问:
A bartender and a customer roll a six-sided, ungalvanized cube. What is his chance of winning?

A bartender and a customer roll a six-sided, ungalvanized cube. If the
visitor throws a higher number than the bartender, he wins. A guest is
a rascal who cheats. He takes advantage of the waiter's inattention by
secretly looking at his first roll under the dice cup and then, if he
throws a 1-3, he throws the dice again, unnoticed.

What is his chance of winning?

Let $k$ be the number the bartender throws:
For every $k$ there are two ways of winning: Either the visitor has a higher number with his first throw or he throws either a $1$, $2$ or $3$ and than has a higher number.

For $k=6$: $0/6$

For $k=5$: $1/6+3/6\cdot 1/6 $

For $k=4$: $2/6+3/6\cdot2/6$

For $k=3$: $3/6+3/6\cdot3/6$

For $k=2$: $4/6+2/6\cdot4/6$

For $k=1$: $5/6+1/6\cdot5/6$

Adding all these up and dividing them by six, gives $p=\frac{121}{216}$

Is that correct?

答: No. The key observation here is that the first roll only counts if it is a 4 or above. This does not have any effect on your $k = 3,4,5,6$ cases, but does on the others: you have counted the case where the customer rolls a $2$ or $3$ on the first roll as a win when it exceeds the bartender's roll, but in fact those will be rerolled, and so have only a $\frac{4}{6}$ or $\frac{5}{6}$ chance of being wins, depending on the bartender's roll.


问:
Coin flipping - conditional expectation

We flip a symmetric coin 6 times. Denote X as "amount of all heads gotten" and Y as "count of all heads gotten in the first two flips". Calculate:

$\mathbb{P}(X=5|Y)$

$\mathbb{E}(Y|X)$

I'm learning to my final exam and banging my head against the wall as I already forgot how would one calculate stuff like that. I appreciate any help.

答:
I'll solve for the expectation as it is more fun (for me). I leave the probability as the exercise for the OP :D

Given $X$, we know that we had $X$ heads and $6-X$ tails. Out of symmetry, all arrangements are equally likely, so the probability that the first toss was heads is $\tfrac{X}{6}$. Same goes for the second toss. Since $Y$ is the sum of two indicators (the first and second toss), then

$$E(Y\vert X)=\tfrac{X}{6}+\tfrac{X}{6}=\tfrac{X}{3}$$

ContentLicense: CC BY-SA 4.0

Title: Coin flipping - conditional expectation


问:
Given 2 points in 3d, how to generate a plane equation in the form $lx+my+nz=d$
I can do this for 2d points by getting the 2 parametric equations for $x$ and $y$ and then solving by eliminating $t$ and getting the 2d equivilent line equation. There is only one line equation in this case though.

In 3d there will be infinitely many planes but I only need one of them for my use case. I can still generate the symmetric line by doing the below

\begin{align*}x & = {x_0} + ta\\ y & = {y_0} + tb\\  z & = {z_0} + tc\end{align*}

which rearranges to

\begin{align*}\frac{{x - {x_0}}}{a} = \frac{{y - {y_0}}}{b} = \frac{{z - {z_0}}}{c}\end{align*}

What I want to do is get one of the planes that does go through the 2 points in  the form

\begin{align*}lx+my+nz = d\end{align*}

The reason for doing this is to solve a plane / Bezier curve intersection (which I have working in 2d)

Example given the following cubic

\begin{align*}\mathbf{X}(t) = (1-t)^3\mathbf{P}_0 + 3t(1-t)^2\mathbf{P}_1 +
               3t^2(1-t)\mathbf{P}_2 + t^3\mathbf{P}_3\end{align*}

And the following line vector equation

\begin{align*}\mathbf{A}\cdot \mathbf{X} = d\end{align*}

with \begin{align*}\mathbf{A} = (l,m,n)\end{align*}

I just need to solve

\begin{align*}(1-t)^3(\mathbf{A} \cdot \mathbf{P}_0) +
3t(1-t)^2(\mathbf{A} \cdot \mathbf{P}_1) +
3t^2(1-t)(\mathbf{A} \cdot \mathbf{P}_2) +
t^3(\mathbf{A} \cdot \mathbf{P}_3) - d = 0\end{align*}

to be able to get the intersection points.

So how do I generate a example plane equation that does go through these 2 points?

Update :

I realise with 2 points there are infinitely many planes, I have rephrased the question to ask how to generate one of them or the formula for generating all of them.

答:
$$\frac{{x - {x_0}}}{a} = \frac{{y - {y_0}}}{b} = \frac{{z - {z_0}}}{c}$$
means
$$
\begin{cases}
\frac{{x - {x_0}}}{a} = \frac{{y - {y_0}}}{b}\\
\frac{{y - {y_0}}}{b} = \frac{{z - {z_0}}}{c}\\
\end{cases}
$$
The line is given as intersection of two planes. Any linear combinations of the two planes gives a plane passing through the two given points.
$$
\begin{cases}
b(x - {x_0}) - a(y - {y_0})=0\to bx-ay+h=0\\
c(y - {y_0}) - b(z - {z_0})=0\to cy-bz+k=0\\
\end{cases}
$$
$$\alpha:\lambda(bx-ay+h)+\mu(cy-bz+k)=0$$

Easier alternative

Suppose we have two points $P=(1,2,3);\;Q=(2,1,1)$

Vector joining the two points is $\vec{QP}=(1,-1,-2)$

A vector $\vec{n}=(a,b,c)$ is normal to $\vec{QP}$ is $\vec{n}\cdot\vec{QP}=0$, that is

$a-b-2c=0\to b=a-2c$

$\vec{n}=(a,a-2c,c)$

The equation of the plane passing through $P$ and normal to $\vec{n}$ is

$$a(x-x_P)+b(y-y_P)+c(z-z_P)=0$$

$$a(x-1)+(a-2c)(y-2)+c(z-3)=0\to a x+(a -2 c) y+c z+c-3a=0$$


答:
Hint

$$\mathbb P\{X=5\mid Y=k\}=\frac{\mathbb P\{X=5, Y=k\}}{\mathbb P\{Y=k\}}.$$
and $$\mathbb E[Y\mid X=k]=\sum_{t\in \{0,1,2\}}t\mathbb P\{Y=t\mid X=k\}.$$

Title: Coin flipping - conditional expectation

Body:
We flip a symmetric coin 6 times. Denote X as "amount of all heads gotten" and Y as "count of all heads gotten in the first two flips". Calculate:

$\mathbb{P}(X=5|Y)$

$\mathbb{E}(Y|X)$

I'm learning to my final exam and banging my head against the wall as I already forgot how would one calculate stuff like that. I appreciate any help.


问:
average score of students
i am a little confused and please help me to figure out how to solve such questions

i know that it could be solved using ven diagram,but i could not  get any correct intersection,now what i have tried,  if question one  solved correctly by $200$ and question $2$  is solved by $304$,in total it is $504$,which means that $4$  solved both,so total ssore would be  $196*25+4*50+296*25$,but  what about third and fourth question?if we do it separately,we get that $100 $ person sold both $3$ and $4$,so only $3$ and $4$ was solved by $250$ and $150$,in total we have $196+4+296+250+150=896$,does it means that  from these  students  all question was done by $896-400=496$?please help me

答:
I don't think you have to do any calculation to get the answer. This can be just estimated.

4 Questions with 25 points each is max 100 points.

400 students answering 4 questions - 200 ~ 350 gets each answer right.

Average must be between 50 and 100.

Only D fits the estimation range

ContentLicense: CC BY-SA 3.0

Title: average score of students


问:
Let X be a non-empty set and U an ultrafilter over X....
Let X be a non-empty set and U an ultrafilter over X. Show that if X = $A_1 \cup A_2 \cup… \cup A_n$, then there is k $\in$ {1,…, n} such that $A_K \in$ U.

X = $A_1 \cup A_2 \cup… \cup A_n$ = $\bigcup_{k=1}^n A_k \in U$, since U is an utrafilter, k $\leq$ n  and n is finite then $a_k \in U$.

I don't think this is good, but I can't think of another way to show it. It is not trivial that $A_1 \cup A_2 \cup… \cup A_n \in U$ then $A_k \in U$ where k $\in$ {1,...,n}.

maybe I got it wrong.

答: Suppose for a contradiction that no $A_k$ belongs to $U$. Then $X\setminus A_k\in U$ for all $k$, since $U$ is an ultrafilter. Filters are closed under finite intersections, so $$\bigcap_{k=1}^n (X\setminus A_k)\in U,$$ but that intersection is empty because the $A_i$ cover $X$, which is a contradiction, since the empty set is not in any ultrafilter.


问:
Set Notation (Axiom of Infinity)
I'm having trouble understanding the notation used in describing the axiom of infinity (which is number 6 in the Wolfram MathWorld page). I understand what the axiom is saying, but I'm trying to understand how the following sentence:

$$\exists  S \space [ \varnothing \in S \wedge (\forall x \in S) [x \cup \{x\} \in S]]$$

describes the existence of an infinity set. If someone could just break this apart for me and describe its pieces and how it all works that would be greatly appreciated.

答:
Using the second part of the definition recursively, we see $S$ must contain

$\varnothing=0$

$0\cup\{0\}=1$

$1\cup\{1\}=2$

$2\cup\{2\}=3$

$\cdots$

which are all distinct. Note that when we look at the set-theoretic construction of the naturals, we have $0=\varnothing$ and $n=\{k:k<n\}=\{k<n-1\}\cup\{n-1\}=n-1\cup\{n-1\}$.

Title: Set Notation (Axiom of Infinity)


问:
If $\lim_{x\to x_{0}} f(x) = L$ and $\lim_{x\to x_{0}}g(x)=\infty $, then $\lim_{x\to x_{0}}(f(x) + g(x)) = \infty$
regarding proving this claim:

If $\lim_{x\to x_{0}} f(x) = L$ and $\lim_{x\to x_{0}}g(x)=\infty $, then $\lim_{x\to x_{0}}(f(x) + g(x)) = \infty$

I am at the end of the proof, right here:

$f(x)+g(x) > L - \square_{1} + \square_{2} \geq  K $

I don't know what to place inseatd of $\square_{1}$ and $\square_{2}$.

I know that $\square_{1}>0$ because it represent $\varepsilon$.

In $\square_{2}$ I tried placing $K$...

Any help will be awesome!

Thank you so much!

答:
For it to tend to $\infty$ we need that $\forall\ K>0, \exists\ \delta>0: x\in B_\delta(x_0)\implies f(x) > K$.

Let K > 0.

If you're close enough to $x_0$ ($\delta_1(\varepsilon)$), then $f(x)> L-\varepsilon$ and if you're close enough to $x_0$ ($\delta_2(R)$) then $g(x)> R$.

so if $x\in B_{min(\delta_1,\delta_2)}(x_0)$ we have that$f(x)+g(x)> L-\varepsilon+R > K \iff R> K-L+\varepsilon$
Note that you can thus choose $R$ so that it fulfills that.

EDIT:
Given that my original answer produced some confusion. Let's do it being a little more careful. Perhaps this way you can see it more clearly.

We need to prove $$\operatorname{lim}_{x\rightarrow x_0} (f(x)+g(x)) = +\infty$$

That means that $\forall \ K>0, \exists\ \delta>0: x\in B_\delta(x_0)\implies f(x) > K$

Someone gives you $K>0$, now you have to find a $\delta$ that works for that K. Take $\varepsilon=1$, and $R=K-L+2$.

Since $g(x)\rightarrow \infty $ then there is a $\delta_1>0: x\in B^*_{\delta_1}(x_0) \implies g(x)>R$, by definition of the limit. Now, since $f(x) \rightarrow L$. Then there is a $\delta_2>0: x\in B^*_{\delta_2}(x_0)\implies |f(x)-L|<\varepsilon \implies f(x)>L-\varepsilon$.

Take $\delta = min \lbrace\delta_1,\delta_2\rbrace$ then if $x\in B^*_{\delta}(x_0) \implies f(x)+g(x)>L-\varepsilon+R = K+1 > K$

So we have what we wanted


问:
Get angle between two scalar inputs
I'm trying to make a game and in m working on the movement system

I'm trying to make the character look the way they are moving

I have two numbers, the horizontal axis and vertical axis; both are between -1 and 1

What I need is the angle that the point (h, v) is at

I would prefer the math to be linear as I am using bolt which is a flow language

答:
The usual function for this is called atan2 in most programming languages. So
$$\theta=atan2(y,x)$$
should give the angle (usually in radians) between the ray going from the origin to the point $(x,y)$ and the positive x-axis.

The function works for any point except of course the origin $(0,0)$, because then the angle is undefined.

Title: Get angle between two scalar inputs


问:
Estimate value of function
I'm asked:

For the function $f(x) = x + 1 - \frac{1}{n} , x \in [\frac{n-1}{n},
> \frac{n}{n+1}]$ in the interval $[0, 1)$.

a) Prove it can be extended by continuity for $x=1$. Call $g$ to that
new functioon.

b) calculate an upper and a lower estimates for $\int_0^1g$

I really don't know how to start point (a) because $x$ is between those intervals, I believe I cannot simply put $x \rightarrow 1^-$.

The second problem I have is with estimates. What am I supposed to do in order to calculate those estimates? Should I use those intervals $[\frac{n-1}{n},
> \frac{n}{n+1}]$ and multiply the value of supremum/infimum of $g(x)$ there in order to calculate the upper/lower estimates respectively? What would that look like?

答: For a you can show that the low points of each interval converge to a certain value as $x \to 1^-$ and that the limits at the high end also converge to that value.  For b I would note that the low points are all on the curve $y=2x$ and use that for my estimate.  The function sags below that line but the wiggles are fairly small.  You were not asked to assess the error in the estimate.  You could do better by explicitly integrating the first few intervals, which is where most of the error will come from.

Title: Estimate value of function


问:
Compactness Theorem (logic) application.
I met a question which says: if we have $A, B$, both are nonempty theories such that every model of $A$ satisfies at least one formula (sentence if they are in the first-order language, the question indicates that there is no difference whether it's propositional or first-order) in $B$. Then it asks me to show there are some formulas/sentences, $a_1....a_k$ from $A$ and $b_1...b_l$ from $B$ such that $(a_1...\land...a_k)\to(b_1...\lor...b_l)$ is a tautology.

The question specifically says I should use the compactness theorem, but I don't see how to apply the theorem here. Any hints?

答: Hint: Notice that the hypothesis says exactly that $\{\varphi,\neg \psi\mid \varphi\in A,\psi\in B\}$ is inconsistent. By compactness, there is a finite inconsistent subset.


