问:
Computing the Euler characteristic of real projective space $\mathbb{R}P^{n}$
I would like to compute the Euler characteristic of $\mathbb{P}^n(\mathbb{R)}$. I do not know if cohomology could help but I should avoid it because I did not studied it yet. I would like to use only basic tools of differential topology.

I though to consider $\mathbb{P}^n(\mathbb{R)}$ as the quotient of $\mathbb{D}^n$ where its edge is identified: $x \sim y$ is and only if $x=y$ or $\|x\|=\|y\|=1$ and $x=-y$. But I am not sure on how to continue.

答:
Hint 1. $\mathbb{R}P^{n} \cong \mathbb{S}^{n}/x \sim -x$. What is the Euler characteristic of $\mathbb{S}^{n}$? Can you deduce $\chi(\mathbb{R}P^{n})$ from there?

Hint 2. $\mathbb{R}P^{n} = \mathbb{R}P^{n-1} \cup_{f} D^{n}$, where $f: S^{n-1} \to  \mathbb{R}P^{n-1}$ is the 2-to-1 covering map and $D^{n}$ is the interior of an $n$-disk.


问:
Sets and set-builder notation
I have few questions about sets and set-builder notations.

How to name the sets correctly ?

I know that typically we use capital letters $A, B, C, ...$ , but I wonder if it is allowed to name sets like $AT, A1, A_1, A^1$ or some of them are reserved for representing some operation on set.

How to show that the set-builder for set B depends on set-builder from set A ?

For example:

$A = \{n, n \in Z\}$

$B = \{\frac{2n + 10}{7}, n \in Z\}$

Where set-builder from set B is connected with set-builder from set A by the function:

$f(n) = \frac{2n + 10}{7}$,

or it is actually shown by naming the variables from both set-builders the same "n" ?

Is it right that if I would like to check if set A and set B have common values I have to compare their set-builders ?

For example:

$A = \{n, n \in Z\}$

$B = \{2n, n \in Z\}$

$n = 2m$

There exists solution for all $n,m \in Z$, so actually set B is subset of set A.

答:
(1) The notation $A, B, C$ are fine. I would only use $AT$ if it can be seen as a mnemonic to describe the set, like in
$$
  AT = \bigl\{ \mathrm{arctan}(x) \mid -1 \leqslant x \leqslant 1 \bigl\}
$$
I would avoid $A1$. The next one, $A_1$, is common if you have a family of sets, like in
$$
\text{For $0 \leqslant r < 5$, let $A_r = \{ n \in \Bbb N \mid n \equiv r \bmod 5\}$. Then $A_1 = \{ 1, 6, 11, 16, \dotsm \}$}
$$
The notation $A^1$ should generally be avoided, since $A^n$ usually denotes the Cartesian product of $n$ copies of $A$, that is $\underbrace{A \times {}\dotsm{} \times A}_{\text{$n$ times}}$.

(2) The correct notation should be
$$
A = \{ n \mid n \in \Bbb Z \} \qquad B = \Bigl\{ \frac{2n + 10}{7} \mid n \in \Bbb Z \Bigr\}
$$
although the set $A$ is simply $\Bbb Z$. If you want to insist that you use different set builders on the same domain, you could first introduce the functions you intend to use, and then define the sets, like in
$$
\text{Let $f_1(n) = n$, $f_2(n) = 2n$ and $f_3(n) = \frac{2n + 10}{7}$. For $1 \leqslant i \leqslant 3$, set $A_i = \bigl\{ f_i(n) \mid n \in \Bbb Z \bigr\}$ }.
$$
(3) To compare the sets $A_1$, $A_2$ and $A_3$, it would indeed be sufficient to compare the images of the functions $f_1$, $f_2$ and $f_3$.

ContentLicense: CC BY-SA 4.0

Title: Sets and set-builder notation


问:
Intuition of the meaning of homology groups
I am studying homology groups and I am looking to try and develop, if possible, a little more intuition about what they actually mean. I've only been studying homology for a short while, so if possible I would prefer it if this could be kept relatively simple, but I imagine it is entirely possible there is no real answer to my query anyway.

As I said above, I want to gain a little deeper understanding of what the n-th homology group actually means: I can happily calculate away using Mayer-Vietoris but it doesn't really give me a great deal of intuition about what the n-th homology group actually means. For example, with homotopy groups, the fundamental group is in some sense a description of how loops behave on the object in question, and it is obvious to me why that is what it is for say, the torus or the circle. However, I have no idea what, if anything, I am actually saying about a triangulable object when I talk about it having 0-th homology group this or 1st homology group that.

The best I have been able to find online or in my limited book selection is the brief description "intuitively, the zeroth homology group counts how many disjoint pieces make up the shape and gives that many copies of $\Bbb Z$, while the other homology groups count different types of holes". What 'different types of holes' are there, roughly speaking? I appreciate that it may often be completely non-obvious what the low-order homology groups are for some complicated construction, but perhaps in simpler examples it might be more explicable. Are there (simple) cases where I could say, just from looking something like e.g. the torus, what its zero-th or first or second etc. homology group was based on the nature of the object? I guess in the zero-th case it is, as my source (http://teamikaria.com/hddb/wiki/Homology_groups) above says, related to the number of disjoint pieces. Can we delve deeper than this for the other homology groups?

Any book/website suggestions would be welcomed (preferably websites as I am nowhere near a library!) - I have Hatcher but not a great deal else, and I haven't gleaned as much as I wish to from that alone. Of course I know that there is a great deal we don't know about homology groups even today, so I don't expect some magical all-encompassing answer, but any thoughts you could provide would be appreciated. I hope this question is appropriate for SE Mathematics, apologies if not! -M

答:
We can read some geometric intuition about homology groups directly from the definition of the singular homology groups. We may also have a look at cellular homology to see what we can extract.

Singular homology. Let $X$ be a space, and let $S:=S_\cdot X$ be its singular chain complex with differential (or boundary map) $d_\cdot$. We call $Z_n(S):=\text{ker}d_n$ the $n$-cycles of $S$, and $B_n(S):=\text{im}d_{n+1}$ the $n$-boundaries of $S$. A pair $f,g$ of $n$-chains of $S$ are homologous if their difference $f-g$ is a boundary. In other words, if there exists an $n+1$-chain $h$ such that $dh=f-g$; we really do mean they are a boundary!

This is analogous to homotopy. A pair $f,g:X\to Y$ are homotopic if there exists a map $h:X\times I\to Y$ such that

commutes (where $i_0$ and $i_1$ are the inclusion maps at $0$ and $1$ respectively). In other words; two maps $f,g:X\to Y$ are homotopic if they are restrictions to the boundary of a full map $h:X\to Y$.

Intuitively, homology finds $n$-holes, and indeed different types of $n$-holes. A way to think about it is that homology finds holes by keeping track of which $n$-chains are the boundary of an $(n+1)$-chain. Indeed, if you cannot fill in your $n$-chain there must be something in your space which is missing.

A thought experiment that helped me: We may try consider various spaces, and try to construct a chain or two which are not the boundary of a higher chain. What can we for instance say about a chain realizing the circle centered at $0$, in the plane punctured at $0$? What happens of we have two such chains? Are they homologous and how?

Cellular homology. We can probably read even stronger geometric intuition from cellular homology. Indeed if $X$ is a CW complex, then it has a filtration $X_0\subset \cdots \subset X$, and it is natural to think that the homology of the $X_i$ together with the data of the inclusion maps (how $X$ is glued together) should determine the homology of $X$.

This is indeed the case, and in fact in a very elegant fashion: The space $X_i/X_{i-1}$ is a wedge of $i$-spheres and as such it has free abelian homology concentrated at $i$ (precisely, if $k$ is the number of $i$-spheres of $X_i/X_{i-1}$, then $\tilde H_j(X_i/X_{i-1})=\mathbb{Z}^k$ for $i=j$ and $0$ otherwise). Using this and the data of the filtration of $X$ to its skeleton, we may tie the groups together in a long exact sequence which produces homology data based on the degrees of gluing maps (in other words, data about how $X$ is glued together from spheres and disks).

This reminds me that I should read more about cellular homology; I'll come back to this answer!

ContentLicense: CC BY-SA 3.0


问:
Find $\lim_{n\to\infty} \frac{n^{1296}}{6^n}$
I need to find $$\lim_{n\to\infty} \frac{n^{1296}}{6^n}$$ I have nothing yet, maybe sandwich theorem but I'm not sure.
Thanks in advance  for help.

答:
Let $$L=\lim_{n\to\infty}\frac{n^{1296}}{6^n}$$
Applying L' Hopital's rule once, we see that
$$L=\lim_{n\to\infty}\frac{1296n^{1295}}{6^n\ln6}$$
We can keep on applying L' Hopital's rule (as we always will have a indeterminate form), until we get
$$L=\lim_{n\to\infty}\frac{1296!~n}{6^n(\ln6)^{1295}}$$
Applying this once more we are left with
$$L=\lim_{n\to\infty}\frac{1296!}{6^n(\ln6)^{1296}}=0$$
So we see that $L=0$.
I hope that was helpful. If you have any questions please don't hesitate to ask :)


问:
Integral involving ratios of polynomial and square roots
Is there a simple way to compute the following integral

$$I(y,\beta,\gamma)=\int_1^y \frac{x^2}{\sqrt{\left(x^2-1\right)\left(x^2-\gamma ^2\right) \left(x^2+\beta ^2\right) }}\mathrm{d}x$$

with $y>1$, $0<\gamma<1$ and $\beta>0$?

答:
Substitute $x=\sqrt t,dx=\frac1{2\sqrt t}\,dt$ to get
$$I=\frac12\int_1^{y^2}\sqrt{\frac t{(t+\beta^2)(t-\gamma^2)(t-1)}}\,dt$$
This is an elliptic integral. Byrd & Friedman 258.13 gives
$$I=\frac1{\sqrt{\gamma^2+\beta^2}}\int_0^{F(\varphi,m)}\frac{\operatorname{dn}^2u}{1-n\operatorname{sn}^2u}\,du$$
where $m=\frac{\gamma^2(1+\beta^2)}{\gamma^2+\beta^2}$, $\sin^2\varphi=\frac{(\gamma^2+\beta^2)(y^2-1)}{(1+\beta^2)(y^2-\gamma^2)}$ and $n=\frac{1+\beta^2}{\gamma^2+\beta^2}$. In turn, B&F 339.01 solves the second integral, yielding
$$I=\frac1{\sqrt{\gamma^2+\beta^2}}\cdot\frac1n(mF(\varphi,m)+(n-m)\Pi(n,\varphi,m))$$
$$=\frac1{\sqrt{\gamma^2+\beta^2}}(\gamma^2F(\varphi,m)+(1-\gamma^2)\Pi(n,\varphi,m))$$


问:
Help! perplexing problem on volume revolution of region about a line
how do we find the volume when the region bounded by $y = x^{\frac 1 2}$ and $y = \frac x 2$ is revolved about the line $y = 1$? I understand how to solve for the regions above and below $y = 1$, however because of the intersection of volumes from $x = 1$ to $x = 2$, the volume of the entire region both above and below $y = 1$ eludes me.

heres the graphs: https://www.desmos.com/calculator/njvae9suit

I would greatly appreciate an explained solution to solving a problem such as this. Thank you!

答:
You should do something like
$$\pi \left[\int_0^1 \left(\left(\frac{x}{2}-1\right)^2-\left(\sqrt{x}-1\right)^2\right) \, dx+ \int_1^4 \left(\left(\sqrt{x}-1\right)^2-\left(\frac{x}{2}-1\right)^2\right) \, dx\right]$$


问:
Proving that the integral closure of $R$ in $S$ is a subring of $S.$
Let $R \subset S$ be commutative rings. Prove that the integral closure of $R$ in $S$ is a subring of $S.$

Note that the integral closure of $R$ in $S$ consists of all $x \in S$ that is integral over $R$. And note that an element $x$ is integral over $R$ if there exists a monic polynomial $f(x) \in  R[x]$ such that $f(x) = 0$.

I was thinking about using the ordinary subring test for doing that but I was given a hint to use the following Lemma: $x$ is integral over $R$ iff $R[x]$ is a finitely generated $R$-module.

My question is:

I do not see why I should use this lemma and how it facilitates the steps of the proof, could anyone explain this to me please?

答: Let $x,y\in S$ be integral over $R$. We wish to show that also $x+y$, $xy$ are integral over $R$. By the lemma $R[x]$ is a finite $R$-module. Note that $y$ is surely also integral over $R[x]$, so $R[x][y]$ is a finite $R[x]$-module. Show that this implies that $R[x][y]$ is a finite $R$-module and use the lemma again (we might need a slightly stronger version of the lemma, namely that if $z$ is contained in a ring that is a finitely generated $R$-module then $z$ is integral over $R$. This is often proved together with the lemma you cite, if not: the proof is almost identical using the Cayley-Hamilton theorem)


问:
Constructing a $4$-state Markov chain model that describes the arrival of customers
The times between successive customer arrivals at a facility are independent and identically distributed random variables with the following PMF:

$$p(k) = 0.2(k = 1)$$
$$p(k) = 0.3(k = 3)$$
$$p(k) = 0.5(k = 4)$$
$$p(k) = 0(k \notin \{1,3,4\})$$

Construct a four-state Markov chain model that describes the arrival process. In this model, one of the states should correspond to the times when an arrival occurs.

Can you please explain in simple words how to construct this markov chain? Because I am totally lost with given distribution and how can I use this in my problem.

答:
The state of the Markov chain the question seems to be asking about is the length of time since the last customer arrived.  It can have any of the four values, $\ 0,$$\ 1,$$\ 2,$ or $\ 3\ $.  The chain is in state $\ 0\ $ at time $\ n\ $ if a customer arrived at that time, and is otherwise in state $\ i\ne0\ $ at that time if the last customer arrived at time $\ n-i\ $.

