问:
Trouble completing derivation using Milne-Thompson Circle Theorem with a Vortex
I am having trouble finishing the derivation using the Milne-Thompson Circle Theorem (MTCT) for a potential flow with a vortex at a location $\zeta_v$.

My understanding of the MTCT is that the complex potential $w_v$ (with subscript $v$ for vortex) transforms accordingly when a circle is present,

$$w_{vc}(\zeta)=w_v(\zeta)+\overline{w_v\left(\frac{a^2}{\overline{\zeta}}\right)}$$ or equivalently $$w_{vc}(\zeta)=w_v(\zeta)+\overline{w_v}\left(\frac{a^2}{\zeta}\right).$$

My complex velocity is $$w_v(\zeta)=\frac{\Gamma}{2\pi i}\ln(\zeta-\zeta_v)$$ where $\zeta_v$ is the position of the vortex. I know the final complex velocity with the circle $w_{vc}$ is $$w_{vc}(\zeta)=\frac{\Gamma}{2\pi i}\left[\ln(\zeta-\zeta_v)-\ln\left(\zeta-\frac{a^2}{\overline{\zeta_v}}\right)+\ln(\zeta)\right].$$ The furthest I've been able to get is the following, $$\frac{\Gamma}{2\pi i}\left[\ln(\zeta-\zeta_v)-\ln\left(a^2\left(\frac{1}{\zeta}-\frac{1}{\zeta_v}\right)\right)\right]$$ and I'm not sure if I made a mistake or if I just can't figure what to do with this form of the equation.

I think some of my confusion may be because of notation. My textbook by Batchelor says the MTCT uses $\overline{w\left(\frac{a^2}{\zeta}\right)}$ where elsewhere it is $\overline{w\left(\frac{a^2}{\overline{\zeta}}\right)}=\overline{w}\left(\frac{a^2}{\zeta}\right)$ which I guess are identical. Although, the answers to these two questions (1, 2) seem to contradict that but I'm probably misunderstanding something.

Thank you very much.

答:
I can't reconcile the last term, but here's what I have so far:

$w_{vc}(\zeta) = w_v(\zeta)+\bar{w}_v(\frac{a^2}{\zeta})$

$=\frac{\Gamma}{2\pi i}[ln(\zeta-\zeta_v)-ln(\frac{a^2}{\zeta}-\bar{\zeta}_v)]$

Since:

$ln(\frac{a^2}{\zeta}-\bar{\zeta}_v)=ln(\zeta-\frac{a^2}{\bar{\zeta}_v})+ln(-\frac{\bar{\zeta}_v}{\zeta})=ln(\zeta-\frac{a^2}{\bar{\zeta}_v})+ln(-\bar{\zeta}_v)-ln(\zeta)$

We have that:

$w_{vc}(\zeta) = \frac{\Gamma}{2\pi i}[ln(\zeta-\zeta_v)-ln(\zeta-\frac{a^2}{\bar{\zeta}_v})+ln(\zeta)-ln(i\bar{\zeta}_v)-\frac{i\pi}{2}]$

Maybe the last terms go away because it's constant, and these are periodic functions? I also could have messed up somewhere; I'm not the best with complex logarithms.


问:
Question Regarding Composition of Functions
Here is an exercise from my textbook that I have been struggling to solve for a while now.

Suppose $f : X \longrightarrow Y$ is called good if for any function $g : Z \longrightarrow Y$ with $|Z| \leq 13$ there exists an injection $h : Z \longrightarrow X$ such that $g = f \circ h$. How can we describe the set of positive numbers $n$ for which there exists a good function $f : [1000] \longrightarrow [n]$?

I would appreciate any help on this!

答:
The first step is to show that

$f$ is good if and only if for every $y \in Y$, the preimage $f^{-1}(y)$ has at least $13$ elements.

Proof: "If" part: let $g:Z \rightarrow Y$ be any function with $|Z| \leq 13$.

For each $y \in Y$, we have $|g^{-1}(y)|\leq 13 \leq |f^{-1}(y)|$, hence there exists an injection $h_y: g^{-1}(y) \rightarrow f^{-1}(y)$.

Since $Z$ is the disjoint union of all $g^{-1}(y)$, the $h_y$'s glue to a well-defined function $h: Z\rightarrow X$. It is easy to see that it satisfies the willing properties.

"Only if" part: if there exists $y\in Y$ such that $|f^{-1}(y)| < 13$, then the constant function $g:[13] \rightarrow Y$ sending every element to $y$ causes a problem: no possible $h$ can exist, as $|f^{-1}(y)| < |g^{-1}(y)|$.

From this, it is clear that

for positive integers $m, n$, there exists a good function $f:[m]\rightarrow [n]$ if and only if $m \geq 13 n$.

And the final answer is "the set of positive integers $\leq 76$".

ContentLicense: CC BY-SA 4.0


问:
Laurent Series at Infinity
I thought that finding the Laurent series was something that was straightforward, however, I am having some difficulty of finding the Laurent series of

$$f(z) = \frac{1}{z(1-z)}$$

for $z= \infty$.  Any suggestions?

答:
Using generalized Laurentseries with two poles
$$\sum _{k=0}^{n-1} x^k-\frac{x^n}{x-1}+\frac{1}{x}$$

ContentLicense: CC BY-SA 3.0

Title: Laurent Series at Infinity


问:
Calculate distance from plane to parallel plane in O using vector and normal
I'm trying to figure out what's the best method to get the distance between two planes where i have the normalized vector of the plane and a point in the plane. What I want to do is to create a parallel plane where 0,0,0 is in and then calculate the distance between the two planes.
My thought was that it's actually the same like the distance between the plane and 0,0,0 which I tried just to calculate as the dot product between the normalized vector and the given point.

e.g.: (The values are that strange because they are from a 3D camera system)

```<code>{ 0.00102464424908972; -1.5806471007608384E-4; -0.0014898099362794807} // normalized vector (plane normal)
{ 1.2168973210867913; -0.2862541104934054; -0.39146720489920095} //given point
0.0018753452463538794  //calculated distance  ?
</code>```

I'm not sure but this looks not like the result I expected. Is it the right formula? I tried to use the Hessesche Normalform

Update: somehow I droped my normalize function and the normal was not normalized. now my results look much better. Thankf for all answers.

答:
Computing the dot product between the plane normal and the position vector of the given point strikes me as the right solution.

If you imagine the plane and the position vector $x$ from the origin to the given point of the plane, you can drop an orthogonal projection of the origin onto the plane to form a triangle (with the given point and the origin.) The length of the side perpendicular to the plane (=parallel to the plane normal $n$) is exactly $|(x\cdot n)|$.


问:
Those fields where all elements can be generated from 1 by addition, subtraction and multiplication.
Consider a field $(F,+,-,*,0,1)$. Let $S$ be the smallest subset of $F$ that is closed under $1$,$+$,$-$, and $*$. Assuming that $S=F$, must $F$ be a finite field? In fact, must $F$ be finite of prime number cardinality?

答:
[Comment promoted to answer, at suggestion of OP]

Consider the sequence $$1,1+1,1+1+1, 1+1+1+1,\dots$$ If zero never shows up in this sequence, then the closure (under addition, subtraction, and multiplication) is the integers, not a field. If zero does show up, let $n$ be the smallest number of $1$s adding up to zero. Then $n$ must be prime (since, if $n=rs$, the distributive law shows that zero is the product of the sum of $r$ ones and the sum of $s$ ones, and the nonexistence of zero divisors in a field implies $\{\,r,s\,\}=\{\,1,n\,\}$), hence the closure is a field with a prime number of elements.


问:
How a linear function is bounded below on $R^n$ only when it is zero and not -$\infty$?
I’ve gone through these two links

Why is a linear function bounded below on $R^n$ only when it is zero?

Bounded linear function implication

But I’m not able to understand why there is a bound. Can’t a function be unbounded from below?
According to 2nd link the author has considered a ray. What if it’s a line with range $R$ , as function is defined as $f : \mathbb{R}^m \to \mathbb{R}$ , hence large -ve input value means large -ve Output, how it’s bounded by zero?

Extremely sorry it’s difficult to digest.

答:
Suppose $f(x)= y \neq 0$ for some $x\in\mathbb{R}^n$.

Now suppose we have a bound $B$ such that we know $f(a) > B$ for all $a\in\mathbb{R}^n$. But by linearity we have $f(\frac{B}{y}x) = \frac{B}{y}f(x) = \frac{B}{y}y = B$, which is a contradiction.


答: I think you have misunderstood the question: the point is that the zero function is the only linear function on $\mathbb{R}^n$ which is bounded below: that is, any non-zero linear function is not bounded below, as you say.

Title: How a linear function is bounded below on $R^n$ only when it is zero and not -$\infty$?

Body:
I’ve gone through these two links

Why is a linear function bounded below on $R^n$ only when it is zero?

Bounded linear function implication

But I’m not able to understand why there is a bound. Can’t a function be unbounded from below?
According to 2nd link the author has considered a ray. What if it’s a line with range $R$ , as function is defined as $f : \mathbb{R}^m \to \mathbb{R}$ , hence large -ve input value means large -ve Output, how it’s bounded by zero?

Extremely sorry it’s difficult to digest.


问:
SVD for Non-Square matrices?
Is QR decomposition a pre-requisite for SVD decompostion of non-square matrices? I have been unable to find a clear cut answer to this question. I will be grateful for a response.</p>

答: If your question is whether in order to compute the singular value decomposition of a given matrix, one needs to compute QR decompositions I believe the answer is no. One can use them to this end but they are not necessary, see here.</p>

Title: SVD for Non-Square matrices?


问:
Prove that $\exists X_0\in \mathbb{R}$ such that $\bigcap_{n=1}^{\infty}I_{n}=\{X_{0}\}$
Let $I_{n}=[a_{n},b_{n}$] for $n\in \mathbb{N}$, such that $I_{n+1} \subseteq I_{n}$ also let $\inf\{b_{n}-a_{n}|n \in \mathbb{N}\}=0.$  Prove that exists $X_0\in \mathbb{R}$ such that $\bigcap_{n=1}^{\infty}I_{n}=\{X_{0}\}.$

I've tried to prove that $\bigcap_{n=1}^{\infty}I_{n} \neq \emptyset$ with the nested interval theorem and then take 2 points and show these points are the same, but I can't. HELP! :(

答: $I_{n+1}\subset I_n$ means $[a_{n+1},b_{n+1}]\subset[a_n,b_n]$ and thus $a_{n+1}\geqslant a_n$ and $b_{n+1}\leqslant b_n$ for all $n\geqslant 1$. $(a_n)_{n\geqslant 1}$ is an increasing sequence, $(b_n)_{n\geqslant 1}$ is a decreasing sequence, thus $(b_n-a_n)_{n\geqslant 1}$ is a decreasing sequence, its limit is $\inf_{n\geqslant 1}(b_n-a_n)=0$. Thus the sequences $(a_n)$ and $(b_n)$ are adjacent and converge towards the same limit $\ell$, and $\ell$ is such that $a_n\leqslant\ell\leqslant b_n$ for all $n\geqslant 1$, this means that $\displaystyle\ell\in\bigcap_{n\geqslant 1}I_n$ and thus the intersection is not empty. Now if $\displaystyle\ell,\ell'\in\bigcap_{n\geqslant 1}I_n$, then $\ell,\ell'\in I_n$ for all $n\geqslant 1$ and thus $|\ell-\ell'|\leqslant b_n-a_n$ for all $n\geqslant 1$. Letting $n\rightarrow +\infty$ gives $\ell=\ell'$.


问:
Convergence in distribution and independence
Consider two sequences of random variables $\{X_n\}_{n \in \mathbb{N}}$ and $\{Y_n\}_{n \in \mathbb{N}}$ such that $X_n \rightarrow_{d} X$ and $Y_n \rightarrow_{d} Y$ and $X_n$ is independent of $Y_n$ for every $n \in \mathbb{N}$. Does $(X_n, Y_n) \rightarrow_d(X,Y)$?

If not can someone suggest a counter example?

答:
As @WoolierThanThou mentioned in the comment, the statement could be wrong without assuming the independence of $X$ and $Y$. Here is a counterexample.

Let $X \equiv Y \sim N(0, 1)$, and $X_n$ be a sequence of $N(0, 1)$ random variables that are independent of $X$, $Y_n$ be a sequence of $N(0, 1)$ random variables that are independent of $\{X_n\}$ and $X$. Then trivially $X_n \to_d X$ and $Y_n \to_d Y$.

Consider the set $A = \{(x, y): x \leq 0, y \leq 0\}$ with $\partial A = \{(x, y): x = 0, y \leq 0\} \cup \{(x, y): x \leq 0, y = 0\}$. Since $P[(X, X) \in \partial A] = 2P[X = 0, X \leq 0] - P[X = 0, X = 0] = 0$, $A$ is a continuity set. On the other hand, since $(X_n, Y_n) \sim N_2(0, I_{(2)})$, we have
\begin{align*}
P[(X_n, Y_n) \in A] = P[X_n \leq 0]P[Y_n \leq 0] = \frac{1}{2}\times\frac{1}{2}
= \frac{1}{4} \not\to P[(X, X) \in A] = P[X \leq 0] = \frac{1}{2},
\end{align*}
showing that $(X_n, Y_n) \not\to_d (X, X)$.


答: We know $\cap_{n=1}^\infty I_n \neq \emptyset$. Suppose there are two points, $x_1$ and $x_2$ in this set. Clearly, $a_n\leq x_1 \leq b_n$ and $a_n\leq x_2 \leq b_n$ for all $n$. Hence, $|x_1-x_2|\leq b_n-a_n$ for all $n$. If $x_1 \neq x_2$, then $|x_1-x_2|>0$. Where's the contradiction? Or you can just prove that $x_1=x_2$ directly using squeeze theorem.

Title: Prove that $\exists X_0\in \mathbb{R}$ such that $\bigcap_{n=1}^{\infty}I_{n}=\{X_{0}\}$

Body:
Let $I_{n}=[a_{n},b_{n}$] for $n\in \mathbb{N}$, such that $I_{n+1} \subseteq I_{n}$ also let $\inf\{b_{n}-a_{n}|n \in \mathbb{N}\}=0.$  Prove that exists $X_0\in \mathbb{R}$ such that $\bigcap_{n=1}^{\infty}I_{n}=\{X_{0}\}.$

I've tried to prove that $\bigcap_{n=1}^{\infty}I_{n} \neq \emptyset$ with the nested interval theorem and then take 2 points and show these points are the same, but I can't. HELP! :(


问:
Ambiguous multiple choice question
So the question is if $f'(c)$ does not exist and $f(x)$ is continuous then $f(x)$ may have

a) a local extremum at $x=c$

b) a point of inflection at $c$

c) neither a nor b is possible

d) both a or b are possible

I know a is not possible because $f'(c)$ would have to equal to $0$. But is it required that the first derivative to exist to have a point of inflection.

答:
Consider the example $$f(x) = |x|, \quad c = 0.$$  Then $f'(0)$ does not exist because $$\lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = 1,$$ but $$\lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = -1.$$  Yet $f(0) = 0$ is a local (and global) extremum because for all $x \in \mathbb R$, $f(x) \ge 0$ with equality attained at $x = 0$.

Now consider the example $$f(x) = x^{1/3}, \quad c = 0.$$  Then $$f'(x) = \frac{1}{3} x^{-2/3}$$ and $f'(0)$ does not exist.  Yet $x = 0$ is an inflection point because $$f''(x) = -\frac{2}{9} x^{-5/3}$$ which is negative for $x > 0$ and positive for $x < 0$.

Title: Ambiguous multiple choice question


问:
CFG for language
I can't find CFG for this language.

$N = \{w ∈ \{0, 1\}^* \mid \text{w contains more } 1\text{s than }0 \text{s}\}$

Basically, it is much easier when order is set (for example, $1$s come after $0$s) but in this situation I got stuck.

I can make a PDA for that but I'm not strong at converting PDA to CFG.

Thanks in advance.

答:
Simple solution

$$S\to E|E1S\\
E \to 0E1E|1E0E|\varepsilon$$

It's clear that anything generated by $E$ will verify $|u|_0 = |u|_1$ and therefore that anything generated by $S$ will verify $|u|_0 \le |u|_1$.

Claim A: $u\in \mathcal L(E)\iff |u|_0 = |u|_1$

Proof A: By induction on $|u|$.

Since $u\in \mathcal L(E)\iff \overline{u}\in \mathcal L(E)$ (where $\overline{u}$ is $u$ with $0$ and $1$ swapped), we can assume WLOG that $u=1v$. Take $w_1$ the greatest prefix of $v$ so that $|w_1|_0=|w_1|_1$. It exists because $\varepsilon$ is one such prefix. Write $v=w_1w_2$ where $w_2$ is just whatever remains. Since we took $w_1$ the greatest with the property, any prefix of $w_2$ has more $0$s than $1s$. In particular, $w_2$ starts with a $0$. So $w_2=0h$. $w_1$ and $h$ both have as many $0$s as $1$s so by the induction hypothesis, $E\to^*w_1$ and  $E\to^*h$. So $E\to0E1E\to^*0w_11h=u$ and we're done.

Claim B: $u\in\mathcal L(S)\iff |u|_0\le |u|_1$

Proof B: Left to the reader. One direction come directly from claim A and the other is by induction on $|u|_1-|u|_0$ (and by taking the greatest prefix for which $|u|_0= |u|_1$ to find the part that comes from $E$).

$\Sigma:= \{0,1\}$

If you want to have more $1$s that $0$s, you just have to make sure that whenever you add a $1$ you add a $0$.

$$S \to E|E1T\\
E \to DE|UE|\varepsilon\\
T \to UT|1T|\varepsilon\\
O \to 0 | 1OO\\
I \to 1|0II\\
\\
D \to 0I\\
U \to 1O$$

This is done with $I$ that means "add one more $1$ than $0$s" (except that it only generates words for which this is false for all their prefixes). Similarly $O$ means "add one more $0$ than $1$s". $U$ therefore means "bounce upwards and come back to where you were" and $D$ "bounce downwards and come back to where you were". $E$ means "bounce upwards or downwards and come back to where you were as many times as you want". $T$ means "(Go one step up or bounce upwards and come back to where you were) as many times as you like". With this interpretation, it feels plausible that $S$ generates what we want because either you take $E$ and it means that you'll have equality, or you take $E1T$ and you'll get strict inequality because once you're done with $E$, you go up once and then $T$ doesn't let you go down so you'll be strictly above the $|v|_1-|v|_0=0$ line.

Here is an example of how $|v|_1-|v|_0$ can evolve with the prefix $v\le u$.

