问:
Topological properties of subspace topology are the same in larger topological space that contains it?
I was originally interested in the statement: suppose $Y$ is a topological space and consider a fixed subset $X\subseteq Y$. If $X$ is compact in its subspace topology, then it is compact as a subset of $Y$.

Proof Let $X\subseteq \bigcup_\alpha U_\alpha$ be an open cover where $U_\alpha$ is open in $Y$. Taking the intersection with $X$ both sides:
$$X=X\cap X\subseteq\bigcup_\alpha (U_\alpha\cap X).$$
Since $X$ is compact in its subspace topology, there exists a finite subcover that
$$X\subseteq\bigcup_{i=1}^m(U_i\cap X)\subseteq \bigcup_{i=1}^m U_i.$$
That is, $X$ is compact as a subset of $Y$.

Assuming that the above proof is correct, this made me wonder about the more general situation.

Problem Suppose $Y$ is a topological space and consider a fixed subset $X\subseteq Y$. If $X$ satisfies some topological property in its subspace topology, then $X$ satisfies the same topological property as a subset of $Y$.

Question Is the above statement true?

答:
This is in general not true, and I will show below that the topological property of being closed (as in $X$ is always closed in $X$ under its subspace topology) can break (in the sense that $X$ need not be a closed subset of $Y$).

Let $Y = \mathbb{R}$ under the usual Archimedean topology and let $X = [0,1) \subseteq Y$. Then $X$ is a closed subset of itself $X$ in the subspace topology (as $X$ is closed in itself) but $X$ is not closed in $Y$ (it does not contain the limit point $x = 1$).


问:
How to find average and power of signal expressed by fourier series
I need to find the average and the power of this signal:

$$x(n)=\sum_{k=1}^{\infty}2^{-k}e^{j2{\pi}kn}$$

The problem is that the summation starts at 1 and not at 0, and a part of that how can I find the average of a signal expressed in this way?

答:
HINT:

Note that we have

$$\begin{align}
x(n)&=\sum_{k=1}^\infty 2^{-k}e^{i2\pi kn}\\\\
&=\sum_{k=1}^\infty\left(\frac{e^{i2\pi n}}{2}\right)^k\\\\
&=\frac{e^{i2\pi n}}{2-e^{i2\pi n}}\\\\
&=\frac{2e^{i2\pi n}-1}{5-8\cos(2\pi n)}
\end{align}$$

Can you proceed now?

ContentLicense: CC BY-SA 4.0


问:
Prove that $-\psi(y) \leq c(x,y) + c(x,\bar y) - \psi(\bar y)$
In the book Optimal Transport for Applied Mathematicians, Theorem 1.37, the authors states an inequality, but I'm having trouble seeing how one shows that the inequality is indeed true.

Definition: Given a function $c:X\times Y \to \mathbb R$, we say that $\Gamma \subset X \times Y$ is $c$-cyclically monotone ($c$-CM) if for every $k \in \mathbb N$, every permutation $\sigma$ and every finite family of points $(x_1,y_1),...,(x_k,y_k) \in \Gamma$ we have
$$
\sum^k_{i=1} c(x_i,y_i) \leq
\sum^k_{i=1} c(x_i,y_{\sigma(i)})
$$

Now, assume that $\Gamma$ is $c$-CM. Then, fix $(x_0, y_0) \in \Gamma$, and define:
$$
-\psi(y) = \inf\{
-c(x_n,y) +c(x_n,y_{n-1})-c(x_{n-1},y_{n-1})+...+ c(x_1,y_0) - c(x_0,y_0) : n \in \mathbb N,
(x_i,y_i) \in \Gamma \quad  \forall i=1,...,n
\}
$$

How does one then shows that fo $(x,y) \in \Gamma$ and $\bar y \in \text{Proj}_y \circ\Gamma$, we can then write:
$$
-\psi(y) \leq - c(x,y) + c(x,\bar y) - \psi(\bar y)
$$

In the proof, the author claims that this inequality follows from the very definition of the function $\psi$. What I found odd was that since $y$ and $\bar y$ are arbitrary, then I could just swap them, obtaining the opposite inequality, and hence, I actually have an equality. Which would be kind of odd. Anyways, how can you prove this inequality? And is it actually an equality?

答:
By assumption $-\psi(\bar{y})>-\infty$. Thus, there is a cycle $\Gamma':=\lbrace (x_{1},y_{1}),\dots,(x_{r},y_{r})\rbrace$ such that $-\psi(\bar{y})=-c(x_{r},\bar{y})+c(x_{r},y_{r-1})+\dots+c(x_{1},y_{0})-c(x_{0},y_{0})$. Then, considering the cylce $\Gamma'':=\Gamma'\cup\lbrace{(x,\bar{y})\rbrace}$ it follows
\begin{equation}
-\psi(y)\le -c(x,y)+\sum_{k=1}^{r+1} c(x_{k},y_{k-1})-c(x_{k-1},y_{k-1})-c(x_{0},y_{0})=-c(x,y)+c(x,\bar{y})-\psi(\bar{y}).
\end{equation}


答: It depends on what topological property you are considering. Let $Y$ be a finite nonempty set and $y_0\in Y$. The open sets of $Y$ are the empty set and each subset which contains $y_0$. Then $X=Y\setminus\{y_0\}$ is discrete and in particular normal Hausdorff, but it is not even $T_0$ in $Y$.

Title: Topological properties of subspace topology are the same in larger topological space that contains it?

Body:
I was originally interested in the statement: suppose $Y$ is a topological space and consider a fixed subset $X\subseteq Y$. If $X$ is compact in its subspace topology, then it is compact as a subset of $Y$.

Proof Let $X\subseteq \bigcup_\alpha U_\alpha$ be an open cover where $U_\alpha$ is open in $Y$. Taking the intersection with $X$ both sides:
$$X=X\cap X\subseteq\bigcup_\alpha (U_\alpha\cap X).$$
Since $X$ is compact in its subspace topology, there exists a finite subcover that
$$X\subseteq\bigcup_{i=1}^m(U_i\cap X)\subseteq \bigcup_{i=1}^m U_i.$$
That is, $X$ is compact as a subset of $Y$.

Assuming that the above proof is correct, this made me wonder about the more general situation.

Problem Suppose $Y$ is a topological space and consider a fixed subset $X\subseteq Y$. If $X$ satisfies some topological property in its subspace topology, then $X$ satisfies the same topological property as a subset of $Y$.

Question Is the above statement true?


问:
Probability of rolling $n$ successes on an open-ended/exploding dice roll
I'm trying to compute the probability of achieving a certain number of successes when rolling a die pool for both open-ended/exploding and closed tests. Success is defined as a roll above a certain (variable) threshold. If a roll is open-ended, any die which rolls the maximum value triggers an additional die to be rolled. This is for a role-playing game (Burning Wheel Gold).

Assume the use of a $n$-sided die. Additionally, we assume $d$ to be the number of dice, $r$ to be the required number of successes, and $c$ to be the minimum value accepted as a success. I'm breaking the question down into two blocks: one for open-ended rolls, and one for closed rolls.

Closed roll: For closed rolls, we can define a probability recurrence as follows:

$$Pr\left[r,d\right] = \begin{cases} \\1-\frac{c}{n}:Pr\left[d-1,r\right]\\\frac{c}{n}:Pr\left[d-1,r-1\right] \end{cases}$$

With probability $\frac{c}{n}$, a success will be met. However, I'm not sure how to solve this recurrence relation for any number of dice/required successes. There's probably an easier way to do this, but I haven't done much discrete probability.

Edit: It looks like a binomial probability distribution works for this part of the problem. This comes in the form of:

$${{d}\choose{r}}\left(\frac{c}{n}\right)^{r}\left(1-\frac{c}{n}\right)^{d-r}$$

Open ended roll: For open-ended rolls, we can find the equivalent number of rolled dice. This is going to be equal to:

$$d+\sum_{j=1}^{\infty}\left(\frac{d}{6^{j}} \right)=\frac{6d}{5}$$

This implies that, given a number of dice $n$, rolling open-ended results in approximately $\frac{6}{5}n$ dice rolled. If we had an easy way to solve the above for any number of dice $n$, it would be easy: we could just input the non-integer value $n$ and get our result. However, since the above <s>is defined with a recurrence relation</s> is a binomial distribution, and thus uses combinations, it won't work for a non-integer number. Thus, we can define a probability recurrence relation as follows:

$$Pr\left[ r,d\right] = \begin{cases} \\1-\frac{c}{n}:Pr\left[d-1,r\right]\\\frac{c-1}{n}:Pr\left[d-1,r-1\right]\\\frac{1}{n}:Pr\left[d,r-1\right] \end{cases}$$

The probability of no significant result is $1-\frac{c}{n}$, the probability of a regular success is $\frac{c-1}{n}$, and the probability of an 'exploding' or open-ended result is $\frac{1}{n}$. However, I don't know how to solve this recurrence relation, either.

So these are my questions:

For closed rolls, how do I find the probability of $n$ successes? I've looked at existing probability tables (PDF), but they don't actually contain the math for their generation, and it's non-intuitive to discern from the table. Edit: See above; the binomial formula applies.

For open rolls, can the closed roll formula be applied to a non-integer number $\frac{6n}{5}$? Edit: No. See above; the binomial formula uses combinations, thus requires integer numbers.

So, how do I solve the closed roll recurrence relation and/or find the open probability?

答:
These can be solved at the same time, the closed roll is just a special case of the open ended.

Let

$$d = \text{Number of sides on each die}$$
$$n = \text{Number of dice rolled}$$
$$s = \text{Score required for success}$$
$$m = \text{number of successes required}$$

Also, let

$$s = n^k+r$$

where
$$k\in 0,1,2,\dots$$
$$r\in 1,2,\dots,d$$

And, let

$$\begin{align}
p_o &= \text{chance of getting open ended}\\
&= \begin{cases}
\frac{1}{d}&,\text{if roll is open}\\
0&,\text{otherwise}\\
\end{cases}
\end{align}$$

So, for 1 die, we have a geometric distribution to get $k$ open ended results followed by a single roll to get $r$ (note that this is not exactly a geometric distribution since we need $k$ failures with $q=1-p_0$ followed by 1 success with a different probability). Also we will require $x^0=1$.

$$p_1=\left(p_0\right)^{k}\left(\frac{d-r+1}{d}\right)$$

To get $m$ success from $n$ dice, we have a binomial distribution as you noticed so

$$\begin{align}
p_{m,n}&=\sum_{i=m}^n\binom{n}{i}p_1^i\left(1-p_1\right)^{n-i}\\
\end{align}$$

Some examples:

$d=6$, $n=1$, $s=4$, $m=1$ & open ended; so $k=0$, $r=4$

$$\begin{align}
p_0=\frac{1}{6}\\
\end{align}$$

$$\begin{align}
p_1&=\left(\frac{1}{6}\right)^{0}\left(\frac{6-4+1}{6}\right)\\
&=1\left(\frac{3}{6}\right)\\
&=1\times\frac{1}{2}\\
&=\frac{1}{2}\\
\end{align}$$

$$\begin{align}
p_{1,1}&=\sum_{i=1}^1\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\
&=\binom{1}{1}\left(\frac{1}{2}\right)^{1}\left(1-\frac{1}{2}\right)^{1-1}\\
&=1\times\frac{1}{2}\times{1}\\
&=\frac{1}{2}\\
\end{align}$$

$d=6$, $n=1$, $s=10$, $m=1$ & open ended; so $k=1$, $r=4$

$$\begin{align}
p_0=\frac{1}{6}\\
\end{align}$$

$$\begin{align}
p_1&=\left(\frac{1}{6}\right)^{1}\left(\frac{6-4+1}{6}\right)\\
&=\frac{1}{6}\times\frac{3}{6}\\
&=\frac{1}{12}\\
\end{align}$$

$$\begin{align}
p_{1,1}&=\sum_{i=1}^1\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\
&=\binom{1}{1}\left(\frac{1}{12}\right)^{1}\left(1-\frac{1}{12}\right)^{1-1}\\
&=1\times\frac{1}{12}\times{1}\\
&=\frac{1}{12}\\
\end{align}$$

$d=6$, $n=3$, $s=4$, $m=2$ & closed; so $k=0$, $r=4$

$$\begin{align}
p_0=0\\
\end{align}$$

$$\begin{align}
p_1&=\left(0\right)^{0}\left(\frac{6-4+1}{6}\right)\\
&=1\left(\frac{3}{6}\right)\\
&=\frac{1}{2}\\
\end{align}$$

$$\begin{align}
p_{3,2}&=\sum_{i=2}^3\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\
&=\binom{3}{2}\left(\frac{1}{2}\right)^{2}\left(1-\frac{1}{2}\right)^{3-2}+\binom{3}{3}\left(\frac{1}{2}\right)^{3}\left(1-\frac{1}{2}\right)^{3-3}\\
&=3\times\frac{1}{4}\times\frac{1}{2}+1\times\frac{1}{8}\times1\\
&=\frac{4}{8}\\
&=\frac{1}{2}\\
\end{align}$$

$d=6$, $n=3$, $s=10$, $m=2$ & open ended; so $k=1$, $r=4$

$$\begin{align}
p_0=\frac{1}{6}\\
\end{align}$$

$$\begin{align}
p_1&=\left(\frac{1}{6}\right)^{1}\left(\frac{6-4+1}{6}\right)\\
&=\left(\frac{1}{6}\right)\left(\frac{3}{6}\right)\\
&=\frac{1}{12}\\
\end{align}$$

$$\begin{align}
p_{3,2}&=\sum_{i=2}^3\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\
&=\binom{3}{2}\left(\frac{1}{12}\right)^{2}\left(1-\frac{1}{12}\right)^{3-2}+\binom{3}{3}\left(\frac{1}{12}\right)^{3}\left(1-\frac{1}{12}\right)^{3-3}\\
&=3\times\frac{1}{144}\times\frac{11}{12}+1\times\frac{1}{1,728}\times1\\
&=\frac{34}{1,728}\\
&=\frac{17}{864}\\
&\approx0.02\\
\end{align}$$

ContentLicense: CC BY-SA 3.0


问:
How to calculate coset intersection in gap
Given a finite group $G$ and two of its subgroups $S_1,S_2$,
I'd like to calculate the intersection of the cosets of $S_1$ and $S_2$.

