问:
Quotient Space and Subspaces
Suppose that we have a vector space $X$ and two vector subspaces $A, B$ such that $A \subset B$. Is the following generally true?
$$
X / A \supset X/B
$$

Can we show this via equivalence classes? By this I mean given some $x \in X$ can we show $[x] \in X/B \implies [x] \in X/A$ where $[x]$ is the equivalence class generated by $x$.

Edit: The motivation for this question was trying to understand a step in the proof in Lax - Functional Analysis on pg. 13. For some operators $M$ and $G$ it claims because $R_M \supset R_{I+G}$ this implies $$codim(R_M) \leq codim (R_{I + G})$$

Here $R_M$ denotes the range of $M$.

答:
No. In the case that $A\subsetneq B$ we cannot say $X/A \supseteq X/B$, because they have different equivalence relations, hence different equivalence classes. This is a set-theoretic obstacle.

Unfortunately the assignment $i:X/B \hookrightarrow X/A$ sending the equivalence class $x+B$ to the equivalence class $x+A$ is not welldefined either. For an  element $b \in B\setminus A$ we would have to have
$$0+A = i(0+B) = i(b+B) =b +A \neq 0+A$$
where the last inequalities holds as $b\notin A$.

Title: Quotient Space and Subspaces


问:
How to calculate the expected distribution of results using random numbers?
See this SO thread: Calculating which item is next in a percentage distribution.

To summarize,

you calculate

```<code>max ((rand1 * 10), (rand2 * 20), (rand3 * 30), (rand4 * 40))
</code>```

for four random numbers in a range (say 1-100).

If the max is

```<code>rand1 * 10, you choose 'A'
rand2 * 20, you choose 'B'
rand3 * 30, you choose 'C'
rand4 * 40, you choose 'D'
</code>```

What is the expected distribution of choosing A, B, C and D?

How to derive it mathematically?

(Using some C code, the percentages come out to be 1, 10, 31 and 56 respectively)

答:
Suppose you only had three random numbers between 0 and 1. Point (x,y,z)=(rand1*10,rand2*20,rand3*30) falls uniformly inside parallelepiped with dimensions 10,20,30.
Sets x>y&&x>z, y>x&&y>z, z>x&&z>y partition this parallelepiped into 3 parts, as pictured below, and their relative volumes represent the fraction of the time that x, y or z are the maximum value, respectively. Working out expression for the volumes should give you the integral expression David gave above.

ContentLicense: CC BY-SA 2.5

ContentLicense: CC BY-SA 4.0


问:
Pullback, 2-form invariant
Let $\mathbb H^2 = \{(x, y) \in \mathbb{R}^2 : y> 0\}$ and consider the $2$-form in $\mathbb H^2$ defined by
$$\varphi = \dfrac{dx \wedge dy}{y^2}.$$

Show that $\varphi$ is invariant $(T^*(\varphi)= \varphi)$ under the transformation $T$ from $\mathbb H^2$ to $\mathbb H^2$ given by
$$T(z) = \dfrac{az + b}{cz + d}$$

where $z \in \mathbb H^2 \subset \mathbb{C}$ and $a, b, c, d \in \mathbb{R}$ with $ad - bc \neq 0$.

My attempt was to try to express $T(x, y)$ and calculate $dx $ and $ dy $ to make $ dx \wedge dy $, and calculate the pullback. I couldn't, the expression for $T$ was strange.

答:
As we know, every such transform is generated by 3 elements:
$$ f_1(z) = z + a$$
$$ f_2(z) = \dfrac{1}{z} $$
$$ f_3(z) =  az $$
So you need to check that you's form is invariant under this 3 transforms, what is not difficult

Title: Pullback, 2-form invariant


问:
Proving compactness in $\mathsf{ZF}$
Consider a family $\{A_i\mid i\in I\}$ of disjoint sets. Let $X$ be the union of all $A_i$, and equip it with the smallest topology $\tau$ making the $A_i$ closed in $X$. How can one prove in $\mathsf{ZF}$ that $(X,\tau)$ is compact?

One may apply the Alexander subbase theorem when working in $\mathsf{ZFC}$, but the theorem's proof is based on some form of choice.

This is part of an exercise in "The Axiom of Choice" by T. Jech, where the goal is to prove that the Axiom of Choice follows from the following assertion: If $(X,\tau)$ is a compact space and $I$ is an arbitrary index set, then $\prod_{i\in I} X$ equipped with the product topology is also compact.

答:
We don't need to look at subbase, because we have a neat description of all open subsets. A subset $U\subseteq X$ is open if and only if $U=X$ or $U=\emptyset$ or $X\backslash U$ is a finite union of some $A_i$-s. You can verify that this is a topology and the smallest topology having $A_i$ as closed subsets. I don't think there's any choice involved here.

Assume a family $\mathcal{U}$ covers $X$. Take a single subset $U\in\mathcal{U}$. It follows that there are at most finitely many $A_1,\ldots, A_n$ not covered by $U$. But each $A_i$ is covered by some $U_i\in\mathcal{U}$. And so $\{U,U_1,\ldots,U_n\}$ is a finite subcovering of $\mathcal{U}$. Here we only utilize the finite choice, no additional axioms required.

Title: Proving compactness in $\mathsf{ZF}$


问:
Distribution of discrete random variable $X$
I am stuck on a question regarding discrete random variable. The question is as follows:

The probability mass function of $X$ with $x \in \mathbb{N}:=\{0,1,2, \ldots\}$ satisfies the following:

\begin{align}
\frac{(=)}{(=+1)}=x+1
\end{align}

I am asked to find $(=0)$ and identify the distribution of $X$

I worked out that $(=0)=(=1)$ from the given PMF but not sure what to do next and how to identify the type of distribution of $X$.

Any help is greatly appreciated.

答:
Consider a poisson distribution $X\sim Po(1)$

then you have that

$$\frac{\mathbb{P}[X=x]}{\mathbb{P}[X=x+1]}=\frac{\frac{e^{-1}}{x!}}{\frac{e^{-1}}{(x+1)!}}=x+1$$

thus

$$\mathbb{P}[X=0]=e^{-1}$$


问:
Construction of triangle
I don't know how to prove or disprove the following problem:
How to construct triangle if elements $a$, $b$, $\beta-\gamma$ are given?
Is it constructible (if not, how to prove it)? Any help is welcome.</p>

答: with $a,\angle B,C$ you can construct a triangle.but the extra b may cause conflict unless you ignore it. or you can use $a,b , \angle C$ to construct a triangle.but you have to omit $\angle B$.</p>

ContentLicense: CC BY-SA 3.0

Title: Construction of triangle


问:
Understanding Artin's derivation of the determinant formula
I'm trying to understand section 1.4 of Artin. I'll mention some context so that the notation he's using is clear, even though it makes sense to me.

He defines a permutation matrix $P$ associated with permutation $p$. Then if $e_{ij}$ are "matrix units," matrices with a $ 1$ in the $ij$th place and $0$'s elsewhere, we have
$$P = e_{p(1)1} + \ldots + e_{p(n), n}.$$
So far, I understand. Then he introduces linearity of the determinant. In the $2 \times 2$ case, it gives:
\begin{align*}
\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \det \begin{pmatrix} a & 0 \\ c & 0 \end{pmatrix} +  \det \begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix} +  \det \begin{pmatrix} 0 & b \\ c & 0 \end{pmatrix} +  \det \begin{pmatrix} 0 & b \\ 0 & d \end{pmatrix}.
\end{align*}
The first and fourth matrices have a column of $0$'s, so there determinant is $0$. So far so good. He calls the remaining matrices $M$ and comments that they have exactly one entry left in each row and each column. I can see this in the $2 \times 2$ case, but don't understand why it is true, generally, though it must be obvious from linearity of the determinant.

Assuming this is true about $M$, $M$ is basically a permutation matrix, but instead of entries of $1$, they're entries $a_{ij}$ coming from some matrix $A$ whose determinant we're taking, so we have
$$M = \sum\limits_{j} a_{p(j)j} e_{p(j)j}.$$
He then states that by linearity, we have:
\begin{align*}
\det M & = (a_{p(1)1} \cdots a_{p(n)n}) \det P \\
& = (\text{sign $p$})(a_{p(1)1} \cdots a_{p(n)n}.
\end{align*}
I don't understand where either of these lines come from, and this may be my greatest source of confusion. I tried to work out a few examples by hand, but it wasn't particularly helpful.

The next observation is there is "one such term for each. permutation," so we can sum over them:
\begin{align*}
\det A & = \sum\limits_{\text{perm $p$}} (\text{sign $p$}) a_{p(1)1} \cdots a_{p(n)n}) \\
& = \sum\limits_{\text{perm $p$}} (\text{sign $p$}) a_{1p(1)} \cdots a_{np(n)})
\end{align*}
By one term for each. permutation, concretely, I assume he means that if we're looking at an $n \times n$ matrix, there are $n!$ permutations of its columns, so there are $n$ matrices $M$ (two above) and we want to sum over each of those. I believe I understand that, but I don't understand where the second, "transpose form" comes from or why it's equivalent. Is this using the fact thatthe determinant of a matrix equals the determinant of its transpose? Even if it's using that fact, I don't understand why the indices shift in this way.

I would appreciate if someone could help to clarify these.

答:
Let's observe that a determinant is a multilinear function of the rows (or columns) of a matrix.

Let's use the above fact on a 2x2 matrix $M$ given by $$M=\begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\end{bmatrix} $$ Also let's note that the unit row vectors can be written as $$e_1=(1,0),e_2=(0,1)$$ and we can write a typical row vector $r=(a, b) $ as $$r=ae_1+be_2$$ Next we can write $$\operatorname {det} (M) =\operatorname {det} \begin{bmatrix} a_{11}e_1+a_{12}e_2\\a_{21}e_1+a_{22}e_2 \end{bmatrix} $$ which further by multilinearity equals $$\operatorname {det} \begin{bmatrix} a_{11}e_1\\ a_{21}e_1 +  a_{22}e_2\end{bmatrix}+\operatorname {det} \begin{bmatrix} a_{12}e_2\\ a_{21}e_1+ a_{22}e_2\end{bmatrix} $$ This can be further split as $$\operatorname {det} \begin{bmatrix} a_{11}e_1\\ a_{21} e_1\end{bmatrix}+\operatorname {det} \begin{bmatrix} a_{11}e_1 \\ a_{22}e_2 \end{bmatrix}+\operatorname {det} \begin{bmatrix} a_{12}e_2 \\ a_{21}e_1\end{bmatrix}+\operatorname {det} \begin{bmatrix} a_{12}e_2\\ a_{22}e_2\end{bmatrix} $$ And twe can take out the matrix entries to get $$\operatorname {det} (M) =a_{11}a_{21}\operatorname {det} \begin{bmatrix} e_1\\  e_1\end{bmatrix}+a_{11}a_{22}\operatorname {det} \begin{bmatrix} e_1 \\ e_2 \end{bmatrix}+a_{12}a_{21}\operatorname {det} \begin{bmatrix} e_2 \\ e_1\end{bmatrix}+a_{12}a_{22}\operatorname {det} \begin{bmatrix} e_2\\ e_2\end{bmatrix} $$ Let's clearly notice that the determinants with at least two equal rows vanish and hence we are left with $$\operatorname {det} (M) = a_{11}a_{22}\operatorname {det} \begin{bmatrix} e_1 \\ e_2 \end{bmatrix}+a_{12}a_{21}\operatorname {det} \begin{bmatrix} e_2 \\ e_1\end{bmatrix}$$ You will thus note that we are left with only those entries which have rows with a permutation of unit row vectors (this happens in exactly the same manner with $n\times n$ matrices and we are left with exactly $n! $ determinants).

If $\sigma$ is a permutation on $n$ symbols (more formally $\sigma$ is a bijection from set $\{1,2,\dots,n\}$ to itself) then we define the sign of permutation $\sigma$ as $$\operatorname {sign} (\sigma) =\operatorname {det} \begin{bmatrix} e_{\sigma(1)}\\e_{\sigma(2)}\\ \dots\\e_{\sigma(n)} \end{bmatrix} $$ The above matrix can always be transformed into the indetity matrix $[e_1,e_2,\dots, e_n] ^{T} $ via a finite number of row exchanges and hence the sign of a permutation $\sigma$ is $(-1)^j$ where $j$ is the number of transpositions which can be applied on $\sigma$ to reduce it to identity permutation (prove that there can be multiple values of $j$ but with same parity so that $(-1)^j$ remains fixed). Using this definition one can show that $\operatorname {sign} (\sigma) =\operatorname {sign} (\sigma^{-1})$.

Coming back to our computation with the $2\times 2$ matrix $M$ we see that $$\operatorname {det} (M) =\sum_{\sigma}\operatorname {sign} (\sigma) a_{1\sigma(1)}a_{2\sigma(2)}=a_{11}a_{22}-a_{12}a_{21}$$ A similar formula holds for matrices of order $n\times n$ and if $M=[a_{ij}] _{n\times n} $ is a square matrix of order $n$ with entries $a_{ij} $ in $i$-th row and $j$-th column then we have $$\operatorname {det} (M) =\sum_{\sigma} \operatorname {sign} (\sigma) a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n\sigma(n)} $$ where $\sigma$ runs over all permutations on symbols $1,2,\dots,n$.

So you can see that we use the following properties of a determinant to prove the desired formula:

it is linear function of each row (multi-linearity)

it is $0$ if two rows are same (this is a consequence of the fact that determinant changes sign if two rows are exchanged so that it is an alternating function of row vectors).

it takes the value $1$ on identity matrix.

The formula above in terms of permutations proves that any such function with those three properties is unique and thus determines the determinant (!!) uniquely.

Next if $\tau$ is the inverse of permutation $\sigma$ and $\sigma(i) =j$ then $\tau(j) =i$ and thus $a_{i\sigma(i)} =a_{\tau(j) j} $. Let $M=[a_{ij}]_{n\times n} $ and $M^{T} =[b_{ij}] _{n\times n} $ be the transpose of $M$ so that $b_{ij} =a_{ji} $.

