问:
What is the analogue to a (directed) path in graph theory called in the theory of relations?
My level is undergrad-master with little or no formal knowledge of relation or order theory, but with some graph theory.

Let $R$ be a binary relation on a finite set $X$. I was looking for terminology and notation for describing a set like

$P=\{(x_1,x_2),(x_2,x_3),(x_3,x_4),\dots,(x_{k-1},x_k)\}$,

i.e. in a sense a directed path if this was graph theory.

For orders I've heard of chains being sets of pairwise comparable elements. But, as I understand it this requires transitivity, e.g. $P$ above is not a chain. I thus see chain mostly (if not only) defined for partial orders, which makes sense.

I also found the definition of a fence in for example Schroeder's Ordered Sets (Def. 2.37). He mentions that they can be seen as analogues to paths in graph theory. However, they are paths with alternating orientation. I guess what I was after was an analogue of a directed path, that does not require transitivity.

Questions:

Is there a terminology and/or notation for sets like $P$ above purely for relations, akin to the one of directed paths in a graph? Preferably, I'd simply denote it as $x_1x_2x_3\dots x_{k-1}x_k$, and even concatenate elements and "paths" into new ones like $yPz$, similar to the notation used in e.g. Diestel's Graph Theory.

Would anyone recommend a good (preferably introductory) treatment on the correspondence between relations, orders and graphs?

Most books for undergrads focus more on order and specifically partial orders. Books I found on more pure relations have either not explicitly contained such definitions, or have been a bit too advanced for me to notice. Also, the abundance of terms for the same things sometimes makes it hard to find what I want, e.g. chain, total order, linear order, simple order, connex order, etc. I looked in e.g. Steven Givant's Relation Algebras, but did not find any such notation.

答:
Sometimes "graph" is used to mean exactly the same thing as a binary relation on a set (a directed graph without multiple edges but where loops are permitted), but of course not always. It seems to make good sense to talk about a "path" in a relation, although "directed path" would be ok for emphasis. Maybe a "path" would be a directed path in the symmetric closure of the relation. While it is important to respect existing terminology, especially when talking to a specific audience, mathematics has a culture of generalizing the meaning of terms. If this is done thoughtfully, the freedom to do so is part of the creativity of the subject. Feel free to develop the terminology and probably writing the precise definitions will be an important part of learning. For example: it's not clear that the set $P$ is quite right in the question (what if the path contains cycles?); should it be a sequence not a set? is it the sequence of pairs in $X \times X$ or the sequence of elements of $X$ that matters most?

From a purely relational point of view (in the Relation Algebra setting of Givant's book you mention) the emphasis is more on the properties of $R$ as a whole, rather than the individual elements (we might call them arrows or pairs) in the relation. In fact, there are models of the relation algebra axioms that cannot be represented as concrete relations on a set (non-representable relation algebras).
Thus if your application needs to talk explicitly about the pairs in the relation there must be something involved that goes beyond the operations in a relation algebra.

Composition of $R$ with itself "forgets" the route used between the two endpoints and only measures the existence of a route. To get involved with different routes between the same endpoints, you might need to bring in category theory. See, for example, the wiring diagrams on p42 of Fong and Spivak An Invitation to Applied Category Theory (2019).

For other sources of information on relations, there is Gunther Schmidt's Relational Mathematics (2011) and possibly Schmidt and Strohlein Relations and Graphs (1993).


问:
For a bounded operator $A$, show that $(fg)(A)=f(A)g(A)$
Let $f$ be an analytic function in the neighbourhood of the spectrum $\sigma(A)$ of a bounded operator $A$. Let $C$ be a contour, counterclockwise around $\sigma(A)$ within the domain of $f$. Finally let $f(A)$ be defined by

$$f(A)=\dfrac{1}{2\pi i}\oint_C f(z) (z-A)^{-1}dz.$$

Now I have to prove that $(fg)(A)=f(A)g(A)$ but I don't know how to start. I do recognize Cauchy's integral and since $f$ is analytic on a neighbourhood of $\sigma(A)$ one could use Cauchy's integral formula but I don't see how this would help.

答:
Let $C'$ be a slightly larger boundary curve within which both $f$ and $g$ are holomorphic. By the holomorphic functional calculus,
\begin{align*}
f(A)&=\frac{1}{2\pi i}\int\limits_{C}f(\zeta)(\zeta-A)^{-1}\, d\zeta\\
g(A)&=\frac{1}{2\pi i}\int\limits_{C'}g(z)(z-A)^{-1}\, dz.
\end{align*} Then (after swapping the order of integration), we get the double integral $$f(A)g(A)=-\frac{1}{4\pi^2}\int\limits_{C'}\int\limits_{C}f(\zeta)g(z)(\zeta-A)^{-1}(z-A)^{-1}\, d\zeta dz.$$ Recall the resolvent identity $$(\zeta-z)(\zeta-A)^{-1}(z-A)^{-1}=(z-A)^{-1}-(\zeta-A)^{-1}.$$ Plugging this into the integral,

$$f(A)g(A)=-\frac{1}{4\pi^2}\int\limits_{C'}\int\limits_{C}f(\zeta)g(z)(\zeta-z)^{-1}\left((z-A)^{-1}-(\zeta-A)^{-1}\right)\, d\zeta dz.$$ If you integrate the $(z-A)^{-1}$ piece in $\zeta$, you get an integral of $0$, by Cauchy's theorem (which is why it helps to take a slightly larger boundary curve). So, this term vanishes, and we only need to look at the $(\zeta-A)^{-1}$ term. Notice that this can be simplified a lot by doing the $z$ integral first using the Cauchy integral formula $$g(\zeta)=\frac{1}{2\pi i}\int\limits_{C'}\frac{g(z)}{z-\zeta}\, dz.$$ Thus,

$$f(A)g(A)=\frac{1}{2\pi i}\int\limits_{C} f(\zeta)g(\zeta)(\zeta-A)^{-1}\, d\zeta=(fg)(A).$$

ContentLicense: CC BY-SA 4.0


问:
Course of traveling
A ship is traveling with speed 12km/h in North direction.A wind with speed 9km/h with direction from West to East changes the direction of ship.
a)Find resultant speed (Answer.15km/h in course 36.9°)
b) The sheep need to traveling back to starting point. In which course need to be her orientation?
(Answer 253,7°)

(I think that a) is easy of course just a Pithagoras theorem and definition of tan.but I not have any idea at point b))

答:
Draw a line segment of length $12$ from the origin to $(0, 12)$ to represent the ship's movement (of its own).

Draw a line segment of length $9$ from $(0,12)$ to $(9, 12)$ to represent the wind.

The length of the line segment connecting the origin $(0,0)$ and $(9, 12)$ can be found by the Pythagorean Theorem, as you noticed. This is a right triangle with legs of length $12$ and $9$.

As for the heading/bearing, you have to understand the convention:

Due North is a bearing of $0^\circ$

Due East is a bearing of $90^\circ$

Due South is a bearing of $180^\circ$

Due West is a bearing of $270^\circ$

Edit: As you have pointed out, the question is likely asking for what course the ship should set, not what it's actual path will be. So the ship will need to aim for $(-9, 0)$ in order to arrive at the origin. This would result in a right triangle with sides $18$ and $12$.

I meant to add earlier - for all such questions, you should definitely draw a triangle (or two)!

Title: Course of traveling


问:
Domain of a binary relationship and cardinality
I need to write the proof of the following exercise: if $A\subset X \times Y$ is a binary relationship and $\operatorname{dom}(A)=\{a \in X \mid \exists b \in Y \; (a,b) \in A\}$ prove that $|\operatorname{dom}(A)|\leq|A|$. My solution is to note that the function $\pi\colon A  \rightarrow \operatorname{dom}(A)$ is surjective and then the statement using AC. I would like to write a proof that doesn't require AC. Any suggestion?

答:
You can't prove this without the axiom of choice.

Suppose that $X$ is an amorphous set, or really just a set such that $[X]^{<\omega}=\{Y\subseteq X\mid Y\text{ is finite}\}$ is itself Dedekind-finite. Then consider the relation $\{(n,Y)\mid Y\in[X]^{<\omega}, |Y|=n\}$, then the domain is $\omega$, but if there is an injection from $\omega$ into the relation, then we can find an infinite sequence of distinct sets in $[X]^{<\omega}$ which is impossible, since it is a Dedekind-finite set.

More generally, if $f\colon X\to Y$ is a surjective function, but there is no injection from $Y$ to $X$, then $f^{-1}$ is an example of a relation whose domain is $Y$, but there is no injection from $Y$ into $f^{-1}$, because that would define an injection into $X$.

So in other words, the statement you're trying to prove is equivalent to the Partition Principle, which states "If there is a surjection from $X$ onto $Y$, then there is an injection from $Y$ into $X$".

PP proves, amongst other things, that every infinite set is Dedekind-infinite, choice from well-ordered families, and a few more things. It is currently an open problem whether or not PP implies AC.


问:
Rotate about a point or an axis?

English is not my native language. Why math text uses the preposition "about" while referring rotation in the context of a point, an axis or a plane? Is not the preposition "around" or "on" more natural to use in this context?

Is matrix multiplication of NON-square matrix and identity matrix commutative?

Scale applies along an axis. For example, if you scale along x-axis, the horizontal length of the shape will increase/decrease in 2D. Why the author of the following para (whole para is pasted below to clear the context) is saying that it is applied about the perpendicular axis? Also, why did he use the preposition "about" here?

Scaling along the Cardinal Axes
The simplest scale operation applies a separate scale factor along each cardinal axis. The scale along an axis is applied about the perpendicular axis (in 2D) or plane (in 3D). If the scale factors for all axes are equal, then the scale is uniform; otherwise, it is nonuniform.

Please help if you can answer any one of the three questions.

Thanks.

答:
"around" is pretty common too; this is more a question for the English language stacks, but I know that "about" was more commonly used in the past the way "around" is now; see for instance this passage from Shakespeare's Macbeth.

O well done! I commend your pains;

And every one shall share i' the gains;

And now about the cauldron sing,

Live elves and fairies in a ring,

Enchanting all that you put in.

A matrix that isn't square can only be multiplied on both sides by another matrix that isn't square: a $2\times3$ matrix can only be multiplied on both sides by a $3\times2$ matrix.  The results of these two multiplications are both square matrices, of different sizes: if $A$ is $2\times3$ and $B$ is $3\times2$, then $C = AB$ is a $2\times2$ matrix, and $D = BA$ is a $3\times3$ matrix.  Because they are different sizes, $C$ and $D$ are obviously not equal; because they are not square, neither $A$ nor $B$ can be an identity matrix.

The final passage is ... not how I would put things, but I think what he means is - a scaling along a particular axis fixes its perpendicular: in 2 dimensions, a scaling along one axis leaves a line perpendicular to the axis unchanged, and in 3 dimensions it similarly leaves a plane perpendicular to the axis unchanged.  This sense of "about" is shared with rotation, in that in both cases these operations leave the things they are "about" unchanged.

Title: Rotate about a point or an axis?


问:
For any integer $a$, there is an integer $k$ such that $a^2=3k$ or $a^2=3k+1$

Let $a$ be an integer. Prove that there exists an integer $k$ such that $a^2=3k$ or $a^2=3k+1$.

Here is what I did:

I said: $a\in\{\ldots,-3,-2,-1,0,1,2,3,\ldots\}$. When I put 2 into $a^2$, I get $4$ but when I put $2=k$, $3k$, I get the number $4$ on the LHS and the number $6$ on the RHS. But $4$ is not equal to $6$. So how can I prove that $a^2=3k$ equivalent?

I've learned what the division algorithm is. However, when I tried to solve this problem, I got confused.

答:
Note that any integer can be written as one of $3j+0$, $3j+1$, or $3j+2$
$$
\begin{align}
(3j+0)^2&=9j^2&=3k+0&\text{for }k=3j^2\\
(3j+1)^2&=9j^2+6j+1&=3k+1&\text{for }k=3j^2+2j\\
(3j+2)^2&=9j^2+12j+4&=3k+1&\text{for }k=3j^2+4j+1\\
\end{align}
$$
Therefore, the square of any integer can be written as $3k$ or $3k+1$.

Since the tag modular-arithmetic appears in the question, it should be said that the preceding can be written as
$$
\begin{align}
0^2\equiv0\pmod{3}\\
1^2\equiv1\pmod{3}\\
2^2\equiv1\pmod{3}
\end{align}
$$

ContentLicense: CC BY-SA 3.0


问:
why $U(P,f) - L(P,f) < \epsilon ?$
Consider the following function on $[0,1]$:
$$f(x)=\begin{cases}1&x\in\{a_1,a_2,\dots,a_n\}\\0 &\text{otherwise}\end{cases}$$
The $a_i$ are fixed and all in $[0,1]$. Determine whether $f$ is Rieman integrable or not.

My attempt :  I found the answer here

Here is the outline of the given answer

Take a partition $P$ of $[0,1]$ that is defined as:
$$
P = \left\{0,a_1-\frac{\epsilon}{2},a_1+\frac{\epsilon}{2},\dots,a_n-\frac{\epsilon}{2},a_n+\frac{\epsilon}{2},1\right\}.
$$
Clearly, $U(P,f) = \epsilon N$ and $L(P,f) = 0$. Since for any given $\epsilon>0$, a partition that could yield
$$
U(P,f) - L(P,f) < \epsilon
$$
can be found, $f$ is Riemann integrable.

My confusion :Here $$U(P,f)=\sum_{i=1}^{n}M_i(x_i- x_{i-1})=  n\epsilon$$ and

$$L(P,f)=\sum_{i=1}^{n}m_i(x_i- x_{i-1})=  0$$

$$\implies U(P,f) - L(P,f)  > \epsilon   \implies  n\epsilon -0 > \epsilon $$

My confusion is that  why  $U(P,f) - L(P,f) < \epsilon ?$

答:
You have the right to be confused here. That answer is misleading (at best).

