问:
Limit of operator norm
Let $n$ be an integer. Let $T : \mathbb{R}^n \to \mathbb{R}^n$ be a linear map.

How to show that
$$
\lim_{p \to 1} \|T: \ell^1_n \to \ell^p_n\|
=\|T: \ell^1_n \to \ell^1_n\| ?
$$

Any reference is welcome.

答:
Indeed the key point is that $p$-norm is a continuous function as stated in the comments. Although, a small argument is needed:

Fix $x=(x_1,...,x_n)\in \mathbb{R^n}$ (i assume that your $p$ is $>1$). Then, by Hölder's Inequality used in the vectors $x,y\in \mathbb{R^n}$ where $y=(1,...,1)$ we have

\begin{align}
\sum_{k=1}^{n}|x_k|&=\sum_{k=1}^{n}|x_k|\cdot 1\\
&\leq \biggl(\sum_{k=1}^{n}|x_k|^p\biggr)^{1/p}\biggl(\sum_{k=1}^n1\biggr)^{1-1/p}\\
&=n^{1-1/p}\cdot ||x||_p
\end{align}

Hence,
$$\tag{1} ||x||_1\leq n^{1-1/p}||x||_p$$

Now, denote by $||T||_{1,1}$ the norm of the operator $T:l_1\to l_1$ and by $||T||_{1,p}$ the norm of the operator $T:l_1\to l_p$. Using $(1)$ we have
$$||Tx||_1\leq n^{1-1/p}\cdot ||Tx||_p\leq n^{1-1/p}\cdot ||T||_{1,p}\cdot ||x||_1$$
Hence, $||T||_{1,1}\leq n^{1-1/p}||T||_{1,p}$ so, taking $\liminf_{p\to 1}$ in the last inequality we get
$$||T||_{1,1}\leq \liminf_{p\to 1}||T||_{1,p}$$

For the other direction we use the continuity of the $p-$norm. Let $\epsilon>0$, by the definition of $||T||_{1,p}$ there is an $||x||_1\leq 1$ such that
$$\tag{2} ||T||_{1,p}-\epsilon<||Tx||_p\leq ||T||_{1,p}$$

Then, using $\lim_{p\to 1}||Tx||_p=||Tx||_1$ in $(2)$ we get $$\tag{3}\limsup_{p\to 1}||T||_{1,p}-\epsilon\leq ||Tx||_1\leq ||T||_{1,1}\cdot ||x||_1=||T||_{1,1}$$

Hence, $\limsup_{p\to 1}||T||_{1,p}\leq ||T||_{1,1}$. Putting all together,

$$\limsup_{p\to 1}||T||_{1,p}\leq ||T||_{1,1}\leq \liminf_{p\to 1}||T||_{1,p}$$
which is equivalent to $\lim_{p\to 1}||T||_{1,p}=||T||_{1,1}$.

Title: Limit of operator norm


问:
What is an example of a bijection from unlabeled graphs to the positive integers?
Suppose $f$ is a bijection from the set of unlabeled graphs--i.e. isomorphism classes of graphs -- to the positive integers $\mathbb{N}$. Due to the isomorphism graph problem, I understand that trying to compute $f(G)$ or $f^{-1}(n)$, where $G$ is a graph, would require a lot of computations, maybe on the order of $O(2^{n^2})$. I was wondering if someone could give me an example of such a function. Clearly such a function exists since there are countably many unlabeled graphs.

答:
Writing such a thing out in full detail will be very annoying, since as you say if you could compute such a thing then you could solve graph isomorphism. Here is a sketch: it's a lot easier to write down such a thing for labeled graphs, since a labeled graph on the vertex set $\{ 1, 2, \dots n \}$ is completely specified by specifying its edges. So we can encode a labeled graph on $\{ 1, 2, \dots n \}$ by writing down a binary string of length ${n \choose 2}$ indicating which edges exist in the graph, where the positions of the characters in the string correspond to the edges in lexicographic order; that is, the edges are indicated in the order $(12), (13), (14), \dots (23), (24), \dots$.

Then we can write down all labeled graphs on $\{ 1, 2, \dots n \}$ by writing down all $2^{ {n \choose 2} }$ such binary strings in lexicographic order. And we can write down all labeled graphs by writing down the strings describing labeled graphs on $0$ vertices, then on $1$ vertex, etc. Writing down the elements of a countable set in a fixed order is equivalent to finding a bijection with $\mathbb{N}$; the bijection sends $n \in \mathbb{N}$ to the $n^{th}$ element in the order.

This bijection begins

$$\emptyset, \emptyset, 0, 1, 000, 001, 010, \dots, 000000, 000001, 000010, \dots$$

where $\emptyset$ denotes the empty string and we need two of them corresponding to the empty graph and the unique graph on $1$ vertex (so the binary string encoding is very slightly non-unique which is annoying but doesn't affect whether this is a bijection).

This bijection can be computed efficiently by subtracting $2^{ {n \choose 2} }$ repeatedly until you can't anymore, then writing down the result in binary.

Given the bijection for labeled graphs, we can construct a bijection for unlabeled graphs as follows: start writing down all labeled graphs in the same order as above, but ignore a graph if it's isomorphic to one you've written down already. Naturally, as expected, this bijection is very hard to compute. It begins

$$\emptyset, \emptyset, 0, 1, 000, 001, 011, 111, 000000, \dots$$

where the $2^3 = 8$ binary strings of length $3$ describe $8$ labeled graphs but only $4$ unlabeled graphs on $3$ vertices, namely the empty graph $000$, the single edge $001$ (isomorphic to $010$ and $100$), the double edge $011$ (isomorphic to $101$ and $110$), and the triangle / complete graph / cycle $111$. So given all binary strings corresponding to different labelings of the same unlabeled graph we write down the first one in lexicographic order only and discard the rest.

ContentLicense: CC BY-SA 4.0


问:
$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right)$, is my solution wrong?
I needed to calculate: $$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right)$$

First of all I saw that it won't be possible to do that in any traditional way and actually calculate the limit, because of the form of expression. I mean - it's a sum with sqares on $n$ so I can't use Stolz lemma that easy. But, I thoght, that the solution is probably $0$, because probably every element of the sum is $0$ when $n \implies \infty$ and the limitation of sum in $\infty$ = sum of limitations in $\infty$. So I just went with that and decided to prove that using induction.

My base is:
$$\lim_{n \to \infty} \frac{1}{\sqrt{n^2 + 1}} = 0$$

My assumption:
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +...+ \frac{1}{\sqrt{n^2 +2n}}\right) = 0$$

My induction:
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right) = 0 + \lim_{n \to \infty} \frac{1}{\sqrt{n^2 +2n + 1}}) = 0$$

So the limit is:
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right) = 0$$

But then my grader at university said that at first sight it looks totaly wrong but he actualy needs to think about it a bit. So here is my question - is that wrong? How is that wrong?

Ok, thank you for your answers. I thought I can solve that exercise that way because I asked not long ago very similar question on that forum: Find the limit of of such a sequence defined by recurrence

Because of your answers I think that the problem is actually that in this case I am dealing with a SUM of elements, am I right (or it the answer that I got in other case wrong?)?

答:
By an induction like this you can correctly prove that for any constant $k$,
$$
   \lim_{n \to \infty} \left(\frac1{\sqrt{n^2+1}} + \frac1{\sqrt{n^2+2}} + \dots + \frac1{\sqrt{n^2+k}} \right) = 0.
$$
But there is no value of $k$ for which this sum of a fixed number of terms will turn into the sum in the question, which has a variable number of terms.

Going in the other direction, you can "bite off" any number of terms, and show that
$$
   \lim_{n \to \infty} \left(\frac1{\sqrt{n^2+1}} + \frac1{\sqrt{n^2+2}} + \dots + \frac1{\sqrt{n^2+2n+1}} \right) =    \lim_{n \to \infty} \left(\frac1{\sqrt{n^2+k}} + \frac1{\sqrt{n^2+k+1}} + \dots + \frac1{\sqrt{n^2+2n+1}} \right).
$$
But there is no value of $k$ which will reduce this to the limit that you've called the base case, because the number of terms will always keep growing with $n$.

In fact, since there are $2n+1$ terms in the sum, and each one of them is at least $\frac1{\sqrt{n^2+2n+1}} = \frac1{n+1}$, that the sum is equal to at least $\frac{2n+1}{n+1} = 2 - \frac1{n+1}$, and therefore the limit is at least $2$, not $0$. Think about what kind of upper bound we can put on the sum in the same way...


Title: $\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right)$, is my solution wrong?

Body:
I needed to calculate: $$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right)$$

First of all I saw that it won't be possible to do that in any traditional way and actually calculate the limit, because of the form of expression. I mean - it's a sum with sqares on $n$ so I can't use Stolz lemma that easy. But, I thoght, that the solution is probably $0$, because probably every element of the sum is $0$ when $n \implies \infty$ and the limitation of sum in $\infty$ = sum of limitations in $\infty$. So I just went with that and decided to prove that using induction.

My base is:
$$\lim_{n \to \infty} \frac{1}{\sqrt{n^2 + 1}} = 0$$

My assumption:
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +...+ \frac{1}{\sqrt{n^2 +2n}}\right) = 0$$

My induction:
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right) = 0 + \lim_{n \to \infty} \frac{1}{\sqrt{n^2 +2n + 1}}) = 0$$

So the limit is:
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right) = 0$$

But then my grader at university said that at first sight it looks totaly wrong but he actualy needs to think about it a bit. So here is my question - is that wrong? How is that wrong?

Ok, thank you for your answers. I thought I can solve that exercise that way because I asked not long ago very similar question on that forum: Find the limit of of such a sequence defined by recurrence

Because of your answers I think that the problem is actually that in this case I am dealing with a SUM of elements, am I right (or it the answer that I got in other case wrong?)?

Body:
Clearly,
$$
\frac{1}{n+1}\le \frac{1}{\sqrt{n^2 + k}} \le \frac{1}{n}
$$
for $k=1,2,\ldots,2n+1$. Hence
$$
\frac{2n+1}{n+1}<\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}<\frac{2n+1}{n}
$$
and
$$
\lim_{n\to\infty}\frac{2n+1}{n+1}=\lim_{n\to\infty}\frac{2n+1}{n}=2.
$$
Therefore
$$
\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\to 2.
$$


答:
The others already told you why your solution is wrong. I am going to show you how to actually compute this limit (and please bear in mind that this is a standard trick for such limits).

Obviously, $\displaystyle n^2+1\le n^2+k \le n^2+2n+1$ for any $n \in \mathbb{N}$, $k= \overline{1,2n+1}$. This is equivalent to saying that $\displaystyle \frac{1}{(n+1)^2}\le \frac{1}{n^2+k}\le \frac{1}{n^2+1}$. After you take the square root and sum up all the inequlities you get that $\displaystyle \frac{2n+1}{n+1}\le \sum_{k=1}^{2n+1}\frac{1}{\sqrt{n^2+k}}\le \frac{2n+1}{\sqrt{n^2+1}}, \forall n\in \mathbb{N}$ and now by the squeeze theorem you may conclude that $\displaystyle \lim_{n\to \infty}\sum_{k=1}^{2n+1}\frac{1}{\sqrt{n^2+k}}=2.$


问:
If $X $ is a random variable and $aX \sim bX $ then $ a = b $
$X $ is a random variable for which $\ P(X \not = 0 ) > 0 $ , and I have some $\ a, b \in [0, \infty) $ such that $\ F_{aX} = F_{bX} $ does it mean that $\ a = b $ ?

My understanding is that since $\ F_{aX}(t) = F_{bX}(t) $ for every $\ t \in \mathbb R $

then $$\mathbb P[(\omega \mid aX(\omega) \in (-\infty, t])]  = \mathbb P [(\omega \mid bX(\omega) \in (-\infty,t])] $$

then $$\ aX(\omega) = bX(\omega) \rightarrow \frac{a}{b} = 1 $$

答:
Your argument is invalid. Just because two random variables have the same distribution you cannot say that the random variables are themselves equal. Here are some steps you can verify to answer this question:

Step I:  Suppose $a >b$. If $aX$ and $bX$ have the same distribution then $X$ has the same as distribution as $\frac b a X$.

Step II. We can iterate the result of Step I to see that $X$ has the same as distribution as $(\frac b a)^{n} X$ for each $n$.

Step III.  Since $(\frac b a)^{n} \to 0$ we can conclude that $X$ has the same distribution as $0$.  But this contradicts the hypothesis.

Finally we have proved that $a \leq b$ and a similar argument gives $b \leq a$ so $a=b$.


问:
What is the solution of $x^3+x=1$?
According to Wolfram|Alpha, the solution of $x^3+x=1$ is approximate $0.68233$ or exactly this monstrosity:

$x_0=\frac{\sqrt[3]{\frac{1}{2}(9+\sqrt{93})}}{3^{\frac{2}{3}}}-\sqrt[3]{\frac{2}{3(9+\sqrt{93})}}$

$x^3+x=1$ is so simple, that I refuse to believe that this ugly construct is the simplest way.  Am I right?

答:
If one re-arranges $x^{3}+x= 1$ as

$x^{2} + 1 = \frac1x$

then the accompanying figure shows an alternative approach to obtaining the (real) solution iteratively, starting with some root approximation (e.g. $x_0=0.5$ in the figure), calculating $x_0^{2} + 1$ to obtain the first iteration $x_1=\frac{1}{x_0^{2}+1}$ and so on. In comparison to the unavoidable ways of expressing the exact solution, the iterative equation for convergence on 0.68233 (5 d.p.) looks rather simple:

$x_{n+1} = \frac{1}{x_n^{2}+1}$

Such is the symmetry of the curves (I've only shown the quadrant with real root), there is no constraint on the choice of initial real value $x_0$ to achieve convergence.

<img src="https://i.stack.imgur.com/6LJoc.jpg" alt="$$x^{2} + 1 = \frac1x$$" />

Title: What is the solution of $x^3+x=1$?


问:
Seat count calculation to fill a row within 50mm
I have 3 sizes of seats measuring 510, 535, and 560. I also have extra seat arms measuring 50.  I want to use combinations of these seats and extra arms to fill a row within 50mm of its total length.  I also have an average seat size that I would like to achieve of 550. Lastly I would like to be able to choose an even or odd total seat count. I assume that most long (ie 10000) row lengths would have multiple solutions. How do I find them?</p>

答:
It depends.  Do you want to find an optimal solution (for example, the solution with the fewest chairs), or do you want to find all solutions?