问:
Will $x+U$ be open set in metric space $X$ if $U$ is open in $X$?
"If $U$ is an open set in normed linear space $X$, then for any $x\in X$, $x+U$ is open". My question is:
What can we say if $X$ is not a normed linear space, will it hold? I tried finding counterexamples but couldn't find it. Any hints?

答: If $X$ is a topological vector space, then by definition the sum operation $+$ is continuous. Since $U$ is open, and $f(y)= y+x$ is a homeomorphism (it is continuous and its inverse $g(z)=z-x$ is also continuous) then also $f(U)=U+x$ is open.


问:
Convergence in distribution of Geometric r.v.s sequence

Let $X_n$ be a sequence of r.v.s with Geometric distribution with parameter $p_n=\frac{5}{n}$. Prove that $Z_n = \frac{X_n}{n}$ converges in distribution. What is its limiting distribution?

I'm studying for my math exam(from all three previous years) but I completely forgot how would one do such task. I guess we'd like to use CLT here but no further progress has been made.

Thanks for any help.

答:
There is no role of CLT here. CLT gives the convergence to the normal distribution, and it is clear here that the limiting distribution must be supported only on the positive half of the real line, it can't be normal.

Since the geometric random variables are easy to analyze, that is, the probability of simple events like $P(X_n/n>x)$ can be computed more or less explicitly. It is a worthwhile idea to try to show the convergence from the definition, that is, by showing the convergence of the distribution function. This is what we do below:

\begin{align}
F_{\frac{X_n}{n}}(t)&=1-P(X_n>tn)\\
&= 1-\left(1-\frac{5}{n}\right)^{tn}\\
&\to 1-e^{-5t}.
\end{align}
This shows that the $\frac{X_n}{n}$ converges to $Exp(5)$ in distribution. (Note that the second equality above is not true, one should be honestly using $\lceil tn \rceil$ but it doesn't matter in the limit, so I cheated a little bit.)

As you mentioned CLT, the usual way of proving CLT is via analyzing the characteristic functions. It is not hard to write the characteristic function of the geometric distribution and showing that the limit indeed is the characteristic function of the exponential distribution. This would give an alternate proof that would be very much in the spirit of (the proof of) CLT.


问:
Prove the convergence in probability of $\bar{X}_{n}=\frac{1}{n} \sum_{i=1}^{n} X_{i}$, with iid X_i
I need to show that: $\bar{X}_{n} \rightarrow^{p} 1+\theta$ and that $min(X_i) = X_{(1)} \longrightarrow^{p} \theta$.

Here $X_{1}, X_{2} \ldots$ is a sequence of iid random variables with p.d.f $f(x)=e^{-x+\theta}, x \geq \theta,\theta>0$, and $\bar{X}_{n}=\frac{1}{n} \sum_{i=1}^{n} X_{i}$.

I tried to apply the definition of convergence in probability, but I didn't know how to move forward.

答:
First observe that the expectation of your rv is

$$\mathbb{E}[X]=e^{\theta}\int_{\theta}^{\infty}xe^{-x}dx=\dots=\theta+1$$

Using the Strong Law of Large Numbers you get that

$$\overline{X}_n\xrightarrow{a.s.}\theta+1$$

Thus also

$$\overline{X}_n\xrightarrow{\mathcal{P}}\theta+1$$

Second, the CDF of $U=X_{(1)}$ is the following

$$F_U(u)=1-e^{n(\theta-u)}$$

Thus

$$\lim\limits_{n \to \infty}\mathbb{P}[\theta<U<\theta+\epsilon]=1-e^{-n\epsilon}=1$$

this proves that

$$X_{(1)}\xrightarrow{\mathcal{P}}\theta$$

Using the facts that the sequence is monotonic (non increasing) and lower bounded to $\theta$ you can also prove that

$$X_{(1)}\xrightarrow{a.s.}\theta$$


答:
To summarize the discussion in the comments:

The stated solution is incorrect because the calculation allows the cheater to (sometimes) consider the bartender's roll, contrary to the specified problem.

To solve the given problem:

We consider probabilities for the various outcomes for the cheater's roll.  For $≤3$ the only way to get that value is to first roll $3$ or less and then roll the specified value.  For $≥4$ the cheater can either get the desired roll initially or get $≤3$ the first time and then roll the desired value.  Thus:

$$ P(X=n) =
\begin{cases}
\frac 1{12}  & \text{if $n≤3$ } \\ \\
\frac 14 & \text{if $n≥4$ }
\end{cases}$$

Where, of course, $X$ is the cheater's roll.

Since the probability of winning if $X=i$ is $\frac {i-1}6$ we see that the probability the cheater wins is $$\frac 1{12}\times \left(\frac 06+\frac 16+\frac 26\right)+\frac 1{4}\times \left(\frac 36+\frac 46+\frac 56\right)=\boxed {\frac {13}{24}}$$

Title: A bartender and a customer roll a six-sided, ungalvanized cube. What is his chance of winning?

Body:
A bartender and a customer roll a six-sided, ungalvanized cube. If the
visitor throws a higher number than the bartender, he wins. A guest is
a rascal who cheats. He takes advantage of the waiter's inattention by
secretly looking at his first roll under the dice cup and then, if he
throws a 1-3, he throws the dice again, unnoticed.

What is his chance of winning?

Let $k$ be the number the bartender throws:
For every $k$ there are two ways of winning: Either the visitor has a higher number with his first throw or he throws either a $1$, $2$ or $3$ and than has a higher number.

For $k=6$: $0/6$

For $k=5$: $1/6+3/6\cdot 1/6 $

For $k=4$: $2/6+3/6\cdot2/6$

For $k=3$: $3/6+3/6\cdot3/6$

For $k=2$: $4/6+2/6\cdot4/6$

For $k=1$: $5/6+1/6\cdot5/6$

Adding all these up and dividing them by six, gives $p=\frac{121}{216}$

Is that correct?


问:
Combinatorics, counting and number theory.
Find number of ordered quadruples (a,b,c,d) [of positive integers] such that $lcm[a,b,c]= lcm[a,b,d]= lcm[a,c,d]= lcm[b,c,d] = 2^r\cdot3^s$

So i approached it like two of a,b,c,d has max power of 2 = r. For rest two, they have r+1 choices each. So it would be $\binom{4}{2} \cdot (r+1)^2$. Same for the power of three.

So as per me answer should be $(\binom{4}{2} \cdot (r+1) \cdot (s+1))^2$ but the answer to this problem is not given in my book.

I would really appreciate if you could tell whether my answer is correct and whether this is the correct approach and if it is not correct approach, then why?

Thanks!

答:
Your approach seems alright but the way you are counting, you will have many duplicates and end up overcounting. We can count each case separately and add.

Let's take $\displaystyle 2^r$ first

Case 1 - two of them have power of $r$ and other two have power of $0 \leq p \lt r$.

Number of ways = ${4 \choose 2} \times r \times r$

Case 2 - three of them have power of $r$ and one has power of $0 \leq p \lt r$.

Number of ways = ${4 \choose 1} \times r$

Case 3 - all of them have power of $r$ for $2$. There is only $1$ way.

That gives total number of ways $ = 6r^2 + 4r + 1$

We similarly find for $s$ and then multiply for all possible quadruples.


问:
Improved Euler's Method 2nd order ODE ,not sure if this is right
SO i have this differential equation and actually i am not sure if i have solved it right with Euler's improved method :

$z'' = f_{z} - C_{z}*|z'|*z'$

$ z' = u$

$u' = z'' = f_{z}-C_{z}*|u|*u$

Improved Euler Method :

$K_{1} = f{z}-C{z}*|u_{n}|*u_{n}$

$K_{2} = f{z}-(C_{z}*|u_{n}+h*K_{1}|)*(u_{n}+h*K_{1})$

so we get :

$z_{n+1} = z{n} + h*u_{n}$

$u_{n+1} = u_{n} + \frac{h}{2}*(K_{1} + h*K_{2})$

Is this right ?

P.S (fz , Cz are just variables with numbers not function inputs)

答:
There are several small errors that would sum up to a more grossly incorrect output. As a system with multiple components, you need to compute the intermediary slopes at all stages for all components, that is, additionally for the slopes in $z$ direction
\begin{align}
L_1&=u\\
L_2&=u+hK_1
\end{align}
Then the equations for the step are
\begin{align}
u_{+1}&=u+\frac h2(K_1+K_2)\\
z_{+1}&=z+\frac h2(L_1+L_2)=z+hu+\frac{h^2}2K_1
\end{align}


问:
The intersection of any family of principal ideals in a UFD is principal
I found this problem from Aluffi's Algebra Chapter 0 and only managed to find one reference (without proof) to this statement online, so I'm posting this here for reference/verification.

Problem

Let $\mathcal{A} = \{(a_i)\}$ be a family of principal ideals in a UFD $R$.
Show that the ideal $I = \bigcap_i (a_i)$ is principal.

Solution (?)

$I$ is an ideal since it's an intersection of ideals.
Pick any $r_0 \in I$. If $(r_0) = I$ we are done, so suppose $(r_0) \subsetneq I$.
Then pick any $s \in I - (r_0)$.
Since $R$ is a UFD, greatest common divisors always exist so we can set $r_1 = \mathrm{gcd}(r_0, s)$.

By definition of gcd, the ideal $(r_1)$ is the smallest principal ideal containing the ideal $(r_0, s)$.
Since $(r_0, s) \subseteq (a_i)$ for each $i$,
we therefore have $(r_1) \subseteq (a_i)$ for each $i$.
Thus $(r_1) \subseteq I$ and we get
$$
(r_0) \subsetneq(r_0, s) \subseteq (r_1) \subseteq I.
$$
As long as $(r_n) \subsetneq I$ this process can be repeated fo find $r_{n+1}$ such that
$(r_n) \subsetneq (r_{n+1}) \subseteq I$, yielding an ascending sequence of principal ideals
$$
(r_0) \subsetneq \cdots \subsetneq (r_{n}) \subsetneq (r_{n+1}) \subseteq I.
$$
But since $R$ is a UFD this sequence must stabilize, which can only happen if we eventually have $(r_n) = I$.

Question

Does my solution look ok?

Isn't the generator of the ideal $I$ the least common multiple of the $a_i$'s? If this is true, does this statement imply that lcms exist for arbitrary families in UFDs? Can we say that gcds of arbitrary families exist in UFDs by the same logic?

答: Your solution looks fine. And you are correct that any generator of the intersection will be an lcm. You can indeed also find arbitrary gcds in a UFD $R$. Take some subset $S\subseteq R$. Let $D$ be the set of principal ideals containing $S$. Of course, the unit ideal $(1)\in D$ so it's nonempty. As you've shown in your proof, the intersection $\bigcap_{(d) \in D} (d)$ has a principal generator, say $e$. Then $S\subseteq (e)$ so for all $s \in S$, $e \mid S$, i.e. $e$ is a common divisor of the elements of $S$. It is the greatest common divisor because if $d$ is another common divisor of the elements of $S$ then $S \subseteq (d)$, i.e. $(d) \in D$. Hence, $(e)\subseteq (d)$ and $d\mid e$.


问:
Coefficients of Newton interpolation polynomial
Given distinct $y_0,...,y_m$ in $\mathbb R$, let $N_m(x)$ be the Newton interpolation polynomial of degree $m$. That is, $N_m(x) = \sum_{n=0}^{m}a_nw_n(x)$ where $w_0 = 1$, $w_n(x) = (x-x_o)(x-x_1)...(x-x_{n-1})$ for $n > 0$ and the coefficients $a_n$ satisfies

$a_0 = y_0$

$a_0 + a_1w_1(x_1) = y_1$

.........................................

$a_0 + a_1w_1(x_m) + ... + a_mw_m(x_m) = y_m$

Prove that $\displaystyle a_n = \sum_{j=0}^n\frac{y_j}{\prod_{k=0, k \neq j}^n(x_j - x_k)}$ for $0 \leq n \leq m$.

Hint is given in the book that $N_n(x) = L_n(x)$, for $0 \leq n \leq m$ where $L_n(x)$ is the Lagrange polynomial that interpolates through $y_o,...,y_m$. However this hint is not helping me much. Any further hints or solution is appreciated.

答:
$$N_n(x) = \sum_{j=0}^n a_jw_j(x) = \sum_{j=0}^n a_j \prod_{k=0}^{j-1}(x-x_k)
= \sum_{j=0}^n y_j \prod_{k=0, k\neq j}^n \frac{(x-x_k)}{(x_j-x_k)}$$
interpolates through the points
$$(x_0,y_0), (x_1,y_1), \ldots, (x_{n-1},y_{n-1}), (x_n,y_n)$$
Note that the polynomial on the right side above is the Lagrange
interpolation polynomial $L_n(x)$ suggested in the hint.
Since $L_n(x)$ also interpolates through these $n+1$ points, it must be
the same as $N_n(x)$.
Note that $w_n(x)$ is the only polynomial of degree $n$ in the sum
on the left while each polynomial in the sum on the right has degree $n$.
Comparing the coefficients of $x^n$ on both sides, we get
$$a_n = \sum_{j=0}^n \frac{y_j}{\prod_{k=0, k\neq j}^{n}(x_j-x_k)}.$$</p>


问:
Coordinate of point, given a line segment
It is given that points A = (−2, 1)  B = (4, −1) form a segment, and it is asked at which point in y axis
(when x = 0) when you "look at" or draw segments to points A and B there is a 90 degree angle.
How do you solve this?
The answer is 0,3 and 0,-3 but I cant understand how to find them.

I added a picture, it is known that x = 0, and is asked y coordinate, and the angle between red lines is 90. what is the coordinate of point c?

答:
One way to do it, might not be the best way, using the fact that the product of the slope of perpendicular line is $-1$.

Let $A(-2,1)$, $B(4,-1)$ and $C(0,c)$ be three points. First, the slope of the line between $A$ and $C$ is
$$m_1=\frac{c-1}{0-(-2)} = \frac{c-1}{2}$$
Second, the slope of the line between $B$ and $C$ is
$$m_2=\frac{c-(-1)}{0-4} = \frac{c+1}{-4}$$
The product of the slopes must be $-1$
$$m_1\times m_2=\frac{c-1}{2}\times\frac{c+1}{-4} = \frac{c^2-1}{-8} = -1$$
This gives a quadratic equation
$$c^2-1=8$$
which gives
$$c=\pm3$$


问:
Side of a square inscribed in regular polygon

I'm trying to derive a general formula for the side of a square inscribed in a regular polygon with $n$ sides.

I know there may be more than one, e.g. infinitely-many for an octagon, but I want to be able to extract the side for at least one of them.

I am aware of Dilworth and Mane's 2010 paper "Inscribing a regular $m$-gon in a regular $n$-gon" (PDF link via sc.edu), but there is no indication on how to calculate the side.

