line. And moreover, $w''$ being on the segment
$[w',w^\dagger]$ (mid point even) in $\Bbb R^2$, makes $b''$ sit
on the segment $[b',b^\dagger]$ (also as the mid point) by linearity via $A$.

Now we can say some words on $U(A)$.

It contains all "boundary parameters" $b$, so that the corresponding system of inequalities
has at least a solution.
In the case of the matrix $A$ from the OP, imagine all possibilities to move the
(hyper)planes delimiting the house $P(A,b)$ to get in the same $\Bbb R^3$ a house (in the sky or underground)
with the same kind of walls, i.e. move them parallely up and down, to the left to the right, in the depth, et caetera.
All possible such movements are parametrized by some $b'$, and if there is a small place for an ant in the house,
well for a point at least, then $b'\in U(A)$.
We have seen above some possibilities for such a $b'$. In some case, the new house has only a thin shape to live in near the wall
$x=3$. In some other cases, we have collapsed the house to one point with all planes delimiting it, the house
can only give place for the point $w'$, resp. $w''$, resp. $w^\dagger$, etc. but to make the house become a point it is
of course enough to move less planes to touch that one point.

We have thus a big freedom to move the roof platform $[1,2]^2\times [2]$ upwards, the ground "face" $[0,3]^2\times[0]$ downwards,
the left wall to the left, etc. from the starting (or some other feasible) $b$,
this means moving from $b$ to an other configuration obtained by going correspondingly in one direction with either $b_1$, $b_2$, ... ,
but when we move in the other direction, there is a point of collapsing the house to something in a plane, where no thin $3$-dimensional ant can live in.
And then, the house vanishes. (Becomes the empty set.)
This was the story of $U(A)$.

(4) What is the support function $u\to h(P,u)$ for some polytope $P$?
We will illustrate this for $P=P(A,b)$ from the OP. By definition, for a general $P$, it is
$$ h(P, u):=\max_{x\in P}x\cdot u\ .
$$
Here, $x\cdot u$ is the scalar product of the vectors $x$ and $u$.
For instance, if $u=(0,0,1)$, the vector pointing to the sky in the lanscape of our house
$P=P(A,b)$, then for some $x=(x_1,x_2,x_3)\in P(A,b)$ the scalar product is $x_3$, its maximal value is $2$,
taken for the points of the roof platform. So having it, and knowing only the house, not the inequalities defining it,
we can already write an inequality, $(0,0,1)\cdot(x_1,x_2,x_3)\le 2$. This means $h(P,(0,0,1))=2$. So (up to norming issues), we
get the corresponding $b$-component from the equation for the line in $A$ which is proportional to this $u$.
Similarly, we may take $u=(0,0,-1)$, then the scalar product is $-x_1$, and the maximum is zero. (On the ground.)
We write this as $h(P, (0,0,-1))=0$.
But there are also some other directions, not the ones perpendicular / normal to the one or other face (e.g. wall, roof, ground, in dimension two,
or vertical line where two walls intersect, in dimension one, or even one corner, dimension zero) of the house.
For instance what happens with $u=(2,3,4)$. Well, we draw the vector as an arrow in the space, and imagine a plane perpendicular / normal to this arrow,
and we push it so far in the direction of the arrow till we have the last point of contact with the house. In this point we get the maximal value for
$h(P, (2,3,4))$. (It is somewhere on the roof. We have to intersect with each plane determined by the faces of the roof.)

(5)
What is the normal cone of a face?
To be specific, let us consider the face $F=[1,2]^2\times[2]$, the roof platform.
A point on this face is of the shape $(x_1, x_2,3)$ with $1\le x_1,x_2\le 2$.

By definition, for this face,
$$
\mathcal N(F,P)
=\{
\ v=(v_1,v_2,v_3)\in\Bbb R^3\ :\
h(P, v):=\max_{y\in P}y\cdot v
\text{ in taken in one and each }x\in F
\
\}
\ .
$$
(It is clear that if some $v$ is in this normal cone, than any positive scalar times $v$ is also in there.)
In other words, some $v$ is in the normal cone $\mathcal N(F,P)$ of this face
if and only if, taking a plane perpendicular / normal on $v$ and pushing it so far as possible in the direction
of the arrow $v$ we touch the house for "the last time" in one and each point in the face $F$.
So $v$ is normal to the plane the face $F$ sits in, and we have to go in its direction (not in the opposite direction).
The immediate guess is that the normal cone contains only the vector $(0,0,1)$, normal to $F$ and making $F$ the
"last face touched in this direction", and those obtained from it by multiplication with a scalar $\ge 0$.
The normal cone is thus a ray, a half-line, starting in $0=(0,0,0)$ and pointing in the direction $(0,0,1)$.

The same can be done with all other $2$-dimensional faces.
To have the confirmation, and the computation for all $2$--faces, the following code is my typist:

Since we can do it, we ask sage for this normal cone.
I am using sage as a typist, there is no need to understand the code.
(But it may be useful if one wants to adapt it to an other situation.)

```<code>for F in P.faces(2):
    # F is a 2-dimensional face of P
    print('Face with vertices {}'.format([tuple(v) for v in F.vertices()]))
    NC = F.normal_cone()    # just ask for it, we know it is a ray starting from the origin
    print(f'Its normal cone has {NC.vertices()} and {NC.rays()}\n')
</code>```

We let this run and get:

```<code>Face with vertices [(0, 3, 1), (0, 0, 1), (1, 1, 2), (1, 2, 2)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (-1, 0, 1),)

Face with vertices [(0, 0, 1), (3, 0, 0), (3, 0, 1), (0, 0, 0)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (0, -1, 0),)

Face with vertices [(0, 0, 1), (2, 1, 2), (1, 1, 2), (3, 0, 1)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (0, -1, 1),)

Face with vertices [(3, 3, 0), (3, 0, 0), (0, 0, 0), (0, 3, 0)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (0, 0, -1),)

Face with vertices [(0, 3, 1), (0, 0, 1), (0, 0, 0), (0, 3, 0)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (-1, 0, 0),)

Face with vertices [(2, 2, 2), (2, 1, 2), (1, 1, 2), (1, 2, 2)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (0, 0, 1),)