To find all solutions, the only real option is to use "brute force"--it would be best to write a computer program for this.

For finding the optimal solution greedy strategy could work (e.g. pick the biggest chair at each step, then proceed).  However, the validity of the greedy approach is dependent on how you define optimal.

For an arbitrary definition of optimal solution, I suggest Dynamic Programming (DP), which is too broad to completely cover here.  The idea of DP is very similar to recursion, but is more time and memory efficient.

ContentLicense: CC BY-SA 3.0


问:
Topology - Every injective quotient map is a homeomorphism
I am going to show that every injective quotient map is a homeomorphism:

Let $(X,\tau_{X})$, $(Y,\tau_{Y})$ be topological spaces.

Definitions:

$q:X \rightarrow Y$ is a quotient map iff $q$ is surjective ($q[X] = Y$) and
$$
\forall V\subseteq Y: V\in \tau_{Y} \iff q^{-1}[V] \in \tau_X
$$
where $[]$ is used to denote the image of a function. $f:X \rightarrow Y$ is a homeomorphism iff $f$ is bijective and
$$
\forall U\subseteq X: U\in \tau_{X} \iff f[U] \in \tau_Y
$$

Lemma:
$$\forall x \forall y: P(x,y) \implies Q(x)$$
is equivalent to
$$\forall x: Q(x) \land \forall x \exists y: P(x,y)$$

Proof of the lemma: proof

Proof:

It is sufficient to show that if $q$ is injective, $\forall V\subseteq Y: V\in \tau_{Y} \iff q^{-1}[V] \in \tau_X$ is equivalent to $\forall U\subseteq X: U\in \tau_{X} \iff q[U] \in \tau_Y$.

Notes:
Injectivity of $q$ ensures $q^{-1}[q[U]] = U$ for all $U \subseteq X$.
For a surjection $q$, $\forall V \subseteq Y: \exists U \subseteq X: q[U] = V$ is a logical necessity.

$$
\begin{align}
&\forall V\subseteq Y: V\in \tau_{Y} \iff q^{-1}[V] \in \tau_X& \\
&\iff
(\forall V\subseteq Y: V\in \tau_{Y} \iff q^{-1}[V] \in \tau_X) \land (\forall V \subseteq Y: \exists U \subseteq X: q[U] = V)& \text{Tautology}\\
&\iff
\forall V\subseteq Y : \forall U \subseteq X: q[U] = V \implies V\in \tau_{Y} \iff q^{-1}[V] \in \tau_X & \text{Lemma}\\
&\iff
\forall U \subseteq X: \forall V\subseteq Y : q[U] = V \implies q[U]\in \tau_{Y} \iff q^{-1}[q[U]] \in \tau_X &p \rightarrow q \iff p\rightarrow p \land q\\
&\iff
\forall U \subseteq X: \forall V\subseteq Y : q[U] = V \implies q[U]\in \tau_{Y} \iff U \in \tau_X & \text{Injectivity}\\
&\iff
(\forall U\subseteq X : q[U]\in \tau_{Y} \iff U \in \tau_X) \land (\forall U \subseteq X:\exists V \subseteq Y: q[U] = V) &\text{Lemma}\\
&\iff
\forall U\subseteq X : q[U]\in \tau_{Y} \iff U \in \tau_X& \text{Tautology}\\
\end{align}
$$

Is this correct?

答:
Your proof is correct, but it is difficult to understand and unnecessarily complicated.

Your lemma is a purely logical statement, and it is unusual to begin a concrete proof with such a statement. In fact, all arguments are based on logic, and we must take that foundation as granted. Otherwise every proof would have to begin with an explanation of all logical, set theoretical  etc. issues.

The essence of your arguments it this:

Any quotient map $q$ is a continuous surjection.

Any injective quotient map $q$ is a continuous bijection. Thus it suffices to show that if $U \in \tau_X$, then $p(U) \in \tau_Y$. But this is obvious since $p^{-1}(p(U)) = U$ for injective maps.


问:
Harmonic Mean: Why does it exist?
I am implementing some descriptive statistics in a program and would like to understand why the Harmonic Mean exists. What, precisely, does it achieve?

I have read this thread, but the answer has not resolved my query.

答:
Actually, in my opinion, the misapplication of statistics is a major issue.

From already presented examples, including the cited importance in the field of finance on calculating price earning ratios, for example, where the harmonic mean apparently can potentially play a key role (as in screening and more generally financial decision making) context is important.

Otherwise, one is just possibly contributing to potentially more misleading and even contextually deceptive statistical-based data.

My suggestion, have an embedded in your descriptive information summary (per links, an appendix, an expanded table, ...) an explanation of the particular statistic of interest along with key areas of its application. Consider also providing examples where it is expected to be likely not advisable.

And yes, the harmonic mean is more rarely referenced and exotic, but it can distinctly be decisively informative. Adding informational value will, in time, be well received.

Title: Harmonic Mean: Why does it exist?


问:
show that if $E$ is a linear space and $L \subset E$ is a subspace such that $codimL=1$ then there is $f \in E^{*}$ with $kerf = L$
Let $E$ be a linear space over $\mathbb{R}$ or $\mathbb{C}$ and let $L \subset E$ be a subspace such that $codimL = 1$.

I want to prove that there is a functional $f \in E^*$ with $kerf = L$.

Since $codimL = 1$ then $E / L = \{\lambda x + L | \lambda \in \mathbb{F} \}$ for some $0 \neq x \in E$.

Equivalently $E = L + W$ where $dimW=1$.

I am not sure how to follow from here, but since $kerF \subset L + W$ then maybe somehow I can take $y \in kerf$ and then $y = v + w$ where $v \in L, u \in w$ and somehow show that $u = 0$ .

Help woukd be appreciated.

答: Note that $L \cap W=\{0\}$. Just define $f(v+\lambda x)=\lambda$. Then the kernel of $f$ is $L$.


问:
Why use Gauss Jordan Elimination instead of Gaussian Elimination, Differences
Why use Gaussian Elimination instead of Gauss Jordan Elimination and vice versa for solving systems of linear equations? What are the differences, benefits of each, etc.?

I've just been solving linear equation systems, of the form Ax = B, by reducing matrix A to a diagonal matrix where every nonzero value equals 1. I'm not exactly sure if this would be considered Gaussian Elimination or Gauss-Jordan Elimination.

答:
To build on the accepted answer, I will add an analysis. So lets say that we wanted to  turn 2 rows into row echelon form.

Using RREF:

Divide all terms in first row by the leading coefficient (n operations) then find out what to multiply the first row by so that you can add it to the second row and make the second rows leading coefficient 0, finding this multiplier will take 1 operation. Then you have to divide/multiply every element of the first row by this number, and add this row to the second row, 2n operations. So out total is n+n+n+1 or around 3n

Using REF:

Find out what to multiply the first row by so that you can add it to the second row and make the second rows leading coefficient 0 (1 operation). Multiply all terms in the first row by this number, then add this row with the second row (2n operations). So our total is 1+n+n or around 2n.


问:
How does one become a mathematician?
I mean how someone goes from a student to become a teacher or a professor ? What is that state feel like ? Are you confident of the knowledge you have conquered over the years ?or there is part of doubt in there ?

I am current in the 3rd year out of 4 to get my degree, and I am struggling to learn the subjects to the fullest, always questions about things that confuse me and  I can't answer , I can't solve problems with confidence (a lot of times I can't solve them at all), I don't have much confidence about my solution etc. I have a feeling that even if I get my degree those doubts and confusion  won't go away like that, if there is someone who felt that way their opinion and point of view are more that welcome . You can imagine you are giving advice to your old self , an advise you never got but you needed.

For example because you did good on your finals doesn't mean you know abstract algebra , it's like the piano, you can't call your self a pianist because you know how to play well 5-10 songs. So when someone can be called a mathematician ? Or even that at least he knows abstract  algebra. Also,  I can observe that my knowledge on statistics and probabilities are a bit weaker over the pure math like real analysis number theory etc(because i study them more and give them more attention). Does every professor know deeply all the subjects ?

答:
Given that I don't know you, nor does anyone else on this forum probably, I cannot say whether your doubts are justified or not. I can only share my limited experience.

Almost my entire (short) mathematical career was centered around algebraic number theory. I've taught a few courses in abstract algebra over the years, from group theory to algebraic geometry. My knowledge of statistics and real analysis is extremely limited, somewhere around second-year level. Certainly nowhere near research level, I have no deep understanding of anything related to those subjects.

I started to gain confidence in my ability as a mathematician once I became a teaching assistant, when I got to see the work of dozens and dozens of other students. Seeing their perspectives, their approaches, and hearing their questions greatly broadened my own mathematical toolbox, and it also made clear to me that my mathematical abilities were at least above average.

At first, teaching (tutorials, as a TA) was quite frightening and I did not think I was fit for the job. After a few months of weekly tutorials I was able to answer most questions quite comfortably, and deliver coherent and somewhat interactive lectures (of 20-30 minutes). After about a year I was comfortable discussing problems instead of simply answering them, and comfortable admitting that I didn't know an answer, and maybe even investigating the question together with the students. Around this time I started considering myself to be an actual teacher.

Title: How does one become a mathematician?


问:
Convert Sums of absolute values in Linear program
Ihave the following problem and it needs to be converted into LP form.

\begin{align}
 \max\> & z = c^T x\\
      & Ax \leq B \\
      & \sum_{i=1}^{n}|x_i|=1
\end{align}

I know that $c^Tx$ and $Ax \leq B$ are matrix dot products.
My question is, How can I convert $\sum_{i=1}^{n}|x_i|=1$ into LP form.

I have read online that if you only have a finite amount of $x$'s, you can simply do one equation for the positive and another one for the negative. For instance, if we have $|x_1| = 1$, the equation will be transformed to $x_1 = 1$ and $-x_1 = 1$. How can we do it for the summation?

I think we need to take the combinations of all the possible $x$'s as maybe $x_1$ is positive and all the rest are negative.

Thanks!

答: Use auxiliary variables. Let $u_i \geq 0, v_i \geq 0$. Then we can write $x_i = u_i - v_i$ and $\vert x_i \vert = u_i + v_i$.


问:
Principal value of integral
How to find the Cauchy principal value of the integral $$\int_0^\infty \left(\frac{1}{x^2}-\frac{\cot(x)}{x} \right) dx?$$</p>

答: See the answer to this question at MO here</p>

Title: Principal value of integral


问:
Is my proof that the Sharkovsky Ordering is a total ordering, correct?
The Sharkovsky ordering is an ordering of the natural numbers $\mathbb{N}$, where

$3$  $\prec$   $5 $      $\prec$      $7 $     $\prec$  $9$      $\prec$   ...

$2*3$  $\prec$  $2*5$   $\prec$      $2*7$     $\prec$  $2*9$      $\prec$  ...

...

$2^n*3$  $\prec$  $2^n*5$   $\prec$ $2^n*7$  $\prec$  $2^n*9$  $\prec$  ...

...

... $2^n$  $\prec$  ...     $\prec$  $2^3$      $\prec$  $2^2$      $\prec$  $2$  $\prec$  $1$

The ordering starts with all odd numbers, except for one, in increasing order, followed by two times the odds, $2^2$ times the odds, $2^3$ times the odds and so one. Finally the powers of two are listed last in decreasing order.

I know what to prove that this order is a total ordering. Thus, I need to show

Antisymmetry: If $a \leq b$ and $b \leq a$ then $a=b$

Transitivity: If $a \leq b$ and $b \leq c$ then $a \leq c$

Connexity: $a\leq b$ or $b \leq a$

I came up with the following mapping from the Sharkovsky ordering to the natural numbers:
\begin{equation}\label{eq1}
  P =
    \begin{cases}
      \mathbb{N}_0 \times \mathbb{N}_0 & \rightarrow \mathbb{N}\\
      (r,p) & \rightarrow 2^r \cdot (2p+1)
    \end{cases}
\end{equation}

As this function is bijective, each natural number appears exactly once in the Sharkovsky ordering.

See answer below for attempt at proof.

答:
I would simply define the relation $\prec$ carefully and then use the definition to show that $\preceq$ is a linear order. Specifically,

$$\begin{align*}
2^r(2k+1)\prec 2^s(2\ell+1)\text{ iff }&r<s\text{ and }k\ne 0\ne\ell,\text{ or}\\
&r=s\text{ and }0<k<\ell,\text{ or}\\
&r>s\text{ and }k=\ell=0,\text{ or}\\
&k>0\text{ and }\ell=0\,.
\end{align*}$$


问:
If $AC=BC$ and $A \cap C = B \cap C$ then $A=B$
This is exercise 2.48 from Rotman's "An Introduction to the Theory of Groups".

I'm having trouble proving $B \subset A$ knowing $A,B,C \leq G$, $A \leq B$, $A \cap C = B \cap C$ and $AC=BC$.

I can't use Dedekind's modular law to prove the proposition.

答:
Let $b\in B$. Since $1\in C$, we have $b=b1\in BC=AC$ thus $b=ac$ for some $a\in A,\,c\in C$.

Then $c=a^{-1}b\in B$ because $A\le B$.

But it was an element of $C$, hence $c\in B\cap C=A\cap C\le A$, so $b=ac\in A$.


问:
Understanding proof of Continuous extension theorem for uniformly continuous function
I was reading, continuous extension theorem for uniformly continuous function Theorem 5.4.8 in "Introduction to real analysis by Robert G. Bartle" fourth edition In the proof, i didn't get one point.

Author has written in end of first para of proof that, if $(u_n)$ is any sequence in $(a,b)$ that converges to $a$ then $lim(u_n-x_n)=0$. So "By uniform Continuity of $f$ we have,

$lim(f(u_n))=lim(f(u_n)-f(x_n))+lim(f(x_n))=0+L=L$

How uniform Continuity of $f$ is used here?

My opinion: as $lim(u_n-x_n)=0$ hence we have $lim(u_n)=lim(x_n)= a$ (since $(x_n)$, $(u_n)$ are sequences in $(a,b)$ both  converging to $a$)

So as $f$ is uniformly continuous on $(a,b)$ and hence it is continuous on $(a,b)$ hence by "Sequential criterion of continuity" we have,

$lim(f(u_n))=lim(f(x_n))=f(a)=L$ (say)

Hence, $lim(f(u_n)-f(x_n))=0$.

So that, $lim(f(u_n))=lim(f(u_n)-f(x_n))+lim(f(x_n))=0+L=L$ as required.