Given $\varepsilon>0$, that answer shows that there is a partition $P$ such that $U(f,P)-L(f,P)=n\varepsilon$ (not $N\varepsilon$; there is no $N$ here). So, given $\varepsilon>0$, take $\varepsilon'<\frac\varepsilon n$ and consider the partition$$P=\left\{0,a_1-\frac{\varepsilon'}2,a_1+\frac{\varepsilon'}2,\ldots,a_n-\frac{\varepsilon'}2,a_n+\frac{\varepsilon'}2,1\right\}.$$Then$$U(f,P)-L(f,P)=n\varepsilon'<\varepsilon.$$

Title: why $U(P,f) - L(P,f) < \epsilon ?$


问:
Finding $n$, $n^n \equiv k \pmod m$
What is the general process of finding $n$ such that
$$n^n \equiv k \pmod m?$$

I tried working with an example, say $k=60$ and $m=120.$ Then, I know that $2,3,5 \mid n.$ So, Letting $n=30w,$ We have
$$30w^{30w} \equiv 30^{30w}(w)^{30w}$$
But continuing this thought leaves me an expression with $w^w$ again. The problem I feel is that the exponent of $n^n$ is varying, so I can't really deduce many properties about it.

答: First lets look at the basics if $n^n=mx+k$ then $\gcd(x,k),\gcd(m,k)\mid n^n$ . Another note is: if $m$ is odd, odd $n$ would produce $k$ with opposite parity to the multiplier they are in. You can use similar tricks modulo small primes.

Title: Finding $n$, $n^n \equiv k \pmod m$


问:
Why is there a Mandelbrot hidden in this sequence?
I was experimenting a bit with tetration escape visualization and tried to visualize the following equation on the complex plane:

$$z_{n+1} = z_{0}^{\sin(z_{n})}$$

This sequence visualized looks like this:

I noticed between $7.8-0.11i$ and $8.03+0.11i$ there was a Mandelbrot set hidden between all the other mess:

Also between $-4.81-0.15i$ and $4.61+0.15i$ there is a Mandelbrot-like structure:

How comes the Mandelbrot set randomly appears in there?

答:
Consider iteration of $z \mapsto c^{\sin(z)}$ near $c$:

$$z \mapsto c^{\sin(c + z)} - c$$

The Taylor series near $z = 0$ is (via Wolfram Alpha):

$$ z \mapsto \left(c^{\sin(c)} - c\right) \\
 + \left(\log(c) c^{\sin(c)} \cos(c)\right) z \\
 + \left(\frac{1}{2} \log(c) c^{\sin(c)} \left(\log(c) \cos(c)^2 - \sin(c)\right)\right) z^2 \\
 + O(z^3)
$$

Locally ($z$ small), the non-linearity is dominated by the $z^2$ term (as higher order terms will be much smaller).  This means miniature quadratic Mandelbrot set copies may appear.  Completing the square to the form $z \mapsto a(z + b)^2 + d$ gives:

$$
ab^2 + d = c^{\sin(c)} - c \\
2ab = \log(c) c^{\sin(c)} \cos(c) \\
a = \frac{1}{2} \log(c) c^{\sin(c)} \left(\log(c) \cos(c)^2 - \sin(c)\right)
$$

Affine conjugation via $z \mapsto z - b$ and $z \mapsto \frac{z}{a}$ gives the form $z \mapsto z^2 + C$ where $C = a (d + b)$ which expands to:

$$ C = \frac{1}{2} \log(c) \left(c^{\sin(c)} - c\right) c^{\sin(c)} \left(\log(c) \cos(c)^2 - \sin(c)\right) \\ - \left(\frac{1}{2} \log(c) c^{\sin(c)} \cos(c)\right)^2 + \frac{1}{2} \log(c) c^{\sin(c)} \cos(c) $$

Solving numerically for $C = 0$ gives (via Wolfram Alpha) a number of solutions including: $$c \approx 7.85398163397448\ldots$$

Solving numerically for $C = 0$ near $c = -4.7$ (via Wolfram Alpha) the solution: $$c \approx -4.71238898038469\ldots$$

These both correspond to mini-Mandelbrot islands with period $1$, there should also be mini-Mandelbrot islands corresponding to higher periods, for example I think there should be a period $3$ near $18.155$ (with $C \approx -1.754\ldots$).


问:
Finding the gradient of $f(x,y)=\max \{|x|,|y| \}$
We have the function $f : \mathbb{R}^2 \to \mathbb{R}$ given by

$$f(x,y) := \max \left\{ |x|, |y| \right\}$$

Let $C := \left\{ (x,y) \mid |x|=|y| \right\}$. Given point $(x,y) \notin C$, how can I find the gradient $\nabla f(x,y)$?

I have to find $\frac{\partial f }{\partial x}$ and $\frac{\partial f }{\partial y}.$ But I am not sure, do we just get the following?

$$\frac{\partial f }{\partial x}=\max \{|1|,|y| \}, \qquad \frac{\partial f }{\partial y}=\max \{|x|,|1| \}$$

I'm not sure, can anyone help me ?

答:
Hint:

$$\max(a, b) = \frac{a + b + |a - b|}{2}$$

Proof:

Assume $a > b$. Then, we have $0.5(a + b + |a - b|) = 0.5(a + b + a - b) = a$. As desired.

Assume $a < b$. Then, we have $0.5(a + b + |a - b|) = 0.5(a + b - a + b) = b$. As desired.

Assume $a = b$. Then, it is obvious.


问:
Lax Milgram and inf-sup
I am trying to understand why the Lax Milgram theorem is a consequence of the following theorem:

Let $H_1, H_2$ be Hilbert spaces and $k : H_1 \times H_2 \rightarrow \mathbb{R}$ be a continuous bilinear form. For $f \in H_2'$ consider the variational problem:
\begin{equation}
\text{find } u \in H_1 \text{ such that } k(u,v) = f(v) \text{ for all }v ∈ H_2.
\end{equation}
The following two statements are equivalent:

For arbitrary $f \in H_2'$ the problem above has a unique solution $u \in H_1$ and $∥u∥_{H_1} ≤ c∥f∥_{H_2'}$ holds with a constant $c$ independent of $f$.

The conditions
\begin{equation}
\exists \varepsilon >0 : \sup_{v \in H_2}{\frac{k(u,v)}{∥v∥_{H_2}}}\geq \varepsilon ∥u∥_{H_1} \text{ for all } u \in H_1,
\end{equation}
\begin{equation}
\forall v \in H_2, \quad  v\neq 0, \quad  \exists u \in H_1: \quad  k(u,v) \neq 0
\end{equation}
hold.

Moreover, for the constants $c$ and $\varepsilon$ one can take $c = \frac{1}{\varepsilon}$ .

Any help is much appreciated.

答:
The Lax-Milgram theorem is stated once again for convenience.

Let $T$ be a bounded coercive bilinear form on a Hilbert space $H$, with coercive constant $C_c>0$ i.e. for all $u \in H, u \neq 0$ we have $T(u,u) \geq C_c\|u\|_H^2$. Then for all $f \in H'$ there exists $u_f \in H$ such that $\|u_f\|_H \leq \frac{\|f\|_{H'}}{C_c}$ and $T(u_f,v) = f(v)$ for all $v \in H$.

Proof : Let $H_1 = H_2 = H$. We will show that the condition $2$ holds in the theorem given, which will imply condition $1$.

For condition $2$ to hold, we need to show that there is an $\epsilon > 0$ such that for all $u \in H$ we have $\sup_{v \neq 0 \in H} \frac{T(u,v)}{\|v\|_{H}} \geq \epsilon \|u\|_H$. However, note that for $u=0$ the condition is trivially satisfied for any $\epsilon$, and for $u \neq 0$ we have :
$$
\sup_{v \neq 0 \in H} \frac{T(u,v)}{\|v\|_{H}} \geq \frac{T(u,u)}{\|u\|_{H}} \geq C_c \|u\|_H
$$
hence the first part holds with $\epsilon = C_c$. For the second part, if $v \neq 0 \in H$ then $T(v,v) \geq C_c \|v\|_H^2 > 0$, so $u = v$ is such that $T(u,v) \neq 0$. Thus, condition $2$ holds.

By the theorem given, condition $1$ holds. Condition $1$ says : for all $f \in H'$ there is a unique $u_f \in H$ such that $\|u_f\|_{H} \leq \frac{\|f\|_{H'}}{C_c}$ and $T(u,v) = f(v)$ for all $v \in H$. But this is precisely the Lax-Milgram theorem!

Note that the second theorem is much stronger for a couple of reasons : one is the fact that it deals with forms which are not just bilinear but also takes arguments from different spaces. This allows us to experiment with PDE where arguments from different spaces (for example in PDE, different Sobolev spaces for two functions which have been differentiated a different number of times) are involved. The second improvement is that this helped tackle boundary problems in PDE where the domain "changes shape" with time, the best example being that of a melting iceberg where of course with time the shape of the iceberg changes according to how heat dissipates in the iceberg, so basically the heat equation is solved with shape changes in the domain.

The second theorem is referred to as the Lions-Lax-Milgram theorem, or just Lions' theorem as well, and is a very vital improvement. Another is the Babushka-Brezzi theorem, these are all fundamental for existence and uniqueness theorems for PDE. More on this can be found in Kesavan's book 'Functional analysis and applications'.

Title: Lax Milgram and inf-sup


问:
show $\lim_{x\rightarrow\infty} x^n\exp(-x^2)=0$
I want to show for all $n\in\mathbb N$
$$\lim_{x\rightarrow\infty}\frac{x^{n}}{\exp(x^2)}=0$$

I am pretty sure that I have to use L'Hospital. I've tried induction:

$n=1$: $$\lim_{x\rightarrow\infty}\frac x{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac1{2x\exp(x^2)}=0$$

And for $n\rightarrow n+1$:
$$\lim_{x\rightarrow\infty}\frac{x^{n+1}}{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac{(n+1)x^n}{2x\exp{x^2}}$$

And now I am stuck. The term $2x$ really annoys my for my induction hypothesis.

Any hints?

答: Note that for positive $x$, $x^{2n}\le e^{x^2}$ (by considering the taylor series), so $\frac{x^n}{e^{x^2}}\le\frac{x^n}{x^{2n}}=\frac{1}{x^n}$, hence $\lim\limits_{x\to\infty}\frac{x^n}{e^{x^2}}\le\lim\limits_{x\to\infty}\frac{1}{x^n}=0$.</p>


问:
If a family has two children, what is the probability that at least one is a boy?
If a family has two children, what is the probability that at least one is a boy, assuming there are only two sexes?

One way can be : The sample space is $\{BB, BG, GB, GG\}$ so that $P = \frac{1}{4}$.

This might also be a way : The sample space is $\{0 \ B, 1 \ B, 2 \ B\}$ so that $P = \frac{1}{3}$.

Are both models correct? Is there a fallacy? I usually see the answers to be $\frac{1}{4}$ so I am confused.

Edit : As I have been pointed out, I did mean $\frac{3}{4}$ and $\frac{2}{3}$.

答:
There are several other assumptions that must be made.

Each child's gender is male or female with probability $1/2$, independent of all other children. That's certainly not quite true: slightly more boys than girls are born, while death rates for males tend to be higher. And I don't know if there is data on how independent they are.

Decisions on whether to have another child are independent of the gender of the existing children.  Again, that's certainly not true.


问:
A corollary of Moser's theorem
In my symplectic geometry class we gave the following result :

Let $M$ be a closed symplectic manifold, $Q\subset M$ a closed submanifold and $\omega_t$ a family of symplectic forms such that $[\frac{d}{dt}\omega_t]=0$ and $(\frac{d}{dt}\omega_t)|_Q=0$ for any $t\in [0,1]$. Then there exists an isotopy $\phi_t :M\rightarrow M$ such that $(\phi_t)_Q=id$ and $\phi_t^*(\omega_t)=\omega_0$ for any $t\in [0,1]$.

Now there's a step on the proof where we realized there is a mistake and so far no one was able to fix it. I have tried to look for this result in symplectic geometry books but I had no sucess so I am wondering if anyone knows if this result is actually true and how one can prove it , or if there is a counterexample ?

Any help is appreciated. Thanks in advance.

答:
If the cohomology class $[\frac{d\omega_t}{dt}]$ is merely assumed to lie in the (absolute) cohomology $H^2_{dR}(M)$, then the result is false. If it is assumed to lie in the relative cohomology $H^2_{dR}(M,Q)$, then the result is true under the following formulation (to be explained below):

Let $M$ be a closed symplectic manifold, $i_Q : Q \subset M$ a closed submanifold and $\omega_t$ a family of symplectic forms such that $\left[ \frac{d \omega_t}{dt} \right] = 0$ in $H^2(M,Q)$ and $i_Q^*\frac{d\omega_t}{dt} = 0$ for any $t \in [0,1]$. Then there exists an isotopy $\phi_t : M \to M$ such that $\phi_t|_Q = id$ and $\phi_t^*\omega_t = \omega_0$ for every $t \in [0,1]$.

Absolute case: For $[\frac{d\omega_t}{dt}] = 0$ in $H^2_{dR}(M)$ , one can produce counter-examples to the statement by taking $Q$ to be a contractible embedded circle in a closed surface $M$. Indeed, pick an embedded smooth disc $D$ bounded by $Q$ and define $\omega_t$ by changing the symplectic form $\omega_0$ away from a neighborhood of the circle in such a way that the area of $D$ disc increases with $t$ while keeping the area of the whole surface fixed. Then any isotopy $\phi_t$ produced from the family $(\omega_t)_{t \in [0,1]}$ by Moser's trick is such that the $\omega_0$-area of $D_t := \phi_t^{-1}(D)$ increases with $t$, which implies $\phi_t|_Q \neq id$.