```<code>111000000110100111111000111001110

                               /\
  /\                /\    /\  /
 /  \              /  \  /  \/
/____\____________/____\/_______________________________
      \          /
       \  /\/\  /
        \/    \/
1I----0I----------1O----11O--111O
U-----D-----------U-----1U---11U-
E-----------------------1T-------
</code>```

I'll remove $U$ and $D$ from the grammar and will prove it for this one but it should be clear that those two grammars are equivalent.

$$S \to E|E1T\\
E \to 0IE|1OE|\varepsilon\\
T \to 1OT|1T|\varepsilon\\
O \to 0 | 1 OO\\
I \to 1|0II$$

Claim 1: Define $H(u)$:$|u|_0 > |u|_1$ and for any strict prefix $v<u$, $|u|_0 \le |u|_1$. Then for any $u\in\Sigma^*$,$u\in \mathcal L(O) \iff H(u)$. Note that $H(u)$ implies $H'(u)$:"$|u|_0=|u|_1+1$" and that the last letter of $u$ is $0$.

Proof 1:

"$\Rightarrow$": Let $P(n)$: "If $O\to^m u\in \Sigma^*$ for some $m\le n$, then H(u)". We prove $P(n)$ by induction on $n\ge 1$.

If $O\to^1 u\in \Sigma^*$, then the production must be $O\to 0$ so that $u=0$ which verifies the properties.

If $0\to^{n+1} u \in \Sigma^*$ for $n>0$, then there is $v\in (\Sigma\sqcup \Gamma)^*$ so that $I\to v \to^n u$. The first production can only be $O\to 1OO$ so that $v=1OO$. By the fundamental lemma, $u$ can be written as $u=u_1u_2u_3$ so that $1\to^{n_1} u_1$, $O\to^{n_2} u_2$, $O\to^{n_3} u_3$, $n=n_1+n_2+n_3$. Then $H(u)$ follows from $H(u_2)$ and $H(u_3)$ (which are given by the induction hypothesis because $\forall i, n_i \le n$) and $u=0u_2u_3$.

"$\Leftarrow$": Let $P(n)$: "For any $u\in \Sigma^*$ so that $H(u)$, $u\in \mathcal L(O)$". We prove $P(n)$ by induction on $n$.

$P(0)$ is vacuously true.

Suppose $P(n)$. Let $u\in \Sigma^*$ so that $|u|=n+1$ and $H(u)$. Then $u=av$ for some $a\in \Sigma$ and $v\in\Sigma^*$.

If $a=0$, then $a|w_2|_1$. Take $p<w_2$. Then $1w_1p<u$ so that $|w_1|_1+1+|p|_0=|w_1|_0+|p|_0=|1w_1p|_0\le |1w_1p|_1=1+|w_1|_1+|p|_1$ and we indeed have $|p|_0\le |p|_1$.

Claim 2: Define $\overline{H}(u)$:$|u|_1 > |u|_0$ and for any strict prefix $v<u$, $|u|_1 \le |u|_0$. Then for any $u\in\Sigma^*$,$u\in \mathcal L(I) \iff \overline{H}(u)$. Note that $\overline{H}(u)$ implies $\overline{H}'(u)$:"$|u|_1=|u|_0+1$" and that the last letter of $u$ is $1$.

Proof 2: Similar to proof 1.

Claim 3: $u\in \mathcal L(T)\iff$ for any prefix $v\le u$, $|v|_1 \ge |v|_0$.

Proof 3:

"$\Rightarrow$": Follows from claim 1 by induction on the size of the derivation.

"$\Leftarrow$": Take $u\in\Sigma^*$ so that for any prefix $v\le u$, $|v|_1 \ge |v|_0$. First note that the first letter of $u$ must be $1$ (if $u$ is not $\varepsilon$). We will prove it by induction on $|u|$.

If $u=\varepsilon$, we have $T\to u$.

If it exists, take $v$ to be the smallest prefix of $u$ so that $|v|_0=|v|_1$. Then $v=1v'$. Since we took the smallest prefix, we have $H(v')$ and we therefore have $O\to^*v'$ by claim 1. Write $u=vw$ for some $w$. Then by the induction hypothesis, $T\to^* w$ and then $T\to^*1OT\to^*1v'w=u$.

If it doesn't exist, write $u=1v$. The fact that the prefix doesn't exist implies that $v$ also verifies the property. By the induction hypothesis, we therefore have $T\to^*v$. And we conclude that $T\to 1T\to^*1v=u$.

Claim 4: $u\in\mathcal L(E) \iff |u|_0=|u|_1$

Proof 4: Left to the reader. One implication follows directly from claim 1 and 2 and the other implication is a case analysis on the first letter.

Claim 5: $u\in \mathcal L(S)\iff |u|_0\le |u|_1$

Proof 5: Left to the reader. One implication follows directly from claims 3 and 4 and the other one has two cases: $|u|_0=|u|_1$ or not, and if not $u$ is factored into $u_11u_2$ where $u_1$ is the greatest prefix so that $|u_1|_0=|u_1|_1$.

Bonus claim: The grammar is non-ambiguous.

Bonus proof: Left to the reader. The idea is that the languages generated by $E$ and $E1T$ will be disjoint so you only have to check that each part is non-ambiguous separately. The part of the grammar used by $E$ is $LL(1)$ so it's non-ambiguous. Then for a word $u$, suppose you can write it as $e_11t_1=u=e_21t_2$ with $E\to^*e_i$ and $T\to^*t_i$. $e_i$ is the longest prefix $v<u=e_i1t_i$ so that $|v|_0=|v|_1$ because $|e_i|_0=|e_i|_1$ and for each prefix $w<t_i$,  $|w|_0\le|w|1$. This implies that $e_1=e_2$ and therefore that $t_1=t_2$. By a previous argument, there is only one derivation tree for $e_1$ so all that remains to be checked is that there is only one for $t_1$, or in other words that the part of the grammar used by $T$ is non-ambiguous. It is because you can only take $T\to1OT$ if you're going to come back to the level you are at and $T\to1T$ if you're not.

Title: CFG for language

Body:
Simple solution

$$S\to E|E1S\\
E \to 0E1E|1E0E|\varepsilon$$

It's clear that anything generated by $E$ will verify $|u|_0 = |u|_1$ and therefore that anything generated by $S$ will verify $|u|_0 \le |u|_1$.

Claim A: $u\in \mathcal L(E)\iff |u|_0 = |u|_1$

Proof A: By induction on $|u|$.

Since $u\in \mathcal L(E)\iff \overline{u}\in \mathcal L(E)$ (where $\overline{u}$ is $u$ with $0$ and $1$ swapped), we can assume WLOG that $u=1v$. Take $w_1$ the greatest prefix of $v$ so that $|w_1|_0=|w_1|_1$. It exists because $\varepsilon$ is one such prefix. Write $v=w_1w_2$ where $w_2$ is just whatever remains. Since we took $w_1$ the greatest with the property, any prefix of $w_2$ has more $0$s than $1s$. In particular, $w_2$ starts with a $0$. So $w_2=0h$. $w_1$ and $h$ both have as many $0$s as $1$s so by the induction hypothesis, $E\to^*w_1$ and  $E\to^*h$. So $E\to0E1E\to^*0w_11h=u$ and we're done.

Claim B: $u\in\mathcal L(S)\iff |u|_0\le |u|_1$

Proof B: Left to the reader. One direction come directly from claim A and the other is by induction on $|u|_1-|u|_0$ (and by taking the greatest prefix for which $|u|_0= |u|_1$ to find the part that comes from $E$).

$\Sigma:= \{0,1\}$

If you want to have more $1$s that $0$s, you just have to make sure that whenever you add a $1$ you add a $0$.

$$S \to E|E1T\\
E \to DE|UE|\varepsilon\\
T \to UT|1T|\varepsilon\\
O \to 0 | 1OO\\
I \to 1|0II\\
\\
D \to 0I\\
U \to 1O$$

This is done with $I$ that means "add one more $1$ than $0$s" (except that it only generates words for which this is false for all their prefixes). Similarly $O$ means "add one more $0$ than $1$s". $U$ therefore means "bounce upwards and come back to where you were" and $D$ "bounce downwards and come back to where you were". $E$ means "bounce upwards or downwards and come back to where you were as many times as you want". $T$ means "(Go one step up or bounce upwards and come back to where you were) as many times as you like". With this interpretation, it feels plausible that $S$ generates what we want because either you take $E$ and it means that you'll have equality, or you take $E1T$ and you'll get strict inequality because once you're done with $E$, you go up once and then $T$ doesn't let you go down so you'll be strictly above the $|v|_1-|v|_0=0$ line.

Here is an example of how $|v|_1-|v|_0$ can evolve with the prefix $v\le u$.

111000000110100111111000111001110

                               /\
  /\                /\    /\  /
 /  \              /  \  /  \/
/____\____________/____\/_______________________________
      \          /
       \  /\/\  /
        \/    \/
1I----0I----------1O----11O--111O
U-----D-----------U-----1U---11U-
E-----------------------1T-------

I'll remove $U$ and $D$ from the grammar and will prove it for this one but it should be clear that those two grammars are equivalent.

$$S \to E|E1T\\
E \to 0IE|1OE|\varepsilon\\
T \to 1OT|1T|\varepsilon\\
O \to 0 | 1 OO\\
I \to 1|0II$$

Claim 1: Define $H(u)$:$|u|_0 > |u|_1$ and for any strict prefix $v<u$, $|u|_0 \le |u|_1$. Then for any $u\in\Sigma^*$,$u\in \mathcal L(O) \iff H(u)$. Note that $H(u)$ implies $H'(u)$:"$|u|_0=|u|_1+1$" and that the last letter of $u$ is $0$.