Then $$\operatorname {det} (M) =\sum_{\sigma} \operatorname {sign} (\sigma) a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n\sigma(n)} $$ We can now rewrite each term in above sum with order of factors such that the second index of $a$ is in order $1,2,\dots,n$ so that $$a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n\sigma(n)}=a_{\tau(1)1}a_{\tau(2)2}\dots a_{\tau(n)n}$$ As $\sigma$ runs over all permutations on $n$ symbols so does its inverse $\tau$ and hence we get $$\operatorname {det} (M) =\sum_{\tau}\operatorname {sign} (\tau) a_{\tau(1)1}\dots a_{\tau(n) n} =\sum_{\tau} \operatorname {sign} (\tau) b_{1\tau(1)}\dots b_{n\tau(n)} =\operatorname {det} (M^T) $$

The above procedure also proves that if $f$ is any alternating and multi-linear function of the rows of $n\times n$ matrices then $$f(M) =(\operatorname {det} (M)) f(I) $$ for any square matrix $M$ of order $n$ and $I$ being the identity matrix of order $n$.


答:
The smallest topology that makes each $A_i$ closed is essentially the following: give $I$ the co-finite topology, then replace $i$ by $A_i$.

It is easy to see why the co-finite topology is compact, even in $\sf ZF$. Given any non-empty open set, then only finitely many points are missing from it, so any cover has a finite subcover: pick some non-empty open set in the cover, then finitely many more are all you need to cover the missing points.

The fact that we replace each $i$ with $A_i$ with completely irrelevant here, since the restriction of the topology to each $A_i$ is indiscrete.

Title: Proving compactness in $\mathsf{ZF}$

Body:
Consider a family $\{A_i\mid i\in I\}$ of disjoint sets. Let $X$ be the union of all $A_i$, and equip it with the smallest topology $\tau$ making the $A_i$ closed in $X$. How can one prove in $\mathsf{ZF}$ that $(X,\tau)$ is compact?

One may apply the Alexander subbase theorem when working in $\mathsf{ZFC}$, but the theorem's proof is based on some form of choice.

This is part of an exercise in "The Axiom of Choice" by T. Jech, where the goal is to prove that the Axiom of Choice follows from the following assertion: If $(X,\tau)$ is a compact space and $I$ is an arbitrary index set, then $\prod_{i\in I} X$ equipped with the product topology is also compact.


问:
Limit and singular point
I have a elementary problem but I don't know how to solve it :

$\lim\limits_{z \to 0} \dfrac{1}{e^z-1}-\dfrac{1}{z}$, it is indetermine because we have :$\infty - \infty$, according to my research on the web I think that is 1/2 but I don't know how to prove it.

singular point of $e^{\frac{1}{z+1}}$

Anybody can help me or give me a clue? Thank you in advance

答:
Regarding 2., $z=-1$ is a singular point of $f(z)=e^{\frac{1}{z+1}}$. It is in fact as essential singularity because the Laurent series has an infinite number of terms with negative index:
$$
e^{\frac{1}{z+1}} = \sum_{n=0}^{+\infty} \frac{1}{n!}\left(\frac{1}{z+1}\right)^n= 1+\frac{1}{z+1} + \frac{1}{2!(z+1)^2} + \frac{1}{3!(z+1)^3} + \cdots
$$

Title: Limit and singular point


答:
The question can be solved with a single applicaton of L'Hospital. As

$$\frac{1+z-e^z}{z(e^z-1)}=\frac{1+z-e^z}{z^2}\frac{z}{e^z-1},$$

the last factor is known to tend to $1$ and can be ignored.

Then by L'Hospital,

$$\lim_{z\to0}\frac{1+z-e^z}{z^2}=\lim_{z\to0}\frac{1-e^z}{2z}=-\frac12.$$

Title: Limit and singular point

Body:
I have a elementary problem but I don't know how to solve it :

$\lim\limits_{z \to 0} \dfrac{1}{e^z-1}-\dfrac{1}{z}$, it is indetermine because we have :$\infty - \infty$, according to my research on the web I think that is 1/2 but I don't know how to prove it.

singular point of $e^{\frac{1}{z+1}}$

Anybody can help me or give me a clue? Thank you in advance


问:
Separation of variables in spherical coordinates issues
I am trying to follow this derivation of the solution of the spherical lapace equation. This is probably very simple but where did $-m^2$ on the bottom of page one?

[

Additionaly, what is happening between the bottom two lines. I know it says diff wrt r or $\theta$ but it doesnt look like the derivative has been taken.

http://www.physics.usu.edu/Wheeler/EM3600/Notes11SeparationOfVariablesSpherical.pdf

答:
If $y(\varphi)=\frac{1}{\Phi}\frac{d^2\Phi}{d\varphi^2}$, then you effectively have an equation $dy/d\varphi = 0$, the solution is $y=\mathrm{const}=C$. However, if $C>0$, then the solution of equation
$$
\frac{d^2\Phi}{d\varphi^2}=C\Phi >0,\qquad \Phi = C_1e^{\varphi \sqrt C}+C_2e^{-\varphi \sqrt C}
$$
is non-periodic. However, we know that $\varphi$ is a cyclic variable, so $\Phi(\varphi)=\Phi(\varphi+2\pi)$. It means that $C\le0$. One of the way how to highlight it is to choose a constant that cannot be positive like $-m^2$.

In your second question, again let's denote functions to easily see what's going on:
$$
\begin{aligned}
f(r) &= \frac{1}{R} \frac{d}{d r}\left(r^{2} \frac{d R}{d r}\right) \\
g(\theta) &=\frac{1}{\Theta} \frac{1}{\sin \theta} \frac{d}{d \theta}\left(\sin \theta \frac{d \Theta}{d \theta}\right)-\frac{m^{2}}{\sin ^{2} \theta}
\end{aligned}
$$
You have an equation $f(r)+g(\theta)=0$, then you differentiate it by $r$, you will get $f'(r)=0$, the solution of this equation is $f(r)=\mathrm{const}=C_f$, by analogy the solution of $g'(\theta)=0$ is $g(\theta)=\mathrm{const}=C_g$. However, if you put those to the initial formula, you will get $C_f+C_g=0$, so clearly those constants are not independent. If you take $C_f=a$, then $C_g=-a$.


问:
Cuboid diagonals
The points $M$ and $N$ are taken to be on the diagonals $AB_1$ and $BC_1$ of the cuboid $ABCD-A_1B_1C_1D_1$. If $MN \parallel A_1C$ , find the $MN:A_1C$ ratio.
Initially I understood what's going on here, but unfortunately I can't solve it... The problem requires vector solution. I have no idea how to do it. Can anyone solve it then explain beautifully?? I will glad...

答:
Here are the hints. Proceed along these lines and in case of doubts, show your work first.

In a $3$-d coordinate system, set $A=(0,0,0)$, $C=(L,W,0)$, $A_1=(0,0,H)$

Using similarity, find coordinates of $M=(0,y,\frac{H}{W}y)$ and $N=(x,W,\frac{H}{L}x)$

The unit vector parallel to $AC$ or $MN$ is $\frac{1}{|A_1C|}(L,W,-H)$. Hence position vectors $M,N$ satisfy
$$N=M+\frac{|MN|}{|A_1C|}(L,W,-H)$$

On equating the components, a numerical value of $MN:A_1C$ is obtained.

 $$\frac{MN}{A_1C}=\frac{1}{3}$$

Title: Cuboid diagonals


问:
Defining a topology in terms of filter neighbourhoods
I am a little confused about the following:

if $X$ is a set for which there is a filter $\mathcal{F}(x)$ of sets containing $x\in X$ assigned to every point $x\in X$, and these filters are such that for all $U \in \mathcal{F}(x)$ there is some set $V \in \mathcal{F}(x)$ such that for all $y \in V$, $U \in \mathcal{F}(y)$, if we say a set $O$ is open if it is empty or such that for all $o \in O,O \in \mathcal{F}(o)$,
then why is it true that the closure of a set $A$ is the set of all points $x$ for which all members of $\mathcal{F}(x)$ have nonempty intersection with $A$?

I can prove this if I know that every member of $\mathcal{F}(x)$ contains an open set containing $x$, because then it just comes down to proving the following statement: 'The closure of a set $A$ in a topological space is the set of all points $x \in X$ such that for all open sets $U$ containing $x$, $U \cap A \neq \emptyset$' which I am familiar with and know how to prove.

The issue is although it seems like intuitively, all members of the filter of 'neighbourhoods' of a point should all themselves contain open sets containing that point, I cannot seem to easily deduce it from the definition of open set in my first paragraph.

This definition is taken from 'Topological Vector Spaces,Distributions and Kernels' by Francois Treves.

Thank you in advance.

答:
In this post I give a full proof of equivalence between neighbourhood spaces and corresponding topological spaces.
Given a neighbourhood space, we can define open set as you describe (the empty set is not a special case BTW, because of void truth) and we get that the neighbourhood system we start with is exactly the one corresponding to that topology.

In particular the axiom: "for all $U \in \mathcal{F}(x)$ there is some set $V \in \mathcal{F}(x)$ such that for all $y \in V$, $U \in \mathcal{F}(y)$" (quoting your post) really "says" that each neighbourhood of $x$ contains an open neighbourhood of $x$.

The above equivalence proves also that the closure of a set is indeed

$$\operatorname{cl}(A) = \{x \in X\mid \forall N \in \mathcal{N}_x: N \cap A \neq \emptyset\}$$

because that's indeed true for neighbourhoods in all topological spaces. We cold have taken that as the definition of closed sets in this correspondence too (though it's slightly more convoluted, unless you're allowed to use the Kuratowski closure axioms as the foundations of topology, and check directly that cl as defined above obeys these).


问:
Conditional probability of 3 continuous random variables
The length of time for events X, Y, Z to occur are independent and have probability density functions
$f(x)$, $f(y)$ and $f(z)$. What is the probability that given the second event occurred at time $t$, find the probability that Y occurred at time $t$.

If it was about the first or last event I would do something like $Pr(Y=t|min(X, Y, Z) = t)$ but not sure about this scenario.

Thank you

答: It's the probability of $Y$ occurring at time $t$ and Y is the second event. Therefore, the probability is $Pr(Y=t|X<Y \land Z>Y) + Pr(Y=t|Z<Y \land X>Y)$.


问:
Calculating a simple limit in $\mathbb{R}^k$
So, I was working on this exercise that required me to prove a certain
function from $\mathbb{R}^k$ to $\mathbb{R}$ is continuous at $0^{[k]}$.
By simply using the infamous triangle inequality I reached the conclusion
that the function (denote by $f:\mathbb{R}^k \rightarrow \mathbb{R}$) is bounded by the following:
\begin{gather*}
    \forall x \in \mathbb{R}^k \\
    \begin{cases}
        -\dfrac{\sum\limits_{i=1}^{k} |x_i|^{5}}{\sum\limits_{i=1}^{k} x_i^4} & x \neq 0 \\
        0 & x = 0
    \end{cases} \leq f(x) \leq
    \begin{cases}
        \dfrac{\sum\limits_{i=1}^{k} |x_i|^{5}}{\sum\limits_{i=1}^{k} x_i^4} & x \neq 0 \\
        0 & x = 0
    \end{cases}
\end{gather*}

So define:
\begin{gather*}
    g: \mathbb{R}^k \rightarrow \mathbb{R} \\
    (x_1, \; ... \, , \: x_k) \rightarrow
    \begin{cases}
        \dfrac{\sum\limits_{i=1}^{k} |x_i|^{5}}{\sum\limits_{i=1}^{k} x_i^4} & x \neq 0 \\
        0 & x = 0
    \end{cases}
\end{gather*}
Thus, if I prove $g(x) \xrightarrow[x \rightarrow 0^{[k]}]{} 0$ I can apply the sandwich rule and the proof is setteled.
The thing is, I know this limit is true, it must be so. I think it is trivial, but I simply can't find the right direction.
I feel as if there is some sort of an algebraic trick I am missing or something.

Any hint or comments would be greatly appreciated.

Edit:

The original function:
\begin{gather*}
f:\mathbb{R}^k \rightarrow \mathbb{R} \\
(x_1, \; ... \, , \: x_k) \rightarrow
    \begin{cases}
        \dfrac{ \sum\limits_{i=1}^{k} (-1)^i \cdot x_i^{2021} }{ \sum\limits_{i=1}^{k} x_i^{2020} } & x \neq 0 \\
        0 & x = 0
    \end{cases}
\end{gather*}
(I changed the powers for simplification purposes)

Thank you so much, have a great weekend.

Body: $ \sum\limits_{i=1}^{k}|x_i|^{5} \leq (\max \{|x_1|,|x_2|,...,|x_k|\} )\sum\limits_{i=1}^{k}|x_i|^{4}$. Can you finish?


问:
Polynomials are continuous with respect to the Zariski topology
Let $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$. Show that if both $\mathbb{F}$ and $\mathbb{F}^n$ have the Zariski topology then all polynomials on $\mathbb{F}^n$
are continuous.</p>

答: Let $p : \mathbb{F}^n \to \mathbb{F}$ be a polynomial. It suffices to show that the preimage of every closed set in $\mathbb{F}$ is closed in $\mathbb{F}^n.$ The closed sets of $\mathbb{F}$ are simply the finite sets and $\mathbb{F}$ itself. The preimage of $\mathbb{F}$ is $\mathbb{F}^n$ which is closed so if we can show the preimage of a point $\alpha\in \mathbb{F}$ is closed then we are done. That preimage consists of the points in $\mathbb{F}^n$ such that $p(x_1,\cdots, x_n) -\alpha =0,$ which is by definition closed.


问:
Prove $X/Y$ is a random variable.
Show that if $X$ and $Y$ are random variables then $X/Y$ is a random variable $(Y\neq 0)$.

