问:
Improving description of vector systems and aggregation rules for social sciences
I am working on a simple individual based model that aggregates information. I am not a mathematician, but I would like to be as precise as possible with the terminology used to describe the system and the aggregation rules, so I thought this forum might be the appropriate place for advice (if you think there is another forum more suitable, please let me know). I try to define things as follows:

Let $X=\{x_{1}, x_{2},\ldots, x_{O}\}$ be a vector of $O$ opinions, and let $A=\{a_{i}, a_{j},\ldots, a_{N}\}$  be the set of agents in a population.
In the initial state, each agent $a_{n} \in A$ is randomly assigned an opinion $x_{o} \in X$ selected from $X$ without replacement. We use $x_{nr}$ to denote the opinion of $a_{n}$ at round $r$.

At each round:

Each agent expresses one opinion from $X$ (according to the probabilistic model MI). Thus, the vector of opinions $X_{r}$ in the population is a vector of $N$ (one per agent) opinions $(x_{ir}, x_{jr},\ldots, x_{Nr})$, where $x_{ir}$ is the opinion produced by agent $a_{i}$ at round $r$, and so on.

Each agent is defined by a vector of preferences $S$, which is a probability distribution on $X$: that is, at each round, an agent $a_{n}$ has a vector of preferences $S_{n}$ = $(s_{n1r}, s_{n2r},\ldots, s_{nOr})$. $S$ assigns a number between $0$ and $1$ to each opinion in $X$, where $s_{n1r}$ is the value assigned by an agent $a_{n}$ to opinion $x_{1}$ at round $r$, and so on. Thus, $s_{nor}$ is the value assigned by a random agent $a_{n}$ to an opinion $x_{o}$ at round $r$. S evolves according to model MII.

I use two aggregation rules to create an emergent vector called $G$ = $(g_{1r}, g_{2r},\ldots, g_{Or})$, which contains one value for each possible opinion in X.

First rule: At each round $r$, the value $g_{or}$ assigned to a given opinion $x_{o}$ is calculated as an arithmetic mean of the $N$ values $s_{ior}, s_{jor},\ldots, s_{Nor}$, where $s_{ior}$ corresponds to the value assigned by agent $a_{i}$ to opinion $x_{o}$, and $s_{jor}$ corresponds to the value assigned by agent $a_{j}$ to opinion $x_{o}$, and so on. That is, the value of $g_{or}$ at round $r$ is:

\begin{equation} \label{Eq2}
g_{or}= \frac{1}{N} \sum_{nor=1}^{N} s_{nor} = \frac{s_{ior}+s_{jor}+\ldots+s_{Nor}}{N}
\end{equation}

Second rule: At each round $r$, the value $g_{or}$ assigned to a given opinion $x_{o}$ is calculated as the relative frequency $f$ of opinion $x_{o}$ in $X_{r}$.

\begin{equation}
g_{or}=  f_{x_{o}} = \frac{n_{x_{o}}}{O} = \frac{n_{x_{o}}}{\sum_{x_{o}}n_{x_{o}}}
\end{equation}

where n stands for the number of the specific opinions $x_{o}$ found in $X_{r}$.

Probabilistic model MI:

At each round, and for each agent, the model yields a probability distribution of opinions for a given history ($h$) (We use the apostrophe ($\prime$) to denote the probabilistic complement):
\begin{equation} \label{eq:3}
 \Pr(x_{nor}\mid h_{nr})= C f(x_{o}\mid h_{nr}) + C's_{nor}
\end{equation}

where $\Pr(x_{nor}\mid h_{nr})$ corresponds to the probability that an agent $a_{n}$ produces opinion $x_{o}$ at round $r$ given the specific history of agent $a_{n}$ by round $r$. The term $f(x_{o}\mid h_{nr})$ stands for the relative frequency of opinion $x_{o}$ in agent's history. $C$ is a constant that takes values form 0 to 1.

Model MII:

At each round, for a given agent $a_{n}$, each value of its vector of preferences $S$ on $X$ is updated as follows:

\begin{equation} \label{eqvalues}
    s_{nor+1}(s_{nor},x_{nor},g_{or},\epsilon)=
\begin{cases}
    \epsilon \cdot g_{or} + (\epsilon'){ |x_{o}|} & \text{if } x_{o} \in h_{nr} \\
    s_{nor}, &
    \text{otherwise}
\end{cases}
\end{equation}

where $s_{nor+1}$ stands for the value that agent $a_{n}$ gives to opinion $x_{o}$ at round $r+1$,  $\epsilon$ is a constant that takes values form 0 to 1. At each round, $|x_{o}|$ is 1 if the specific opinion $x_{o}$ is produced by the agent in the current round $r$, 0 otherwise.

Questions:

Is this description mathematically acceptable and clear?

Is there a better way to describe this system more precisely mathematically? If so, suggestions and rewriting of the system are greatly appreciated.

Regarding the rules of opinion aggregation, is anyone aware of classical literature where these methods (simple arithmetic mean) and (relative frequency) have been used?

Thank you very much.

答:
Some parts are unclear to me

I cannot confidently say whether these parts are incorrect, or whether I simply don't understand them well enough. Specifically, there are a lot of variables all indexed quite a few times. It's difficult to remember what they all stand for.

I'll first describe the system as I understand it broadly, followed by some notes on the details and mathematical notations.

Broad overview

Now then, as I understand it, you have some agents $A$ and some opinions $X$. At the very start, each agent gets a single opinion (but later agents have a vector of opinions. Is the initial state a one-hot vector for each agent?). The system then progresses and, at each round, each agent broadcasts a single opinion, which is a sample from the distribution from MI. From these opinions we can obtain an aggregate opinion, which can be seen as the opinion of the group, as opposed to the opinion of an individual agent. Then, the opinion distribution of each agent changes, taking into account its own previous opinions as well as the group's. Together, this forms a model for the evolution of opinions.

Mathematical details

As for the mathematics, we have four clear parts: the two aggregation rules and the two models. Let's go over them one by one. First off:

\begin{equation}
g_{or}= \frac{1}{N} \sum_{nor=1}^{N} s_{nor} = \frac{s_{ior}+s_{jor}+\ldots+s_{Nor}}{N}
\end{equation}

As I understand it, the aggregate support for any opinion should be the mean of the support over all agents. In the second part of the equation, you seem to be summing over $nor$, but those are all supposed to be different things. I recommend only summing over agents. In the third part you unnecissarily introduce $i$ and $j$, which I believe should just be 1 and 2. So I recommend:

\begin{equation}
g_{or}= \frac{1}{N} \sum_{i=1}^{N} s_{ior} = \frac{s_{1or}+s_{2or}+\ldots+s_{Nor}}{N}
\end{equation}

For the second aggregate rule, I believe you just want the value for each opinion to be the proportion of that opinion in the expressed opinions of that round. I believe you mean to be dividing by $N$, not $O$, as that will be the amount of opinions expressed (one for each agent). I would also further clarify what you mean by $n_{x_{o}}$.

For the third equation MI, I don't know what you mean by probabilistic complement, but it seems like you simply mean $1-C$. If this is the case I would recommend writing it as such, which prevents the introduction of yet more notation. As I understand it, the probability of an agent emitting an opinion is a linear combination of the proportion of previous emissions of that opinion by that agent, and their current value for that opinion.

Fourth, MII:

\begin{equation}
    s_{nor+1}(s_{nor},x_{nor},g_{or},\epsilon)=
\begin{cases}
    \epsilon \cdot g_{or} + (\epsilon'){ |x_{o}|} & \text{if } x_{o} \in h_{nr} \\
    s_{nor}, &
    \text{otherwise}
\end{cases}
\end{equation}

I think the value for an opinion for a specific agent is a linear combination of the group's value for that opinion and the boolean value indicating whether the opinion was expressed in that round by the agent. Except if the agent has never expressed this opinion, in which case the value remains constant. Here, $|x_{o}|$ is again entirely new notation, which I believe to be confusing. $x_o$ is an opinion, yet the absolute value or norm of this opinion $|x_o|$ is a binary value indicating information about some unnamed agent. I don't have a clear opinion on how to better express this.

In summary, I think some of the notation is unclear, and I think the root cause might be the definition of the vectors. The vector $X$ of opinions leads to some confusion as it is unclear what these opinions are (numbers? Complex objects?) and yet we want to be able to use them in equations.

Other research in this area

I think this would fall under computational sociology. This field is highly multidisciplinary, with social scientists, economists, computer scientists, AI researchers, logicians and even the occasional physicist all publishing in this same field. (Your fondness of functions, vectors and indices leads me to believe that you are more on the computer science side.) In these diverse fields it is imperative to create a shared language, so I would recommend reading some papers within the field and copying their mathematical style. Not just in the method they use to define the objects they work with, but also the specific variables they use to do so. For example, I expect that an opinion vector might be called $O$ instead of $X$, and the amount of opinions to be $m$ instead of $O$, with the amount of agents to be $n$ instead of $N$.

Specifically for the modelling of the evolution of opinions, lots of work has been done. For a good overview, I recommend the table in the appendix of "Modelling Opinion Formation with Physics Tools: Call for Closer Link with Reality", which contains a systematic overview of lots of papers on opinion modelling and their approaches.

Sobkowicz, Pawel. 'Modelling Opinion Formation with Physics Tools: Call for Closer Link with Reality'. Journal of Artificial Societies and Social Simulation 12(1)11 http://jasss.soc.surrey.ac.uk/12/1/11.html.


ContentLicense: CC BY-SA 3.0

Title: How to prove a generalized integral identity

Body:
$$
\int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)(1+t^{2})}dt=-\frac{1}{4}+\frac{\gamma}{2}
$$
where $\gamma$ = Euler Gamma
$$
\int_{0}^{\infty }\frac{t}{( e^{2\pi t}-1)(1+t^{2}) ^{2}}dt=\frac{\pi^2}{24} -\frac{3}{8}
$$
$$
\int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)(
1+t^{2})^{3}}dt=\frac{\pi^2}{96} +\frac{\zeta(3)}{8} -\frac{7}{32}
$$</p>

Body:
By
$$
\int_{0}^{\infty }\frac{\arctan \frac{t}{z}}{e^{2\pi t}-1}dt=\frac{z-z\ln z}{2}-\frac{1}{4}\ln \frac{2\pi }{z}+\frac{1}{2}\ln \Gamma (z)      [{Re}z>0]
$$
we have
$$
\int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left(
z^{2}+t^{2}\right) }dt=\frac{\ln z}{4}-\frac{1}{4z}-\frac{1}{2}\psi \left(
z\right),
$$
$$
\int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left(
z^{2}+t^{2}\right) ^{2}}dt=\frac{1}{4z}\psi ^{\prime }\left( z\right) -\frac{%
1}{4z^{2}}-\frac{1}{8z^{3}} ,
$$
$$
\int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left(
z^{2}+t^{2}\right) ^{3}}dt=\frac{1}{16z^{3}}\psi ^{\prime }\left( z\right) -
\frac{1}{16z^{2}}\psi ^{\prime \prime }\left( z\right) -\frac{1}{8z^{4}}-
\frac{3}{32z^{5}}
$$
and the recurrence formula
$$
I_{p}(z)=\int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left(
z^{2}+t^{2}\right)^p }dt=\frac{-1}{2pz}I_{p-1}^{\prime }(z).
$$
For $\psi ^{(k)}(z)$ we have the following identities
\begin{equation}
\begin{array}{c}
\psi ^{(k)}(n+z)=\psi ^{(k)}(z)+k!(-1)^{k}\sum\limits_{l=1}^{n-1}\frac{1}{%
\left( {l+z}\right) ^{k+1}}, \\
\psi ^{(k)}(z-n)=\psi ^{(k)}(z)+k!\sum\limits_{l=1}^{n}\frac{1}{\left( {l-n-z%
}\right) ^{k+1}}.%
\end{array}
\end{equation}
and
\begin{equation}
\begin{array}{c}
\psi ^{(k)}(0)=\left\{
\begin{array}{c}
-\gamma ,k=0 \\
k{{!(-1)^{k+1}}\zeta (k+1)},k>0%
\end{array}%
\right. , \\
\psi ^{(k)}(\frac{1}{2})=\left\{
\begin{array}{c}
-\gamma -2\ln 2,k=0 \\
k{{!(-1)^{k+1}}}\left( 2^{k+1}-1\right) {\zeta (k+1)},k>0%
\end{array}
\right. , \\
\psi ^{(k)}(\frac{1}{4})=\left\{
\begin{array}{c}
-\gamma -2\ln 2,k=0 \\
-\frac{k{!}}{2\pi }\left( \left( k+2+4^{k+2}\right) {\zeta }%
(k+2)-2\sum\limits_{l=0}^{k/2-1}4^{k-2l}{{{\zeta }(k-2l)\zeta }}(2l+2)\right)
\\
-k{!}2^{k}(2^{k+1}-1){\zeta }(k+1),k\text{ is even}%
\end{array}
\right. , \\
\psi ^{(k)}(\frac{3}{4})=\left\{
\begin{array}{c}
-\gamma -2\ln 2,k=0 \\
\frac{k{!}}{2\pi }\left( \left( k+2+4^{k+2}\right) {\zeta }%
(k+2)-2\sum\limits_{l=0}^{k/2-1}4^{k-2l}{{{\zeta }(k-2l)\zeta }}(2l+2)\right)
\\
-k{!}2^{k}(2^{k+1}-1){\zeta }(k+1),k\text{ is even}%
\end{array}
\right.
\end{array}
\end{equation}


问:
Can $R$ be isomorphic to $R[X]/\langle X^2-a\rangle$ where $a\in R$ and $R$ is an integral domain?
I have proved that if $a=0$, then $R[X]/\langle X^2\rangle\not\cong R$

If they are isomorphic then suppose $f:R[X]/\langle X^2\rangle\to R$ is an isomorphism. Then let $c\in R,c\ne 0$ such that $f(\overline{X})=c$. Again we have $\overline{X}\ \overline{X}=\overline{X^2}=0\implies 0=f(0)=c^2$. As $R$ is integral domain, we have $c=0$, contradiction.