Relative case: (Preliminary remark: Let $i_Q : Q \subset M$ denote the inclusion map. Given a differential form $\alpha$ on $M$, we are going to consider two types of 'restrictions' to $Q$, (i) the pullback form $i_Q^*\alpha$ on $Q$ and (ii) the restriction $\alpha|_{Q}$ of $\alpha$ to $Q$, by which I mean the form induced on the pullback bundle $i_Q^*TM = T_{Q}M$. The pullback $i_Q^*\alpha$ only captures the value of $\alpha|_Q$ on the subbundle $(i_Q)_*TQ \subset T_Q M$.)

The desired property $\phi_t|_Q = id$ amounts to $X_t|_Q = 0$, where $X_t$ is the vector field produced by Moser's trick via the equation $\lambda_t = \iota_{X_t}\omega_t$ where $d\lambda_t = \frac{d \omega_t}{dt}$. Hence, for $\phi_t$ to have the desired property, we need to be able to find for every $t \in [0,1]$ a primitive 1-form $\lambda_t$ for $\frac{d \omega_t}{dt}$ whose restriction $\lambda_t|_Q$ to $Q$ vanishes. This is only possible if $\frac{d \omega_t}{dt}$ is 'well-behaved relative to $Q$', in a sense we now explore.

The counter-examples given in the absolute case suggest that for the statement to hold, one should further require that all members of the family $(\omega_t)_{t \in [0,1]}$ give the same area to every 2-cycle with boundary in $Q$. To satisfy this requirement, it suffices to ask that for every $t \in [0,1]$, the exact two-form $\frac{d\omega_t}{dt}$ admits a primitive 1-form $\lambda_{t}$ whose pullback $i_Q^*\lambda_{t}$ to $Q$ is exact. (Note: The condition $i_Q^*\frac{d\omega_t}{dt} = 0$ already implies that $i_Q^*\lambda_{t}$ is closed.)

This last condition can be expressed in relative cohomological terms as defined in Bott's & Tu's Differential forms in algebraic topology. Namely, we define the de Rham complex of $M$ relative to $Q$ as the cone of $i_Q$, i.e. as the complex $\Omega^*(M, Q) := \Omega^*(M) \oplus \Omega^{*-1}(Q)$ with differential given by $d(\alpha, \eta) = (d\alpha, i_Q^*\alpha - d\eta)$. Then the condition stated in the previous paragraph is precisely that for every $t \in [0,1]$, the relative 2-form $\left( \frac{d \omega_t}{dt}, 0 \right) \in \Omega^2(M, Q)$ should be exact; In other words, the relative cohomology class $\left[ \frac{d \omega_t}{dt} \right] := \left[ \left( \frac{d \omega_t}{dt}, 0 \right) \right] \in H^2(M, Q)$ should be required to vanish.

Under this assumption, we can find for every $t \in [0,1]$ a primitive 1-form $\lambda_t$ to $\frac{d\omega_t}{dt}$ whose pullback to $Q$ is exact, say $i_Q^*\lambda_t = df_t$ with $f_t : Q \to \mathbb{R}$. Now, by Whitney's extension theorem, there exists $F_t : M \to \mathbb{R}$ such that $i_Q^* F_t = f_t$ and $dF_t|_Q = \lambda_t|_Q$. Let's define $\lambda'_t := \lambda_t - dF_t$; This is another primitive 1-form for $\frac{d\omega_t}{dt}$ which satisfies $\lambda'_t|_Q = 0$. Since all the above choices can be made to depend smoothly on $t$, we can apply Moser's trick with this family $\lambda'_t$ to obtain the desired isotopy.

Title: A corollary of Moser's theorem


问:
What is the maximum number of the pieces that a watermelon can be cut into 10 knife cuts? (circle cutting problem extension)
I saw this problem in a game "Brain out" level 85. In this game, the answer is 1024, simply 10th power of 2. But I learned circle cutting problem in school and by solving recurrence formula, I could get explicit formula for the problem.
I think this problem also require recurrence formula similarly and the answer is not just 1024. What is the true answer about this game stage?

答: i think that answer's actually right since it doesn't have any limitations like the watermelon cannot be rearranged. But if you weren't allowed to rearrange it and some force were holding the watermelon pieces together as you cut, the answer would be $176$. It's the same as the cake number problem and follows the formula $(n^3+5n+6)/6$


问:
Is it true that any set which contains elements with contradictory properties is the empty set?
Here's what I've "discovered" but can't verify.

Let's take a set $A$ defined as "the set of all $x$ which don't belong to $A$".

Obviously this definition wouldn't hold in particular axiomatic systems, but it would in others.

Let's analyze this in NST. $X\in A$ iff $X\not\in A$.

This is a logical contradiction, but it still doesn't prove that $A$ doesn't exist. In fact, I would argue that $A$ is the empty set.

If put all of the elements of $A$ into the empty set, we get $(X∈A \text { iff } X∉A) \text { iff } X\in \emptyset$.

This is a tautology for whatever $X$ and $A$. Now, if we switch two of the atomic sentences, we get $X\not\in A \text { iff } (X\in A \text { iff } X\in\emptyset$). We know this to also always be true, which means that either $A$ doesn't exist or $A$ is the empty set.

Therefore, the empty set does contain every element belonging to any set it doesn't belong to.

Is there some error in my proof?

答:
The statement $$(*)\quad(X∈A \text { iff } X∉A) \text { iff } X\in \emptyset$$ does not mean what you seem to think it means (though I'm not sure exactly what you think it means).  If $A=\{X:X\not\in A\}$, then $(X∈A \text { iff } X∉A)$ must be true for all $X$.  When you assert $(*)$ you are asserting that $(X∈A \text { iff } X∉A)$ is true if $X\in\emptyset$, but also that it is false if $X\not\in\emptyset$.  So if $A$ actually is equal to $\{X:X\not\in A\}$, then this statement (with an implicit universal quantifier on $X$) is false, since $(X∈A \text { iff } X∉A)$ would be true for all $X$, even $X$ such that $X\not\in\emptyset$.

Again, I'm not entirely sure what your train of thought is, but it sounds like you are confusing $$A=\{X:X\not\in A\}$$ with $$A=\{X:X∈A \text { iff } X∉A\}.$$  If $A$ was supposed to satisfy the latter equation, then $A=\emptyset$ would be equivalent to $(*)$ (and indeed, your argument then correctly shows that $A=\emptyset$ satisfies the latter equation).  But the first equation is quite different.


问:
Determining if a model is fully specified by given information
I have a zero-parameter model $M_1$ which predicts a parameter $x = 0$ and a model $M_2$ with one parameter $b$ which predicts $x=b$. Then, the sampling distribution of the data $d$ is

$$P(d|x,σ) = N=(d;x,σ^2) = \frac{1}{\sqrt2\sqrtπσ }* exp[\frac{-1}{2} (\frac{d-x}{σ})^2]$$

How can I determine if the models are fully specified by the given information?

答: M1 is fully specified but M2 isn’t (because we don’t know the distribution of b)</p>


问:
Find expectation $E(X)$ in a probability problem

At work they play secret friend (secret santa gift exchange). Everyone sits in a circle randomly and each
one gives a gift to the one on the right or the one on the left, without knowing who is on the right and
who to the left (is to avoid preferences). For each, the choice of giving it to the
the right or the left is random with probability $0.5$ of giving it to the right and $0.5$
to give it to the one on the left.  Find the expectation of the number of participants that end without a gift, if in total there are $200$ people.

Idea: Is not hard to imagine cases where people end wihout gift, for example if circle starts in $A$ and ends in $B$, then anyone in one side gives a gift to the next, but the last participant $B$ do not gives a gift to $A$ who is the next, instead he gives the gift to the the previous person, so $A$ ends without gift. Diagram: A---->O----->O---->O----><B.
How to extend this idea to find the answer?

答:
Let $X_i$ be an indicator random variable that equals $1$ if the $i_{th}$ person doesn't receive a gift, and $0$ otherwise.

Then $\Bbb {P}(X_i)$

$= \Bbb P$[not receive gift from both $(i-1) \;and\; (i+1)$ person]

$ = \frac12\cdot\frac12 = \frac14$

Now the expectation of an indicator variable is just the probability of the event it indicates,

thus $\Bbb E[X_i] = \Bbb P(X_i) = \frac14$

and by linearity of expectation, which applies even if the variables are not independent,

$\Bbb E[X] = \Bbb E[X_1] + \Bbb E[X_2] + ... + \Bbb E[X_{200}] = 200\cdot\frac14 = 50$


问:
Multiple choice questions in exam
In a country-level placement exam, there are 10 multiple-choice questions, each with 4 equivalent possible replies.
What is the minimum number of students taking this exam, so that it is always possible to find at least 2 sets with at least 9 same replies? All questions must be answered (and we assume that they are answered at random, that is, without any previous knowledge of the subjects).

My thoughts:
All possible different sets of replies are $4^{10}$.
If we get $4^{9}+1$, do we ensure we will have at least 9 same sets?

答:
Indeed, the minimum number is $4^9+1$. You can prove this works using the pigeonhole principle, with $4^9$ holes corresponding to the answers to the first $9$ questions.

To prove this is the minimum, you need to find a set of $4^9$ tests where no two agree on $8$ answers. From the pigeonhole argument, you know that every possible string of $4^9$ answers for the first $9$ questions must appear exactly once. You just need to choose the last answer for each of these in such a way that when any two answer sequences agree in $8$ places, their $10^{th}$ answers do not agree. Addition $\pmod 4$ is one way to accomplish this...

Title: Multiple choice questions in exam


问:
Finding a basis with Change of Basis

Find the coordinate vector for v relative to the basis S = {v1, v2, v3} for $R^3$.
  $$v = (2,-1,3);$$
  $$v1 = (1,0,0); v2 = (2,2,0); v3 = (3,3,3);$$

So I did and I got the coordinate vector space as (v)s = (3, -2, 1). I do not believe that this answer is wrong. So now I am at the next question.

Consider the coordinate vector:
  $$ [w]s =
\left[ \begin{array}{c}
6 \\
-1 \\
4
\end{array} \right]$$
  Find w if S is the basis of the first question.

I have no idea how to do this as my textbook is not accurate. Can someone please help me.

答:
$$\begin{pmatrix}6\\\!\!-1\\4\end{pmatrix}=a\begin{pmatrix}1\\0\\0\end{pmatrix}+b\begin{pmatrix}2\\2\\0\end{pmatrix}+c\begin{pmatrix}3\\3\\3\end{pmatrix}\iff$$

$$\begin{align*}a+2b+3c&=6\\
2b+3c&=-1\\
3c&=4\end{align*}$$

Well, now just find $\,a,b,c\,$ and these will be your coordinates for your vector wrt the given basis. The inverse process you already did, just as John pointed out.

Title: Finding a basis with Change of Basis


问:
Can an open ball be a closed set and cannot be interpreted as a closed ball at the same time?
For example, in the metric space $E=(0,1)\cup(1,2)\cup(2,3)$ with normal distance, $(1,2)=\{x:d(x,\frac{3}{2})<\frac{1}{2}\}$ is an open ball, and $(1,2)$ is also a closed set. However, we also have $(1,2)= \{x:d(x,\frac{3}{2})\leq \frac{1}{2}\}$, which is a closed ball, so $(1,2)$ cannot be served as a supportive example.

I would also like to know the answer of the mirror question: Can a closed ball be an open set and cannot be interpreted as an open ball at the same time?

Thanks!

答:
Yes. Consider the set$$[2,\infty)\cup\left\{1,0,-2(=-1-1),-1-\frac12,-1-\frac13,-1-\frac14,\ldots\right\}.$$Then the closed ball with center $0$ and radius $1$ is an open set, but not an open ball.

Now, fro the question from the title, consider$$(-\infty,-2)\cup(-2,-1)\cup(-1,0]\cup\{2\}.$$Then the open ball with center $0$ and radius $2$ is a closed set which is not an closed ball.


问:
Bonus scheme calculation
Problem Background:

We are a project based company. Employees work on project basis and each project duration can be between 3 days to 30 days. At the end of each project, the employee will get rated from a score of 1 (worst) to 3 (best) across 5 different categories, let's call it the project rating.

We have a year end bonus scheme and the project component weightage is 30%. Each employee can get up to 30% and it's not shared between all employees.