If the state at time $\ n\ $ is $\ 0\ $ then there is a probability of $\ 0.2\ $ that the next customer will arrive at time $\ n+1\ $, in which case the chain will remain in state $\ 0\ $, and a probability of $\ 0.8\ $ that the customer will arrive at either time $\ n+3\ $ or $\ n+4\ $, in which case the state at time $\ n+1\ $ will be $\ 1\ $.  Thus we have
\begin{align}
p_{00}&=0.2\\
p_{01}&=0.8
\end{align}

If the chain is in state $\ 1\ $ at time $\ n\ $, then the last customer arrived at time $\ n-1\ $ and none arrived at time $\ n\ $, so the next one won't arrive until either time $\ n+2\ $ or $\ n+3\ $, and the state at time $\ n+1\ $ will be $\ 2\ $ with probability $\ 1\ $.  That is,
$$
p_{12}=1\ .
$$
If the chain is in state $\ 2\ $ at time $\ n\ $, then the last customer arrived at time $\ n-2\ $ and none arrived at time $\ n\ $.  There is then a (conditional) probability of $\ \frac{0.3}{0.8}\ $ that the next customer will arrive at time $\ n+1\ $, in which case the state at time $ n+1\ $ will be $\ 0\ $, and a probability of $\ \frac{0.5}{0.8}\ $ that the customer will arrive at time $\ n+2\ $, in which case the state at time $\ n+1\ $ will be $\ 3\ $.  Thus we have
\begin{align}
p_{20}&=\frac{0.3}{0.8}\\
p_{23}&=\frac{0.5}{0.8}\ .
\end{align}
Here, we must condition our probabilities on the event that the time between the arrival of the last customer and the next was not $\ 1\ $ time unit, because we know that that event has occured.

If the chain is in state $\ 3\ $ at time $\ n\ $, then the last customer arrived at time $\ n-3\ $ and none arrived at time $\ n\ $, so the next one must arrive at time $\ n+1\ $, and the chain will then be in state $\ 0\ $. Thus we have
$$
p_{30}=1\ .
$$
Putting all this together, we get the following transition matrix for the Markov chain.
$$
P=\pmatrix{0.2&0.8&0&0\\
           0&0&1&0\\
           \frac{0.3}{0.8}&0&0&\frac{0.5}{0.8}\\
           1&0&0&0}\ .
$$


问:
Logic math question?
How do I shorten "(a and b and c) or (a and negation b and c)"? I feel like the answer is easy but I cant seem to grasp which rule I should use.</p>

答: Hint. Writing $d$ for ($a$ and $c$), your formula becomes ($d$ and $b$) or ($d$ and not $b$). Can you now simplify this expression?

Title: Logic math question?


答:
You prove that it is a subring by ordinary "subring test" but you use the lemma while doing so. Although in our commutative algebra course we used also the following version of lemma (we actually prove this first to get your lemma but I am not sure whether it is necessary to do it in this order. )

Lemma 1 $R\subset S$,  $x\in S $ then $x$ is integral over $R$ if $R[x]$ is finitely generated $R-$module.

Lemma 2 $x$ is integral over $R$ iff there exists finitely generated $R-$module $R'$ such that $R\subset R[x]\subset R'\subset S$

Take $a,b\in S$ integral over $R$. By your lemma $R[a]$ is finitely generated $R$-module. But we can view $R[a]$ simply as a ring so $b$ is integral also over $R[a]$. Hence $R[a][b]$ is finitely generated module over $R[a]$.

Now $R[a][b]=R[a,b]$ as subrings of $S$ (they consist of the same elements) so $R[a,b]$ is also finitely generated as $R-$module.

Using lemma 2: Now set $x:=a+b$ and $R':=R[a,b]$. $R'$ is finitely generated $R-$module and $R\subset R[x]\subset R'\subset S$ so by lemma 2 we conclude that $x$ is integral.

Commentary I think that the reason we need to use lemma 2 instead of ypur version (lemma 1) is that you can't conclude that $R[a+b]$ is finitely generated from the fact that it is $R-$submodule of finitely generated $R-$module $R[a,b]$. there are submodules of fin. gen. modules that are not finitely generated (take ring $R$ as a module over itself it is generated by $1$ but unless it is Noetherian it has ideals that are not finitely generated)

However, in this particular setting, lemma 2  gives you  that $a+b$ is integral so by lemma 1 is $R[a+b]$ finitely generated $R-$module after all.

Title: Proving that the integral closure of $R$ in $S$ is a subring of $S.$

Body:
Let $R \subset S$ be commutative rings. Prove that the integral closure of $R$ in $S$ is a subring of $S.$

Note that the integral closure of $R$ in $S$ consists of all $x \in S$ that is integral over $R$. And note that an element $x$ is integral over $R$ if there exists a monic polynomial $f(x) \in  R[x]$ such that $f(x) = 0$.

I was thinking about using the ordinary subring test for doing that but I was given a hint to use the following Lemma: $x$ is integral over $R$ iff $R[x]$ is a finitely generated $R$-module.

My question is:

I do not see why I should use this lemma and how it facilitates the steps of the proof, could anyone explain this to me please?


问:
How is this value of a slope for a parabola
My teacher wrote  that $$y + \frac{D}{4a} =a*(x+\frac  {b}{2a})^2$$ is the value of slope . I thought it could be instantaneous slope then since slope is of a line and not a parabola. Then , my question is how did he derive this as value of slope.

My teacher said that equate the slope to 0.
that is $dy/dx=0$

$dy/dx= 2a(x+\frac{b}{2a})$

so
$2a(x+\frac{b}{2a})=0$

so $x=-b/2a$

答: As @Raffaele pointed out, what your teacher is attempting to do is to write the quadratic equation in vertex form, or to "Complete the square". Now at the vextex $\left(-\frac{b}{2a},-\frac{D}{4a}\right)$ the slope is $0$. I believe the goal is to show that you can find the $x$ value of the turning point either by completing the square or by setting the derivative to $0$ and solving for $x$.


问:
Proof that a linear transformation is continuous
I got started recently on proofs about continuity and so on. So to start working with this on $n$-spaces I've selected to prove that every linear function $f: \mathbb{R}^n \to \mathbb{R}^m$ is continuous at every $a \in \mathbb{R}^n$. Since I'm just getting started with this kind of proof I just want to know if my proof is okay or if there's any inconsistency. My proof is as follows:

Since $f$ is linear, we know that there's some $k\in \mathbb{R}$ such that $|f(x)|\leq k|x|$ for every $x\in \mathbb{R}^n$, in that case let $a\in \mathbb{R}^n$ and let $\varepsilon >0$. Consider $\delta = \varepsilon /k$ and suppose $|x-a|<\delta$, in that case we have:

$$|f(x)-f(a)|=|f(x-a)|\leq k |x-a|<k \frac{\varepsilon}{k}=\varepsilon$$

And since $|x-a|<\delta$ implies $|f(x)-f(a)|<\varepsilon$ we have that $f$ is continuous at $a \in \mathbb{R}^n$. Since $a$ was arbitrary, $f$ is continous in $\mathbb{R}^n$. Is this proof fine? Or there was something I've missed on the way?

答:
This proof is correct modulo result you stated in the begining, i.e.
$$
\text{there exist $k\in\mathbb{R}$ such that $|f(x)|\leq k|x|$ for all $x\in\mathbb{R}^n$}
$$
Proof of this fact is much more interesting and uses compactness of unit ball in $\mathbb{R}^n$.</p>


问:
A simple method to calculate minimal area enclosed between a tangent to $f(x)$ and coordinate axes
Given a function $y=f(x)$, take a tangent to the function at some point. We are to find the minimum area between this line and the coordinate axes. An example question is the coordinate axes have minimum area.

I faced two different algorithms to find the solution. The first one is straight-forward:

Pick a point on the function: $(x_1, y_1) = (x_1, f(x_1))$.

Find the derivative of the function at $x_1$ to calculate the slope of the line. ($m=f'(x_1)$)

Derive the tangent-line formula. $y-y_1=m(x-x_1)$

Find the formulations of intersections with coordinate axes $(0, y_0)$, $(x_0, 0)$.

Calculate the formulation of the area of triangle as $A = x_0 y_0/2$.

$A$ shall be a function of $x_1$, minimize that to calculate minimum area.

Second algorithm is very short compared to that:

Take the function $g(x) = 2 x f(x)$.

Minimize that function to calculate the result.

I cannot figure out how the second algorithm works, or when it works. I checked both against the following family of functions, and both algorithms give the same result:

$f(x) = ax+b$

$f(x) = k/x$

$f(x) = 3b^2 - a^2 x^2$

$f(x) = \frac{b}{a} \sqrt{a^2-x^2}$

Question is: Can we prove/disprove the second algorithm? If disproved, under what conditions does the second algorithm work?

答:
$\def\x{\tilde x}$
Observe that all your examples except for $f(x)=ax+b$ have the following common properties. In the first quadrant they are:

$$\begin{align}&\text{positive and decreasing}\tag1\\
 &\text{either concave or convex}\tag2
\end{align}$$

Let $(x,f(x))$ be the tangency point. Then the area of the triangle in question is:
$$
A(x)=-\frac{(xf'(x)-f(x))^2}{2f'(x)}.\tag3
$$
Hence the equation $A'(x)=0$ reads:
$$
\frac{[xf'(x)-f(x)]f''(x)[xf'(x)+f(x)]}{2(f'(x))^2}=0.\tag4
$$

Since due to properties (1) and (2) neither the first nor the second factor in the numerator can be $0$, we are left with the equation:
$$
0=xf'(x)+f(x)=[xf(x)]',\tag5
$$
so that the extrema of the functions $A(x)$ and $xf(x)$ are attained at the same point $\x$. Substituting $\x f'(\x)=-f(\x)$ into equation (3) one finally obtains that the extreme area is:
$$
A(\x)=2\x f(\x)\tag6
$$
in agreement with the claim.


答:
Using the Distributive Property we get: $$(a \land b \land c) \;\lor\; (a \land \neg b \land c) \;=\; (a \land c) \;\land\; (b \lor \neg b)$$
Where $(b \lor \neg b)$ is a tautology, meaning it always evaluates to 1.

Because 1  is the neutral element of an 'and' operation, we can leave it out, giving us: $$a \land c$$

Title: Logic math question?

Body: How do I shorten "(a and b and c) or (a and negation b and c)"? I feel like the answer is easy but I cant seem to grasp which rule I should use.</p>


问:
Lebesgue measure of a 1D set over 2D space
I'm trying to prove mathematically that if two continuous real independent random variables $X$ and $Y$ have different densities then they are different.

This is what I did :

$P(X(w)=Y(w))$=$P((X,Y)$$\in A)$ with $A=${$(X(w),Y(w))/X(w)=Y(w)$}

Now I calculate $P((X,Y)\in A)$$=\int_{R^2}$$  _A$$d\lambda_{R^2}$ = $\int_{A}$$d\lambda_{R^2}$$=0$

is this right? since the dimension of A is 1 and the the lebesgue measure is in $R^2$

答:
That is too vague an argument.

$P(X=Y)=\int_{\mathbb R^{2}} 1_A d\mu$ where $A=\{(x,x):x \in \mathbb  R\}$ and $\mu (E)=P((X,Y) \in E)$ for all Borel sets $E$ in $\mathbb R^{2}$. By Fubini's Theorem $P(X=Y)=\int \int 1_{\{x\}}(y) f_Y(y)dy f_X(x)dx =\int 0  f_X(x)dx =0$.

I have used the fact that for fixed $x$, $(x,y) \in A$ iff $y=x$.

Incidentally, the result is true even if $X$ and $Y$ do not densities provided $P(X=x)=0$ for all $x$ and $P(Y=y)=0$ for all $y$.


问:
Transforming an operation involving the components of matrixes, ${C}_{i j}=B_{i k} B_{j l} A_{k l}$, into a matrix multiplication
The components of a matrix ${\mathbf{C}}=({C}_{i j})_{n\times n}$ are obtained from the components of two matrices ${\mathbf{A}}=({A}_{i j})_{n\times n}$ and ${\mathbf{B}}=({B}_{i j})_{n\times n}$ in the following way:

$${C}_{i j}=B_{i k} B_{j l} A_{k l} \tag{1}$$

Using Einstein's notation. I need to express $(1)$ as matrix multiplication.

My attempt

I know that for square matrices $n \times n$, if $\mathbf{c}=\mathbf{a}·\mathbf{b}$, then their components will satisfy that
$$c_{i j}= a_{i k} b_{k j}$$

So, in $(1)$, $B_{i k} A_{k l}=X_{il}$ will correspond to the matrix multiplication $\mathbf{B}·\mathbf{A}=\mathbf{X}$, getting that

$${C}_{i j}= B_{j l} X_{il}$$

However, since the second index in $B_{j l}$ and the first index in $X_{il}$ doesn't match, how could this be expressed in terms of a matrix multiplication?

Body: Dovetail indices. Since $C_{\color{red}{i}\color{limegreen}{j}}=B_{\color{red}{i}\color{blue}{k}}A_{\color{blue}{k}\color{brown}{l}}B^T_{\color{brown}{l}\color{limegreen}{j}}$, $C=BAB^T$.


答:
Using your sketch, the point to note is that for $1 \leq x \leq 2$, one of the curves is below $y = 1$ and one of the curves is above $y=1$.

The radius of revolution will be defined by which vertical distance is more from $y = 1$. As you can see from the sketch that for $x = 1$, the distance to $y = \frac{x}{2}$ is more whereas for $x = 2$, distance to $y = \sqrt{x}$ is more. Say that the distance  from $y = 1$ to both curves equal at $x = a$ then the volume of revolution can be written as

$V = \displaystyle \int_0^1 \int_{1-\sqrt x}^{1-\frac{x}{2}} 2 \pi r \ dr \ dx +
\int_1^a \int_{0}^{1-\frac{x}{2}} 2 \pi r \ dr \ dx + \int_a^2 \int_{0}^{\sqrt x - 1} 2 \pi r \ dr \ dx + \int_2^4 \int_{\frac{x}{2} - 1}^{\sqrt x - 1} 2 \pi r \ dr \ dx$

Now to find $a$, equate $1 - \frac{a}{2} = \sqrt a - 1$. Solving it we get $a = 6 - 2 \sqrt5$.

