答:
Great question! I was just pondering this recently. It turns out that $B \backslash G / B$ is just another way to write $G \backslash (G \backslash B \times G \backslash B)$

Check out here for how these are the same thing: A bijection between between the orbits $G$ on $G \backslash B \times G \backslash B$ and orbits of $B$ on $G \backslash B$?

Also, what Qiaochu Yuan linked is of course exactly what you are looking for if you want to understand what's happening on a deeper level. As of right now, my favorite post on MSE is Emertons post here we he explains the Bruhat Decomposition in a very intuitive way: https://mathoverflow.net/questions/15438/a-slick-proof-of-the-bruhat-decomposition-for-gl-nk

Many groups besides $SL(n)$ have Bruhat decompositions. What is required for a group $G$ to have a Bruhat decomposition is it needs to have what is called an $(B,N)$ pair. Fo $SL(n)$, $B$ (the Borel subgroup) will be the collection of upper triangular matrices and $N$ (the Weyl group) will be the normalizer of the Diagonal subgroup, whose quotient with the diagonal subgroup can be seen to be isomorphic to $S_n$.

If you read Emertons post, you will eventually read where he claims how any two Borel subgroups contain a common torus, and here is an arguement for showing that that is true: https://wildonblog.wordpress.com/2015/03/20/intersections-of-borel-subgroups/

Title: Simple description of double coset

Body:
Let $G=SL_2(\mathbb{F}_p)$, two by two matrices over $\mathbb{F}_p$ whose determinant is one. Let $H$ be the subgroup of $G$ that is upper triangular matrices.
Is there any simple way to describe ${H}\backslash^{G}/_{H}$?

Context is that in applying Mackey's irreducibility creterion, I am trying to find an element in ${H}\backslash^{G}/_{H}$ that is not identical to identity, but I have trouble understanding this double coset.


问:
Properties of P-matrices?
I have two $n \times n$ matrices satisfying $0 \leq A \leq B$, where $\leq$ denotes element-by-element comparison. I'm trying to show that if $(I-B)^{-1}$ exists and all of its elements are non-negative, then $(I-A)^{-1}$ exists and all its elements are non-negative. Not sure where to start. I'm deliberately not assuming that $\sum_{j=1}^{n} B_{ij} < 1$, which would imply that $(I-B)^{-1} = \sum_{k=0}^{\infty} B^k \geq \sum_{k=0}^{\infty} A^k =  \geq 0$.

答:
The spectral radius of $B$ is necessarily $<1$.

Let $v$ be a Perron vector of $B$. As $(I-B)^{-1}$ is nonnegative, we have $0\le (I-B)^{-1}v=(1-\rho(B))^{-1}v$. Hence $(1-\rho(B))^{-1}\ge0$ and $\rho(B)<1$.

It follows that $\rho(A)\le\rho(B)<1$ and the Neumann series $I+A+A^2+\cdots$ converges to $(I-A)^{-1}$. Hence $(I-A)^{-1}$ is nonnegative.

ContentLicense: CC BY-SA 4.0

Title: Properties of P-matrices?


问:
Find area of right triangle △ with circumcircle $O$ and tangent circle $K$.
The triangle $ABC$, with $\angle A = 90^\circ$,
is inscribed in the circle $\mathcal{O}$.
The circle $\mathcal{K}$ is tangent to the circle $\mathcal{O}$,
the sides $AB$ and $AC$.
Suppose that the radius of $\mathcal{O}$ is $3$
and the radius of $\mathcal{K}$ is $2$. Find the area of $\triangle ABC$.

I drew this problem out by making circle $\mathcal{K}$
internally tangent to circle $\mathcal{O}$.
I was able to make it tangent to $AB$ and $AC$.
However, I have gotten nowhere using trigonometry to solve this.

I also tried to name the sides and angles
to find the area, which did not work, either

答:
Note that BK $=2\sqrt2$, the diagonal of the square of side length $2$ and $\triangle$BOK is a right triangle because of its sides, $1$, $2\sqrt2$ and $3$, satisfying the Pythagorean rule. Then, $\sin\theta =\frac{KO}{BO}= \frac13$ and
$$\sin\alpha \cos\alpha = \sin(45-\theta)\cos(45-\theta)=\frac{1}2\cos2\theta=\frac12-\sin^2\theta=\frac7{18}
$$
Thus, the area of $\triangle$ABC is
$$Area_{ABC}=\frac12AB\cdot BC=\frac12AC^2\sin\alpha\cos\alpha = \frac12\cdot6^2\cdot\frac7{18}=7
$$


问:
What type of stochastic process satisfy $\mathrm{Var}[X_t]\mathrm{Var}[X_s] = \mathrm{Cov}[X_t,X_s]$ for all $t,s \in \mathbb R^+$?
Let $X=(X_t)_{t\in \mathbb R^+}$ be an $L^2$ stochastic process. What does it say about $X$ if $\mathrm{Var}[X_t]Var[X_s] = \mathrm{Cov}[X_t,X_s]$ for all $t,s \in \mathbb R^+$ ? What does it says about $X$ if $\mathrm{Var}[X_t]\mathrm{Var}[X_s] \neq \mathrm{Cov}[X_t,X_s]$ for all $t,s \in \mathbb R^+$ ?

Is there a special class of processes that satisfy one of the above ?

Now we repeat the same questions, but we suppose that $X$ is a Gaussian process. Do we learn something new ?

答:
With $s=t$ the condition is
$$\mathrm{Var}(X_s)=\mathrm{Var}(X_s)^2,$$
which forces, that
$\mathrm{Var}(X_s)=1$ for all $s$. And thus
$$\mathrm{Cov}(X_s,X_t) = 1^2 = \sqrt{\mathrm{Var}(X_s)}\sqrt{\mathrm{Var}(X_t)},$$
which implies, that the correlation between $X_s$ and $X_t$ is $1$ for all $s$ and $t$, and therefore $X_t$ is almost surely a linear function of $X_s$, that is
$$X_t = aX_s + b$$
for some $a$ and $b$. It is clear from the covariance condition, that $a=1$ and we can see, that $b=\mathbb{E}[X_t - X_s]$. Thus we may write
$$X_t = X_0 + f(t),$$
where $f(t)$ is the deterministic function $f(t)=\mathbb{E}[X_t-X_0]$. Also any proces defined as $X_t := X_0 + f(t)$ with $\mathrm{Var}(X_0)=1$ and $f$ some arbitrary function, will satisfy the given condition.


问:
How to prove that two sequence formulas are the same
I derived the formula for the nth term of a sequence to be $$\frac{1-\displaystyle\sum_{i=1}^{n-1} a_i}{2},$$ but upon inspection this appeared to be the same as $$\left(\frac{1}{2}\right)^n.$$ I was wondering how to prove that these are the same more rigorously?

答:
Since $\left( \frac 12 \right)^n $ is a geometric sequence you may want to prove $a_n$ is one too. Indeed
$$
a_{n+1}=\frac{1-\sum_{i=1}^n a_i}{2}=\frac{1-\sum_{i=1}^{n-1} a_i}{2}-\frac{a_n}{2}=a_n-\frac{a_n}{2}=\frac{1}{2} a_n.
$$

Of course you need $a_1=\frac 12$.


问:
Is the set of all natural numbers acctually a proper class?
I have been searching about the difference between a set and a class. The main definitions I found can be resumed in “all sets are classes, but not all classes are sets. If a class is not a set, then it is a proper class”.

When I was looking at ZF axioms (https://plato.stanford.edu/entries/set-theory/ZF.html), I saw the axiom of union, and then start thinking if the set of natural numbers is a set, since 1, for example, is not a set and it looks to be impossible to apply it in the axiom of union, because $\bigcup\mathbb{N}$ doesn't look to make any sense.

So I ask: is $\mathbb{N}$ a set or a proper class?

P.S.: I also don't doubt that one of my premisses is wrong.

答:
The usual approach in ZF set theory is to implement the natural numbers inductively as sets: $0$ is defined to be the empty set, and for each natural number $n$, we define $n + 1 = n \cup \{n\}$. (So, for example, $2 = \{0, 1\} = \{\{\}, \{\{\}\}\}$.)

The axiom of infinity says that there is a set $S$ that contains the empty set and such that, for all $x \in S$, we have $x \cup \{x\} \in S$. This tells us that the class $\mathbb{N}$ of all natural numbers is in fact a set. So, essentially, $\mathbb{N}$ is a set because one of the axioms of ZF says it's a set. (If we omit the axiom of infinity, there are models of the resulting set theory where all sets are finite, because none of the other axioms requires the existence of an infinite set.)


答:
There's a gap between how we normally do mathematics as an "informal-but-rigorous" activity and how we formalize mathematics inside $\mathsf{ZF}$ or similar. To get there from here we have to choose some method of implementation of our standard mathematical ideas into the set-theoretic framework. Specifically, $\mathsf{ZF}$ adopts the position that everything is a set, and that means we have to find ways to "encode" the not-really-set-flavored mathematics we do day-to-day in terms of just $\in$.

For the natural numbers this is done via the finite (von Neumann) ordinals - specifically, a "natural number" in the sense of $\mathsf{ZF}$ is an ordinal which is not in bijection with any of its proper subsets. The first few natural numbers in this sense are $$0=\emptyset, 1=\{0\}=\{\emptyset\}, 2=\{0,1\}=\{\emptyset,\{\emptyset\}\}, ..., n+1=\{0,1,...,n\}=n\cup\{n\},...$$

From this we can directly calculate $\bigcup\mathbb{N}$:

 It's exactly $\mathbb{N}$ itself! Neat, huh?

There are a few tricks to "set-ifying" mathematics. The above is one of these: the idea to represent natural numbers by finite ordinals "gets us off the ground." Another fundamental point is the observation that we can talk about ordered pairs - and hence relations, functions, etc. - using set theory alone: see here. Note that these aren't axioms (or theorems), but rather methodological choices we make about how to translate mathematics into the narrow language of set theory; the $\mathsf{ZF}$ axioms come into the picture when we prove the desired facts about these implementations. E.g. it's not at all obvious at first how to define addition of natural numbers in terms of set theory alone, but there's a $\mathsf{ZF}$-theorem which does the job (the recursion theorem).

Now one might reasonably object to this "all-things-are-sets" framework for mathematics; see e.g. the discussion here, and more tangentially here. Opinions on the matter vary.

So what does make a class fail to be a set?

Well, in $\mathsf{ZF}$ it's ultimately all down to size. For a class $C$, exactly one of the following holds:

There is a surjection from $C$ to the proper class $Ord$ of all ordinals (put another way, $Ord$ is the smallest proper class in a precise sense).

$C$ is a set.

(Actually the above is a bit slippery since $\mathsf{ZF}$ can't talk about classes directly, but I'm ignoring this point.)

Title: Is the set of all natural numbers acctually a proper class?

Body:
I have been searching about the difference between a set and a class. The main definitions I found can be resumed in “all sets are classes, but not all classes are sets. If a class is not a set, then it is a proper class”.

When I was looking at ZF axioms (https://plato.stanford.edu/entries/set-theory/ZF.html), I saw the axiom of union, and then start thinking if the set of natural numbers is a set, since 1, for example, is not a set and it looks to be impossible to apply it in the axiom of union, because $\bigcup\mathbb{N}$ doesn't look to make any sense.

So I ask: is $\mathbb{N}$ a set or a proper class?

P.S.: I also don't doubt that one of my premisses is wrong.


问:
Let $f(x)$ be a differentiable function satisfying $f'(x)=f(x)+1$
I need help with this exercise.

Let $f(x)$ be a differentiable function satisfying $f'(x)=f(x)+1$

a) Let $g(x)=f(x)e^{-x}$. Find $g'(x)$.

So this is what I did

$$g'(x)=f'(x)\cdot e^{-x}+e^{-x}\cdot -1 \cdot f(x)$$

$$g'(x)=f'(x)\cdot e^{-x}-e^{-x} \cdot f(x)$$

$$g'(x)=e^{-x}[f'(x)-f(x)]$$

$$g'(x)=e^{-x}[f(x)+1-f(x)]$$

$$g'(x)=e^{-x}[1]$$

$$g'(x)=e^{-x}$$

b) Find all the differentiable functions $f$ satisfying $f'(x)=f(x)+1$

I'm stuck with this one. Please help!

答:
You should really have explained in your question that you needed to use (a) to solve (b).

Spoilers ahead: click each one for a hint. (In the following, DE stands for Differential Equation)

 Can you rewrite the DE for f using what you got in (a)?

 $f(x)= g(x)e^x$, so what is f' using the product rule?

 Anyway, have you learned about "integrating factor", or "exact differentials"?


问:
Proving Monotone of $\cos(x)$.
How can we show that $\cos(x)\geq0$ whenever $x\in(-\pi/2,\pi/2)$ where $\cos(x)$ is represented by its infinite series representation?

答: $\cos$ is continuous and $\cos(x)\neq 0$ for all $x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ using your definition of $\pi$, thus $\cos(x)>0$ for all $x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ or $\cos(x)<0$ for all $x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Since $\cos(0)=1>0$, we have $\cos(x)>0$ for all $x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

Title: Proving Monotone of $\cos(x)$.


问:
Why is the cross product is alternating?
In the book An introduction to manifolds by Tu, Loring W, it says that

The cross product $v\times w$ on $\mathbb R^3$ is alternating.

However in the definition of alternating, it says that

A $k$-linear function $f:V^k\to\mathbb R$ is symmetric, if
\begin{align*}
    f\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right)=f\left(v_{1}, \ldots, v_{k}\right)
\end{align*}
for all permutations $\sigma\in S_k$; it is alternating, if
\begin{align*}
    f\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right)=(\operatorname{sgn} \sigma) f\left(v_{1}, \ldots, v_{k}\right)
\end{align*}
for all $\sigma\in S_k$.

However, I can't understand why the cross product is alternating. I think that the cross product $\times$ can be viewed as a function $\mathbb R^3\times \mathbb R^3\to\mathbb R^3$, but not any function that maps to $\mathbb R$. So why this is alternating?

答:
If you expand the definition of alternating to linear functions from $\mathbb{R}^n$ to $\mathbb{R}^m$, one has the following.