I read this in some notes I found online from a university stating that $X+Y,XY,X/Y$ are all random variables. I know the proof for $X+Y$ which looks like,

$\forall r \in R,\ \exists q \in Q,  \{X<r,Y>c-r\} \subset \{X<q, Y>c-q\}$
$\cup_{r\in R} (\{X<r\}\cap\{Y>c-r\})\subset \cup_{q\in Q} (\{X<q\}\cap\{Y>c-q\})$
$\cup_{q\in Q} (\{X<q\}\cap\{Y>c-q\}) \subset \cup_{r\in R} (\{X<r\}\cap\{Y>c-r\})$

Is it possible the proof for $X/Y$ is similar? I cant seem to fix a constant ($r$).

答:
Hint: $$(\frac X Y <z)=(Y>0, X <zY) \cup (Y<0, X >zY)$$ $$ =\bigcup_{q\in \mathbb  Q} (Y>0, X <q<zY) \cup \bigcup_{r \in \mathbb Q} ((Y<0, X >r>zY).$$

[$(Y>0,X<q<zY)=(Y>0)\cap (X<q)\cap (zY>q)$ etc].

Title: Prove $X/Y$ is a random variable.


问:
On the Continuity of Sobolev Functions
The statement is simple. Further discussion follows. Does there exist a proof of the following lemma which doesn't make use of the fact that Schwartz functions or test functions are dense in the Sobolev space?

Lemma: Let $d$ be a positive integer, $\Omega$ a measurable subset of $\mathbb{R}^d$, and $p \geq 1$ a real number. If $f \in W^{k,p}(\Omega)$ for every $k = 0,1,\ldots,$ then $f$ is almost everywhere equal to a continuous function on $\Omega$.

If needed, $\Omega$ can be whatever you want it to be, but I would prefer to not infringe upon $\Omega$'s free will, if possible.

Apparently, this fact is at least true for $\Omega = \mathbb{R}^d$. According to Exercise 33 in Terrence Tao's blog post (link), we can find $k$ great enough that $W^{k,p}(\mathbb{R}^d)$ embeds into $W^{1,2d}(\mathbb{R}^d)$. Then, by Hölder-Sobolev embedding, $W^{1,2d}(\mathbb{R}^d)$ embeds into $C^{0,\alpha}(\mathbb{R}^d)$ for some $\alpha$, the exact value of which is not consequential. Overall,
$
    W^{k,p}(\mathbb{R}^d)
\hookrightarrow
    W^{1,2d}(\mathbb{R}^d)
\hookrightarrow
    C^{0,\alpha}(\mathbb{R}^d)
$
for some $k$ which has to be chosen arbitrarily large given arbitrarily large $d$. In fact, if I'm not mistaken, this reasoning generalizes to show that $W^{\infty,p}(\mathbb{R}^d) \hookrightarrow C^\infty(\mathbb{R}^d)$ where $W^{\infty,p}(\Omega) := \bigcap_{k=0}^\infty W^{k,p}(\Omega).$

I admit that I, myself, still can't articulate the proof of either the Sobolev embedding theorem or the aforementioned Exercise. However, assuming I haven't misstated them, that isn't the present concern. In summary, those embedding theorems both feel like they might be overkill, don't they? Is there not a direct proof?

References are never a bad thing.

Thank you.

答:
As mentioned in the comments I don't think there's a way to avoid density in some form, but there is a relatively simple and self-contained way to prove this result. Broadly one proceeds as follows:

Localise by multiplying by a cut-off, and show the cutoff also lies in $W^{k,1}$ for all $k \geq 0.$

By mollifying we can assume this localised function is smooth.

Iteratively using the fundamental theorem of calclus prove the embedding $W^{d,1} \hookrightarrow C.$

Pass to the limit to conclude.

The idea is that continuity is a local condition, which allows us to localise. In this case the approximation argument follows by a straightforward mollification. Finally I use a relatively simple variant of the Sobolev inequality which can be proven without reference to the general inequality. I will make this construction precise below; here $Q_r(x)$ denotes the open cube in $\Bbb R^d$ centered at $x$ with side-length $2r.$

Step 1: Given any $Q_{3\rho}(x) \subset \Omega,$ choose a cutoff $\chi \in C^{\infty}_c(\Omega)$ such that $\chi \equiv 1$ on $Q_{\rho}(x)$ and vanishes outside $B_{2\rho}(x).$ Then $v = \chi u$ can be checked to lie in $W^{k,q}(\Omega)$ for all $k \geq 0$ and $1 \leq q \leq p;$ the case $q=p$ follows using the Leibniz rule and the fact that all derivatives of $\chi$ are bounded, and in general we can use Hölder's inequality since $v$ is supported in the bounded domain $B_{2\rho}(x).$ Also note $v = u$ on $Q_{\rho}(x)$ still.

Step 2: Now if $\rho_{\varepsilon}$ is a standard mollifier, we have $v_{\varepsilon} = \rho_{\varepsilon} \ast (\chi u)$ is smooth and is supported in $\Omega$ provided $\varepsilon < 1.$ By standard properties of mollifiers we know that $v_{\varepsilon} \to v$ in $W^{k,q}(\Omega)$ for all $k \geq 0$ and $1 \leq q \leq p.$ We will prove a uniform estimate for $v_{\varepsilon}$ and use this convergence to pass to the limit.

Step 3: Since each $v_{\varepsilon}$ is smooth, observe by the fundamental theorem of calculus that for any $z \in Q_{3\rho}$ we have
$$ v_{\varepsilon}(x+z) = \int_0^{z_d} \int_0^{z_{d-1}} \dots \int_0^{z_1} \partial_{x_d}\partial_{x_{d-1}} \dots \partial_{x_1}v_{\varepsilon}(x_1+t_1,\dots,x_d+t_d) \,\mathrm{d}t_1\dots \,\mathrm{d}t_{d-1}\,\mathrm{d}t_d. $$
Hence taking absolute values we get
$$ |v_{\varepsilon}(x+z)| \leq \int_{-3\rho}^{3\rho} \dots \int_{-3\rho}^{3\rho} |\nabla^du(x_1+t_1,\dots,x_d+t_d)| \,\mathrm{d}t_1\dots \,\mathrm{d}t_d = \int_{Q_{3\rho}(x)} |\nabla^du(y)| \,\mathrm{d}y.$$
This establishes the estimate
$$ \sup_{y \in Q_{3\rho}(x)} |v_{\varepsilon}(y)| \leq \lVert v_{\varepsilon} \rVert_{W^{d,1}(Q_{3\rho}(x))}. $$

Step 4: Sending $\varepsilon \to 0$ and applying the above inequality to $v_{\varepsilon} - v_{\delta}$ since the derivation is linear, we deduce that since $v_{\varepsilon}$ is Cauchy in $W^{d,1}$ it is Cauchy in the uniform norm. Hence $v_{\varepsilon} \to v_*$ uniformly on $B_{3\rho}(x),$ which gives a continuous representative of $v.$ Hence restricted to $B_{\rho}(x)$ this gives a local continuous representative of $u,$ and doing this at all points establishes the result.

Title: On the Continuity of Sobolev Functions


问:
Is there something wrong with this explanation of Chebyshev's Inequality?
So I was revising (more like re-learning coz I suck) Chebyshev's Inequality using this document. Specifically, I am referring to Example $7$ on Page $3$, which I shall reproduce below.

Question

A coin is weighted so that the probability of landing on heads is $0.2$. Suppose the coin is flipped $20$ times. Using Chebyshev's Inequality, find a bound for the probability it lands on heads at least $16$ times.

Answer

\begin{align}
P(X \geq 16) & =
P(0 \leq X \leq 16)
\\[5 mm] & =
P(-8 \leq X \leq 16)
\\[5 mm] & =
P(|X - 4| \geq 12)
\\[5 mm] & \leq
\frac {Var(X)} {12^2}
\\[5 mm] & =
\frac {(20)(0.2)(0.8)} {144}
\\[5 mm] & =
\frac 1 {45}
\end{align}

I understand if we skip the first jump from $P(X \geq 16)$ to $P(|X - 4| \geq 12)$ because, as suggested in the document, $X \geq 0$. However, what I cannot get is how the first equality makes sense. In particular, how does $P(X \geq 16)$ equal $P(0 \leq X \leq 16)$? Also, how does $P(-8 \leq X \leq 16)$ equal $P(|X - 4| \geq 12)$?

I know the answer is correct but I cannot help but feel that the proof is not entirely mathematically right. Anyone who can explain why the first $3$ equalities make sense please do tell.

答:
I think it should be
$$
P(X>16)=P(X-4>12)
12)<\frac{20 \cdot 0.2 \cdot 0.8}{12^2}
$$
This is because $|x|>x$ always

Body:
I think it should be
$$
P(X>16)=P(X-4>12) < P(|X-4|>12)<\frac{20 \cdot 0.2 \cdot 0.8}{12^2}
$$
This is because $|x|>x$ always


问:
Finding maximum volume
A box has corner (0,0,0) and all edges parallel to the axes. If the opposite corner (x,y,z) is on the plane $$ 3x+2y+z=1 $$, what position gives maximum volume? Show first that the problem maximizes
$$
xy-3x^2y-2xy^2
$$

Can somebody please explain how this should be solved?

答: You want to maximize $|xyz|$ w.r.t. constraint $3x+2y+z=1.$  The hint given in the problem is eliminating variable $z.$</p>

Title: Finding maximum volume


问:
Are there integers $a,b,c,d,e,n$ such that $a^3-nb^3=c^3-nd^3=e^3$?
Can anyone help me find (with a computer search) nonzero integers $a,b,c,d,e,n$ such that
$$a^3-nb^3=c^3-nd^3=e^3$$
where $(a,b,e)$ and $(c,d,e)$ are pairwise coprime and $n^2\ne 1$

One solution set will do. Thanks.

答:
TL;DR

Computer search gives us these solutions
$$\begin{cases}a = -5 \\ b = -2 \\ c = 2 \\ d = -1 \\ e = 3 \\ n = 19\end{cases} \
\begin{cases}a = -5 \\ b = 2 \\ c = 2 \\ d = 1 \\ e = 3 \\ n = -19\end{cases} \
\begin{cases}a = -2 \\ b = -1 \\ c = 5 \\ d = -2 \\ e = -3 \\ n = -19\end{cases} \
\begin{cases}a = -2 \\ b = 1 \\ c = 5 \\ d = 2 \\ e = -3 \\ n = 19\end{cases}
$$
after removing those obtained by permuting $(a, b)$ and $(c, d)$.

Method

We first use the diophantine function from SymPy to solve the diophantine equation $A - n B = C - n D$, which gives us the parametric solution $$\begin{cases}A = t_0 \\ B = t_1 \\ C = t_0 + n t_2 \\ D = t_1 + t_2 \\ E = t_0 - n t_1\end{cases}$$ (Actually, diophantine doesn't support variable coefficient. But we can let $n = 2, 3, 4, \dots$ and guess the correct  parametric solution.)

Then, we substitute $t_0 = a^3, t_1 = b^3$ and $t_2 = t$ to get the parametric solution $$\begin{cases}a = a \\ b = b \\ c = \sqrt[3]{a^3 + n t} \\ d = \sqrt[3]{b^3 + t} \\ e = \sqrt[3]{a^3 - n b^3} \end{cases}$$

Finally, we use Python 3 to check all integers in $[[-M, M]] = [[-32, 32]]$ for the parameters $a, b, n, t$ and obtain a solution.

Python 3 script for the final step:

```<code># To the extent possible under law, the author(s) have dedicated all copyright
# and related and neighboring rights to this software to the public domain
# worldwide. This software is distributed without any warranty.

# You should have received a copy of the CC0 Public Domain Dedication along
# with this software.
# If not, see <https://creativecommons.org/publicdomain/zero/1.0/>.

import itertools
import math

def sgn(x):
    return math.copysign(1, x)

def sym_itv(M):
    """return the integer interval [[-M, M]]"""
    return range(-M, M + 1)

def sym_itv_wo_0(M):
    "return the union of integer intervals [[-M, -1]] and [[1, M]]"
    return itertools.chain(range(-M, 0), range(1, M + 1))

def sym_itv_wo_0_pm_1(M):
    "return the union of integer intervals [[-M, -2]] nad [[2, M]]"
    return itertools.chain(range(-M, -1), range(2, M + 1))

def cubert(x):
    return round(sgn(x) * abs(x)**(1/3))

def is_cube(x):
    return cubert(x) ** 3 == x

M = 32

for a, b, n, t in itertools.product(sym_itv_wo_0(M),
                                    sym_itv_wo_0(M),
                                    sym_itv_wo_0_pm_1(M),
                                    sym_itv(M)):

    c_cube = a ** 3 + n * t
    d_cube = b ** 3 + t
    e_cube = a ** 3 - n * b ** 3

    if any(not is_cube(x) for x in [c_cube, d_cube, e_cube]):
        continue

    c = cubert(c_cube)
    d = cubert(d_cube)
    e = cubert(e_cube)

    if (a, b) != (c, d) and c != 0 and d != 0 and \
       math.gcd(a, b) == math.gcd(a, e) == math.gcd(b, e) == 1 and \
       math.gcd(c, d) == math.gcd(c, e) == math.gcd(d, e) == 1:
        print('a = {a}, b = {b}, c = {c}, d = {d}, e = {e}, n = {n}'.format(
            a=a, b=b, c=c, d=d, e=e, n=n))
</code>```

Python 3 script for the first step:

```<code># To the extent possible under law, the author(s) have dedicated all copyright
# and related and neighboring rights to this software to the public domain
# worldwide. This software is distributed without any warranty.

# You should have received a copy of the CC0 Public Domain Dedication along
# with this software.
# If not, see <https://creativecommons.org/publicdomain/zero/1.0/>.

import sympy

A, B, C, D = sympy.symbols('A B C D', integer=True)

for n in range(2, 17):
    sympy.pretty_print(sympy.diophantine(A - n * B - C + n * D))
</code>```

Body:
TL;DR

Computer search gives us these solutions
$$\begin{cases}a = -5 \\ b = -2 \\ c = 2 \\ d = -1 \\ e = 3 \\ n = 19\end{cases} \
\begin{cases}a = -5 \\ b = 2 \\ c = 2 \\ d = 1 \\ e = 3 \\ n = -19\end{cases} \
\begin{cases}a = -2 \\ b = -1 \\ c = 5 \\ d = -2 \\ e = -3 \\ n = -19\end{cases} \
\begin{cases}a = -2 \\ b = 1 \\ c = 5 \\ d = 2 \\ e = -3 \\ n = 19\end{cases}
$$
after removing those obtained by permuting $(a, b)$ and $(c, d)$.