Proof 1:

"$\Rightarrow$": Let $P(n)$: "If $O\to^m u\in \Sigma^*$ for some $m\le n$, then H(u)". We prove $P(n)$ by induction on $n\ge 1$.

If $O\to^1 u\in \Sigma^*$, then the production must be $O\to 0$ so that $u=0$ which verifies the properties.

If $0\to^{n+1} u \in \Sigma^*$ for $n>0$, then there is $v\in (\Sigma\sqcup \Gamma)^*$ so that $I\to v \to^n u$. The first production can only be $O\to 1OO$ so that $v=1OO$. By the fundamental lemma, $u$ can be written as $u=u_1u_2u_3$ so that $1\to^{n_1} u_1$, $O\to^{n_2} u_2$, $O\to^{n_3} u_3$, $n=n_1+n_2+n_3$. Then $H(u)$ follows from $H(u_2)$ and $H(u_3)$ (which are given by the induction hypothesis because $\forall i, n_i \le n$) and $u=0u_2u_3$.

"$\Leftarrow$": Let $P(n)$: "For any $u\in \Sigma^*$ so that $H(u)$, $u\in \mathcal L(O)$". We prove $P(n)$ by induction on $n$.

$P(0)$ is vacuously true.

Suppose $P(n)$. Let $u\in \Sigma^*$ so that $|u|=n+1$ and $H(u)$. Then $u=av$ for some $a\in \Sigma$ and $v\in\Sigma^*$.

If $a=0$, then $a<u$ would lead to $1=|0|_0\le |0|_1=0$. So we have $0=a=u$ the derivation $O\to u$ works.

If $a=1$, then we take $w_1$ to be the shortest prefix of $v$ so that $H(w_1)$ and we take $w_2$ so that $u=1w_1w_2$ holds. Since $H(u)$ and $H(w_1)$, we also have $H'(u)$ and $H'(w_1)$. This gives us $H'(w_2)$ and so $|w_2|_0>|w_2|_1$. Take $p<w_2$. Then $1w_1p<u$ so that $|w_1|_1+1+|p|_0=|w_1|_0+|p|_0=|1w_1p|_0\le |1w_1p|_1=1+|w_1|_1+|p|_1$ and we indeed have $|p|_0\le |p|_1$.

Claim 2: Define $\overline{H}(u)$:$|u|_1 > |u|_0$ and for any strict prefix $v<u$, $|u|_1 \le |u|_0$. Then for any $u\in\Sigma^*$,$u\in \mathcal L(I) \iff \overline{H}(u)$. Note that $\overline{H}(u)$ implies $\overline{H}'(u)$:"$|u|_1=|u|_0+1$" and that the last letter of $u$ is $1$.

Proof 2: Similar to proof 1.

Claim 3: $u\in \mathcal L(T)\iff$ for any prefix $v\le u$, $|v|_1 \ge |v|_0$.

Proof 3:

"$\Rightarrow$": Follows from claim 1 by induction on the size of the derivation.

"$\Leftarrow$": Take $u\in\Sigma^*$ so that for any prefix $v\le u$, $|v|_1 \ge |v|_0$. First note that the first letter of $u$ must be $1$ (if $u$ is not $\varepsilon$). We will prove it by induction on $|u|$.

If $u=\varepsilon$, we have $T\to u$.

If it exists, take $v$ to be the smallest prefix of $u$ so that $|v|_0=|v|_1$. Then $v=1v'$. Since we took the smallest prefix, we have $H(v')$ and we therefore have $O\to^*v'$ by claim 1. Write $u=vw$ for some $w$. Then by the induction hypothesis, $T\to^* w$ and then $T\to^*1OT\to^*1v'w=u$.

If it doesn't exist, write $u=1v$. The fact that the prefix doesn't exist implies that $v$ also verifies the property. By the induction hypothesis, we therefore have $T\to^*v$. And we conclude that $T\to 1T\to^*1v=u$.

Claim 4: $u\in\mathcal L(E) \iff |u|_0=|u|_1$

Proof 4: Left to the reader. One implication follows directly from claim 1 and 2 and the other implication is a case analysis on the first letter.

Claim 5: $u\in \mathcal L(S)\iff |u|_0\le |u|_1$

Proof 5: Left to the reader. One implication follows directly from claims 3 and 4 and the other one has two cases: $|u|_0=|u|_1$ or not, and if not $u$ is factored into $u_11u_2$ where $u_1$ is the greatest prefix so that $|u_1|_0=|u_1|_1$.

Bonus claim: The grammar is non-ambiguous.

Bonus proof: Left to the reader. The idea is that the languages generated by $E$ and $E1T$ will be disjoint so you only have to check that each part is non-ambiguous separately. The part of the grammar used by $E$ is $LL(1)$ so it's non-ambiguous. Then for a word $u$, suppose you can write it as $e_11t_1=u=e_21t_2$ with $E\to^*e_i$ and $T\to^*t_i$. $e_i$ is the longest prefix $v<u=e_i1t_i$ so that $|v|_0=|v|_1$ because $|e_i|_0=|e_i|_1$ and for each prefix $w<t_i$,  $|w|_0\le|w|1$. This implies that $e_1=e_2$ and therefore that $t_1=t_2$. By a previous argument, there is only one derivation tree for $e_1$ so all that remains to be checked is that there is only one for $t_1$, or in other words that the part of the grammar used by $T$ is non-ambiguous. It is because you can only take $T\to1OT$ if you're going to come back to the level you are at and $T\to1T$ if you're not.


问:
Matrix sends a vector to a scalar multiple of its complex conjugate
Does anyone know any methods for solving the following for $\mathbf{v}$:

$\mathbf{A\bar{v} = \lambda v}$

Ideas:

Maybe multiplying by some matrix takes the complex conjugate

This is like a weird eigenvector problem where instead of sending the vector to a scalar multiple of itself, it sends it to a scalar multiple of its complex conjugate

What if we force $\mathbf{v} = e^{i\mathbf{k}}$ for some $\mathbf{k}$? Does this help?

答: $e^{i \mathbf k}$ doesn't make sense if $\mathbf k$ is a vector. $e^x$ is defined for rings, but unless there is some multiplicative operator defined on the vector space that $\mathbf k$ is in, that space is not a ring. We can define $e^{iA}$, since matrices are a ring, but I don't think that helps. Now, if we take $\mathbf v = e^{ix}\mathbf u $ for some real-valued vector $\mathbf u$ and scalar $x$, we can say $\mathbf A e^{-ix}\mathbf u = e^{ix}\lambda \mathbf u$. Scalars commute with matrices, so we have $ e^{-ix}\mathbf A\mathbf u = e^{ix}\lambda \mathbf u$ and therefore $ \mathbf A\mathbf u = e^{2ix}\lambda \mathbf u$, which shows that $\mathbf u$ is an eigenvector of  $\mathbf A$ with eigenvalue $e^{2ix}\lambda $. So that implies that there is a real-valued eigenvector $ \mathbf u$, and we need to take its eigenvalue, write it as $e^{2ix}\lambda $, and then take $\mathbf v = e^{ix}\lambda \mathbf u $. But if an eigenvalue can be written as $e^{2ix}\lambda $,  then it also can be written as $e^{2(x+\pi)i}\lambda $, giving $\mathbf v = e^{(x+\pi)i}\lambda \mathbf u $.


问:
Example of point where f'(x) <0 and f''(x) = 0 but not point of inflection
$x^4=0$ is a example of a curve where there is a point that has $f''(x)=0$ but is not a point of inflection (POI). This point is a stationary point with $f'(x)=0$.

I am looking for a example of a curve where there is a point where $f''(x)=0$ and $f'(x) < 0$ (non stationary point) and the point is not a POI.

答:
Take $f(x) = x^4-x$, at $x=0$. We have $f’(x)=4x^3-1$, so $f’(0)=-1\lt 0$. And $f’’(x)=12x^2$, which is nonnegative everywhere and $f’’(0)=0$. It is not a point of inflection, since the function is always concave up.

Of course, $f(x)=x^4+x$ will give you a similar example with $f’(0)\gt 0$.

Yves Daoust’s suggestion in the comments of $f(x)=-x$ is of course also an example that meets your required conditions, though I thought you might object to that since it is “flat”...


答:
The function $f(x)=x^4-x$ has a second derivative of zero at $x=0$. However only having the second derivative being zero is not sufficient to be an inflection point. In order for the point to be an inflection point the sign of the second derivative must change when passing through the point. If we take our example function, it's second derivative is:

$$f''(x) = 12x^2$$

And this function is always positive, which means that the second derivative never crosses the $x$ axis, and only touches it at $x=0$.

Title: Example of point where f'(x) <0 and f''(x) = 0 but not point of inflection

Body:
$x^4=0$ is a example of a curve where there is a point that has $f''(x)=0$ but is not a point of inflection (POI). This point is a stationary point with $f'(x)=0$.

I am looking for a example of a curve where there is a point where $f''(x)=0$ and $f'(x) < 0$ (non stationary point) and the point is not a POI.


问:
Differential equations of first order?
I have the equation y' - 2xy =2x*e^(x2) (1)
To solve it,I do y' -2xy=0 then lny=x^2 +c so e^(x^2 +c)=y
I find y' here so I have y'=(2x+c') * e^(x^ 2 +c)
I replace this in equation (1) and I have 2x*e^(x^2+c) +c'* e^(x^2+c)-2x*e^(x^2+c)=2x*e^(x^2).. the problem is that now I have to find c' here but when I do,c doesnt simplify..where am I wrong?</p>

Title: Differential equations of first order?

Body: $y'-2xy=2xe^{x^2}$   multiply with $m=e^{\int -2xdx}=e^{-x^2}$ now you have $e^{-x^2}(y'-2xy)=2xe^{x^2}e^{-x^2}$ or $(ye^{-x^2})'=2x$  or $ye^{-x^2}=x^2+K$ which gives $y=(x^2+K)e^{x^2}$


答:
Consider everything to be over a $2n$-dimensional real vector space, in the obvious way, ordering the entries as (real part of the first entry), (imaginary part of the first entry), (real part of the second entry), etc.

Then we can replace $\mathbf{A}$ with a new matrix $r(\mathbf{A})$ in the obvious way, $\mathbf{v}$ by $r(\mathbf{v})$ (in the same obvious way), $\bar{\mathbf{v}}$ by $\mathbf{B}r(\mathbf{v})$, where $$\mathbf{B} = \bigoplus^n \left(\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right),$$ and $\lambda$ by a matrix $$r(\lambda) = \bigoplus^n\left(\begin{array}{cc}\mathfrak{Re}(\lambda) & \mathfrak{Im}(\lambda)\\-\mathfrak{Im}(\lambda) & \mathfrak{Re}(\lambda)\end{array}\right).$$

That then gives us a new equation of the form $(r(\mathbf{A})\mathbf{B})r(\mathbf{v}) = r(\lambda)r(\mathbf{v})$, which you can then solve in the usual way (getting eigenvalues/eigenvectors for $r(\mathbf{A})\mathbf{B}$). The eigenvectors so produced are the $r(\mathbf{v})$ for $\mathbf{v}$ solving the original equation, which you can recover by simply taking each $k$th pair $(x_k,y_k)$ of non-overlapping consecutive entries of each $r(\mathbf{v})$ and letting your $k$th entry of $v$ be $x_k + iy_k$.

Title: Matrix sends a vector to a scalar multiple of its complex conjugate

Body:
Does anyone know any methods for solving the following for $\mathbf{v}$:

$\mathbf{A\bar{v} = \lambda v}$

Ideas:

Maybe multiplying by some matrix takes the complex conjugate

This is like a weird eigenvector problem where instead of sending the vector to a scalar multiple of itself, it sends it to a scalar multiple of its complex conjugate

What if we force $\mathbf{v} = e^{i\mathbf{k}}$ for some $\mathbf{k}$? Does this help?


问:
Proving the matrix $A$ is diagonizable without finding its eigenvectors
I need to show that $A$ is a diagonizable matrix:
$$A \in M_{n\times n}(\mathbb{R}) = \begin{bmatrix}0 & a & b\\ a & 0 & b\\ b & a & 0\end{bmatrix}; \space a\neq b; \space a,b \neq 0$$

Since I can't find the eigenvectors I think the best way to approach this is to see if, with the help of the characteristic polynomial, I can prove that $A$ has three different eigenvalues.

$$|A - \lambda I_{n}| = 0 \Leftrightarrow$$
$$\Leftrightarrow -\lambda (\lambda^{2} - ab) -a (-a \lambda -b^{2}) + b(a^{2} + b\lambda) = 0 \Leftrightarrow$$
$$ \Leftrightarrow -\lambda^{3} + (ab + a^{2} + b^{2})\lambda + ab^{2} + a^{2}b = 0$$

But I didn't come to any conclusion... Maybe I'm missing on how to factorize this polynomial, or maybe there is a better approach to the problem... Can some one guide me?

答:
The statement is wrong with $a = 1, b = -2$, whose Jordan form is
\begin{align*}
\begin{pmatrix}
-1 & 1 & 0 \\
0  & -1 & 0 \\
0  & 0  & -2
\end{pmatrix},
\end{align*}
hence it's not diagonalizable.


问:
If $A$ is a $2\times 2$ non-diagonal diagonalizable matrix then have two distinct eigenvalues
Let $A$ be a $2\times 2$ non-diagonal real matrix. I know that if it has two distinct real eigenvalues then it is diagonalizable. I would like to know if the reverse is also true, that is, matrix $A$ is digitizable if it has two distinct (real) eigenvalues.

Obviously if I remove the hypothesis to be non-diagonal, then it is not true, as can be seen in these other questions. Q1 and Q2

I tried to create an example of a non-diagonal matrix with equal eigenvalues that is diagonalizable, but I couldn't. But, in this comment, it is said that my statement is true,but I couldn't prove it.

I thought this theorem could help me, but I was unable to apply it in the direction I want.

