问:
Why does the group of units functor respect products? That's I believe we have $(X×Y)^×\cong X^××Y^×$, in general. Why?
Perhaps someone can explain this category theoretic fact.  I must have read somewhere that it follows from the fact that there's a left adjoint for this functor.  That would be the group ring construction.

So I guess my question boils down to: why does the existence of a left adjoint for a functor guarantee that it respects (preserves) products?

答:
Here's a low-tech answer before answering about adjoints (I'm not sure whether that was part of your question, or whether your question was purely about the category-theoretic proof, but let me do it anyway for potential passer-by's):

a unit in $X\times Y$ is a couple $(x,y)$ with an inverse, that is, a couple $(x',y')$ with $(x,y)(x',y') = (1,1)$. Since multiplication is defined pointwise, this is equivalent to $xx' = 1, yy' = 1$ (and symmetrically for $x'x, y'y$) so that if $x,y$ are invertible, so is $(x,y)$.

Now here's the category theoretic version (this is a summary from the comments essentially): $R\mapsto R^\times$ is right adjoint to $G\mapsto \mathbb Z[G]$ as is easily checked, so that it must commute with all limits : this is a general fact, right adjoints always preserve limits.

Moreover, products are limits: $A\times B$ is the limit of the functor from the category $\bullet \space\bullet$ with values $A$ and $B$

The proof in that particular case could go as follows: $\hom(G,(X\times Y)^\times) \cong\hom(\mathbb Z[G],X\times Y)\cong \hom(\mathbb Z[G],X)\times \hom(\mathbb Z[G],Y)\cong \hom(G,X^\times)\times \hom(G,Y^\times) \cong \hom(G,X^\times\times Y^\times)$ and so $(X\times Y)^\times \cong X^\times \times Y^\times$ by the Yoneda lemma.


问:
Characterising inverses of $g'\in G'$ with the right cosets of $\ker \phi$ where $\phi:G\to G'$ is a homomorphism

$\phi: G\to G'$ is a homomorphism, with $\ker\phi = K$. The set of all inverse images of $g'\in G'$ under $\phi$ in $G$ is given by $Kx$, where $x$ is any particular inverse image of $g'\in G'$, under $\phi$.

My work:

Let $e,e'$ be the identity elements in $G,G'$ respectively. Let $x_0$ be a particular inverse image of $g'$, i.e. $\phi(x_0) = g'$. By definition, $K = \{x\in G: \phi(x) = e'\}$. Consider two sets, $Kx_0$ and $S$, where:
$$Kx_0 = \{kx_0: k\in K\}$$
$$S = \{x\in G: \phi(x) = g'\}$$
First we show $Kx_0 \subseteq S$. This is easy! Take $kx_0 \in Kx_0$ for some $k\in K$. Note that $\phi(kx_0) = \phi(k)\phi(x_0) = e'g' = g'$. So, $kx_0\in S$. This direction is done.

I am slightly suspicious of my proof of $S\subseteq Kx_0$. I take $x\in S$, so $\phi(x) = g'$.
$$g' = \phi(x) = \phi(x_0) = e'\phi(x_0) =  \phi(k)\phi(x_0) = \phi(kx_0)$$ for some $k\in K$. Now, $x = kx_0$ certainly satisfies the above equality (though there may be other possibilities). So, $x\in Kx_0$.

Is this last step correct? If yes, how do I make it more rigorous? If not, please suggest alternatives/improvements. Thank you!

答:
What you have done is not correct. Starting with $x \in S$, you must show that $x \in Kx_0$. All you have done is shown that "it's possible" for that to happen. What you have to actually do is show that that must happen.

Here's how you can prove $S \subseteq Kx_0$. Let $x \in S$. Then, $\phi(x) = g' = \phi(x_0)$ and hence,
$$e' = \phi(x)\phi(x_0)^{-1} = \phi(xx_0^{-1})$$
and hence, $xx_0^{-1} \in \ker \phi = K$. This finishes the proof.

ContentLicense: CC BY-SA 4.0


答:
Let $x\in\phi^{\leftarrow}(\{g'\})$. Then:

\begin{alignat}{1}
Kx &= \{gx\in G\mid \phi(g)=e'\} \\
&= \{y\in G\mid \phi(yx^{-1})=e'\} \\
&= \{y\in G\mid \phi(y)\phi(x)^{-1}=e'\} \\
&= \{y\in G\mid \phi(y)=\phi(x)\} \\
&= \varphi^{\leftarrow}(\{\phi(x)\}) \\
&= \varphi^{\leftarrow}(\{g'\}) \\
\end{alignat}

Title: Characterising inverses of $g'\in G'$ with the right cosets of $\ker \phi$ where $\phi:G\to G'$ is a homomorphism

Body:
$\phi: G\to G'$ is a homomorphism, with $\ker\phi = K$. The set of all inverse images of $g'\in G'$ under $\phi$ in $G$ is given by $Kx$, where $x$ is any particular inverse image of $g'\in G'$, under $\phi$.

My work:

Let $e,e'$ be the identity elements in $G,G'$ respectively. Let $x_0$ be a particular inverse image of $g'$, i.e. $\phi(x_0) = g'$. By definition, $K = \{x\in G: \phi(x) = e'\}$. Consider two sets, $Kx_0$ and $S$, where:
$$Kx_0 = \{kx_0: k\in K\}$$
$$S = \{x\in G: \phi(x) = g'\}$$
First we show $Kx_0 \subseteq S$. This is easy! Take $kx_0 \in Kx_0$ for some $k\in K$. Note that $\phi(kx_0) = \phi(k)\phi(x_0) = e'g' = g'$. So, $kx_0\in S$. This direction is done.

I am slightly suspicious of my proof of $S\subseteq Kx_0$. I take $x\in S$, so $\phi(x) = g'$.
$$g' = \phi(x) = \phi(x_0) = e'\phi(x_0) =  \phi(k)\phi(x_0) = \phi(kx_0)$$ for some $k\in K$. Now, $x = kx_0$ certainly satisfies the above equality (though there may be other possibilities). So, $x\in Kx_0$.

Is this last step correct? If yes, how do I make it more rigorous? If not, please suggest alternatives/improvements. Thank you!


问:
Probability Confusion/Clarification
Two pizzas are each divided into eight slices and placed inside separate boxes with the
lids closed, and left in an empty room. People randomly arrive at the room to take a slice.
Each person who arrives randomly chooses a box, opens it, takes a slice of pizza and closes
the box. If all eight slices from a box are taken then the lid is not closed.

The first person to discover an empty box opens the other box. Calculate the probability
that it contains all eight slices.

The answer says $\left(\dfrac{1}{2}\right)^8$. This is what I intuitively thought, until my friend posed the solution $\left(\dfrac{1}{2}\right)^7$ as he is looking at the problem from the standpoint that all the others have to do is pick the same box as the first person, and it doesn't matter which box the first person chooses. I have a feeling that perhaps his situation is more conditional probability, but I'm struggling to get my head around what is correct! Can anyone clarify?

答: The only possible way for a person to discover an empty box and a full one, is if  eight slices have been taken (not more). The second condition is, that every slice was taken from the same box. This adds up to your result $(\frac{1}{2})^8$ since every person must take a slice from the same box.

Title: Probability Confusion/Clarification


问:
Expectation of sample variance
Let $s^2$ be sample variance, $\sigma^2$ be population variance

$E[\frac{(n-1)s^2}{\sigma^2}] = E[\chi^2_{n-1}] = (n-1) \implies \frac{(n-1)E[s^2]}{\sigma^2} = n-1 \implies E[s^2]=\sigma^2$

But if i do in followng way, i am getting wrong answer

$E[s^2]=E[\frac{1}{n-1} \sum_{i=1}^n (x_i-\bar x)^2] \\ = \frac{1}{n-1} \sum_i (E[x_i^2] + E(\bar x^2) - 2E[\bar x]E[x_i])$

I know that

$E[\bar x] = \mu$, population mean

$E[x_i^2] = var(x_i)+E[x_i]^2 = \sigma^2+\mu^2 \\ E[\bar x^2] = var(\bar x)+E[\bar x]^2 = \frac{\sigma^2}{n}+\mu^2$

Plugging these in above boldface equation

$E[s^2] = \frac{1}{n-1} (n(\sigma^2+\mu^2)+n(\frac{\sigma^2}{n}+\mu^2)-2\mu^2) \\ \frac{n+1}{n-1} \sigma^2$

I know this is wrong. But i donot know where i made the mistake.

Title: Expectation of sample variance

Body:
$$\mathbb{E}[S^2]=\frac{1}{n-1}\mathbb{E}[\Sigma_i(X_i-\overline{X}_n)^2]=$$

$$=\frac{1}{n-1}\mathbb{E}[\Sigma_i(X_i-\mu)^2-n(\overline{X}_n-\mu)^2]=$$

$$=\frac{1}{n-1}\left\{\Sigma_i\underbrace{\mathbb{E}[(X_1-\mu)^2]}_{\text{var pop}=\sigma^2}-n\underbrace{\mathbb{E}[(\overline{X}_n-\mu)^2]}_{\mathbb{V}[\overline{X}_n]=\sigma^2/n}\right\}=$$

$$=\frac{1}{n-1}\left[n\sigma^2-n\frac{\sigma^2}{n}\right]=\sigma^2$$


问:
"Introduction to Smooth Manifolds" - differential of a function query
I'm trying to understand the differential $df$ as an approximation to $\Delta f$, in Lee's Introduction to Smooth Manifolds (p282-283 - see image below). He says “let $p$ be a point on $M$” but then, in the diagram, labels $p$ in $U$ but not on $M$. Why does he do that?

答: Lee  uses a chart $\phi : V \to U \subset \mathbb R^n$ around $p$ and considers $f \circ \phi^{-1} : U \to \mathbb R$. In Fig. 11.2 he should have written $\phi(p)$ instead of $p$, but working with a chart means that we may consider without loss of generality the special case $M = U$ and $p \in U$. Lee explicitly says "we can think of $f$ as a function on an open $U$". This is what Fig. 11.2 shows.


问:
What is the mathematical foundation of Control Theory?
There is a question which I'm wondering again and again in recent months. I have taken courses on elementary differential equations, signals and systems, linear control systems, general theory of circuits and networks but I still do not know how can a simple linear integral transformation (Laplace transform) be so much helpful in analyzing and designing linear systems. For example, I think I somewhat know what "time domain" is, but how about the "frequency domain"?

Consider a block diagram in the frequency domain. Conceptually, what do the blocks represent? What is traveling(?) on the connections between blocks? First, I have thought that instead of taking care of the statement "when does some phenomenon happens" we consider "how many times does a phenomenon oscillate", but how can this approach be more helpful? Is it the best (optimal) way of treating such problems?

I think control systems and even systems have a more general meaning in mathematics. I even think that the very problem of "controlling a system" does have multiple "paradigms" — A.I., neural networks, state space approach, fuzzy logic, robust control, adaptive control, etc.

What I'm looking for is a book (preferred) or paper discussing the very beginning notions of system, control system, its history, and more importantly, what is the field or language in mathematics to study systems rigorously? Dynamical systems? Game theory? Differential equations? Linear algebra? A.I?

Any help is appreciated. Thanks.

P.S.: I've already studied these books:

Alan Oppenheim's Signals and Systems (I think it is the best in the field)

Ogata's Modern Control Engineering

Desore's Basic Circuits Theory

Charles A - Kuh and Ernest S's A rather brief book on the history of control systems ($H_\infty$ methods and so on, but I didn't get much; at least, this book induced some information on me that nobody actually knows what "feedback" is, why is it so helpful and why we should take samples of the output! It wasn't a precise book, I admit)

26 May 2013:

As I said, I am looking for a book or article discussing the origins of Control Theory and Systems. Seven days are passed since I started the bounty, yet I didn't find an answer. I'll get to the bottom of this issue as soon as I get some free time. Thanks.

答:
All mathematics, functions, transforms and ..., can just model a physical system and sometimes all these models dont work properly. The blocks in linear control systems, are just models to let you understand better about the physics, say voltages, currents or maybe  fields, that travel(!) from one par of the system to another… The reality of a control system is the physics not the blocks in frequency domain, but these block just try to let you understand it better and let you control it when its going to be unstable.  Think of Fourier series which is indeed, expansion of a function to an infinite sum of complex exponentials, sometimes we dont need all of these exponential   to model a physical waveform  and a prat of them work, and sometimes the whole series fail to model a specific waveform.

A little bit Electronics (active circuts with feedback), Linear algebra and measure theory help you to understand better about the physics and foundation of Control theory...

ContentLicense: CC BY-SA 3.0


答:
The difference between $\left(\frac{1}{2}\right)^8$ versus $\left(\frac{1}{2}\right)^7$ depends on how we interpret the problem.

Problem 1:

We have two pizzas, each with eight pizza slices.  Each time someone enters the room, they open one of the two boxes uniformly at random, independent of each other persons' choices, and will take a slice if able and then close the lid except in the case where they have taken the last slice in which case they throw out the box.  We ask what the probability is that the first person to come in the room after a box has been thrown out will find a full eight slices of pizza in the remaining box.

For this problem, let us label the two pizza boxes $H$ and $T$ and let the people make the decision which box to eat from based on a coin flip.  Here, we have the two sequences HHHHHHHH and TTTTTTTT both result in the ninth person coming in and seeing one of the boxes missing and the remaining box having all of the slices of pizza in it, with a probability of $\left(\frac{1}{2}\right)^8 + \left(\frac{1}{2}\right)^8 = \left(\frac{1}{2}\right)^7$

The calculation could have been made simpler as your friend had noted by not caring which of the two boxes was chosen by the first person and then for the next seven all of them choosing the same box, yielding the same $\left(\frac{1}{2}\right)^7$

Problem 2:

We have two pizzas, each with eight pizza slices.  Each time someone enters the room, they open of of the two boxes uniformly at random, independent of each other persons' choices, and will take a slice if able and then close the lid... even in the case that they closed the lid on a now empty box.  We ask what the probability is that the first time that someone comes into the room and opens an empty box (that they did not cause to be empty themselves), that the remaining box has all eight slices still in it.

For this problem, similarly to the last, we describe via a sequence of coin flips.  Here however, after the first box becomes empty, we require that the empty box be discovered before slices from the still full box be taken.  In this case we required nine heads in a row or nine tails in a row.  This occurs with probability $\left(\frac{1}{2}\right)^9 + \left(\frac{1}{2}\right)^9 = \left(\frac{1}{2}\right)^8$

Which of these was the intended problem?  That is up to the person who asked the problem, but I interpret it as the first as per the included phrase "If all eight slices from a box are taken then the lid is not closed." which implies to me that no random event needs to occur for the ninth person in the event that the first eight people all ate from the same box since the contents of the empty box are plainly visible due to the lid remaining open.