Second part:

Is it true that for an odd $n$ there is no inscribed square that is concentric with the polygon?

答:
Dilworth and Mane publish another paper
https://link.springer.com/article/10.1007/s00022-011-0065-3
This gives the maximal regular m-gon in a regular n-gon.
If gcd(m,n) > 1 the problem is completely solved.
If gcd(m,n)=1 the result is stated as a conjecture (which passes numerical checks).
Fixing m=4 gives the answer, except for n=5 and n=9 the maximal square has only three points of contact (= not inscribed).
For all other n \ge 3 the maximal square has four points of contact.
The formula for the edge length is given.

Second part:
For gcd(m,n)=1 the maximal m-gon is never concentric with the n-gon.
Concentric inscribed square for odd n --- see Prop 4.1?


问:
Functions in $H^s$ vanish at infinity
I am learning about sobolev spaces and try to prove some simple properties. Our definition of $H^s(\mathbb{R}^n)=W^{k,2}(\mathbb{R}^n)$ is the following

$$
f \in H^s \text{ if and only if there exists a sequence } f_k \in C_0^{\infty}(\mathbb{R}^n), \text{ such that } \Vert f_k-f\Vert_{H^s} \rightarrow 0
$$
We defined the norm for the space $H^s$ to be $$ \Vert f \Vert_{H^s} = \Big(\int_{\mathbb{R^n}}{(1+\vert y \vert^2)^s \vert \hat f(y) \vert^2dy}\Big)^{1\diagup2},$$ where $\hat f$ is the fourier transform of $f$.

For $s > \frac{n}{2}.\; $Let $f \in H^s(\mathbb{R}^n)$, than $f$ vanishes at $\infty$

I know that if $f_k \in C_0^{\infty}(\mathbb{R}^n)$ converges uniformly to $f$, than $f$ vanishes at $\infty$. So my "idea" would be to show that convergence in $\Vert \cdot \Vert_{H^s}$ for $s > \frac{n}{2}$ implies uniform convergence. Would appreciate any hint/help  to prove the statement above using only the definition I have given.

答:
This is very much related to a common Sobolev embedding theorem, namely that $H^s$ embeds continuously into $C^k\cap L^\infty$ if $s>k+n/2$ (for you, $k=0$).

Observe that if $u\in H^s$, then
$$\int|\hat{u}(\xi)|\, d\xi=\int|\hat{u}(\xi)|(1+|\xi|^2)^{-s/2}(1+|\xi|^2)^{s/2}\, d\xi\leq \left(\int (1+|\xi|^2)^{-s}\, d\xi\right)^{1/2}\|u\|_{H^s}<\infty.$$ Hence, $\hat{u}\in L^1,$ then an application of the Riemann-Lebesgue lemma yields that $u$ vanishes at infinity.

Title: Functions in $H^s$ vanish at infinity


问:
To show a subspace is dense in a normed linear space.
Let $A$ be a normed linear space and $B$ be a subspace of $A$. In order to show $B$ is dense in $A$ it is enough to show that for any $T\in A^*$(dual space of $A$), if $T$ vanishes on $B$ then $T$ is identically zero.

My questions:

Let $C$ be dense subspace of $A^*$. Then is the statement :
"In order to show $B$ is dense in $A$ it is enough to show that for any $T\in C,$ if $T$ vanishes on $B$ then $T$ is identically zero" true?

If (1) is true then can it be generalized further in some sense?

答:
I believe the claim is false.

Let $A = \ell^2$ and let $B$ be the orthogonal complement of the sequence $\left(\frac1n\right)_n$ in $\ell^2$. Then $A^* = \ell^2$ as well and set $C = c_{00}$, the subspace of all finitely-supported sequences.

Then $B$ is not dense in $A$, but for every $f \in C$ holds $f|_B = 0 \implies f=0$. Indeed, let $f =(\alpha_n)_n \in C$ be such that $f|_B=0$. For every $n\in\Bbb{N}$ we have that $$e_1-ne_n \perp \left(\frac1n\right)_n \implies e_1-ne_n \in B \implies f(e_1-ne_n) = 0 \implies \alpha_1 - n\alpha_n=0 \implies \alpha_n = \frac1n\alpha_1.$$
But if $\alpha_1 \ne 0$, we would have $\alpha_n \ne 0$ for all $n \in \Bbb{N}$ contradicting the fact that $f \in c_{00}$. Hence $\alpha_1 = 0$ and consequently $\alpha_n = 0$ for all $n\in\Bbb{N}$. Therefore $f=0$.


问:
If $A + \Delta A$ has eigenvalue $\lambda$, show that $A$ has eigenvalue $\lambda + \Delta \lambda$.
$A$ is symmetric and can be diagonalised as $A = V\Lambda V^{-1}$, where $\Lambda$ is a diagonal matrix with eigenvalues of $A$ and $V$ is an orthogonal matrix with column vectors equal to eigenvectors of $A$.

If $A + \Delta A$ has eigenvalue $\lambda$,  show that $A$ has eigenvalue $\lambda + \Delta \lambda$ with

$$\Delta \lambda = 0 \qquad\text{or}\qquad |\Delta \lambda| = ||(\lambda I - \Lambda)^{-1}||_2^{-1}.$$

My attempt:

$$A u = (\lambda + \Delta \lambda)u$$
$$A u = \lambda u + \Delta \lambda u$$
$$- \Delta \lambda u = (\lambda I - A)u$$
$$-(\lambda I - A)^{-1} \Delta \lambda u = u$$
$$||u||_2 = ||(\lambda I - A)^{-1} \Delta \lambda u||_2$$

Where do I go from here?

答:
The key to this problem, I think, is to realize that $A + \Delta A$ doesn't tell us anything about $A$.

By replacing $A$ with $(A - \lambda I)$, suppose without loss of generality that $\lambda = 0$.

We are given that $A + \Delta A$ has eigenvalue $0$, and we want to show that it must either hold that $A$ has eigenvalue $0$ or $A$ has eigenvalue $\Delta \lambda$ with $|\Delta \lambda| = \|\Lambda^{-1}\|^{-1}$. Equivalently, if $A$ does not have $0$ as an eigenvalue (i.e. $A$ is invertible), then $A$ has an eigenvalue $\Delta \lambda$ with $|\Delta \lambda| = \|\Lambda^{-1}\|^{-1}$.

This is easy to see: suppose that $A$ is invertible. We write
$$
\Lambda = \pmatrix{\lambda_1 \\ & \ddots \\ && \lambda_n},
$$
where each $\lambda_i$ is an eigenvalue of $A$. We note that
$$
\Lambda^{-1} = \pmatrix{\lambda_1^{-1}\\ & \ddots \\ && \lambda_n^{-1}},
$$
from which it follows that
$$
\|\Lambda^{-1}\|_2 = \max_{j=1,\dots,n} |\lambda_j|^{-1} = (\min_{j=1,\dots,n} |\lambda_j|)^{-1}.
$$
Thus, we have $\|\Lambda^{-1}\|_2^{-1} = \min_{j=1,\dots,n}|\lambda_j|$. So, it is indeed the case that $A$ has an eigenvalue $\Delta \lambda = \lambda_j$ for which $|\lambda_j| = \|\Lambda^{-1}\|^{-1}$, which is what we wanted to show.

A less "matrix-dependent" approach: again, suppose WLOG that $\lambda = 0$, and suppose that $A$ is invertible. We note that for $z \in \Bbb C$,
$$
A - zI \text{ is invertible }\iff I - z A^{-1} \text{ is invertible }.
$$
For $z$ with $|z|$ sufficiently small, we see that $I - zA^{-1}$ must be invertible because the inverse can be expressed via the convergent Neumann series
$$
(I - zA^{-1}) = \sum_{k=0}^\infty A^{-k} z^k.
$$
The radius of convergence of this series is equal to $\rho(A^{-1})^{-1}$, which is equal to $\|\Lambda^{-1}\|_2^{-1}$ (since $A$ is real and symmetric, this is in turn equal to $\|A^{-1}\|_2^{-1}$). By the analyticity of holomorphic functions, there exists a $z \in \Bbb C$ with $|z| = \|\Lambda^{-1}\|^{-1}$ at which the function $(I - zA^{-1})^{-1} = A^{-1}(A - zI)^{-1}$ has a singularity, which is to say that $A - zI$ fails to be invertible. In other words, such a $z$ must be an eigenvalue of $A$.


问:
Exponential Models Help

The volume of a substance, A, measured in cubic centimeters increases according to the exponential growth model $\frac{\mathrm{d}A}{\mathrm{d}t} = 0.3A$, where $t$ is measured in hours.

The volume of another substance, B, also measured in cubic centimeters increases at a constant rate of
$1\ \mbox{cm}^{3}$ per hour according to the linear model $\frac{\mathrm{d}B}{\mathrm{d}t} = 1.$

At $t = 0$, substance A has a volume $A\left(0\right) = 3$ and substance B has size $B\left(0\right) = 5$.

I got $3.4531$ using the equations $\frac{1}{0.3}log|\frac{A}{3}|= t$ and $B-5=t,$ is that correct?

答:
Starting from @RyanK's answer, we have
$$t=-\frac{10}{3}\, W_{-1}\left(-\frac{9}{10\, e^{3/2}}\right)-5$$ and this is the only analytical solution.

If you cannot use Lambert function, you will need a numerical method and then a reasonable guess of the root.

Consider that you are looking for the zero of function
$$f(t)=3 e^{3 t/10}-t-5$$ for which
$$f'(t)=\frac{9}{10}\, e^{3 t/10}-1 \qquad \text{and}\qquad f''(t)=\frac{27}{100}\, e^{3 t/10} \quad >0 \quad \forall t$$
The first derivative cancels at
$$t_*=\frac{10}{3} \log \left(\frac{10}{9}\right)$$ which is a minimum; since
$$f(t_*)=-\frac{5}{3}-\frac{10}{3} \log \left(\frac{10}{9}\right) <0$$ build a Taylor series around $t_*$. This will give, as an approximation*,
$$f(t)=-\left(\frac{5}{3}+\frac{10}{3} \log \left(\frac{10}{9}\right)\right)+\frac{3}{20}
   \left(t-\frac{10}{3} \log
   \left(\frac{10}{9}\right)\right)^2+O\left(\left(t-\frac{10}{3} \log
   \left(\frac{10}{9}\right)\right)^3\right)$$ solve the quadratic and keep the largest root. This will give
$$t_0=\frac{10}{3} \left(\log \left(\frac{10}{9}\right)+\sqrt{1+2 \log
   \left(\frac{10}{9}\right)}\right)\approx 4.01896$$ which does not look too bad.

Now, using Newton method, we should have the following iterates
$$\left(
\begin{array}{cc}
n & t_n \\
 0 & 4.01896 \\
 1 & 3.52114 \\
 2 & 3.45418 \\
 3 & 3.45305
\end{array}
\right)$$

Title: Exponential Models Help


问:
A problem in limits to show that $\lim_{n\to\infty}$ $p_n$/$q_n$ = $π/4$
Define two sequences of real numbers $p_n$ and $q_n$ with $n = −1, 0, 1, 2, \ldots$ as follows:
$$p_n = 2p_{n−1} + (2n−1)^2p_{n−2}$$ and
$$q_n = 2q_{n−1} + (2n−1)^2q_{n−2}$$
for every $n \geq 1$, and starting with $p_{−1}= 0, q_{−1}= 1, p_0 = q_0 = 1$.
Show that
$$\lim_{n \to \infty}\frac{p_n}{q_n} = \frac{\pi}{4}.$$

I did:
Suppose $a_n$ is a sequence that satisfies $a_n = 2a_{n−1} + (2n − 1)^2a_{n−2}$ for all
$n\geqslant1$. Subtracting $(2n + 1)a_{n−1}$ from both sides we derive a simple recursion for a new
sequence $b_n = a_n − (2n + 1)a_{n−1}$, namely $b_n = −(2n − 1)b_{n−1}$.

What to do after this... Is it correct?
If not kindly provide a detailed answer in the answer section.. As it helps a new beginner like me to understand who has just started to solve advanced problems.

答:
Let $r_n := p_n/q_n$. We can compute the first few times to observe that: $r_0 = 1$, $r_1 = \frac{2}{3} = 1-\frac{1}{3}$, $r_2 = \frac{13}{15} =1-\frac{1}{3} + \frac{1}{5}$, $r_3 = \frac{76}{105} = 1-\frac{1}{3} + \frac{1}{5} -\frac{1}{7}$. So this gives us enough motivation to conjecture that: $$\frac{p_n}{q_n} = 1 -\frac{1}{3} + \frac{1}{5} - \cdots + (-1)^n \frac{1}{2n+1} $$
for all $n \ge 0$. The first case is done, as stated above. Suppose the hypothesis for $n-1$ is true. Then \begin{align*}
\frac{p_n}{q_n} & = \frac{2 p_{n-1}}{q_n} + \frac{(2n-1)^2 p_{n-2}}{q_n} \\
& = 2\bigg(1-\frac{1}{3} + \cdots (-1)^{n-1}\frac{1}{2n-1}\bigg)\frac{q_{n-1}}{q_n} +(2n-1)^2 \bigg(1 - \frac{1}{3} + \cdots + (-1)^{n-2} \frac{1}{2n-3}\bigg) \frac{q_{n-2}}{q_n} \\
&=  \bigg(1 - \frac{1}{3} + \cdots + (-1)^{n-2} \frac{1}{2n-3}\bigg) + 2\frac{q_{n-1}}{q_n} (-1)^{n-1}\frac{1}{2n-1} \\
&= \bigg(1 - \frac{1}{3} + \cdots + (-1)^{n-2} \frac{1}{2n-3} +(-1)^{n-1}\frac{1}{2n-1}\bigg) + (2n-1)\frac{q_{n-2}}{q_n} (-1)^n \quad -\star
\end{align*}

Where I applied IH to both terms $p_{n-1}$ and $p_{n-2}$ in the second line. I now claim that $\frac{q_{n-2}}{q_n} =\frac{1}{(2n-1)(2n+1)}$ for all $n$. As usual, the base case is easily verified. Now suppose the hypothesis for $n-1$ is true. Then \begin{align*}
q_{n-2} &= 2q_{n-3} + (2n-5)^2 q_{n-4} \\
& = \frac{2q_{n-1}}{(2n-3)(2n-1)} + (2n-5)\frac{q_{n-2}}{2n-3} \quad \text{Apply IH to both terms} \\
& =\frac{1}{(2n-3)(2n-1)}\bigg(2 q_{n-1} + (2n-5)(2n-1) q_{n-2} \bigg) \\
&= \frac{1}{(2n-3)(2n-1)} \bigg(q_n - 4(2n-1)q_{n-2}\bigg)
\end{align*}
You can check that this easily simplifies to $\displaystyle \frac{q_{n-2}}{q_n}=\frac{1}{(2n-1)(2n+1)}$, which is what we want! Now substituting this back into the $(\star)$ we have proven by induction that:

$$\frac{p_n}{q_n} = 1 -\frac{1}{3} + \frac{1}{5} - \cdots + (-1)^n \frac{1}{2n+1} $$

So if we let $n\rightarrow \infty$, we see that this term is just equal to the series $$1-\frac{1}{3}+\frac{1}{5} - \frac{1}{7} + \cdots $$ This is precisely equal to the $\arctan 1$! (Search up the Taylor series expansion for arctan for this.) So this value is just $\displaystyle \frac{\pi}{4}$, which is what we desired.