Let $T:V\to V$  be a linear operator, in a space of finite dimension and let $\lambda_1,\cdots \lambda_t$ be its distinct eigenvalues. The following statements are equivalent:

A is diagonalizable

$p_T(x)=(x-\lambda_1)^{n_1}\cdots (x-\lambda_t)^{n_t},n_i\ge 1$ and $\gamma_{T}(\lambda_i)=\mu_{T}(\lambda_i)$

$\dim V=\sum_{i=1}^{t}\dim Aut_A(\lambda_i)$

where $\gamma_{T}(\lambda_i)$ and $\mu_{T}(\lambda_i)$ are geometric
multiplicity and algebraic multiplicity, respectively.

Is it possible to prove that if $A$, (not diagonal) is diagonalizable, then it has two distinct real eigenvalues with the theorem above? Or does anyone have any suggestions for another useful result.

答:
Suppose $A$ is non-diagonal and diagonalizable. By contradiction, suppose also that it has only one eigenvalue $\lambda$ repeated two times.

Since $A$ is diagonalizable, you have a base-change matrix $M$ such that $A=MDM^{-1}$ with $D$ a diagonal matrix containing the eigenvalue. Since $A$ has only one eigenvalue with algebraic multiplicity 2, then $D=\lambda I$. As a consequence,
$$A = M \lambda IM^{-1} = \lambda MM^{-1} = \lambda I$$
so $A$ is diagonal. Contradiction.


问:
Is there a general notion of nonabelian cech cohomology?
Diving into the theory of vector bundles and characteristic classes, I've often come across the introduction of $\check{H}^1(\mathcal U, Gl(n))$, where $\mathcal U = \{U_i\}_i$ is an open cover of a space $X$, as the orbit space of coboundaries $(\delta_i)_i$ acting on cocycles $(g_{ij})_{ij}$ by morphing each $g_{ij}$ to $\delta_i \,g_{ij}\, \delta_j^{-1}$.  However, this seems like an isolated workaround to the fact that the coboundary map $(\delta_i)_i\mapsto (\delta_i\delta_j^{-1})_{ij}$ may not be a group homomorphism in general, which is due to $GL(n)$ being nonabelian for $n>1$.

Of course, the outlined approach works to construct $\check H^1(\mathcal U, G)$ for any nonabelian group $G$.

But is there a reasonable definition for the higher cohomology classes?

Please bear in mind that I know nothing about algebro-geometrical methods such as sheaf cohomology, if that's relevant.

答:
There are no higher cohomology classes if $G$ is nonabelian. $H^1(X, G)$ computes (under mild hypotheses) the set of homotopy classes of maps $X \to BG$ where $BG$ is the classifying space of $G$. When $G$ is abelian $BG$ can be modeled as a topological abelian group so itself has a classifying space $B^2 G$, so this construction can be iterated, which is the abstract reason why higher cohomology exists; under mild hypotheses $H^n(X, A)$ computes the set of homotopy classes of maps $X \to B^n A$.

When $G$ is nonabelian this is (generally) impossible, morally because of the Eckmann-Hilton argument, although there are subtleties involving the possibility that, say, a nonabelian group is contractible and so "homotopy abelian." In order for $G$ to have a double classifying space $B^2 G$, $G$ must be a double loop space, or equivalently what is called a grouplike $E_2$-algebra, and most nonabelian topological groups won't have this property.


问:
Convergent sequence bounded by other sequence
I never know what to do (or how to start) in proofs.

"Let $u_n, v_n$ be two sequences in $\mathbb{R}$ such that $u_n < v_n < u_{n+1}$. Show that if $v_n$ is bounded, then it converges."

答:
Start by writting down the given info and have a feel of what it happening.

If $v_n$ is bounded then there must be $A, B \in \mathbb{R}$ such that $A \leq v_n \leq B$. But for all $n \in \mathbb{N}$ it is true that $u_n < v_n$, therefore $u_n < v_n \leq B$ and $u_n$ is also upper bounded.

Now see that if $u_n < v_n < u_{n+1}$ then $u_n < u_{n+1}$ meaning $u_n$ is increasing - so it has a lower bound: $u_1$. By the same triangular inequality you have that $v_{n-1} < u_n < v_n$ and $v_n$ is also increasing. This would complete the proof because $v_n$ is both bounded and monotonic - so it must converge.

Another way to finish it would be to state that since $u_n$ is bounded and monotonic (as we showed), then $u_n$ converges and has a limit $L$ such that:
$$\lim u_n = \lim u_{n+1} = L$$

Hence
$$u_n < v_n < u_{n+1}$$
$$\lim u_n < \lim v_n < \lim v_{n+1}$$
$$L < \lim v_n < L$$

You can see that $\lim v_n = L$ which is a finite number, so $v_n$ must converge.


问:
Suppose $n\mathbb{Z} = \{0, n,−n, 2n,−2n, . . .\}$ and $I_n = n\mathbb{Z} \cap [1, 1000]$.
Question: Suppose $n\mathbb{Z} = \{0, n,−n, 2n,−2n, . . .\}$ and $I_n = n\mathbb{Z} \cap [1, 1000]$. How many elements are contained in the
set $I_6 \cup I_{15} \cup I_{25}$?

Answer: The answer is $226$.

Here is what I tried. Is this correct?

$\left \lfloor{\frac{1000}{6}}\right \rfloor=166$

$\left \lfloor{\frac{1000}{15}}\right \rfloor=66$

$\left \lfloor{\frac{1000}{25}}\right \rfloor=40$

$166+66+40=272$

$\left \lfloor{\frac{1000}{\text{lcm}(6,15)}}\right \rfloor=\left \lfloor{\frac{1000}{30}}\right \rfloor=33$

$\left \lfloor{\frac{1000}{\text{lcm}(6,25)}}\right \rfloor=\left \lfloor{\frac{1000}{150}}\right \rfloor=6$

$\left \lfloor{\frac{1000}{\text{lcm}(15,25)}}\right \rfloor=\left \lfloor{\frac{1000}{75}}\right \rfloor=13$

$33+6+13=52$

$\left \lfloor{\frac{1000}{\text{lcm}(6,15,25)}}\right \rfloor=\left \lfloor{\frac{1000}{150}}\right \rfloor=6$

$272-(52-6)=272-46=226$.

答: The number of elements in $I_n$, which we will denote by $|I_n|$, is uniquely determined by the value of its largest element, which in turn is the largest integer multiple of $n$ not exceeding $1000$.  In particular, $$|I_n| = \left\lfloor \frac{1000}{n} \right\rfloor$$  so that we have $$|I_6| = 166, \\ |I_{15}| = 66, \\ |I_{25}| = 40.$$  But we would be double-counting some of these elements since for instance $(6)(15) = 90$ is included in both $I_6$ and $I_{15}$.  Adjusting for these leads us to the inclusion-exclusion formula $$|I_6 \cup I_{15} \cup I_{25}| = |I_6| + |I_{15}| + |I_{25}| - |I_6 \cap I_{15}| - |I_{15} \cap I_{25}| - |I_{25} \cap I_6| + |I_{6} \cap I_{15} \cap I_{25}|.$$  Since $$I_a \cap I_b = I_c$$ where $c = \operatorname{lcm}(a,b)$, we also need to compute $$I_{6} \cap I_{15} = I_{30}, \\ I_{15} \cap I_{25} = I_{75}, \\ I_{25} \cap I_6 = I_{150}, \\ I_{6} \cap I_{15} \cap I_{25} = I_{150}.$$  Consequently $$|I_6 \cup I_{15} \cup I_{25}| = 166 + 66 + 40 - (33 + 13 + 6) + 6 = 226.$$


问:
Homeomorphism betwen Unit circle in the Taxicab Geometry to the Unit circle in the Euclidean Geometry
I need make a homeomorphism to the unit circle in the Taxicab Geometry to the Unit circle in the Euclidean Geometry, but I can't find information about a parametrization to the unit circle in the Taxicab Geometry</p>

答:
Take a point $x$ on the boundary of the taxicab's unit ball. Draw a line from the origin through $x$. It meets the boundary of the circle at some unique point $f(x)$. Prove that $x\to f(x)$ is a homeomorphism.

Added: Actually $f(x)=\frac{x}{\|x\|}$ where $\|x\|$ is the Euclidean norm, and this same formula provides a homeomorphism between the boundary of the unit ball of any norm and the corresponding boundary of the Euclidean unit ball.


问:
How to write a compact list terms through set notation?
I would like to ask a question about a mathematical notation used in economics, because I'm not an expert on it. I'm studying an economic model in which I understood measure theory is used, but I'm not sure because it has not been told. There are definitions in which there are summations for every element of a particular set.

For example, the definition of the composite good, one of the goods which are consumed in the model, which has n regions, is the following, in the slides

$$M_{r}=\Big [\sum_{s\in R} \int_{i \in N_s} [q_{sr}(j)]^{(\sigma -1)/\sigma} dj\Big ]^{\sigma/(\sigma-1)}$$

r, is the region in which the good is consumed, s is the region in which the good is produced, j is the variety of the good, R is the set of regions, $N_s$ is the mass of firms - I know there is a continuum of firms - and I have a question: if I want to write all the composite goods consumed in all regions, can I write under the M  $r \in R $ and under the summation sign $s,r \in R$? I mean, am I writing all the composite goods with this notation, or am I doing an exception with mathematics just to get to the point?

I'd like just to write all the R composite goods in the correct way.

Thanks in advance

答:
Personally, I'd write $$\sum\limits_{r \in R}M_r =\sum\limits_{r \in R}\left[\sum_{s \in R}\int\limits_{i \in N_s}[q_{sr}(j)]^{(\sigma - 1)/\sigma}dj\right]^{\sigma/(\sigma - 1)}$$
but I don't know if there's some other standard notation used in the field - it may well be that there's some other name given to this thing (and if I was writing it a lot, I'd probably come up with a name for it just to avoid having to deal with $\sum_{r\in R}M_r$ every time).

However, there are a couple of things which I can say for certain:

$M_{r\in R}$ doesn't work unless you define it first: what does that mean? How are we combining all of these things? I don't know, unless you tell me (unless this is standard in the field, in which case just follow the standard).

Combining the two sums definitely doesn't work, even if you define it: this is a problem of mathematics, rather than notation: if we throw out the extraneous part by  assuming $\sigma/(\sigma - 1) = 2$ and that $R$ only has two elements (neither of these actually matter - the problem appears regardless, except in a handful of essentially silly cases), then what you're after is something of the form

$$(a + b)^2 + (c + d)^2$$ (where $a$ and $b$ are the values of the two integrals with $r$ equal to our first value, and $c$ and $d$ the other two).

However, if you write the sums combined, what you've written down is instead $$(a + b + c + d)^2,$$ which is not the same thing at all (for example, if $a = b = c = 1$, then the first gives us $2^2 + 2^2 = 8$, whereas the second gives us $4^2 = 16$).


问:
What is the probability of a sampled election to be wrong?
Suppose that only a subset of people who wants to vote is allowed to. The sampling of voters is fair, and anyone has the same chance of being selected to vote.

Suppose $n$ people are allowed to vote, sampled from a very large population $P$ of interested voters such as $|P| \gg n$. There are only two candidates. For a given election result, what is the probability of it being correct? I.e. what is the probability the result would be the same if everybody was allowed to vote?