Method

We first use the diophantine function from SymPy to solve the diophantine equation $A - n B = C - n D$, which gives us the parametric solution $$\begin{cases}A = t_0 \\ B = t_1 \\ C = t_0 + n t_2 \\ D = t_1 + t_2 \\ E = t_0 - n t_1\end{cases}$$ (Actually, diophantine doesn't support variable coefficient. But we can let $n = 2, 3, 4, \dots$ and guess the correct  parametric solution.)

Then, we substitute $t_0 = a^3, t_1 = b^3$ and $t_2 = t$ to get the parametric solution $$\begin{cases}a = a \\ b = b \\ c = \sqrt[3]{a^3 + n t} \\ d = \sqrt[3]{b^3 + t} \\ e = \sqrt[3]{a^3 - n b^3} \end{cases}$$

Finally, we use Python 3 to check all integers in $[[-M, M]] = [[-32, 32]]$ for the parameters $a, b, n, t$ and obtain a solution.

Python 3 script for the final step:

# To the extent possible under law, the author(s) have dedicated all copyright
# and related and neighboring rights to this software to the public domain
# worldwide. This software is distributed without any warranty.

# You should have received a copy of the CC0 Public Domain Dedication along
# with this software.
# If not, see <https://creativecommons.org/publicdomain/zero/1.0/>.

import itertools
import math

def sgn(x):
    return math.copysign(1, x)

def sym_itv(M):
    """return the integer interval [[-M, M]]"""
    return range(-M, M + 1)

def sym_itv_wo_0(M):
    "return the union of integer intervals [[-M, -1]] and [[1, M]]"
    return itertools.chain(range(-M, 0), range(1, M + 1))

def sym_itv_wo_0_pm_1(M):
    "return the union of integer intervals [[-M, -2]] nad [[2, M]]"
    return itertools.chain(range(-M, -1), range(2, M + 1))

def cubert(x):
    return round(sgn(x) * abs(x)**(1/3))

def is_cube(x):
    return cubert(x) ** 3 == x

M = 32

for a, b, n, t in itertools.product(sym_itv_wo_0(M),
                                    sym_itv_wo_0(M),
                                    sym_itv_wo_0_pm_1(M),
                                    sym_itv(M)):

    c_cube = a ** 3 + n * t
    d_cube = b ** 3 + t
    e_cube = a ** 3 - n * b ** 3

    if any(not is_cube(x) for x in [c_cube, d_cube, e_cube]):
        continue

    c = cubert(c_cube)
    d = cubert(d_cube)
    e = cubert(e_cube)

    if (a, b) != (c, d) and c != 0 and d != 0 and \
       math.gcd(a, b) == math.gcd(a, e) == math.gcd(b, e) == 1 and \
       math.gcd(c, d) == math.gcd(c, e) == math.gcd(d, e) == 1:
        print('a = {a}, b = {b}, c = {c}, d = {d}, e = {e}, n = {n}'.format(
            a=a, b=b, c=c, d=d, e=e, n=n))

Python 3 script for the first step:

# To the extent possible under law, the author(s) have dedicated all copyright
# and related and neighboring rights to this software to the public domain
# worldwide. This software is distributed without any warranty.

# You should have received a copy of the CC0 Public Domain Dedication along
# with this software.
# If not, see <https://creativecommons.org/publicdomain/zero/1.0/>.

import sympy

A, B, C, D = sympy.symbols('A B C D', integer=True)

for n in range(2, 17):
    sympy.pretty_print(sympy.diophantine(A - n * B - C + n * D))


答:
Express this as a recurrence:
$$
\pi_1 = \pi_0\\
\pi_2 = \frac{\pi_1}{2}=\frac{\pi_0}{1 \cdot 2}\\
\pi_3 = \frac{\pi_0}{1 \cdot 2 \cdot 3}\\
$$
Hence
$$
\pi_k = \frac{\pi_0}{k!}
$$
Since probabilities sum to $1$, you get $1$ on LHS, and $\exp(1) \pi_0$ on RHS and the result follows.

Title: Distribution of discrete random variable $X$

Body:
I am stuck on a question regarding discrete random variable. The question is as follows:

The probability mass function of $X$ with $x \in \mathbb{N}:=\{0,1,2, \ldots\}$ satisfies the following:

\begin{align}
\frac{(=)}{(=+1)}=x+1
\end{align}

I am asked to find $(=0)$ and identify the distribution of $X$

I worked out that $(=0)=(=1)$ from the given PMF but not sure what to do next and how to identify the type of distribution of $X$.

Any help is greatly appreciated.


答:
You can easily prove a much more general statement using continuity and the fact that composition of measurable functions remain measurable.

From your attempt one can conclude that you are considering the canonical sigma topology on $\mathbb R$ and thus the Borel sigma algebra as canonical sigma algebra.

A simple exercise is now to show that addition, multiplication, etc are continuous functions $\mathbb R\times\mathbb R\rightarrow\mathbb R$. By definition of continuity, it is straightforward to see that continuous functions are measurable. Finally, use the fact that compositions of measurable functions are measurable (which is also easy to prove) to conclude that all elementary operations are measurable.

Title: Prove $X/Y$ is a random variable.

Body:
Show that if $X$ and $Y$ are random variables then $X/Y$ is a random variable $(Y\neq 0)$.

I read this in some notes I found online from a university stating that $X+Y,XY,X/Y$ are all random variables. I know the proof for $X+Y$ which looks like,

$\forall r \in R,\ \exists q \in Q,  \{X<r,Y>c-r\} \subset \{X<q, Y>c-q\}$
$\cup_{r\in R} (\{X<r\}\cap\{Y>c-r\})\subset \cup_{q\in Q} (\{X<q\}\cap\{Y>c-q\})$
$\cup_{q\in Q} (\{X<q\}\cap\{Y>c-q\}) \subset \cup_{r\in R} (\{X<r\}\cap\{Y>c-r\})$

Is it possible the proof for $X/Y$ is similar? I cant seem to fix a constant ($r$).


问:
Basis funcions Discontinuous Galerkin
I am learning about Disconinuous Galerkin methods. I fail to understand how he basis funcions are constructed. I understand that typically Legendre polynomial are used, but I can't see how they relate to the nodes.

In Continuous Galerkin methods, the basis funcions for a linear triangle (3 nodes) are:

$N_1(\xi, \eta) = 1 - \xi - \eta\\
N_2(\xi, \eta) = \xi\\
N_3(\xi, \eta) = \eta\\$

What are the basis functions for a linear triangle used in Discontinuous Galerkin methods? Say we choose Legendre polynomials of order 2, then I would have one basis function of order 0, one of order 1, and one of order 2 for each node?

答:
It is good to hear you figured out your problem!

My response to your questions is thus. There is nothing stopping you using those basis functions in DG - you will just be considering a broken Sobolev space. By this we simply mean that each triangle is independent of each other.

Title: Basis funcions Discontinuous Galerkin


问:
Solving equation of the type $ -1 = \cos x - a x \sin x $
I have the equation
$$ -1 = \cos x - a x \sin x $$
where, $a$ is a constant, $x$ is a real variable and taken in the range $0<x<\pi$. When I solve it by simply rearranging the terms, I end up with
$$ \frac{1+\cos x}{\sin x}=\cot \frac{x}{2}=a x. $$

However, the correct answer is supposed to give both
$$\cot \frac{x}{2}=a x,$$
$$\cos \frac{x}{2}=0.$$

I am not sure how the latter was found. I must be missing something. Any advice is appreciated.

答:
If we discrad the trivial solution (as explained in previous answers), the problem is effectively to find the zero of function
$$f(x)=\cot \left(\frac{x}{2}\right)-ax$$ which will not show any explicit solutions. There will not be any problem with numerical method.

But we can build approximations. For example, using Taylor expansion
$$\frac{\cot \left(\frac{x}{2}\right)}{x}=\frac{2}{x^2}-\frac{1}{6}-\frac{x^2}{360}+O\left(x^4\right)$$ which is a more than decent approximation (just plot the lhs and rhson the same graph for $0\leq x\leq \pi$).

Neglecting the higher order terms
$$a=\frac{2}{x^2}-\frac{1}{6}-\frac{x^2}{360}\implies x^4+60(1+6a)x^2-720=0$$ which is a quadratic in $x^2$ and then

$$x_{\text{est}}=\sqrt{6 \left( \sqrt{15(60 a^2+20 a+3)}-30 a-5\right)}$$

For example, using $a=12.345$, this would give $x_{\text{est}}= 0.399806370$ while the "exact" solution, obtained by Newton method, is $x=0.399806343$.

Using $a=0.12345$, this would give $x_{\text{est}}= 2.54763$ while the "exact" solution, obtained by Newton method, is $x=2.53501$.

Now, trying to have better estimates, using the first iterate of an high order iterative method starting with $x_0=0$, one can propose
$$x_{\text{est}}=\sqrt{\frac{220 \left(22680 a^4+15120 a^3+4158 a^2+558 a+31\right)}{2494800 a^5+2079000
   a^4+748440 a^3+145200 a^2+15202 a+691}}$$ which seems to be decent as shown below
$$\left(
\begin{array}{ccc}
a & x_{\text{est}} & x_{\text{exact}}\\
 1 & \color{red} {1.3065423}80219720 &
                  1.306542374188806\\
 2 & \color{red} {0.96018887}4044812 &
                  0.960188873914783 \\
 3 & \color{red} {0.7944961284}50111 &
                  0.794496128438359 \\
 4 & \color{red} {0.69270825340}9536 &
                  0.692708253407499 \\
 5 & \color{red} {0.62210569640}1108 &
                  0.622105696400595 \\
 6 & \color{red} {0.569452885113}318 &
                  0.569452885113153 \\
 7 & \color{red} {0.5282417875158}91 &
                  0.528241787515829 \\
 8 & \color{red} {0.494850930368852} &
 0.494850930368852
\end{array}
\right)$$


问:
Probability to reach $a>0$ before returning to the origin in a one dimensional random walk.
Given a one dimensional random walk where I have $p$ probability to go forward and $q$ probability to go backwards.

I have to prove that, starting from the origin, the probability of reaching $a>0$ before returning to the origin is $p(1-q_1)$

Where

$$q_1=\frac{\left(\frac{q}{p}\right)^a-\left(\frac{q}{p}\right)}{\left(\frac{q}{p}\right)^a-1}$$

答:
After what Ian's response.

Ian (https://math.stackexchange.com/users/83396/ian), Probability to reach $a>0$ before returning to the origin in a one dimensional random walk., URL (version: 2021-03-25): https://math.stackexchange.com/q/4076368

I thought that there was a paralelism between this problem and the gambler's ruin:

Let player A have a capital of $z$ and player B have a capital of $b$. If we define $a=z+b$. And the probability for the player A to win the game is A and the probability for player B to win the game is q. We know that the probability for player A to lose  all his money is:

if $p\neq q$

$$q_z=\frac{\left(\frac{q}{p}\right)^a-\left(\frac{q}{p}\right)^z}{\left(\frac{q}{p}\right)^a-1}$$
if $p=q$
$$q_z=1-\frac{z}{a} $$

So in our model we need to redefine the $p_0$ because in the gambler's ruin it's $0$ (If you don't have money you can't gamble). But in our game we can actually go forward with a probability of $p%$.

In order to achieve the point $a$ you first have to start from the point $z=1$ for the probability to be defined in the gambler's ruin model. So, as Ian said, we need to go forward with a probability of $p$. Once we are in $z=1$, we can apply the gambler's ruin model so we have a probability of $p_1$ to reach $a$.

Finally we multiply all this two probabilities and we have the probability to reach $a$ starting from $0$.
$$pp_1=p(1-q_1)$$


问:
How to calculate a result based on an exponentially changing pattern/sequence.
Given any number (including decimals) between 12.5 and 87.5 I need to calculate another number based on these results:

 Input = 12.5 | 31.25 | 50 | 68.75 | 87.5

Result = 12.5 | 64.5  | 78 | 84       | 87.5

So, for example, if "input = 25" or "input = 58", what can I use to calculate a result?

The only thing I can determine from looking at these numbers is that the closer the input is to 87.5, the smaller the increment in the result. (The result of 12.5 to 13.5 would have a much greater difference than 86.5 to 87.5).

答: You are looking for an interpolation function or a fitted curve that comes close to the points.  It is a huge field in numerical analysis, and routines can be found in any text.  If you plot the data in Excel, "Add Trendline", and "Show Equation on Graph" you can see the function and how well it fits.  You can choose various functional forms to see what fits your needs.</p>


问:
Direct integration of a complex integral
In my homework there is a integral looks so strange, this is the first time that I see this kind of thing. So I have no idea what to do.

$\int_{0}^{-1+i\infty}(z+1)e^{iz}dz$

What does it mean the upper part of integral $-1+i \infty$? And how can i solve it?

I tried to take integral respect to $z$. Then i wrote $z=x+iy$ and put from $0$ to $-1$ for all x and put from $0$ to $\infty$ for all y. Am I right?

答:
It is better to consider first
$$I_p=\int_{0}^{-1+ip}(z+1)\,e^{iz}\,dz$$ Expandin the integrand, one integration by parts gives the antiderivative and using the bounds
$$I_p=e^{-(p+i)} (p+1)-(1-i)$$ Now, take the limit when $p\to \infty$.


答:
You may assume that the big box is the unit cube $[0,1]^3$. You then have
$M=(0,\mu,\mu)$,$\>N=(\nu,1,\nu)$. You now have to determine $\mu$ and  $\nu$, such that the vector $\vec{MN}=(\nu,1-\mu,\nu-\mu)$ is parallel to $\vec{A_1C}=(1,1,-1)$. This fortunately has a solution.