Title: Help! perplexing problem on volume revolution of region about a line

Body:
how do we find the volume when the region bounded by $y = x^{\frac 1 2}$ and $y = \frac x 2$ is revolved about the line $y = 1$? I understand how to solve for the regions above and below $y = 1$, however because of the intersection of volumes from $x = 1$ to $x = 2$, the volume of the entire region both above and below $y = 1$ eludes me.

heres the graphs: https://www.desmos.com/calculator/njvae9suit

I would greatly appreciate an explained solution to solving a problem such as this. Thank you!


答: Using a few laws of boolean algebra, specifically the distributive law and complement law, we get:

Title: Logic math question?


问:
Working out efficiently if points are within polygons for large datasets
Perhaps this question should be elsewhere but as its mainly a mathematical problem Im posting here.

I have a dataset of which looks like this:

```<code>[
  [ 242841.86914496083,
  1090.0027001134586,
  11711.344635666988,
  142639.20305005432 ],

  [ 212841.86914496083,
  128007.0027001134586,
  11711.344635666988,
  142639.20305005432 ]
  ...
]
</code>```

Each array is called an "event" and in this case have 4 "parameters". I want to plot these events on multiple scatter charts. For example, on the first scatter chart, i'll plot the first and second parameter of each event, so the (x, y) point for the first event is (242841.86914496083, 1090.0027001134586), for the 2nd event is (212841.86914496083, 128007.0027001134586) etc. On the second chart, i'll plot the second and third parameter of each event, so the (x, y) point for the first event is (11711.344635666988, 142639.20305005432), for the 2nd event is (11711.344635666988, 142639.20305005432) etc.

In the example Im giving the range is 0 to 262144;

The scatter charts are 200px x 200px but can be any pixel size up to 2000px. It's always square.

When plotting, I want to work out which points are within a polygon. Here is a sample polygon:

```<code>[
    [
        200000,
        100000
    ],
    [
        230000,
        180000
    ],
        [
        200000,
        200000
    ],
    [
        200000,
        230000
    ],
    [
        110000,
        220000
    ],
    [
        100000,
        180000
    ],
]
</code>```

The points within the polygon are to be plotted in a different color to the points outside it. The polygon is to be applied to scatter chart 1 and 2, but it's the first and second parameter of each event that is to be used to figure out if the event is within the polygon and what color it should be on both charts. E.g. say the point for event 2 (212841.86914496083, 128007.0027001134586) are within the polygon, this mean that event will appear in red in each chart.

I have a simple algorithm which runs through each point and checks if it's within the polygon. I use https://www.npmjs.com/package/point-in-polygon.

Keep in mind, there are often multiple polygons (n), and these can be applied to different combinations of parameters. So I need to work out if an event is in all n polygons, or if its in n-1 polygons etc.

To sum up, what I need to know is:

percentage of total events within each polygon

number of events within a polygon

number of events within n, n-1 etc polygons

x value mean of event with polygon

y value mean of event with polygon

My algorithm performs well for up to 10,000 events. But then it hits issues. Some of my users have files that contain 1 millions events, and my algorithms take minutes.

How can I work out what I need more efficiently? Binning data is an option but then I lose accuracy in results. E.g. for a 200 by 200 pixel image I could create 200 bins for each axis. Problem is a then lose accuracy on the x and y mean values. Also, users can change to 201 x 201 pixel images, now I need to work out 201 bins etc.

After file upload, these files are on the server and can be pre-processed before graphs are generated. Is there anything that can be done mathematically to improve performance for large number of events?

EDIT:
To be clear - figuring out if a point is in a polygon is not the most performance intensive. The most performance intensive part is looping through 1 million events in order to plot them on a, say, 400px by 400px image. The solution, probably lies somewhere in binning data - but if I make 400 bins, then the user - on the fly - changes the image size to 410px, I now need to make 410 bins from the ORIGINAL dataset. Is there some way around this?

EDIT
So the data is known in advance. This sits on my server and can be manipulated in anyway. The size of the image and the polygons are now knows in advance - users draw the polygons and the image must change on the fly. With, say, 1 million events, this is not possible - the browser usually crashes...I can bin the data, but if I create 200 bins (standard image is 200 x 200), then the user makes the image bigger e.g. 210 x 210, now I have to cycle through 1 million events and create 210 bins, which is obviously a big performance issue

答:
If the polygon is convex then there is a simple solution.

Every convex polygon is created by the intersection of half-planes.

A point can be determined to lie on a particular side of a half-line (the bounding line of a half-plane) if its line segment with a point known to be on the 'outside' fails or succeeds to intersect with the half-line.

If a point is 'inside' all the half-lines of the bounding (convex) polygon, it is inside the polygon.


问:
Proof the Existence of an Integer given $a^2 x \equiv a \pmod m$
In my final abstract algebra exam, I had the following problem I can't solve completely yet. It states the following:

Given $a \in \mathbb Z$, (a) given p is prime, if there exists $x \in \mathbb Z$ such that $a^2 x \equiv a \pmod  p$ then there exists $z \in \mathbb Z$ such that $a^2 z \equiv a \pmod p$ and $az^2 \equiv z \pmod  p$. (b) Same but substituting $p$ with $p^r$. (c) Same but substituting $p$ with any arbitrary number $m$.

I had (a) correct cause is almost trivial; as we are working in $\mathbb F_p$, then $a$ is either unity or zero. If $a=0$, $z=0$ would do the job. Otherwise, we get that $ax \equiv 1 \pmod p$. From this expression, we can just multiply by $x$ and we get $ax^2 \equiv x \pmod  p$. The first equation($a^2 x \equiv a \pmod p$ is given by hypothesis).

For (b), I followed a similar argument. If $a$ is not a power of $p$, as $p$ is prime then $\gcd(p^r,a)=1$ hence $a$ is unity and we can proceed like in (a). If $a$ is not unity, then $a=p^s$. I'm not sure of the next steps; I see then that $p^{2s} x \equiv p^s \pmod  {p^r}$. As $p$ is prime, $x$ needs to be a power of $p$ hence $p^{2s} x \equiv p^s \pmod  {p^r}$. I deduce that $x=p^{rk - s}$. From here, you get that $p^s * p^{2rk-2s} \equiv p^{2rk-s} \equiv p^{-s} \equiv x$. However, I'm not sure of the argument. Moreover, I don't know how to proceed to an arbitrary integer. I would thank any suggestion or hint.

答:
for answer (a): when ${ax=1 (mod p)}$,  where ${x \equiv a^{-1} (mod p)}$, now say ${a^{-1}=t}$, now the set ${t+np}$, contains the required set, and ${z}$ can be any element from this set.

for ex. p=5 and then a=3 , ${a^{-1}=2}$, so now, x= 2+5n contains from a set {2,7,12,17,22,27,32,37,42,47,52,57,62,67,72,77...}  = {(n2,n7)...}  nice pair of 2 and 7!

now for (b) , ${p^r}$you will still have the subset from above bigger set. for ex.. for ${p^2}$ , the set is {27,52,77..}

for (c) : As ${p^r}$ is composite for r>1, then, it can be prove for any composite m.


问:
Solve $\int_0^\infty f(x) \cos(kx) \, dx=\frac {sin(ak)} {k}$ with using Fourier transform
$$\int_0^\infty f(x) \cos(kx) \, dx=\frac {\sin(ak)} {k}$$

From the table fourier cos transform of Heasivide function is $\frac {\sin(ak)}{k}$

Let $f(x)\begin{cases}1, &  0\leq x \leq1
\\
0 & x > 1 \end{cases}$

$$\ F_c(f(x))= \int_0^\infty f(x)\cos(kx) \, dx =  \int_0^1 1\ \cos(kx)\ \text{d} x + 0 = \left[\frac{\sin(kx)}k\right]^1_0 = \frac{\sin(ak)}{k}$$

(not sure of this steps).

without using transform table, How can we get $f(x)=H(a-x)$?

答:
Note that your last calculation is wrong, since you pop out a factor $a$ which as not present before.

Heaviside Theta in your case is defined as

$$\theta(a - x) = \begin{cases} 1 & 0 \leq x \leq a \\
0 & x > a
\end{cases}$$

Whence

$$\int_0^{+\infty} \theta(a-x)\cos(kx) \text{d}x = \int_0^a \cos(kx)\ \text{d}x = \frac{\sin (a k)}{k}$$


问:
How to derive the cartesian equation of cycloid expressed for y?
I found an expression for x in wiki: https://proofwiki.org/wiki/Equation_of_Cycloid_in_Cartesian_Coordinates but I need it expressed for y. How can I do that?</p>

答:
It cannot be done with only elementary functions. The closest you can do is define function
$$ F(\theta) = \theta - \sin\theta $$
and use its inverse $F^{-1}$ (but that inverse cannot be expressed using only elementary functions and algebraic operations, see Inverse of $f(x)=\sin(x)+x$ ). We have then
$$ y(x) = a\Big(1-\cos \big( F^{-1}(\frac{x}{a}) \big)\Big) $$
It may be still useful, depending on what do you want to obtain from it, because although $F^{-1}$ cannot be expressed using elementary functions, it can be still examined using the tools of mathematical analysis.


问:
Question about a step in Prop 7.50 in Milne's Algebraic Number Theory

This question talks about the correspondence between finite unramified extensions of $K$, a complete field with a discrete, nonarchimedean absolute value, that are contained in an algebraic extension $L$ of $K$ and finite subextensions of the tower of residue fields $k \subset l$.

So Milne starts by considering a finite subextension $k' \supset k$ of the residue field of $k$ and writes it as $k[a]$ for some $a$. Now he uses Newton's Lemma (which was proved along with Hensel's Lemma) to argue that there should be an $\alpha \in L$ such that $f(\alpha)=0$ and $\alpha \equiv a \mod \mathfrak{p}$.

My issue is this: why should such an $\alpha$ exist in L? Since L is an arbitrary algebraic extension (not necessarily finite), it doesn't have to be complete under the extension of the absolute value and so I'm not sure why Newton's Lemma applies. Also, since L is not necessarily algebraically closed, why should the extension $K[x]/(f)$ embed into L (where f is a lift of the minimal polynomial of the primitive generator k')?

Am I missing something obvious or is there a mistake here?

You can look at the proposition and its proof in page 127 in Milne's notes.

EDIT:

I also took a look at these notes from 18.785 (page 1-2) where they consider $K[X]/(f)$ but they say that by construction it is true that the valuation ring of this is $A[\alpha]$. I don't see why this should be so. Could someone help with this as well?

答:
My issue is this: why should such an $\alpha$ exist in L? Since L is an arbitrary algebraic extension (not necessarily finite), it doesn't have to be complete under the extension of the absolute value and so I'm not sure why Newton's Lemma applies.

I agreed with you that there seems to be something wrong when $L$ is infinite.

However, the argument for existence of unramified $K'\subset L$ whose residue is $k'=k(a)$ can be rewritten following MIT course notes: Pick a lift $\alpha$ of $a\in k'$ to $B$, valuation ring of $L$. Let $g\in A[x]$ be minimal polynomial of $\alpha$ over $L$ then $K(\alpha)\cong K[x]/(g)$ is a separable finite subextension of $L$. Then you proceed to show $K'=K(\alpha)$ is unramified as in the MIT notes.

Also, since L is not necessarily algebraically closed, why should the extension $K[x]/(f)$ embed into L (where f is a lift of the minimal polynomial of the primitive generator k')?

When people write it like this, $K[x]/(f)\subset L$ would mean $K(\alpha)\subset L$ where $\alpha\in L$ with minimal polynomial $f\in K[x]$. I think they
do this in this situation because it is safe to do so.

I also took a look at these notes from 18.785 (page 1-2) where they consider $K[X]/(f)$ but they say that by construction it is true that the valuation ring of this is $A[\alpha]$. I don't see why this should be so. Could someone help with this as well?

Once we view $K[x]/(f)$ and $K(\alpha)$ to be the same object $K'$, as discussed above, then its valuation ring is $A[\alpha]$ or $A[x]/(f)$ because it is algebraic extension of $A$ in $K'$, and as it is of degree $n=[K':K]$, so it is precisely algebraic closure of $A$ in $K'$, i.e. valuation ring of $K'=K[x]/(f)$.


问:
Using the function $f'\left( x \right) = \left( {{e^{\left( {f\left( x \right) - g\left( x \right)} \right)}}} \right)g'\left( x \right)$
Let $f:R \to R\ $ and  $g:R \to R$ be two non-constant differentiable function. If $f'\left( x \right) = \left( {{e^{\left( {f\left( x \right) - g\left( x \right)} \right)}}} \right)g'\left( x \right)$  for all $x \in R$ and $f(1)=g(2)=1$, then which of the following statement is(are) true

(A) $f\left( 2 \right) < 1 - {\log _e}2$

(B) $f\left( 2 \right) > 1 - {\log _e}2$

(C) $g\left( 2 \right) > 1 - {\log _e}2$

(D) $g\left( 2 \right) < 1 - {\log _e}2$

My approach is as follow

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

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

$\int {{e^{ - f\left( x \right)}}f'\left( x \right)dx}  = \int {{e^{ - g\left( x \right)}}g'\left( x \right)dx}  + c$

$t = {e^{ - f\left( x \right)}} \Rightarrow dt =  - {e^{ - f\left( x \right)}}f'\left( x \right)dx\& u = {e^{ - g\left( x \right)}} \Rightarrow du =  - {e^{ - g\left( x \right)}}g'\left( x \right)dx$

$ - \int {dt}  =  - \int {du}  + c \Rightarrow  - t =  - u + c$

$ \Rightarrow  - {e^{ - f\left( x \right)}} =  - {e^{ - g\left( x \right)}} + c \Rightarrow  - {e^{ - f\left( 1 \right)}} =  - {e^{ - g\left( 1 \right)}} + c \Rightarrow  - {e^{ - 1}} =  - {e^{ - g\left( 1 \right)}} + c \Rightarrow c = {e^{ - g\left( 1 \right)}} - {e^{ - 1}}$

$ \Rightarrow  - {e^{ - f\left( 2 \right)}} =  - {e^{ - g\left( 2 \right)}} + c \Rightarrow c = {e^{ - g\left( 2 \right)}} - {e^{ - f\left( 2 \right)}} \Rightarrow c = {e^{ - 1}} - {e^{ - f\left( 2 \right)}}$

On comparing we get $ \Rightarrow {e^{ - g\left( 1 \right)}} - {e^{ - 1}} = {e^{ - 1}} - {e^{ - f\left( 2 \right)}} \Rightarrow {e^{ - g\left( 1 \right)}} + {e^{ - f\left( 2 \right)}} = 2{e^{ - 1}}$.