Consider the following example:

```<code>Employee A
| Project Name | Project rating (across 5 categories) |
| ------------ | ------------------------------------ |
| One  | 3,2,2,2,2 |
| Two  | 2,2,2,2,2 |
| Three| 2,2,2,2,2 |

Employee B
| Project Name | Project rating (across 5 categories) |
| ------------ | ------------------------------------ |
| One  | 3,2,2,2,2 |
| Two  | 2,2,2,2,2 |
| Three| 2,2,1,1,2 |
| Four | 1,1,2,3,2 |
</code>```

Questions:

Is there a fair good way translate the project ratings into the weightage given the following limitations:

The number of projects completed between different employees differs (some did more while some did lesser)

Not all projects have the same level of difficultly, this data wasn't captured in the project rating as well. Employees dont get to choose the project either.

There must be a way to get full 30% weightage without compromising the fairness criteria, if possible

答:
With the data provided there is no way to distinguish work on hard or long projects from work on easy or short ones.
Perhaps the employees who did fewer projects did so because those were long or hard.

You ask for a "fair" bonus calculation, but "fair" is not a mathematically well defined term. I think that given the data the "fair" assumption is that each of the five ratings on each employee's projects contributes the same weight to that employee's bonus, independent of what the other employees have done.

The easiest calculation is to find the average rating for each employee. For the first one in the question that would be
$$
\frac{3 + 14 \times 2}{15} = 2.07.
$$
Then you could award that employee a bonus of $2.07 \times 10\% = 20.7\%$.

With this algorithm the minimum bonus is $10\%$ and the maximum is $30\%$.

You could do a little more work  to count ratings of $3$ as more than three times as good as ratings of $1$ or to scale the result so that the bonus range was from $0\%$ to $30\%$. I suspect that any scheme would look "unfair" to someone. The simplest one would be the easiest to defend.

Title: Bonus scheme calculation


问:
Books for proving the hopf invariant is a homomorphism.
Books for proving that the Hopf invariant is a homomorphism. I need to see many proofs of this, could someone recommend some books to me please?</p>

答: A somewhat recent question asked about the proof of this fact in I.M. James' "Note on cup-products".  In one of the comments to my answer to that question I pointed out that Mosher-Tangora's book also contains a proof.</p>


问:
Arbitrage opportunity
Given odds $o_i$ for $i=1,2,\dots,n$ and the possibility to bet the amount $b_i \in \mathbb{R}$ on each event such that if event $i$ occurs you receive $b_i o_i$ and if it doesn't you receive $-b_i$. I am trying to find out the condition for arbitrage. My immediate thoughts are that $1/o_i$ represents probability, and since these events are independent then $1/o_1+1/o_2+\dots+1/o_n=1$ has some significance and that if $$\sum\limits_{i=1}^n o_i^{-1}\neq1$$ then arbitrage is possible.

In a specific example, $n=3$, $o_1=1,o_2=2,o_3=3$ I have worked out that if  $b_1=-5,b_2=-4,b_3=-3$, then the profit is always greater or equal to zero and positive with finite probability. (i.e. $\{1\}\rightarrow 2,\{2\}\rightarrow 0,\{3\}\rightarrow 0$)

How do I show this without trial and error? I want a way to find the bids, maybe even for general $n$? Also, I get the feeling that if the sum is less than unity you need to back all possibilities for a sure profit ($b_i>0$) and if it is greater than unity you need to lay all possibilities ($b_i<0$) How would I show this?

答:
Your odds are what I would call odds against.

If you want a Dutch Book (a guaranteed outcome) then you can bet $\dfrac{k}{1+o_i}$ on each option $i$. Then if possibility $j$ wins you will receive $\dfrac{k}{1+o_j}o_j-\displaystyle\sum_{i\not =j} \dfrac{k}{1+o_i}$, so whatever happens you will end up with $k\left(1-\displaystyle\sum_{i=1}^n\dfrac{1}{1+o_i}\right)$.

If you want to come out ahead, make sure that $k$ has the same sign as $\left(1-\displaystyle\sum_i\dfrac{1}{1+o_i}\right)$.

In your example $\displaystyle\sum_i\dfrac{1}{1+o_i}=\dfrac{13}{12}$, you could let $k=-120$ and make bets of $-60,-40,-30$ for a guaranteed outcome of $+10$.

Title: Arbitrage opportunity


问:
Pattern in dropping time of Collatz iteration
I've noticed a linear pattern in the dropping time $C(n)$ of the Collatz iteration (A122437) when plotted against the number of digits $d(n)$ in the base-2 representation of $3^n$ (A020914).

Collatz iteration dropping time (vertical) versus the number of digits in the base-2 representation of $3^n$ (horizontal).

For this limited data set the connection can be described by something like
$$C(n)=\lfloor1.635\cdot d(n)\rfloor$$
Is this connection known and if not would it be worth investigating further?
Thanks in advance for sharing your ideas.

Here's how I found the pattern.

Given $n,b\in\mathbb{N}$ let $c(n,b)$ be the dropping time of the set $\{2^n\cdot k+b|k\in\mathbb{N}\}$ if it exists. E.g. $c(1,0)=1$ and $c(4,3)=6$ while $c(3,1)$ is not defined. We don't consider multiples of 2, because $c(3,2)$ is the same as $c(2,1)$, for instance.

From numerical observation it becomes apparent that $\{c(n,b)|b\in\mathbb{N}\wedge(n,b)\in\textrm{dom}(c)\}$ is a singleton, say $\{e(n)\}$. Finally, $e$ only seems to be defined on $\textrm{cod}(d)$ which yields the relation:

$$\forall n\in\mathbb{N}:C(n)=e(d(n))$$

Here's the C++ code I did the analysis with.

```<code>#include <fstream>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>

template <typename T>
class residue_class {
 public:
  residue_class(T divisor, T remainder)
      : divisor{divisor},
        remainder{remainder},
        divisor_orig{divisor},
        remainder_orig{remainder} {}

  std::vector<residue_class> solve(std::ostream& stream,
                                   std::map<T, T>& steps_by_divisor) {
    return solve_light(stream, steps_by_divisor)
               ? std::vector<residue_class>{}
               : std::vector<residue_class>{create_even(), create_odd()};
  }

  bool solve_light(std::ostream& stream, std::map<T, T>& steps_by_divisor) {
    while (step()) {
      if (is_smaller()) {
        print(stream);
        auto found{steps_by_divisor.find(divisor_orig)};
        if (found == steps_by_divisor.end()) {
          steps_by_divisor.emplace(divisor_orig, steps);
        } else if (found->second != steps) {
          std::cout << "Found inconsistency in number of steps. Expected "
                    << found->second << ", but found " << steps
                    << " for divisor " << divisor_orig << '\n';
        }
        return true;
      }
    }
    return false;
  }

 private:
  // a | b | n | a*n+b (mod 2)
  // -----------------
  // 0 | 0 | * | 0
  // 0 | 1 | * | 1
  // 1 | 0 | * | ?
  // 1 | 1 | * | ?
  bool step() {
    bool success{false};
    if (divisor % 2 == 0) {
      if (remainder % 2 == 0) {
        divisor /= 2;
        remainder /= 2;
      } else {
        detect_overflow(3);
        divisor *= 3;
        remainder *= 3;
        ++remainder;
        if (divisor == remainder) {
          remainder = 0;
        }
      }
      ++steps;
      success = true;
    }
    return success;
  }

  bool is_smaller() {
    return divisor < divisor_orig ||
           (divisor == divisor_orig && remainder < remainder_orig);
  }

  residue_class create_even() const {
    residue_class even{*this};
    detect_overflow(2);
    even.divisor *= 2;
    even.divisor_orig *= 2;
    return even;
  }

  residue_class create_odd() const {
    residue_class odd{*this};
    odd.remainder += odd.divisor;
    odd.remainder_orig += odd.divisor_orig;
    detect_overflow(2);
    odd.divisor *= 2;
    odd.divisor_orig *= 2;
    odd.remainder %= odd.divisor;
    odd.remainder_orig %= odd.divisor_orig;
    return odd;
  }

  void print(std::ostream& stream) const {
    stream << divisor_orig << "n+" << remainder_orig << "->" << divisor << "n+"
           << remainder << " in " << steps << "\n";
  }

  void detect_overflow(std::int8_t modulus) const {
    if (divisor >= std::numeric_limits<T>::max() / modulus) {
      std::cout << "Overflow detected: ";
      print(std::cout);
    }
  }

 private:
  T divisor;
  T remainder;
  T steps{0u};
  T divisor_orig;
  T remainder_orig;
};

int main() {
  std::ofstream file{"out.txt"};
  using T = std::int64_t;
  std::map<T, T> steps_by_divisor;
  std::queue<residue_class<T>> todo;
  todo.emplace(2, 0);
  todo.emplace(2, 1);

  std::size_t solved{0u};
  for (auto j{1}; j < 7; ++j) {
    for (auto i{0}; i < std::pow(10, j) && !todo.empty(); ++i) {
      auto modular{std::move(todo.front())};
      todo.pop();
      auto followup{modular.solve(file, steps_by_divisor)};
      if (followup.empty()) {
        ++solved;
      }
      for (auto& task : followup) {
        todo.emplace(std::move(task));
      }
    }
    std::cout << "Summary: " << todo.size() << " unknowns and " << solved
              << " solved with ratio "
              << static_cast<double>(solved) / static_cast<double>(todo.size())
              << '\n';
  }

  std::size_t unknowns{0u};
  while (!todo.empty()) {
    auto modular{std::move(todo.front())};
    todo.pop();
    if (modular.solve_light(file, steps_by_divisor)) {
      ++solved;
    } else {
      ++unknowns;
    }
  }

  std::cout << "Summary: " << unknowns << " unknowns and " << solved
            << " solved with ratio "
            << static_cast<double>(solved) / static_cast<double>(unknowns)
            << '\n';

  std::cout << "log2(a)\t#steps\n";
  for (auto const& pair : steps_by_divisor) {
    std::cout << std::log2(pair.first) << "\t" << pair.second << '\n';
  }
  return 0;
}
</code>```

Body:
Mostly to get this out of the unanswered queue but this should be fairly simple to show:

$$r\bmod 2= 2^nx+r\bmod 2$$ so until all the factors of 2 are gone, the two have the same pattern in dropping. The residues $r$ that survive $n=11$ without dropping below their starting points are:

$$27, 31, 47, 63, 71, 91, 103, 111, 127, 155, 159, 167, 191, 207, 223, 231, 239, 251, 255, 283, 303, 319, 327, 359, 383, 411, 415, 447, 463, 479, 487, 495, 511, 539, 543, 559, 603, 615, 623, 639, 667, 671, 679, 703, 719, 743, 751, 763, 767, 795, 799, 831, 839, 859, 871, 879, 895, 927, 935, 959, 991, 1007, 1019, 1023, 1051, 1055, 1071, 1087, 1095, 1115, 1127, 1135, 1151, 1179, 1183, 1191, 1215, 1231, 1247, 1255, 1263, 1275, 1279, 1307, 1327, 1343, 1351, 1383, 1407, 1435, 1439, 1471, 1487, 1503, 1511, 1519, 1535, 1563, 1567, 1583, 1627, 1639, 1647, 1663, 1691, 1695, 1703, 1727, 1743, 1767, 1775, 1787, 1791, 1819, 1823, 1855, 1863, 1883, 1895, 1903, 1919, 1951, 1959, 1983, 2015, 2031, 2043, 2047$$

These are the only residues that could be minimal escapes or minimum in a loop.


问:
Almost sure linear independence of random vectors
Draw $m$ vectors of $n$ entries at random, $m \leq n$, all the entries are i.i.d. Gaussian random variables with mean zero and unit variance. How to show that these $m$ vectors are linearly independent with probability one?

答:
Say $m \leq n$ and we draw $m$ vectors in $\mathbb{R}^n$ according to any distribution which is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}^n$.  We will inductively on $m$ show that the probability these vectors are linearly dependent is zero.

The probability that the first vector is the zero vector is obviously zero. Now suppose the result true for $m-1$. Then the probability that $m$ vectors are linearly dependent is equal to the probability that the $m$th is a linear combination of the first $m-1$ (all the other failures occur with probability zero by the inductive hypothesis). The set of vectors in the span of the first $m-1$ vectors has Lebesgue measure zero, so you are done.

(Induction is a bit of a crutch here: the idea of the proof is that any proper subspace has Lebesgue measure zero, so the vectors should fall into it with zero probability, but we have to get around the fact that there are infinitely many proper subspaces. A prettier proof would use Lebesgue measure on a Grassmannian.)


答:
NO. If we assume $\exists x: [x\in A \land [P(x) \land \neg P(x)]]$ we can easily obtain a contradiction. Therefore, such a set $A$ cannot exist.

EDIT: In your example, if we assume $\exists x: [x\in A \land [x\in A \iff x \notin A]]$ we can similary obtain a contradiction. Therefore, in this case, too, such a set $A$ cannot exist.

BTW, when you say that a set contains any elements, you rule out that it is empty.

EDIT: Also, if you assume $x\in A \iff x\notin A$, this is already a contradiction. We can only conclude that there cannot exist such a set $A$ and element/object $x$.

We can, however, define an empty set $E$ given set A as follows:

$\forall x: [x \in E \iff x \in A \land x \notin A]$

Title: Is it true that any set which contains elements with contradictory properties is the empty set?

Body:
Here's what I've "discovered" but can't verify.

Let's take a set $A$ defined as "the set of all $x$ which don't belong to $A$".

Obviously this definition wouldn't hold in particular axiomatic systems, but it would in others.

Let's analyze this in NST. $X\in A$ iff $X\not\in A$.

This is a logical contradiction, but it still doesn't prove that $A$ doesn't exist. In fact, I would argue that $A$ is the empty set.

If put all of the elements of $A$ into the empty set, we get $(X∈A \text { iff } X∉A) \text { iff } X\in \emptyset$.

This is a tautology for whatever $X$ and $A$. Now, if we switch two of the atomic sentences, we get $X\not\in A \text { iff } (X\in A \text { iff } X\in\emptyset$). We know this to also always be true, which means that either $A$ doesn't exist or $A$ is the empty set.

Therefore, the empty set does contain every element belonging to any set it doesn't belong to.

Is there some error in my proof?


问:
How to prove that characteristic polynomial divides the $n$-th power of the minimal polynomial?
Let $M \in\textrm{Mat}_{n\times n}(\mathbb{C})$ be a matrix with complex coefficients, $\textrm{char}_M(X)$ its characteristic polynomial and $m_M(X)$ its minimal polynomial.

How do I prove that char$_M(X)$ divides $m_M(X)^n$ using that in $\mathbb{C}$, every monic polynomial of degree $d$ factors as $\prod_{i=1}^d(X-a_i)$?

答: Hint: You should use the fact that if $\lambda$ is an eigenvalue of $M$, then $(X - \lambda)$ divides $m_M(X)$. In other words, every linear factor of $\operatorname{char}_M$ divides $m_M$.


