问:
Proof Verification: $T$ is complete if and only if $\text{Mod}(T)$ satisfies the joint embedding property
This answer has an example exercise to show a relationship between the joint embedding property of the category of models of a given theory and the completeness of that theory, paraphrased in the section below with links to definitions.

I tried to prove it. I'm pretty sure my proof attempt works but I'm not totally certain.

Does the proof attempt work?

Is there a more elegant way to prove this?

Am I assuming any nontrivial results improperly (e.g. without naming the theorem when it would be idiomatic to name it)?

A theory $T$ is complete if and only if its category of models with arrows given by elementary embeddings $\text{Mod}(T)$ satisfies the joint embedding property.

Suppose that $T$ is complete.

For any pair of objects $x$ and $y$ in $\text{Mod}(T)$, $x$ and $y$ are elementarily equivalent because they satisfy the same sentences due to the completeness of $T$. Suppose without loss of generality that $|x| \le |y|$, then there exists an elementary embedding $f : x \to y$. The cardinals are totally ordered and every object has an identity arrow going to itself, therefore both $x$ and $y$ have arrows going to $\max(x, y)$, the larger of the two by cardinality, or $x$ if they have the same cardinality. Since $x$ and $y$ were chosen arbitrarily, $\text{Mod}(T)$ has the joint embedding property.

Suppose $T$ satisfies the joint embedding property.

For any objects $x$ and $y$ in $\text{Mod}(T)$, there exists an object $z$ so that there exist arrows $f : x \to z$ and $g : y \to z$. By definition of $\text{Mod}(T)$, if an arrow exists between two objects, then they satisfy the same sentences. Therefore, $x$ satisfies the same sentences as $z$ and $y$ satisfies the same sentences as $z$. Therefore, $x$ satisfies the same sentences as $y$. Since $x$ and $y$ were chosen arbitrarily, this means that all models satisfy the same sentences. Thus, $T$ is complete.

答:
As others have pointed out in the comments, your proof of the implication "complete$\implies$ JEP" is not quite right. In particular, the claim that $|M|\leqslant |N|$ implies the existence of an elementary embedding $M\to N$ is not true in general. Alex gives a nice example of this, but here is perhaps a simpler one:

Let $\mathcal{L}=\{P_1,P_2\}$ consist of two unary predicate symbols, and let $T$ assert that $P_1\cap P_2=\emptyset$, that each $P_i$ is infinite, and that every element belongs to one of the $P_i$. Then $T$ is $\aleph_0$-categorical (why?), and is hence in particular complete. However, consider a model $M$ of $T$ where $|P_1^M|=\aleph_1$ and $|P_2^M|=\aleph_1$, and a model $N$ of $T$ where $|P_1^N|=\aleph_0$ and $|P_2^N|=\aleph_2$. Then $|M|=\aleph_1<\aleph_2=|N|$, but there is no elementary embedding $M\to N$,  since such an embedding would induce an injection $P_1^M\to P_1^N$.

There is a relevant notion you might find interesting, of a "prime model" of a complete theory, the defining property of which is that they elementarily embed into every other model. You can read about these in, eg, chapter 4 of Marker.

Here's an alternative proof.
Let $\mathcal{L}'=\mathcal{L}\cup\{c_a\}_{a\in M}\cup\{d_b\}_{b\in N}$, where the $c_a$ and $d_b$ are constant symbols corresponding to the elements of $M$ and $N$, respectively. We can naturally consider $M$ as an $\mathcal{L}\cup\{c_a\}_{a\in M}$-structure, by realizing each $c_a$ as $a$; let $T_M$ be the set of all $\mathcal{L}\cup\{c_a\}_{a\in M}$ sentences satisfied by $M$. (This is known as the "elementary diagram" of $M$.)

Similarly, let $T_N$ be the elementary diagram of $N$ in the language $\mathcal{L}\cup \{d_b\}_{b\in N}$, and let $T'=T\cup T_M\cup T_N$. Can you use completeness of $T$ to show that $T'$ is finitely satisfiable?

 Since $M\models T\cup T_M$, it suffices to show that $T_N$ is finitely satisfiable in $M$. By taking conjunctions, we therefore just need to to show that $\psi(d_{b_1},\dots,d_{b_n})$ is satisfiable in $M$ for any $\mathcal{L}$-formula $\psi(\overline{w})$ with $\psi(\overline{d})\in T_N$. By construction, we know $$N\models\psi(d_{b_1},\dots,d_{b_n}),$$ and so in particular $N\models\exists w_1,\dots,w_m\ \psi(w_1,\dots,w_n)$. Moreover, the sentence $\exists \overline{w}\ \psi(\overline{w})$ is an $\mathcal{L}$-sentence (ie, it contains no symbols outside $\mathcal{L}$). Since $T$ is complete, this means we must have $\exists \overline{w}\ \psi(\overline{w})\in T$, and hence $M\models\exists\overline{w}\ \psi(\overline{w})$. Interpreting the $d_{b_i}$ as witnesses of this sentence in $M$ then gives the desired result.

Thus $T'$ is finitely consistent, and so – by compactness – there exists an $\mathcal{L}'$-structure $O$ modeling $T'$. Now, consider the functions $f:M\to O$ and $g:N\to O$ given by $f(a)=c_a^O$ and $g(b)=d_b^O$. Can you show that this map is an elementary $\mathcal{L}$-embedding?


问:
limit of series exponential

Compute the limit of the series $$\sum\limits_{n=4}^\infty 3\frac{2^{n+1}}{5^{n-2}}$$

How do you approach these types of problems?

I'm thinking that this one is in indeterminate form, is that correct?

答:
The biggest hint lies on the fact that you have a $2^n$ in the numerator and $5^n$ on the denominator.

So if you are able to manipulate the series so that it kind of looks like $$\sum_{n=0}^\infty ({2\over5})^n$$, then you can use the fact that this is an infinite geometric sequence with the common ratio having an absolute value less than 1.

Thus you can find the infinite sum to be $$1 \over {1 - {2\over 5}}$$

Now let's move on to how to manipulate it. First we want to get rid of the numbers around $({2 \over 5})^n$, so it will go as follows. $$3 {{2^{n+1}}\over{5^{n-2}}}
 = 3 {2 \over 5^{-2}}・{{2^n}\over{5^n}}$$ $$ = 150 ・ ({2 \over5})^n$$

so the original series can be expressed as $$150\sum_{n=4}^{\infty}({2\over 5})^n $$

Now what we need to do is to deal with the index of the sum, which oddly starts from 4. The following is what we usually do $$\sum_{n=4}^{\infty}({2\over 5})^n $$ is basically the same thing as the geometric sum, but it's "missing" the first four terms, as in $$\sum_{n=0}^{\infty}({2\over 5})^n = 1 + {2 \over 5} + ({2\over 5})^2 + ({2\over 5})^3 + \sum_{n=4}^{\infty}({2\over 5})^n$$

So calling the first four terms $S$ for convenience, the following is what you will need.

$$\sum_{n=4}^{\infty}3 {{2^{n+1}}\over{5^{n-2}}} = 150 \left[ \sum_{n=0}^{\infty}({2\over 5})^n - S \right]$$

Most problems that seems like it's related to geometric series goes like this.

ContentLicense: CC BY-SA 3.0

Title: limit of series exponential


问:
Singular Value Decomposition gives a row permutated matrix
I want to find the SVD of the following matrix.
$$A = \begin{bmatrix}1&2\\2&2\\2&1\end{bmatrix}$$

The singular values of the above matrix are $\sqrt{17}$,$1$. The following results are posted here in order to make this question shorter.

$$AA^T=\begin{bmatrix}5&6&4\\6&8&6\\4&6&5\end{bmatrix}$$
Eigen values of $AA^T$=$\{0,1,17\}$

Eigen vectors of $AA^T$=$\{\begin{bmatrix}-2\\3\\-2\end{bmatrix},\begin{bmatrix}1\\0\\-1\end{bmatrix},
\begin{bmatrix}3\\4\\3\end{bmatrix}\}$

$$A^TA=\begin{bmatrix}9&8\\8&9\end{bmatrix}$$
Eigen values of $A^TA$=$\{1,17\}$

Eigen vectors of $A^TA$=$\{\begin{bmatrix}1\\-1\end{bmatrix},\begin{bmatrix}1\\1\end{bmatrix}\}$

Therefore, we can write the U,V,W matrices as follows.

$$U=\begin{bmatrix}\frac{3}{\sqrt{34}}&\frac{1}{\sqrt{2}}&\frac{-2}{\sqrt{17}}\\
\frac{4}{\sqrt{34}}&0&\frac{3}{\sqrt{17}}\\
\frac{3}{\sqrt{34}}&\frac{-1}{\sqrt{2}}&\frac{-2}{\sqrt{17}}\end{bmatrix}$$

$$W=\begin{bmatrix}\sqrt{17}&0\\0&1\\0&0\end{bmatrix}$$
$$V=V^T=\begin{bmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\end{bmatrix}$$

My problem is that why in here, $$A\neq UWV^T$$
$$UWV^T=\begin{bmatrix}2&1\\2&2\\1&2\end{bmatrix}$$
which is a row permuted version of A.

This problem won't occur if I multiplied the eigenvector of $AA^T$ corresponding to the eigenvalue of $1$ by $(-1)$.(i.e; by using $\begin{bmatrix}-1\\0\\1\end{bmatrix}$ instead of $\begin{bmatrix}1\\0\\-1\end{bmatrix}$). Both of those eigenvectors are correct for the eigenvalue $1$ of that matrix. So, my problem is how do we identify such occasions when obtaining the SVD?

答:
With the help of Lutz Lehmann's comment above I figured out the way to obtain the SVD without worrying about the problem I previously faced. I am posting my answer here so that it would be a help to anyone who has the same problem.

In the SVD, we express the matrix A as a product of 3 matrices which I am denoting as U, W, $V^T$. In this less problematic method, what we have to first do is obtaining the $A^TA$.

$$A = \begin{bmatrix}1&2\\2&2\\2&1\end{bmatrix}$$
$$\implies A^TA=\begin{bmatrix}9&8\\8&9\end{bmatrix}$$

Then we obtain the eigenvalues and corresponding eigenvectors of $A^TA$ as follows.

$det(A^TA-\lambda I)=det(\begin{bmatrix}9-\lambda&8\\8&9-\lambda\end{bmatrix})=0$

$\implies (1-\lambda)(17-\lambda)=0$

$\therefore \lambda=\{17,1\}$

When $\lambda = 17,$

eigenvector = $\begin{bmatrix}1\\1\end{bmatrix}$

$\therefore$ normalized eigenvector when ($\lambda = 17$) = $\begin{bmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{bmatrix}=v_1$

When $\lambda = 1,$

eigenvector = $\begin{bmatrix}1\\-1\end{bmatrix}$

$\therefore$ normalized eigenvector when ($\lambda = 1$) = $\begin{bmatrix}\frac{1}{\sqrt{2}}\\\frac{-1}{\sqrt{2}}\end{bmatrix}=v_2$

The corresponding singular value would be $\{\sigma_1=\sqrt{17},\sigma_2=1\}$.

Using the above results, we can write W and V as follows.

$$W=\begin{bmatrix}\sqrt{17}&0\\0&1\end{bmatrix}$$
$$V=\begin{bmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\end{bmatrix}$$

Now we can obtain the columns of U without explicitly calculating the eigenvectors of $AA^T$ as follows.

Simply use the following relation,

$$Av_i=\sigma_i u_i, \forall i$$

$$\therefore Av_1=\sigma_1 u_1$$
$$\begin{bmatrix}1&2\\2&2\\2&1\end{bmatrix} \begin{bmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{bmatrix}= \sqrt{17} u_1$$

$$\therefore \begin{bmatrix}\frac{3}{\sqrt{2}}\\\frac{4}{\sqrt{2}}\\\frac{3}{\sqrt{2}}\end{bmatrix}=\sqrt{17} u_1\implies u_1=\begin{bmatrix}\frac{3}{\sqrt{34}}\\\frac{4}{\sqrt{34}}\\\frac{3}{\sqrt{34}}\end{bmatrix}$$

Also, $Av_2=\sigma_2 u_2$

$$\therefore \begin{bmatrix}1&2\\2&2\\2&1\end{bmatrix} \begin{bmatrix}\frac{1}{\sqrt{2}}\\\frac{-1}{\sqrt{2}}\end{bmatrix}= 1 u_2 \implies u_2 = \begin{bmatrix}\frac{-1}{\sqrt{2}}\\0\\\frac{1}{\sqrt{2}}\end{bmatrix}$$

$$\therefore U = \begin{bmatrix} \frac{3}{\sqrt{34}}&\frac{-1}{\sqrt{2}}\\
\frac{4}{\sqrt{34}}&0\\
\frac{3}{\sqrt{34}}&\frac{1}{\sqrt{2}}\end{bmatrix}$$

So, the SVD of A is,

$$\begin{bmatrix}1&2\\2&2\\2&1\end{bmatrix}=
\begin{bmatrix} \frac{3}{\sqrt{34}}&\frac{-1}{\sqrt{2}}\\
\frac{4}{\sqrt{34}}&0\\
\frac{3}{\sqrt{34}}&\frac{1}{\sqrt{2}}\end{bmatrix}
\begin{bmatrix}\sqrt{17}&0\\0&1\end{bmatrix}
\begin{bmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\end{bmatrix}$$

The above is the reduced SVD. In any case if you want to obtain the full SVD, all you have to do is obtain the normalized eigenvector of $AA^T$ corresponding to the eigenvalue of $0$ and, put that as the third column of the above U matrix. Also, remember to add another row of zeros to the W matrix as its third row.

The above method can be applied to any given matrix to obtain the singular value decomposition.