As for why the linked Matchbox Problem gives the other answer, that is because it follows the style of problem 2 where we do not remove the newly emptied box once it becomes empty and we leave it until we attempt to retrieve a match from it and discover there are none to be had.

Title: Probability Confusion/Clarification

Body:
Two pizzas are each divided into eight slices and placed inside separate boxes with the
lids closed, and left in an empty room. People randomly arrive at the room to take a slice.
Each person who arrives randomly chooses a box, opens it, takes a slice of pizza and closes
the box. If all eight slices from a box are taken then the lid is not closed.

The first person to discover an empty box opens the other box. Calculate the probability
that it contains all eight slices.

The answer says $\left(\dfrac{1}{2}\right)^8$. This is what I intuitively thought, until my friend posed the solution $\left(\dfrac{1}{2}\right)^7$ as he is looking at the problem from the standpoint that all the others have to do is pick the same box as the first person, and it doesn't matter which box the first person chooses. I have a feeling that perhaps his situation is more conditional probability, but I'm struggling to get my head around what is correct! Can anyone clarify?


问:
A specific type of orthogonal bases
In $\mathbb{R}^n$, can we find an orthogonal basis $v_1,v_2,\ldots,v_n$ with $\|v_i\|=1$ such that
$$\|\bar{v}_1\|=\|\bar{v}_2\|= \cdots =\|\bar{v}_n\|$$
where $\bar{v}_i\in\mathbb{R}^m$ is the vector formed by the first $m$ coordinates of $v_i$ with respect to the standard basis and $m$ is a given integer $m<n$.</p>

答:
I think there's a trick that can make joriki's idea work in any case (I hope I'm not mistaken). Let's denote $A_{m,n}$ the assertion that there exists a solution to the problem for the values $m \le n$ (we take the convention that $A_{0,n}$ is always true). Obviously, $A_{m,m}$ is true for all $m$. By flipping rows, we see that

$A_{m, n} \Longleftrightarrow A_{n-m, n}$

And joriki showed that $A_{m,n}$ is equivalent to the fact that there exist a (non zero) $m$ by $n$ matrix whose rows are orthonormal, and columns have same norm. So by concatenation of such matrices (with suitable scaling factor so that the rows remain of norm 1, and column of both matrices have same norm), we get

$(A_{m, n_1} \textrm{ and } A_{m, n_2} ) \Longrightarrow A_{m, n_1 + n_2}$

Let $m < n$ be two integers, and $r$ the remainder of $m$ modulo $n$. Using the properties above, we see that

$A_{r, m} \Longrightarrow A_{r, m+r} \Longrightarrow A_{m, m+r} \Longrightarrow A_{m, n}$

So checking if $A_{m,n}$ is true can be reduced to checking if $A_{r, m}$ is. Applying Euclide's algorithm, we can reduce this to a point where the remainder is $0$, in which case there is a solution.

Title: A specific type of orthogonal bases


问:
Sequences with the following properties...
Suppose that $\lim_{n\to\infty}\frac{a_n}{b_n}$=1. And $\sum^{\infty}_{n=1}a_n$ converges, $\sum^{\infty}_{n=1}b_n$ diverges. Are such sequences $(a_n)$, $(b_n)$ exist? </p>

答: You may consider, for example, $a_n=\frac{(-1)^{n-1}}{\sqrt{n}}$ and $b_n=a_n+\frac{1}{n}$.</p>


问:
Differentiation for 2nd order ODE
Having an issue differentiating the following equation for purposes of reducing into two 1st order ODE's. Where: $y(1)=0$, $y'(1) = 2/3$ and $h = 0.5$

$$3x^2y'' - 5xy'+5y = 0$$

when making $y'=z$.

Trying to find $dy/dx$ and $dz/dx$, then put it into a 4th order Runge-Kutta where:

$$dy/dx=f(x,y,z)$$ $$dz/dx=g(x,y,z)$$

This is my process (roughly):
Help with using the Runge-Kutta 4th order method on a system of 2 first order ODE's.

答:
If $z = y'$ then $z' = y''$, and you get
\begin{cases}
3x^2z' - 5xz+5y = 0\\
\phantom{3x^2z' - 5xz+5}y' = z.
\end{cases}

We can rewrite this by solving for $z'$ and $y'$ as
\begin{cases}
z' = \frac{5xz - 5y}{3x^{2}}\\
y' = z.
\end{cases}

Title: Differentiation for 2nd order ODE


问:
Prove that $\Sigma\Vdash^g_M \phi$ iff $\{\square^n\sigma: \sigma\in \Sigma,n\in\omega\} \Vdash_M\phi$

(Modal Logic, Blackburn, Ex 1.5.3): Let $\Sigma$ be a set of formulas in the basic modal language, and let $M$ denote the class of all models. Prove that $$\Sigma\Vdash^g_M \phi \Longleftrightarrow \{\square^n\sigma: \sigma\in \Sigma,n\in\omega\} \Vdash_M\phi$$

My work:

For the $[\Leftarrow]$ direction: Assume $\{\square^n\sigma: \sigma\in \Sigma,n\in\omega\} \Vdash_M\phi$, i.e. $\forall \mathfrak M\in M,\forall w\in W, \mathfrak M,w\Vdash \{\square^n\sigma: \sigma\in \Sigma,n\in\omega\} \implies \mathfrak M,w\Vdash\phi$. We want to show that $\forall \mathfrak M\in M, \mathfrak M\Vdash\Sigma \implies \mathfrak M\Vdash\phi$. Consider arbitrary $\mathfrak M$, and suppose $\mathfrak M\Vdash\Sigma$. So, $\mathfrak M,w\Vdash\Sigma$ for all $w\in W$ (some would like to write $\mathfrak M,w\Vdash\sigma$ for all $w\in W,\sigma\in\Sigma$ - but that notation doesn't add any value). To show $\mathfrak M\Vdash\phi$, we take arbitrary $w\in W$ and show $\mathfrak M,w\Vdash\phi$. Since we have $\forall w\in W, \mathfrak M,w\Vdash\Sigma, $ we conclude $\mathfrak M,w\Vdash \{\square^n\sigma: \sigma\in \Sigma,n\in\omega\}$. By assumption, $\mathfrak M,w\Vdash \phi$.

I'm stuck with the $[\Rightarrow]$ direction. Assume $\Sigma\Vdash^g_M \phi$ holds, i.e. $\forall \mathfrak M\in M, \mathfrak M\Vdash\Sigma \implies \mathfrak M\Vdash\phi$. We want to show that $\forall \mathfrak M\in M, \mathfrak M,w\Vdash \{\square^n\sigma: \sigma\in \Sigma,n\in\omega\} \implies \mathfrak M,w\Vdash\phi$. Let us argue by contradiction. Suppose, there exists $\mathfrak M\in M$ and some $w\in W$, for which $\mathfrak M,w\Vdash \{\square^n\sigma: \sigma\in \Sigma,n\in\omega\}$ but $\mathfrak M,w\nVdash\phi$. What do I do next?

I really need to be able to use the $\mathfrak M$ and $w$ above to construct some model $\mathfrak M'$ wherein $\mathfrak M'\Vdash\Sigma$ holds so I can use the assumption. I don't know how exactly that would help though.

Definitions if you need them:

$\Sigma\Vdash^g_M \phi$ - For all $\mathfrak M \in M$ (class of all models), $\mathfrak M\Vdash\Sigma \implies \mathfrak M\Vdash\phi$. This is the global semantic consequence relation.

$\Sigma\Vdash_M \phi$ - For all $\mathfrak M \in M$ (class of all models), $\forall w\in W$, $\mathfrak M,w\Vdash\Sigma \implies \mathfrak M,w\Vdash\phi$. This is the local semantic consequence relation.

Please help, thanks a lot! Hints are okay, and solutions are good too!

答:
Here is a proof $[\Rightarrow]$ with some details omitted. Let me know if this helps.

Assume you have a countermodel to the right hand side of the equivalence, i.e. a pair $(\mathfrak M,w)$ such that $\mathfrak M,w\Vdash \{\square^n\sigma: \sigma\in \Sigma,n\in\omega\}$ and $\mathfrak M,w\nVdash\phi$.

In the next step, you should consider the model $\mathfrak M'$ which arises by removing from $\mathfrak M$ all worlds which cannot be reached from $w$ via a directed path using the accessibility relation. Note that $\mathfrak M'\in M$ as $M$ is the class of all models.

(Maybe you want to do the proof from here for yourself. Otherwise, the rest goes as follows:)

Convince yourself that removing such unreachable worlds does not have an effect on which modal formulas are true at $w$. So we still have (i) $\mathfrak M',w\Vdash \{\square^n\sigma: \sigma\in \Sigma,n\in\omega\}$ and (ii) $\mathfrak M',w\nVdash\phi$. Let $w'$ be any world in $\mathfrak{M}'$. By construction $w'$ is reachable from $w$ in $\mathfrak{M}'$ and so from (i) it follows (check!) that $\mathfrak{M}',w'\Vdash \sigma$ for all $\sigma\in\Sigma$. Therefore $\mathfrak M'\Vdash\Sigma$, and from this and (ii) you get a countermodel to the left hand side of the equivalence.


问:
Suppose $n \geq a_1>a_2>...>a_k\geq1$ and $lcm(a_i,a_j) \leq n$ for $i,j=1,2,...,k$. Show that $ia_i \leq n$.
Could I receive some hints on the problem. I do not know how to touch it since index $i$ seems like it does not depend on any of the other conditions.

答:
Suppose that $ia_i>n$.

Let the LCM of $a_{i-1}$ and $a_i$ be given by $$sa_{i-1}=ta_i$$ where $s$ and $t$ are positive integers. These integers must be different and less than $i$ and so $$a_{i-1}>\frac{i-1}{i-2}a_i>\frac{i-1}{i-2}\frac{n}{i}.$$ Then  $$(i-1)a_{i-1}>\frac{i^2-2i+1}{i^2-2i}n>n.$$  We can now repeat this argument for $(i-1)a_{i-1}$, and we get:   $$a_1>2a_2> ...>(i-1)a_{i-1}>n.$$ A contradiction.


问:
The usual metric on $\mathbb N$ is equivalent to the discrete metric, and any metric on a finite set is equivalent to the discrete metric

Show that the usual metric on $\mathbb N$ is equivalent to the discrete metric, and any metric on a finite set is equivalent to the discrete metric.

First of all, two metrics $d,\rho$ are equivalent if they generate the same convergent sequences, i.e. $d(x_n,x) \to 0$ iff $\rho(x_n,x)\to 0$.

The usual metric on $\mathbb N$ is the one it inherits from $\mathbb R$. Let's denote this metric by $d$, and the discrete metric by $d_0$. My proof is as follows:

$[\Rightarrow]$ Let $d(x_n,x) \to 0$, i.e. $|x_n-x| \stackrel{n\to\infty}{\longrightarrow} 0$. In $\epsilon$ notation, $$\forall \epsilon > 0 \exists N\in\mathbb N\ \forall n\ge N (|x_n - x| < \epsilon)$$
Take $\epsilon = 1$, and find suitable $N$. For all $n\ge N$, we have no choice but $x_n = x$. This is immediate from the fact that $x_n$ is a sequence on $\mathbb N$. $x_n$ is thus, eventually constant. This tells us that $d_0(x_n,x) = 0$ for $n\ge N$. We conclude $$\forall \epsilon > 0 \exists N\in\mathbb N\ \forall n\ge N (d_0(x_n,x) < \epsilon)$$
because no matter what $\epsilon$, we can always just pick $N$ as the same one picked above to get $d_0(x_n,x) = 0$.

$[\Leftarrow]$ Let $d_0(x_n,x)\to 0$. So, $$\forall \epsilon > 0 \exists N\in\mathbb N\ \forall n\ge N (d_0(x_n,x) < \epsilon)$$
Take $\epsilon = 1$, and find suitable $N$. For $n\ge N$, we have $d_0(x_n,x) = 0$ so $x_n = x$, i.e. $x_n$ is eventually constant. It is easy to see that $$\forall \epsilon > 0 \exists N\in\mathbb N\ \forall n\ge N (|x_n - x| < \epsilon)$$ holds, since for whatever $\epsilon$ we have, we can always choose $N$ as the same one we chose in the previous step. The proof ends here.

Question 1: Is the above proof alright?

Question 2: I have no clue how to begin the second half, i.e. any metric on a finite set is equivalent to the discrete metric. How do I do this? It is a really strong statement!

Thanks a lot for your help and attention.

P.S. A follow-up question:

If $M$ is a countable set and $d$ is a metric on $M$, then is $d$ equivalent to the discrete metric on $M$?

答:
Answer 1. The proof looks correct to me.

Answer 2. The idea is the same. (A sequence converges in the discrete metric iff it is eventually constant.)

Let $A = \{a_1, \ldots, a_n\}$ be a finite set and let $d_0$ be a metric on $A$. Consider $$r = \min_{1 \le i < j \le n}d(a_i, a_j).$$
The $\min$ makes sense because it is over a finite set. Moreover, $r > 0$. (I am assuming that the listed quantities are distinct.)

Now, can you imitate your proof of part 1 to show that convergent sequences in $(A, d)$ are eventually constant? (Note that $\epsilon = 1$ was super useful there for some reason. Can $\epsilon = r$ fulfil the same purpose here?)

As earlier, can you now conclude that $d$ must be equivalent to the discrete metric?

For the follow-up: Consider the countable set $\Bbb Q$ with the standard metric. Note that $\dfrac{1}{n} \to 0$ in this metric. Is the same true in the discrete metric?


问:
Concrete Bezout's Theorem Calculation
For context: I'm taking an algebraic geometry class and TAing for a calculus class. My students had the classic problem of finding the area between $x^3$ and $x^2$, but primed by algebraic geometry, I had a thought while I was grading:

Shouldn't $x^2$ and $x^3$ intersect in $2 \cdot 3 = 6$ points by Bezout's Theorem?

Obviously for the purpose of the calculus class the answer is "no", the only intersections are $0$ and $1$, but there are 3 reasons why this might be the case:

We're working over $\mathbb{R}$ not $\mathbb{C}$

We're not working in a projective setting

$x^2$ and $x^3$ actually intersect twice at $0$, so our curves aren't suitably "generic" (whatever that means).

So I tried to solve these problems by considering (in $\mathbb{C}$) the intersection of

$yz = x^2 + \alpha z^2$

$yz = x^3 + \beta z^3$

Here we've projectivized, and the $\alpha$ and $\beta$ were added to try and solve the genericness issue while keeping the equation somewhat simple.