问:
What happens to the trace if you multiply with orthogonal matrices
Suppose we are given matrices $A$ and $Q$. Furthermore denote by $U$ and $V$ orthogonal matrices. For a matrix $Q'$ the equality $Q'=UQV^T$ holds. Since $U$ and $V$ are orthogonal matrices, the matrices $Q$ and $Q'$ have the same singular values. Denote by $\langle A, B \rangle = \mbox{tr} (A^T B)$ the Frobenius inner product for two matrices $A$ and $B$. A fellow student now claims that $\langle Q, A \rangle = \langle Q', A \rangle$, but unfortunately I do not see why. Does somebody know if this is true?

答:
No, the statement is not true. As a counterexample, consider
$$
A = Q = V = \pmatrix{1&0\\0&1}, \quad U = \pmatrix{0&1\\1&0}.
$$


问:
Invertible operator
Let $K:V\to W$ such that $Kf = k$, where $V,W$ are infinite-dimensional Banach spaces. Is it correct to say that in general $f = (K^*K)^{-1}k$, however, when $V=W$, then $f = K^{-1}k$. $T^*$ here denotes the adjoint operator. In other words if $V\ne W$, we can't really talk about $K^{-1}$. I'm just trying to draw a parallel with $K$ being a real-valued matrix.</p>

答: Inverse operators are a bit more abstract than what one may first think. The notation and ease of defining inverses from linear algebra is misleading. Really, the inverse operator (when it is well-defined) takes $w\in W$ and gives $v\in V$, with the added condition that $Kv = w$. This is the definition of the inverse operator. What the operator looks like really depends on the context. For example, the inverse operator to a differential operator is a Green's function which you can't nicely represent in terms of the PDE itself in the way you're suggesting. You could, in fact, define an inverse for a linear transformation from $V$ to $W$ if $\dim(W) < \dim(V)$ but you would need to make some specifications on the kinds of output you can have because the system is underdetermined. The way linear algebra is taught at the introductory level is actually a bit detrimental, I think. The notation and somewhat lack of clarity makes functional analysis a lot more difficult than it needs to be.

Title: Invertible operator


问:
Number of possible solutions to $2 \le |x-1||y+3| \le 5$, where $x$ and $y$ are negative integers

If $$2 \le |x-1||y+3| \le 5$$ and both $x$ and $y$ are negative integers, find the number of possible combinations of $x$ and $y$ .

Below is my solution approach :-

As $x$ is a negative integer, hence $|x-1|$ in the $2 \le |x-1||y+3| \le 5$ will be come $-(x-1)$ or $(1-x)$ and the main equation will transform into $2 \le (1-x)|y+3| \le 5$.

$1st$ case when $y+3 \ge 0 \Rightarrow y \ge -3 \Rightarrow y \in \{{-3,-2,-1}\} $ as $y$ is a negative integer :

For $y=-3$ we can see that there is no valid solution for $x$ as $|y+3|$ part will become $0$, hence this case is invalid.

For $y=-2$ we get the solution set for $x$ to be $x \in \{{-4,-3,-2,-1}\} $  and total number of solutions possible in this case is $4$.

For $y=-1$ we get the solution set for $x$ to be $x \in \{{-1}\} $ and total number of solutions possible in this case is $1$.

So for this $1st$ case when $y+3 \ge 0$ we have in total 5 solutions.

$2nd$ case when $y+3 \lt 0 \Rightarrow y \lt -3$ and in this case $y$ will have infinite values and I am not able to proceed from here.

The answer for the total number of solutions provided is $10$ and you can see that I've been able to find out the $5$ solutions in my $1st$ case. Can someone please guide or help me about how to proceed in the 2nd case?

Thanks in advance !

Body:
If $x,y$ are integers then $|x-1|$ and $|y+3|$ are non negative integers.

So just rewrite them as $m,n$ and we have $1 \le mn \le 5$.

Those are small and simple numbers.

We can have $nm = 1,2,3,4,5$

But you said $x,y$ are both *negative integers.  So $x \le -1$ and so $x -1 \le -2$ so $|x- 1| \ge 2$.

so we can have:

$nm = 2$ and $|x-1| =2$ and $|y + 3| = 1$ so $x-1= -2$ and $x =-1$ while $|y+3| = 1$ so $y+3 = \pm 1$ so $y = 1-3=-2$ or $y = -1 -3 = -4$.

or

$nm =3$ and $|x-1| = 3$ and $|y+3| = 1$ so $x-1 =-3$ and $x = -2$ while, as above, y=-2, -4$.

or $nm = 4$ and either $|x-1|=2$ and $|y+3|=2$.  Or $|x-1| =4$ and $|y+3| =1$.

If $|x-1|=|y+3|= 2$ then $x = -1$ and $y+3 =\pm 2$ so $y = -1; -5$.

And if $|x-1|=4$ and $|y+3| =1$ then $x-1 = -4$ and $x =-3$ and $y= -2,4$ (as above).

And if $nm = 5$ then $|x-1| = 5$ so $x-1 = -5$ and $x =-4$ and $|y+3| = 1$ and $y=-2,-4$.

So all solutions are.

$(x,y) = \{$

$(-1, -2),(-1, -4), (-2,-2),(-2,-4), (-1,-1),(-1,-5), (-3,-2)(-3,-4), (-4,-2),(-4,-4)$

$\}$.


问:
Estimating the remainder term in $(1+x)^s$, $s\in\mathbb R^+$
If $s \in \mathbb R^+$ and $-1<x<1$, Then we have:

$$(1+x)^s=1+{s\choose 1}x+{s\choose 2}x^2+...+{s\choose n}x^n+R_{n+1}$$
Where $R_{n+1}$ is the remainder term, and $${s\choose k}=\prod_{i=0}^{k-1}\frac{s-i}{k}, k>0$$

I'm having a hard time trying to estimate the remainder term, it should be the usual integral (Taylor Series) but I don't know how to estimate it.

Edit:

since $\mid(1+x)\mid^s<2^s$, $\forall s\in \mathbb R^+$ so $$\mid R_n\mid\le 2^s \frac{\mid x\mid^n}{n!}$$
is that correct?

答:
We can write our formula succinctly in terms of the Gamma function. Given $|x|<1$ we can write
$$(1+x)^p=\sum_{k=0}^\infty \frac{\Gamma(p+1)}{\Gamma(p-k+1)}\frac{x^k}{k!}$$
Let's stop the sum at some positive integer $n$, in particular, let $n=\lfloor p\rfloor$.
$$(1+x)^p=\sum_{k=0}^n \frac{\Gamma(p+1)}{\Gamma(p-k+1)}\frac{x^k}{k!}+\sum_{k=n+1}^\infty \frac{\Gamma(p+1)}{\Gamma(p-k+1)}\frac{x^k}{k!}$$
The $\Gamma(p+1)$ is a constant. Let's write this as
$$(1+x)^p=\sum_{k=0}^n \frac{\Gamma(p+1)}{\Gamma(p-k+1)}\frac{x^k}{k!}+\Gamma(p+1)\sum_{k=n+1}^\infty a_kx^k$$
With $a_k=\left(\Gamma(p-k+1)k!\right)^{-1}$. Our task now is to show that $|a_k|$ is decreasing.
We have
$$a_{k+1}=\frac{1}{\Gamma(p-k-1+1)(k+1)!}$$
$$=\frac{1}{\Gamma(p-k)k!}\cdot\frac{1}{k+1}=\frac{1}{\Gamma(p-k+1)k!}\cdot\frac{p-k}{k+1}=\frac{p-k}{k+1}a_k$$
Since $k\geq n+1=\lfloor p\rfloor+1$, $|(p-k)/(k+1)|\leq 1$. Hence $|a_k|$ is decreasing. So, we can bound the sum by the first coefficient:
$$\left|\sum_{k=n+1}^\infty a_kx^k\right|\leq\left|\sum_{k=n+1}^\infty a_{n+1}x^k\right|$$
But this sum is now easy, and can be evaluated with a geometric series (again remember $|x|<1$). So
$$\left|\sum_{k=n+1}^\infty a_{n+1}x^k\right|=\left|a_{n+1}x^{n+1}\sum_{k=0}^\infty x^k\right|=\left|\frac{x^{n+1}}{\Gamma(p-k)(n+1)!(1-x)}\right|$$
So going back to our original expansion
$$(1+x)^p=\sum_{k=0}^n \frac{\Gamma(p+1)}{\Gamma(p-k+1)}\frac{x^k}{k!}+\underbrace{\Gamma(p+1)\sum_{k=n+1}^\infty a_kx^k}_{R_{n+1}}$$
We have
$$|R_{n+1}|\leq\left|\frac{\Gamma(p+1)x^{n+1}}{\Gamma(p-n)(n+1)!(1-x)}\right|$$
You can check this numerically on Desmos.


答: Since every polynomial splits, this is evident given that all the zeros of the characteristic polynomial are zeros of the minimal polynomial.  Note that the algebraic multiplicity of a zero of the characteristic polynomial can't exceed $n$.  So the $n$-th power of the minimal polynomial has all the factors of the characteristic polynomial in powers of greater than or equal order.

Title: How to prove that characteristic polynomial divides the $n$-th power of the minimal polynomial?

Body:
Let $M \in\textrm{Mat}_{n\times n}(\mathbb{C})$ be a matrix with complex coefficients, $\textrm{char}_M(X)$ its characteristic polynomial and $m_M(X)$ its minimal polynomial.

How do I prove that char$_M(X)$ divides $m_M(X)^n$ using that in $\mathbb{C}$, every monic polynomial of degree $d$ factors as $\prod_{i=1}^d(X-a_i)$?


问:
How can I calculate the size of a square block and the number of rows and columns needed to fit a known number of blocks on a page of known size?
I am a web developer and have a problem to solve that I thought might be suitable on here...

I am developing an app that will be used on a range of phones and tablets all with different screen sizes. One of the pages of this app will show a grid full of square buttons. The number of buttons to show will change each time the app is used. I want to fill the page as much as possible with the buttons and space them evenly on the page in a grid. The page will always be in landscape format (wider than it is high - sorry to state the obvious).

So my question is this - how can I calculate

1) The size of the square buttons
2) The number of rows
3) The number of columms
4) The distance between each button so they are spaced evenly

given that when the app loads i will know

a) The width of the screen
b) The height of the screen
c) The number of buttons to show

Until the app loads I will not know any of these values.

The number of buttons is likely to be in the region of 60 or so. SO for example if I have 60 buttons to show on an phone with width of 480 pixels and height of 270 pixels, I will show 10 columms and 6 rows, each block will be about 35 pixels square and I will space them out by 10 pixels. My grid will thus be (10*35)+(10*10) = 450 wide and (6*35)+(6*10) = 270 high.

答:
As long as there is no analytical solution, you may just enumerate possibilities and test them.

For a number of rows $r$ from $1$ to a square root of a number of buttons $\sqrt N$ (rounded down to the nearest integer), divide the screen's height by $r$ to calculate a button's size: $s=H/r$ (rounded down to a whole number of pixels). Then divide the screen's width by the button's size (and round down) to determine a number of columns that will fit: $c=W/s$. Multiply them to see if the resulting grid has enough cells to accomodate all your buttons: $c\cdot r \ge N.$

From all results satisfying the final inequality choose the one with a biggest $s$.

For $N$ up to $120$ it will take up to $\left\lfloor\sqrt{120}\right\rfloor = \sqrt{100} = 10$ tests which seems pretty affordable. :)


问:
Joint density function of $f_\theta(x) = \theta x^{\theta-1}$
Let $f_\theta (x) = \theta x^{\theta-1}$, where $0<x<1$, $x_1, x_2$ are iid

Find $P(x_1x_2 \geq 0.75)$

I used double integration to obtain the joint probablity,

$$\begin{align*}
P(x_1x_2 \geq 0.75) &= \int_{0.75}^1 \theta x_1^{\theta - 1} \int_{0.75/x_1}^1 \theta x_2^{\theta - 1} dx_2 dx_1 \\
&= \dots \text{skip the steps of integration} \\
&= 1 - 0.75^\theta + 0.75^\theta \cdot ln0.75
\end{align*}$$

However, refer to For the density function $f_\theta (x) = \theta x^{\theta-1}$, how can we find the product distribution, $\prod_{i=1}^n X_i$?

Should $P(x_1x_2 \geq 0.75) = P(Gamma(2,\theta) \geq 0.75)$?

The answers are different using two approaches.

Update

I think I got it,

$$\begin{align*}
P(x_1x_2 \geq 0.75) &= P(exp(-Gamma(2,\theta)) \geq 0.75) \\
&= P(Gamma(2,\theta) \leq -ln0.75)
\end{align*}$$
Is this right?

答:
to get a gamma distribution you transformed $X_1X_2$ in a certain way...you have to do the same with the requested quantile...

$$P(G< -\log0.75)$$


问:
Differential equation with start condition and maximum capacity
The house with 1000 people have got 1 infected with an unknown virus.
The number of students who are newly infected is proportional to the number of healthy product of current and the current number of infected students. After 2 days is 5 people infected. How many of them will be infected in a week?

I have calculated this with diferencial equation and i have got 1000 people. Is that OK??

