问:
If $X$ and $Y$ are Lévy processes with $X_t\sim Y_t$ for all $t$, can we infer that $X\sim Y$?
Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$, $E$ be a $\mathbb R$-Banach space and $(X_t)_{t\ge0}$ be an $E$-valued process on $(\Omega,\mathcal A,\operatorname P)$. Remember that $X$ is called $\mathcal F$-Lévy if

$X$ is $\mathcal F$-adapted;

$X_0=0$;

$X_{s+t}-X_s$ and $\mathcal F_s$ are independent for all $s,t\ge0$;

$X_{s+t}-X_s\sim X_t$ for all $s,t\ge0$.

Assume $X$ is $\mathcal F$-Lévy and let $(Y_t)_{t\ge0}$ be another $E$-valued $\mathcal F$-Lévy process on $(\Omega,\mathcal A,\operatorname P)$ (or possibly on another probability space) with $X_t\sim Y_t$ for all $t\ge0$.

Are we able to show that $X$ and $Y$ have the same (finite-dimensional) distribution(s)?

EDIT: To be precise, the question is whether we can show that $$\operatorname P\left[X_{t_1}\in B_1,\ldots,X_{t_k}\in B_k\right]=\operatorname P\left[Y_{t_1}\in B_1,\ldots,Y_{t_k}\in B_k\right]\tag1$$ for all $B_1,\ldots,B_k\in\mathcal B(E)$, $k\in\mathbb N$ and $0\le t_1<\cdots<t_k$ or even $$\operatorname P\left[X\in B\right]=\operatorname P\left[Y\in B\right]\;\;\;\text{for all }B\in\mathcal B\left(E^{[0,\:\infty)}\right)\tag2.$$

答: I think the answer whether $(1)$ holds is rather trivial by observing that $X$ is a time-homogeneous Markov process with transition semigroup $$\kappa_t(x,B):=\operatorname P\left[X_t\in B-x\right]\;\;\;\text{for }(x,B)\in E\times\mathcal B(E)\text{ and }t\ge0.$$ From this we know that, for all $k\in\mathbb N$ and $0\le t_0<\cdots<t_k$, $$\left(X_{t_0},\ldots,X_{t_k}\right)\sim\mathcal L\left(X_{t_0}\right)\otimes\bigotimes_{i=1}^k\kappa_{t_i-t_{i-1}}\tag3.$$ So, unless I'm missing something, it's only the question left whether $(2)$ holds as well. But this should be a general fact for any processes over an arbitrary index set.


问:
Find $\displaystyle\sum_{r=0}^n\binom nr\cos(r\theta)$using$1+\cos(\theta)+i\sin(\theta)=2\cos(\frac\theta2)(\cos(\frac\theta2)+i\sin(\frac\theta2))$
I have proven the identity shown as the textbook requests, and made attempts on manipulating the identity. The most I have procured is $$(1+(\cos(\theta)+i\sin(\theta))^n=2^n\cos^n(\frac\theta2)(\cos(n\frac \theta2))$$Recognising that $$2^n=\sum_{r=0}^n \binom {n}{r}$$ But this is as far as I can get as I cannot recognise any geometric series or other techniques I could use. The book provides an answer but if someone could provide a nudge in the right direction or processes/techniques I could use it would be greatly appreciated.

ContentLicense: CC BY-SA 4.0

Body: $\cos(r\theta)=\mathfrak{Re}(e^{ir\theta})$ so you want to find$$\begin{align*}\mathfrak{Re}\left(\sum_{r=0}^n\binom nre^{ir\theta}\right)&=\mathfrak{Re}[(1+e^{i\theta})^n]\\&=\mathfrak{Re}[(1+\cos\theta+i\sin\theta)^n]\\&=\mathfrak{Re}[2^n\cos^n(\theta/2)e^{in\theta/2}]\\&=2^n\cos^n(\theta/2)\cos(n\theta/2)\end{align*}$$


问:
Does there exist a matrix $\mathbf{A}\in\mathbb{R}^{3\times3}$ such that $\mathbf{A}^{2}=-\mathbf{I}$?
Is it possible for a matrix $\mathbf{A}\in\mathbb{R}^{3\times3}$,

$$\mathbf{A}^2=-\mathbf{I}$$

I know that It is possible for $2\times2$ matrix, but is it possible for $3\times3$ matrix ?