This works but it's taking too long as the group size gets larger :

G:=(group);

S1:=Subgroup(G,...);Cosets1:=RightCosets(G,S1);

S2:=Subgroup(G,...);Cosets2:=RightCosets(G,S2);

R:=List(Cosets1,x->List(Cosets2,y->Size(Intersection(x,y))));

Are there faster ways to do this.

If it simplifies the problem, instead of the actual size of the intersection
I can also use a binary indicator whether the two cosets intersect or not.

答:
There is no special routine for the intersection of cosets - it computes elements lists and intersects these. Thus the inefficiency when working in larger groups.
In general, in GAP, working with transversals instead of cosets is faster, and the command PositionCanonical finds the transversal position corresponding to a given element.

The intersection of cosets of subgroups $S$, respectively $T$, is a coset of $S\cap T$ (Intersection of cosets). To find which cosets of $T$ a coset $Sx$ intersects, it is thus sufficient to split $Sx$ into cosets of $S\cap T$ (representatives are $cx$ where $c$ runs through a transversal of $S\cap T$ in $S$), and for each such coset $(S\cap T) cx$ find the coset of $T$ in which the representative $cx$ lies.

You can do so in GAP as follows (example, but this should work quickly for much larger sizes):

```<code>gap> G:=SymmetricGroup(6);;
gap> S:=Stabilizer(G,[1,2],OnSets);;T:=Stabilizer(G,[1,2,3],OnSets);;
gap> Ttrans:=RightTransversal(G,T);
RightTransversal(Sym( [ 1 .. 6 ] ),Group([ (4,5,6), (4,5), (1,2,3), (1,2) ]))
gap> ST:=Intersection(S,T);;
gap> ST:=Intersection(S,T);
Group([ (5,6), (4,5,6), (1,2)(5,6) ])
gap> STt:=RightTransversal(S,ST);
RightTransversal(Group([ (3,4,5,6), (3,4), (1,2) ]),Group(
[ (5,6), (4,5,6), (1,2)(5,6) ]))
gap> x:=(2,4,5);; # whatever element you want
gap> List(STt,c->PositionCanonical(Ttrans,c*x));
[ 5, 9, 8, 2 ]
</code>```

So the coset $Sx$ intersects the cosets of $T$ with representatives in Positions 2,5,9,8. A brute-force check (not recommended in general) for this is:

```<code>gap> Filtered([1..Length(Ttrans)],y->Length(Intersection(S*x,T*Ttrans[y]))>0);
[ 2, 5, 8, 9 ]
</code>```

(This approach assumes that the group is given as a permutation group (or a PCGroup). And - unavoidably - this becomes more expensive if the indices $[S:S\cap T]$ or $[G:T]$ become larger. But for example I tried itin $M_{24}$ (order ~$2\cdot 10^8$), it just took seconds. )

Body:
There is no special routine for the intersection of cosets - it computes elements lists and intersects these. Thus the inefficiency when working in larger groups.
In general, in GAP, working with transversals instead of cosets is faster, and the command PositionCanonical finds the transversal position corresponding to a given element.

The intersection of cosets of subgroups $S$, respectively $T$, is a coset of $S\cap T$ (Intersection of cosets). To find which cosets of $T$ a coset $Sx$ intersects, it is thus sufficient to split $Sx$ into cosets of $S\cap T$ (representatives are $cx$ where $c$ runs through a transversal of $S\cap T$ in $S$), and for each such coset $(S\cap T) cx$ find the coset of $T$ in which the representative $cx$ lies.

You can do so in GAP as follows (example, but this should work quickly for much larger sizes):

gap> G:=SymmetricGroup(6);;
gap> S:=Stabilizer(G,[1,2],OnSets);;T:=Stabilizer(G,[1,2,3],OnSets);;
gap> Ttrans:=RightTransversal(G,T);
RightTransversal(Sym( [ 1 .. 6 ] ),Group([ (4,5,6), (4,5), (1,2,3), (1,2) ]))
gap> ST:=Intersection(S,T);;
gap> ST:=Intersection(S,T);
Group([ (5,6), (4,5,6), (1,2)(5,6) ])
gap> STt:=RightTransversal(S,ST);
RightTransversal(Group([ (3,4,5,6), (3,4), (1,2) ]),Group(
[ (5,6), (4,5,6), (1,2)(5,6) ]))
gap> x:=(2,4,5);; # whatever element you want
gap> List(STt,c->PositionCanonical(Ttrans,c*x));
[ 5, 9, 8, 2 ]

So the coset $Sx$ intersects the cosets of $T$ with representatives in Positions 2,5,9,8. A brute-force check (not recommended in general) for this is:

gap> Filtered([1..Length(Ttrans)],y->Length(Intersection(S*x,T*Ttrans[y]))>0);
[ 2, 5, 8, 9 ]

(This approach assumes that the group is given as a permutation group (or a PCGroup). And - unavoidably - this becomes more expensive if the indices $[S:S\cap T]$ or $[G:T]$ become larger. But for example I tried itin $M_{24}$ (order ~$2\cdot 10^8$), it just took seconds. )


问:
Math Self Learning
Is it possible to learn math all on your own and reach the same level as a person who has gone through a four years undergrad math specialist study in a university?</p>

答:
The short answer is Yes for sure.

However, you need a plan and you need to be honest with yourself.

Regarding the plan:

You need to know what are the courses that you have to follow. From A to Z. The resources are countless on the MOOCS (MIT, HARVARD, YOUTUBE...). However, knowing what to study is key for success.

Regarding honesty:

When you understand something, you need to find a way to test your understanding similar to the exams at school. You can't just claim that you have learned it by yourself, if you can't test yourself in regards to the course/chapter.

Suggestions

If you are a visual person, watch the courses on MIT website. Else, buy or download books from the internet and start learning them from page 1 to the end.

Motivation

I wanted to point out, that in the past, people used to be multidisciplinary, they used to be philosophers, mathematicians, physicians etc... Therefore, if they were able to do it without electricity/internet/community help ... You, I and everyone can do it with all the resources we have in hand. Stay consistent!

Good luck!

Title: Math Self Learning


问:
Primary School fractions question
My daughter asked me for help with the following question. While it's fairly easy to work out with a bit of common sense and trial and error, I was wondering if there is a simple way to work mathematically that you'd expect an 8 year old to understand?

答:
Perhaps something like the following:

$$ \frac{2}{3} \times \frac{\text{top}}{\text{bottom}} \;\; =  \;\; \frac{3}{10} \times \frac{\text{top}}{\text{bottom}}   $$

If the left side "top" is $3$ and the right side "top" is $2,$ then the numerators on both sides will be the same, because $2 \times 3 = 3 \times 2.$ And if the left side "bottom" is $10$ and the right side "bottom" is $3,$ then the denominators on both sides will be the same (because $3 \times 10 = 10 \times 3),$ and we'll have equality. Note that this solution is a sight-recognition task.

However, in showing this to the student you'd want to use an empty box for "top" and an empty triangle for "bottom", or something along these lines. The fact that the two boxes (top's here) can represent different numbers should not be made an issue of. For the student, the boxes simply represent places to put numbers and not variables representing numbers (the latter of which they don't know about, so shouldn't be brought up).

Title: Primary School fractions question


问:
Can a densely defined closed operator have a closed extension?
Let $X$ be a separable Banach space with two densely defined, closed unbounded operators $A_1 \subset A_2$ on it:
$$
  A_1: D(A_1) \to X, \quad A_2: D(A_2) \to X \\[8pt]
  D(A_1) \subseteq D(A_2) \subseteq X \\[8pt]
  \overline{D(A_1)} = \overline{D(A_2)} = X \\[8pt]
  \overline{A_1} = A_1, \quad \overline{A_2} = A_2 \\[8pt]
  A_1(x) = A_2(x) \quad \forall x \in D(A_1)
$$
Is it possible that $A_1 \neq A_2$?

I know that closed operators are defined by this condition:
$$
  x_n \to x \;\wedge\; Ax_n \to x \implies x\in D(A) \;\wedge\; Ax=y
$$
It seems to me, that $A_1 = A_2$ it doesn't follow from this condition. On the other hand, I couldn't find a counterexample and in my Quantum Mechanics class, we're always satisfied when we find any closed densely defined extension, so I always assumed other extensions just don't exist.

答:
In fact it is possible that $A_1\neq A_2$.  Here is an example.

Lemma.  Let $H=\ell ^2({\mathbb N})$.  Then there exists a bounded  operator $T:H\to H$, such that

$T$ is injective,  and $T(H)$ is a proper dense subspace of  $H$,

there exists a proper closed subspace $K\subseteq H$, such that $T(K)$ is also dense.

Proof.  Setting  $T((x_n)_n) = (x_n/n)_n$,  it is clear that $T$ satisfies all of the properties listed in (1).
Notice also that $T$ is self-adjoint.

Now let $y$ be any vector in $H\setminus T(H)$ and put $K=\{y\}^\perp$.  To see that $T(K)$ is dense, suppose by
contradiction that there is a nonzero vector  $z\in T(K)^\perp$.  This means that, for every
$x\in  K$  one has that
$$
  0 = \langle T(x), z\rangle  =   \langle x, T(z)\rangle ,
  $$
so
$$
  T(z) \in  K^\perp = (\{y\}^\perp)^\perp  = {\mathbb C}y.
  $$
Since $z$ is nonzero and $T$ is injective, $T(z)$ is nonzero and we deduce that $y$ is a multiple of $T(z)$,
contradicting the choice of $y$.  QED.

This said, let $A_2$ be the inverse of $T$, defined on $T(H)$, and let $A_1$ be the restriction of $A_2$ to $T(K)$.
Then $A_2$ is a closed operator since the its graph is (up to a flip of coordinates) the same as the
graph of the bounded operator $T$, and the same goes for $A_1$, being the inverse of  the restriction of $T$ to $K$.


问:
Is there notation for the logical and/or of lots of items?
You can add lots of things together using summation notation $$\sum_{n=1}^5 n = 15$$

Is there a similar operator for logical "and" and "or"?

I considered $\exists$/$\forall$, but that doesn't allow me to easily specify how to determine the answer in some cases.

答: You mentioned in a comment that this came up in an algorithmic context. In programming terms, the general expression for this is reduce(operation, iterable) (of course, this implies that you're dealing with an associative binary operator; reduce(mean, iterable) isn't going to get the mean of the iterable), e.g. reduce(or, (f(_) for _ in range(k))). There's also any(f(_) for _ in range(k)). Those are the Python syntax, but if you're doing psuedo code, this should be clear even to CS people who aren't familiar with Python. In mathematical logic terms, it would be $\exists n:f(n)$.


问:
After using the Deferred Acceptance algorithm, is this matching stable?
All profiles are strictly preferred. I have a set of students (s) and teachers (t). I have to come up with a construct where the matching is stable and the algorithm gives two outcomes that are different. The students in S should also prefer the outcome of the student-proposal and the teachers in T should prefer the outcome of the teacher-proposal. Have I done this correctly?

答: Yes both of the matching are stable. Instability requires both of the participants to be happier with an alternative match, but from the perspective of the students as proposers $s2$ was already rejected by $t3$, so they can not be happier with anyone other than $t1$ and $t1$ had no other offers. It shows that there can be more than one stable matching and as can be the case the matches are optimal from the proposers prospective.


答:
I might try drawing some blank lines to represent the situation:

$\underline{\hspace{.35 in}} \hspace{.2in} \underline{\hspace{.35 in}} \hspace{.2in}\underline{\hspace{.35 in}} \hspace{.3 in}$

$\underline{\hspace{.35 in}} \hspace{.2in} \underline{\hspace{.35 in}} \hspace{.2in}\underline{\hspace{.35 in}}
\hspace{.2in} \underline{\hspace{.35 in}}\hspace{.2in} \underline{\hspace{.35 in}}\hspace{.2in} \underline{\hspace{.35 in}}\hspace{.2in} \underline{\hspace{.35 in}}\hspace{.2in} \underline{\hspace{.35 in}}\hspace{.2in} \underline{\hspace{.35 in}}\hspace{.2in} \underline{\hspace{.35 in}}\hspace{.3 in}$

And then give a hint/explanation like this:
The line of three boxes each contain the same amount. And the line of ten boxes each contain the same amount.  The first two boxes of the three-box line contain the same total amount as the first three boxes of the ten-box line. We are interested in the grand total of each line (two-thirds of the first grand total is equal to three-tenths of the second grand total).

Title: Primary School fractions question

Body: My daughter asked me for help with the following question. While it's fairly easy to work out with a bit of common sense and trial and error, I was wondering if there is a simple way to work mathematically that you'd expect an 8 year old to understand?


问:
Can $p$-adic numbers be interpreted as the result of some summation method?
Perhaps an example of why I might want this information would clarify the question.

I know that Hensel's Lemma allows one to lift to roots of a polynomial to roots in $\mathbb{Q}$ given a value for $p$ and an initial guess.  My question is whether we can come up with a general rule that, say, limits the primes $p$ for which a rational polynomial can have roots in $\mathbb{Z}_p$.  If I were to try, e.g., substituting the series

$$\sum_{k\geq 0} a_kp^k$$

into a polynomial expression, then is it a simple matter of applying formal power series arithmetic, combining like powers of $p$, then forming a system of equations where the resulting coefficients for each $p^k$ must cancel?  Or am I missing some subtlety?

答:
Your question is completely unclear.

Hensel lemma is saying that from $F\in \Bbb{Q}[x]$, that we transform in $f\in \Bbb{Z}[x]$ monic  squarefree, there is some $N$ (easily computable but don't ask me its value) such that $f$ has a root in $\Bbb{Z}_p$ iff $f\bmod p^N$ has a root in $\Bbb{Z}/p^N\Bbb{Z}$. The correspondence is $a_N\to \lim_{n\to \infty } a_n$ where $a_{n+1}=a_n-f(a_n)/f'(a_n)$.