How do we procced this question

答:
$2e^{-1}=e^{-g(1)}+e^{-f(2)}>e^{-f(2)}$. Hence, $\ln 2 -1 >-f(2)$ which gives $f(2) >1-\ln 2$. So (B) is true which makes (A) false.For (C) and (D) just use the hypothesis that $g(2)=1$.

[I have written $\ln x$ for $log_2(x)$].


问:
Unable to understand option 2 of this question
Let $A$ and $B$ be events satisfying $P(A∩B)$ = $P(A)$ + $P(B)$.

Which of the following statements is/are true?

If $A$≠ $B$, then $A$ and $B$ are mutually exclusive.

If $A$ and $B$ are independent, then $P(A) = 0 $ or $P(B) = 0$.

If $P(A) > P(B) > 0$, then $A$ and $B$ are not independent.

If $A$ = $S$, the sample space, then $P(B) = 0$.

By process of elimination:

Option $1$ is correct since if $A$ and $B$ are mutually exclusive, $P(A ∩ B) = 0$.

Option $3$ is correct as if $P(A)>0$ and $P(B)>0$, $P(A)*P(B)>0$ meaning $P(A ∩ B) >0$. This is a contradiction thus not independent.

Option $4$ is correct as $P(B) > 0$ the initial statement given would be false.

Option $2$ is the one I'm having difficulty understanding, is it vacuously true(is logic language even used here or is it a normal if-then statement?) since it's established that $A$ and $B$ are exclusive?

答:
This seems more like an exercise to check the knowledge of definitions. I hope I am using the same definitions as you.

Option 1
Firstly, I believe that your argument for option 1 is not valid. Events are mutually exclusive if the intersection of both of them (as a subset of your probability space)  is empty. This indeed implies $P(A\cap B)=0$. But you need for your argument the opposite implication.

Now if you work over countable probability space, what I wrote in the previous paragraph is just some unnecessary technicality. But if you are in uncountable probability space things get more funnier.  Because intersection can be non-empty but still of measure zero.

Option 2 This option is indeed correct. But I would not say it is vacuously correct. Though if you have a mutual exclusiveness it is indeed easy, because you have $0=P(A\cap B)=P(A)+P(B)$.

Without the assumption of mutual exclusiveness, you go by $P(A\cap B)=P(A)P(B)$ by assumption of independence. Hence you have $ P(A)P(B)=P(A)+P(B)$.

Then you have two options. Firstly $P(A)=P(A)P(B)=P(A)+P(B)$ this directly implies $P(B)=0$ and also $P(A)=0$.  Or $P(A)<P(A)P(B)=P(A)+P(B)$ but this is contradiction.


问:
find the maximum value of $m$ where $3^m$ be a factor of $A_{1395}$.

Consider the number $A_{1395}=\underbrace{333\cdots3}_{1395}$. what is
the maximum value of $m$ where $3^m$ be a factor of $A_{1395}$.

$a) 2\quad\quad b)3\quad\quad c)4\quad\quad d)5\quad\quad e)6$

Here is my attempt :

We have $A_{1395}=3\times \underbrace{111\dots1}_{1395}$ sum of the digits of $\underbrace{111\cdots1}_{1395}$ is $1395$ which is a multiply of $9$. therefor this number is dividable to $9$. so we conclude $A_{1395}$ has the factor $3^3$ . to see whether it has more factor of $3$ or not I divided it to $27$. after doing some steps of long division I got:

$$\underbrace{11\cdots1}_{1395}÷27=4115226\cdots$$
Unfortunately I couldn't see any pattern as the result. so I don't know how to check this number has more factor of $3$ or not.

答:
We write the number as ${10^{1395}-1\over 3}$

By Euler Thorem,
$10^{162}\equiv1\pmod{243}$ so $10^{1395}\equiv10^{99}\pmod{243}$

Since $10^{99} \equiv 28^{33}\equiv82^{11}\equiv(-80)^5\times82\equiv82^2\times(-80)\times82\equiv(-80)^2\times82\equiv-80\equiv163\pmod{243}$

We know $10^{99}-1\equiv162\pmod{243}$ which implies $10^{1395}-1\equiv 162\pmod{243}$

Therefore $10^{1395}-1$ is not a multiple of $243$ but is a multiple of $81$.

So ${10^{1395}-1\over 3}$ is a not multiple of $81$ but is a multiple of $27$


问:
Stereographic projection of "unit sphere" in a Banach space
So I was wondering, if it is possible to extend the idea that a sphere (minus one point) is isomorphic to a plane to the case of an infinite dimensional Banach (or Hilbert if needed) space.
Suppose we have a $L^2$ space, then consider the unitary sphere $U$ , namely the family of function in $L^2$ that have unitary norm. Is there an analog of the Stereographic projection in this case?

Thanks in advance!

答:
Hilbert seems to be needed.  Consider a unit vector $n$ and let $S$ be the sphere with center $n$ and radius $1$.  The straight line through the north pole $2n$  and a point $f$ on the sphere is given by
$$x=2n+t(f-2n).$$
To calculate the intersection of the straight line and the orthogonal complement of $\{n\}$ solve $\langle x,n\rangle=0$ for $t$.
Hence the stereographic projection $P$ from $S\setminus\{2n\}$ to the orthogonal complement of $\{n\}$ is given by
$$
P(f)=\frac{2}{2-\langle n,f\rangle}(f-\langle n,f\rangle n).
$$
Furthermore, the inverse mapping is easily computed to
$$
P^{-1}(f)=\frac{2}{4+\|f\|^2}(\|f\|^2n+2f)
$$
again using a straight line through $2n$ and $f$; this time we assume $n\perp f$.
Works in any Hilbert-space.


答:
Is (b)  (so also (c)) false? Take $a=2, p=2$ and $r=2.$ The equation  $$a^2x\equiv a \mod p^r$$ would be $$4x\equiv 2 \mod 4,$$ but $4x$ is just $0$ so there's no solution... or am I missing something?

Also, I think your solution for (a) is correct.

Title: Proof the Existence of an Integer given $a^2 x \equiv a \pmod m$

Body:
In my final abstract algebra exam, I had the following problem I can't solve completely yet. It states the following:

Given $a \in \mathbb Z$, (a) given p is prime, if there exists $x \in \mathbb Z$ such that $a^2 x \equiv a \pmod  p$ then there exists $z \in \mathbb Z$ such that $a^2 z \equiv a \pmod p$ and $az^2 \equiv z \pmod  p$. (b) Same but substituting $p$ with $p^r$. (c) Same but substituting $p$ with any arbitrary number $m$.

I had (a) correct cause is almost trivial; as we are working in $\mathbb F_p$, then $a$ is either unity or zero. If $a=0$, $z=0$ would do the job. Otherwise, we get that $ax \equiv 1 \pmod p$. From this expression, we can just multiply by $x$ and we get $ax^2 \equiv x \pmod  p$. The first equation($a^2 x \equiv a \pmod p$ is given by hypothesis).

For (b), I followed a similar argument. If $a$ is not a power of $p$, as $p$ is prime then $\gcd(p^r,a)=1$ hence $a$ is unity and we can proceed like in (a). If $a$ is not unity, then $a=p^s$. I'm not sure of the next steps; I see then that $p^{2s} x \equiv p^s \pmod  {p^r}$. As $p$ is prime, $x$ needs to be a power of $p$ hence $p^{2s} x \equiv p^s \pmod  {p^r}$. I deduce that $x=p^{rk - s}$. From here, you get that $p^s * p^{2rk-2s} \equiv p^{2rk-s} \equiv p^{-s} \equiv x$. However, I'm not sure of the argument. Moreover, I don't know how to proceed to an arbitrary integer. I would thank any suggestion or hint.


答: Answer for (2): The given equation gives $(1-P(A)) (1-P(B))=1$. If either $P(A) \neq 0$ or $P(B) \neq 0$ then the left is a product of two numbers one of which is less than $1$ and the other $\leq 1$ leading to a contradiction. Hence we must have $P(A)=P(B)=0$.

Title: Unable to understand option 2 of this question

Body:
Let $A$ and $B$ be events satisfying $P(A∩B)$ = $P(A)$ + $P(B)$.

Which of the following statements is/are true?

If $A$≠ $B$, then $A$ and $B$ are mutually exclusive.

If $A$ and $B$ are independent, then $P(A) = 0 $ or $P(B) = 0$.

If $P(A) > P(B) > 0$, then $A$ and $B$ are not independent.

If $A$ = $S$, the sample space, then $P(B) = 0$.

By process of elimination:

Option $1$ is correct since if $A$ and $B$ are mutually exclusive, $P(A ∩ B) = 0$.

Option $3$ is correct as if $P(A)>0$ and $P(B)>0$, $P(A)*P(B)>0$ meaning $P(A ∩ B) >0$. This is a contradiction thus not independent.

Option $4$ is correct as $P(B) > 0$ the initial statement given would be false.

Option $2$ is the one I'm having difficulty understanding, is it vacuously true(is logic language even used here or is it a normal if-then statement?) since it's established that $A$ and $B$ are exclusive?


Title: find the maximum value of $m$ where $3^m$ be a factor of $A_{1395}$.

Body:
Consider the number $A_{1395}=\underbrace{333\cdots3}_{1395}$. what is
the maximum value of $m$ where $3^m$ be a factor of $A_{1395}$.

$a) 2\quad\quad b)3\quad\quad c)4\quad\quad d)5\quad\quad e)6$

Here is my attempt :

We have $A_{1395}=3\times \underbrace{111\dots1}_{1395}$ sum of the digits of $\underbrace{111\cdots1}_{1395}$ is $1395$ which is a multiply of $9$. therefor this number is dividable to $9$. so we conclude $A_{1395}$ has the factor $3^3$ . to see whether it has more factor of $3$ or not I divided it to $27$. after doing some steps of long division I got:

$$\underbrace{11\cdots1}_{1395}÷27=4115226\cdots$$
Unfortunately I couldn't see any pattern as the result. so I don't know how to check this number has more factor of $3$ or not.

Body:
For $x\in\Bbb N$, letr $v_3(x)$ be the exponent of the maximal power of $3$ dividing $x$.

If $n=3^ka+1$ with $k\ge1$ and $3\nmid a$, then
$$ n^3=3^{3k}a^3+3^{2k+1}a^2+3^{k+1}a+1
=3^{k+1}(a+3^k(\ldots))+1$$
i.e.,

If $v_3(n-1)>0$ then $v_3(n^3-1)=1+v_3(n-1)$.

Note that $A_{1395}=\frac{10^{1395}-1}3$, so what we are looking for is $m=v_3(10^{1395}-1)-1$.
By the above remark,
$$m=v_3(10^{1395}-1)-1=v_3(10^{465}-1) = v_3(10^{155}-1)+1. $$
Also, as $v_3(10^3-1)=1+v_3(10-1)=3$, so that
$$ 10^{155}=(10^3)^{51}\cdot 100\equiv 100\equiv 19\pmod{27}$$
(but $\equiv 1\pmod 9$). We conclude
$$v_3(10^{155}-1)=2 $$ and ultimately
$$ v_3(A_{1395})=3.$$


问:
Is this possible to solve?
Is this possible to solve the length of $AB$ while only knowing that $l_1$ and $l_2$ are two parallel lines?

Example with geogebra

答:
This question is tricky!

While it is tempting to use the Intercept Theorem, and calculate $AB$ from the following equation:

$$\frac{DB}{DE}=\frac{AB}{FE}$$

This would yield a wrong answer. Raffaele's answer explains why very clearly.

Title: Is this possible to solve?


问:
A question about a limit
I have the next $\lim_{x \to \frac{\pi}{4}} \frac{f(x)-f(\frac{\pi}{4})}{x-\frac{\pi}{4}}$ when the function is $$f(x)=\frac{x+\sin(x)}{\tan(x)}.$$
I don't know how to even start. I am sorry that this is short but I really don't know.

答:
$f$ is said to be differentiable at $c$, if there exists a real number $\alpha$ such that
$$f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=\alpha.$$
Now see your question.

Find $f'$ and then put $x=\frac{π}{4}.$

Note that $\frac{d}{dx}f=f'.$
If $$f(x)=\frac{h(x)}{g(x)}$$, then $$f'(x)=\frac{g(x)\frac{d}{dx}h(x)-h(x)\frac{d}{dx}g(x)}{(g(x))^2}.$$

Title: A question about a limit


答:
Write using the binomial theorem,
\begin{align}
999\cdots 9 &= 10^{1395} -1 \\
&= (1+9)^{1395} -1 \\
&= \left(\array{1395\\1}\right)\cdot 9 + \left(\array{1395\\2}\right)\cdot9^2 + \cdots 9^{1395}
\end{align}
But $1395 = 3^2 \cdot 5 \cdot 31$.  So, the first term is divisible by $3^4$ and each term thereafter is divisible by $3^5$ meaning the sum is divisible $3^4$ and not by $3^5$.  Accordingly, $333\cdots 3$ is divisible by $3^3$ but not $3^4$.  The answer is therefore $3$.