Thanks for answers

My solution:

MAX people: 1000

day: 1 infected

day: 5 infected

$P (7) = (1000* (\frac{2}{998} e^{10})*e^{5*7}) / ( 1 + (\frac{2}{998} e^{10}) * e^{5*7}) = 999,99 = 1000$

答:
Let $y$ be the number infected at time $t$, where $y(0)=1$. The model we are asked to use is
$$y'=ky(1000-y).$$
This is a separable differential equation, which can be rewritten as
$$\frac{dy}{y(1000-y)}=k\,dt,$$
and then, using partial fractions, as
$$\frac{1}{1000}\left(\frac{1}{y}+\frac{1}{1000-y}\right)\,dy=k\,dt$$
and then as
$$\left(\frac{1}{y}+\frac{1}{1000-y}\right)\,dy=1000k\,dt.$$
Integrate. We get, since $y$ will stay between $1$ and $1000$,
$$\ln\left(\frac{y}{1000-y}\right)=1000kt+C.\tag{1}$$
Since $y(0)=1$, we get $C=\ln(1/999)=-\ln 999$.
We have $y(2)=5$. Substituting in our equation we get
$$\ln(5/995)=2000k -\ln 999,$$
and therefore
$$k=\frac{1}{2000}\ln(999/199).$$
Finally, take the exponential of both sides of (1). We get
$$\frac{y}{1000-y}=\frac{e^{kt}}{999},$$
where $k$ is known. Set $t=7$. With some manipulation, we get a linear equation for $y(7)$. Solve.

Remark: The model should not be taken entirely seriously. One can expect at best a rough fit with reality.


问:
Characterizing Totally Ramified Extensions
I am trying to understand two portions of the proof of Theorem 2.4 here.

Let $L/K$ be an extension of nonarchimedean fields. Say the valuation on $K$ is $v$ extending to $w$ on $L$. A polynomial $f = a_nx^n + ... + a_1x + a_0 \in K[x]$ is called Eisenstein if $v(a_0) = 0$, $v(a_i) \geq 1$ for all $1 \leq i < n$, and $v(a_n) = 1$.

Let $\pi_L$ be a uniformizer of $L$. First, why does every root $\alpha_i$ of a degree $n$ Eisenstein polynomial over $K$ having extended valuation $w(\alpha_i) = 1/n$ force the ramification index $e = (w(L^*):v(K^*)) = v(\pi_L)$ to be at least $n$?

Second, suppose you have $\alpha \in L$ such that $w(\alpha) = 1$ and that $L/K$ is totally ramified; that is $e := v(\pi_L) = [L:K]$.  It follows that the $1,\alpha,...,\alpha^{n-1}$ form a $K$-linear independent set, so

$$a_0\alpha^n + a_1\alpha^{n-1} ... + a_{n-1}\alpha + a_n = 0$$

The proof then claims that it must be the case that there exist $i,j$ such that $w(a_i\alpha^i) = w(a_j\alpha^j)$ are of maximal valuation. Why? And why does this then force $f = \sum a_ix^i$ to be Eisenstein?

答:
It can be a bit confusing with various different normalizations of the valuation, but let's agree to say that $v(\pi_K) = 1$.

If $L/K$ is a finite extension, and $\pi_L$ is a uniformizer of $L$, then since $\pi_K \in L$ you can write

$$\pi_K = u \cdot \pi^e_L$$

for some unit $u \in \mathcal{O}^{\times}_L$. The number $e$ here the ramification index. The valuation $v$ extends to $L$ and you see you have the formula

$$1 = v(\pi_K) = e v(\pi_L)$$

so $v(\pi_L) = 1/e$.

Now suppose that $L/K$ is Eisenstein, so $L = K(\alpha)$. From the Newton Polygon, you can deduce that $v(\alpha) = 1/n$. But you can also write $\alpha = u \pi^r_L$ in $L$ for some $r$, and then you see that

$$1/n = v(\alpha) = r v(\pi_L) = r/e,$$

so $e = nr$. But you also know that $e \le n$, because otherwise $\pi_L$ could not satisfy a degree $\le n$ polynomial over $K$. So you deduce not only that $e = n$ but also that $r = 1$, that is, that $\alpha$ is a uniformizer of $L$.

For the second question, the $p$-adic valuation has the following behavior: if $\alpha$ and $\beta$ are elements of $L$, then $v(\alpha + \beta) \ge \min\{v(\alpha),v(\beta)\}$. But it actually satisfies a stronger property: if $v(\alpha)$ and $v(\beta)$ are different, then $v(\alpha+\beta) = \min\{v(\alpha),v(\beta)\}$. By induction, that means that if you add together $p$-adic numbers all with different valuation then the valuation of the sum is the smallest valuation of any term. But the sum in this case is $0$, which has infinite valuation, so this can't happen, that is, there must be two terms with the same minimal (not maximal) valuation.

Now back to the particular sum. First of all, we want to make sure we don't get confused about valuations. The implicit normalization here is that $\alpha$ is a uniformizer of $L$ and $v(\alpha)=v(\pi_L)=1$. Then this implies that $e=n=[L:K]$ and then $v(\pi_K)=n$. That is, the valuation of any element of $K$ is divisible by $n$. But now look at the sum.

One can certainly assume that that $a_i$ are all in $\mathcal{O}_K$ and are not all divisible by $\pi_K$. That means they all have valuations which are divisible by $n$. Moreover, there is at least on $k$ with $a_k$ a unit. Consider the two terms of minimal valuation: since $v(\alpha) = 1$ and $v(a_i) \in n \mathbf{Z}$, the valuation of the $k$th term is $k \bmod n$. So the only way that two terms can have equal valuation is if they are the first and last terms. Moreover, these are the terms of smallest valuation. The valuation of $a_0 \alpha^n$ is at least $n$ since $a_0$ is assumed to be in $\mathcal{O}_K$. But we are assuming that there is a $k$ with $a_k$ a unit; the corresponding term must have valuation exactly $v(a_k \alpha^k) = k \le n$. So the only way the first and last terms can have the smallest equal valuation and the

$a_0$ is a unit, so $v(a_0 \alpha^n) = n$.

$a_n$ is exactly divisible by $\pi_K$, so $v(a_n) = n$.

$a_k$ for $1 \le k \le n-1$ are all divisible by $\pi_K$ --- otherwise $v(a_k \alpha^k) = k < n$.

These together say exactly that the polynomial is Eisenstein.


问:
Measure-valued function as transition kernel
Let $X$ and $Y$ be Polish spaces. Consider $\mathcal{M}(Y)$, the space of (nonnegative) measures on $Y$, as a Polish space, equipped with the topology of weak convergence of measures.

Let $\mu_x:X\rightarrow \mathcal{M}(Y)$ be a Borel function. Is it then automatically the case that $f$ induces a transition kernel? In other words, does it follow that for any Borel subset $B$ of $Y$, we have that $ x \mapsto \mu_x (B)$ is measurable?

答: At least for probability measures, the answer is yes. This is a consequence of the result in this math stackexchange article, and is also stated and proved explicitly as Proposition 7.26 in Bertsekas and Shreve's stochastic control book.</p>


问:
Finding extremum of quadratic function using factorisation as symmetry operation.
In this book in Section 3.2.3, the author shows how to find extremum of a quadratic function using an invariant quality and symmetry operation, which preserves it.

E.g. to find the extremum of $f(x)=6x-x^2$ we can notice that $f(x)=6x-x^2 = x(6-x) = (6-x)x$. This operation maps every value $x$ to $6-x$ through the axis of symmetry and vice-versa (e.g. 0 is mapped to 6 and 6 mapped to 0). This operation preserves symmetry for $x=3$ - the axis of symmetry and hence the extremum of the function.

What about the function $f(x)=6x+x^2 = x(6+x) = (6+x)x$?
We know that the minimum of this function is at $x=-3$ but $6+(-3)=3$.
Also this operation maps e.g $4$ to $10$, but $10$ to $16$! Hence, this approach fails with this example (symmetry is not preserved).

Why is that so? Why is this approach not working for all the quadratic functions?

答:
If $f(x)=f(c-x)$ for all $x$, then $x=\frac{c}2$ is an axis of symmetry since $$f\left(\frac{c}2+\epsilon\right)=f\left(c-\left(\frac{c}2+\epsilon\right)\right)=f\left(\frac{c}2-\epsilon\right)$$

Since $f(x)=x(6-x)=f(6-x)$, the axis of symmetry is $x=3$.

Since $f(x)=6x+x^2=-(-6x-x^2)=-x(-6-x)=f(-6-x)$, the axis of symmetry is $x=-3$.

Now, let's consider a general quadratic equation $f(x)=ax^2+bx+c$.

We have

$$f(x)=-ax\left( -\frac{b}a-x \right)+c=f\left( -\frac{b}{a}-x\right),$$

The axis of symmetry is $x=-\frac{b}{2a}$.


问:
show $\lim_{x\rightarrow\infty} x^n\exp(-x^2)=0$
I want to show for all $n\in\mathbb N$
$$\lim_{x\rightarrow\infty}\frac{x^{n}}{\exp(x^2)}=0$$

I am pretty sure that I have to use L'Hospital. I've tried induction:

$n=1$: $$\lim_{x\rightarrow\infty}\frac x{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac1{2x\exp(x^2)}=0$$

And for $n\rightarrow n+1$:
$$\lim_{x\rightarrow\infty}\frac{x^{n+1}}{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac{(n+1)x^n}{2x\exp{x^2}}$$

And now I am stuck. The term $2x$ really annoys my for my induction hypothesis.

Any hints?

答:
Applying L'Hospital $n$ times would do the trick too (assuming $n$ is fixed):

$$
\lim_{x\to\infty}\frac{x^{n}}{e^{x^2}} = \lim_{x\to\infty}\frac{(x^{n})'}{(e^{x^2})'} = \lim_{x\to\infty}\frac{nx^{n-1}}{e^{x^2} 2x},
$$
again we have an $\infty/\infty$ type limit, repeating L'Hospital until we get a constant in the denominator:
$$
\lim_{x\to\infty}\frac{(x^{n})^{(n)}}{(e^{x^2})^{(n)}} =  \lim_{x\to\infty}\frac{n!}{e^{x^2} p(x)} = 0,
$$
where $p(x)$ is a degree $n$ polynomial in $x$.


问:
Find numbers from GCD and LCM
two numbers gcd and lcm are respectively 6 and 600. What is the possible pairs of two numbers?</p>

答:
If $(a,b)$ is the pair then $a=2^x3^y5^z$ and $b=2^u3^v5^t$ and the requirements are: $\min(x,u) = 1, \max(x,u) = 3, \min(y,v) = 1, \max(y,v) = 1, \min(z,t) = 0, \max(z,t) = 2$.

Therefore there are $2\times 1 \times 2=4$ possibilities for $x,y,z,u,v,t$ and they turn into the following pairs for $(a,b)$: $(600,6),(150,24),(24,150),(6,600)$

Title: Find numbers from GCD and LCM


问:
Does the matrix product $X^T X$ have a special meaning?
I have come across this specific matrix product several times, lately in the context of stochastic linear models where it is an integral part of the Least Squares Estimator (LSE). Often times in linear algebra there is some beautiful intuition hidden behind recurring formulae and since I don't see the one behind this one I'm asking for help.

Is there a geometric interpretation or special meaning for the matrix product

$$X^T X$$

for a matrix $X \in \mathbb{R}^{n \times p}$ with $p \leq n$?

答:
Suppose you have an inner product space, but what you care about is not the dot product of the two vectors, but the dot product of their images under $X.$ Well,

$$\langle X a, X b\rangle = a^t X^t X b.$$

Note that the correct way to think of $X^t X$ is as a quadratic form, not so much a linear transformation.


答: Hint: Setting $a=6a',\;b=6b'$,  you know that $$a'b'=\frac{ab}{36}=\frac{6\cdot 600}{36}=100$$ and $a', b'$ are coprime.

Title: Find numbers from GCD and LCM

Body: two numbers gcd and lcm are respectively 6 and 600. What is the possible pairs of two numbers?</p>


问:
Real characters of odd degree
Suppose that $G$ is a finite group and $\chi$ is an irreducible real character, namely that $\chi(g) \in \mathbb{R}$ for every $g \in G$. Is it true that if $\chi(1)$ is an odd number, then $\chi$  is afforded by a representation over real numbers?

Motivation, let $m_{\mathbb{R}}(\chi)$ the Schur index of $\chi$ as defined in $10.1$ of Isaacs' "Character Theory of Finite Groups". If $\mathcal{X}$ is a complex representation that affords $\chi$, then there is an $\mathbb{R}$-representation $\mathcal{Y}$ of twice the dimention of $\mathcal{X}$ such that, viewed as complex, its character is $2\chi$. Hence the Schur $m_{\mathbb{R}}(\chi)$ should be $\le 2$. Nevertheless, by point $h)$ of Corollary $10.2$ of Isaacs' book, we have that $m_{\mathbb{R}}(\chi)$ divides $\chi(1)$ that is odd.

答: You don't need Brauer-Speiser or anything complicated, just the Frobenius-Schur indicator and some linear algebra. Since $\chi$ is a real character, the indicator is either $+1$ or $-1$, not $0$. $+1$ means it is a real representation, $-1$ means it is a quaternionic representation. But quaternionic representations preserve an alternating form, and this requires the dimension to be even. So all odd-dimensional (irreducible) representations are of $+$ type, i.e., are real representations.

Title: Real characters of odd degree


问:
Multi-linearity of determinants
Given a set $S = \{(a,b,c) \in \mathbb{R^3} : \det A = 0\}$, where $A = \begin{pmatrix} a & b & c \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}$. How do we show it is a subspace and find its dimension? For example I thought I could use multi-linearity of determinants to show its a subspace but I'm not sure how the proof would look like to be complete. Also with multi-linearity, I'm not a 100% sure i understand it correctly. This is what I understood for multi-linearity  $$ \det \begin{pmatrix} c_1a & c_2b & c_3c \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix} = c_1 \det \begin{pmatrix} a & c_2b & c_3c \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix} + c_2 \det \begin{pmatrix} c_1a & b & c_3c \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix} + c_3\det \begin{pmatrix} c_1a & c_2b & c \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}$$

This doesn't seem right to me.