Face with vertices [(0, 3, 1), (3, 3, 1), (3, 3, 0), (0, 3, 0)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (0, 1, 0),)

Face with vertices [(0, 3, 1), (2, 2, 2), (3, 3, 1), (1, 2, 2)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (0, 1, 1),)

Face with vertices [(3, 3, 1), (3, 3, 0), (3, 0, 0), (3, 0, 1)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (1, 0, 0),)

Face with vertices [(2, 2, 2), (3, 3, 1), (2, 1, 2), (3, 0, 1)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (1, 0, 1),)
</code>```

Let us consider now a $1$-dimensional face.
To make things complicated, i will take an edge of the roof platform,
$F=[1,2]\times[2]\times[2]$. Is is determined by two vertices, $(1,2,2)$ and $(2,2,2)$.
Now imagine again a vector $v$, together with its normal plane, and we push as far as possible this plane in the $v$ direction,
till it touches for the last time the house. When is this the case for the entire face $F$?
It is clear that the last touch is this whole face $F$, so the "last plane" contains the face $F$, and thus also the line supporting it.
Now we look at all planes "rotating" around this line parametrized by $(t,2,2)=(0,2,2)+t(1,0,0)$ or given by the equations $x_2=x_3=2$.
Among all these planes two of them are "interesting", the plane of the roof platform $R$, which contains the $2$--face $[1,2]^2\times [2]$
and the oblique roof wall $W$ behind, of trapezoidal shape with parallel bases/edges $F=[1,2]\times[2]\times[2]$ and
$F'=[0,3]\times[3]\times[1]$. We start with this plane $W$ and rotate it around the face $F$ till it becomes $R$, such that the house
remains touched only in $F$. It is clear that for each plane in this family the face $F$ is the "last touch".
But we are not interested in these planes, but rather in the vectors $v$ perpendicular to the planes.
Which are these vectors. First of all, they are perpendicular to (the direction of) $F$ (and of the line supporting it),
so to $(1,0,0)$ (see the $t$-parametrization). So for such a $v$, from $v\cdot(1,0,0)=0$ we get $v_1=0$. Then in the other two remained coordinates,
$v_2,v_3$, we move from the perpendicular to the roof $R$ to the perpendicular to the roof wall $W$.
This set is indeed a cone, and we can draw it.

Well, when we draw, we have to draw a cone based in $(0,0)$, but in our thoughts, the $1$-dimensional face $F$ and the
$2$-dimensional faces containing it were geometrically the essence of the construction.
So by convention, and by easy identification of the normal cones, when we draw more of them, it is usual to draw
a translated version of them, namely the version moved to the face.

Here is the code and the shape of the normal cones for the $1$-faces, i assume that P is as above.

```<code>P = Polyhedron(ieqs=b_A)
for F in P.faces(1):
    # F is a 2-dimensional face of P
    print('Face with vertices {}'.format([tuple(v) for v in F.vertices()]))
    NC = F.normal_cone()    # just ask for it, we know it is a ray starting from the origin
    print(f'Its normal cone has\n- {NC.vertices()} and\n- {NC.rays()}\n')
</code>```

It gives:

```<code>Face with vertices [(0, 0, 1), (1, 1, 2)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 1), A ray in the direction (0, -1, 1))

Face with vertices [(0, 3, 1), (0, 0, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 0), A ray in the direction (-1, 0, 1))

Face with vertices [(1, 1, 2), (1, 2, 2)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 1), A ray in the direction (0, 0, 1))

Face with vertices [(0, 3, 1), (1, 2, 2)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 1), A ray in the direction (0, 1, 1))

Face with vertices [(0, 0, 1), (3, 0, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, -1, 0), A ray in the direction (0, -1, 1))

Face with vertices [(3, 0, 0), (0, 0, 0)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, -1, 0), A ray in the direction (0, 0, -1))

Face with vertices [(0, 0, 1), (0, 0, 0)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 0), A ray in the direction (0, -1, 0))

Face with vertices [(3, 0, 0), (3, 0, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, -1, 0), A ray in the direction (1, 0, 0))

Face with vertices [(2, 1, 2), (1, 1, 2)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, -1, 1), A ray in the direction (0, 0, 1))

Face with vertices [(2, 1, 2), (3, 0, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, -1, 1), A ray in the direction (1, 0, 1))

Face with vertices [(0, 0, 0), (0, 3, 0)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 0), A ray in the direction (0, 0, -1))

Face with vertices [(3, 3, 0), (0, 3, 0)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, 0, -1), A ray in the direction (0, 1, 0))

Face with vertices [(3, 3, 0), (3, 0, 0)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, 0, -1), A ray in the direction (1, 0, 0))

Face with vertices [(0, 3, 1), (0, 3, 0)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 0), A ray in the direction (0, 1, 0))

Face with vertices [(2, 2, 2), (1, 2, 2)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, 0, 1), A ray in the direction (0, 1, 1))

Face with vertices [(2, 2, 2), (2, 1, 2)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, 0, 1), A ray in the direction (1, 0, 1))

Face with vertices [(0, 3, 1), (3, 3, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, 1, 0), A ray in the direction (0, 1, 1))

Face with vertices [(3, 3, 1), (3, 3, 0)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, 1, 0), A ray in the direction (1, 0, 0))

Face with vertices [(2, 2, 2), (3, 3, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, 1, 1), A ray in the direction (1, 0, 1))

Face with vertices [(3, 3, 1), (3, 0, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (1, 0, 0), A ray in the direction (1, 0, 1))
</code>```