问:
Testing to see if triangles intersect

I’m curious, because it was an interesting question, and would like someone to please clarify / help me determine if my statement is actually true.

So by subtraction of co-ordinates and applying Pythagoras, they could use the co-ordinate values to ensure the distances are greater than s. Thanks. Are there any better ways of doing this?

答:
As rightly pointed out by @TobyMak one condition is to determine if any vertex of a triangle is, or is not, inside another triangle.

However, if all vertices of a triangle are outside another, this will also include the case that the two triangles intersects.

So we shall counter-test also for this condition, if overlapping shall be excluded.

Now to test if a point is external to a given triangle, we have better use the barycentric coordinates
approach.

Given the triangle $ABC$ with vertices $(x_A, y_A),\, (x_B, y_B), \, (x_C, y_C)$ we construct the matrix
$$
{\bf T} = \left( {\matrix{
   {x_A } & {x_B } & {x_C }  \cr
   {y_A } & {y_B } & {y_C }  \cr
   1 & 1 & 1  \cr
 } } \right)
$$

A point $P=(x,y)$ will be internal to the triangle if its barycentric coordinates wrt $\triangle ABC$ are all positive, i.e.
iff
$$
\left( {\matrix{   {\lambda _{\,1} }  \cr    {\lambda _{\,2} }  \cr    {\lambda _{\,3} }  \cr } } \right)
 = {\bf T}^{\, - \,{\bf 1}}
 \left( {\matrix{   x  \cr    y  \cr    1  \cr  } } \right)
\quad  \Rightarrow \quad 0 < \lambda _{\,1} ,\lambda _{\,2} ,\lambda _{\,3}
$$

The $\lambda _k$ sum to $1$, so they cannot be all negative.

If one at least is negative, then the point $P$ is external.

And if it is $\lambda_1$ to be negative, then you know that $P$ is on the side of the edge $BC$ opposite to $A$.

So, if you take the $\triangle XYZ$ and modify its coordinates by a translation and a rotation
$$
\eqalign{
  & {\bf X} = \left( {\matrix{
   {x_X } & {x_Y } & {x_Z }  \cr
   {y_X } & {y_Y } & {y_Z }  \cr
   1 & 1 & 1  \cr
 } } \right)\quad  \Rightarrow   \cr
  & {\bf X}' = \left( {\matrix{
   {\cos \alpha } & { - \sin \alpha } & 0  \cr
   {\sin \alpha } & {\cos \alpha } & 0  \cr
   0 & 0 & 1  \cr
 } } \right)\left( {\left( {\matrix{
   {x_X } & {x_Y } & {x_Z }  \cr
   {y_X } & {y_Y } & {y_Z }  \cr
   1 & 1 & 1  \cr
 } } \right) + \left( {\matrix{
   u & u & u  \cr
   v & v & v  \cr
   0 & 0 & 0  \cr
 } } \right)} \right) \cr}
$$
then by taking the product
$$
{\bf \Lambda } = {\bf T}^{\, - \,{\bf 1}} \;{\bf X}'
$$
you can adjust the parameters $\alpha, u, v$ in such a way that at least one component in every column
of $\Lambda$ be negative, to ensure that all the vertices of $\triangle X'Y'Z'$ are external to $\triangle ABC$.

Finally, to test that the two triangles do not overlap,  we take a generic point $Q$ internal to $\triangle X'Y'Z'$
$$
Q = {\bf X}'\,
 \left( {\matrix{   {\mu _{\,1} }  \cr    {\mu _{\,2} }  \cr    {\mu _{\,3} }  \cr } } \right)
 \quad \left| \matrix{
  \,0 \le \mu _{\,1} ,\mu _{\,2} ,\mu _{\,3} \left( { \le 1} \right) \hfill \cr
  \,\mu _{\,1}  + \mu _{\,2}  + \mu _{\,3}  = 1 \hfill \cr}  \right.
$$
and test that not all the components of
$$
{\bf T}^{\, - \,{\bf 1}} Q = {\bf T}^{\, - \,{\bf 1}} {\bf X}'\left( {\matrix{
   {\mu _{\,1} }  \cr
   {\mu _{\,2} }  \cr
   {\mu _{\,3} }  \cr
 } } \right) = {\bf \Lambda }\left( {\matrix{
   {\mu _{\,1} }  \cr
   {\mu _{\,2} }  \cr
   {\mu _{\,3} }  \cr
 } } \right)
$$
are positive for any of the allowed values of $\mu _k$.

Title: Testing to see if triangles intersect


问:
Prove that from $\mathbb E(X\mid Y) \ge Y, \mathbb E(Y\mid X) \ge X$ follows that $X = Y$.
Let $X$ and $Y$ discrete random variables. Prove that from $\mathbb E(X\mid Y) \ge Y, \mathbb E(Y\mid X) \ge X$ follows that $X = Y$.

I tried to prove it by contradiction. If the statement is not true, then without loss of generality $X \lt Y.$ Then from $X \le \mathbb E(Y\mid X)$ we have that $Y \lt \mathbb E(Y\mid X)$.

As a result, we have $Y \lt \mathbb E(Y\mid X)$ and $Y \le \mathbb E(X\mid Y)$, what can we do next?

答:
"If the statement is not true, then without loss of generality $X < Y$."

This is wrong.  What you can say is that without loss of generality $\mathbb P(X < Y) > 0$.


问:
A counter example which shows that $\Omega(f|_{\Omega}) \neq \Omega(f)$
I'm looking for an example :

A counter example which shows that $\Omega(f|_{\Omega}) \neq \Omega(f)$

$(X,f)$ is a Dynamical System if $f:X \to X$ is a homeomorphism and $X$ is a compact space. $\Omega(f)$ is the set of non wandering points, i.e. all $x$ such that $\forall$ U open containing $x$ and $\forall$ $N>0$ there exists some $n>N$ such that $f^n(U) \cap U \ne \emptyset$.

I think we have to take $f$ on a circle but I don't know how to compute the non wandering set.

答:
My previous answer have an error. I will try differrently.

We will work on polar coordinate.

Our space will be the union of the sets $C_{n} = \{(\frac{n}{n+1},\frac{\pi}{k}) , k \in \{-n,...,-1,1,...,n \}\}$ for $n \in \mathbb{N}^* $ with $C_{\infty} = \{(1,\frac{\pi}{k}) k \in \mathbb{Z}^* \} \cup \{(1,0)\}$ and with $D=\{(n,0) ,n \geq 2 \}$ with the topology induced by $\mathbb{C}$.

Here a picture of the situation, the blue points are in the sets $C_k$ and the red one in $C_\infty$

We will consider the function $f$ defined by :

On $D$

$f(n,0)=(n-1,0)$ if $n \ne 2$

$f(2,0)=(1/2,\pi) \in C_{1}$

On $C_k$

$f(\frac{n}{n+1},\frac{\pi}{k})=(\frac{n}{n+1},\frac{\pi}{k+1})$ if $k \ne n$ and $k \ne -1$

$f(\frac{n}{n+1},\pi)=(\frac{n}{n+1},\frac{\pi}{2})$ if $k=-1$

$f(\frac{n}{n+1},\frac{\pi}{n})=(\frac{n+1}{n+2},\frac{-\pi}{n}) \in C_{k+1}$ if $k =n $

And on $C_{\infty}$

$f(1,\frac{\pi}{k})=(1,\frac{\pi}{k+1})$ if $k \ne -1$

$f(1,\pi)=(1,\frac{\pi}{2})$ if $k =-1$

$f(1,0) = (1,0)$

We have that $f$ is bijective. $\underset{k}{\cup} C_k \cup D$ and $C_{\infty}$ are dynamically speaking two bi-infinite shift and $(1,0)$ is a fixed point.

Let's show that $f$ is a homeomorphism.

$f$ is continuous and with continuous inverse at every point of $\underset{k}{\cup} C_k \cup D$ because the topology there is discrete.

Now on $C_{\infty}$, take a sequence $(\frac{n_i}{n_i+1},\frac{\pi}{k_i}) \underset{i \to \infty}{\to} (1,\frac{\pi}{k})$. We should have that $n_i \underset{i \to \infty}{\to} \infty $ and $k_i \underset{i \to \infty}{\to} k $ so for every $i>I$ for $I$ large enough, $k_i \ne n_i $ and  $f(\frac{n_i}{n_i+1},\frac{\pi}{k_i}) = (\frac{n_i}{n_i +1},\frac{\pi}{k_i +1}) \underset{i \to \infty}{\to} (1,\frac{1}{k +1})= f(1,\frac{1}{k})$. We avoid the difficulties $k =-1 $ because $-\pi = \pi$ on polar coordinate.

If $(\frac{n_i}{n_i+1},\frac{\pi}{k_i}) \underset{i \to \infty}{\to} (1,0)$ then we could have for infinitely many i, $k_i = n_i$ but then $f(\frac{n_i}{n_i+1},\frac{\pi}{k_i}) = (\frac{n_i+1}{n_i + 2},\frac{-\pi}{n_i +1 }) \underset{i \to \infty}{\to} (1,0)= f(1,0)$

So $f$ is continuous.

The same can be done for $f^{-1}$ it "just" rotate conterclockwise and sometimes can "gain" a level in $C_k$.

Now we have that $\underset{k}{\cup} C_k \cup D$ is not in $\Omega(f)$ since the topology there is discret, you can just take a singleton wich won't intersept itself after iteration of $f$.

$C_{\infty}$ is in $\Omega$, indeed for every $(1,\frac{\pi}{k})$, every open set which contain it, should contain  $ \{ (\frac{n}{n+1},\frac{\pi}{k}) , n \geq N\}$ for a $N$ large enough. This subset intersect itself many times.

So $\Omega(f)=C_{\infty}$, but on $C_{\infty}$ $f$ is just a shift and a fixed point so $\Omega(f_{| \Omega}) = \{ (0,1) \}$


问:
Given 2 points in 3d, how to generate a plane equation in the form $lx+my+nz=d$
I can do this for 2d points by getting the 2 parametric equations for $x$ and $y$ and then solving by eliminating $t$ and getting the 2d equivilent line equation. There is only one line equation in this case though.

In 3d there will be infinitely many planes but I only need one of them for my use case. I can still generate the symmetric line by doing the below

\begin{align*}x & = {x_0} + ta\\ y & = {y_0} + tb\\  z & = {z_0} + tc\end{align*}

which rearranges to

\begin{align*}\frac{{x - {x_0}}}{a} = \frac{{y - {y_0}}}{b} = \frac{{z - {z_0}}}{c}\end{align*}

What I want to do is get one of the planes that does go through the 2 points in  the form

\begin{align*}lx+my+nz = d\end{align*}

The reason for doing this is to solve a plane / Bezier curve intersection (which I have working in 2d)

Example given the following cubic

\begin{align*}\mathbf{X}(t) = (1-t)^3\mathbf{P}_0 + 3t(1-t)^2\mathbf{P}_1 +
               3t^2(1-t)\mathbf{P}_2 + t^3\mathbf{P}_3\end{align*}

And the following line vector equation

\begin{align*}\mathbf{A}\cdot \mathbf{X} = d\end{align*}

with \begin{align*}\mathbf{A} = (l,m,n)\end{align*}

I just need to solve

\begin{align*}(1-t)^3(\mathbf{A} \cdot \mathbf{P}_0) +
3t(1-t)^2(\mathbf{A} \cdot \mathbf{P}_1) +
3t^2(1-t)(\mathbf{A} \cdot \mathbf{P}_2) +
t^3(\mathbf{A} \cdot \mathbf{P}_3) - d = 0\end{align*}

to be able to get the intersection points.

So how do I generate a example plane equation that does go through these 2 points?

Update :

I realise with 2 points there are infinitely many planes, I have rephrased the question to ask how to generate one of them or the formula for generating all of them.

答:
Suppose given $\dfrac{{x_1 - {x_0}}}{a} = \dfrac{{y_1 - {y_0}}}{b} = \dfrac{{z_1 - {z_0}}}{c}$. Where $x_1 ,y_1 ,z_1$ is any point on the line and $x_0 ,y_0 ,z_0$ is the fixed point on the line.

Imagine another line going through the same fixed point $x_0 ,y_0 ,z_0$. We write this equation as $\dfrac{{x_2 - {x_0}}}{a_1} = \dfrac{{y_2 - {y_0}}}{b_1} = \dfrac{{z_2 - {z_0}}}{c_1}$

Now we consider any line which lies on the plane created by this two line. Consider a point $x,y,z$ on this plane so

$x-x_0 = \alpha(x_1 - {x_0})+\beta(x_2-x_0) = \alpha at+\beta a_1k\\
y-y_0 = \alpha(y_1 - {y_0})+\beta(y_2-y_0) = \alpha bt+\beta b_1k\\
z-z_0 = \alpha(z_1 - {z_0})+\beta(z_2-z_0) = \alpha ct+\beta c_1k\
$

Solve for $\alpha t$ and $\beta k$ from first two equations and replace those in third equation.


问:
For all $m,n \in \Bbb{N}$, $n>m$, there is a unique $k \in \Bbb{N}$ such that $km \le n < (k+1)m$
For all $m,n \in \Bbb{N}$, $n>m$, there is a unique $k \in \Bbb{N}$ such that $km \le n < (k+1)m$

My attemt for existence (is this approach correct?)

Let $m,n \in \Bbb{N}$ and consider the set $S = \{q \in \Bbb{N} | qm \le n \}$

Since $1 \in S$, $S$ is nonempty, so by the Well Ordering Principle there is a least element $k \in S$.

Then, suppose $(k+1)m \le n$

Then $km \le n - m = 0$, which is absurd, as $km \in \mathbb{N}$. So $n < (k+1)m$ and then $km \le n < (k+1)m$.

Is this correct?

For the uniqueness, I'm stuck. I tried supposing there are different $k,k'$ and using the two inequalities for each, but can't get to a contradiction...