Hence $R[X]/\langle X^2\rangle\not\cong R$.

But I am not sure about if $a\ne 0$. By divison algorithm, we can say that $R[X]/\langle X^2-a\rangle=\{\overline{\alpha X+\beta}|\alpha,\beta\in R\}$.

Can anyone help me in this regard?

答:
It is possible that $R[X]/\langle X^2-a\rangle\cong R$. Take for example $R=k[t]$ and $a=t$. Then $$R[X]/\langle X^2-t\rangle\cong k[\sqrt{t}]\cong k[t]=R$$
This doesn't work if we would ask if $R$ and $R[X]/\langle X^2-a\rangle$ are isomorphic as $R$-algebras (which is in general more interesting). For example $R$ is a free $R$-module of rank $1$ and $R[X]/\langle X^2-a\rangle$ has rank $2$, hence they cannot be isomorphic as modules and therefore also not as $R$-algebras (assuming that $R$ is commutative).  (See also Tsemo's answer for another argument for the $R$-algebra case)

ContentLicense: CC BY-SA 4.0


问:
Extended Bertrand's Boxes Problem - 3 boxes (2 coins per box)
Apologies for the long text, but it is to give context to the exercise.

I am having some trouble understanding an extended version of Bertrand's paradox. It comes in a few questions I have answered most, but the last one, I couldn't determine if my answer is the correct one.

Basically, you have 3 boxes, each containing 2 coins, the first one contains 2 gold coins, the second contains one gold coin and one silver coin, and the third one contains 2 silver coins.

Question 1 : you pick one box at random, and take a coin (and DON'T LOOK) what is the probability that the remaining coin is a gold coin ?  Answer :  all events have the same probability so here it is 1/2

Question 2 : you open your palm, you see that you have a gold coin, what is the probability that the remaining coin is a gold coin ?  Answer : 2/3 (out of the 3 gold coins you could get in your palm only 2 are in a box where the remaining would be a gold coin).

Question 3 : you put the gold coin back, and IN THE SAME BOX, you pick a coin, and don't look again,  what is the probability that the remaining coin is a gold coin ?
Answer : Here you operate under the condition that you have picked a gold coin before, so 2/3 probability of having chosen the first box, and 1/3 probability of having chosen the second box, so the probability that the remaining coin is a gold coin is : $ 1* \frac{2}{3} +\frac{1}{2}*\frac{1}{3} = \frac{5}{6} $

Question 4: you open your palm and you see that you had a gold coin, what is the probability that the remaining coin is a gold coin ? Answer : Here you operate under the condition that you have picked a gold coin before, AND that you picked in the same box a gold coin, in the question 3, having a gold coin remaining, or picking a gold coin are equivalent, so here there can only be a gold coin remaining if you picked from the first box, so the answer is : $ \frac{\frac{2}{3}}{\frac{5}{6}} = \frac{4}{5} $

Question 5 : you get rid of the coin that you picked (there only remains 5 coins now), and pick a box (out of the 3) at random, pick a coin, what is the probability that the remaining coin is a gold coin ?

My Answer is : $  \frac{3}{5} $, basically, I compute conditional on the fact that we eliminated a gold coin from the first box, and conditional on the fact that we eliminated a gold coin from the second box.

First box case, the probability of eliminating that coin is $ \frac{4}{5} $, which as computed before, is the probability of having picked the "eliminated" gold coin from box #1. That said, let $B_{i} , i = 1,2,3$ be the event of picking a coin from box #$i$, we call $O$ the event of having a gold coin remaining in the chosen box, and $G_{i}, i = 1,2$ the event that the eliminated coin is coming from box #$i$ , we have , $P(G_{1}) = \frac{4}{5} $, and $P(G_{2}) = \frac{1}{5} $ and :

$ P(O) = P(O,G_{1}) + P(O,G_{2}) $, with :  $P(O,G_{1}) = P(O|G_{1}) P(G_{1}) = P(O,B1|G_{1}) P(G_{1}) + P(O,B2|G_{1}) P(G_{1})  + P(O,B3|G_{1}) P(G_{1})  =  0 * \frac{4}{5} + \frac{1}{2} * \frac{4}{5}  + 0* \frac{4}{5}  = \frac{2 }{5}$

And :  $P(O,G_{2}) = P(O|G_{2}) P(G_{2}) = P(O,B1|G_{2}) P(G_{2}) + P(O,B2|G_{2}) P(G_{2})  + P(O,B3|G_{2}) P(G_{2})  =  1 * \frac{1}{5} +0*\frac{1}{5}  + 0* \frac{1}{5}  = \frac{1 }{5}$

Total is  $\frac{3}{5}$, but the solution says $\frac{2}{5}$, because there are 2 gold coins out of 5, am I missing something ?

Thank you, and sorry again for the long text.

答:
Your first 4 answers look right to me.

In the last question, you have two possible scenarios to begin with. The first is

```<code> 1     2     3
G X   G S   S S
</code>```

with probability 4/5. In this case the probability that the remaining coin in the box is gold is the average over the three boxes:

$$    \frac{1}{3}(0  + \frac12 + 0)=\frac16$$

The second scenario, with probability 1/5 is

```<code> 1     2     3
G G   X S   S S
</code>```

In this case the probability is

$$    \frac{1}{3}(1  + 0 + 0)=\frac13$$

Then the global probability is

$$ \frac45 \frac16 + \frac15 \frac13 = \frac15$$

So, now we have three answers :-)

Body:
Your first 4 answers look right to me.

In the last question, you have two possible scenarios to begin with. The first is

 1     2     3
G X   G S   S S

with probability 4/5. In this case the probability that the remaining coin in the box is gold is the average over the three boxes:

$$    \frac{1}{3}(0  + \frac12 + 0)=\frac16$$

The second scenario, with probability 1/5 is

 1     2     3
G G   X S   S S

In this case the probability is

$$    \frac{1}{3}(1  + 0 + 0)=\frac13$$

Then the global probability is

$$ \frac45 \frac16 + \frac15 \frac13 = \frac15$$

So, now we have three answers :-)


问:
Proof: if $f(x)\ge c$ on $[a,b]$, then $\int_{a}^{b}{f(x)dx} \ge c(b-a)$,where $a,b,c \in{R}$ and $a<b$, for some real-valued continuous function $f$.
This proof is very simple but I have a little problem:
I start with $f(x) \geq c$, which is obvious. But in order to turn the left side ($f(x)$) into the form of the limit of a Riemann sum, I need to change $x$ to $x_{i}$. How do I achieve this?

答:
$[\inf_{x \in [a,b]} f(x) ] \cdot (b-a)$ is a lower Darboux-sum for the partition $x_0=a,x_1=b$.

Then

$c(b-a) \leq [\inf_{x \in [a,b]} f(x) ] \cdot (b-a) \leq \int_a^b f(x) dx$


问:
A basic doubt on independence of events in probability
Let $X_1$ and $X_2$ be i.i.d random variable. Now, in a book I see the following steps to calculate $P(X_1 < X_2 < x)$

$P(X_1 < X_2 < x)$

= $P(X_1 < x, X_2 < x, X_1 < X_2)$

Now it is argued that $X_1 < x, X_2 < x$ and $X_1 < X_2$ are independent events. Hence we can factor them. I feel that this is wrong. Am I correct ?

答:
Let $A=[X_1\lt x,X_2\lt x]$ and $B=[X_1\lt X_2]$. If $X_1$ and $X_2$ are i.i.d. and with no atoms then indeed $A$ and $B$ are independent.

To show this, consider $C=[X_2\lt X_1]$ and $D=[X_1=X_2]$ and note that, since $(X_1,X_2)$ and $(X_2,X_1)$ have the same distribution,
$$
P[B]=P[C],\qquad P[A\cap B]=P[A\cap C].
$$
Since $(B,C,D)$ is a partition, $2P[B]=1-P[D]$ and $2P[A\cap B]=P[A]-P[A\cap D]$. Thus, $A$ and $B$ are independent if and only if $P[A\cap D]=P[A]P[D]$. If the common distribution of $X_1$ and $X_2$ has no atom, $P[D]=0$ hence the assertion holds..


问:
Partial derivatives and orthogonal relations
I am studying some stuffs relative to PDEs, and appear the following sentences that for are very artificial and I do not understand.

By considering the relation on $x=a$
$$-A (\partial_x f - \partial_x g)  = \partial_t f - \partial_t g + h, $$
by decomposing the relation before into the direction $(\partial_x f - \partial_x g)$ and its orthogonal direction, we conclude that
$$
A=-\frac{(\partial_x f - \partial_x g)}{|(\partial_x f - \partial_x g)|^2} \left(
\partial_t f - \partial_t g + h
\right) \left|_{x=a}\right.
$$
and
$$
(\partial_x f - \partial_x g)^{\bot}
 \left(
\partial_t f - \partial_t g + h
\right) \left|_{x=a}\right. =0.
$$

How is possible deduce the last two equalities ?
Which means the last two equalities ? which method are considering to obtain it ?

答:
$\require{amsmath}$

Call $\pmb{\phi} = \pmb{f} - \pmb{g}$, it is important to realize here that $\pmb{\phi}$ is a vector quantity, so let's go back to your original equation

$$
-A\partial_x\pmb{\phi} = \partial_t \pmb{\phi} + \pmb{h} \tag{1}
$$

Evaluate $\partial_x \pmb{\phi}$ at $x=a$, remember that is a vector, you can multiply (inner product) both sides of equation (1) by that vector

$$
-A(\partial_x \pmb{\phi}) \cdot (\partial_x \pmb{\phi}) = (\partial_x \pmb{\phi})\cdot(\partial_t \pmb{\phi} + \pmb{h})= -A |\partial_x \pmb{\phi}|^2 \tag{2}
$$

from here you can get $A$

$$
A = -\frac{\partial_x \pmb{\phi}}{|\partial_x\pmb{\phi}|^2}\cdot (\partial_t\pmb{\phi} + \pmb{h}) \tag{3}
$$

Let's go back to the statement that $\partial_x \pmb{\phi}$ is a vector, as such you can define a normal vector to it, call it $(\partial_x \pmb{\phi})^\perp$, multiply both sides of equation (1) by this quantity

$$
-A \underbrace{(\partial_x \pmb{\phi})\cdot (\partial_x\pmb{\phi})^\perp}_{=0, ~~\text{because they are perpendicular}} = (\partial_t \pmb{\phi} + h) \cdot (\partial_x \pmb{\phi})^\perp = 0
$$


问:
Are differences of open balls open?
In the metric topology the open sets of a metric space $X$ are union of open balls, but if $r>r'$ then $\forall x_0 \in X$

$B(x_0, r) = B(x_0, r') \cup(B(x_0, r) - B(x_0, r'))$

is open, that in turn means that $B(x_0, r) - B(x_0, r')$ is open.

And that is not an open ball, so why is it open?

答:
An open set can be written as a union of an open set and a non-open set. For example, $(0, 2) = (0, 1) \cup [1, 2)$.

So your argument after "in turn" is incorrect.

Title: Are differences of open balls open?


问:
How can we make four distinct lists of books so that the total number of pages in the books on each list are equal?
I have $15$ books each with between $300$ to $400$ pages. How can we make four distinct lists of books  so that the total number of pages in the books on each list are equal ?

Can anyone please give me a hint to proceed ?

答: Choosing 7 books gives the largest number of lists since the binomial coefficients are unimodal, so the largest is $\binom{15}{7}$ or $\binom{15}{8}$. A list of 7 books has from $2107$ to $2793$ (inclusive) total pages, and so there are less than $700$ possibilities. We can show that $\binom{15}{7}>9\cdot700$ using Stirling's approximation, hence we can find $10$ distinct lists of $7$ books each with the same total number of pages.


答: By no means does that imply that $B(x_0,r)\setminus B(x_0,r')$ is open: see for instance $X=\Bbb R$, $x_0=0$, $r'=1/2$ and $r=1$. $(-1,-1/2]\cup[1/2,1)$ isn't open.

Title: Are differences of open balls open?

Body:
In the metric topology the open sets of a metric space $X$ are union of open balls, but if $r>r'$ then $\forall x_0 \in X$

$B(x_0, r) = B(x_0, r') \cup(B(x_0, r) - B(x_0, r'))$

is open, that in turn means that $B(x_0, r) - B(x_0, r')$ is open.

And that is not an open ball, so why is it open?


问:
What manifolds $M$ have a $CW-$structure so that the $n-$skeleton, $M_n$, is a manifold for all $n$ aswell?
If you have a $CW-$structure on a connected manifold $M$ we obtain a filtration $M_n$ of $M$ where $M_n$ is the $n-$skeleton. If in addition we have that the $M_n$ are also all manifolds (like with the standard $CW-$decomposition of $\mathbb RP^n$ or $\mathbb CP^n$) what can we say about the manifold $M$?