ContentLicense: CC BY-SA 4.0


问:
In the binomial distribution, why does order not matter?
(For context): $X$ is the random variable for number of successes ($x$) out of $n$ independent trials with probability of success $p$ and probability of failure $1-p = q$

The formula for mass my book gives me for the binomial distribution is this:
$P(X=x) = {n \choose x}p^xq^{n-x}$

My question is this. How come order doesn't matter? Why do we use ${n \choose x}$ in the above to calculate all the ways we can get $x$ successes out of $n$ independent trials instead of using $nPx$? I.E. Why do we use combination instead of permutation? Can someone explain this to me because it's been really bothering me.

Thank You!

答: If we have three events $A,B,C$ where $A,B$ are successes and $C$ is a failure, then $A,B,C$ and $B,A,C$ both represent two successes in the first two trials and a failure in the last trial, so they are considered to be the same.


答:
Look at a small example. In a sequence of $4$ trials, what are the ways to get $2$ successes? With $S$ for success and $F$ for failure they are $SSFF$, $SFSF$, $SFFS$, $FSSF$, $FSFS$, AND $FFSS$; there aren’t any more. Specifically, there is one way to get $2$ successes for every pair of trials, and there are $\binom42$ pairs of trials.

The successes and failures don’t have individual identities: their only identities are as success or failure and where in the sequence of trials they occur. You might as well imagine that you have $x$ white and $n-x$ black stones, identical in all respects save color, and want to know in how many ways you can line them up in a row. The stones aren’t labelled; all that matters is the color of each stone and its position in the row. For instance, there is only one distinguishable arrangement with all $x$ of the white stones together at the head of the line.

It would be another story if the white stones were labelled $1$ through $x$: then there would be $k$ distinguishable arrangements with all of white stones together at the head of the line. That’s the model that your proposed calculation would be following, and it clearly doesn’t fit what you’re trying to model: the individual successes don’t come with individual labels beyond success in trial number so-and-so.

Title: In the binomial distribution, why does order not matter?

Body:
(For context): $X$ is the random variable for number of successes ($x$) out of $n$ independent trials with probability of success $p$ and probability of failure $1-p = q$

The formula for mass my book gives me for the binomial distribution is this:
$P(X=x) = {n \choose x}p^xq^{n-x}$

My question is this. How come order doesn't matter? Why do we use ${n \choose x}$ in the above to calculate all the ways we can get $x$ successes out of $n$ independent trials instead of using $nPx$? I.E. Why do we use combination instead of permutation? Can someone explain this to me because it's been really bothering me.

Thank You!


问:
Counting the number of zeros yielded by modulo arithmetic functions
I have 4 functions

\begin{align}
f_1(x) &= x \text{ mod } 1\\
f_2(x) &= x \text{ mod } 3 \\
f_3(x) &= x \text{ mod } 4 \\
f_4(x) &= x \text{ mod } 12\\
\end{align}

for $x \in [0,\infty) $

I want to create a function $g(x)$ that counts the number of zeroes at $x$ for those four functions.

For example, $x=6$ would have

\begin{align}
f_1(6) &= 0\\
f_2(6) &= 0 \\
f_3(6) &= 2 \\
f_4(6)& = 6 \\
\end{align}

Since only two of these are zero, then $g(6) = 2$

Another example is $x=24$

\begin{align}
f_1(24) &= 0\\
f_2(24) &= 0 \\
f_3(24) &= 0 \\
f_4(24)& = 0\\
\end{align}

Since all four of these are 0, then $g(24) = 4$

If $x =0.5$, then

\begin{align}
f_1(0.5) &= 0.5 \\
f_2(0.5)& = 0.5 \\
f_3(0.5) &= 0.5\\
f_4(0.5) &= 0.5 \\
\end{align}

Since none of these are zero, then $g(0.5)=0$

How can I define $g(x)$ elegantly?

Inelegant method I've found

One inelegant way I thought of was basically count them for each possible value.

\begin{align}
g(0)& = 4 \\
g(1) &= 1 \\
g(2)& = 1\\
g(3)& = 2 \\
g(4)& = 2 \\
g(5)& = 1\\
g(6)& = 2 \\
g(7)& = 1\\
g(8)& = 2\\
g(9) &= 2 \\
g(10) &= 1\\
g(11)& = 1\\
\end{align}

Then for any $x$, I could do $g(x \text{ mod } 12)$ and it would either match one of these or else $g(x) = 0$.

This gives me the correct result. However, this seems very inelegant and I'm hoping for a nicer solution.

答:
For non-integers $x$, $g(x)=0$ clearly.

If $x$ is an integer not divisible by both $3$ and $4$, we have $f_1(x)=0$, but the values of $f_2(x),f_3(x),f_4(x)$ is not zero. (Try to figure why by yourself) So $g(x)=1$ in this case.

If $x$ is an integer divisible by $3$ but not $4$, $f_1(x)=f_2(x)=0$ and the values of $f_3(x),f_4(x)$ are not zero. (Why?) So $g(x)=2$ in this case.

If $x$ is an integer divisible by $4$ but not $3$, $f_1(x)=f_3(x)=0$ and the values of $f_2(x),f_4(x)$ are not zero. (Why?) So $g(x)=2$ in this case.

If $x$ is an integer divisible by both $3$ and $4$, then it is divisible by $12$. (Why?) So $f_1(x)=f_2(x)=f_3(x)=f_4(x)=0$, so $g(x)=4$ in this case.


问:
How to complete the square for this?
How to express $2x^2 - 3x$ in the form $a(x+b)^2 + c$ where $a$, $b$ and $c$ are constants?

Title: How to complete the square for this?

Body:
$$a(x+b)^2 +c= ax^2+ 2abx + ab^2+c=2x^2 -3x$$
Comparing the coefficients,
$$a=2$$
$$2ab=-3$$
$$2(2)b=-3$$
$$b=\frac{-3}{4}$$
$$ab^2 +c=0$$
$$c=-ab^2=-(2)(\frac{-3}{4})^2= \frac{-9}{8}$$


问:
If $E \lvert X_1 \rvert = \infty$, $\sum_{n=1}^{\infty} P(\lvert X_1 \rvert > a_n) <\infty $and $a_n/n$ increase, then $a_n/n \rightarrow \infty$
I want to prove followings:

If $E \lvert X_1 \rvert = \infty$, $\sum_{n=1}^{\infty} P(\lvert X_1 \rvert > a_n) <\infty $and $a_n/n$ increase, then $a_n/n \rightarrow \infty$

It's described in Durrett's Probability theory and examples (5th ed) Thm 2.5.13

He said its obvious, but how can I prove it?

Could you explain details of it?

答:
Suppose $\frac {a_n} n$ increase to a finite number $C$. Then $a_n <2C n$ for $n$ sufficiently large. Hence $P(|X_1| >a_n) \geq P(|X_1|>2Cn)$ for such $n$. It follows that $\sum P(|X| >n) <\infty$, where $X=\frac {X_1 } {2C}$. But this implies that $E|X| <\infty$ and hence $E|X_1| <\infty$.

I have assumed the following well known fact: $E|X| <\infty$ iff $\sum P(|X| >n) <\infty$. You can find  a proof of this on this site as well as ion Wikipedia.


问:
When is a uniformly dense family of functions dense in L^p?
Suppose $\mathcal{A}\subset L^p(\mathbb{R})$ is an algebra of functions with the following property:

For every compact $K\subset\mathbb{R}$, $\mathcal{A}$ is dense in $\mathcal{C}(K)$ with respect to the uniform norm $\|\cdot\|_{\infty}$, where $\mathcal{C}(K)$ is the collection of real continuous functions on $K$. The uniform norm I'm referring to is
$$\|f\|_{\infty}=\sup_{t\in K}|f(t)|.$$
(See http://en.wikipedia.org/wiki/Uniform_norm.)

Can we conclude that $\mathcal{A}$ is dense in $L^p(\mathbb{R})$ (with respect to the $L^p$ norm)?

I became interested in this question while investigating a special case, the $L^2$-density of finite linear combinations of Gaussians:

$$\sum_{i=1}^n\alpha_ie^{-k_i(x-x_i)^2},\qquad\alpha_i,k_i,x_i\in\mathbb{R},k_i>0.$$

The question above occurred to me because I can imagine it being useful in cases like this to, say, verify the hypotheses of the Stone-Weierstrass theorem for a given family of functions rather than to explicitly approximate functions in $L^p$.

答:
I think you can pull an argument like this : define the sequence of compacts $[-n, n]$ ; since $\mathcal A$ is dense in each of them, for each function $f \in L^p(\mathbb R)$ and for all $n$, there exists a sequence $f_{n,m}(x)$ such that $f_{n,m} \to f$ in $[-n,n]$ with respect to the norm you've chosen as $m \to \infty$.

You could extract a subsequence out of the sequences $f_{n,m}$ by the following argument. Choose a sequence $\varepsilon_k \to 0$. We will find $g_k$ such that
$$
\|g_k - f\|_{\infty} \overset{def}{=} \inf \{ K > 0 \, | \, \mu(\{x \in \mathbb R \, | \, |g_k(x) - f(x)| > K\} ) = 0 \} < \varepsilon_k.
$$
(I believe this is the detailed definition of your norm? Correct me if I'm wrong.)

with $\mu$ being the Lebesgue measure.

Now $f \in L^p$, so that for $n$ big enough, $|f(x)| < \varepsilon_k/2$ almost everywhere outside $[-n,n]$. Fix this $n$.

This is the point where I am stuck at. Since $f_{n,m} \to f$ in $[-n,n]$, there exists $m$ such that $\|f_{n,m}-f \|_{\infty} < \varepsilon_k$ inside $[-n,n]$ and $|f_{n,m}(x)| < \varepsilon_k/2$ outside$^{*}$ $[-n,n]$ ($f_{n,m} \in L^p$ also, hence goes to $0$ at infinity.)

This guy ($^{*}$) is the problem ; I don't have any control over what happens to $f_{n,m}$ outside $[-n,n]$ and that is what gives me intuition that it might be false in the general case. Although your choice of functions exhibit more properties than the abstract one... but this is now at a state of intuition only. Let me finish just to see where I froze in my argument.

With those properties summed up, outside $[-n,n]$, $|f_{n,m}(x) - f(x)| \le |f_{n,m}(x)| + |f(x)| < \varepsilon_k$ almost everywhere outside $[-n,n]$ and $|f_{n,m}(x) - f(x)| < \varepsilon_k$ inside $[-n,n]$, so that $|f_{n,m}(x)-f| < \varepsilon_k$ almost everywhere over $\mathbb R$, that is, we have found $f_{n,m}$ such that $\|f_{n,m} - f\|_{\infty} < \varepsilon_k$. Just let $g_k = f_{n,m}$.

I hope it helped in some way to at least give you some ideas.

EDIT : I forgot the "$ = 0$" in my definition of the norm. Typo error.


问:
Searching for unbounded, non-negative function $f(x)$ with roots $x_{n}\rightarrow \infty$ as $n \rightarrow \infty$
If a function $y = f(x)$ is unbounded and non-negative for all real $x$,
then is it possible that it can have roots $x_n$ such that $x_{n}\rightarrow \infty$ as $n \rightarrow  \infty$.</p>

答: The function $ y = |x \sin(x)|$ has infinitely many roots $x_n$ such that $x_{n}\rightarrow \infty$ as $n \rightarrow  \infty$.


答:
Instead of equating coefficients, you can also follow this straightforward procedure.

First factor out the leading coefficient, so that you can work with something of the form $x^2+bx+c$, and start completing the square by writing this as $\left(x+\frac{b}2\right)^2+d$ for some $d$ that remains to be determined:

$$\begin{align*}
2x^2-3x&=2\left(x^2-\frac32x\right)\\
&=2\left[\left(x-\frac34\right)^2+d\right]\,.
\end{align*}$$

Now

$$\left(x-\frac34\right)^2=x^2-\frac32x+\frac9{16}\,,$$

and what we want inside the square brackets is just $x^2-\frac32x$, so clearly $d$ must be $-\frac9{16}$ to get rid of the unwanted $\frac9{16}$ term:

$$\begin{align*}
2x^2-3x&=2\left[\left(x-\frac34\right)^2-\frac9{16}\right]\\
&=2\left(x-\frac34\right)^2-\frac98\,.
\end{align*}$$

Title: How to complete the square for this?

Body: How to express $2x^2 - 3x$ in the form $a(x+b)^2 + c$ where $a$, $b$ and $c$ are constants?


问:
Examples of Galois Correspondence being inapplicable with non-normal groups and fields
In ``Fields and Galois Theory '' by JM Howie, I cannot think of any example to illustrate strict inclusion for the Galois correspondence in the case of non-normal groups and fields. In particular Howie uses symbols such as $\mathbf{K}, \mathbf{E}, \mathbf{L} $ for fields where often $\mathbf{K} \subset \mathbf{E} \subset \mathbf{L} $ and often uses $\mathbf{H}, \mathbf{G}$ for groups where often $\mathbf{H} \subset \mathbf{G}$. The fixed field  of a subgroup  $\mathbf{H}$  of a Galois group is denoted $\Phi(\mathbf{H})$ and the subgroup of the Galois group which fixes $\mathbf{E}$ is denoted $\Gamma (\mathbf{E})$ I note Howie's theorem 7.11 which shows $\mathbf{E} \subseteq \Phi (\Gamma (\mathbf{E}))$ and $\mathbf{H} \subseteq \Gamma (\Phi (\mathbf{H}))$. and theorem 7.28 which in effect shows that these inclusions are strict for non normal fields and subgroups.
I was then looking at the 8 automorphisms of $D_4$ (Galois group of $x^4-2=0$) as an example. In particular there is a normal and a non-normal subgroup each with one automorphism other than the identity. The normal subgroup's non-identity automorphism doesn't move $i$ but takes $\sqrt[4]{2}$ to its negative and the non normal subgroup's non-identity automorphism takes $i$ to $-i$ as well as taking $\sqrt[4]{2}$ to its negative (This is Howies subgroup $H_7$). I could not use this to find an example to illustrate strict inclusion $\mathbf{E} \subset \Phi (\Gamma (\mathbf{E}))$ and $\mathbf{H} \subset \Gamma (\Phi (\mathbf{H}))$ for non normal fields and subgroups. This makes me suspect that I am missing something important in the whole concept of the Galois correspondence. I would be grateful for an example using this $H_7 \subset D_4$ and the corresponding fields. Thanks David Kault

答:
Through his unintuitive and idiosyncratic notation, Howie is trying to explain that if the subgroup isn't normal as a subgroup of the Galois group of the splitting field over the rational numbers, then the fixed field of that subgroup isn't normal as a field extension of the rational numbers.