Similarly, when we consider a vertex, a $0$-dimensional face, then its cone
starts in zero and is delimited by three rays, those normals of the $2$--faces determining it.
For instance:

```<code>sage: P = Polyhedron(ieqs=b_A)
....: for F in P.faces(0):
....:     # F is a 2-dimensional face of P
....:     print('Face with vertices {}'.format([tuple(v) for v in F.vertices()]))
....:     NC = F.normal_cone()    # just ask for it, we know it is a ray starting from the origin
....:     print(f'Its normal cone has\n- {NC.vertices()} and\n- {NC.rays()}\n')
....:     break
....:
Face with vertices [(0, 0, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 0),
   A ray in the direction (-1, 0, 1),
   A ray in the direction (0, -1, 0),
   A ray in the direction (0, -1, 1))
</code>```

(The result was manually rearranged.)

(6) Complete fan.
We collect all the cones above.
(In my convention, the origin is part of these cones. But tomorrow it may be different. One should read the definitions
to see if this is the case, also if the cones in the fan are containing the boundary or not.)

Note that in each direction $v$ we go, there is a last touch of the house when moving a $v$-normal plane (from far away, $-\infty$
in its direction, in the direction of $+\infty$ on it line $\Bbb R v$), this last touch is a vertex, or an edge, or a $2$-face,
at any rate, it comes somewhere in the one or the other cone.

So the union of all these fans is $\Bbb R^3$, the entire space. (Without zero, depending on convention.)
This means that the fan is complete.

(7) Combinatorially equivalent polytopes.
We start with our house, $P(A,b)$.
And move the "walls" ($2$-faces) parallely. (The roof platform, and the other oblique roof parts are now walls.)
This is parametrically done by using some $b'$ instead of $b$.

If the "shape" of the house remains, i.e. we have in the new house $P(A,b')$
the same number of faces, thus in correspondence $W\to W'$, wall $W$ of the house $P(A,b)$ is bijectively associated to wall $W'$
of $P(A,b')$, it is clear that the normal cone for $W$ is the same as for $W'$.
And the same for some $1$--face, i.e. edge $F=W_1\cap W_2$, the normal cone is the same as the one for $F':=W_1'\cap W_2'$ in the new house.

(So please do not move the left wall to the right so far, that it eliminated the roof platform from the house.)

So the normal fans of the old and the new house coincide.
This can be safely taken as definition for the combinatorial equivalence.

(8) How to see the decomposition of the house as "sum" of some pieces?

Let us take the (solid) "roof" $R$ only, it is in the picture Figure 1 from the link the (solid) pyramid, that roof placed on the ground.

Consider also the cube $C=[0,1]^3$. What is $R+C$?

It is the set obtained as the union of all $R+c$ for $c$ in the cube $C$.
So we take $R$ and move it in all $x,y,z$ directions, at most one unit.
Some points will be taken more than once in this movement.
But note that the upper vertex of the pyramid can reach at most the height $2$, and in this case it
can realize exactly the points on the roof platform.

This illustrates that our house is decomposed as a sum in the shape $R+C$.

Now $C$ further decomposes as the sum of elements on the three segments $[0,1]\times[0]\times[0]$, it corresponds to the red segment in
Figre one, but i want the opposite orientation, $[0]\times[0,1]\times[0]$, the green segment with opposite orientation,
and $[0]\times[0]\times[0, 1]$, the blue segment.

(I have to submit... almost too long to be shown, there is a limit of characters, i often hit it. For the rest, how to get the $b$-inequalities, a new post is needed. If there is a real interest, the post will be here as soon as my typing skills allow it.)

Title: Meaning cone, ray, fan for polytopes

Body:
This answer is rather a story, not a pointed answer.
The intuition will be supported by sage code and examples. Sage is free, installed by one click on linux boxes.

The definitions are already given, so there is nothing to do in this regard,
i will only try to digest them, based on the example. And i will  add a 2D-example.

Since $P(A,b)$ looks like a house, i'll call it so.

(1) First of all, we consider the polytope, (solid) polyhedral set from the OP given by
$$
\underbrace{
\begin{bmatrix}
0 & 1 & 1 \\
1 & 0 &1 \\
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1 \\
0 &  0& 1\\
0 &  -1& 0\\
0 &  -1& 1\\
-1 &  0& 0\\
-1 &  0& 1
\end{bmatrix}}_{=:A}
\underbrace{
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}}_{=:v}
\leq
\underbrace{
\begin{bmatrix}
4\\
4\\
3\\
3\\
0\\
2\\
0\\
1\\
0\\
1
\end{bmatrix}}_{=:b}\ .
$$
This is explicitly the set of all $v=(x,y,z)\in\Bbb R^3$ satisfying all the following conditions:
$$
\left\{
\begin{aligned}
y+z  & \le 4\ ,\\
x+z  & \le 4\ ,\\
x    & \le 3\ ,\\
y    & \le 3\ ,\\
-z   & \le 0\ ,\\
 z   & \le 2\ ,\\
-y   & \le 0\ ,\\
-y+z & \le 1\ ,\\
-x   & \le 0\ ,\\
-x+z & \le 1\ .
\end{aligned}
\right.
$$
(It is convenient to swicht tacitly between $v=(x,y,z)$, simpler when writing humanly the inequations,
and $x=(x_1,x_2,x_3)$, the sage preference.)
Note that each equation is an equation determining a half-space,
delimited by a (hyper)plane in the 3D-space.
(In 2D, the space is a plane, a hyperplane is an affine set of codimension one, so it is a line, et caetera.)
Each such half-space is a convex set,
i.e. together with two points in it, the segment joining them is also in it.

The intersection of convex sets is a convex set.
So we expect that the intersection of the half-planes determined by the above inequalities,
i.e. the polytope
$$
P(A,b):=\{\ v\in\Bbb R^d=\Bbb R^3\ :\
\underbrace{Av}_{\in\Bbb R^m=\Bbb R^{10}}
\ge
\underbrace{0 }_{\in\Bbb R^m=\Bbb R^{10}} \ \}
$$
is also a convex set.