If $u$ and $v$ are linearly independant vectors, then $u\times v$ is defined to be the unique vector such that

$\|u\times v\| = \|u\|\|v\||\sin(u,v)|$

$u\times v$ is orthogonal to the plane generated by $u$ and $v$

$(u,v,u\times v)$ is a positively oriented basis

Thus, if $(u,v,u\times v)$ is positively oriented, then $(v,u,u\times v)$ is negatively oriented. Hence, $(v,u,-(u\times v))$ is again positively oriented. The norm of $-(u\times v)$ is the good one, and it is orthogonal to $u$ and $v$. By uniqueness, $-(u\times v) = v\times v$.

As the cross product is bilinear, and the only permutations on a set of two elements are the identity or the transposition, it follows that the cross product is alternating.


问:
Probability of a number N being reduced to a nonpositive value after K trials
Let's say we have an integer $N$. In each of $K$ trials, that number is reduced by a random integer number from the uniform interval $[0, M]$ (so if we had $M = 5$, then a number $N$ in each trial could be reduced by either $0$, $1$, $2$, $3$, $4$ or $5$, each with a probability $\frac{1}{6}$). What is the probability that the number $N$ will be less than or equal to zero after $K$ trials? As an example, for $N=2$, $M=1$ and $K=3$ the answer is $0.5$.

I have to write the solution as an algorithm in some programming language, so the brute force solution would be to simply enumerate every permutation for a total of $(M+1)^K$ and count cases when $N \le 0$ at the end. But for this problem, $M$ and $K$ could be up to 1000, and then this complexity becomes $1000^{1000}$ which is intractable.

So I was wondering if there is some math formula that could help me avoid generating all the permutations?

答:
I'll think about the problem as adding from 0 to N, in steps of less than or equal to M.

Calculate the distribution of the integers you can hit, after 1 trial, and then, using that distribution, calculate the distribution after 2 trials, and so on. At trial n, you're doing at most order (Mn) calculations, so the whole thing can be done in quadratic complexity. There are probably better ways, and perhaps you can ask stack overflow.

EDIT: (With an example). (I'm on a phone so pardon the lack of formatting

If N = 1000, and M = 3, at each trial n, you should come up with such a table:

First trial (n=1)

0 1 2 3

0.25 0.25 0.25 0.25

For n=2, Consider each cell in the previous trial. For each cell, where can it hit?

For the 0 cell

0 1 2 3

0.25 0.25 0.25 0.25

For the 1 cell

1 2 3 4

0.25 0.25 0.25 0.25

For the 2 cell

2 3 4 5

0.25 0.25 0.25 0.25

For the 3 cell

3 4 5 6

0.25 0.25 0.25 0.25

Now add up all these probabilities and divide by M+1 = 4, to make sure the resulting sum is still 1.


问:
CDF of the sum of 2 random variables uniformly distributed
I'm trying to understand the solution to a homework problem and am getting nowhere.  The explanation given is really unspecific and I have no idea what the thought process was behind solving this problem.  Any insight into what this solution is supposed to mean would be greatly appreciated!

Thank you so much!

答:
$F_Z(z)=P(Z \le z) = P(X+Y \le z)$

Because $(X,Y)$ is uniformly distributed in the triangle $D = \{0 \le y \le x \le 1\}$, for any subset $A$ of the plane, we have $P(X+Y \in A) = \frac{\text{area}(A \cap D)}{\text{area}(D)}$.

If you can draw a picture of $A = \{(x,y) : x + y \le z\}$, you can use what the solution calls "simple geometry" to compute $\text{area}(A \cap D)$.


问:
A function is given by $f(x)=x^4−2x^3−2x^2+6x−3$. What is the solution set for $f(x)\geqslant 0$?
A function is given by  $f(x)=x^4−2x^3−2x^2+6x−3$. What is the solution set for  $f(x)\geqslant 0$?

答:
Hint

$f(x) = (x^4 - 2x^2 - 3) - (2x^3 - 6x).$

$(x^2 - 3)$ is a factor in both of the above terms.

Solution

$$f(x) = (x^2 - 3) \times [(x^2 + 1) - 2x] ~=~ (x^2 - 3) \times (x - 1)^2.$$

Therefore, $f(x)$ hits the $x$-axis at $~-\sqrt{3}, ~1, ~$ and $~+\sqrt{3}.$

Since the tag is pre-calculus, I will complete the solution without Calculus.

To determine for which regions $f(x)$ is non-negative, explore a specific value inside of the pertinent regions.  The idea is that if a specific value inside of a region is negative, the entire region must be negative.  This follows because otherwise $f(x)$, being a continuous function, would have to hit the $x$-axis at some location other than $~-\sqrt{3}, ~1, ~$ and $~+\sqrt{3}.$

By a symmetrical argument, if a specific value inside of a region is positive, the entire region must be positive.

The exploration can be done without a calculator by employing the following shortcut:

$f(x) = (x^2 - 3) \times (x - 1)^2.$

When $x \neq 1,~$ the 2nd factor above (being the square of a number) will always be positive.  Therefore, for any value of $x$ that is not a root, the sign of $f(x)$ will be completely determined by the sign of $(x^2 - 3).$  Therefore

$$f(x) > 0 \iff (x^2 - 3) > 0.$$

Therefore

$$f(x) \geq 0 \iff x~ \text{is in one of the following regions:}$$

$$[x \leq -\sqrt{3}] ~\text{or} ~[x = 1] ~\text{or} ~[x \geq +\sqrt{3}].$$


问:
How many zeros does $3z^5+21z^4+5z^3+6z+7$ have in $D$?
How many zeros does $f(z)=3z^5+21z^4+5z^3+6z+7$ have in closed unit disc $\overline{D}$?

I tried using Rouche's theorem which states that for $\gamma$, a closed curve in a region $\Omega$, if $f$ and $g$ are analytic in $\Omega$ and satisfy $|f(z)+g(z)|<|f(z)|+|g(z)|$ for all $z\in \gamma$ then $f$ and $g$ have the same number of zeros enclosed in $\gamma$.
I let $f(z)=3z^5+21z^4+5z^3+6z+7$ and $g(z)=-21z^4$ then $|f(z)+g(z)|=|3z^5+5z^3+6z+7| \leq 3+5+6+7=21 \leq |-21z^4|=21$ on $\gamma$ function $|z|=1$. However, we need strictly less than relation; ie. $|f(z)+g(z)|<|f(z)|+|g(z)|$ for $f$ and $g$ to have same number of of zeros.

So I can't figure out how to find the number of zeros for the function.

答:
Taking David's hint, we want to show that $f(z) \neq 0$ whenever $|z| = 1$. Suppose by way of contradiction that this could be the case. Then
$$3z^5 + 21z^4 + 5z^3 + 6z + 7 = 0 \implies |-21z^4| = 21 = |3z^5 + 5z^3 + 6z + 7|$$
and here we may appeal to the triangle inequality, which says that
$$|3z^5 + 5z^3 + 6z + 7| \leq |3z^5| + |5z^3| + |6z| + 7 = 21$$
which equality if and only if all of the addends are each non-negative (real) multiples of some complex number $z$. So this would force $z = 1$ (as $6z$ must be a non-negative multiple of $7$). But you can verify $f(1) \neq 0$, so you can conclude that $f(z)$ is never $0$ whenever $|z| = 1$.


答:
You're correct that using the given definition strictly, this doesn't make sense. However, there are a couple small changes in interpretation you could choose to make it work:

Extend the definition of a symmetric and alternating functions to apply to $k-$linear functions $V^k \to W$ for any vector space $W$. (You can use the same defining formulae.) In the case of the cross product, it's alternating in this sense simply because $u \times v = - v \times u.$

"Lower an index" on the cross product via the metric, i.e. consider $f(u,v,w) = (u \times v) \cdot w.$ This is then a 3-tensor on $\mathbb R^3$ that co-incides with the determinant.

Title: Why is the cross product is alternating?

Body:
In the book An introduction to manifolds by Tu, Loring W, it says that

The cross product $v\times w$ on $\mathbb R^3$ is alternating.

However in the definition of alternating, it says that

A $k$-linear function $f:V^k\to\mathbb R$ is symmetric, if
\begin{align*}
    f\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right)=f\left(v_{1}, \ldots, v_{k}\right)
\end{align*}
for all permutations $\sigma\in S_k$; it is alternating, if
\begin{align*}
    f\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right)=(\operatorname{sgn} \sigma) f\left(v_{1}, \ldots, v_{k}\right)
\end{align*}
for all $\sigma\in S_k$.

However, I can't understand why the cross product is alternating. I think that the cross product $\times$ can be viewed as a function $\mathbb R^3\times \mathbb R^3\to\mathbb R^3$, but not any function that maps to $\mathbb R$. So why this is alternating?


问:
How to find roots of $X^5 - 1$?
How to find roots of $X^5 - 1$? (Or any polynomial of that form where $X$ has an odd power.)</p>

答: Hint: Use De Moivre's theorem and note that $$1^{1/5} = \Bigl( \cos 2m\pi + i \sin{2m\pi} \Bigr)^{1/5} \ ;\qquad m \in \mathbb{Z}$$</p>

ContentLicense: CC BY-SA 3.0

Title: How to find roots of $X^5 - 1$?


问:
Transitivity of union of two transitive relations
I have a question concerning proving properties of Relations. The question is this: How would I go about proving that, if R and S (R and S both being different Relations) are transitive, then R union S is transitive?

The answer is actually FALSE, and then a counter example is given as a solution in the book.

I understand how the counterexample works as explained in the book, but what I don't understand is, how exactly they arrive to the conclusion that the statement is actually false.

Basically I can see myself giving a proof that if that for all values (x,y,z) in R and S, if (x,y) is in R and (y,z) is in R, (x, z) is in R since R is transitive. And if (x,y) is in S and (y,z) is in R, (x,z) is in S since S is transitive. Since (x,z) is in both R and S, the intersection is true. But why wouldn't the union of R and S be true as well?

Is it because that the proof cannot be ended with "since (x,z) is in both R and S, (x,z) can be in R or S"?

Basically, that a proof can't be ended with an OR statement at the end?

Hope that all makes sense.

答:
The existence of any counterexample is a proof that the statement is false. The statement is of the form

"For all transitive relations, if $S$ and $R$ are transitive relations, then $S\cup R$ is transitive." $\quad (*)$

Hence, to prove this statement is false, it suffices to prove the existence of two transitive relations whose union is not transitive. This is what a counterexample proves.

That is, the negation of $(*)$ is

$\lnot (*)$ "It is NOT THE CASE that for all transitive relations, S, R, that $S \cup R$ is transitive."

which is equivalent to

$\lnot (*)$"There exists transitive relations $S, R$ such that $S \cup R$ is NOT transitive."


问:
Is it possible to construct matrices $A$ and $B$ such that $AB$ is invertible?
Let $A$ be an $n \times m$ matrix and let $B$ be an $m \times n$ matrix. Also, let $n$ be less than $m$. Is it possible to construct $A$ and $B$ such that $AB$ is invertible?

I know that $AB$ must have dimensions $n \times n$. If both $A$ and $B$ are full rank, then $AB$ should also be full rank. Thus, $AB$ should be invertible. However, I can't find a line of reason for why $AB$ should also be full rank.

Any guidance is greatly appreciated!

答:
Edit: My answer adresses the original version of the question.

A short answer: no.  $A$ and $B$ have $m$ columns and $n$ rows respectively, and you let $n < m$ in the question, so the matrix product $AB$ can't even be defined.


问:
Converting statements to propositional logic and conjunctive normal form (CNF)
I have the following task:

Defne the following statements in propositional logic:

The system heats if the cover is closed and the power is on.

If the lamp shines, the power is on.

The lamp shines.

The cover is closed.

Rewrite above statements to conjunctive normal form and prove by contradiction and resolution that the system heats.

My propositional logic:

```<code>systemHeats → coverIsClosed ∧  powerIsOn
lampShines → powerIsOn
lampShines
coverIsClosed
</code>```

My CNF:

```<code>(¬ systemHeats ∨ coverIsClosed) ∧ (¬ systemHeats ∨ powerIsOn)
¬ lampShines ∨ powerIsOn
lampShines
coverIsClosed
</code>```

Is this correct?

I'm supposed to prove by contradiction and resolution that the system heats. Can you link me to an explanation on how to do so?

答:
No it is not correct.  You need to revise the first premise.  "This if that" means $\textsf{that}\to\textsf{this}$.

To use resolution, you add an additional premise --the negation of the aledged conclusion-- aiming to derive a contradiction.