Background: Let $F = \Bbb{Q}( \sqrt[4]{2}, i )$ be the splitting field of $x^4 - 2$ over $\Bbb{Q}$. For visualization purposes, it's most natural to understand $\operatorname{Gal}(F, \Bbb{Q})$, the group of $\Bbb{Q}$-preserving field automorphisms of $F$, as being generated by two elements $a, b$, where $a$ maps $\sqrt[4]{2}$  to  $i \sqrt[4]{2}$, but fixes $\pm i$, and $b$ is complex conjugation. Clearly $a$ permutes the four roots of $x^4 - 2$ via $90°$ rotation CCW in the complex plane, and $b$ is reflection in the $x$-axis, which analogizes nicely with our ordinary geometric understanding of $D_4$ (the symmetry group of a square).

Having named these two elements, and understanding the expression $xy$ as performing transformation $y$, followed by transformation $x$, we find that $$D_4 = \{ e, a, a^2, a^3, b, ab, a^2 b, a^3 b \}.$$ In this notation, Howie's non-normal order $2$ subgroup $H_7$ (why $7$? Was he just numbering the subgroups of $D_4$ randomly?) is then $$H_7 = \langle a^2 b \rangle,$$ while the normal subgroup of order $2$ is $\langle a^2 \rangle$. Here $\langle x_1, ..., x_n \rangle$ indicates the subgroup generated by the elements $x_1, ..., x_n$.

The fixed field of the normal subgroup $\langle a^2 \rangle$ is $\Phi( \langle a^2 \rangle ) = \Bbb{Q}(i, \sqrt{2})$, which is indeed a normal extension, as it's the splitting field of $x^4 - 4$ over $\Bbb{Q}$. The Galois group $\operatorname{Gal}(\Bbb{Q}(i, \sqrt{2}) : \Bbb{Q})$ is the Klein four-group $$V = \{e, x, y, xy \},$$ where $x$ swaps $i$ with $-i$ and $y$ swaps $\sqrt{2}$ with $-\sqrt{2}$. $V$ is commutative and each element is self-inverse, so the subgroups $\langle x \rangle$, $\langle y \rangle$, $\langle xy \rangle$ are all normal. These subgroups correspond to the subfields $\Bbb{Q}(\sqrt{2})$, $\Bbb{Q}(i)$, $\Bbb{Q}(i \sqrt{2})$. We can directly find that
\begin{align*}
\Gamma(\Bbb{Q}) = V &\text{ and } \Phi(V) = \Bbb{Q}, \\
\Gamma(\Bbb{Q}(\sqrt{2}) = \langle x \rangle &\text{ and } \Phi(\langle x \rangle) = \Bbb{Q}(\sqrt{2}), \\
\Gamma(\Bbb{Q}(i) = \langle y \rangle &\text{ and } \Phi(\langle y \rangle) = \Bbb{Q}(i), \\
\Gamma(\Bbb{Q}(i\sqrt{2}) = \langle xy \rangle &\text{ and } \Phi(\langle xy \rangle) = \Bbb{Q}(i\sqrt{2}), \\
\Gamma(\Bbb{Q}(\sqrt{2}, i)) = \{ e \} &\text{ and } \Phi(\{ e \}) =
\Bbb{Q}(\sqrt{2}, i). \
\end{align*}
In other words, because the subgroup was normal, the corresponding field extension is normal as well.

By contrast, the fixed field of $H_7$, in Howie's notation, is $\Phi(H_7) = \Bbb{Q}(i\sqrt[4]{2})$.

Just as the group $H_7 = \langle a^2 b \rangle$ is non-normal as a subgroup of $D_4 = \langle a, b \rangle$, the field $\Bbb{Q}(i\sqrt[4]{2})$ is non-normal as an extension of $\Bbb{Q}$.

Indeed, $x^4 - 2$ itself fails to split over $\Bbb{Q}(i\sqrt[4]{2})$: when we try to factor it, we get $$x^4 - 2 = (x - i\sqrt[4]{2})(x + i\sqrt[4]{2})(x^2 - 2),$$ and here we are stuck, because $\Bbb{Q}(i\sqrt[4]{2})$ contains a square root of $-2$, but no square root of $2$.

How can we use this to answer [part of] Howie's question?

Let $E = \Bbb{Q}$ considered as a subfield of the field extension $\Bbb{Q}(i\sqrt[4]{2})$.

The automorphism group $\operatorname{Aut}(\Bbb{Q}(i\sqrt[4]{2}):\Bbb{Q})$ has 2 elements $\{ e, g \}$: the identity $e$, and the map $g$ which swaps $i\sqrt[4]{2}$ with its negative.

Then $$\Gamma(E) = \Gamma(\Bbb{Q}) = \{ e, g \},$$ but $$\Phi(\Gamma(E)) = \Phi(\Gamma(\Bbb{Q})) = \Phi(\{ e, g \}) = \Bbb{Q}(\sqrt{2}),$$ which strictly contains $E = \Bbb{Q}$.

However, there are only two subgroups of $\{ e, g \}$: the trivial subgroup $\{ e \}$ and the entire group $\{ e, g \}$. We check that both of these subgroups satisfy $H = \Gamma(\Phi(H))$:
\begin{align*}
\Gamma(\Phi( \{ e \} )) = & \Gamma(\Bbb{Q}(i\sqrt[4]{2})) = \{ e \}, \\
\Gamma(\Phi( \{ e, g \} )) = & \Gamma(\Bbb{Q}(\sqrt{2})) = \{ e, g \}. \
\end{align*}
So no subgroup $H < \operatorname{Aut}(\Bbb{Q}(i \sqrt[4]{2}): \Bbb{Q})$ can give us an example where $H$ is strictly contained in $\Gamma(\Phi(H))$.


问:
Is limit in indeterminate form?
Quick question. I was wondering whether the following limit is simply equal to zero or it is rather indeterminate:

$$\lim_{x\to0}\frac0x$$

For more detail: I was studying the differentiability of the function $\lfloor x\rfloor\sin(\pi x)^2$ at $\frac12$. When you evaluate the function at $\frac12+h$ you get $\left\lfloor\frac12+h\right\rfloor\times1$ up in the denominator divided by $h$ as $h$ goes to zero. Since $\left\lfloor\frac12+h\right\rfloor$ should be $0$, I guess we would end up having a situation similar to the one in the picture above. But I'm not sure how to tackle it.

答: It is $0$, since, for each $x\ne0$, $\frac0x=0$. So, $\lim_{x\to0}\frac0x=\lim_{x\to0}0=0$.

Title: Is limit in indeterminate form?


答:
The notion of an "indeterminant form" follows moreso from taking the limit of a ratio as the ratio of limits:

$$\lim_{x \to a} \frac{f(x)}{g(x)} \stackrel{\text{becomes}}{\longrightarrow} \frac{\lim \limits_{x \to a} f(x)}{\lim \limits_{x \to a} g(x)}$$

...or, at least it does in your case. Of course, these limits are not always equal, but they are within certain conditions. If the individual limits on the right-hand side are both infinite in magnitude or both zero, then we can call that an indeterminant form. Of course, it is clear that your limit satisfies this condition.

Of course, that doesn't mean the limit cannot be evaluated. For instance, your limit evaluates to $0$, because $f(x)=0/x$ is $0$ for all $x \ne 0$. From the definition of a limit (like in terms of $\varepsilon$-$\delta$ for instance), you can show that this means your limit is $0$.

The notion of a limit being indeterminant is moreso a statement that, on the face of it, the limit will be a bit harder to solve than, say, naive substitution. For instance, for ratios which have indeterminant forms, L'Hopital's rule is a common means of finding the limit despite the indeterminant form.

In short: there is a difference between a limit not existing, and the limit simply having an indeterminant form.

Title: Is limit in indeterminate form?

Body:
Quick question. I was wondering whether the following limit is simply equal to zero or it is rather indeterminate:

$$\lim_{x\to0}\frac0x$$

For more detail: I was studying the differentiability of the function $\lfloor x\rfloor\sin(\pi x)^2$ at $\frac12$. When you evaluate the function at $\frac12+h$ you get $\left\lfloor\frac12+h\right\rfloor\times1$ up in the denominator divided by $h$ as $h$ goes to zero. Since $\left\lfloor\frac12+h\right\rfloor$ should be $0$, I guess we would end up having a situation similar to the one in the picture above. But I'm not sure how to tackle it.


问:
Converting a LP into standard form
I am trying to convert this LP into standard form and I am stuck on the part with $-1\leq x_1\leq 4$. I am unsure how to turn this into variables that are all greater than 0.

$\begin{array}{l@{}l}\text{minimize} & 4x_1+x_2+ 5x_3\\
    \text{subject to}& x_1+x_3 \geq 10 \\
    &2x_1+x_2+3x_3 \geq 15 \\
    &-1 \leq x_1 \le 4, x_2 \leq 0, x_3 \geq 0 \\
\end{array}$

My approach so far:

Let $x_2'=-x_2$

$\begin{array}{l@{}l}\text{maximize} & -4x_1+x_2'- 5x_3\\
    \text{subject to}& -x_1-x_3 \leq -10 \\
    &-2x_1+x_2'-3x_3 \leq -15 \\
    &-1 \leq x_1 \le 4,~x_2',~x_3 \geq 0 \\
\end{array}$

but then I don't know what to do with the $-1\leq x_1\leq 4$

答:
Substitute $x_1'=x_1+1\in[0,5]$ to get
$$\begin{array}{l@{}l}\text{maximize} &Z-\color{red}4=-4x_1'+x_2'- 5x_3\\
    \text{subject to}& -x_1'-x_3 \leq -11 \\
    &-2x_1'+x_2'-3x_3 \leq -17 \\
    &x_1' \le 5\\
    &x_{1,2}',~x_3 \geq 0 \\
\end{array}$$and note that $Z$ is maximized iff $Z-4$ is maximized.

Title: Converting a LP into standard form


问:
Prove that if $p(A)=0$ where A is a matrix of a linear operator ($A \in L(V)$), $p(\lambda)=0$ if $\lambda \in \sigma(A)$
I think it's all in the title. $p$ is some random polynomial.
I don't know how to approach this one. I've tried taking the roots of $p$, placing them on the diagonal of a new matrix and reasoning that the linear operator of the new matrix has a common specific value (if it's called that in English) with A, but I have no idea how to prove that that specific value is also a root of $p$.</p>

答:
I'll give you an example ($p(x)=x^2+x+1$) and leave the general case to you.
Suppose that $\lambda\in\sigma(A)$ with corresponding eigenvector $u\neq0$. Then

$$
0=p(A)u=A^2u+Au+u=A(\lambda u)+\lambda u+u=
\\
\lambda Au+\lambda u+u=\lambda^2u+\lambda u+u=(\lambda^2+\lambda+1)u=p(\lambda)u.
$$
Since $u\neq0\Rightarrow p(\lambda)=0$.


问:
Prove or disprove: If $H$ is a normal subgroup and cyclic and $G/H$ are cyclic, then $G$ is cyclic.

Prove or disprove: If $H$ is a normal subgroup and cyclic and $G/H$ are cyclic, then $G$ is cyclic.

I don't understand the quotient group $G/H$ being cyclic. What does it mean? From what I understand, $G/H$ is a group that includes groups and not elements, right?

I know that the reverse is right (If $G$ is cyclic then $G/N$ is cyclic.)

How do I disprove it?

答: The smallest counterexample is Klein's $4$-group, $G=\{e,a,b,ab\}$. Take e.g. $H=\{e,a\}$; then, $H\unlhd G$, $H\cong C_2$ and $G/H\cong C_2$ (it's worthwile to explicitly write the two elements of the quotient, and squaring the nonidentity one), but $G$ is not cyclic.


答: Another way to look at this problem is noticing that $p$ has to be a (polynomial) multiple of the minimal polynomial of $A$. Now what happens to the minimal polynomial if you apply it on an Eigenvalue of $A$? What happens consequently to $p$ ?</p>

Title: Prove that if $p(A)=0$ where A is a matrix of a linear operator ($A \in L(V)$), $p(\lambda)=0$ if $\lambda \in \sigma(A)$

Body:
I think it's all in the title. $p$ is some random polynomial.
I don't know how to approach this one. I've tried taking the roots of $p$, placing them on the diagonal of a new matrix and reasoning that the linear operator of the new matrix has a common specific value (if it's called that in English) with A, but I have no idea how to prove that that specific value is also a root of $p$.</p>


问:
Is A Purely Reflexive Relation Automatically Transitive?
For a given set $A$, such that $A = \{a, b, c, d\}$, and a relation on $A$, $R=\{(a,a),(b,b),(c,c),(d,d)\}$

Since transitivity is defined as:
$$\forall x,y,z \in A : (x R y \land y R z)\to x R z$$

And if $x, y, z = a$ for instance, then we have all $(x,y), (y,z),(x,z)$ and so on for all other elements in $A$.