问:
Solving two algebraic equations
I'm trying to prove an identity from physics. I have the following two equations ($M$ is a constant):

$$e^2 = \left(1-\frac{2M}{r}\right)\left(1+\frac{l^2}{r^2}\right)$$ and $$r = \frac{l^2}{2M}\left[1 - \sqrt{1 - 12\left(\frac{M}{l}\right)^2}\right]$$  and I need to show that $$\frac{l}{e} = \sqrt{Mr}\left(1 - \frac{2M}{r}\right)^{-1}$$

I've tried rearranging the second eqation to get $l^2$ on its own and then dividing this by $e^2$ from equation 1 to get an expression for $l^2/e^2$ but this is quite messy and I can't see a clear way to simplify it. Can anyone see a straightforward way of doing this?

答:
let $x=\dfrac{l^2}{M^2},a=\dfrac{2r}{M} \to a=x-\sqrt{x^2-12x} \to x^2-12x=(x-a)^2 \to x=\dfrac{a^2}{2a-12}$

$\sqrt{1-\dfrac{12}{x}}=\sqrt{\dfrac{a^2-12*2a+12*12}{a^2}}=\dfrac{|a-12|}{a}$

now you can simplify more easy.

ya,if 12 is wrong, you give a right number but the middle result is same.

But it seems the last one is not correct,there should a factor of $\sqrt{\left(1 - \dfrac{2M}{r}\right)^{-1}}$

edit: I am interesting to go further after seeing the source poster.

$l^2=\dfrac{(aM)^2}{2a-12}=\dfrac{4r^2}{\dfrac{4r}{M}-12}=\dfrac{r^2M}{r-3M}=\dfrac{rM}{1-\dfrac{3M}{r}}$
if $a>12$, we have :

$r=\dfrac{l^2}{2M}\left[1-\dfrac{a-12}{a} \right]=\dfrac{6l^2}{aM}=\dfrac{3l^2}{r} \to \dfrac{l^2}{r^2}=\dfrac{1}{3}\to e^2 = \left(1-\dfrac{2M}{r}\right)\left(1+\dfrac{1}{3}\right)=\left(1-\dfrac{2M}{r}\right)\dfrac{4}{3}$

$\dfrac{l^2}{e^2}=\dfrac{3l^2}{4}\left(1 - \dfrac{2M}{r}\right)^{-1}=\dfrac{3rM}{4}\left(1 - \dfrac{2M}{r}\right)^{-1}\left(1 - \dfrac{3M}{r}\right)^{-1}$

if $a<12$ ,we have :

$r=\dfrac{l^2}{2M}\left[1-\dfrac{12-a}{a} \right]=\dfrac{l^2}{2M}\dfrac{2a-12}{a}=\dfrac{a-6}{2r} \to \dfrac{l^2}{r^2}=\dfrac{2}{a-6}=\dfrac{M}{r-3M} \to$
$1+\dfrac{l^2}{r^2}=\dfrac{r-2M}{r-3M} \to e^2=\left(1 - \dfrac{2M}{r}\right)\dfrac{r-2M}{r-3M} \to $

$\dfrac{l^2}{e^2}=\dfrac{r-3M}{r-2M}*rM*\left(1 - \dfrac{2M}{r}\right)^{-1}\left(1 - \dfrac{3M}{r}\right)^{-1}=rM\left(1 - \dfrac{2M}{r}\right)^{-2}$

so that is the answer!

Title: Solving two algebraic equations


问:
Technique for evaluating Limits of recursive sequences
I was trying to evaluate the limit of a Recursive Sequence, and I found the following website on how to do so: Evaluating Limits of Recursive Sequences.

They used one strange trick in order to evaluate the limit. The process basically is the following:
Let's define the following sequence:

$$\begin{cases}
a_1=k \\
a_{n+1}=f(a_n)
\end{cases}$$

And they basically did the following: $L = \lim a_{n+1} =\underbrace{ \lim f(a_n)= f(\lim a_n) }_{(1)}= f(L)$.

And so the limit would be the solution of the equation $L = f(L)$.

My question is, is this a correct way of evaluating these kinds of limits? Does it always work? If this is correct what are the cases where we can use this? What if this equation has multiple solutions? If the equation does not have any solutions does it mean that the limit does not exist? I'm looking for an analysis of this technique.

I think that in order to do step $(1)$ we need to make sure that $f$ is continuous at $L$, but I'm not sure.

答:
So I think that I sorta figured it out:

$$\begin{cases}
a_1 = k \\
a_{n +1} = f(a_n)
\end{cases}$$

Is actually the orbit of $k$ by $f$. This means that if it converges, then it's limit $L$ is a fixed point of $f$, meaning that:

$$L = f(L)$$

If this equation has multiple solutions then, if the sequence converges, one of them should be the limit, and if this has no solutions, then it's impossible for the sequence to converge. This equation having solutions does not imply that the sequence converges, it just means that if it does, then it's limit is one of those values.

Is this correct?


问:
If the sets defined by a formula are all trivial then what can we conclude from that?
Let $\mathcal L$ be a language, $n$ a positive integer and $\phi=\phi(x_1,\dots,x_n)$ an $\mathcal L$-formula having $x_1,\dots,x_n$ as free variables.

Then for every $\mathcal L$-structure $\mathfrak A$ there is a set $\phi^{\mathfrak A}$ defined by $\phi$ as: $$\phi^{\mathfrak A}:=\{(a_1,\dots,a_n)\in |\mathfrak A|^n: \mathfrak A\vDash\phi[a_1,\dots,a_n]\}$$where $|\mathfrak A|$ denotes the domain of structure $\mathfrak A$.

Now let it be that for every $\mathcal L$-structure $\mathfrak A$ we have: $$\phi^{\mathfrak A}=\varnothing\text{ or }\phi^{\mathfrak A}=|\mathfrak A|^n$$

Can we conclude from this that one of the following statements must be true?

$\phi^{\mathfrak A}=\varnothing$ for every $\mathcal L$-structure $\mathfrak A$.

$\phi^{\mathfrak A}=|\mathfrak A|^n$ for every $\mathcal L$-structure $\mathfrak A$.

答: No, we cannot; let $\phi'$ be any $\mathcal{L}$-sentence (so $\phi'$ contains no free variables), and let $\phi$ be the $\mathcal{L}$-formula $\phi'\wedge\bigwedge_{i=1}^nx_i=x_i$. (Note that $\phi$ does indeed contain every $x_i$ as a free variable, albeit in a vacuous way.) Then if $\mathfrak{A}\models\phi'$ we have $\phi^\mathfrak{A}=|\mathfrak{A}|^n$, and if $\mathfrak{A}\models\neg\phi'$ we have $\phi^\mathfrak{A}=\emptyset$. Since every $\mathcal{L}$-sentence either holds or does not hold in an $\mathcal{L}$-structure, one of these two cases must apply. Thus, chosing $\phi'$ so that there are some structures in which it holds and some structures in which it does not provides a countexample.


问:
Prove that the subspace topology is the coarsest topology on $Y$ for which the inclusion function is continuous

Prove that the  subspace topology  is the coarsest topology on $Y$ for which the inclusion function $i:Y \longrightarrow X$ given by $i(y)=y$ is continuous.

Attempt

Define $T^{\prime}=\lbrace V\subset X \mid i^{-1}(V) \text{ is open in $Y$} \rbrace$
since $Y$ have the subspace topology $ i^{-1}(V)$ should look like $Y \cap U$ where $U$ is open in $X$.
By definition $i^{-1}(V)=\lbrace y\in
Y \mid i(y)\in V\rbrace=\lbrace y\in Y \mid y\in V\rbrace$ since $V\subset X$ and not necessary in all $Y$ we conclude that $T^{\prime}=\lbrace  Y\cap V  \mid  V\subset X \rbrace$ and since the subspace topology in $Y$ is given by $T=\lbrace U\cap Y \mid U\,  \text{is open in $X$} \rbrace$
And  obviously $T\subset T^{\prime}$ and therefore the subspace  topology on $Y$ is coarsest topology for which the function $i$ is continuous.

答: Hint: Let $U\subset X$ be open.  Then $i^{-1}(U)=U\cap Y$.


问:
connection between affine map and differential
Assume that exists  $T\in\hom\left(\mathbb{R}^{k},\mathbb{R}^{m}\right)$ such that  $\left(Df\right)_{a}=T$ forall $a\in\mathbb{R}^{K}$. Prove that exists $T\in\hom\left(\mathbb{R}^{k},\mathbb{R}^{m}\right)$ such that $f\left(x\right)=Tx+f\left(0\right)$ forall $x\in\mathbb{R}^{k}$.

can anyone give me any clue?
thanks!

答:
If I understood correctly, you have, by hypothesis, that there exists a linear map $T$ such that, for every $a\in\mathbb R^k, Df_a=T$; and what you have to prove is that, there exists another linear map $S$ such that $f=S+f(0)$, right?

So, for the solution take the function $g=f-T$ which has derivative identically zero, that is, for every $a\in\mathbb R^k, D(f-T)_a=0$. By the connectedness of the domain of $f-T$, we get that $f-t$ is constant. Evaluating at $x=0$ we get $(f-T)(0)=f(0)$. Since $f-T$ is constant, we get, for every $x\in\mathbb R^k$, $f(x)=T(x)+f(0)$. So you have to take $S=T$.


问:
Prove that an infinite number of triangles can be inscribed in the hyperbola xy=c^2 whose sides touch the parabola y^2=4ax.
So I tried writing 3 equations of tangents to the parabola:

$t_1y+x=at_1^2$,  $t_2y+x=at_2^2$ and $t_3y+x=at_3^2$

The respective points of intersection are:

$(at_2.t_3,a(t_2+t_3))$, $(at_1.t_3,a(t_1+t_3))$ and $(at_1.t_2,a(t_1+t_2))$

Since these points lie on the hyperbola, Putting these points in $xy=c^2$:

$t_2.t_3(t_2+t_3)=\frac{c^2}{a^2}$ ...(1)

$t_3.t_1(t_3+t_1)=\frac{c^2}{a^2}$ ...(2)

$t_1.t_2(t_1+t_2)=\frac{c^2}{a^2}$ ...(1)

I am not able to figure out how to proceed after this... Can someone please help me out?

答:
I haven't checked the derivation of your three equations, but
it is not difficult to see that they are equivalent to this system of only TWO equations:
$$
_1+_2+_3=0 \\
_1_2_3=−{^2\over ^2}.
$$
Hence there are infinitely many solutions: for instance you can choose $t_1$ at will and find $t_2$, $t_3$ from the above equations.


问:
Matrix Algebra Over the Opposite Ring
I'm trying to show that $M_n(R^{opp}) \cong M_n(R)^{opp}$ as rings. My thought was that $\varphi : M_n(R)^{opp} \to M_n(R^{opp})$ defined by $\varphi (A) = A^{T}$ is the desired isomorphism. But I'm a little confused as to what there is to verify...

Clearly $\varphi^{-1} = \varphi$, so it is bijective, $\varphi (A+B) = (A+B)^{T} = A^{T} + B^{T}$, and $\varphi (A \star B) = \varphi (BA) = (BA)^{T} = A^{T} B^{T} = \varphi (A) \varphi (B)$, where $\star$ is the opposite product of $M_n(R)^{opp}$.

Is that it? Seems like a cheesy proof. I tried consulting some of the users in the chatroom. The only one to proffer a critique was Thorgott. He said that "I don't think $\varphi^{-1} = \varphi$, given that codomain and domain are different." After thinking about this for some time, I don't think this is a problem, because although $M_n(R^{opp})$ and $M_n(R)^{opp}$ denote different rings, the underlying sets are the same.

答:
The entire argument is hidden in the step $(BA)^T = A^T B^T$ and the notation obscures what's going on slightly; on the LHS the matrices are being multiplied in $M_n(R)$ and on the RHS the matrices are being multiplied in $M_n(R^{op})$ (and the statement is false if $R$ is noncommutative and both matrix multiplications are interpreted in the same matrix ring). Once you write out what's going on here in coordinates it should be clear what's going on and what you have to prove.

Maybe the cleanest way to write this proof is to introduce explicit notation for the "identity" map $(-)^{op} : M_n(R) \to M_n(R^{op})$. $\varphi(A)$ should really be defined as $(A^{op})^T = (A^T)^{op}$ (the "opposite transpose"). Now you're no longer sweeping anything under the rug: you need to prove that

$$((BA)^T)^{op} = (A^T)^{op} (B^T)^{op}$$

(now the op's have been properly inserted!) which will then give $\varphi(A \star B) = \varphi(A) \varphi(B)$.

As for Thorgott's objection, this is a notational issue again; write $\varphi_R$ for the transpose map $M_n(R)^{op} \to M_n(R^{op})$ and you'll see that the inverse of $\varphi_R$ is $(\varphi_{R^{op}})^{op}$, where the "op" here has yet another meaning: if $f : R \to S$ is a ring homomorphism then it has an opposite ring homomorphism $f^{op} : R^{op} \to S^{op}$ which on underlying sets is just $f$ again.

Title: Matrix Algebra Over the Opposite Ring


问:
Sigma value for defect ratio
What is the smallest sigma value (out of $1\sigma$, $2\sigma$, $3\sigma$, $4\sigma$, $5\sigma$, $6\sigma$) we can use when we want a less than 0.3% defect ratio?

My idea is to use $68-95-99.7$ rule, and obtain $3\sigma$ as an answer. Is it correct approach, or has missing parts?

答:
Allow me to ask you here (since I can't comment yet) why would you choose $6σ$.

Since 99.7% means less than 0.3% defect ratio, maybe if we take an additional $σ$ only we would be in a safe place, in my opinion. Therefore, my answer would be $4σ$ (3 is already safe for your question).

New Information

As I mentioned, since we are talking about defects, it's more about the Six Sigma and the Sigma Level. Therefore, if we go to the wikipedia page, we can see that the value should be between 4 and 5 Sigma Levels.