Is my opinion (reason) is correct? Please help.....

答:
$f$ being continuous on $(a,b)$ does not mean that $f$ is continuous at $a$, because $a \notin (a,b)$. $f$ is not even defined at $a$, so there is no "$f(a)$" for those two limits to converge to.

The whole point of this theorem is that when $f$ is uniformly continuous on $(a,b)$ you can extend the domain of $f$ to include $a$ in such a way that $f$ will be continuous at $a$. You do not get to assume that the continuous extension has already happened.

Because $x_n$ and $u_n$ converge to point outside the domain of $f$, you need uniform continuity to assure $$\lim_{n\to\infty} f(u_n) - f(x_n) = 0$$

Since you cannot assume a common limit for $f(u_n)$ and $f(x_n)$, you will need to examine the definition of uniform continuity to see why else it indicates they converge to the same value.

To see that plain continuity is not sufficient, consider $(a,b) = (0,1), f(x) = \sin 1/x$.


问:
Surface element of inverted cone in cylindrical coordinates.
I have to integrate over the curved surface of a inverted cone, since the surface is of a cone I think all three $s, \phi$ and $z$ are varying under the integral sign.

Now this causes the problem, I don't know surface element when all three are varying nor I can find a solution on internet,  I thought I will take $\displaystyle|d\vec a| = s\ ds\ d\phi\ dz$ but that is volume element.

What should I do ? Drop $s$ from $\displaystyle |d\vec a| = s\ ds\ d\phi \ dz$ ?

答:
From [1], consider a right solid circular cone with height $h$ and aperture $2\, \phi $, whose axis is the $z$ coordinate axis and whose apex is the origin.  By $\omega$, I denote the differential area element of this cone in cylindrical coordinates. I find that

$$
\omega = z\,\tan {(\phi)} \,\sqrt{\tan^2{(\phi)}  + 1}\,dz\,d\theta
$$

Following from the sphere example in [2],  this cone can be parametrized using cylindrical coordinates with the map
$$\Phi{(\theta, z)} = \left( z\,\tan{(\phi)}\,\cos{(\theta)}, z\,\tan{(\phi)}\,\sin{(\theta)}, z \right).$$

Then,
\begin{align}
g_{11}
&=
\sum\limits_{k=1}^{3}
\frac{\partial{\Phi_k}}{\partial{\theta}}
\frac{\partial{\Phi_k}}{\partial{\theta}}
\\
&=
\frac{\partial{\Phi_1}}{\partial{\theta}}
\frac{\partial{\Phi_1}}{\partial{\theta}}
+
\frac{\partial{\Phi_2}}{\partial{\theta}}
\frac{\partial{\Phi_2}}{\partial{\theta}}
+
\frac{\partial{\Phi_3}}{\partial{\theta}}
\frac{\partial{\Phi_3}}{\partial{\theta}}
\\
&=
\left[-z\,\tan{(\phi)}\,\sin{(\theta)}\right]^2
+
\left[+z\,\tan{(\phi)}\,\cos{(\theta)}\right]^2
+ 0^2
\\
&=
\left[z\,\tan{(\phi)} \right]^2
\\
g_{12}
&=
\sum\limits_{k=1}^{3}
\frac{\partial{\Phi_k}}{\partial{\theta}}
\frac{\partial{\Phi_k}}{\partial{z}}
\\
&=
\frac{\partial{\Phi_1}}{\partial{\theta}}
\frac{\partial{\Phi_1}}{\partial{z}}
+
\frac{\partial{\Phi_2}}{\partial{\theta}}
\frac{\partial{\Phi_2}}{\partial{z}}
+
\frac{\partial{\Phi_3}}{\partial{\theta}}
\frac{\partial{\Phi_3}}{\partial{z}}
\\
&=
\left[-z\,\tan{(\phi)}\,\sin{(\theta)}\right]\,\left[\tan{(\phi)}\,\cos{(\theta)}\right]
+
\left[+z\,\tan{(\phi)}\,\cos{(\theta)}\right]\,\left[\tan{(\phi)}\,\sin{(\theta)}\right]
+ (0)\,(1)
\\
&=
z\,\left[ - \,\tan^2{(\phi)}\,\sin{(\theta)} \,  \cos{(\theta)} +
 \tan^2{(\phi)}\,\cos{(\theta)} \, \sin{(\theta)}\right]
\\
&=
0
\\
g_{21}
&=
0
\\
g_{22}
&=
\sum\limits_{k=1}^{3}
\frac{\partial{\Phi_k}}{\partial{z}}
\frac{\partial{\Phi_k}}{\partial{z}}
\\
&=
\frac{\partial{\Phi_1}}{\partial{z}}
\frac{\partial{\Phi_1}}{\partial{z}}
+
\frac{\partial{\Phi_2}}{\partial{z}}
\frac{\partial{\Phi_2}}{\partial{z}}
+
\frac{\partial{\Phi_3}}{\partial{z}}
\frac{\partial{\Phi_3}}{\partial{z}}
\\
&=
\left[\tan{(\phi)}\,\cos(\theta)\right]^2
+
\left[\tan{(\phi)}\,\sin(\theta)\right]^2
+
1^2
\\
&=
\tan^2{(\phi)}  + 1
\end{align}
Now the Jacobian, $g$, is
\begin{align}
g
&=
\begin{vmatrix}
z^2\,\tan^2{(\phi)}
&
0
\\
0
&
\tan^2{(\phi)}  + 1
\end{vmatrix}
\\
&=
\left|z^2\,\tan^2{(\phi)} \,\left(\tan^2{(\phi)}  + 1\right)
\right|
\\
&=
z^2\, \tan^2{(\phi)} \,\left(\tan^2{(\phi)}  + 1\right)
\end{align}
Finally, I write the differential area element as
\begin{align}
\omega = \sqrt{g} \,dz\,d\theta
\end{align}

Billiography

[1] https://en.wikipedia.org/wiki/Cone#Equation_form

[2] https://en.wikipedia.org/wiki/Volume_element


问:
Conjecture: Are there infinitely many triangular numbers which are of the form $qp$ , where $p$,$q$ are distinct primes?

Show that there are infinitely many positive integers $n$ such that the number of distinct odd prime factors of $n(n + 3)$ is a multiple of $3$.

I couldn't get much progress, I took $n= 3k$, and then was trying to show, that there are infinitely many  many positive integers $k$ such that the number of distinct odd prime factors of $k(k + 1)$ is $1\mod 3$.

So if I can show that

There are infinitely many triangular numbers which are of the form $qp$ , where $p,q$ is a prime

This looks true seeing the OEIS Link , the first term  is $55$, then $91$ , then $231$ and so on ..
then I will be done.

However, I think I am in a wrong path, because it's a contest problem.

Thanks in advance!

Here's the link of the question

答:
Suppose that $\frac{n(n + 1)}{2}$ is a product of $2$ primes where $n > 2$. If $n$ is even, this means that both $\frac{n}{2}$ and $n + 1$ are primes, and if $n$ is odd, then both $n$ and $\frac{n + 1}{2}$ are primes.

So we find that there are infinitely many triangular numbers that are a product of $2$ primes if and only if either there are infinitely many primes $p$ such that $2p + 1$ is a prime, or there are infinitely many primes $p$ such that $2p - 1$ is a prime. Both of these are unsolved problems.

Primes $p$ such that $2p + 1$ is also a prime are called Sophie Germain primes. Primes $p$ such that $2p - 1$ is also a prime don't have a special name. In both cases it is conjectured but not known that there are infinitely many such primes.


问:
Prove if $G$ is a finite group and $H\unlhd G$ that there is a composition series of $G$ one of whose terms is $H$

Prove if $G$ is a finite group and $H\trianglelefteq G$ that there is a composition series of $G$ one of whose terms is $H$

So I previously proved every finite group has a composition series. Which is maybe useful here.

So I use two cases, either $G/H$ is simple or it is not. If $G/H$ is simple, then since $H$ has a composition series as it is a finite group, then $1\trianglelefteq H_1\trianglelefteq ...\trianglelefteq H\trianglelefteq G$ is a composition series of $G$ containing $H$.

If $G/H$ is not simple, then there exists some $K/H$ in $G/H$ such that $K/H\trianglelefteq G/H$. By correspondence theorem there exists a $K\trianglelefteq G$ containing $H$. I believe since $G$ is finite, I can also say that I can choose $K/H$ to have minimal index in $G/H$, since there can at most be finitely many $K_i/H\trianglelefteq G/H$ so there exists a minimal index. This ensures that $G/K$ is simple as if not then there exists an $N\trianglelefteq G$ containing $K$ which has $\vert G:N\vert\vert N:K\vert=\vert G:K\vert=\vert G/H:K/H\vert$, so then $\vert G/H:N/H \vert=\vert G:N\vert<\vert G:K\vert$ a contradiction.

Then since $H$ has a composition series, we get that $1\trianglelefteq H_1\trianglelefteq...\trianglelefteq H\trianglelefteq K\trianglelefteq G$ is a compositions series of $G$.

I think what I'm missing here however is that $K/H$ needs to be simple and I haven't shown that, I've only got that $G/K$ is simple. I don't for instance see anything wrong with $K/H$ not being simple, since I would get some $N$ with the property that $H\trianglelefteq N\trianglelefteq K$ where $K/N$ is simple but $N/H$ may not be and this would repeat again. And I believe since the group is finite this must terminate at some point.

答:
Okay, you are getting tied up in knots, and comment thread is getting too long.

Let $G$ be a group. By “a composition series” I understand a subnormal series
$$ 1=N_0\triangleleft N_1\triangleleft N_2\triangleleft\cdots \triangleleft N_{m-1}\triangleleft N_m=G$$
such that for all $i=0,\ldots,m-1$, $N_{i+1}/N_{i}$ is simple.

You say you already know that every finite group has a composition series.

Let $G$ be a finite group, and $H\triangleleft G$. We want a composition series for $G$ with $N_i=H$ for some $i$.

Since $H$ is finite, $H$ has a composition series,
$$1=N_0\triangleleft N_1\triangleleft\cdots \triangleleft N_r=H.$$
Now, $G/H$ is also finite, so it has a composition series, which for ease I will number starting at $r$:
$$ 1 = M_r\triangleleft M_{r+1}\triangleleft \cdots \triangleleft M_{r+s}=\frac{G}{H}.$$
By the correspondence theorem, each $M_i$ corresponds to a subgroup $N_i\leq G$ (I will show shortly that the “new” $N_r$ is the same as the old one) that contains $H$, and moreover, $N_i\triangleleft N_{i+1}$ because the correspondence preserves normality. Note that $N_r/H = M_r=\{1_{G/H}\}$, so both subgroups called $N_r$ are actually equal and I’m not abusing notation. Now we have a subnormal series
$$1 = N_0\triangleleft\cdots\triangleleft N_r =H \triangleleft N_{r+1}\triangleleft\cdots \triangleleft N_{r+s}=G.$$
It only remains to show this is a composition series.

For $i=0,\ldots,r-1$, the quotient $N_{i+1}/N_i$ is simple, since the $N_i$ are a composition series for $H$. For $i=r,\ldots,r+s-1$, we have
$$\frac{N_{i+1}}{N_i} \cong \frac{N_{i+1}/H}{N_i/H} \cong \frac{M_{i+1}}{M_i}.$$
The first isomorphism follows from the Isomorphism Theorems (usually second or third, depending on how you number them, which tells you that a quotient of a quotient is a quotient of the original group).

Now, we know $M_{i+1}/M_i$ is simple, because the $M_i$ form a composition series for $G/H$. Therefore, $N_{i+1}/N_i$ is simple for $i=r,\ldots,r+s-1$ as well.

This shows that in our subnormal series, all quotients are simple. Thus, it is a composition series. And $N_r=H$, so it is a composition series one of whose terms is $H$, as desired. $\Box$


问:
Lindeberg-Feller alternative formulation in Durrett
I feel a little confused and need help in understanding equivalent formulations of Lindeberg-Feller CLT given in different sources.

Let $X_{n,k}$ to be the triangular array, where
$$\forall n: $X_{n,1}, ..., X_{n, n} \text { are independent}$$
$$\forall n, m: EX_{n, m} = 0$$

Then in Durrett, the Lindeberg-Feller condition for is formulated as:

$s_n^2 = \sum_{m=1}^{n}EX_{n,m}^2 \to \sigma^2$

$\forall \epsilon > 0:\lim_{n\to \infty}\sum_{m = 1}^nE(|X_{n,m}|^2;|X_{n,m}>\epsilon|) = 0$

However, in other textbooks, for example Shiryaev, the Lindeberg-Feller condition is given by:

$EX_{n,m}^2 < \infty $, i.e., variance exists and is finite

$\forall \epsilon > 0: \lim_{n\to \infty} \frac{1}{s_n^2}\sum_{m = 1}^nE(|X_{n,m}|^2;|X_{n,m}>\epsilon s_n|) = 0$, where $s_n^2 = \sum_{m=1}^nVarX_{n, m} = \sum_{m=1}^n EX_{n,m}^2$

After playing with this expressions for a while, I still fail to see how are they equivalent. More specifically, I think I could see how Durrett's formulation implies the other one, however, I am having difficulty to conclude the reverse.

答:
To show the Shiryaev conditions imply the Durrett conditions, assume that $(X_{n,m})$ satisfy (S1) and (S2). Define $s_n$ as in (S2), and define $(Y_{n,m})$ by
$$Y_{n,m}:=\frac{X_{n,m}}{s_n}.$$
The $(Y_{n,m})$ satisfy (D1), since
$$\sum_{m=1}^n E(Y_{n,m}^2)=\frac 1{s_n^2}\sum_{m=1}^n E(X_{n,m}^2)=1,$$
so we can take $\sigma^2:=1$ in (D1). And the $(Y_{n,m})$ satisfy (D2), since
$$
E( Y_{n,m}^2 ; |Y_{n,m}|>\epsilon) = E\left(\frac{X_{n,m}^2}{s_n^2} ; |X_{n,m}|>\epsilon s_n\right).
$$
So the conclusion of Durrett applies to the $(Y_{n,m})$, not to the $(X_{n,m})$.


问:
Counting couples of numbers
I have no trouble believing that, if $|n| \leq J$, then $$\#\{ (j_1,j_2) \in \{ 1,...,J \} \, | \, j_1-j_2 = n \} = J-|n|,$$

but can anyone explain it a little more formally?

Thank you in advance for any help.