答:
Outline:

Note that $\det A=-(b-c)$. So your set is rewritten as $$\{(a,b,c) \in \Bbb R^3: b-c=0\}$$

This set is just the homogeneous solution of $Ax=0$ where $A= \begin{pmatrix} 0&1&-1\end{pmatrix}$ and $x=\begin{pmatrix} a&b&c\end{pmatrix}^t$, which is clearly a subspace of dimension $2!$

Title: Multi-linearity of determinants


答:
Your understanding of multilinearity is incorrect. For the context of this problem, the important thing to understand is that for vectors $(a_1,b_1,c_1),(a_2,b_2,c_2)$ and scalars $k_1,k_2$, we have
$$
\det \pmatrix{k_1 a_1 + k_2 a_2 & k_1b_1 + k_2 b_2 & k_1c_1 + k_2 c_2\\
1&1&1\\
0&1&1} =\\
k_1\cdot \det \pmatrix{a_1& b_1& c_1\\
1&1&1\\
0&1&1} +
k_2 \cdot \det
\pmatrix{a_2 & b_2 & c_2\\
1&1&1\\
0&1&1}.
$$
More generally: if $v,w$ are (constant) row vectors and $x = (x_1,x_2,x_3)$ is a row-vector, then the function $T:\Bbb R^3 \to \Bbb R$ defined by
$$
T(x) = \det \pmatrix{x\\v\\w}
$$
is linear. This is true for each of the rows of the determinant, which is to say that the functions
$$
x \mapsto \det \pmatrix{v\\x\\w}, \quad x \mapsto \det \pmatrix{v\\w\\x}
$$
are also linear. The fact that all these maps are linear is what is meant by the "multilinearity" of the determinant.

Title: Multi-linearity of determinants

Body:
Given a set $S = \{(a,b,c) \in \mathbb{R^3} : \det A = 0\}$, where $A = \begin{pmatrix} a & b & c \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}$. How do we show it is a subspace and find its dimension? For example I thought I could use multi-linearity of determinants to show its a subspace but I'm not sure how the proof would look like to be complete. Also with multi-linearity, I'm not a 100% sure i understand it correctly. This is what I understood for multi-linearity  $$ \det \begin{pmatrix} c_1a & c_2b & c_3c \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix} = c_1 \det \begin{pmatrix} a & c_2b & c_3c \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix} + c_2 \det \begin{pmatrix} c_1a & b & c_3c \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix} + c_3\det \begin{pmatrix} c_1a & c_2b & c \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}$$

This doesn't seem right to me.


问:
How can I show this implication for $\frac x{xy+x+1} + \frac y{yz+y+1} +\frac z{zx+z+1} = 1$?
$$(xy+x\ne -1 \land yz+y\ne-1 \land zx+z\ne -1 \land xyz=1)\\
\implies \frac x{xy+x+1} + \frac y{yz+y+1} +\frac z{zx+z+1} = 1$$

I tried finding each variables value using $xyz=1$ $(ex.x=1/yz)$ , and then replacing them but didnt really get anything. Then i tried writing the $1$s as $xyz$ and i could simplify some of them but still nothing. Thank you in advance

答:
Hint:
$$
\frac x{xy+x+1} + \frac y{yz+y+1} +\frac z{zx+z+1}\\
\stackrel{z=\frac1{xy}}=\frac x{xy+x+1}+\frac{xy}{xy+x+1}+\frac{1}{xy+x+1}.
$$

Can you take it from here?


问:
I don't understand how implication works in logic
The statement $(p \wedge q ) \Rightarrow (p \Rightarrow q)$ is a tautology, i.e., always true.

Now, imagine I am writing a proof and I want to show that $p \Rightarrow q$. I show that statement $p$ is always true, then I show that statement $q$ is always true, then I can end my proof and claim $p \Rightarrow q.$

This doesn't make sense to me. If I prove that $\sqrt2$ is irrational and then I prove that $7$ is odd, I don't understand how "$\sqrt2$ is irrational" $\Rightarrow$ "$7$ is odd".

Can someone help this disconnect I am having? Perhaps, you can provide information about what these logical statements actually mean and how they are used.

答:
I show that statement $p$ is always true, then I show that statement
$q$ is always true, then I can end my proof and claim $p \Rightarrow q.$ This doesn't make sense to me.

Using a simple form of natural deduction, we have:

(Screen shot from my proof checker)


问:
Is my proof of irrationality correct?
There is a lot of proof on this site about different numbers being irrational. This is just a generalized version.

THE STATEMENT:-

If $\forall c,k\in\mathbb{N},N\in\mathbb{N}:N\neq c^k$

then $\sqrt[k]{N}$ is irrational.

MY PROOF:-

Let $a,b\in\mathbb{Z}:\gcd(a,b)=1$

$\implies \gcd(a^k,b^k)=1$, call it $eq(1)$

Assume $\sqrt[k]{N}=\frac{a}{b}$

$\implies N=\frac{a^k}{b^k}$

$\implies a^k=Nb^k$

$\implies \gcd(a^k,b^k)=\gcd(Nb^k,b^k)=b^k$, call it $eq(2)$

Now, combining $eq(1)$ and $eq(2)$ would give

$b^k=1$

$\implies b=1$

$\implies \sqrt[k]{N}=a$

$\implies non-integer=integer$, which is a Contradiction

Which means our assumption was wrong and $\sqrt[k]{N}$ is irrational.$QED$

MY QUESTION:-

Is my proof entirely valid or is there a problem and if there is a problem then what is it?

答: The proof is entirely valid. Although you are using the fact that $(a,b) = 1 \implies (a^k,b^k) = 1$. Which of course is easy to prove with prime factorization.

Title: Is my proof of irrationality correct?


答:
@Jorge's answer should give you the way to exhaust your solutions. But I want to help you gain some insight as to why the formulas he gave actually work.

You can use a couple of interesting facts to get an idea. The first is the fundamental theorem of arithmetic, which allows you to decompose your integers into a unique product of prime numbers.

Secondly, the GCD is the "meet" of two numbers, and the LCM is the "join" of two numbers. Anytime you have a meet or join of $a$ and $b$ in mathematics, you have

$$meet(a,b) \times join(a,b) = a \times b$$

So:

we have $a \times b = 3600 = 36 \times 100 = 6^2 \times 10^2 = 2^4 \times 3^2 \times 5^2$;

we have $meet(a, b) = 2 \times 3$;

and we have $join(a, b) = 2^3 \times 3 \times 5^2$,

by the fundamental theorem of arithmetic.

For all your solutions, both $a$ and $b$ will have the factors of the meet in common, and the factors of the join-divided-by-the-meet be unique to one of the factors. Also, the sum of powers of each factors of $a$ and $b$ must sum to the exponent for the same factor in $a \times b$ (because $a^b a^c = a^{b+c}$). So for example, one such pair is $a = 2 \times 3 \times 5^2 = 150$ and $b = 2^3 \times 3 = 24$.

Title: Find numbers from GCD and LCM


问:
Asymptotic of series
Let $x_n$ be a sequence of positive numbers with the property that

$$\sum_n \frac{x_n}{n \log(n)^3}< \infty.$$

Can we show that

$$\limsup_{\varepsilon \downarrow 0}\frac{\sum_{i=1}^{\infty} \exp(-\varepsilon x_n)}{\sum_{i=1}^{\infty} \exp(-\varepsilon \log(i)^2)}>0?$$

答:
This isn't a complete solution, but it shows that the limit is $+\infty$ for "most" $t$ (the meaning will be made precise near the end -- here my $t$ is your $\varepsilon$). The proof shows that the "large" terms of $x_n$ can be chosen so that the sum of the $e^{-t(\ln{n})^2}$ (with $n$ running over the indices of the large terms) can be (for some $t$, relatively few) the most significant part of the sum $\sum_{n \geq 1}{e^{-t(\ln{n})^2}}$.

First, it's important to get a good idea of how the series $F(t)=\sum_{n \geq 1}{e^{-t(\ln{n})^2}}$ behaves for small $t > 0$. We'll be very crude in the estimate, but the result will be more than sufficient.

By series-integral comparison, it's easy to see that $|F(t)-F_i(t)| \leq 2$, where $F_i(t) = \int_0^{\infty}{e^{-t(\ln{u})^2}du}$, because the part from $0$ to $1$ is bounded in modulus by $1$, and the general term is a (smooth) decreasing function of $n$ bounded by $1$.

Writing $u = e^v$, we see that $F_i(t) = \int_{-\infty}^{\infty}{e^{v-tv^2}dv} = \frac{\sqrt{\pi}}{\sqrt{t}}\exp\left(\frac{1}{4t}\right)$ by completing the square.

We can obviously assume $x_1=x_2=0$ (that's not very important but it makes the computations simpler).

So, let now $S = \{n \geq 3,\, 2x_n \geq \ln^2{n}\}$: we know that $\sum_{n \in S}{\frac{1}{2n\ln{n}}} < \infty$.

For each $p \geq 1$, let $N_p = \{n \in S,\, e^p < n \leq e^{p+1}\}$. That the previous series converges can be rewritten as $\sum_{p \geq 1}{\frac{N_p}{pe^p}} < \infty$. Write $u_p = N_pe^{-p}$: it is a bounded nonnegative sequence (by $e-1$) such that $\sum_{p \geq 1}{p^{-1}u_p} < \infty$.

In particular, let $T$ be the set of indices $p$ such that $u_p > 1/10$, and let $b_n$ be an increasing sequence of image $T$. We know that $\sum_{n \geq 1}{\frac{1}{b_n}} \leq \sum_{p \in T}{\frac{10u_p}{p}} < \infty$. By a standard analysis lemma, it follows that $n = o(b_n)$.

Then $$\sum_{n \in S}{e^{-t(\ln{n})^2}} = \sum_{p \geq 1}{\sum_{n \in S \cap (e^p,e^{p+1}]}{e^{-t(\ln{n})^2}}} \leq \sum_{p \geq 1}{u_pe^{p-tp^2}} \leq \frac{1}{10}\sum_{p \geq 1}{e^{p-tp^2}}+(e-1)\sum_{n \geq 1}{e^{b_n-tb_n^2}}.$$

So, define $G(t) = \sum_{n \geq 1}{e^{n-tn^2}}$ for each $t > 0$. We want to show an explicit inequality between $G(t)$ and $t^{-1/2}e^{1/4t}$.

As $x \longmapsto x-tx^2$ is increasing for $0 \leq x \leq 0.5/t$, and decreasing for larger $x$, and as this function is bounded by $1/4t$, it follows that $$0 \leq G(t) \leq 2e^{0.25/t}+\int_0^{\infty}{e^{x-tx^2}dx} \leq 2e^{0.25/t}+e^{.25/t}\int_{\mathbb{R}}{e^{-t(x-1/2t)^2}dx} \leq \left(2+\frac{\sqrt{\pi}}{\sqrt{t}}\right)\exp{\frac{1}{4t}}.$$

Let's study $H(t)=\sum_{n\geq 1}{e^{b_n-tb_n^2}}$ -- we want to show that this quantity is, for small enough $t$, bounded by some $B(1+o(1))\frac{e^{0.25t^{-1}}}{t^{1/2}}$, where $B$ is a constant.

First, let's assume this is the case. Then, for small enough $t$, we thus have $\sum_{n \in S}{e^{-t(\ln{n})^2}} \leq \frac{1}{\sqrt{t}}\exp\left(\frac{1}{4t}\right)(c+o(1))$, where $c < \sqrt{\pi}$ (here $c = B(e-1)+\frac{\pi^{1/2}}{10}$). It follows that there exists some $k > 1$ such that for small enough $t >0$, $F(t) \geq k\sum_{n \in S}{e^{-t(\ln{n})^2}}$, and thus that $\sum_{n \notin S}{e^{-t(\ln{n})^2}} = \Theta(F(t))$. From this, it follows that for small enough $t > 0$,

$\frac{\sum_{n \geq 1}{e^{-tx_n}}}{\sum_{n\geq 1}{e^{-t(\ln{n})^2}}} \geq \Theta\left(\frac{F(t/2)}{F(t)}\right) = \Theta\left(\frac{F_i(t/2)}{F_i(t)}\right) = \Theta(e^{1/4t})$, which obviously concludes.

Now, we still have to compare $H(t)$ to $G(t)$. We split the sum in three:

$\sum_{1 \leq p \leq 0.5t^{-1}-t^{-1/2}}{e^{p-tp^2}} \leq e^{1/4t}+\int_{-\infty}^{1/2t-1/\sqrt{t}}{e^{x-tx^2}\,dx} = e^{1/4t}+e^{1/4t}\int_{-\infty}{-1/\sqrt{t}}{e^{-tu^2}dx} \leq e^{1/4t}+e^{1/4t}\times \frac{0.14}{\sqrt{t}}$.

Similarly, $\sum_{p \geq 0.5/t+t^{-1/2}}{e^{p-tp^2}} \leq e^{1/4t}+e^{1/4t}\int_{1/\sqrt{t}}^{\infty}{e^{-tx^2}dx} <e^{1/4t}(1+0.14t^{-1/2})$.

(Note that, if we had chosen $t^{-2/3}$ instead of $t^{-1/2}$, then the two sums would have been negligible before $\frac{e^{1/4t}}{\sqrt{t}}$ -- so if $T$ contains the whole interval $[1/2t-t^{-2/3},1/2t+t^{2/3}]$, then $H(t)$ is very large and we won't be able to say anything. Fortunately, as explained below, this is very rare).

Thus, if $T_t$ is the cardinality of $T \cap [1/2t-1/\sqrt{t},1/2t+1/\sqrt{t}]$, then $H(t) < (2+T_t+0.28/\sqrt{t})e^{1/4t}$. So if $|T_t| \leq 0.648t^{-1/2}$ (the $0.648$ comes from the formulas for $B$ and $c$ above), we are done.