答:
For (b): If $s=r$ we get that $a\equiv0$ and we can take $z=0$ as you noted for the first part. Now assume that $1<s<r$. Then as $p^{2s}x\equiv p^s\mod p^r$ we get $p^r\mid (p^{2s}x-p^s)=p^s(p^sx-1)$. Since $s<r$ this implies $p\mid p^sx-1$ which is a contradiction as $p\nmid 1$. So if $\gcd(p^r,a)\ne1$ then the only possibility for $a^2x\equiv a$ to be satisfied is if $a\equiv0\mod p^r$.

For (c): Let $m=p_1^{r_1}\dots p_k^{r_k}$ and let $a^2x\equiv a\mod m$. This implies $a^2x\equiv a\mod p_i^{r_i}$ for all $i$. Hence there are $z_i$ that satisfy

$a^2z_i\equiv a\mod p_i^{r_i}$

$az_i^2\equiv z_i\mod p_i^{r_i}$

By the Chinese remainder theorem there is some integer $z$ such that $z\equiv z_i\mod p_i^{r_i}$. Now the congruences $a^2z\equiv a$ and $az^2\equiv z$ are satisfied modulo the $p_i^{r_i}$. It follows that they are also satisfied mod $m$.


问:
$\cot 30^\circ = \cot 20^\circ - \operatorname{cosec} 80^\circ$ (?)
Prove that,
$\cot 30^\circ = \cot 20^\circ - \operatorname{cosec} 80^\circ$

I could not do that by trying heart and soul. Please solve it.

Body:
$$\cot20-\cot30=\dfrac{\sin(30-20)}{\sin20\cdot\sin30}$$

Use $\sin20=\sin(2\cdot10)=?$

and $\cos10=\sin(?)$


答:
This is just a comment, but too long to be written in the comments section.

As I like to proceed when meeting those nasty things, I would suggest you to use Dirac's Elegance.

What I mean is that the integrand function is full of large numbers and patterns which could be written in a shorter way. For example, if you just name $a_0, \ldots, a_{19}$ the first terms and $b_0, \ldots, b_{23}$ the next terms, the first two parts of the integrand function can be written as

$$a_0 + a_1q + a_2q^2 + \ldots + a_{19}q^{19} = \sum_{k = 0}^{19} a_kq^k$$

$$b_0\log(q) + b_1q \log(q) + \ldots + b_{23}q^{23}\log(q) = \log(q)\sum_{k = 0}^{23} b_kq^k$$

The other parts I leave them to you to be written in a compact elegant way.

When you will have finished, please rewrite the whole question by making use of this form so we could go on by investigating the eventual absolute convergence and split into $N$ parts the integrand and make try to get out of this tunnel!

Title: Nasty integrals involving polylogarithms

Body:
Is there a general method how to solve integrals including polylogarithms? I have, for example, a following peculiar integral:

$$\int_0^1 \frac{\log (1-q)}{(1-q)^{19}} \big[19122427017242392409179196696957 -2335312823676226739258108857200 \pi ^2-290355604888434431371406862640901 q+32737624286330384009900385643800 \pi ^2
   q+2089178147864338015378359942349146 q^2-217908181258375090989657061033200 \pi ^2 q^2-9513735107311961105716228372823832 q^3+923058652929127659169477549327800 \pi ^2
   q^3+30860609840601979343569796243664360 q^4-2809234642150007839007261169271200 \pi ^2 q^4-75991613070307679674084684299763488 q^5+6560138082212689647470512926380400 \pi ^2
   q^5+147604422934589574241111368769031514 q^6-12225064353406525792601847884728800 \pi ^2 q^6-231661749521735051694587401454488374 q^7+18614430483751971866343908315754000 \pi
   ^2 q^7+298158588677298781555571530676473698 q^8-23467372646806215931232745831715200 \pi ^2 q^8-317258462849638359532191702700757050 q^9+24645135112158586773581379231285600
   \pi ^2 q^9+279929745859987197112521991075600990 q^{10}-21578983058687084626530891022466400 \pi ^2 q^{10}-204525346750399538308156387121441190
   q^{11}+15702582554471282285512462526560800 \pi ^2 q^{11}+123049277403587430345156505609140720 q^{12}-9430354354307541344225967691260000 \pi ^2
   q^{12}-60342329966544036497302398449157240 q^{13}+4622406501952432638430248014413200 \pi ^2 q^{13}+23738252991464115594141538592264430
   q^{14}-1818869679546039116051640669453600 \pi ^2 q^{14}-7312550577333389628853566957018834 q^{15}+560636064795229527948416643193200 \pi ^2
   q^{15}+1699354671638585927650255051925577 q^{16}-130382484017845412425019267350800 \pi ^2 q^{16}-280173803150789489806907495704617 q^{17}+21513060264068052978453696664200 \pi
   ^2 q^{17}+29212691355365195449956724538624 q^{18}-2244803977761215443782290650800 \pi ^2 q^{18}-1448394095369360889798167890490 q^{19}+111380159353759803311707564200 \pi ^2
   q^{19}-864839351159251344060440256000 \pi ^2 \log (q)+9614284037777341680385275648000 \pi ^2 q \log (q)-48596947298066064189458694528000 \pi ^2 q^2 \log
   (q)+149016194597643828944407014096000 \pi ^2 q^3 \log (q)-311515153604740784777334800544000 \pi ^2 q^4 \log (q)+472522854921446523354235979904000 \pi ^2 q^5 \log
   (q)-540131604846004142202578675328000 \pi ^2 q^6 \log (q)+476835369490723415021630300520000 \pi ^2 q^7 \log (q)-330431763957397218290451028752000 \pi ^2 q^8 \log
   (q)+181570501304738210584142319096000 \pi ^2 q^9 \log (q)-79469952891751512800087794416000 \pi ^2 q^{10} \log (q)+27618312749376667202083351152000 \pi ^2 q^{11} \log
   (q)-7469792604527284717674468480000 \pi ^2 q^{12} \log (q)+1441811254291252370952714288000 \pi ^2 q^{13} \log (q)-47548846798971176802583440000 \pi ^2 q^{14} \log
   (q)-242992965510346357364274216000 \pi ^2 q^{15} \log (q)+315652971889245500921659248000 \pi ^2 q^{16} \log (q)-286658885174869914970791528000 \pi ^2 q^{17} \log
   (q)+190547414679020492013416496000 \pi ^2 q^{18} \log (q)-91723410310948362363008592000 \pi ^2 q^{19} \log (q)+31251841079492830435486848000 \pi ^2 q^{20} \log
   (q)-7178196236973842421012672000 \pi ^2 q^{21} \log (q)+1000683546793492378825008000 \pi ^2 q^{22} \log (q)-64152027504348590478672000 \pi ^2 q^{23} \log (q)+116396280 \log
   (1-q) \left((-1+q)^2 \left(-257586022317616038442244+3541119036337161972270259 q-22985344007557567401249245 q^2+94080232161007231115696965 q^3-272947169011867587813604265
   q^4+596855442237497586986976649 q^5-1019309093239565006695750169 q^6+1388155171326881304329317345 q^7-1524633136813377308169137885 q^8+1356305480530438789865344115
   q^9-975804898738457785419837691 q^{10}+563741882769447393607104851 q^{11}-257984481217331023678963375 q^{12}+91454555996476460965835495 q^{13}-24229819048292743306257775
   q^{14}+4517255786475215854979111 q^{15}-528777243429852450045991 q^{16}+29245925849336575939850 q^{17}\right)+17463600 \left(2552782515875592-28378884658059936
   q+143445643657566096 q^2-439857339564018222 q^3+919512319237712508 q^4-1394765491161108528 q^5+1594329068484771696 q^6-1407494920941509015 q^7+975349269884284614
   q^8-535949249427887597 q^9+234574786644003962 q^{10}-81522129873123914 q^{11}+22048899523933360 q^{12}-4255854634983666 q^{13}+140351921539830 q^{14}+717252508230687
   q^{15}-931726090681386 q^{16}+846143031204471 q^{17}-562446780417522 q^{18}+270743829850494 q^{19}-92247367548336 q^{20}+21188182319304 q^{21}-2953759514706
   q^{22}+189360225054 q^{23}\right) \log ^2(q)\right)+2032698075408000 \left(2552782515875592-28378884658059936 q+143445643657566096 q^2-439857339564018222
   q^3+919512319237712508 q^4-1394765491161108528 q^5+1594329068484771696 q^6-1407494920941509015 q^7+975349269884284614 q^8-535949249427887597 q^9+234574786644003962
   q^{10}-81522129873123914 q^{11}+22048899523933360 q^{12}-4255854634983666 q^{13}+140351921539830 q^{14}+717252508230687 q^{15}-931726090681386 q^{16}+846143031204471
   q^{17}-562446780417522 q^{18}+270743829850494 q^{19}-92247367548336 q^{20}+21188182319304 q^{21}-2953759514706 q^{22}+189360225054 q^{23}\right) \log (q) \text{Li}_2(1-q)\big] \mathrm dq$$


问:
Topology-Continuous functions and compact spaces

If $f$ is a continuous function from $X$ to $Y$ and $A$ is a subset of $X$ then is it true that $x \in A^d$ implies $f(x) \in (f(A))^d$? Here $A^d$ is the derived set of $A$.

If $f$ is a continuous function from $X$ to $Y$ and $X$ is a compact space,then show that $f$ is bounded and attains its bounds.

$A$ and $B$ are two compact subsets of a Hausdorff space $Y$.

i. Union of $A$ and $B$ is compact in $Y$:

$A$ and $B$ are closed and hence the union is closed,so it is compact.

ii. Intersection of $A$ and $B$ is compact in $Y$:

$A$ and $B$ are closed and hence the intersection is closed,so it is compact.

iii. $\operatorname{fr}(A)$ is compact:

$\operatorname{fr}(A)$ is the intersection of $\operatorname{cl}(A)$ and $\operatorname{cl}(Y-A)$. Intersection of two closed sets, hence compact.
    Am I correct?

答:
1.) If $A$ is a subset of $X$ which is mapped to a single closed point in $Y$, then $(f(A))'$ is empty.

Question 2.) only makes sense for a function $f:X\to\mathbb R$. If $X$ is compact, then by continuity of $f$ the image is also compact, hence closed and bounded. So the supremum of $f(X)$ is actually the maximum and there is an $x\in X$ which is mapped to this maximum.

3.) The union of finitely many compact sets is always compact.

In a Hausdorff space compact sets are closed, so the intersection of $A$ and $B$ is a closed subset of the compact set $A$, thus compact.

The boundary is the intersection of cl$(A)$ and cl$(Y-A)$, hence a closed subset of the compact set $A$ and therefore also compact.


问:
Neat proof to show that $\sup\{x\; : P(X<x)< \lambda\}=\inf\{x\; : P(X\leq x)\geq \lambda\}$
Consider a probability space $(\Omega,\mathcal{F},P)$ and some real-valued random variable $X$. I want to show that $\sup\{x\; : P(X<x)< \lambda\}=\inf\{x\; : P(X\leq x)\geq  \lambda\}$ for any $\lambda \in (0,1)$.

My idea:

By monotonicity of the probability measure: for any $x \in \mathbb R$ we have:

$P(X<x)\leq P(X\leq x)$ and consider that sets

$$ A:=\{x:P(X< x)< \lambda\}\;\;, B:= \{x:P(X\leq x)\geq \lambda\}$$

For any $x^{*}\in A\cap B$, we have:

$$ P(X<x^{*})< \lambda \leq P(X\leq x^{*})$$ such that$P(X=x^{*})>0$ and $x^{*}$ satisfies:

$$\sup\{x\; : P(X<x)< \lambda\}=x^{*}=\inf\{x\; : P(X\leq x)\geq  \lambda\} $$

Now assume that $A\cap B= \varnothing$, then $x^{a} \in A$ and $x^{b}\in B$ such that $x^{a}<x^{b}$ such that $\sup A\leq \inf B$. Furthermore, since $A\cap B= \varnothing$, we can say that there exists $x$ such that $P(X\leq x)=\lambda$

I feel like I may be making this over-complicated looking at different cases.

答:
For $x<\sup A$, there exists $x'\in A$ with $x'>x$ and hence $P(X\le x)\le P(X<x')<\lambda$, i.e., $x\notin B$.
For $x>\sup A$, there exists $x'<x$ with $x'\notin A$, hence $P(X\le x)\ge P(X<x')\ge \lambda$, i.e., $x\in B$. Thus with $s:=\sup A$, we find that either $B=(s,\infty)$ or $B=[s,\infty)$. At any rate $\inf B=s$.


问:
Time complexity for checking if a matching is a maximum matching in bi-partite graph
Let $\ M $ be a matching in a simple bi-partite undirected graph $\ G = (V,E) $ where $\ |V| = n $ and $\ |E| = m $ . The graph is represented by adjacency list. I need to prove or disprove that it can be checked wether $\ M $ is a maximum matching in $\ \Theta(n+m) $

I was thinking of turning the graph into a network with source and sink and then find minimum cut but it will be a problem computing the time complexity of such algorithm (if there's any?)

答:
I'll give some hints without details. A matching $M$ is a maximum matching if and only if the graph has no alternating paths (with respect to $M$) ending in unmatched vertices. You can find such a path in linear time. If you prefer to think in terms of networks, you should use the fact that the flow is maximum if and only if there are no augmenting paths. I.e. you only have to run one iteration of the Edmonds-Karp algorithm to check whether a given flow is maximum.

EDIT: A standard way of finding an augmenting path in a bipartite graph $G = (A,B,E)$ is to orient the edges in $M$ from $A$ to $B$ and the rest of the edges from $B$ to $A$. Let $X,Y$ denote the set of unmatched verties in $A$ and $B$, respectively. Then the problem is to decide whether any vertex in $X$ is reachable from a vertex in $Y$ by a directed path. You can solve this by using standard graph traversal algorithms (e.g. BFS). A small trick to make this run in linear time is to add a new root vertex $v$ to the graph, along with edges $vy$ for every $y \in Y$ and start the search from this vertex.