I wrote a python simulator that says that (using $n = 10000$) if one candidate wins with 52% of the votes, the chance of the election being right is 99.98%. Get better if the margin is bigger, worse if the margin is smaller. This is the output of one execution:

```<code>50 % : 7076 / 10181 , ( 69.50201355466064% )
51 % : 18932 / 19755 , ( 95.83396608453556% )
52 % : 20104 / 20108 , ( 99.98010741993237% )
53 % : 19913 / 19913 , ( 100.0% )
54 % : 20101 / 20101 , ( 100.0% )
  ... (lines omitted, all the ~20000 executions for each result nailed with 100%)
99 % : 19576 / 19576 , ( 100.0% )
100 % : 10191 / 10191 , ( 100.0% )
</code>```

I am pretty sure there is formula to calculate it: for a given $n$ and result, what is the probability the result is correct?

What about considering $|P|$ as well?

EDIT:
It was implicit, but I want to make it explicit: every voter supports one or the other candidate, thus will not abstain.

答:
After briefly reviewing sampling distribution, as K.defaoite suggested, I dismissed it because I got the feeling that it either was not going the right way to answer my question, or the road to get there would be too long. But it pointed me to binomial distribution, what pointed to hypergeometric distribution, and after some thinking, I believe I got the right answer. I would very much appreciate if you could verify it.

I know I framed the question in $n$ and victory margin $m$, but it is easier to answer considering the number of votes in favor (call it $x$) and the number of votes against (call it $y$), so:
$$
x = n * m \\
y = n - x \\
x \ge y
$$

A voter drawn at random from $P$ has a probability $p$ of supporting the winner. This means that the result is correct only if $p > 0.5$, otherwise the other candidate would have won if everybody was allowed to vote.

We don't know the probability $p$, but we know, from the election result, that out of $n$ voters drawn at random, $x$ voters supports the winner and $y$ voters do not.

Let $f_{x,y}(p)$ be the probability of a given $p$ yielding the known $x$ and $y$ election result. For every possible $p$, the ones with greatest $f_{x,y}(p)$ are the most likely to be the real proportion of voters in $P$ that supports the winner.

If we assume $f_{x,y}(p)$ is proportional to the likelihood of $p$ being the real support for the winner (and I think it is, I am just not sure how to prove it), then the probability of the election being wrong is given by:

$$
l(x,y) = \frac{\int_{0}^{0.5}f_{x,y}(p)\, dp}{\int_{0}^{1}f_{x,y}(p)\, dp}
$$

Which is to say the more cases of $p < 0.5$ are capable of producing the know result of $x$ and $y$, bigger the chance of the election results being wrong, because it is only correct if $p > 0.5$.

Now we only need to know $f_{x,y}(p)$ to be able to calculate $l(x,y)$.

The case for $|P| \gg n $:

It wont make much of a difference for the proportion of supporters if  10 or 10,000 voters are removed out of 300,000,000, thus, for these cases, we can approximate the chance of each one of the $n$ voters to be a supporter of the correct winner as independent of each other, and use the binomial distribution, in which case $f_{x,y}(p)$ is given by:

$$
f_{x,y}(p) = \frac{(x+y)!}{x!y!} p^x (1-p)^y
$$

By plugging into the definition of $l(x,y)$, we get:

$$
l(x,y) = \frac{\int_{0}^{0.5} p^x (1-p)^y\, dp}{\int_{0}^{1} p^x (1-p)^y \, dp}
$$

The numerator bears striking resemblance for the incomplete beta function, as the denominator to the full beta function, so much that it can be written as:
$$
l(x,y) = \frac{B(0.5; x+1,y+1)}{B(x+1,y+1)}
$$

Now we can see a striking resemblance to the regularized incomplete beta function, so much that it can be written as:
$$
l(x,y) = I_{0.5}(x+1, y+1)
$$

Turns out that the regularized incomplete beta function is the CDF for the beta distribution, thus I think we can say the probability of a given result in a sampled election have its voters chosen from a population with some support level to the winner is given by the beta distribution.

In practice, $l(x,y) = 0.5$ for $x = y$, and quickly drops to 0 as the difference between $x$ and $y$ and their magnitude increases.

The case for small $|P|$

If you want to consider the statistical dependency between draws from $P$, you can instead use the PMF of the hypergeometric distribution as $f_{x,y}(p)$:

$$
f_{x,y}(p) = \frac{\binom{p|P|}{x} \binom{(1-p)|P|}{y}}{\binom{|P|}{n}}
$$

Notice that this function is now discrete, and is only defined for where values of $p$ where $p|P| \in \mathbb{N}$. This means that you'll need to replace the integrals for summation over every valid discrete value of $p$ within the integration interval.


问:
Similar triangles in Ptolemy and inversion
In this video at 30:56 seconds an equation is given.

https://youtu.be/bJOuzqu3MUQ?t=1856

How could that equation achieved?
So, are those triangles similar?

答:
By definition of inversion, the points $A_1$ and $B_2$ verify
$$r^2=OA\cdot OA_1=OB\cdot OB_1$$
This implies that
$$\frac{OA}{OB}=\frac{OB_1}{OA_1}$$
Since the angle
$$\angle AOB=\angle B_1OA_1$$
This indeed implies that the triangles $\triangle AOB$ and $\triangle B_1OA_1$ are similar!


问:
Looking for an equation that roughly generates the given graph
I would like to precursor this with the fact that I am in no way a mathematician. Anyways, I am looking help finding an equation, series of equations, or any information on a good starting point for generating a graph roughly like the one in the picture below. Also, I'd like to be able to rotate it around 0,0. Any help is greatly appreciated.

Example of what I would roughly like to generate:

答:
The parametric equations
$$x=3sin(t-\frac{\pi}{4})+2sin(3t-\frac{\pi}{4})$$
$$y=3cos(t+\frac{\pi}{6})+2cos(3t+\frac{\pi}{6})$$
$$0≤t≤2\pi$$
gets you

I arrived at the numbers largely through trial and error, but started from this observation: from your sketch, if you look at the movement along just, say, the $x$-axis it describes a sine wave modulated by another sine. In other words, two sine functions added together. Using Geogebra, it was then a matter of playing with the various parameters to shift the sine curves with respect to one another until getting something approximating your sketch. No doubt there are more formal approaches.


问:
If $A$ and $B$ each have at least two elements, then not every element of $P(A × B)$ has the form $ A_1 \times B_1$
The full definition is:

If $A$ and $B$ each have at least two elements, then not every element of
$P(A \times B)$ has the form $A_1 \times B_1$ for some $A_1 \in P(A)$ and $B_1 \in P(B)$.

Well In my opinion it is not difficult to understand why this is true, If I have for example $A=\{A_1,A_2\}$ and $B=\{B_1,B_2\}$ then I will have an element in $A\times B$ that is $A_1\times B_2$ that would result in $(A_1,B_2)$ and then $\{(A_1,B_2)\}\in P(A\times B)$.

But this definition is so obvious I am having trouble writing up a formal proof. Any guidance would be appreciated!

答:
Let $a_1$ and $a_2$ be distinct elements of $A$ and let $b_1$ and $b_2$ be distinct elements of $B$. Then $C := \{(a_1,b_1), (a_2,b_1), (a_1,b_2)\} \in P(A \times B)$. To see that it is not of the form $A_1 \times B_1$, note that if that were the case, then:

$A_1$ must contain both $a_1$ and $a_2$ (else we miss either $(a_1,b_1)$ or $(a_2,b_1)$ from the product).

$B_1$ must contain both $b_1$ and $b_2$ (else we miss either $(a_1,b_1)$ or $(a_1,b_2)$ from the product).

Thus, $A_1 \times B_1$ must contain $(a_2, b_2)$, which $C$ does not.

Thus, $C$ cannot be such a product, and we win.


问:
Rewrite $\int\limits_{y=0}^p{2{{({\frac{a}{y}})}^{\frac{M}{2}}}{K_M}({2\sqrt{\frac{b}{y}}})\exp({-\frac{y}{c}})dy}$ as non-elementary function?
How to rewrite this integral as some non-elementary function or any possible close-form $?$:
$$
I \equiv
\int_{0}^{p}2\left(a \over y\right)^{M/2}\
\operatorname{K}_{M}\left(2\,\sqrt{\,{b \over y}}\,\right)
\exp\left(-\,{y \over c}\right)\,\mathrm{d}y
$$
where $a,b,c,p$ are all positive integer and
$\operatorname{K}_{M}\left( {} \right)$ is the modified Bessel function of the second kind with $M$ is a positive integer that can be $1,2\ \mbox{or}\ 3$.

Currently, I am not considering the general case for all positive integer $M$ but any contribution is great.

答:
The closest identity I could find that might be useful is:

$$\int_0^\infty t^{\mu-1}~e^{-at}~K_{\nu}(t)~\mathrm{d}t=\begin{cases}
\sqrt{\frac{\pi }{2}} \Gamma ( \mu -\nu ) \Gamma ( \mu +\nu )\left( 1-a^{2}\right)^{-\mu /2+1/4} P^{-\mu +1/2}_{-\nu -1/2}( a) & -1< a< 1\\
\sqrt{\frac{\pi }{2}} \Gamma ( \mu -\nu ) \Gamma ( \mu +\nu )\left( 1-a^{2}\right)^{-\mu /2+1/4} P^{-\mu +1/2}_{\nu -1/2}( a) & \operatorname{Re}( a) \geq 0\ ;\ a\neq 1
\end{cases}$$

With $P^\alpha_\beta$ the associated Legendre function.

REFERENCE: https://dlmf.nist.gov/10.43


问:
Are there any trivial examples of languages that cannot be produced by formal grammars?
Since the cardinality of the set of all languages that can be produced by a grammar is smaller (countably infinite) than the cardinality of the set of all languages (which is uncountable infinite) I'm wondering if there is an easy example for a language that cannot be produced by a grammar.</p>

答:
If you mean grammars with a finite number of substitution rules I think these are equivalent in power to outputs of computer programs, so that the question asks for a set (of words) that is not recursively enumerable.  Examples are as easy or hard or explicitly describable as for non-r.e. sets.

examples: The set of strings whose shortest generating program in C has an odd number of characters.  The set of true sentences of first-order Peano arithmetic.  The set of programs that do not halt.  Any set higher in the arithmetic hierarchy than r.e. sets.  The set of $n$ for which $n$-th bit of Chaitin's constant is $1$. The set of ...

None of these sets have any kind of explicit description, they are constructs that can be named in a theory of computer programs, or arithmetic, or set theory.


问:
Bijection between Power Set of Natural numbers and Power Set of integers
There is for sure more than one method to do this , however I want to understand the hint provided in my book .The book says to use the bijection between N and Z given by f(n)=n/2 if n is even and f(n)= (1-n)/2 if n is odd .I think  I can prove one to one , but why is f(n) onto ?? More importantly how does this translate into a bijection between P(N) and P(Z)?? is it automatic? I believe some explanation is required but I cannot figure it out...............</p>

答:
Thinking in terms of the specific sets may make this look more mysterious than it actually is: what we're seeing here is an instance of a more general phenomenon.

Namely:

Suppose there is a bjiection between $A$ and $B$. Then there is also a bijection between $\mathcal{P}(A)$ and $\mathcal{P}(B)$.

To see this, suppose $f:A\rightarrow B$ is a bijection and I have a set $S$ of elements of $A$. The function $f$ only takes as input individual elements of $A$, so something like "$f(S)$" doesn't make sense; nonetheless, do you see a way to use $f$ to turn $S$ into a subset $\hat{S}$ of $B$?

 Just apply $f$ to each of the elements in $A$: that is, set $$\hat{S}=\{f(x): x\in S\}.$$

Let's give the construction above a name, call it "$F$." That is, $F$ is the function

 $$F:\mathcal{P}(A)\rightarrow\mathcal{P}(B): S\mapsto\{f(x): x\in S\}.$$

Can you see how to use the fact that our original function $f$ is a bijection to show that $F$ is also a bijection between $\mathcal{P}(A)$ and $\mathcal{P}(B)$?

 I'll sketch how to prove injectivity (proving surjectivity is similar). Suppose $F(S_1)=F(S_2)$; we want to show that $S_1=S_2$. Well, by definition of $F$ we have $$(*)\quad\{f(x): x\in S_1\}=\{f(x): x\in S_2\}.$$ To show that $S_1\subseteq S_2$ (the converse containment is proved identically), note that $(*)$ implies that given $s\in S_1$ there is some $t\in S_2$ such that $f(s)=f(t)$. What does this tell you about $s$ and $t$, and what properties of $f$ are you using to show this? And do you see why this means $S_1\subseteq S_2$?


问:
How can we define $\mathbb{R}^0$?
I'm reading a paper which asks to consider some box B $\subset \mathbb{R}^{v-1}$, for all $v\geq 1$. So for example, if we let $v=2, $ we would have that B $\subset \mathbb{R}$, so B would just be some interval $I$. I'm wondering what the case would be for $v=1$. Then we would be considering some box $B \subset \mathbb{R}^0$. I'm not sure what to make of this. Would it be empty...or just a singular point?

I can link the paper if its helpful. Thanks in advance.

答: The way I think of $\mathbb R^0$ is as a $0$-dimensional vector space, so it is a singleton $\{0\}$ consisting of the $0$ element.

Title: How can we define $\mathbb{R}^0$?


问:
First Order formula Validity?
The given first order formula is valid or not?

[ β→ ∃ₓ α(x) ]→ [ ∀ₓ(β→α(x)) ]

Someone please answer this, I have been Struggling to know the correct answer to this Question?

Intuitively it seems to be a valid formula. However I am not sure!

答:
As was discussed in the comments, the formula in question is not valid. Any model in which $\beta$ is true and $\alpha$ holds of some but not all elements will witness this.

Let me say a bit about how one might search for such a countermodel, since this is a skill it can take some time to develop. We start thinking about the simplest component of our formula, namely $\beta$ (it's just true or false, full stop - it doesn't involve thinking about elements of our model). So we think for a moment about the two possibilities for $\beta$ and how they would impact the model. Since it's easy to see that making $\beta$ false makes the whole formula true (since both antecedent and consequent will be vacuously true), we know we want $\beta$ to be false in our countermodel.

Now what about $\alpha$? Unlike with $\beta$, this is a bit more complicated. Essentially we have three choices: $\alpha$ holds always, $\alpha$ holds never, or $\alpha$ holds sometimes but not always.$^*$ We could whip up and check examples of all three possibilities separately, but the third should stand out as particularly promising. This is because the formula involves comparing an "$\exists_x$" and a "$\forall_x$." In the first two cases for $\alpha$, these don't behave differently since all elements of the domain will look alike; in the third case however there is differentiation between them (e.g. "$\exists_x\alpha(x)$" is true but "$\forall_x\alpha(x)$" is false). So the third seems to have the most promise, and we try it and see that it works.

$^*$OK, that's not totally true. Each of those cases is itself a rather wide umbrella: e.g. "$\alpha$ holds sometimes and fails sometimes" includes "$\alpha$ holds of one element and fails of two," and "$\alpha$ holds of forty elements and fails of seventeen," or ... However, the formula doesn't suggest that this will be important: since we never consider two variables, so multiplicity doesn't matter. For this reason we should look at the simplest possible examples of each case: for the first two cases this is a model with one element, and for the third case this is a model with two elements.

Title: First Order formula Validity?


问:
Distribution of waiting time when lifetime is exponentially distributed
Suppose the lifetime of a PC-hard-drives is exponentially distributed with average lifetime $\tau$. Now, $N$ hard drives are switched on simultaneously. Question: How is the waiting time until the first hard drive fails distributed?

I'm not especially adept at statistics, so I suppose my approach is quite wrong. But here is my idea:

I want to know the probability of one hard-drive failing within the interval $[t,t+dt]$. This means, that in the interval $[0,t]$ none of the $N$ hard drives are "allowed" to fail. The probability for this is

$$\int_0^t f(t') dt' = F(t)$$

where $f(t)$ is the (exponential) density function. Now the probabilty of one hard drive failing (within the mentioned interval) is

$$ 1 - f(t) dt$$

However, since the others still have to be functioning in that very same interval, the probability for exactly one malfunction is