Not all manifolds $M$ have such a CW structure.

Like $S^1 \times S^1 = M$ for example, $M_1$ has to be either a point or a circle since it can't be a wedge of circles since that's not a manifold, it also has to be connected since if $M_1$ is disconnected $M_2$ will always be disconnected but that's impossible since $M_2$ is supposed to be $S^1 \times S^1$, and if we denote by $m_i$ the number of $i-$cells in $M$ we have that

$\text{rank } H_i(M) \leq m_i$ (by a $CW-$homology argument) which is not true for $i = 1$ in $S^1 \times S^1$ since $H_1(S^1 \times S^1) = \mathbb Z^2$ and $m_1 = 1$ or $0$.

How far can we go with this? Is there some easy way to see through invariants like homology or cohomology wether such a filtration $M_n$ exists or not? All of the spaces $S^n, \mathbb RP^n, \mathbb CP^n, \mathbb HP^n$ are examples of such manifolds. Don't know of any else from the top of my head.

答:
I believe you gave a homological classification.

Let's work with $\mathbb{Z}/2$ coefficients; then Poincare duality tells us that if our lowest homology is in degree $k$, to obtain a manifold from $M^n$ by attaching higher degree cells we must be attaching degree $(n+k)$-cells. Again by Poincare duality (obvious in the case $k>1$) is that we can only attach a single $(n+k)$-cell because the differential from the $(n+k)$-cells is homologically trivial since otherwise it would destroy Poincare duality.

Let us call the generator of $H^k(M)$ the element $x$, we first show that there is a m so $x^m=g$ generates $H^n (M)$. We know that there is a $y$ so $xy=g$ and by Poincare duality $x^2 y=g'$ where $g'$ generates $H^{n+k}(M')$. Hence $x^2$ is nontrivial. Repeating this, we know there is a $y'$ so $x^2 y'=g$ and $x^3 y'=g'$, so $x^3$ is nontrivial. Repeating this we conclude that if $|x^{m'}| \leq n$ then $x^{m'+1}$ is nontrivial, so by degree considerations and Poincare duality we know there is a $m$ so $x^m =g$, the generator of $H^n (M)$.

Hence, the cohomology of $M'$ contains a truncated polynomial algebra on $x$ with highest degree element $n+k$. Could there be other elements? No, since for any element $y$, we have some equation $yy'=g=x^m$ and so $yy'x=g'$, but by Poincare duality $yx=0$ if $y$ is not a power of $x$. Of course this is rather sloppy, to write this down formally we need to pick a basis of the cohomology and talk about basis elements $y \neq x^l$, but this is possible because the pairing by the cup product is nonsingular.

Hence the cohomology algebra of $M'$ is a truncated polynomial algebra. Hopf classified up to $\mathbb{Z}/2$ homology equivalence all spaces which have truncated polynomial $\mathbb{Z}/2$ cohomology algebras and they are exactly what you listed. It is also known that there are many manifolds homotopy equivalent but not homeomorphic to projective spaces. Obviously this is only the algebraic obstructions, but I suspect it will not be possible to give an actual local analysis. There are many types of these questions where giving a homological answer is rather easy, but strengthening this to homeomorphism is almost impossible. For example, the topological Poincare conjecture is equivalent to the statement "The suspension of the manifold $M$ is a manifold, if and only if, $M$ is a sphere." Whereas it is extremely easy to see $M$ must be a homology sphere.


问:
Find the lyapunov function to prove the asymptotic stability
Let's the following non linear system:
$\begin{cases}
\dot{x_1}=x_2&\\
\dot{x_2}=-x_1^3&\\
\end{cases}$
determine if the origin is asymptotically stable and in this case if it is globally asymptotically stable.
I have tried to linearized the system but I have obtained two eigenvalues equal to $0$. I have tried to find out an opportune $V$ to prove that the origin is AS but I can't.

Anyone can help me?

答: Hint: Try something with $x_1^4$ instead of $x_1^2$.


问:
Can we apply the component test for the given field?

Consider the following vector field $\mathbf F$, where
$$\mathbf F=x^3\sin\left(\frac1x\right)\mathbf{i}+(2x+e^y)\mathbf{j}+\left(\frac{x^2-y^2}{x^2+y^2}\right)\mathbf{k}$$
defined on its domain. Can you apply the Component Test for Conservative Fields to check whether the vector field is conservative or not. If yes, then use it to check whether $\mathbf F$ is conservative or not. If not, explain.

I understand that to apply the component test for any field, the domain must be simply connected. But for such a field, how can we determine whether the domain is simply connected or not?

答: Well, first of all, you should say what the domain actually is. I assume it's all the purely imaginary quaternions $x \,\boldsymbol i + y \,\boldsymbol j+ z \,\boldsymbol k$ for which the formula is defined. That would be $Im\mathbb H$ minus the plane spanned by $\boldsymbol j$ and $\boldsymbol k$, which is homeomorphic to $\mathbb R^3$ minus the $(y,z)$-plane, which is homotopy equivalent to the $x$-axis minus the origin by projecting, which has two simply connected components. That should be enough for your component test.


问:
Basis vectors question
Suppose the columns of a $5\times 5$ matrix $A$ are a basis for $\mathbb R^5$

If $b$ is in $\mathbb R^5$ then $Ax = b$ is solvable because the basis vectors span $\mathbb R^5$
.

Why is this the case? I don't get how the basis vectors spanning $\mathbb R^5$ says anything about it being solvable.

答: It is solvable by the definition of basis. Multiplying A with x represents a linear combination of these basis vectors. Since $b \in \mathbb{R}^5$ it is possible to linearly combine it from the columns of A. And the matrix vector  product $Ax$ is exactly a linear combination of the columns of A.

Title: Basis vectors question


答: Let $f(x)=c+g(x)$ where $g(x)\ge0$ on $[a,b]$. Therefore $\int_a^bf(x)dx=\int_a^b(c+g(x))dx=c(b-a)+\int_a^bg(x)dx$, Well, if $g(x)$ is non-negative then $\int_a^bg(x)dx$ is non-negative also.

Title: Proof: if $f(x)\ge c$ on $[a,b]$, then $\int_{a}^{b}{f(x)dx} \ge c(b-a)$,where $a,b,c \in{R}$ and $a<b$, for some real-valued continuous function $f$.

Body:
This proof is very simple but I have a little problem:
I start with $f(x) \geq c$, which is obvious. But in order to turn the left side ($f(x)$) into the form of the limit of a Riemann sum, I need to change $x$ to $x_{i}$. How do I achieve this?


问:
Calculate $\iint_D \frac{y}{x}  $
I am trying to calculate $\iint_D \frac{y}{x} $ where D is restricted by 3<_3x+y<_5 and by 4x<_y<_7x. I started by switching variables so that s=3x+y and t=$\frac{y}{x}$. However that gives me a very weird Jacobian determinant; $\frac{x^3}{3x^3+y}$, so maybe I shouldn't set t to $\frac{y}{x}$? Does anyone have any suggestion on how to go about this? Here are my calculations this far:

Thanks in advance!

/Nick

答:
What about defining $s = 3x + y$ and $t = x$? You don't get constant integration limits, but it is simple enough, as the integral becomes

$$
\int_3^5  \int_{s/10}^{s/7} |J| \dfrac{s- 3t}{t} dt ds,
$$

with $J=\begin{pmatrix}  1 & 0 \\ -3 & 1 \\  \end{pmatrix}$. The integral then becomes

$$
\int_3^5 \int_{s/10}^{s/7} \left(\frac st -3 \right) dt\,ds = 8 \left(\log \left(\frac{10}{7}\right)-\frac{9}{70}\right).
$$

obs: Final result computed with wolfram, but the primitive functions are immediate.

Title: Calculate $\iint_D \frac{y}{x}  $


问:
Characterization of linear independence by wedge product
Let $V$ be a vector space of finite dimension. Show that $x_1,...,x_k$ is linearly independent iff  $x_1\wedge ... \wedge x_k \neq 0$.</p>

答:
Hint for one direction: if there is a linear dependence, one of the $x_i$ is a linear combination of the others. Then substitute into $x_1\wedge\cdots \wedge x_k$.

Hint for the other direction:  You can do row operations $x_i\mapsto x_i+rx_j$ for $i\neq j$ without affecting the wedge $x_1\wedge\cdots\wedge x_k$. Similarly you can divide any $x_j$ by a scalar without affecting whether $x_1\wedge\cdots\wedge x_k$ is nonzero. I'm not sure what properties you already know about the wedge. If you know that wedges $e_{i_1}\wedge\cdots \wedge e_{i_k}$, $i_1<i_2<\cdots<i_k$ form a basis for $\wedge^kV$, when $e_i$ is a basis for $V$, then you're home free.


问:
If $V$ is an irreducible representation then is $S(V\otimes V)$?
Let $V$ be an irreducible $FG$-module of dimension $2$. Is $S(V\otimes V)$ irreducible? Why?

$G$ is a finite group.
$F$ is a field, its order is unspecified.

$S(V\otimes V)=\{x \in V\otimes V : xT=x\}$

$A(V\otimes V)=\{x \in V\otimes V : xT=-x\}$

答:
No, there's no reason for this to be true, e.g. there exist groups which only have irreducible representations of dimension at most $2$, but if $\dim V = 2$ then $\dim S^2(V) = 3$, so it can't be irreducible. The smallest such group is $S_3$.

In fact $S^2(V)$ frequently has a trivial direct summand; this happens iff $V$ admits a $G$-invariant symmetric bilinear form iff $V$ is the complexification of a real $G$-representation (see Frobenius-Schur indicator).


问:
Convert cubic spline to Bézier curve and get control points
There is a cubic spline represented by the standard equation:
$$
    f(x) = a + b  (x - x_0) + c  (x - x_0)^2 + d  (x - x_0)^3
$$
and 2 endpoints:

$P_0~ [x, y]$ - starting point

$P_1~ [x, y]$ - end point

Is it possible to convert it to a cubic Bézier curve
and get all control points (CV0 and CV1).

What I was thinking of is to
build a system of 2 parametric equations of the Bezier curve and 2 points
$$
    P_i(t) = (1 - t_i)^3 \cdot P_0 + 3t(1 - t_i)^2 \cdot \text{CV}_0 + 3t^2(1 - t_i) \cdot \text{CV}_1 + t_i^3 \cdot  P_1
$$
I can calculate Pi[x, y] that belongs to a cubic spline. But how to get the t_i value for the current point? Or there is different approach?

Body:
Let us assume a curve defined on $t\in[0,1]$

$$x(t)=a+bt+ct^2+dt^3$$

with the boundary conditions

$$x_0=a,\\x'_0=b,\\x_1=a+b+c+d,\\x'_1=b+2c+3d.$$

From this we draw

$$a=x_0,\\b=x'_0,\\c=3(x_1-x_0)-x'_1-2x'_0,\\d=-2(x_1-x_0)+x'_1+x'_0.$$

Now the Bezier polynomial is

$$w_0(1-t)^3+3w_1(1-t)^2t+3w_2(1-t)t^2+w_3t^3=
\\w_0+3(w_1-w_0)t+3(w_2-2w_1+w_0)t^2+(w_3-3w_2+3w_1-w_0)t^3.$$

We can identify to the standard form and solve the triangular system. This gives

$$w_0=a=x_0,\\w_1=\frac b3+a=\frac{x'_0}3+x_0,\\w_2=\frac c3+\frac{2b}3+a=-\frac{x'_1}3+x_1,\\w_3=d+c+b+a=x_1.$$


答:
There is no asymptotic stability. Taken the system

$$
\cases{
\dot x_1 = x_2\\
\dot x_2 = -x_1^3
}
$$

and multiplying by $x_1^3, x_2$ as

$$
\cases{
x_1^3\dot x_1 = x_1^3x_2\\
x_2\dot x_2 = -x_1^3x_2
}
$$

and adding we have

$$
\frac 12x_1^4+x_2^2 = C
$$

so those orbits characterize a center around the origin.

Title: Find the lyapunov function to prove the asymptotic stability

Body:
Let's the following non linear system:
$\begin{cases}
\dot{x_1}=x_2&\\
\dot{x_2}=-x_1^3&\\
\end{cases}$
determine if the origin is asymptotically stable and in this case if it is globally asymptotically stable.
I have tried to linearized the system but I have obtained two eigenvalues equal to $0$. I have tried to find out an opportune $V$ to prove that the origin is AS but I can't.

Anyone can help me?


问:
Inner products equality for one of vectors fixed
Is is true that
$$ z \in \mathbb{R}^n, \forall u,v \in \mathbb{R}^n, \langle u,z\rangle = \langle v,z\rangle \implies u = v $$ i.e. if two inner products with fixed vector $ z $ are equal so that $ u $ and $ v $ are equals.</p>

答:
No, they may not equal. Since vectors have directions. If the angle between u and z is $\theta_1$
$\langle u, z\rangle=|u||z|cos(\theta_1)$
the angle between v and z is $\theta_2$
$\langle v, z\rangle=|v||z|cos(\theta_2)$

Any two vectors satisfy $|u|cos(\theta_1)=|v|cos(\theta_2)$ will satisfy $\langle u, z\rangle=\langle v, z\rangle$


问:
Why is a triangle a rigid shape?
Triangles are used in structures because they are the most rigid shape. And they are many descriptions of this on network and, if I am not mistaken, this all comes down to the law of cosines.But why this holds? Is it maybe because if we connect three random points on a planes their internal angles will add to half revolution? Is there a more fundamental reason for the rigidness of triangles?