So surely these equations will have 6 solutions, right? Well...

If we look in the $x=1$ plane, then we see $1 + \alpha z^2 = yz = 1 + \beta z^2$, so either $z=0$ or $z = \frac{\alpha}{\beta}$. I think the solution $z=0$ doesn't actually exist, though, since
$y0 = 1 + \alpha 0^2$ has no solutions for $y$. So we pick up 1 solution.

If we look in the $y=1$ plane, nothing interesting happens since we can always just eliminate $y$, so any solutions will already by in the $x=1$ or $z=1$ planes. Let's move on.

If we look in the $z=1$ plane, then we get $x^2 + \alpha = y = x^3 + \beta$, so we pick up 3 complex solutions.

So then we end up with 4 solutions instead of the expected 6. Where did I go wrong? I suspect the issue is either

adding $\alpha$ and $\beta$ alone don't make these curves "generic enough", whatever that means (so the double root $z=0$ that we killed off in step 1 should really be counted, bringing our total to 6)

I've misunderstood something about solving equations where we can eliminate some variables, and there are some extra roots in the $y=1$ plane that I shouldn't have ignored.

Any clarification is appreciated! Thanks in advance!

答:
KReiser has given a nice explanation of what's going on with your original system of equations.  Let me discuss how to correctly investigate the perturbation with $\alpha$ and $\beta$ that you tried.

First, you made a mistake when homogenizing the second equation: it should be $$yz^2 = x^3 + \beta z^3,$$ not $$yz = x^3 + \beta z^3,$$ since every term needs to have degree $3$.

Now let's look at what happens when $x=1$.  The equations become $$yz=1+\alpha z^2$$ and $$yz^2=1+\beta z^3.$$  These can be combined to get $$z+\alpha z^3=1+\beta z^3$$ which for generic $\alpha$ and $\beta$ will have $3$ different solutions for $z$, all of which are nonzero.  For each value of $z$ you can then solve for $y$ (since $z$ is nonzero), so this gives $3$ solutions to the system.

In your attempt, you proceeded to consider the case $y=1$, and discarded it as redundant because you can "eliminate" $y$.  This doesn't make sense, though.  Remember that a solution to the original homogeneous system of equations is a triple $[x:y:z]$ which is defined up to scaling, and you are putting them in one of your affine planes by scaling so that one variable is $1$.  So, a solution $[x:y:z]$ can be scaled to be in the $y=1$ plane as long as its $y$-coordinate is nonzero.  This solution will also show up in the $x=1$ plane as long as its $x$-coordinate is nonzero, since you could instead scale it to make the $x$-coordinate nonzero.  This has nothing to do with whether $y$ can be "eliminated" from the system; rather, what is relevant is whether the resulting values of $x$ and $z$ you get are $0$ or not.

Here's how I would think about it instead.  We've already found the solutions where $x$ is nonzero (so we can scale to make $x=1$), so let's consider solutions where $x=0$.  As long as $y$ is nonzero, we can scale such a solution to make $y=1$, so let's set $x=0$ and $y=1$ in the equations to get $$z=\alpha z^2$$ and $$z^2=\beta z^3.$$  For generic $\alpha$ and $\beta$, this system has only $1$ solution, $z=0$.

The only remaining case is if $x$ and $y$ are both $0$, in which case $z$ must be nonzero (since we are not allowed to have all three variables $0$), and thus we can scale so $z=1$.  For generic $\alpha$ and $\beta$ this does not solve the equations, so we get no solutions in this case.

So in total, we got only $4$ solutions, not $6$!  What's wrong?  Well, it turns out that even with your $\alpha$ and $\beta$ perturbation, these polynomials are not sufficiently "generic", i.e. they still intersect with multiplicity.  Specifically, the intersection $[x:y:z]=[0:1:0]$ that we found in the second case actually has multiplicity $3$.  To get a sense of why this is the case, let's look at it in the $y=1$ plane, in which our equations become $$z=x^2+\alpha z^2$$ and $$z^2=x^3+\beta z^3.$$  Isolating $x$ in the first equation and cubing, we write $x^6$ as some degree $6$ polynomial in $z$.  Isolating $x$ in the second equation and squaring, we write $x^6$ as some other degree $6$ polynomial in $z$.  Setting these equal, we get a degree $6$ equation in $z$ which ought to have $6$ solutions.  Actually, though, it has only $4$, since $z=0$ is a triple root.  We can see this quickly by observing that both of the expressions we get for $x^6$ will be divisible by $z^3$.  On the other hand, if we perturbed the equations a little more (say by adding $\gamma y^2$ and $\delta y^3$ terms which would become constant terms when we set $y=1$), the degree $6$ polynomial in $z$ that we would get would generically have no repeated roots, so we would get $6$ different solutions for $z$.  For each one, we could then solve uniquely for $x$ (though it takes a bit of work to see we really do get solutions that work in the original system), giving $6$ different intersection points of our curves.

Title: Concrete Bezout's Theorem Calculation


问:
If $17 \mid \frac{n^m - 1}{n-1}$ find the values of $n$ where $m$ is even but not divisible by $4$

Let $m, n \in \mathbb{Z}_+$ with $n > 2$, and let $\frac{n^m-1}{n-1}$ be divisible by $17$. Show that either $m$ is even:$ m \equiv 0 \mod 17$ and $n \equiv 1 \mod 17$. Find all possible values of $n$ in the cases when $m$ is even but not divisible by $4$, or divisible by $4$ but not divisible by $8$.

So far, I have done this: First, let's assume that $n \equiv 1 \mod 17$. Also, we know that we can write

$$\frac{n^m-1}{n-1} = \sum_{i = 0}^{m-1} n^i.$$

As $n \equiv 1 \mod 17 \implies n^i \equiv 1 \mod 17$ and so $\sum_{i = 0}^{m-1} n^i \equiv \underbrace{1 + 1 + \cdots + 1} \equiv  m$.

Also, we know that $\sum_{i = 0}^{m-1} n^i \equiv 0 \mod 17 \implies m \equiv 0 \mod 17$.

Now, let's assume that $n \not\equiv 1 \mod 17$. Then, $\frac{n^m - 1}{n - 1}$ is divisible by $17$ if $n^m - 1$ is divisible by $17$, i.e $n^m - 1 \equiv 0 \mod 17 \implies n^m \equiv 1 \mod 17$. Fermat's little theorem tells us that $n^{p-1} \equiv 1 \mod p \implies n^{16} \equiv 1 \mod 17$. From here, we have that $m \mid 16$ which tells us that $m$ must be even.

However I am now stuck on the next bit, I'm not sure how to find those values of $n$. Can someone help me please. Thank you.

答:
As the first case is perfect, let me start with the second.

As $17$ is  prime and hence $\phi(17)=16$

As Primitive Root is a Generator of Reduced Residue System, let's find a primitive root of $17$

$2^4\equiv-1, 2^8\equiv(-1)^2\equiv1\pmod{17} \implies \text{ord}_{17}2=8<\phi(17)$

$3^2\equiv9, 3^4\equiv-4,3^8=(3^4)^2\equiv(-4)^2\equiv-1\pmod {17}$
$\implies \text{ord}_{17}3=16$ i.e., $3$ is  a primitive root of $17$

From this, if $d=$ord$_ma,$ ord$_m(a^r)=\frac d{(r,d)}$ where $m,d$ are positive integers  and $a,r$  are integers.

So, ord$_{17}(3^n)=\frac{16}{(16,n)}$

Case $1:$ If ord$_{17}(3^n)\equiv2\pmod 4, n=8s$ where $s$ is odd $\implies 3^{8s}=(3^8)^s\equiv(-1)^s\equiv-1\pmod{17}\implies $ord$_{17}(-1)=2$

Case $2:$ If ord$_{17}(3^n)\equiv4\pmod 8, n=4s$ where $s$ is odd
$\implies n\equiv4,12\pmod{16}$

Now, $3^4\equiv-4\pmod{17}$ and $3^{12}=3^8\cdot3^4\equiv(-1)(-4)\equiv4\pmod {17}$
$\implies $ord$_{17}(\pm4)=4$

Case $3:$ If ord$_{17}(3^n)\equiv8\pmod {16}, n=2s$ where $s$ is odd
$\implies n\equiv2,6,10,14\pmod{16}$


问:
Interpreting domain notation
I've come across a phrase in a paper with notation I don't know how to interpret. The phrase is this:

...given a signal $f:\Bbb R \rightarrow \Bbb R$ a function $L:\Bbb R \times \Bbb R_+ \rightarrow \Bbb R$ is defined by ...

My question is how to interpret  $\Bbb R \rightarrow \Bbb R$ and $\Bbb R \times \Bbb R_+$.

答:
$\mathbb{R}\to\mathbb{R}$ means the function's domain and codomain are both $\mathbb{R}$.

$\mathbb{R}\times\mathbb{R}_+$ is the cartesian product of $\mathbb{R}$ and $\mathbb{R}_+$. I believe $\mathbb{R}_+$ is the set of positive real numbers.

Title: Interpreting domain notation


问:
$y''+p(t)y'+q(t)y=0$ has finite number of zeros

Given $y''+p(t)y'+q(t)y=0$ and $p,q$ are continuous on $\mathbb{R}$ and let $y$ be a non-trivial solution of the system.

Prove that in any finite interval $[a,b]$ exists at most a finite number of zeros $\{t_k\}_{k=1}^{n}\subset[a,b]$ such that $y(t_k)=0$ for every $k$

I've thought transforming into S-L form by multiply the equation with $e^{\int p(s)}$ and then I get :

$(y'e^{\int p(s)})'+e^{\int p(s)}q(t)y=0$ but I don't know how can I say something about the number of zeros from this equation. any hint?

答:
Any solution with $y(c)=y'(c)=0$ is the zero function.

Thus any non-zero solution has only isolated roots (non-zero derivative at root and thus a root-free neighborhood).

There can only be a finite number of isolated roots in a finite interval.


问:
Why is the kernel of a group homomorphism so called?
I took a course in Linear Algebra last fall. I have come to associate the word kernel with the nullspace of matrices/linear transformations - and now that I am studying Group Theory, similar intuition doesn't seem to carry forward.

Why would anyone name it kernel (in group theory) if it really isn't the same thing as that in linear algebra? Perhaps there is a relation or connect between the two definitions that I am missing. Could someone help me better understand why the kernel of a homomorphism is so defined?

To document the definitions:

(Linear Algebra): Consider a linear transformation $T:V\to W$. $$\ker T = \{x\in V: T(x) = 0\}$$
(Group Theory): Let $\phi:G\to H$ be a group homomorphism. $$\ker\phi = \{x\in G: \phi(x) = 1_H\}$$ where $1_H$ is the identity in $H$.

Thanks!

答:
It might be worth noting that it appears that the use of "kernel" in group theory precedes the use in Linear Algebra. According to the Earliest known uses of some of the words in mathematics website, though the term "kernel" was already in use in Fourier analysis and the theory of integral equations, it first appeared in the algebraic context in Pontrjagin's 1931 paper, Über den algebraischen Inhalt topologischer Dualitätssätze in Math. Annalen 105. The quotation, translated, reads "The set of all the elements of the group $G$ which go into the identity of the group $G^*$ under the homomorphism $g$ is called the kernel of this homomorphism." The use in linear algebra came from Birkohff and Mac Lane's "Survey of Modern Algebra", 3rd edition, where they write:

Since $\mathbf{0}$ is the group identity, it follows that that the null-space of $T$ is precisely the kernel of $T$ regarded as a group-homomorphism.

"Kernel" can either mean "the inner, softer part of a seed, fruit stone, or nut", or "the central or essential part". I suspect, but this is mere surmise, that the use in algebra is related to the first meaning, thinking about the things that get "squished" into the trivial element, at the "center" or "middle" of the group.


问:
Normal vector of an implicit curve
We consider the equation $y^2-4y+4+x^2y+\arctan(x)=0$, the exercice of my book say that the normal vector $n$ of the cartesian curve implicitally defined by the previous equation is $n=l(1,0)$ for $l \neq 0$. I don't understand the reason of this affermation, because I think that if $y(x)$ is the implicit function obtained form the equation, the parametric equation of the curve is $(t, y(t))$, so the first coordinate of the normal vector is zero.
I would therefore expect that the normal vector is $n=l(0,1)$. Is it right?

答:
That's a little odd.  The tangent vectors and normal vectors to a curve typically vary in direction but the result you give doesn't!  IF the normal vector is always in the direction (1, 0) the tangent vector is always in direction (0, 1) (or (0, -1)) so is a straight line of the form x= constant.

The curve is given by $y^2- 4y+ 4+ x^2y+ a tan(x)= 0$. The derivative is given by $2yy'- 4y'+ 2xy+ x^2y'+ a sec^2(x)= 0$.
$(2y- 4+ x^2)y'= -2xy- asec^2(x)$

The slope of the tangent line at point $(x_0, y_0)$ is $-\frac{2x_0y_0+ a sec^2(x_0)}{2y_0- 4+ x_0^2}$ and the slope of the normal line at that point is $\frac{2y_0- 4+ x_0^2}{2x_0y_0+ a sec^2(x_0)}$.  The normal line at that point is given by $y= \frac{2y_0- 4+ x_0^2}{2x_0y_0+ a sec^2(x_0)}(x- x_0)+ y_0$

Title: Normal vector of an implicit curve


答:
For $n \ge 3$, let $a_n= -\frac{1}{\log n}$ if $n$ is odd, and $\frac{1}{\log(n-1)}$ if $n$ is even. The series converges to $0$, consecutive terms cancel.

Let $b_n=-\frac{2}{\log 2n+\log n}$ if $n$ is odd, and $b_n=a_n$ if $n$ is even.

The sum of two consecutive terms, the first of which has odd index, of the series $\sum b_n$ is $\frac{1}{\log(n-1)}-\frac{2}{\log(2n)+\log(n)}$. This simplifies to
$$\frac{\log(2n)+\log n-2\log(n-1)}{(\log(2n)+\log n)\log(n-1)}.$$

The top has limit $\log 2$. The bottom is slow-growing, so the series diverges.

Note that $\lim_{n\to\infty}\frac{a_n}{b_n}=1$.

Remark: The example by Landscape is much nicer.

Title: Sequences with the following properties...