Title: Cuboid diagonals

Body:
The points $M$ and $N$ are taken to be on the diagonals $AB_1$ and $BC_1$ of the cuboid $ABCD-A_1B_1C_1D_1$. If $MN \parallel A_1C$ , find the $MN:A_1C$ ratio.
Initially I understood what's going on here, but unfortunately I can't solve it... The problem requires vector solution. I have no idea how to do it. Can anyone solve it then explain beautifully?? I will glad...


问:
Proving that if $r$ is rational, then $\frac27(3r+5)-9$ is also rational
I'm trying to write a proof but I'm getting stuck. I have to prove that

For all real numbers $r$, if $r$ is rational, then $\frac27(3r+5)-9$ is also rational.

I let $r = a/b$ by definition of rational. Then I substituted $r$ for $a/b$. After doing some algebra, I got down to $$\frac27\cdot\frac{3a+5b}{b}-9$$ but I'm not sure if this is correct

答:
Continue:
$$\underbrace {\frac {6a+10b}{7b}}_m-9$$
Here, $m$ is a rational as $a $ and $b $ are integers and multiplication and addition are closed under integers. Further, $b\neq0$ implies $7b\neq0$. So, $m$ is rational.

Now, combine the fact that difference of two rationals is a rational.

Hope this helps. Ask anything if not clear :)


问:
Can any polynomial in $\Bbb R [x]$ be written as the difference of two sum-of-squares polynomials in $\Bbb R [x]$?
Let $f \in \Bbb R[x_1,\dots,x_n]$ be an arbitrary polynomial, not necessarily non-negative. Are there always two sum-of-squares (SOS) polynomials $g, h \in \Bbb R[x_1,\dots,x_n]$ such that $f=g-h$?

If so, to what extent will $\max\{\deg(g),\deg(h)\}$ increase, compared to $\deg(f)$? Linearly? Polynomially? Exponentially?

答:
I claim that any $f(x_1,...,x_n)$ polynomial is a difference $g-h$ where $g,h$ are sums of square polynomials.  To see this let $S$ be the set of all polys that can be written in this form.  Note that

(1) For each $f\in S$, $c f\in S$ for all $c\in\Bbb R$.

(2) For $f_1,f_2\in S$, $f_1+f_2\in S$.

(3) For $f_1,f_2\in S$, $f_1f_2\in S$.

(1) and (2) are obvious.  To see (3), let $f_1=g_1-h_1$ and $f_2=g_2-h_2$, $g_i,h_i$ are SOS.  Then
$$f_1f_2=(g_1g_2+h_1h_2)-(g_1h_2+g_2h_1).$$
But $g_1g_2,h_1h_2,g_1h_2,g_2h_1$ are SOS.

But $1=1^2-0^2$, $x_i=(x_i+1/4)^2-(x_i-1/4)^2$.  So $S$ is generated by $1,x_1,x_2,...,x_n$.  Therefore $S$ is the whole $\Bbb R[x_1,...,x_n]$.  We can easily see that if $\deg f=d$ then we can take $g,h$ to have degrees $\le 2d$.

In fact you can also see that $x_i^{2k}=(x_i^k)^2-0^2$ and $x_i^{2k+1}=(x_i^k(x_i+1/4))^2-(x_i^k(x_i-1/4))^2$.  So you can take $g,h$ to have degrees $\le$ the minimum of $2d$ and $d+n$.

This answer is possible because of Arthur's comment.


问:
How to construct likelihood ratio tests when the null hypothesis is of this form?
For example, I know how to get the likelihood ratio when $H_o: \theta = 3$ against $H_1: \theta > 3$. We derive some rule involving $x$ and a constant from the likelihood ratio, and then find the probability of type I error. But when the null hypothesis is of the form $H_o: \theta \leq 3$, how do I do this? Assume that $X$ is from an exponential distribution with mean $\frac{1}{\theta}$ and I have to construct a LRT with $\alpha=0.05$.

答:
$H_0=3$ and $H_0\leq 3$ are equivalent. This because of the definition of $\alpha$

$$\alpha= \sup_{\theta \in \Theta_0} \pi(\theta)$$

where $\pi(\theta)$ is the power function

thus if you know how to verify this system

$$\begin{cases}
H_0: \theta=3\\
H_1: \theta>3
\end{cases}$$

verify this one in the same way.

$$\begin{cases}
H_0: \theta\leq3\\
H_1: \theta>3
\end{cases}$$

in an informal way: The null hypotesis is always considered TRUE until you have enough data to reject it...thus you assume as $H_0$ the best value you can choose...so if $H_0$ is $\theta \leq 3$ you choose $\theta=3$ as the best value to be verified

If your model belongs to the exponential family (as it is the example you posted) this test is well explained here, on page 422, theorem 5

If you read next page, 423 the example 13 is exactly yours! What the book does not state is that the resulting integral can be solved via chi-squared distribution, but I think you perfectly know that.


问:
What happens after the cardinality $\mathfrak{c}$?
While having measure theory this year the following came in my mind:

  When we go from finite objects to infinite we "lose" a lot of properties. For example
  the summation isn't well defined anymore, when the sum doesn't converge, and operations
  which are abelian in the finite case are not abelian in the infinite case (think of conditional convergent series).

When we go from countable to uncountable we have things like that too. Measures are most time countable additive but not uncountable additive (at least the lebesgue measure). Sums of uncountable indexing sets only can converge when they are in fact countable (meaning nearly all are 0).

So my question is after Arthur Comments slightly rephrased

What happens when you go from the cardinality $\mathfrak{c}$ to a higher cardinality? Are there more intermediate steps, meaning having a cardinality strictly greater than $\mathfrak{c}$ but not "to big"?

答:
Allow me to share a small piece of my philosophical about the uncountable. It may not actually answer your question, but in a way it does.

Mathematics is created by humans. Humans are essentially finite beings, and they communicate in a finite matter. This is why we base things on finite intuition, which is the leading cause of confusion when people first deal with infinite sets.

Countable sets are sets that can be enumerated, and approximated by finite means. This is exactly the idea behind Cauchy's definition of the limit (and even more so if we only use rational numbers). We can list the rational numbers, for example, in a way that is assuring us that every number is met at a finite stage.

For this reason separability is a useful property. It allows us to approximate every element of the space using countable means. This is visible in metric space more than it is in non-metrizable spaces, but it's still true. That's the idea behind separability. So in some sense we may have an uncountable set, but we can approximate it using its structure and finite means.

But the uncountable, the truly uncountable, is not describable by such means. It's too big for us to assure that finite approximations can eventually swallow the set whole. We can't describe its initial segments in a finite fashion. We can't even assure that we can order the set nicely. This is why the axiom of choice is so important to us in the modern era of mathematics. We deal with abstract objects and we usually don't pose much limitations on their size (even though we usually care for objects quite small), and the axiom of choice gives us reasonable means to handle their sizes and structure.

But the uncountable is still uncountable. Note that even if we assume that $2^{\aleph_0}=\aleph_2$, without further assumptions such as Martin's Axiom, then we don't even know what's going on at the $\aleph_1$ level. All we know is that we can find a subset of the real numbers which has this cardinality. This is why the continuum hypothesis - in my opinion - is unprovable. The uncountable is indeed too large for us to comprehend (and manage in full) using our essentially finite means.

So what does happen beyond the continuum? A lot that we can't get a good grip on. The Lowenheim-Skolem theorem tells us that given a first-order structure which has an infinite model, it has an infinite model of every cardinality. This means that everything that happens at one infinite cardinality happens at the others, at least first-order-wise (e.g. there aren't many Archimedean fields, but there are real-closed fields). So we have a lot of fields, rings, measure algebras, Banach spaces, ordered sets, and so on.

What properties do they have? The answer often depends on set theoretic assumptions now. Much like many things at the countable level can fail without some axiom of choice (which assures the transition from finite to countable is smooth), many more can fail at the uncountable level. Assumptions such as $\lozenge_\kappa$, Martin's Axiom, cardinal arithmetics (e.g. $\kappa^{<\kappa}=\kappa$), and many many more set theoretical assumptions can - and will - tell you about the behavior of very large objects, but you will need to know about them first.


答:
First fix some index $i\in[n]$.
For even $k\in\mathbb{N}$, $x_i^{k}$ is already an SOS.
For odd $k\in\mathbb{N}$, $x_i^{k} = (x_i^{k}+1)^2/4 - (x_i^{k}-1)^2/4$.

Now fix some monomial $x_1^{d_1} \cdots x_n^{d_n}$.
For each $i\in[n]$, $x_i^{d_i} = g_i-h_i$ where $g_i,h_i$ are SOS and $\max\{\deg(g_i),\deg(h_i)\} \leq 2d_i$.
Thus, $$ \prod_{i\in[n]} x_i^{d_i} = \prod_{i\in[n]}(g_i-h_i)$$
which is a difference of two SOS and the degree of RHS is at most $2(d_1+\cdots+d_n)$.

Title: Can any polynomial in $\Bbb R [x]$ be written as the difference of two sum-of-squares polynomials in $\Bbb R [x]$?

Body:
Let $f \in \Bbb R[x_1,\dots,x_n]$ be an arbitrary polynomial, not necessarily non-negative. Are there always two sum-of-squares (SOS) polynomials $g, h \in \Bbb R[x_1,\dots,x_n]$ such that $f=g-h$?

If so, to what extent will $\max\{\deg(g),\deg(h)\}$ increase, compared to $\deg(f)$? Linearly? Polynomially? Exponentially?


答:
What you need to do here is show that $(2/7)(3r+5)-9$ can be written in the form $A/B$, where $A$ and $B$ are integers. To do this, keep going with what you started:

$${2\over7}\left({3a\over b}+5\right)-9={2\over7}\cdot{3a+5b\over b}-9={6a+10b\over7b}-9={6a+10b-63b\over7b}={6a-53b\over7b}$$

and now note that this is of the form $A/B$ where $A=6a-53b$ and $B=7b$ are integers.

Title: Proving that if $r$ is rational, then $\frac27(3r+5)-9$ is also rational

Body:
I'm trying to write a proof but I'm getting stuck. I have to prove that

For all real numbers $r$, if $r$ is rational, then $\frac27(3r+5)-9$ is also rational.

I let $r = a/b$ by definition of rational. Then I substituted $r$ for $a/b$. After doing some algebra, I got down to $$\frac27\cdot\frac{3a+5b}{b}-9$$ but I'm not sure if this is correct


问:
Is $[0,1]\cdot A$ a convex set?
Let $A\subset\mathbb{R}^d$ be a convex set. We define \begin{align}[0,1]\cdot A:=\{x\in\mathbb{R}^d:x=\lambda a \text{ for a }\lambda\in[0,1] \text{ and }a\in A\}.\end{align} Is this set convex?

I was thinking that in the case $d=1$ the statement is true, because $A$ is an interval and, hence, $[0,1]\cdot A$ is an interval too. But is it also true in higher dimension $d>1$?

I want to say, this is not a homework. I want to show that a specific problem in the theory of optimal control can be solved by considering a replacement problem.

答:
Let $x=\lambda a$ and $y=\mu b$ with $a,b\in A$ and $\lambda,\mu\in[0,1]$.

You want to prove that $tx+(1-t)y\in [0,1]A$ for all $t\in[0,1]$.

Now set $$\eta=\lambda t+\mu(1-t).$$
$\eta$ is  in $[0,1]$.
and set $$s=\frac{\lambda t}{\lambda t+\mu(1-t)}$$
also $s$ is in $[0,1]$.

With these constants you have

$tx+(1-t)y=s\eta a+(1-s)\eta b$

since $\eta A$ is convex then $s\eta a+(1-s)\eta b\in \eta A$ for all $s\in[0,1]$, so $tx+(1-t)y=s\eta a+(1-s)\eta b\in [0,1]A$, as wanted.

Title: Is $[0,1]\cdot A$ a convex set?


答:
For a geometric approach, let $x = \lambda a \in [0,1] \cdot A$ and $ y = \mu b \in [0,1] \cdot A$.

Consider the triangle $T$ determined by $a$, $b$ and the origin. As $A$ is convex, the edge between $a$ and $b$ lies completely in $A$. Hence for any point $z$ on the edge between $a$ and $b$ the line between $z$ and the origin lies completely in $[0,1] \cdot A$. Therefore, the whole triangle T lies in $[0,1] \cdot A$, and in particular the line between $x$ and $y$ lies in $[0,1] \cdot A$, which means that $[0,1] \cdot A$ is convex.

Title: Is $[0,1]\cdot A$ a convex set?

Body:
Let $A\subset\mathbb{R}^d$ be a convex set. We define \begin{align}[0,1]\cdot A:=\{x\in\mathbb{R}^d:x=\lambda a \text{ for a }\lambda\in[0,1] \text{ and }a\in A\}.\end{align} Is this set convex?

I was thinking that in the case $d=1$ the statement is true, because $A$ is an interval and, hence, $[0,1]\cdot A$ is an interval too. But is it also true in higher dimension $d>1$?

I want to say, this is not a homework. I want to show that a specific problem in the theory of optimal control can be solved by considering a replacement problem.


问:
An interesting problem using Pigeonhole principle
I saw this problem: Let $A \subset \{1,2,3,\cdots,2n\}$. Also, $|A|=n+1$.
Show that There exist $a,b \in A$ with $a \neq b$ and $a$ and $b$ is coprime.
I proved this one very easily by using pigeon hole principle on partition on $\{1,2\},\{3,4\},\dots,\{2n-1,2n\}$.

My question is How can I prove or disprove that:

Let $A \subset \{1,2,3,\cdots,2n\}$. Also, $|A|=n+1$.
Show that There exist $a,b \in A$ with $a \neq b$ and $a|b$.

I can't make suitable partition. Is this true?

答:
Any number from the set $A$ is of the form $2^{k}l$ where $k\ge 0,0\le l\le (2n-1)$ and $l$ is odd.

Number of odd numbers $l\le (2n-1) $ is $n$.

Now if we select $(n+1)$ numbers from the set $A$ then there must be two numbers(among the selected numbers) with the same $l$.

That is, we must get $a,b$ with $a=2^{k_1}l$ and $b=2^{k_2}l$ now as $a\ne b$ so $k_1\ne k_2$.Now if $k_1>k_2$ then $b|a$ else $a|b$.