Thus your initial list should be:

```<code>(coverIsClosed ∧ powerIsOn) → systemHeats
lampShines → powerIsOn
lampShines
coverIsClosed
~systemHeats
</code>```

Now, put these in CNF (which you've shown that you can do okay) and resolve.

Body:
No it is not correct.  You need to revise the first premise.  "This if that" means $\textsf{that}\to\textsf{this}$.

To use resolution, you add an additional premise --the negation of the aledged conclusion-- aiming to derive a contradiction.

Thus your initial list should be:

(coverIsClosed ∧ powerIsOn) → systemHeats
lampShines → powerIsOn
lampShines
coverIsClosed
~systemHeats

Now, put these in CNF (which you've shown that you can do okay) and resolve.


问:
How to "vertically" scale an arbitrary polar equation?
Given an arbitrary polar equation $r = P(\theta)$, which represents some graph, when plotted, is it possible to write a new $r = Q(\theta) = f(P(g(\theta)))$, such that plotting $Q(\theta)$ produces a new graph, which is exactly the same as $P(\theta)$, except that it is scaled along one axis?

答:
No, not in the way you describe. Your $f$ takes radial values as inputs, and it cannot distinguish between a radial value when $\theta$ is "horizontal" (a multiple of $\pi$) and when it is "vertical". For example, if $P$ is constant (so the graph is a circle) then $g$ is irrelevant, and $f$ would have no way to stretch vertically without also stretching horizontally.

If you want to scale vertically by $k$, then first assume we have an "old" point (before the stretch transformation) where the angle is in $[0,\pi/2)$. And after the stretch, it is at a "new" location, but still it would have angle in $[0,\pi/2)$.

$$\begin{align}
y_{\text{new}}&=k\cdot y_{\text{old}}&x_{\text{new}}=x_{\text{old}}\\
r_{\text{new}}\sin\left(\theta_{\text{new}}\right)&=k\cdot r_{\text{old}}\sin\left(\theta_{\text{old}}\right)\\
r_{\text{new}}\sin\left(\arctan\frac{y_{\text{new}}}{x_{\text{new}}}\right)&=k\cdot r_{\text{old}}\sin\left(\theta_{\text{old}}\right)\\
r_{\text{new}}\frac{\frac{y_{\text{new}}}{x_{\text{new}}}}{\sqrt{1+\frac{y_{\text{new}^2}}{x_{\text{new}^2}}}}&=k\cdot r_{\text{old}}\sin\left(\theta_{\text{old}}\right)\\
r_{\text{new}}\frac{k\frac{y_{\text{old}}}{x_{\text{old}}}}{\sqrt{1+k^2\frac{y_{\text{old}^2}}{x_{\text{old}^2}}}}&=k\cdot r_{\text{old}}\sin\left(\theta_{\text{old}}\right)\\
r_{\text{new}}\frac{\tan\left(\theta_{\text{old}}\right)}{\sqrt{1+k^2\tan^2\left(\theta_{\text{old}}\right)}}&= r_{\text{old}}\sin\left(\theta_{\text{old}}\right)\\
r_{\text{new}}&=P\left(\theta_{\text{old}}\right)\cos\left(\theta_{\text{old}}\right)\sqrt{1+k^2\tan^2\left(\theta_{\text{old}}\right)}\\
r_{\text{new}}&=P\left(\theta_{\text{old}}\right)\sqrt{\cos^2\left(\theta_{\text{old}}\right)+k^2\sin^2\left(\theta_{\text{old}}\right)}\\
\end{align}$$

So multiplying $P(\theta)$ by $\sqrt{\cos^2\left(\theta\right)+k^2\sin^2\left(\theta\right)}$ (not composing by some function) will have the desired effect.

To get a plottable curve for the full demonstration:

$$\begin{align}
\theta_{\text{old}}&=\arctan\left(\frac{y_{\text{old}}}{x_{\text{old}}}\right)\\
&=\arctan\left(\frac{y_{\text{new}}}{kx_{\text{new}}}\right)\\
&=\arctan\left(\frac{\tan\left(\theta_{\text{new}}\right)}{k}\right)\\
\end{align}$$

Compose this into the above for something like $r=P(g(\theta))\cdot f(\theta)$. Here is what it looks like in a Desmos plot.

I'll leave it to you to explore what happens for a point with angle outside of $[0,\pi/2)$. The details with $\arctan$ and the sign on trig functions need modifications.


答:
If $A$ is $n\times m$ and $B$ is $m\times n$, with $m\lt n$, then $AB$ cannot be invertible. If $n\lt m$, then yes, it can be, but it doesn’t have to be.

If $n\lt m$, one example would be to let $A$ have first $n$ columns be the $n\times n$ identity and then fill in with zero columns; and let $B$ have first $n$ rows be the $n\times n$ identity, and the rest of the rows be zero. i.e.
$$\begin{align*}
 A &= \left(\begin{array}{ccccccc}
1 & 0 &\cdots &0 & 0 & \cdots &0\\
0 & 1 & \cdots & 0 & 0 & \cdots &0\\
\vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 1 & 0 & \cdots &0\end{array}\right)\\
B&=\left(\begin{array}{cccc}
1 & 0 & \cdots& 0\\
0 & 1 & \cdots& 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 1\\
0 & 0 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 0\end{array}\right).
\end{align*}$$
Then $AB$ is the $n\times n$ identity.

So it is possible for $AB$ to be invertible. However, it need not be invertible necessarily. For instance, just move the identity matrix in $A$ one column to the right, and the product of $A$ and $B$ will not be full rank. For  a worst case scenario, take $m=2n$, and let $A$ have $n$ columns of zeros followed by the identity matrix, and leave $B$ the same. Then $A$ and $B$ both have full rank, but $AB$ is the zero matrix.

On the other hand, if $m\lt n$, then note that the rank of $B$ is at most $m$; that means that the rank of $AB$ is at most $m$, and hence cannot equal $n$ (which you would need for $AB$ to be invertible). Alternatively, because $B$ has more columns than rows, there exists a nonzero vector $\mathbf{v}$ such that $B\mathbf{v}=\mathbf{0}$. But then $(AB)\mathbf{v}=\mathbf{0}$, with $\mathbf{v}\neq\mathbf{0}$, which again tells you $AB$ cannot be invertible.

Title: Is it possible to construct matrices $A$ and $B$ such that $AB$ is invertible?

Body:
Let $A$ be an $n \times m$ matrix and let $B$ be an $m \times n$ matrix. Also, let $n$ be less than $m$. Is it possible to construct $A$ and $B$ such that $AB$ is invertible?

I know that $AB$ must have dimensions $n \times n$. If both $A$ and $B$ are full rank, then $AB$ should also be full rank. Thus, $AB$ should be invertible. However, I can't find a line of reason for why $AB$ should also be full rank.

Any guidance is greatly appreciated!


答:
Suppose $(x,y)$ is in $R$, and $(y,z)$ is in $S$, but $(x,z)$ is in neither.

Then $R$ and $S$ could both be transitive, but  their union wouldn't be, because it would be missing $(x, z)$.

So you can't prove that the union of two transitive relations is also transitive, because it is not always true.

Title: Transitivity of union of two transitive relations

Body:
I have a question concerning proving properties of Relations. The question is this: How would I go about proving that, if R and S (R and S both being different Relations) are transitive, then R union S is transitive?

The answer is actually FALSE, and then a counter example is given as a solution in the book.

I understand how the counterexample works as explained in the book, but what I don't understand is, how exactly they arrive to the conclusion that the statement is actually false.

Basically I can see myself giving a proof that if that for all values (x,y,z) in R and S, if (x,y) is in R and (y,z) is in R, (x, z) is in R since R is transitive. And if (x,y) is in S and (y,z) is in R, (x,z) is in S since S is transitive. Since (x,z) is in both R and S, the intersection is true. But why wouldn't the union of R and S be true as well?

Is it because that the proof cannot be ended with "since (x,z) is in both R and S, (x,z) can be in R or S"?

Basically, that a proof can't be ended with an OR statement at the end?

Hope that all makes sense.


问:
Extending Vectors to Invertible Matrices
Let $\mathbb{F}$ be a field and suppose $a=(a_1,\ldots,a_n)\in\mathbb{F}^n$ and $b=(b_1,\ldots,b_n)\in\mathbb{F}^n$ are such that $\sum_{k=1}^na_kb_k=1$. Is it always possible to find $n\times n$ matrices $A$ and $B$ such that the first row of $A$ is $a$, the first column of $B$ is $b$, and $AB=I$?

To me the answer is clearly yes but I am having some trouble proving this. I am fairly certain it is not needed but $\mathbb{F}$ is algebraically closed.

For those wondering what the context is for this problem, I am trying to determine whether two subalgebras are conjugate and I can construct an inner automorphism provided this fact always holds.

答:
Yes, it is always possible. It should be straightforward that with the given information, we can always complete $(a_1, a_2, \cdots, a_n)$ to a full-rank matrix $A$ with $A b = \hat{e}_1$ (where $\hat{e}_1$ is the first column of the identity, and is often referred to as the first standard basis vector). Indeed, we just need to find $n - 1$ vectors $\{v_2, \cdots, v_n\}$ that are linearly independent in the space $b^\perp = \{x: x^T b = 0\}$ to serve as the remaining $n - 1$ rows. This will always be possible as $\dim(b^\perp) = n - 1$. Since $a^T b = 1 \neq 0$, it will follow that $\{a, v_2, \cdots, v_n\}$ is linearly independent (why).

From here, we just need to choose the columns $b_i$ of $B$ so that $A b_i = \hat{e}_i$. As $A$ is invertible, this equation always has a solution.


答:
Alternative:

Let $h(z)=21z^4+7$, and $g(z)=3z^5+5z^3+6z$.

So $|h(z)|>14$ for all $z \in \{z \in \mathbb{C} : |z|=1\} \setminus \{e^{i\pi/4},e^{3i\pi/4}\}$, and $|g(z)| \leq 14$ for all $z \in \{z \in \mathbb{C} : |z|=1\}$. If $z=e^{\pm i\pi/4}$, then $|h(z)|=14$ but $|g(z)|=|3z^4+5z^2+6|=|-3 \pm 5i+ 6|=|3 \pm 5i|=\sqrt{34} <14$. Thus $|h(z)|>|g(z)|$ for all $z \in \{z \in \mathbb{C} : |z|=1\}$. By Rouche’s Theorem, $h$ and $h+g=f$ have the same number of zeros in the unit disc.

Note: $h(z)=7(3z^4+1)$.

Though I think you should use the idea you started with combined with the advice others have provided here for your solution.

Title: How many zeros does $3z^5+21z^4+5z^3+6z+7$ have in $D$?

Body:
How many zeros does $f(z)=3z^5+21z^4+5z^3+6z+7$ have in closed unit disc $\overline{D}$?

I tried using Rouche's theorem which states that for $\gamma$, a closed curve in a region $\Omega$, if $f$ and $g$ are analytic in $\Omega$ and satisfy $|f(z)+g(z)|<|f(z)|+|g(z)|$ for all $z\in \gamma$ then $f$ and $g$ have the same number of zeros enclosed in $\gamma$.
I let $f(z)=3z^5+21z^4+5z^3+6z+7$ and $g(z)=-21z^4$ then $|f(z)+g(z)|=|3z^5+5z^3+6z+7| \leq 3+5+6+7=21 \leq |-21z^4|=21$ on $\gamma$ function $|z|=1$. However, we need strictly less than relation; ie. $|f(z)+g(z)|<|f(z)|+|g(z)|$ for $f$ and $g$ to have same number of of zeros.

So I can't figure out how to find the number of zeros for the function.


问:
Supose $N$ is a normal subgroup of order $p^n$. Then show that $\sigma (N)=N$
Let $G$  be a group of order $p^nm, \ p \nmid m$. Supose $N$ is a normal subgroup of order $p^n$. Then show that $\sigma (N)=N$ for any  automorphism $\sigma \in \text{Aut}(G)$.

I have tried in the way-

Suppose $H$ be the smallest subgroup of $G$ containing both $\sigma(N)$ and $N$.

My target is to show $|H|=p^n$.

One way to show this, we need to show $H\subset N$.

Suppose $H \nsubseteq N$, then  there exists at least an element, say, $g \in G \setminus \{H\}$  such that order of $g$  doesn't divide $p^n$.

How to show contradiction here ?

Edit:

I got the following answers but if we can prove the question in my direction of approach.

答:
Ok, here is a proof/hint which avoids using the Sylow theorems (again, not sure why this is desirable). We have that $$\lvert \sigma(N) N \rvert = \frac{\lvert \sigma(N) \rvert \lvert N \rvert}{\lvert \sigma(N) \cap N \rvert} = \frac{p^{2n}}{\lvert \sigma(N) \cap N \rvert}.$$

Since $N$ is normal, $\sigma(N)N$ is a subgroup of $G$, so its order divides $p^n m$. What does this tell you about the order of $\sigma(N) \cap N$?


答:
$AB$ need not be invertible. To see this, take\begin{equation*}
A =
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
\end{pmatrix}
\end{equation*} \begin{equation*}
B =
\begin{pmatrix}
1 & 2\\
2 & 4\\
3 & 1
\end{pmatrix}
\end{equation*} Notice $A$ and $B$ both have full rank, but \begin{equation*}
AB =
\begin{pmatrix}
1 & 2 \\
2 & 4 \\
\end{pmatrix}
\end{equation*} does not.


问:
Finding the largest of $S_1,S_2,S_3,\cdots,S_{13}$
Let $S_n$ be the sum of the first $n$ terms of a arithmetic sequence $\{a_n\}.$ If $a_3 = 12,  S_{12} > 0$ and $S_{13} <0,$ then find the largest out of $S_1, S_2, S_3, \ldots, S_{13}.$

I first let $a$ be the first term and $d$ be the common difference. Then, I got that