Does this mean that the relation satisfies all requirements for transitivity?

答:
Yes. Such a relation is indeed a transitive relation, since the only relevant cases for the premise "$xRy \land yRz$" are $x=y=z$ in such relations. Since the premise never holds for cases where $x,y,z$ are not all the same, there is no need to consider them.

Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.


问:
Find the function p(x)
Solve for $p(x)$:

$$(x-1)\cdot p(x)=x^3+x-2$$

I made this expression:

$$p(x)=\frac{x^3+x-2}{x-1}$$

But no matter what I tried I couldn't simplify this further. I sat with this problem for over 2 hours and could not find a solution. I looked online for an explanation but could not find it. Can you please explain  in detail how and why I can simplify this further. I found the solution but it didn't explain why that was the solution. Please note that this is not my math homework, I just want to know how to solve this problem.

答:
Notice that $p(x)$ is of the form $p(x)=x^2+ax+b$

So we have just two coefficients to find (the one of $x^2$ being trivially $1$) and we can do it by simply plugging values for $x$.

$x=0$ you get $(0-1)p(0)=0^2+0-2\iff (-1)b=-2\iff b=2$

$x=1$ gives nothing

$x=2$ gives $p(2)=8\iff 6+2a=8\iff a=1$

Title: Find the function p(x)


问:
Kac-Moody algebras/groups for reductive groups
The question in brief is: can $\mathfrak{gl}_n$ be constructed as a Kac-Moody algebra, and can $\operatorname{GL}_n$ be constructed as a Kac-Moody group in a compatible way? In general, given any root datum, can we construct the corresponding Lie algebra and reductive group?

Background: In the books I've read on Kac-Moody algebras and groups (Infinite Dimensional Lie Algebras, by Kac, and Kac-Moody Groups, their Flag Varieties and Representation Theory by Kumar), it seems that the option of constructing a reductive (not just semisimple) Lie algebra or group is ruled out, since the only input for the construction is a generalised Cartan matrix rather than a root datum. For example, all of the reductive groups $\operatorname{SL_3}, \operatorname{GL}_3, \operatorname{PSL}_3$ have the Cartan matrix
$$ A = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} $$
however they are all different reductive groups. Can these all be constructed as Kac-Moody groups associated to that Cartan matrix? Looking at the corresponding Kac-Moody algebra, we have $\mathfrak{g}(A) \cong \mathfrak{sl}_3 \cong \mathfrak{psl}_3$, however it seems that $\mathfrak{gl}_3$ cannot be constructed since these books require that the dimension of the Cartan subalgebra is $n + \operatorname{corank}(A) = 2$, while $\mathfrak{gl}_3$ has a three-dimensional subalgebra.

It seems that as input to the Kac-Moody construction, one could instead put a root datum. For example, to construct $\operatorname{GL}_n$ or $\mathfrak{gl}_n$, we would use the root datum
$$\begin{aligned}
\text{coweight lattice} = Y &= \mathbb{Z}\{\varepsilon_1^\vee, \ldots, \varepsilon_n^\vee\}, \\
\text{weight lattice} = X &= \mathbb{Z}\{\varepsilon_1, \ldots, \varepsilon_n\}, \\
\text{perfect pairing}: \quad& \langle \varepsilon_i^\vee, \varepsilon_j \rangle \mapsto \delta_{ij}, \\
\text{coroots} = \Pi^\vee &= \{\alpha_1^\vee = \varepsilon_1^\vee - \varepsilon_2^\vee, \ldots, \alpha_{n-1}^\vee = \varepsilon_{n-1}^\vee - \varepsilon_n^\vee \}, \\
\text{roots} = \Pi &= \{\alpha_1 = \varepsilon_1 - \varepsilon_2, \ldots, \alpha_{n-1} = \varepsilon_{n-1} - \varepsilon_n \}.
\end{aligned}$$
Then the matrix $a_{ij} = \langle \alpha_i^\vee, \alpha_j \rangle$ is the type $A_{n-1}$ Cartan matrix, and we should be able to define the Cartan subalgebra $\mathfrak{h} = Y \otimes_{\mathbb{Z}} \mathbb{C}$, making it $n$ (rather than $n-1$) dimensional, and identifying $\mathfrak{h}^*$ with $X \otimes_{\mathbb{Z}} \mathbb{C}$ by extending the perfect pairing up to $\mathbb{C}$. Would most of the theory of Kac-Moody algebras and groups go through if the construction allowed this? If not, what would fail? The most extreme example of this would be constructing the Lie algebra corresponding to a torus, which would simply be the $n$-dimensional abelian Lie algebra for any $n \geq 0$ fixed in advance.

Update: I should point out some of the reasons for asking this. Firstly, there is a very good off-the-self theory for Kac-Moody algebras, and it would be nice to know if this theory could be applied to any root datum (for example, to $\mathfrak{gl}_n$ and $\hat{\mathfrak{gl}_n}$). Secondly, it seems that some authors believe that Kac-Moody algebras are defined in this generality (or perhaps to them, the term "Kac-Moody algebra" means something more general). For example, in Lusztig's book Introduction to Quantum Groups, he refers to "the Kac-Moody Lie algebra attached to the root datum".

答:
Joppy said in comment:

The answer is yes, you can use pretty much whatever realisation of the root data you want and have the Kac-Moody construction go through with minor changes (dimension of the centre, etc). The most important thing for the theory is that the simple roots and simple coroots each remain linearly independent: without this, significant parts of the theory change.


问:
For a filter on set, ultrafilters are exactly prime filters.
I'm following through a book, and I'm understanding the proof of the proposition that ultrafilters are exactly prime filters.

The proof that ultrafilters are prime goes as the following.

Suppose $F$ is ultrafilter on a set $X$ that is not prime. Then there are $A, B \subseteq X$ with $A \cup B \in F$ but $A, B \notin F$. We have proved before in the text that a filter $U$ on set $X$ is an ultrafilter if and only if for every $A \subseteq X$, $A \notin U$ if and only if there is $B \in U$ with $A \cap B = \emptyset$.

So we may find $A', B' \in F$ with $A \cap A' = \emptyset = B \cap B'$. The proof says that this implies $(A \cup B) \cap (A' \cup B') = \emptyset$. I'm having trouble with this line as $A \cap A' = \emptyset = B \cap B'$ does not imply $(A \cup B) \cap (A' \cup B') = \emptyset$. For example, take $A = \{0,1\}$ $A' = \{2,3\}$, $B = \{2,4\}$, $B' = \{1,5\}$.

So with what condition do we know that $(A \cup B) \cap (A' \cup B') = \emptyset$?

After this line, the proof is easy as $(A \cup B) \cap (A' \cup B') = \emptyset$ if and only if $A \cup B \notin F$.

Thanks!

答:
Another approach: Let $F$ be an ultrafilter on $X$ and let $A\cup B=C\in F.$

We have

$(i).\;\, \forall E\in F\,(A\cap E\ne \emptyset),$ or

$(ii).\, \forall E\in F\,(B\cap E\ne \emptyset).$

Proof: If $\neg (i),$ let $D\in F$ with $A\cap D =\emptyset.$ So $C\cap D=B\cap D.$ So for any $E\in F$ we have $$\emptyset \ne \;C\cap D\cap E\;=\;  B\cap D\cap E\;\subseteq B\cap E.$$ Now if $(i)$ holds: Then $F_A=\{(A\cap E)\cup G: E\in F\land G\subseteq X\}$ is a filter on $X$ with $A\in F_A\supseteq F.$ But $F\supseteq F_A$ by the maximality of $F,$ so $A\in F_A=F.$

Or if  $(ii)$ holds, replace $A$ with $B$ in the above paragraph to conclude $B\in F.$

Remarks: $A=(A\cap X)\cup \emptyset\in F_A.$ And if $E\in F$ then $E= (A\cap E)\cup (E\setminus A)\in F_A.$ I leave it to the reader to confirm that if $(i)$ holds then $F_A$ is a filter. A further remark: $(i)\lor (ii)$ depends only on $F$ being a filter.


问:
Why is the relation $R = \{(a, b), (a, c)\}$ over $X = \{a, b, c\}$ transitive?
I know it is not complete and I know it is asymmetric. But how is it transitive? The textbook says it is transitive...

Instead, I used a different example of transitive, asymmetric, and not complete, which is $X = \{a, b, c, d\}$ with $R = \{(a, b), (b, c), (a, c)\}$.

答:
As noted in the comments, this is indeed a transitive relation. This is because the definition of transitivity requires "$xRy \land yRz$" to hold; however, for the given relation, this never holds, and thus via a vacuous argument the relation is transitive.

Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.


问:
Question about absolute value of the addition of vectors
I would say the answer is "The angle between A and B is 120" because it |A| describes the L2 norm and every other option could not yield a correct result
1

Am I right? If so, explain why. The friend that asked this said it's wrong and it was because of a Chegg study post so I want to know why he was right/wrong.

答:
$|A + B|^2 = |A|^2 + |B|^2 +2(A\cdot B)$ which yields
$$
\begin{align}
A\cdot B
&= \frac{1}{2}\{|A + B|^2 - (|A|^2 + |B|^2)\}\\
&= -\frac{1}{2}|A|^2
\end{align}
$$
Since $A\cdot B = |A|^2\cos\theta$, we have

$|A| = |B| = 0$
or

$|A| = |B| \not= 0$ and $\cos\theta = -1/2$, which gives one of the solutions $\theta = 120\deg$.

The problem missed case 1 though.


问:
Double Integral with Trig Function
The question is:
$$\int_{1/2}^1\left( \int_{\pi/y}^{2\pi}\sin(xy) \ dx\right) \ dy+\int_1^2\left(\int_\pi^{2\pi/y}\sin(xy) \ dx\right) \ dy$$
I've tried this but I'm stuck at
$$\int_{1/2}^1\dfrac{-\cos(2\pi y)}{y} \ dy + \int_1^2\dfrac{\cos(\pi y)}{y} \ dy-2\ln2 $$

How should I proceed from here?

Thanks in advance

答: Juts put $x=\frac y 2$ in the second integral in your attempt. You will see that it cancels with the first term!

Title: Double Integral with Trig Function


问:
Inequalities considering surfaces
I have one general question considering the surface area.
If we have for example body bounded with $z^2\le x^2+y^2, z\le 2-x^2-y^2$.
The first inequality tells us we are looking outside of the cone, and the second inside of paraboloid.
But in solution they are first looking at the part that is inside of cone and paraboloid between $z=0,z=1$(and I don't understand why because we should only look outside of cone?). Second area according to solutions is symmetric to this one just below $z=0$. And the third(one that I would only consider if I had to work on this) is outside of cones and inside of paraboloid.
So, my main question is why are we looking those two areas that are inside cones, when the given inequality should only be outside of cone?

答:
We are talking about surface area so it is always on the surface of at least one of the intersecting solids. Now take the surfaces in your question with given inequalities,

$z^2\le x^2+y^2$

$z\le 2-x^2-y^2$

Equating both, we have $z^2 +z - 2 = 0 \implies z = 1, -2$.

First let's consider only the cone above $z-$ plane and its intersection with the paraboloid. Also the sketch shows you that you have part of the paraboloid inside the cone and part of the cone inside the paraboloid. So which surface area do you find? For that, let's see the inequalities.

You have surface of paraboloid inside the cone for $z \geq 1$. But at any given $z$, if you are inside the cone, clearly, $x^2 + y^2 \lt z^2$ which is not what the inequality suggests.

Now let's take the surface of the cone inside the paraboloid which is for $0 \leq z \leq 1$. As you are on the surface of the cone, the cone inequality (read equality) is obviously met and as we are inside the paraboloid surface, at any given $z$, clearly $x^2 + y^2 \lt 2 - z$ which meets the other inequality as well.

The same logic applies for the cone below $z-$plane. But you must note that for this cone, while there is finite surface of cone inside the paraboloid, for $z \leq -2$, the paraboloid is completely inside the cone surface and hence there is unbound surface area of the paraboloid inside the cone.

Last but not least the bound surface area above $z-$plane and below are not equal and it is not symmetric with respect to the $z-$ plane.

Title: Inequalities considering surfaces


问:
How to find the p-value in the chi-square test
I'm reading Probability and Statistical Inference, ninth edition by Hogg and on page 436, he gives an example using the chi-square goodness-of-fit test. How did he find this p-value?

答:
P-value is a value of quantile function at 14.4.

Quantile function may be found using special tables.


问:
Is every convergent sequence Cauchy?
Wikipedia: "Every convergent sequence (with limit s, say) is a Cauchy sequence, since, given any real number ε > 0, beyond some fixed point, every term of sequence is within distance ε/2 of s, so any two terms of the sequence are within distance ε of each other." http://en.wikipedia.org/wiki/Cauchy_sequence

Cauchy sequence in $ L^p(\mu) : \forall \epsilon>0 \space{ } \exists N \in \mathbb{N}$ such that $\forall n,m>N \space{} ||f_n(x)-f_m(x)||_p<\epsilon $

Pointwise convergent sequence: $\forall \epsilon>0 \space{ } \exists N \in \mathbb{N}$ such that $\forall n>N \space{} |f_n(x)-f(x)|<\epsilon $

Uniformly convergent sequence (on set $E$): $\forall \epsilon>0 \space{ } \exists N \in \mathbb{N}$ such that $\forall x \in E, \forall n>N \space{} |f_n(x)-f(x)|<\epsilon $

However $L^p$ norm $(\int_X|f_n-f_m|^p  d{\mu})^{\frac{1}{p}}$  is not the same as the absolute value in 2. and 3. So is the Wikipedia claim true?