This completes the proof.


问:
Does this series converge to zero?
Does $|x_k|$ $\to$ $0$ and $\sum_{k=1}^{\infty} |x_k| < \infty$ $\implies$ $\sum_{k=m+1}^{\infty} |x_k| \to 0$ as $m\to\infty$ ?

If so, how would one formally prove this?

Any help would be greatly appreciated.

Title: Does this series converge to zero?

Body:
Hint:
$$\sum_{k=m+1}^{\infty} |x_k| = \sum_{k=1}^{\infty} |x_k| - \sum_{k=1}^{m} |x_k| $$


问:
Finding Laurent series around each singularity
I have this function $f(z)=\frac{1-iz}{1+iz}$ and I would like to find the Laurent expansion around the singularities of it.

So first I know there is a singuality at $1+iz=0$ so $z=i$. Next rationalising the denominator I get:

$$f(z)=\frac{1-z^2-2iz}{z^2+1}$$.

This has singularities at $z=±i$. Then I am not sure how to further. Next say we look at $z=i$, so:

$$f(z)=\frac{1-z^2-2iz}{(z-i)(z+i)}$$

Then about our singularity:

$$f(z)=\left(\frac{1}{z-i}\right)\frac{1-z^2-2iz}{(z+i)}$$

$$f(z)=\left(\frac{1}{z-i}\right)\frac{(1-iz)(1+iz)}{(z+i)}$$

However I am not sure how to continue this? Could anyone help me out?

答: The only singularity is located at $z=i$ and you have$$\frac{1-iz}{1+iz}=\frac{z+i}{-z+i}=-1-2i(z-i)^{-1}.$$


问:
Minimisation with combinatorics

Suppose Dianne has a number of friends. She wants to invite three
to her house for dinner every day of the year. What is the minimum
number of friends she can have if she does not want to invite the same
three friends more than once?

I reason that the number of friends can be obtained by $ {n \choose 3} = \frac{n!}{3!(n-3)!} = 365$

but this gives me a third-order polynomial which is unsolvable without a computer. I know guess and check works but I am looking for a "more correct" solution that uses mathematics. I know this si also technically incorrect because you cant "choose" out of numbers other then integers. I know the solution but I'm having trouble framing my answer using combinatorics

I also tried applying pigeonhole principle to this question (initially) but that didn't get me anywhere.

答:
It is interesting to notice that you face the same problem as Newton before Newton method !

Simpliying all numbers, you need to solve for $n$ the cubic equatio
$$n^3-3 n^2+2 n-2190=0$$

First guess : $n^3-2190=0 \implies n=\sqrt[3]{2190}=12.9862$; say $n=13$.

Now, make one single iteration of Newton method to get $n=\frac{6077}{431}=14.0998$. So $n=\lceil 14.0998\rceil=15$.

Title: Minimisation with combinatorics


问:
Recommended (free) software to plot points in 3d
I am looking for (preferably free) software to:

1) plot 3d points read from a file. A scatter plot would be fine.

2) Optionally color the points by a property - also read from the file

It would be terrific if this program could also compute and display the best fit plane through these points.

答:
There are quite a number of freewares which do that.

Gnuplot, with a very interesting not so frequently asked questions here (archive version), fit is easily obtained. Nice outputs to PDF and LaTeX

Matplotlib: requires Python, so that's probably not your best option if you need to produce graphs quickly, but if you are thinking about a long term solution, I would go for this one,

R, normally for statistical computations, but quite nice plotting possibilities. Also, it is a software which is intended to read data in files, so it has very powerful and easy to use functions to import CSV files and the likes,

Octave not a big fan. Don't know the new major release, though. Previous releases were based on Gnuplot for the graphic part,

Scilab has continuously grown througout the years, so as to become a monster now. Not a big fan either.

Hope you'll find your best choice here !!!
Good luck,
Sébastien


问:
How to solve this system of trigonometric equations?
I wanto to express $x$ and $y$ as a function of the known parameters $a, b, c, d, p, q$. The angles $\phi_1$ and $\phi_2$ are unknowns, but not of interest for me:

$$
\begin{align}
a &= x \cdot \cos(\phi_1) + y\cdot \cos(\phi_2) \\
b &= -x \cdot \sin(\phi_1) -y\cdot \sin(\phi_2) \\
c &= -x \cdot p \sin(\phi_1) - y \cdot q\sin(\phi_2) \\
d &= -x \cdot p \cos(\phi_1) - y \cdot q \cos(\phi_2)
\end{align}
$$
All variables are real, $p$ and $q$ are positive.

How do I do this? Even though I had a couple of higher math courses in college, I didn't encounter trigonometric systems of equation once.

An additional piece of information that might be relevant: I am actually only looking for $x^2$ and $y^2$, so I don't care about any ambiguities because of a $\pm$-solution.

答:
You can treat the middle two equations as a system of linear equations in the two unknowns $x\sin\phi_1$ and $y\sin\phi_2$. So with just those two equations you can find the values of $x\sin\phi_1$ and $y\sin\phi_2$.

Similarly, use the other two equations to solve for $x\cos\phi_1$ and $y\cos\phi_2$.

With those four values you can calculate $x^2$ and $y^2$.


问:
A consequence of the comprehension schema is that the intersection of two sets is a set
I am currently doing a project on set theory and at the minute i am studying ZF axioms. A consequence of the comprehension schema is that the intersection of two sets is a set. Hnece i have come up with the following proporsition.

Proposition: Let $x$ and $y$ be two sets. Then the intersection of $x$ and $y$, $x\cap y$ is a set.

Proof: Define $P(z) = z \in Y$. Then  $x\cap y = \{ z \in x \wedge P(z)\} =\{ z \in x \wedge z \in y \}$. Hence, by the comprehension schema  $x\cap y$ is a set.

I would greatly appreciate if someone could let me know if there are any mistakes in the proof.

答: Yes not only is your proof correct for the intersection of two sets but with a minor adjustment you can prove it for any non empty class of sets (defined by a first order formula). This is intuitively true because the intersection is a subset of any of the sets partaking the intersection. The way you prove it is similar. Given a class (a collection of sets that may or may not be a set itself) defined by the formula $\varphi(x)$ then we can choose a set $a$ in the class (this is where you need it to be non empty) and we define the intersection to be $\{x\in a:\forall y (\varphi(y)\rightarrow x\in y)\}$ The new first order formula which uses comprehension is now $\psi(x)\equiv \forall y (\varphi(y)\rightarrow x\in y)$.


问:
Control Design Problem with continue and discrete signals;
I met a real application problem within my research.

The control signal will be three dimension, $u_1$,$u_2$,and $u_3$.

$u_1$ and $u_2$ are continous. Unfortuately, $u_3$ is disrete with the 6 selections with {1,2,3,4,5,6}. I have to design this control from optimal perspective.

Is there anyone who has done this situation before? Could you give me a hint.

thanks

Jie

答: Model predictive control is a general framework for optimization-based control, and having discrete control inputs does not make the control structure any more complicated conceptually, although the actual computation of control inputs might become prohibitively expensive as it will lead to mixed-integer optimization problems.</p>


问:
Problem with a definition of $\succsim$-maximal element and $\succsim$-maximum
I found in a book the following definition of $\succsim$-maximal element and $\succsim$-maximum, but I really don't see the difference between what is written in the book (I know the difference between those two concepts). Am I wrong?

Let $(X,\succsim )$ be a preordered set and $∅ = Y ⊆ X$. An element $x$ of $Y$ is said to be $\succsim$-maximal in $Y$ if there is no $y \in Y$ with $y \succ x$. If $x \succsim y $ for all $y \in Y$ , then $x$ is called the $\succsim$-maximum of $Y$.

答:
The difference in the negation:

Maximal element is one that there does not exist anyone strictly larger.

Maximum element is one that every element is smaller or equivalent to.

We have $\lnot\exists y\varphi(y,x)$ and $\forall y\varphi(x,y)$, where $\varphi(x,y)$ states $y\precsim x$.


问:
Solve the equation $2 \rvert 3x + 4y - 2 \lvert + 3\sqrt{25 - 5x + 2y} = 0$
I have been trying to solve the following equation but have not been able to figure out the solution. My textbook says that $x=4$ and $y=\frac{-5}{2}$. On inspection, I noticed that $25 - 5x + 2y \geqslant 0$ and also figured out that $y=\frac{1}{64}(-48x\pm 3\sqrt{1673-416x}+41)$ and hence $1673-416x \geqslant 0$. However, I feel like these inequalities are getting me nowhere since it still feels like there are infinite solutions. I was curious about how to continue (if I should continue at all); or is there a quicker approach to take which I have clearly ignored?

答:
If the equation has a solution, then it must be if and only if

$$\begin{cases} 3x + 4y-2  =0 \\ 25 - 5x + 2y = 0\end{cases}$$

This is the simple linear system of equation.


问:
What does the derived functor send roofs of one derived category to?
For reference, I am working out of Pramod Achar's notes here, and my confusion centres around definition 7 on page 15.

For $F: \mathcal A \to \mathcal B$ a left exact functor, and $\mathcal R$ an adapted class for $F$ (ie $F$ is exact on $\mathcal R$ and $\mathcal R$ is large enough in $\mathcal A$), the derived functor $$ RF: D^+(\mathcal A) \to D^+(\mathcal B)$$ is defined by sending $A^\bullet$ to the complex $F(R^\bullet)$, where $R^\bullet$ is an $\mathcal R$-resolution of $A^\bullet$ (a quasi-isomorphism $t : A^\bullet \to R^\bullet$ exists).
I am hoping to understand the corresponding morphism of $\text{Hom}_{D^+(\mathcal A)}(A^\bullet, B^\bullet)$ is under $RF$.

I know that a morphism in $\text{Hom}_{D^+(\mathcal A)}(A^\bullet, B^\bullet)$ is a roof $$A^\bullet \xleftarrow{q} X^\bullet \rightarrow B^\bullet$$ where $q$ is qis, so my guess would be that (with $s: B^\bullet \to Q^\bullet$ an $\mathcal R$-resolution of $B^\bullet$) the roof is sent to $$F(R^\bullet) \xleftarrow{F(t \circ q)} F(X^\bullet) \rightarrow F(Q^\bullet).$$
However I don't think this is very well-defined, unless I add the constraint that the $X^\bullet$ is also in $\mathcal R$ (if this were the case, then the fact that $F$ is exact on $\mathcal R$ tells me that $F(t \circ q)$ is qis, from which I get an equivalence class for the roof in $D^+(\mathcal B)$).

Since in the definitions of the derived categories I don't get such a constraint, my suspicion is that I am slightly wrong here, but I can't think of how to fix this. I have tried further unpacking the equivalence classes on the roofs of $D^+(\mathcal B)$ to show that we don't need to take $F(X^\bullet)$, but I haven't managed to show that sending the roof $A^\bullet \xleftarrow{q} X^\bullet \rightarrow B^\bullet$ to any equivalence class $F(R^\bullet) \leftarrow Y^\bullet \rightarrow F(Q^\bullet)$ is well-defined, at least functorially-speaking.

答:
First of all you can use the roof in the opposite direction $A^\bullet \rightarrow X^\bullet \xleftarrow{q} B^\bullet$ where $q$ is a quasi isomorphism. now you have a quasi-isomorphism from $X\to R$ so you get an equivalent  roof such that the middle term is in $R$ and basically your idea work. but to made it precise you have two choice:

you can consider the subcategory $\mathcal R\to \mathcal A$ and use the previous paragraph to show that $S^{-1}\mathcal R\to S^{-1}\mathcal A$ is an equivalence of categories(where $S$ is the set of quasi isomorphism). then as you said it is easy to define a map $S^{-1}\mathcal{R}\to \mathcal B$ which is enough fo most practical purposes.

or you can first choose a complex in $\mathcal R$ which is quasi-isomorphism to each element of $\mathcal A$ and then using this choices define the $RF$ it is easy to check that this map is well defined when you work with only one functor $F$ and every two choices give equivalent functors. of course, when you want to talk about the combinacation of two different derived functor things become uglier.

Or you can use the language of higher category theory. any way the point is that you can make these things precise so usually people just accept this things in practice.


问:
Intuition behind Definition of Nilpotent Groups

Definition: A group $G$ is called nilpotent if there exists a chain of subgroups $N_0, N_1,\ldots, N_k$ such that $$\{e\} = N_0 \le N_1 \le N_2 \le ... \le N_k = G$$ and for $0\le i\le k-1$,

$N_i \vartriangleleft G$, i.e. $N_i$ is normal in $G$,

$N_{i+1}/N_i \subset Z(G/N_i)$

where $Z(G)$ denotes the center of some group $G$, and $e$ is the identity of $G$.

What motivated this definition? What is so special about $N_{i+1}/N_i \subset Z(G/N_i)$? It feels so random and out of the blue!

答:
I like the nLab description, which phrases nilpotent groups in terms of towers of central extensions. In this guise it's a generalization of being abelian.

An extension of a group $G$ is a group $E$ such that $G$ is a quotient of $E$. It's actually slightly more--it's a short exact sequence $1 \to A \to E \to G \to 1$, effectively a "witness" to the fact that $G$ is a quotient of $E$.

A central extension is a group extension as above where additionally $A$ is central in $E$, i.e. (the image of) every element of $A$ commutes with everything in $E$, including all of $A$ itself, so $A$ in particular is abelian. As an example, the central extensions of $G=1$ are precisely the abelian groups.

What are the central extensions of abelian groups? What about the central extensions of central extensions of abelian groups, etc.? Well, whatever they are, we call them nilpotent groups.

The central series can be thought of as a "witness" to this iterated construction. The successive quotients are the central subgroups we're extending by.

Another reason this is a nice definition is that solvable groups are pretty transparent: they're iterated abelian extensions, rather than central extensions, a somewhat weaker constraint. That tracks back to solvability in Galois theory.