Engineer here, please don't mind the naive question

答:
If three lengths can form a triangle at all, that is, if the longest is less that the sum of the other two, there is only one triangle that they can form. You can uniquely solve for the angles, and the only way for them to change would be for (at least one of) the lengths to change.

If you have four lengths, from which you can assemble a quadrilateral, there are infinitely many sets of angles that could work. Just consider how a cardboard box collapses flat when the top and bottom are removed: you can see that the shape moves smoothly through various angle solution sets with no change in side lengths, starting at 90,90,90,90, and ending at 180,0,180,0.

Title: Why is a triangle a rigid shape?


问:
Evaluate sum of tangents $\tan^4{10°}+\tan^4{50°}+\tan^4{70°}$
How to evaluate $$\tan^4{10°}+\tan^4{50°}+\tan^4{70°} = ?$$

It should equals 59.

But I don't know how?
I also think that we should use power-reduction formulas.

答:
Per the triple-angle identity
$$\tan 3x = \frac{3\tan x-\tan^3x}{1-3\tan^2x}$$
$\tan 10^\circ$, $-\tan50^\circ$ and $\tan 70^\circ$ are the three roots of
$$ \sqrt3 (3t- t^3)=1-3t^2$$

Square to get $3t^6+33t^2 = 27t^4+1$,
and square again to get the cubic equation
$$9t^{12} -531t^8 +1035t^4-1=0$$
in $t^4$, with roots $\tan^410^\circ$, $\tan^450^\circ$ and $\tan^470^\circ$ . Thus
$$\tan^4{10°}+\tan^4{50°}+\tan^4{70°} =\frac{531}9=59$$


问:
Volume between paraboloid and cone

Find the volume blocked between $z^2=x^2+y^2$ and $z=x^2+y^2$.

So one way is to use cylindrical coordinates, which actually worked for me, but I'm wondering why this solution doesn't work:

Our shapes intersect at $(0,0,0)$ and the plane $z=1$.

So we want to find the volume of: $\Omega = \left \{ (x,y,z) : z^2 \leq x^2+y^2 \leq z, 0 \leq z \leq 1\right \}$.

Using Fubini's Theorem: $$V = \int_{\Omega}1dV = \int_0^1 \int_{z^2 \leq x^2+y^2 \leq z} 1 dV^2dz$$

Now we can notice the inner integral is the area between 2 circles, so we have:

$$\int_0^1 \int_{z^2 \leq x^2+y^2 \leq z} 1 dV^2dz = \int_0^1 \pi z^2-\pi z^4dz= \frac{2 \pi}{15}$$

But the answer should be $$\frac{2 \pi}{12}$$

Where am I wrong?

答:
$\displaystyle V = \int_0^1 (\pi z^2-\pi z^4dz) \ $ step has a mistake.

Please note that the area is $(\pi z - \pi z^2)$ and not $(\pi z^2-\pi z^4)$.

So the integral should be $V = \displaystyle \int_0^1 \pi (z - z^2) \ dz = \frac{\pi}{6}$

Title: Volume between paraboloid and cone


答: The maximal domain of the given vector field is $ D= \mathbb R^3\setminus \{(x,y,z):x=0\}$. This is not simply connected, since a closed curve around the $x$-axis can not be contracted to a point. However, if you restrict the domain to a simply connected set, e.g., the half space $D_+ = \{(x,y,z):x>0\} $, then you can apply the component test.

Title: Can we apply the component test for the given field?

Body:
Consider the following vector field $\mathbf F$, where
$$\mathbf F=x^3\sin\left(\frac1x\right)\mathbf{i}+(2x+e^y)\mathbf{j}+\left(\frac{x^2-y^2}{x^2+y^2}\right)\mathbf{k}$$
defined on its domain. Can you apply the Component Test for Conservative Fields to check whether the vector field is conservative or not. If yes, then use it to check whether $\mathbf F$ is conservative or not. If not, explain.

I understand that to apply the component test for any field, the domain must be simply connected. But for such a field, how can we determine whether the domain is simply connected or not?


问:
Proof verification of a sequence limit
Show that $$\lim_{n\rightarrow \infty} \frac{6n-5\cos(n)}{2n-1}=3 $$

We need to find such $n_0\in \text{(Natural numbers)}$ that for every $\epsilon \in \text{(Real numbers)}$ there exists $n_0$ such that $n_0 < n$ for every $n \in \text{(Natural numbers)}$ and that $|a_n - a|<\epsilon$

$$|\frac{6n-5\cos(n)}{2n-1} - 3|<\epsilon$$
$$|\frac{6n-5\cos(n) -6n + 3}{2n-1}|<\epsilon$$
$$|\frac{-5\cos(n)+3}{2n-1}|<\epsilon$$
$$|\frac{-5\cos(n)}{2n-1}|+|\frac{3}{2n-1}|<\epsilon$$
$$ |\frac{-5*1}{2n-1}|+\frac{3}{2n-1}\leq|\frac{-5\cos(n)}{2n-1}|+\frac{3}{2n-1}<\epsilon$$
$$\frac{|-5*1|}{|2n-1|}+\frac{3}{2n-1}< \epsilon$$
$$\frac{5}{2n-1}+\frac{3}{2n-1}< \epsilon$$
$$8<\epsilon*2n-\epsilon$$
$$\frac{8+\epsilon}{2\epsilon} < n \implies n_0 = [\frac{8+\epsilon}{2\epsilon}] + 1$$

答:
Observe that
$$
\left|\frac{3-5\cos(n)}{2n-1}\right|\le \frac{3+|5\cos(n)|}{2n-1}\le \frac{8}{2n-1}\tag{1}
$$
In order the find $n_0$, work with
$$
\frac{8}{2n-1}<\epsilon\tag{2}
$$

Note that (2) implies
$$
\left|\frac{3-5\cos(n)}{2n-1}\right|<\epsilon
$$

Title: Proof verification of a sequence limit


问:
Diamond diagram for Correspondence Theorem
This paragraph appears in Isaacs' Algebra (chapter on homomorphisms).

We comment briefly on the interpretation of the Correspondence theorem in terms of lattice diagrams, at least in the case where $\phi$ is the canonical homomorphism $G\rightarrow G/N$. If we have a lattice diagram for some of the subgroups of $G$, including $N$, then the part of the diagram above $N$ is a valid lattice diagram for $G/N$. In fact, diamonds go over to diamonds, since if $N\subseteq U$ and $N\subseteq V$ and $UV$ is a group, then in $G/N$ we have $(U/N)(V/N)=(UV)/N$.

I get the last part that if $N\subseteq U$ and $N\subseteq V$ and $UV$ is a group, then in $G/N$ we have $(U/N)(V/N)=(UV)/N$, but I don't understand what he means in terms of the diagram and diamonds, and that diamonds go over to diamonds. Maybe this part of algebra is too "abstract" for me to understand...