答:
I believe you need the greatest $k\in S$. Try, for instance, $n=5$, $m=1$. Then the smallest element in $S$ is $k=0$ and ... you see where this is going. Fortunately, you can show that $S$ is bounded, and hence that it achieves a maximum. This is also a consequence of the well ordering principle. The fact that $S$ is bounded means $\mathbb{N}-S$ is non-empty. The maximum of $S$ is one less than the minimum of $\mathbb{N}-S$.

If $k$ is the max, then $k+1$ is not in $S$ by definition. Note that this approach kind of hints at the uniqueness argument ... what's wrong with choosing a different $k$? Can you take it from here / fill in the uncited principles?

Edit: Also, $n-m>0$.

Edit: I am adding some comments on uniqueness. For $k=max(S)$, $k+1\in \mathbb{N}-S$ by definition. Therefore, $(k+1)m\leq n$ is not true, and the negation of that inequality is $n< (k+1)m$. So, in this situation, $km\leq n<(k+1)m$.

If we choose $k$ such that $k>max(S)$, then we lose the $km\leq n$ since $k$ is not in $S$.

If we choose $k$ such that $k<max(S)$, then $k+1\leq max(S)$ and $$(k+1)m\leq max(S)m\leq n$$
and we lose $n<(k+1)m$.

Does this help?


问:
Convergence in distribution of Geometric r.v.s sequence

Let $X_n$ be a sequence of r.v.s with Geometric distribution with parameter $p_n=\frac{5}{n}$. Prove that $Z_n = \frac{X_n}{n}$ converges in distribution. What is its limiting distribution?

I'm studying for my math exam(from all three previous years) but I completely forgot how would one do such task. I guess we'd like to use CLT here but no further progress has been made.

Thanks for any help.

答:
Using MGF you get

$$MGF\Big(\frac{x_n}{n}\Big)=\frac{\frac{5}{n}}{1-\Big(1-\frac{5}{n}\Big)e^{\frac{t}{n}}}\xrightarrow{n\to \infty}=\frac{0}{0}\xrightarrow{H}\frac{5}{5e^{\frac{t}{n}}-\Big(1-\frac{t}{n}\Big)e^{\frac{t}{n}}t}=\frac{5}{5-t}$$

Thus

$$\frac{x_n}{n}\xrightarrow{d}Exp(5)$$

Because we recognize the MGF of an exponential distribution with parameter $\lambda=5$


问:
Derived category and so on
I am looking for an introductive reference to the theory of derived categories. Especially I need to start from the very beginning and I need to know how to use this in examples which comes from algebraic geometry. I don't want a too rigorous approach, made of a lot of definition and propositions but instead I would like to find an introduction which gives the main ideas and many examples.
Thank you</p>

答:
You'll find Bernhard Keller's notes for a short course on the subject in his web page. His exposition is characteristically lucid and clear. His focus is representation theory, so they may not match your interests, though.

I also like a lots the to-the-pointness approach taken by Dieter Happel in his book about triangulated categories.

Title: Derived category and so on


问:
Solve a System with Variable
Given these matrices, how does one find two real solutions?

$dx/dt$ =
$\begin{bmatrix}
3 & -5\\
5 & 3
\end{bmatrix}x$

with $x(0) = \begin{bmatrix}
2\\
-3
\end{bmatrix}$

答:
Hints:

We are given the system:

$$x' = Ax = \begin{bmatrix}
3 & -5\\
5 & 3
\end{bmatrix}x$$

with IC:

$$x(0) = \begin{bmatrix} 2\\ -3 \end{bmatrix}$$

The general solution will be:

$$x_1(t) = c_2 \sin 4 t + c_1 (3 \sin 4 t + 4 \cos 4 t)$$

$$x_2(t) =  c_1 \sin 4 t + c_2 (4 \cos 4 t - 3 \sin 4 t)$$

Using the IC, will yield the final solution of:

$$x_1(t) = \frac{21}{4} \sin 4 t + 2 \cos 4 t$$

$$x_2(t) = \frac{19}{4} \sin 4 t - 3 \cos 4 t$$

Title: Solve a System with Variable


问:
Definition of a random sequence of random variables
The following is a passage from Durrett's book on probability.

''If $X_n \in \{0,1\}$ are independent with $P(X_n=1)=a_n \to 0$ and $\sum a_n = \infty$, then $X_n \to 0$ in probability, but if we let $N(n)=\text{inf}\{m\geq n: X_m=1\}$, then $X_{N(n)}=1$ a.s.''

The statement itself is not important, but I'm having trouble understanding the sequence $X_{N(n)}$ in a rigorous way. Can someone explain how the random variables $\{X_{N(n)}\}$ look like as measurable functions and also as a subsequence of the original sequence $\{X_n\}$?

答:
It's all complicated in the beginning. I personally feel that it is much easier to understand it in terms of measure theory.

Recall that $X_n$ is just a measurable function on some probability space (about which we usually don't care). But let us fix this probability space $(\Omega, \mathcal{F}, \mathbb{P}).$ So $X_n:(\Omega, \mathcal{F}, \mathbb{P})\to \{0, 1\}$ is a measurable function with some properties (but I will ignore them as you said the exact statement is not important).

Now if you have another measurable function $N:(\Omega, \mathcal{F}, \mathbb{P})\to \mathbb{N},$ then you can define a compostie function say $Y:(\Omega, \mathcal{F}, \mathbb{P})\to \{0, 1\}$ by $Y(\omega)=X_{N(\omega)}(\omega).$ This is what we are talking about.

Probabilistically, think of the situation as follows: Think of $X_n$ as being representing head/tail in a coin toss. But let us say you are interested in some particular $X_n,$ for example, you might be interested in what happens in the $10$th toss, that is, you are interested in $X_{10}.$ This is simple. But you might be interested in something more complicated. Imagine that you are playing a game where you win if you see $5$ consecutive zeroes and then $1.$ Naturally, if you look at the time of your first win, it is a random time, you can call it $N.$

So fundamentally, at the level of measure theory, we are just talking about the composition of measurable functions. In terms of probability, we have a sequence but we are now chosing one of the term in the sequence randomly.


问:
multivariable convex function
Consider the following two-variable function:
\begin{equation}
f(x,t)=h(x)+g(t)
\end{equation}
assume the following relation holds for $0\leq\lambda\leq 0$:
\begin{align}
&h(\lambda x +(1-\lambda) y)\leq \lambda h(x)+(1-\lambda)h(y)\notag\\
&g(\lambda t +(1-\lambda) s)\leq \lambda g(t)+(1-\lambda)g(s)
\end{align}
can we conclude that the function $f(x,t)$ is also convex?

more generally if we have an arbitrary two-variable function $k(x,t)$ which is convex w.r.t its individual arguments, can we conclude that it is convex as a multivariable function as well? i.e.

if
\begin{align}
&k\Big((\lambda x +(1-\lambda) y),t\Big)\leq \lambda k(x,t)+(1-\lambda)k(y,t)\notag\\
&k\Big(x,(\lambda t +(1-\lambda) s)\Big)\leq \lambda k(x,t)+(1-\lambda)k(x,s)
\end{align}
Can we conclude that
\begin{align}
&k\Big((\lambda x +(1-\lambda) y),(\lambda t +(1-\lambda) s)\Big)\leq \lambda k(x,t)+(1-\lambda)k(y,s)
\end{align}

答:
The answer to the first question is “yes”: Both functions
$$
 (x, t) \mapsto h(x) \\
 (x, t) \mapsto g(t)
$$
are convex, and the sum of convex functions is convex.

The answer to the second question is “no”: A counterexample is $f(x, t) = xt$, which is linear (and therefore convex) in both variables separately, but not convex:
$$
 1 = f(1, 1) \le \frac 12 \left( f(2, 0) + f(0, 2) \right) = 0
$$
does not hold.

Title: multivariable convex function


问:
Estimate value of function
I'm asked:

For the function $f(x) = x + 1 - \frac{1}{n} , x \in [\frac{n-1}{n},
> \frac{n}{n+1}]$ in the interval $[0, 1)$.

a) Prove it can be extended by continuity for $x=1$. Call $g$ to that
new functioon.

b) calculate an upper and a lower estimates for $\int_0^1g$

I really don't know how to start point (a) because $x$ is between those intervals, I believe I cannot simply put $x \rightarrow 1^-$.

The second problem I have is with estimates. What am I supposed to do in order to calculate those estimates? Should I use those intervals $[\frac{n-1}{n},
> \frac{n}{n+1}]$ and multiply the value of supremum/infimum of $g(x)$ there in order to calculate the upper/lower estimates respectively? What would that look like?

答:
To have continuity at $x=1$ we set $g(x)=x+1$ so that when $n\to\infty$ we have $x\to 1$ and $f(x)\to 2$.

The width of intervals is $$\frac{n}{n+1}-\frac{n-1}{n}=\frac{1}{n^2+n}$$
We can define Riemann sums considering the left value and the right value of the interval. Left sums give $$s_n=\sum _{k=1}^{n } \left(\frac{k-1}{k}+1\right)\frac{1}{k^2+k}=\sum _{k=1}^{n } \frac{2 k-1}{k^2 (k+1)}$$
partial fractions $$\frac{2 k-1}{k^2 (k+1)}=3\left(\frac{1}{k}-\frac{1}{k+1}\right)-\frac{1}{k^2}$$
As $n\to\infty$ the first series is a telescoping one and gives $3$, the second is the famous $\zeta(2)=\frac{\pi^2}{6}$, therefore
$$s=\sum _{k=1}^{\infty } \frac{2 k-1}{k^2 (k+1)}=3-\frac{\pi^2}{6}\approx 1.355$$
The right sums are
$$S_n=\sum _{k=1}^{n } \left(\frac{k}{k+1}+1\right)\frac{1}{k^2+k}=\sum _{k=1}^{n } \frac{2 k+1}{k (k+1)^2}$$
like the series above we get
$$S=\sum _{k=1}^{\infty } \frac{2 k+1}{k (k+1)^2}=\frac{\pi ^2}{6}\approx 1.645 $$

Title: Estimate value of function


问:
Dynamic dice game: optimal price to enter
Here's a dynamic dice game. We can roll dice as many times as we want unless a 6 appears. After each roll, if 1 appears, we win \$1, if 2 appears we win \$2 etc. If 6 appears, we lose all money and game ends. We can decide if we want to continue the game after witnessing anything between 1 and 5. How much would you be willing to pay to play the game?

My approach:
I first computed the expected payoff of the game. Sa $x$ is the expected payoff. Then we have
$$x = \frac{1}{6}(1+x)+\frac{1}{6}(2+x)+\frac{1}{6}(3+x)+\frac{1}{6}(4+x)+\frac{1}{6}(5+x) $$
This gives me $x=15$.

What next? How to find the optimal price to play the game before it begins? Thanks.

答:
A strategy is simply given by a set $S\subset\Bbb N$ and amount to "If we have collected $x$ then stop of $x\in S$ and continue if $x\notin S$". We assume there is some optimal strategy $S_0$.

Let $E(x)$ be the expected final payout if we have already collected $x>0$ and make optimal decisions from now on.
Clearly, we have
$$\tag1 E(x)=\begin{cases}x&x\in S_0\\
\frac{E(x+1)+E(x+2)+E(x+3)+E(x+4)+E(x+5)}6&x\notin S_0\end{cases}$$
With strategy $S=\Bbb N$, we will of course stop immediately and have payout $x$. Therefore,
$$\tag2 E(x)\ge x.$$
With $(1)$, this gives us the lower estimate
$$ E(x)\ge \frac{E(x+1)+\cdots+E(x+5)}6\ge \frac{(x+1)+\cdots+(x+5)}6=\frac {5x+15}6,$$
which is certainly $>x$ if $x<15$. In other words, $\{1,\ldots,14\}\cap S_0=\emptyset$.

If $x>k$, let $S=S_0+k=\{\,s+k\mid s\in S_0\,\}$. Then strategy $S$ shows $$E(x)\le E(x-k)+k$$ (namely, we make our decision as if there were $k$ less, hence gain $k$ more than "expected" if we win something, or end up with the same $0$ if we lose)

Combining with $(1)$, we find for $x\in S_0$ that
$$ E(x)\le\frac{5E(x)+15}6$$
and hence
$$ E(x)\le 15.$$
As also $E(x)\ge x$, we conclude that $x\in S_0$ if $x>15$. So
$$ E(x)=x\quad\text{if }x\ge 15.$$
This makes
$$ E(14)=\frac{E(15)+\cdots+E(19)}6=14\tfrac16$$
$$ E(13)=\frac{E(14)+\cdots+E(18)}6=13\tfrac{13}{36}$$
$$ E(12)=\frac{E(13)+\cdots+E(18)}6=12\tfrac{127}{216}$$
and so on until
$$ E(0)=\frac{E(1)+\cdots+E(5)}6=13\tfrac{13}{36}=6\tfrac{ 72285265495}{470184984576}\approx 6.1537.$$

Hence the fair price for the game is $\approx 6.15$ and the best strategy is to keep playing until one reaches $15$ or more.


问:
Basic Series Question-Euler
I'm studying Calc 2 and I have a basic series question.

Leonhard Euler discovered that $\sum _{n=1}^{\infty }\:\frac{1}{n^2}=\frac{\pi ^2}{6}$.

Given $\sum \:_{n=1}^{\infty \:}\:\frac{1}{\left(2n\right)^2}$, what does the series equal?

My work:

Let k=2n

$\sum \:_{k=2}^{\infty \:}\:\frac{1}{k^2}=\sum \:_{n=1}^{\infty \:}\:\frac{1}{\left(2n\right)^2}$.

Now $\sum \:_{k=2}^{\infty \:}\:\frac{1}{k^2}=-1+\sum \:_{k=1}^{\infty \:}\:\frac{1}{\left(k\right)^2}$

Finally: $\frac{\pi ^2}{6}-1=-1+\sum \:\:_{k=1}^{\infty \:\:}\:\frac{1}{\left(k\right)^2}$

My textbook says that $\sum \:_{n=1}^{\infty \:}\:\frac{1}{\left(2n\right)^2}$=$\frac{\pi ^2}{24}$ though I cannot see how, based on my work above.

Does anyone see an error?

Thanks in advance!

答:
As a commenter already pointed out - your $k$ sum is over even numbers. It excludes more than just the first term.

If you'd like a hint, see if you can use this: $(2n)^{2}=4n^{2}$.

Title: Basic Series Question-Euler


问:
Is the subset $X:=\{x \in C^2([0,1]) | x(0)=x(1)=0 \}$ of the Banach space $C^m ([0,1])$ also a Banach space?
To prove this statement I planned to use the fact that every closed linear subspace of a Banach space is also a Banach space. Therefore I only have to show that $X$ is a linear subspace, which it was easy to see, and that $X$ is closed (i.e, if $(x_n)_n$ is a sequence in $X$ such that $x_n \to x$, then $x \in X$).