Now, we go back to our original SIGMAs,  which are easier to understand.
Our goal is to get 0.3% defect, or 99.7% good.
Therefore, we take our table of values we know that within 3σ we get 0.9973. Therefore, the real value should be a bit less than $3σ$. But we can go with $3σ$

Being Very Precise

Table for Z-value if you read the table you will find 2.75 Sigma for 0.99702.

Reference

Quora Difference between Sigma and Sigma Level

Title: Sigma value for defect ratio


问:
Integral equation to determine equilibrium
I have the integral equation
$$\left(1-\frac{3x}{2}\right)f(x) + (1-x)\int_0^1 x f(x) dx + (1-x) \int_0^x f(x) dx = 0$$

with boundary conditions:
$f(x) > 0$ for all $x$ and $\int_0^1 f(x) = 1$, or with other words, $f(x)$ is a probability density function on the domain $[0,1]$.

I am wondering (1) is there a solution? (2) can we find it numerically or exactly by some procedure?

I was trying to turn this into a differential equation but I couldn't manage to do that. I also have some doubts that the system may be underspecified... ?

Background: I believe this integral equation is modelling some behavior of a system I am studying, and the integral equation would be this system in equilibrium. Successive approximations of the state of this system show that the system is converging and thus equilibrium seems likely, but I wonder if I can prove equilibrium via this way. The partial solutions of the system not in equilibrium (but almost) seem to indicate that $f(x)$ is biased to small $x$, where $f(0)$ is quite large and rapidly decreases to zero for $x > 0$, and looks similar to an exponential distribution with $\lambda \approx 8$.

答:
Rewrite

$$\frac{2-3x}{2(1-x)}f(x) + \int_0^1 x f(x) dx + \int_0^x f(x) dx = 0.$$

Then after differentiation,

$$\frac{2-3x}{2(1-x)}f'(x)+\left(1-\frac1{2(1-x)^2}\right)f(x)=0$$

which is a separable ODE.


问:
Are there single words for being within two bounds exclusive and non-exclusive?
When $a <= b <= c$ is there a single word that describes $b$ being between $a$ and $c$ (inclusive)? Is $b$ within $a$ and $c$?

Similarly, when $a < b < c$ is there a single word that describes $b$ being between the bounds $a$ and $c$a (exclusive)? Is $b$ between $a$ and $c$?

It seems like we can say $a$ to $c$ to mean $[a, c]$ and $a$ until $c$ to mean $[a, c)$ so I am curious if there are succinct ways of describing a value within open and closed intervals (until in this example is half-open so is not exactly the same).

答:
You might say that b is bounded by a and c.

Extending this a little, and possibly creating some terminology, you might say the bound is inclusive if equality is possible and exclusive if equality is not.

So you might say that b is exclusively lower bounded by a and inclusively upper bounded by c.

Or, to be traditional, a less than b less than or equal to c.


问:
Preimages of a function: Is the following proposition true or false?
Let $g: ℤ \times ℤ → ℤ \times ℤ$ be defined by $g(m,n) = (2m, m – n)$.

Is the following proposition true or false? Justify your conclusion.
For each $(s, t) ∈ ℤ \times ℤ$, there exists an $(m, n) ∈ ℤ \times ℤ$ such that $g(m,n) = (s, t)$.

I understand the question, and I believe it is true. But I have no idea of what the steps are on how to prove it is true. I guess I should not have skipped the past couple of days. Thanks for any help.

答:
Hint: the first coordinate of the image point $(2m,m-n)$ is always even.

What happens if $s$ is odd?


问:
Riemann integral in two variables
Let $f:[0,1]\times[0,1]\to \mathbb{R}$, be defined by:
\begin{equation}
    f(x,y)=\begin{cases}
0 & \text{ if  }\; 0\leq x<\frac{1}{2},\\
1 & \text{ if  }\; \frac{1}{2}\leq x \leq 1.
\end{cases}
\end{equation}
Show that $f$ is integrable and $\displaystyle\int_{[0,1]\times [0,1]}f=\frac{1}{2}$.

Attempt. Let $P$ a partition of $[0,1]\times[0,1]$, such that:
$\mathcal{P}=(\mathcal{S}_{1},\mathcal{S}_{2})$, where :

\begin{align}
    \begin{split}
        \mathcal{S}_{1}&=\left[0,\frac{1}{2}\right]\times \left[0,1\right]\\
        \mathcal{S}_{2}&=\left[\frac{1}{2},1\right]\times \left[0,1\right]\\
    \end{split}
\end{align}

Define
\begin{align}
   \begin{split}
     m_{\mathcal{S}_{1}}(f)&=\inf_{x\in \mathcal{S}_{1}}f(x)=0\\
   m_{\mathcal{S}_{2}}(f)&=\inf_{x\in \mathcal{S}_{2}}f(x)=1\\
   \end{split}
\end{align}
and
\begin{align}
      \begin{split}
   M_{\mathcal{S}_{1}}(f)&=\sup_{x\in \mathcal{S}_{1}}f(x)=0\\
   M_{\mathcal{S}_{2}}(f)&=\sup_{x\in \mathcal{S}_{2}}f(x)=1\\
      \end{split}
  \end{align}
The lower and upper sums:
\begin{align*}
    L(f,\mathcal{P})& =\sum_{\mathcal{S}_{1}}m_{\mathcal{S}_{1}}(f).\upsilon(\mathcal{S}_{1})+\sum_{\mathcal{S}_{2}}m_{\mathcal{S}_{2}}(f).\upsilon(\mathcal{S}_{2})\\
    &=\sum_{\mathcal{S}_{1}}0\cdot\frac{1}{2}+\sum_{\mathcal{S}_{2}}\frac{1}{2}\\
    &=\frac{1}{2}
\end{align*}
and
\begin{align*}
    U(f,\mathcal{P})& =\sum_{\mathcal{S}_{1}}M_{\mathcal{S}_{1}}(f).\upsilon(\mathcal{S}_{1})+\sum_{\mathcal{S}_{2}}M_{\mathcal{S}_{2}}(f).\upsilon(\mathcal{S}_{2})\\
    &=\sum_{\mathcal{S}_{1}}0\cdot\frac{1}{2}+\sum_{\mathcal{S}_{2}}\frac{1}{2}\\
    &=\frac{1}{2}
\end{align*}

Then
\begin{align*}
\sup{L(f,\mathcal{P})}=\frac{1}{2}=\inf{U(f,\mathcal{P})}.
\end{align*}
Hence $f$ is integrable on $[0,1]\times [0,1]$ and  $\displaystyle\int_{[0,1]\times [0,1]}f=\frac{1}{2}$.

I have doubts about my solution since my instructor told me that the conclusion is not correct since I calculate the lowest of the upper sums on a particular partition, and by definition of the upper integral it must be calculated on the set of all partitions.
He suggested using the definition of infimum $ \varepsilon $. I appreciate if someone could help me or give an indication.

答:
The Riemann integral can be shown to exist, by the Riemann criterion, if for any $\epsilon > 0$ it is possible to produce a partition $\mathcal{P}$ such that $U(f, \mathcal{P}) - L(f, \mathcal{P}) < \epsilon$.

Your choice does not help when $\epsilon < \frac{1}{2}$ because both subrectangles include points where $f(x) = 1$ and $U(f, \mathcal{P})- L(f, \mathcal{P})= 1- \frac{1}{2} = \frac{1}{2}$.

A better choice is the partition with subrectangles

$$\left[0,\frac{2-\epsilon}{4} \right]\times [0,1], \,\, \left[\frac{2-\epsilon}{4},\frac{2+\epsilon}{4} \right]\times [0,1], \,\, \left[\frac{2+\epsilon}{4},1 \right]\times [0,1]$$

We then have

$$L(f,\mathcal{P}) = 1 \cdot \left(1 - \frac{2+\epsilon}{4}\right) = \frac{2-\epsilon}{4}, \\ U(f,\mathcal{P}) = 1 \cdot \left(1 - \frac{2+\epsilon}{4}\right) + 1 \cdot \left(\frac{2+\epsilon}{4}- \frac{2-\epsilon}{4} \right) = \frac{2-\epsilon}{4} + \frac{\epsilon}{2},$$

and $U(f, \mathcal{P}) - L(f, \mathcal{P})= \frac{\epsilon}{2}  < \epsilon$.

It should now be easy for you to show that the value of the integral must be $\frac{1}{2}$.

Hint: $L(f, \mathcal{P}) \leqslant \int_{[0,1]\times[0,1]} f \leqslant U(f, \mathcal{P}) $ for every partition $\mathcal{P}$.

Title: Riemann integral in two variables


问:
Number of ways to assign scores
A test consisting of $12$ questions is scored with $200$ points. How many ways can the $200$ points be assigned if each question must be worth at least $10$ points but no more than $25$ and the scores must be a multiple of $5$?

I'm thinking about a possible combination with repetition. $15$ points ($80/5$) out of $12$ questions. But I don't know how to apply the maximum restriction of $25$.$3$ to each question $\frac{25-10}{3}$.

But it may not be a combination with repetition either, as I interpret the statement as meaning that it doesn't matter if the first question is worth $30$ points and the second is worth $10$, which is the other way around. I understand that this specific example is a unique way, isn't it?

答:
Since each score must be a multiple of $5$, we might as well divide all of the point values by $5$ and have $12$ questions worth a total of $40$ points, each question being worth at least $2$ and at most $5$ points. If $p_k$ is the point value of the $k$-th question, we’re looking for the number of solutions to

$$\sum_{k=1}^{12}p_k=40\tag{1}$$

in integers $p_k$ satisfying the condition that $2\le p_k\le 5$ for $k=1,\ldots,12$. Let $x_k=p_k-2$ for $k=1,\ldots,12$; then the number of solutions to $(1)$ subject to the stated restriction is the same as the number of solutions to

$$\sum_{k=1}^{12}x_k=16$$

in non-negative integers $x_k$ satisfying the condition that $x_k\le 3$ for $k=1,\ldots,12$. If it weren’t for the upper limit on the numbers $x_k$, this would be a standard stars-and-bars problem, and there would be $\binom{16+12-1}{12-1}=\binom{27}{11}$ of them. Unfortunately, many of those solutions violate the upper limit on one or more of the numbers $x_k$, so $\binom{27}{11}$ is a significant overestimate. To correct this you will need to make an inclusion-exclusion calculation. My answer to this question includes such a calculation; try to use it as a model to complete the solution of your problem.

Title: Number of ways to assign scores


问:
Problem computing an angle
Consider the following drawning:

sorry if I did it by hand. In this case, how do I compute the angle $\gamma$?

If I have understood well, it should be:

$\gamma=atan2(y,x)-\theta$

but I am not sure.

答: Note that the angle between the line from the point to the origin and the $x$-axis is $\arctan\big(\frac{y}{x}\big)$. Thus, $\boxed{\gamma = \theta-\arctan\big(\frac{y}{x}\big)}$. If you are working on the whole coordinate plane, you should use the $\text{atan2}$ function instead of $\arctan$, which gives the correct angle for all $4$ quadrants.

Title: Problem computing an angle


问:
What's the value of the inverse square and inverse of an unknown exponential?
The problem is as follows:

Given:

$$x^{-x}=(-8)^{-3^{-1}} \cdot (-8^{0})^{3^{4^{5}}}$$

Find the result of

$$(x^{-2}-x^{-1})$$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\frac{3}{4}\\
2.&\frac{1}{4}\\
3.&-\frac{1}{4}\\
4.&\frac{1}{8}\\
5.&\frac{3}{8}\\
\end{array}$

What I attempted to do was as follows:

From what I can understand this whole expression equals to $-1$ as the sign is taken out of the exponential and the $8$ powered to the rest becomes just $1$.

$(-8^{0})^{3^{4^{5}}}=-1$

Then if I'm not mistaken the thing is with the rest:

$(-8)^{-3^{-1}}=(-8)^{-\frac{1}{3}}=-\left(\frac{1}{2^3}\right)^{\frac{1}{3}}$

$-\left(\frac{1}{2^3}\right)^{\frac{1}{3}}=-\frac{1}{2}$

Then this means:

$\left(-\frac{1}{2}\right)\times (-1)=\frac{1}{2}$

Then this means:

$x^{-x}=2^{-1}$

But this doesn't seem right. I'm stuck there. Can someone help me here?. Maybe is my interpretation of the problem not accurate?. The official answer is the first option but I don't know how to get there.

答:
x^x = 2
Equalize the options in the book (1-x)/(x^2) and solve the equation.
And put the obtained answers in the equation x^x that the answer should be equal to 2.

check https://www.wolframalpha.com/input/?i=x%5Ex%3D2


问:
Computing the limit $\lim_{z\to 0}\frac{\operatorname{Arg}(1+z)}{z}$
For every $z\in\mathbb C\setminus\{0\}$ we define $\operatorname{Arg}(z)$ to be the principal argument of $z$, that is $\operatorname{Arg}(z)=\theta$, where $z=re^{i\theta }$ and $\theta\in(-\pi,\pi]$. Does the following limit exists?

$$
\lim_{z\to 0}\frac{\operatorname{Arg}(1+z)}{z}
$$
I guess that if it exists, then it must be $0$ since if $z\to0$ along the real line, then the limit is $0$.

答:
$$ \frac{\log (1+z)}{z} = \frac{\log |1+z|}{z}+ i \frac{\arg{(1+z)}}{z}$$

$$\lim_{z\to 0} \frac{\log(1+z)}{z}=\lim_{z\to 0}\frac{1}{1+z}=1.$$

Consider $z\to 0$ and $z=x$ ($z$ approaches zero along the real axis).

Then $$\lim_{z\to 0} \frac{\arg(1+z)}{z} = \Im \lim_{z\to 0} \frac{\log(1+z)}{z} =0.$$

Consider $z\to 0$ and $z=iy$ ($z$ approaches zero along the imaginary axis).
$$ \frac{\log (1+z)}{z} = \frac{\log |1+iy|}{iy}+ i \frac{\arg{(1+iy)}}{iy}$$

Then $$\lim_{y\to 0} \frac{\arg(1+iy)}{y} = 1$$

and  $$\lim_{z\to 0} \frac{ \arg(1+z)}{z} = -i \ne 0.$$

Since the limit obtained from different directions is different, the limit does not exist.