问:
Solving $\;x+y+z =8 ; \;\;\sqrt{x^2+1}+\sqrt{y^2+4}+\sqrt{z^2+9}=10 $
Solve the problem

\begin{cases}x+y+z =8 \\ \\
\sqrt{x^2+1}+\sqrt{y^2+4}+\sqrt{z^2+9}=10
\end{cases}

with $(x,y,z) \in \mathbb R^3$

I have already solved it, but I'd like to see others creative solutions and before all, share this funny problem with the community.

答: Since $\sqrt{x^2+1}+\sqrt{y^2+4}+\sqrt{z^2+9}\ge \sqrt{(x+y+z)^2+(1+2+3)^2}= 10$ by the Minkowski inequality, now check when the equal sign hold.</p>


问:
Commutant built from the Unilateral Shift
Let $T: \ell^2 \rightarrow \ell^2$ be the unilateral shift, and let $\mathcal{T}= \{ I, T, T^* \}$.

Show that $\mathcal{T}' = \{\lambda I_{\ell^2}  : \lambda \in \mathbb{C} \}$ and $\mathcal{T}'' = B(\ell^2)$.

Attempt:

Since $T$ is the unilateral shift, we have that $Te_n = e_{n+1}$.

Also, by definition, we have $\mathcal{T}' = \{ S \in B(\ell^2) : ST =TS, \forall T \in \mathcal{T}\}$.

Let's look at the each element in $T$:

We know that the identity $I$ always has the property $IS=SI$ for $S \in B(\ell^2) $.

Also, let $S \in B(\ell^2)$. Then we have  $STe_n= Se_{n+1} =e_{n+1 }S= Te_n S = TSe_{n}$.

For $T^*$, I know that $T^*e_n= e_{n-1}$, for $n>1$. So don't we get a similar result as above, i.e $ST^*e_n =T^* S e_n$?

So I'm a bit confused on how to get the results we want. I feel like I got the definitions of $\mathcal{T}'$ and $\mathcal{T}''$ mixed up somewhere...

Thank you for your help!

答:
Let's rename $E=\{I,T,T^*\}$. If we show that $E'=\mathbb{C}I$, then the claim that $E''=\mathbb{B}(\ell^2)$ follows trivially, since $E''=(E')'=(\mathbb{C}I)'=\mathbb{B}(\ell^2)$.

Lemma: Let $H$ be a Hilbert space and $\mathbb{K}(H)$ denote the compact operators over $H$. Then $\mathbb{K}(H)''=\mathbb{B}(H)$.

Proof: $\mathbb{K}(H)$ is a $C^*$-algebra that has an approximate unit of the form $\{p_\lambda\}$ where $p_\lambda$ are finite rank projections and $p_\lambda\to\text{id}_{H}$ in the strong operator topology. In other words, the inclusion $\mathbb{K}(H)\subset\mathbb{B}(H)$ is a non-degenerate representation, so, by von Neumann's double commutant theorem we have that
$$\mathbb{K}(H)''=\overline{\mathbb{K}(H)}^{sot}$$
But, in general, if $I$ is a closed, self-adjoint two sided ideal in $\mathbb{B}(H)$ then so is $\overline{I}^{sot}$: it is a $C^*$-algebra as we know that the strong closures of $C^*$-subalgebras are again $C^*$-algebras and it is an ideal since if $x\in \overline{I}^{sot}$ and $y\in\mathbb{B}(H)$ then we take a net $\{x_\mu\}\in I$ with $x_\mu\to x$ strongly and simply observe that $x_\mu y\in I$ for all $\mu$ and that $x_\mu y\to xy$ strongly, since multiplication is separately strongly continuous. Since $\mathbb{K}(H)$ is an ideal, its strong closure is an ideal and it contains the identity operator, since $p_\lambda\to\text{id}_{H}$. Thus it is equal to the entire algebra $\mathbb{B}(H)$, q.e.d.

Now it is elementary to check that $\mathbb{B}(\ell^2)'=\mathbb{C}I$, so $\mathbb{K}(\ell^2)'=\mathbb{K}(\ell^2)'''=\mathbb{B}(\ell^2)'=\mathbb{C}I$. It is also elementary to check that if $x\in E'$ then $x\in C^*(E)'$. So, if we show that $\mathbb{K}(\ell^2)\subset C^*(E)$, then $E'\subset C^*(E)'\subset\mathbb{K}(\ell^2)'=\mathbb{C}I$ and we will be done. Let's show this inclusion:

For $n\geq1$ define the projection $p_n\in\mathbb{B}(\ell^2)$ by
$$p_n(\xi_1,\xi_2,\dots)=(\xi_1,\dots,\xi_n,0,0,\dots)$$
The projections $\{p_n\}$ form an approximate unit for $\mathbb{K}(\ell^2)$, so every $x\in\mathbb{K}(\ell^2)$ satisfies $x=\lim_{n\to\infty}p_nxp_n$. So it suffices to show that $p_nxp_n\in C^*(E)$ for all $x\in\mathbb{K}(\ell^2)$.

More generally, if $\{e_k\}$ is the canonical orthonormal basis of $\ell^2$, we define $y_{i,j}:\ell^2\to \ell^2$ by $y_{i,j}(\xi)=\langle \xi,e_j\rangle e_i$. Observe that $p_n=\sum_{i=1}^ny_{i,i}$. Moreover, $$p_1=y_{1,1}=I-TT^*\in C^*(E)$$ and also $$y_{i,j}=T^{j-1}\cdot p_1\cdot (T^*)^{i-1}\in C^*(E)$$
for all $i,j$. Now it is easy to verify that $p_n\cdot\mathbb{B}(\ell^2)\cdot p_n=\text{span}\{y_{i,j}: 1\leq i,j\leq n\}$, so $p_n\cdot\mathbb{B}(\ell^2)\cdot p_n\subset C^*(E)$ for all $n$. This shows that $p_nxp_n\in C^*(E)$ for all $x\in\mathbb{K}(\ell^2)$ and all $n\geq1$, so, since $C^*(E)$ is closed, we have that $x=\lim_{n\to\infty}p_nxp_n\in C^*(E)$ and the claim is proved.

Comment: My answer is basically the details to what Mike F describes in their answer. I am posting this because OP asks for "a more detailed answer".


问:
Conditional joint CDF
Is this the definition of conditional probability of 2 continuous random variables conditioned on 1 continuous random variable? Please provide a reference.

(I forgot if any 2 continuous random variables necessarily have a well-defined joint pdf.)

Instead of the usual $$P(z_1 < Z < z_2 | B=b) := \int_{z_1}^{z_2} f_{Z|B=b}(z) dz,$$ with $f_{Z|B=b}(z):=\frac{f_{Z,B}(z,b)}{f_B(b)},$ it looks like we'll have like $$P(z_1 < Z < z_2, u_1 < U < u_2 | B=b) := \int_{u_1}^{u_2} \int_{z_1}^{z_2} f_{(Z,U)|B=b}(z,u) dz du,$$ with $f_{(Z,U)|B=b}(z,u) := $, I think, $\frac{f_{Z,U,B}(z,b,u)}{f_B(b)}$

Note: if any of the definitions ':=' are in fact not definitions, then you'll have to explain conditioning on an event of probability zero to me please.

Wiki just says for 'conditional cumulative distribution function' or conditional CDF of $X$ and $Y$ on $Z=z$ is $F_{(X,Y)|Z=z}(x,y):=P(X \le x, Y \le y|Z=z)$, but it doesn't quite define $P(X \le x, Y \le y|Z=z)$.

For just 1 continuous random variable conditioned on 1 continuous random variable, it's $P(X \le x, |Z=z) := \int_{-\infty}^{x} f_{X|Z=z}(t) dt$, where $f_{X|Z=z}(t) := \frac{f_{(X,Z)}(x,z)}{f_{Z}(z)}$.

For 2 continuous random variables conditioned on 1 continuous random variable, I think it's $P(X \le x, Y \le y|Z=z) := \int_{-\infty}^{x} \int_{-\infty}^{y} f_{(X,Y)|Z=z}(t,u) du dt$, but then...

What's $f_{(X,Y)|Z=z}(t,u)$? (I guess we do the elementary probability way of thinking: define the pdf before the cdf...) According to this site (see problems 1 and 16), it's $f_{(X,Y)|Z=z}(t,u): = \frac{f_{(X,Y,Z)}(t,u,z)}{f_Z(z)}$. So, I guess I'm right about joint cdf/pdf stuff. I'm just hoping for a reference please.

答:
Found in Larsen and Marx

unfortunately this is for discrete. it seems there's no mention of a continuous BUT THEN there's an exercise THAT ASSUMES a definition for continuous.

Also found in

Walpole, Myers, Myers and Ye

I believe it applies for both continuous and discrete.

Title: Conditional joint CDF


问:
Help with density functions in economics - easy double integration
I've already posted in the econ.stack, since the model I’m having trouble with is an econ one (link). However, I feel the nature of my doubt is more about math than the actual econ model behind it, so I decided to post also here changing my question accordingly.

The author identifies two product characteristics (x,y) that are orthogonal and are uniformly distributed over the unit square [0,1]x[0,1].

I understand the mechanics of the model, but I have a problem with the math behind some profit functions and consumer surplus functions. In particular, the integrals over the pdf/cdf really bug me.

Take eq. (3) for example:

$$CS^{fp} = 2 \int_0^{1-x} \int_0^{1} (v-t-t(x+y)) f(x) f(y) dx dy = v - \frac{5t}{3} $$

given $f(x)$ is the std. uniform I know that this density over the 0-1 interval is just 1 and I can basically ignore it, correct?
An alternative view should be to consider $\int f(x)f(y) dx = \int f(x,y) dx = f(y)$ given independence. Correct?

If so, I can write

$$CS^{fp} = 2 \int_0^{1-x} f(y) \left[ v -t -t \int_0^{1} (x+y) dx  \right]dy =$$

$$CS^{fp} = 2 \int_0^{1-x} f(y) \left[ v -ty - \frac32 t  \right]dy =$$

If this is correct what is then $f(y)$ in this interval? How do I proceed?

My guess is...
$$CS^{fp} = 2 \int_0^{1-x} v -ty - \frac32 t  dy = 2\left[ vy - \frac{t y^2}{2} - \frac34 ty^2 \right]_0^{1-x} = 2\left[ vy - ty^2\frac{5}{4} \right]_0^{1-x}$$

$$ = 2v(1-x) - \frac52 t (1-x)^2 = $$

答:
Yes, being $f(x)=f(y)=1$ you can write them off.

Your integral becomes, differently written:

$$2\int_0^1\left[\int_0^{1-x}(v-t-tx-ty)dy    \right]dx=$$

$$=2\int_0^1\left[vy-ty-txy-\frac{t}{2}y^2\right]_{y=0}^{y=1-x}dx=\dots=v-\frac{5}{3}t$$


答:
A maximum element is necessarily comparable with all the other elements. It is possible to have two different maximal elements which cannot be compared with each other.

To illustrate take the set of integers greater than $1$, and $a \succsim b$ defined as $b=ra$ for some positive integer $r$.

Then $2$ and $3$ are both maximal, but neither is a maximum.

Add $1$ to the set, and $1$ becomes the unique maximum.

Title: Problem with a definition of $\succsim$-maximal element and $\succsim$-maximum

Body:
I found in a book the following definition of $\succsim$-maximal element and $\succsim$-maximum, but I really don't see the difference between what is written in the book (I know the difference between those two concepts). Am I wrong?

Let $(X,\succsim )$ be a preordered set and $∅ = Y ⊆ X$. An element $x$ of $Y$ is said to be $\succsim$-maximal in $Y$ if there is no $y \in Y$ with $y \succ x$. If $x \succsim y $ for all $y \in Y$ , then $x$ is called the $\succsim$-maximum of $Y$.


问:
How to factorise $x^2 + 1$
Keep in mind that I am not asking how to find the solutions to the equation $x^2 + 1 = 0$. I know that there will be imaginary answers In this case. I just want to know how to factor this using a logical method and obtaining 2 factors. Thanks for answering.

答:
Although you are not asking how to solve the equation $x^2+1=0$, usually the easiest way of factorising a polynomial is by solving the equation when the polynomial is equal to $0$. This is useful as a result of the factor theorem, which states:

Let $f(x)$ be a polynomial in $x$. If $f(a)=0$ then $(x-a)$ is a factor of $f(x)$ (and if $(x-a)$ is a factor of $f(x)$ then $f(a)=0$).

In our case then, solving the equation $x^2+1=0$ yields the solutions $x=\pm i$; hence the polynomial factorises as $(x+i)(x-i)$.

I hope that was helpful. If you have any questions please don't hesitate to ask :)

Title: How to factorise $x^2 + 1$


问:
Determining flux through a paraboloid.
I am a bit confused about how to solve this exercise. I can't use Stokes Theorem since both sides are open right?

Let S be the surface of the paraboloid:        $ z=3-x^2-y^2 $ for $-1\leq z\leq 1$
oriented with the outward unit normal. Let

$\vec F(x,y,z)=x\vec i+z\vec j-y\vec k $ be the velocity field of the fluid

答:
The paraboloid is $x^2 + y^2 = 3-z$, so you always must have $z\leq 3$ (the vertex is at $z=3$).

So I wonder if you can do something like $$\int_{z\in[-1,1]} = \int_{z\in[-1,3]} - \int_{z\in[1,3]}$$

so that in the two integral on the RHS the boundary is connected?

Title: Determining flux through a paraboloid.


问:
Describe $\operatorname{Spec} R/fR$ with respect to prime decomposition of $f$ in a PID.
This is an exercise from Goertz/Wedhorn: Let $R$ be a PID and $f\in R$ with prime-decomposition $f=\epsilon f_1^{n_1}\ldots f_k^{n_k}$. Analyse the affine scheme $X=\operatorname{Spec} R/fR$, so the topological space, stalks and $\mathcal{O}_X(U)$ for $U$ open.