Body: Suppose that $\lim_{n\to\infty}\frac{a_n}{b_n}$=1. And $\sum^{\infty}_{n=1}a_n$ converges, $\sum^{\infty}_{n=1}b_n$ diverges. Are such sequences $(a_n)$, $(b_n)$ exist? </p>


问:
Applying the exterior differential to the first law of thermodynamics
I'm working on an exercise for an advanced statistical physics course. The question I'm struggling with is this:

$$TdS=dE+PdV-\mu dN\tag{1}$$

Write the first law $(1)$ as $dS=....$ Applying the exterior differential $d$ to the resulting equation and using $d(dS)=0$, derive the relation

$$\frac{\partial T}{\partial V}\Bigg|_{E,N}-P\frac{\partial T}{\partial E}\Bigg|_{V,N}+T\frac{\partial P}{\partial E}\Bigg|_{V,N}=0\tag{3}$$

(Hint: Keep in mind that $dEdV=-dVdE$.)

So I've written down the first law of thermodynamics of the form $S(E,V,N)$ in terms of dS: $$dS=\frac{1}{T}dE+\frac{P}{T}dV-\frac{\mu}{T}dN$$ I can also write dS in terms of its partial derivatives: $$dS=\frac{\partial S}{\partial E}dE+\frac{\partial S}{\partial V}dV+\frac{\partial S}{\partial N}dN$$

Now I'm being asked to apply the exterior differential, to calculate $d(dS)$, and then use $d(dS)=0$ to prove the statement from the exercise.

I thought I needed to calculate the exterior differential as follows: $$ d(dS)=\frac{-1}{T^2}dTdE+\frac{1}{T}d(dE)-\frac{P}{T^2}dTdV+\frac{1}{T}dPdV+\frac{P}{T}d(dV)-\frac{1}{T}d\mu dN+\frac{\mu}{T^2}dTdN-\frac{\mu}{T}d(dN) $$

where I would then use that $d(dE)=d(dV)=d(dN)=0$ to eliminate those terms. But then I'm stuck. Nothing seems to get me to the required equation. Could anyone point me in the right direction?

edit: thanks to the hint from Michael Albanese i've made some progress.

I've written $dT$ in terms of $dE,dV,dN$ which gives me:

$$ dT=\frac{1}{S}dE+\frac{P}{S}dV-\frac{\mu}{S}dN $$
and
$$ dT=\frac{\partial T}{\partial E}dE+\frac{\partial T}{\partial V}dV+\frac{\partial T}{\partial N}dN $$

So that i now have that $\frac{\partial T}{\partial V}=\frac{P}{S}$ and $\frac{\partial T}{\partial E}=\frac{1}{S}$.

Then I tried writing $dP$ in terms of $dE,dV,dN$ and I'm getting stuck again.

I started with

$$ P = \frac{TS+\mu N-E}{V} $$

and then took the derivative:

$$ dP = \frac{1-\mu -TS}{V^2}dV+\frac{\mu}{V}dN-\frac{1}{V}dE $$
along with
$$ dP = \frac{\partial P}{\partial V}dV+\frac{\partial P}{\partial N}dN+\frac{\partial P}{\partial E}dE $$

to give me $\frac{\partial P}{\partial E}=-\frac{1}{V}$

Now the equation that i'm supposed to derive turns into:

$$ \frac{P}{S}-P*\frac{1}{S}+T*\frac{-1}{V} \neq 0 $$

am i missing a very obvious step? I also haven't figured out how to use $d(dS)$ up until this point...

答:
If a 1-form has the expression $\omega =\sum_i \alpha_i d x_i$, then the expression of its exterior derivative is
\begin{equation}
d\omega =\sum_i\sum_j \frac{\partial \alpha_i}{\partial x_j} d x_j\wedge d x_i
\end{equation}
Use this with $E, V, N$ as the $x_i$'s and $d S$ as $\omega$ (which implies that $d\omega = d(dS) = 0$).


答:
Note that the zero solution is always a solution.
Also note that any solution given a value and derivative in a point is unique.
If for a given solution $y$ there is an infinite number of zeros in $[a, b]$, then there is a convergent subsequence of zeros in $[a, b]$, let's say this limit is $c$. What is the value and derivative of $y$ in $c$? If you can prove this means $y$ has to be the zero solution itself, this shows the only solution with an infinite number of zeros in $[a, b]$ is $0$.

Title: $y''+p(t)y'+q(t)y=0$ has finite number of zeros

Body:
Given $y''+p(t)y'+q(t)y=0$ and $p,q$ are continuous on $\mathbb{R}$ and let $y$ be a non-trivial solution of the system.

Prove that in any finite interval $[a,b]$ exists at most a finite number of zeros $\{t_k\}_{k=1}^{n}\subset[a,b]$ such that $y(t_k)=0$ for every $k$

I've thought transforming into S-L form by multiply the equation with $e^{\int p(s)}$ and then I get :

$(y'e^{\int p(s)})'+e^{\int p(s)}q(t)y=0$ but I don't know how can I say something about the number of zeros from this equation. any hint?


问:
Prove That $\sin(1)+\sin(2)+...+\sin(n)=\frac{\sin(\frac{n+1}{2})\sin(\frac{n}{2})}{\sin(\frac{1}{2})}"$
I need help Proving this: "For Every $n\in \mathbb{N-{0}}$ Prove the following:

$$\sin(1)+\sin(2)+...+\sin(n)=\frac{\sin(\frac{n+1}{2})\sin(\frac{n}{2})}{\sin(\frac{1}{2})}"$$

I assume we have to use Induction On this but I failed to do so

If anyone can provide a simple and clear proof I would be grateful!

答:
Hint:

You don't need induction. Simply use that

$$\sin(1)+\sin(2)+\dots+\sin(n)=\operatorname{Im}\bigl(\mathrm e^{i}+\mathrm e^{2i}+\dots+\mathrm e^{ni}\bigr),$$
which is the sum of a geometric series with common ratio $\mathrm e^{i}$.


问:
Differentiation, how do I find the turning points of this curve? also how do I determine whether the turning point is a maximum or a minimum?

Given $$y=\frac{\ln(x)}{x}$$
what are the coordinates of the turning point?

I believe that when you differentiate it you get $x=e$, but I need help finding both coordinates.

How do I determine whether the turning point is a maximum or a minimum?

Another part of the question is : The points A and B lie on the curve and have x-coordinates 2 and 4. show that the line AB is parallel to the x axis.

how do I go about this

答: Differentiating $y = \frac{\ln x}{x}$ gives you $$y' = \frac{1 - \ln x}{x^2}$$ which, as you seem to have determined, is zero when $x = e$.  To get the $y$-coordinate we just need to plug $x = e$ back into the original function, so the $y$-coordinate is $\frac{\ln e}{e} = \frac{1}{e}$.  That tells us that the turning point is at $(e,1/e)$. To determine whether this is a maximum or minimum, we notice that because $x^2$ is always non-negative, the sign of $y'$ is determined by the sign of $1 - \ln x$.  Now, if $x < e$ then $1 - \ln x > 0$ and if $x > e$ then $1 - \ln x < 0$.  This tells us that $y$ is increasing on $(0,e)$ and decreasing on $(e,\infty)$, and so $(e,1/e)$ is a maximum.


问:
Let $a_1=a_2=1$, $a_3=2$ and $a_n=a_{n-1}+a_{n-2}$ for $n\ge3$. Prove that $a_1 + a_2 + \ldots + a_n = a_{n+2} - 1$.
I think that this needs the use of a strong form of induction.

So, I tried it with $k$

$3\le k\le n$, where $n$ is arbitrarily chosen.

But I can't proceed beyond this.

答:
You have :

$$a_n=a_{n-1}+a_{n-2}\Leftrightarrow a_n-a_{n-1}=a_{n-2} \Rightarrow \sum_{k=3}^{n+2}(a_k-a_{k-1})= \sum_{k=3}^{n+2}a_{k-2} \\
\Leftrightarrow a_{n+2}-1=\sum_{k=1}^{n}a_k \text{ using telescopic sums}$$


答:
The integer $17$ is prime,

The integer $3$ appears in your sequence,

Also

$2^3+2^3+1=17$

Title: If $x$ and $y$ are in this given sequence, can $2^x+2^y+1$ be prime?

Body:
The sequence:

$3, 11, 13, 17, 19, 29, 37, 41, 53, 59, 61, 67, 83, 97, 101, 107, 113, 131, 137, 139, 149, 163, 173, 179, 181, 193, 197, 211, 227, 257, 269, 281, 293, 313, 317, 347, 349, 353, 373, 379, 389, 401, 409, 419, 421, 443, 449, 461, 467, 491, 509, 521, 523, 541, 547, 557, 563, 569, 577, 587, 593, 613, 617, 619, 653, 659, 661, 677, 701, 709, 757, 761, 769, 773, 787, 797, 809, 821, 827, 829, 853, 857, 859, 877, 883, 907, 929$...

$n = 2^x + 2^y + 1$

$x,y$ in this sequence.

Is there  case that $n$ is prime with this sequence?


问:
Non-central chi-squared distribution, scaled or not?
I am a bit confused as to the non-central chi-squared scaling... It may sound like a basic question to most, not for me!

Suppose we have a variable (for the sake of argument, that has only one degree of freedom) that is defined as

$Y = \lambda + (\sigma X + \mu)^2$, with $X\sim\mathcal{N}(0,1)$.

We have $(\sigma X + \mu)^2 \sim \mathcal{X'}^2_1(\mu^2/\sigma^2)$ and thus, we can say

$(Y - \lambda) \sim \mathcal{X'}^2_1(\mu^2/\sigma^2)$.

Alternatively, we can reformulate the variable as

$Y = \lambda + \sigma^2(X + \mu/\sigma)^2$, with $X\sim\mathcal{N}(0,1)$.

We have $(X + \mu/\sigma)^2 \sim \mathcal{X'}^2_1(\mu ^2/\sigma^2)$ and thus, we can say

$(Y - \lambda)/\sigma^2 \sim \mathcal{X'}^2_1(\mu^2/\sigma^2)$.

Which one is correct?

答:
The second expression is correct, the first is incorrect since

$$
\sigma X + \mu \sim N(\mu, \sigma^2)
$$

and squaring this random variable does not give a non-central chi-square - by definition a non-central chisq is the square of a normal random variable with mean $\mu$ and variance $1$.

It follows that

$$
\frac{Y-\lambda}{\sigma^2} \sim \chi^2_1(\mu^2/\sigma^2)
$$


问:
Lebesgue integration : Inequality
Let $f\in C^1([-2,2],\mathbb{R})$, and $t\in (-1,1)$ and for all $n\geq 1$, $t_n=t+(-1)^n$. I want to prove the following inequality :
$$\Bigg|\int_t^{t_n}f(s)ds\Bigg|\leq c|t_n-t|,$$
where $c=\displaystyle\sup_{x\in[0,2]}|f(x)|$.

My idea using Lebesgue's theory : we can rewrite
$\int_t^{t_n}f(s)ds$ as follows :
$$ \int_{\{s\in\mathbb{R}, \; t\in (t-1,t)\}}f(s)ds+ \int_{\{s\in\mathbb{R}, \; t\in (t,t+1)\}}f(s)ds,$$
which implies that
$$\Bigg|\int_t^{t_n}f(s)ds\Bigg|\leq c(\mu\{s\in\mathbb{R}, \; t\in (t-1,t)\}  + \mu \{s\in\mathbb{R}, \; t\in (t,t+1)\}),$$
but I could not reappear the quantity $|t_n-t|$ again.

答:
If the exercise explicitly says that $c=\sup\{|f(x)| : x\in [0,2]\}$, then the inequality you are looking for is not true. For a counterexample, consider $f:[-2,2] \to \mathbb{R}$ given by $$ f(x)= \begin{cases} x^2, & \text{if} \hspace{2mm} -2\leq x \leq 0, \\
0, & \text{if} \hspace{2mm} 0\leq x \leq 2.
  \end{cases} $$
It is easy to see that $f\in C^{-1}([-2,2],\mathbb{R})$. Now take $t=-\frac{1}{2}$ and consider $t_2=\frac{1}{2}$. Then, on the one hand $$\left| \int_{t}^{t_n} f(x)\ dx \right|=\left| \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x)\ dx \right|= \left| \int_{-\frac{1}{2}}^{0} x^2\ dx \right|=\frac{1}{24}.$$
However, on the other hand, $c=0$ and so $c|t_2-t|=0$

If it was a typo and $c$ is actually equal to $\sup\{|f(x)| : x\in [-2,2]\}$ then the proof is quite easy. Recall that for any integrable function $g:[a,b]\to \mathbb{R}$ the following inequality holds $$\left| \int_{a}^{b} g(x)\ dx \right|\leq  \int_{a}^{b} \left|g(x) \right|\ dx.$$

Having this in mind, notice that for an even $n$ we get  $$\left| \int_{t}^{t_n} f(x)\ dx \right| \leq  \int_{t}^{t_n} \left|f(x)\right|\ dx \leq \int_{t}^{t_n} c\ dx = c|t_n-t|.$$

Edit: Adding a separate case to avoid confusion. For an odd $n$ we get

$$\left| \int_{t}^{t_n} f(x)\ dx \right| = \left|- \int_{t_n}^{t} f(x)\ dx \right| =  \left| \int_{t_n}^{t} f(x)\ dx \right| \leq $$ $$ \int_{t_n}^{t} \left|f(x)\right|\ dx \leq \int_{t_n}^{t} c\ dx = c|t_n-t|.$$

Hope this helps.

Title: Lebesgue integration : Inequality


答:
Such pairs do exist, and they are precisely the columns of the following matrix:

$$\left(
\begin{array}{ccccccccccccccc}
 3 & 3 & 11 & 37 & 53 & 59 & 179 & 179 & 197 & 227 & 353 & 421 & 449 & 467 & 853 \\
 37 & 317 & 17 & 67 & 83 & 797 & 293 & 509 & 227 & 509 & 587 & 787 & 659 & 653 & 907
   \\
\end{array}
\right)$$

The minimal prime obtained in this way is, by inspection, $2^{11}+2^{17}+1=133121$.

Edit: I forgot to check the diagonal!  That adds the single pair $(3,3)$, corresponding to $$2^3+2^3+1=17.$$
This brings us to a total of $16$ solutions.


问:
Hypotheses for Gödel's condensation lemma
Gödel's condensation lemma is that given a suitable $L_{\alpha}$, and given $X \prec L_{\alpha}$, then upon taking the Mostowski collapse on $X$ we are returned with $L_{\beta}$ for some $\beta < \alpha$.