答: Well, the definition of Cauchy sequence can be given in any metric space $(X,d)$ as Wikipedia points out, while the notion of converging sequence requires only a topology on a set to be well-defined (see here). So, if you can understand the sketch of proof given by Wikipedia and write it down rigourously, you'll see that it works for every metric space: you just have to substitute the absolute value of the difference of two real numbers, which is the standard metric on $\mathbb{R}$, with the given distance for an arbitrary metric space. $L^{p}$ does not make any difference, since it is a metric space with the distance induced by $\vert\vert\cdot\vert\vert_{p}$ norm. </p>

Title: Is every convergent sequence Cauchy?


问:
Convergence of series in p-adic norm
I want to know the convergence of the series

$\sum_{n\geq 0} {{p^{n+1}}\choose{p^n}}$

My idea is to show the convergence of the partial sum $|s_n-s_{n-1}|_p$ but I am stucked in expressing the terms properly.

Is there any hint to this question?

答:
Let me tell you a cute identity for binomial coefficients:
$$
\binom{a}{b} = \frac{a}{b}\binom{a-1}{b-1}.
$$
Therefore
$$
\binom{p^{n+1}}{p^n} = p\binom{p^{n+1}-1}{p^n-1}.
$$

Now let me tell you a cute congruence for binomial coefficients modulo $p$, going back to Kummer:
$$
\binom{a_0 + a_1p + \cdots + a_rp^r}{b_0 + b_1p + \cdots + b_rp^r} \equiv \binom{a_0}{b_0}\binom{a_1}{b_1}\cdots  \binom{a_r}{b_r} \bmod p
$$
where $a_j$ and $b_j$ are base $p$ digits.  The base $p$ digits of $p^{n+1}-1$ and $p^n-1$ are all $p-1$, so $\binom{p^{n+1}-1}{p^n-1} \bmod p$ can be computed as follows:
$$
\binom{(p-1) + \cdots + (p-1)p^{n-1} +(p-1)p^n}{(p-1) + \cdots + (p-1)p^{n-1} + (0)p^n}
\equiv \binom{p-1}{p-1} \cdots \binom{p-1}{p-1}\binom{p-1}{0} \bmod p,
$$
so $\binom{p^{n+1}-1}{p^n-1} \equiv 1 \bmod p$.

Since $\binom{p^{n+1}-1}{p^n-1}$ is not divisible by $p$, ${\rm ord}_p\binom{p^{n+1}}{p^n} = {\rm ord}_p(p\binom{p^{n+1}-1}{p^n-1}) = 1$. Therefore every term in the series you wrote about has the same $p$-adic absolute value, so the series does not converge (terms do not tend to $0$).

Title: Convergence of series in p-adic norm


问:
In the binomial distribution, why does order not matter?
(For context): $X$ is the random variable for number of successes ($x$) out of $n$ independent trials with probability of success $p$ and probability of failure $1-p = q$

The formula for mass my book gives me for the binomial distribution is this:
$P(X=x) = {n \choose x}p^xq^{n-x}$

My question is this. How come order doesn't matter? Why do we use ${n \choose x}$ in the above to calculate all the ways we can get $x$ successes out of $n$ independent trials instead of using $nPx$? I.E. Why do we use combination instead of permutation? Can someone explain this to me because it's been really bothering me.

Thank You!

答: This is tautological, or definitional. You only use the binomial distribution when the order of the successes doesn’t matter because the successes are all interchangeable and independent. If the order of the successes matters, or the successes aren’t interchangeable, or they’re not independent, then you shouldn’t be using the binomial distribution, lol.</p>


问:
Graph $|x| + |y| \leq 1$
I need to graph the inequality in the title but I'm doing something wrong...
so far I have:

$$|y| \leq -|x| + 1 \Leftrightarrow$$

$$\Leftrightarrow |y| \leq -x + 1, \space \forall \space x > 0 \space \vee |y| \leq x + 1, \space \forall \space x < 0 $$

This gives me four different conditions, depending on the value of $x$:

$$y \leq -x + 1, \space \forall \space (x > 0 \space \wedge -x + 1 < 0)$$
$$-y \leq -x + 1, \space \forall \space (x > 0 \space \wedge -x + 1 > 0)$$
$$y \leq x + 1, \space \forall \space (x < 0 \space \wedge x + 1 > 0)$$
$$-y \leq x + 1, \space \forall \space (x < 0 \space \wedge x + 1 < 0)$$

The resulting graph is not the correct one.

答:
A little higher perspective about this graph, this is the unit ball of radius $1$ for the norm $||\cdot||_1=|x|+|y|$ it is a rotated square.

Note that for $||\cdot||_\infty=\max(|x|,|y|)$ the unit ball is a square  with sides // to the coordinate axis.

This is easy to see $|x|\le 1$ is a vertical band and $|y|\le 1$ is an horizontal band, intersection of both conditions gives a square.

Now $|x|+|y|\le 1$ is either $\{x+y,x-y,-x+y,-x-y\}\le 1$ depending on signs of $x,y$

so we can also rewrite it $\begin{cases}|x+y|\le 1\\|x-y|\le 1\end{cases}\quad$ or equivalently $\max(|x+y|,|x-y|)\le 1$

If you do the change of variable $\begin{cases}u=x+y\\v=x-y\end{cases}\quad$ you'll notice $$||(x,y)||_1\le 1\iff ||(u,v)||_\infty\le 1$$

This is why you get a rotated square.

Note:

The change of variable is not a proper rotation, it needs to be normalized with a factor $\sqrt{2}$ (because for a rotation of angle $45^\circ$ both are equal to $\cos(45^\circ)=\sin(45^\circ)=\frac 1{\sqrt{2}})$

Title: Graph $|x| + |y| \leq 1$


问:
Pedantic question about symmetric/non-symmetric bilinear forms: is symmetry of the bilinear form redundant?
A quadratic form $q$ is defined in terms a symmetric bilinear form $f$ by $q(\mathbf{v}):=f(\mathbf{v},\mathbf{v}).$ However, this definition seems too restrictive. Why must $f$ be symmetric? Even if $f$ were a non-symmetric bilinear form, $q$ could still be defined in terms of $f$ and would still have the defining property of quadratic forms, namely that,

In mathematics, a quadratic form is a polynomial with terms all of degree two.

Any clarifications are welcomed, thank you.

答:
Actually it's the opposite: the non-symmetric part of the bilinear form is redundant. Any bilinear form $B(u, v)$ has a canonical decomposition (over a field of characteristic $\neq 2$) into a symmetric part and a skew-symmetric part

$$B(u, v) = \frac{B(u, v) + B(v, u)}{2} + \frac{B(u, v) - B(v, u)}{2}$$

and the skew-symmetric part of $B$ does not contribute to $Q(v) = B(v, v)$ at all, so can be safely ignored. That is, if we write $S(u, v)$ for the symmetric part, then $B(v, v) = S(v, v)$. This is why we define quadratic forms using only symmetric bilinear forms. The symmetric part can be recovered from the quadratic form $Q$ as the polarization $\frac{Q(u + v) - Q(u) - Q(v)}{2}$ so the quadratic form and the symmetric bilinear form are two different ways of talking about the same thing.

You can work this out quite explicitly for two variables, for example: here the vector space of quadratic forms is $3$-dimensional with basis $\{ x^2, xy, y^2 \}$ and the vector space of symmetric bilinear forms is also $3$-dimensional and can be identified with the vector space of symmetric matrices, with basis

$$\left\{ \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right] \right\}$$

and these correspond to the quadratic forms $x^2, 2xy, y^2$ respectively. Meanwhile the vector space of skew-symmetric forms is $1$-dimensional with basis $\left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right]$ and the corresponding quadratic form is identically zero.

(There are additional subtleties if the field has characteristic $2$ because we can no longer perform the division by $2$ necessary for the formulas above, but on a first pass we can safely ignore them. Most people will never need to deal with them.)


问:
Convergence of sequence in p-adic integers
Let $a\in Z_p^{*}$. I need to prove the sequence $\{a^{p^n}\}_{n\geq 0}$ converges in $\mathbb{Z}_p$ to $w$, where $$a=wb$$ for $w$ is the $p-1$ root of unity in $\mathbb{Z}_p^*$ and $b\in 1+p\mathbb{Z}_p$.

I feel like I need to use Hensel's Lemma for this question but I do not know how to start it. Is there any particular hint for this question?

答:
Don't use Hensel's lemma: use the contraction mapping theorem.
For $a, b \in \mathbf Z$, show
$$
a \equiv b \bmod p^k \Longrightarrow a^p \equiv b^p \bmod p^{k+1}.
$$
Since $p$-adic integers modulo $p^k$ are "just like" integers modulo $p^k$, the same implication holds for $a, b \in \mathbf Z_p$. Therefore when $a$ and $b$ are in $\mathbf Z_p$,
$$
|a^p - b^p|_p \leq \frac{1}{p}|a-b|_p
$$
Indeed, if $a = b$ there is nothing to check, and if $a \not= b$ let $p^k$ be the power of $p$ where $a - b = p^ku$ for $u \in \mathbf Z_p^\times$.  That makes $|a - b|_p = 1/p^k$, so from the above implication,
$$
|a^p - b^p|_p \leq \frac{1}{p^{k+1}} = \frac{1}{p}|a-b|_p.
$$

Now use the contraction mapping theorem for the $p$th power map on the coset $C = a + p\mathbf Z_p$. This is a complete metric space and the $p$th power map sends $C$ to itself since $x^p \equiv x \bmod p\mathbf Z_p$ for all $x \in \mathbf Z_p$.  The limit coming from this contraction is a number $w \in C$ such that $w^p = w$. If $a \in \mathbf Z_p^\times$ then $w \not= 0$, so $w^{p-1} = 1$. (If $a \in p\mathbf Z_p$ then $w = 0$.)


答: If $X\sim \chi ^2(3)$, then the p-value is given by $\mathbb P(X>14.4)$. With a table or with software it is easy to obtain the probability $0.0024$.

Title: How to find the p-value in the chi-square test

Body: I'm reading Probability and Statistical Inference, ninth edition by Hogg and on page 436, he gives an example using the chi-square goodness-of-fit test. How did he find this p-value?


问:
Finding surface area and volume of a sphere using only Pappus' Centroid Theorem
I wonder if it is possible to derive surface area and volume of a sphere seperately using techniques involving pappus' theorem.

I did some calculation and found out the ratio of surface area and volume. Here is my work,

My key observation is finding out that the centroid of a semidisk of raidus $r$ is also the centroid of a semicircle of raidus $2r/3$ when the centers coincide. (By "centers" I mean the centers of full circle and disc induced by half of them.)

I sliced the semidisk identical pieces of triangles(infinitely many) and by locating each triangle's centroid I form a semicircle of radius $2r/3$ which must have the same centroid with semidisk.

Then let's say our centroid is located $h$ distance above the center. We still don't know it.

Using the theorem, the circular path taken by the centroid of the semidisk times the area of the centroid should give the volume of the sphere of radius $r$. And similarly, the circular path taken by the same centroid of the semicircle times the arc length of semicircle should give the surface area of a sphere of radius $2r/3$.

We know that when the radius increases with a proportion, corresponding surface area will also increase with the square of that proportion. Thus, we need to multiply the surface area of sphere of radius $2r/3$ by the factor $9/4$ to get the surface area of the sphere of radius $r$.

Here is the calculations,

$$V = 2\pi h * \pi r^2/2=\pi^2r^2h$$
$$S = 9/4*2\pi h * \pi (2r/3) = 3\pi^2rh$$

Since we don't know $h$, I simply divide them to cancel it out and get,
$$V/S = r/3.$$

答:
Comment

It is insightful to consider center of mass in general after integration ; from a sector in full $ \alpha= \pi/2$ towards a narrow triangle $\alpha\to 0 $ in limit:

For semi-disc/volume (blue)
$$ y_{semi-disc}= \dfrac{Vol}{\text{area before sweep}}=\dfrac{2 r\sin \alpha}{3\alpha}\to\dfrac{2r}{3}$$

For arc/area (red)
$$ y_{semicircle-arc}== \dfrac{Area}{\text{arc length  before sweep}}\dfrac{r\sin \alpha}{\alpha}\to r$$


问:
Sum of unitary transformation
I am having struggle with this question.

suppose I have two unitary matrices.

Is their sum is  normal ?

I am try to give an example to show it is not true and I can not find.

I try to proof and I reach this results:

$T$,$S$ are unitary then:

$(T+S)(T+S)^{*}=(T+S)^{*}(T+S)$

$(T+S)(T^{*}+S^{*})=(T^{*}+S^{*})(T+S)$

$(T+S)T^{*}+(T+S)S^{*}=(T^{*}+S^{*})T+(T^{*}+S^{*})S$

$TT^{*}+ST^{*}+TS^{*}+SS^{*}=T^{*}T+S^{*}T+T^{*}S+S^{*}S$

$ST^{*}+TS^{*}=S^{*}T+T^{*}S$

and I can not continue from here.

I can not tell is this equation is true or not.

and BTW

matrix is unitary iff matrix is symmetric ?

Thanks in advanced !!

答:
First, symmetric matrix and unitary matrix are two unrelated concepts. It is easy to find symmetric matrices that are not unitary, and vice versa.

For an example where $T+S$ is not normal, we can take $T=\begin{bmatrix} 0 & i \\ -i & 0\end{bmatrix}$, and $S=\begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix}$ with $\lambda_i\bar{\lambda}_i=1$ for $i=1,2$. The calculations are omitted.

It is interesting to note however, that if both $S$ and $T$ are taken in $SU(2)$ (i.e., $2\times 2$ unitary matrices whose determinant is one), then indeed $S+T$ is normal.  One way to see this is to identify $SU(2)$ with the norm 1 elements in the Hamilton quaternion algebra ( then the *-operator corresponds to taking conjugate).  Then $$ST^*+TS^*=ST^*+(ST^*)^*=\mathrm{Tr}(ST^*),$$ and similarly $$T^*S+S^*T=\mathrm{Tr}(T^*S).$$
Now $\mathrm{Tr}(ST^*)=\mathrm{Tr}(T^*S)$.