Among the above inequalities some are "very simple", for instance from $x\le 3$ and $-x\le 0$
we obtain $x\in[0,3]$. Similar restrictions appear for $y,z$, explicitly
$y\in[0,3]$ and $z\in [0,2]$, so the whole "solid polyhedral set" $P(A,b)$, the whole house is at any rate
contained in the "box" $[0,3]\times[0,3]\times[0,2]$.
The ground of the house is the (solid) square $[0,3]\times[0,3]\times[0]$, and it is a $2$--dimensional (or $1=3-2$ codimensional) face.
The roof has an upper platform, the intersection with the house with the (hyper)plane $z=2$,
and it is the (solid) square $[1, 2]\times[1,2]\times[2]$.
We have side walls, which are also faces of dimension $2$ (i.e. codimension $1$), for instance $[0,3]\times[0]\times[0,1]$.
And on the roof there are also four dimension $2$ faces that let the water flow downwards.

Think about $d=3$ in the example as the dimension of the space containing the polyhedral set.
And $m=10$ is the number of conditions (inequalities) used as plane delimiters for some half-spaces.
The intersection of these half-spaces is our set of interest, it is a convex set, a (solid) polyhedral set. It may be that such a set is either bounded or unbounded. Our set in the example is bounded. So it is determined by its vertices (=extremal points). (In the sense that the convex hull of the set of these vertices is $P(A,b)$. In less mathematical terms, somehow intuitively...) One can put a mass (a real quantity $\ge 0$ in this physical world i accept today) in each such extremal point. Then this physical system has a mass center. All such possible mass centers are the given polyhedral set.

(1-2D) It may be good (at least for drawing) to also have a similar 2D-example.
I will use the following set of inequalities, also allowing only $\le$ (so no $\ge$) while writing them:
$$
\left\{
\begin{aligned}
-x   & \le 0\ ,\\
x    & \le 3\ ,\qquad\text{ so }x\in[0,3]\\
-y   & \le 0\ ,\\
y    & \le 2\ ,\qquad\text{ so }y\in[0,2]\\
-x+y & \le 1\ ,\\
 x+y & \le 4\ .
\end{aligned}
\right.
$$
I hope that this is simpler to spot. From the first four inequalities
we obtain $(x,y)\in[0,3]\times[0,2]$. And the other two conditions
are "biting" from two corners. Is the whole system of inequalities "feasible",
i.e. can we find some point satisfying all these conditions?
Yes, $(0,0)$ is one such point. Some of the conditions are satisfied with the equality.
But we can even do "better", the point $(1,1)$ satisfies them all with strict inequalities.
So it is an "interior point", and we have indeed a $d=2$--dimensional "polyhedron",
well a "solid polygon". The corresponding $A,b$ data can be extracted from:
$$
\underbrace{
\begin{bmatrix}
-1 & 0 \\
1 & 0 \\
0 & -1\\
0 & 1\\
-1 &  1\\
 1 &  1
\end{bmatrix}}_{=:A}
\underbrace{
\begin{bmatrix}
x\\
y
\end{bmatrix}}_{=:v}
\leq
\underbrace{
\begin{bmatrix}
0\\
3\\
0\\
2\\
1\\
4
\end{bmatrix}}_{=:b}\ .
$$

(2-2D)
It may be a good idea to get computer support. For (1-2D) first.
In sage, the above set can be declared as follows. The sage convention wants from us to rewrite the above,
using $\ge$, and moving all terms on one side, so we pass from $Av\le b$ to $0\le b-Av$ and finally to $b-Av\ge 0$:
$$
\underbrace{
\begin{bmatrix}
0 & 1 & 0 \\
3 & -1 & 0 \\
0 & 0 & 1\\
2 & 0 & -1\\
1 & 1 & -1\\
4 & -1 &  -1
\end{bmatrix}}_{=[b, -A}
\begin{bmatrix}
1\\
x\\
y
\end{bmatrix}
\ge
\begin{bmatrix}
0\\
0\\
0\\
0\\
0\\
0
\end{bmatrix}\ .
$$
Now we extract the coefficients and ask for...

b_A = matrix(QQ, 6, 3, [[0,  1,  0],
                        [3, -1,  0],
                        [0,  0,  1],
                        [2,  0, -1],
                        [1,  1, -1],
                        [4, -1, -1],])
P = Polyhedron(ieqs=b_A)

Now we can interactively ask for:

sage: P.inequalities()
(An inequality (-1, -1) x + 4 >= 0,
 An inequality (-1, 0) x + 3 >= 0,
 An inequality (0, -1) x + 2 >= 0,
 An inequality (1, 0) x + 0 >= 0,
 An inequality (0, 1) x + 0 >= 0,
 An inequality (1, -1) x + 1 >= 0)
sage: P.vertices()
(A vertex at (3, 0),
 A vertex at (2, 2),
 A vertex at (3, 1),
 A vertex at (1, 2),
 A vertex at (0, 1),
 A vertex at (0, 0))
sage: P.show()

Here, sage uses $x$ instead of $(x_1,x_2)$.
Yes, the picture shown is

and we have the claimed vertices.
And the show can go on with the polyhedral set in the OP:

We can ask for the faces in each dimension,
and for the corresponding normal cones. But first we will introduce $h(P, u)$ in (4).

(2) The sage code and dialog for the set $P(A,b)$ from the OP is:

A = matrix( QQ, 10, 3,
        [[0, 1, 1], [1, 0, 1], [1, 0, 0], [0, 1, 0], [0, 0, -1],
        [0, 0, 1], [0,-1, 0], [0,-1, 1], [-1,0, 0], [-1,0,  1]])

b = matrix( QQ, 10, 1,
        [4, 4, 3, 3, 0, 2, 0, 1, 0, 1])

b_A = block_matrix([[b, -A]])
P = Polyhedron(ieqs=b_A)

This initializes P in sage.
(And use it also below.)
Then we can interactively ask for:

sage: P.inequalities()
(An inequality (-1, 0, -1) x + 4 >= 0,
 An inequality (-1, 0, 0) x + 3 >= 0,
 An inequality (0, -1, -1) x + 4 >= 0,
 An inequality (0, -1, 0) x + 3 >= 0,
 An inequality (0, 0, -1) x + 2 >= 0,
 An inequality (1, 0, 0) x + 0 >= 0,
 An inequality (0, 0, 1) x + 0 >= 0,
 An inequality (0, 1, -1) x + 1 >= 0,
 An inequality (0, 1, 0) x + 0 >= 0,
 An inequality (1, 0, -1) x + 1 >= 0)
sage: P.vertices()
(A vertex at (0, 3, 1),
 A vertex at (0, 0, 1),
 A vertex at (2, 2, 2),
 A vertex at (3, 3, 1),
 A vertex at (3, 3, 0),
 A vertex at (2, 1, 2),
 A vertex at (1, 1, 2),
 A vertex at (3, 0, 0),
 A vertex at (3, 0, 1),
 A vertex at (1, 2, 2),
 A vertex at (0, 0, 0),
 A vertex at (0, 3, 0))
sage: P.show()
Launched html viewer for Graphics3d Object

Then i get the image in the browser. It is a 3D-image, and we can rotate it.

(3) After these preliminaries, let us understand the set $$U(A)\ .$$