I'm concerned about the exact requirements on $\alpha$ or $L_{\alpha}$. The way I learnt it was that $L_{\alpha} \vDash T$ is required, where $T$ is the finite fragment of $ZF-P$ that is used to formulate model theory, definability and construct the $L$-hierarchy. Then $L_{\alpha} \vDash V = L$, and is correct about the levels of the $L$ hierarchy so that $(L)^{L_{\alpha}} = L_{\alpha}$ and $ (L_{\gamma})^{L_{\alpha}} = (L_{\gamma})$ for $\gamma < \alpha$. Then by elementarity + isomorphism the same holds for $\pi(X)$ ($\pi$ being the Mostowski collapse) and from this the result easily follows.

However it seems that we can loosen the requirements beyond $L_{\alpha} \vDash T$. I've seen weakenings to $\alpha$ being a limit or $\alpha$ being such that $\alpha = \omega \cdot \alpha$. I'm not sure how this is supposed to work. If $\alpha$ is simply a limit, it seems to me that it might not satisfy a good chunk of $ZF-P$ and potentially $T$. For instance the proof of $(Comprehension)^L$ uses the reflection theorem scheme, and simply assuming $\alpha$ a limit, it seems that $(Comprehension)^{L_{\alpha}}$ may not work. Is Comprehension then not needed to formulate model theory, the $L$-hierarchy, etc needed in the proof above? If $L_{\alpha} \not \vDash T$, then it seems that the traditional argument (or the one  I used above) doesn't follow through.

So how far can the hypotheses on $\alpha$ be weakened? Can $\alpha$ really just be a limit ordinal or that $\alpha = \omega \cdot \alpha$? Does this follow simply from being very careful in observing what exactly is used in $T$ above and coming to the conclusion that limits are enough? If that was the case, then I was hoping to see/understand this analysis and find out what exactly is needed, and what is the complexity of this when put as a sentence, etc. Or is it a different proof entirely?

答:
In fact, condensation is very strong:

Suppose $\alpha$ is a limit ordinal and $X\prec_1L_\alpha$. Then $X\cong L_\beta$ for a unique ordinal $\beta$, and the Mostowski collapse provides the unique isomorphism.

Here "$\prec_1$" refers to elementarity for $\Sigma_1$ formulas only: $A\prec_1 B$ iff $A$ is a substructure of $B$ and for every $\Sigma_1$ formula $\varphi(\overline{x})$ and every $\overline{a}\in A$ we have $A\models\varphi(\overline{a})\iff B\models\varphi(\overline{a})$.

The proof isn't really different from the proof of the weaker argument that this holds with $\prec$ in place of $\prec_1$ and restricting attention to $\alpha$s such that $L_\alpha$ satisfies a reasonably strong set theory, it's just much more tedious. Part of this is essentially the realization that arbitrary limit levels of $L$ satisfy a larger fragment of set theory than one might expect; the other part is showing that the construction of the $L$-hierarchy is a $\Sigma_0$ proccess in a precise sense. These fine-grained analyses are moot if we assume that $L_\alpha$ satisfies a reasonable fragment of set theory.

Devlin's presentation (page $80$) of this is quite good in my opinion. Devlin's book has serious flaws elsewhere, but this material is solid.


问:
Determinant equals 1 whereas Inverse unequal to the transpose?
So briefly: Given the matrix $A$ with det$(A) = 1$:

$\left(\begin{array}{ccc} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{array}\right)$

Its inverse however is set by

$\left(\begin{array}{ccc} 1 & -a & a\,c-b\\ 0 & 1 & -c\\ 0 & 0 & 1 \end{array}\right)$

If I'm not mistaken, that's not $A^T$. Also I hold the opinion that this kind of matrices with the property of  $\det= \pm 1$ belong to the orthogonal group, which is why I came for the conclusion.

答: You are right about $A^{-1}$. Since it is distinct from $A^T$, the conclusion is that $A$ is not an orthogonal matrix. Although the determinant of every orthogonal matrix is indeed $\pm1$, most matrices whose determinant is $\pm1$ are not orthogonal.


问:
Are the Fourier coefficients of smooth functions on torus absolutely summable?
Suppose $T: L^p(\mathbb{T}^n) \to L^q(\mathbb{T}^n)$ is linear and bounded, and for each $m\in \mathbb{Z}^n$, $T(e^{2m\pi ix}) = a_m e^{2m\pi ix}$, where $\{a_m\}_{m\in \mathbb{Z}^n}$ is bounded. How does this imply the following?

\begin{equation}
Tf(x) = \sum_m a_m \hat f(m) e^{2m\pi ix}, ~\text{for every}~ f\in C^\infty(\mathbb{T}^n).
\end{equation}

I know the above equation holds for trigonometric polynomials. I tried to use the density argument to extend to all $C^\infty(\mathbb{T}^n)$ functions. But I need that the Fourier coefficients of $f$ are absolutely summable. Is this true that the Fourier coefficients of a $C^\infty(\mathbb{T}^n)$ function are absolutely summable? If not, does the Fourier series of a $C^\infty(\mathbb{T}^n)$ function always converge to itself? And what is the right way to complete this density argument?

答: I summarize reuns' comments here. By smoothness of $f$, van der Corput's lemma implies that for every $k$, there exists a constant $C_k>0$, such that for all $m\in \mathbb{Z}^n$, $|\hat f(m)| \leq C_k m^{-k}$. Note $\widehat f(m)$ are also bounded. Then by taking $k>n$, we obtain the absolute summability.


Title: Let $a_1=a_2=1$, $a_3=2$ and $a_n=a_{n-1}+a_{n-2}$ for $n\ge3$. Prove that $a_1 + a_2 + \ldots + a_n = a_{n+2} - 1$.

Body:
I think that this needs the use of a strong form of induction.

So, I tried it with $k$

$3\le k\le n$, where $n$ is arbitrarily chosen.

But I can't proceed beyond this.

Body:
Here is a slightly different take.

Let $b_n = 1 + a_1 + a_2 + \cdots + a_n$.

Then

$b_{n-1} = 1 + a_1 + a_2 + a_3 + \cdots + a_{n-1}$

$b_{n-2} = \hspace{9mm} 1 + a_1 + a_2 + \cdots + a_{n-2}$

gives

$b_{n-1}+b_{n-2} = 1 + (a_1 + 1) +(a_2+a_1) + (a_3+a_2) + \cdots + (a_{n-1}+ a_{n-2})$

$\hspace{22mm}= 1 + (a_1 + a_2) + a_3 + a_4 + \cdots + a_{n} = b_{n}$

Since $b_1 = 2 = a_3$ and $b_2=3=a_4$, we have $b_n=a_{n+2}$ for all $n$.


问:
What is the meaning of $\cap_{n=1}^\infty \cup_{i=n}^\infty A_i=\{A_n \text{ infinitely often}\}$ mean?
The second Borel-Cantelli lemma refers to a sequence of independent events $A_n$ such that $\sum_{n=1}^\infty \Pr(A_n)=\infty,$ and says that in this case infinitely many $A_n$ events occur almost surely.

$$\bigcap_{n=1}^\infty \bigcup_{i=n}^\infty A_i=\{A_n \text{ i.o.}\}$$

The indexing in the union considers the union of only events beyond a certain event $n$ in the sequence. So it would comprise the occurrence of any events after $n.$

But what is the intersection? Does it simply mean that all these composite events after every single $n$ do occur?

答: Just read the intersection as a "for all", and the union as a "there exists". The event $\bigcap_{n=1}^\infty \bigcup_{i=n}^\infty A_i$ then means : "for all $n \geq 1$, there exists $i \geq n$ such that the event $A_i$ occurs". In other words, this means indeed that infinitely many events $A_i$ occur.


答:
Let $Y=\lambda+(\sigma X+\mu)^2$. We have

$$\begin{split}X&\sim N(0,1)\\
\sigma X&\sim N(0, \sigma^2)\\
\sigma X+\mu&\sim N(\mu, \sigma)^2\text{ (*)}\\
\left(\frac{\sigma X+\mu}{\sigma}\right)&\sim N\left(\frac \mu \sigma, 1\right)\\
\left(\frac{\sigma X+\mu}{\sigma}\right)^2&\sim\chi^2_{1, \frac{\mu^2}{\sigma^2}}\end{split}$$

Therefore

$$\frac{Y-\lambda}{\sigma^2}=\left(\frac{\sigma X+\mu}{\sigma}\right)^2\sim\chi^2_{1, \frac{\mu^2}{\sigma^2}}$$

the second is correct. Notice that $\sigma X+\mu$ does not have unit variance (*) so squaring it does not create a non-central chi squared random variable, which requires it be the sum of independent, normally distributed variables with unit variance. Therefore the first is not correct.

Title: Non-central chi-squared distribution, scaled or not?

Body:
I am a bit confused as to the non-central chi-squared scaling... It may sound like a basic question to most, not for me!

Suppose we have a variable (for the sake of argument, that has only one degree of freedom) that is defined as

$Y = \lambda + (\sigma X + \mu)^2$, with $X\sim\mathcal{N}(0,1)$.

We have $(\sigma X + \mu)^2 \sim \mathcal{X'}^2_1(\mu^2/\sigma^2)$ and thus, we can say

$(Y - \lambda) \sim \mathcal{X'}^2_1(\mu^2/\sigma^2)$.

Alternatively, we can reformulate the variable as

$Y = \lambda + \sigma^2(X + \mu/\sigma)^2$, with $X\sim\mathcal{N}(0,1)$.

We have $(X + \mu/\sigma)^2 \sim \mathcal{X'}^2_1(\mu ^2/\sigma^2)$ and thus, we can say

$(Y - \lambda)/\sigma^2 \sim \mathcal{X'}^2_1(\mu^2/\sigma^2)$.

Which one is correct?


问:
Help with an expression of the Schur polynomial
I am reading: https://www2.math.upenn.edu/~peal/polynomials/schur.htm where it says:

The Schur polynomial is defined as $$s_\lambda(x_1,...,x_n)=\prod_{1\le i<j\le n}(x_i-x_j)^{-1}\det(x_j^{\lambda_i+n-i})$$ And this can also be phrased as $$ s_\lambda(x_1,...,x_n)=\sum_{\sigma\in S_n}\sigma\left(\frac{x^\lambda}{\prod_{i<j}1-x_j/x_i}\right) $$ where $\sigma$ act by permuting the indices of the variables.

I am confused with the second expression of the Schur polynomial. For example, if we are given $n=2$ and $\lambda=(3,1)$, then we have
$$ s_{(3,1)}(x_1,x_2)=\frac{1}{x_1-x_2}\det\begin{bmatrix}x_1^4&x_2^4\\x_1&x_2\end{bmatrix}=\frac{x_1^3x_2-x_2^4}{1-x_2/x_1} .$$
But this obviously does not agree with the second expression since the second term in the numerator of the last fraction above has partition type $(4)$ which is not $(3,1)$. What's wrong with my understanding? And how to prove the second expression is equal to the first expression?

答:
The second expression gives $s_\lambda(x)=\sum_{w\in S_2}\dfrac{x_1^3x_2}{1-x_2/x_1}$, so $s_\lambda(x)=\dfrac{x_1^3x_2}{1-x_2/x_1}+\dfrac{x_2^3x_1}{1-x_1/x_2}=\dfrac{x_1^4x_2-x_1x_2^4}{x_1-x_2}$ which is what you got from the first formula.

To see the equivalence of the two definitions:
Note that
$$ \sum_{\sigma \in S_n} \sigma \bigg(\dfrac{x^\lambda}{\prod_{i<j} 1- x_j/x_i}\bigg)=\sum_{\sigma \in S_n} \sigma \bigg(\dfrac{x^\lambda x_1^{n-1}x_2^{n-2}...x_{n-1}}{\prod_{i<j} (x_i- x_j)} \bigg)$$.

Now $\Delta= \prod_{i<j} (x_i- x_j)$ is skew-symmetric, i.e, $w \Delta = \epsilon(w) \Delta$ for $w \in S_n$. Therefore, the right hand side of the above expression simplifies to $$ \Delta^{-1} \sum_{\sigma \in S_n} \epsilon (\sigma) (x^{\lambda+\rho}) $$ where $\rho=(n-1,n-2,...,1,0)$.

Now I leave you to check that the sum in the above expression is the determinant of the matrix $(x_j^{\lambda_i+n-i})$.


问:
How is the Algebra $C_{p}^{\infty}(U)$ of germs of $C^\infty$ functions in $U$ at $p$ is the Same as $C_{p}^{\infty}(M)$
Hi i am reading An introduction to manifolds by Loring and have some doubts in remark 8.2. It is written that

If $U$ is an open set containing $p$ in $M$ then the algebra $C_{p}^{\infty}(U)$ of germs of $C^\infty$ functions in $U$ at $p$ is the same as $C_{p}^{\infty}(M)$.Hence, $T_pU=T_pM$.

My first question is:

How do we know that the algebra $C_{p}^{\infty}(U)=C_{p}^{\infty}(M)$. I know that, the equivalence class of $(f,U)$ is called the germ of $f$ at $p$.

I have a second question which is related to the differential of a map. It is written that:

The equation $(F_{*}(X_p))f=X_p(f\circ F)$ is independent of the representative of the germ.

My second question is that

here in the above equation $f$ is the representative of a germ and it is appearing inside the equation then how this equation is independent of the representative $f$?

For reference i am attaching the screenshots where i have highlighted the part where these two statements are mentioned.

答:
For $X= M, U$ let $G_p^\infty(X)$ denote the set of
all pairs $(f ,V)$, where $U$ is an open neighborhood of $p$ in $X$ and $f : V \to \mathbb R$ is a $C^\infty$-function. Then $C_p^\infty(X) = G_p^\infty(X)/\sim_X \phantom{.}$ where $(f,V) \sim_X (g,W)$ if there exists an open neighborhood $W$ of $p$ in $X$ such that $W \subset U \cap V$ and  $f \mid_W = g \mid_W$.

Since $U$ is open in $M$, we have $G_p^\infty(U) \subset G_p^\infty(M)$, but if $U \subsetneqq M$, then certainly $G_p^\infty(M)$ has elements $(V,f)$ such that $V \not\subset U$, i.e. $G_p^\infty(M)$ has more elements than $G_p^\infty(U)$. For $(V,f), (W,g) \in G_p^\infty(U)$ we have $(V,f) \sim_U (W,g)$ if and only if $(V,f) \sim_M (W,g)$, i.e. we get an injective function
$$i : C_p^\infty(U) \to C_p^\infty(M), i([(V,f)]_U) = [(V,f)]_M $$
which is easily seen to be an algebra homomorphism. This map is in fact a bijection: Given $(V,f) \in G_p^\infty(M)$, we have $(V,f) \sim_M (V \cap U, f \mid_{V \cap U})$. But clearly $(V \cap U, f \mid_{V \cap U}) \in G_p^\infty(U)$.

