问:
Bolzano-Weierstrass and zeros of complex analytic function
I am working on a textbook exercise. A similar question: An analytic function in a compact region has finitely many zeros, but it's not quite clear to me and I also have possibly another approach? I want to prove basically the same question, that if $f$ is analytic inside and on a simple closed contour $C$ (except possibly for poles inside $C$), and if all zeros of $f$ are inside $C$ and of finite order, then the zeros must be finitely many.

Hopefully my attempt below can be verified or corrected.

My attempt:

Suppose otherwise. Then by Bolzano-Weierstrass, the set $S$ of all zeros of $f$ (which is infinite) contains an accumulation point inside $C$. Let's say it is $z_0$. This $z_0$ is also a zero of $f$ since it is the limit a subsequence of zeros in $S$ and $f$ is analytic (hence continuous too). By assumption, it is a zero of finite order, say $m$.

I claim that in any neighborhood $N$ of $z_0$, $f$ cannot be identically zero. To see this, write $f(z)=(z-z_0)^mg(z)$ where $g$ is nonzero and analytic at $z_0$. Hence by these properties of $g$, there is a neighborhood around $z_0$ (intersected with $N$) where $g$ is nonzero. However, this neighborhood contains another (different) zero, say $z'$, of $f$ by definition of accumulation point. Hence, $0=f(z')=(z'-z_0)^mg(z')$, implying that $g$ can be zero in this neighborhood, a contradiction.

Now by a theorem in the textbook, since $f$ is analytic and zero at $z_0$, but not identically zero in any neighborhood of $z_0$, there must be a deleted neighborhood of $z_0$ where $f$ is identically nonzero. But again, in this deleted neighborhood contains a zero of $f$, say $z''$, by definition of accumulation point, contradicting $f$ being identically nonzero there. QED.

So my questions would be:

Is the above valid? If not, which part should be improved?

Are there any other approaches?

Usually Q2 is more interesting, but I highly appreciate if Q1 is answered too. Thanks a lot!

EDIT: Now that I think about it after some comment inputs:

My first paragraph should be fine.

As for my second paragraph until conclusion, I should do it like this:

As $z_0$ is of order $m$, we can write $f(z) = (z-z_0)^m g(z)$ where $g$ is analytic and nonzero at $z_0$. By continuity of $g$ and being nonzero at $z_0$, there is a neighborhood at $z_0$ where $g$ is identically nonzero. Deleting $z_0$ there, $f$ is then nonzero in that deleted neighborhood. However, this contradicts the fact that $z_0$ is an accumulation point of zeros. Done?

OR

Another method, I can also say: Either $f$ is not identically zero in any neighborhood $N$ of $z_0$ , or $f$ is identically zero in some neighborhood $N$ of $z_0$ . For the former, my original third paragraph follows to conclude. For the latter, by identity theorem $f$ must be identically zero inside $C$. By analyticity, their derivatives of all order are zero, showing infinite order. Done?

答:
I propose the following: let us prove that  if a function $f$ is analytic in the region $R$ consisting of all points inside and on a simple closed contour $C$, except possibly for poles inside $C$, and if all the zeros of $f$ in $R$ are interior to $C$ and are of ﬁnite order, then those zeros must be ﬁnite in number.  I think we must add the condition that $\;f\;$ isn't identically equal to zero in any non-trivial open, connected subset of $\;R\;$ . This is from a book (I already found a paper about this from 1981...) which I still cannot locate and it seems to be something very close to what you actually want. Observe that the conditions above for the function $\;f\;$ actually say the function's meromorphic on the domain enclosed by $\;C\;$ .

Proof: Suppose there are infinite zeros $\;\{z_1,z_2,...\}\;$ of $\;f\;$ inside $\;C\;$ . Then by Bolzano-Weierstrass, there exists $\;z_0\;$ on $\;R\;$ s.t. $\;\lim\limits_{n\to\infty}z_n=z_0\;$ . By continuity of $\;f\;$ , we get that $\;f(z_0)=0\;$ , too.

Since we're assuming all the zeros of $\;f\;$ on $\;R\;$ are of finite order and isolated, there exists $\;m\in\Bbb N\;$ s.t. $\;f(z)=(z-z_0)^mg(z)\;$ , in some open neighborhood $\;U\;$  of $\;z_0\;$ and for some meromorphic function $\;g\;$ s.t. $\;g(z)\neq0\;\;\forall\,z\in U\;$ . Since the possible poles of $\;f\;$ inside $\;C\;$ are isolated, we can take a neighborhood $\;V\;$  of $\;z_0\;$ where there are no poles of $\;f\;$  inside $\;V\;$ , and take the above relation $\;f(z)=(z-z_0)^mg(z)\;$ in $\;U':=U\cap V\;$, and this time $\;g\;$ is non-zero and analytic in $\;U'\;$ .

Thus we are almost thru, since then by the identity theorem of analytic functions we'd get that $\;f\;$ would be identically zero in some connected neighborhood of $\;z_0\;$ , since this point is an accumulation point of a set where $\;f\;$ and the zero function coincide, and this contradicts the further condition added above.


问:
$X,Y \sim Geo(p)$,what is $\frac{X}{Y}$ and $X-Y$ distribution?
let $X,Y\sim Geo(p)$, $X,Y$ are i.i.d

i know that $X+Y \sim NB(2,p)$

im trying to find how $Z= X-Y, W= \frac{X}{Y}$ are distributed

my attemp:
$P(Z\le k) = \sum_{i=-\infty}^k\sum_{j=-\infty}^\infty P(Y=j)P(X=i+j)=\sum_{i=-\infty}^k\sum_{j=-\infty}^\infty p(1-p)^{j-1}(1-p)^{i+j-1}p= \sum_{i=-\infty}^k p^2(1-p)^{i-2}\sum_{j=-\infty}^\infty(1-p)^{2j} = \sum_{i=-\infty}^k \frac{p^2(1-p)^{i-2}}{1-(1-p)^2}+\sum_{i=-\infty}^k\frac{p^2(1-p)^{i-2}}{(1-(1-p)^{-2})^2} = \frac{p^2}{(1-p)^2} \left (\frac{1}{1-(1-p)^2} + \frac{1}{(1-(1-p)^{-2})^2} \right ) \sum_{i=-\infty}^k (1-p)^i = \frac{p^2}{(1-p)^2} \left (\frac{1}{1-(1-p)^2} + \frac{1}{(1-(1-p)^{-2})^2} \right ) \left( \frac{1}{1-(1-p)^{-1}} + \frac{1-(1-p)^k}{1-(1-p)}\right)$

from here i really struggled with the fractions and didnt get any known distirbution or something that i think is correct, the same goes for $W$ as well, did i get something wrong along the way? how do i continue from here and also how to i calculate $W$ distirbution?

答:
For $\frac{X}{Y}$ to be finite this would need $Y\not=0$ and so suggests the "trials until success" version of the geometric distribution with the support $\{1,2,3,\ldots\}$.

So $\mathbb P(X=n) =\mathbb P(Y=n) = p(1-p)^{n-1}$ for $n\ \in \{1,2,\ldots\}$ and $0$ otherwise.

For $z$ an integer: $$\mathbb P(X-Y=z) = \sum\limits_{n=|z|+1}^\infty  p^2(1-p)^{2n-2+|z|} = \frac{p}{2-p} (1-p)^{|z|}$$

For $w=\frac{a}{b}$ a positive rational in lowest terms: $$\mathbb P\left(\frac XY=\frac{a}{b}\right) = \sum\limits_{n=1}^\infty  p^2(1-p)^{an+bn-2} = p^2 \frac{(1-p)^{a+b-2}}{1-(1-p)^{a+b}}$$

As a check, the peak occurs when $X=Y$, with $\mathbb P(X-Y=0) =\mathbb P\left(\frac XY=1\right)=\frac{p}{2-p}$

ContentLicense: CC BY-SA 4.0


问:
Difference between normal convergence, pointwise convergence and uniform convergence
Let be $\left(f_k\right)_{k\in\mathbb{N}}$ a sequence of functions, $f_k:M\to Y$, where $(Y,\Vert\cdot\Vert_Y)$ is a normed space.

Why do we sometimes talk about "normal" convergence and sometimes we refer to uniform convergence or pointwise convergence? Is it correct if I explain like this:

If we talk about $\left(f_k\right)_{k\in\mathbb{N}}$ as a sequence that lives in a function space then we simply refer to convergence and say $\left(f_k\right)_{k\in\mathbb{N}}$ converges to some element $f$ from this function space and write $\lim\limits_{k\to\infty}f_k=f$. However, if we look at a sequence of functions regardless its function space then we make the distinction between the two notions of convergence. So the distinction between convergence, uniform convergence and pointwise convergence comes from the different levels of abstractions.

EDIT:

Here an example to illustrate the idea:

Let's say the function space is equipped with the supremum norm, then in this case convergence of $\left(f_k\right)_{k\in\mathbb{N}}$ means that it is uniformly convergent (and vice versa). If the function space is equipped with some other norm then it could be the case that $\left(f_k\right)_{k\in\mathbb{N}}$ is uniformly convergent (and hence pointwisely convergent) but not convergent in the function space.

Is this explanation correct?

Maybe you have some other explanations which help to grasp the idea :)

答:
I believe when you say "normal" convergence that it actually is the same thing as pointwise convergence. What would it mean for a sequence of functions to converge to a function in a function space? Functions are defined by acting on elements of a given set, so the only way to check convergence would be to check how they act on their domain, but this is just how they act for each point, hence pointwise convergence.

So really there only is a distinction between pointwise and uniform convergence. This difference can best be summarized by the following: pointwise convergence is concerned with a single point at a time. Uniform convergence is concerned with ALL points in the domain at the SAME time. This is important because while a sequence of functions may converge pointwise, if it is converging at different "rates" at each point in may not converge uniformly.

For example,$f_n:(0,\infty)\to \mathbb{R}$ given by $f_n(x) = \frac{1}{nx}$ is a sequence of functions that converges pointwise to 0. This is clear because if you pick any $x_0\in (0,\infty)$, then $f_n(x_0) = \frac{1}{nx_0}$ is just a sequence of real numbers clearly converging to $0$. But $f_n$ does not converge uniformly to $0$, because you can always find, for some $\varepsilon$, a $\delta>0$ such that for all $x\in (0,\delta)$, $f_n(x) \geq \varepsilon$. The idea is when $x$ is large then clearly $f_n(x)=\frac{1}{nx}$ will be small for all values of $n\in\mathbb{N}$. But when you shrink $x$ you have to increase $n$ for $f_n(x)$ to remain small and this relationship between $x$ and $n$ implies that the convergence is not uniform as it depends on what point we pick for us to determine the convergence, whereas if it were uniform then for large enough $n$, EVERY $x$ would satisfy $f_n(x) <\varepsilon$


问:
Limit for $n\to\infty$ with the sine
It is true that $\lim\limits_{n\to\infty} \left(n\pi\sin{n\pi}\right)=0\,$? I thought that $\sin{n\pi}=0$ and so $\lim\limits_{n\to\infty} (n\pi\cdot0)=0\;.$

答: Yes, since $\;\sin(n\pi)=0\;\;\forall n\in\mathbb{N}\;,\;$ every term of the sequence is $\;0\;$ and so the limit is $\;0\;.$

Title: Limit for $n\to\infty$ with the sine


问:
When/why is a point on a curve an effective Cartier divisor?
The definition of effective Cartier divisor that I'm using is: a closed immersion whose corresponding quasicoherent sheaf of ideals is an invertible sheaf.

Let $X$ be a scheme that is dimension 1 and locally of finite type over a field $k$ (but can be singular or have multiple components). Let $p$ be a closed point on $X$. Then why is the canonical morphism $f : Spec(k(p)) \rightarrow X$ is an effective Cartier divisor?

This is my attempt so far:

Since $p$ is a closed point, $f : Spec(k(p)) \rightarrow X$ is a homeomorphism onto its range and the range is closed. Also, the pullback morphism $f^\# : O_X \rightarrow f_* O_{Spec(k(x))}$ is surjective (as on an affine open neighbourhood $U = Spec(A)$ of $p$, we have $A \rightarrow Quot(A/p) = A/p$ since $p$ is a closed point so it's a maximal ideal, so pullback is surjective). Therefore it's a closed immersion.

Let $\mathcal{I}$ be the corresponding quasicoherent sheaf of ideals. For open sets U that don't contain $p$, $\mathcal{I} \sim O_X|_{U}$ so is an invertible sheaf on there. But now I'm stuck on the part where open sets $U$ that contain $p$.

Is the statement even true? If so, how do I prove $\mathcal{I}$ is invertible on some open set $U$ that contains $p$?

答: This statement is true precisely when $p$ is a regular point of $X$. If $p$ is regular, then $\mathcal{O}_{X,p}$ is a regular local ring of dimension one and hence a DVR, so we can select a uniformizer which gives a generator on some neighborhood of $p$. Conversely, if the ideal sheaf of $X$ is locally principal then its stalk at $p$ must be a principal ideal of $\mathcal{O}_{X,p}$. This suffices to show that $\mathcal{O}_{X,p}$ is a DVR, which is a regular local ring.