$$ \big(1-f(t)dt \big) \cdot \big( F(t+dt)-F(t) \big)^{N-1} $$

Finally, the overall probability is the product of these:

$$\big( F(t) \big)^N \cdot \big( 1 - f(t) dt \big) \cdot \big( dF dt \big)^{N-1} \cdot N$$

The factor $N$ is there because of all possible combinations of one hard drive failing out of $N$.

As I said, I suppose this approach is not only incorrect, but also very wrong. I would appreciate some help in solving this problem.

答:
Let's label the $N$ drives with numbers $i \in \{1, 2, \ldots, N\}$ and denote the random lifetimes of each drive as $T_1, T_2, \ldots, T_N$.  Each one is independent and identically distributed as an exponential random variable with mean $\tau$.  Let $$F_{T_i}(t) = \Pr[T_i \le t]$$ be the cumulative distribution function that gives the probability that drive $i$ has failed by time $t$.

Now let $T_{(1)}$ represent the failure time of the first drive to fail when all $N$ drives are operated simultaneously.  Then we have $$T_{(1)} = \min_i (T_1, T_2, \ldots, T_N);$$ that is to say, it is the minimum of the set of random failure times of all the drives.  So for instance, if $N = 5$ and we ran each drive until failure and observed $(T_1, T_2, T_3, T_4, T_5) = (10, 25, 34, 15, 9)$, then we have $T_{(1)} = 9$, the smallest observed failure time.

The question you are interested in is, what is $$F_{T_{(1)}}(t) = \Pr[T_{(1)} \le t]?$$  Well, it is easier to work with the complementary probability--the survival function  $$S_{T_{(1)}}(t) = \Pr[T_{(1)} > t] = 1 - \Pr[T_{(1)} \le t] = 1 - F_{T_{(1)}}(t).$$  The survival function of the first/minimum failure time is $$\Pr[T_{(1)} > t] = \Pr[\min(T_1, T_2, \ldots, T_N) > t] = \Pr[(T_1 > t) \cap (T_2 > t) \cap \cdots \cap (T_N > t)],$$ because if the smallest of the $T_i$ exceeds $t$, we know that all of the $T_i$ exceed $t$; and vice versa--if all of the $T_i$ exceed $t$, then the smallest also exceeds $t$.  This is why we switched from the CDF to the survival, because this logic doesn't work properly if we used the CDF, since $T_{(1)} \le t$ does not guarantee that the other $T_i$ are also $t$ or less; they can be greater.

Now because each $T_i$ is independent, the probability of the intersection of the events $(T_1 > t) \cap (T_2 > t) \cap \cdots \cap (T_N > t)$ is simply the product of the probability of each event; i.e., $$\Pr[T_{(1)} > t] = \Pr[T_1 > t] \Pr[T_2 > t] \cdots \Pr[T_N > t].$$  And because each of the $T_i$ are identically distributed, the LHS is simply the $N^{\rm th}$ power of any single probability of a single drive surviving past time $t$:  $$\Pr[T_{(1)} > t] = (\Pr[T_1 > t])^N.$$  Written in terms of the CDF, we then have $$F_{T_{(1)}}(t) = 1 - S_{T_{(1)}}(t) = 1 - (\Pr[T_1 > t])^N = 1 - (1 - F_{T_1}(t))^N. \tag{1}$$

Note that our derivation does not use the fact that the $T_i$ are exponentially distributed, so formula $(1)$ is distribution-free.  The only requirement is that the $T_i$ are independent and identically distributed.


问:
Real life usage of Benford's Law
I recently discovered Benford's Law. I find it very fascinating. I'm wondering what are some of the real life uses of Benford's law. Specific examples would be great.</p>

答:
From a physics perspective you can look at how Benford's law (or devations from it) can arise from the different stastical mechanics distributions (Boltzmann, Bose-Einstein, etc...). There is a good paper on the arVix that can get you started on this topic:

Abstract:

The occurrence of the nonzero leftmost digit, i.e., 1, 2, ..., 9, of numbers from many real world sources is not uniformly distributed as one might naively expect, but instead, the nature favors smaller ones according to a logarithmic distribution, named Benford's law. We investigate three kinds of widely used physical statistics, i.e., the Boltzmann-Gibbs (BG) distribution, the Fermi-Dirac (FD) distribution, and the Bose-Einstein (BE) distribution, and find that the BG and FD distributions both fluctuate slightly in a periodic manner around the Benford distribution with respect to the temperature of the system, while the BE distribution conforms to it exactly whatever the temperature is. Thus the Benford's law seems to present a general pattern for physical statistics and might be even more fundamental and profound in nature. Furthermore, various elegant properties of Benford's law, especially the mantissa distribution of data sets, are discussed.

The Significant Digit Law in Statistical Physics

ContentLicense: CC BY-SA 2.5

Title: Real life usage of Benford's Law


答: For $n \in \mathbb N$ we can think $A^n$ as the set of all functions from $n = \{0,\dots,n-1\}$ to $A$. An element $f \in A^n$ can be denoted as $f = (f(0),\dots,f(n-1))$. Thus, $A^0$ only consist of the empty function $\varnothing \to A$, and then $A^0 = \{\varnothing\} = \{0\} = 1$.

Title: How can we define $\mathbb{R}^0$?

Body:
I'm reading a paper which asks to consider some box B $\subset \mathbb{R}^{v-1}$, for all $v\geq 1$. So for example, if we let $v=2, $ we would have that B $\subset \mathbb{R}$, so B would just be some interval $I$. I'm wondering what the case would be for $v=1$. Then we would be considering some box $B \subset \mathbb{R}^0$. I'm not sure what to make of this. Would it be empty...or just a singular point?

I can link the paper if its helpful. Thanks in advance.


答:
$\mathbb R^{v-1}$ is the set of all $v-1$-tuples of real numbers, i.e. all functions from
$\{1,2,\ldots, v-1\}$ to $\mathbb R$.  In the case $v=1$, that's all functions from the empty set to $\mathbb R$.  There is one function from the empty set to anything, namely a function that doesn't do anything.

I'd have to look closer at the paper you're referring to, to see what is done with $B$, but my guess is that the author didn't really bother much about the case $v=1$.  Either didn't think about it at all, or is leaving this up to the reader.

Title: How can we define $\mathbb{R}^0$?


问:
Doubtful solution to an exercise on faithful flatness in Matsumura's Commutative Algebra
The exercise on page 30. It says that:

Let $A, B$ be integral domain having the same field of fractions, $B \supseteq A$. Prove that $B$ is faithfully flat over $A$ only when $B = A$.

My solution seems to be easy, which makes me doubt its correctness. My idea is to prove that for any $x \in B$, $x \in A$ so that $B \subseteq A$. By assumption that $A, B$ have the same field of fraction, any element of $B$ is of the form $a/b$ for $a, b \in A$, $b \not= 0$. By faithful flatness, the sequence of $A$-modules
$$0 \rightarrow A \rightarrow A[X]/(bX - a) \rightarrow 0$$
is exact if and only if the sequence of $B$-modules
$$0 \rightarrow \underbrace{A \otimes_A B}_{B} \rightarrow A[X]/(bX - a) \otimes_A B \rightarrow 0$$
is exact. We have
\begin{align*}
A[X]/(bX - a) \otimes_A B &\cong A[X]/(bX - a) \otimes_{A[X]} A[X] \otimes_A B\\
&\cong A[X]/(bX - a) \otimes_{A[X]} B[X]\\
&\cong B[X]/(bX - a)\\
&\cong B &\text{ because }a/b \in B
\end{align*}
So the second sequence is exact: $0 \rightarrow B \rightarrow B \rightarrow 0$. This implies the first sequence is exact and that can only occurs when $a/b \in A$.

答:
There is fatal flaw towards the end of your proof when you say $B[X]/(bX - a) \cong B$ because $a/b \in B$.

For example, let $A$ be a domain and $a \in A$.  Then $a^2 / a = a \in A$ but $A[X]/(aX - a^2)$ is not even a domain, so clearly is not isomorphic to $A$.

A correct solution might look like:

Let $A \subseteq B$ and $a/b \in B$.  Then $(b :_A a) = (b :_A a)B \cap A = (bB :_A aB) \cap A = B \cap A = A$.  The first equality uses that $I = IB \cap A$ for all $A$-ideals under faithfully flat extension.  The second equality is a fundamental property of colon ideals under flat extensions.  So $a = bc$ for some $c \in A$, i.e. $a/b \in A$.


问:
For a continuous function f on open interval $(a,b)$, if $\lim_{a^{+}}f(x)=\infty$ and $\lim_{b^{-}}f(x)=-\infty$, prove it has at least one zero.
This is quite an obvious consequence to the intermediate value theorem, but I　was wondering if there was a more concrete way to prove it?(by using the definition of limits, and choosing some  $(c,d)$ in $(a,b)$? Just a hunch--it might not work)
I also have another generalization (an unproved hypothesis, actually)

For a continuous function $f:I\rightarrow R$, where I is an open interval $(a,b)$, if $\lim_{x\to a^{+}}f(x)=\infty$ and $\lim_{x\to b^{-}}f(x)=-\infty$, then the equation $f(x)=x$ has at least one solution.

答:
Any proof of your result will either use the IVT or essentially prove the IVT.  In fact you can derive the IVT from your result.  Namely, suppose $g$ is a continuous function on a closed interval $[c,d]$, and $v$ any real number between $g(c)$ and $g(d)$.  We may assume wlog $g(c) > v > g(d)$ (otherwise replace $g$ by $-g$). Let $f(x) = g(x) - v$.  Extend $f$ continuously to the interval $(c-1,d+1)$ by taking $f(x) = f(c)/(x-c+1)$ for $c-1 < x < c$ and $f(x) = f(d)/(d+1-x)$ for $d < x < d+1$.  Your result implies there is some $x \in (c-1,d+1)$ with $f(x)=0$,
and thus $g(x) = v$, and it is easy to see $x \in [c,d]$.


问:
Hardy-Littlewood inequality in $L^p$ space.
Let $p \in (1, \infty]$ and $f \in L^p(\mathbb{R}^n)$.
I would like to show that if $Mf(x)=\sup_{r>0}\frac{1}{m(B(x,r))} \int_{B(x,r)}|f(y)|df$, then for all $f \in L^p(\mathbb{R}^n)$,
$$m(\{x:Mf(x)> \lambda\}) \leq  \frac{3^n}{\lambda^p}||f||^p_p, $$
where $m$ denotes the Lebesgue measure.
We know that for $f\in L^1(\mathbb{R}^n)$,
$$m(\{x:Mf(x)> \lambda\}) \leq  \frac{3^n}{\lambda}||f||_1, $$ I thought I would be able to use this one but I could not figure out how.

Body:
First, by Jensen's inequality:
$$(Mf(x))^p=\sup_{r>0}(\frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)|\,dy)^p\leq \\\sup_{r>0}\frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)|^p\,dy=M|f|^p(x)$$
Thus:
$$m(\{x:Mf(x)>\lambda\})=m(\{x:(Mf(x))^p>\lambda^p\})\leq m(\{x:M|f|^p(x)>\lambda^p\})\leq \\\frac{3^n}{\lambda^p}|||f|^p||_1=\frac{3^n}{\lambda^n}||f||_p^p$$


问:
Linear Algebra Done Right: Approach for 1a doesn't make much sense (to me)
Let $F$ either be $\mathbb R$ or $\mathbb Z/2\mathbb Z$. For each of the following subsets of $F^3$, determine whether it is a subspace. Be sure to explain your answer.

$U_1=\{(x_1,x_2,x_3)\in F^3: x_1+2x_2+3x_3=0\}$;
$U_2=\{(x_1,x_2,x_3)\in F^3: x_1+2x_2+3x_3=4\}$;

For 1a, what does the solution mean by "properties of addition/multiplication for any field"? It's puzzling because not every field has 2,3 as elements in it. Multiplication/addition are not defined for ANY field; the field itself has its own rule for those 2 operations, right?

For 1b, I notice that 1a uses "properties ... for any field." Why can't you do the same general approach for 1b like so.

Choose an arbitrary $(x_1,x_2,x_3) \in U_1$. We know $(0,0,0) \notin U_1$, since $0 + 2(0)+3(0) \neq 4$. Since we have used properties of addition/multiplication for any field, this statement holds over $\mathbb R$ and $\mathbb Z/2\mathbb Z$.

Here, I used the 1a approach by taking an approach for any field. But this doesn't work for $\mathbb Z/2\mathbb Z$. In short, I'm confused when I should use operations for ANY field vs a specific field; basically, I'm confused on the why the 1a approach is justified.

答:
Any field contains $1$, and hence contains elements $n := 1 + ... + 1$, where there are $n$ copies of $1$, for all $n$.

By "properties of addition/multiplication for any field", they mean "those properties that are true in all fields - to wit, the field axioms and their consequences.

In 1b, you're using properties that are not true in all fields: "$0 \neq 4$" is not a field axiom, nor a consequence of the field axioms.


问:
Determine limit if the function diverges
How do we prove $$\lim_{x\rightarrow \infty}f(x)=\infty\Rightarrow\lim_{x\rightarrow \infty}\frac{1}{f(x)}=0$$
I'm thinking of using the definition of a diverging limit to prove it: I will pick a random $m$, that $$f(x) > m$$
Then what do I do next? Is there anything else that I can know from the fact the limit of $f(x)$ diverges?

答:
Yes your idea is fine, proceeding by definition we have that

$$\lim_{x\rightarrow \infty}f(x)=\infty \iff \forall m>0 \quad \exists x_0 \quad \forall x\ge x_0 \quad f(x)\ge m$$

and then

$$ \forall \varepsilon = \frac 1m>0 \quad \exists x_0 \quad \forall x\ge x_0 \quad \frac1{f(x)}\le \frac 1m=\varepsilon$$


答:
It depends what you mean by "concrete", but you can do it in a way that constructs such a zero:

We shall define two sequences $a_n$ and $b_n$, which shall converge exponentially quickly to a zero.

To do this, define $a_0 = a$ and $b_0 = b$, and for each $n$, define $c_n$ to be the midpoint of $a_n$ and $b_n$. Then we have three cases:

$f(c_n) = 0$. In this case, we have found a zero, and can set $a_k = b_k = c_n$ for all $k > n$.

$f(c_n) < 0$. Then set $a_{n+1} = c_n$ and $b_{n+1} = b_n$.

$f(c_n) > 0$. Then set $a_{n+1} = a_n$ and $b_{n+1} = c_n$.

In this way, we note that:

We always have $f(a_n) \leq 0$ (or $a_n = a$).

We always have $f(b_n) \geq 0$ (or $b_n = b$).

Both $(a_n)$ and $(b_n)$ converge (they're bounded and monotonic).

They converge to the same limit (call it $c$), since$(a_n - b_n) = 2^{-n}(b - a)$.

By (1), $f(c) \leq 0$.

By (2), $f(c) \geq 0$.

Hence, $f(c) = 0$.

And so we have found our zero, and have an exponentially fast algorithm for doing so.

For your generalisation (and any other generalisation of the form mentioned in the comments): just apply the result directly to $h: x \mapsto f(x) - x$ (or $f(x) - g(x)$ for the generalised version in the comments): $h$ is continuous since $f$ and $g$ are, $\lim\limits_{x \to a^+}h(x) = \lim\limits_{x \to a^+}f(x) - \lim\limits_{x \to a^+}g(x) = -\infty$ (since the latter limit is not $\infty$), and similarly $\lim\limits_{x\to b^-}h(x) = \infty$. The above then gives some $c$ such that $h(c) = 0$, and then $f(c) = g(c)$.

Title: For a continuous function f on open interval $(a,b)$, if $\lim_{a^{+}}f(x)=\infty$ and $\lim_{b^{-}}f(x)=-\infty$, prove it has at least one zero.

Body:
This is quite an obvious consequence to the intermediate value theorem, but I　was wondering if there was a more concrete way to prove it?(by using the definition of limits, and choosing some  $(c,d)$ in $(a,b)$? Just a hunch--it might not work)
I also have another generalization (an unproved hypothesis, actually)

For a continuous function $f:I\rightarrow R$, where I is an open interval $(a,b)$, if $\lim_{x\to a^{+}}f(x)=\infty$ and $\lim_{x\to b^{-}}f(x)=-\infty$, then the equation $f(x)=x$ has at least one solution.


问:
Characteristic function of a gaussian
There is a claim in my book, that the characteristic function of a Gaussian with mean $0$ and  variance $\frac{1}{\sigma^2}$ is

$$ \phi(x)=\int_{\mathbb{R}} e^{ixt}e^{-\frac{\sigma^2}{2}t^2} \frac{\sigma dt}{\sqrt{2\pi}} $$

I just don't understand the presence of that $\frac{\sigma}{\sqrt{2\pi}}$ multiplying the measure. I would perfectly agree  with the result without that factor. What am I missing?

答: The density function of $\mathcal N(0,\sigma^{2})$ is $\frac  1 {\sqrt {2\pi} \sigma} e^{-x^{2}/2\sigma^{2}}$. For $\mathcal N(0,1/\sigma^{2})$ you just have to change $\sigma $ to $\frac  1 {\sigma}$ in this.

Title: Characteristic function of a gaussian


答:
Use argument by contradiction. Suppose $f(x)=x$ has no solution in $(a,b)$. Then either $f(x)<x$ or $f(x)>x$ for $x\in(a,b)$. If $f(x)<x$ for $x\in(a,b)$, one has
$$ f(x)<x<b, x\in(a,b)$$
which is against $\lim_{x\to b^-}f(x)=\infty$. If $f(x)>x$ for $x\in(a,b)$, one has
$$ f(x)>x>a, x\in(a,b)$$
which is against $\lim_{x\to a^+}f(x)=-\infty$. Thus $f(x)=x$ must have a solution in $(a,b)$.


问:
Formula for difference between two numbers on a wall clock in clokwise direction
It's my first post in math.stackexchange.com.
I got a necessity to find out the clockwise difference between two numbers on a wall clock. For example, difference between 12 and 1 is 1 where as the difference between 1 and 12 is 11. Actually what I need is number of clock wise steps required reach number b from number a on a clock.
While writing an application, I struck writing this function.

Please help me.
Thanks in advance.

答:
i am writing a C function

Code