I can prove that $x(0)=x(1)=0$, but I am stuck at showing that $x$ is twice differentiable. I would appreciate a hint for solving this.

edit: This has to be shown for any integer $m \geq 0$.

edit2: This is respect to the norm $||x||_{C^m} := \sum_{k=0}^m \text{sup}_{t\in[0,1]}|x^{(k)}(t)|$, where $x^{(k)}$ means the $k$-th derivative of $x \in C^m([0,1])$.

答: I don't think $X$ must be closed, since uniform convergence does not respect differentiability. More formally, take some $0<a<b<c<d<1$ and let $f$ be a smooth bump function that is 1 on $[b,c]$ and 0 outside of $[a,d]$. Now, by the Stone Weierstrass theorem, we can find a sequence of polynomials $p_n$ converging uniformly to any continuous function $g: [a,d]\longrightarrow \mathbb R$. For all $n$, $p_n f\in X$. In fact, they are all smooth. Furthermore, $p_n f \to gf$, and on $[b,c]$ we have $gf=g$. If we take $g$ to be $C^1$ but not $C^2$ on $[b,c]$, then we have exhibited a sequence $p_n f \in X$ that converges in $C^1([0,1])$ to a limit not in $X$.


问:
Determine the constants C and K so that ${Ce^{iK\pi}}/{2e^{i\pi/4}} = 1+i$
Determine the constants C and K so that
$$
\frac{Ce^{iK\pi}}{2e^{i\pi/4}} = 1+i
$$

Pretty sure you need to use De Moivres formula and Eulers formula (complex numbers in notation)
Need help!!

答:
Hint: Simplify the left side and write the right side in polar form to get
$$
\frac{C}{2}e^{i \pi [K - 1/4]} = \sqrt{2} e^{i \pi/4}.
$$


问:
Gradient in coordinates of function in 2-sphere
Let $\mathbb{S}^{2}$ a 2-sphere that is subset of $\mathbb{R}^{3}$ and let a function $f:\mathbb{S}^{2}\longrightarrow \mathbb{R}$ defined by $f(x,y,z)=z$. Compute in coordinates the gradient $\nabla f$ with a metric of $\mathbb{S}^{2}$. I am unable to calculate the metric and inverse matrix coefficients using stereographic projection and I am a little confused in this step. A helping hand is very grateful.

答:
Let $i : \mathbb{S}^2 \to \mathbb{R}^3$ be the inclusion. It is an embedding and its differential at a point $p \in \mathbb{S}^2$ is just the inclusion of the tangent space $\mathrm{d}i_p : T_p\mathbb{S}^2=p^{\perp} \to \mathbb{R}^3$.

The metric on the sphere is defined to be the pullback metric $g=i^*g_0$ where $g_0$ is the euclidean metric.

Let $h : \mathbb{R}^3 \to \mathbb{R}$ be defined by $h(x,y,z) = z$ and let $f$ be its restriction to the sphere. We have
$$
f = h\circ i
$$
Thus, we have
$$
\forall p \in \mathbb{S}^2,~ \mathrm{d}f_p = \mathrm{d}h_{f(p)}\circ \mathrm{d}i_p = \mathrm{d}h_p\circ \mathrm{d}i_p
$$
and if $v$ is a tangent vector at $p$, then
$$
\mathrm{d}f_p(v) = \mathrm{d}h_p \circ \mathrm{d}i_p(v) = \mathrm{d}h_p(v)
$$
Now, the gradient of $f$ at $p$, say $\mathrm{grad}f_p$, is defined to be the unique tangent vector in $T_p\mathbb{S}^2$ such that
$$\forall v \in T_p\mathbb{S}^2,~ \mathrm{d}f_p(v) = g_p\left(\mathrm{grad}f_p,v \right)
$$

Let $\partial_z=(0,0,1)$ be the third coordinate vector in the canonical basis. It is the usual gradient of $h$ in $\mathbb{R}^3$, so that
$$
\mathrm{d}h_p(v) = \langle \partial_z,v\rangle
$$

If $\pi^\perp_p$ is the orthogonal projection from $\mathbb{R}^3$ to $T_p\mathbb{S}^2$, $\pi^{\perp}_p\partial_z \in T_p\mathbb{S}^2$ and for all $v$ tangent to the sphere at $p$:
$$
\mathrm{d}f_p(v) = \langle\partial_z,v \rangle = \langle \pi^{\perp}_p\partial_z,v\rangle = \langle \mathrm{d}i_p(\pi^{\perp}_p\partial_z),\mathrm{d}i_p(v) \rangle = g_p\left( \pi^{\perp}_p\partial_z,v\right)
$$
And this shows that $\mathrm{grad}f_p = \pi^{\perp}_p\partial_z$. Thus,
$$
\mathrm{grad}f_p = \partial_z - \langle \partial_z,p\rangle p =(-zx,-zy,1-z^2)
$$
if $p=(x,y,z) \in \mathbb{S}^2 \subset \mathbb{R}^3$. As a consequence, one can see that at the north and south poles (where $p = \pm \partial_z$), the gradient of $f$ is zero, which correspond to extrema of the function $f$.

More generally, let $M \hookrightarrow N$ be an embedded submanifold, $g_0$ a riemannian metric on $N$, and $g$ the induced metric on $M$. Let $h : N \to \mathbb{R}$ be a smooth real function and $f$ its restriction of $M$. Then
$$\mathrm{grad}f_p = \pi^{\perp}_p (\mathrm{grad}h_p)$$
where $\pi^{\perp}_p : T_pN \to T_pM$ is the orthogonal projection with respect to $g_0$. The proof is the same as above.


问:
Finding height of the tower when the angles of elevation of its top from two points are given

The angle of elevation of the top of a vertical tower from a point $A$, due east of it is $45^{\circ}$. The angle of elevation of the same tower from a point $B$, due south of $A$ is $30^{\circ}$. If the distance between $A$ and $B$ is $54\sqrt2$m the find the height of the tower.

My attempt:

Let the height of tower be $h$ meters. So, the distance of the foot of the tower from $A$ is $h$ and from $B$ is $h\sqrt3$. So, $h^2+(54\sqrt2)^2=(h\sqrt3)^2\implies h=54m$.

Answer is given as $108$m.

答:
As already mentioned in the comment, your solution $h=54 m$ is correct.

Below is a 3d-diagram depicting the situation.

Btw., the distance from $B$ to the top of the tower is $108 m$.


问:
Let T1 ⊂ T2 be two topologies on X. For a subset A⊂X, compare the closures of A according to T1 and T2.
Let T1 ⊂ T2 be two topologies on X. For a subset A⊂X, compare the closures of A according to T1 and T2.

Can you help me please?

答:
Any set that is closed with respect to $T_1$ must also be closed with respect to $T_2$. However, the closure is the "smallest" closed set containing $A$. It follows that
$$
\operatorname{cl}_{T_1}(A) \supseteq  \operatorname{cl}_{T_2}(A),
$$
where $\operatorname{cl}_{T}(A)$ denotes the closure of $A$ with respect to the topology $T$.


问:
Analysing ways to choose $m$ naturals from set of first $n$, s.t. they differ by $\ge k$.
Ways to choose $m$ distinct natural numbers from set of first $n$ natural numbers, s.t. the chosen natural numbers differ by at least $k$.

Request vetting for my approach:

The numbers to be chosen are $m$, with at least $k$ 'buffer' contiguous numbers.

Say, if $n=25, m=4, k=3$, then few possible choices are stated below with chosen $m$ elements 'outside of brackets', while $\ge (k-1)$ buffers in brackets. :

#1: $1,(2,3), 4, (5,6), 7, (8,9), 10, (11,12,13,14,15,16,17,18,19,20,21, 22,23,24,25)$

#2: $(1),2,(3,4), 5, (6,7,8,9), 10, (11,12), 13, (14,15,16,17,18,19,20,21, 22,23,24,25)$

#3: $1,(2,3,4,5), 6, (7,8,9,10,11,12,13,14,15,16,17,18,19), 20, (21,22,23),24,(25)$

#4: $(1),2,(3,4,5), 6, (7,8,9,10,11,12,13,14,15,16,17,18,19), 20, (21,22,23,24),25$

As can be seen, the $m$ chosen ones form $m+1$ slots, with two 'border ones' (one before the first marker element, & one after the last marker) possibly being empty or can have $\lt k-1$ buffer elements.

Need choose $m$ markers defining boundary of $m+1$ slots, with at least $k-1$ elements in non-empty slots.

So, need reserve for non-empty slots (all apart from 'border ones') at least $k-1$ elements, leading to reserving $(m-1)(k-1)$ elements from $n$. From the left out elements, need choose $m$ elements. Simply, need to subtract from the $n$ elements, $(m-1)(k-1)$ elements. This makes available needed buffer elements.

(Kindly note that it covers cases where the first (or, any other element) is a marker, as well as the cases where the first (or, any other element) element is a buffer.)

Then, choose $m$ markers. So, need choices for that first.

Need choose later from the free elements.

So, get in the first step ${n-(m-1)(k-1) \choose m}$ choices.

To see, a small example, for $n=6, m=2,k=3$, get the #choices as :  ${6-(2-1)(3-1) \choose 2}= {4 \choose 2}= 6$. These choices are:

#1. $1(23)4(56)$,

#2. $1(234)5(6)$,

#3. $1(2345)6$,

#4. $(1)2(34)5(6)$,

#5. $(1)2(345)6$,

#6. $(12)3(45)6$,

Also, in second step need place $n - m - (n-1)(m-1)$ elements in their proper slots.

But, there are no choices available for the second step, as the values of elements (due to some sort of ordering, which indicates the need of ordering, or classification for elements) dictate their slot. In other words, the second step is completed in the first one.

Hence, the choices of the first step dictate the answer.

This gels with the multisets based star-and-bars approach's answer.

Edit:

But, the elements are indistinguishable for that, so indistinguishability (as in answer by @ChristianBlatter, by taking one set of selection as $1$, others as $0$) is w.r.t. the possibility of elements being in a class. Here, the elements are ordered (as in the selected answer), yet indistinguishable in a class! The abstraction of class is marked by slots' value, and the slot is chosen based on any left $m$ choices after taking $(m-1)(k-1)$ values from $n$, with no arrangements as ordering cannot be changed for elements chosen as slots ($m$).

The formula for multisets will be ${(n'+m-1) \choose m}$ choices.

This can be viewed as $n' = (n - (m-1)(k-1))+1$, and the size of multiset =$m$, which is fixed. The #of multisets gives all possible choices of getting a given size multiset for a given $n'$.

The factor $n'-1$ in the formula for multisets : ${(n'-1+m) \choose m}$, shows that we take/adapt multisets' formula to mean that the roles of marker is being taken by free elements, and the slots formed by the free elements is given by $n'$. So, the actual markers/ #free elements are given by: $n'-1$.

Taking the above small example, to show the adaptation of logic to make free elements act as markers is:

$n=6, m=2,k=3, n'= 6-(2)(1)= 4.$

#1. Taking $2,5$ as buffer, get free elements as: $1,3,4,6$.

$3$ forms the first boundary, and $6$ the second one. The chosen elements in set $m= 1,4$: $1(23)4(56)$,

#2. Taking $2,3$ as buffer, get free elements as: $1,4,5,6$.

$4$ forms the first boundary, and $6$ the second one. The chosen elements in set $m= 1,5$: $1(234)5(6)$,

and so on by taking suitably buffer elements, free elements as markers, and chosen elements; for the rest three cases.

答:
An admissible choice can be encoded as a binary word $w$ of length $n$, satisfying certain conditions. In particular this word $w$ contains exactly $m$ ones. Furthermore, immediately after each of the first $m-1$ of these ones there have to be $k-1$ zeros.

When we omit these $(m-1)(k-1)$ zeros from $w$ we obtain a binary word $w'$ of length $n':=n-(m-1)(k-1)$, containing exactly $m$ ones, and there are no other constraints about $w'$. Conversely: Given such a $w'$ of length $n'$, introduce $k-1$ zeros immediately after the first $m-1$ ones, and you obtain an admissible word $w$.

It follows that the number $N$ of admissible words $w$ is given by
$$N={n'\choose m}={n-(m-1)(k-1)\choose m}\ .$$
You have obtained this value, and wanted to add "a second step" (which I have not understood). But no such step is needed.


问:
Prove that $\log X < X$ for all $X > 0$
I'm working through Data Structures and Algorithm Analysis in C++, 2nd Ed, and problem 1.7 asks us to prove that $\log X < X$ for all $X > 0$.

However, unless I'm missing something, this can't actually be proven. The spirit of the problem only holds true if you define several extra qualifiers, because it's relatively easy to provide counter examples.

First, it says that $\log_{a} X < X$ for all $X > 0$, in essence.

But if $a = -1$, then $(-1)^{2} = 1$. Therefore $\log_{-1} 1 = 2$. Thus, we must assume
$a$ is positive.

if $a$ is $< 1$, then $a^2 < 1$.  Therefore we must assume that $a \geq 1$.

Now, the book says that unless stated otherwise, it's generally speaking about base 2 for logarithms, which are vital in computer science.

However, even then - if $a$ is two and $X$ is $\frac{1}{16}$, then $\log_{a} X$ is $-4$. (Similarly for base 10, try taking the log of $\frac{1}{10}$ on your calculator: It's $-1$.) Thus we must assume that $X \geq 1$.

...Unless I'm horribly missing something here. The problem seems quite different if we have to prove it for $X \geq 1$.

But even then, I need some help solving the problem. I've tried manipulating the equation as many ways as I could think of but I'm not cracking it.

答:
I believe I managed to answer this question:

Proof is by contradiction:

Let us assume that Log X >= X for all X > 0.

Then we define Log X to be equal to Y.
X must be <= Y, by definition.

By the definition of logarithms, 2^Y = X.
Using simple substitution, we can say that 2^Y <= Y, since X is <= Y.

This inequality only holds true for Y < 0.  Since X <= Y, then the inequality also holds true only for X <= 0, via syllogism.  Thus Log X >= X if and only if X <= 0, thus Log X < X for all X > 0.

Title: Prove that $\log X < X$ for all $X > 0$


问:
Three-term recurrence for $A$-orthogonal vectors obtained by Gram-Schmidt procedure.
I would be glad for any help on how to approach the following problem.

Let $A$ be a positive definite $n\times n$ real matrix, $p_0 \in \mathbb{R^n}$.

Put $p_i = A^ip_0$.