Hmm, I guess I am only using the fact that $X+X^{-1}=\mathrm{Tr}(X)$ for any $X\in SL_2(\mathbb{C})$ using the formula $\begin{bmatrix} a & b \\ c & d\end{bmatrix}^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a\end{bmatrix}$.

Title: Sum of unitary transformation


答: Hint. When you see problems like this (that is, division problems with a polynomial of higher degree in the numerator and a linear polynomial in the denominator), let the denominator be equal to a new variable. In this case let $x-1=u$ and see what you obtain.

Title: Find the function p(x)

Body:
Solve for $p(x)$:

$$(x-1)\cdot p(x)=x^3+x-2$$

I made this expression:

$$p(x)=\frac{x^3+x-2}{x-1}$$

But no matter what I tried I couldn't simplify this further. I sat with this problem for over 2 hours and could not find a solution. I looked online for an explanation but could not find it. Can you please explain  in detail how and why I can simplify this further. I found the solution but it didn't explain why that was the solution. Please note that this is not my math homework, I just want to know how to solve this problem.


问:
Context in context of Type Theory
I am reading a book on  homotopy type theory  or let me better say 'the' book of HoTT.  On page 416, the author
intruduces a judgment that includes the ctx (context) formalism. The
judgment

$$(x_1 : A_1, x_2 : A_2, ... , x_n : A_n) \text{ ctx }$$

expresses that each $A_i$ is a type in the context
$x_1 : A_1, x_2 : A_2, ... , x_{i-1} : A_{i-1}$. The next statement
I not understand:

In particular, therefore, if $\Gamma \vdash a : A $ and
$\Gamma \text{ ctx } $, then we know that each $A_i$ contains only
the variables $x_1, ... , x_{i-1}$, and that $a$ and $A$ contain
only the variables $x_1, ... , x_n $.

As far I not understand this conclusion. By definition the judgement
$\Gamma \vdash a : A $ means just that $a: A$
in context of $\Gamma $. Literally: $a$ has type $A$
under the assumptions listed in $\Gamma$.

The judgment $\Gamma \text{ ctx } $ says literally that the context $\Gamma$
is well-formed. As far as I understand it correctly, formally this
is by definition above the statement that if
$\Gamma = (x_1 : A_1, x_2 : A_2, ... , x_n : A_n)$ then that just
means recursively that $A_i$ is a type in the context
$x_1 : A_1, x_2 : A_2, ... , x_{i-1} : A_{i-1}$. Can it be token as
general 'definition' of $\Gamma \text{ ctx } $, am I right?

But then, why from $\Gamma \vdash a : A $ and
$\Gamma \text{ ctx } $  we conclude that $A_i$ contains only
the variables $x_1, ... , x_{i-1}$ and that $a$ and $A$ contain
only the variables $x_1, ... , x_n $? This 'conclusion' isn't clear
for me.

答:
The sentence you are quoting is part of an introduction, meant to give you some intuition behind the formalism of the theory. It does not pretend to give you the meta-theoretical proof that you are looking for: it says "then we know ...", not "this is a proof that ...".

In other words, the formal rules of the theory are designed so that in a context $\Gamma$:

the type $A_i$ may refer only the the variables $x_0, ..., x_{i-1}$

the terms $a$ and $A$ may refer only to the variables $x_0, ..., x_n$

To have a proof of it, you need to check -all- formal rules of the theory, verify that the rule "Vble" in A.2.2 is the only one to introduce fresh variables, and does it only in a context where this variable is already defined by its type.

Title: Context in context of Type Theory


问:
prove $x_n(t) =(\prod_{k=0}^{n-1}a(k))y_0+\sum_{k=0}^{n-1}(\prod_{j=k+1}^{n-1}a(j))c(k)$..
Prove the following : Let $a(n)$ and $c(n)$,$n ∈ N$, be real sequences. Then the linear first order difference
equation $x(n + 1) = a(n)x(n) + c(n)$ with initial condition $x(0) = y_0$ has the following solution:
$x_n(t) =(\prod_{k=0}^{n-1}a(k))y_0+\sum_{k=0}^{n-1}(\prod_{j=k+1}^{n-1}a(j))c(k)$..

I've managed to show that $x_H=(t)=(\prod_{k=0}^{n-1}a(k))y_0$ but i couldn't figure how to get the non-homogenous part.

any hint?

答:
We consider the recurrence relation
\begin{align*}
\color{blue}{x_{n+1}}&\color{blue}{=a(n)x_n+c(n)\qquad\qquad n\geq 0}\tag{1}\\
\color{blue}{x_0}&\color{blue}{=y_0}
\end{align*}
where $\left(a(n)\right)_{n\geq 0},\left(c(n)\right)_{n\geq 0}$ are real-valued sequences and the initial value $x_0=y_0$ is given. We do not assume that $x_n=x_n(t), n\geq 0$ is a function of $t$. This is not needed, as it is not part of the right-hand side of OPs solution.

Method: The idea is to simplify the recurrence relation to
\begin{align*}
z_{n+1}&=z_n+d(n)\qquad\qquad n\geq 0
\end{align*}
by multiplication with a convenient factor. This way we get rid of $a(n)$ and it can then be solved easily via telescoping.

We assume $a(n)\ne 0, n\in\mathbb{N}$ and multiply (1) with
\begin{align*}
\color{blue}{A(n+1)}&\color{blue}{=\frac{1}{\prod_{j=0}^n a(j)}\qquad\qquad n\geq 0}\tag{2}\\
\end{align*}
We set $A(0)=1$.

We multiply (1) with $A(n+1)$ and obtain:
\begin{align*}
A(n+1)x_{n+1}&=A(n+1)a(n)x_n+A(n+1)c(n)\\
&=A(n)x_n+A(n+1)c(n)\tag{3}
\end{align*}

From (3) we get by summing up from $k=0,\ldots,n$ and telescoping:
\begin{align*}
\sum_{k=0}^nA(k+1)x_{k+1}&=\sum_{k=0}^nA(k)x_k+\sum_{k=0}^nA(k+1)c(k)\tag{4}\\
A(n+1)x_{n+1}&=A(0)x_0+\sum_{k=0}^nA(k+1)c(k)\tag{5}\\
\color{blue}{x_{n+1}}&=\frac{A(0)x_0+\sum_{k=0}^nA(k+1)c(k)}{A(n+1)}\\
&=\left(x_0+\sum_{k=0}^n\frac{c(k)}{\prod_{j=0}^{k}a(j)}\right)\prod_{j=0}^{n}a(j)\tag{6}\\
&\,\,\color{blue}{=y_0\prod_{j=0}^n a(j)+\sum_{k=0}^nc(k)\prod_{j={k+1}}^na(j)}\tag{7}
\end{align*}
and the claim follows.

Comment:

In (4) we sum up (3) from $k=0,\ldots,n$.

In (5) we apply telescoping and cancel common summands.

In (6) we use the definition (2) of $A(k)$.

In (7) we simplify and use the initial condition $x_0=y_0$.


答:
How did he find this p-value?

Using a standard chi-square table you can say that the p-value is less than $0.005$ as $14.4 >12.84$ tabulated.

This result is enough to take any decision, as it is very very low.

If you want a precise result  you have to use a calculator. For example, with Excel, typing +DISTRIB.CHI(14.4;3) you get $p=0.002408$


问:
Is $a$ bigger than $0$ or not?
Consider Goldbach's original conjecture (no need worry, because we don't talk about "Goldbach" itself):

every integer $n> 2$ could be detached as a sum of three primes. (In Goldbach's age, $1$ is regarded as a prime.)

For any $n>2$, we denote the statement above as $p(n)$. Then define

$$a_n:=\begin{cases}
0, &\text{if $p(n)$ is true},\\
1, &\text{otherwise}.
\end{cases}$$

Note that $a(n)$ is definite, since every $p(n)$ could be verified, at least by machine, as long $n$ is given. However, consider the decimal number $$a:=0.00a_3a_4a_5\cdots.$$

What's $a$? Is it a real number?

If $a$ is a real number, then we can compare the magnitude of $a$ and
$0$, since the real number field is ordered.

But we can't in fact. If $a=0$, then "Goldbach" is proven, and if not, then "Goldbach" is overturned. However, it's in suspense hitherto whether Goldbach is true or not.

Probably, we may say $a$ is indefinite, that's to say, we do not know what $a$ is yet. But a new question is coming about:

Why we know any digit but do not know the number itself ? In order
to know the number, what else we need ?

Further, let's compare $a$ with another classic number
$$\pi=3.1415926\cdots.$$

Indeed, we do not know all the digits of $\pi$ untill we compute every one. But we can safely say it's $\pi$. What's the exact difference between $a$ and $\pi$ on earth? Why we think we know $\pi$ but do not know $a$? In another word, what does "knowing a number" mean exactly? And in what sense can we say we know a number?

答:
We can know any digit of $a$ (after a possibly very long but finite computation), but we do not currently know every digit. There is a difference: if you wanted to know the $n$th digit of $a$, give me some long but finite time, and I can tell you. But that is not the same as saying that I can tell you, currently, off the back of my mind, every single digit of $a$, for arbitrary $n$.

There is no paradox, you just need to be precise about the meaning of "know". You could say, for instance, that we can know any digit of $a$, but do not currently know every digit.

Title: Is $a$ bigger than $0$ or not?


问:
Evaluate $ \lim_{n \to \infty} \int_{-\infty}^\infty \frac{1}{1+x^2e^{nx^2}}\,dx$
I am trying to evaluate the following limit:

$$ \lim_{n \to \infty} \int_{-\infty}^\infty \frac{1}{1+x^2e^{nx^2}}\,dx$$

Now seems we can use the fact that $$ \int_\mathbb{R}\frac{1}{1+x^2}dx = \pi$$ since this is the definite integral of $\arctan(x)$.

I tried finding a clever way to re-write the denominator but I am not getting anywhere.

答:
One does not need any measure theory. It suffices to study $\int_0^\infty \frac{dx}{1+x^2e^{nx^2}}$.  Observe that for $x>0$,
\begin{align}
0< \frac1{1+x^2e^{nx^2}}
&=\frac{e^{-nx^2}}{e^{-nx^2}+x^2}
\le \frac{e^{-nx^2}}{x^2}
\end{align}
Also, $\frac1{1+x^2e^{nx^2}}\le 1$.
So for any $\epsilon>0$,
\begin{align}
0&\le \int_0^\infty \frac{dx}{1+x^2e^{nx^2}}
\\&
\le \int_0^\epsilon dx + \int_{\epsilon}^\infty \frac{e^{-nx^2}}{x^2}dx
\\&
= \epsilon + {\sqrt n}\int_{\epsilon\sqrt n}^\infty \frac{e^{-y^2}}{y^2}dy
\\&
\le \epsilon + \frac1{\epsilon^2\sqrt n}\int_{\epsilon\sqrt n}^\infty e^{-y^2}dy
\\&
\le \epsilon + \frac1{\epsilon^2\sqrt n}\int_{0}^\infty e^{-y^2}dy    \end{align}
Its well known that $\int_{0}^\infty e^{-x^2}dx $ is finite, but a good enough elementary bound follows from $e^t \ge 1+t$, or rather $e^{-t} \le \frac1{1+t}$, and the fact that $\int_0^\infty \frac{dy}{1+y^2} =\frac\pi2$ which is assumed in the OP. In addition, the integrand $\frac1{1+x^2e^{nx^2}}$ is decreasing as $n\to\infty$. So, the desired limit exists, and
$$0 \le \lim_{n\to\infty} \int_0^\infty \frac{dx}{1+x^2e^{nx^2}} \le \epsilon$$
Since $\epsilon$ was arbitrary, we conclude that $ \lim_{n\to\infty} \int_0^\infty \frac{dx}{1+x^2e^{nx^2}}=0$.


问:
Approximation of the Gamma function for small value
It is well known that $\Gamma(u) \stackrel{u \to 0}{\sim} \frac{1}{u}$.

I am looking for more precise information on the behavior of $\Gamma(x)$ when $x$ is small, ie: $x\to 0$.

My question is then, are there accurate (for small value) inequalities for the Gamma function ?

Any other information on the behavior is also very welcome.

答:
I'm a little late to the party but I derived something for this exact purpose.

Published here: http://albert.life/Math/FactorialApproximation/

Wolfram Alpha compatible: x!≈(0.44265*x+e^-eulergamma)^(0.986261*x) for -1/e<x<0

For applying above to Gamma, shift one, via Γ(x+1)==x!


问:
Substitution principle example? (for ring homomorphisms $R[x]\to S$)

I have to know the substitution principle for an upcoming exam. This is the definition given in my book. I understand the notation I believe. However, I am having a very difficult time understanding what this might look like. Can anyone give me a simple example that would display the principle pretty well? I'm having difficulty understanding the point or when/how this would be used or constructed. Any other forms of explanation are also welcome. Thank you.

答:
Given $f$ is a ring homomorphism from $R$  to $S$. We can fine an evaluation map $\psi_\alpha:R[x]\to R$,$\sum_i r_ix^i\mapsto \sum_ir_i\alpha^i$(which is also a ring homomorphism).
Now, we have
$R[x]\to^{\psi_{\alpha}} R\to^f S$.
Thus we have desire map $\Phi=fo\psi_\alpha:R[x]\to S$ that agrees on the coefficients of the polynomial and send $x\mapsto \alpha$.