```<code> int clockstep(int a, int b){
    int steps =0;
    if (a<=b)
    {
       steps = b-a;
     }
     else
    {
       steps = 12-b+a;
     }
  return steps
 }
</code>```

means you take input from user than compare it if 1st $<=$ 2nd then simply (2nd-1st) is your answer otherwise do (12-2nd+1st).

Body:
i am writing a C function

Code

 int clockstep(int a, int b){
    int steps =0;
    if (a<=b)
    {
       steps = b-a;
     }
     else
    {
       steps = 12-b+a;
     }
  return steps
 }

means you take input from user than compare it if 1st $<=$ 2nd then simply (2nd-1st) is your answer otherwise do (12-2nd+1st).


问:
Inverting the vorticity via Biot-Savart in Navier Stokes
Given the Navier Stokes equation $\partial_t u+u\cdot \nabla u+\nabla p=\nu \Delta u$ in $\mathbb{R}^3$ with $u$ divergence-free, one is often interested in the vorticity $\omega=\text{curl} \ u$. In Majda Bertozzi (Chapter 2.4, page 73) it is claimed that one has the Biot-Savart Law
$$u(t,x)=\dfrac{1}{4\pi} \int_{\mathbb{R}^3 } \dfrac{(x-y) \times \omega (t,y)}{|x-y|^3} dy $$

To my understanding, they way they arrive at this solution is by showing that the system $\text{curl} \ u =\omega,\ \text{div}\ u=0$ has the above solution with which I agree (they don't do brute force computation, they take $\psi$ such that $\Delta \psi=\omega$ and it turns out that $-\text{curl} \ \psi$ has the desired properties and is of the above form).

What is not clear to me is why $u$ should be exactly that;  the system $\text{curl} \ u =\omega,\ \text{div}\ u=0$ clearly does not have a unique solution since the system $\text{curl} \ u =0,\ \text{div}\ u=0$ does not have a unique solution. Should there not be an additional freedom given by some $\nabla f$ where $f:\mathbb{R}^3\rightarrow \mathbb{R}$ is harmonic? Freedom in the sense that
$$u(x)=\dfrac{1}{4\pi} \int_{\mathbb{R}^3 } \dfrac{(x-y) \times \omega (t,y)}{|x-y|^3} dy\ + \ \nabla f $$
This would make intuitive sense since $\omega$ depends only on the derivatives of $u$ so it "loses" information.

I've seen the formula from Majda Bertozzi even in papers and it's not clear to me why nobody talks about the extra term $\nabla f$. If anybody has any input I'd appreciate it.

答:
The Biot-Savart law only determines the velocity field up to an arbitrary irrotational component $\nabla f$.  Since $\nabla \times \nabla f= 0$, the presence of this term does not change the vorticity.

The function $f$ is required to enforce a no-flux condition at a bounding surface $S$.  For incompressible flow, we have the divergence-free condition $\nabla \cdot \nabla f = \nabla^2f= 0$, and $f$ can be determined uniquely as the solution to the potential problem

$$\nabla^2 f = 0, \quad \frac{\partial f}{\partial n} + \mathbf{u}_\omega\cdot \mathbf{n}= U_n \,\text{ for } \mathbf{x} \in S. \quad f(\mathbf{x}) \to 0 \, \text{ as } \, \mathbf{x} \to \infty ,$$

where $\mathbf{u}_\omega$ is the Biot-Savart velocity field and $U_n$ is a specified surface velocity (zero for no flux through a stationary surface).


问:
Calculating expected value of infimum of two point distrubution
Let $(X_k)$ be a sequence of random variables as following : $P(X_i=-1)=q$ and $P(X_i=1)=1-q$. I want to calculate $$E[\inf\{n:X_1+X_2+...+X_n>0\}]$$

My work so far

Let's denote $Y(x)=\inf\{X_1+X_2+...+X_n>x\}$

Then we want to calculate $E[Y(1)]$

$$E[Y(x)]=E[Y(x)|X_1=-1]P(X_1=-1)+E[Y(x)|X_1=1]P(X_1=1)=$$
$$=E[Y(x)|X_1=-1]q+1-q$$

And now I have a little troubles with calculating $E[Y(x)|X_1=-1]$. Could you please tell me if I'm going in right direction ?

答:
Let $S_n:=k+\sum_{i=1}^n X_i$ and $T_z:=\inf\{n\ge 1: S_n=z\}$. Then
$$
\mathsf{P}_0(T_1=1)=1-q,
$$
and for $n\ge 1$,
\begin{align}\label{1}\tag{1}
\mathsf{P}_0(T_1=2n+1)&=q\mathsf{P}_{-1}(T_1=2n)=\frac{q}{n}\mathsf{P}_{-2}(S_{2n}=0) \\
&=\frac{q}{n}\binom{2n}{n-1}q^{n-1}(1-q)^{n+1},
\end{align}
where $\mathsf{P}_k$ is the law of $S_n$ starting at $k$. Therefore, for $q<1/2$,
\begin{align}
\mathsf{E}T_1&=\sum_{n\ge 0}(2n+1)\mathsf{P}_0(T=2n+1) \\
&=(1-q)+q\sum_{n\ge 1}\frac{2n+1}{n}\binom{2n}{n-1}q^{n-1}(1-q)^{n+1} \\
&=\frac{1}{1-2q}.
\end{align}
Obviously, for $q\ge 1/2$, $\mathsf{E}T=\infty$.

For the second equality in \eqref{1} we used the fact that
$$
\mathsf{P}_{-1}(T_1=k)=\mathsf{P}_{-2}(T_0=k)=\frac{2}{k}\mathsf{P}_{-2}(S_k=0).
$$


问:
Almost Surely discontinuous path
For $t\in\mathbb{R}$ let $\{X_t\}$ be a family of random variables independent and with the same distribution. Assume the distribution is absolutely continuous with respect to Lebesgue measure. I want to show that the paths of the process $X_t$ are discontinuous almost surely. Does anyone have a hint for this problem?

答:
Lemma

Let $(Y_n)$ be i.i.d. If $P(Y_n \to Y)>0$ for some random variable $Y$ then $Y$ is degenerate.

Proof of the lemma: $P(Y_{2n+1}-Y_{2n} \to 0)$ is $0$ or $1$ by Kolmogorov's $0-1$ law. If it is $1$ then $Y_{2n+1}-Y_{2n} \to 0$ in distribution. Since this sequence is i.i.d. it follows that $|\phi (t)|=1$ for all $t$ where $\phi$ is the common characteristic function. This implies that $Y_{2n+1}-Y_{2n}$ is degenerate and hence $Y_n$ is degenerate for each $n$. The lemma follows.

Now consider $P(X_{1/n} \to X_0)$. If this probability is $>0$ then the lemma shows that $X_0$ has a degenerate distribution contradicting absolute continuity.

Title: Almost Surely discontinuous path


问:
Given two sequences of positive numbers with $\sum_{n\ge1}\frac{x_n}{y_n}$ and $\sum_{n\ge1}y_n$ convergent, is $\sum_{n\ge1}\sqrt{x_n}$ convergent?
Consider two sequences of positive numbers $(x_n)$ and $(y_n)$. Suppose that the series

$$\sum_{n \ge 1} \frac{x_n}{y_n}$$

and the series

$$\sum_{n \ge 1} y_n$$

are both convergent. Is the series

$$\sum_{n \ge 1} \sqrt{x_n}$$

also convergent?

This is what I tried:

We know that the series:

$$\sum_{n \ge 1} \frac{x_n}{y_n} \hspace{4cm} \sum_{n \ge 1} y_n$$

are both convergent. By the $n^{th}$ Term Test, we can conclude that:

$$\lim_{n\to\infty} \frac{x_n}{y_n} = 0$$

and

$$\lim_{n\to\infty} y_n = 0$$

Now, since we know that the series $\sum_{n \to \infty} y_n$ is convergent and that

$$\lim_{n \to \infty} \frac{x_n}{y_n} = 0 \in [0, \infty) \cup \{\infty \}$$

we can conclude by the Second Comparison Test that the series

$$\sum_{n \to \infty} x_n$$

is convergent.

Now, using the $n^{th}$ Term Test again, we get that

$$\lim_{n \to \infty} x_n = 0$$

But this is about as far as I got. I don't really see how to draw any conclusion about the series $\sum_{n \to \infty} \sqrt{x_n}$. I don't know how to take into consideration that square root.

Body: Hint: $$\sqrt{x_n}\leqslant\frac12\left(\frac{x_n}{y_n}+y_n\right).$$


答:
By Cauchy–Schwarz inequality we have

$$\sum_{n \ge 1} \sqrt{x_n}=\sum_{n\ge1}\left(\frac{ \sqrt{x_n} }{ \sqrt{y_n}}\cdot \sqrt{y_n}\right)\le \left(\sum_{n\ge1}\frac{x_n}{y_n}\right)^2 \cdot \left(\sum_{n\ge1}y_n\right)^2 $$

Title: Given two sequences of positive numbers with $\sum_{n\ge1}\frac{x_n}{y_n}$ and $\sum_{n\ge1}y_n$ convergent, is $\sum_{n\ge1}\sqrt{x_n}$ convergent?

Body:
Consider two sequences of positive numbers $(x_n)$ and $(y_n)$. Suppose that the series

$$\sum_{n \ge 1} \frac{x_n}{y_n}$$

and the series

$$\sum_{n \ge 1} y_n$$

are both convergent. Is the series

$$\sum_{n \ge 1} \sqrt{x_n}$$

also convergent?

This is what I tried:

We know that the series:

$$\sum_{n \ge 1} \frac{x_n}{y_n} \hspace{4cm} \sum_{n \ge 1} y_n$$

are both convergent. By the $n^{th}$ Term Test, we can conclude that:

$$\lim_{n\to\infty} \frac{x_n}{y_n} = 0$$

and

$$\lim_{n\to\infty} y_n = 0$$

Now, since we know that the series $\sum_{n \to \infty} y_n$ is convergent and that

$$\lim_{n \to \infty} \frac{x_n}{y_n} = 0 \in [0, \infty) \cup \{\infty \}$$

we can conclude by the Second Comparison Test that the series

$$\sum_{n \to \infty} x_n$$

is convergent.

Now, using the $n^{th}$ Term Test again, we get that

$$\lim_{n \to \infty} x_n = 0$$

But this is about as far as I got. I don't really see how to draw any conclusion about the series $\sum_{n \to \infty} \sqrt{x_n}$. I don't know how to take into consideration that square root.