Title: Counting couples of numbers

Body:
We can restrict to the case $n \ge 0$, since if $n \le 0$ then
$$\{ (j_1,j_2) \in \{ 1,...,J \} \, | \, j_1-j_2 = n \} =
\{ (j_2,j_1) \in \{ 1,...,J \} \, | \, j_2-j_1 = -n \},$$
with $-n \ge 0$.

Write
$$S = \{ (j_1,j_2) \in \{ 1,...,J \} \, | \, j_1-j_2 = n \},$$
with $n \ge 0$.

Now if $(j_1, j_2) \in S$, we have $j_1 = n + j_2 > n$. Conversely, if $J \ge j_1 > n$, then take $j_2 = j_1 - n \in \{ 1, \dots , J \}$, and $(j_1, j_2)\in S$.

So there are as many pairs in $S$ as the $J - n$ values of $j_1$ from $n+1$ to $J$.


问:
Inserting zeroes - is this map continuous?
The map $[0,1) \to [0,1)$ which inserts a $0$ between every decimal, e.g. $0.1234567\ldots \mapsto 0.010203040506070\ldots$

Body:
No, it's not, because the sequence
$$\begin{gather}
0.09, \\
0.099, \\
0.0999, \\
0.09999, \\
\vdots
\end{gather}$$ converges to $0.1$, while the sequence
$$\begin{align}
f(0.09) &= 0.0009, \\
f(0.099) &= 0.000909, \\
f(0.0999) &= 0.00090909, \\
&\,\,\,\vdots
\end{align}$$
converges to $0.00\overline{09} \neq 0.01 = f(0.1)$.


问:
Proving $f(x) = 0$ for almost every $x\in \mathbb{R}$
Can someone help me with the following, please? I am strugging proving this.

Let $f: \mathbb{R} \to \mathbb{R}$ be a Lebesgue integrable function such that $$\int_a^b f(x) dx=0 \hspace{20pt} \text{ for every } a<b.$$

Show $f(x) = 0$ for almost every $x\in \mathbb{R}$

答:
Let $\mathcal{C}=\{(a,b]\mid a<b\}$ and $\mathcal{L}=\{A\in\mathcal{B}(\mathbb{R})\mid\int_{A}f(x)dx=0\}.$
Clearly $\mathcal{C}$ is a $\pi$-class and $\sigma(\mathcal{C})=\mathcal{B}(\mathbb{R})$.
Also, it is given that $\mathcal{C}\subseteq\mathcal{L}$. We go to
check that $\mathcal{L}$ is a $\lambda$-class. Clearly $\emptyset\in\mathcal{L}$.
By Dominated Convergence Theorem, $\int f(x)dx=\lim_{n\rightarrow\infty}\int_{-n}^{n}f(x)dx=0$.
We have that $\int_{A^{c}}f(x)dx+\int_{A}f(x)dx=\int f(x)dx=0$,
so $\int_{A^{c}}f(x)dx=-\int_{A}f(x)dx=0$. That is, $A\in\mathcal{L}\Rightarrow A^{c}\in\mathcal{L}$.
Let $A_{1},A_{2},\ldots\in\mathcal{L}$ by pairwisely disjoint. Let
$A=\cup_{n}A_{n}$. By Dominated Convergence Theorem, we have
\begin{eqnarray*}
\int_{A}f(x)dx & = & \int\lim_{n\rightarrow\infty}\sum_{k=1}^{n}1_{A_{k}}f(x)dx\\
 & = & \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\int_{A_{k}}f(x)dx\\
 & = & 0.
\end{eqnarray*}
Therefore, $\cup_{n}A_{n}\in\mathcal{L}$. Hence, $\mathcal{L}$ is
a $\lambda$-class. By Dynkin's $\pi$-$\lambda$ Theorem, we have
$\sigma(\mathcal{C})\subseteq\mathcal{L}$. It follows that $\mathcal{L}=\mathcal{B}(\mathbb{R}).$

Let $A\subseteq\mathbb{R}$ be an arbitrary Lebesgue measurable set.
Choose a Borel set $B$ such that $m(A\Delta B)=0$. Observe that
$B=(A\cap B)\cup(B\setminus A),$ $m(B\setminus A)\leq m(A\Delta B)=0,$
and the union is disjoint, so
\begin{eqnarray*}
0 & = & \int_{B}f(x)dx\\
 & = & \int_{A\cap B}f(x)dx+\int_{B\setminus A}f(x)dx\\
 & = & \int_{A\cap B}f(x)dx.
\end{eqnarray*}
Next, $A=(A\cap B)\cup(A\setminus B),$ $m(A\setminus B)\leq m(A\Delta B)=0$,
and the union is disjoint, so
\begin{eqnarray*}
\int_{A}f(x)dx & = & \int_{A\cap B}f(x)dx+\int_{A\setminus B}f(x)dx\\
 & = & \int_{A\cap B}f(x)dx\\
 & = & 0.
\end{eqnarray*}

Let $A=\{x\in\mathbb{R}\mid f(x)>0\}.$ We prove by contradiction
that $m(A)=0$. Suppose the contrary that $m(A)>0$. For each $n$,
let $A_{n}=\{x\mid f(x)>\frac{1}{n}\}$, then $A=\cup_{n}A_{n}$.
Therefore, there exists $n$ such that $m(A_{n})>0$. Since $A_{n}$
is Lebesgue measurable, so $\int_{A_{n}}f(x)dx=0$. On the other hand,
$\int_{A_{n}}f(x)dx\geq m(A_{n})\cdot\frac{1}{n}>0$, which is a contradiction.
Similarly, we can prove that the set $\{x\mid f(x)<0\}$ has Lebesgue
measure 0. Therefore $f=0$ a.e..


答:
Is my proof correct?

1. Antisymmetry

First, $$2^r (2k+1) \preceq 2^s (2l+1)   \text{(1)}$$
$\text{ if } r < s \text{ and } k \neq 0 \neq l, \text{or}$

$\text{ if } r = s \text{ and } 0 < k < l, \text{or}$

$\text{ if } r > s \text{ and } k = l = 0, \text{or}$

$\text{ if } k > 0 \text{ and } l = 0,\text{or}$

$\text{ if } r = s \text{ and } k = l,$

Second, $$2^s (2l+1) \preceq 2^r (2k+1)   \text{(2)}$$
$\text{ if } s < r \text{ and } l \neq 0 \neq k, \text{or}$

$\text{ if } r = s \text{ and } 0 < l < k, \text{or}$

$\text{ if } s > r \text{ and } l = k = 0, \text{or}$

$\text{ if } l > 0 \text{ and } k = 0,\text{or}$

$\text{ if } r = s \text{ and } k = l,$

Thus, if both (1) and (2) are true, $r = s$ and $k = l$, because all other options lead to contradictions. Hence $2^r (2k+1) = 2^s (2l+1)$, which implys Antisymmetry.

2. Transitivity

First, $$2^r (2k+1) \preceq 2^s (2l+1)   \text{(1)}$$
$\text{ if } r < s \text{ and } k \neq 0 \neq l, \text{or}$

$\text{ if } r = s \text{ and } 0 < k < l, \text{or}$

$\text{ if } r > s \text{ and } k = l = 0, \text{or}$

$\text{ if } k > 0 \text{ and } l = 0,\text{or}$

$\text{ if } r = s \text{ and } k = l,$

Second, $$2^s (2l+1) \preceq 2^t (2m+1)   \text{(2)}$$
$\text{ if } s < t \text{ and } l \neq 0 \neq m, \text{or}$

$\text{ if } t = s \text{ and } 0 < l < m, \text{or}$

$\text{ if } s > t \text{ and } l = m = 0, \text{or}$

$\text{ if } l > 0 \text{ and } m = 0,\text{or}$

$\text{ if } t = s \text{ and } m = l,$

Than,

$r<s<t$ and $k \neq 0 \neq l \neq 0 \neq m \Rightarrow r<t$ and $k \neq 0 \neq m$

$r = s = t$ and $0 < k < l<m$ $\Rightarrow r = t$ and $0 < k < m$

$r > s > t$ and $k = l = m = 0$ $\Rightarrow r > t$ and $m = 0$

$r = s = t$ and $k = l = m$ $\Rightarrow r = t$ and $k = m$

Hence,  $2^r (2k+1) \preceq 2^t (2m+1)$.

Title: Is my proof that the Sharkovsky Ordering is a total ordering, correct?

Body:
The Sharkovsky ordering is an ordering of the natural numbers $\mathbb{N}$, where

$3$  $\prec$   $5 $      $\prec$      $7 $     $\prec$  $9$      $\prec$   ...

$2*3$  $\prec$  $2*5$   $\prec$      $2*7$     $\prec$  $2*9$      $\prec$  ...

...

$2^n*3$  $\prec$  $2^n*5$   $\prec$ $2^n*7$  $\prec$  $2^n*9$  $\prec$  ...

...

... $2^n$  $\prec$  ...     $\prec$  $2^3$      $\prec$  $2^2$      $\prec$  $2$  $\prec$  $1$

The ordering starts with all odd numbers, except for one, in increasing order, followed by two times the odds, $2^2$ times the odds, $2^3$ times the odds and so one. Finally the powers of two are listed last in decreasing order.

I know what to prove that this order is a total ordering. Thus, I need to show

Antisymmetry: If $a \leq b$ and $b \leq a$ then $a=b$

Transitivity: If $a \leq b$ and $b \leq c$ then $a \leq c$

Connexity: $a\leq b$ or $b \leq a$

I came up with the following mapping from the Sharkovsky ordering to the natural numbers:
\begin{equation}\label{eq1}
  P =
    \begin{cases}
      \mathbb{N}_0 \times \mathbb{N}_0 & \rightarrow \mathbb{N}\\
      (r,p) & \rightarrow 2^r \cdot (2p+1)
    \end{cases}
\end{equation}

As this function is bijective, each natural number appears exactly once in the Sharkovsky ordering.

See answer below for attempt at proof.


问:
Is Newton-Raphson the best we can do if we only know the derivative?
I was wondering if NR is the fastest method to find a root if all we know about a function is how to evaluate it and its derivative at any point.

Since you can use the first derivative to approximate the second I was wondering if this lets you converge on a root faster (as in $\epsilon_{n+1} \approx \epsilon_n ^3$)?

答:
About the only thing you can do to improve on Newton-Raphson is to abandon its habit of forgetting every point you've calculated except the last.

For example, instead of assuming at each step that the function is linear and using that assumption to find the next estimate of the root, you find $x_0$ by guessing, $x_1$ by assuming the function is linear and using the values $f(x_0), f'(x_0)$, $x_2$ by assuming the function is a cubic and using $f(x_0), f(x_1), f'(x_0), f'(x_1)$, then find $x_3$ by assuming the function is a quintic, etc.

But this gives only very minor improvements in convergence speed. It has been too long for me to remember, but I think converge is still quadratic in the limit.

Alternatively, instead of polynomials, you can use other families of functions in hopes of improved performance. Polynomials are not always that great at interpolating or extrapolating. For example Brent's Method, a widely-used root-finding algorithm when you don't even have derivatives, relies on inverse quadratic functions based on three previous values of the function.


问:
Can the Leibniz integral rule be applied to an integral whose integrand does not contain the variable of derivation?
To my understanding, the Leibniz Integral lets
$$\frac{d}{dx}\int_{a(x)}^{b(x)} f(x,t)dt = \int_{a(x)}^{b(x)}\frac{d}{dx}f(x,t)dt$$
under certain restrictions. (Pretend the second $\frac{d}{dx}$ is a partial derivative, since I can't figure out how to do it).

I haven't seen the proof and the Wikipedia page is a bit dense for me, so I'm asking here. Can this rule also be applied in the case that $f(x,t) = f(t)$? My initial reaction makes me think "why not?", but if so, then any integral of that form would simply be $0$:
$$
\begin{align*}
\frac{d}{dx}\int_{a(x)}^{b(x)} f(t)dt &= \int_{a(x)}^{b(x)}\frac{d}{dx}f(t)dt\\
&= \int_{a(x)}^{b(x)}0dt\\
&= 0\Big|_{a(x)}^{b(x)}\\
&= 0
\end{align*}
$$

So simply, can the Leibniz integral be applied to integrals of this form? And if so, why is the result valid? And if not, what assumption does the proof rest on that makes it so?

答:
What I think you are asking is:
$$\frac{d}{dx}\int_{a(x)}^{b(x)}f(t)dt=\frac d{dx}\left[F(b(x))-F(a(x))\right]=b'(x)F'(b(x))-a'(x)F'(a(x))$$
$$=b'(x)f(b(x))-a'(x)f(a(x))$$


问:
Error in Spivak's proof that an odd degree polynomial has a root?
I'm reading Spivak's Calculus 3rd edition. On page 123 (chapter 7, Three Hard Theorems) he gives a proof of the theorem that states if $n$ is odd, then any equation $x^n+a_{n-1}x^{n-1}+...+a_0=0$ has a root.

Essentially he's showing that there's an interval $[a, b]$ such that $f(a)<0<f(b)$ where  $f(x)=x^n+a_{n-1}x^{n-1}+...+a_0$, concluding that there's $x \in [a,b]$ such that $f(x)=0$.

The proof begins as follows:

$$f(x)=x^n\left(1+\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n}\right)$$

$$\left|\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n}\right|\leq\frac{|a_{n-1}|}{|x|}+...+\frac{|a_0|}{|x^n|}$$

If we choose $x$ satisfying

$$|x|>1, 2n|a_{n-1}|,...,2n|a_0|$$
then
$$\mbox{ (1) } |x^k|>|x|$$
and
$$\mbox{ (2) }\frac{|a_{n-k}|}{|x^k|}<\frac{|a_{n-k}|}{|x|}<\frac{|a_{n-k}|}{2n|a_{n-k}|}=\frac1{2n}$$

Now he doesn't say anything about $k$. Since he stated that $|x^k|>|x|$, I assumed that $k\in[2,n]$ because if $k$ were in $[1,n]$ then it wouldn't have been true that $|x^k|>|x|$ - we'd have had $|x^1|>|x|$ which is false. But if $k\in[2,n]$, how does it follow that $\frac{|a_{n-1}|}{|x|}<\frac1{2n}$? If this doesn't, does (3) still holds?

In the next step he says that

$$\mbox{ (3) } \left|\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n}\right|\leq\sum_{1}^n\frac1{2n}=\frac12$$

where "less than or equals" doesn't really make sense - it should have been "strictly less" since we have "strictly less" in (2).