问:
Does the Probability holds the distributivity?
I am doing a basic probability problem from a textbook exercise. The question is as follows:

Consider two events A and B such that P(A) = 1/3 and P(B) = 1/2 and P(AB) = 1/8. Determine the value of P(BA^c). The way I did the problem is as,
       P(BA^c) =  P(B(1-A))  --> step 1
               =  P(B - BA)  --> step 2
               = P(B) - P(BA)  --> step 3
               = (1/2) - (1/8) = 3/8.

Answer from the book is same as mine.Is my approach correct? Does the Probability holds the distributivity as I did it from step 2 to step 3?

Please correct me if I were wrong with my terminology.

Thanks

答: A way to see that your method works is to observe that for any two events $A$ and $B$ the probability of $B$ is the probability of "$B$ and $A$" plus the probability of "$B$ and not-$A$." Subtracting the "$B$ and $A$" term from both sides gives what you got. </p>


答:
Since all scores are a multiple of $5$ we can divide by it, resulting in $40$ points total and question scores from $2$ to $5$ points. Since each question must be at least $2$ points, we can think of the questions as holding $24$ "base points", leaving the problem of how many ways there are to distribute $16$ "extra points" to the $12$ questions with no question having more than $3$ of them.

As a generating function problem, this is the $x^{16}$ coefficient of $(1+x+x^2+x^3)^{12}$. And the answer turns out to be $1501566$.

Title: Number of ways to assign scores

Body:
A test consisting of $12$ questions is scored with $200$ points. How many ways can the $200$ points be assigned if each question must be worth at least $10$ points but no more than $25$ and the scores must be a multiple of $5$?

I'm thinking about a possible combination with repetition. $15$ points ($80/5$) out of $12$ questions. But I don't know how to apply the maximum restriction of $25$.$3$ to each question $\frac{25-10}{3}$.

But it may not be a combination with repetition either, as I interpret the statement as meaning that it doesn't matter if the first question is worth $30$ points and the second is worth $10$, which is the other way around. I understand that this specific example is a unique way, isn't it?


问:
Tensor product of submodules is a submodule of the tensor product?
If $M',N'$ be submodules of $M$ and $N$ respectively. Then is it true that $M'\otimes N'$ is a submodule of $M\otimes N$?

I think it should not be true.
I considered $2\mathbb{Z}$ and $\mathbb{Z_2}$ as submodules of $\mathbb{Z}$ and $\mathbb{Z_2}$ then $2\otimes 1=0 \in \mathbb{Z}\otimes \mathbb{Z_2}$ and $2\otimes 1 \neq 0 \in \mathbb{2Z}\otimes \mathbb{Z_2}$.

I think elements of $M' \otimes N'$ are very different from the elements $M\otimes N$, just like elements of $\mathbb{Z_3}$ is very different from elements of $Z_4$ although both contains $\bar{1},\bar{2}$.

Is my thinking correct?

答:
It's not true and your counterexample is correct. The problem is that tensoring by $M$ or $N$ can fail to respect injections if $M$ or $N$ aren't flat, and in your example $N$ is not flat.

It is true if $M$ and $N'$ are both flat (in which case we have injections $M' \otimes N' \hookrightarrow M \otimes N' \hookrightarrow M \otimes N$) or if $M'$ and $N$ are both flat (in which case we have injections $M' \otimes N' \hookrightarrow M' \otimes N \hookrightarrow M \otimes N$). Notably this means it's true if $M$ and $N$ are torsion-free modules over a Dedekind domain (which implies that $M, N, M', N'$ are all flat), which is a useful special case.


问:
WQO - why transitivity?
This article says that $\sf WQO$s are a strenghtening of the definition of well-founded relations for the purpose of having the property be preserved by operations. However, the few operations and proofs (minimal bad sequence arguments) that I know about don't seem to need the extra assumption of transitivity. The only exception is the product of relations/posets but it can be done with infinite Ramsey's theorem. Is transitivity used in other parts of $\sf WQO/BQO$ theory?

答: This likely isn't the greatest theorem of WQO theory but for an example, a qoset is WQO iff it has no infinite decreasing chains nor infinite antichains. This fails for WQO-minus-transitivity, consider the relation $\preceq$ on $\mathbb{N}$ such that $a \preceq b \iff a=b \lor a=b+1 \lor (a<b \land a\equiv b \mod 2)$. There is an infinite decreasing chain $0\succ1\succ2\ldots$ but in an infinite sequence we will eventually have two numbers of same parity in increasing order.

Title: WQO - why transitivity?


问:
Deriving iterative applications of an operator using recursion
In this answer by Angina Seng, they start with the proposition that:

$$ P^n = \sum_{k=1}^n a_{n,k} x^k D^k$$

Why were they able to do this? Is there any easy justification to give for it?

答:
It's a general fact about the Weyl algebra $k[x, D]$ (where $D = \partial_x$) that every element can be written this way. The proof is just by induction; you apply the canonical commutation relation $D x - x D = 1$ a lot to move $x$'s past $D$'s. As a simple example,

$$(xD)^2 = x(Dx)D = x(xD + 1)D = x^2 D^2 + xD.$$

In general, it suffices to show that $D^i x^j$ can be written as a sum of terms of the form $x^k D^{\ell}$, which you do by commuting the $k$ copies of $x$ past the $j$ copies of $D$ using the canonical commutation relation. It will be convenient to adopt the notation $[X, Y] = XY - YX$ and to prove lemmas about it, e.g. that it's a derivation in each variable, and a lemma about what $[X, Y^n]$ looks like.

Abstractly you can get this out of the Poincare-Birkhoff-Witt theorem, applied to the universal enveloping algebra of the Heisenberg algebra $\text{span}(x, D, 1)$. (You then need to take a quotient of the universal enveloping algebra given by setting $1$ to the identity to get the Weyl algebra.)


问:
How is $ B_n = 1- \sum_{k=0}^{n-1} \binom{n}{k} \frac{B_k}{n-k+1} $ Where $B_n$ are the Bernoulli Numbers with $B_1 = \frac{1}{2}$
So I was browing Wikipedia just looking at Bernoulli Number identities and I stumbled across this
$$ B_m^+ = 1- \sum_{k=0}^{m-1} \binom{m}{k} \frac{B_k^+}{m-k+1} $$
The Wikipedia page said that this identity was derived from the identity :
$$
\sum_{k=0}^{m}\left(\begin{array}{c}
m+1 \\
k
\end{array}\right) B_{k}^{+}=m+1
$$
My question is, how are these identities derived? Wikipedia gives the reference for these theorems to be from the page on Bernoulli Numbers on WolframMathWorld but I can't find the proof of any of these identities there. Can anyone guide me to or state a proof of either identity? (From the definition of Bernoulli Numbers stated below)
$$
\frac{t}{1-e^{-t}}=\frac{t}{2}\left(\operatorname{coth} \frac{t}{2}+1\right) \quad=\sum_{m=0}^{\infty} \frac{B_{m}^{+} t^{m}}{m !}
$$

答:
As Wikipedia says, the first identity can be obtained from the second by "solving" the latter for $\ B_m^+\ $:
\begin{align}
m+1&=\sum_{k=0}^m{m+1\choose k}B_k^+\\
&=(m+1)B_m^++\sum_{k=0}^{m-1}\frac{(m+1)!}{k!\,(m+1-k)!}B_k^+\\
&=(m+1)\left(B_m^++\sum_{k=0}^{m-1}\frac{m!}{k!\,(m-k)!}\left(\frac{B_k^+}{m-k+1}\right)\right)\\
&=(m+1)\left(B_m^++\sum_{k=0}^{m-1}{m\choose k}\frac{B_k^+}{m-k+1}\right)\ .
\end{align}
Dividing through by $\ m+1\ $ and subtracting $\ \displaystyle\sum_{k=0}^{m-1}{m\choose k}\frac{B_k^+}{m-k+1}\ $ from both sides of this equation gives the required first identity.

Here is a version of reuns's derivation of the second identity which works directly from the identity
$$
\frac{t}{1-e^{-t}}=\sum_{k=0}^\infty\frac{B_k^+t^k}{k!}\ .
$$
Multiplying both sides of this identity by $\ e^t-1\ $ gives
\begin{align}
\sum_{m=0}^\infty\frac{t^{m+1}}{m!}&=
te^t\\
&=\big(e^t-1\big)\sum_{k=0}^\infty\frac{B_k^+t^k}{k!}\\
&=\sum_{n=0}^\infty\sum_{k=0}^\infty\frac{B_k^+t^{n+k+1}}{(n+1)!k!}\\
&=\sum_{m=0}^\infty t^{m+1}\sum_{k=0}^m\frac{B_k^+}{(m+1-k)!k!}\ .
\end{align}
Equating coefficients of $\ t^{m+1}\ $ gives
$$
\frac{1}{m!}=\sum_{k=0}^m\frac{B_k^+}{(m+1-k)!k!}\ ,
$$
and multiplying this equation by $\ (m+1)!\ $ gives the required identity.


问:
Trigonometric doubt on squaring both sides
I have the equation: $$\tan(x)+\sec(x)=1$$ What I tried was moving the $\sec(x)$ to the right side $$\sec(x)=1-\tan(x)$$ and here's the doubt, I decided to try squaring both sides, as after I will be able to apply $\tan^2(x)-\sec^2(x)=-1$, but when I squared both sides, the solution changed when I plugged it into Wolfram, $$\sec^2(x)=(1-\tan(x))^2$$ gives as solution from Wolfram $x=\pi k, \quad k\in \mathbb{Z}$ while the original equation gives $x=2\pi k, \quad k\in \mathbf{Z}$. Why is this?

答: Squaring an equation $f(x)=g(x)$ introduces all the solutions of $-f(x)=g(x)$. You need to filter out the solutions to this second, "unwanted" equation once you have worked out the solutions to the squared equation.


问:
Probability that $3$ hands of a clock in the same semi-circle?
I am wondering how to calculate the probability that $3$ hands of a clock are in the same semi-circle ? I know a similar question is that if we randomly choose n points in a circle, then the probability that all of them in the same semi-circle will be $n/2^{(n-1)}$ with $n = 3$.

But when it comes to the dial hands, where I guess there are correlations between them, I am not sure whether my question is equivalent to the previous one.
If they are not equivalent, and this question will be hard to calculate, then whether we can determinate this probability will be greater or smaller than $3/4$?

Anyone can help? Thanks.

答:
Let's denote by $\omega$ the angular speeds, so
$$
\left\{ \matrix{
  \omega _{\,h}  = {{2\pi } \over {12h}} = {{2\pi } \over {12 \cdot 3600}}s^{\, - 1}  \hfill \cr
  \omega _{\,m}  = {{2\pi } \over h} = {{2\pi } \over {3600}}s^{\, - 1}  \hfill \cr
  \omega _{\,s}  = {{2\pi } \over m} = {{2\pi } \over {60}}s^{\, - 1}  \hfill \cr}  \right.
\quad  \Rightarrow \quad \left\{ \matrix{
  \omega _{\,m}  - \omega _{\,h}  = 11\;\omega _{\,h}  \hfill \cr
  \omega _{\,s}  - \omega _{\,h}  = 719\;\omega _{\,h}  \hfill \cr}  \right.
$$

Clearly we are interested in the relative angles
$$
\left\{ \matrix{
  \alpha _{\,m}  = \left( {\omega _{\,m}  - \omega _{\,h} } \right)t\,
\quad \left( {\bmod 2\pi } \right)\quad  \Rightarrow \quad
 \rho _{\,m}  = {{\alpha _{\,m} } \over {2\pi }}
 = \left\{ {11{{\omega _{\,h} \,t} \over {2\pi }}} \right\}\, \hfill \cr
  \alpha _{\,s}  = \left( {\omega _{\,s}  - \omega _{\,h} } \right)t\,
\quad \left( {\bmod 2\pi } \right)\quad  \Rightarrow \quad
 \rho _{\,s}  = {{\alpha _{\,s} } \over {2\pi }}
 = \left\{ {719{{\omega _{\,h} \,t} \over {2\pi }}} \right\} \hfill \cr}  \right.
$$

The hands will be on the same half circle when
$$
\left\{ \matrix{
  0 \le \alpha _{\,m}  < \pi  \hfill \cr
  0 \le \alpha _{\,s}  < \pi \; \vee  \hfill \cr
  \;\pi  + \alpha _{\,m}  \le \alpha _{\,s}  < 2\pi  \hfill \cr}  \right.\;\;
 \vee \;\;\left\{ \matrix{
  \pi  \le \alpha _{\,m}  < 2\pi  \hfill \cr
  \pi  \le \alpha _{\,s}  < 2\pi \; \vee \; \hfill \cr
  0 \le \alpha _{\,s}  < \alpha _{\,m}  - \pi  \hfill \cr}  \right.
$$
or more simply they will not be so when
$$
\left\{ \matrix{
  0 \le \alpha _{\,m}  < \pi  \hfill \cr
  \pi  \le \alpha _{\,s}  < \pi  + \alpha _{\,m} \; \hfill \cr}  \right.\;\;
 \vee \;\;\left\{ \matrix{
  \pi  \le \alpha _{\,m}  < 2\pi  \hfill \cr
  \alpha _{\,m}  - \pi  \le \alpha _{\,s}  < \pi \; \hfill \cr}  \right.
$$

The cycle will endure $12$ h, that is
$$
0 \le {{\omega _{\,h} \,t} \over {2\pi }} < 1
$$
and let's put
$$
11{{\omega _{\,h} \,t} \over {2\pi }} = k + x\quad \left| \matrix{
  \;0 \le k \in \mathbb Z \le 10 \hfill \cr
  \;0 \le x \in \mathbb R < 1 \hfill \cr}  \right.
$$