问:
Completing a proof for a pigeonhole principle question
Seventeen people correspond by mail with one another - each one with all the rest. In their letters only three different topics are discussed. Each pair of correspondents deals with only one of these topics. Prove that there are at least three different people who write to each other about the same topic.

Work I have done so far

Each person can talk to maximum $16$ people on $3$ topics $\implies$ at least one topic will be spoken about $6$ times by Pigeonhole principle.

This part is clear to me. I do not understand how to complete the proof by showing that there are at least three different people who write to each other about the same topic.

答:
So there is someone who corresponds with $6$ other people on topic $1$. If any pair of these $6$ people correspond with one another on topic $1$, then we done. So suppose that each of these $6$ people only correspond on topics $2$ or $3$ ... Well that's a $2$ coloring of $K_6$, which must have a $K_3$, so we are done.

For more details see https://en.wikipedia.org/wiki/Ramsey%27s_theorem#Examples


问:
Definition(s) of rational numbers
The definitions of rational numbers are somewhat confusing for me. The definition of rational numbers on wikipedia and most other sites is:

In mathematics, a rational number is any number that can be expressed as the quotient or fraction $\frac pq$ of two integers, with the denominator $q$ not equal to zero.

All integers can be expressed as the quotient $\frac pq$. So if they are not written in the fraction form, will they still be called rational numbers?

Another definition which I've found is:

A rational number is a quotient $\frac mn$ where $m$ and $n$ are integers and $n \neq 0$.

These definitions might be the same. However, I'm not able to figure out how! Are rational numbers those numbers which can be expressed as the quotient $\frac mn$ of two integers or those which are directly the quotient of two integers? For example, if $n$ is an integer, so is $n$ a rational number? Or is $\frac n1$ a rational number?

答: I hope this helps clarifing things a little bit. $\mathbb{Z}$ is built from $\mathbb{N}\times \mathbb{N}$ introducing an equivalence relation on it and $\mathbb{Q}$ is gained from $\mathbb{Z}\times (\mathbb{Z}\setminus\{0\})$ considering its quotient by another appropriete equivalence relation (details can be found everywhere, for example in Wikipedia). So, strictly speaking, it is not right to say that $\mathbb{Z}\subseteq \mathbb{Q}$ because it is not true that each element of the LHS is also an element of RHS. But we have a canonycal function $i\colon\mathbb{Z}\longrightarrow \mathbb{Q}$, mapping $n\in\mathbb{Z}$ to $\frac{n}{1}$, which is injective (therefore a bijection onto its image) and also a ring (monotone) homomorphism. Hence we identify, reasoning up to isomorphism, $\mathbb{Z}$ with $i(\mathbb{Z})$ when we want to think to $\mathbb{Z}$ as contained ("embedded") in $\mathbb{Q}$. </p>

ContentLicense: CC BY-SA 3.0

Title: Definition(s) of rational numbers


问:
Cannot Solve Tangent Plane Equation to Parametric Surface
I have a problem while finding the equation of a tangent plane to a parametric surface. The surface is given by $$\pmb r = \begin{bmatrix} s^2 + t^2 \cr 2s+2t \cr 2 \end{bmatrix}$$

The point at which plane is tangent $$ \pmb r_o = \begin{bmatrix} 2 \cr 4 \cr 2 \end{bmatrix} (s=1, t=1)$$

I know that for tangent-plane equation, I need to find a normal vector $\pmb n$ by doing a cross-product of $\frac{\partial \pmb r}{\partial t}$ at $r_o$ and $\frac{\partial \pmb r}{\partial s}$ at $r_o$.

My problem is that both $\frac{\partial \pmb r}{\partial t}$ and $\frac{\partial \pmb r}{\partial s}$ are being computed as the same vector $\begin{bmatrix} 2 \cr 2 \cr 0 \end{bmatrix}$. So my normal vector will always be zero, and I can't use it to find the equation of the tangent.

How to get around this problem?

Also, in general, how do we solve tangent-plane equations to parametric surfaces in such cases?

答: Your "surface" $S$ is the subset $x\geq y^2/8$ inside the plane $z=2$. As such, it is not regular along the curve $(\tau, \tau^2/8, 2)$ and, according to most standard definitions, one cannot talk about tangent planes at points along that curve. Since $\pmb r(1,2)$ lies on this curve, we have no well-defined tangent. Of course, since $S$ is entirely contained in the plane $z=2$, at any point not on its boundary curve the tangent plane is $z=2$.


问:
How good function approximation can be achieved with N bits?
Alice has a function $f: [0,1]\rightarrow [0,1]$ that is piecewise continuous. Bob wants to approximate $f$, but can only get $N$ bits of information. How good can his approximation $\hat{f}$ be?

The question needs some refinement since the answer may depend on the protocol: In case 1 Bob asks $N$ yes/no questions about $f$ to Alice she answers truthfully, in case 2 he only gets to ask about function values and get yes/no answers back (e.g. "Is $f(0.3)>0.5$?), and in case 3 Alice gives Bob $N$ bits of information for yes/no questions of her choosing.

We also need to settle on how to measure the fit; a reasonable choice may be to use the norm $\int_0^1 |f(x)-\hat{f}(x)|dx$ but I am open for alternatives.

One upper bound on the approximation in case 2 is if Bob asks $N=M^2$ questions that fill in a binary grid: "Is $f(i/M))\leq j/(M+1)$?" (for $i,j$ from 1 to $M$). This produces a piecewise constant approximation where the total error is maximally 1 for wildly varying functions: not so good. In this case just assuming $\hat{f}(x)=1/2$ is better.

For "nice" functions that do not vary wildly the total error will be $\approx 1/2M$. There Bob can ask $M$ questions to do interval halving for the best approximation in $[j/(M+1),(j+1)/(M+1)]$, arriving at a total error scaling as $2^{-M}$: much better.

In case 3, if Alice is feeling unhelpful, she will just tell Bob approximations to one single value $x^*$ and Bob's $\hat{f}$ will presumably have to be just $\hat{f}(x)=1/2, x\neq x^*$ and $f(x^*)$ otherwise. The error will be 1/2. But if she is helpful she might use the algorithm in (Konno & Kuno 1988) or something like it to give approximations to an optimal answer. I am however uncertain of what bounds on how good the answer would be.

In summary, it seems that the interactive questions in case 1 and 2 are exponentially more valuable than following a pre-committed list of questions, and a helpful Alice is equally good or better. But can Bob ask more clever questions in case 1? Or do we need to place more constraints on $f$ to make the problem well-posed?

答:
In case 1, Bob's procedure is some function from strings $s$ of $N$ bits to functions $\hat{f}_s$. Fixing any such mapping, we can WLOG make his questions
"What is the $k$th binary digit of the string $s$ for which $\hat{f}_{s}$ achieves the best fit to your function?"

With this framing, it's clear that the problem is just a restriction of the general problem "Bob can precommit to a list of $n$ functions and wants to minimize the error of the best function $\hat{f}_i$ to an adversarially-selected $f$" to the case where $n$ is a power of two; I expect that looking at $n=1,2,3,\ldots$ is the more natural way to investigate the question.

However, without more restrictions on $f$ I think Bob can't learn anything, even in Case 1: the idea is basically "Alice picks $0$ or $1$ at random for the value of her function on each of many small intervals". Then any specific $\hat{f}$ will have expected integral difference $\epsilon/2$ over a given interval of length $\epsilon$, so by picking enough intervals, we can make the distribution of expected error be so tight around $1/2$ that the odds any one $\hat{f}$ goes below, say, $0.49$ in error is less than 1. Hence, by the probabilistic method, some concrete choice on Alice's part must evade good approximation for Bob's specific list of $n$ functions. (The same sort of idea should work even if we require Alice's function to be continuous and differentiable, or a lot of other "niceness" conditions - we just make the jump from $0$ regions to $1$ regions very steep.)

I think a natural restriction on Alice to make this problem interesting would be to require that $f$ be differentiable, and bound the absolute value of the derivative of $f$. (Or perhaps relax the differentiability constraint and just say that $\left|\frac{f(y)-f(x)}{y-x}\right|<c$ for some $c>0$.)

Under these restrictions, one can make nontrivial improvements: when $n=2$ and $c=1$, taking $\hat{f}_1=1/3$ and $\hat{f}_2=2/3$ guarantees an error of at most $4/9$, though I think this isn't tight even for this specific choice of $f_1$ and $f_2$.


问:
Prove $\{X_k\}_{k=1}^{n}$ are independent iff $F_{X_1,X_2,\ldots,X_n}(x_1,x_2,\ldots,x_n)=\prod_{k=1}^{n}F_{X_k}(x_k)$

Let $\{X_k\}_{k=1}^{n}$ be RVs

Prove $\{X_k\}_{k=1}^{n}$ are independent iff for every $(x_1,x_2,\ldots,x_n)\in\mathbb{R}^n$ :$F_{X_1,X_2,\ldots,X_n}(x_1,x_2,\ldots,x_n)=\prod_{k=1}^{n}F_{X_k}(x_k)$

I've managed to prove the first part where $X_k$ are independent.

Any help for the second will be welcome.

答:
Let $\mathcal{I}$ be the set of intervals of the form $(-\infty, x]$ with $x\in\mathbb{R}$, and $\mathcal{B}$ be the Borel $\sigma$-algebra on the real line. Fix $I_2,...,I_n\in\mathcal{I}$. Then the class
$$\mathcal{G}_1=\left\{B_1\in\mathcal{B} : \mathbb{P}(X_1\in B_1, X_2\in I_2, ..., X_n\in I_n)=\mathbb{P}(X_1\in B_1)\mathbb{P}(X_2\in I_2)\cdots \mathbb{P}(X_n \in I_n) \right\}$$
is a monotone class. Moreover, $\mathcal{I}\subset\mathcal{G}_1$ is stable under finite intersections so the monotone class lemma ensures that $\mathcal{M}(\mathcal{I})=\sigma(\mathcal{I})\subset\mathcal{G}_1$. Since $\sigma(\mathcal{I})=\mathcal{B}$, we have shown that for all $B_1\in \mathcal{B}$ and $I_2,...,I_n\in\mathcal{I}$
$$\mathbb{P}(X_1\in B_1, X_2\in I_2,..., X_n\in I_n)=\mathbb{P}(X_1\in B_1)\mathbb{P}(X_2\in I_2)\cdots\mathbb{P}(X_n\in I_n)$$
Proceeding by induction with the same reasoning, we can replace $I_2,...,I_n\in\mathcal{I}$ by $B_2,...,B_n\in\mathcal{B}$ one by one. Once we are done, we can simply notice that we have arrived to the definition of independence.


问:
How can I solve C(n,2)+2×C(n,3)+3×C(n,4)+...+(n-1)×C(n,n)=?
'm able to find an answer but I don't think it's the right one. I tested it for n=2 and it worked but not for n=3.Here's my answer:

$$C(n,2)+2×C(n,3)+3×C(n,4)+...+(n-1)×C(n,n)=[C(n,2)+C(n,3)+...+C(n,n)] + [0×C(n,2)+1×C(n,3)+2×C(n,4)+...+
(n-2)×C(n,n)] =
2^{(n-2)} + 2^{(n-2)}×(n-2)$$

I'm either forgetting something or it's just fully wrong. Sorry for the bad formatting and thank you in advance

答:
Note that
\begin{align*}
n(1 + x)^{n - 1}  - (1 + x)^n & = \left( {\frac{d}{{dx}}(1 + x)^n } \right) - (1 + x)^n  = \sum\limits_{k = 0}^n {\binom{n}{k}kx^{k-1} }  - \sum\limits_{k = 0}^n {\binom{n}{k}x^k } \\ & = \sum\limits_{k = 0}^n {\binom{n}{k}(k - x)x^{k-1} } .
\end{align*}
Now just substitute $x=1$.


问:
Finding a normal subgroup of a fiber product giving a prescribed quotient.
Let $D_{2^{n-1}}$ be the Dihedral group of order $2^n$. Letting $s_1:D_{2^{n-1}}\rightarrow \mathbb{Z}_2$ be determined by one of its subgroups isomorphic to $D_{2^{n-2}}$. The book I'm reading says that the pullback given by this map has a normal subgroup $N$ such that when quotiend by it, the resulting group is isomorphic to the semi-dihedral group $QD_{2^{n-1}}$.

What I'm assuming this means, is that we get a fiber product $D_{2^{n-1}}\times_{\mathbb{Z}_2} \mathbb{Z}_4$ where $\mathbb{Z}_4\rightarrow \mathbb{Z}_2$ is the canonical epimorphism.

(Here $\mathbb{Z}_n$ denotes the cyclic group of order $n$.)

What I am wondering is how to find this subgroup $N$. I think one could try to construct an explicit surjection from the fiber product to $QD_{2^{n-1}}$, and then we have the result by the first isomorphism theorem. But this seems like a lot of work. The book usually does some calculations in e.g. Maple. Is there an easy way to see this? My main problem being that while I know explicitly how the fiber product looks like, I find it computationally hard to work with.