But everything would've made sense if he did the following:

$$\mbox{ (1*) } |x^k|\geq|x|, k\in [1,n]$$
$$\mbox{ (2*) }\frac{|a_{n-k}|}{|x^k|}\leq\frac{|a_{n-k}|}{|x|}\leq\frac{|a_{n-k}|}{2n|a_{n-k}|}=\frac1{2n}$$
$$\mbox{ (3*) } \left|\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n}\right|\leq\sum_{1}^n\frac1{2n}=\frac12$$

Am I missing something, or it's indeed an error in the book?

答:
You are right when you state

But everything would've made sense if he did the following: ....

Note that your approach requires the general assumption that $x$ is chosen such that

$$(0^*) \quad |x| \ge 1, 2n|a_{n-1}|,...,2n|a_0| .$$

Spivak's proof is nevertheless correct although it has little shortcomings. In fact, $(1)$ is true only for $k \ge 2$, thus also $(2)$ is true only for $k \ge 2$. However, the essence of $(2)$ is
$$(2') \quad \frac{|a_{n-k}|}{|x^k|} < \frac1{2n} ,$$
the intermediate inequalities are irrelevant. And this is true also for $k = 1$ because $|x| > 2n|a_{n-1}|$. The condition $|x| > 1$ is only needed to obtain $(1)$, but in fact it would also suffice to require  $|x| \ge 1$ to obtain
$$(2'') \quad \frac{|a_{n-k}|}{|x^k|} \le \frac{|a_{n-k}|}{|x|} < \frac{|a_{n-k}|}{2n|a_{n-k}|} = \frac1{2n} .$$

Anyway, in my opinion it is unnecessary to work with $(1)$ and $(2)$. We have
$$f(x)=x^n\left(1+\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n}\right)$$
and clearly
$$1+\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n} \to 1$$
as $\lvert x \rvert \to \infty$. Thus
$$f(x) \to \pm \infty$$
as $x \to \pm \infty$.
Therefore we find $a < 0 < b$ such that $f(a) < 0 < f(b)$.


问:
Determining the largest interval of definition of a Cauchy problem
I have some problems trying to solve this exercise:

Consider the following Cauchy problem: $\left\{\begin{matrix}x'=t+\frac{y}{1+x^2}
\\ y'=txe^{-ty^2}
\\ x(0)=y(0)=1
\end{matrix}\right.$

Discuss the existence of a solution and its uniqeness. Then I'm asked to prove or to disprove the fact that the supremum of the max interval on which the solution is definted is $[0,+\propto)$.

I've defined a function as following $F:\mathbb{R}^3 \rightarrow \mathbb{R}^2$, $F(t,**z**=(x,y))=\begin{bmatrix}t+\frac{y}{1+x^2}
\\ txe^{-ty^2}
\end{bmatrix}$

Since $F$ is infinitely differentiable (which implies it's continuous and locally lipschitz) on $\mathbb{R}^3$, we can conclude that $\exists \delta>0, \ \exists \ \phi:[-\delta,\delta] \rightarrow \mathbb{R}^2$ which is a unique solution of the Cauchy problem on that interval. I have some troubles while trying to compute the max interval of definition of the solution. Can someone suggest me how to procede?

Thanks in andvance for the help.

答:
Lutz Lehmann gives a good solution. Here is another one which uses argument by contradiction. Suppose there is $t_0\in(0,\infty)$ such that
$$ \lim_{t\to t_0^-}\|(x(t),y(t))\|=\infty. $$
Let $u=\|(x,y)\|^2$ and then (1) becomes
$$ \lim_{t\to t_0^-}\|u(t)\|=\infty. \tag1$$
Multiplying both sides of the firs, second equations by $x$ and $y$, respectively, one has, for $t\in[0,t_0)$,
\begin{eqnarray}
\frac12(x^2+y^2)'&=&tx+\frac{xy}{1+x^2}+txye^{-ty^2}\\
&\le& t_0|x|+\frac12|y|+t_0|x||y|\\
&\le&\frac12(t_0^2+x^2)+\frac14(1+y^2)+\frac{t_0}2(x^2+y^2) \tag2
\end{eqnarray}
Let
$$ a=t_0^2+\frac12, b=\frac12+t_0$$
and then (2) becomes
$$ u'\le a+bu. $$
It is easy to see
$$ u(t)\le u(0)e^{bt}+\frac{a}{b}(e^{-bt}-1). $$
Letting $t\to t_0^-$ gives
$$ \limsup_{t\to t_0^-}u(t)<\infty $$
which is against (1). So $t_0=\infty$.


问:
Topologies and manifolds
This question might seem philosophical a bit:

in a standard manifolds introductory course. when one talks about open , closed sets in $\mathbb{R}^n$ it's always the standard euclidian topology that we think about. Is there a more general framework in which one defines manifolds in different topologies in a way that generalize the standard setting that is closer to the physical world of course.

Thank you in advance for your answers

答:
Algebraic geometry essentially uses the same construction to define varieties — I won't go into details, but basically we develop the theory for the special case of affine varieties, then build more general algebraic varieties by gluing affine varieties together along open sets. The topology (the Zariski topology) used here is markedly different — for example, it's rarely ever Hausdorff because the open sets are far too large.

However, in a sense, it's not really a generalisation of the concept of manifold — we still have a canonical class of spaces which we glue together, and this class of spaces is fixed in advance for the whole subject. Moreover, although there is a way to make some varieties into manifolds, not every variety is a manifold, and not every manifold is a variety.

Title: Topologies and manifolds


问:
can we divide by any term when we have an differential homogeneous equation?
I am asking because i think we divided by x here for whatever reason since the other side is equal to 0 and it wont affect the equation in any meaningful way.

Letting $y=ux$ we have
  $$\begin{align}
(x-ux) dx + x(udx + x du) &= 0 \\
dx+ x du &= 0\\
\frac{dx}x+du&=0\\
\ln |x| + u &= c\\
x\ln |x|+ y &= cx.
\end{align}$$

I got $x/2 = y + c$ instead

The thing I don't understand though is how this is legal, because if it is it means there is an infinite number of solutions possible as we could also multiply both sides by anything.

答:
Yes, provided that term is non-zero.

Notice that after division by $x$, and a non-explicit integration, we had a $\ln |x|$ term which, just like $\frac{1}{x}$, is not well-defined when $x=0$.

Notice that several lines have been missed out. Going from the third to the fourth line, we had:

\begin{array}{ccc}
\frac{\operatorname{d}\!x}{x} + \operatorname{d}\!u &=& 0 \\ \\
\frac{\operatorname{d}\!x}{x} &=& -\operatorname{d}\!u \\ \\
\int \frac{\operatorname{d}\!x}{x} &=& -\int \operatorname{d}\!u \\ \\
\ln|x| &=& -u + c \\ \\
\ln|x| + u &=& c
\end{array}

The original question was in terms of $y$ and $y$ was swapped for $ux$. If $y=ux$ then $u=\frac{y}{x}$, so lets make the swap:

$$\ln|x| + \frac{y}{x} = c$$.

Since $x \neq 0$, we can multiply through by $x$ to give:

$$x\ln|x| + y = cx$$


问:
$ \sum \limits_{k=n+2}^{2n+7} \frac{1}{k}$ Check if there is a limit to that sum. Is my solution correct?
I needed to check if a given sequence has a limit (there is no need to find that limit)

$$ \sum \limits_{k=n+2}^{2n+7} \frac{1}{k}$$

I saw that it is impossible to evaluate the limit using Stolz lemma (since I don't have any sequence in the denominator. So I did the test of value-change in $\infty$ (I don't know how it's called, I am sorry):

$a_{n-1} = \frac{1}{n+2} + \frac{1}{n+3} + \frac{1}{n+4} + ... + \frac{1}{2n+6}$

$a_{n} = \frac{1}{n+2} + \frac{1}{n+3} + \frac{1}{n+4} + ... + \frac{1}{2n+7}$

I kewn that in nominators I have arithmetic sequences that are always $\geq0$ and growing.

So I got:

$$\lim_{n \to +\infty} \frac{a_{n}}{a_{n-1}} = \frac{\frac{1}{n+2} + \frac{1}{n+3} + \frac{1}{n+4} + ... + \frac{1}{2n+7}}{\frac{1}{n+2} + \frac{1}{n+3} + \frac{1}{n+4} + ... + \frac{1}{2n+6}} \implies \frac{0}{0}$$

And now I could use Stolz lemma:
$$\lim_{n \to +\infty} \frac{a_{n}}{a_{n-1}} = \frac{\frac{1}{2n+7}}{\frac{1}{2n+5}} = \frac{2}{2}=1$$

The idea standing by that solution was that if in $\infty$ the difference between 2 elements is equal to $0$, then the sequence limited (because 'new' elements are equal).

However, my friend told me that this solution is wrong because there are plenty of seqences with $\lim_{n \to +\infty} \frac{a_{n}}{a_{n-1}} = 0$ that are growing and are not limited. One of thouse would be $a_n = n$. So now I don't know if my solution is valid. I am actually worried that it's not.

But if it was wrong, then that test of $\lim_{n \to +\infty} \frac{a_{n}}{a_{n-1}} $ would be totally useless in limit calculations.

答:
You can use the theorem: every monotone bounded sequence converges.

First, this sequence is bounded: $$0<a_n=\sum_{k=n+2}^{2n+7}{\frac{1}{k}}<\frac{n+6}{n+2}=1+\frac{4}{n+2}<3.$$
Besides, it is decreasing:
\begin{align*}
a_{n+1}-a_n=\sum_{k=n+3}^{2n+9}{\frac{1}{k}}-\sum_{k=n+2}^{2n+7}{\frac{1}{k}}&=\frac{1}{2n+8}+\frac{1}{2n+9}-\frac{1}{n+2} \\
&=\left(\frac{1}{2n+8}-\frac{1}{2n+4}\right)+\left(\frac{1}{2n+9}-\frac{1}{2n+4}\right)<0.
\end{align*}
Then it must converge to some limit.


问:
How does one solve $\lim_{n\to\infty} \int_0^\infty \frac{\sin(x/n)}{x+x^2}dx$
I believe this converges point wise but I am not sure how to evaluate $$\lim_{n\to\infty} \int_0^\infty \frac{\sin(x/n)}{x+x^2}dx.$$ Can someone please help?

答:
So maybe this is another example of tricking the integral to allow the limit inside itself. We can do this in a lot of ways, but I'm very fond of leveraging dominated convergence theorem. Observe, for any $n,$ we see:
$$\int_0^{\infty} \frac{\sin(x/n)}{x + x^2} dx = \underbrace{\int_0^{\epsilon} \frac{\sin(x/n)}{x + x^2} dx}_{I} + \underbrace{\int_{\epsilon}^{\infty}  \frac{\sin(x/n)}{x + x^2} dx}_{II}$$

Consider the first integral. We know within an $\epsilon$-ball of the origin, we see:
$$\sin(x/n) = \frac{x}{n} + \frac{x^3}{3! n^3 } + \ldots $$
we can then evaluate
$$\int_0^{\epsilon} \frac{\sin(x/n)}{x + x^2} dx = \int_0^{\epsilon} \frac{\frac{x}{n} + \frac{x^3}{3! n^3 } + \ldots}{x + x^2} dx = \int_0^{\epsilon} \frac{\frac{1}{n} + \frac{x^2}{3! n^3}}{1 + x^2} dx \leq \frac{\epsilon}{n}$$
Clearly, as $n \rightarrow \infty,$ the integral decays to zero. So $I \rightarrow 0.$ Now consider the other integral.

$$II = \int_{\epsilon}^{\infty}  \frac{\sin(x/n)}{x + x^2} dx$$

Clearly $\left | \frac{\sin(x/n)}{x + x^2}\right| \leq \frac{1}{x+x^2} $ for all choices of $n$. Further, we know
$$\int_{\epsilon}^{\infty}  \frac{1}{x + x^2} dx < \infty$$
Therefore, by the dominated convergence theorem, we know $II$ converges. Since both converge uniformly, then we can see:
$$\lim_{n \rightarrow \infty} II =\lim_{n \rightarrow \infty} \int_{\epsilon}^{\infty}  \frac{\sin(x/n)}{x + x^2} dx = \int_{\epsilon}^{\infty} \lim_{n \rightarrow \infty}  \frac{\sin(x/n)}{x + x^2} dx =  \int_{\epsilon}^{\infty}  0 \ dx = 0$$

Therefore,
$$\lim_{n \rightarrow \infty} \int_{0}^{\infty}  \frac{\sin(x/n)}{x + x^2} dx =\lim_{n \rightarrow \infty} I + II = \lim_{n \rightarrow \infty} \frac{\epsilon}{n} + 0 = 0$$


问:
How to handle endpoint constraints in optimal control problem?
I am have problems trying to solve a question in my assignment with endpoint constraints on the state, after googling this problem I was unable to find any examples in optimal control theory dealing with endpoint constraints in a clear enough manner. To make my question more general as to not have my assignment solved directly I have reformulated it into the optimal control problem with two states defined below:

\begin{gather}
\max \int_0^Tf(u(t))\,dt \nonumber \\
\begin{aligned}[t]
&\text{subject to:}\quad x'(t)=y(t),\quad x'(0) = 0 \\
&\phantom{\text{subject to:}\quad} y'(t) = u(t),\quad y'(0) = 0\\
&\phantom{\text{subject to:}\quad}x(T)+y(T) \geq 2
\end{aligned}
\end{gather}

My approach:

I started by writing out the Hamiltonian:

$$ H(u(t),x(t),y(t),\lambda_1(t), \lambda_2(t))= f(u(t))+\lambda_1(t)y(t) + \lambda_2(t)u(t)$$

It is at this step that I'm stuck, I understand that it may be necessary to introduce Lagrange multipliers for the endpoint constraint but I don't know how.

If I do it directly I have:

$$ L(u(t),x(t),y(t),\lambda_1(t), \lambda_2(t), l(T))= f(u(t))+\lambda_1(t)y(t) + \lambda_2(t)u(t) + l(T)(x(T)+y(T)-2)$$

and I get a set of equations that are unrelated to each other and can't be solved. How should I deal with the endpoint constraint $x(T)+y(T) \geq2$ in this case?