答:
Here is an argument without using determinant or eigenvalues (yes, this is a deliberately complicated answer). We prove by contradiction. Suppose $A^2=-I_3$. Take $x\neq0$ and let $y=Ax$. If $y=kx$ for some $k\in\mathbb{R}$, then
$$k^2x=ky=kAx=A(kx)=Ay=A^2x=-x\ \Rightarrow\ k^2=-1,$$
which is impossible. Therefore $x$ and $y$ are linearly independent. Extend them to a basis $\{x,y,z\}$ of $\mathbb{R}^3$. Let $w=Az$. By similar reasoning, $z$ and $w$ are linearly independent and hence $w\neq0$. Therefore
$$w=ax+by+cz\tag{1}$$
for some $(a,b,c)^T\in\mathbb{R}^3\setminus0$. Then $-z=Aw=A(ax+by+cz)=ay-bx+cw$, i.e.
$$cw=bx-ay-z.\tag{2}$$
As $x,y,z$ are linearly independent, $c\neq0$. Therefore $(2)$ implies that
$$w=\frac{b}{c}x-\frac{a}{c}y-\frac{1}{c}z.\tag{3}$$
Comparing the coefficients of $z$ in $(1)$ and $(3)$, we get $c=-\frac1c$, which is impossible. Hence there does not exist a real matrix $A$ such that $A^2=-I_3$.</p>

ContentLicense: CC BY-SA 3.0


问:
The density of M = Median($U_1, U_2, U_3$)
$U_1, U_2, U_3$ are identically independent distributed from Unif(0,1). Let M = Median($U_1, U_2, U_3$). What is the density and cdf of M?

答:
Using the theory of Order Statistics, we can easily find

$$f_{U_{(2)}}(u)=3!u(1-u)=6u(1-u)$$

$u \in (0;1)$

We recognize a known density: $U_{(2)}\sim Beta(2;2)$


问:
Find the residue of $\frac{1 - \cos z}{z^{3} (z-3)}$
Is my solution correct? Also, are there removable singularities? Im having trouble classifying singularities

答:
Yes......you can also see that around $0$ the series expansion looks like
$$ -1/6z -1/18 - z/216 \ - .... etc $$
So at zero you have a simple pole of order 1. This clearly follows as for $z \rightarrow 0 $ you have
$$  z(1-\cos(z))/z^3(z-3) =  2\sin^2(z/2)/z^2(z-3)\sim 1/2(z-3) \rightarrow -1/6$$
Hence $ z^k(1-\cos(z))/z^3(z-3) \sim z^{k-1}/2(z-3) \rightarrow 0 $ for all $ k \geq 2 $. Thus the Laurent series starts from $1/z $.</p>


问:
Variation of Birthday problem - Group of n people
I know this has been posted several times and I have gone through most of the relevant posts.
Here is one which I am having a difficult time to solve:

There are 450 people in a room; (1) how many of them are expected to have the same birthday with some other person in the room, (2) with at least 2 other people in the room and (3) with at least 3.

(1) is easy - by the pigeonhole principle, 450-365 (or 366) = 85 people are expected to have the same birthday.

How do we do (2) and (3)?

I am thinking that in 85 people we have $\frac {85*84} {2} = 3570$ possible pairs so the probability for a 3rd person to share one of their birthdays is $1-\frac {364}{365}^{85}$. And then how do we find the expected number of people for each case?

Any help is greatly appreciated! Thank you!

答:
Hints:

Picking a particular individual, what is the probability that person shares their birthday:

with no other people

with exactly one person

with exactly two people?

So for that individual,  what is the probability that person shares their birthday:

with at least one person

with at least two people

with at least three people?

And using the linearity of expectation, what is the expected number of people that share their birthday

with at least one person  (much more than or $85$ or $86$)

with at least two people

with at least three people?