答:
The direct product can be defined by the presentation
$$\langle x,y,z \mid x^{2^{n-1}}=y^2=z^4=1, y^{-1}xy=x^{2^{n-2}-1}, [x,z]=[y,z]=1 \rangle,$$
and the fibre product is the subgroup $\langle u,y \rangle$ of index $2$, where $u=xz$.

There are three normal subgroups of index $2$, and (as you might expect) factoring out either of the two contained in one of the original direct factors does not give what you want, so it must the third of these, whichn is $\langle x^{2^{n-2}}z^2 \rangle = \langle u^{2^{n-2}}z^2\rangle$.

Now  $y^{-1}uy = x^{-1}z$, whereas $$u^{2^{n-2}-1} = (xz)^{2^{n-2}-1} = x^{-1}u^{2^{n-2}}z^{-1} = x^{-1}z(u^{2^{n-2}}z^2) = y^{-1}uy(u^{2^{n-2}}z^2),$$ we see that the quotient group is indeed isomorphic to ${\rm QD}_{2^{n-1}}$.


问:
Find sum of $5$ digit number that can be formed using digits $0,0,1,2,3,4$
I'm attaching my approach in image format. I'm new to mathematics stack exchange, so I'm not able to write in latex format properly.
I'm getting different answers. what going wrong in my logic.
I've solved every problem related to sum by this way.

Given answer is $6399960$.
strong text

答:
Observe that a digit of $0$ does not contribute anything to the sum, so you do not need to count them.

For the first digit, you should have $\frac{120}{2!} = 60$, as both zeroes are indistinguishable, so they give the same number when interchanged. For the other digits, you should have $\frac{3 \times 4 \times 3 \times 2}{2!} = 36$ instead of $80$ for the same reason, as the factor of $1$ for the chosen digit does not affect the calculation  (the first digit cannot be $0$ as you have noticed).

Hence the answer is ($1+2+3+4=10$):

$$60 \times 10 \times 10^4 + 36 \times 10 \times 10^3 + 36 \times 10 \times 10^2 + 36 \times 10 \times 10^1 +  36 \times 10 \times 10^0$$
$$=6399 960.$$

or around $6$ million.


答:
Consider the person that talks about a topic with other $6$. Take that group of people, and consider that if some pair talks about that same topic then we're done. So consider they don't talk about that topic. They're now a group of $6$ people talking about $2$ topics.

If we apply the same process you did at the start, we have that anyone talks about one of the two remaining topics at least $3$ times. If any pair of those three remaining people talk about that topic, say topic number two, we're done. If there is no pair of those three remaining people talking about  topic number two, then they're all talking about topic number three. But then we're done too, because they're a triangle of people talking about the same topic.

It's basically applying what you did three times, until you have a trivial case of a triangle.

Title: Completing a proof for a pigeonhole principle question

Body:
Seventeen people correspond by mail with one another - each one with all the rest. In their letters only three different topics are discussed. Each pair of correspondents deals with only one of these topics. Prove that there are at least three different people who write to each other about the same topic.

Work I have done so far

Each person can talk to maximum $16$ people on $3$ topics $\implies$ at least one topic will be spoken about $6$ times by Pigeonhole principle.

This part is clear to me. I do not understand how to complete the proof by showing that there are at least three different people who write to each other about the same topic.


问:
Prove $\lim\limits_{x\to -\infty} \frac{\cos(\tan x)}{x+1} = 0$ using definition

Problem

Prove  $\lim\limits_{x\to -\infty} \frac{\cos(\tan x)}{x+1} = 0$ using definition

proof

$\forall \varepsilon>0$ take $M=-\frac{1}{\varepsilon}-1$ if $x<M<0$, implies

$$x<-\frac{1}{\varepsilon}-1<0$$ it  follows that  $$-\varepsilon <\frac{1}{x+1}<0<\varepsilon \iff \bigg|\frac{1}{x+1}\bigg|<\varepsilon$$

since $|\cos(\tan x)|\leq 1$ it follows
$$f(x)=\bigg|\frac{\cos(\tan x)}{x+1}\bigg|<\varepsilon$$
so
$$\lim\limits_{x\to -\infty} \frac{\cos(\tan x)}{x+1} = 0$$

Please verify the proof above.

答: Your proof is fine, albeit only on $\{x<-1|\tan x\in\Bbb R\}$, a space where the limit makes sense. Alternatively (again subject to the same constraint @TheSilverDoe notes), squeeze between $\frac{\pm1}{x+1}$, each of limit $0$.


问:
Why does $\mathbb{P}(x)=\frac{3}{x^2\pi^2}$ have no expectation?
The textbook answer to $\mathbb{E}(X)$ for $\{\mathbb{P}(X=x)=\frac{3}{x^2\pi^2}, x\in \mathbb{Z}\setminus 0 \}$ is it doesn't exist because of subtraction of $\infty$. But in my calculation I find that $\mathbb{E}(X)=\sum_{-\infty}^{-1}{\frac{3}{x\pi^2}}+\sum_{1}^{\infty}{\frac{3}{x\pi^2}}=\frac{3}{\pi^2}\sum_1^\infty{1/x}-\sum_1^\infty{1/x}=0$. What's wrong with factorizing the negative out?

答:
We define $\mathbb{E}[X] := \mathbb{E}[X^+]-\mathbb{E}[X^-]$ where $X^+$ is the positive and $X^-$ the negative part of $X$, whenever the two expectations exist.

But as you probably (unconsciously) noted, $E[X^+] = E[X^-] = \sum_{n=1}^\infty \frac{3}{n\pi^2} = \infty$, so $\mathbb{E}[X]$ is undefined.


问:
What does the Lattice actually means in the Small World Propensity formula?
I am studying small-world networks and I came across the formula of Small World Propensity (reference: https://doi.org/10.1371/journal.pone.0216146.s001). And I am having trouble understanding the Lattice in the formula. Please find the detailed problem after the definition.

Definition: The small-world propensity measures the extent to which a network is characterized by high levels of local clustering and low average path length. It is defined as follows:

$\Phi = 1-\sqrt{\frac{{\Delta_T}^2+{\Delta_L}^2}{2}}$

where

$\Delta_T = \frac{T_{lattice}-T_{observed}}{T_{lattice}-T_{random}}$

and

$\Delta_L = \frac{L_{observed}-L_{random}}{L_{lattice}-L_{random}}$

with

$L$ being the path length of a network, i.e. the average shortest path length between all node pairs, and $T$ representing the network clustering (transitivity) coefficient, defined as the average of node-specific clustering coefficient values.

Problem: I am trying to code this using the python-igraph library and my understanding is that the we have to calculate $T$ and $L$ values using -

random graph for $T_{random}$ and $L_{random}$

k-regular graph for $T_{lattice}$ and $L_{lattice}$

but when I checked the documentation of igraph I noticed that apart from K_Regular(n, k) function (ref. page no. 123  in the pdf) there is a separate function called Lattice(dim) (ref. page no. 124 in the pdf) that generates a regular lattice. Screenshots attached below, in case this pdf is removed in the future.

Now, I am confused that should I use K_Regular(n, k) or Lattice(dim) to calculate $T_{lattice}$ and $L_{lattice}$? And what is the difference between these two functions?

答:
Apparently, my understanding of creating the regular graphs was wrong in the case of Small World Propensity.

I did a discussion with people working in the field of brain connectivity and with the help of this paper (in this Small World Propensity is originally discussed), I was able to figure out the meaning of lattice used in this context.

Below I am quoting a few points from the paper that helped me figure it out.

In section Generating Weighted Small-World Networks

Begin with a network of N nodes that are arranged on a lattice and connected to all neighbors within a radius, r.

In Figure 3a shows that the initial lattice is a circular graph in which every node is connected to n neighbours both on left and right.

Now, this confirmed that I needed to use Lattice(dim) for generating the initial lattice, but now a second question arose  - what should be the value of dim?

This was answered in this paper, Figure 2a has explained that such a lattice is A periodic one-dimensional ( D = 1 ) lattice (i.e. a ring) where each vertex is connected to its n neighbours on left and right.

and the documentation of igraph for Lattice confirmed that I need to put a list containing 1 element (1 dimension), and the value of that element should be the number of nodes to be generated, further, I need to put the number of neighbours on either side to directly connect.