答:
It's possible this approach will lead you to the same issue, but here's something that you might try.  Try a change of variables $z = x + y$.  This gives you
\begin{align}
z' &= x' + y' = y + u\\
z(0) &=x'(0) + y'(0) = 0\\
z(T) &=x(T) + y(T) \ge 2
\end{align}
and so your optimal control problem can be restated as
\begin{gather}
\max \int_0^Tf(u(t))\,dt \nonumber \\
\begin{aligned}[t]
&\text{subject to:}\quad z'(t) = y + u,\quad z'(0) = 0 \\
&\phantom{\text{subject to:}\quad} y'(t) = u(t),\ \ \quad y'(0) = 0\\
&\phantom{\text{subject to:}\quad}z(T) \geq 2
\end{aligned}
\end{gather}
Again, you might run into the same issue as before, but at least now your terminal constraint only includes one state variable.


答:
To evaluate the limit (more than is asked):
$$
\lim_{n\to\infty} \sum_{k=n+2}^{2n+7}\frac{1}{k}
= \lim_{n\to\infty} \int_n^{2n}\frac{dx}{x} = \ln 2 .
$$

Title: $ \sum \limits_{k=n+2}^{2n+7} \frac{1}{k}$ Check if there is a limit to that sum. Is my solution correct?

Body:
I needed to check if a given sequence has a limit (there is no need to find that limit)

$$ \sum \limits_{k=n+2}^{2n+7} \frac{1}{k}$$

I saw that it is impossible to evaluate the limit using Stolz lemma (since I don't have any sequence in the denominator. So I did the test of value-change in $\infty$ (I don't know how it's called, I am sorry):

$a_{n-1} = \frac{1}{n+2} + \frac{1}{n+3} + \frac{1}{n+4} + ... + \frac{1}{2n+6}$

$a_{n} = \frac{1}{n+2} + \frac{1}{n+3} + \frac{1}{n+4} + ... + \frac{1}{2n+7}$

I kewn that in nominators I have arithmetic sequences that are always $\geq0$ and growing.

So I got:

$$\lim_{n \to +\infty} \frac{a_{n}}{a_{n-1}} = \frac{\frac{1}{n+2} + \frac{1}{n+3} + \frac{1}{n+4} + ... + \frac{1}{2n+7}}{\frac{1}{n+2} + \frac{1}{n+3} + \frac{1}{n+4} + ... + \frac{1}{2n+6}} \implies \frac{0}{0}$$

And now I could use Stolz lemma:
$$\lim_{n \to +\infty} \frac{a_{n}}{a_{n-1}} = \frac{\frac{1}{2n+7}}{\frac{1}{2n+5}} = \frac{2}{2}=1$$

The idea standing by that solution was that if in $\infty$ the difference between 2 elements is equal to $0$, then the sequence limited (because 'new' elements are equal).

However, my friend told me that this solution is wrong because there are plenty of seqences with $\lim_{n \to +\infty} \frac{a_{n}}{a_{n-1}} = 0$ that are growing and are not limited. One of thouse would be $a_n = n$. So now I don't know if my solution is valid. I am actually worried that it's not.

But if it was wrong, then that test of $\lim_{n \to +\infty} \frac{a_{n}}{a_{n-1}} $ would be totally useless in limit calculations.


问:
Inner product space - simple proof
This seems like a simple proof but I haven't been able to do it.

Problem: Suppose $V$ is an inner-product space with $u, v \in V$, with $v \neq 0$. Provided that $u_{perp} = u - u_{par}$ and that $u_{par} = \frac{}{||v||^2}v$, prove that $<u_{perp}, u_{par}> = 0$.

My attempt:

Note that $<u_{perp}, u_{par}>$ = $$ = $ - <u_{par}, u_{par}>$ = $$

In a similar fashion, $<u_{perp}, u_{par}>$ = $<u_{perp}, u - u_{perp}>$ = $<u_{perp}, u> - <u_{perp}, u_{perp}>$ = $<u_{perp}, u>$

Therefore: $<u_{perp}, u_{par}>$ = $$ = $<u_{perp}, u>$

$$\implies  = <u_{perp}, u>$$

$$\implies  - <u_{perp}, u> = 0$$

$$\implies  = 0$$

$$\implies <u_{parr}, -u_{perp}> = 0$$

$$\implies -<u_{parr}, u_{perp}> = 0$$

$$\implies <u_{parr}, u_{perp}> = 0$$

$$\implies -<u_{perp}, u_{parr}> = 0$$

$$\implies <u_{perp}, u_{parr}> = 0$$

QED.

I can't help but think that I'm using wrong reasoning or my proof has some flaw. In fact, I'm pretty sure there is a flaw when expanding the inner products. In another attempt, I tried to use the fact that $||u_{perp} + u_{par}||^2 = ||u_{perp}||^2 + ||u_{par}||^2$ implies orthogonality, but the RHS here is $||u||^2$ whereas the LHS after expanding is $2<u_{par}, u_{par}> - 2 + ||u||^2$, which I'm unable to simplify further.

Any assistance or even hints are much appreciated.

答: $\langle u_{perp},u_{par} \rangle = \langle (u - \frac{\langle u,v \rangle}{\| v \|^{2}}\cdot v),\frac{\langle u,v \rangle}{\| v \|^{2}}\cdot v \rangle = \langle u,\frac{\langle u,v \rangle}{\| v \|^{2}}\cdot v \rangle - \langle \frac{\langle u,v \rangle}{\| v \|^{2}}\cdot v, \frac{\langle u,v \rangle}{\| v \|^{2}}\cdot v \rangle = \frac{\langle u,v \rangle}{\| v \|^{2}}\langle u,v \rangle - \frac{\langle u,v \rangle^{2}}{\| v \|^{4}}\langle v,v \rangle = \frac{\langle u,v \rangle}{\| v \|^{2}}\langle u,v \rangle - \frac{\langle u,v \rangle^{2}}{\| v \|^{4}}\|v\|^{2} = \frac{\langle u,v \rangle^{2}}{\| v \|^{2}} - \frac{\langle u,v \rangle^{2}}{\| v \|^{2}} = 0$

Title: Inner product space - simple proof


答:
The problem here is:
$$L=\lim_{x\to 0}\lim_{n\to\infty}\frac{\sin(x/n)}{x+x^2}$$
if we look at:
$$\lim_{n\to\infty}\frac{\sin(x/n)}{x+x^2}=\lim_{n\to\infty}\frac{-x\cos(x/n)}{x(x+1)n^2}=\lim_{n\to\infty}\frac{-\cos(x/n)}{(x+1)n^2}=-\frac{1}{x+1}$$
and so:
$$L=\lim_{x\to 0}\frac{-1}{x+1}=-1$$
so the integrand converges to a finite value for $x=0$ now if we look at the integral:
$$\int_0^\infty\frac{\sin(x/n)}{x+x^2}dx<\int_0^\infty\frac{x/n}{x+x^2}dx=\lim_{h\to\infty}\frac1n\int_0^h\frac{1}{1+x}dx$$
and so our integral, $I$, satisfies:
$$I\le\lim_{(n,h)\to(\infty,\infty)}\frac{\ln(1+h)}{n}=0$$

Title: How does one solve $\lim_{n\to\infty} \int_0^\infty \frac{\sin(x/n)}{x+x^2}dx$

Body: I believe this converges point wise but I am not sure how to evaluate $$\lim_{n\to\infty} \int_0^\infty \frac{\sin(x/n)}{x+x^2}dx.$$ Can someone please help?


问:
Matrix Norm Inequality $\lVert A\rVert_\infty \leq \sqrt{n} \lVert A\rVert_2$
So I'm trying to prove that

$\lVert A\rVert_\infty \leq \sqrt{n} \lVert A\rVert_2$.

I've written the right hand side in terms of rows, but this method doesn't seem to be getting me anywhere.

Where else should I go?

答:
Let $x\in \mathbb{R}^n_\infty$. We know that $\Vert y\Vert_\infty\leq \Vert y\Vert_2\leq\sqrt{n}\Vert y\Vert_\infty$ for all $y\in\mathbb{R}^n$, so
$$
\Vert A(x)\Vert_\infty\leq\Vert A(x)\Vert_2\leq\Vert A\Vert_2\Vert x\Vert_2\leq\Vert A\Vert_2\sqrt{n}\Vert x\Vert_\infty
$$
Since $x\in \mathbb{R}^n_\infty$ is arbitrary
$$
\Vert A\Vert_\infty\leq\sqrt{n}\Vert A\Vert_2
$$</p>


问:
What is the solution of $x^3+x=1$?
According to Wolfram|Alpha, the solution of $x^3+x=1$ is approximate $0.68233$ or exactly this monstrosity:

$x_0=\frac{\sqrt[3]{\frac{1}{2}(9+\sqrt{93})}}{3^{\frac{2}{3}}}-\sqrt[3]{\frac{2}{3(9+\sqrt{93})}}$

$x^3+x=1$ is so simple, that I refuse to believe that this ugly construct is the simplest way.  Am I right?

答:
Alternatively
$$x=\frac2{\sqrt3}\sinh\left( \frac13\sinh^{-1}\frac{3\sqrt3}2\right)$$
which may be less monstrous.

Title: What is the solution of $x^3+x=1$?


问:
Conditional probability and independent sigma algebras
Exercise: Given a random variable $X$ , sigma algebras $\mathcal{G},\mathcal{H}$ such that $\mathcal{H}$ is independent from both $\mathcal{G}$ and $X$ prove that $\operatorname{E}[X\mid\sigma(\mathcal{G},\mathcal{H})]=\operatorname{E}[X\mid\mathcal{G}]$.

I've proven that $\sigma(\mathcal{G},\mathcal{H})=\sigma(S)$ where $S=\{ H\cap G \vert H\in\mathcal{H},G\in\mathcal{G}\}$ .

Obviously $\operatorname{E}[X\mid\mathcal{G}]$ is $\sigma(\mathcal{G},\mathcal{H})$-mesurable.  I've also proven that $\forall A\in S ,\ \operatorname{E}[X\mid\mathcal{G}]$ satisfies the conditional expectation property, i.e. $\int_AX=\int_A\operatorname{E}[X\mid\mathcal{G}]$

I would like to know if i can conclude using a criteria for coincidence of finite measures:

given $A\in\sigma(\mathcal{G},\mathcal{H})$ the maps that send $A \rightarrow \int_AX$ and $A \rightarrow \int_A\operatorname{E}[X\mid\mathcal{G}]$ are well defined finite measures on $\sigma(\mathcal{G},\mathcal{H})$ and both coincide on $S$ which is a set of generators for $\sigma(\mathcal{G},\mathcal{H})$ and it is also stable under finite intersection, thus such measures coincide on $\sigma(\mathcal{G},\mathcal{H})$ which is equivalent with what i want to show.

Thank you

答:
Yes, the criterion you are applying is sometimes known as the Uniqueness Lemma:

Let $\mathcal I$ be a $\pi$-system. If two finite measures agree on $\mathcal I$, then they agree on $\sigma({\mathcal I})$.

A $\pi$-system is a collection of sets that is closed under finite intersection, which is true for your family $S$.


问:
Uniqueness of plurisubharmonic functions
In complex analysis there is a uniqueness theorem as,

For holomorphic functions $f,g: \Omega \to \mathbb C$ and domain $\Omega$ in $\mathbb C^n$, if $f=g$ on $D$; then $f=g$ on $\Omega$, where $D$ open in $\Omega$.

Is preceding theorem true for real valued $\mathscr{C}^2$ plurisubharmonic functions instead of holomorphic functions?

答:
Think about the special case that $\Omega = \mathbb{C}$ and $D = \{z: |\textbf{Re}(z)| < 1 \}$.  Take $f$ to be the constant zero function.

In this context, a function is plurisubharmonic iff it is subharmonic, which is to say that it has a nonnegative Laplacian.  Can you rig up a subharmonic function on $\mathbb{C}$ which is zero on $D$, but not elsewhere?  Hint:  if $z = x+iy$, try to find one which is a function of $x$ only.


问:
Show that isometries of $\mathbb{S^2}$ are determined by the image of three points $A,B,C$ not in a "line".
Given information: If points $P,Q \in \mathbb{S^2}$ have the same distace from three points $A,B,C \in \mathbb{S^2}$ not in a "line", then $P=Q$.

Deduce from the given information that an isometry of $\mathbb{S^2}$ is determined by the images of the three points $A,B,C$ not in a "line".

My understanding of the given information is that the points that are equidistant from $A,B,C$ are given by the intersection of three great circles. These three great circles intersect at two points, one is between the triangle spherical triangle $A,B,C$ and the other is it's antipodal point on the other side of the sphere. So we know that any point that is equidistant from $A,B,C$ by a given distance $k$ is unique. But how does this relate to isometries of $\mathbb{S^2}$?

答:
Let $d(X,Y)$ be the distance between points $X$ and $Y$.

Then the given information is saying that $d(P,A)=d(Q,A), d(P,B)=d(Q,B),$ and $d(P,C)=d(Q,C)$.  It's not saying that $P$ is equidistant from $A,B,C$ (and same for $Q$).

You are being asked to show that a point is uniquely determined by its distances from three points not in a line.

(Hint: assume that two distinct points $P$ and $Q$ have the same distances to three points not on a line, and see if you can get a contradiction).

(Hint 2: Draw the line that is equidistant from $P$ and $Q$.  Where are $A,B,C$ relative to that line?)


问:
Remainders problem
What will be the reminder if $23^{23}+ 15^{23}$ is divided by $19$?
 Someone did this way:

$15/19 = -4$ remainder and $23/19 = 4$ remainder So $(-4^{23}) + (4^{23}) =0$ but i didn't understand it

Title: Remainders problem

Body:
$$\forall\,\text{prime}\;p\;\wedge\;\forall\,a\in\Bbb Z\;,\;(a,p)=1\implies a^{p-1}=1\pmod p$$

$$23=4\pmod{19}\;\wedge\;23^{19}= 4^{19}=4\pmod{19}\;,\\\;15=-4\pmod{19}\implies 15^{19}=15=-4\pmod {19}$$

Thus (all the following is onde modulo $\,19\,$ ):

$$23^{23}+15^{23}=4^5+(-4)^5=0$$


问:
showing $\mu(\lim\sup_{n\to\infty} E_n) = 0.$
Let $(X,\mathcal{M},\mu)$ be a measure space. Suppose $E_n\in \mathcal{M}$ such that $$\sum_{n=1}^\infty \mu(E_n) < \infty$$ show $\mu(\lim\sup_{n\to\infty} E_n) = 0.$

Also, how can I prove or give a counterexample of the conclusion if the hypothesis is replaced with $$\sum_{n=1}^\infty \mu(E_n)^2 < \infty$$

Can I prove this by Fatou's Lemma? Thank you

答:
It is a very general and very useful theorem of calculus that if $\sum a_i < \infty$ then it must be that $\lim\sup a_i = 0$. To get your result from this theorem, simply apply subadditivity.