\begin{align*}
S_1 &= a \\
S_2 &= 2a + d \\
S_3 &= 3a + 3d \\
S_4 &= 4a + 6d \\
S_5 &= 5a + 10d \\
S_6 &= 6a + 15d \\
S_7 &= 7a + 21d \\
S_8 &= 8a + 28d \\
S_9 &= 9a + 36d \\
S_{10} &= 10a + 45d \\
S_{11} &= 11a + 55d \\
S_{12} &= 12a + 66d \\
S_{13} &= 13a + 78d.
\end{align*}
However, I am unsure how to proceed from here. Can someone help please?

Body:
By symmetry
$$S_{13}=13a_7<0 \implies a_7<0$$
$$S_{12}=12\frac{a_6+a_7}{2} >0 \implies a_6+a_7>0 \implies a_6 > -a_7>0$$

Therefore $$a_1 > a_2 > \cdots > a_6 > 0 > a_7 > \cdots > a_{13}\\
\implies S_1 < S_2 < \cdots < S_6 \text{ and } S_6 > S_7 > \cdots >S_{13}. \blacksquare$$


问:
“Proof” that zero is equal to one by infinitely subtracting numbers
Recently, I came across a “proof” that $0=1$. Here is how it goes:

Let $x = 1-1-1-1-1-1-1-\cdots$. Since $1-1=0$, $x=0-1-1-1-1-1-1-\cdots$. Now, we bracket the $1-1-1-1-1-1-\cdots$ on both sides and we get $x=1-(1-1-1-1-1-1\cdots)=0-(1-1-1-1-1-1-\cdots)$. Then, we get $1-x=0-x$. So, $1-x+x=0-x+x$. Hence, $1+0=0+0$ and so $1=0$.

I could not figure out what went wrong in this proof. The result is clearly not true but the proof seems to be true. I then asked a few people and they all could not figure out what went wrong. Can someone come please help me to identify what went wrong? Thank you.

答:
So called infinite sums in mathematics have formal definition as series and is based on concept of partial sum:
$$a_1+a_2+ \cdots =\sum\limits_{n=1}^{\infty}a_n=\lim\limits_{n \to \infty}\sum\limits_{i=1}^{n}a_i$$
where $S_n=\sum\limits_{i=1}^{n}a_i$ is partial sum.

Now let's come to your example: if you consider  $1-1-1-1-1-1-1-...$, then we should construct partial sum for it
$$\begin{array}{}
S_1=1 \\
S_2=1-1=0 \\
S_3=1-1-1=-1 \\
S_4=1-1-1-1=-2 \\
S_5 =1-1-1-1-1=-3 \\
\cdots \\
S_n=2-n \\
\cdots
\end{array}$$
As you see partial sum have no finite limit, which means, that expression $1-1-1-1-1-1-1-...$ is not finite number and cannot be used as such.

Funny example of such "proof" can be obtained if you consider expression $1-1+1-1+1-1+1-...$ and do not investigate convergence:
$$0=(1-1)+(1-1)+\cdots= 1+(-1+1)+(-1+1)+\cdots=1$$


答:
If $\langle x,y\rangle$ and $\langle y,z\rangle$ belong to $R\cap S$, then both belong to $R$, so you can apply transitivity of $R$, and both belong to $S$, so you can apply transitivity of $S$. Thus, you get $\langle x,z\rangle$ in both $R$ and $S$ and therefore in they’re intersection.

$R\cup S$ doesn’t behave nearly so nicely. The problem is that $\langle x,y\rangle$ and $\langle y,z\rangle$ might belong to $R\cup S$ because $\langle x,y\rangle\in R\setminus S$ and $\langle y,z\rangle\in S\setminus R$. Then transitivity would require $\langle x,z\rangle$ to be in $R\cup S$, i.e., in $R$, in $S$, or in both, but there is nothing in the hypotheses that requires $\langle x,z\rangle$ to belong to $R$ (since $\langle y,z\rangle\notin R$) or to $S$ (since $\langle x,y\rangle\notin S$). And in fact it’s this problem with the straightforward attempt to prove that $R\cup S$ is transitive that suggests how to construct a counterexample.

All you need is three elements, so let $A=\{0,1,2\}$, say. If $R=\{\langle 0,1\rangle\}$ and $S=\{\langle 1,2\rangle\}$, then $R$ and $S$ are both (vacuously) transitive, but $R\cup S=\{\langle 0,1\rangle,\langle 1,2\rangle\}$ fails to be transitive, since it doesn’t contain $\langle 0,2\rangle$. It shows that the potential difficulty that we spotted in trying to carry out the proof is a real difficulty: it actually can happen.


问:
Conditional Probability on Joint Uniform Distribution
This is a very basic question, but I somehow manage to confuse myself all the time. So any help is greatly appreciated. Suppose we have two random variables $X$ and $Y$ with joint distribution function $f$ where $f$ is uniform. That is

$$
f(x,y)= \begin{cases} 1 & \text{if $0\leq x\leq 1$ and $0\leq y \leq 1$}, \\ 0 & \text{elsewhere}. \end{cases}
$$

If I want to determine the conditional probability $\Pr(Y>q\mid X=p)$ I get confused since

$$
\Pr(Y>q\mid X=p)=\frac{\Pr((Y>q)\cap (X=p))}{\Pr(X=p)}
$$
but then isn't the denominator a zero probability event? In general what is the answer of this conditional probability?

答:
You should use the probability density functions.

$$\begin{align}\mathsf P(Y>q\mid X=p) &= \int_{y>q} f_{\small Y\mid X}(y\mid p)~\mathsf d y\\[2ex]&= \dfrac{\int_{y>q}f_{\small X,Y}(p,y)\,\mathsf d y}{\int_{\Bbb R}f_{\small X,Y}(p,y)\,\mathrm d y}\\[2ex]&=\left(\dfrac{\int_q^1\mathrm d y}{\int_0^1\mathrm dy}\,\mathbf 1_{0\leqslant q\lt 1}+\mathbf 1_{1\leqslant q}\right)\,\mathbf 1_{0\leqslant p\leqslant 1} \\[2ex]&=\left((1-q)\,\mathbf 1_{0\leqslant q\lt 1}+\mathbf 1_{1\leqslant q}\right)\,\mathbf 1_{0\leqslant p\leqslant 1}\end{align}$$


答:
When you write out an infinite series, you should first check if it converges. If not, then normal procedures like bracketing doesn't work anymore.

For example, here's a similar (false) proof that all integers are $0$:
Let $x = 1 + 1 + 1 +  \cdots $. For any integer $n > 0$, bracket the first $n$ terms so that $x = (1+1+\cdots+1) + 1 + 1 + 1+ \cdots = n + x$. Hence $n=0$.

Title: “Proof” that zero is equal to one by infinitely subtracting numbers

Body:
Recently, I came across a “proof” that $0=1$. Here is how it goes:

Let $x = 1-1-1-1-1-1-1-\cdots$. Since $1-1=0$, $x=0-1-1-1-1-1-1-\cdots$. Now, we bracket the $1-1-1-1-1-1-\cdots$ on both sides and we get $x=1-(1-1-1-1-1-1\cdots)=0-(1-1-1-1-1-1-\cdots)$. Then, we get $1-x=0-x$. So, $1-x+x=0-x+x$. Hence, $1+0=0+0$ and so $1=0$.

I could not figure out what went wrong in this proof. The result is clearly not true but the proof seems to be true. I then asked a few people and they all could not figure out what went wrong. Can someone come please help me to identify what went wrong? Thank you.


问:
Application of Rouche's theorem in complex analysis
Let $p(z)=\sum_{j=0}^np_jz^j\in\mathbb{C}[z]$. Define
\begin{equation}
\tilde{p}(z)=z^n\overline{p(1/\overline{z})}=\sum_{j=0}^n\overline{p}_{n-j}z^j.
\end{equation}

Assume $p$ and $\tilde{p}$ have no common factors and $|p_0|>|p_n|$. Then $p$ and $\overline{p_0}p-p_n\tilde{p}$ have the same number of zeros in $\mathbb{D}.$

My guess is that since the statement is related to the number of zeros, I think we could use the Rouche's theorem. Then can we first show $p$ and $\overline{p_0}p$ has same number of zeros and then show $\overline{p_0}p$ and $\overline{p_0}p-p_n\tilde{p}$ has same number of zeros? I also notice that $\deg(\overline{p_0}p-p_n\tilde{p})<n$ if that helps.

答:
The result follows by an easy application of Rouche. First wlog assume $p_n \ne 0$ as otherwise there is nothing to prove since $|p_0|>0$ so $\bar p_0p$ has same zeroes as $p$.

The hypothesis implies that there is no root of $p$ on the unit circle and numbering them $a_1,..a_k, b_1,..b_m, |a_l|<1, |b_l|>1, k+m=n$, we also get that $|a_1..a_kb_1..b_m|>1$ (that is $|p_0/p_n|$ by the usual considerations), so in particular $m \ge 1$ and at least a root is outside the unit disc.

Clearly $p(z)=p_n(z-a_1)..(z-a_k)(z-b_1)..(z-b_m)$) , while the roots of $\tilde{p}$ are $1/\bar a_l, 1/\bar b_l$ so $\tilde{p}(z)=\bar p_0(z-1/\bar a_1)..(z-1/\bar a_k)(z-1/\bar b_1)..(z-1/\bar b_m)$

But now notice that on the unit circle $|\frac{z-a_l}{1-\bar a_lz}|=1$ so $|\frac{z-a_l}{z-1/\bar a_l}|=|\bar a_l|=|a_l|$ and similarly for $b_l$ we have $|\frac{z-1/\bar b_l}{1- z/b_l}|=1$ so $|\frac{z-b_l}{z-1/\bar b_l}|=|b_l|$

But this means that on the unit circle $|\frac{\bar p_0p(z)}{p_n\tilde{p}(z)}|=|\frac{\bar p_0p_n}{p_n \bar p_0}||\frac{(z-a_1)..(z-a_k)(z-b_1)..(z-b_m)}{(z-1/\bar a_l)..(z-1/\bar a_k)(z-1/\bar b_1)..(z-1/\bar b_m)}|=|a_1..a_kb_1..b_m|>1$, hence indeed $|p_n\tilde{p}(z)|<|\bar p_0p(z)|$ and Rouche applies, so we are done!


问:
$f(x_0)=\sup\limits_{x\in{X}}\, f(x)$ in a compact metric space
Let $X$ be a compact metric space. Suppose that $f:X\rightarrow \mathbb{R}$ is such that for all convergent sequences $(x_n)_{n=1}^\infty\subseteq X$ , $f(\lim\limits_{n\rightarrow \infty}\, x_n)\geq \lim\limits_{n\rightarrow \infty}\, \sup\, f(x_n)$ . Show that there is an $x_0\in{X}$ such that $f(x_0)=\sup\limits_{x\in{X}}\, f(x)$.

How do we proceed on this?

答:
Set $M= \sup_{x\in X} f(x)$. Then there exists a sequence $\{x_n\}$ in $X$ such that $M = \lim_{n\to\infty} f(x_n)$. Since $X$ compact, we may assume that $x_n$ converges to some $x_0\in X$. We have

$$M\ge f(x_0)=\lim_{n\to\infty} f(x_n) \ge \limsup_{n\to\infty} f(x_n) = M$$
That is $f(x_0) =M$.


问:
Relationship between 2 Dimensional Quadratic systems and roots
Given four points

$(x_1, y_1)
(x_2, y_2)
(x_3, y_3)
(x_4, y_4)$

How does one construct a system of two equations:

$a_1x + a_2x^2 + a_3y + a_4y^2 + a_5xy = c_1$

$b_1x + b_2x^2 + b_3y + b_4y^2 + b_5xy = c_2$

such that the set of solutions of this system is the four original points?

Solving the general systems is growing massively complicated.

答:
Here's one way to do it, at least when the points are in a general position.

Call the four given points $P_1, P_2, P_3, P_4$.  Let $\ell_{ij}$ denote the line connecting $P_i$ and $P_j$, and $L_{ij}$ a linear equation in $x$ and $y$ satisfied by the points on $\ell_{ij}$.  Then you can use the system of equations:
\begin{equation}
L_{12} L_{34} = 0, \\
L_{13} L_{24} = 0.
\end{equation}

Indeed, it is easy to see that each of the four points $P_i$ satisfy both equations, so they lie on the intersection.  Provided none of the lines are identical (the condition that the points are in general position), then there are exactly four solutions to this system of equations, which you've already found.


问:
Without Gabriel's theorem, how do you classify indecomposable reps of $\mathbb{A}_n$?
Given the Dynkin quiver of type $\mathbb{A}_n$ without orientation,
$$
\begin{aligned}
  1\!-\!2\!-\dotsb-\!n
\end{aligned}
$$
let $M(i,j)$ be a representation of $\mathbb{A}_n$, such that $V_l=K$ for all $i \leq l \leq j$ and $0$ otherwise.

It is a well know result that $\mathbb{A}_n$ is representation finite. In particular every indecomposable representation of $\mathbb{A}_n$ is isomorphic to some $M(i,j)$. This means that up to isomorphism there are $n(n+1)/2$ indecomposable representations of $\mathbb{A}_n$, which are given by the $M(i,j)$.

My question is, how can one show this without using Gabriels Theorem? Is there an author who has done that already? Gabriels Theorem is very long and involves reflection functors, but at the same time its statement is very general and applies to different kinds of Dynkin quivers.

答:
Since your looking at representations of $A_n$ first pick some orientation of the edges as arrows. And if you'd like, without loss of generality choose the orientation where they're all pointing in the same direction, to the right.

Sketch. Think in terms of strings. In a representation $M$ let $i$ be the left-most vertex such that $\dim M_i >0$. Pick a basis for $M_i$ and let $x$ be an element of that basis. Let $M_{f_i}, M_{f_{i+1}}, \dotsc, M_{f_n}$ be the linear maps to the right of the vertex $i$. Follow $x$ along this chain of maps, forming a string of one-dimensional vector spaces spanned by the successive images of $x$. There must be some left-most $k$ between $i$ and $n$ such that you get $$M_{f_k} \circ\dotsb \circ M_{f_{i+1}} \circ M_{f_{i}} (x) = 0$$ This is where the string stops. The idea is that you can peel off this string of one-dimensional spaces
$$ 0 \rightarrow \dotsb \rightarrow \langle x \rangle \rightarrow \langle M_{f_i}(x)\rangle \rightarrow \langle M_{f_{i+1}}(M_{f_i}(x))\rangle \rightarrow \dotsb \rightarrow \langle M_{f_{k-1}} \circ\dotsb \circ M_{f_{i+1}} \circ M_{f_{i}} (x)\rangle \rightarrow 0 \rightarrow \dotsb \rightarrow 0$$
as a direct summand of $M$. The details of are too annoying for me to want to type up here, but I hope this conveys the idea well enough.


问:
Simplification of complex rational?
I was looking at a problem in a textbook and found the following simplification:

$$\frac{1}{x+i\omega} * \frac{2y}{y^2 + \omega^2} = \frac{2}{xy} * \frac{1}{1+i\frac{\omega}{x}}*\frac{1}{1 + (\frac{\omega}{y})^2}$$