Then using  $[  ]$ to denote the Iverson bracket
and $\overline K (k,x)$ to denote the indicator function of the "forbidden" area, we get
$$
\eqalign{
  & \overline K (k,x)\quad \left| \matrix{
  \;0 \le k \le 10 \hfill \cr
  \;0 \le x < 1 \hfill \cr}  \right.\quad  =   \cr
  &  = \left[ {0 \le x < {1 \over 2}} \right]\left[ {{1 \over 2}
 \le \left\{ {{{719} \over {11}}k + {{719} \over {11}}x} \right\} < {1 \over 2} + x} \right] +   \cr
  &  + \left[ {{1 \over 2} \le x < 1} \right]\left[ {x - {1 \over 2}
 \le \left\{ {{{719} \over {11}}k + {{719} \over {11}}x} \right\} < {1 \over 2}} \right] =   \cr
  &  = \left[ {0 \le x < {1 \over 2}} \right]\left( \matrix{
  \left[ {{1 \over 2} \le \left\{ {{{719\left( {2k} \right)} \over {22}}
 + {{719} \over {11}}t} \right\} < {1 \over 2} + x} \right] +  \hfill \cr
   + \left[ {x \le \left\{ {{{719\left( {2k + 1} \right)} \over {22}} + {{719} \over {11}}x} \right\}
 < {1 \over 2}} \right] \hfill \cr}  \right) =   \cr
  &  = \left[ {0 \le x < {1 \over 2}} \right]\left( \matrix{
  \left[ {{1 \over 2} \le \left\{ {{{15\left( {2k} \right)} \over {22}} + {{719} \over {11}}x} \right\}
 < {1 \over 2} + x} \right] +  \hfill \cr
   + \left[ {x \le \left\{ {{{15\left( {2k + 1} \right)} \over {22}} + {{719} \over {11}}x} \right\}
 < {1 \over 2}} \right] \hfill \cr}  \right) \cr}
$$

Therefore the probability $\overline P$ of not having the hands on the same half circle
will be
$$
\eqalign{
  & \overline P  = {1 \over {11}}\sum\limits_{k = 0}^{10} {\int_{\,x\, = \,0}^{\,1}
 {\overline K (k,x)dx} }  =   \cr
  &  = {1 \over {11}}\sum\limits_{k = 0}^{10} {\int_{\,x\, = \,0}^{\,1/2} {\,\left( \matrix{
  \left[ {{1 \over 2} \le \left\{ {{{15\left( {2k} \right)} \over {22}}
 + {{719} \over {11}}x} \right\} < {1 \over 2} + x} \right] +  \hfill \cr
   + \left[ {x \le \left\{ {{{15\left( {2k + 1} \right)} \over {22}}
 + {{719} \over {11}}x} \right\} < {1 \over 2}} \right] \hfill \cr}  \right)dx} }  \cr}
$$

$\overline K (k,x)dx $ is represented by the segments of the two sawtooth waves intercepted

inside the respective forbidden areas like those shown in the following sketch.

Note that for better clarity  the inclination $m$ of the lines is much lower than $719/11$.

The integral translate into the sum of the $\Delta x$ of each segment.

To compute them let's start with computing the intercepts with $y=1/2$.

The minimum intercept $s_{\,0}$ will be
$$
\eqalign{
  & y = \left\{ {q + m\,x} \right\} = \left\{ {\left\{ q \right\} + m\,x} \right\}
 = {1 \over 2}\quad  \Rightarrow   \cr
  &  \Rightarrow \quad \left\{ q \right\} + m\,x = n + {1 \over 2}\quad  \Rightarrow   \cr
  &  \Rightarrow \quad m\,x = n - \left( {\left\{ q \right\} - {1 \over 2}} \right)  \cr
  &  \Rightarrow \quad m \, s_{\,0}  = \left\lceil {\left\{ q \right\} - {1 \over 2}} \right\rceil
  - \left( {\left\{ q \right\} - {1 \over 2}} \right) = \left( {{1 \over 2} - \left\{ q \right\}} \right)
 - \left\lfloor {{1 \over 2} - \left\{ q \right\}} \right\rfloor  =   \cr
  &  = \left\{ {{1 \over 2} - q} \right\} \cr}
$$
For the lower triangle we can simply replace $q$ with the negative of the intercept with $x=1/2$ , i.e.:
$$
m\,s_{\,0}  = \left\{ {{1 \over 2} + {m \over 2} + q} \right\}
$$

Next we take the upper triangle and slant it as shown

where
$$
\Delta y(x) = \left\{ {\matrix{
   {{m \over {m - 1}}x}
 & {\left| {\;0 \le x < {1 \over 2}\left( {1 - {1 \over m}} \right)} \right.}  \cr
   {m\left( {{1 \over 2} - x} \right)}
 & {\left| {\;{1 \over 2}\left( {1 - {1 \over m}} \right) \le x < {1 \over 2}} \right.}  \cr
 } } \right.
$$
so that we will compute $\Delta x$ as $ 1/m \, \Delta y$.

Finally, we are sampling the slanted triangle at points starting from $s_{\, 0}$  (which depends on $q$ and thus on $k$) and progressing with $\Delta s = 1/m$.
$$
s(k,j) = s_{\,0}  + {j \over m} = \left\{ {\matrix{
   \matrix{
  {1 \over m}\left\{ {{1 \over 2} - q} \right\} + {j \over m}
 = {{11} \over {719}}\left( {\left\{ {{1 \over 2}
 - {{15\left( {2k} \right)} \over {22}}} \right\} + j} \right) =  \hfill \cr
   = {{11} \over {719}}\left( {\left\{ {{{11 - 8k} \over {22}}} \right\} + j} \right)
 = s_{\,u} (k,j) \hfill \cr}  \hfill & {{\rm upper}\;{\rm Tr}{\rm .}} \hfill  \cr
   \matrix{
  {1 \over m}\left\{ {{1 \over 2} + {m \over 2} + q} \right\} + {j \over m}
 = {{11} \over {719}}\left( {\left\{ {{1 \over 2} + {{719} \over {22}}
 + {{15\left( {2k + 1} \right)} \over {22}}} \right\} + j} \right) =  \hfill \cr
   = {{11} \over {719}}\left( {\left\{ {{{19 + 8k} \over {22}}} \right\} + j} \right)
 = s_{\,d} (k,j) \hfill \cr}  \hfill & {{\rm lower}\;{\rm Tr}{\rm .}} \hfill  \cr
 } } \right.
$$

Since
$$
\eqalign{
  & {{11} \over {719}}\left( {\left\{ {{{11 - 8k} \over {22}}} \right\} + j} \right)
 = {1 \over {2 \cdot 719}}\left( {22\left\{ {{{11 + 14k} \over {22}}} \right\} + 22j} \right)  \cr
  & 22\left\{ {{{11 + 14k} \over {22}}} \right\}\quad  \Rightarrow \quad 11 + 14k
 = 2n + 1\quad \left( {\bmod 22} \right)\quad  \Rightarrow   \cr
  &  \Rightarrow \quad 5 + 7k = n\quad \left( {\bmod 11} \right)\quad
  \Rightarrow \quad 7k \equiv n + 5\quad \left( {\bmod 11} \right) \cr}
$$

and
$$
\eqalign{
  & {{11} \over {719}}\left( {\left\{ {{{19 + 8k} \over {22}}} \right\} + j} \right)
 = {1 \over {2 \cdot 719}}\left( {22\left\{ {{{19 + 8k} \over {22}}} \right\} + 22j} \right)  \cr
  & 22\left\{ {{{19 + 8k} \over {22}}} \right\}\quad  \Rightarrow \quad 19 + 8k
 = 2n + 1\quad \left( {\bmod 22} \right)\quad  \Rightarrow   \cr
  &  \Rightarrow \quad 9 + 4k = n\quad \left( {\bmod 11} \right)\quad  \Rightarrow
 \quad 4k \equiv n + 2\quad \left( {\bmod 11} \right) \cr}
$$

then the {Independent Residues theorem](https://en.wikipedia.org/wiki/Residue_number_system)

assures us that the the two relations above are bijections between

$$\left[ {0 \le k \le 10} \right] \Leftrightarrow \left[ {0 \le n \le 10} \right]$$

We have that the set of the sampling points are the same for both the triangles, and we can simply put

$$
\left\{ \matrix{
  s_u (n) = s_d (n) = {{2n + 1} \over {2 \cdot 719}} \hfill \cr
  0 \le s(n) < {1 \over 2}\left( {1 - {1 \over m}} \right)
 = {1 \over 2}\left( {{{708} \over {719}}} \right)\quad
  \Rightarrow \quad 0 \le n \le 353 \hfill \cr
  {1 \over 2}\left( {1 - {1 \over m}} \right) \le s(n) < {1 \over 2}\quad
  \Rightarrow \quad 354 \le 2n \le 358 \hfill \cr}  \right.
$$

And finally obtain

$$
\eqalign{
  & \overline P  = {1 \over {11}}\sum\limits_s {\Delta x(s)}
  = {1 \over {11m}}\sum\limits_s {\Delta y(s)}
  = {2 \over {11m}}\sum\limits_{\,0 \le n \le 358} {\Delta y(s_u (n))}  =   \cr
  &  = 2\left( {{1 \over {708}}\sum\limits_{\,0 \le n \le 353} {{{2n + 1} \over {2 \cdot 719}}}
  + {1 \over {11}}\sum\limits_{\,354 \le n \le 358}
 {\left( {{1 \over 2} - {{2n + 1} \over {2 \cdot 719}}} \right)} } \right) =   \cr
  &  = {1 \over {719}}\left( {{1 \over {708}}\sum\limits_{\,0 \le n \le 353} {\left( {2n + 1} \right)}
  + {1 \over {11}}\sum\limits_{\,354 \le n \le 358} {\left( {718 - 2n} \right)} } \right) =   \cr
  &  = {1 \over {719}}\left( {{{354 + 354 \cdot 353} \over {708}}
 + {{5 \cdot 718 - 2 \cdot 5 \cdot 356} \over {11}}} \right) =   \cr
  &  = {1 \over {719}}\left( {{{354} \over 2} + {{30} \over {11}}} \right)
 = {{1977} \over {7909}} = {1 \over 4}\left( {1 - {1 \over {7909}}} \right) \cr}
$$


问:
Prove that the subspace topology is the coarsest topology on $Y$ for which the inclusion function is continuous

Prove that the  subspace topology  is the coarsest topology on $Y$ for which the inclusion function $i:Y \longrightarrow X$ given by $i(y)=y$ is continuous.

Attempt

Define $T^{\prime}=\lbrace V\subset X \mid i^{-1}(V) \text{ is open in $Y$} \rbrace$
since $Y$ have the subspace topology $ i^{-1}(V)$ should look like $Y \cap U$ where $U$ is open in $X$.
By definition $i^{-1}(V)=\lbrace y\in
Y \mid i(y)\in V\rbrace=\lbrace y\in Y \mid y\in V\rbrace$ since $V\subset X$ and not necessary in all $Y$ we conclude that $T^{\prime}=\lbrace  Y\cap V  \mid  V\subset X \rbrace$ and since the subspace topology in $Y$ is given by $T=\lbrace U\cap Y \mid U\,  \text{is open in $X$} \rbrace$
And  obviously $T\subset T^{\prime}$ and therefore the subspace  topology on $Y$ is coarsest topology for which the function $i$ is continuous.

答:
Your attempt contradicts itself: you have two descriptions of $T'$ that clearly cannot be equal. On the one hand you let

$$T'=\{V\subseteq X:i^{-1}[V]\text{ is open in }Y\}\,;\tag{1}$$

clearly $X\in T'$, since $i^{-1}[X]=Y$. But later you conclude that

$$T'=\{Y\cap V:V\subseteq X\}\,,\tag{2}$$

which contains only subsets of $Y$; if $Y\subsetneqq X$, then $X$ is not in this version of $T'$, so $(1)$ and $(2)$ cannot describe the same family of sets. And there is a further problem with $(2)$: it is simply $\wp(Y)$, since every subset of $Y$ has the form $Y\cap V$ for some $V\subseteq X$.

Let $\tau_X$ be the topology on $X$, $\tau_Y$ the subspace topology on $Y$, and $\tau$ the coarsest topology on $Y$ for which $i$ is continuous; by definition

$$\tau=\left\{i^{-1}[V]:V\in\tau_X\right\}\,,$$

and you want to show that $\tau=\tau_Y$. You know that $\tau_Y=\{V\cap Y:V\in\tau_X\}$, so you want to show that

$$\left\{i^{-1}[V]:V\in\tau_X\right\}=\{V\cap Y:V\in\tau_X\}\,.$$

The natural way to try to do this is to try to prove that $i^{-1}[V]=V\cap Y$ for each $V\in\tau_X$. This is true and is not hard to prove; see if you can manage it.


问:
Map Quadrant Conformally onto the Unit Disc and find $|g'(1+i)|$.
If $w = g(z)$ maps the quadrant $\{z = x + iy \; : \; x,y >0\}$ conformally onto $|w|<1$ with $g(1) = 1$, $g(i) = -1$, and $g(0) = -i$, find $|g'(1+i)|$.

My initial reaction was to directly compute the transform and then take the derivative, i.e. set $g(z) = \frac{az + b}{cz + d}$ and use the given data to solve for the coefficients. I was expecting something similar to $\frac{z^2 - i}{z^2 + i}$ but I ended up getting $g(z) = (1+i)z - i$. This satisfies the given data but doesn't seem to map the quadrant to the unit disc...

I'm not sure if I am making some kind of algebra mistake or my approach is wrong.

答:
I figured I should answer this myself so this question can be closed.

The first step is to map the quadrant to the upper half-plane via the map $T_1(z) = z^2.$  This fixes $0$ and $1$, while mapping $i$ to $-1$.

Now we want to find a map of the form $T_2(z)=\frac{az+b}{cz+d}$ such that $T_2(0) = -i$, $T_2(-1) = -1$ and $T_2(1) = 1$.  Given the conditions we find $T_2(z) = \frac{z - i}{-iz + 1}$.

Thus, composing the above two functions we see that $g(z) = \frac{z^2 - i}{-iz^2 + 1}$.  Then $g'(z) = \frac{4z}{(-iz^2 + 1)^2}$ so that $|g'(1+i)| = \frac{4}{9}\sqrt{2}.$


问:
Is it always true that intersection of two Sylow p-subgroups is trivial?
I would like to investigate a few properties of Sylow p-subgroup, especially, when $p^n \;|\; |G|$ for $n > 1$.