For the counter example, can you think of a series such that $\sum a_i^2$ converges
and $\sum a_i$ diverges?


问:
Is $\sqrt{n}(\bar X_n - \bar X_m)$ tight?
My question is a bit more specific than the title.

Suppose $\{X_i\}$ is iid with finite second moments, and suppose we select some random subset $D_n \subset \{1, \dots, n\}$ such that $n_d = |D_n|$, and suppose that $D_n$ is independent of $\{X_i\}_{i=1}^n$. Moreover, suppose that $n_d/n \xrightarrow{p} c$ for some constant $c \in (0,1)$.

Now, is it true that
$$\sqrt{n}\left(\frac{1}{n}\sum_{i=1}^n X_i - \frac{1}{n_d}\sum_{i \in D_n} X_i\right)= O_p(1)$$

I know that $\frac{1}{n}\sum_{i=1}^n X_i - \frac{1}{n_d}\sum_{i \in D_n} X_i = o_p(1)$, and I basically want to know if $\frac{1}{n}\sum_{i=1}^n X_i - \frac{1}{n_d}\sum_{i \in D_n} X_i = O_p(n^{-1/2})$

答:
Clarification I understand your big O notation as probabilistic big O notation

Short Answer

Yes, you have the right speculation.

And, if my calculations are right, your speculation is still valid if $(n_d;n \ge 1)$ is a sequence of random variables such that :

$n_d/n$ is bounded below by a real number $a>0$

$n_d/n$ converges in law (as a sequence of real random variables)

Main idea

We'll prove the result the same way that we've done for the central limit theorem. That is, by characteristic functions and Levy's continuity theorem.

Long answer
WLOG: $\mathbb{E}(X_1)=0, \mathbb{E}(X^2)=1$

Let $\phi$ be the characteristic function of $X$, that is:
$$\phi(\lambda) = \mathbb{E}( e^{i\lambda X_1})$$

The finiteness of the second moment of $X$ implies that:
$$\phi(\lambda) = exp ( \lambda^2(-1+o_{\lambda}(1)))$$
where $|o_{\lambda}(1)| \le h( |\lambda|)$ for some increasing function $h$ bounded around $0$ and $h(\lambda) \xrightarrow{\lambda \rightarrow 0^+} 0$

($o_\lambda$  can be complex but doesn't matter)

Let

$d^n_i:= 1_{ i \in \mathbb{D_n}}$

When there is no confusion, by abuse of notation, we'll let $d_i= d^n_i$

$Y_n$ is the random quantity that we want to study, i.e
$$Y_n:= \sqrt{n}\left( \dfrac{1}{n}\left(\sum_{k=1}^n X_k\right) - \dfrac{1}{n_d}\left(\sum_{k \in D_n} X_k\right) \right)$$
So clearly, $Y_n$ can be written as
$$Y_n = \sqrt{n} \left(\sum_{k=1}^n \left(\dfrac{1}{n}-\dfrac{d_k}{n_d}\right)X_k\right)$$
So, by indepedence between $D_n$ and $(X_i)$, for all $\lambda$ real, we have:
$$\phi_{Y_n}(\lambda) = \mathbb{E}\big( \prod_{k=1}^n \phi\underbrace{\left( \lambda \dfrac{1}{\sqrt{n}}-\lambda \sqrt{n}\dfrac{d_k}{n_d} \right)}_{ =o(1/\sqrt{n}) \text{regardless of the value of} d_i} \big)$$

Thus, by the remark we have to $\phi$, we imply that
$$\phi_{Y_n}(\lambda)  = \mathbb{E}\left( exp\left[ \sum_{k=1}^n \left(\lambda \dfrac{1}{\sqrt{n}}-\lambda \sqrt{n}\dfrac{d_k}{n_d} \right)^2(-1+ o(1)) \right] \right)$$

where $|o(1)| \le h( \frac{b|\lambda|}{\sqrt{n}})$ for every $o(1)$ and some $b$ depending only on $a$.

As $ \left(\lambda \dfrac{1}{\sqrt{n}}-\lambda \sqrt{n}\dfrac{d_k}{n_d} \right)^2 \ge 0$, we have:

$$ \sum_{k=1}^n \left(\lambda \dfrac{1}{\sqrt{n}}-\lambda \sqrt{n}\dfrac{d_i}{n_d} \right)^2(-1+ o(1)) = (-1+o(1))\underbrace{\left[ \sum_{k=1}^n \left(\lambda \dfrac{1}{\sqrt{n}}-\lambda \sqrt{n}\dfrac{d_k}{n_d} \right)^2 \right]}_{ = \lambda^2(\dfrac{n}{n_d}-1)} $$

Thus
$$ \phi_{Y_n}(\lambda) = \mathbb{E}\left[ e^{ -(1+o(1))\lambda^2( n/n_d-1)} \right] \xrightarrow{n \rightarrow +\infty} e^{-\lambda^2(1/c-1) }$$

Therefore, by Levy's theorem, we have:
$$ \mathcal{L}( Y_n ) \longrightarrow \mathcal{N}( 0, 1/c-1)$$

Hence the conclusion.
$\square$


问:
Geometry - Equilateral triangle covered with five circles
I have to cover an equilateral triangle (whose sides are 1m long) with 5 identical circles: what's the minimum radius of the circles?</p>

答: Here is a solution where $r \approx 0.251$. This is not optimal, but is very close. Finding a precise, or non-numerical, solution of this sort is an exercise in trigonometry and algebra.


问:
What is an example of a bijection from unlabeled graphs to the positive integers?
Suppose $f$ is a bijection from the set of unlabeled graphs--i.e. isomorphism classes of graphs -- to the positive integers $\mathbb{N}$. Due to the isomorphism graph problem, I understand that trying to compute $f(G)$ or $f^{-1}(n)$, where $G$ is a graph, would require a lot of computations, maybe on the order of $O(2^{n^2})$. I was wondering if someone could give me an example of such a function. Clearly such a function exists since there are countably many unlabeled graphs.

答:
Here's another way of thinking about it: with orderings, rather than encodings.

First, define a well-ordering of all labeled graphs on vertex sets of the form $\{1, 2, \dots, n\}$. This well-ordering will say that $G \prec H$ if

$G$ has fewer vertices than $H$, or

$G$ has the same number of vertices as $H$, and if we sort the edges in $E(G) \mathbin{\Delta} E(H)$ in lexicographic order, the first edge will be an edge of $G$.

Second, use $\prec$ to get another well-ordering of unlabeled graphs, which we'll also call $\prec$. To compare two unlabeled graphs $G$ and $H$:

Let $\mathcal G$ be the set of all labeled graphs on vertex set $\{1, 2, \dots, |V(G)|\}$ isomorphic to $G$; let $\mathcal H$ be the corresponding set for $H$.

Let $\min \mathcal G$ be the minimum element of $\mathcal G$ with respect to $\prec$, and the same for $\min \mathcal H$.

We say $G \prec H$ if $\min \mathcal G \prec \min \mathcal H$.

Finally, define a map $f$ from the set of all unlabeled graphs to $\mathbb Z^+$ by letting $f(G)$ be the number of unlabeled graphs that precede $G$ with respect to $\prec$, plus $1$.


问:
Can an isomorphism of subgroups of a finite group always be extended to an automorphism on the group?
Consider the problem:

Let $G$ be a finite group. Let $H, K$ be arbitrary subgroups of $G$. Let $f: H\to K$ be an isomorphism on groups. Can we always construct $F\in\text{Aut}\,(G)$ such that $F_{|H}=f$?

I am quite sure the answer is no. To this end, I am seeking the following:

A concrete example of when this fails.

A characterization of groups where this condition fails/does not fail from a representation-theoretic perspective (if it exists).

I have searched around a bit and not come up with much else other than this thread on MO and this thread on MSE from ten years ago, so links to (recent?) research papers are very much appreciated!

Edit: I am now aware of a paper by Pettet and another one by Schupp both of which assert that an automorphism $\phi: G\to G$ is extendable to an automorphism $\psi: E\to E$ where there is a natural inclusion $\iota: G\to E$ if and only if $\phi$ is inner. I suppose what I am seeking now is an equivalent formulation and proof of this problem (if one exists) from a representation-theoretic perspective. I would also be interested in approaches making use of homological algebra.

答: The cyclic subgroups generated by $(12)$ and $(12)(34)$ in $S_4$ are isomorphic, but there is no automorphism of $S_4$ sending one to another: every automorphism of $S_4$ is inner and $(12)$ and $(12)(34)$ are not conjugate, being of different cycle types.


问:
Norm of a linear form in a function space
I would appreciate some help with the following problem. Let $u, u_n:C([0,1])\to\mathbb{R}$ the linear forms defined by $$u(f)=\int_0^1f(x)dx\quad\&\quad u_n(f)=\frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n}\right).$$ I'm trying to prove that $$||u_n-u||=\sup_{||f||_\infty=1}|u_n(f)-u(f)|=2.$$

I already have that $||u_n-u||\leq2$ and I've tried to show that the $\sup$ is attained (I'm not sure that it is true) by using trigonometric functions but I haven't been able to get anything. Could anyone give me a hint?

Thanks in advance.

答:
You indeed can do it using trigonometric functions. Following the suggestion of @Berci, define
$$f_k(x) = -1+2|\sin(n\pi x)|^{1/k}$$
so that $u_n(f_k) = -1$ for all $k \in \Bbb{N}$. On the other hand, $f_k \to 1$ almost everywhere so by the Lebesgue dominated convergence theorem we have
$$\lim_{k\to\infty} u(f_k) = \lim_{k\to\infty}\int_0^1 f_k(x)\,dx = \int_0^1 \lim_{k\to\infty} f_k(x) = 1.$$
Hence $$\|u_n-u\| \ge \lim_{k\to\infty} \frac{|u_n(f_k)-u(f_k)|}{\|f_k\|_\infty} = 2.$$


问:
Discrete Valuation Ring and Subring of the Fractions Field
Let $R$ be a Discrete Valuation Ring, and $K$ its fractions field. Now if $B\subseteq K$ is a subring with $R\subseteq B$ then we have $$B=R \text{ or } B=K.$$

Now this seems to be a very basic fact however I am just unable to prove it.  Well after a hint from YACP: we can look at the at the uniformizer $p$ i.e. $v(p)=1$ - where $v$ is the valuation. We know for every element $\alpha \in K $ exists an $i\in \mathbb{Z}$ so that $ \alpha = u \cdot p^{i}$. Also we have that $K=S^{-1}R$ where $S=\{1,p,p^2,...\}$. I just can not say anything about $B\neq R$.:(

That $R$ is principal ideal domain with maximal ideal $m=(p)$ keeps coming up but I am unable to use it.

答: As $R$ contains exactly the elments $a\in K$ with $|a|\le 1$, the ring $B$ must contain some element $|b|$ with $|b|>1$. If $x\in K$ is arbitrary, therefore $|x|<|b|^n$ for some $n$ and the element $\frac{x}{b^n}$ is in $R$, hence  $x=\frac x{b^n}\cdot b^n$ is in $B$.</p>


问:
Does the strong law of large numbers imply the convergence of moments of multivariate empirical distribution?
Suppose, $X_1, \dots, X_n$ are iid random vectors with some distribution $P$ on $\mathbb{R}^d$ and suppose $E_P(|X_1^k|)$ exists for all $k \leq K$ for some $K > 1$. Now, let $\hat P_n$ be the empirical distribution of $X$, constructed by putting weight $1/n$ on each observed $X_i$; let $F_n$ be its cdf.

If $\dim(X) = 1$, then by Glivenko-Cantelli, $\Vert F_n - F\Vert_\infty \rightarrow 0$ almost surely, and so, $E_{\hat P_n}(X^k) \rightarrow E_P(X^k)$, almost surely, for all $k < K$.

My question is, does this generalize to multiple dimensions? I know that the Glivenko-Cantelli result does not generalize directly without additional restrictions on $P$, but does almost sure convergence of moments hold for general $P$? If not, what kind of conditions are sufficient?

What about coordinate product moments? i.e. do we have $E_{\hat P_n}(X_{J}^kX_{J}^{k'}) \rightarrow E_{P}(X_{J}^kX_{J}^{k'})$ almost surely, where now $X_{J}$ represents the the sub-vector $(X^{(j)})_{j \in J}$ for some $J \subset \{1, \dots, \dim(X)\}$

EDIT: Does this simply follow from the Strong Law of Large numbers?

We should have
$$E_{P_n}(X^k) = \sum_{i=1}^n X_i^k \xrightarrow{a.s.} E_{P}(X^k)~,$$
since $\{X_i\}_{i=1}^n$ is an iid sequence from $P$ and $E(|X^{k}|)< \infty$. Is this correct? I think this also directly generalizes to the coordinate product moments, no?

Changed title to reflect this.

答:
While the almost sure uniform convergence for empirical distribution functions is not clear.

The almost sure convergence for moments is pretty straightforward.

Let $\alpha$ be any multi-index in $\mathbb{N}^d$ such that $|\alpha|\le K$

You can see that:

$$\mathbb{E}_{\hat{P}_N}(X^{\alpha}) = \dfrac{1}{N}\left( \sum_{n=1}^N X^{\alpha}_n \right)$$

Clearly, $(X^{\alpha} ,n \ge 1)$ is a sequence of integrable real random variables.

Hence, by the strong law oflarge number, we have:
$$ \dfrac{1}{N}\left( \sum_{n=1}^N X^{\alpha}_n \right) \longrightarrow \mathbb{E}_P(X^{\alpha}) \quad \text{a.e}$$

Hence, your conclusion. $\square$


问:
Pythagorean Triples, $a^2+b^2=c^2$ and $b^2+c^2=d^2$. Is it possible?
I'm trying to solve a problem, and it would be solved if I could find a solution to the above system, $a^2+b^2=c^2$ and $b^2+c^2=d^2$  for natural numbers.

答: This is impossible by Fermat's right triangle theorem.</p>


问:
smoothness of an inclusion map proof verification
I know this problem was asked before. My solution is different. I want to verify my solution.