Following the hints:

Picking a particular individual, the probability that person shares their birthday:

with no other people is $\frac{364^{449}}{365^{449}}$

with exactly one person is ${449 \choose 1}\frac{364^{448}}{365^{449}}$

with exactly two people is ${449 \choose 2}\frac{364^{447}}{365^{449}}$

So for that individual,  the probability that person shares their birthday:

with at least one person is $1-\frac{364^{449}}{365^{449}}$

with at least two people is $1-\frac{364^{449}}{365^{449}}-{449 \choose 1}\frac{364^{448}}{365^{449}}$

with at least three people is $1-\frac{364^{449}}{365^{449}}-{449 \choose 1}\frac{364^{448}}{365^{449}} - {449 \choose 2}\frac{364^{447}}{365^{449}}$

And using the linearity of expectation, the expected number of people that share their birthday

with at least one person  is $450\left(1-\frac{364^{449}}{365^{449}} \right)$

with at least two people is $450\left(1-\frac{364^{449}}{365^{449}}-{449 \choose 1}\frac{364^{448}}{365^{449}} \right)$

with at least three people  is $450\left(1-\frac{364^{449}}{365^{449}}-{449 \choose 1}\frac{364^{448}}{365^{449}} - {449 \choose 2}\frac{364^{447}}{365^{449}} \right)$

and these values are about $318.7$  (much more than or $85$ or $86$) and $156.8$ and $57.1$


问:
What's the best way to think about the Hessian?
I've always thought about the Hessian like this:

Let $f:\mathbb R^n \to \mathbb R$ be smooth.  Let $g:\mathbb R^n \to \mathbb R^n$ such that $g(x) = \nabla f(x)$.  (I am using the convention that $\nabla f(x)$ is a column vector.)  Then the Hessian of $f$ at $x$ is, by my definition, the $n \times n$ matrix $H(x) = g'(x)$.

However, with this way of looking at the Hessian, I'm not thinking of $H(x)$ as being a quadratic form.  I'm worried that there is a "quadratic form" viewpoint of the Hessian that I am missing.  There is a rule of thumb I've heard that when a matrix (such as the Hessian) is automatically symmetric, then it's often most natural to think of it as defining a quadratic form.  I realize that you can define a quadratic form at $x_0$ by $x \mapsto \langle x, H(x_0) x \rangle$, and that this quadratic form appears in Taylor's formula.  But I still think I'm missing something, because I don't see why it is fundamentally most natural to think of the Hessian as being a quadratic form.

Is it true that there is a quadratic form viewpoint that I'm missing out on?  If so, what is it?  More generally, what do you think is the best way to think about the Hessian?

答:
When I happen to work in analysis, I tend to work in the part that's covered by limits of (possibly multivariable, possibly formal) power series, which you can get quite a lot of mileage from as far as intuition goes. As a basic example, the independence of the order of partial derivatives is obvious in this setting. So, I think of the Hessian in those terms first, and of the rest as generalizations of this setting. Those generalizations are appropriate for when you have less structure, but they aren't the right way to introduce the idea.

Anyway, take a multivariable polynomial $f(x_1, \ldots, x_n)$. We can write it as

$$f(x_1, \ldots, x_n) = f(0, \ldots, 0) + a_1 x_1 + \cdots + a_n x_n + \sum_{i,j=1}^n a_{i,j} x_i x_j + \cdots.$$

The constant bit is clear, and the linear bit is just $\nabla f(0) \cdot (x_1, \ldots, x_n)$. Since $x_i x_j = x_j x_i$, it seems clean to pick $a_{i,j} = a_{j,i}$. The quadratic term is

$$\sum_{i,j=1}^n a_{i,j} x_i x_j = \mathbf{x}^T A \mathbf{x}$$

--oh look, a symmetric quadratic form popped out. Pick off the coefficients with $a_{i,j} = f_{ij}(0)/2 = H_{i,j}/2$ so $A = H/2$ and
$$f(x_1, \ldots, x_n) = f(0) + \nabla f(0) \cdot \mathbf{x}^T + \frac{1}{2} \mathbf{x}^T H \mathbf{x} + \cdots$$

In this way, the Hessian obviously encodes the best quadratic approximation to the local behavior of $f$, after dealing with the constant and linear behavior. Make this precise with the formalism of your choosing--Laithy has provided one such--and bake until golden brown.

Connecting this to your definition, note that your definition of the Hessian is basically the best linear approximation to the local behavior of $\nabla f$. Since $\nabla f$ itself is the best linear approximation to the local behavior of $f$, it's at least intuitively plausible that the best linear approximation of $\nabla f$ gives the best quadratic approximation to the local behavior of $f$. So, I'd say at an intuitive level your definition is explained by my approach.