答:
In the original diagram with $G$ at the top, the groups $UV,U,V$, and $N$ form a diamond with $UV$ at the top and $N$ at the bottom. The map $\varphi$ carries $UV$ to $(U/N)(V/N)$, $U$ to $U/N$, $V$ to $V/N$, and $U\cap V$ to $(U\cap V)/N$, which form a diamond in the lattice diagram for $G/N$. More generally, every diamond in the part of the $G$-lattice lying above $N$ is carried in similar fashion to a diamond in the $(G/N)$-lattice. Writing $W$ for $U\cap V$, we have this correspondence of diamonds:</p>

```<code>                 UV                                 UV/N
                /  \                                /  \
               U    V                             U/N  V/N
                \  /                                \  /
                  W                                  W/N
                  |                                   |
                  N                                   1
</code>```

Body:
In the original diagram with $G$ at the top, the groups $UV,U,V$, and $N$ form a diamond with $UV$ at the top and $N$ at the bottom. The map $\varphi$ carries $UV$ to $(U/N)(V/N)$, $U$ to $U/N$, $V$ to $V/N$, and $U\cap V$ to $(U\cap V)/N$, which form a diamond in the lattice diagram for $G/N$. More generally, every diamond in the part of the $G$-lattice lying above $N$ is carried in similar fashion to a diamond in the $(G/N)$-lattice. Writing $W$ for $U\cap V$, we have this correspondence of diamonds:

                 UV                                 UV/N
                /  \                                /  \
               U    V                             U/N  V/N
                \  /                                \  /
                  W                                  W/N
                  |                                   |
                  N                                   1


问:
Prove that $C=\bigcup_{n=1}^{\infty}I_n$ is a compact subset of a space with a rail metric

Let $I_n$ - line segment connecting the points $(0,0)$ and $\Big( \frac{\cos (\frac{2\pi}{n})}{n}, \frac{\sin (\frac{2\pi}{n})}{n} \Big)$ for $n\in \mathbb N \setminus \{0\}$.
Prove that $C=\bigcup_{n=1}^{\infty}I_n$ is a compact subset of a space with a rail metric.

I think $C$ is a compact in rail metric because we can cover each of the line segments with a railway ball starting at $(0,0)$ and radius $1$. There will be countabless infinite numbers of these balls, because $|C|=|\mathbb N|=\aleph_0$. That's why $C$ is compact.

However I think that my proof is not very formal and results more from graphic intuition. Anyone would like to say how to prove it "elegantly"?

答:
Note that the sequence of end points converges to $(0,0)$ and in any space, a convergent sequence with its limit together form a compact set, as any neighbourhood of the limit already contains all but finitely many terms of the sequence by convergence, and this keeps on being true with the extra line segments because with the endpoint a neighbourhood will also contain the line segment. And the finitely many segments not in the neighbourhood of the origin form a finite union of compact segments so can also by covered by finitely many members of the cover.

Note that your proof is totally wrong: you have to show that every open cover of the set has a finite subcover. Showing that one finite cover exists proves nothing at all.


问:
Do theoreticians check proof of all lemmas and theorems proposed by others while being used in their own proof?
I am trying to figure out how theoreticians work. In my understanding, theoreticians work to find new theorems from previously established theoretical systems. To prove new theorems, they have to read a lot of previous papers to find paths for their own proofs. I am wondering, in the process, do they have to read and understand all proofs of theorems that build their own construction carefully? Simply using an existing lemma or theorem is convenient, but the intricate construction process will be ignored, which may contain valuable hints for the problem of the theoretician himself/herself.

Although not working as a pure theoretician, from my research experience in parallel algorithms of computational physics, I can imagine the difficulty of fully digesting theoretical papers, especially when a systematic understanding of a whole theoretical system is needed to proceed. The general structure of a system is more important than the path to build it, for most scholars. But for theoreticians, the building path may be of more interest.

P.S. I purposely avoided using the word "mathematician". The reason is, I'd like to incorporate researchers not only in pure math, but also other areas, e.g., theoretical computer science, theoretical neuroscience, theoretical physics, etc. I guess anyone who works on establishing, or expanding a theoretical system (in the sense of formal logic) should be counted as working in theory. These researchers may not directly contribute to the mainstream of mathematics, but they also rely on constructing proofs in their career.

My question is mainly from my observation of theoretical computer science, especially from many theoretical works for understanding deep learning. Many excellent young scholars publish several (around five) theoretical papers on top venues every year, full of proofs for lemmas and theorems. I was astonished by their publication speed, and was wondering if they really completed understanding all previous relevant papers to write their own. For me (who is not working directly in theory, but has to read some theoretical papers for implementations), understanding each of these papers thoroughly could take at least weeks (considering reading reference papers). Honestly, I am not a theoretician, but this publication speed really exceeded my imagination, and I am figuring out if I misunderstood theoreticians' work.

答:
It depends, but I'd say: to a large extent yes.

There are certainly some theorems which are simultaneously extremely useful but have very difficult proofs, which get used as blackboxes at least some of the time. At the same time I can confidently say that every result I've used is one whose proof I've understood completely, at least at some point in time (I may have forgotten it since then), and at present at least I have no interest in using a result without fully understanding its proof first. Certainly this is made easier by the nature of my subfield (mathematical logic is relatively young, and reverse math and computable structure theory in particular don't really have "big theorems" as such), but I don't really think it's particularly rare.

There's certainly no rule saying that you can't use existing results without knowing their proofs, and in principle one could build a career based entirely on combining known theorems in new ways (although I don't think this is actually possible in practice). However, I'd strongly warn against that, partly because of the lack of insight, partly because of the drastically increased possibility for error (referees aren't perfect and there are plenty of results which are snappy-sounding but have delicate hypotheses), and partly because to be honest taking too much on faith will make it harder, not easier, to see how to apply or generalize the results you already know.


答:
The idea of your proof is ok, but you should put but implications between inequalities so everyone keeps track of what's going on. Also, you can make it simpler if you observe right from the start that it is equivalent to show that
$$
\lim_{n \to \infty} \dfrac{3- 5 \cos n}{2n -1} = 0.
$$

Also, if you note that $\left| \dfrac{3- 5 \cos n}{2n -1}\right| \leq \frac{8}{2n-1}$, you can pull this off just by showing that $\lim_{n\to \infty} \dfrac{8}{2n-1} = 0$.

Title: Proof verification of a sequence limit

Body:
Show that $$\lim_{n\rightarrow \infty} \frac{6n-5\cos(n)}{2n-1}=3 $$

We need to find such $n_0\in \text{(Natural numbers)}$ that for every $\epsilon \in \text{(Real numbers)}$ there exists $n_0$ such that $n_0 < n$ for every $n \in \text{(Natural numbers)}$ and that $|a_n - a|<\epsilon$

$$|\frac{6n-5\cos(n)}{2n-1} - 3|<\epsilon$$
$$|\frac{6n-5\cos(n) -6n + 3}{2n-1}|<\epsilon$$
$$|\frac{-5\cos(n)+3}{2n-1}|<\epsilon$$
$$|\frac{-5\cos(n)}{2n-1}|+|\frac{3}{2n-1}|<\epsilon$$
$$ |\frac{-5*1}{2n-1}|+\frac{3}{2n-1}\leq|\frac{-5\cos(n)}{2n-1}|+\frac{3}{2n-1}<\epsilon$$
$$\frac{|-5*1|}{|2n-1|}+\frac{3}{2n-1}< \epsilon$$
$$\frac{5}{2n-1}+\frac{3}{2n-1}< \epsilon$$
$$8<\epsilon*2n-\epsilon$$
$$\frac{8+\epsilon}{2\epsilon} < n \implies n_0 = [\frac{8+\epsilon}{2\epsilon}] + 1$$


问:
Volume and orientation application -Calculus of manifolds
If $\omega$ is the volumen element of $V$ determined by $T$ and $\mu$, and $f\colon \mathbb{R}^n\to V$ is an isomorphism such that $f^*T=\langle,\rangle$, and such that $[f(e_1),\ldots,f(e_n)]=\mu$, show that $f^*\omega=det$

I am trying to do exercise 4-4 of calculus of manifolds Spivak, The hypothesis of the exercise is very similar to Theorem 4-2.(Calculus of manifolds - Spivak)

what I have tried to give an ortnormal base $[f(e_1),\ldots,f(e_n)]$, such that the pullback $f^*T(e_i,e_j)=\delta_{ij}$, and $f^*\omega(e_1,\ldots,e_n)=\omega(e_1,\ldots,e_n)$.

They are only ideas, since I have not been able to solve it, any help or suggestion is well received

答:
The thorem you attached isn't really what you should be looking at. This exercise amounts to reading carefully the pages on how an inner product and orientation determines a volume element.

Let us briefly recall how this goes. First a definition: if $W$ is an $n$-dimensional real vector space then a volume element is by definition a non-zero alternating $(0,n)$ tensor  over $W$; i.e an element $\eta \in \mathcal{A}^n(W)\setminus\{0\}$ (recall that this space is $1$-dimensional so we're essentially choosing a basis for this space).

On an arbitrary vector space $W$ we don't have a specific way of deciding which volume element to take: if we choose one, then any non-zero scalar multiple will do. However, if we have an oriented inner-product space $(W,g,\nu)$ then there is a unique choice, because there is a unique $\eta \in \mathcal{A}^n(W)\setminus\{0\}$ such that for some (and hence any) ordered orthonormal basis $\{w_1,\dots, w_n\}$ of $W$ with the correct orientation $[w_1,\dots, w_n]=\nu$, we have
\begin{align}
\eta(w_1,\dots, w_n)&=+1.
\end{align}

I intentionally avoided the notation $V$ in this discussion because for your question we'll be taking $W=\Bbb{R}^n$, $g=\langle \cdot,\cdot\rangle$ the standard inner product, and the usual orientation $\nu=[e_1,\dots, e_n]$.

For your question, you're given an isomorphism $f$ along with a volume element $\omega$ on $V$. You should know (pretty much by basic definition unwinding) that $f^*\omega$ is an alternating $(0,n)$ tensor on $\Bbb{R}^n$, i.e $f^*\omega\in \mathcal{A}^n(\Bbb{R}^n)$. But of course we already know that $\det$ spans this $1$-dimensional vector space (because as mentioned in the comments this is usually how one defines the determinant... or if you take a more abstract definition of the determinant this also follows easily). Hence, there is a $c\in\Bbb{R}$ such that
\begin{align}
f^*\omega &= c\cdot \det
\end{align}
Ideally, we'd like $c=1$, so how do we show that? Simple. Take the standard basis $\{e_1,\dots, e_n\}$ of $\Bbb{R}^n$ then we have
\begin{align}
c&= c\cdot 1\\
&= c\cdot \det(e_1,\dots, e_n)\\
&=(f^*\omega)(e_1,\dots, e_n)\\
&= \omega(f(e_1),\dots, f(e_n))
\end{align}
Now, based on everything I've mentioned above can you justify why this last line equals $1$? (Of course you need to use your assumptions on $f$ at this stage).


问:
Does there exist a way to compute matrix "properties" (eigenvalues, determinant, etc.) from its vectorized representation?
I am wondering whether there are some formulas that compute matrix eigenvalues, determinant, rank, etc. directly from its vectorized representation.

More formally: suppose $\bf S \in \mathbb{S}^N$, i.e. square and symmetric. We can represent any matrix with its vectorized form $\bf s= \operatorname{vec(\bf S)} \in \mathbb{R}^{N^2}$, i.e. by storing all its entries in a vector. At this point we can perform operations involving $\bf S$, by equivalently using $\bf s$. Simple example:
$$\operatorname{trace}(\bf A^\top \bf S) = \sum_{i,j}A_{ij} S_{ij}= \operatorname{vec(\bf A)}^\top \operatorname{vec(\bf S)}= \bf a ^\top \bf s$$

This makes sense clearly.

Question 1: My first question stems from the fact that we have clearly the same information in $\bf s$ as in $\bf S$, we are not losing anything if not the structure. Can we then compute, say, the eigenvalues of $\bf S$ from $\bf s$ without reshaping it? At the end, they are function of the entries of $\bf S$.

Question 2: If the answer to the above question is yes, then another fundamental question is whether we can perform the same operation in the half-vectorization space of the matrix, i.e. the vector storing only its lower (or upper) triangular part, for a symmetric matrix. Again, we are not losing any information.

An affirmative answer to the above question is important, for instance, in algorithms where the use of vectors reduces the dimensionality of the problem at hand, but where matrix properties are necessary.
Thank you

EDIT: to make the context more clear. I am implementing an iterative algorithm working with the independent variables of the matrix only. I need to  perform an SVD to shrink the singular values at every iteration. I am wondering whether I can perform such operation without reshaping the matrix and then reshape back to a vector.

答:
As you note, the same information is there in either representation, so any property can be computed from either representation.

You are asking about whether there are efficiencies to be had choosing one representation over the other. I think the answer is "it depends". In algorithm design and implementation there are always tradeoffs to be made between work done on the data structure and work done in the procedural code. The balance will be different in different computations.


问:
Help with functions
How many functions  $f\ne0$, $f:\mathbb{Z}\to\mathbb{C}$

Periodic in an integer $a$ are there so that

$$f(x+y)=f(x)f(y)$$

What I have sofar is that
$$f(a)=f(2a)=f(a)f(a)$$
So that $f(a)=1$,

Also if I multiply $f(x)$ together with itself $a$ times I get that $$f(x)^a=f(x)f(x)f(x)....f(x)=f(ax)=1$$

So that $f(x)^a=1$

I would appreciate any help

答:
From the equality
$$f(a)=\left(f(a)\right)^2$$
we have $f(a)=0$ or $f(a)=1$.

If $f(a)=0$ and if we suppose $f(1)=w$ then $w^a=0$ and then $w=0$ so $f(n)=0$ for all $n\in\mathbb Z$ is a solution.

If $f(a)=1$ then $w^a=1$ and then $w$ is a root of the unity so $w=e^{2ik\pi/a}$ for $k\in\mathbb Z$ and since $f$ is $a$-periodic so $k$ is  relatively prime to $a$ hence there are $\varphi(a)+1$ solutions

Title: Help with functions


问:
Inverse of an integral function
$$f(x) =\int_0^x  \sin(t)\,dt.$$

I have to compute $(f^{-1})'(0)$ and I know it involves the formula for the derivative of an inverse function $1/f'(f^{-1}(x))$ and that $f' = \sin(x)$ but I do not know how to find $f^{-1}$ of an inverse function.

答:
if you have:
$$y=f^{-1}(x)\Rightarrow x=f(y)$$
$$\frac{dx}{dy}=f'(y)$$
$$\frac{dy}{dx}=\frac{1}{f'(y)}=\frac{1}{f'(f^{-1}(x))}$$
so:
$$\left[f^{-1}\right]'=\frac{1}{f'\circ f^{-1}}$$
Just in case you were interested in where that formula comes from.

$$f(x)=\int_0^x\sin(t)dt=\left[-\cos(t)\right]_0^x=1-\cos(x)$$
$$f'(x)=\sin(x)$$
now to find to inverse function solve the following for $y$:
$$x=1-\cos(y)$$
$$\cos(y)=1-x$$
$$y=\arccos(1-x)=f^{-1}(x)$$
now combine the two and you are done :)

Title: Inverse of an integral function


答:
If you're talking about theoretical mathematics, it depends on the subject.  There are some areas of mathematics that have advanced so quickly and build on such a huge foundation that most of the people working in them do not understand much of the context or proofs for many results they make use of.

To understand every single detail of every lemma or theorem you make use of would require you to read tens of thousands of pages of technical details.  Instead of doing that, many mathematicians find it easier to keep a collection of working examples in their minds.  The examples are not themselves proofs, but in understanding the examples, one is sufficiently convinced of the truth of many theorems which in full generality are difficult to understand why they are true.

If you work in the Langlands program of algebraic number theory, for example, there is a good chance you are using a lot of algebraic geometry, and that is a huge timesink to try and work out the details.  Stacksproject, which includes detailed proofs of a lot of foundational algebraic geometry, is 7000 pages long.  Even the foundational papers in the Langlands program themselves are long and incredibly difficult to read.  Langlands' paper on Eisenstein series is notoriously difficult, and many people who work with Eisenstein series just take those results for granted.

Title: Do theoreticians check proof of all lemmas and theorems proposed by others while being used in their own proof?

Body:
I am trying to figure out how theoreticians work. In my understanding, theoreticians work to find new theorems from previously established theoretical systems. To prove new theorems, they have to read a lot of previous papers to find paths for their own proofs. I am wondering, in the process, do they have to read and understand all proofs of theorems that build their own construction carefully? Simply using an existing lemma or theorem is convenient, but the intricate construction process will be ignored, which may contain valuable hints for the problem of the theoretician himself/herself.

Although not working as a pure theoretician, from my research experience in parallel algorithms of computational physics, I can imagine the difficulty of fully digesting theoretical papers, especially when a systematic understanding of a whole theoretical system is needed to proceed. The general structure of a system is more important than the path to build it, for most scholars. But for theoreticians, the building path may be of more interest.

P.S. I purposely avoided using the word "mathematician". The reason is, I'd like to incorporate researchers not only in pure math, but also other areas, e.g., theoretical computer science, theoretical neuroscience, theoretical physics, etc. I guess anyone who works on establishing, or expanding a theoretical system (in the sense of formal logic) should be counted as working in theory. These researchers may not directly contribute to the mainstream of mathematics, but they also rely on constructing proofs in their career.

My question is mainly from my observation of theoretical computer science, especially from many theoretical works for understanding deep learning. Many excellent young scholars publish several (around five) theoretical papers on top venues every year, full of proofs for lemmas and theorems. I was astonished by their publication speed, and was wondering if they really completed understanding all previous relevant papers to write their own. For me (who is not working directly in theory, but has to read some theoretical papers for implementations), understanding each of these papers thoroughly could take at least weeks (considering reading reference papers). Honestly, I am not a theoretician, but this publication speed really exceeded my imagination, and I am figuring out if I misunderstood theoreticians' work.


问:
Baby Rudin Problem 2.17: Homeomorphism to a Cantor Set?
Problem 2.17 asks us to consider some properties of numbers in $[0,1]$ in base $10$ with decimal expansions consisting only of $4$ and $7$.  Let's call the set of all these numbers $E$.

The question asks us something different, but is this set homeomorphic to the Cantor set?  In particular, if we write each element in the Cantor set in base $3$, and send $0$s to $4$s and $2$s to $7$s, will we get a homeomorphism with $E$?

Also, what about the set of base-$10$ numbers with decimal expansions consisting only of $3$ or more separate digits, for example?  Will these be homeomorphic to the Cantor set, or to something else?

答:
Every metric space that is compact, totally disconnected (the only non-empty connected subsets are singletons) and has no isolated points is homeomorphic to the Cantor set. This is a classical result due to Brouwer.

Your set fulfills these criteria: it's a closed subset of $[0,1]$ (so compact metric), contains no non-trivial intervals (which implies totally disconnected) and every neighbourhood of a point in it contains (many) other points from the set (so no isolated points).

The homeomorphism you suggested also works.

Other applications: the Cantor set $C$ is homeomorphic to $\{0,1\}^\mathbb{N}$, homeomorphic to any of its finite or countable powers, etc. etc.

Also, the final question is answered as well: the same arguments apply and we also get a Cantor set, whether we restrict to 2 or 3 digits (up to 9 I think will work). It's pretty ubiquitous.


问:
Can the exterior algebra be seen as a "model for area/volume"?
I am tying to understand the exterior product. Unfortunately I am not sure what it is supposed to describe. I found a wikipedia article, which I think says that the exterior algebra is sort of a formalisation/model of the notion of (signed) area. https://en.wikipedia.org/wiki/Exterior_algebra