and I have no clue how they did this simplification? Is this like some well known formula? I tried multiplying by the complex conjugate but that doesn't seem to help so any guidance/formulas they used would be much appreciated.

Thanks

答: Divide top and bottom by $y^2$, and $x+i\omega=x\left(1+i\frac\omega x\right)$

Title: Simplification of complex rational?


问:
how to calculate $\int_{0}^{\infty}\frac{x}{\sqrt{e^x-1}}\mathrm{d}x$
I was trying to solve another integral when then I reached this, I've no idea of how to select the contour for the integration.</p>

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{t \equiv \expo{x} - 1 \implies x = \ln\pars{1 + t}}$:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
{x \over \root{\expo{x} - 1}}\,\dd x} =
\int_{0}^{\infty}
t^{-1/2}\,\,\,{\ln\pars{1 + t} \over 1 + t}\,\dd t
\\[5mm] = &\
\int_{0}^{\infty}
t^{\color{red}{1/2} - 1}\bracks{%
-\sum_{k = 0}^{\infty}H_{k}\,\pars{-t}^{k}}\,\dd t
\\[5mm] = &\
-\int_{0}^{\infty}
t^{\color{red}{1/2} - 1}
\bracks{%
\sum_{k = 0}^{\infty}
\color{red}{H_{k}\,\Gamma\pars{1 + k}}
{\pars{-t}^{k} \over k!}}\dd t
\end{align}
With the Ramanujan's Master Theorem:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
{x \over \root{\expo{x} - 1}}\,\dd x}
\\[5mm] = &\
-\ \underbrace{\Gamma\pars{\color{red}{1 \over 2}}}
_{\ds{\root{\pi}}}\
\underbrace{\quad H_{\color{red}{-1/2}}\quad}
_{\ds{\int_{0}^{1}{1- t^{\color{red}{-1/2}} \over 1 - t}
\,\dd t}}\
\Gamma\pars{1 \color{red}{- {1 \over 2}}}
\\[5mm] = &\
-\pi\int_{0}^{1}{1- t^{-1} \over 1 - t^{2}}\,2t\,\dd t =
2\pi\int_{0}^{1}{\dd t \over 1 + t}
\\[5mm] = &\
\bbx{2\pi\ln\pars{2}} \approx 4.3552 \\ &
\end{align}


问:
Is it true that any continuous injective map from a circle to circle is surjective?
Here is my idea:

Suppose $f : \mathbb S^1 \to \mathbb S^1$ is continuous and injective.

If $f$ is not surjective, its image is homeomorphic to a subset of $\mathbb S^1 \setminus \{ c\} \cong \mathbb R$ (by considering the projection).

Then, the restriction of $f$'s codomain $g : \mathbb S^1 \to f(\mathbb S^1)$ is injective and surjective, so it is bijective.

Also, since $\mathbb S^1$ is compact and connected, its image under the continuous map is also compact and connected, so $f(\mathbb S^1)$ is homeomorphic to some closed interval $[a, b]$, which is simply connected.

Noting $f(\mathbb S^1)$ is Hausdorff and $\mathbb S^1$ is compact, by the closed map lemma, it follows $g$ is a homeomorphism. This contradicts that $\mathbb S^1$ is not simply connected, since $[a, b]$ is simply connected. Thus, $f$ must be surjective.

Does it work?

答:
May assume that our map $f$ takes $1$ to $1$. Consider the  map $p\colon \mathbb{R}\to \mathbb{S}^1$, $t \mapsto \exp 2\pi i t $.  There exists a "covering"  $F\colon \mathbb{R} \to \mathbb{R}$  of $f$, such that  $p\circ F= f \circ p$, and $F(0) = 0$. Now, since $p(F(t+1)) = p(F(t))$ for all $t\in \mathbb{R}$, we have
$$F(t+1) - F(t) \equiv k \ \textrm {(an integral constant)}$$

In particular, $F(1) =k$. ( $k$ is the index of $f$).

If $k\ne 0$, then the image of $F$ contains the interval $[0,k]$ ( or $[k,0]$),
and $p$ applied to this interval is the whole $\mathbb{S}^1$. Using $f\circ p = p\circ F$, we conclude $f$ is surjective.

If $k=0$, then $F(0)=F(1)=0$.  Then it is easy to see that there exist $0<t_1<t_2<1$ such that $F(t_1) = F(t_2)$.  It follows that $f(p(t_1)) = f(p(t_2)$, and so $f$ is not injective.


问:
Proving the existence of a basis
Suppose that $V$ is a non-zero vector space and $T: V \rightarrow V$. Show that there is an ordered basis for $V$ s.t. $A$, the matrix representation of $T$ in the basis has a first column that is either $(c\space0 ...\space0)^T$ for some $c \in F$, or $(0\space 1\space 0 ...\space 0)^T$.

My attempt:
Suppose that $B = {v_1, ..., v_n}$ is a basis for $V$ s.t. $T(v_j) \in$ span$(v_1, ..., v_{j+1})$. Then, the first column of A is of the form:

$$A = (a_{11}\space a_{21}\space 0 ...\space0)^T$$

where $aij$ is the $ij$ entry of A.

The first column is given by $T(v_1)$ and $T(v_1) = a_{11}v_1 + a_{21}v_2$. Therefore, there is an ordered basis for $V$ s.t. $A$ has a first column that is either $(c\space0 ...\space0)^T$ where $c = a_{11}$ or $(0\space 1\space 0 ...\space 0)^T$.

I don't think what I've done above is right. Nevertheless, I haven't done a proof of showing the existence of a basis before, so I was hoping someone can go over the proof above and let me know if and where I've made a mistake(s). Any assistance is much appreciated.

答:
If the first column of $A$ in that basis is of the form you claim, then you need either $a_{21} = 0$ or $a_{11}=0$ and $a_{21}=1$ to solve the problem.

Here's how I would do this problem. There are two cases: $T$ has an eigenvalue or it does not. If $T$ has an eigenvalue $c$ then by definition there is some $0 \neq v \in V$ such that $Tv = c v$. Extend $v$ to an ordered basis where it is the first element, so $(v=v_1, \dots, v_n)$. As $Tv_1=cv_1$, we therefore have that the first column of $A$ is $(c, 0, \dots, 0)^t$.

On the other hand, suppose $T$ does not have an eigenvalue. Then for any $0 \neq v \in V$, $Tv$ is not a scalar multiple of $v$. Thus, $\{v, Tv\}$ is linearly independent. It can therefore be extended to an ordered basis $(v=v_1,Tv=v_2, \dots, v_n)$. Then as $Tv_1=v_2$, the first column of $A$ is $(0, 1, 0, \dots, 0)^t$.

By the way, it's a bit pedantic but you do need to specify that $\dim V > 1$ otherwise this makes no sense.

Title: Proving the existence of a basis


问:
Proof $(X×Y)-(A×B)$ is connected [Contradiction]
Let $A$ be a proper subset of $X$, and let $B$ is a proper subset of $Y$. If $X$ and $Y$ are connected, show that

$$(X×Y)-(A×B)$$

is connected.

I know there's already a question answering this problem, but as I was working on this problem, I did it from another way and wanted to make sure is right.

Proof Idea:

Suppose by contradiction that $H=(X×Y)-(A×B)$ is not connected. Therefore There exist a separation of H from two opens $U,V$.

$$(X×Y)-(A×B)=U\cup V = U_1\times V_1 \cup U_2 \times V_2$$

Where $U_1,U_2$ are open in $X$ and $V_1,V_2$ are open in Y. This way

$U_1\times V_1 \cap U_2 \times V_2 = (U_1 \cap U_2) \times (V_1 \cap V_2) = \emptyset$. So there are two cases. I will work only on the first case, the second is similar.

$U_1 \cap U_2 = \emptyset$, note that $U_1, U_2 \notin A$, as consequence $U_1 \cap U_2 \cap A = \emptyset$. Consider in particular $U_c=U_1 \cup U_2$. Observe that $U_c \cup A = X$, is a separation of $X$, but $X$ is connected. $\Rightarrow\!\Leftarrow$.

Let me know if there's any mistake, will help me prepare for my exam.

答:
We adopt the following definition for connected topological space.

Let $X$ be a topological space. We say that $X$ is connected if
there does not exist non-empty open sets $U$, $V$ such that $U\cup V=X$
and $U\cap V=\emptyset$. Let $Y\subseteq X$ be a subset. We say
that $Y$ is a connected subset of $X$ if $Y$ is connected with
respect to the relative topology. Note that the emptyset $\emptyset$
is connected by definition.

Proposition 1: Let $X$, $Y$ be topolocal spaces. Let $f:X\rightarrow Y$
be a continuous function. If $X$ is connected, then $f(X)$ is a
connected subset of $Y$.

Proof of Propositon 1: Prove by contradiction. Suppose the contrary
that there exist open subsets $U$, $V$ of $Y$ such that $f(X)\cap U$,
$f(X)\cap V$ are non-empty, $[f(X)\cap U]\bigcup[f(X)\cap V]=f(X)$
and $[f(X)\cap U]\bigcap[f(X)\cap V]=\emptyset$. We have that
\begin{eqnarray*}
X & = & f^{-1}\left([f(X)\cap U]\bigcup[f(X)\cap V]\right)\\
 & = & f^{-1}\left(f(X)\cap U\right)\bigcup f^{-1}\left(f(X)\cap V\right)\\
 & = & f^{-1}(U)\bigcup f^{-1}(V).
\end{eqnarray*}
That $f(X)\cap U$ and $f(X)\cap V$ are non-empty implies that $f^{-1}(U)$
and $f^{-1}(V)$ are non-empty. Finally,
\begin{eqnarray*}
\emptyset & = & f^{-1}\left([f(X)\cap U]\bigcap[f(X)\cap V]\right)\\
 & = & f^{-1}(U)\bigcap f^{-1}(V).
\end{eqnarray*}
This shows that $X$ can be written as a disjoint union of two non-empty
open sets, which is a contradiction.

Proposition 2: Let $X$ be a topological space. Then the following
are equivalent:

(a) $X$ is connected.

(b) If $f:X\rightarrow\{0,1\}$ is continuous, then $f$ is a constant
function.

Proof of Proposition 2: $(a)\Rightarrow(b)$: Suppose that $X$ is
connected. Let $f:X\rightarrow\{0,1\}$ be a continuous function.
By Proposition 1, $f(X)$ is a non-empty connected set. Note that
non-empty connected subsets of $\{0,1\}$ are $\{0\}$ and $\{1\}$.
Hence, $f(X)=\{0\}$ or $f(X)=\{1\}$. In another word, $f$ is a
constant function.

$(b)\Rightarrow(a)$: Prove by contradiction. Suppose that $X$ is
disconnected. Choose non-empty open sets $U,V$ such that $U\cup V=X$
and $U\cap V=\emptyset$. Define $f:X\rightarrow\{0,1\}$ by $f(x)=\begin{cases}
0, & \mbox{ if }x\in U\\
1, & \mbox{ if }x\in V
\end{cases}.$ It is trivial to verify that $f$ is continuous but $f$ is not a
constant function. A contradiction!

Proposition 3: Let $X$ be a topological space. Let $\{A_{i}\mid i\in I\}$
be a family of connected sets such that for any $i_{1},i_{2}\in I$,
there exists a fintie sequence $(j_{k})_{k=1}^{n}$ in $I$ such that
$j_{0}=i_{1}$, $j_{n}=i_{2}$ and $A_{j_{k-1}}\cap A_{j_{k}}\neq\emptyset$
for $k=1,\ldots,n$. Then $A=\cup_{i\in I}A_{i}$ is connected.

Proof of Proposition 3: Let $f:A\rightarrow\{0,1\}$ be a continuous
function. For each $i\in I$, let $f_{i}=f\mid_{A_{i}}$, the restriction
of $f$ on $A_{i}$, which is continuous with respect to the relative
topology. By Proposition 2, $f_{i}$ is a constant function. Let $x,y\in A$
be arbitrary. Choose $i_{1},i_{2}\in I$ such that $x\in A_{i_{1}}$
and $y\in A_{i_{2}}$. By assumption, we may choose a sequence $(j_{k})_{k=0}^{n}$
such that $j_{0}=i_{1}$, $j_{n}=i_{2}$ and $A_{j_{k-1}}\cap A_{j_{k}}\neq\emptyset$
for $k=1,\ldots,n$. For each $k$, choose $z_{k}\in A_{j_{k-1}}\cap A_{j_{k}}.$
We have that
\begin{eqnarray*}
 &  & f(x)\\
 & = & f_{j_{0}}(x)\\
 & = & f_{j_{0}}(z_{1})\\
 & = & f_{j_{1}}(z_{1})\\
 & = & f_{j_{1}}(z_{2})\\
 & = & f_{j_{2}}(z_{2})\\
 & = & \cdots\\
 & = & f_{j_{k}}(z_{k})\\
 & = & f_{j_{k}}(y)\\
 & = & f(y).
\end{eqnarray*}
This shows that $f$ is a constant function. By Proposition 2, $A$
is connected.

Proposition 4: Let $X,Y$ be topological spaces. If $X$ and $Y$
are connected, then $X\times Y$ is connected (with respect to the
product topology).

Proof: Let $f:X\times Y\rightarrow\{0,1\}$ be a continuous function.
Let $(x_{1},y_{1}),(x_{2},y_{2})\in X\times Y$ be arbitrary. We go
to show that $f(x_{1,}y_{1})=f(x_{2},y_{2})$. Define function $\iota_{y_{2}}:X\rightarrow X\times Y$
by $\iota_{y_{2}}(x)=(x,y_{2})$. Define function $\iota^{x_{1}}:Y\rightarrow X\times Y$
by $\iota^{x_{1}}(y)=(x_{1},y)$. Clearly $\iota_{y_{2}}$ and $\iota^{x_{1}}$
are continuous. Therefore $f\circ\iota_{y_{2}}:X\rightarrow\{0,1\}$
and $f\circ\iota^{x_{1}}:Y\rightarrow\{0,1\}$ are continuous. Since
$X$ and $Y$ are connected, by Proposition 2, $f\circ\iota_{y_{2}}$
and $f\circ\iota^{x_{1}}$ are constant functions. Therefore
\begin{eqnarray*}
f(x_{2},y_{2}) & = & f\circ\iota_{y_{2}}(x_{2})\\
 & = & f\circ\iota_{y_{2}}(x_{1})\\
 & = & f(x_{1},y_{2})\\
 & = & f\circ\iota^{x_{1}}(y_{2})\\
 & = & f\circ\iota^{x_{1}}(y_{1})\\
 & = & f(x_{1},y_{1}).
\end{eqnarray*}
It follows that $f$ is a constant function. By Proposition 2, $X\times Y$
is connected.

Definition 5: Let $X$ be a topological space. We say that $A\subseteq X$
is a connected component of $X$ if:

(a) $A$ is connected.

(b) If $B\subseteq X$ is connected and $A\subseteq B$, then $A\subseteq B$.

Proposition 6: Let $X$ be a non-empty topological space. Let $\mathcal{C}$
be the collection of all connected components of $X$. Then

(a) $\mathcal{C}$ is a partition of $X$, i.e., $X=\cup\mathcal{C}$
and $A\cap B=\emptyset$ whenever $A,B\in\mathcal{C}$ and $A\neq B$.

(b) For each $A\in\mathcal{C}$, $A$ is closed.

Proof of Proposition 6: Let $x\in X$. We show that $x$ is contained