问:
If $A^2 = I$, $B^2=I$ and $(AB)^2=I$, then $AB = BA$
Matrix Question

If $A^2 = I$, $B^2=I$ and $(AB)^2=I$, then $AB = BA$

Basically, got up to $A(BA-AB)B = 0$ (by cancelling and equating terms from $I^2 = I$ and to $A^2B^2 = A^2B^2$ and using distributive laws), but that doesn't work out too well!

Thanks for help in advance!

答:
Multiplying $A(BA-AB)B$ on the left by $A$ and the right by $B$ gives
$$AB-BA=A^2(BA-AB)B^2=A0B=0$$
as desired.</p>


问:
Question about inertia groups and unramified extensions
Let $K$ be a number field, and $v$ a finite place. If $\bar{K}$ is a separable closure of $K$, then in $G_K=\text{Gal}(\bar{K}/K)$ we can find the decomposition group of (a place over) $v$, which is isomorphic to the Galois group of $\bar{K_v}/K_v$, with $K_v$ the completion at $v$.

It is well know that the fixed field of the inertia $I_v$ in $\bar{K_v}$ is the maximal unramified extension of $K_v$. Is it also true that the fixed field of the inertia in $\bar{K}$ is the maximal extension of $K$ unramified at $v$? I think this is true, since we can easily move to the finite case where is true, but also I could appreciate a check.

After this, if we consider the maximal extension of $K$ unramified at $v$ and $v'$, with $v\ne v'$, then it is the intersection of the maximal extension unramified at $v$ with the one unramified at $v'$ (it is true? It seems to me obvious), therefore, by Galois correspondence, the product of the inertia $I_vI_{v'}$ is the group corresponding to that field.

But what happens if we consider the maximal extension unramified outside a finite set of places, so unramified at an infinite set of places? The infinite intersection would correspond to an infinite product of subgroups, which of course make no sense. So have we to compute it with, maybe, inverse limit, or something like this?

(My final goal is to understand a proof in Rubin's book Euler systems: he proved that, given a Galois representation $T$ with coefficients in the valuation ring $O$ of a finite extension of $\mathbb{Q}_p$, and a finite set of primes $\Sigma$ containing all primes where $T$ ramifies, primes above $p$ and infinite places, then the Selmer group $S^{\Sigma}(K,T)$ is equal to $H^1(K_{\Sigma}/K,T)$, where $K_{\Sigma}$ is the maximal extension unramified outside $\Sigma$.

The proof is the following: $$\begin{split}S^{\Sigma}(K,T)&\overset{(1)}{=}\ker \left(H^1(K,T)\to \prod_{v\not\in\Sigma} H^1(K_v,T)/H_f^1(K_v,T)\right)=
\\&\overset{(2)}{=}\ker\left(H^1(K,T)\to\prod_{v\not\in\Sigma}\text{Hom}(I_v,T)\right)=\\&\overset{(3)}{=}\ker\left(H^1(K,T)\to H^1(K_{\Sigma},T)\right){=}H^1(K_{\Sigma}/K,T).
\end{split}$$
(1) is the definition. In (2) we need $\text{Hom}(I_v,T)^{Fr}$, the fixed points of Frobenius, but not a big deal, since we want the kernel,so we can enlarge the codomain. But my big problem is in (3), with which the question is related).

答:
Pick a place $w$ of $\overline{K}$ above $v$. The absolute Galois group acts transitively on those places so we don't care which $w$ we chose.

By density of $\overline{K}$ in $\overline{K}_w\cong \overline{K_v}$ then $Gal(\overline{K}_w/K_v)$ is the $\sigma\in Gal(\overline{K}/K)$ that are continuous for $w$, ie. $D_w$.

$$I_w = Gal(\overline{K}_w/K_v^{nr}), \qquad K_v^{nr}=\bigcup_{n\ge 1} K_v(\zeta_{p^n-1})$$
For a finite Galois extension $L/K$, iff $L$ is fixed by $I_w$ then $L\subset K_v^{nr}$ to that $L/K$ is unramified at $w|_L$. And since $L/K$ is Galois we have unramified at $w|_L$ iff unramified at all the places above $v$.

$\overline{K}^{I_w}$ is the largest extension of $K$ unramified at $w$, not the same as unramified at all the places above $v$.

When considering the largest extension unramified at a set of places, consider the subfield fixed by the all corresponding intertia groups, ie. the subfield fixed by the subgroup of $Gal(\overline{K}/K)$ generated by those intertia groups.


问:
Why is the character value sum of roots of unity?
I know that every finite representation is equivalent to a unitary representation if we choose a proper inner product("averaging") for the underlying vector space. Also, each element of unitary representation is a unitary matrix and the unitary matrix is unitarily diagonalizable.
But what does this tell us about the character value(trace of this matrix), in particular, why does it have to be the sum of roots of unity?</p>

答:
Let $\rho: G \rightarrow GL(V)$ be a complex representation of a finite group $G$.  Then if $g \in G$, there is some $k$ (the order of $g$) that $g^k = e$, the group identity.  Then if $M = \rho(g)$, then $M^k = \rho(g)^k = \rho(g^k) = \rho(e) = I$ where $I \in GL(V)$ is the identity.  So the minimal polynomial of $M$ must divide $x^k - 1$.  It follows that every eigenvalue of $M$ is a root of unity, and the trace of $M$ is the sum of the eigenvalues, hence a sum of roots of unity.

Another, more direct proof:

Let $\lambda$ be an eigenvalue of $\rho(g)$ and let $k$ be the order of $g$.  Then there is $v$ so that $\rho(g)v = \lambda v$.  Then $\rho(g^k)(v) = \rho(e)(v) = Iv = v$, and also $\rho(g^k)(v) = \rho(g)^k v = \lambda^k v$.  It follows that $\lambda^k = 1$.  So all eigenvalues of $\rho(g)$ are roots of unity, and hence the trace of $\rho$ is a sum of roots of unity.


问:
If a polynomial $f(x)$ has a positive leading coefficient then $\lim_{x\to\infty}f(x)=\infty$
This seems pretty intuitive-- but is there a rigorous $\epsilon-\delta$ proof for it?
I could prove $\lim_{x\to\infty}x^n=\infty$ quite easily- $x^n\geq x$ would complete the proof nicely- but when other terms are added then I don't know what to do-- thanks!

Body:
$$P_n(x)=a_nx^n + a_{n-1} x^{n-1} + ... + a_1x + a_0 = a_nx^n( 1 + \frac{a_{n-1}}{a_nx} + ... + \frac{a_1}{x^{n-1}} + \frac{a_0}{x_n} - a_nx^n f(x).$$
We have
$$\lim_{x \to \infty} f(x) = 1.$$
Hence
$$\lim_{x \to \infty } P_n(x)= \lim_{x \to \infty} a_nx^n = a_n\lim_{x \to \infty } x^n.$$


Body:
Matrix Question

If $A^2 = I$, $B^2=I$ and $(AB)^2=I$, then $AB = BA$

Basically, got up to $A(BA-AB)B = 0$ (by cancelling and equating terms from $I^2 = I$ and to $A^2B^2 = A^2B^2$ and using distributive laws), but that doesn't work out too well!

Thanks for help in advance!

Body: We know $AA = I, BB= I, ABAB = I$. Thus $A^{-1} = A, B^{-1} = B$. Now $$ABAB = I \\ AB = IB^{-1}A^{1} = BA$$


问:
Assigning 20 tennis players to 14 games
$20$ tennis players are scheduled to play exactly $14$ two-person games among themselves, with each player playing at least one game. Prove there must be a set of $6$ games with $12$ distinct players.

To set it up, I let $G$ be a graph on $20$ vertices (players) and $14$ edges (games). Two vertices are adjacent if and only if they play each other in a game. Note that since each player plays a game, each vertex has degree at least $1$.

Using $\displaystyle{\sum_{i=1}^{20} deg(v_i) = 28} $, we can show that at least $12$ vertices have degree $1$. Also, we can show that the largest degree any vertex can have is $8$.

I believe we should show there is a matching of size $6$ on $G$.

Is this a good method? If so, any suggestion on how to show this would be helpful. Thank you.

答: Suppose you have atmost $5$ games with distinct players. Let the set of these distinct players be $S$. Then the remaining $10$ players can't play any game among themselves because if they did, you would have $6$ games with distinct players. Thus each of these $10$ players must play a game with someone from $S$. This will make $5+10=15$ games which is a contradiction.

Title: Assigning 20 tennis players to 14 games


答: One tactic is to simply factor out the largest power of $x$. Then we have $x^n(1+c_1x^{-1}+c_2x^{-2}...)$. Then argue that the part in the parentheses goes to $1$. From there, you can make an argument along the lines of "the limit of the product is the product of the limits", but you have to make sure that you are doing so under conditions that this is valid. Another way to progress is to argue that since the sum goes to $1$, there is some $x_0$ such that if $x>x_0$, then the sum is greater than $0.5$. Then you have $f(x) > 0.5x^n$.

Title: If a polynomial $f(x)$ has a positive leading coefficient then $\lim_{x\to\infty}f(x)=\infty$

Body:
This seems pretty intuitive-- but is there a rigorous $\epsilon-\delta$ proof for it?
I could prove $\lim_{x\to\infty}x^n=\infty$ quite easily- $x^n\geq x$ would complete the proof nicely- but when other terms are added then I don't know what to do-- thanks!


问:
Evaluation of the Lebesgue Integral
I'm having some trouble with the second point of this question and I'm not completely sure that I made the first right so:

For $x\in\mathbb{R} - \{0\}$, let $f_n(x) = \frac{1}{(1 + n^2 + x^2)(\arctan (x^2))^{1/n}}$, $n \in\mathbb{N} - \{0\}$

(i) determine for which $n ∈ \mathbb{N}, f_n$ ∈ $L^1(\mathbb{R})$

(ii) evaluate  $\lim_{n\to \infty}$ $\int_{\mathbb{R}} f_n \,dm$;

For the fist part I notice that the $f_n$ are a.e. continuous so that implies that they are measurable, and that the functions are even, so I can narrow my analysis to $[0, \infty )$.

Then when $x\rightarrow\infty, \, f_n \sim \frac1{x^2\bigl({\pi\over2} \bigr)^{1 \over n}} \le {1 \over {x^2}}$ so I can say that the integral converges for every $n\in\mathbb{N} - \{0\}$

Instead when $x \rightarrow 0$,  $f_n \sim \frac1{(1+x^2)x^{2 \over n}} \le {1 \over {x^{2\over n}}}$ that converges for every $n \gt 2
$.
Then we came to the second point, here I could not find a an integral majorant to use Lebesgue's dominated convergence theorem and I have not even been able to determine if there is a chain of the type $f_1 \le f_2 \le f_3 \le \cdots \le f_n \le \cdots$ in order to use the monotone convergence theorem in order to pass the limit under sign of integral and evaluate it.

As an addition I think that the integral in the end is zero because $\lim_{n\to \infty} f_n = 0$.

Thank you very much.

答:
Just note that for any fixed $\epsilon \in(0,1)$ then $0\leqslant f_n(x)\leqslant \frac1{(1+x^2)\arctan (\epsilon ^2) }$ for all $x\in \mathbb{R}\setminus (-\epsilon ,\epsilon )$ and all $n\in \mathbb{N}$ so the dominated convergence theorem show us that
$$
\lim_{n\to\infty}\int_{\mathbb{R}\setminus (-\epsilon ,\epsilon )}f_n\mathop{}\!d \lambda =0
$$
Now note that $|x|/2\leqslant |\arctan x|$ for all $x\in (-\epsilon ,\epsilon )$. Therefore
$$
0\leqslant \int_{(-\epsilon ,\epsilon )}f_n\mathop{}\!d \lambda \leqslant \int_{(-\epsilon ,\epsilon )}\frac2{n^2|x|^{2/n}}\mathop{}\!d x =\int_{(0,\sqrt[n]{\epsilon })}\frac4n y^{n-3}\mathop{}\!d y\leqslant \int_{(0,1)}\frac4n\epsilon ^{1-3/n}\mathop{}\!d y
$$
Then the dominated convergence theorem show us that
$$
\lim_{n\to\infty}\int_{(0,\sqrt[n]{\epsilon })}\frac4n y^{n-3}\mathop{}\!d y=0\implies \lim_{n\to\infty}\int_{(-\epsilon ,\epsilon )}f_n\mathop{}\!d \lambda =0
$$
so we conclude that $\lim_{n\to\infty}\int_{\mathbb{R}}f_n\mathop{}\!d \lambda =0$.

Title: Evaluation of the Lebesgue Integral


问:
How to find the length of one segment in a triangle when it is crossed by another triangle?
The problem is as follows:

The figure from below shows a parallelogram $ABCD$ and
$\triangle\,APD$ and $QR=3\,cm\,RD=4\,cm$. Using this information find
the value of PQ.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\frac{3}{4}\,cm\\
2.&\frac{7}{3}\,cm\\
3.&\frac{10}{3}\,cm\\
4.&\frac{7}{10}\,cm\\
\end{array}$

So far the only relationship which I was able to spot was that since there's a parallelogram involved this meant.

$\angle BCA = \angle DAC.$

But that's where I'm still stuck. Is there any other relationship which can be obtained from what is given?.

Does this problem require construction or similarity or something like that?.

Please include some drawing in the answer. How, exactly, relying only in Euclidean geometry postulates can this be solved?.

答:
Triangles $\triangle QRC$ and $\triangle DRA$ are similar to each other by A.A.A

$$\frac{QC}{AD}=\frac{QR}{RD}=\frac{3}{4}$$

Hence

$$\frac{BQ}{AD}=\frac{1}{4}$$

And triangles $\triangle PBQ$ and $\triangle PAD$ are similar to each other by A.A.A

$$\frac{PQ}{PD}=\frac{PQ}{PQ+7}=\frac{BQ}{AD}=\frac{1}{4}$$

$$4PQ=7+PQ$$

$$PQ=\frac{7}{3}$$


答: Hint: Is $BA$ the inverse of $AB$. How to check that?</p>

Title: If $A^2 = I$, $B^2=I$ and $(AB)^2=I$, then $AB = BA$