Using the above two examples. In the definition of the (solid) polyhedral set
$P(A,b)$ we use $A$, $b$ as constants, they are previously fixed.
And we have a non-trivial, non-empty set $P(A,b)$ in some cases, the input data
$(A,b)$ is called feasible then, and $\emptyset$ else.

Now we fix $A$, and let $b$ run among all $b\in\Bbb R^m$. Think at the corresponding
sets $P(A,b)$ as "deformations" of the given special situation of some $b_0$.
I will try to illustrate this for the 2D-case.

Imagine we change the two $x$-conditions, and the two $y$-conditions,
and the two remaining conditions "slightly".
$$
\left\{
\begin{aligned}
-x   & \le 0\ ,\\
x    & \le 3\ ,\qquad\text{ so }x\in[0,3]\\
-y   & \le 0\ ,\\
y    & \le 2\ ,\qquad\text{ so }y\in[0,2]\\
-x+y & \le 1\ ,\\
 x+y & \le 4\ .
\end{aligned}
\right.
\qquad\text{ into }\qquad
\left\{
\begin{aligned}
-x   & \le -0.1\ ,\\
x    & \le 3.2\ ,\qquad\text{ so }x\in[0.1\ ,\ 3.2]\\
-y   & \le -0.3\ ,\\
y    & \le 1.89\ ,\qquad\text{ so }y\in[0.3\ ,\ 1.89]\\
-x+y & \le 1.1\ ,\\
 x+y & \le 4.1\ .
\end{aligned}
\right.
$$
Then the corresponding picture is:

What happened? Every delimiting line was "slighlitly" moved parallely to itself
corresponding to the deformations of the corresponding conditions.
Some half-spaces are now "bigger" (w.r.t. the inclusion), other are "smaller".
So chosing other $b$-values means moving the delimiting lines (in 2D, planes in 3D) in a parallel way, and then $P(A,d)$ is the new intersection of half-planes (in 2D, half-spaces in 3D) given by the new fronteers.

Now we are playing the following games starting with the original $P(A,b)$,
$b=(0,3,0,2,1,4)\in\Bbb R^m=\Bbb R^6$, which implements the $m=6$ delimiting fronteers.
We move the $\color{red}0$ from
$b_0=(\color{red}{0},3,0,2,1,4)$ towards $3$, getting closer and closer to $3$.
This moves the delimiting rectangular box
$[\color{red}{0},3]\times[0,2]$ till it becomes smaller and smaller, of the shape
$[\color{red}{3-\epsilon},3]\times[0,2]$, $\epsilon>0$.
Yes, some conditions become superfluous now, but still, we have some thin (solid) trapezium
left as $P(A,b)$. If we move this first component $b_1$ of $b=(b_1,b_2,b_3,b_4,b_5,b_6)\in\Bbb R^6$ beyond $3$, then there is nothing left, $P(A,b)=\emptyset$.

So far we have moved only one boundary condition. We can move two of them, but let us
move all of them, we try to get closer to the situation where the set $P(A,b)$ becomes
smaller and smaller, becomes a point (or something living in less dimensions) and "vanishes".
A way of thinking of a six-dimensional move that does this is as follows.
Let us focus on some fixed point say $(1,1)$. Now we move all boundaries as we would move the lens of a camera (parallely),

lens camera movement

so that we have one point when all boundaries touch $w'=(1,2)$, say.
Which is the value for this $b$-parameter. Well we plug in this value $w'=(x,y)=(1,2)$ into
the function
$v\to Av$, and get $w'\to Aw'=b'=(-1,1,-2,2,1,3)$.

Note that when we plug in the $b$-value as it was in the 2D-example,
$b=(0,3,0,2,1,4)$, we can always slightly change each component, still having a feasible
$P(A,b)$. But for the above $b'$ we have to be careful. There is a strict movement
in each component, either in direction "bigger" or in direction "smaller", but not both.
We can do the same with $w''=(1,1)$ and $w^\dagger=(0,1)$, getting
$b''=Aw''=(-1,1,-1,1,0,2)$ and $b^\dagger=Aw^\dagger=(0,0,-1,1,1,1)$.

And observe that parallely to the fact that $w', w'', w^\dagger$ are on a line, also
the points $b',b'',b^\dagger$ are on a line. And moreover, $w''$ being on the segment
$[w',w^\dagger]$ (mid point even) in $\Bbb R^2$, makes $b''$ sit
on the segment $[b',b^\dagger]$ (also as the mid point) by linearity via $A$.

Now we can say some words on $U(A)$.

It contains all "boundary parameters" $b$, so that the corresponding system of inequalities
has at least a solution.
In the case of the matrix $A$ from the OP, imagine all possibilities to move the
(hyper)planes delimiting the house $P(A,b)$ to get in the same $\Bbb R^3$ a house (in the sky or underground)
with the same kind of walls, i.e. move them parallely up and down, to the left to the right, in the depth, et caetera.
All possible such movements are parametrized by some $b'$, and if there is a small place for an ant in the house,
well for a point at least, then $b'\in U(A)$.
We have seen above some possibilities for such a $b'$. In some case, the new house has only a thin shape to live in near the wall
$x=3$. In some other cases, we have collapsed the house to one point with all planes delimiting it, the house
can only give place for the point $w'$, resp. $w''$, resp. $w^\dagger$, etc. but to make the house become a point it is
of course enough to move less planes to touch that one point.

We have thus a big freedom to move the roof platform $[1,2]^2\times [2]$ upwards, the ground "face" $[0,3]^2\times[0]$ downwards,
the left wall to the left, etc. from the starting (or some other feasible) $b$,
this means moving from $b$ to an other configuration obtained by going correspondingly in one direction with either $b_1$, $b_2$, ... ,
but when we move in the other direction, there is a point of collapsing the house to something in a plane, where no thin $3$-dimensional ant can live in.
And then, the house vanishes. (Becomes the empty set.)
This was the story of $U(A)$.

(4) What is the support function $u\to h(P,u)$ for some polytope $P$?
We will illustrate this for $P=P(A,b)$ from the OP. By definition, for a general $P$, it is
$$ h(P, u):=\max_{x\in P}x\cdot u\ .
$$
Here, $x\cdot u$ is the scalar product of the vectors $x$ and $u$.
For instance, if $u=(0,0,1)$, the vector pointing to the sky in the lanscape of our house