Let $M$ and $N$ be manifolds and let $q_0$ be a point in $N$. Prove that the inclusion map $i_{q_0} : M \to M×N : p \mapsto (p,q_0)$, is $C^\infty$.

Solution attempt:

Let $(U,\phi)$ and $(V,\psi^{\prime})$ be charts on M and $M \times N$ respectively. Charts on $M \times N$ are of the form $(U_1\times V_1, \psi_1 \times \psi_2)$. Therefore we can take the following chart $(U \times V_1, \phi \times \psi_2) = (U \times V_1,\psi^{\prime})$.

It follows that:

$(\psi^{\prime} \circ F \circ \phi^{-1})(\phi(p)) = (\psi^{\prime} \circ F)(p) = \psi^{\prime}((p,q_0)) = (\phi(p),\psi_2(q_0))$.

Since components $\psi_1$ and $\psi_2$ are $C^{\infty}$ it follows F is $C^{\infty}$.


答:
Example 1. A normal subgroup isomorphic to a nonnormal subgroup.

$H=\{(1),\ (1\ 2)(3\ 4),\ (1\ 3)(2\ 4),\ (1\ 4)(2\ 3)\}$ and $K=\{(1),\ (1\ 2),\ (3\ 4),\ (1\ 2)(3\ 4)\}$ are isomorphic subgroups of $S_4$, but $H$ is a normal subgroup of $S_4$ and $K$ is not. An automorphism of a group must take normal subgroups to normal subgroups.

Example 2. An Abelian example.

$G=\langle(1\ 2\ 3\ 4),\ (5\ 6)\rangle$ is an Abelian group of order $8$. The subgroups $H=\langle(1\ 2\ 3\ 4)^2\rangle$ and $K=\langle(5\ 6)\rangle$ are isomorphic subgroups of order $2$. No automorphism of $G$ maps $(1\ 2\ 3\ 4)^2$ to $(5\ 6)$, since the odd permutation $(5\ 6)$ is not a square; also, the quotient groups $G/H$ and $G/K$ are nonisomorphic.

Title: Can an isomorphism of subgroups of a finite group always be extended to an automorphism on the group?

Body:
Consider the problem:

Let $G$ be a finite group. Let $H, K$ be arbitrary subgroups of $G$. Let $f: H\to K$ be an isomorphism on groups. Can we always construct $F\in\text{Aut}\,(G)$ such that $F_{|H}=f$?

I am quite sure the answer is no. To this end, I am seeking the following:

A concrete example of when this fails.

A characterization of groups where this condition fails/does not fail from a representation-theoretic perspective (if it exists).

I have searched around a bit and not come up with much else other than this thread on MO and this thread on MSE from ten years ago, so links to (recent?) research papers are very much appreciated!

Edit: I am now aware of a paper by Pettet and another one by Schupp both of which assert that an automorphism $\phi: G\to G$ is extendable to an automorphism $\psi: E\to E$ where there is a natural inclusion $\iota: G\to E$ if and only if $\phi$ is inner. I suppose what I am seeking now is an equivalent formulation and proof of this problem (if one exists) from a representation-theoretic perspective. I would also be interested in approaches making use of homological algebra.


问:
effective way to get the integer sequence A181392 from oeis
the sequence A181392 are perfect squares and any digit in the sequence says
"I am part of an integer in which you'll find d digits "d"" (see A108571, How can we call them? "digit-valid"?)

How to get all these number in sequence A108571?
This is maybe (hopefully) more a theoretical problem instead of using brute-force.
I know two approaches:

compute sequence A108571 and check if the number is a perfect square (use this topic)

test every perfect square if it has the correct "format"

The first one seems to be very slow (there are many numbers in A108571= multinom.coeff. (45,9,8,7,6,5,4,3,2,1)). Of course we can create a suffixset containing the last 5 digits numbers of any squares, and put digits before the suffix such that the number is in sequence A108571, permute the all digits except the last 5 digits and test whether it is a perfect square.

The second approach has the advantage that there are "only" $\sqrt{999999999888888887777777666666555554444333221}$ numbers to test.

Are there some theorems I can use? Hensel's lemma gives me a statement about the three last digits of every odd square. So it would be more effective to use store the last 5 digits and start from there.

Is there another tricky theorem or statement I can use?

答:
Define a string of nonzero digits to be $n$-consistent if when read in base 10, for all $1 \leq d \leq 9$ it has at most $d$ digits equal to $d$ and is congruent to a square mod $10^n$.

Build a tree whose root is the empty string and whose edges represent extensions of a $k$-consistent number to a $(k+1)$-consistent number, recording square integers (in A181392) as they appear.  The tree can be built depth-first and fully explored branches deleted. It is useful to keep track of the powers of $2$ and $5$ dividing each number and maybe the part prime to $10$, and probably also to have computed the nontrivial 2-adic square roots of 1 to mod $2^{45}$ precision.

Some thought may be needed on how to do the mod $10^k$ liftings using the base 10 representation without recomputing in base $2$ and $5$.


问:
Is $\mathbb R$ with usual topology a Hausdorff space?
$\mathbb R$ with usual topology is a Hausdorff space. By the definition of Hausdorff space, for any two distinct point in $\mathbb R$ we can find  disjoint neighbourhoods... Now consider the set $A$=$\{{{1}\over{n}}\}$  where  $n\in \mathbb N$. We know that the sequence $\{{{1}\over{n}}\}$ converges to "$0$". Clearly $0$ does not belong to $A$, that is every point in $A$ and the point $0$ are distinct. Hence by definition of Hausdorff space we can find a neighbourhood of $0$ which does not intersect $A$. This shows that $0$ is not a limit point. This is contradiction to ${{1}\over {n}}\rightarrow 0$. Is my opinion is correct? please spot where i make mistake?
my thought: consider every point in $A$ and the point $0$, we can get disjoint neighbourhoods because they are disjoint. In this way of thinking I am getting lot of complications.. what can I do?</p>

答:
given $a, b \in \mathbb{R}$, where $a<b$, one can find a middle point so that

$$
a< \frac{a+b}{2} <b
$$

so

$( 2a - \frac{a+b}{2}, \frac{a+b}{2} )$ is centered at $a$

$( \frac{a+b}{2}, 2b-\frac{a+b}{2} )$  is centered at $b$.

the two intervals are open in order topology (the standard topology defined in $\mathbb{R}$) in $\mathbb{R}$.

the two intervals are disjoint.

For $0$ and $1/n$, the two intervals are $( -\frac{1}{2n},\frac{1}{2n} )$ and $( \frac{1}{2n},\frac{3}{2n} )$.

I suppose your mental struggling is whether $\frac{1}{2n} < \frac{1}{n}  $ when $n \rightarrow  \infty$ so both terms "become" zero. Probably this has puzzled crazy people like Dedekind or Cantor.


问:
Eigenvalues of laplacean matrix of directed graph are all non-zero when calculated in mathematica
I have a problem where I am given the incidence matrix of a directed graph, and need to calculate its Laplacean matrix and respective eigenvalues. Since it is a relatively large graph, it's reasonably suggested to do this using software. The incidence matrix I am given, formatted for input in mathematica, is:

```<code>{{0,0,0,0,0,0,1,0,0,0},{1,0,0,0,0,0,0,0,0,0},{0,1,-1,0,-1,0,0,1,0,0},{-1,0,0,0,0,1,0,0,0,1},
{0,0,0,0,0,-1,-1,0,1,0},{0,-1,0,0,0,0,0,0,0,0},{0,0,0,-1,0,0,0,0,0,0},{0,0,1,0,0,0,0,0,0,0},
{0,0,0,1,1,0,0,0,0,-1},{0,0,0,0,0,0,0,0,-1,0},{0,0,0,0,0,0,0,-1,0,0}}
</code>```

It's my understanding that the Laplacean matrix should have at least one zero eigenvalue. But when I calculate in Mathematica, the set of eigenvalues I get is {4,3,3,3,1,1,1,1,1,1,1}. I do this by:

```<code>m = {{0,0,0,0,0,0,1,0,0,0},{1,0,0,0,0,0,0,0,0,0},{0,1,-1,0,-1,0,0,1,0,0},{-1,0,0,0,0,1,0,0,0,1},
    {0,0,0,0,0,-1,-1,0,1,0},{0,-1,0,0,0,0,0,0,0,0},{0,0,0,-1,0,0,0,0,0,0},{0,0,1,0,0,0,0,0,0,0},
    {0,0,0,1,1,0,0,0,0,-1},{0,0,0,0,0,0,0,0,-1,0},{0,0,0,0,0,0,0,-1,0,0}}
g = IncidenceGraph[m]
k = KirchoffMatrix[g]
Eigenvalues[k]
</code>```

Some of my colleagues have tried this in different mathematical software (Sagemath, numpy) and we have all arrived at different results for the eigenvalues, despite checking that we didn't have any typographical errors. What is behind this discrepancy, and what is the correct method to approach this problem using software?

答:
You are arriving at different results that don't have the properties you expect because the Laplacian matrix of a directed graph

does not have a standard definition, and

does not have all the nice properties we expect from the Laplacian matrix of an undirected graph.

For an undirected graph, the definition of the Laplacian matrix is straightforward: its $(i,i)$ entry is the degree of vertex $i$, and its $(i,j)$ entry for $i\ne j$ is $-1$ if there is an edge $ij$ and $0$ otherwise. There are lots of ways we can try to generalize this to directed graphs:

When we have an edge $i \to j$, should we put a $-1$ in the $(i,j)$ entry and a $0$ in the $(j,i)$ entry? Or a $+1$ in the $(i,j)$ entry? Or maybe $-1$ in both? (Mathematica makes the first choice.)

Should the $(i,i)$ entry record the out-degree of vertex $i$? Or the in-degree? Or maybe their sum, or their difference? (Mathematica takes the sum.)

In the case of the undirected Laplacian, there is an eigenvalue of $0$ because the all-$1$ vector is an eigenvector: to put it differently, the sum of every row is $0$. If each edge $i \to j$ puts a $-1$ in entry $(i,j)$ and a $0$ in entry $(j,i)$, then the diagonal should record only the out-degree for this to work.

But we might have to make different design decisions to preserve different properties of the undirected Laplacian matrix. So you should first decide what you want the Laplacian matrix to do, and then figure out if there is a way to define it so that it has those properties.


问:
Algebraically Evaluate Limit
The following function is half the perimeter of a regular $n$-gon with circumradius $1$ where $x$ is the number of subdivisions in half following a hexagon (when $x=1$, $n$ is $12$):

$p(x)=6 (2^{x-1})\sqrt{2-2\cos(\frac{\pi}{3(2^x)})}$

Geometrically and numerically, it is obvious that $\lim \limits_{x \to \infty} p(x) = \pi$.

How can you prove this algebraically?

As a bonus question, is there any closed form expression with no trig functions that approaches $\pi$? This came up in an attempt to find one. It failed, but I did find out an interesting recurrence relation:

$q(x)=\sqrt{2-2\cos(\frac{\pi}{3(2^x)})}$

satisfies:

$q(x+1)=\sqrt{2-\sqrt{4-q(x)^2}}$

at least when $x$ is a non-negative integer.

答:
Let's do a couple algebraic manipulations to get: $p(x)=6\sqrt{2}\cdot 2^{x-1}\sqrt{1-\cos\left(2\cdot \frac\pi{12\cdot2^{x-1}}\right)}$

Now, if we let $y=2^{x-1}$, then $y\to\infty$ as $x\to\infty$ and so our limit is equal to $L=\lim_{y\to\infty}6\sqrt 2\cdot y\sqrt{1-\cos\left(2\cdot\frac\pi{12y}\right)}$

But $\cos(2\theta)=1-2\sin^2(\theta)$ and so $L=\lim_{y\to\infty}6\sqrt{2}\cdot y\cdot\sqrt{2\sin^2\left(\frac\pi{12y}\right)}=\lim_{y\to\infty}12y\sin\left(\frac\pi{12y}\right)$

One last substitution in case it isn't clear yet, $t=\frac1y$ so that $t\to0$ as $y\to\infty$:

$$L=12\lim_{t\to0}\frac{\sin\left(\frac\pi{12}t\right)}t=12\cdot\frac\pi{12}=\boxed{\pi}$$

As for your bonus question, it really does depend on what you mean by "closed form." I would argue that $\pi=\frac{\ln(-1)}i$ is a closed form expression. $\pi=2\displaystyle\int_{-1}^1\sqrt{1-t^2}dt$ is also perfectly valid and can be converted into a limit of a Riemann sum. Generally, however, the fundamental property of $\pi$ as $\frac1{2i}$ of the period of $\exp$ means that you'll probably want to invoke the complex exponential (or the trig functions that are just the complex exponential in disguise) in your "closed-form" limits for $\pi$.

Hope this helps!

Title: Algebraically Evaluate Limit


问:
Prove the sum of integrals of function equals to the integrals of sum of functions
Let $(f_n)$ be a countable sequence of functions in $M^+(S,S)$, and let $g=\sum_nf_n$. Prove $\int gd\mu=\sum_n\int f_nd\mu$

答:
For $N \in \mathbb{N}$, define $g_N = \sum\limits_{i=1}^{N} f_i$. Then since each $f_i \geq 0$, we have $g_N \nearrow \sum\limits_{i=1}^{\infty}f_i$, so by the monotone convergence theorem,
\begin{align}
   \int \sum_{i=1}^{\infty} f_i d \mu &= \int \lim_{N \to \infty} g_N d \mu = \lim_{N \to \infty} \int g_N d \mu = \lim_{N \to \infty} \int \sum_{i=1}^{N} f_i d\mu \\
   &= \lim_{N \to \infty} \sum_{i=1}^{N} \int f_i d\mu = \sum_{i=1}^{\infty} \int f_i d\mu
  \end{align}


问:
Find the area of the region enclosed by the curves $y = \tan(x)$, $y = -3\tan(x)$ and $x = \pi/4$
Find the area enclosed by the curves $y = \tan(x)$, $y = -3\tan(x)$ and $x = \pi/4$.

What can possibly be the area, $\tan(x)$ and $-3\tan(x)$ never intersect unless it is $\frac{\pi}{2}k$. There is no upper bound to find because the two curve goes to different direction, so confused about this question.

答:
HINT

The integral you are interested in can be written as
\begin{align*}
I = \int_{0}^{\pi/4}(\tan(x) - (-3\tan(x)))\mathrm{d}x = \int_{0}^{\pi/4}4\tan(x)\mathrm{d}x
\end{align*}

Can you take it from here?