in a connected component. Let $\mathcal{F}=\{A\subseteq X\mid x\in A\mbox{ and }A\mbox{ is connected.}\}.$ Note that $\{x\}\in\mathcal{F}$, so $\mathcal{F}$ is non-empty.
Clearly the condition in Proposition 3 is satisfied, so $A=\cup\mathcal{F}$
is a connected set. If $B\subseteq X$ is a connected and $A\subseteq B$,
then $x\in B$ and hence $B\in\mathcal{F}$. Therefore $B\subseteq\cup\mathcal{F}=A$,
so $A=B$. This shows that $A$ is a connected component. In particular,
we have proven that $\cup\mathcal{C}=X$. Next, let $A,B\in\mathcal{C}$
with $A\neq B$. Suppose the contrary that $A\cap B\neq\emptyset$.
Then $\{A,B\}$ satisfies the condition in Proposition 3, so $A\cup B$
is also a connected set. Clearly $A\subseteq A\cup B$. By maximality
of $A$, we have $A=A\cup B$. Similarily, $B\subseteq A\cup B\Rightarrow B=A\cup B$.
Therefore $A=B$, which is a contradiction. Finally, we go to show
that for each $A\in\mathcal{C}$, $A$ is closed. Prove by contradiction.
Assume that $A\neq\bar{A}$, then there exists $x\in\bar{A}\setminus A$.
Since $\mathcal{C}$ is a partition of $X$, there exists $B\in\mathcal{C}$
such that $x\in B$. Note that $A$ is a proper subset of $A\cup B$,
by maximality of $A$, $A\cup B$ must be disconnected. Choose open
subsets $U,V$ of $X$ such that $U\cap\left(A\cup B\right),$ $V\cap\left(A\cup B\right)$
are non-empty, disjoint, and whose union is $A\cup B$. Without loss
of generality, assume that $x\in V\cap(A\cup B)$. Observe that $U\cap A$,
$V\cap A$ are open in $A$, disjoint, and $\left(U\cap A\right)\cup\left(V\cap A\right)=A$.
Since $A$ is connected, we must have $U\cap A=\emptyset$ or $V\cap A=\emptyset$.
Note that $x\in V$ and $x\in\bar{A}$, so $V\cap A\neq\emptyset$.
It follows that $U\cap A=\emptyset$ and $A\subseteq V$. Now, $U\cap\left(A\cup B\right)=U\cap B$
and $V\cap\left(A\cup B\right)=A\cup(V\cap B)$. Therefore,
\begin{eqnarray*}
B & = & B\cap\left(A\cup B\right)\\
 & = & B\cap\left\{ \left[U\cap\left(A\cup B\right)\right]\cup\left[V\cap\left(A\cup B\right)\right]\right\} \\
 & = & B\cap\left\{ \left[U\cap B\right]\cup\left[A\cup(V\cap B)\right]\right\} \\
 & = & \left[U\cap B\right]\cup\left[V\cap B\right].
\end{eqnarray*}
Note that $U\cap B=U\cap\left(A\cup B\right)$ is non-empty, $V\cap B$
is also non-empty because it contains $x$. $U\cap B$ and $V\cap B$
are disjoint because $U\cap\left(A\cup B\right)$ and $V\cap\left(A\cup B\right)$
are disjoint. We arrive a contradiction because $B$ is connected.

Proposition 7: Let $X$ and $Y$ be connected topological spaces.
Let $A\subset X$ and $B\subset Y$ be proper subsets. Let $H=(X\times Y)\setminus(A\times B)$,
then $H$ is connected.

Proof of Proposition 7: Note that $H=(X\times B^{c})\bigcup(A^{c}\times Y)$.
Let $\mathcal{F}_{X}=\{C_{i}\mid i\in I\}$ be the family of non-empty
connected components of $A^{c}$ and $\mathcal{F}_{Y}=\{D_{j}\mid j\in J\}$
be the family of non-empty connected components of $B^{c}$. These
families are non-empty because $A^{c}$ and $B^{c}$ are non-empty sets.
Then $A^{c}=\cup_{i}C_{i}$ and $B^{c}=\cup_{j}D_{j}$. It follows
that
\begin{eqnarray*}
H & = & (X\times B^{c})\bigcup(A^{c}\times Y)\\
 & = & \cup_{j}(X\times D_{j})\bigcup\cup_{i}(C_{i}\times Y)\\
 & = & \cup\mathcal{G},
\end{eqnarray*}
where $\mathcal{G}=\{X\times D_{j}\mid j\in J\}\cup\{C_{i}\times Y\mid i\in I\}$.
By Proposition 4, each element in $\mathcal{G}$ is connected. We
go to check that $\mathcal{G}$ satisfies the condition in Proposition
3. Let $E_{1},E_{2}\in\mathcal{G}$. Consider the following few cases.

Case 1: $E_{1}=X\times D_{j_{1}}$ and $E_{2}=X\times D_{j_{2}}$
for some $j_{1},j_{2}\in J$. Choose any $i\in I$. Note that $(X\times D_{j_{1}})\cap(C_{i}\times Y)=C_{i}\times D_{j_{1}}\neq\emptyset$
and $(C_{i}\times Y)\cap(X\times D_{j_{2}})=C_{i}\times D_{j_{2}}\neq\emptyset$.

Case 2: $E_{2}=C_{i_{1}}\times Y$ and $E_{2}=C_{i_{2}}\times Y$
for some $i_{1},i_{2}\in I$. This case can be treated similarly as
Case 1.

Case 3: $E_{1}=X\times D_{j}$ and $E_{2}=C_{i}\times Y$. Note that
$E_{1}\cap E_{2}=C_{i}\times D_{j}\neq\emptyset$.

By Proposition 3, $H=\mathcal{\cup G}$ is connected.

We remark that Proposition 4 can be generalized for arbitrary product.

Theorem 8: Let $I$ be a non-empty index set (which may be uncountable).
Let $\{X_{i}\mid i\in I\}$ be a family of connected topological spaces.
Let $X=\prod_{i\in I}X_{i}$, equipped with product topology. Then
$X$ is connected.

Proof of Theorem 8: Prove by contradiction. Suppose the contrary that
$X$ is disconnected. By Proposition 2, there exists a non-constant
continuous function $f:X\rightarrow\{0,1\}$. Let $U=f^{-1}(\{0\})$
and $V=f^{-1}(\{1\})$, which are non-empty, disjoint open subsets
of $X$. Choose non-empty open sets $U_{i}\subseteq X_{i}$ such that
$\prod_{i}U_{i}\subseteq U$ and $U_{i}=X_{i}$ except finitely many
$i$. Similarly, we can choose non-empty open sets $V_{i}\subseteq X_{i}$
such that $\prod_{i}V_{i}\subseteq V$ and $V_{i}=X_{i}$ except finitely
many $i$. Let $I_{0}\subseteq I$ be a finite set such that $U_{i}=V_{i}=X_{i}$
for all $i\in I\setminus I_{0}$. For each $i\in I\setminus I_{0}$, fix $a_{i}\in X_{i}$. Observe
that $\prod_{i\in I_{0}}U_{i}$ and $\prod_{i\in I_{0}}V_{i}$ are
non-empty. Choose $u'\in\prod_{i\in I_{0}}U_{i}$ and $v'\in\prod_{i\in I_{0}}V_{i}$.
Define $u\in X$ by $u(i)=\begin{cases}
u'(i), & \mbox{ if }i\in I_{0}\\
a_{i}, & \mbox{ if }i\in I\setminus I_{0}
\end{cases}.$ Define $v\in X$ by $v(i)=\begin{cases}
v'(i), & \mbox{if }i\in I_{0}\\
a_{i}, & \mbox{if }i\in I\setminus I_{0}
\end{cases}.$ Then, $u\in\prod_{i\in I}U_{i}$ and $v\in\prod_{i\in I}V_{i}$.
In particular $f(u)=0$ and $f(v)=1$. Define $X'=\prod_{i\in I_{0}}X_{i}$.
Define a map $\theta:X'\rightarrow X$ by $\theta(x')(i)=\begin{cases}
x'(i), & \mbox{if }i\in I_{0}\\
a_{i}, & \mbox{if }i\in I\setminus I_{0}
\end{cases},$ $x'\in X'$. Note that $\theta$ is continuous because for each $i\in I$, $\pi_{i}\circ\theta=\begin{cases}
\pi'_{i}, & \mbox{if }i\in I_{0}\\
\bar{a}_{i}, & \mbox{if }i\in I\setminus I_{0}
\end{cases}$ is continuous, where $\pi_{i}:X\rightarrow X_{i}$, $\pi_{i}':X'\rightarrow X_{i}$
are canonical projections while $\bar{a}_i:X'\rightarrow X_{i}$
is the constant function: $x'\mapsto a_{i}$, $x'\in X'$. Therefore
$f\circ\theta:X'\rightarrow\{0,1\}$ is continuous. Note that $X'$,
being a finite product of connected spaces, is connected. Therefore
$f\circ\theta$ is a constant function. On the other hand, $0=f(u)=f\circ\theta(u')=f\circ\theta(v')=f(v)=1$,
which is a contradiction.


问:
Solving $\lim_ {a\to \frac{\pi}{4}} f(a)$ where $f$ is a definite integral.
$f:(0,\frac{\pi}{2})\to$$\mathbb{R}$,  $f(a) = \int_0^1\frac{dx}{x^2+2x\tan a + 1}$, $\forall a\in(0,\frac{\pi}{2})$

What is $\lim_ {a\to \frac{\pi}{4}} f(a)$?

I am confused. Function $f$'s domain is given at the beginning but then this  $\forall a\in(0,\frac{\pi}{2})$ is given . Isn't $a$ the variable with which $f$ is defined?

The only thing i can extract from here is $\tan a$ being positive always but aside from that i do not know what to do. I tried writing the denominator by "completing the square" and then i used u-substitution but in the end i get $\frac{1}{2\sqrt {\tan^2{a -1}}}$ multiplied by some logarithms but this is defined only for ${\tan^2a} \gt 1$.

How do i solve it?

Body:
Consider first
$$x^2+2k x+1=(x+k)^2+(1-k^2)$$
$$I=\int \frac {dx}{x^2+2k x+1} =\int \frac {dx}{(x+k)^2+(1-k^2)}$$ Let $x+k=t\sqrt{1-k^2} $
$$I=\frac 1 {\sqrt{1-k^2}} \int \frac {dt}{t^2+1}=\frac{\tan ^{-1}(t)}{\sqrt{1-k^2}}$$

Back to $x$
$$I=\frac{\tan ^{-1}\left(\frac{x+k}{\sqrt{1-k^2}}\right)}{\sqrt{1-k^2}}$$
Using the bounds
$$J=\int_0^1 \frac {dx}{x^2+2k x+1}=\frac{\tan ^{-1}\left(\frac{k+1}{\sqrt{1-k^2}}\right)-\tan
   ^{-1}\left(\frac{k}{\sqrt{1-k^2}}\right)}{\sqrt{1-k^2}}$$ Recombine the arctangents
$$J=\frac{1}{\sqrt{1-k^2}}\tan ^{-1}\left(\frac{\sqrt{1-k}}{\sqrt{1+k}}\right)$$

Now, using Taylor series around $k=1^\pm$, in both cases
$$J=\frac{1}{2}-\frac{1}{6}(k-1)+\frac{1}{15} (k-1)^2+O\left((k-1)^{3}\right)$$


问:
Better than Runge-Kutta-Fehlberg 4(5) at high order?
I wonder what are currently the best numerical solvers of ODE for high-accuracy computations. I need an efficient and accurate method to solve ODE that are not pathological (all is smooth) using $128$-bit floating-point number (quad arithmetics).

So, what would be good choices (some references comparing different algorithms would be appreciated) ?

答:
Do you something about the nature of your differential equation? One that involves exponential forms may be very efficiently approximated using exponentials in a way similar to Taylor series...

Utilizing logarithms, xlog(x)... And mixtures of different elementary functions may ask prove useful too... Constructing the method isn't significantly difficult you just need to use regression/basic slope and sum analysis to fit a particular type of function to your curve, rinse and repeat with higher powers.

I would recommend this approach of customizing your own approximation method, especially, if you know something about the nature/behavior of your solution (which you appear to do since you can confidently say its smooth) and want high performance as well as high accuracy.


问:
What is the proof that the total number of subsets of a set is $2^n$?
What is the proof that given a set of $n$ elements there are $2^n$ possible subsets (including the empty-set and the original set).</p>

答: To write out some details of the discussion under this answer. Consider the two element set $\{0,1\}$ where you think of $1$ as meaning in the subset and $0$ as meaning not in the subset. In this way any function $f\colon S \to \{0,1\}$ defines a subset of $S$, where $s$ is in that subset if and only if $f(s)=1$. So then our question is how many such functions are there? For each of the $n$ elements of $S$ there are two choices for what $f(n)$ can be, so making these $n$ choices independently, there are $2^n$ options.


问:
vector identities in solving wave equations with different speeds of propagation

Assume that $u = (u^1, u^2, u^3)$ solves the evolution equations of linear elasticity: $$u_{tt}-µ \Delta u − (λ + µ) D (\nabla\cdot u) = 0$$  in $\mathbf{R}^3 × (0, ∞)$.
Show that $w := \nabla \cdot u $ and $w := \nabla \times u$ each solve wave equations, but with differing speeds of propagation.

This is problem 21 in chapter 2 of Evan's PDE.

I am able to do this problem when $ w := \nabla \times u$ to obtain $w_{tt} = \mu \Delta w$. For $ w:= \nabla \cdot u$, I am not recognizing how to proceed from

$$ w_{tt} = \mu \Delta w + (\lambda + \mu) \nabla (\nabla \cdot w) $$

to $$w_{tt} = \mu (\Delta w) + (\lambda + \mu)(\Delta w)$$

Since $ w = \nabla \cdot u$ is a scalar, I am not sure how the divergence of $w$ is defined here. The identity $\Delta w = \nabla(\nabla \cdot w) - \nabla \times \nabla \times w $ would be useful here but I can't see why the curl of curl of $w$ would be zero in this case or even defined when $w$ is a scalar. Maybe I am misunderstanding something in the statement of the problem? Any help will be appreciated.

答: Evans means that $\nabla\cdot u$ solves a scalar-valued wave equation and $\nabla\times u$ solves a vector-valued wave equation. Your issue is that you are overloading your derivative operators, in particular $\Delta$. There is a vector Laplacian and a scalar Laplacian in play here. The identity you stated for $\Delta w$ is for a vector Laplacian and only makes sense for a vector-valued function $w$. Since $w = \nabla\cdot u$ is a scalar function, one should use the definition of the scalar Laplacian $\Delta w = \nabla\cdot(\nabla w)$ instead.


问:
Pairwise non-isomorphic graphs on n vertices
Show that there are at least $\frac {2^{n\choose 2}}{n!}$ pairwise non-isomorphic graphs on $n$ vertices

答: HINT: Explain why there are $2^{\binom{n}2}$ different graphs on $n$ vertices labelled $1$ through $n$. HINT: Think about the possible edges. Now let $G$ be a graph on $n$ unlabelled vertices, and explain why there are $n!$ different ways to label the vertices of $G$ with the numbers $1$ through $n$. Now put these two results together.