What I have so far:$X=\operatorname{Spec} R/fR$ is homeomorphic to $V(f)\subseteq \operatorname{Spec} R$, and the prime ideals in the letter set are precisely $\{(f_1)=\mathfrak{p}_1,\ldots,(f_k)=\mathfrak{p}_k\}$. Thus, as a set we have $\operatorname{Spec} R/fR=\{\overline{\mathfrak{p}_1},\ldots,\overline{\mathfrak{p}_k}\}.$ For the stalk we have $$\mathcal{O}_{X,\overline{\mathfrak{p}_s}}=(R/fR)_{\overline{\mathfrak{p}_s}}\cong (R/(f_1^{n_1}))_{\overline{\mathfrak{p}_s}}\times\ldots\times (R/(f_k^{n_k}))_{\overline{\mathfrak{p}_s}}=(R/(f_s)^{n_s})_{\overline{\mathfrak{p}_s}},$$ because for $l\neq s$ we have $(f_l^{n_l})\cap(R\setminus\mathfrak{p}_s)\neq\emptyset$, as otherwise we would have $(f_l)^{n_l}\subseteq \mathfrak{p}_s$ and then also $(f_l)\subseteq\mathfrak{p}_s$ because the latter is prime. But this would force $l=s$ since both ideals are maximal. Now the topology on $\operatorname{Spec} R/fR$ is the discrete topology, because it's finite and the closed sets are precisely the finite subsets. Given an open - so arbitrary - $U=\{\overline{\mathfrak{p}_{m_1}},\ldots,\overline{\mathfrak{p}_{m_r}}\}\subseteq X,$ we consider $g=\prod _{i\neq m_j}f_i\in R$ and then have $D(\overline{g})=U$. This gives $$\mathcal{O}_X(U)=(R/fR)_{\overline{g}}\cong (R/f_1^{n_1})_{\overline{g}}\times\ldots\times (R/f_k^{n_k})_{\overline{g}}=\prod_j (R/f_{m_j}^{n_{m_j}})_{\overline{g}}.$$

My questions:

Is this correct?

Am I done with this exercise? I have kind of a bad feeling regarding if I'm done with such vaguely formulated exercises. I don't know if I could simplify a bit more, for example. Maybe a bit of life advice concerning this would also be appreciated. I'm grateful for any sort of help.

答:
The "chinese remainder theorem" says the following: If $I_1,..,I_l$ are coprime ideals in a commutative ring $A$ ($I_i +I_j=(1)$ for $i \neq j$) let $I:=I_1\cdots I_l$.
It follows there is an isomorphism of rings

$$A/I \cong A/I_1\times \cdots \times A/I_l:=A_1\times \cdots \times A_l$$

If your ideals $\mathfrak{p}_i$ are coprime you get such a decomposition. This gives a "decomposition"

$$Spec(A/I) \cong Spec(A_1)\cup \cdots \cup Spec(A_l)$$.

of $Spec(A/I)$ into a disjoint union of the schemes $Spec(A_i)$.

If your ring $R$ is  PID and your ideals $(f_i)\subseteq R$ are maximal, it follows your ideals are coprime, meaning  $(f_i)+(f_j)=(1)$ for $i \neq j$. Hence you get if $f:=f_1\cdots f_k$ an isomorphism of commutative rings

$$\phi: R/(f) \cong R/(f_1)\times \cdots \times R/(f_k) .$$

The map $\phi$ induce an isomorphism of schemes $Spec(R/(f)) \cong \cup_i Spec(R/(f_i)$
where the union is disjoint. The subscheme $X_i:=Spec(R/(f_i))$ is open and closed in $X:=Spec(R/(f))$. Since $(f_i)$ is a maximal ideal it follows $X_i$ is a reduced and irreducible scheme of dimension zero: $X_i\cong \{(0)\}$ where $(0)\subseteq R/(f_i)$ is the zero ideal. $X$ is a reduced scheme since it is a disjoint union of reduced schemes.

More generally: If $(f_i)+(f_j)=(1)$ for $i \neq j$ it follows $(f_i^{n_i})+(f_j^{n_j})=(1)$ and hence the ideals $(f_i^{n_i})$ are coprime for any $n_i\geq 1$. You get similarly using the crt an isomorphism of rings

$$R^*:=R/(f_1^{n_1}\cdots f_k^{n_k})\cong R/(f_1^{n_1})\times \cdots \times R/(f_k^{n_k}):=R_1\times \cdots \times R_k.$$

The ring $R_i$ is reduced iff $n_i=1$. Let  $X:=
Spec(R^*)$ and $X_i:=Spec(R_i)$. It follows $X\cong \cup X_i$ is a disjoint union and $X_i \subseteq X$ is open and closed. $X$ is reduced iff $n_i=1$ for all $i$.

Note that if $I,J \subseteq R$ are coprime ideals with $a\in I, b\in J$ with $a+b=1$ we get the following:

$$1=1^{k+l+1}=(a+b)^{k+l+1} =\sum_{u+v=k+l+1} \alpha_{u,v}a^ub^v$$

and for all $u,v$ it follows $\alpha_{u,v}a^ub^v \in I^k+J^l$ hence $1\in I^k+J^l$ and it follows $I^k$ and $J^l$ are coprime for any integers $k,l \geq 1$. Hence in your case if the ideals $(f_1),..,(f_k)$ are coprime it follows for any integers $n_i \geq 1$ the ideals $(f_i^{n_i})$ are coprime.

Your ring $R$ is zero dimensional and you will find a similar discussion here:

Is the ring of the polynomials on a finite algebraic set always a product of local rings?


问:
Tensor product of $\partial_i$ and $dx^j$
In general the tensor product is not commutative. But I have a doubt:

in defining the absolute derivative (or covariant derivative) of a vector field $X$, as $\nabla X\in \cal{T_{1}^{1}(\cal{M})}$ (tensor field of type $(1,1)$) I have read that it is defined by:
$$\nabla X=\xi_{;j}^{i}dx^j\otimes \partial_i$$ with $X=\xi^i\partial_i$.

But in defining  a vector field of type $(1,1)$ in coordinates I know that it can be written as
$$t=t_i^j\partial_i\otimes dx^j$$
So it seems that $\partial_i$ and $dx^j$ are shifted in the order in defining $\nabla X$ with respect to the general definition of a tensor field of type $(1,1)$.

How it is possible? Maybe it is true that $\partial_i\otimes dx^j=dx^j\otimes \partial_i$?

答:
It really depends on your conventions. Some books define a $(1,1)$ tensor to be a section of $T^{*}M \otimes TM$ (the cotangent bundle appears first) while others define a $(1,1)$ tensor to be a section of $TM \otimes T^{*}M$ (the tangent bundle appears first). The bundles are of course isomorphic but still one makes a certain choice.

Let's say your convention is $TM \otimes T^{*}M$. Then the expression $dx^i \otimes \partial_j$ is not a "legal" $(1,1)$ tensor since it is not a section of $TM \otimes T^{*}M$. Once a choice is made, one uses certain isomorphisms to identify $(1,1)$-tensors with bilinear or linear maps. This choice is also somewhat arbitrary but most people try to be consistent. In the case of $TM \otimes T^{*}M$, one can identify sections of this bundle with smoothly varying bilinear maps $T_p^{*}M \times T_pM \rightarrow \mathbb{R}$ by declaring that
$$ (\alpha \otimes X)(Y, \omega) = \alpha(Y) \cdot \omega(X) $$
(and extending linearly). Note that the first argument $Y$ is a vector field and it "acts" on the first factor $\alpha$ (which is a one-form) and similarly the second argument $\omega$ is a one-form and it acts on the second factor $X$ (which is a vector field). One could do it the other way around but this is not common.

So in your case, the notation $(\nabla X)(Y,\omega)$ suggests that $(1,1)$-tensors are considered as sections of $T^{*}M \otimes TM$ and the formula $\nabla X = \xi^i_{;j} dx^j \otimes \partial_i$ is consistent with this guess.


问:
How many numbers are with $5$ different digits in decreasing order?
I mean that $54321$ is a valid number, but $16755$ is not.

I was thinking that there are $6^5$ possible numbers because there are 6 possibilities for each position in the number, but I don't know exaclty.

答:
Each number between $0$ and $9$ can appear exactly once or not at all (since there can be no repetitions). Given a set of five such numbers (say $\{1, 3, 0, 5, 8\}$) there is exactly one corresponding decreasing sequence (85310). So the number of decreasing sequences is the number of ways to choose 5 elements from a set of 10 digits. This is just
$$ {10 \choose 5} = 252 \,. $$


问:
Compute eigenvalues and eigenvectors problem
I really don't know how solve this problem:

Let $V$ be the space of real functions spanned by $\cos(x)$, $\cos(2x)$ and $\cos(3x)$. Let $T\in\mathcal{L}(V,V)$ con $T(\cos(x)) = 3\cos(x) + 2\cos(2x) - \cos(3x)$, $\;T(\cos(2x)) = 3\cos(2x) + \cos(3x)$ and $T(\cos(3x)) = \cos(3x)$. Find the eigenvalues and eigenvectors of $T$.

Please, I need help

Thanxs in advance.

答:
It's clear that
$$T\left(\begin{array}{c}\cos(x)\\\cos(2x)\\\cos(3x)\end{array}\right)\ =\ \underbrace{\left(\begin{array}{ccc}3 & 2 & -1 \\0 & 3 & 1 \\0 & 0 & 1\end{array}\right)}_{A}\left(\begin{array}{c}\cos(x)\\\cos(2x)\\\cos(3x)\end{array}\right)$$
where $A$ have the eigenvalues 1, 3 and 3. Later,

You have that $T(\cos(3x)) = \cos(3x)$, and then $1$ is an eigenvalue of $T$ with eigenvector $v_1 = \cos(3x)$.

Now, if $v_2 = \alpha\cos(x) + \beta\cos(2x) + \gamma\cos(3x)$ is an eigenvector of $T$ for eigenvector $3$, then
\begin{eqnarray*}
T(v_2) & = & 3v_2,\\
\alpha T(\cos(x)) + \beta T(\cos(2x)) + \gamma T(\cos(3x)) & = & 3\alpha\cos(x) + 3\beta\cos(2x) + 3\gamma\cos(3x),\\
3\alpha\cos(x) + (2\alpha + 3\beta)\cos(2x) - (\alpha - \beta - \gamma)\cos(3x) & = & 3\alpha\cos(x) + 3\beta\cos(2x) + 3\gamma\cos(3x),\\
2\alpha\cos(2x) - (\alpha - \beta + 2\gamma)\cos(3x) & = & 0.
\end{eqnarray*}
Now, $\cos(2x)$ and $\cos(3x)$ are linear independent, then you have
$2\alpha = 0 \Rightarrow \alpha = 0$ and $\alpha - \beta +2\gamma = 0 \Rightarrow \beta = 2\gamma$. Therefore
$$v_2\ =\ \gamma\left[2\cos(2x) + \cos(3x)\right],\quad \gamma \in \mathbb{K}$$
is an eigenvector for 3.


问:
Matrix Derivative of Log Gaussian with respect to Reparameterized Covariance
The traditional multivariate Gaussian density is given by:

$$p(x; \mu, \Sigma) := (2 \pi \det(\Sigma))^{-1/2} \exp(-\frac{1}{2} (x-\mu)^T (\Sigma)^{-1} (x-\mu)) $$

Suppose I reparameterize the density with another matrix $A$, such that $\Sigma := A^T A$:

$$p(x; \mu, A) := (2 \pi \det(A^T A))^{-1/2} \exp(-\frac{1}{2} (x-\mu)^T (A^T A)^{-1} (x-\mu)) $$

What is the matrix derivative of the log multivariate Gaussian PDF with respect to $A$? That is, what is

$$ \nabla_A \log p(x; \mu, A)$$

答:
$\def\S{\operatorname{Sym}}\def\p#1#2{\frac{\partial #1}{\partial #2}}$Let
$\,w=(x-\mu),\;$ then the function of interest is
$$\eqalign{
\lambda &= \log(p)
  \;=\; -\tfrac 12 w^T\Sigma^{-1}w - \tfrac 12\log\det(\Sigma)
  - \tfrac 12\log(2\pi) \\
}$$
First expand the differential in terms of $\Sigma$
$$\eqalign{
d\lambda
 &= \tfrac 12\Sigma^{-1}ww^T\Sigma^{-1}:d\Sigma - \tfrac 12\Sigma^{-1}:d\Sigma \\
}$$
Then substitute $\;\Sigma = A^TA$
$$\eqalign{
d\Sigma &= dA^TA+A^TdA \\&= 2\operatorname{Sym}(A^TdA) \\\\
d\lambda
 &= \Big(\tfrac 12\Sigma^{-1}ww^T\Sigma^{-1} - \tfrac 12\Sigma^{-1}\Big)
  : 2\operatorname{Sym}(A^TdA) \\
 &= \operatorname{Sym}\Big(\Sigma^{-1}ww^T\Sigma^{-1} - \Sigma^{-1}\Big)
  : \Big(A^TdA\Big) \\
 &= A\,\Big(\Sigma^{-1}ww^T\Sigma^{-1} - \Sigma^{-1}\Big)
  : dA \\\\
\p{\lambda}{A}
 &= A\,\Big(\Sigma^{-1}ww^T\Sigma^{-1} - \Sigma^{-1}\Big) \\\\
}$$
If you wish, you can eliminate the remaining $\Sigma$ variables in favor of $A$

$$\eqalign{
A\Sigma^{-1} &= A\left(A^TA\right)^{-1} = \left(A^+\right)^T \\
\Sigma^{-1} &= A^+\left(A^+\right)^T \\
}$$
where $A^+$ denotes the Moore-Penrose inverse of $A$.

In some steps above, a colon is used to denote the trace/Frobenius product, i.e.
$$\eqalign{
A:B &= {\rm Tr}(A^TB) \\
A:A &= \big\|A\big\|_F^2 \\
}$$


答: Alternatively: these numbers correspond one-to-one to $5$-element subsets of $\{0,\dots,9\}$ (the set of the digits), so their number is $\binom{10}{5}=252$.

Title: How many numbers are with $5$ different digits in decreasing order?

Body:
I mean that $54321$ is a valid number, but $16755$ is not.

I was thinking that there are $6^5$ possible numbers because there are 6 possibilities for each position in the number, but I don't know exaclty.