问:
Searching for unbounded, non-negative function $f(x)$ with roots $x_{n}\rightarrow \infty$ as $n \rightarrow \infty$
If a function $y = f(x)$ is unbounded and non-negative for all real $x$,
then is it possible that it can have roots $x_n$ such that $x_{n}\rightarrow \infty$ as $n \rightarrow  \infty$.</p>

答:
For an analytic example, you can take $f : \left\{ \begin{array}{ccc} [0,+\infty) & \to & \mathbb{R} \\ x & \mapsto & x + x \sin(x) \end{array} \right.$.

Then $f(x)=0$ iff $x=0$ or $\sin(x)=-1$ (ie. $x=-\pi/2[2\pi]$) and $f(x) \geq x-x=0$.

More generally, you can take $f : x \mapsto g(x)(1+\sin(x))$ where $g(x) \geq 0$. Then the graph is a sinusoid bounded above by $2g(x)$. For example:


问:
Write a linear program for the following model
My approach is at the bottom

The table at the bottom gives a composition of various foods used in making cereals. The other material in each food is fiber, water, etc. The company blends these food materials and makes two kinds of cereals. In the process of blending, 3% of protein, 5% of starch, and 10% of minerals and vitamins are completely lost from the mix. For each 100 kg of foods added in the blend, the blending process adds 5 kg of other material (mainly water and fat).
Cereal type 1 sells for \$1.50/kg. It should contain at least 22% protein, 2% of minerals and vitamins, and at most 30% of starch by weight. Cereal type 2 sells for \$1.00/kg. It should contain at least 30% starch by weight. The company would like to determine the most profitable product mix that meets the requirements described above.

What I tried:
Let $x_1=$ food 1, $x_2=$ food 2, $x_3=$ food 3, $x_4=$ food 4
For cereal 1:

minimize $0.68x_1+0.27x_2+0.31x_3+0.45x_4$

subject to $45x_1+7x_2+12x_3+27x_4-3\geq22$ (protein)

$12x_1+38x_2+25x_3+40x_4-5\geq 30$ (starch)

$4x_1+x_2+2x_3+3x_4-10\geq 2$ (minerals)

For cereal 2:

minimize $0.68x_1+0.27x_2+0.31x_3+0.45x_4$

subject to $12x_1+38x_2+25x_3+40x_4-5\geq 30$ (starch)

$x_1\leq 1500, x_2\leq 500, x_3\leq 1000, x_4\leq 2000$

Then overall, we want to maximize $1.5C_1+C_2$

I don't think that I am doing this correctly but I can't see any other way to do it.

答:
$$\bf\text{Review of your solution}$$
Looking at the objective functions, it seems $x_i$ represents the weight of different foods in $kg$ used to make the cereals per day.

You can't use $x_i$ to denote the amount of food $i$ in both cereal $1$ and cereal $2$ because both cereals have different nutritional requirements and you might obtain an optimum at different values of $x_i$ for both cereals. Use $x_i$ to denote the amount of food $i$ in cereal $1$ and $y_i$ to denote the corresponding amount for food $2$.

Your existing constraints are incorrect as you disregarded the percentages. For example, in the first constraint, the coefficients should be $0.45,0.07,...$. You subtracted $3$ from the amount of protein. Note that the question mentions $3\color{red}\%$ of the protein is lost and the remaining protein should be at-least $22\color{red}\%$ of the total weight of the cereal.

You didn't take into account the availability of different foods per day.

I am not sure what $C_{1,2}$ represent. But according to me, the one and only objective of this linear program should be to maximize the profit of the company assuming all cereal is sold.

$$\bf\text{My solution}$$

The amount of protein added into cereal $1$ is $0.45 x_1+0.7x_2+0.12x_3+0.27x_4$ out of which $3\%$ is lost, so protein present in cereal $1$ is $p_1=0.97[0.45 x_1+0.7x_2+0.12x_3+0.27x_4]$.

Similarly starch and minerals present in cereal $1$ are $s_1=0.95[0.12x_1+0.38x_2+0.25x_3+0.40x_4]$ and $m_1=0.90[0.04x_1+0.01x_2+0.02x_3+0.03x_4]$.

The combined weight of the foods added to the blend is $x_1+x_2+x_3+x_4$ and to this water and fat are added to make the total weight $w_1=1.05[x_1+x_2+x_3+x_4]$.

The cost of making the cereal is $c_1=0.68x_1+0.27x_2+0.31x_3+0.45x_4$ and the income from total sale is $1.50w_1$ so the profit from sale is $P_1=1.5w_1-c_1$.

The constraint on the amount of protein is thus $p_1\ge0.22w_1$. Similarly $s_1\le0.30w_1$ and $m_1\ge0.02w_1$.

For cereal $2$, we only have a constraint on the amount of starch: $s_2\ge0.30w_2$. The profit from sale of cereal $2$ is $P_2=w_2-c_2$. Thus we have the following linear program:

\begin{align*}
\text{maximize}&~~\mathbf{\text{Profit} P}=P_1+P_2\\
\text{subject to}&~~\bf\text{Cereal 1 nutritional constraint(s)}\\
&~~\color{blue}{p_1\ge0.22w_1}\\
&~~\color{blue}{s_1\le0.30w_1}\\
&~~\color{blue}{m_1\ge0.03w_1}\\\\
&~~\bf\text{Cereal 2 nutritional constraint(s)}\\
&~~\color{green}{s_2\ge0.30w_2}\\\\
&~~\bf\text{Food availability constraint(s)}\\
&~~\color{red}{x_1+y_1\le1500}\\
&~~\color{red}{x_2+y_2\le500}\\
&~~\color{red}{x_3+y_3\le1000}\\
&~~\color{red}{x_4+y_4\le2000}\\\\
&~~\bf\text{Non-negative constraint(s)}\\
&~~x_i,y_i\ge0\\\\
\text{where}&~~p_1=0.97[0.45 x_1+0.7x_2+0.12x_3+0.27x_4]\\
&~~s_1=0.95[0.12x_1+0.38x_2+0.25x_3+0.40x_4]\\
&~~m_1=0.90[0.04x_1+0.01x_2+0.02x_3+0.03x_4]\\
&~~w_1=1.05[x_1+x_2+x_3+x_4]\\
&~~c_1=0.68x_1+0.27x_2+0.31x_3+0.45x_4\\
&~~P_1=1.5w_1-c_1\\\\
&~~s_2=0.95[0.12y_1+0.38y_2+0.25y_3+0.40y_4]\\
&~~w_2=1.05[y_1+y_2+y_3+y_4]\\
&~~c_2=0.68y_1+0.27y_2+0.31y_3+0.45y_4\\
&~~P_2=w_2-c_2
\end{align*}

Now it is your job to bring all the constraints and objective function in terms of $x_i$ and $y_i$ only.


问:
finding a four elements group and binary operation upon it that has closure, associativity, an identity element and an inverse to each element.
the question is in the heading. I tried using the group {0,1,2,3} and the binary operation $*$ defined as $a*b=|a-b|$ which I proved to have closure, an identity and an inverse, but am unable to prove associativity; $(a*b)*c=|a-b|*c=||a-b|-c|$ and $a*(b*c)=a*|b-c|=|a-|b-c||$. I cant seem to show they are equal (maybe I could do it by separating into cases, $x>y$, $x<y$, $x=y$ so that $x,y\in \{0,1,2,3\}$, but there seems to be a lot of cases to check, and I'm sure there's an easier solution).

答:
If you have a $2$-element group, you can take the Cartesian product of it with itself to get a four-element group. This gives rise to a group called the Klein four-group, which is (up to isomorphism) the only group with four elements other than Hank Igoe’s sensible and obvious suggestion in the comments, $\Bbb{Z}/4\Bbb{Z}$. Hank’s is probably better if you insist on labeling the elements of your group as “$0$”, “$1$”, “$2$”, “$3$”.

The number of elements is small enough that you could verify by hand if your operation gave rise to a group. I think that your operation does not, because it lacks associativity. If it had associativity, I should be able to find an element $b$ so that $1 \star b = 3$ by left-starring by the inverse of $1$. But clearly there’s no element in $\{ 0, 1, 2, 3 \}$ at a distance of $3$ from $1$.


问:
Proof that each well ordered set is similar to a unique ordinal number (according to Halmos)
I'm trying to follow the proof that is given in the "Naive Set Theory" by Halmos that each well ordered set is similar to a unique ordinal number (which Halmos calls the Counting Theorem):

Counting Theorem. Each well ordered set is similar to a unique ordinal number.

Since for ordinal numbers similarity is the same as equality, uniqueness is obvious. Suppose now that $X$ is a well ordered set and suppose that an element $a$ of $X$ is such that the initial segment determined by each predecessor of $a$ is similar to some (necessarily unique) ordinal number. If $S(x, \alpha)$ is the sentence that says "$\alpha$ is an ordinal number and $s(x)$
is similar to $\alpha$", then, for each $x$ in $s(a)$, the set $\{\alpha: S(x, \alpha)\}$ can be formed; in fact, that set is a singleton. The axiom of substitution implies the existence of a set consisting exactly of the ordinal numbers similar to the initial segments determined by the predecessors of $a$. It follows, whether $a$ is the immediate successor of one of its predecessors or the supremum of them all, that $s(a)$ is similar to an ordinal number. This argument prepares the way for an application of the prinicple of transfinite induction; the conclusion is that each initial segment in $X$ is similar to some ordinal number. This fact, in turn, justifies another application of the axiom of substitution, just like the one made above; the final conclusion is, as desired, that $X$ is similar to some ordinal number.

I cannot understand what is the justification of the step: "It follows, whether $a$ is the immediate successor of one of its predecessors or the supremum of them all, that $s(a)$ is similar to an ordinal number."

I understand the construction of the set $\{\alpha: S(x, \alpha)\}$, but I only see that each element of that set is an ordinal number which is similar to some of the initial segments of $s(a)$. I also see that they are ordered by continuation (but so is any set of ordered numbers). From the step that puzzles me, I expect that the said set has to be an ordinal number, or at least to be similar to an ordinal number, but I can't see how this can be shown (or how this can be obvious).

PS.
As far as I know, some of the terms that Halmos uses have modern equivalents:
similar = order-isomorphic
Axiom of Substitution = Axiom of Replacement

Halmos also has proven a fact hat every set of ordinal numbers has a supremum in the text just before that. This might be helpful, as $\{\alpha: S(x, \alpha)\}$ is definitely a set of ordinal numbers. So it must have a supremum, but what is it to do next, I don't understand.

UPDATE

After asking the question I think I pinpointed the part which is difficult to me.

Halmos asserts the fact that $s(a)$ is similar to an ordered number whether $a$ is the immediate successor of one of its predecessors, or the supremum of them.

In case $a$ is the immediate successor, it's easy to see that $s(a)$ is similar to some ordered number. There should be some $b \in s(a)$ such that $a$ is the immediate successor of $b$. Then $s(b)$ is similar to some ordinal number, say $\beta$ and $s(a)$ is similar to $\beta^+$. But I don't understand how to show the same if $a$ is the supremum of $s(a)$.

答:
Been reading this book to fill some knowledge gaps on set theory, so here's my solution (taking uniqueness of ordinal equivalence for granted):

Claim: Suppose that $X$ is a well-ordered set such that every initial segment of $X$ is similar to an ordinal number. Then $X$ is itself similar to an ordinal number.

Proof: For any $x \in X$, we know that there exists an ordinal number similar to $x$ by hypothesis; the singleton set containing this ordinal is then precisely the set $\{ \alpha : \alpha \text{ is an ordinal and } \alpha \cong s(x) \}$ by uniqueness of ordinal similarity. By the axiom of substitution, there exists a function $F$ with domain $X$ such that $F(x)$ is the aforementioned set for each $x \in X$. Since each element $F(x)$ is a (non-empty) singleton set, we can form the unique function $f$ of domain $X$ satisfying $F(x) = \{ f(x) \}$ (i.e. $f$ removes the curly braces from the result of $F$). The result is a function $f$ with the property that, for all $x \in X$, $f(x)$ is an ordinal and $f(x) \cong s(x)$.

Denote by $S$ the range of $f$. We claim that $S$ is an ordinal. As established, $S$ is a set of ordinals, so it is well ordered under set inclusion/membership. Suppose now that $\alpha \in S$. The initial segment of $\alpha$ in $S$ is included in $\alpha$ by definition; it remains to be shown that every element of $\alpha$ is included in $S$ (and therefore also in its initial segment). Suppose $\beta \in \alpha$. Then if $\alpha$ is the image of $x \in X$, it is similar to $s(x)$, say through the similarity $\varphi$. Now it is easy to show that $\varphi$ restricts over the initial segment of $\beta$ in $\alpha$ to a similarity. Noting that $\varphi(s(\beta)) = s(\varphi(\beta))$ as a consequence of the fact that $\varphi$ is a similarity, $\left.\varphi\right|_{s(\beta)}$ is a similarity between $s(\beta) = \beta$ (taking $\beta \in \alpha$, an ordinal) and $s(\varphi(\beta))$. Since $\varphi(\beta) \in s(x) \subseteq X$, we have $\beta \in S$ as desired.

Now the sets $X$ and $S$ are each well-ordered, and so there are three cases. If they are similar, we are done. If $X$ is similar to an initial segment of $S$, then since an initial segment of an ordinal is again an ordinal, we are done. Finally, suppose $S$ is similar to an initial segment $s(x)$ of $X$. We then have that $s(x)$ is similar to the ordinals $f(x)$ and $S$, so that $f(x) \cong S \implies f(x) = S$. However, $f$ has range $S$, so $f(x) \in S \implies f(x) \in f(x) \implies f(x) < f(x)$, a contradiction. $\blacksquare$

Now the proof in the book goes as follows:

Since for the ordinal numbers similarity is the same as equality, uniqueness is obvious. Suppose now that $X$ is a well ordered set and suppose that an element $a$ of $X$ is such that the initial segment determined by each predecessor of $a$ is similar to some (necessarily unique) ordinal number. By the claim, $s(a)$ itself is then similar to some ordinal number. Applying transfinite induction, all of the initial segments of $X$ are similar to some ordinal number. Applying the claim now to $X$, $X$ itself is similar to some ordinal number.