I find this point of view really convincing. However what makes me wonder is that I haven't seen any such explanation before. Neither in the lecture nor in any book I looked at. As a matter of fact all I found regarding the exterior algebra are abstract definitions and not a single note to how one might come up with it or some other "intuitive" explanation.

Is this a valid way of looking at the exterior algebra?

The article then defines the exterior algebra in terms of a quotient.

From what I understand the above construction is/can be viewed as a realisation of a "model/formalisation of area". Is that correct?

Furthermore I would like to make sense of why we "mod out" by $I$.

If we only look at $v\otimes w$ it does not satisfy all of properties 1 to 5. In fact only the first and the 4th property is met only partially. In order to have property 4 we would need $w\otimes w$ to be zero. $w\otimes w=0$ is property 2. I take this as a hint to consider $(V\otimes V)/I$, since for the equivalence class $[v\otimes v]=\pi(v\otimes v)=[0]$, since $v\otimes v\in\ker\pi$. The other properties follow if we look at $\pi(v\otimes w)$.
Hence, it is reasonable to consider it a model of area/volume.

It is all a bit vague, but I hope it still makes some sense. Many thanks in advance.

答:
The first wikipedia excerpt is a good explanation of why the signed area of a parallelogram is an antisymmetric bilinear function of the two vectors giving two adjacent sides. But I consider the second excerpt irrelevant.

First, given a vector space $V$, which we visualize as a set of arrows, we always, in practice, use only the exterior algebra of the dual vector space, $V^*$, and almost never the exterior algebra of $V$ itself.

So here's how it goes: Given an abstract vector space $V$, there is no way to define the area of a parallelogram, if $\dim V=2$, or the volume of a parallelotope in higher dimensions, because there are no units of measurement yet. For example, the length of a vector or the angle between two vectors is undefined.

The first cool thing is that you can define the concept of volume without using lengths or angles. What you do is choose a parallelotope in $V$, spanned by a basis $(e_1,\dots, e_n)$ of $V$, and declare it to have volume $1$. We can now try to define a function $f(v_1, \dots, v_n)$ to be the volume of the parallelotope spanned by the vectors $v_1, \dots, v_n$.

You do this by studying what properties you want a volume function $f$ to have. After some effort (as described in the first wikipedia excerpt), you find that it should be the absolute value of an antisymmetric multilinear function of the vectors $v_1, \dots, v_n$. The absolute value is hard to work with both algebraically and analytically, so we dump it. This requires allowing volume to be negative, so we call it signed volume. So the set of signed volume functions is the set of antisymmetric multilinear functions of an ordered set of $n$ vectors. This space is written as $\bigwedge^nV^*$.

The space $\bigwedge^nV$ is not as useful in geometry or differential geometry and, in general, should be avoided. That also means you can ignore its definition as the quotient of the tensor product by the ideal $I$.

Instead, you should think of $\Lambda^nV^*$ as the subspace of $\bigotimes^nV^*$ (which is the space of all multilinear functions of $n$ vectors) containing only multilinear functions that satisfy the antisymmetry property.

There is also a subtlety in the definition of the wedge product. Let me restrict to $n=2$; it is easily generalized to higher dimensions. Suppose you have $\theta^1, \theta^2 \in V^*$ and want to define their wedge product $\theta^1\wedge\theta^2 \in \bigwedge^2V^*$. What should it be? Well, if the forms are linearly independent and $(e_1,e_2)$ is the basis of $V$ dual to $(\theta^1,\theta^2)$, then you want
$$
(\theta^1\wedge\theta^2)(e_1,e_2) = 1.
$$
So, given an arbitrary ordered pair of vectors $(v_1,v_2)$, we define
$$
(\theta^1\wedge\theta^2)(v_1,v_2) = \theta^1(v_1)\theta^2(v_2)-\theta^1(v_2)\theta^2(v_1).
$$
Note the lack of a normalizing constant like $\frac{1}{2!}$. When $(v_1,v_2)=(e_1,e_2)$, the second term is zero, and we get what we want.

Finally, I advise against ever writing something like $(\theta^1\wedge\theta^2)(e_1\wedge e_2)$. You run into a lot of confusion about when to include a normalizing constant and when not to. Use only the exterior algebra of $V^*$ and avoid using the exterior algebra of $V$ unless there is somehow no way to avoid it. I've never seen any need for $\bigwedge^nV$ in differential geometry.


问:
Given $A \in \mathbb{R}^{n \times n}$ such that $y^T A y \geq 0$ for any $y \in \mathbb{R}^n$, does $x^TAx=0$ imply $Ax=0$?
Given $A \in \mathbb{R}^{n \times n}$ such that $y^T A y \geq 0$ for any $y \in \mathbb{R}^n$, does $x^TAx=0$ imply $Ax=0$?

I think i have proven it in the case where $A$ is symmetric (so $A$ is positive semidefinite), but does it hold in this more general setting?

答:
it is only true for positive definite matricies, whose definition are exactly :
$$x^TAx>0 \text{ if }x\neq 0 $$

This is explained by the fact that if the matrix is only positive semi definite, you only have : $$\lambda\in Sp(A)\Rightarrow \lambda \geq 0 $$


答:
If I may, a few observations:

Merely finding a suitable $n_0$ is not a proof of the claim, keep in mind the logic: for all $\epsilon \gt 0$ there is an $n_0$ such that, for all naturals $n$:

IF $n \gt n_0$ THEN $|a_n -3| \lt \epsilon$.

So what you must do in order to prove the claim is (after finding a suitable $n_0$): assume that $n \gt n_0$ and then deduce that $|a_n -3| \lt \epsilon$.

also:

By triangle inequality we have: $|\frac{-5\cos(n)+3}{2n-1}| \le |\frac{-5\cos(n)}{2n-1}|+|\frac{3}{2n-1}|$ and not the other way around. But it's true that $|\frac{3-5\cos(n)}{2n-1}| \ge |\frac{3}{2n-1}|-|\frac{5\cos(n)}{2n-1}|$.

Lastly: $|-5|$ is bigger than $|-5\cos n|$, not smaller.

Title: Proof verification of a sequence limit


问:
Evaluate $\cos 10^{\circ}+\cot 23^{\circ}-\sin 12^{\circ} + cos 12^{\circ}$
Evaluate

$$\cos 10^{\circ}+\cot 23^{\circ}-\sin 12^{\circ} + cos 12^{\circ}$$

Can you help me please!

Body:
Per the identity $4\cos^3t-3\cos t=\cos 3t$, verify that $\cos20$, $-\cos40$ and $-\cos80$ are the three roots of
$$4x^3-3x=\frac12 $$
Square to get $64x^6-96x^4+36x^2-1=0$, or the cubic equation for $\frac1{x^2}$

$$\frac1{x^6}-\frac{36}{x^4} +\frac{96}{x^2} -64=0$$

Thus
$$
\csc^2 10+\csc^2 50+\csc^2 70= \frac1{\cos^280}+ \frac1{\cos^240}+ \frac1{\cos^220}=36
$$


问:
What is $\sum_{3\leq p\leq x} \pi(\sqrt{p})$?
What is the $\displaystyle \sum_{3\leq p\leq x} \pi(\sqrt{p})$? I thought about starting from  $\displaystyle  2\sum_{3\leq p\leq x}\frac{\sqrt{p}}{\log p}$.

答:
You may want to look at Abel's lemma. For a sequence of complex numbers $\{a_n\}$ and $f(t)$ a differentiable function for $t\geq 0$,
$$\sum_{n\leq x} a_nf(x) = A(x)f(x) - \int_{1}^x A(t)f'(t)dt$$

where $A(t) = \sum_{n\leq x} a_n$. In your case, $a_n = 1$ for all $n$ (so, $A(x) = [x]$ where [$\cdot$] is the greatest integer function of $x$), and $f(t) = \frac{2\sqrt{t}}{\log(t)}$

Of course, this only gives the Asymptotic behavior, because $\pi(t)$ itself is not differentiable for all $t\geq 0$

Edit: Since we know that this is only going to give us the asymptotic behavior anyway (based on our approximation of $\pi$ by $f$) you might as well take $A(x) = x$ to simplify the integral computation.

Edit 2: In this case, your $A(x) = x$ if you are looking for the upper bound, you can bound the second ugly integral by
$$-\int_1^x \frac{t(\log(t)-2)}{2\sqrt{t}\log^2(t)} \leq -\int_{1}^x t\frac{(\log(1)-2)}{t\sqrt{t}}dt$$
Because for $t \geq 1$, $\log(t) \geq \log(1)$ and $\frac{1}{2\sqrt{t}\log^2(t)} \geq \frac{1}{t\sqrt{t}}$ as this plot shows

Edit 3: $-\int_{1}^x t\frac{(\log(1)-2)}{t\sqrt{t}}dt = \int_{1}^x \frac{(2)}{\sqrt{t}}dt = \int_{1}^x 2t^{-1/2} = 4t^{1/2}]_1^x = 4\sqrt{x} - 4$

So, by Abel, $\sum_{n\leq x} a_nf(x) = x\frac{2\sqrt{x}}{\log(x)} + 4\sqrt{x} - 4$


问:
Is there a function that grows asymptotically faster than the Busy Beaver numbers?
Is there a function that grows asymptotically faster than the Busy Beaver numbers?

That is, I know that BB(n)^n grows faster than BB(n), but is there another sequence that can be shown to grow faster than BB(n) without itself referencing the busy beaver numbers in its definition?

答:
Sure. You just need to define the analogue of the busy beaver numbers with respect to a strictly stronger model of computation. For example, let a super Turing machine be a Turing machine equipped with a halting oracle. Super Turing machines are strictly more powerful than Turing machines because they can solve a problem (namely the halting problem) that Turing machines can't. In particular, they can compute the Busy Beaver numbers.

Then define the super Busy Beaver numbers to be the Busy Beaver numbers of super Turing machines. Since super Turing machines can't solve the super halting problem (for the same reason that Turing machines can't solve the halting problem), super Turing machines can't compute the super Busy Beaver numbers, so they must grow much faster than the Busy Beaver numbers.

And of course you can iterate this process. You might enjoy reading Scott Aaronson's Who Can Name the Bigger Number? for more along these lines, and this MO question for an update.


问:
I need to find a sum expressed as a function of $\sum_{n=1}^\infty -\frac{1}{2^n \cdot n}$
$$\sum_{n=1}^\infty -\frac{1}{2^n \cdot  n}$$

I also got the first $4$ terms of this, which are pretty obvious, but I've written them down below anyways. I'm supposed to find a sum expressed as a function based on known Maclaurin series.

I really suck at manipulating known series, or seeing what series I can manipulate to get what I need. Anything that can get me started is appreciated!

First four terms, just for practicing MathJax:
$-\frac{1}{2}-\frac{1}{2^2\cdot 2}-\frac{1}{2^3 \cdot 3}-\frac{1}{2^4 \cdot 4}$

答:
You have, for $|x|<1$ :

$$\int\sum_{n=0}^\infty x^n~dx =\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$$

So you get :

$$\sum_{n=0}^\infty x^n=\frac{1}{1-x} \\
\sum_{n=0}^\infty \frac{x^{n+1}}{n+1}=-\log(1-x)$$

So you can just take $x=\frac{1}{2}$ and make a change of variable $k=n+1$, and you get the result :

$$\sum_{n=1}^\infty -\frac{1}{2^n\times n}=-\log2 $$


问:
Prove that solution of following equations are integer

If $a$ and $b$ are integers and solutions of equation $y-2x-a=0$ and $y^2-xy+x^2-b=0$ are rational, then prove that the solutions are integers.

My Approach:

$y=2x+a$ and I put $y=2x+a$ in second equation from that i obtained a result $3x^2+3xa+(a^2-b)=0$

Then i considered the roots of above equations $\alpha$ and $\beta$

so sum of roots = $\alpha+\beta=-a$ and products of roots $\alpha\cdot\beta=\frac{(a^2-b)}{3}$

Now i can clearly see that roots are rations but cannot prove  that roots are integer.

答:
Write the whole formula out: $$x={-3a\pm\sqrt{12b-3a^2}\over 6}$$

The whole thing is rational, and $a,b$ are integers, means $12b-3a^2=c^2$ for some integer $c$.

First, it's easy to see $c$ is a multiple of $3$ because the left side is a multiple of $3$.

(1) If $a$ is odd, then $12b-3a^2$ is odd so $c$ is odd.

(2) If $a$ is even, then $12b-3a^2$ is even so $c$ is even.

Therefore $a$ and $c$ always has same parity and $-3a\pm c$ is always even and hence is a multiple of $6 \implies x$ is integer. $y$ is therefore integer from the first original equation.


问:
Find all critical points of the function $f(x, y) =\vert x\vert + (x−1)^2−y^2$. Classify them using the second derivative test.
this is what I have so far:

$$\frac{\mathrm{d}f}{\mathrm{d}x} = x/\vert x\vert+2(x-1)=0$$
$$\frac{\mathrm{d}f}{\mathrm{d}y} = -2y=0$$

$y=0$

$x=2$ or $\vert x\vert = 1/2$

What do I do with $\vert x\vert$? Am I on the right track?

Thanks

答:
Hint.

Consider $f(x,y) = \sqrt{x^2}+(x-1)^2-y^2$ then

$$
\nabla^2 f = \left(
\begin{array}{cc}
 2 & 0 \\
 0 & -2 \\
\end{array}
\right)
$$


答:
In other words, is it possible for a linear map to transform a nonzero vector $x$ into a vector that is orthogonal to $x$? Asked this way, the answer is trivially yes (that is, the answer to your initial question is no).

It's even possible for the linear map to transform every nonzero vector into an orthogonal one: for instance in two dimensions, a rotation of angle $\pi/2$.

$$A=\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}$$

Then

$$x^TAx=\begin{pmatrix}x_1 &x_2\end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix} \begin{pmatrix}x_1 \\ x_2\end{pmatrix}=\begin{pmatrix}x_1 &x_2\end{pmatrix}\begin{pmatrix}-x_2 \\ x_1\end{pmatrix}=0$$

Body:
Given $A \in \mathbb{R}^{n \times n}$ such that $y^T A y \geq 0$ for any $y \in \mathbb{R}^n$, does $x^TAx=0$ imply $Ax=0$?

I think i have proven it in the case where $A$ is symmetric (so $A$ is positive semidefinite), but does it hold in this more general setting?


问:
Why is a triangle a rigid shape?
Triangles are used in structures because they are the most rigid shape. And they are many descriptions of this on network and, if I am not mistaken, this all comes down to the law of cosines.But why this holds? Is it maybe because if we connect three random points on a planes their internal angles will add to half revolution? Is there a more fundamental reason for the rigidness of triangles?

Engineer here, please don't mind the naive question

答:
For simplicity, let's think about shapes made up of line segments, which can't bend or stretch in any way, and "joints" between them, which can rotate but can't pull apart. (You can imagine solid metal bars with pins put through them, for example.)

First, imagine a unit-length line segment in a plane. We can say that this segment has three degrees of freedom: you can't fully specify it with less than three numbers (two for the position, one for the rotation).

This is a reasonable definition of "rigidity" within this model. So let's define it: a shape is rigid if it has no more than three degrees of freedom. In other words, it can be moved around as a whole, and it can be rotated as a whole, but there's nothing else that can change about it.

Now imagine two unit-length line segments. There are now six degrees of freedom: each segment needs two numbers to define its position, and one for its rotation.

But if you fasten them together at one end, that takes away two degrees of freedom. The position of the second line segment is now completely determined by the position and rotation of the first one (and the length of the first one, but we've said the segments can't stretch or shrink). So now we only need four numbers: the position of the first one, the rotation of the first one, and the rotation of the second one.

This turns out to be true in general: a joint between two segments, in this model, removes two degrees of freedom. If you want to join more than two segments at a single point, it fixes the position of each one, so a joint between $n$ segments removes $2\times(n-1)$ degrees of freedom. (You can imagine needing to push more pins through the same point to hold more bars in place.)

So now imagine a triangle. It's made up of three line segments (nine degrees of freedom), with three joints (removing six degrees of freedom). Three degrees of freedom remain, so the triangle is rigid.

Then imagine a square (or, more generally, any sort of quadrilateral). It's made up of four line segments (twelve degrees of freedom), with four joints (removing eight degrees of freedom). Four degrees of freedom remain, so it's not rigid: the additional degree of freedom is that it can flex and change its angles.

Finally, imagine adding another segment across the square, connecting two opposite corners together. This is an additional line segment (adding three more degrees of freedom), and two additional joints (removing four more degrees of freedom). This brings the total back down to three, so this structure is rigid.

(This particular explanation isn't at all formal, but this type of analysis can be made rigorous; you just need to be more careful in your definitions.)

Title: Why is a triangle a rigid shape?


答:
You have a rational solution $x=p/q$ where $p, q$ are co-prime and we can take $q >0$ without loss of generality.  Then $p/q$ satisfies
$$
3\left(\frac{p}{q}\right)^2+3\left(\frac{p}{q}\right)a+(a^2-b)=0.
$$
Multiply by $q^2$ to give you
$$
3p^2 +3pqa+(a^2-b)q^2=0.$$
Now $q$ divides the middle and right hand term so $q$ divides $3p^2$.  But $p$ and $q$ are relatively prime, which means $p$ and $q^2$ are also relatively prime and therefore by Euclid's lemma $q$ must divide $3$.  Accordingly $q=1$ or $3$.

If $q=3$, the quadratic becomes
$$
\frac{p^2}{3}+ap+(a^2-b) = 0
$$
implying $p^2/3$ is an integer and so $3$ now divides $p$ which contradicts the requirement $p$ is co-prime to $q = 3$.  Thus $q \neq 3$ and so $q=1$.  Thus, $x$ is an integer and so is $y$.

Title: Prove that solution of following equations are integer

Body:
If $a$ and $b$ are integers and solutions of equation $y-2x-a=0$ and $y^2-xy+x^2-b=0$ are rational, then prove that the solutions are integers.

My Approach:

$y=2x+a$ and I put $y=2x+a$ in second equation from that i obtained a result $3x^2+3xa+(a^2-b)=0$

Then i considered the roots of above equations $\alpha$ and $\beta$

so sum of roots = $\alpha+\beta=-a$ and products of roots $\alpha\cdot\beta=\frac{(a^2-b)}{3}$

Now i can clearly see that roots are rations but cannot prove  that roots are integer.


问:
Sufficient condition for measurability
Let $f$ be a real-valued function defined on a measurable set $E$. Show that the measurability of set {$x$ : $f(x)$ = c} , where c $\in$ $\Bbb R$, is not sufficient for $f$ to be measurable.

答:
Let $E=[0,1]$ and let $\mathcal{A}$ be the sigma-algebra of the subsets of $E$ that are either countable or their complement in $[0,1]$ is countable. Take the function $f:E\to\mathbb{R}$ to be $f(x)=x$. Then $f$ is non-measurable: Indeed, if $c\in(0,1)$ then $f^{-1}(-\infty,c)=[0,c)$ which is not measurable (not an element of $\mathcal{A}$). However, $\{x\in E: f(x)=c\}=\{c\}$ when $c\in[0,1]$ and $\{x\in E: f(x)=c\}=\emptyset$ when $c\not\in [0,1]$ which is measurable in any case (because $\mathcal{A}$ contains singletons and the empty set).

Edit

A second example, just in case OP refers to Lebesgue measurability only.

Take $E=\mathbb{R}$ and let $N$ be a non-measuralbe subset of $[0,1]$. Such subsets exist, due to the axiom of choice. Now take $f(x)=x+\chi_N(x)$, where $\chi_N$ is the characteristic function of $x$, i.e. $\chi_N(x)=1$ when $x\in N$ and $\chi_N(x)=0$ when $x\not\in N$. The function $\chi_N$ is a non-measurable function obviously. Now $f$ is non-measurable: the identity function $x\mapsto x$ is of course Lebesgue measurable and it is very easy to prove that the sum of a measurable and a non-measurable function is non-measurable.

On the other hand, if $c\in\mathbb{R}$ then $\{x\in\mathbb{R}: f(x)=c\}=\{x\in N: x+1=c\}\cup\{x\in \mathbb{R}\setminus N: x=c\}$. Now we distinguish cases:

Case $1$: if $c\in \mathbb{R}\setminus N$ and $c-1\in N$, then $\{x\in \mathbb{R}: f(x)=c\}=\{c\}\cup\{c-1\}$.

Case $2$: if $c\in \mathbb{R}\setminus N$ and $c-1\in \mathbb{R}\setminus N$ then $\{x\in \mathbb{R}: f(x)=c\}=\{c\}$

Case $3$: if $c\in N$ and $c-1\in N$ then $\{x\in \mathbb{R}: f(x)=c\}=\{c-1\}$.

Case $4$ if $c\in N$ and $c-1\in\mathbb{R}\setminus N$ then $\{x\in \mathbb{R}: f(x)=c\}=\emptyset$.

So in any case, i.e. for any $c\in\mathbb{R}$, this is a Lebesgue-measurable set.

Title: Sufficient condition for measurability


答:
$y-2x=a$   $\Rightarrow  y^{2}-4xy+4y^{2}=a^{2}$

$4b=4y^{2}-4xy+4x^{2}$

Subtracting will give, $3y^{2}=4b-a^{2}$.

$y$ can't have any integer apart from $1$ as a denominator, hence $y$ is an integer. Write down the formula for quadratic formula for $x$ and then apply this and the answer will follow.


问:
Helicoid in $\mathbb{R}^4$
I'm considering a parametric equation of helicoid in $\mathbb{R}^4$ is $X(u,v)=(u \cos v, u \sin v,v,v).$
Now I can calculate two unit tangent vectors $t_1=\dfrac{X_u}{\vert X_u\vert}=(\cos u, \sin u, 0,0) $ and
$t_2=\dfrac{X_v}{\vert X_v \vert}=\dfrac{1}{\sqrt{u^2+2}}(-u \sin v, u \cos v, 1, 1)$. The problem is how to find 2 unit normal vectors $n_1, n_2$? I tried and find that $n_1= \dfrac{(0,0,1.-1)}{\sqrt{2}}$ is ok, but the second unit normal vector is more difficult.

答: Of course, this is just "the usual" helicoid sitting in the hyperplane $x_3=x_4$. The normal vector you found is the normal vector to that hyperplane. If you review the computation for "the usual" helicoid in $\Bbb R^3$, you have the (non-unit length) normal $(\sin v,-\cos v, u)$. So you might guess the vector $(\sin v,-\cos v, u/2, u/2)$. Note that this is orthogonal to both $t_j$ and to $n_1$.

Title: Helicoid in $\mathbb{R}^4$


问:
Is it possible with Linear Regression that the p-value is low, but the coefficient is high when the explanatory variables are standardized?
Let's suppose that the explanatory variables are standardized, so they are on the same scale. In this case, if I understand correctly, the coefficients of each feature determine whether the attribute is relevant or not.
Is my conclusion true? Or can it happen that a coefficient of a variable is high but based on its p-value the variable is not relevant?</p>

答:
It can get much more complicated than that in cases where you almost have colinearity.  There are probably better examples, but try this in R with some numbers I just invented