答:
The answer is affirmative for the unit sphere of a Hilbert space. The point I would like to make is that the stereographic projection is essentially a one-dimensional mapping, which only involves one polar coordinate. Thus, dimensionality does not really matter.

Remark. On the other hand, for a general Banach space, non-Hilbert, the stereographic projection is not generally available. I have in mind the example of $\mathbb R^2$ with the $\lVert\cdot\rVert_\infty$ norm, for which the unit sphere is the unit square. I don’t see how to stereographically project that; the straight faces of the square are surely going to pose problems. These straight faces are a manifestation of the lack of strict convexity of $(\mathbb R^2, \lVert\cdot\rVert_\infty)$.

Consider the usual stereographic projection of $P\in \mathbb S^2$ onto $\mathbf x\in \mathbb R^2$, but let us reason in polar coordinates, with the physicist’s convention that $\theta\in [0, \pi]$ and $\phi\in [0, 2\pi)$. The point
$$
P=(\cos\phi\sin \theta, \sin \phi \sin \theta, \cos \theta)$$
is mapped onto the point
$$
\mathbf x=(r\cos \phi, r \sin \phi, 0),\quad r\in [0,\infty),$$
where
$$\tag{1}
\tan\frac\theta2 = r.$$
See picture.

Writing the stereographic projection in this way, it becomes manifest that the mapping is essentially one-dimensional.

So let us replace $\mathbb S^2$ with the unit sphere $S$ of a Hilbert space $H$, where we choose a unit vector $e_1$, which will play the role of $(0,0,1)$ in the picture above.

We define a mapping of $P\in S$ onto $\mathbf x\in e_1^\bot$ by mimicking the above; for
$$
\cos \theta := \langle P, e_1\rangle, $$
we let
$$
r= \tan\frac{\theta}2\in [0, \infty), $$
and we denote
$$\tag{2}
\mathbf x = r\frac{P’}{\lVert P’\rVert},\quad \text{where }P’=P-\langle P, e_1\rangle e_1.$$

To compare this mapping with Michael Hoppe’s one, we remark that
$$
\tan \frac\theta 2 = \frac{\sin \theta}{1+\cos \theta},$$
(this trig formula can be proved by inspection of the picture above), and that
$$
\lVert P\rVert^2 = 1 = \cos^2 \theta + \lVert P’\rVert^2 \quad \Rightarrow \lVert P’\rVert =\sin \theta.$$
Thus, (2) reads
$$\tag{3}
\mathbf x = \frac{P’}{1+\langle P, e_1\rangle}, $$
where Michael’s formula reads, in our notation,
$$
\tag{4}
\mathbf x = \frac{2P’}{2-\langle P, e_1\rangle}.$$

The two mappings are not exactly the same, which corresponds to the different choices of the projection pole. Here we projected from the South Pole of $S$, which is the point $-e_1$. Micheal projected from $2e_1$.

Title: Stereographic projection of "unit sphere" in a Banach space

Body:
So I was wondering, if it is possible to extend the idea that a sphere (minus one point) is isomorphic to a plane to the case of an infinite dimensional Banach (or Hilbert if needed) space.
Suppose we have a $L^2$ space, then consider the unitary sphere $U$ , namely the family of function in $L^2$ that have unitary norm. Is there an analog of the Stereographic projection in this case?

Thanks in advance!


问:
Example 15.9 in Loring Tu's An Introduction to manifolds
Example 15.9 is showed as follows in Loring Tu's An Introduction to manifolds:

Example 15.9 (Lines with irrational slope in a torus). Let $G$ be the torus $\mathbb{R}^2/\mathbb{Z}^2$ and $L$ a line through the origin in $\mathbb{R}^2$. The torus can also be represented by the unit square with the opposite edges identified. The image $H$ of $L$ under the projection $\pi:\mathbb{R}^2\longrightarrow\mathbb{R}^2/\mathbb{Z}^2$ is a closed curve if and only if the line $L$ goes through another lattice point, say $(m,n)\in\mathbb{Z}^2$. This is the case if and only if the slope of $L$ is $n/m$, a rational number or $\infty$; then $H$ is the image of finitely many line segments on the unit square. It is a closed curve diffeomorphic to a circle and is a regular submanifold of $\mathbb{R}^2/\mathbb{Z}^2$ (Figure 15.1).

If the slope of $L$ is irrational, then its image $H$ on the torus will never close up. In this case the restriction to $L$ of the projection map, $f=\pi|_L:L\longrightarrow\mathbb{R}^2/\mathbb{Z}^2$, is a one-to-one immersion. We give $H$ the topology and manifold structure induced from $f$.

It can be shown that $H$ is a dense subset of the torus [3, Example 111.6.15, p. 86]. Thus, $H$ is an immersed submanifold but not a regular submanifold of the torus $\mathbb{R}^2/\mathbb{Z}^2$.

Whatever the slope of $L$, its image $H$ in $\mathbb{R}^2/\mathbb{Z}^2$ is an abstract subgroup of the torus, an immersed submanifold, and a Lie group. Therefore, $H$ is a Lie subgroup of the torus.

I see that $H$ is an immersed submanifold, but I cannot understand how $H$ is an immersed submanifold but not a regular submanifold of the torus $\mathbb R^2/\mathbb Z^2$ is deduced from previous context. Can anyone explain the details for me?

答: In order for $H$ to be a regular submanifold, the natural map from $H$ into $\Bbb R^2/\Bbb Z^2$ should be a homeomorphism onto its image. But it is not, since the image is not locally connected. On the other hand, every manifold is locally connected, since it is locally homeomorphic to some $\Bbb R^n$ (in this case, $n=1$).


问:
Separating the variables possible?
Given the differential equation $$ \frac{dy}{dx} = y+x$$

I am told this differential equation is separable. Meaning I need to rewrite the RHS into a product of two variables depending on y and x.

I've tried for some time now but I simply cannot figure out how this is separable. I'm able to solve it using the method of "integrating factor" and so I know the solution should be $$ y = Ce^x-x-1 $$

Any ideas?

答:
Perhaps with $z:=y+x+1$,
$$ \frac{\mathrm dz}{\mathrm dx}= \frac{\mathrm dy}{\mathrm dx}+1=y+x+1=z.$$

Title: Separating the variables possible?


问:
Double Integrals. Simple question, don't understand their wording.
Make a sketch of the region over which

$$\int_0^{\pi/2} dx \int_0^{\sin(x)} dy$$

Would this be the same as

$$\int_0^{\pi/2} 1 dx \int_0^{\sin(x)} 1 dy$$

Which simply evaluates to $(\pi/2)\sin(x)$.

Is this correct?

答: Yup, that's right: $\int_a^b dx$ is a shorthand for $\int_a^b1dx$.</p>


问:
Can we prove that odd and even numbers alternate without using induction?
It is a simple exercise to prove using mathematical induction that if a natural number n > 1 is not divisible by 2, then n can be written as m + m + 1 for some natural number m.  (Depending on your definition of odd number, this can either be stated as "every number is even or odd", or as "every odd number is one greater than an even number".  My question is, can this be proven without induction?  In other words, can this be proven in Robinson arithmetic?  If not, what is example of a nonstandard model of Robinson arithmetic that doesn't have this property?

The reason I'm asking this is that the proof of the irrationality of the square root of 2 is usually presented with only one use of induction (or an equivalent technique).  But the proof depends on the fact if k^2 is even, then k is even, and that fact in turn depends on the fact that the square of a number not divisible by 2 is a number not divisible by 2.  And that fact is, as far as I can tell, is a result of the proposition above.  (I'm open to correction on that point.)  So if that proposition depended on induction, then the proof that sqrt(2) is irrational would depend on two applications of induction.  (The reason that the ancient Greeks wouldn't have been aware of this is that Euclid implicitly assumes the proposition above, when he defines an odd number as "that which is not divisible into two equal parts, or that which differs by a unit from an even number".)

Any help would greatly appreciated.

Thank You in Advance.

答:
I know this is a pretty old post, but I want to share some of my own insights.

First, I want to highlight a slight error in the answer by mercio. It is a correct model of PA - induction, but Robinson Arithmetic is PA - induction + $\forall n (n = 0 \vee \exists m (n = S(m)))$. Their model does not satisfy this additional axiom, as $X$ does not have a predecessor. Now that I think of it, I am not sure if this is also addressed in the model in the Elements?

Let me add a model of my own. Take $N$ together with two non-standard elements $a$ and $b$. Extend $S$ by $S(a) = a$ and $S(b) = b$, addition by $a + n = a$ for $n \in N$ and all other new combinations result in $b$, and multiplication by $n * 0 = 0$ always, and any other new combination results in $b$.

It is some annoying work to check that this satisfies all the axioms (which I formalized in Agda, for good measure), and it can be verified that $a$ is neither even nor odd (i.e. there does not exist $n$ such that any of $n + n = a$, $2*n=a$, $n + n + 1 = a$ or $2*n+1=a$ are true).


问:
About sum of binomial coefficients
Suppose you are a cashier of a cinema, the ticket is $\$5$, there are N people in the queue, each of them are buying one ticket, $x$ of them have $\$5$ notes, and others have $\$10$ notes. These people are arranged randomly. If you start with no money in your hand, what is the probability of you failing to provide changes to any of them?

To solve this problem, I need to use the following identity:

$$\sum\limits_{n=0}^{a}\frac{{a\choose n}{b-a\choose n}}{ (n+1){b\choose 2n}}=\frac{b+1}{b-a+1}, b \geq 2a \geq 0$$

How to prove it?

答:
Note that
$$\frac{{a\choose n}{b-a\choose n}}{{b\choose 2n}}=\frac{{2n\choose n}{b-2n\choose a-n}}{{b\choose a}}.$$
Therefore
$$
\begin{align}\sum_{n=0}^{a}\frac{{a\choose n}{b-a\choose n}}{ (n+1){b\choose 2n}}
&=\frac{1}{{b\choose a}}\sum_{n=0}^{a}\frac{{2n\choose n}{b-2n\choose a-n}}{n+1}=
\frac{{b\choose a}+\sum_{n=1}^{a}\frac{{2n\choose n}{b-2n\choose a-n}}{n+1}}{{b\choose a}}\\
&=\frac{{b\choose a}+\color{blue}{{b\choose a-1}}}{{b\choose a}}=\frac{{b+1\choose a}}{{b\choose a}}=\frac{b+1}{b-a+1}
\end{align}$$
where to obtain $\color{blue}{{b\choose a-1}}$ we applied the identity shown at A Vandermonde-like identity, new or existing?

Title: About sum of binomial coefficients


问:
When is floor function homogeneous?
I am trying to figure out, when is $\lfloor ax \rfloor = a\lfloor x \rfloor$ for $x,a \in \mathbb{R}$, where $\lfloor x \rfloor$ is floor function, and I'm completely stuck.

Is there a general rule, for which pairs of $a$ and $x$ is the equation above true? It's obvious that $a \in \mathbb{Q}$ but I couldn't get any further than that. Clearly, for fixed $a={b\over c}$ we have at least one $x$, namely $x=c$ and every multiple of $c$. But how to  find all such $a$ and $x$?

答:
First note that if $a \lfloor x \rfloor \notin \Bbb{Z}$, then $\lfloor ax \rfloor \in \Bbb{Z}$ so it can't be equal. Now let's suppose that $a \lfloor x \rfloor \in \Bbb{Z}$.

Now let's mark $r = x \mod 1$. Then $x = \lfloor x \rfloor + r$. And we have:$$
\lfloor ax \rfloor = a \lfloor x \rfloor\\
\lfloor a\lfloor x\rfloor + ar \rfloor = a \lfloor x \rfloor
$$
And because $a\lfloor x\rfloor\in\Bbb{Z}$:$$
a\lfloor x\rfloor + \lfloor ar \rfloor = a \lfloor x \rfloor\\
\lfloor ar \rfloor = 0
$$
And this holds only for $ar \in [0;1)$.

Title: When is floor function homogeneous?


答:
I think this was originally
$$\int_0^1\int_0^{\sin(x)} dy\, dx.$$

In that even, you are doing the double integral over the region $G = \{(x,y)|0\le y \le \sin(x), 0\le x \le 1\}$.

Title: Double Integrals. Simple question, don't understand their wording.

Body:
Make a sketch of the region over which

$$\int_0^{\pi/2} dx \int_0^{\sin(x)} dy$$

Would this be the same as

$$\int_0^{\pi/2} 1 dx \int_0^{\sin(x)} 1 dy$$

Which simply evaluates to $(\pi/2)\sin(x)$.

Is this correct?


问:
What's $x^n = n!$ approximately
What is the approximate solution to:

$x ^{1000} = 1000 !$

How can you solve for x?

More generally, for some constant $k$, how can you solve:

$x^k = k!$

答:
Notice that by the Hierarchy of limits, $\frac{x^n}{n!}\rightarrow 0$ as $n\rightarrow\infty$, so $n$ would have to be 'quite small'. In response to your first problem, $1000\log{x}=log{1000!}$ so $\log{x}=\frac{log{1000!}}{1000}$. Can you see a way to complete the problem? If not, here is a hint:

What can you deduce about the function $f(x)=\frac{log{x!}}{x}$? Can you sketch a graph of it? How does its derivative compare with $\log{x}$? You should be able to use these to find an approximate intersection point.

[To differentiate ${\log{x!}}$, consider the Gamma function definition of factorials, then apply the Fundamental Theorem of Calculus to differentiate.]

Title: What's $x^n = n!$ approximately


答:
You can use stirling's approximation when $n \rightarrow +\infty$:
$$\ln(n!) = n\ln(n)-n+O(\ln(n))$$

So,
\begin{align}
x^k = k! & \Leftrightarrow  k\ln(x) =\ln(k!) \\
& \Leftrightarrow  k\ln(x) =k\ln(k)-k+O(\ln(k)) \\
& \Leftrightarrow  \ln(x) =\ln(k)-1+O(\frac{\ln(k)}{k}) \\
& \Leftrightarrow  \ln(x) =\ln(k)-1+O(\frac{\ln(k)}{k}) \\
& \Leftrightarrow  x =\frac{k}{e} \times\exp{(O(\frac{\ln(k)}{k}))} \approx \frac{k}{e}\\
\end{align}

Title: What's $x^n = n!$ approximately

Body:
What is the approximate solution to:

$x ^{1000} = 1000 !$

How can you solve for x?