问:
How can I prove that this map is 1-1 and onto?
If I have the following homomorphism $f: \mathbb Z_{p^a} \rightarrow \Bbb Z_{p^b}$ defined by multiplication by $n/d$ where $n = p^b$ and $d = \gcd(p^a, p^b).$

How can I prove the following:

if $a\geq b$ then $f$ is onto.

if $a \leq b$ then $f$ is 1-1.

My intuition is that I will use the fact that I have cyclic groups but how can I relate that to the relation between $a$ and $b,$ could anyone give me a hint please?

答:
To expand on my comment, I'll do $a)$ and leave $b)$ for you to do. The map
$$f: \mathbb{Z}/p^a \mathbb{Z} \longrightarrow \mathbb{Z}/p^b \mathbb{Z}$$
is given explicitly by $f(x)=\frac{n}{d}x$. In the first case, this simply sends a number mod $p^a$ to itself mod $p^b$. To see that this is surjective, note that the quotient map $\mathbb{Z} \to \mathbb{Z}/p^b \mathbb{Z}$ factors through $f$ since $a \geq b$.

To do $b)$, you should first ask yourself why $f$ is well-defined. In other words, if $n=m+kp^a$, then why do we have $f(n)=f(m)$? Then try to show that $f(n)=f(m)$ implies $n=m$ in $\mathbb{Z}/p^a \mathbb{Z}$.


问:
If $X_i$ is not iid from normal a distribution, whether $(n-1)S^2$ is still follow chi-square distribution with $df = n-1$?
If $X = (X_1,...,X_n)$ is a random vector from $N(\mu, \Sigma)$, where $$ \Sigma =
\begin{bmatrix}
    1     & \rho   & \dots  & \rho \\
   \rho   & 1      & \dots  & \rho \\
   \vdots & \vdots & \ddots & \vdots \\
   \rho   & \rho   & \dots  & 1
\end{bmatrix},
$$
we can get that $S^2$ is independent with $\bar{X}$, since $\Sigma = \sigma^2[(1-\rho)I + \rho11^T]$. If we set
$$ A =
\begin{bmatrix}
    1-1/n    & 1/n    & \dots   & 1/n \\
    -1/n   & 1-1/n  & \dots   & -1/n \\
   \vdots  & \vdots & \ddots  & \vdots \\
   -1/n    & -1/n   & \dots   & 1-1/n
\end{bmatrix},
$$ then we can get that $AX \sim N(A\mu, A\Sigma A^T)$ and then we can verify $S^2$ and $\bar{X}$ are independent because of the independence between $\bar{X}$ and $X_i - \bar{X}$. However, do we know that $(n-1)S^2/\sigma^2$ still follows a $\chi^2$ distribution with $df = n-1$? Since from the covariance\variance matrix of $AX$ we know the $X_i - \bar{X}$ and $X_j - \bar{X}$ are not independent, where $i$ is not equal to $j$.

If yes, how can we prove it? If no, which distribution does it follow?

答:
Recall that $A$ is a symmetric idempotent matrix, i.e. $A = A^T = A^2.$ And $A1= 0$ if $1$ denotes the column vector whose every entry is the scalar $1.$

You have $\Sigma = (1-\rho)I + \rho 1 1^T$ but also $\Sigma = \sigma^2[(1-\rho)I + \rho 1 1^T].$ I will assume the latter was intended.

Then you have
$$
\operatorname{var}(AX) = \sigma^2 A[(1-\rho)I + \rho 11^T] A^T = \sigma^2 (1 - \rho ) A.
$$
Recall next that $(n-1)S^2 = \|AX\|^2.$

And recall that a suitable rotation of $\mathbb R^n$ transforms $X$ to a vector whose expected value is $(\mu\sqrt n,0,0,\ldots,0)^T$ and transforms $\sigma^2(1-\rho)A$ to
$$
\sigma^2(1-\rho)\left[ \begin{array}{ccc} 0 \\ & 1 \\ & & 1 \\ & & & \ddots \\ & & & & 1 \end{array} \right].
$$
Hence we have
$$
\frac{(n-1)S^2}{\sigma^2(1-\rho)} \sim \chi^2_{n-1}.
$$


问:
When is an Integer a Rational Number, and are All Ratios Rational, Even $\frac{\sqrt{7}}{2}$?

$$\Bbb{Q} = \left\{\frac ab \mid \text{$a$ and $b$ are integers and $b \ne 0$} \right\}$$

In other words, a rational number is a number that can be written as one integer over another.

For an integer, the denominator is $1$ in that case. For example, $5$ can be written as $\dfrac 51$.

Is $5$ a rational number? Or is $\dfrac 51$  a rational number? I'm not able to figure out what the definition is actually saying. What are the numbers that cannot be written as one integer over another?

Irrational numbers are the numbers that cannot be written as one integer over another. Roots of numbers that are not perfect squares are examples of irrational numbers.

However, what is this then: $\dfrac {\sqrt 7} {2}$?

答:
Rational numbers are usually defined in terms of the integers.

I would define a rational number
as an ordered pair of integers
$(a, b)$ such that $b \ne 0$.
I would then define equality for rational numbers
by $(a, b) = (c, d)$ if and only if
$ad = bc$.

With the interpretation of $(a, b)$
as the rational number $a/b$,
this means that $a/b = c/d$
if and only if $ad = bc$.
In other words,
our previous knowledge of how
rational number should behave
drives all our definitions relating to
these rational numbers.

For example, since we want
$a/b + c/d = (ad+bd)/bd$,
we would $define$ addition by
$(a, b)+(c,d) = (ad+bc, bd)$.
We would then show that
these "rational numbers"
behave exactly like we want rational numbers to behave.

Finally, we would say
"From now on, we will write
$(a, b)$ as $a/b$",
and then forget all the constructions
since we have laboriously constructed
something we we already quite familiar with.

Another example of this is defining all integers
in terms of positive integers.
We can write an integer as
an ordered pair of positive integers
$(a, b)$
with equality being defined by
$(a, b) = (c, d)$ if and only if
$a+d = b+c$.
This is because we want $(a, b)$ to mean $a-b$,
and $a-b = c-d$ is the same as
$a+d = b+c$.
This reason we use this definition of equality
is that, in our definition
of "integers", we can only use positive integers,
where addition is always possible, but subtraction is not always possible.

Similarly, in our definition of
rational numbers in terms of integers,
multiplication is always possible,
but division is not always possible.

For more than you probably want to know
about this process,
read "Foundations of Analysis" by Edmund Landau.
This starts with constructing the integers
and ends up with the complex numbers.


问:
proving that $f'(x) > g'(x)$ implies $f$ and $g$ cannot intersect more than once.
I'm not completely certain if the statement is true, but intuitively I think it is. My sort of intuitive explanation would be if there is a point where $f$ and $g$ are equal, then the fact that $f'$ is greater than $g'$ would prevent them from intersecting again.

I was thinking of doing something with the mean value theorem and a proof by contradiction:

If we assume $f$ and $g$ intersect at two points, say $a$ and $b$, then $f(a)=g(a)$ and $f(b)=g(b)$. By the MVT, there must be points $c_f$ and $c_g$ such that $f'(c_f)=\frac{f(b)-f(a)}{b-a}$ and $f'(c_g)=\frac{g(b)-g(a)}{b-a}$, which implies $f'(c_f)=g'(c_g)$. But I'm not sure how much this would help since it would only lead to a contradiction if $c_f=c_g$.

Is there a way to continue the proof from here or some other way to prove this statement?

答:
Okay, I think I got the answer.

Since $f'(x)>g'(x)$, that means $f'(x)-g'(x)>0$ or the derivative of $f(x)-g(x)$ is greater than $0$. Using Rolle's theorem, if $f(x)-g(x)=0$ at two or more points, then there's a point where $f'(x)-g'(x)=0$, which is a contradiction since $f'(x)-g'(x)>0$. This means that $f(x)-g(x)=0$ at at most one point. Or, $f$ and $g$ can intersect at most once.


问:
How to determine the most probable value of the perimeter of a quadrilateral?
The problem is as follows:

The diagonals of a quadrilateral measure $8\,cm$ and $10\,cm$. Using
this information find the most probable perimeter of this
quadrilateral indicated as a range.

The alternatives given in my book are:

$\begin{array}{ll}
1.&\textrm{Between 14 cm and 16 cm}\\
2.&\textrm{Between 18 cm and 36 cm}\\
3.&\textrm{Between 38 cm and 40 cm}\\
4.&\textrm{Between 42 cm and 50 cm}\\
5.&\textrm{Between 12 cm and 14 cm}\\
\end{array}$

In this problem I don't know what sort of strategy can be used?. Does it exist a relationship between the diagonals of a quadrilateral?.

So far what it come to my mind was to use the triangle inequality which states that the sum of the two sides of a triangle is greater than the third side.

Given a quadrilateral ABCD then this means the diagonals are from $AC$ and from $BD$.

$AC<AB+BC$

$BD<BC+CD$

$AC<CD+AD$

$BD<AB+AD$

Given that the diagonals are AC and BD then its sum must be less than:
By summing the expressions given:

$2AC+2BD<2AB+2BC+2CD+2AD$

$AC+BD<AB+BC+CD+AD$

This must be the upper bound in the perimeter of the quadrilateral. But how about the lower bound?. Well, for this part I'm using the reverse triangle inequality which states that any side is greater than the difference of the other two sides provided one is greater than the other.

In this case I'm assuming $AB>BC$ and $BC>CD$ and $CD>AD$ and $AB>AD$

Then this would mean:

$AC>AB-BC$

BD>BC-CD

$AC>CD-AD$

$BD>AB-AD$

The thing here is that you may not get the perimeter directly.

Suming these:

$2AC+2BD>2AB-2AD$

$AC+BD>AB-AD$

But we know that we're assuming $BD>AC$.

Then summing both gets:

$2BD>AB-AD$

The perimeter is $AB+BC+CD+AD$

But that's where I'm stuck. Can someone help me here to finish this proof?. I think I'm near but I don't know how to use those inequalities to get to the answer. Can someone help me?.

By the way, the official answer from my book states to be the second option. Which it kinds of confirms my initial finding for the upper bound. But it doesn't make sense for the lower bound. Shouldn't it be less?. Therefore, can someone help me?

答:
What you found is actually the lower bound for the perimeter:

$$18 = AC + BD < AB + BC + CD + DA$$

To find the upper bound, we must assume that $ABCD$ is convex. Otherwise the perimeter can be as large as we want (consider a very long arrow-shape.)

If $ABCD$ is convex, the diagonals $AC$ and $BD$ intersect within the quadrilateral at some point $E$. Then we have, by triangle inequality:

$$AE + BE > AB$$$$BE + CE > BC$$$$CE + DE > CD$$$$DE + AE > DA$$

Summing them up, we have:

$$2 (AE + BE + CE + DE) = 2(AC + BD) = 36 > AB + BC + CD + DA$$

hence the perimeter will be bounded between $18$ cm and $36$ cm (or the sum and twice the sum of the diagonals).


问:
what is $\int_{|z|=3}\frac{1}{(z-2)^n(z-4)^m}dz$ for $m,n\in\mathbb{N}$?
I saw many examples how to calculate integrals with the residue theorem. But now I'm stuck with this integral: $$\int_{|z|=3}\frac{1}{(z-2)^n(z-4)^m}dz,$$where $m,n\in\mathbb{N}$ and $z=2$ and $z=4$ are the poles. Here we are only interested in $z=2$, because $4$ is not in $B_3(0)$.

For n=1 it is $2\pi if(z_0)=\int_{|z|=3}\frac{\frac{1}{(z-4)^m}}{(z-2)^1}dz$ with $f(z)=\frac{1}{(z-4)^m}$ and $z_0=2$ I would say, i.e. the integral is $2\pi i\frac{1}{(-2)^m}$ for $n=1$. But in genereal case I don't know what to do here, how to prove it ith the residue theorem.

I appreciate your help.

Body:
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 \newcommand{\dd}{\mathrm{d}}
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\begin{align}
&\bbox[5px,#ffd]{2\pi\ic\,{1 \over \pars{n - 1}!}
\lim_{z \to 2}\,\,\totald[n - 1]{}{z}\bracks{\pars{z - 2}^{n}\,
{1 \over \pars{z - 2}^{n}\pars{z - 4}^{m}}}}
\\[5mm] = &\
2\pi\ic\,{1 \over \pars{n - 1}!}
\lim_{z \to 0}\,\,\totald[n - 1]{\,\,\pars{z - 2}^{-m}}{z}
\\[5mm] = &\
2\pi\ic\,{\pars{-1}^{m}\,2^{-m} \over \pars{n - 1}!}\,
\lim_{z \to 0}\,\,\totald[n - 1]{\,\,\pars{1 - z/2}^{-m}}{z}
\\[5mm] = &\
2\pi\ic\,\pars{-1}^{m}\,2^{-m}
\bracks{z^{n - 1}}\pars{1 - {z \over 2}}^{-m}
\\[5mm] = &\
2\pi\ic\,\pars{-1}^{m}\,2^{-m}\
{-m \choose n - 1}\pars{-1}^{n - 1}\,\,{1 \over 2^{n - 1}}
\\[5mm] = &\
\bbx{2\pi\ic\,{\pars{-1}^{m} \over 2^{m + n - 1}}\
{m + n - 2\choose n - 1}} \\ &
\end{align}


问:
Quadratic form Gauss sum and quadratic residue
Let $Q(n_1,n_2,\ldots,n_r)$ be a quadratic form of several variables with integral coefficients. Let $q$ be a prime and let $0<a<q$. We are interested in the associated Gauss sum
$$S(a,q)=\sum_{n_1,n_2,\ldots,n_r=0,\ldots,q-1}e\left(\frac{aQ(n_1,n_2,\ldots,n_r)}{q}\right).$$

Now suppose $a$ is a quadratic residue mod $q$. Writing $a\equiv b^2$ for some $b$ coprime to $q$, we see that
$$S(a,q)=\sum e\left(\frac{Q(bn_1,bn_2,\ldots,bn_r)}{q}\right)=S(1,q).$$

What happens when $a$ is a quadratic nonresidue? When the quadratic form $Q$ is a sum of squares, we do know that