If $n = 1$ and $P_1, P_2$ are Sylow $p$-subgroups, then if $id \neq g \in P_1 \cap P_2$, then $\{g, \cdots, g^p\} = P_1 = P_2$, so the proof is done. I would like to know what happens if $n \geq 2$; I cannot use the same trick as before since there's is no guarantee that a Sylow p-subgroup $P \cong \mathbb Z/p^n$

(I am aware of the result that if $P_1, P_2$ are Sylow $p$-subgroup, then $P_1$ and $P_2$ are conjugate to each other. If I were to use it, then it is sufficient to show that if $H_1, H_2 < G$ such that $H_1, H_2$ are conjugates, then $H_1 \cap H_2 = \{id\}$ or $H_1 = H_2$. But how, and is it even true? So really, I am asking two separate questions here.)

答: No, there's no reason this should be true. A simple conceptual reason is that the center $Z(G)$ may have nontrivial elements of order $p$, in which case the Sylow $p$-subgroup of $Z(G)$ is a subgroup of every Sylow $p$-subgroup. So we can consider, for example, the Sylow $2$-subgroups of $C_2 \times S_3$, which has center $C_2$, as ancientmathematician says in the comments. More generally the same argument works if $G$ has a normal $p$-subgroup.


问:
If $X\sim f$, what is the variable $f(X)$?
For some real random variables $X$ with density $f\in C^\infty(\mathbb R)$, I am looking to find the distribution of the random variable $f(X)$.
I found that
$$\mathbb P(t< f(X)\le t+h)=\mathbb E[1_{(t,t+h]}(f(X))]=\int_{-\infty}^\infty  1_{(t,t+h]}(f(x))f(x)\ dx$$
$$\approx f(t) \int_{-\infty}^\infty  1_{(t,t+h]}(f(x))\ dx$$
But now I don't know how to proceed.

Is there a way the distribution of $f(X)$ in a simple form?
Is there someone who has already studied this type of variables?

答:
From what you have already found, the probability density would then be $$\lim_{h \to 0^+}\mathbb{P}\left( t < f(X) \le t+h \right) = \lim_{h \to 0^+}\frac{1}{h}\int_{-\infty}^\infty  1_{(t,t+h]}(f(x))f(x)\ dx$$

Now because $h$ is approaching $0$, consider it to be small enough that each region such that $t < f(x) \le t+h$ is disjoint and that $f(x)$ is monotone (the second condition will be true for all $x$ where $f(x)$ isn't a local extrema or constant). Let $x_n$ be the $n$th value of $x$ such that $f(x) = t$ and $f^{-1}_n(x)$ be the inverse of $f(x)$ near $x_n$. Then the limit of the integral is $$\lim_{h \to 0^+}\frac{1}{h}\left( \sum_{n} \left| \int_{f_n^{-1}(t)}^{f_n^{-1}(t+h)}f(x)dx \right| \right) = \sum_{n}\lim_{h \to 0^+}\frac{1}{h}\left(  \left| \int_{f_n^{-1}(t)}^{f_n^{-1}(t+h)}f(x)dx \right| \right)$$

This evaluates to $$\sum_n \lim_{h \to 0^+} \frac{\left| F\left(f_n^{-1}(t+h)\right)-F\left(f_n^{-1}(t)\right) \right|}{h}$$

Then using L'Hopital's, this becomes $$\sum_n \lim_{h \to 0^+} \left|f(f_n^{-1}(t+h))f_n^{-1}{'}(t+h)\right|$$

This simplifies to $$\sum_n \left|\frac{t}{f'(x_n)}\right| = |t|\sum_n \left|\frac{1}{f'(x_n)}\right|$$


问:
Complete Noetherian local ring with residue field $\mathbb{F}_p$ is an algebra over the $p$-adic integers $\mathbb{Z}_p$
Let $R$ be a complete, Noetherian, local ring with maximal ideal $\mathfrak{m}$ and residue field $R/\mathfrak{m} \cong \mathbb{F}_p$. Note that by "complete", we mean that
$$
R \cong \varprojlim_n R/\mathfrak{m}^n.
$$

I am trying to show that $R$ is an algebra over the $p$-adic integers $\mathbb{Z}_p$.

My attempt

Since we want to construct a ring homomorphism $\mathbb{Z}_p \to R$, I would like to construct compatible homomorphisms
$$
R \to \mathbb{Z}/p^n\mathbb{Z}.
$$
which would give a cone on the inverse system used to define
$$
\mathbb{Z}_p = \varprojlim_n \mathbb{Z}/p^n\mathbb{Z},
$$
and hence induce a homomorphism $\mathbb{Z}_p \to R$ be the universal property of limits. For $n=1$, there is an obvious candidate, namely the quotient map $\varphi_1:R \to R/\mathfrak{m} \cong \mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$.

Now we would like a homomorphism $\varphi_2:R \to \mathbb{Z}/p^2\mathbb{Z}$ such that the composition
$$
R \overset{\varphi_2}{\to} \mathbb{Z}/p^2\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}
$$
is the same as $\varphi_1$. Now, I would like it to be the case that $R/\mathfrak{m}^n =\cong \mathbb{Z}/p^n\mathbb{Z}$, which would allows us to define all higher maps the same way we definded $\varphi_1$. However, I don't see why this would be the case, and I'm not sure how to proceed.

Any help would be much appreciated.

答:
As @DanielHast hints in his comment, you’re looking for a nonexistent thing. There’s no ring homomorphism from $R=\Bbb Z_p[\sqrt p\,]$ to $\Bbb Z_p$, for instance, even though this $R$ definitely is the kind of complete ring you’re considering. The maximal ideal of $R$ is $\mathfrak m=\langle\sqrt p, p\rangle$, where the angle brackets denote an enumeration of the $\Bbb Z_p$-basis. And it is not true that $R/\mathfrak m^2\cong \Bbb Z_p/(p)^2$.

What you want is a map $\Bbb Z_p\to R\,$; that’s what makes $R$ a $\Bbb Z_p$-algebra. You can construct it by taking the necessary map $\Bbb Z\to R$, and then showing that $p$-adically convergent sequences of elements of $\Bbb Z$ get carried to $\mathfrak m$-adically convergent sequences of elements of $R$. As I recall, it may be rather tiresome to verify that you’ve defined a good ring morphism $\Bbb Z_p\to R$.


问:
Proving that $f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ixy}\int_{-\infty}^{\infty}e^{-ixy}f(x)dxdy$.
I need to show that if $f\in L^2(\mathbb{R})$ then $$\displaystyle f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ixy}\left(\int_{-\infty}^{\infty}e^{-ixy}f(x)dx\right)dy$$

答:
The standard proof uses the idea of gaussian regularization. First, write

$$ \mathcal{F}\{f\}(\xi) = \hat{f}(\xi) = \int_{-\infty}^{\infty} f(x)e^{-i\xi x} \, \mathrm{d}x. $$

Since $\mathcal{F}$ is a continuous linear operator on $L^2(\mathbb{R})$, it suffices to establish the Fourier inversion formula on a dense subset  $\mathcal{D}$ of $L^{2}(\mathbb{R})$. For the purpose of  our proof, we will choose $\mathcal{D}$ such that $f \in C_b(\mathbb{R}) \cap L^{1}(\mathbb{R})$ and $\hat{f} \in L^1(\mathbb{R})$ for each $f \in \mathcal{D}$. (Note that such $\mathcal{D}$ indeed exists. For instance, we may choose $\mathcal{D}$ as the Schwartz space.)

Let $f \in \mathcal{D}$. Since $\hat{f} \in L^1(\mathbb{R})$, the dominated convergence theorem tells that

$$ \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} e^{ix\xi} e^{-\epsilon \xi^2} \hat{f}(\xi) \, \mathrm{d}\xi = \int_{-\infty}^{\infty} e^{ix\xi} \hat{f}(\xi) \, \mathrm{d}\xi. $$

On the other hand, by the Fubini's Theorem and the gaussian integral,

\begin{align*}
\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix\xi} e^{-\epsilon \xi^2} \hat{f}(\xi) \, \mathrm{d}\xi
&= \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix\xi} e^{-\epsilon \xi^2} \left( \int_{-\infty}^{\infty} e^{-i\xi y}f(y) \, \mathrm{d}y \right) \, \mathrm{d}\xi \\
&= \frac{1}{2\pi} \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} e^{i(x-y)\xi} e^{-\epsilon \xi^2} \, \mathrm{d}\xi \right) f(y) \, \mathrm{d}y \\
&= \frac{1}{\sqrt{4\epsilon\pi}} \int_{-\infty}^{\infty} e^{-\frac{(y-x)^2}{4\epsilon}} f(y) \, \mathrm{d}y
\end{align*}

Substituting $y=x+\sqrt{4 \epsilon \pi} \, t$, it follows that

\begin{align*}
\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix\xi} e^{-\epsilon \xi^2} \hat{f}(\xi) \, \mathrm{d}\xi
&= \int_{-\infty}^{\infty} e^{-\pi t^2} f(x+\sqrt{4 \epsilon \pi} \, t) \, \mathrm{d}t
\end{align*}

Since $f$ is bounded and continuous, this converges to

$$ \int_{-\infty}^{\infty} e^{-\pi t^2} f(x) \, \mathrm{d}t = f(x) $$

as $\epsilon \to 0^+$ by the dominated convergence theorem. This proves that

$$ \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix\xi} \hat{f}(\xi) \, \mathrm{d}\xi = f(x) $$

for all $f \in \mathcal{D}$ and $x \in \mathbb{R}$, hence establishing the Fourier inversion formula over $\mathcal{D}$. Now as mentioned in the beginning, this extends to all of $L^2(\mathbb{R})$ via the continuity of $\mathcal{F}$.


问:
Partitioning a compact and totally disconnected space
Let $X$ be a compact, Hausdorff, totally disconnected space, i.e. a compact space that has a basis consisting of clopen subsets. Let also $F\subset C(X)$ be finite and $\varepsilon>0$. I am trying to show that there exist disjoint, clopen subsets $X_1,\dots, X_r$ such that $X=\cup_{j=1}^rX_j$ and such that for each $j=1,\dots,r$ there exists a point $x_j\in X$ so that $|f(x)-f(x_j)|<\varepsilon$ for all $x\in X_j$ and all $f\in F$.

I am not sure on how to do this. On the one hand, we may set $U_x=\{y\in X: |f(x)-f(y)|<\varepsilon \text{ for all }f\in F\}$. Obviously $U_x$ are open and $X=\cup_{x\in X}U_x$, so by compactness we may reduce the cover to a finite subcover. But then I do not know how to bring into play the total disconnectedness. On the other hand, starting from total disconnectedness, I can write $X$ as a finite, disjoint union of clopen subsets. But then I do not know how prove the existence of these magical points. Any ideas?

答: Start with your sets $U_x$; since $X$ is zero-dimensional, $U_x$ contains a clopen nbhd of $x$, so you might as well assume that the sets $U_x$ themselves are clopen. Now take a finite subcover $\{U_1,\ldots,U_n\}$ and refine it to a disjoint clopen cover of $X$: let $V_1=U_1$, $V_2=U_2\setminus V_1$, $V_3=U_3\setminus(V_1\cup V_2)$, and so on.


问:
Probability of picking a real number randomly
If we randomly pick a real number from the number line, the probability of picking a number (say x) is 0. This is true for all real numbers x and it makes sense to me why this must be true. But seemingly there is a paradox lurking here. Suppose we pick the number y. This number also had probability 0 but was still chosen in the random process. If it was chosen, it means the probability of it being chosen was finite. Isn't this a paradox? What am I missing here?</p>

答:
Probability zero does not mean impossible. Just as probability 1 doesn't mean guaranteed to happen.

A different example of the same phenomenon is this: Flip a coin until you get a heads. The probability that you get a heads at some point and therefore stop is 1. But it clearly isn't completely, absolutely guaranteed to happen. The probability that you keep getting tails indefinitely and never stop is 0. But it clearly isn't impossible. (This is actually almost equivalent to picking a real number uniformly at random from the interval $[0,1)$ and asking whether the chosen number was $0$.)


问:
Special numbers in patterns and the reasons they are special
I know there are several big list questions out there (e.g. Patterns that break down at certain numbers) that touch on classifications of mathematical structures where certain numbers don't fit in, but I'd like to find more explanations on why these numbers are special.

For instance, all automorphisms of symmetric groups are conjugacies, except for $n=6$, where the conjugacies form a subgroup of index two. The reason 6 is special is that the number of partitions of six distinct objects into two sets of three is the same as the number of partitions into three sets of two.

As another example, 3-manifold groups are special because in higher dimensions any group can be a manifold group (by Dehn surgery), but we can still embed all graphs into all 3-manifolds (which makes them more interesting than 2-manifolds).

Some special numbers I'd like to see explanations for are:

Fermat's last theorem: where does the proof break down for $n=2$? (I know there are obvious counterexamples, but why do the techniques used require $n>2$)?

Exotic smooth structures in dimension 4.

The finitely many quadratic extensions of the rationals that are not Euclidean or ufd's, etc.

答: There's a question over at MathOverflow in this vein, focusing on why characteristic $2$ is so special.  I think that you might like some of the answers there.</p>


问:
vertices on a path
Let $G = (V,E)$ be a graph on at least $3$ vertices so that for every vertex $v \in V, G\backslash v$ is connected. Prove that for any vertices $u$ and $v,$ there is a cycle containing $u$ and $v$.

I would prefer not to use Menger's Theorem (if so, then I could claim that since the minimum separator for $u$ and $v$ has length at least $2$, there are $2$ vertex-disjoint disjoint paths from $u$ to $v$ and hence $u$ and $v$ are contained in the same cycle). I think I can show that each edge of $G$ is contained in a cycle. I'm thinking of showing by induction on the length of a shortest path $P$ from $u$ to $v$ that every vertex $P$ in the path is in a cycle containing $u$. For the inductive step, I suppose the shortest path would have length $0$. Then $u=v,$ which is part of an edge in $G$ and thus contained in a cycle. Assume that for all shortest paths between $u$ and $v$ of length less than $k\geq 1,$ every vertex in the path is contained in a cycle that contains $u$. I'm not sure how to proceed with the inductive step.