So, If I need 20 nodes and connect 3 neighbours in each direction (so total of 6 neighbours will be connected), function should be:

```<code>G = ig.Graph().Lattice(dim=[20],nei=3)
</code>```

This generates the following graph.

Body:
Apparently, my understanding of creating the regular graphs was wrong in the case of Small World Propensity.

I did a discussion with people working in the field of brain connectivity and with the help of this paper (in this Small World Propensity is originally discussed), I was able to figure out the meaning of lattice used in this context.

Below I am quoting a few points from the paper that helped me figure it out.

In section Generating Weighted Small-World Networks

Begin with a network of N nodes that are arranged on a lattice and connected to all neighbors within a radius, r.

In Figure 3a shows that the initial lattice is a circular graph in which every node is connected to n neighbours both on left and right.

Now, this confirmed that I needed to use Lattice(dim) for generating the initial lattice, but now a second question arose  - what should be the value of dim?

This was answered in this paper, Figure 2a has explained that such a lattice is A periodic one-dimensional ( D = 1 ) lattice (i.e. a ring) where each vertex is connected to its n neighbours on left and right.

and the documentation of igraph for Lattice confirmed that I need to put a list containing 1 element (1 dimension), and the value of that element should be the number of nodes to be generated, further, I need to put the number of neighbours on either side to directly connect.

So, If I need 20 nodes and connect 3 neighbours in each direction (so total of 6 neighbours will be connected), function should be:

G = ig.Graph().Lattice(dim=[20],nei=3)

This generates the following graph.


问:
Homotopy Class of Maps from Euclidean Space
Let $n,m\in \mathbb{Z}^+$, and let $\pi(\mathbb{R}^n,S^m)$ be the set of free homotopy classes from $\mathbb{R}^n$ to $S^m$.  Is there more than one homotopy class in $\pi(\mathbb{R}^n,S^m)$?  In other words, is every continuous function $f:\mathbb{R}^n\rightarrow S^m$ null-homotopic?

Edit: What about the converse; i.e.: $\pi(S^m,\mathbb{R}^n)$?

答:
Let $f: \mathbb{R}^n \rightarrow X$ be any continuous map where $X$ is any topological space. Then $(t,x) \in [0,1] \times \mathbb{R}^n \longmapsto f(tx) \in X$ is a homotopy between $f$ and the constant map to $f(0)$, so $f$ is null homotopic.

Similarly, let $f: X \rightarrow \mathbb{R}^n$ any continuous map, then $(t,x) \in [0,1] \times X \longmapsto tf(x)\in \mathbb{R}^n$ is a homotopy between $f$ and the constant map to $0$. So $f$ is null homotopic.


问:
Solve the linear congruences
To solve $\,26x\equiv 6\pmod{\!110},\,$ note $\,\gcd(26,110)=2\,$ so there are $\,2\,$ solutions.

Next, $\ 104x\equiv 24\pmod{\!110}\ $ follows by scaling the above by $\,4.$

i.e. $\,\ \ \ \ \ {-}6x\equiv 24\pmod{\!110}$

Hence $\ \ \ \ \,x\equiv -4\pmod{\!55}\ $ by dividing prior by $\,6$.

What is the final answer?

答:
Avoid scaling by a factor that is not comprime to the modulo; you may create/miss solutions.

From $26x \equiv 6 \pmod {110}$, we can divide everything by $2$ to reach $13 x \equiv 3 \pmod {55}$.

Now scaling by $4$ is valid, since $4$ and $55$ are coprime. We have:

$$12 \equiv 52 x \equiv -3 x \pmod {55}$$

Dividing both sides by $-3$, which is also coprime with $55$:

$$x \equiv -4 \pmod {55}$$

which is the solution you somehow managed to arrive at, despite the questionable method.

Title: Solve the linear congruences


答:
Note $\,x\equiv -4\pmod{\!55}\iff x\equiv 51, 106\pmod{\!110},\,$ so those are the sought $2$ solutions.

However you got lucky. The method you used will generally introduce extraneous solutions because you scaled by a number $(4)$ this is noninvertible $\bmod{110}$. To remedy that in general you need to test which of the derived possible solutions are actual solutions. This is not needed when we scale by invertibles since that yields equivalent congruences, i.e. the congruences are connected by $(\!\!\iff\!\!)$ arrows, vs. $(\Rightarrow)$.

For example if we do the same problem mod $220$ and scale by $8$ we deduce that any solution again satisfies $\,x\equiv -4\pmod{\!55},\,$ but only two of those $4$ possible solutions are actually solutions, viz. $\,x\equiv 51\pmod{\!110}\iff x\equiv 51,\,161\pmod{\!220}$.

You can avoid such problems by cancelling $\,2 =\gcd(26,110)\,$ at the start (vs. end), as in the general solution method explained in this answer.

Another handy way to solve such congruences is to use the fractional extended Euclidean algorithm, which is closely related to the method that you (implicitly) employed.

Title: Solve the linear congruences

Body:
To solve $\,26x\equiv 6\pmod{\!110},\,$ note $\,\gcd(26,110)=2\,$ so there are $\,2\,$ solutions.

Next, $\ 104x\equiv 24\pmod{\!110}\ $ follows by scaling the above by $\,4.$

i.e. $\,\ \ \ \ \ {-}6x\equiv 24\pmod{\!110}$

Hence $\ \ \ \ \,x\equiv -4\pmod{\!55}\ $ by dividing prior by $\,6$.

What is the final answer?


问:
Probability newspaper company
I'm trying to solve this task but I'm obviously stuck. Will anyone help me solve it?

In a certain company three newspapers are published. 20% of the population read A, 16% read B, 14% read C, 8% read A and B, 5% read A and C, 4% read B and C, and 2% read all 3 newspapers. A person is selected at random.

a) none of the papers
b) at least one of the papers
c) exactly one of the papers
d) A or B (or both)
e) A, given that the person reads at least one paper
f) C, given that the person reads either A or B (or both).

答:
I am not sure what kind of answer you are looking for. The easiest way to solve this is to make a Venn Diagram and fill in the values as required. This gives you probabilities quickly.

To solve this you need to look at intersection and union of events, and need to know the general formula
$$ \mathbb{P}(A\cup B\cup C) = \mathbb{P}(A) + \mathbb{P}(B) + \mathbb{P}(C) - \mathbb{P}(A\cap B) -+ \mathbb{P}(B\cap C)- \mathbb{P}(C\cap A)+ \mathbb{P}(A\cap B\cap C). $$

Let $A$ denote the event that a person reads newspaper $A$. Set similar defintions for $B$ and $C$. It is given to you that,

$\mathbb{P}(A) = 0.20,\ \mathbb{P}(B) = 0.16,\ \mathbb{P}(C) = 0.14,\ \mathbb{P}(A\cap B) = 0.08,\ \mathbb{P}(A\cap C) = 0.05,\ \mathbb{P}(B\cap B) = 0.04,\ \mathbb{P}(A\cap B\cap C) = 0.02$

Now, calculate these by yourself.

a) Probability that the person reads none of the newspapers is the even given by complement of the event of reading any one of the news papers. Thus, if $E_1$ is this event, then
$$ \mathbb{P}(E_1) = 1 - \mathbb{P}(A\cup B\cup C) = 1 - (0.20+0.16+0.14-0.08-0.05-0.04+0.02) = 1 - 0.35 = 0.65$$ from the definition given above.

I will only mention the events now.

b) At least one of the newspapers means, either A or B or C, is the event $$\mathbb{P}(A\cup B \cup C) = 0.35.$$ Thus, answer is which you calculated above.

c) Exactly one of the papers implies the event $$\mathbb{P}(A\cap B^c\cap C^c) + \mathbb{P}(A^c\cap B^c\cap C) + \mathbb{P}(A^c\cap B\cap C^c) = 0.22$$

d) $A$ or $B$ ( or both) is the event $A\cup B$. Thus, the probability is
$$ \mathbb{P}(A\cup B) = 0.20 + 0.16 - 0.08 = 0.28.$$ This is a little ambiguous.

e) Conditional probability. This is
$$ \mathbb{P}(A| A\cup B\cup C) = \frac{\mathbb{P}(A\cap (A\cup B\cup C)}{\mathbb{P}(A\cup B\cup C)} = \frac{\mathbb{P}(A)}{\mathbb{P}(A\cup B\cup C)} = 0.20 / 0.35 = 4/7 = 0.571428.$$

f) Similar to above you have the probability,
$$ \mathbb{P}(C | A\cup B) = \frac{\mathbb{P}(C)}{\mathbb{P}(A\cup B)} = 0.14/0.28 = 1/2 = 0.5.$$

Title: Probability newspaper company


问:
Let all zeros of $f(z)=a_0 z^n + ... + a_n$ be inside c. Find $\int_{c} \frac{z f'(z)}{f(z)}$
Let all zeros of $f(z)=a_0 z^n + ... + a_n$ be inside c. Find
$\int_{c} \frac{z f'(z)}{f(z)}$

I dont know how to approach this problem

答:
Hint: Observe that $f(z)$ is a polynomial and can be expressed as
$$f(z)=a_0(z-z_1)...(z-z_n)$$
where $z_k,k=1..n$ are zeros. Then, it is not difficult to see (left as an exercise) that
$$\frac{f'(z)}{f(z)}=\frac{1}{z-z_1}+\frac{1}{z-z_2}+...+\frac{1}{z-z_n}$$
Now, taking the integral should be really easy. And don't forget about Vieta's!


问:
Given $A\cap B$, $A\cap C$ and $B\cap C$, what is $A\cap (B\cap C)=$?
Suppose we know about $A\cap B$, $A\cap C$, and $B\cap C$.
Here, each set is a binary representation of a decimal number, hence each set includes 0's and 1's with specific positions or order. The intersection here means the 1's in these binary numbers that overlap at some positions. We are interested in the size of the set, meaning the number of overlapping positions with value 1.
Hence, the set is defined as the coordinate of 1's in each binary number.
Note that the size of these sets is actually Hamming weight or the number of 1's.

We have some extra information as well . We know that $|A\cap B|=x$,  $|A\cap C|=x$ or $x+1$,  $|B\cap C|=x$ or $x+1$ if $|A\cap C|=x+1$. You can consider all $x$.
How can we find $|A\cap (B\cap C)|=$?

Example 1: When $|A\cap B|=x$,  $|A\cap C|=x$,  $|B\cap C|=x$

A = 1 0 1 1 = {0, 1, 3}

B = 1 1 0 1 = {0, 2, 3}

C= 1 1 1 0 = {1, 2, 3}

$A\cap B = \{0, 3\}$

$A\cap C = \{1, 3\}$

$B\cap C = \{2, 3\}$

$A\cap B\cap C = \{3\}$

Here, 0, 1, 2, and 3 refer to the position of every bit where 0 refers to the rightmost bit (LSB) and 3 refers to the leftmost bit (MSB)

For many examples like this one when $|A\cap B|=x$,  $|A\cap C|=x$,  $|B\cap C|=x$, I am getting  $|A\cap (B\cap C)|=x-1$.

Example 2: When $|A\cap B|=x$,  $|A\cap C|=x+1$,  $|B\cap C|=x+1$

A = 1 0 1 1 = {0, 1, 3}

B = 1 1 0 1 = {0, 2, 3}

C= 1 1 1 1 = {0, 1, 2, 3}

$A\cap B = \{0, 3\}$

$A\cap C = \{0, 1, 3\}$

$B\cap C = \{0, 2, 3\}$

$A\cap B\cap C = \{0, 3\}$

For many examples like this one when $|A\cap B|=x$,  $|A\cap C|=x+1$,  $|B\cap C|=x+1$, I am getting  $|A\cap (B\cap C)|=x$.

I would appreciate any tip or help to solve or understand this problem better. Thank you so much in advance.

答:
Let $|A\cap B| = a,$ $|B\cap C| = b,$ $|C\cap A|=c$ and $|A\cap B\cap C|=x$.

Also, let
$$A = \{1,2,3,4\}, \ \ B = \{2,3,4,5\}, \ \ C = \{\color{red}4,6,7,8\} $$
Then
$$
A\cap B = \{2,3,4\}, \ \ B\cap C = \{4\}, \ \ A\cap C = \{4\} \\
a=3, \ \  b=c=1 \\
A\cap B\cap C = \{4\} \implies \boxed{x = 1}$$

Now, let
$$A = \{1,2,3,4\}, \ \ B = \{2,3,4,5\}, \ \ C = \{\color{red}5,6,7,1\} $$
Then,
$$
A\cap B = \{2,3,4\}, \ \ B\cap C = \{5\}, \ \ A\cap C = \{1\} \\
a=3, \ \  b=c=1 \\
A\cap B\cap C = \emptyset \implies \boxed{x=0}$$

Notice that the same triplets $(a,b,c)$ led to two different results ($x=1$ and $x=0$) from which we can deduce that there is no such formula in general.

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$


答:
In a certain company three newspapers are published. 20% of the population read A, 16% read B, 14% read C, 8% read A and B, 5% read A and C, 4% read B and C, and 2% read all 3 newspapers.

Hint

First of all, Gerry Myerson's comment hits the nail on the head.

Consider that with 3 newspapers, there are $2^3$ categories, re a person does or does not read each of the 3 newspapers.  To solve, you must assign an explicit percentage to each of the 8 categories.  Then, everything falls into place and you can easily answer all of the questions.

You can actually construct the arithmetic equivalent of the Venn diagram merely by taking the information and working backwards.

2% read all three newspapers.

8% read A and B, 5% read A and C, 4% read B and C.

Therefore

(8% - 2% = 6%) read A and B and don't read C.

You can similarly compute :

how many read A and C and don't read B

how many read B and C and don't read A

Assume that you have done this:

20% of the population read A, 16% read B, 14% read C

At this point, you should have been able to determine:

how many read all three

how many read A and B but not C

how many read A and C but not B

Taking the above 3 percentages and deducting them from the 20% that read A, will give you the percentage of people that read A but do not read B or C.

You can similarly compute the number of people who read B but do not read A or C.

You can similarly compute the number of people who read C but do not read A or B.

At this point, you will have computed the percentage for 7 of the 8 categories.  Then the 8th category, how many people read none of the newpapers is simply 1 - (the sum of the 7 categories).

At this point it is game over.

Title: Probability newspaper company

Body:
I'm trying to solve this task but I'm obviously stuck. Will anyone help me solve it?

In a certain company three newspapers are published. 20% of the population read A, 16% read B, 14% read C, 8% read A and B, 5% read A and C, 4% read B and C, and 2% read all 3 newspapers. A person is selected at random.

a) none of the papers
b) at least one of the papers
c) exactly one of the papers
d) A or B (or both)
e) A, given that the person reads at least one paper
f) C, given that the person reads either A or B (or both).


问:
Consequences of a rectangular matrix being of maximal rank
I have a real matrix $A$, $(m+1) \times m$ and a vector $b  \in \mathbb R^{m+1}$ such that $b_{m+1}=0$. For any vector $u\in \mathbb R^m$, $Au=0 \Rightarrow u=0$. This means that $A$ is a rectangular matrix of maximal rank, i.e. of rank $m$.