$P=P(A,b)$, then for some $x=(x_1,x_2,x_3)\in P(A,b)$ the scalar product is $x_3$, its maximal value is $2$,
taken for the points of the roof platform. So having it, and knowing only the house, not the inequalities defining it,
we can already write an inequality, $(0,0,1)\cdot(x_1,x_2,x_3)\le 2$. This means $h(P,(0,0,1))=2$. So (up to norming issues), we
get the corresponding $b$-component from the equation for the line in $A$ which is proportional to this $u$.
Similarly, we may take $u=(0,0,-1)$, then the scalar product is $-x_1$, and the maximum is zero. (On the ground.)
We write this as $h(P, (0,0,-1))=0$.
But there are also some other directions, not the ones perpendicular / normal to the one or other face (e.g. wall, roof, ground, in dimension two,
or vertical line where two walls intersect, in dimension one, or even one corner, dimension zero) of the house.
For instance what happens with $u=(2,3,4)$. Well, we draw the vector as an arrow in the space, and imagine a plane perpendicular / normal to this arrow,
and we push it so far in the direction of the arrow till we have the last point of contact with the house. In this point we get the maximal value for
$h(P, (2,3,4))$. (It is somewhere on the roof. We have to intersect with each plane determined by the faces of the roof.)

(5)
What is the normal cone of a face?
To be specific, let us consider the face $F=[1,2]^2\times[2]$, the roof platform.
A point on this face is of the shape $(x_1, x_2,3)$ with $1\le x_1,x_2\le 2$.

By definition, for this face,
$$
\mathcal N(F,P)
=\{
\ v=(v_1,v_2,v_3)\in\Bbb R^3\ :\
h(P, v):=\max_{y\in P}y\cdot v
\text{ in taken in one and each }x\in F
\
\}
\ .
$$
(It is clear that if some $v$ is in this normal cone, than any positive scalar times $v$ is also in there.)
In other words, some $v$ is in the normal cone $\mathcal N(F,P)$ of this face
if and only if, taking a plane perpendicular / normal on $v$ and pushing it so far as possible in the direction
of the arrow $v$ we touch the house for "the last time" in one and each point in the face $F$.
So $v$ is normal to the plane the face $F$ sits in, and we have to go in its direction (not in the opposite direction).
The immediate guess is that the normal cone contains only the vector $(0,0,1)$, normal to $F$ and making $F$ the
"last face touched in this direction", and those obtained from it by multiplication with a scalar $\ge 0$.
The normal cone is thus a ray, a half-line, starting in $0=(0,0,0)$ and pointing in the direction $(0,0,1)$.

The same can be done with all other $2$-dimensional faces.
To have the confirmation, and the computation for all $2$--faces, the following code is my typist:

Since we can do it, we ask sage for this normal cone.
I am using sage as a typist, there is no need to understand the code.
(But it may be useful if one wants to adapt it to an other situation.)

for F in P.faces(2):
    # F is a 2-dimensional face of P
    print('Face with vertices {}'.format([tuple(v) for v in F.vertices()]))
    NC = F.normal_cone()    # just ask for it, we know it is a ray starting from the origin
    print(f'Its normal cone has {NC.vertices()} and {NC.rays()}\n')

We let this run and get:

Face with vertices [(0, 3, 1), (0, 0, 1), (1, 1, 2), (1, 2, 2)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (-1, 0, 1),)

Face with vertices [(0, 0, 1), (3, 0, 0), (3, 0, 1), (0, 0, 0)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (0, -1, 0),)

Face with vertices [(0, 0, 1), (2, 1, 2), (1, 1, 2), (3, 0, 1)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (0, -1, 1),)

Face with vertices [(3, 3, 0), (3, 0, 0), (0, 0, 0), (0, 3, 0)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (0, 0, -1),)

Face with vertices [(0, 3, 1), (0, 0, 1), (0, 0, 0), (0, 3, 0)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (-1, 0, 0),)

Face with vertices [(2, 2, 2), (2, 1, 2), (1, 1, 2), (1, 2, 2)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (0, 0, 1),)

Face with vertices [(0, 3, 1), (3, 3, 1), (3, 3, 0), (0, 3, 0)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (0, 1, 0),)

Face with vertices [(0, 3, 1), (2, 2, 2), (3, 3, 1), (1, 2, 2)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (0, 1, 1),)

Face with vertices [(3, 3, 1), (3, 3, 0), (3, 0, 0), (3, 0, 1)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (1, 0, 0),)

Face with vertices [(2, 2, 2), (3, 3, 1), (2, 1, 2), (3, 0, 1)]
Its normal cone has (A vertex at (0, 0, 0),) and (A ray in the direction (1, 0, 1),)

Let us consider now a $1$-dimensional face.
To make things complicated, i will take an edge of the roof platform,
$F=[1,2]\times[2]\times[2]$. Is is determined by two vertices, $(1,2,2)$ and $(2,2,2)$.
Now imagine again a vector $v$, together with its normal plane, and we push as far as possible this plane in the $v$ direction,
till it touches for the last time the house. When is this the case for the entire face $F$?
It is clear that the last touch is this whole face $F$, so the "last plane" contains the face $F$, and thus also the line supporting it.
Now we look at all planes "rotating" around this line parametrized by $(t,2,2)=(0,2,2)+t(1,0,0)$ or given by the equations $x_2=x_3=2$.
Among all these planes two of them are "interesting", the plane of the roof platform $R$, which contains the $2$--face $[1,2]^2\times [2]$
and the oblique roof wall $W$ behind, of trapezoidal shape with parallel bases/edges $F=[1,2]\times[2]\times[2]$ and
$F'=[0,3]\times[3]\times[1]$. We start with this plane $W$ and rotate it around the face $F$ till it becomes $R$, such that the house
remains touched only in $F$. It is clear that for each plane in this family the face $F$ is the "last touch".
But we are not interested in these planes, but rather in the vectors $v$ perpendicular to the planes.