答:
This is a nice proof due to the (undergraduate) book by Bondy and Murty (Theorem 3.2)

I will use the fact that for a nontrivial graph, if the graph is $2$-connected, then the graph is at least $2$-edge-connected; this means that if the graph cannot be disconnected by removing a single vertex, then it cannot be disconnected by removing a single edge either (this is good to prove as an exercise).

We will use induction on $d(u,v)$. I will use the base case of $d(u,v) = 1$; if $d(u,v) = 1$, then if $e$ is the edge joining $u$ and $v$, $G-e$ is connected; so there is a $(u,v)$-path in $G-e$ which, together with $e$, gives a cycle containing $u$ and $v$.

Now we will assume that this holds for any two vertices at distance less than $k$, and assume that $d(u,v) = k \geq 2$. Consider a $(u,v)$-path of length $k$, and let $w$ be the vertex that preceeds $v$ on this path. Now $d(u,w) = k-1$, so there is a cycle containing $u$ and $w$; we will consider this cycle $P \cup Q$ where $P$ and $Q$ are internally disjoint $(u,w)$-paths. Furthermore, $G-w$ is connected so there is a $(u,v)$ path $P^{\prime}$ not containing $w$. Let $x$ be the last vertex in $P^{\prime}$ that is also in $P \cup Q$; without loss of generality we can assume that $x \in P$.

Now we get a cycle containing $u$ and $v$ by following $P$ from $u$ to $x$; following $P^{\prime}$ from $x$ to $v$; the edge $vw$; and then $Q$ from $w$ to $u$ (technically I guess we are traversing the path $Q$ backwards).

Title: vertices on a path


问:
A holomorphic function which vanishes at a point can be written as a product
This is from a proof in a textbook but the authors don't justify this particular step (restated):

Let $f(z)$ be a holomorphic function in some disk $D$ with $f(a) = 0$ for some $a \in D$.
Then there exists a holomorphic function $g(z)$ on $D$ such that $f(z) = (z-a)g(z)$. Why?

答: Since $f$ is holomorphic around $a$, it equals its own Taylor series at $a$. Since $f(a)=0$, the $(x-a)^0$ or constant term is zero. Dividing the Taylor series by $(x-a)$, then, yields the necessary $g(x)$.


答: $f(x)/(x-a)$ has a singularity at $x=a$, but that singularity is removable.  Removing it gives you $g$.

Title: A holomorphic function which vanishes at a point can be written as a product

Body:
This is from a proof in a textbook but the authors don't justify this particular step (restated):

Let $f(z)$ be a holomorphic function in some disk $D$ with $f(a) = 0$ for some $a \in D$.
Then there exists a holomorphic function $g(z)$ on $D$ such that $f(z) = (z-a)g(z)$. Why?


答:
Infinity doesn't work like you think it does.  There are several ways to interpret probability.  One of those ways is the "frequentist" interpretation - you divide the number of outcomes you're looking for by the number of outcomes possible, and that's the probability.  The issue is that, although that's fine for finite sets of outcomes, it doesn't work when there are infinitely many outcomes (or, more accurately, it always produces zero).

In your interpretation, you are really asking "what is the probability that the number is $y$ now that I've already selected it, and the answer is 100%.  That isn't surprising - there is one outcome and you've chosen it.

However, from the frequentist perspective, what was the probability that you would select $y$ and not, say, $y + \delta$, for any real $\delta$?  Well, there are two issues:

There are infinitely many reals on any interval, so technically this is $\frac{y}{\infty}$, which is always zero

Worse, the size of the interval doesn't seem to matter.  Whether we choose some extremely small interval or an extremely large one, the size is still infinite.

It is #2 above that breaks the frequentist interpretation entirely unless the probability of a particular outcome is always zero.  Basically, the probability of a particular outcome in the frequentist interpretation must change (or stay at zero and only zero) if the set of possible outcomes expands.  Specifically it must reduce, presuming that the outcomes are independent.  You cannot have a probability less than 0, so it works out.

So instead we don't talk about probability of a given number.  We talk about the probability that the outcome will be within some interval, which does have a very well defined interpretation from the perspective of the size of the various sets.

Title: Probability of picking a real number randomly

Body: If we randomly pick a real number from the number line, the probability of picking a number (say x) is 0. This is true for all real numbers x and it makes sense to me why this must be true. But seemingly there is a paradox lurking here. Suppose we pick the number y. This number also had probability 0 but was still chosen in the random process. If it was chosen, it means the probability of it being chosen was finite. Isn't this a paradox? What am I missing here?</p>


问:
Show that a set is not a complex vector space
Let vector space $W$ be $\{p(z)\in P_{3}(\mathbb{C}):p(a)=0 \Rightarrow a\in \mathbb{R}\}$ under usual addition and usual scalar multiplication.

How to show that W is not a vector space over $\mathbb{C}$?

It should violate one or more of the 10 axioms of complex vector space, but I am stuck here, I tried and it all seems that 10 axioms holds, how can W not be an complex vector space?

答: Both $p(x)=x^2$ and $q(x)=1$ belong to $W$. However, $p(x)+q(x)=x^2+1\notin W$, since its roots are $\pm i$, none of which is real.


问:
Are vectors equal when their product is equal?
When we have such a simple outer product on two sides of the equation as below:

$aa^T$ = $bb^T$

where a and b are vectors, does it necessarily means a=b?

答: No.  Here is a counter example: $a = (1,-1)$ and $b = (-1,1)$.  Then $aa^T = bb^T$ but $a \ne b$.


问:
Flux of a vector field through a sphere
What is the flux of the vector field $\mathbf{f}(\mathbf{r})=(x- y^2z, ysinz, cosz)$ through the sphere $x^2+y^2+z^2=4$?

What I have done:

$\nabla\cdot \mathbf{f}= 1+ sinz -sinz= 1 $

I know that the flux is $\iiint\limits_{V}\nabla\cdot \mathbf{f} dxdydz= \iiint\limits_{V}1 dxdydz $. But I don't know what limits to use.

答:
hint

$$\iint_S \vec{f}\cdot \vec{ds}=$$
$$\iiint_V \nabla\cdot \vec{f} dv=$$
$$\iiint_Vdxdydz=$$
volume of the sphere
$$=\frac 43 \pi (\;radius\;)^3$$

but, the equation of your sphere is
$$x^2+y^2+z^2=4=R^2$$

Title: Flux of a vector field through a sphere


问:
Rolling dice until 6 appears, X is a number of trials, while Y is the number of odds in the resulting sequence.
The main question, which I do not understand is that how we can get conditional expectation of $E(Y \mid X)$ and what is the probability $P(Y=y \mid X)$.
Idea to use binomial for the probability $Y \sim \text{Binom}(X,3/6)$

答:
Given that $X$ was realised as a natural number $x$, you rolled the die $x-1$ times and failed to get a six those $n-1$ times, i.e. all those rolls were from one to five. In that range are three odd and two even numbers, so $Y$ follows the binomial distribution with $x-1$ trials and success probability $\frac35$:
$$P(Y=y\mid X=x)=\binom{x-1}y(3/5)^y(2/5)^{x-y-1}$$
The conditional expectation is $E(Y\mid X=x)=\frac35(x-1)$.


问:
Number of ways for making a set of signature list
A meeting is being held in a big hall which has $7$ doors by which the mourners enter the hall. One can use any one of the $7$ doors to enter and can come at any time during meeting.At each door a register is kept in which a mourner has to put down his signature while entering

If $200$ people attend the meeting the no. Of different sets of $7$ lists that can arise

The problem can be solved by considering that each mourner has 7 options to choose from but the rank or order in which $200$ people sign also matters . What is an easy way to solve both the tasks I.e check the order of every mourner eith condition that each has $7$ options.

答:
We can arrange the mourners in a list in $200!$ ways.  Now divide the list into $7$ sections, and assign the first section to the first guest book, the second section to the second guestbook, and so on.  To split the list into $7$ sections, we need to insert $6$ dividers into the list, and there are $201$ places we can put a divider -- to the left of any person, or to the right of all the persons.  If none of the dividers could be placed in the same spot, we could say that there are $$200!\binom{201}{6}$$ possibilities, this isn't so.  We can choose the same position for a divider more than once, creating an empty list.

So, it comes down to this: how many ways are there to choose $6$ items from $201$ if the order of choosing doesn't matter and we can choose the same item more than once?  But this is stars and bars: how many ways are there to place $6$ identical balls in $201$ distinct buckets?  Therefore, the solution is $$200!\binom{206}6$$


问:
Use Lambert W Function to Solve $\frac\alpha\beta\ln(x)-x+c=0$
I need to solve this equation for x:

$\frac{α}{β}\ln(x)-x+c=0$

Apparently it has to be done with the Lambert W function. I think the answer is

$x=\frac{α\operatorname{W}(-\frac{βe^-\frac{Cβ}{α}}{α})}{β}$

But I don't know how to get to the answer.

Thanks in advance!

答:
$ \frac{\alpha}{\beta}\ln{x}-x+c=0 $

$ \frac{\alpha}{\beta}\ln{x} = x-c $

$ \ln{x} = \frac{\beta}{\alpha}(x-c) $

$ x = e^{\frac{\beta}{\alpha}(x-c)} $

$ xe^{-\frac{\beta}{\alpha}x} = e^{-\frac{\beta}{\alpha}c} $

$ -\frac{\beta}{\alpha}xe^{-\frac{\beta}{\alpha}x} = -\frac{\beta}{\alpha}e^{-\frac{\beta}{\alpha}c} $

$ W(-\frac{\beta}{\alpha}xe^{-\frac{\beta}{\alpha}x}) = W(-\frac{\beta}{\alpha}e^{-\frac{\beta}{\alpha}c}) $

$ -\frac{\beta}{\alpha}x = W(-\frac{\beta}{\alpha}e^{-\frac{\beta}{\alpha}c}) $

$ x = -\frac{\alpha}{\beta}W(-\frac{\beta}{\alpha}e^{-\frac{\beta}{\alpha}c}) $


问:
How to rewrite the lower bound and upper bound of the double sum to accommodate a constraint?
How to rewrite the upper bound and lower bound of this double sum:

$$\sum\limits_{K = 0}^{K = M - 1} {\sum\limits_{L = 0}^{L = M - 1} {f\left( {{x_L},{x_K}} \right)} }$$

When the summation process can only be proceeded under the constraint $K + 1 > L$ ?

Since $M$ could be very large, for example the number of antenna in Massive MIMO system dropping the conditional checking of $K + 1 > L$ can save a little bit time.

Thank you for your enthusiasm !

答:
You can write it as

$$\sum_{L=0}^{M-1}\sum_{K=L}^{M-1}f(x_L,x_K)$$

or as

$$\sum_{K=0}^{M-1}\sum_{L=0}^Kf(x_L,x_K)\,;$$

both are equivalent to

$$\sum_{0\le L\le K\le M-1}f(x_L,x_K)\,.$$


问:
Box Topology Name
To put it simply, why is the box topology on an infinite product of topological spaces called the box topology? What is "boxy" about it?</p>

答:
Say we have a product $\prod_{a\in A}X_a$ of topological spaces $X_a$ for some index set $A$. Take open subsets $U_a\subseteq X_a$ for all $a\in A$. Then the "box" $\prod_{a\in A} U_a$ is open in the box topology (in fact, these sets form the standard basis for the box topology).

Think $X_a=\Bbb R$ with standard topology and $U_a$ being intervals, and the geometric analogy is complete.

Title: Box Topology Name


问:
Prove that the dimension of $X$ is finite

Let $X$ be a normed space and suppose there are $n$ points $\lbrace x_{k} : 1\leqslant k\leqslant n\rbrace$
and a number $r>1$ such that
$$B(0,r) \subset \bigcup_{k=1}^n B(x_{k},1)$$
Prove that the dimension of $X$ is finite.

I have tried using Riesz's theorem, we know that if the closed unit ball is compact by this theorem we have that the dimension of $X$ is finite, but I don't know how to prove the compacity of the closed unit ball.

答:
Let $$r'=\frac{1+r}{2}$$
so that $1 < r' < r$, and let $0 < \varepsilon < r'-1$.

Let $V = \mathrm{Vect}(x_1,..., x_n)$, and suppose that $X \neq V$.

Then there exists $y \in X \setminus V$. Let $d=d(y,V)$ the distance from $y$ to the subspace $V$ : because $V$ is finite-dimensional, hence it is closed, hence $d>0$. By definition of $d$, there exists $z \in V$ such that
$$d \leq ||y-z|| \leq (1+\varepsilon)d$$

Let $$z_0 = r'\frac{y-z}{||y-z||}$$

Now, for all $x \in V$, one has
$$\left|\left| x-z_0\right|\right| = \left|\left| x-r'\frac{y-z}{||y-z||}\right|\right|   = r'\frac{\left|\left|(x\frac{||y-z||}{r'}+z)-y\right|\right|}{||y-z||}$$

But $$x\frac{||y-z||}{r'}+z \in V$$ so the numerator is greater than $d$ ; and the denominator is less than $(1+\varepsilon)d$ by definition : you deduce $$\left|\left| x-z_0\right|\right| \geq \frac{r'}{1+\varepsilon}$$

In particular, $$d(z_0, V) \geq \frac{r'}{1+\varepsilon} > 1$$ by definition of $\varepsilon$. But by hypothesis, because $||z_0||=r' < r$, then $z_0 \in B(0,r)$, so there exists $i \in \lbrace 1, ..., n \rbrace$ such that $z_0 \in B(x_i, 1)$, i.e. such that $d(z_0, x_i) < 1$. So one must also have $d(z_0, V) < 1$. Contradiction.

So the hypothesis $X \neq V$ is not true, so $X=V$ is a finite-dimensional vector space.