More generally, for some constant $k$, how can you solve:

$x^k = k!$


答:
Consider an extension 9f the familiar sum of digits test for divisibility by $9$.

The number $999$ factors as $27×37$ or with prime factors, $3^3×37$.  Thus if you add three-digit blocks together to cast out multiples if $999$, you are also casting out multiples of $27$ and $37$ and thus the sum of tyree-digit blocks tests for divisibility by these factors.

This can be extended to higher powers of $3$.  The sum of nine-digit blocks tests for divisibility by $81$, and more generally adding blocks of $3^k$ digits tests for divisibility by $3^{k+2}$

In your case, then, first sum the three-digit blocks to get  $333×465$ where each of $465$ blocks is $333$.  Since $333$ is a multiple of $9$ and $465$ is a multiple of $3$, your three-digit block sum passes for divisibility by $27$.

Next try the nine-digit block sum getting $333333333×155$.  The three-digit blocks of $333333333$ add up to a multiple of $27$, but dividing out a factor of $3$ gives $111111111$ which fails that test.  So $333333333$ is a multiple of $27$ but not $81$, and $155$ is not a multiple of $3$, so the nine-digit block sum which equals their product fails divisibility by $81$.

Thus the maximum power of $3$ that divides $A_{1395}$ is $3^3=27$.


答:
Sine law in triangle $ABD$

$\gamma=\beta=37°$ because $l_1\parallel l_2$
$$\frac{AB}{\sin 55^{\circ}}=\frac{25}{\sin (180^{\circ}-55^{\circ}-37^{\circ})}\to AB\approx 20.4913$$

edit

The segment $FE$ in the original drawing is a deception. As you can see from the calculation above, $FE$ is not necessary to find $AB$. Furthermore $FE=39$ would make points $D,A,F$ not aligned because $F'E$ is larger than $FE$.

Hope this is clear

$$...$$

Title: Is this possible to solve?

Body:
Is this possible to solve the length of $AB$ while only knowing that $l_1$ and $l_2$ are two parallel lines?

Example with geogebra


问:
For discrete group $G$ and $H\leq G$. Show that $G$ also satisfies the Folner condition if $H$ satisfies it and $[G:H]<\infty$.
A finitely generated group $G=\langle S \rangle$ is said to have the Folner condition if $\forall \varepsilon>0$, there exists a finite subset $F\subset G$ such that
$$\#((S\cup S^{-1})F\setminus F)<\varepsilon\# F,$$
where $\#S$ is the cardinality of a set $S$. Or equivalently, $\forall\varepsilon>0$, for any finite $T\subset G$, there exists a finite $F\subset G$ such that
$$\#(TF\setminus F)<\varepsilon\#F.$$

Let $G$ be a discrete group and let H be a subgroup of $G$. Suppose that the index $[G:H]$ is finite and that $H$ satisfies the Folner condition. Prove that $G$ also satisfies the Folner condition.

答:
Here are some ideas. Let  $S$ be a generating set of $G$, and let  $\{t_i: 1 \le i \le n \}$ be a set of left coset representatives of $H$ in $G$. For $g \in G$, denote the unique $t_i$ with $gH = t_iH$ by $\bar{g}$.

Now let $T = \{\overline{st_i}^{-1}st_i : s \in S \cup S^{-1}, 1 \le i \le n \}$. Then $T$ generates $H$ by a result of Schreier.

Now, given $\epsilon > 0$, choose $F$ as a corresponding Fölner set for $H$ with respect to the generating set $T$. Then $\cup_{i=1}^n t_iF$ is a Fölner set for $G$ with respect to $S$.


答:
It would evaluate to $1$. ${}{}{}{}{}{}{}{}$

Integrate first with respect to $y$. We get $y$. Do the limit points substitution. We get $\sin x-0$. Now integrate $\sin x\,dx$, from $0$ to $\pi/2$.

An antiderivative of $\sin x$ is $-\cos x$. So our final result is $-(\cos(\pi/2))-(-\cos(0))$, which is $1$.

The double integral we are evaluating is $1$ over the region below $y=\sin x$, above the $x$-axis, from $x=0$ to $x=\pi/2$. This is half an arch of the sine curve, and has area $1$.

For the picture they are asking for, draw the sine curve, and shade in the region described above.


问:
The Hausdorff Condition in the Smooth Manifold Chart Lemma
Here is the lemma stated in Lee's book, the second edition.

LEMMA. Let $M$ be a set and $\{U_\alpha\}_{\alpha\in J}$ be a collection of subsets of $M$, along with maps $\varphi_\alpha:U_\alpha\to\mathbf R^n$, such that the following properties are satisfied:

(i) $\forall \alpha\in J$: $\varphi_\alpha$ is a bijection between $U_\alpha$ and an open subset $\varphi_\alpha(U_\alpha) \subset\mathbf R^n$.

(ii) $\forall \alpha,\beta\in J$: the sets $\varphi_\alpha(U_\alpha\cap U_\beta)$ and $\varphi_\beta(U_\alpha\cap U_\beta)$ are open in $\mathbf R^n$.

(iii) $\forall\alpha,\beta\in J$: $U_\alpha\cap U_\beta\neq \emptyset
   \quad
   \Rightarrow
   \quad \varphi_\beta\circ\varphi_\alpha^{-1}:\varphi_\alpha(U_\alpha\cap U_\beta)\to \varphi_\beta(U_\alpha\cap U_\beta)$ is smooth.

(iv) Countably many of the sets $U_\alpha$ cover $M$.

(v)
$
\left.
\begin{array}{c}
p,q\in M\\
p\neq q
\end{array}
\right\}
\quad
\Rightarrow
\quad
\left\{
\begin{array}{c}
\exists \alpha\in J\text{ such that } p,q\in U_\alpha,\quad\text{ or}\\
\exists \alpha,\beta\in J\text{ such that } p\in U_\alpha, q\in U_\beta \text{ and } U_\alpha\cap U_\beta=\emptyset
\end{array}
\right.
$

Then $M$ has a unique manifold structure such that each pair $(U_\alpha,\varphi_\alpha)$ is a smooth chart.

We define the topology on $M$ by taking all sets of the form $\varphi_\alpha^{-1}(V)$, with $V$ an open subset of $\mathbf R^n$.

I feel like the condition (v) in the lemma is redundant but I am not sure about that. The condition (v) is used to provide the Hausdorff property of $M$. But it seems like we can show that $M$ is Hausdorff even without the condition (v). Here is my reasoning.

Let $p,q\in M$ and $p\neq q$.

By the condition (iv), there exist some $\alpha,\beta \in J$ s.t. $p\in U_\alpha$ and $q\in U_\beta$. If $\alpha = \beta$, then by the condition (i), we can find two disjoint open subsets of $U_\alpha$ to separate $p$ and $q$ by the Hausdorff property of $\mathbf R^n$. So we may assume that $\alpha \neq \beta$ and $U_\alpha \, \cap U_\beta \neq \emptyset$.

Let $A:= U_\alpha \setminus \overline{U_\beta}$, $B:=U_\beta \setminus \overline{U_\alpha}$, $C:= U_\beta \,\cap U_\alpha$.

Note that $A,B,C$ are all open by the definition of the topology on $M$ and the condition (i).

Case $1$: $p\notin B$ and $q\notin B$.

In this case, $A$ and $C$ separate $p$ and $q$. We are done.

Case $2$: $p\in B$ and $q\in B$.

In this case, it reduces to the situation that $p,q \in U_\alpha$. And we are done again.

Case $3$: One of $\{p,q\}$ is in $B$ and the other one is not.

W.L.O.G, assume that $p\in B$ and $q\notin B$.

In this case, we must have $q\in C$. But $B$ and $C$ are disjoint, which implies that $B$ and $C$ separate $p$ and $q$. So we obtain the Hausdorff property of $M$ again.

$\tag*{Q.E.D}$

I think there may be something wrong in my proof but I can't see it. Thanks for help.

答: In a manifold like the line with two origins (classic non-Hausdorff example), we can have that for $0$ and $0'$ $\alpha \neq \beta$ and $0 \in U_\alpha$ and $0' \in U_\beta$ with $U_\alpha \cap U_\beta$= $U_\alpha \setminus \{ 0 \} = U_\beta \setminus \{0'\}$. So $U_\alpha \setminus \overline{U_\beta}= \emptyset$ and vice versa. So $A=B= \emptyset$ but $C$ contains neither origin, in your notation.


问:
Question about random samplings and estimations.
100  students  were  selected  to  participate  in  a  survey  by  the  school  in  May  2013.  The  next  month,  another  100 students  were  randomly  selected  to  participate  in  the  same  survey,  out  of  which  2  had  already  participate  in the  survey  the  previous  month.  If  the  number  of  students  did  not  change  from  May  to  June  then  what  is  the approximate number of students in the school?

Came across this question. Is the question weird or am I just dumb? Can anyone suggest a way to solve this or an approach on how it can be solved. Thank you.

答:
Suppose there are $n$ students.

Number them with $i=1,2,\dots,n$ and let $S_i$ take value $1$ if student $i$ is selected twice, and value $0$ otherwise.

Note that: $$\mathbb ES_i=P(S_i=1)=\left(\frac{100}{n}\right)^2$$for every $i$.

If we set:$$S:=\sum_{i=1}^nS_i$$then $S$ is exactly the number of students that are chosen twice.

Taking its expectation by applying linearity of expectation we find:$$\mathbb ES=\mathbb E\sum_{i=1}^n S_i=\sum_{i=1}^n\mathbb ES_i=n\cdot\left(\frac{100}{n}\right)^2=\frac{10000}{n}$$

The fact that exactly $2$ students were chosen twice makes it reasonable to go for $\mathbb ES=2$ which leads to $n=5000$.


问:
ratios: why is the ratio of this equation like so?
The equation is w = 3m

and the ratio of w is to m is 3 : 1

Shouldn't it be 1 : 3?

could you please explain the reasoning?

答: The ratio of w to m is $w : m = \frac{w}{m} = \frac{3m}{m} = \frac{3}{1} = 3:1.$


Title: ratios: why is the ratio of this equation like so?

Body:
The equation is w = 3m

and the ratio of w is to m is 3 : 1

Shouldn't it be 1 : 3?

could you please explain the reasoning?

Body:
$$a:b=\frac{a}{b}$$
$$w=3m$$
$$\frac{w}{m}=\frac{3m}{m}=3$$
$$\frac{w}{m}=w:m=3$$


问:
What is the purpose to define different moments on a distribution?
What is the purpose to define different moments on a distribution?

The first moment is the expectation value of a function, what about the other?

答:
The second moment is a close relative of the variance, a very important measure of variability. More precisely, $\text{Var}(X)=E(X^2)-(E(X))^2$.

Higher moments tend to be of much less importance than the first two.


答:
I believe what is needed is more arithmetic than pure algebra. So here we go:

If $w=3m$ then when $m = 1, w=3(1)=3$; therefore, the ratio of $w$ to $m$ is 3:1.
Likewise, when $m=2, w=3(2)=6$. Note that the value of $w$ is always three times the value of $m$.


问:
Probability of getting two consecutive tails on the $n^{th}$ and $(n-1)^{th}$ toss for a biased coin.
The experiment goes like this.

The coin is tossed repeatedly and independently until two successive tails are obtained. And the probability of getting a head is $p$.
Let $q_n$ be the probability that we stop at $n^{th}$ coin toss.

What is the recursion for $q_n$?

答:
Let $\{X_n:n=0,1,2,\ldots\}$ be a Markov chain on $\{0,1,2\}$ with initial distribution $\alpha = \begin{pmatrix}1&0&0\end{pmatrix}$ and transition matrix
$$
P = \begin{pmatrix}
1-q & q & 0\\
1-q & 0 & q\\
0 & 0 & 1
\end{pmatrix}.
$$
This is a terminating Markov chain, i.e. one in which all states are transient except for one which is absorbing. We can write $P$ as
$$
P = \begin{pmatrix}
T & \mathbf T^0\\
\mathbb 0 & 1
\end{pmatrix},
$$
where $T$ is the substochastic matrix corresponding to transitions between transient states and $\mathbf T^0$ the (row) vector corresponding to transitions from transient states to the absorbing states. Let $\tau = \inf\{n:X_n=2\}$, then the distribution of $\tau$ is
\begin{align}
\mathbb P(\tau = k) &= \alpha T^{k-1}\mathbf T^0\\
&= \frac{q^2 \left(\left(1-q+\sqrt{q (2-3 q)+1}\right)^k-\left(1-q-\sqrt{q (2-3 q)+1}\right)^k\right)}{2^k\sqrt{q (2-3 q)+1}}.
\end{align}


问:
Trigonometric definite integral $\int ^\frac{\pi}{2} _0 \cos^4(s)\mathrm{ds}-\int^\frac{\pi}{2}_0 \cos^6(s)\mathrm{ds} $
$$\int ^\frac{\pi}{2} _0 \cos^4(s)\mathrm{ds}-\int^\frac{\pi}{2}_0 \cos^6(s)\mathrm{ds}  $$

the answer is $\dfrac{3\cdot1}{4\cdot2}-\dfrac{5\cdot3\cdot1}{6\cdot4\cdot2}$

I really don't know why

There may be some tricks in this, but I have not seen it in the textbook

答:
Integrate-by-parts below to obtain the recursive formula

\begin{align}
I_n=\int ^\frac{\pi}{2} _0 \cos^n s \>{ds}&= \int ^\frac{\pi}{2}_0 \frac{\cot^{n-1} s }{n}d(\sin^{n}s) {ds}=\frac{n-1}nI_{n-2},\>\>\>I_0=\frac\pi2
\end{align}

and then apply it to

$$\int^\frac{\pi}{2}_0 \cos^4s \> {ds}  -\int^\frac{\pi}{2}_0 \cos^6s \>{ds}  =I_4-I_6=\frac34\frac12I_0-\frac56 \frac34\frac12I_0$$


问:
compact metric spaces

Let $(X,d_X)$ and $(Y, d_Y)$ be compact metric spaces.