```<code>> x1 <- c(-1, -0.51, -0.11, 0.11, 0.52, 1.0)
> x2 <- c(-1, -0.51, -0.12, 0.12, 0.51, 1.0)
> y <-  c(-1, -0.50, -0.10, 0.10, 0.50, 1.0)
> fit <- lm(y ~ x1+x2)
> summary(fit)

Call:
lm(formula = y ~ x1 + x2)

Residuals:
        1         2         3         4         5         6
-0.005201  0.008268  0.009765 -0.006011 -0.015776  0.008955

Coefficients:
             Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.001877   0.005720  -0.328    0.764
x1           1.126269   0.812855   1.386    0.260
x2          -0.133347   0.813761  -0.164    0.880

Residual standard error: 0.01361 on 3 degrees of freedom
Multiple R-squared:  0.9998,    Adjusted R-squared:  0.9996
F-statistic:  6798 on 2 and 3 DF,  p-value: 3.277e-06
</code>```

Each of $x_1$ and $x_2$ is very close to $y$ and on virtually the same scale, so on its own would make a very good predictor of $y$ with a coefficient very near $1$.

But crudely looking at the $p$ values on the right might suggest neither coefficient is significantly different from zero in the multiple regression, while the sign on the coefficient of $x_2$ is in fact negative when its relationship with $y$ looks obviously positive.

Body:
It can get much more complicated than that in cases where you almost have colinearity.  There are probably better examples, but try this in R with some numbers I just invented

> x1 <- c(-1, -0.51, -0.11, 0.11, 0.52, 1.0)
> x2 <- c(-1, -0.51, -0.12, 0.12, 0.51, 1.0)
> y <-  c(-1, -0.50, -0.10, 0.10, 0.50, 1.0)
> fit <- lm(y ~ x1+x2)
> summary(fit)

Call:
lm(formula = y ~ x1 + x2)

Residuals:
        1         2         3         4         5         6
-0.005201  0.008268  0.009765 -0.006011 -0.015776  0.008955

Coefficients:
             Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.001877   0.005720  -0.328    0.764
x1           1.126269   0.812855   1.386    0.260
x2          -0.133347   0.813761  -0.164    0.880

Residual standard error: 0.01361 on 3 degrees of freedom
Multiple R-squared:  0.9998,    Adjusted R-squared:  0.9996
F-statistic:  6798 on 2 and 3 DF,  p-value: 3.277e-06

Each of $x_1$ and $x_2$ is very close to $y$ and on virtually the same scale, so on its own would make a very good predictor of $y$ with a coefficient very near $1$.

But crudely looking at the $p$ values on the right might suggest neither coefficient is significantly different from zero in the multiple regression, while the sign on the coefficient of $x_2$ is in fact negative when its relationship with $y$ looks obviously positive.


答:
Let $S$ be a non-measurable subset of $\mathbb{R}$ and let $S^c=\mathbb{R}{\setminus}S$.

Let $g$ be an injective function from $S$ into $(0,1)$, and let $h$ be an injective function from $S^c$ into $(-1,0)$.

Now define $f:\mathbb{R}\to\mathbb{R}$ by
$$
f(x)
=
\begin{cases}
g(x)&\text{if}\;x\in S\\[4pt]
h(x)&\text{if}\;x\not\in S\\
\end{cases}
$$
Then $f$ is injective, hence the inverse image of any singleton is measurable (since it has cardinality at most one).