Except that there's no guarantee that this happens for all $t$ small enough:  if $T = \bigcup_{n \geq 5}{[2^{2n-1}-3^n,2^{2n-1}+3^n]}$, then $|T_{4^{-n}}| = 2^{n+1}+1 > 1.49 \sqrt{4^n}$.

However, what we can say is that, if $0 < t_0 < 1/2$,

$\int_{t_0}^{1/2}{T_t\sqrt{t}\frac{dt}{t}} = 2\sum_{n \in T}{\int_{\sqrt{2}}^{\sqrt{1/t_0}}{1(2n \in [u^2-2u,u^2+2u])\frac{du}{u^2}}}$. Solving the quadratic equations -- since $u^2 = \Theta(n)$ when $u^2-u \leq n \leq u^2+u$ -- yields that $$\int_{t_0}^{1/2}{T_t\sqrt{t}\frac{dt}{t}} \leq C\sum_{n \in T}{n^{-1}} < \infty.$$

In other words: there exists a set $U \subset (2,\infty)$ (here $U$ is the set of $t$ such that $|T_{2/t}|/\sqrt{t/2}$ or $|T_{1/t}|/\sqrt{t}$ is greater than the $0.648$ from above) such that $\int_U{\frac{dt}{t}} < \infty$ and such that if $t_n \rightarrow 0$ and $1/t_n \notin U$, then $\frac{1}{F(t_n)}\sum_{q \geq 1}{e^{-t_n(\ln{q})^2}} \rightarrow \infty$.

This, of course, doesn't settle the problem. There is the question of these "exceptional" $t$. It's surprising that this arises because there's no obvious reason why the behavior should be so "erratic" when $t$ changes.

Note that I haven't used the full power of the original assumption (I think the second one doesn't matter -- we can just make a few terms huge, say $a_n = n(\ln{n})^{5/2}$, at $n=2^{k^k}$) about the convergent series. I just used that most terms would be smaller than $(\ln{n})^2/2$. I'm not sure if this is helpful, because it could be that there are still too many "large" $x_n$ for the contribution of the $e^{-tx_n}$ with $x_n$ "small" to matter (at least for some $t$), or that the $x_n$ aren't small enough to make a difference.

Title: Asymptotic of series


问:
Diagonalization of the following Toeplitz complex symmetric matrix
There is a Toeplitz matrix of the following form:

\begin{equation}
M = \begin{pmatrix}
1 & e^{i\phi} & e^{2i\phi} & \ldots & e^{(N-1)i\phi} \\
e^{i\phi} & 1 & e^{i\phi} &  \ldots & e^{(N-2)i\phi} \\
e^{2i\phi} & e^{i\phi} & 1 &  \ldots & e^{(N-3)i\phi} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
e^{(N-1)i\phi} & e^{(N-2)i\phi} & e^{(N-3)i\phi} & \ldots & 1
\end{pmatrix}.
\end{equation}

where $\phi$ is just some real number. This is a complex symmetric Toeplitz matrix. There is no general algorithm to diagonalize it analytically, and I am not absolutely sure that this is possible. Right now I am in the process of obtaining the eigenvalues (or at least a simple form of the eigenvalue equation).

However, what I do know is the answer in the case of $\phi=0$, then it is just a matrix of all $1$'s, which is a circulant type, and can be diagonalized by the DFT matrix. So, I wonder, may be there is some sort of simple transformation between the $\phi=0$, and $\phi \ne 0$ cases so that I can find the eigenvectors without explicitly finding the Kernel for each $M-\lambda_jI$?

Background

Matrix $M$ describes a certain periodic quantum-mechanical system. Physically, it is somewhat similar to what is called a photonic crystal, where each subsystem interacts with others by means of plane waves (hence the constant amplitude of all entries, and an acquired multiple of $\phi$
in phase).

Update 23.03.2021
So, basically, I was able to obtain an explicit form of the eigenvalue equation:
\begin{equation}
\det \left(M - \lambda I \right) = \dfrac{\left( \beta_+^N - \beta_-^N \right) (1-\lambda) + (\beta_-^N \beta_+ - \beta_+^N \beta_-)}{\beta_+-\beta_-} = 0, \\
\beta_{\pm} = \frac{1}{2} \left( 1-\lambda - e^{i2\phi}(1+\lambda) \pm \sqrt{ (1-\lambda - e^{i2\phi} (1+\lambda))^2 - 4 e^{i2\phi} \lambda^2 } \right),
\end{equation}
which is... fairly complicated. I believe this can not be solved explicitly, and, probably, the eigenvectors can not be expressed exactly. However, there is also a way to solve it by using a physical rationale, mb I will share some materials on that later.

答: Computing the eigensystem with Mathematica indicates that there is nothing particularly enlightening going on. Perhaps in the case where $\exp(i \theta)$ is a root of unity, something may be doable, but in general...


问:
Is the limit function continuous?

Is this function continuous or not? $$y\ =\ \lim _{a\to +\infty }\bigg[\frac{\ln \left(1+e^{\alpha \cdot x}\right)}{\ln \left(1+e^{\alpha }\right)}\bigg]$$

I computed that limit of function as $a\to+\infty$ is $x$. How I proceed from here for proving the continuous part?

答:
Calculating the limit gives
$$y = \lim _{a\to +\infty }\frac{\ln \left(1+e^{a \cdot x}\right)}{\ln \left(1+e^{a }\right)} = \lim_{a\to+\infty}\frac{xe^{ax}(1+e^a)}{e^a(1+e^{ax})} = x$$
and $y=x$ is continuous.

Title: Is the limit function continuous?


问:
Problem regarding functions and sequences

There is just so much going on in this question that I feel kind of lost.
How should I connect all the information into something that will help me? I've tried opening up the definition of limit of f(x), using continuity, using differentiability to no vail.. I'm having a hard time grasping the combination of functions and sequences into one problem..

答:
Let's summarize the most important points:

$f:[a,b]\to[a,b]$ is differentiable and $f(\alpha)=\alpha$

$(x_n)$ is a sequence in $[a,b]$ with $f(x_{n})=x_{n+1}$ for all $n\in\mathbb N$.

Assume the derivative of $f$ is bounded less than one: $|f'(x)|\leq q<1$.

Show that $x_n$ converges to $\alpha$.

from the hint, we need to compare $|x_{n+1}-\alpha|$ and $|x_n-\alpha|$. From points (1) and (2), we have $|x_{n+1}-\alpha|=|f(x_n)-f(\alpha)|$. By the Mean Value Theorem (which I am guessing is Lagrange's theorem), we have if $$\left|\frac{f(x_n)-f(\alpha)}{x_n-\alpha}\right|>q,$$ then there must be a point $y$ between $\alpha$ and $x_n$ such that $|f'(y)|>q$, but this contradicts point (3). Therefore, $$|x_{n+1}-\alpha|=|f(x_n)-f(\alpha)|<q|x_n-\alpha|.$$ We can apply this fact inductively to see that $|x_n-\alpha|<q^n|x_1-\alpha|.$ Thus, since $q<1$, $|x_n-\alpha|\to 0$ as $n\to\infty$, so $x_n$ converges to $\alpha$.


问:
Solve equation involving matrices and Kronecker product
I have what appears to be a simple equation:

$$
C_u = H(C_y \otimes R)H^T
$$

where the matrix dimensions are compatible, i.e. $\dim(C_u)=\dim(C_y)=n \times n$, $\dim(H)=n \times np$, $\dim(R)=p \times p$, and the unknown matrix is $C_y$.  ($\otimes$ denotes the Kronecker product.)

What is the closed-form solution for $C_y$?  I am expecting that operations such as matrix multiplication, $\otimes$, and $vec$ should be enough.

Many thanks in advance.

答:
For convenience, I rename $C_u$ and $C_y$ as $U$ and $Y$. So, we have the equation $H(Y\otimes R)H^T=U$, where $Y$ is the unknown. Partition $H$ as $(H_1,H_2,\ldots,H_n)$, where each $H_i$ is an $n\times p$ matrix. Then our equation can be rewritten as
\begin{align*}
\sum_{i,j} y_{ij} H_i R H_j^T &= U,\\
\sum_{i,j} y_{ij} \operatorname{vec}(H_i R H_j^T) &= \operatorname{vec}(U).
\end{align*}
This is a usual linear system of equations in the form of $A\operatorname{vec}(Y)=\operatorname{vec}(U)$, where
\begin{align*}
A &= (v_{11},v_{21},\ldots,v_{n1},v_{12},v_{22},\ldots,v_{n2},\ldots,v_{1n},v_{2n},\ldots,v_{nn}),\\
v_{ij} &= \operatorname{vec}(H_i R H_j^T).
\end{align*}</p>


问:
Diffusion process with "centering" drift
Consider a diffusion process which has a drift that pushes it towards some fixed value.

Specifically, consider the example of a MLE estimator for a Bernoulli random variable, given by $\mu_t = \frac{C_t}{t}$ where $C_t$ is the count of successes. We know $\mu_t \rightarrow p$ where $p$ is the Bernoulli parameter, i.e. $C_t \sim \text{Bernoulli}(p)$.

The drift of a process is defined as $\lim_{dt\rightarrow 0} \frac{1}{dt}\mathbb{E}[\mu_{t+dt} - \mu_t | \mu_t] $.

Then, whenever $\mu_t < p$, the drift would be positive, pushing $\mu_t$ toward $p$, and vice versa.

In the continuum limit of this process, i.e. $dC_t = \sqrt{dt}$ and $dt \rightarrow 0$, this is a diffusion process. How can we define the drift? I believe it should be piecewise constant, with value depending on whether $\mu_t > p$ or $\mu_t < p$.

答:
So first off you must work with the Poisson process with jump rate $p$ instead of directly with $C_t$. This is a reasonable assumption for large time, by basically the same logic as the Poisson approximation to the binomial. I'll call that Poisson process $C_t$ from here out.

The main idea is that the Poisson process $C_t$ itself cannot be described by a diffusion process directly at all. One must instead work with a process with at least small drift if not zero drift, because just $C_t$ when converted to a continuum limit in a naive way would either be deterministic (because $O(t/dt)$ steps of magnitude $dt$ occur and the law of large numbers kicks in) or diverge instantly (because $O(t/dt)$ steps of magnitude $(dt)^{1/2}$ occur and they tend not to cancel out because of the drift).

In this context it makes sense to achieve the approximation by defining a time scale $T \gg 1$ and examining $c_\tau=C_{T \tau}=T \mu_\tau + T^{1/2} \sigma_\tau$ where $\mu_\tau$ is deterministic and $\sigma_\tau$ is random. Neither can depend explicitly on $T$ (if they do then our scaling was wrong).

This is a kind of analogue of van Kampen system size expansion where the large parameter is forced in by stretching out the time scale. (Note that this focus on large time is necessary, there is no way that the short time dynamics can be resolved by a diffusion process in any reasonable sense.)

In this case $\mu_\tau=p\tau$ necessarily, and then you are left to perform Taylor expansion on the master equation for the evolution of $\sigma_\tau=T^{-1/2} \left ( c_\tau-T\mu_\tau \right )=T^{-1/2} c_\tau - T^{1/2} p \tau$ in powers of the small parameter $T^{-1/2}$ in order to isolate the "microscopic drift" (if there is any) and the diffusion inside of $c_\tau$. It looks like you get that the first moment is just zero while the second moment is $T^{-1}$ times the second moment of the increment distribution of $c_\tau$ which is $pT$. So for $T \gg 1$ and $t=T\tau$ where $\tau$ is of order $1$, we have that the PDF of $\sigma_\tau$ asymptotically exists and evolves as

$$\frac{\partial f}{\partial \tau}=\frac{p}{2} \frac{\partial^2 f}{\partial x^2}.$$

Consequently the overall process $C_t$ behaves like $pt+\sqrt{p} B_t$ where $B_t$ is a Brownian motion, for large time.

Remarks:

This implies that the one-dimensional distribution of $C_t$ is asymptotically $N(pt,pt)$ distributed. This is consistent with what we know for CLT for the Poisson distribution. It's not exactly consistent with what we know for CLT for the binomial (which would tell us the variance should be $pqt$ not $pt$), but that error was already committed in passing to the Poisson process approximation in the first place.

You could do the procedure over by writing $C_t=t-(t-C_t)$ and then starting from scratch with that. This will be more accurate than what I wrote here if $p>1/2$. (This is the analogue of the Poisson approximation to the binomial in the $p \to 1^-$ limit.)

Keep in mind that it does not make sense to say that the one dimensional distribution of $C_t$ converges to $N(pt,pt)$ as $t \to \infty$; this is the same situation as with CLT, if you want convergence you need to shift and rescale.

My main source here is Gardiner Handbook of Stochastic Methods Chapter 7.


问:
Homeomorphism from the whole plane to the open upper half plane.

Find a homeomorphism from the whole plane to the open upper half plane.

So I need to find a bicontinuous map $f:\mathbb{R^2} \to \mathbb{R} \times[0, \infty)$. If I consider $f(x,y)=(x,e^y)$ then certainly $f$ is continuous since this is a $C^\infty$ function? Both component functions also have a continuous inverse, but neither one is defined at $(x,y)=(0,0)$? Is this a valid candidate or am I just mistaken here?

答: suggest sending both to the open unit disc. The upper half plane goes by a Moebius transformation, one direction is $ \frac{z+i}{iz+1},$  the other way is just the reciprocal $ \frac{iz+1}{z+i}.$  The entire plane  just needs radial direction shrunk,  $(r,\theta )  \mapsto (\frac{r}{1+r}, \theta)$ works in one direction