Show that for a nonempty set $A\subseteq X,$ the function $f : X\to \mathbb{R}, f(x) = \inf \{d_X(x,a) : a\in A\}$ is continuous.

Show that if a map $f : X\to Y$ is both continuous and bijective, then its inverse $f^{-1} : Y\to X$ is also continuous.

Let $C$ be the Cantor set with metric from $(\mathbb{R},|\cdot|)$ and $P = \prod_{k=1}^\infty \{0,\frac{1}{2^k}\}$ with metric from $(\ell_1, \lVert \cdot \rVert_1).$ Show that there exists a continuous bijection $f : C\to P.$

For $1),$ let $x\in X$ and let $\epsilon > 0.$ Choose $\delta = \epsilon.$ Then if $y\in X$ is such that $d(x,y) < \epsilon, we $ have \begin{align}d(f(x), f(y))
&= \inf\{d(x,a) | a \in A \} - \inf\{d(y,a) | a\in A\}\\
&\leq \inf \{d(x,a) | a\in A\} - (\inf \{d(x,a) - d(x,y) | a \in A\})\\
&= \inf \{d(x,a) | a\in A\} - (\inf \{d(x,a) | a \in A\} - d(x,y)\}\\
&= d(x,y) < \epsilon.\end{align}

Is this part correct?

For $2),$ suppose $f:X\to Y$ is continuous and bijective. Let $y\in Y$ and $\epsilon > 0.$ We want to show that there exists $\delta > 0 $ so that if $d_Y(x,y) < \delta$ then $d_X(f^{-1}(x), f^{-1}(y)) < \epsilon$. Let $(y_n)$ be a sequence in $Y$ such that $y_n\to y.$ Then $(f^{-1}(y_n))\subseteq X$ and since $X$ is compact, it has a convergent subsequence, say $(f^{-1}(y_{n_k}))_k.$ Since $f$ is bijective, we can write $y_n = f(x_n)$ for each $n\in\mathbb{N},$ where $x_n \in X.$ Let $x = \lim_{k\to\infty} f^{-1}(y_{n_k}) = \lim_{k\to \infty} x_{n_k}.$ Since $f$ is continuous, $f(x) = \lim_{k\to\infty} f(x_{n_k}).$ But here I'm stuck as I'm not sure how to show $f^{-1}$ is continuous from here.

As for the third part, every element of the Cantor set can be uniquely written in the form $\sum_{k=1}^\infty \frac{a_k}{3^k}, a_k \in \{0,2\}\forall k.$ Then it suffices to show that the function $f : C\to P, f(\sum_{k=1}^\infty \frac{a_k}{3^k}) = \prod_{k=1}^\infty \{\frac{a_k/2}{2^k}\}$ is continuous. I think one could show it's sequentially continous, but I'm not really sure about the details.

答:
I think (1) is correct, but I wouldn't bother with the infima: let $a,b\in A$. Then

$$|d(x,a)-d(y,b)|\leq |d(x,a)-d(a,y)|+|d(a,y)-d(y,b)|\leq d(x,y)+d(a,b)$$
where we used the triangle inequality. Take infimum on both sides as $a\in A$, while $b$ is (for now) fixed, so $|f(x)-d(y,b)|\leq d(x,y)+0=d(x,y)$. Now this is true for all $b\in A$, so take infimum over $b$ and conclude that $|f(x)-f(y)|\leq d(x,y)$, I think this is more neat.

For (2) this is a standard topological fact: If $f:X\to Y$ is a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism. Indeed, all we need to do is show that $f^{-1}$ is continuous: if $E\subset X$ is a closed subset, we want to show that the inverse image of $E$ through $f^{-1}$ is closed in $Y$, which is the same as $f(E)$ is closed in $Y$. Now $X$ is compact and $E$ is a closed subset, so $E$ is also compact. Now $f$ is continuous, so $f(E)$ is a compact subset of $Y$. Now $Y$ is Hausdorff, so all compact sets are closed, thus $f(E)$ is closed in $Y$. Sometimes it helps viewing things in a more abstract frame; for example the metrics here are not essentially needed, so the extra information causes extra confusion.

For (3) you have a correct guess: Indeed, every element of the Cantor set $C$ can be written uniquely as $\sum_{k=1}^\infty\frac{a_k}{3^k}$, where $a_k$ is either $0$ or $2$. But it is important to note that the converse is also true: Any sum $\sum_{k=1}^\infty\frac{a_k}{3^k}$ where $a_k\in\{0,2\}$ is an element of the cantor set. Now define $f:C\to P$ as
$$f\big(\sum_{k=1}^\infty\frac{a_k}{3^k}\big)=(\frac{a_k}{2^{k+1}})_{k=1}^\infty$$

$f$ is well-defined: since $a_k$ is either $0$ or $2$, we have that $a_k/2^{k+1}$ is either $0$ or $1/2^k$, so each $a_k/2^{k+1}$ does indeed belong to $\{0,\frac{1}{2^k}\}$, so $f$ does indeed take values in $\prod_{k=1}^\infty\{0,\frac{1}{2^k}\}$.

$f$ is one to one: this is obvious.

$f$ is surjective: indeed, let $x=(x_k)_{k=1}^\infty\in P$, so each $x_k$ is either $0$ or $1/2^k$. Set $a_k=2^{k+1}\cdot x_k$. Then $a_k$ is either $0$ or $2$, so if $t=\sum_{k=1}^\infty\frac{a_k}{3^k}\in C$ then $f(t)=x$.

$f$ is continuous: Let $\varepsilon>0$ and fix a point $x\in C$, say $x=\sum_{k=1}^\infty\frac{a_k}{3^k}$, where $a_k\in\{0,2\}$.
Let $y\in C$ be another element, say $y=\sum_{k=1}^\infty\frac{b_k}{3^k}$ where $b_k\in\{0,2\}$. Note that
$$d_P(f(x),f(y))=\|\{\frac{a_k}{2^{k+1}}\}_{k=1}^\infty-\{\frac{b_k}{2^{k+1}}\}_{k=1}^\infty\|_{\ell^1}=\|\{\frac{a_k-b_k}{2^{k+1}}\}_{k=1}^\infty\|_{\ell^1}=\sum_{k=1}^\infty\frac{|a_k-b_k|}{2^{k+1}} $$

Claim: Let $m\geq1$ be an integer. Then if $|x-y|<\frac{1}{3^m}$, then $a_k=b_k$ for all $k=1,\dots,m$.

Proof of the claim: I will only describe the idea here. Note that the construction of the Cantor set is done in steps: in each step we divide the segments in 3 pieces and throw out the middle part. In the expression $x=\sum_{k=0}^\infty\frac{a_k}{3^k}$ with $a_k=0$ or $2$, we have the following interpretation: in the first step of the construction, we divide $[0,1]$ in three segments and we only keep $[0,1/3]$ and $[2/3,1]$. If $a_1=0$, then $x$ lies in $[0,1/3]$. If $a_1=2$, then $x$ lies in $[2/3,1]$. So if $|x-y|<1/3$ where $y\in C,y=\sum_{k=1}^\infty\frac{b_k}{3^k}$ with $b_k=0$ or $2$, then $b_1=a_1$: otherwise we would have $x\in[0,1/3]$ and $y\in[2/3,1]$ or $x\in [2/3,1]$ and $y\in[0,1/3]$, which contradicts the fact that $|x-y|<1/3$.
Let's assume that $a_1=0$, so $x\in[0,1/3]$. Now again, in the second step of the construction, we cut each of $[0,1/3]$ and $[2/3,1]$ in three pieces each and throw out the middle third of each. Let's look at $a_2$. If $a_2=0$, then $x$ lies in the left segment of $[0,1/3]$ that is left after the second step, i.e. $x\in[0,1/3^2]$. If $a_2=2$, then $x$ lies in the right segment of $[0,1/3]$ that is left after the second step, i.e. $x\in[2/3^2,1/3]$. Now if $|x-y|<1/3^2$ then we also have $|x-y|<1/3$, so $y$ lies in $[0,1/3]$ as we explained for the first step, i.e. $b_1=0$. Now if $b_2\neq a_2$, we have that $x,y$ are in different thirds of $[0,1/3]$ after the application of the second step of the construction, so $|x-y|\geq 1/3^2$, a contradiction. I believe that you get the idea.

Now let $m$ be so large that $\sum_{k={m+1}}^\infty\frac{1}{2^k}<\varepsilon$. Taking $\delta=\frac{1}{3^m}$; from the claim we have that if $y\in C$ satisfies $|x-y|<\delta$, then $a_k=b_k$ for all $k=1,\dots,m$, so
$$d_P(f(x),f(y))=\sum_{k=m+1}^\infty\frac{|a_k-b_k|}{2^{k+1}}\leq\sum_{k=m+1}^\infty\frac{2}{2^{k+1}}<\varepsilon$$
and continuity follows. Observe that $\delta$ is independent of $x$, so $f$ is uniformly continuous.

Is this good news? No, this is expected, since $f$ is continuous on a compact space, thus it must be uniformly continuous. Also, by (2) this gives us a homeomorphism of the cantor set and $P$.

Title: compact metric spaces


问:
compute $\iint_Y F.N\ dS $ with Gauss
The question is:
$$ \iint_Y F.N\ dS \quad Y=(x-z)^2+(y-z)^2=1+z^2,\ \  0\leq z\leq 1 \quad N=\text{pointing outward}
$$ $$ F=(y,x,1+x^2z)$$
Here is how I have tried to solve it: I tried to solve it with Gauss theorem like this, $\gamma=\text{upper lid}$, $\sigma=\text{lower lid} $
$$ \iint_{Y+\sigma+\gamma}F.N\ dS=\iiint_K \text{div}f\  dA \iff \iint_Y F.N\ dS=\iiint_k-\iint_\sigma-\iint_\gamma$$
$$ \iint_\sigma=-\pi, \iint_\gamma=2+\pi$$
But I have trouble to integrate this triple integral
$$ \iiint_kx^2 dxdydz$$
I don't know if I have been  calculating right so far, and how should I proceed? Any suggestion would be great, thanks

答:
You can use transformation $f(x,y,z)=(x+z,y+z,z)$. The jacobian of this transform is 1. You get:$$
\iiint_{k'}(x+z)^2dxdydz
$$
where $x^2+y^2-z^2=1$.

Using cylindric coordinates, you can transform it to:
$$
\int_{0}^{1}\int_0^{2\pi}\int_0^{\sqrt{1+z^2}}(\rho\cos\phi+z)^2\rho d\rho d\phi dz
$$
Does this help you?

Title: compute $\iint_Y F.N\ dS $ with Gauss


问:
Proving Expectation law for $E[\max{X,Y}]$
Prove:
$$E[\max{X,Y}] = E[X] + E[Y] - E[\min{X,Y}]$$ where $X, Y$ are random variables.

Attempt:
$\max{X,Y}= X,$ $X$ greater than or equal to $Y$
$Y, X$ less than or equal to $Y$

$\min{X,Y}= Y,$ $X$ greater than or equal to $Y$
$X$, $X$ less than or equal to $Y$

Do I start with using the definition of Expectation?:
$$E[X] = \sum_i (i\cdot \operatorname{Pr}(X=i))$$

答:
It can easily be shown that for $a,b\in\mathbb R$,
$$
\min\{a,b\} + \max\{a,b\} = a+b.
$$
If $\mathbb E[|X|]<\infty$ and $\mathbb E[|Y|]<\infty$ then
$$
\mathbb E[\max\{X,Y\}] \leqslant \mathbb E[|X|] + \mathbb E[|Y|]<\infty
$$
and similarly we see that $\min\{X,Y\}$ has finite expectation. Since
$$
\min\{X,Y\} + \max\{X,Y\} = X+Y
$$
with probability one, taking expectations yields
$$
\mathbb E[\min\{X,Y\} + \max\{X,Y\}] = \mathbb E[X+Y],
$$
from which we conclude.


问:
Is there a special term for an array consisting only of ones?
Is there a special term for an array consisting only of ones?

Sorry for the rather elementary question. I am getting into MapReduce programming and am trying to frame my code to be nice and neat.

答: Not sure what you mean by "array". The $m\times n$ matrix whose entries are all ones could be denoted $1_{m\times n}$. Sometimes the $n\times n$ all-ones matrix is called $J_n$, or just $J$ if $n$ is understood from context. </p>


问:
How to study mathematics (Graduate level)
I know that normally, after attending class and understand the theorem and proofs, we will go directly to the problem and solve a couple of them, strengthening the concepts. Clearly, this is the best way to study and this is why in the US they always give a homework assignments to keep track of the study.

But it seems this way doesn't really work for me. The main problem is always, I found it's hard for me to do the problem directly after the class as I need more time to absorb the materials (In some sense, I might need to prove a theorem 5 times on my own to really understand what that theorem is all about.) So I don't usually do any homework on my own, instead, I focus on understanding the material first and do extra homework 2 weeks before exams.

This causes a big problem, that I don't have enough time to think about the homework assignment and end up "discussing problems" (mimicking solutions) with the others during the week. Also, I feel more comfortable just having a solid understanding of the whole course before moving on to doing questions, say two weeks before final exams.

Is there any way to solve this problem? Due to US system, there are always homework assignment and mid term and I cannot avoid them.

答: One recommendation is to get a study partner or form a study group of 2 or 3 dedicated individuals. Working with others can get you unstuck quicker and reveal more incites into the problems from different angles; hence allowing for conceptual understanding rather than rote learning.</p>