I still unfortunately can not answer what the hell the author intended with the phrase "whether $a$ is the immediate successor of one of its predecessors or the supremum of them all, that $s(a)$ is similar to an ordinal number."


问:
For all topologies of R, are the open sets of R union of open intervals?
My textbook gave the following proposition which followed the section on Euclidean topology:

"A subset S of R is open if and only if it is a union of open intervals"

but did not state whether this proposition is true only for the Euclidean topology or any topology on R. There are topologies on R that can be only closed intervals, but since infinite union of open interval can be closed intervals, I think the proposition is true for any topology on R right?

答:
The book is defining the "standard" or "usual" topology on $\Bbb R. $

A topological space is a pair $(X,T)$ where $T$ is a collection of some or all of the subsets of $X,$ such that

(i). $X\in T$ and $\emptyset\in T,$ and

(ii). If $S$ is any sub-collection of $T$ then the common union $\bigcup S=\cup_{U\in S}\,U$ is in $T,$ and

(iii). If $S$ is any $finite$ subset of $T$ then the common intersection $\cap S$ is in $T.$

One example is $T=\{X, \emptyset\},$ called the coarse or anti-discrete topology. The opposite extreme is to have every subset of $X$ belong to $T,$ called the discrete or fine topology. Or, if you want, if $U$ is any  subset of $X,$ you can let $T=\{X,U, \emptyset\}.$

$T$ is called a topology on $X$ and the members of $T$ are called the open sets. "Open" is not an intrinsic property of any set. When you speak of an open set, your audience should already be aware of the topology that you are referring to.


问:
Incomplete space
I must show that $C_0(\mathbb{R})$ space of all continuous real functions $f:\mathbb{R} \longrightarrow \mathbb{R}$ with compact support is not a complete space endowed by norm
$\|f\|=\sup\limits_{t \in \mathbb{R}}|f(t)|$.

Thank for any help.

答: Hint: You can use indicator functions.

Title: Incomplete space

Body: Hint: You can use indicator functions.


问:
Is vector dot product and vector cross product allowed in Vector space?
As per the definition of Vector space, for a set of vectors and scalars, the only operations allowed are the 2 closure properties(for addition and scalar multiplication) and 10 axioms(for associativity commutativity, inverse and identity) for the vectors, scalars involved.
If so, where does dot product and cross product(which is vector time vector multiplication) fit in?
These 2 vector times vector multiplication operations are not defined and hence not allowed as per  the definition of Vector space. Is this understanding correct or am I missing something here?</p>

答:
It depends on your structure.
For example, I prefer to define vector space with a single axiom

A vector space is an abelien group $V$ on which a field $k$ acts

Let's look at this for the vector space $V=\Bbb R^3$. Initially, we can see that $\Bbb R^3$ is an abelien group under addition; this is so because of course $\Bbb R$ is an abelien group under addition and for every abelien group $A$, we can naturally define an abelien group structure on $A\times A$ or $A\times A\times A$ or ... component-wise. Viewed as an abelien group, we only have the addition operation at our disposal. Of course $\Bbb R$ itself is much more than an abelien group. We might view $\Bbb R^3$ as a non-unitary ring by also defining multiplication component-wise from the underlying ring $\Bbb R$. But that is of no interest when we want to have a vector space. The ring $\Bbb R$ happens to be a field, but that property does not transfer to a product. And $\Bbb R$ is also a topological space (or even a metric space), which does allow us to define a natural structure of topological space on $\Bbb R^3$. This does not give us a new (binary) operation on $\Bbb R^3$, but allows us to speak og convergence and limits. In fact, the topology is compatible with the group law, i.e., $\Bbb R^3$ is a topological abelien group. But in the context of vector space, we care only for the abelien group structure.

As with every abelien group, the set $\operatorname{End}(V)$ of endomorphisms $A\colon V\to V$ is endowed with a natural ring structure by defining $(A+B)(v)=A(v)+B(v)$ and $(A\cdot B)(v)=A(B(v))$. In particular, the identity endomorphism is the $1$-element of this ring.
To make $\Bbb R^3$ a vector space, we need a base field (and decide to use $k=\Bbb R$) and an action of $k$, i.e., a  ring homomorphism $\phi\colon k\to \operatorname{End}(V)$ and for convenience write $x\cdot v$ for $\phi(x)(v)$, giving us a new operation $\cdot\colon k\times V\to V$ that automatically has some nice properties: E.g., $1\cdot v=v$ follows because in a homomorphism between unitary rings, we postulate that the 1-element is mapped to the 1-element. And things like $(x+y)\cdot v=x\cdot v+y\cdot v$ or $x\cot(v+w)=x\cdot v+x\cdot w$ or $(x\cdot y)\cdot v = x\cdot(y\cdot v))$ also follow immediately from the definition of ring homomorphism. Everything acts together so nicely that using the same operation symbols $+$ and $\cdot$ for various different operations does not confuse us (and in fact is so suggestive that it simplifies our life a lot).

But how do we actually get such a ring homomorphism? There is one natural way, using the fact that $V$ is a direct product of copies of $k$: We map $x\in k$ to the endomorphism of $k^3$ that multiplies component-wise by $x$ (using the fact that each factor $k$ is our field).
This is how we could define a (natural) structure of a real vector space on $\Bbb R^3$, arriving at the usual vector space operations. What we do not have in general is additional structure beyond that of a vector space.

I already mentioned that $\Bbb R^3$ is a topological group. All operations including multiplication with a scalar are continuous under this topology and that allows us to view $\Bbb R^3$ as something more than just a vector space: It is a topological vector space. Not every vector space is also a topological vector space.

I also noted that we can consider component-wise multiplication to view $\Bbb R^3$ as a (non-unitary) ring; this operation is also compatible with the other operations and allows us to view $\Bbb R^3$ as a (non-unitary) algebra over $\Bbb R$. But we cannot do the same for every vector space.

Next, in our specific example we have a nice bilinear map $V\times V\to \Bbb R$ given by multiplying component-wise and adding the results - the dot product. This turns $\Bbb R^3$ into an inner product space. Again, not every vector space is an inner product space, and even if we have an inner product, it is not uniquely determined. But in the case of $\Bbb R^3$ (or more generally, $\Bbb R^n$), we have a natural way to define an inner product. But to repeat: A vector space is in general not endowed (or even endowable) with such an inner product.

Finally, the cross product. To give the broader picture about this, I'd have do dive quite deeply into abstract structures (even compared to what I have done so far). Let's leave it at this: Just as not every vector space is a topological space, or an algebra, or an inner product space, it is also not the case that every vector space allows a cross product. To keep things simple, you may rest assured for the moment that this is only available in a natural way for the specific vector space $\Bbb R^3$. In particular, there is not even a cross product in $\Bbb R^2$ or $\Bbb R^4$.

Summary: If be begin with "Let $V$ be a real vector space ..." then all we have are the vector space operations and are not "allowed" to use dot product or cross product. If we begin "Let $V$ be an inner product space ..." we have the dot product at our disposal. This is also implicitly the case if be begin "Let $V$ be the vector space $\Bbb R^n$ for some $n\in\Bbb N$ ...". But we can use the cross product only  in a context of "Let $V=\Bbb R^3$ ...".


问:
Some questions on Proof of Structure Theorem
I understand the general idea of the proof of Structure Theorem for finitely generated modules over a principal ideal domain but I found it quite difficult to follow some lines of reasoning in the proof. I have spent hours tried to think about it and look for other sources but still unable to fully convince myself. Any detailed helps are truly appreciated.

The Theorem is stated as follows:

Let $M$ be a finitely generated module over a principal ideal domain $R$. Then there exist elements $d_1, d_2, ..., d_k \in R$ satisfying $d_1\mid d_2\mid\cdots\mid d_k$ such that $$M\cong R/(d_1)\oplus R/(d_2)\oplus \cdots\oplus R/(d_k).$$

Proof:

Since $M$ is finitely generated (by $k$ elements say), then $M\cong R^k/N$. But $N\le R^k\Rightarrow N\cong R^s$ for some $s\le k$. Let $\zeta:R^s\to N$ be an isomorphism. This gives a homomorphism $\varphi:R^s\to R^k$ with $Im(\varphi)= N, ker(\varphi)=\{0\}$. Let $A\in M_{k\times s}(R)$ be the matrix of $\varphi$ with respect to the standard bases for $R^s$ and $R^k$. Then A is equivalent to a matrix
$$D=\begin{bmatrix}d_1 & 0 & ... & & 0\\0 & d_2 & 0 & ... & 0\\ & &.& & \\ & &:& & \\0 & 0 & 0 & ... & d_s\\0 & 0 & 0 & ... \\0 & 0 & 0 & ... \end{bmatrix},$$ where $D = XAY$, for $X\in M_{k\times k}(R)$ invertible and $Y\in M_{s\times s}(R)$ invertible. So $D=[\varphi]_{C,B}$ for some bases $B$ for $R^s$, $C$ for $R^k$. Let $C=\{f_1,f_2,...,f_k\}$ be a basis for $R^k$, then $\varphi(B)=\{d_1f_1,d_2f_2,...,d_kf_k\}$ is a basis for $N\subseteq R^k$. So we have $$R^k=<f_1>\oplus<f_2>\oplus...\oplus<f_k>,$$ $$N=<d_1f_1>\oplus<d_2f_2>\oplus...\oplus<d_sf_s>\oplus<0f_{s+1}>\oplus...\oplus<0f_k>$$ where $d_i:=0$ for $s<i\le k$. So,
$$R^k/N\cong\frac{<f_1>}{<d_1f_1>}\oplus\frac{<f_2>}{<d_2f_2>}\oplus...\oplus\frac{<f_k>}{<d_kf_k>}.$$But for any $v\in R^k\setminus \{0\}$, by First Isomorphism Theorem, $<v>/<dv>\cong R/(d)$. So $R^k/N\cong R/(d_1)\oplus R/(d_2)\oplus ...\oplus R/(d_k)$, $d_1|d_2|...|d_{s}|d_{s+1}=0|...$. (QED)

My questions are:

1. Why is $ker(\varphi)=\{0\}$? I could not find any convincing explanation.

2. I have been told that $C$ is given by columns of $Y$ and $B$ is given by columns of $X^{-1}$. I have some understanding of change of bases but how can we show it in a more rigorous way?

3. The part that I struggle the most are $R^k=<f_1>\oplus<f_2>\oplus...\oplus<f_k>$ and $N=<d_1f_1>\oplus<d_2f_2>\oplus...\oplus<d_sf_s>\oplus<0f_{s+1}>\oplus...\oplus<0f_k>$, what does <> represent, why are we taking their direct sum and how do we derive them?

4. Lastly, why can we just divide $R^k/N\cong\frac{<f_1>}{<d_1f_1>}\oplus\frac{<f_2>}{<d_2f_2>}\oplus...\oplus\frac{<f_k>}{<d_kf_k>}$ on both sides, how can I explain/convince myself rigorously.

5. I have consulted other sources, in the final step, it is given that the map $\psi:R^k\to R/(d_1)\oplus R/(d_2)\oplus ...\oplus R/(d_k)$ given by $\psi(\sum_{i=1}^{k}r_if_i)=(r_1+(d_1),...,r_k+(d_k))$ is a homomorphism of R-modules. The result follows from FIT since $\psi$ is surjective and $ker(\psi)=N$. How can we show $\psi$ is surjective and $ker(\psi)=N$?

6. Why if $d_i=0$, then $d_i=0$ for all $j\ge i$? And why if $d_i$ is a unit, then $d_j$ is a unit for all $j\le i$?

Many thanks in advance!

答:
Some ideas:

The homomorphism $\,\phi\,$ described there could be thought of as an embedding one: note that $\,N\cong R^s\le R^k\,$ , so one can embed the former in the latter.

Of course, this embedding is usually far from being unique. From here  $\,\ker\phi=0\,$ is immediate.

The notation $\,\langle x\rangle\,$ usually denotes the cyclic (module, group, etc. ) generated by $\,x\,$ . In this case for example, $\,\langle f_i\rangle\,$ denotes the cyclic $\;R$-module  generated by $\,f_i\,$ in $\,R^k\,$ , i.e.:

$$\langle f_i\rangle:=\{rf_i\;;\;r\in R\}$$

Observe that any finitely generated $\,R$-module $\;M\;$ is the set of all the $\,R$-linear combinations of a finite number of elements of $\,M\,$ , and from here the term "basis" .

The dividing thing is just taking the quotient module...Apparently you already proved before this that the quotient takes that form (this is (5) in your questions):

$$\bigoplus_{i=1}^k\langle f_i\rangle/\bigoplus_{i=1}^k\langle d_if_i\rangle \cong\bigoplus_{i=1}^k \langle f_i\rangle/\langle d_if_i\rangle$$

Your questions about $\,\psi\,$ could probably be partially answered if you read carefully about the CRT = Chinese Remainder Theorem , which has a very similar homomorphism there, yet directly: clearly $\,\psi\,$ is surjective since any element in the direct sum of quotient modules on the right is precisely of the form $\;r_j+(d_j)\;$ , for some $\,r_j\in R\,$ , and you can take an element in $\,R^k\,$ with these $\,r_j$-s as corresponding coefficients of the basis $\,\{f_1,...f_k\}\,$ . That $\,N=\ker\psi\,$ is even more trivial (Sorry, I'm not teasing you, honest) if one understand what's the role of each thing (and this is why I thought subdividing the question in several, focused parts could help...)