Which are these vectors. First of all, they are perpendicular to (the direction of) $F$ (and of the line supporting it),
so to $(1,0,0)$ (see the $t$-parametrization). So for such a $v$, from $v\cdot(1,0,0)=0$ we get $v_1=0$. Then in the other two remained coordinates,
$v_2,v_3$, we move from the perpendicular to the roof $R$ to the perpendicular to the roof wall $W$.
This set is indeed a cone, and we can draw it.

Well, when we draw, we have to draw a cone based in $(0,0)$, but in our thoughts, the $1$-dimensional face $F$ and the
$2$-dimensional faces containing it were geometrically the essence of the construction.
So by convention, and by easy identification of the normal cones, when we draw more of them, it is usual to draw
a translated version of them, namely the version moved to the face.

Here is the code and the shape of the normal cones for the $1$-faces, i assume that P is as above.

P = Polyhedron(ieqs=b_A)
for F in P.faces(1):
    # F is a 2-dimensional face of P
    print('Face with vertices {}'.format([tuple(v) for v in F.vertices()]))
    NC = F.normal_cone()    # just ask for it, we know it is a ray starting from the origin
    print(f'Its normal cone has\n- {NC.vertices()} and\n- {NC.rays()}\n')

It gives:

Face with vertices [(0, 0, 1), (1, 1, 2)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 1), A ray in the direction (0, -1, 1))

Face with vertices [(0, 3, 1), (0, 0, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 0), A ray in the direction (-1, 0, 1))

Face with vertices [(1, 1, 2), (1, 2, 2)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 1), A ray in the direction (0, 0, 1))

Face with vertices [(0, 3, 1), (1, 2, 2)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 1), A ray in the direction (0, 1, 1))

Face with vertices [(0, 0, 1), (3, 0, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, -1, 0), A ray in the direction (0, -1, 1))

Face with vertices [(3, 0, 0), (0, 0, 0)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, -1, 0), A ray in the direction (0, 0, -1))

Face with vertices [(0, 0, 1), (0, 0, 0)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 0), A ray in the direction (0, -1, 0))

Face with vertices [(3, 0, 0), (3, 0, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, -1, 0), A ray in the direction (1, 0, 0))

Face with vertices [(2, 1, 2), (1, 1, 2)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, -1, 1), A ray in the direction (0, 0, 1))

Face with vertices [(2, 1, 2), (3, 0, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, -1, 1), A ray in the direction (1, 0, 1))

Face with vertices [(0, 0, 0), (0, 3, 0)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 0), A ray in the direction (0, 0, -1))

Face with vertices [(3, 3, 0), (0, 3, 0)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, 0, -1), A ray in the direction (0, 1, 0))

Face with vertices [(3, 3, 0), (3, 0, 0)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, 0, -1), A ray in the direction (1, 0, 0))

Face with vertices [(0, 3, 1), (0, 3, 0)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 0), A ray in the direction (0, 1, 0))

Face with vertices [(2, 2, 2), (1, 2, 2)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, 0, 1), A ray in the direction (0, 1, 1))

Face with vertices [(2, 2, 2), (2, 1, 2)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, 0, 1), A ray in the direction (1, 0, 1))

Face with vertices [(0, 3, 1), (3, 3, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, 1, 0), A ray in the direction (0, 1, 1))

Face with vertices [(3, 3, 1), (3, 3, 0)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, 1, 0), A ray in the direction (1, 0, 0))

Face with vertices [(2, 2, 2), (3, 3, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (0, 1, 1), A ray in the direction (1, 0, 1))

Face with vertices [(3, 3, 1), (3, 0, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (1, 0, 0), A ray in the direction (1, 0, 1))

Similarly, when we consider a vertex, a $0$-dimensional face, then its cone
starts in zero and is delimited by three rays, those normals of the $2$--faces determining it.
For instance:

sage: P = Polyhedron(ieqs=b_A)
....: for F in P.faces(0):
....:     # F is a 2-dimensional face of P
....:     print('Face with vertices {}'.format([tuple(v) for v in F.vertices()]))
....:     NC = F.normal_cone()    # just ask for it, we know it is a ray starting from the origin
....:     print(f'Its normal cone has\n- {NC.vertices()} and\n- {NC.rays()}\n')
....:     break
....:
Face with vertices [(0, 0, 1)]
Its normal cone has
- (A vertex at (0, 0, 0),) and
- (A ray in the direction (-1, 0, 0),
   A ray in the direction (-1, 0, 1),
   A ray in the direction (0, -1, 0),
   A ray in the direction (0, -1, 1))

(The result was manually rearranged.)

(6) Complete fan.
We collect all the cones above.
(In my convention, the origin is part of these cones. But tomorrow it may be different. One should read the definitions
to see if this is the case, also if the cones in the fan are containing the boundary or not.)

Note that in each direction $v$ we go, there is a last touch of the house when moving a $v$-normal plane (from far away, $-\infty$
in its direction, in the direction of $+\infty$ on it line $\Bbb R v$), this last touch is a vertex, or an edge, or a $2$-face,
at any rate, it comes somewhere in the one or the other cone.

So the union of all these fans is $\Bbb R^3$, the entire space. (Without zero, depending on convention.)
This means that the fan is complete.

(7) Combinatorially equivalent polytopes.
We start with our house, $P(A,b)$.
And move the "walls" ($2$-faces) parallely. (The roof platform, and the other oblique roof parts are now walls.)