Let $d_i$ be $A$-orthogonal vectors obtained by Gram-Schmidt procedure from $p_i.$

Show that $d_i$ can be obtained using three-term recurrence, that is $d_{i+1}$ can be obtained from $Ad_i, d_i$ and $d_{i-1}.$

答:
If the matrix is positive definite then it's symmetric. So we know that $A^{T} = A$. For a general matrix the process is called the Arnoldi iteration. Given a set of vectors $\{ p_{1}, p_{2}, \cdots p_{k}\}$  then we know the Gram-Schmidt procedure computes an orthonormal basis $\{ d_{1}, d_{2}, \cdots , d_{k} $} so that for all $ j \leq k$ we know

$$ \text{Span}( p_{1},p_{2}, \cdots p_{k}) = \text{Span}(d_{1}, d_{2}, \cdots d_{k}) $$

If you try to compute the set $\{p , Ap , A^{2}p ,\cdots  \}$ you'll note that it will be close to linearly dependent so it won't be numerically stable. Since the span of $\{d_1, d_2, \cdots d_{k-1}\}$ is equal to the span of $\{ p, Ap , \cdots , A^{k-2}p \}$ then $d_{k-1}$ can be written as a linear combination of $\{ p, Ap , \cdots , A^{k-2}p \}$. That is there exists coefficients $c_i$ such that (Note I'm going to use the indexing you have so $p_0$ is there instead.

$$ d_{k-1}  = c_1 p_0 + c_2Ap_0 + \cdots + c_k A^{k-2} p_0$$

Then if we multiply by $A$ we have

$$ Ad_{k-1}  = c_1 A p_0 + c_2 A^{2}p_0 + \cdots + c_k A^{k-1} p_0$$

Now since each of $\{ p_0 ,Ap_0, \cdots ,A^{k-2} p_0 \} $ are in the span of $\{d_1, d_2, \cdots d_{k-1} \}$ they will disappear when we orthogonalize $Ad_{k-1}$ against them. We get an kth term recurrence relation from this

$$ AD_{k} =  D_k H_k  + h_{k+1, k} d_k \xi_k^{T} $$

Where $D_k = [d_1, d_2, \cdots , d_k]$ is the $ n \times k$ matrix with columns $\{ d_1, d_2, \cdots d_k \}$ and $H_k$ is a $k \times k$ Upper Hessenberg matrix and $\xi_{k}^{T} = [0, \cdots , 0, 1]^{T}$ is the k-th unit vector.

When $A$ is symmetric then $D_k^{T}AD_k  = H_k$ is also symmetric (instead of Upper Hessenberg). An upper Hessenberg matrix a upper Triangular matrix but with non-zero entries on the sub-diagonal. In this case if it's symmetric then it's tridiagonal. This gives us the 3-term recurrence relation. Then the $d_j$ orthogonal vector satisfies

$$ Ad_j = \beta_{j-1}d_{j-1}  + \alpha_{j}d_{j} + \beta_{j}d_{j+1} $$

this gives the matrix equation

$$ AD_k = D_k T_k + \beta d_{k+1}\xi_{k}^{T} $$

Where $T$ is the tridiagonal matrix formed from these coefficients at the kth step

$$ T = \begin{bmatrix}
\alpha_1 & \beta_2 &  &  &  & 0\\
\beta_2 & \alpha_2 & \beta_3  &  &  \\
 & \beta_3 & \alpha_3  &  \ddots & \\
  &  & \ddots & \ddots  &  \beta_{m-1} &  \\
 &  &  & \beta_{m-1}  &  \alpha_{m-1}  &  \beta_{m}   \\
 0 &  &  &   &  \beta_{m}  &  \alpha_{m}   \\  \end{bmatrix}  $$


问:
one parameter-family of unitaries lie in the commutant of a von Neuamann algebra

The screenshot is from Herman's paper. I have two questions in the proof of Prolosition 3.

Question 1: If $M_{\varphi}\subset M_{\psi}$, then $v_t\in M_{\psi}'\cap M$. For any $x\in M_{\psi}$, we want to prove $xv_t=v_tx$. Since $x\in M_{\psi}$, we have $x\psi=\psi x$, then $\psi(v_t x)=\psi(x v_t)$. How to conclude $xv_t=v_tx$?

Question 2. How to prove $\sigma_t^{\psi}=v_t\sigma_t^{\psi}(y)v_t^{*}$? Is $\sigma_t^{\psi}(y)\in M_{\psi}$?

答:
Let $x\in M_\varphi$. Then
$$
x=\sigma_t^\psi(x)=v_t^*\sigma_t^\varphi(x)v_t=v_t^*xv_t.
$$
So $x\in M_\varphi'\cap M=M_\psi'\cap M$.

If $y\in M_\psi$, since $v_t\in M_\psi'\cap M$,
$$
\sigma_t^\varphi(y)=v_t\sigma_t^\psi(y)v_t^*=v_tyv_t^*=y.
$$


问:
Solve $\displaystyle\lim_{x \to \frac{\pi}{2}}{\biggl[(1+\cot x)^{\tan x}\biggr]^3}$

Evaluate the limit: $$\displaystyle\lim_{x \to \frac{\pi}{2}}{\biggl[(1+\cot x)^{\tan x}\biggr]^3}$$

Could someone help me to solve this limit?

I found the answer in symbolab, however I could not understand how the steps are explained.

Please explain this to me. Thank you.

答:
Let $y = \tan x$. Then:

$$\displaystyle\lim_{x \to \frac{\pi}{2}}{\biggl[(1+\cot x)^{\tan x}\biggr]^3} = \displaystyle\lim_{y \to \infty}{\biggl[\left(1+\frac{1}{y}\right)^{y}\biggr]^3} = e^3.$$


问:
Given $a_n$ converges to $a$ how to find this limit?

If the terms of a sequence $a_n$ converge to $a$, show that $\frac{a_1+a_2+...+a_n}{n}$ also converges to $a$. Also show that the converse isn't necessarily true.

Please help me with this- I am unable to approach this problem, and don't even know where to begin looking for a solution.

答:
Hint:

Just remind that if $a_n$ converges to $a$, as of a certain $N$, the terms $a_n$ remain at distance at most $\epsilon$ of $a$. Now if you take the average, you will have a finite number of terms with arbitrary values, but an infinity of them at distance $\epsilon$ or less. So with sufficiently many terms, the average will stay close to $a$.


问:
Combinatorics, counting and number theory.
Find number of ordered quadruples (a,b,c,d) [of positive integers] such that $lcm[a,b,c]= lcm[a,b,d]= lcm[a,c,d]= lcm[b,c,d] = 2^r\cdot3^s$

So i approached it like two of a,b,c,d has max power of 2 = r. For rest two, they have r+1 choices each. So it would be $\binom{4}{2} \cdot (r+1)^2$. Same for the power of three.

So as per me answer should be $(\binom{4}{2} \cdot (r+1) \cdot (s+1))^2$ but the answer to this problem is not given in my book.

I would really appreciate if you could tell whether my answer is correct and whether this is the correct approach and if it is not correct approach, then why?

Thanks!

答:
For power of $2$, consider the complement: how many ways are there at most one of $a,b,c,d$ has power $r+1$? That's $(r+1)^4 - r^4-{4 \choose 1} r^3=6r^2+4r+1$.

The total number of ways is then $(6r^2+4r+1)(6s^2+4s+1).$


问:
Eilenberg-Watt's theorem reference.
I'm looking for a reference (with proof!) for the following result:

Let $A$ and $B$ be unital, associative rings.

(Eilenberg-Watt) Let $F: {}_{A}\mathrm{Mod} \to {}_{B}\mathrm{Mod}$ be a functor that is
additive and preserves direct sums and quotients. Then there is a
natural isomorphism $$\alpha: P \otimes_A \bullet \to F$$ where
$P=F(A)$ has the natural right $A$-module structure.

In short, an additive functor that preserves direct sums and quotients coincides with the functor $P \otimes_A \bullet$.

Please check if all assumptions match, because I have seen plenty of variations of this theorem but not this exact one.

答:
See this question and answer for a proof. The idea is exactly the one described in the second part of Andrea's answer: for any $A$-module, you have a presentation $A^{(J)}\to A^{(I)}\to M\to 0$, which you can then use together with your assumptions to deduce that $F(M)\cong F(A)\otimes_A M$.

Note that $F(A)$ has a $B$-module structure by definition, and also a right $A$-module structure given by $F(r_a), a\in A$ where $r_a : A\to A$ is given by $x\mapsto xa$ (this is a morphism of left $A$-modules).

To get naturality of the isomorphism $F(M)\cong F(A)\otimes_A M$, the trick is to define, for any $F$ and any $M$ (no assumptions so far) a natural morphism $F(A)\otimes_A M\to F(M)$. One then uses the assumptions to prove that this natural morphism is an isomorphism.

This natural morphism is defined by the following composite: $M\to \hom(A,M) \to \hom(F(A),F(M))$ and then the universal property of the tensor product. There is one thing to note here : $\hom(A,M)\to \hom(F(A),F(M))$ is a morphism of $A$-modules because of the definition of the right $A$-module structure on $F(A)$.

This gives you a natural morphism, which is obviously an isomorphism when $M=A$, and since both sides preserve direct sums and cokernels, it is an isomorphism for any $M$.

Let me mention two additional things with respect to some points in Andrea's answer :

1- In the example he gives in the first part of the question, there is a mistake : we never consider $P\otimes_\mathbb C X$, but $P\otimes_A X$ - in particular the dimension is not the product of the dimensions.

2- There is no difference between preserving cokernels of general maps and preserving cokernels of injections: indeed if you preserve the latter, then you preserve epimorphisms, and then for a general cokernel sequence $M\to N\to C\to 0$, you can separate it as $M\to \mathrm{im}(M\to N)\to 0$ and $0\to \mathrm{im}(M\to N)\to N\to C\to 0$.
This gives two a priori different definitions of right exact functors which end up agreeing.

Title: Eilenberg-Watt's theorem reference.


问:
Equation of hyperbola with eccentricity $\frac32$

A hyperbola whose transverse axis is along the major axis of the conic $\frac{2x^2}{3}+\frac{y^2}{4}=4$ and has vertices on the foci of this conic. If the eccentricity of the hyperbola is $\frac32$ then which of the following points does NOT lie on it? $(0,2)/(\sqrt5,2\sqrt2)/(\sqrt{10},2\sqrt3)/(5,2\sqrt3)$

Ellipse is $\frac{x^2}{6}+\frac{y^2}{16}=1$. Its eccentricity $\frac{\sqrt{10}}4$. Foci $(0,\pm\sqrt{10})$.

Let the equation of hyperbola be $-\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Vertex$=b\frac32=\sqrt{10}\implies b=\frac{2\sqrt{10}}{3}$. And $a^2=b^2(e^2-1)=\frac{50}9$.

Thus, hyperbola is $-\frac{9x^2}{50}+\frac{9y^2}{40}=1$.

With this, no point is matching.

答:
Equation of rotated hyperbola is$$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$$
Vertices of the hyperbola are on the foci of the ellipse, therefore
$a=\sqrt{10}$

Eccentricity is $e=\frac{c}{b}=\frac32\to c=\frac32 b$

$b^2=c^2-a^2\to b^2=\frac{9}{4}b^2-10\to b^2=8$

Hyperbola equation is $$\frac{y^2}{10}-\frac{x^2}{8}=1$$

It is true that none of the points lie on the hyperbola. See the graph below.

The red hyperbola is what they had in mind (blue points are the given points and one of them does not lie on the red hyperbola), but its equation doesn't fit the given informations


问:
Solving $x^2 \equiv 12 \pmod{13}$

Solve $x^2 \equiv 12 \pmod{13}$.

I was instructed on defining $12$ using primitive roots and then Lagrange's theorem.

It seems that ord$_{13}(12) = 2$, but this isn't a primitive root since $\phi(13) = 12$ and ord$_{13}(12) \ne \phi(13)$. Any hints to proceed would be appreciated. I feel like I missing something here...

答:
$2$ is a primitive root mod $13$, so $2^{12}\equiv1$, $2^6\equiv-1$,

and the solutions to $x^{12}\equiv12\equiv-1$ are $2^3\equiv8$ and $2^9\equiv5\equiv-8$.

(You can see that $8^2=64=65-1\equiv12\bmod13$, and $5^2=25=26-1\equiv12\bmod13$.)

These are the primitive fourth roots of unity mod $13$.

$4\times3\equiv4\times9\equiv0\bmod12$, and $2\times3\equiv2\times9\not\equiv0\bmod12$.

Title: Solving $x^2 \equiv 12 \pmod{13}$


问:
Proof of Möbius function on subset poset
Prove that Möbius function of subset poset of $[n]$ is following, given $A,B \subseteq [n]$.

$\mu (A,B)=\left\{
\begin{array}{l} (-1)^{|B|-|A|},& \text{if}\,\, A \subseteq B\\
0,& \text{otherwise}\end{array} \right.$

I searched for the proof and in one paper they use induction. $A = B$ this is obvious. Assume this holds up to $|B|= |A| + n, A\subset B$. Then for $|B|=n+1$

$\mu (A,B)=- \sideset{}{_{A \subseteq C \subset B} (-1)^{|C|-|A|}} \sum = (-1)^{|B|-|A|}$

But I do not understand the last step. How

$\sideset{}{_{A \subseteq C \subset B} (-1)^{|C|-|A|}} \sum = (-1)^{|B|-|A|}$

Can anyone explain this or point me to a more detailed proof? Thanks

答:
$\mathsf{Hint}$ (use the substitution $D=C\setminus A$):

$$\sum_{A\subseteq C\subseteq B}(-1)^{|C|-|A|}=\sum_{D\subseteq B\setminus A}(-1)^{|D|}=\sum_{\ell=0}^{|B\setminus A|}\binom{|B\setminus A|}{\ell}(-1)^\ell=(1-1)^{|B\setminus A|}=0.$$

Notice $C\color{Red}{\subseteq}B$ is used above, as opposed to $\color{Blue}{\subset}$ as used in the original expression.


问:
Motivation behind the sheaf of relative Kähler differentials
I'm interested in the geometric motivation behind the bundle (or in more general framework the sheaf) of relative differentials $\Omega_{X/Y}$
of a morphism $f: X \to Y$ smooth $k$-varieties. Differential geometry provides following picture of relative differentials:

Let $f: X \to Y$ a equidimensional surjective map between connected manifolds $Y, X$
and moreover assume that every fiber $F:= f^{-1}(y) \subset X$ for $y \in Y$ is also a connected
submanifold of same dimension. We obtain an exact sequence of
tangent spaces

$$  0 \to T_{X/Y} \to T_X \to f^*T_Y \to 0  $$
where $T_{X/Y}$ is the kernel of induced map of tangent bundles. intuitively what is really going on there is that for every $x \in f^{-1}(y)$, the $(T_{X/Y})_x$ is the tangent space of the fiber at $x$. The relative space of Kähler differentials
$\Omega_{X/Y}$ defined as the dual of $T_{X/Y}$ and sits in the sequence which we will
obtain if we dualize the sequence above of tangent spaces:

$$ 0 \to f^*\Omega_Y \to \Omega_X \to \Omega_{X/Y} \to 0  $$

Another definition of relative Kähler differentials which is more common to use in modern algebraic geometry works as follows:

Let us embedd $X$ as the image of the diagonal $\Delta: X \to X \times_Y X$
and assume that the ideal sheaf $I \subset O_{X \times_Y X}$ defines the closed image
$\Delta(X) \subset X \times_Y X$.