More formally, let $\phi : N \to M$ be smooth. Then $\phi$ induces a function $G^\infty_p \phi : G^\infty_{f(q)} M \to G^\infty_q N, G^\infty_p \phi (U,f) = (\phi^{-1}(U), f \circ \phi\mid_{\phi^{-1}(U)})$. Clearly $(U,f) \sim_M (V,g)$, then $G^\infty_p \phi (U,f) \sim_N G^\infty_p \phi (V,g)$. Therefore we get an induced
$$C^\infty_p \phi : C^\infty_{f(q)} M \to C^\infty_q N$$

which is easily seen to be an algebra homomorphism.

Now let $\iota :  U \to M$ denote the inclusion map which is clearly smooth. Then $C^\infty_p \iota : C^\infty_p M \to C^\infty_p U$ is easily seen to be the inverse algebra isomorphism of $i : C^\infty_p U \to C^\infty_p M$.

In that sense $C_p^\infty(U)$ is the same as $C_p^\infty(M)$. But be aware that, strictly speaking, we do not have $C_p^\infty(U) = C_p^\infty(M)$, instead we have the canonical algebra isomorphism $i : C_p^\infty(U) \to C_p^\infty(M)$ with inverse $C^\infty_p \iota : C^\infty_p M \to C^\infty_p U$.

Similarly it is an abuse of notation to write $T_pU = T_pM$, but again we have a canonical  isomorphism $T_pU \to T_pM$.

In fact, each algebra homomorphism (isomorphism) $\mathfrak h :  \mathfrak A \to \mathfrak B$ induces a vector space homomorphism (isomorphism)
$$\mathscr D(\mathfrak h) :  \mathscr D(\mathfrak B) \to \mathscr D(\mathfrak A), \mathscr D(\mathfrak h)(D)  = D \circ \mathfrak h .$$
If we take our above canonical algebra isomorphism $i : C_p^\infty(U) \to C_p^\infty(M)$ we get a canonical vector space isomorphism $\mathscr D(i) : T_p M =  \mathscr D(C_p^\infty(M)) \to T_pU = \mathscr D(C_p^\infty(U))$ with inverse
$$\mathscr D(C^\infty_p \iota) : T_pU \to T_pM .$$
The isomorphism $\mathscr D(C^\infty_p \iota)$ is nothing else than the differential $d_p\iota$ of $\iota$ at $p$. So writing $T_pU = T_pM$ is a sloppy way to say that $d_p\iota : T_pU \to T_pM$ is a canonical isomorphism.

Concerning your second question:

$X_p$ is a derivation on $C_p^\infty(N)$. Given $(V,f) \in G_{F(p)}^\infty(M)$, the map $f \circ F : F^{-1}(V) \to \mathbb R$ gives us an element $(F^{-1}(V),f \circ F) \in G_p^\infty(N)$. It is easy to see that if $(V,f) \sim_M (W,g)$, then $(F^{-1}(V),f \circ F) \sim_N (F^{-1}(W),g \circ F)$. This means that the map $C_{F(p)}^\infty(M) \to C_p^\infty(N), [(V,f)]_M \mapsto [(F^{-1}(V),f \circ F)]_N $ is well-defined.

Instead of $(F_{*}(X_p))f=X_p(f\circ F)$ Tu should have written
$$(F_{*}(X_p))([(V,f)]_M) = X_p([(F^{-1}(V),f\circ F)]_N) .$$
But I think his little abuse of notation is acceptable.

Update:

The above algebra isomorphism $i : C_p^\infty(U) \to C_p^\infty(M)$ has the special feature that its definition is completely obvious and natural. In fact, it does not involve any choices, but is induced by the set inclusion $G_p^\infty(U) \hookrightarrow G_p^\infty(M)$. If you consider germ algebras $C_p^\infty(M), C_q^\infty(N)$ for manifolds $M, N$ of dimension $n
$, you will see that they are always isomorphic, but in general all isomorphisms $\beta: C_p^\infty(M) \to C_q^\infty(N)$ involve choices which do not make $\beta$ "natural" in any sense. In fact, you can choose charts $\phi : U \to B^n$ around $p$ on $M$ and $\psi : V \to B^n$ around $q$ on $N$ (where $B^n$ is an open ball in $\mathbb R^n$ with center $0$) such that $\phi(p) = \psi(q) = 0$. Then $C_p^\infty(M) \approx C_p^\infty(U)$ and  $C_q^\infty(N) \approx C_q^\infty(V)$ in a natural way and $C_p^\infty(U) \approx C_0^\infty(B^n), C_q^\infty(V) \approx C_0^\infty(B^n)$, but the last two isomorphisms are induced by $\phi, \psi$ (which have be chosen arbitarily), i.e. are not natural in  our sense.

For that reason one frequently writes $C_p^\infty(U) = C_p^\infty(M)$ which would be inadequate in general for $C_p^\infty(M)$ and $C_q^\infty(N)$.

This transfers to tangent spaces: $T_pU$ and $T_pM$ are isomorphic in a unique obvious way, the vector space isomorphism being induced by the algebra isomorphism $i: C_p^\infty(U) \to C_p^\infty(M)$ and its inverse $C^\infty_p \iota : C_p^\infty(M) \to C_p^\infty(U)$. For that reason one frequently writes $T_pU  = T_pM$. Note that all tangent spaces $T_pM$ and $T_qN$ are isomorphic, but finding an isomorphism involves choices which prevents writing $T_pM = T_qN$.


答:
I believe we can reduce this to Wallis's product.  I haven't written out the full justification, but it ought to be straightforward.

Assume that we can rearrange the product arbitrarily.  This is the step I haven't justified, but it ought t be easy, since everything in sight is positive.  In particular, we could take the base $2$ logarithm of the partial products to get the partial summand of a sequence of positive terms.

Now if we pull out the expressions with a factor of $$\frac{(2n)^{2n}}{(2n-1)^{2n-1}},$$ we get $$\left(\frac1{2n}\right)^{2/(2n-1)}\prod_{k=n}^\infty\left(\frac{(2n)^{2n}}{(2n-1)^{2n-1}}\right)^{4/((2k-1)(2k+1))}\tag1$$  where I've also associated each of the terms of the form $\left(\frac1{2n}\right)^{2/(2n-1)}$ with one such expression.

The logarithm of the product in $(1)$ is  $$
\log\left(\frac{(2n)^{2n}}{(2n-1)^{2n-1}}\right)\sum_{k=n}^\infty\frac4{(2n-1)(2n+1)}=
\frac2{2n-1}\log\left(\frac{(2n)^{2n}}{(2n-1)^{2n-1}}\right)
$$
because the sum telescopes.

Now $(1)$ becomes $$\left(\frac1{2n}\right)^{2/(2n-1)}\left(\frac{(2n)^{2n}}{(2n-1)^{2n-1}}\right)^{2/(2n-1)}=
\left(\frac{2n}{2n-1}\right)^2,$$ and taking the product over all $n$ gives $$\prod_{n=1}^\infty\left(\frac{2n}{2n-1}\right)^2,$$
which is clearly equivalent to Wallis's product for $\frac\pi2$.

Title: An infinite product for $\frac{\pi}{2}$


问:
How to sum $\frac{1}{(3n+2)(3n+3)}$
Is there analytic way of summing this series? (I have used Wolfram)

$$\sum_{n=0}^{+\infty} \dfrac{1}{(3n+2)(3n+3)}=\dfrac{1}{18}\left(9\log(3)-\sqrt{3}\pi\right)$$

If I decompose it, it is going to be $\dfrac{1}{3n+2}-\dfrac{1}{3(n+1)}$. I don't think this is a telescopic series.

If I convert this into integration, then I will have

$$\displaystyle\int_{0}^{\infty} \dfrac{1}{(3n+2)(3n+3)}=\dfrac{1}{3}\log(\dfrac{3}{2})$$

Can I use a power series to sum this? I don't think it is easy to find a power series, and even if we can, we cannot easily find the limiting function to which this series converge.

Title: How to sum $\frac{1}{(3n+2)(3n+3)}$

Body: You can do it as follows:\begin{align}\sum_{n=0}^\infty\frac1{(3n+2)(3n+3)}&=\sum_{n=0}^\infty\left(\frac1{3n+2}-\frac1{3n+3}\right)\\&=\sum_{n=0}^\infty\int_0^1x^{3n+1}-x^{3n+2}\,\mathrm dx\\&=\sum_{n=0}^\infty\int_0^1x^{3n}(x-x^2)\,\mathrm dx\\&=\int_0^1\sum_{n=0}^\infty x^{3n}(x-x^2)\,\mathrm dx\\&=\int_0^1\frac{x-x^2}{1-x^3}\,\mathrm dx\\&=\int_0^1\frac x{x^2+x+1}\,\mathrm dx\\&=\frac12\left(\int_0^1\frac{2x+1}{x^2+x+1}\,\mathrm dx-\int_0^1\frac1{\left(x+\frac12\right)^2+\frac34}\,\mathrm dx\right)\\&=\log\left(\sqrt3\right)-\frac\pi{6\sqrt3}.\end{align}


问:
Let $f$ be analytic on $D = \{z \in \Bbb{C} : |z| < 1\}$ and $f(0) = 0$.
I was stuck on the following problem:

Let $f$ be analytic on $D = \{z \in \Bbb{C} : |z| < 1\}$ and $f(0) = 0$.

Define
  $$g(z) = \begin{cases} \displaystyle \frac{f(z)}{z} & z \neq 0, \\ f'(0) & z = 0. \end{cases}$$

Then which of the following option(s) is/are correct?

$g$ is discontinuous at $z = 0$ for all $f$.

$g$ is continuous, but not analytic, at $z = 0$ for all $f$.

$g$ is analytic at $z = 0$ for all $f$.

$g$ is analytic at $z = 0$ only if $f'(0) = 0$.

Can someone point me in the right direction with some explanation? Thanks in advance for your time.

EDIT: I have posted an answer .Feel free to  comment if I missed anything in my answer.

答:
We have,
$$g(z) = \begin{cases} \displaystyle \frac{f(z)}{z} & z \neq 0, \\ f'(0) & z = 0. \end{cases}$$.

Since $f$ is analytic in some neighbourhood of $0$, so I can write it as a power series in $z$. So,$$f(z)=a_0+a_1z+ \cdots \cdots +a_nz^n $$ for $|z|<1$. Now,$f(0)=0$ gives $a_0=0.$ Hence  $$ \begin{align}f(z) &=a_1z+ \cdots \cdots +a_nz^n \\ \implies {f(z) \over z} &=g(z) =a_1+a_2z+ \cdots +a_nz^{n-1} \\ \lim_{z \to 0} g(z) &= a_1=f'(0)=g(0)\\ \text{Since},f'(z) &=a_1+ \cdots +na_nz^{n-1} \implies f'(0)=a_1\end {align}$$  Hence $g$ is continuous at $z=0.$ This eliminates option 1.

Also,we notice $$\lim_{z \to 0}{g(z)-g(0) \over z-0}=a_2$$, which exists for all $f$. This eliminates options 2 and 4.

So, option 3 is the right choice.

In this context  Riemann's Removable Singularity Theorem is useful.


问:
Is this a mistake in the definition of Liouville's theorem?
In my book the theorem of Liouville is defined as follows:

Let $f\in H(\mathbb{C})$ and $f(z)=O(|z|^m)$ with $m\in \mathbb{N}_0$ then $f$ is a polynomial with $deg(f) \leq m$.

Shouldn't it be $|f(z)|= O(|z|^m)$ ?

答: For a positive function $g$, the expression $f(z)=O(g(z))$ means $|f(z)|\le Mg(z)$ for sufficiently large $|z|$. So the absolute value on the function $f$ would be redundant.


问:
Expectation of an non-negative integer-valued random variable
I'm aware that this is in regards to the same question set as Find the Mean for Non-Negative Integer-Valued Random Variable and Expected value equals sum of probabilities

my issue is understanding and unfortantly either of those two help my understanding.

Let T be a non-negative integer-valued random variable with $E(T) < \infty$. prove that
  $$E(T) = \sum_{k=1}^{\infty} P(T \geq k)$$

so my understanding of this question is shocking infact it wasn't untill i had read the above that i relised $P(T \geq k) = \sum_{i=k}^{\infty}P(T=i)$ and why this made sense. (completely slipped my mind)

i've tried to follow one of the aboves answers and i need to make sure my understanding of whats going on is concrete and correct.

from my understanding the key to this is remembering that infinite double series can be written as a single series which is going to be the expectation. in all honesty its the reworking of the lower and upper limits which has me concerned.

we start with $$E(T) = \sum_{k=1}^{\infty} P(T \geq k) = \sum_{k=1}^{\infty}\sum_{i=k}^{\infty}P(T=i)= \sum_{k=0}^{\infty}\sum_{i=k}^{\infty}P(T=i)$$

this is just straight substitution of the above identity and shifting the lower index by 1 which if i remember rightly we can do because this is a convergent series. from reading the above the next step i believe is.

$$\sum_{k=0}^{\infty}\sum_{i=k}^{\infty}P(T=i) = \sum_{i=0}^{\infty}\sum_{k=0}^{i}P(T=i)$$ which looks like a rearrangement of the series. i will tend to infinity anyway so this looks like we're just taking particular values at a time in order to end up with i inside the summation. but this is the step in which im unsure about and so any explination would be great. from here it looks like

$$\sum_{i=0}^{\infty}\sum_{k=0}^{i}P(T=i) = \sum_{i=0}^{\infty} iP(T=i) = E[T]$$

which answers the set question but not my question as to why this is allowed.

again any and all help would be great. thanks in advance.

答:
What really helps me is to see that

$$
E[X] = \sum_{k=1}^\infty P(X \geq k) = \sum_{k=1}^\infty [P(X = k) +P(X = k+1) + \cdots] = \sum_{k=1}^\infty\sum_{j=k}^\infty P(X = j).
$$

Now, listing the values that $j$ can take for every value of $k$ (you can think the vertical columns here represent the inner summation)

$k=1$,  $j$ = 1 2 3 4 ...

$k=2$, $j$ = x 2 3 4 ..

$k=3$, $j$ = x x 3 4 ..

so you can se how in the sum the $P(X=1)$ appears once, $P(X=2)$ appears twice and the generic $P(X=j)$ has to appear $j$ times, which justifies the rewriting as

$$
\sum_{k=1}^\infty\sum_{j=k}^\infty P(X = j)=\sum_{j=1}^\infty\sum_{k=1}^j P(X = j)=\sum_{j=1}^\infty j P(X = j) = E[X].
$$


问:
Connectedness between subsets
I know a theorem:

If in a topological space $(X,\tau)$  we have: $S\subset T\subset \overline{S}$ and $S$ is connected set then also $T$ is connected. In particular, the closure of a connected space is connected.