This is parametrically done by using some $b'$ instead of $b$.

If the "shape" of the house remains, i.e. we have in the new house $P(A,b')$
the same number of faces, thus in correspondence $W\to W'$, wall $W$ of the house $P(A,b)$ is bijectively associated to wall $W'$
of $P(A,b')$, it is clear that the normal cone for $W$ is the same as for $W'$.
And the same for some $1$--face, i.e. edge $F=W_1\cap W_2$, the normal cone is the same as the one for $F':=W_1'\cap W_2'$ in the new house.

(So please do not move the left wall to the right so far, that it eliminated the roof platform from the house.)

So the normal fans of the old and the new house coincide.
This can be safely taken as definition for the combinatorial equivalence.

(8) How to see the decomposition of the house as "sum" of some pieces?

Let us take the (solid) "roof" $R$ only, it is in the picture Figure 1 from the link the (solid) pyramid, that roof placed on the ground.

Consider also the cube $C=[0,1]^3$. What is $R+C$?

It is the set obtained as the union of all $R+c$ for $c$ in the cube $C$.
So we take $R$ and move it in all $x,y,z$ directions, at most one unit.
Some points will be taken more than once in this movement.
But note that the upper vertex of the pyramid can reach at most the height $2$, and in this case it
can realize exactly the points on the roof platform.

This illustrates that our house is decomposed as a sum in the shape $R+C$.

Now $C$ further decomposes as the sum of elements on the three segments $[0,1]\times[0]\times[0]$, it corresponds to the red segment in
Figre one, but i want the opposite orientation, $[0]\times[0,1]\times[0]$, the green segment with opposite orientation,
and $[0]\times[0]\times[0, 1]$, the blue segment.

(I have to submit... almost too long to be shown, there is a limit of characters, i often hit it. For the rest, how to get the $b$-inequalities, a new post is needed. If there is a real interest, the post will be here as soon as my typing skills allow it.)


问:
Logical consequence using Replacement Axiom
Let $\mathfrak{A}=$ be a countable model of $\mathrm{ZF}$. And $\mathfrak{A}_{A}=(\mathfrak{A},a)_{a\in A}$ is defined to be the model for the language $\{\in\}\cup \{c_{a}:a\in A\}$ by interpreting each new constant $c_{a}$ by the element $a$.

How can we show that for every $\phi$, if $\mathfrak{A}_{A}\vDash \forall y\exists z\exists x[z\not\in y\land\phi(x,z)\land x\in c_{a}]$, then $\exists x\forall y\exists z[z\not\in y\land\phi(x,z)\land x\in c_{a}]$ is also true? In Chang and Keisler's model theory, they says it can be shown by using the axiom of replacement in $\mathrm{ZF}$. But there is no further argument. I have no idea..

答:
Here's the idea of what's going on: The first sentence essentially says "the collection of all $z$ which are related to some $x\in a$ by $\phi$ is a proper class". The second sentence essentially says "there is some $x\in a$ such that the collection of all $z$ which are related to $x$ by $\phi$ is a proper class". Prove the contrapositive: If for every $x\in a$, the collection of all $z$ which are related to $x$ by $\phi$ is a set, then we can gather up all these $z$'s into a single set by Replacement and Union, i.e., the collection of all $z$ which are related to some $x\in a$ by $\phi$ is a set.

More formally: Suppose $$\mathfrak{A}_A\models \forall x\exists y\forall z[(x\in c_a\land \phi(x,z))\rightarrow z\in y].$$ Then for each $x\in a$, there is some set $b_x$ which contains all $z$ such that $\phi(x,z)$. Let $b_x' = \{z\in b_x\mid \phi(x,z)\}$. Then $b_x'$ is a set by Separation, and we can define the function $x\mapsto b_x'$ by $\forall z\,(z\in y \leftrightarrow \phi(x,z))$. By Replacement, there is a set $c = \{b_x'\mid x\in a\}$. Let $d = \bigcup c$. Then $d$ witnesses $$\mathfrak{A}_A\models \exists y \forall z\forall x[(x\in c_a\land \phi(x,z))\rightarrow z\in y].$$


问:
Find the level of confidence given that standard deviation is 12 and there is sample size of 16:
A confidence interval for the mean price shows a $9.30 margin of error. What was the level of confidence?
A. 99.9%
B. 99.8%
C. 99.7%
D. 99.6%
E. 99.5%

I know that z(a/2) (std dev)/(sqrt n)=9.3

z(a/2)=3.1

And the corresponding z score is 0.9987, but I am struggling because of the (a/2), and I feel like an extra step is required.

答:
The correct answer is Option B: 99.8% Below is the explanation in the image.

Standard deviation $\sigma=\$12$, $n=16$, $E=\$9.30$

\begin{align*}
E=Z_{\alpha/2}\cdot\frac\sigma{\sqrt n}
&\implies 9.30=Z_{\alpha/2}\cdot\frac{12}{\sqrt{16}}
\\
&\implies Z_{\alpha/2}=9.30/3=3.10
\end{align*}

As
$$P(2<3.10)=0.9990\implies\alpha/2=1-0.9990=0.001$$
it follows that $\underline{\alpha=0.002}$. So, for the level of confidence $c$
$$c=1-\alpha=1-0.002=0.998\,\therefore\, c=99.8\%$$


问:
How to solve $\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4 d\theta $?
How to solve $\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4 d\theta  $?

I have tried substitution of cosine and sine, dividing and multiplying by Sine and cosine, converting everything to tan. It can probably be solved using beta/gamma function but I can't figure it out.

答: The integral is equal to $$\frac12\int_{-\pi}^\pi\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^2\left(1+\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^4d\theta=-\frac{1}{128}\int_{-\pi}^\pi e^{-6i\theta}(e^{2i\theta}-1)^2(e^{i\theta}+1)^8\,d\theta,$$ and if we imagine the integrand written as a linear combination of exponentials, then all the exponentials vanish after the integration, and only its constant term "survives". It is equal to $$[z^6](z^2-1)^2(z+1)^8=\binom{8}{2}-2\binom{8}{4}+\binom{8}{6}=28-2\cdot70+28=-84,$$ so the integral is equal to $\dfrac{84}{128}\cdot2\pi=\dfrac{21}{16}\pi$.