Since $m< \infty$, I'm told this means there is a solution $u\in \mathbb R^m$ to $Au=b$. I don't understand why: if $\operatorname{rk}(A)=m$, then the dimension of the image of the linear application defined by $A$ in $\mathbb R^{m+1}$ is $m$. So for $b\in\mathbb R^m$ we could guarantee a solution, but how can we guarantee the solution satisfies our $b\in\mathbb R^{m+1}$?

Is there something simple I'm missing? (I've asked a couple of classmates who didn't know either, and my professor who told me this in the first place told me to look it up in a book that's in a library that's closed until next week.)

答: You're right, your professor is wrong. In such a case, it's usually a good idea to look for a simple example. For $m=1$, the matrix $A=\pmatrix{1\\1}$ has full rank $1$. For any vector $u\in\mathbb R^1$, $Au=0$ implies $u=0$. Yet there is clearly no $\mathrm u\in\mathbb R^1$ such that $Au=\pmatrix{b_1\\0}$ for any $b_1\ne0$.</p>


问:
$\operatorname{vol}(\operatorname{int}(Z+K))=\operatorname{vol}(Z+\operatorname{int}K)$ when $Z$ is a finite set
I want to prove that if $K$ is a compact and convex set of $\mathbb{R}^n$ and $Z=\{z_1, \cdots, z_n\}$ is a finite set of points of $\mathbb{R}^n$ then
$$\operatorname{vol}(Z+K)=\operatorname{vol}(\operatorname{int}(Z+K))=\operatorname{vol}(Z+\operatorname{int}K)$$
I know the first equality is true in general so I think I just need to prove the second one, but I don't really know how to do it.

At first I thought that $\operatorname{int}(Z+K)$ was equal to $Z+\operatorname{int}K$, but as it is shown here, this is not true in general but it suggested me that whenever this is not true, it seems that the difference between the sets is of measure zero so if this was true, and then these two volumes would be indeed equal, but again I don't know if this is true or how to prove it.

Any ideas?

答:
$Z+\text{int}(K)\subset Z+K$, so any difference between the first and third terms would come from points in the latter set but not the former.

But observe that $Z+\text{int}(K)$ is the translates of points in the interior of $K$, while $Z+K$ is the translates of points anywhere in $K$, so $(Z+K)\setminus Z+\text{int}(K)\subset Z+(K\setminus\text{int}(K))$. But the RHS of this consists of finitely many copies of a set of measure $0$ (namely, the boundary of $K$), so it is also of measure $0$.

More precisely:

$$\text{Vol}(Z+(K\setminus \text{int}(K)))\le\sum_{i=1}^n\text{Vol}(z_i+(K\setminus \text{int}(K)))=\sum_{i=1}^n\text{Vol}(K\setminus \text{int}(K))=\sum_{i=1}^n0=0$$


问:
Help with $\int_{0}^{\pi /2}\arccos\frac{\cos(x)}{1+2\cos(x)}dx$
I've been struggling with this integral for a few days.

$$\int_{0}^{\pi /2}\!\arccos\left(\frac{\cos(x)}{1+2\cos(x)}\right)dx$$

Any help would be appreciated. Cheers.

答:
This is the so-called Coxeter integral. It is related to Ahmed integrals. You can refer to this and this for calculating it. The result is
$$
\int_{0}^{\frac{\pi}{2}}\arccos \left (\frac{ \cos(x)}{1+2 \cos(x)}\right )dx=\frac{5\pi^2}{24}
$$

Also you can find more on this page.


问:
Geometric interpretation of differentiability
I know that the geometric interpretation of differentiability for a function $f:\mathbb{R}^2\to \mathbb{R}$ in a point $(x_0,y_0)$ is that it admits a tangent plane in the point $P=(x_0,y_0,f(x_0,y_0))$. Thinking about the concept geometrically, I came up with the following ""conjecture"". Let for simplicity $\ell_{v}$ the tangent line in the point $P$ to the graphic of $f$ in the direction of the versor $v$, and let's suppose the function $f$ admits directional derivatives in every direction.

$f \text{ differentiable in } (x_0,y_0)$ $\iff$ $\forall v,u \ \ \text{versors}$, $\ell_{v}$ $\text{and}$ $\ell_{u}\ \text{are coplanar}$

I tried proving this using affine geometry but calculations get pretty messy. I would like if it's true or not(in the former case I would like a proof, in the latter a counterexample)

答:
This is borderline duplicate of Continuous function with linear directional derivatives=>Total differentiability?, but restated in geometric language. The fact that $l_v$ exist is equivalent to directional derivatives $f_v$ existing. The fact that they are coplanar is equivalent to the function $L(v)=f_v$ being linear. So the counterexamples in that question are counterexamples for your question as well.

Geometrically, the tangent plane has to approximate the graph "uniformly well" over a disc in $\mathbb{R}^2$, while tangent line each approximates the graph over a segment; when, for a "given quality of approximation", these segments fail to assemble into a disc (as in the counterexample given), the resulting plane will fail to approximate, and will not be the tangent plane in the usual sense (of total differentiability).


问:
Is there an uncomputable number between any two real numbers?
I know close to nil about uncomputable numbers, so perhaps it doesn't even make sense to ask this question. All the information I can find about them is unaccessible with my level of education, but I have heard that "most numbers are uncomputable."</p>

答:
Assume $\ a<b\ $ are real numbers and define $\ d:=b-a\ $ .

Case $1$ : $\ a\ $ is uncomputable

Choose positive integer $\ n\ $ so large that $\ \frac{1}{n}<d\ $ holds, then $\ a+\frac{1}{n}\ $ is an uncomputable number in the interval $\ (a,b)\ $.

Case $2$ : $\ a\ $ is computable

Let $\ r\ $ be an arbitary positive non-computable number and choose $\ n\ $ so large that $\ \frac{r}{n}<d\ $ holds. Then, $\ a+\frac{r}{n}\ $ is an uncomputable number in the interval $\ (a,b)\ $.


问:
Unifying abstraction of duality between $A - B$ and $A + B$
I'm wondering whether there's an abstraction that unifies the special cases of dual or complementary equations of the form $A - B$ and $A + B$ that I've seen in math.

Here are some examples:

1: Even and Odd Parts of Functions

If $A = f(x)$ and $B = f(-x)$ then $\frac{A + B}{2}$ is the even part of $f$ and $\frac{A - B}{2}$ is the odd part of $f$.

For instance if $f(x) = e^x$ then $\frac{f(x) + f(-x)}{2} = \cosh x$ and $\frac{f(x) - f(-x)}{2} = \sinh x$.

2: Real and Imaginary Parts of Complex Numbers

If $A$ and $B$ are conjugate complex numbers, then $\frac{A + B}{2}$ is the real part of $A$ and $\frac{A - B}{2 i}$ is the imaginary part of $A$.

3: Lagrangian and Hamiltonian

This example is from physics, but I'm posting on this stackexchange (not physics) because I'm looking for a mathematical answer.

In classical mechanics, two useful formulations are the Lagrangian and the Hamiltonian.

The Lagrangian of a given system is the kinetic energy of the system minus the potential energy, $L = T - V$.

The Hamiltonian is the kinetic energy plus the potential energy, $H = T + V$.

So I'm wondering if anybody knows of any unifying abstractions of which some of the above are special cases.  Or if anybody has more diverse examples of dualities between $A - B$ and $A + B$ to add, please comment.

答: I don't know if this is general enough but if you take an abelian group such that the square roots exist (that is for any $a\in G$ there is $b\in G$ such that $b^2=a$) and an automorphism $h$ such that $h^2(g)=g$, then any element $g\in G$ can be expressed as $g=AB$ where $h(A)=A$ and $h(B)=B^{-1}$, this can be proven by taking $b^2=g$ and letting $A=b\,h(b)$ and  $B=b\,h(b)^{-1}$. All the examples above (except for the third one which i assume is just a coincidence) can be explained by this.


问:
prove that $\lim_{x \to 1^{-}}\frac{(x+2)}{2x^{2}-3x+1} = -\infty$
Prove that $\lim_{x \to 1^{-}}\frac{(x+2)}{2x^{2}-3x+1} = -\infty$

Without loss of generality, assume tha $M<0$ and $\delta>0$,

choose $0<\delta<\frac{1}{2}$ such that $1-\delta<x<1$ implies   $\frac{3}{M}<{2x^{2}-3x+1}<0$ for  that is $\frac{M}{3}>\frac{1}{2x^{2}-3x+1}$

since $\frac{1}{2}<x<1$ also implies $\frac{5}{2}<x+2<3$ it follows that $f(x)<M$ for all $1-\delta<x<1$

I don't know whether this is correct or not if there is any easy method to prove it using definition show it. Also tell me is this correct?

答:
There is definitely an easier way, without going into th rigourous definition of a limit.
Try factorising the denominator.
Then, substiute x=1-h in the eqn, with h tending to zero.
Manipulate the eqn a bit, and you would get your desired result.</p>


问:
Infinite union of interval
For each $n\in \Bbb{N}$, let $A_n=[0,n)$. Find

$\bigcup_{n\in \Bbb{N}}A_n$ .

Attempt: Let $x\in\bigcup [0,n)$. Then $x\in A_n$ for some $n\in\Bbb{N}$.Then $0\leq x<n.$
therefore, $x\in [0,\infty)$. Thus $\bigcup_{n\in \Bbb{N}}[0,n)\subset [0,\infty)$.

Let $x\in [0,\infty)$. Then $0\leq x<\infty$. As $\Bbb{N}$ is unbounded, $x<n$ for some $n\in\Bbb{N}$. Thus $x\in [0,n)$ for some $n\in\Bbb{N}$implies$ [0,\infty)\subset \bigcup_{n\in\Bbb{N}}A_n$.

Hence $\bigcup_{n\in\Bbb{N}}A_n=[0,\infty)$.

Is it correct?

I already assumed answer. I need a way to do it without assuming the answer.

答:
For the first inclusion, you've proven $x\in [0,\infty)$ but that doesn't need the fact that $\mathbb{N}$ is unbounded. $x$ is in $[0,\infty)$ because all numbers in $A_n$ are nonnegative, for all $n$. Remember, $[0,\infty)$ can be written as the set of $x$ such that $x \geq 0$. So for that you don't need the unboundedness.

When you say "without loss of generality" in the second inclusion is exactly where you need to write "Due to $\mathbb{N}$ being unbounded", etc. Because it's there where you're using it.

Title: Infinite union of interval


问:
How does a calculator calculate the sine, cosine, tangent using just a number?

Sine $\theta$ = opposite/hypotenuse

Cosine $\theta$  = adjacent/hypotenuse

Tangent $\theta$  = opposite/adjacent

In order to calculate the sine or the cosine or the tangent I need to know $3$ sides of a right triangle. $2$ for each corresponding trigonometric function. How does a calculator calculate the sine, cosine, tangent of a number (that is actually an angle ?) without knowing any sides?

答:
I always wondered about the same thing, till I attended my first calculus class.

As Julien rightly noted. It uses power series of $\sin x, \cos x$ etc, to only approximately calculate the value of angles(in radians) you put in. You can read more about it here. And power series of $\tan x, \sec x$ and $\text{cosec } x$ is given here.


问:
Can it be that the only way to prove a theorem is based on generation of new tautologies?
We can use a language of a given logic (for example propositional logic or predicate logic) to formulate statements. We can also use rules of inference to reformulate single statements (derive "new" statements from "old" ones) or to combine two existing statements to generate a new statement.

We can also say that we might have a minimal complete set of inference rules (we do not need more rules and we cannot remove any rules).

Can it be the case for a given statement (or a set of statements) none of the inference rule is applicable but a combination of them is applicable?

Here is what I mean in a bit extended form. Inference rules are tautologies likee: $A \land B \iff B \land A$, $A \land (B \lor C) \iff (A \land B) \lor (A \land C)$. We can combine these tautologies to create more complex (larger) tautologies. Now, I can imagine that one of these "larger" tautologies can be applied to one of my statements (or a combination of two statements) to generate a new statements that follows (in a logical sense) from the "old" statement(s).

Can it be the case that it would be impossible to go from the "old" statement(s) to the "new" one just by applying the original (basic, atomic) tautologies but it is possible to get the "new" statement if we combine the original (basic, atomic) tautologies to get a more complex one and then we apply it to derive a "new" statement from the old one?

ADDED

I feel like I need to clarify my question. In a logic (for example propositional logic) we have a small and limited set of inference rules (let's call the "basic rules") that we can use to derive "new" statements (theorems) from "old" statements (old theorems and axioms). The whole process of derivations can be imagined as applying "basic rules" to existing statements to generate new statements until we reach the statement we need.

However, alternatively, one can combine "basic rules" with each other to derive new rules (let's call them "derived rules"). For example, if the first basic rules is "a + b = b + a" and the second basic rule is "a * (b + c) = (a * b) + (a * c)", then we can combine these two rules to generate a new "derived" rule: "a * (b + c) = (a * c) + (a * b)". And then, as soon as we have a new rule, we can apply it to a statement, to transform it. For example if we have a "statement" "2 * (3 + 4)" we can transform it using our new "derived rule" to get "(2 * 4) + (2 *3)".

So, my question was, if it might be the case that some statements ("theorems") are unreachable by "basic rules" only. So, in that case, instead of applying our "basic rules" to the statements, we need first to combine our "basic rules" to get new rules ("derived") and then apply the derived rules to our statements.

答:
This is somewhat ambiguous. An inference rule is just a rule that tells you what sentences you can derive from other sentences. A tautology is a sentence that is true on every interpretation. The term usually refers to sentences in propositional logic that are true for any assignment of truth values to atoms. In this sense, using the standard axiomatisation of propositional logic, we can prove any tautology. This isn't very dramatic: truth tables are themselves a quasi-formal system, just not one consisting of strings of symbols. For this reason propositional logic is called "complete". We might also wonder about things true on every interpretation in predicate logic (although tautology has a stricter meaning here). It turns out we have a positive result here called Godel's Completeness Theorem. This says that a statement in predicate logic is provable if and only if there it has a model, or interpretation. However, models are a tricky business. Peano Arithmetic, the standard first-order formalism for arithmetic, has non-standard models, which behave nothing like we imagine numbers working.

You might ask, can every true statement in a theory be proven. If you're bound by the apparent laws of physics, then the answer is no. Godel's First Incompleteness Theorem states that no sufficiently expressive computable (emulatable on a computer) formal system can be both consistent (not prove contradictions) and complete (in this context, give every sentence of the theory a true or false value). This matters because there is no evidence that physics itself can do anything more than a computer, with the (possible) exception of randomisation, which wouldn't help you to make a formal logic that you actually believed in. So, there will always be sentences about certain things that you can't decide are true or false using your favourite formal system. So you would have to add new rules to any formal system you could actually use to derive them. Of course, this would include the formal system you're using in your brain (if your the human brain actually has consistent fundamental rules of reason), so after a certain point you wouldn't be able to work out whether or not you should add them.


问:
Show that $\exists \eta \in (a,b) ~s.t.~\frac{f(\eta)}{\int_a^{\eta} f(t){\rm d}t}-\frac{g(\eta)}{\int_{\eta}^b g(t){\rm d}t}=2021$.
Suppose $f(x),g(x) \in C[a,b]$ and $f(x),g(x)>0$ for every $x \in (a,b)$. Show that there exists a $\eta \in (a,b)$ such that $$\frac{f(\eta)}{\int_a^{\eta} f(t){\rm d}t}-\frac{g(\eta)}{\int_{\eta}^b g(t){\rm d}t}=2021.$$

This problem comes from here, a website of China.

答:
HINT. We can generalize $2021$ to arbitrary $k$. Let
$$
F(x)=\int_{a}^{x} f(t)\mathrm{d}t ,G(x)=\int_{x}^{b} g(t)\mathrm{d}t
$$
And the original claim is equivalent to
$$
\frac{F'\left( \eta \right)}{F\left( \eta \right)}+\frac{G'\left( \eta \right)}{G\left( \eta \right)}=k
$$
Then
$$
F'\left( \eta \right) G\left( \eta \right) +G'\left( \eta \right) F\left( \eta \right) =kF\left( \eta \right) G\left( \eta \right) \\
F'\left( \eta \right) G\left( \eta \right) +G'\left( \eta \right) F\left( \eta \right) -kF\left( \eta \right) G\left( \eta \right) =0
$$
Let
$$
H (x)=C e^{-kx}F(x )G(x )
$$
What can we say about $H(x)$?

$$
H'(x)=C (F'\left( \eta \right) G\left( \eta \right) +G'\left( \eta \right) F\left( \eta \right) -kF\left( \eta \right) G\left( \eta \right) )
$$
Since $H(a)=H(b)=0$, we can apply Rolle's Theorem.


问:
prove that $v$ is not a Linear combination of $u,w$
Let $U,W$ be subspaces of $V$, Let $u,v,w$ $\in V-{0}$

Given that

$u\in U,u\notin W$

$w\in W$

$v\in W$ and ($v$ is not a scalar multiplication of $w$)

prove that $v$ is not a Linear combination of $u,w$

答: I think that the correct question is: prove that $\;v\;$ is not a linear combination of $\;u,w\;$. In that case $\;v\;$ and $\;w\;$ are linearly independent by hypothesis. Moreover $\;u\notin W \supseteq Span(v,w)\;.\;$ For that reason $\;u,v,w\;$ are linearly independent and so $\;v\;$ is not a linear combination of $\;u,w\;$


问:
Find a cyclic sum
Given that

$$ \frac{a}{bc-a^2}+\frac{b}{ca-b^2}+\frac{c}{ab-c^2}=0,\quad a,b,c\in \mathbb{R}, $$

find the value of

$$ \frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}. $$

I solved it via direct computation which is clearly not the best way to do. Is there any simple transformation I could perform or geometical meaning of this equation?

答:
Hint:
$$\left(\dfrac a{bc-a^2}+\dfrac b{ca-b^2}+\dfrac c{ab-c^2}\right)\cdot{\underbrace{\left(\dfrac1{bc-a^2}+\dfrac1{ca-b^2}+\dfrac1{ab-c^2}\right)}_{\text{assuming this to be finite}}}$$

$$= \underbrace{\dfrac a{(bc-a^2)^2}+\dfrac b{(ca-b^2)^2}+\dfrac c{(ab-c^2)^2}}_{\text{this is to be calculated}}+\dfrac{(a+b)(ab-c^2)+(b+c)(bc-a^2)+(c+a)(ca-b^2)}{(bc-a^2)(ab-c^2)(ca-b^2)}$$

Now $(a+b)(ab-c^2)+(b+c)(bc-a^2)+(c+a)(ca-b^2)=\cdots=0$

We only need $(bc-a^2)(ab-c^2)(ca-b^2)\ne0$

${\quad a,b,c\in \mathbb{R}, {\text { is not actually required}}}$

Title: Find a cyclic sum


问:
What do we mean when we say that $x \in \mathbb R$
When we say that $x \in \mathbb R$ does here $x$ represents all real numbers at the same time or $x$ represents only a single real number at one time. How can a single variable($x$) represent all the real numbers at the same time ? when we write $x \in \mathbb R$ it means that $x$ is an element of the set of real numbers, this means that $x$ represents a single real number but then why we start to treat it as if $x$ represents all the real numbers as in inequality suppose we have $x>-2$ this means that $x$ can be any real number greater than $-2$ but then why we say that all the real numbers greater than $-2$ are the solutions of the inequality. $x$ should represent a single real number right ? then how can $x$ represents all the real number.Please help me....

答:
A sentence like

Let $x \in E$. Suppose $x$ satisfies $blahblahblah$. Then $P$ is true.

means

$\forall x,\quad (x \in E \mbox{ and } blahblahblah) \Rightarrow $P$.

The rest is irrelevant: "represent" is not a mathematical word, so it doesn't make sense to wonder about how a single letter could represent all real numbers.

I think that sometimes, the teachers express themselves in a strange way, using the word "variable" which has no formal sense either, etc. and it confuses students; in any case, I think it is a healthy reaction to get confused by these things!


答: When we say $x\in \mathbb R$, we mean that $x$ is an arbitrary element in $\mathbb R$. That doesn't mean to "represent all the real numbers at the same time", but to present any numbers. Usually we will have a proposition including $x$, and it is true if and only if you choose any real values for $x$, the proposition always holds.

Title: What do we mean when we say that $x \in \mathbb R$

Body: When we say that $x \in \mathbb R$ does here $x$ represents all real numbers at the same time or $x$ represents only a single real number at one time. How can a single variable($x$) represent all the real numbers at the same time ? when we write $x \in \mathbb R$ it means that $x$ is an element of the set of real numbers, this means that $x$ represents a single real number but then why we start to treat it as if $x$ represents all the real numbers as in inequality suppose we have $x>-2$ this means that $x$ can be any real number greater than $-2$ but then why we say that all the real numbers greater than $-2$ are the solutions of the inequality. $x$ should represent a single real number right ? then how can $x$ represents all the real number.Please help me....


问:
limits of continuous probability density function
hello i have to find a limit of a probability distribution function. the FX(x) are not  informed in the assignment, as how large or anything else.

and there are noting in my textbook about it.

my 2 problems are

$\lim\limits_{x \to -\infty} \frac{F_X(x)^2-1}{F_X(x)+1}$ negative infinity

and

$\lim\limits_{x \to \infty} \frac{F_X(x)^2-1}{F_X(x)+1}$ positive infinity

my solution are there are rules for limits of a density function as
$\lim\limits_{x \to -\infty}=0$ and $\lim\limits_{x \to \infty}=1$

so would a solution not be

$\lim\limits_{x \to -\infty} \frac{F_X(x)^2-1}{F_X(x)+1}=\frac{0^2-1}{0+1}=-1$

and

$\lim\limits_{x \to \infty} \frac{F_X(x)^2-1}{F_X(x)+1}=\frac{1^2-1}{1+1}=\frac{0}{2}=0$

答: Note that for the $-\infty$ you could expand the numerator:$ (F-1)(F+1)$ and then cancel the denominator.


答:
Your answer seems consistent with

$\frac{a^2-1}{a+1}=a-1$ if $a\not=-1$.  In particular this is $-1$ when $a=0$ and $0$ when $a=1$

for a cumulative distribution function (not a density): $\lim\limits_{x \to -\infty} F_X(x) = 0$ and $\lim\limits_{x \to +\infty} F_X(x) = 1$

Title: limits of continuous probability density function

Body:
hello i have to find a limit of a probability distribution function. the FX(x) are not  informed in the assignment, as how large or anything else.

and there are noting in my textbook about it.

my 2 problems are

$\lim\limits_{x \to -\infty} \frac{F_X(x)^2-1}{F_X(x)+1}$ negative infinity

and

$\lim\limits_{x \to \infty} \frac{F_X(x)^2-1}{F_X(x)+1}$ positive infinity

my solution are there are rules for limits of a density function as
$\lim\limits_{x \to -\infty}=0$ and $\lim\limits_{x \to \infty}=1$

so would a solution not be

$\lim\limits_{x \to -\infty} \frac{F_X(x)^2-1}{F_X(x)+1}=\frac{0^2-1}{0+1}=-1$

and

$\lim\limits_{x \to \infty} \frac{F_X(x)^2-1}{F_X(x)+1}=\frac{1^2-1}{1+1}=\frac{0}{2}=0$


Body:
Notice that
$$
\frac{F_X(x)^2-1}{F_X(x)+1} = \frac{(F_X(x)-1)(F_X(x)+1)}{F_X(x)+1}=F_X(x)-1
$$


问:
Roots of a function despite first derivative imaginary regions.
If a function has roots which satisfy it, yet some of those roots fall in a region that the first derivative is imaginary/unknown, then will those roots count as part of the solution?

Example: The below equation has roots: 2,3,4,5, and 6.  (Note the exponent can be factored as (x-6)(x-5))
Now the first derivative (Used Mathematica) does not have solutions of x^2-7x+11<=0 (due to both the denominator and Log components). And the region which that is true for is between (7-sqrt(5))/2 and (7+sqrt(5))/2 , ~(2.38,4.62) - Hence this will result in discarding both 3 and 4 as part of the solution, since they both fall in that region. Is this correct? (if so, why do both 3 and 4 nicely satisfy the original equation, yet the graph is unknown for that region).

Does this mean we always have to evaluate the first derivative of a function all the time, in order to verify discontinuity, despite having correct roots? Also: If that is the case, wouldn't that mean that we have to in-turn evaluate the 2nd derivative also, in order to ensure that the first derivative itself will not have the same issue? ...and Nth derivative, by the same logic?

Function:

$$y=-1+(x^2-7x+11)^{(x^2-11x+30)}$$

First derivative:

$$ (11 - 7 x + x^2)^{(
 30 - 11 x +
  x^2)} (((-7 + 2 x) (30 - 11 x + x^2))/(
   11 - 7 x + x^2)) + (-11 + 2 x) Log[11 - 7 x + x^2] $$

Function

First Derivative Unknown region - x^2-7x+11 <=0

答:
3 and 4 ARE ROOTS on the graph (they are 0), and there are a few sporadic points that will be real as well (at like 3.4 for example), but since they are infinitely small and surrounded by nonreal points, you will likely not see them on a graphing utility. If you see below, the only parts that will show are the parts where the imaginary graph is at 0.

Here is Re[F(x)]

And Here is the Im[F(x)]

In conclusion: No. Your roots are still your roots regardless of how wacky the graph might look.


问:
Attaching two manifolds along their boundary
I have a question about a proof in John Lee's Introduction to Topological Manifolds.

Suppose $M$ and $N$ are two topological $n$-manifolds with nonempty boundary (for reference, the definition I am using is: $M$ and $N$ are second countable Hausdorff spaces such that each point has an open neighborhood homeomorphic to either an open set in $\mathbf{R}^n$ or an open set in $\mathbf{H}^n = \{ (x_1,\ldots,x_n) \in \mathbf{R}^n: x_n \geq 0\}$). If $\partial M$ and $\partial N$ denote the manifold boundaries of $M$ and $N$, suppose $h: \partial N \to \partial M$ is a homeomorphism. We may attach $M$ and $N$ along their boundaries by forming the adjuction space

$$ M \cup_h N = \frac{M \sqcup N}{\sim}$$

where $\sim$ is the relation generated by identifying $x\in \partial N$ with $h(x)$. The theorem is:

With $M$, $N$, and $h$ as above, $M \cup_h N$ is an $n$-manifold (without boundary).

The proof first proceeds to show that each point of the adjuction space has a euclidean neighborhood. If $q: M \sqcup N \to M \cup_h N$ denotes the quotient map, then Lee puts $S=q(\operatorname{Int} M \cup \operatorname{Int} N)$ (where $\operatorname{Int}$ denotes the manifold interior) and shows that each point of $M \cup_h N -S$ has a euclidean neighborhood. I am OK with this argument.

Now, I quote:

Suppose $s\in S$, and let $y \in \partial N$ and $x=h(y) \in \partial M$ be two points in the fiber $q^{-1}(s)$. We can choose coordinate charts $(U,\phi)$ for $M$ and $(V, \psi)$ for $N$ such that $x\in U$ and $y\in V$, and let $\hat{U}=\phi(U)$, $\hat{V} = \psi(V) \subseteq \mathbf{H}^n$. It is useful in this proof to identify $\mathbf{H}^n$ with $\mathbf{R}^{n-1}\times [0,\infty)$ and $\mathbf{R}^n$ with $\mathbf{R}^{n-1} \times \mathbf{R}$. By shrinking $U$ and $V$ if necessary, we may assume $h(V \cap \partial N) = U \cap \partial M$, and that $\hat{U} = U_0 \times [0,\epsilon)$ and $\hat{V} = V_0 \times [0,\epsilon)$ for some $\epsilon>0$ and some open subsets $U_0,V_0 \subseteq \mathbf{R}^{n-1}$.