I have two question:

Is there a similar claim: we have $A\subset B\subset C$ and $A,C$ are connected, then $B$ is connected?

$S$ is connected $\Rightarrow$ $\overline{S}$ is connected, but $\Leftarrow$ is not truth?

答:
No, that's far too weak! Take $A=\{0\}$ and $C=\mathbb R$. Both are connected, but obviously there are subsets like $B=(0,1)\cup(2,3)$ which aren't.

Yes, consider $S=(0,1)\cup(1,2)$ in $\mathbb R$. This is obviously disconnected, but its closure is $[0,2]$, which is connected.

Title: Connectedness between subsets


答:
(1) is obviously false. For example, consider $X=\mathbb{R}$, equipped
with the usual topology. Let $A=\emptyset$, $B=(-2,-1)\cup(1,2)$,
and $C=\mathbb{R}$. Clearly $A$ and $C$ are connected, $A\subseteq B\subseteq C$
but $B$ is not connected.

Title: Connectedness between subsets

Body:
I know a theorem:

If in a topological space $(X,\tau)$  we have: $S\subset T\subset \overline{S}$ and $S$ is connected set then also $T$ is connected. In particular, the closure of a connected space is connected.

I have two question:

Is there a similar claim: we have $A\subset B\subset C$ and $A,C$ are connected, then $B$ is connected?

$S$ is connected $\Rightarrow$ $\overline{S}$ is connected, but $\Leftarrow$ is not truth?


答:
For 1.: no, the similar claim doesn't hold: take for instance the segment $A = [-1/2, 1/2] \times \{ 0 \}$ inside the plane. Let $B = B((0,0), 1) \cup B((2,2),1)$, $B(x,r)$ being the ball of center $x$ and radius $r$. Take $C$ any connected set which contains $B$ (for example $\mathbb R^2$ itself) and you have a counterexample.

For 2.: The "$\Leftarrow$" is indeed not true: $\overline {\mathbb Q} = \mathbb R$, so $\overline {\mathbb Q}$ is connected. But $\mathbb Q$ is not connected: the sets $A = (-\infty, \sqrt 2)$, $B = (\sqrt 2,\infty)$ disconnect it.

Title: Connectedness between subsets


问:
Calculate Galois group of finite degree extension of the rationals in GAP
is there a way to calculate the Galois group of finite field extension in GAP?

I've tried the following but it doesn't work.

```<code>x:=Indeterminate(Rationals,"x");
q:=x^4-2;
f:=AlgebraicExtension(Rationals, q);
g:=GaloisGroup(f);
</code>```

答:
GAP does not have a built-in function for the actual Galois group. The reason is that the Galois group consists of automorphisms of the splitting field, and we thus need to construct the splitting field first. This gets messy quickly.

In your example, let us verify first that $f$ is not a splitting field:

```<code>gap> GaloisType(q);
3
gap> Size(TransitiveGroup(4,3));
8
</code>```

So the splitting field has degree 8 (the Galois group is dihedral), we need another extension of degree 2.
We can deduce easily that for this we do not just need a root of $q$ (say $\sqrt[4]{2}$), but also a root of $x^2+1$. Using the primitive element theorem, we make an attempt to construct a polynomial that has $\sqrt[4]{2}+i$ as a root:

```<code>gap> y:=X(Rationals,"y");
y
gap> r:=Resultant((x-y)^4-2,y^2+1,y);
x^8+4*x^6+2*x^4+28*x^2+1
gap> Factors(r);
[ x^8+4*x^6+2*x^4+28*x^2+1 ]
</code>```

(If the polynomial was not irreducible, we would have tried Resultant((x-2*y)^4-2,y^2+1,y) instead and so on.) Construct the splitting field, and express the roots of $q$ in terms of the primitive element of the splitting field:

```<code>gap> e:=AlgebraicExtension(Rationals,r);
<algebraic extension over the Rationals of degree 8>
gap> Factors(Value(q,X(e)));
[ x_1+(-5/24*a^7-19/24*a^5-5/24*a^3-151/24*a),
  x_1+(5/24*a^7+19/24*a^5+5/24*a^3+151/24*a),
  x_1+(1/24*a^6+5/24*a^4+13/24*a^2+29/24),
  x_1+(-1/24*a^6-5/24*a^4-13/24*a^2-29/24) ]
</code>```

So if $\alpha$ (printed a) is a root of $r$, then one root of $q$ is $\frac1{24}\left(5\alpha^7+19\alpha^5+5\alpha^3+151\alpha\right)$.

The elements of the Galois group then are maps that map $\alpha$ to other roots of $r$:

```<code>gap> roots:=RootsOfUPol(Value(r,X(e)));
[ 5/12*a^7+19/12*a^5+5/12*a^3+139/12*a, -a,
  5/24*a^7-1/24*a^6+19/24*a^5-5/24*a^4+5/24*a^3-13/24*a^2+127/24*a-29/24,
  5/24*a^7+1/24*a^6+19/24*a^5+5/24*a^4+5/24*a^3+13/24*a^2+127/24*a+29/24, a,
  -5/12*a^7-19/12*a^5-5/12*a^3-139/12*a,
  -5/24*a^7-1/24*a^6-19/24*a^5-5/24*a^4-5/24*a^3-13/24*a^2-127/24*a-29/24,
  -5/24*a^7+1/24*a^6-19/24*a^5+5/24*a^4-5/24*a^3+13/24*a^2-127/24*a+29/24 ]
</code>```

So for example there is an automorphism that will map $\alpha\mapsto-\alpha$, and another one that will map $\alpha\mapsto\frac1{12}\left(5\alpha^7+19\alpha^5+5\alpha^3+139\alpha\right)$, and so on. They compose by composing polynomials modulo $r$. For example, taking three automorphisms to start with:

```<code>gap> au1:=-x;;
gap> au2:=1/12*(5*x^7+19*x^5+5*x^3+139*x);;
gap> au3:=1/24*(5*x^7-x^6+19*x^5-5*x^4+5*x^3-13*x^2+127*x-29);;
gap> Value(au2,au2) mod r;
x
gap> Value(au3,au3) mod r;
x
</code>```

we see they all three have order 2, but the product of two of them has order 4:

```<code>gap> au4:=Value(au1,au3) mod r;
-5/24*x^7+1/24*x^6-19/24*x^5+5/24*x^4-5/24*x^3+13/24*x^2-127/24*x+29/24
gap> Value(au4,au4) mod r;
-5/12*x^7-19/12*x^5-5/12*x^3-139/12*x
gap> Value(Value(au4,au4),au4) mod r;
-5/24*x^7-1/24*x^6-19/24*x^5-5/24*x^4-5/24*x^3-13/24*x^2-127/24*x-29/24
gap> Value(Value(Value(au4,au4),au4) mod r,au4) mod r;
x
</code>```

We can use the same idea to get permutations for the action on the roots of q (i.e.we get for each of teh 8 automorphisms a permutation of degree 4:

```<code>gap> rootpol:=[ 5/12*x^7+19/12*x^5+5/12*x^3+139/12*x, -x,
>   5/24*x^7-1/24*x^6+19/24*x^5-5/24*x^4+5/24*x^3-13/24*x^2+127/24*x-29/24,
>   5/24*x^7+1/24*x^6+19/24*x^5+5/24*x^4+5/24*x^3+13/24*x^2+127/24*x+29/24, x,
>   -5/12*x^7-19/12*x^5-5/12*x^3-139/12*x,
>   -5/24*x^7-1/24*x^6-19/24*x^5-5/24*x^4-5/24*x^3-13/24*x^2-127/24*x-29/24,
>   -5/24*x^7+1/24*x^6-19/24*x^5+5/24*x^4-5/24*x^3+13/24*x^2-127/24*x+29/24 ];;
gap> qrpol:=[ 5/24*x^7+19/24*x^5+5/24*x^3+151/24*x, -5/24*x^7-19/24*x^5-5/24*x^3-151/24*x,
>   -1/24*x^6-5/24*x^4-13/24*x^2-29/24, 1/24*x^6+5/24*x^4+13/24*x^2+29/24 ];;
gap> perms:=List(rootpol,x->PermList(List(qrpol,y->Position(qrpol,Value(y,x) mod r))));
[ (3,4), (1,2), (1,3)(2,4), (1,4)(2,3), (), (1,2)(3,4), (1,3,2,4), (1,4,2,3) ]
gap> Size(Group(perms));
8
</code>```

While this gives you the element arithmetic, I am not sure that it is immensely useful, in particular if examples get larger.

Body:
GAP does not have a built-in function for the actual Galois group. The reason is that the Galois group consists of automorphisms of the splitting field, and we thus need to construct the splitting field first. This gets messy quickly.

In your example, let us verify first that $f$ is not a splitting field:

gap> GaloisType(q);
3
gap> Size(TransitiveGroup(4,3));
8

So the splitting field has degree 8 (the Galois group is dihedral), we need another extension of degree 2.
We can deduce easily that for this we do not just need a root of $q$ (say $\sqrt[4]{2}$), but also a root of $x^2+1$. Using the primitive element theorem, we make an attempt to construct a polynomial that has $\sqrt[4]{2}+i$ as a root:

gap> y:=X(Rationals,"y");
y
gap> r:=Resultant((x-y)^4-2,y^2+1,y);
x^8+4*x^6+2*x^4+28*x^2+1
gap> Factors(r);
[ x^8+4*x^6+2*x^4+28*x^2+1 ]

(If the polynomial was not irreducible, we would have tried Resultant((x-2*y)^4-2,y^2+1,y) instead and so on.) Construct the splitting field, and express the roots of $q$ in terms of the primitive element of the splitting field:

gap> e:=AlgebraicExtension(Rationals,r);
<algebraic extension over the Rationals of degree 8>
gap> Factors(Value(q,X(e)));
[ x_1+(-5/24*a^7-19/24*a^5-5/24*a^3-151/24*a),
  x_1+(5/24*a^7+19/24*a^5+5/24*a^3+151/24*a),
  x_1+(1/24*a^6+5/24*a^4+13/24*a^2+29/24),
  x_1+(-1/24*a^6-5/24*a^4-13/24*a^2-29/24) ]

So if $\alpha$ (printed a) is a root of $r$, then one root of $q$ is $\frac1{24}\left(5\alpha^7+19\alpha^5+5\alpha^3+151\alpha\right)$.

The elements of the Galois group then are maps that map $\alpha$ to other roots of $r$:

gap> roots:=RootsOfUPol(Value(r,X(e)));
[ 5/12*a^7+19/12*a^5+5/12*a^3+139/12*a, -a,
  5/24*a^7-1/24*a^6+19/24*a^5-5/24*a^4+5/24*a^3-13/24*a^2+127/24*a-29/24,
  5/24*a^7+1/24*a^6+19/24*a^5+5/24*a^4+5/24*a^3+13/24*a^2+127/24*a+29/24, a,
  -5/12*a^7-19/12*a^5-5/12*a^3-139/12*a,
  -5/24*a^7-1/24*a^6-19/24*a^5-5/24*a^4-5/24*a^3-13/24*a^2-127/24*a-29/24,
  -5/24*a^7+1/24*a^6-19/24*a^5+5/24*a^4-5/24*a^3+13/24*a^2-127/24*a+29/24 ]

So for example there is an automorphism that will map $\alpha\mapsto-\alpha$, and another one that will map $\alpha\mapsto\frac1{12}\left(5\alpha^7+19\alpha^5+5\alpha^3+139\alpha\right)$, and so on. They compose by composing polynomials modulo $r$. For example, taking three automorphisms to start with:

gap> au1:=-x;;
gap> au2:=1/12*(5*x^7+19*x^5+5*x^3+139*x);;
gap> au3:=1/24*(5*x^7-x^6+19*x^5-5*x^4+5*x^3-13*x^2+127*x-29);;
gap> Value(au2,au2) mod r;
x
gap> Value(au3,au3) mod r;
x

we see they all three have order 2, but the product of two of them has order 4:

gap> au4:=Value(au1,au3) mod r;
-5/24*x^7+1/24*x^6-19/24*x^5+5/24*x^4-5/24*x^3+13/24*x^2-127/24*x+29/24
gap> Value(au4,au4) mod r;
-5/12*x^7-19/12*x^5-5/12*x^3-139/12*x
gap> Value(Value(au4,au4),au4) mod r;
-5/24*x^7-1/24*x^6-19/24*x^5-5/24*x^4-5/24*x^3-13/24*x^2-127/24*x-29/24
gap> Value(Value(Value(au4,au4),au4) mod r,au4) mod r;
x

We can use the same idea to get permutations for the action on the roots of q (i.e.we get for each of teh 8 automorphisms a permutation of degree 4:

gap> rootpol:=[ 5/12*x^7+19/12*x^5+5/12*x^3+139/12*x, -x,
>   5/24*x^7-1/24*x^6+19/24*x^5-5/24*x^4+5/24*x^3-13/24*x^2+127/24*x-29/24,
>   5/24*x^7+1/24*x^6+19/24*x^5+5/24*x^4+5/24*x^3+13/24*x^2+127/24*x+29/24, x,
>   -5/12*x^7-19/12*x^5-5/12*x^3-139/12*x,
>   -5/24*x^7-1/24*x^6-19/24*x^5-5/24*x^4-5/24*x^3-13/24*x^2-127/24*x-29/24,
>   -5/24*x^7+1/24*x^6-19/24*x^5+5/24*x^4-5/24*x^3+13/24*x^2-127/24*x+29/24 ];;
gap> qrpol:=[ 5/24*x^7+19/24*x^5+5/24*x^3+151/24*x, -5/24*x^7-19/24*x^5-5/24*x^3-151/24*x,
>   -1/24*x^6-5/24*x^4-13/24*x^2-29/24, 1/24*x^6+5/24*x^4+13/24*x^2+29/24 ];;
gap> perms:=List(rootpol,x->PermList(List(qrpol,y->Position(qrpol,Value(y,x) mod r))));
[ (3,4), (1,2), (1,3)(2,4), (1,4)(2,3), (), (1,2)(3,4), (1,3,2,4), (1,4,2,3) ]
gap> Size(Group(perms));
8

While this gives you the element arithmetic, I am not sure that it is immensely useful, in particular if examples get larger.