This gives us another sequence of tagent spaces

$$ 0  \to T_{\Delta(X)} \to T_{X \times_Y X} \to N_{X \times_Y X/X} \to 0 $$

with normal bundle $N_{X \times_Y X/X}$. It's a basic fact that the dual
of $N_{X \times_Y X/X}$ is $I/I^2$ and most books on algebraic
geometry define the sheaf relative Kähler differentials by

$$\Omega_{X/Y} := I/I^2$$

Since this definition not uses that $f$ is a map of smooth maps this
is a far generalization of the old fashion setting from
differential geometry. Now, if there is any justice n this world then these two definitions of relative differentials should coinside if we deal with $X, Y$ and $fY$ nice enough.

Therefore, if $f$ a surjective map between conneted manifolds such that
every fiber is a connected submanifold, why the tangent bundle
$T_{X/Y}$ and the pullback of normal bundle $\Delta^* N_{X \times_Y X/X}$ of
$\Delta(X) \subset X \times_Y X$ are canonically isomorphic?

Can we write down an explicit isomorphism and understand what is geometrically going on there?

答:
In general if $k \rightarrow A \rightarrow^f B$ is an arbitrary sequence of maps of commutative unital rings, there is a canonical right exact sequence of $B$-modules

S1. $B\otimes \Omega^1_{A/k} \rightarrow \Omega^1_{B/k} \rightarrow \Omega^1_{B/A} \rightarrow 0$.

In Matsumura, Commmutative ring theory Thm 25.1 you may find this construction. If $B$ is $0$-smooth over $A$ it follows when you add a $0$ on the left it follows S1 is split exact. When you dualize S1 you get the sequence

S2. $0 \rightarrow Der_A(B) \rightarrow Der_k(B) \rightarrow^{Tf} B\otimes_A Der_k(A)$

where $Tf$ is the "tangent morphism" of the map $f$. If $X:=Spec(B)$ and $Y:=Spec(A)$
it follows the relative tangent bundle $T_{X/Y}$ is the sheafification of the module $Der_A(B)$.  When you sheafifiy S2 you get the sequence

S3. $0 \rightarrow T_{X/Y} \rightarrow T_{X/k} \rightarrow^{Tf} f^*T_{Y/k}$

and S3 is an exact sequence of $\mathcal{O}_X$-modules in general.

If $k$ is the field of complex numbers and $X,Y$ are finitely generated and regular algebras over $k$ it follows $T_{X/k}$ and $f^*T_{Y/k}$ are finite rank locally free sheaves on $X$. The "relative tangent sheaf" $T_{X/Y}$ is related to properties of the morphism $f$.

Definition. Assume $X,Y$ are smooth irreducible schemes of finite type over $k$ the field of complex numbers (this means the local rings of $X,Y$ are regular at all points). The morphism $f$ is "smooth of relative dimension $n$" iff

$f$ is flat

$dim(X)=dim(Y)+n$

The cotangent sheaf $\Omega^1_{X/Y}$ is locally free of rank $n$.

By Hartshorne, Prop III.10.2 this implies that the fibers $f^{-1}(y):=X_y$ of $X$ at any point $y\in Y$ are regular. Hence the cotangent sheaf $\Omega^1_{X/Y}$ "measures" when the fibers of $f$ are regular/smooth of the same dimension $n$.

Example: Let $\pi:X:=\mathbb{V}(\mathcal{E}^*)\rightarrow Y$ be a finite rank geometric vector bundle on $Y$. Locally it follows the sheaf $\mathcal{E}^*$ is a free $A$-module $E:=A\{x_1,..,x_n\}$ on the elements $x_i$ hence locally the map $\pi$ is the following map:

S4. $A \rightarrow B:=A[x_1,..,x_n]$

where the latter is a polynomial ring on $n$-variables. In this case the relative module of Kahler differentials $\Omega^1_{B/A}$ becomes

S5. $\Omega^1_{B/A}\cong B\{dx_1,..,dx_n\}$

which is the free $B$-module of rank $n$ on the elements $dx_i$. Hence the relative
cotangent sheaf $\Omega^1_{X/Y}$ is in this case locally trivial of rank $n$ and the morphism $\pi$ is smooth of relative dimension $n$. The fibers are affine spaces and they are smooth. Hence the relative cotangent sheaf is related to properties of the morphism. The relative tangent sheaf is the dual of the relative cotangent sheaf, but when you dualize you loose information. Hence the cotangent sheaf is more fundamental.

Note: When you write down the relative tangent sequence, this is not a sequence of "tangent spaces", it is an exact sequence of coherent sheaves on $X$. The relative "tangent bundle" is not a vector bundle in general. In your situation it is a "coherent sheaf". If the relative cotangent sheaf is locally trivial it follows the fibers of your morphism are smooth manifolds (if you consider holomorphic maps of complex projective manifolds).


问:
Is it independent?
Let there be two tests for high blood sugar performed by two different
hospitals, A and B. Blood samples are taken from people are first sent to Hospital A and then Hospital B. Hospital A does not reveal the outcome of its tests to Hospital B.

Suppose that the tests are based on similar biological mechanisms. Let
A+ = Event that A declares blood sample has high blood sugar and
B+ = Event that B declares blood sample has high blood sugar

Are A+ and B+ independent or dependent?

I think that it's independent because Hospital A does not reveal the outcome of its tests to Hospital B.

However, does the fact that blood samples are taken from people are first sent to Hospital A and then Hospital B affect its independence?

答:
Two events are said to be independent when the outcome of one event does not affect the outcome of the other event

A+ and B+ are independent of each other

Reason : It is given that hospital A does not reveal any test result  to hospital B so the decision made by hospital B is not dependent on the decision made by hospital A . The fact that the test results are given to A first is immaterial as decision made by B is not influenced by the decision made by A.

Title: Is it independent?


问:
For all $m,n \in \Bbb{N}$, $n>m$, there is a unique $k \in \Bbb{N}$ such that $km \le n < (k+1)m$
For all $m,n \in \Bbb{N}$, $n>m$, there is a unique $k \in \Bbb{N}$ such that $km \le n < (k+1)m$

My attemt for existence (is this approach correct?)

Let $m,n \in \Bbb{N}$ and consider the set $S = \{q \in \Bbb{N} | qm \le n \}$

Since $1 \in S$, $S$ is nonempty, so by the Well Ordering Principle there is a least element $k \in S$.

Then, suppose $(k+1)m \le n$

Then $km \le n - m = 0$, which is absurd, as $km \in \mathbb{N}$. So $n < (k+1)m$ and then $km \le n < (k+1)m$.

Is this correct?

For the uniqueness, I'm stuck. I tried supposing there are different $k,k'$ and using the two inequalities for each, but can't get to a contradiction...

答:
You were on the right track in your last sentence.

"For the uniqueness, I'm stuck. I tried supposing there are different $k,k′$ and using the two inequalities for each, but can't get to a contradiction..."

Ok, so suppose $k \neq k'$. Then we may assume without loss of generality that $k < k',\ $ and so $\exists i \in \mathbb{N}$ such that $k' = k+i$. So:

$(1) \quad km \leq n < (k+1)m,\ $ and

$(2) \quad (k+i)m \leq n < (k+i+1)m . $

Taking $m(i-1)$ away from $(2)$ gives:

$(2b) \quad (k+1)m \leq n+(1-i)m < (k+2)m$.

Putting $(1)$ and $(2b)$ together:

$km \leq n < (k+1)m \leq n+(1-i)m \leq n,\quad$ since $1-i \leq 0$.

This says that $n<n$ which is impossible.

Therefore, our assumption that $k \neq k'$ has lead to a contradiction: this proves uniqueness.


问:
Trace of a matrix
What is the trace of $e^{A}$ where A is a $4 \times4$ matrix

$$\begin{bmatrix}0 & 0 & 0 & t\\
 0 & 0 &-t & 0\\
 0 & t & 0 & 0\\
-t & 0 & 0 & 0 \end{bmatrix}$$

where $t=\dfrac{\pi}{4}$.

答:
By definition:

$$ Tr(e^A)=\sum_{k=1}^4 e^{\lambda_k},$$

where, $\lambda_k$ are eigenvalues of $A$, which solve

$$
\det(A-\lambda I)=\lambda^4+2t^2\lambda^2+t^4=(\lambda^2+t^2)^2=0.
$$

So,

$$ Tr(e^A)=2(e^{it}+e^{-it})=4\cos t|_{\pi/4}=2\sqrt{2}.$$

Btw,

$$ \det(e^A)=\prod_{k=1}^4 e^{\lambda_k}=e^{\sum_{k=1}^4\lambda_k}=e^{Tr(A)}=e^0=1.$$

Title: Trace of a matrix


答:
i think its correct,however we could rewrite diffrently

Suppose we select the $m$ numbers  $x_1<x_2......<x_m$,

let $y_{i}$ denote the number of integers between $x_{i},x_{i+1}$ Easy to see $y_i\ge k-1$ for $1\le i\le m-1$ Also let $y_0,y_{m}$ be the number of integers before $x_1$ and after $x_m$.Again $y_{0},y_{m+1}\ge 0$.

Hence we have to find number of solutions of $$y_0+y_1..+y_m=n-m$$ with the condition $y_0,y_m\ge 0$ and $y_i\ge k-1 $ for $1\le i\le m-1$.We sub $y_0=t_0,y_m=t_m,y_i=t_i-k+1$ for $1\le i\le m-1$

$$t_0+t_1+t_2....t_m=n-m-(m-1)(k-1)$$ where $t_i\ge 0$ which is just the stars and bars problem to get $$\binom{n-(k-1)(m-1)}{m}$$

Title: Analysing ways to choose $m$ naturals from set of first $n$, s.t. they differ by $\ge k$.

Body:
Ways to choose $m$ distinct natural numbers from set of first $n$ natural numbers, s.t. the chosen natural numbers differ by at least $k$.

Request vetting for my approach:

The numbers to be chosen are $m$, with at least $k$ 'buffer' contiguous numbers.

Say, if $n=25, m=4, k=3$, then few possible choices are stated below with chosen $m$ elements 'outside of brackets', while $\ge (k-1)$ buffers in brackets. :

#1: $1,(2,3), 4, (5,6), 7, (8,9), 10, (11,12,13,14,15,16,17,18,19,20,21, 22,23,24,25)$

#2: $(1),2,(3,4), 5, (6,7,8,9), 10, (11,12), 13, (14,15,16,17,18,19,20,21, 22,23,24,25)$

#3: $1,(2,3,4,5), 6, (7,8,9,10,11,12,13,14,15,16,17,18,19), 20, (21,22,23),24,(25)$

#4: $(1),2,(3,4,5), 6, (7,8,9,10,11,12,13,14,15,16,17,18,19), 20, (21,22,23,24),25$

As can be seen, the $m$ chosen ones form $m+1$ slots, with two 'border ones' (one before the first marker element, & one after the last marker) possibly being empty or can have $\lt k-1$ buffer elements.

Need choose $m$ markers defining boundary of $m+1$ slots, with at least $k-1$ elements in non-empty slots.

So, need reserve for non-empty slots (all apart from 'border ones') at least $k-1$ elements, leading to reserving $(m-1)(k-1)$ elements from $n$. From the left out elements, need choose $m$ elements. Simply, need to subtract from the $n$ elements, $(m-1)(k-1)$ elements. This makes available needed buffer elements.

(Kindly note that it covers cases where the first (or, any other element) is a marker, as well as the cases where the first (or, any other element) element is a buffer.)

Then, choose $m$ markers. So, need choices for that first.

Need choose later from the free elements.

So, get in the first step ${n-(m-1)(k-1) \choose m}$ choices.

To see, a small example, for $n=6, m=2,k=3$, get the #choices as :  ${6-(2-1)(3-1) \choose 2}= {4 \choose 2}= 6$. These choices are:

#1. $1(23)4(56)$,

#2. $1(234)5(6)$,

#3. $1(2345)6$,

#4. $(1)2(34)5(6)$,

#5. $(1)2(345)6$,

#6. $(12)3(45)6$,

Also, in second step need place $n - m - (n-1)(m-1)$ elements in their proper slots.

But, there are no choices available for the second step, as the values of elements (due to some sort of ordering, which indicates the need of ordering, or classification for elements) dictate their slot. In other words, the second step is completed in the first one.

Hence, the choices of the first step dictate the answer.

This gels with the multisets based star-and-bars approach's answer.

Edit:

But, the elements are indistinguishable for that, so indistinguishability (as in answer by @ChristianBlatter, by taking one set of selection as $1$, others as $0$) is w.r.t. the possibility of elements being in a class. Here, the elements are ordered (as in the selected answer), yet indistinguishable in a class! The abstraction of class is marked by slots' value, and the slot is chosen based on any left $m$ choices after taking $(m-1)(k-1)$ values from $n$, with no arrangements as ordering cannot be changed for elements chosen as slots ($m$).

The formula for multisets will be ${(n'+m-1) \choose m}$ choices.

This can be viewed as $n' = (n - (m-1)(k-1))+1$, and the size of multiset =$m$, which is fixed. The #of multisets gives all possible choices of getting a given size multiset for a given $n'$.

The factor $n'-1$ in the formula for multisets : ${(n'-1+m) \choose m}$, shows that we take/adapt multisets' formula to mean that the roles of marker is being taken by free elements, and the slots formed by the free elements is given by $n'$. So, the actual markers/ #free elements are given by: $n'-1$.

Taking the above small example, to show the adaptation of logic to make free elements act as markers is:

$n=6, m=2,k=3, n'= 6-(2)(1)= 4.$

#1. Taking $2,5$ as buffer, get free elements as: $1,3,4,6$.

$3$ forms the first boundary, and $6$ the second one. The chosen elements in set $m= 1,4$: $1(23)4(56)$,

#2. Taking $2,3$ as buffer, get free elements as: $1,4,5,6$.

$4$ forms the first boundary, and $6$ the second one. The chosen elements in set $m= 1,5$: $1(234)5(6)$,

and so on by taking suitably buffer elements, free elements as markers, and chosen elements; for the rest three cases.