But $f$ is not measurable since the inverse image of $(0,1)$ is $S$.

Title: Sufficient condition for measurability

Body: Let $f$ be a real-valued function defined on a measurable set $E$. Show that the measurability of set {$x$ : $f(x)$ = c} , where c $\in$ $\Bbb R$, is not sufficient for $f$ to be measurable.


问:
Find a measurable function such that $f(x)\le \alpha$ for $x\in E_\alpha$
Theorem: Given $\{ E_\alpha \}_{\alpha \in \mathbb{R}}\subset \mathcal{M}$ such that $E_\alpha \subset E_\beta$ for $\alpha < \beta$. We have also that $\bigcup_{\alpha \in \mathbb{R}}E_\alpha=X$ and $\bigcap_{\alpha \in \mathbb{R}}E_\alpha=\emptyset$. Then there exists a measurable function $f:X\rightarrow \mathbb{R}$ such that $f(x)\le \alpha$ if $x\in E_\alpha$ and $f(x)\ge \alpha$ if $x\notin E_\alpha$.

I proved this by defining, $$f(x) = \inf \{ q\in \mathbb{Q} : x\in E_q \}$$

Then it is easy to show that $f$ has the desired properties and $f(x) = \inf_{q \in \mathbb{Q}} f_q(x)$. Where $f_q(x)$ is $q$ if $x\in E_q$ and is $\infty$ otherwise. Then since each $f_q$ is measurable we have that $f$ is measurable.

I don't feel very confident about this proof because I don't use the fact that $\bigcap_{\alpha \in \mathbb{R}}E_\alpha=\emptyset$ and the book hints that I should use the fact that $g:X\rightarrow \overline{\mathbb{R}}$ is measurable if $g^{-1} \left( (r,\infty] \right)\in\mathcal{M}$ for $r\in \mathbb{Q}$. And I don't use this fact, so I'd appreciate if someone could tell me if I'm making a bad mistake here.

答:
Stuart, since your question calls for a $f:X\rightarrow \mathbb{R}$, you need to ensure that $f$, as you have defined it, is never $\pm \infty$.  This is where your two conditions come in. ( $\cup_{\alpha} E_\alpha=X , \cap_{\alpha} E_\alpha = \emptyset$)

To use the fact about $g^{-1}((r,\infty])$ being measurable for $r \in \mathbb{Q}$, just show that $f$ satisfies that condition, and then show that $f$ takes only real values.


问:
A formula satisfiable from an infinite sequence of formulas

Let $\lbrace\varphi_n\rbrace_{n\geq 1}$ be a sequence of semantically inequivalent formulas s.t. $\vDash \varphi_{n+1} \to \varphi_{n}$. Is there a formula $\psi$ s.t. $(\lbrace\varphi_n\rbrace_{n\geq 1}\vDash \psi) \land (\forall n\geq 1.\psi\vDash \varphi_{n})$?

I think there is no such $\psi$ but I can't formulate the proof quite yet.
By assuming such $\psi$ exists and using compactness theorem we can have a finite subset $\Delta \subset \lbrace\varphi_n\rbrace_{n\geq 1}$ s.t. $\forall n\geq 1.\Delta \vDash \varphi_{n}$ but I can't see where does a contradiction suppose to come from.

答:
I believe this counts as a counterexample, although things will probably become clearer if the original poster clarifies some minor points I asked about in the comments.

So, take the set of all propositional variables to be $\{p_{n} : n\in\mathbb{N}\setminus\{0\}\}$ and define
$$\varphi_{n}=\bigwedge_{i=1}^{n}p_{i};$$
they are certainly inequivalent, and also satisfy $\vDash\varphi_{n+1}\rightarrow\varphi_{n}$ (actually $\varphi_{m}\vDash\varphi_{n}$ for every $m\geq n$). But there can not be any formula $\psi$ satisfying the desired properties: if $\psi\vDash\varphi_{n}$, for every $n\geq 1$, let $p_{i_{1}}, ... , p_{i_{n}}$ be the variables in $\psi$; if $\psi$ is not a contradiction, take a valuation $\nu$ such that $\nu(\psi)=1$ and define $\nu'$ such that $\nu'(p_{i})=\nu(p_{i})$, for $i\leq M=\max\{i_{1}, ... , i_{n}\}$, and $\nu'(p_{i})=0$ for $i>M$. Then $\nu'(\psi)=1$ but $\nu(\varphi_{i})=0$, for $i>M$, so we must have that $\psi$ is a contradiction.

But then you cannot have $\{\varphi_{n} : n\in\mathbb{N}\setminus\{0\}\}\vDash\psi$: take the valuation $\nu$ such that $\nu(p_{i})=1$, for every $i\in\mathbb{N}\setminus\{0\}$, and it is clear that $\nu(\psi)=0$ but $\nu(\varphi_{i})=1$, for every $i\geq 1$.


问:
A standard variant of the Lefschetz principle
I am studying this paper, and I am a little confused about the terminology "a standard variant of the Lefschetz principle".

Assume I have a smooth and proper morphism of schemes $X\longrightarrow\operatorname{Spec}(k)$, where $k$ is a field. Assume that $X$ is connected. The author writes:

A standard variant of the ''Lefschetz principle'' guarantees the
existence of the following very noncanonical data: a finitely
generated $\mathbb{Z}$-Algebra $A$, a ring homomorphism $A\longrightarrow k$, a smooth proper $A$-scheme $\mathfrak{X}$, and an isomorphism $\mathfrak{X}_{k}\cong X$.

What exactly is he talking about here? Is $\mathfrak{X}$ some sort of a Néron model, and how do I construct it (or prove its existence)? Why does (as is claimed by Eric Wofsey's answer) Chow's Lemma help me here?

Thanks in advance.

答:
This is just saying that defining the scheme $X$ involves only finitely many elements of $k$.  Cover $X$ with finitely many affine $k$-schemes.  Each of these affine $k$-schemes are actually defined over some finitely generated subring of $k$, since the corresponding $k$-algebras have a presentation that involves only finitely many specific elements of $k$.  To glue these affine $k$-schemes together, you just have to cover their pairwise intersections with finitely many affine open subschemes, which again is witnessed by finitely many elements of $k$.  To witness that these $k$-algebras are smooth over $k$ uses finitely many more elements of $k$.  To witness that the scheme you get by gluing them together is proper again needs only finitely many elements of $k$ (using Chow's lemma).

Taken together, this gives a finitely generated subring $A\subseteq k$ which contains all the elements of $k$ needed to construct the scheme $X$ and prove it is smooth and proper over $k$.  So, you can carry out this construction and obtain a smooth and proper scheme $\mathfrak{X}$ over $A$ which becomes $X$ after base-changing to $k$.


问:
Checking non-negativity over an interval
I have been working on non-negative univariate polynomials and I found the following equivalent relationship to check if a polynomial is non-negative or not:

The polynomial $g(x) = \sum_{r=0}^k y_rx^r$ satisfies $g(x) \geq 0$ for all $x \in [0, a]$ if and
only if there exists a positive semidefinite matrix $X = [x_{ij}]_{i,j=0,...,k}$, such that
\begin{equation*}
    \sum_{\substack{i,j=0,\ldots,k \\ i+j=2l-1}}x_{ij}=0, \qquad l=1,\ldots, k
\end{equation*}
\begin{equation*}
    \sum_{\substack{i,j=0,\ldots,k \\ i+j=2l}}x_{ij}=\sum_{m=0}^l y_r {k-r\choose l-m}a^{r}, \qquad l=0,\ldots, k.
\end{equation*}

This result is found in
Bertsimas, D., & Popescu, I. (2005). Optimal inequalities in probability theory: A convex optimization approach. SIAM Journal on Optimization, 15(3), 780-804.

In the proof, found in the Appendix of the paper, it says:

We observe that $g(x)\geq 0$ for $x\in [0,a]$ if and only if
$$(1+t^2)^k g\left(\frac{at^2}{1+t^2}\right) \geq 0 \qquad \text{for all }t.$$

Why does this equivalence hold? I see that if $g(x)\geq 0$, then the composed polynomial must be non-negative too, but I am unable to see why the other implication is also true.

Moreover, I am curious if similar results hold for higher dimensions. In particular, given a bivariate polynomial $f(x,y)=\sum_{r=0}^k\sum_{s=0}^l y_{r,s}x^rz^l$, can we find a characterization of non-negativity over a rectangle $[0,a]\times [0,b]$ using similar arguments? Maybe using a function like
$$(1+t^2)^{k} (1+u^2)^{l}g\left(\frac{at^2}{1+t^2}, \frac{bu^2}{1+u^2}\right)$$
or something similar...

Any help would be appreciated.

答:
Direct implication:
You can easily see that $0 \leq \dfrac{a t^2}{1+t^2} < a$ for $t \in \mathbb{R}$,  in particular, $\dfrac{a t^2}{1+t^2} \in [0, a]$ and, therefore
$$
g\left( \dfrac{a t^2}{1+t^2} \right) \ge 0, \quad t \in \mathbb{R}.
$$

On the other hand, since $(1+t^2)^k > 0$, this is equivalent to
$$
(1+t^2)^k  g\left( \dfrac{a t^2}{1+t^2} \right) \ge 0, \quad t \in \mathbb{R}.
$$

The reverse implication follows trivially because if $ x \in [0, a)$, then $x = \frac{a t_0^2}{1+t_0^2}$ for some $t_0 \ge 0$. The case $x = a$ can be obtained by continuity of $g$.

The multivariable extension you propose works in the exact same way.


问:
Prove: $ \forall x\in\mathbb{R}$, there exists a unique $n\in\mathbb{Z}$ such that $n-1\leq x<n$.
For all $x\in\mathbb{R}$, there exists a unique $n\in\mathbb{Z}$ such that $n-1\leq x<n$.

I know Archimedean Property gives $m\in\mathbb{N}$ such that $x<m$, but what then can we say about $m-1$?

答:
Let $x\in\mathbb{R}$. Then by the Archimedean Property, there is some $m\in\mathbb{N}$ such that $x<m$. Let $M$ be the set of all such $m$, then $M$ has a lower bound of $x$, and thus it must have a smallest element $n$. Since $n\in\mathbb{N}$, $n-1\in\mathbb{Z}$, and $n-1\leq x<n$.

(Should I add more detail to that last part?)


问:
Levi decomposition of $\mathfrak{gl}_n(\mathbb{K})$
Im trying to find the Levi decomposition of $\mathfrak{gl}_n(\mathbb{K})$ where $\mathbb{K}$ has characteristic zero. By Levi's theorem $\mathfrak{gl}_n(\mathbb{K})=Rad(\mathfrak{gl}_n(\mathbb{K} )+S$ where $Rad$ is the solvable radical and S is semisimple.

I know that $Rad{$\mathfrak{gl}_n(\mathbb{K})$ is the set of scalar matrix, but how I can find this set $S$?

答:
For a reductive Lie algebra $L$ we have the Levi decomposition $L=[L,L]\oplus Z(L)$. Here the commutator subalgebra is semisimple, i.e., a Levi subalgebra, and the center $Z(L)$ is abelian, i.e., the solvable radical of the Levi decomposition. For $L=\mathfrak{gl}_n(K)$ we have
$$
S=[L,L]=\mathfrak{sl}_n(K), \quad Z(L)=K\cdot I_n.
$$


问:
Find all $x$ for which the derivative of $f(x)=\sqrt[3]{x}|\sin{x}|$ doesn't exist
The task is to find all $x$ for which the derivative of $f(x)=\sqrt[3]{x}|\sin{x}|$ doesn't exist. $\sqrt[3]{x}$ doesn't cause any problems. The derivative can be problematic because of $|\sin{x}|$. I think that the values for which the derivative doesn't exist are $x=k\pi, k=\pm1, 2,...$. However, I came up with my solution by intuition and I haven't proven that no other $x$, which fit the condition, exist.
If I try to write my thoughts, they would be something like this:
$f'(x)=\frac{|\sin{x}|}{3\sqrt[3]{x^2}}+\sqrt[3]{x}\cos{x}$ and $\lim\limits_{x\rightarrow \pi+}{f'(x)}\neq \lim\limits_{x\rightarrow \pi-}{f'(x)}$. But then comes the problem about the zero. I think the derivative at that points exists because $f'(0)=\frac{|\sin{x}|x}{3x\sqrt[3]{x^2}}+\sqrt[3]{x}\cos{x}=\frac{|\sin{x}|\sqrt[3]{x}}{3x}+\sqrt[3]{x}\cos{x}=0$.

But I think that what I've written is very vague and not enough. What is the formal way of solving the problem? How sould I write the solution on a test?

答:
You approached the problem correctly.

Hint

$$f(x)= \sqrt[3]{x}|sin{x}|$$
$$\implies f'(x)=\frac{|\sin{x}|\sqrt[3]{x}}{3x} + \frac{|\sin{x}|*\cos{x}}{\sin{x}}*\sqrt[3]{x}$$

Wondering from where the $ \frac{|\sin{x}|*\cos{x}}{\sin{x}}$ came?

Well you made a mistake while differentiating $f(x)$. The correct differentiation of $|x|$ is $\frac{|x|}{x}$.

So apply chain rule to get $f'(x)$.

The rest approach would remain same. Just see when the function shows different values when approaching from $x^+$ and $x^-$.