My question: How does one choose the coordinate domains $U$ and $V$ to satisfy these two conditions simultaneously? I know how to choose $U$ and $V$ such that $h(V \cap \partial N) = U \cap \partial M$, and I know how to choose $U$ and $V$ such that their images under the coordinate maps are equal to sets of the form $U_0 \times [0,\epsilon)$ and $V_0 \times [0,\epsilon)$, but I don't know how to pick $U$ and $V$ such that both of these conditions are satisfied at the same time.

EDIT: Let me be a bit more specific. We are to choose the coordinate domains (by shrinking if necessary) to satsify the following two conditions simultaneously:

(a) $h(V \cap \partial N) = U \cap \partial M$,

(b) $\hat{U} = U_0 \times [0,\epsilon)$ and $\hat{V} = V_0 \times [0,\epsilon)$.

Now, assuming that I have adjusted $U$ and $V$ to satisfy (a), then I may adjust them further to satisfy (b). But then how do I know that $U$ and $V$ still satisfy (a)? I can be more specific by what I mean by "adjust" if neeeded.

Am I misreading the proof, or am I missing something glaringly obvious?

答:
EDIT 6/17/19: Another reader recently pointed out that the argument I sketched below is not appropriate for this place in the book, since it uses properties of compactness that are not introduced until Chapter 4. I'll leave my original answer here for those who are curious; but a much better solution is to just abandon the idea of arranging for the images of the coordinate maps to be product open sets, because that was not really needed anyway. I've added a correction to my online list.

Original Answer:

Here's a way to see it.  Choose charts $(U,\varphi)$ and $(V,\psi)$ such that $x\in U$ and $y\in V$, and let  $U_1 = U\cap \partial M$ (which is a  neighborhood of $x$ in $\partial M$) and   $V_1 = V\cap \partial N$ (a neighborhood of $y$ in $\partial N$).  After shrinking $U$ if necessary, we may assume that $h(U_1)\subset V_1$. (Because $U_1\cap h^{-1}(V_1)$ is a neighborhood of $x$ in the subspace topology $\partial M$ inherits from $M$, there is an open subset $\widetilde U\subset M$ such that $U_1\cap h^{-1}(V_1) = \widetilde U\cap\partial M$.  Replacing $U$ by $U\cap \widetilde U$ does the trick.)   Let $U_0\subset U_1$ be a neighborhood of $x$ that's precompact in $U_1$, and let $V_0 = h(U_0)\subset V_1$, a precompact neighborhood of $y$ in $V_1$.

Now let $\widehat U=\varphi(U)$, $\widehat V=\psi(V)$, considered as subsets of $\mathbb H^n  = \mathbb R^{n-1}\times [0,\infty)$.  Since $\varphi(\overline {U_0})$ is a compact subset of $\mathbb R^{n-1}\times \{0\}$,  there is some $\varepsilon_1>0$ such that $\varphi(\overline {U_0})\times[0,\varepsilon_1)\subset \widehat U$.  Similarly, there's $\varepsilon_2>0$ such that $\psi(\overline {V_0})\times[0,\varepsilon_2)\subset \widehat V$.  Now just take $\varepsilon = \min(\varepsilon_1,\varepsilon_2)$, and replace $U$ by $\varphi^{-1}\big(\varphi(U_0)\times[0,\varepsilon)\big)$ and $V$ by $\psi^{-1}\big(\psi(V_0)\times[0,\varepsilon)\big)$.


问:
How can I solve C(n,2)+2×C(n,3)+3×C(n,4)+...+(n-1)×C(n,n)=?
'm able to find an answer but I don't think it's the right one. I tested it for n=2 and it worked but not for n=3.Here's my answer:

$$C(n,2)+2×C(n,3)+3×C(n,4)+...+(n-1)×C(n,n)=[C(n,2)+C(n,3)+...+C(n,n)] + [0×C(n,2)+1×C(n,3)+2×C(n,4)+...+
(n-2)×C(n,n)] =
2^{(n-2)} + 2^{(n-2)}×(n-2)$$

I'm either forgetting something or it's just fully wrong. Sorry for the bad formatting and thank you in advance

答:
$$\sum_{r=2}^n(r-1)\binom nr=\sum_{r=2}^nr\binom nr-\underbrace{\sum_{r=2}^n\binom nr}_{(1)}$$

Now for $r\ge1,$ $$r\binom nr=r\cdot\dfrac{n\cdot(n-1)!}{r\cdot(r-1)!\cdot(n-1-(r-1)))!}=n\binom{n-1}{r-1}$$

$$\implies\sum_{r=2}^nr\binom nr=n\sum_{r=2}^n\binom{n-1}{r-1}$$

$$\sum_{r=2}^n\binom{n-1}{r-1}=-\binom{n-1}0+\underbrace{\sum_{m=0}^{n-1}\binom{n-1}m}_{(2)}$$

From $(1),$  $$\sum_{r=2}^n\binom nr=-\binom n0-\binom n1+\underbrace{\sum_{r=0}^n\binom nr}_{(3)}$$

Use $\displaystyle(1+1)^m=\sum_{k=0}^m\binom mk$ for $(2),(3)$


问:
Find the Expectation and Variance of 4 Independent Dice
We were given this seatwork:

With four independent dice:

a) the expected value of the sum of the rolls,

b) the expected value of the product of the rolls, and

c) the variance of the sum of the rolls .

I was able to answer a and b, but I don't know how to get the variance. Here's my attempt:

\begin{align*}
S = \dfrac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5
\end{align*}
This is the expectation of one die. For the expectation of four dice, we could assume the expectation of the sum four dice is equal to the sum of the expectations of a die:
\begin{align*}
&= S + S + S + S\\
&= 4S\\
&= 4(3.5)\\
&= 14
\end{align*}

Similarly, we could also do this for the products. The expected product of four dice rolls is:
\begin{align*}
&= S \cdot S \cdot S \cdot S\\
&= S^4\\
&= 3.5^4\\
&= 150.06
\end{align*}
Are these assumptions correct?

答:
You have found the expected value of a dice roll to be $E(X)=3.5$, thus the variance of one roll is equal to:

\begin{align}
  \operatorname{Var}(X) &= \sum_{k=1}^6 \frac{1}{6}\left(k - 3.5\right)^2 \\
    &= \frac{35}{12} \approx 2.92.
\end{align}

Since the rolls of the 4 dies are independent the variance of their sum is equal to $4 \cdot \frac{35}{12}\approx 11.66$


问:
Definition(s) of rational numbers
The definitions of rational numbers are somewhat confusing for me. The definition of rational numbers on wikipedia and most other sites is:

In mathematics, a rational number is any number that can be expressed as the quotient or fraction $\frac pq$ of two integers, with the denominator $q$ not equal to zero.

All integers can be expressed as the quotient $\frac pq$. So if they are not written in the fraction form, will they still be called rational numbers?

Another definition which I've found is:

A rational number is a quotient $\frac mn$ where $m$ and $n$ are integers and $n \neq 0$.

These definitions might be the same. However, I'm not able to figure out how! Are rational numbers those numbers which can be expressed as the quotient $\frac mn$ of two integers or those which are directly the quotient of two integers? For example, if $n$ is an integer, so is $n$ a rational number? Or is $\frac n1$ a rational number?

答:
I think the problem is that if you simply state those definitions exactly as they are you'll fall in the problem of not having defined the notion of "a quotient of integers". So the good and cool definition of rationals that solve all of these problems is to introduce one equivalence relation in a certain set. I don't know if you are used to equivalence relations (or even relations at all), so I'll talk about that first.

If you have two sets $A$ and $B$ you can create a relation $R$ between them which is a subset of the cartesian product $R\subset A\times B$. Think for a minute, elements of $A\times B$ are pairs $(a,b)$ with $a\in A$ and $b\in B$, so if $(a,b)\in R$ we are telling that $a$ is in some way related to $b$ and in this case we write $aRb$. Now, an equivalence relation can be introduced between a set and itself to mimic equality, it is usually denoted $\sim$ and satisfies those properties:

$a\sim a$ (Reflexivity)

$a \sim b \Longrightarrow b \sim a$ (Symmetry)

$a \sim b \wedge b \sim c \Longrightarrow a\sim c$ (Transitivity)

In the third one the $\wedge$ symbol means AND. Look now that equality always obeys those three properties. So when we have a set and we want to construct a notion of the objects being equivalent without being equal we use an equivalence relation. Now, given a set $A$, an equivalence relation $\sim$ in $A$ and some element $a \in A$ the set of all other elements of $A$ equivalent to $a$ by $\sim$ is called equivalence class and denoted $\left[a\right]$. The set of all equivalence classes is called the quotient set and denoted $A/\sim$ and although the elements are sets of elements of $A$ we can usually think of $A/\sim$ as just the elements of $A$ with $\sim$ imposed on them.

Now returning to your problem! Given the set $\mathbb{Z}\times (\mathbb{Z}\setminus\{0\})$, in other words, ordered pairs of integers without any pair with $0$ in the second element we introduce the following equivalence relation $\sim$ on the set:

$$(a,b)\sim(a',b') \Longleftrightarrow ab'=a'b$$

Now stop for a while and look what we did! We are almost defining the quocient of integers. If we introduce the notation:

$$\frac{a}{b}=\left\{(a',b') \in \mathbb{Z}\times (\mathbb{Z}\setminus\{0\}) : (a',b')\sim (a,b)\right\}$$

We have exactly what the quotient is: the equivalence class of all those elements $(a,b)$ with the relation imposed. Proving this is really an equivalence relation is a good exercise. Now, look what I've said before: formally we define the quotient using equivalence classes but in practice we simply think of it as the element $(a,b) \in \mathbb{Z}\times(\mathbb{Z}\setminus\{0\})$ with the relation $\sim$ imposed.

Now, we define the set of rationals by:

$$\mathbb{Q}=(\mathbb{Z}\times(\mathbb{Z}\setminus\{0\}))/\sim$$

In other words, the set of rationals is the set of all quotients of integers, recalling that we defined the quotient as that equivalence class. With this your first definition is obviously equivalent (indeed we just made it formal using this thought) and the second is the exact same thing.

I hope this helps you somehow! Good luck!

Title: Definition(s) of rational numbers


问:
Prove $\lim\limits_{x\to -\infty} \frac{\cos(\tan x)}{x+1} = 0$ using definition

Problem

Prove  $\lim\limits_{x\to -\infty} \frac{\cos(\tan x)}{x+1} = 0$ using definition

proof

$\forall \varepsilon>0$ take $M=-\frac{1}{\varepsilon}-1$ if $x<M<0$, implies

$$x<-\frac{1}{\varepsilon}-1<0$$ it  follows that  $$-\varepsilon <\frac{1}{x+1}<0<\varepsilon \iff \bigg|\frac{1}{x+1}\bigg|<\varepsilon$$

since $|\cos(\tan x)|\leq 1$ it follows
$$f(x)=\bigg|\frac{\cos(\tan x)}{x+1}\bigg|<\varepsilon$$
so
$$\lim\limits_{x\to -\infty} \frac{\cos(\tan x)}{x+1} = 0$$

Please verify the proof above.

答:
I make an answer because this will be too long for a comment.

Your proof is actually almost correct, except that you did not take into account the fact that your function is never defined on an interval of the form $(-\infty,a)$, with $a \in \mathbb{R}$. Indeed the function $$f(x)= \frac{\cos(\tan(x))}{x+1}$$ is not-defined in every $\frac{\pi}{2}-k\pi$ where $k\in \mathbb{Z}$ (because of the $\tan$), so there is no interval $(-\infty,a)$ on which $f$ is everywhere defined. So when you say "for every $x < M$, etc.", this is not really correct, because $x<M$ can be a point where the function is not-defined and in that case, what you write then has no sense.

What you must have in mind is that the definition of the limit takes this issue into account. For a function $g$ defined on a domain $\mathcal{D}$, the definition of $\lim_{x \rightarrow -\infty} g(x)=L$ is

$$\forall \varepsilon > 0, \exists M\in \mathbb{R}, \forall x \in (-\infty,M)\cap \mathcal{D}, |f(x)-L|< \varepsilon$$

The important point is "$\forall x \in (-\infty,M)\cap \mathcal{D}$", which is not just "$\forall x \in (-\infty,M)$". You only have to consider the $x$'s which belong to the domain of $g$. With this small precision, you can make an absolutely correct proof of your limit.

I just also want to add an often forgotten point about this definition of the limit in $\pm \infty$ : this definition has sense only if $\mathcal{D}$ is not bounded below (which is the case in your question, so no problem here). Consider for example the function $g : [0,1] \rightarrow \mathbb{R}$ defined by $g(x)=x^2$ on the domain $\mathcal{D}=[0,1]$. Then let $\varepsilon > 0$ and $M = -1$. You have $(-\infty,M) \cap \mathcal{D}= \emptyset$, so it is true that $\forall x \in (-\infty,M) \cap \mathcal{D}$, one has $|g(x)-42|<\varepsilon$, and also $|g(x)-127|<\varepsilon$ ! Therefore, just with the definition, you get that
$$\lim_{x \rightarrow - \infty} g(x) = 42 \quad \quad \text{and} \quad \quad \lim_{x \rightarrow - \infty} g(x) = 127$$

ie $g$ has two distinct limits, which is a thing we don't want (an important property is that the limit, when it exists, must be unique). This happens here because $g$ is only defined on $[0,1]$ which is bounded below.


问:
Determinant of a matrix without actually expanding it.
Given a Matrix $$A =  \begin{vmatrix}
a^2 & (s-a)^2 & (s-a)^2\\
(s-b)^2 & b^2 & (s-b)^2 \\
(s-c)^2 & (s-c)^2 & c^2\\
\end{vmatrix}$$

and $$a+b+c = 2s$$ , how can I find its determinant without expanding?

I've tried finding all factors which make two columns (/rows) similar or proportional. Only with s=0 all the rows and columns becomes 1 (after taking out the common value). With s=a $$\left|\matrix{
a^2 & 0 & 0\\
(a-b)^2=c^2 & b^2 & (a-b)^2=c^2 \\
(a-c)^2=b^2 & (a-c)^2=b^2 & c^2\\
}\right|=0$$

Here is a list of trivial factors
(s-a), (s-b), (s-c), (s-0). I'm given a hint that the determinant is a degree 6 polynomial, which I don't understand cause max degree of s is 4, please do explain how degree is 6.

I think det(A) = (s-a)(s-b)(s-c)(s-0).

Are there any more factors? What are they?

Edit: Please also explain how to arrive at the solution.

答:
Not a 'real' answer, but it was too big for a comment.

I wrote and ran some Mathematica-code:

```<code>In[1]:=Clear["Global`*"];
FullSimplify[
 Det[{{a^2, (s - a)^2, (s - a)^2}, {(s - b)^2,
    b^2, (s - b)^2}, {(s - c)^2, (s - c)^2, c^2}}],
 Assumptions -> a + b + c == 2 s]
</code>```

Running the code gives:

```<code>Out[1]=-2 (b - s) (b + c - s) s^3 (-c + s)
</code>```

If you're not only looking for the solution, it is not very hard to show that this is correct.

Body:
Not a 'real' answer, but it was too big for a comment.

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
FullSimplify[
 Det[{{a^2, (s - a)^2, (s - a)^2}, {(s - b)^2,
    b^2, (s - b)^2}, {(s - c)^2, (s - c)^2, c^2}}],
 Assumptions -> a + b + c == 2 s]

Running the code gives:

Out[1]=-2 (b - s) (b + c - s) s^3 (-c + s)

If you're not only looking for the solution, it is not very hard to show that this is correct.


问:
Mistake in number of ways in which we can assign $7$ projects to $4$ people such that each person get at least one project
Number of ways in which we can assign $7$ projects to $4$ people such that each person get at least one project:

My approach to this question was:

First assign $1$ project to everyone.

For this I first choose $4$ jobs out of $7$ and count every way i.e.$\binom{7}{4}4!$.

Next thing I did was applying the combinations with replacement as

J # J # J #

J J J # # #

J # # # J J

.

.

.

where 2 # are kind of boundaries and 3rd is also a boundary which also covers one position.

(Sorry but I don't remember the formula and always use it like this)

I got the solution $\binom{6}{3}$.

Now I thought I got all the cases and by product rule
$$\binom{7}{4}4!\binom{6}{3}$$

would give me answer. But the answer that I got was $16800$ while the answer in my book is $8400$. So where did I go wrong and why do I have to divide with $2$?

答:
What you are looking for is number of surjective mappings between set of project to set of people(functions onto). There is formula for it:
$$ m!S(n,m)$$
Where :

S(n,m) is stirling number of second kind

n,m are numbers of elements in domain and codomain

4!S(7,4) = 8400


问:
Why do I get a math error for this when trying to compute $\int_{-1}^1(x^2-\frac{1}{x^2}+3) \, dx$?
I tried to evaluate
$$
\int_{-1}^1(x^2-\frac{1}{x^2}+3) \, dx
$$
in the following way:
$$
\left[\frac{x^3}{3}+\frac{1}{x}+3x\right]_{-1}^1=\frac{26}{3} \, .
$$
But when I typed the integral into a calculator I got a math error. Why did this happen?

答:
Well, what happends to:

$$f(x)=\frac{1}{x^2}$$

When $x\in\left[-1,1\right]$?


问:
Count the number of sequences possible which have no adjacent element?
Length of the sequence = $N$

Each element of the sequence can be in the range $[1,M]$

$A[1]=1$

$A[N]=1$

Find the number of ways to assign values to remaining elements such that , $A[i]!=A[i+1]$, for any $'i'$

答:
Just an attempt. I might be wrong.

$A[1] = 1 = A[N]$ is fixed.

Each of the odd positions can be filled in  $M$ ways. The number of odd positions $k$ can be given by the following formula $k = N/2-2$ if $N$ is even and $k=(N+1)/2 - 2$ if $n$ is odd. So, now we have filled these odd positions in $M^k$ ways.

Now each of these intermediate positions can be filled two ways as follows:

If the previous odd position $\alpha-1$ has the same value as the current odd position $\alpha+1$, then the postion $\alpha$ can be filled in $M-1$ ways, i.e., if $A[\alpha-1] = A[\alpha+1]$, then $A[\alpha]$ can be filled in $M-1$ ways. If $A[\alpha-1] != A[\alpha+1]$, then $A[\alpha]$ can be filled in $M-2$ ways.

Assuming $A[\alpha-1]$ is fixed, $A[\alpha-1] != A[\alpha+1]$ can occur in $M-1$ ways and $A[\alpha-1] = A[\alpha+1]$ can occur in exactly $1$ way.

Combining the above two arguments, each even space can now be filled in
$$ (M-1).(M-2) + 1.(M-1) = (M-1)^2$$
ways. The total number of solutions should thus be something of the form
$$ M^k(M-1)^{2(N-k-2)}.$$


答:
The equation $a+b+c=2s$ has a geometric interpretation: $s$ is the semiperimeter of a triangle of sides $a,\,b,\,c$. If this triangle is degenerate, say with $a=b+c$, then $s=a$ and$$A=\left|\begin{array}{ccc}
\left(b+c\right)^{2} & 0 & 0\\
c^{2} & b^{2} & c^{2}\\
b^{2} & b^{2} & c^{2}
\end{array}\right|=0,$$so $(s-a)(s-b)(s-c)|A$. A similar treatment of $-a=b+c\implies s=0$ can prove $s|A$. Since $A$ is of dimension $6$ and is homogeneous when all sides are rescaled, the symmetries require a result proportional to $s^3(s-a)(s-b)(s-c)$. An equilateral triangle with $a=b=c=2,\,s=3,\,A=54$ proves the coefficient is $2$, as in @JanEerland's answer.

Title: Determinant of a matrix without actually expanding it.

Body:
Given a Matrix $$A =  \begin{vmatrix}
a^2 & (s-a)^2 & (s-a)^2\\
(s-b)^2 & b^2 & (s-b)^2 \\
(s-c)^2 & (s-c)^2 & c^2\\
\end{vmatrix}$$

and $$a+b+c = 2s$$ , how can I find its determinant without expanding?

I've tried finding all factors which make two columns (/rows) similar or proportional. Only with s=0 all the rows and columns becomes 1 (after taking out the common value). With s=a $$\left|\matrix{
a^2 & 0 & 0\\
(a-b)^2=c^2 & b^2 & (a-b)^2=c^2 \\
(a-c)^2=b^2 & (a-c)^2=b^2 & c^2\\
}\right|=0$$

Here is a list of trivial factors
(s-a), (s-b), (s-c), (s-0). I'm given a hint that the determinant is a degree 6 polynomial, which I don't understand cause max degree of s is 4, please do explain how degree is 6.

I think det(A) = (s-a)(s-b)(s-c)(s-0).

Are there any more factors? What are they?

Edit: Please also explain how to arrive at the solution.


问:
How do I turn this sequence into a sigma function?
$$3\times2+1$$
$$ 3\times3+1$$ $$3\times3+2$$
$$3\times4+1$$ $$3\times4+2$$ $$3\times4+3$$
$$3\times5+1$$$$3\times5+2$$$$3\times5+3$$$$3\times5+4$$

And so it goes on for a number of terms dependent on the n-value, though exactly how it is dependent I'm not sure. You can leave the number of terms blank, as I'll figure that out myself. However, how I make this sequence into a sigma function I have no idea. Note that something like $(3\times k)(k-1)+\frac{(k-1)k}{2}$ will not work, because the terms are the individual operations, and not the bigger groupings: i.e., the last term doesn't need to be $3\times k + (k-1)$. The last term can be anything from $3\times k + (k-(k-1))$ all the way to the aforementioned one.

As such, the sequence becomes: 7, 10, 11, 13, 14, 15, 16, 17, 18, 19, 19, 20, 21, 22, 23, 22, 23, 24, 25, 26, 27, and so on.
So, any thoughts?

EDIT:

$$\sum_k^? \lceil \frac{n-x}{2} \rceil$$

"?" is something related to the n-value, and "x" is related to the k-value, and is somehow supposed to give the sequence. However, it appears that the current answers to this post give the answer for the objective of subtracting from "n" the entire sequence-parts, instead of the individual terms. As in, those answers make it so that "x" does this: $n - ((3\times4+1) + (3\times4+2) + (3\times4+3))$. However, "x" doesn't need to be all of that. The last "x" can be $3\times4+2$, the number added doesn't need to be k-1.

答:
You can express this as a double summation:

$$\sum_{1≤n<m}^{n} \sum_{i=2}^m 3m+n$$

For example, when $m=4$, you get $3 \times 4 + 1, 3 \times 4 + 2$, and $3 \times 4 + 3$.


问:
What do we mean when we say "Let $x$ be an element of the set $\mathbb{R}$"?

What do we mean when we say "Let $x$ be an element of the set $\mathbb{R}$"?

Does $x$ represents only a single  element of set $\mathbb{R}$? or does $x$ represent all the element of set $\mathbb{R}$ simultaneously at the same time?

People say that if $x\in\mathbb{R}$ then $x$ is any real number; that means $x$ represents all real numbers. But if $x$ is any real number then let's say $x=1$; so $x$ is one, then how it can represent all real numbers?

Please help me I am very confused.

答:
There are already good answers, but I really struggled with this when first encountering logic and proofs, so I hope my own perspective is useful to someone.

Often we want to to prove something about all elements of a set S. That is, we want to prove that every element in $S$ satisfies some property, call it $P$. For example, say $S = \{1,2,...\}$. Now say $P(x)$ stands for the statement "$x$ is greater than $0$". Clearly $P(1)$ is true, and so is $P(2)$ is true, and indeed it should be obvious that no matter what number we choose from $S$, $P(\text{ 'that number' )}$ will be true.

We want a way to express the truth of the statement in the above example. We know that no matter what object I choose from $S$, the statement $P$ will be true for that object. We do this by reasoning about an arbitrary object from that set. We say $x \in S$, and specify nothing more. Now the key bit here is that $x$ is not simultaneously "all objects in $S$". $x$ represents a specific number from the set $S$, we just don't say which one.

Going back to our example, say I tell you $x \in S$. Now $x$ represents some specific number in $S$, but we don't know which. Suppose $x$ actually represented the number $20$. Well of course $P(20)$ is true, because $20>0$. Similarly, if $x$ actually represented the number $1727361$, then $P(1727361)$ would be true too. The pattern here is clear: no matter what number $x$ actually represents, $P(x)$ is true. It is in this sense that $x$ can represent any/all elements of $S$.

Ultimately, our example can be written as $\forall x \big(x \in S \rightarrow P(x)\big)$. This statement is really saying, "for every object $x$, if this object is a specific a number in the set $S$, then the statement $P(x)$ is true." Note the key point here is that I have said $x$ is specific, yet I have not told you what it is. The meaning here is that in the statement, we treat $x$ as a specific number. After all, we say that $x \in S$ and $P(x)$, which only make sense if $x$ is just one number. But $x$ could represent any specific number in $S$, because every number I use makes $P(x)$ true.

The above is an intuitive and informal approach; but fundamentally this question can be answered formally. This question revolves around notions of variables, quantifiers, and logic. Answers to my own questions and others' here on stack exchange do a very good job of explaining this more formal approach. For example, see this. The answer and Noah Schweber's comment are incredibly useful.

Response to comments by OP

When we say $x>-2$, yes, it is correct to say that $x$ can be any real number. $x$ could also be a function, a group, or any kind of mathematical object. $x$ is simply a symbol. There is no reason to say that $x$ must be a real number larger than $-2$. If I were to then tell you $x$ was the number $-3$, that's perfectly fine, it just means the statement is false for that 'value' of $x$.

Now, if you were to say that $x$ is an arbitrary real number such that $x>-2$, then yes, now $x$ represents a specific (yet unspecified) real number larger than $-2$. In essence we are saying $x$ is an arbitrary element of the set $\{y \in \mathbb{R}: y > -2\}$. Now it makes sense to say that $x$ is an arbitrary element greater than $-2$, because I have told you that is the case.

To summarise, the statement $\forall x \in \{y \in \mathbb{R}: y > -2 \} \big(x > -2 \big)$ is true, but the statement $\forall x \in \mathbb{R} \big(x > -2 \big)$ is not. And furthermore, $x > -2$ by itself is not comparable to the other two. This is a well formed formula, and whether it is true or not depends on what $x$ is. Of course, we may restrict $x$ to only take on those values which make $x>-2$ true. In this case, then $x$ represents some specific yet unspecified value from $\{y \in \mathbb{R}: y > -2 \}$.