$$S(a,q)=\left(\frac{a}{q}\right)^rS(1,q)$$
where $\left(\frac{a}{q}\right)$ stands for the Legendre symbol. My question is, is it true for a general $Q$?

答:
Yes https://en.wikipedia.org/wiki/Quadratic_form#Equivalence_of_forms

So $$S(a,q)=\left(\frac{a}{q}\right)^R S(1,q)$$ with $R$ the number of non-zero terms in the sum of squares, it is $r$ when the symmetric matrix of the quadratic form has full rank.

When $R$ is odd and $a$ is not a square, together with $S(0,q)+ \frac{q-1}{2} S(1,q)+\frac{q-1}{2}S(a,q)= \sum_{k=0}^{q-1} S(k,q)=q f(0)$ it gives $f(0)=\# \{ n \in F_q^r, Q(n)=0\}=q^{r-1}$.


问:
The set of complex numbers of modulus $1$ is a group under multiplication
Show that $C=\{z\in \mathbb{C} \mid  |z|=1\}$ is a group under complex multiplication.

I'm a little confused because isn't the identity the only element with order $1$?  What is this set?

答:
Hint: prove that if you multiply two unitary complex numbers then the result is also an unitary number.

I suggest you to learn about polar representation of a complex number. This could make the solution easier.


问:
Fisher-Yates algorithm proof
I have an array of numbers $[a_0, \ldots, a_n]$ and want to shuffle it fairly. The Fisher-Yates algorithm proposes the following.

from $i = 0$ to $n$:

$j$ = randomly pick an index in $(0,  i)$,

swap $a_i$ with $a_j$.

The claim is that $P(x_0 = a_0, \ldots, x_n = a_n) = \frac{1}{n!}$. I am looking for a way to prove this rigorously.

答:
Proof by induction.

Suppose $A_i$ is the set of possible arrangement after the step i.

When $i = 1$, we have two possibles arrangements,
$A_1 = \{(x_0=a_0, x_1=a_1,\dots) ; (x_0=a_1, x_1=a_0,\dots)\}$

for $i = 2$, each of preceding posibilities, alow $3$ times mores, because
$a_2$ can switch with any of the three first variables $x_0, x_1, x_2$

$(x_0=a_0, x_1=a_1, x_2=a_2, \dots) $
$(x_0=a_2, x_1=a_1, x_2=a_0, \dots)$
$(x_0=a_0, x_1=a_2, x_2=a_1, \dots)$

$(x_0=a_1, x_1=a_0, x_2=a_2, \dots)$
$(x_0=a_2, x_1=a_0, x_2=a_0, \dots)$
$(x_0=a_1, x_1=a_2, x_2=a_1, \dots)$

then $|A_i| = 2 \times 3 = 3!$

Suppose when $i = n-1$, we have $|A_{n-1}| = n!$, then
for $i = n$, $a_n$ can switch with any of the $n+1$ first variables $x_0,\dots, x_{n}$, hence we count $$|A_n| = (n+1) \times |A_{n-1}| = (n+1)! $$ possibilities. Hence $P(x_0=a_0,\dots, x_n=a_n) = 1/|A_{n}| = 1 / (n+1)!$

Title: Fisher-Yates algorithm proof


问:
Why isn't the Euler characteristic equal to 2 for this polyhedron?
I've managed to confuse myself on a topic I thought I understood.

To set up, if a surface $M$ is a compact Riemannian manifold with boundary $\partial M$, then the Gauss-Bonnet theorem states $$\int_{M}KdS+\int_{\partial M}kds=2\pi \chi(M)$$ where $K$ is the Gaussian curvature of $M$, $k$ is the geodesic curvature of $\partial M$, and $\chi(M)$ is the Euler characteristic of the surface. Now, if our surface was $\textit{almost}$ smooth everywhere, except at some finite set of points $\left\{ x_{k}\right\}_{k\le n}\subset M$ where there were vertex singularities with angular defects $\left\{ \theta_{k}\right\}_{k\le n}$ at these points, then the Gauss-Bonnet theorem generalizes to $$\int_{M}KdS+\int_{\partial M}kds+\sum_{k\le n}\theta_{k}=2\pi \chi(M).$$ This is intuitively very pleasing, because it's essentially the statement that a surface's curvature, when viewed as a measure, can either live on the subset of the surface that's Riemannian, the boundary, or at vertices. This generalizes both Gauss-Bonnet for smooth surfaces and Descarte's theorem for the total angular defect of polyhedra.

In particular, if $M$ is a polyhedron with $n$ vertices, then $K=k=0$, and the theorem reduces to Descarte's theorem $$\boxed{\sum_{k\le n}\theta_{k}=2\pi \chi(M)}$$ This formula works beautifully for convex or toroidal polyhedra. But for some nonconvex polyhedra, I have no idea what's going on and am completely lost. To illustrate, I'll focus my attention particularly on two examples.

First, the $\textit{pentakis dodecahedron}$ is simply a regular dodecahedral surface with (short) pentagonal pyramids attached to each face. From reading its Wiki page, its Euler characteristic is $2$. No surprises here, since it has no holes or handles.

Then we come to the $\textit{small stellated dodecahedron}$, which is very similar, but with pentagonal pyramids substantially taller. It's Wiki page, Wolfram page, and this source all state, quite strangely, that it has twelve pentagrams as faces with thirty edges and only twelve vertices (which I would never even consider being the case). Its Euler characteristic then drops to $-6$. The same peculiar counting method seems to be implemented for several other nonconvex polyhedra.

I tested this by counting the small stellated dodecahedron's total angular defect, as shown here. Since each triangular face is isosceles with one angle of $36^{\circ}$ at the top and two angles of $72^{\circ}$ at the base, this is straightforward. There are twelve vertices (the pyramidal vertices) with a positive defect of $$360^{\circ}-5\cdot36^{\circ}=180^{\circ}$$ and twenty vertices (the original dodecahedron's vertices) with a defect of $$360^{\circ}-6\cdot72^{\circ}={-72^{\circ}}.$$ The total angular defect then comes to $$12\cdot 180^{\circ}-20\cdot72^{\circ}=720^{\circ}.$$ This is exactly what we would expect given that the small stellated dodecahedron has no holes or handles. This would also suggest that it has the same topology as a sphere and has an Euler characteristic of $2$. Moreover, whenever I count the vertices, edges, and faces, I get $V-E+F=32-90+60=2,$ which I'd expect anyways.

My question is, how could this $\textit{not}$ be the correct answer? My background is in analysis and differential geometry, not topology. If I'm wrong, I have no idea how. I also doubt all these sources are simultaneously wrong. But nowhere in the Gauss-Bonnet/Descarte theorem is convexity assumed. Moreover, it sounds preposterous from an analysis perspective that increasing the heights of the pentagonal pyramids to a certain point suddenly causes the total curvature to discontinuously become negative.

The sources for the stellated dodecahedron all state that this polyhedron confused early topologists. If this is just a funny way of counting faces, edges, and vertices, then who decided this was the $\text{"correct"}$ way to count them, and which way is $\text{"correct"}$ for the Gauss-Bonnet/Descarte theorem?

答:
You are treating the small stellated dodecahedron as a collection of triangles glued together along their edges -- just the exposed shapes that you can see in the figure.
You are also treating each intersection of line segments as a vertex.
As a result, you end up with something that is topologically equivalent to a sphere.

But when someone says the small stellated dodecahedron is made of $12$ pentagrams,
the pentgram they refer to is the star polygon $\left\{\frac52\right\}$.
Although this figure looks something like a pentagon with a triangle glued to each edge, it has only $5$ vertices and $5$ edges: each edge connects one of the "points" of the "star" to another "point", and in between these two vertices is crosses two other edges without creating any additional vertices.
The pentagram is a self-intersecting polygon.

So we have edges that don't end when they intersect another edge, and faces that don't end when they intersect another face. This is getting weird already, even before we calculate the Euler characteristic.

In short, the small stellated dodecahedron is not just a pentakis dodecahedron with the pyramids stretched outward to make certain faces coplanar. It's a completely different construction.

It you try to project the pentagram faces of the small stellated dodecahedron onto a concentric sphere, they will overlap each other.
I find it hard to see a spherical topology in this object--if it's there, it is strangely twisted around itself, just as the pentagram is a pentagon twisted around itself--so it should be no surprise really that it does not have the same Euler characteristic.


问:
What is the correct notation for binary digital image and dice score?
I wanted to ensure the correctness of notation regarding some concepts of digital images. If I have a binary image $X$ of width $W$ and height $H$ is the correct notation for describing image $X$:
$$X \in \{0, 1\}^{W\text{x}H}$$

Also, I have seen the definition of the Dice Score using the set notation for binary images but I still cant find the sense of it using sets:

$$\frac{2 \mid X \cap Y \mid}{\mid X \mid +
\mid Y \mid}$$

Since $\mid X \mid$ and $\mid Y \mid$ express the cardinality, as I understand, these will be the total number of pixels and not the non-zero pixels. Note: The other definition posted in the link makes absolute sense; I just want to understand if the set definition can also, in some way, be extended for images.

答:
There is no single correct notation. Often images are defined as a function on a discrete domain $D \in \mathbb{Z}^2$.

But when talking about binary images as sets, only the true (foreground) pixels are considered part of the set. So the cardinality of the binary image set is the number of true (foreground) pixels. This explains the Dice score definition.


问:
Linear transformations: left and right inverses
I want to prove the following statement.

Let $V$ and $W$ be $n$-dimensional vector space and $T \in \mathcal{L}(V,W)$. Prove that $S \in \mathcal{L}(W,V)$ is a left inverse if and only if it is a right inverse.

My first thought was that I can use the following facts:

(1) A function $f: A \to B$ is injective if and only if it possesses a left inverse and is surjective if and only if it possesses a right inverse.

(2) A linear map $T \in \mathcal{L}(V)$ is injective if and only if it surjective if and only if it is invertible.

The first statement seems valid, but the 'link' between surjectivity and injectivity breaks down, I believe, because $V$ and $W$ only have the same dimension, but are not necessarily the same vector space. Finite-dimensional doesn't imply finite, so I cannot use the usual fact for functions that if $f: A \to B$ where $|A| = |B| = n$ is injective, then it is surjective (and vice-versa). I also can't use the fact that a function is bijective if and only if it is invertible because having a left or a right inverse says nothing about invertibility (with the key word being 'or').

Any help would be appreciated. I'd like to think I'm on the right track.

答:
If $S$ is a left-inverse, then $T$ must be injective.

You say that you know that a linear map $L : V \rightarrow V$ from a finite-dimensional vector space to itself is injective iff it is surjective. But $V, W$ are both $n$-dimensional. So they are in fact the same space; more precisely, they're isomorphic. We can (by sending a basis of $V$ to a basis of $W$) find a bijective linear map $\phi : V \rightarrow W.$ Then $\phi^{-1} \circ T$ is still an injective linear map, but now it is from $V$ to $V.$ Thus $\phi^{-1} \circ T$ is surjective... but this means $T$ is surjective. Now concluding is straightforward: $T$ has a right inverse. But if a function has a left- and right-inverse, those inverses must be equal.


问:
Prove that if $f$ is differentiable at 0, then the function $f(|x|)$ is differentiable at 0 if and only if $f'(0)=0$
I have the following exercise:

Prove that if $f$ is differentiable at 0, then the function $f(|x|)$ is differentiable at 0 if and only if $f'(0)=0$

This problem is somewhat similar and the accepted solution seems correct to me. But I would like to know if there is a different way to solve it, also, the difference to the exercise here is that we do not know that $f(0) = 0$.
I hope you can help me, thank you.

答:
Let $g(x) = f(|x|)$.

If $f'(0) = 0$ then ${f(|x|) - f(0) \over |x|} \to 0$ so  ${f(|x|) - f(0) \over x} \to 0$ and hence $g$ is differentiable at $x=0$ (and $g'(0) = 0$).

If $g$ is differentiable at $x=0$ then ${f(|x|) - f(0) \over x} \to g'(0)$ and ${f(|x|) - f(0) \over -x} \to g'(0)$ and so $g'(0) = 0$. Taking the limit with $x>0$ shows that
$f'(0) = g'(0) = 0$.