问:
Why is the equivalence relation in $L^p$ space of equal almost everywhere valid?
I know that on $L^p(\mathbb{R}^n)$, two functions $f$, $g$ are regarded as the same, denoted $f\sim g$, if $f=g$ almost everywhere.
Transitivity of equivalence relation is satisfied, i.e. if $f\sim g$ and $g\sim h$, then $f\sim h$. And "countable transitivity" also holds, i.e., if $f \sim r_1$, $r_1 \sim r_2 \sim r_3 \sim \cdots \sim g$, then $f\sim g$. This is because a countable union of measure zero sets is still of measure zero. Now my question is why "uncountable transitivity" doesn't hold? Functions $\chi_{[0,s]}$ where $s$ ranges over $[0,1]$, cannot be all in the same equivalence class. But it seems plausible to connect any two such functions $\chi_{[0,s]}, \chi_{[0,t]}$by uncountably many equivalence relations. I think it might be because the equivalence relation cannot be used infinitely many times. But I'm still confused and I guess it is a little related to set theory. Could anyone say more about this?

答:
I think you need to be more careful when talking about "countable transitivity".

Let's suppose we have a sequence of functions $f_n$, all belonging to the same equivalence class. If I understand correctly, you want to claim that the limiting function $g = \lim_{n} f_n$ is also in the same equivalence class. This is no longer an assertion about the equivalence relation. It depends on how the notion of convergence interacts with the equivalence relation.

Suppose you are working in the space of sequences of natural numbers, i.e. functions from $\mathbb{N} \to \mathbb{N}$. Define an equivalence relationship on such sequences that says that $a(k) \sim b(k)$ if $\exists M$ such that $\forall m \geq M$, $a(m) = b(m)$. It's easy to find sequences of such sequences, $a_{n}(k)$, all belonging to the same equivalence class, where the pointwise limit $\lim_{n} a_{n}(k)$ converges and is no longer in the same equivalence class, e.g. by recursively adjusting the values of the sequence at higher and higher inputs. In this example, the convergence does not play well with the equivalence relation.

Returning to $L^p$ space, you do have "countable transitivity" because the condition underlying the equivalence relation, disagreeing on sets of measure zero, plays well with countable sequences. However, if you were to define an uncountable sequence of functions, each differing from the last at only a single point, and if you could make sense of the limit as a function, you would have a process that could transform any function into any other function. However, this kind of limit does not play nicely with the equivalence relation, because uncountable unions of measure zero sets may have positive measure.

In summary: the best you can get from transitivity is "finite transitivity", i.e. a finite sequence of objects, each in the same equivalence class as the last object, never leaves the class. Any kind of infinite transitivity depends on how the limits you are taking interact with the specific equivalence relation in question.


答:
The transitivity of the equivalence relation is defined as $a \sim b$, $b\sim c$ implies $a \sim c$. This, by induction, only implies "finite transitivity", i.e. $a_1 \sim a_2 \sim \cdots \sim a_n$ implies $a_1 \sim a_n$. The "infinite transitivity" has nothing to do with the equivalence relation.

By "countable transitivity" I mean, $f_1$ differs from $f_2$ on a measure zero set $E_1$, $f_2$ differs from $f_3$ on a measure zero set $E_2$, and so on, and let $f_0$ differs from $f_1$ on the union $E_1 \cup E_2 \cup \cdots$, then we still have $f_0 \sim f_1$. This actually follows from the fact that a union of countably many measure zero sets is still of measure zero, and thus $f_0$ only differs from $f_1$ on a measure zero set. Then by how we define two functions are equivalent, we have $f_1 \sim f_0$. There is nothing to do with the transitivity property of the equivalence relation.

"uncountable transitivity" does not hold at all.

Title: Why is the equivalence relation in $L^p$ space of equal almost everywhere valid?

Body:
I know that on $L^p(\mathbb{R}^n)$, two functions $f$, $g$ are regarded as the same, denoted $f\sim g$, if $f=g$ almost everywhere.
Transitivity of equivalence relation is satisfied, i.e. if $f\sim g$ and $g\sim h$, then $f\sim h$. And "countable transitivity" also holds, i.e., if $f \sim r_1$, $r_1 \sim r_2 \sim r_3 \sim \cdots \sim g$, then $f\sim g$. This is because a countable union of measure zero sets is still of measure zero. Now my question is why "uncountable transitivity" doesn't hold? Functions $\chi_{[0,s]}$ where $s$ ranges over $[0,1]$, cannot be all in the same equivalence class. But it seems plausible to connect any two such functions $\chi_{[0,s]}, \chi_{[0,t]}$by uncountably many equivalence relations. I think it might be because the equivalence relation cannot be used infinitely many times. But I'm still confused and I guess it is a little related to set theory. Could anyone say more about this?


问:
Cube question without coordinates, cross sections and planes
We have a cube with corners $ABCDA_1B_1C_1D_1$. The points $A_1B_1C_1D_1$ lie above the square $ABCD$. We also have a plane that goes through the middle of line $BC$, through the middle of square $ABA_1B_1$, through the middle of $A_1B_1C_1D_1$. At what ratio does the plane cut line $AB$.

What I did, I've put the cube into the coordinate system $B(0,0,1), A_1 (0,1,0) \text{ and } A(0,0,0)$, $D(1,0,0)$ and then it was pretty easy to solve. I just got the equation of the plane and then solved for intersection of the line that goes through AB, to get the point. Then I knew the ratio.

My question is, how to solve this problem without putting the cube into the coordinate system?

SOLUTION: solution should be that the plane cuts AB in ratio 3:1

答:
In response to the question how I solved with coordinate system:

Let's say:

$$AA_1 = (0,0,1) = c$$
$$AB = (0,1,0) = a$$
$$AD = (1,0,0) = b$$

Now we get all three points as mentioned above:

$$P_1 = c + (1/2)a + (1/2)b = (1/2,1/2,1/2)$$
$$P_2 = (1/2)c + (1/2)b = (0,(1/2),(1/2))$$
$$P_3 = a + (1/2)b = ((1/2),1,0)$$
Now we get two vectors $P_2P_3$ and $P_2P_1$. When we do cross product we get: $$n = (1,-2,-1)$$

The equation of the plane is thus: $$x - 2y - z = \frac{-3}{2}$$
Because: $\text{We use the second point and get:}0*1 + (1/2)*(-2)+(1/2)(-1) = \frac{-3}{2}$

Now we know that $AB$ is on  $y$ axis , or the second coordinate as we may call it is a parameter $t$. So the equation of the line on which $AB$ is on is $y = t, x =0, z=0$, where $t$ is the parameter.

Now we input $y =t$ into the plane equation and get $$(-2t = \frac{-3}{2}\iff t =\frac{3}{4} )\implies \text{ratio is: } 1:3$$


答:
We assume the following fact without proof: Let $X$ be a topological
space. Then the following conditions are equivalent:

(a) $X$ is connected.

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

We go back to your question. Let $f:T\rightarrow\{0,1\}$ be an arbitrary
continuous function. We go to show that $f$ is a constant function.
Let $g=f|_{S}$, the restriction of $f$ on $S$. Clearly $g:S\rightarrow\{0,1\}$
is also continuous. By the above fact, $g$ is a constant function.
Without loss of generality, we assume that $g(x)=0$ for all $x\in S$.
Let $x\in T$ be arbitrary. Since $x\in\bar{S}$, there exists a net
$(x_{\alpha})$ in $S$ such that $x_{\alpha}\rightarrow x$. By continuity
of $f$, we have that $f(x_{\alpha})\rightarrow f(x).$ However, $f(x_{\alpha})=g(x_{\alpha})=0$
for each $\alpha$, so $f(x)=0.$ This shows that $f$ is a constant
function.

Title: Connectedness between subsets


问:
Special case of Elliott's Theorem
Let $A$ and $B$ be unital $AF$-algebra. By Elliott's theorem we know that if there an order isomorphism $\psi: K_0(A) \rightarrow K_0(B)$ with $\psi([1_{A}]) = [1_{B}]$, then there exists an isomorphism $h : A \rightarrow B$ such that $h_{*} = \psi$.

My question is this is true for homomorphism? In other words, Let $A$ and $B$ be unital $AF$-algebras. Suppose that there is an order homomorphism $\alpha: K_0(A) \rightarrow K_0(B)$ such that $\alpha([1_{A}]) = [1_{B}]$. Is there homomorphism $h: A \rightarrow B$ such that $h_{*} = \alpha$?

答: The result is true: See this paper https://www.researchgate.net/publication/264425120_The_Category_of_Bratteli_Diagrams by Elliot, Golestani and Amini, theorem 6.1. What you are asking is whether the $K_0$ functor is full when restricted to the subcategory of unital AF algebras and is co-restricted to the category of ordered abelian groups with unital, positive group homomorphisms as morphisms. This is exactly what the above paper shows in theorem 6.1. I think that it can also be proved by more elementary means, but I need to think a bit more about this.

Title: Special case of Elliott's Theorem


问:
How to find remainder?
$$a=r\mod (r+1) \ \ \forall r\in\{2,3,4,\dots,9\}$$

Then how do we find $'x'$ if $$a=x\mod 11$$

I get $$2a=9\mod11$$ but that does not help.

Please keep solution simple , i don't now number theory.

The above is the crux what I got from the question:

Let $n_1,n_2,... $ be an increasing sequence of natural numbers each of which leaves remainder $r+1$ when divided with $r\in\{2,3,....9\}$. Find the remainder when $n_{2008}$ is divided with $11$.

答:
$$2a\equiv9\pmod {11}\equiv -2\implies a\equiv-1\pmod {11}\equiv10$$

Alternately, as $2\cdot 6=12\equiv1\pmod{11}\implies 2^{-1}\equiv6$

$2a\equiv 9\pmod{11}\implies a\equiv2^{-1}\cdot9\equiv 6\cdot9\equiv 10\pmod {11} $

If $a\equiv r\pmod{r+1}\equiv-1\implies (r+1)$ divides $a+1 \ \ \forall r\in\{2,3,4,\dots,9\} $

If $m_i$ divides $b,$ we know lcm$(m_i)$ will divide $b$

EDIT:  to answer the changed question

Here lcm$(3,4,\cdots,9,10)=5\cdot 8\cdot7\cdot 9=2520$

So, $n_i\equiv-1\pmod{2520}=2520c_i-1$ for some integer $c_i$s

For $n_i$ to be natural number $n_i=2520c_i-1>0\implies c>0\implies c_{\text{min}}=1$

So, if we choose, any arbitrary set of increasing positive integers for $c_i,$ we shall get an increasing sequence of natural numbers satisfying the given condition and solution will depend on the choice of $c_i$s

Now if we take $c_i=i$ for $i\ge1,$

$n_{2008}=2520\cdot 2008-1$

$\equiv1\cdot6-1\pmod {11}$ as $2520\equiv1\pmod {11}$ and $2008\equiv6\pmod {11}$

$\implies n_{2008}\equiv5\pmod{11}$

Title: How to find remainder?


问:
The expectation of $e^X \left(1-(1-e^{-X}\right)^n)$ when $X$ has Exponential Distribution
To my surprise, I was able to evaluate the following expression in Mathematica:

$$E\left[e^X \left(1-(1-e^{-X}\right)^n) \right] = \frac{y}{y-1} \left(1-\frac{1}{\binom{n+y-1}{y-1}}\right)\quad X\sim\text{Exp}(y)$$
with the right hand side being equal to $\text{HarmonicNumber}(n)$ in the particular case $y=1$, and equal to $n$ when $y=0$.

If we define $\binom{n}{k} = \frac{\Gamma(n+1)}{\Gamma(n-k+1)\Gamma(k+1)}$ the results seems to hold for all $x,y\in\mathbb R$, though I mostly care about $n$ and $y$ as positive integers.

I have no idea how to prove this by hand.
I tried a series expansion of the exponentials without realizing much.
I also tried rewriting in terms of the uniform distribution, since $e^{-X}$ has the same distribution as $U^{1/y}$ for $U\sim\text{Uniform}(0,1)$.

Are there any tricks or properties I'm missing?

Body: This expectation is an integral,$$\begin{align}\int_0^\infty ye^{-(y-1)x}(1-(1-e^{-x})^n)dx&=\frac{y}{y-1}-\int_0^\infty ye^{-(y-1)x}(1-e^{-x})^ndx\\&\stackrel{z:=e^{-x}}{=}\frac{y}{y-1}-\int_0^1yz^{y-2}(1-z)^ndz\\&\stackrel{\star}{=}\frac{y}{y-1}-y\operatorname{B}(y-1,\,n+1)\\&=\frac{y}{y-1}\left(1-\frac{\Gamma(y)\Gamma(n+1)}{\Gamma(n+y)}\right)\\&=\frac{y}{y-1}\left(1-\frac{1}{\binom{n+y-1}{y-1}}\right),\end{align}$$with $\star$ using the Beta function.


问:
Prove that the power set of an $n$-element set contains $2^n$ elements

Theorem. Let $X$ denote an arbitrary set such that $|X|=n$. Then $|\mathcal P(X)|=2^n$.

Proof. The proof is by induction on the numbers of elements of $X$.

For the base case, suppose $|X|=0$. Clearly, $X=\emptyset$. But the empty set is the only subset of itself, so $|\mathcal P(X)|=1=2^0$.

Now, the induction step. Suppose $|X|=n$; by the induction hypothesis, we know that $|\mathcal P(X)|=2^n$. Let $Y$ be a set with $n+1$ elements, namely $Y=X\cup\{a\}$. There are two kinds of subsets of $Y$: those that include $a$ and those that don't. The first are exactly the subsets of $X$, and there are $2^n$ of them. The latter are sets of the form $Z\cup\{a\}$, where $Z\in\mathcal P(X)$; since there are $2^n$ possible choices for $Z$, there must be exactly $2^n$ subsets of $Y$ of which $a$ is an element. Therefore $|\mathcal P(Y)|=2^n+2^n=2^{n+1}$. $\square$

Image that replaced text.

From the above explanation, I don't understand why the set that contains $\{a\}$ will contain $2^{|n|}$ elements when it should clearly be $2^{|1|}$.

The construction of a new set $S$ is the union of the old set with cardinality $n$ and a new element $\{a\}$, therefore the set that does not contain $\{a\}$ still has cardinality $n$ and the set that contains $\{a\}$ is just $\{a\}$, one element.

Can someone please elucidate?

答:
Alternative proof:

Represent a subset of $X$ as a binary number such that its $k^{th}$ bit is set if and only if the $k^{th}$ element is taken.

E.g., $\{a,b,c\}\to111$, $\{a,c\}\to101$, $\{c\}\to001$, ...

You can convince yourself that the two representations are equivalent, and that the second includes all binary numbers from $0$ to $2^{|X|}-1$.

Note that if you add an element, the subset of $X$ with signature $m$ generates two new subsets, with signatures $2m$ and $2m+1$.