问:
Find the probability of the event that we run out of white balls first
I have $a$ white balls, $b$ black balls, $c$ red balls in a basket.
We take balls one by one and do not return.
I have to find the probability of the event that we run out of white balls first.
If I define:

$A_1$ - The event that we run out of white balls first.

$B_2$ - The event that we run out of black balls second.

$B_3$ - The event that we run out of black balls third.

$C_2$ - The event that we run out of red balls second.

$C_3$ - The event that we run out of red balls third.

I get that:

$P(A_1)=P(B_2 \cap C_3) + P(C_2\cap B_3)$

$P(B_2 \cap C_3) = P(B_2|C_3) * P(C_3)$

$P(C_2\cap B_3) = P(C_2|B_3) * P(B_3)$

I also found out that:

$P(C_3)=\frac{c}{a+b+c}$

$P(B_3)=\frac{b}{a+b+c}$

I need help finding $P(B_2|C_3)$ and $ P(C_2|B_3$

Thanks in advance!

答:
Here's an unusual way using stars and bars to solve it.

Let us say the order in which the balls are exhausted is white, black and red.

Lay out all the white balls with one black and one red at the end as shown.
$-w-w-w-w-w-w-w-b-r$
There will be $(a+1)$ interstices including the start of the line where black balls can be placed in $\binom{a+b-1}{b-1} = X, say$

There will now be $(a+b+1)$ interstices for the red balls to be placed in $\binom{a+b+c-1}{c-1}$ = Y,say

$X\cdot Y= F_1$ gives total arrangements where white balls, black balls and red balls get exhausted in that order.

Do a similar exercise  for arrangements where white,red,black are exhausted in that order and get $F_2$

Unconstrained arrangements = $\frac{(a+b+c)!}{a!b!c!} = Z,say$

$Pr = \frac{F_1+F_2}{Z}$

FURTHER THOUGHTS:

A much simpler approach emerges using direct Pr,sparked by @Lulu remarks.

P(ends first) = (red ends last and black ends  second last) + P(black ends  last and red ends second last)

= $\frac{c}{a+b+c}\times \frac{b}{a+b}+ \frac{b}{a+b+c}\times \frac{c}{a+c}$


问:
Check the continuity and differentiability of $f(x)= \sin^{-1}(\cos x)$ at $x=0$

Check the continuity and differentiability of  $f(x)= \sin^{-1}(\cos x)$  at $x=0$

This is how I tried to solve the problem:

$$f(x)= \sin^{-1}(\cos x)=???$$

$$\lim_{x \to 0^+}f(x)=\lim_{x \to 0^-}f(x)=f(0)=???$$
Therefore the function is continuous at $x=0$
$$\lim_{x \to 0^+}\frac{f(x)-f(0)}{x-0}=???$$and
$$\lim_{x \to 0^-}\frac{f(x)-f(0)}{-x}=???$$

答:
In this question it's important to note that $\sin^{-1}(c)$ is (customarily) defined as the unique value of $x \in [-\pi/2,\pi/2]$ such that $\sin x = c$.  In particular, $f(x) = \sin^{-1}(\cos x)$ can never take values outside the range $[-\pi/2,\pi/2]$, contradicting the claim in another answer that $f(x) = \pi/2 - x$.

If you graph the function carefully near $x=0$, you'll find it simplifies to $f(x) = \pi/2 - |x|$ for $-\pi \le x \le \pi$.  I think you already know something about whether $|x|$ is differentiable at $x=0$.


问:
Quadrics and a theorem of Springer
Let $F$ be a complete discretely valued field with ring of valuation $R$, uniformizer $\pi$, and residue characteristic $\neq 2$. A theorem of Springer says that a quadratic form $q=q_1 \bot \pi q_2$, where $q_1,q_2$ have coefficients in $R^\times$, is isotropic over $F$ if and only if at least one of the images of $q_1,q_2$ is isotropic over the residue field $R/\pi$.

I am interested in knowing whether:

a similar result holds for other varieties than quadrics (this is a bit vague, but anything that comes to mind might be useful),

can anything remotely similar be said for quadratic forms in the dyadic case (i.e. when char $R/\pi=2$)?

(Note: I first asked the question on mathoverflow, but did not get an answer. That post is now deleted.)

答:
For Q2, the answer is yes. See Section 19 of " the algebraic and geometric theory of quadratic forms", Elman, Karpenko, Merkurjev, AMS Coll. Pub.56,2008.

For Q1: You can generalize the fact that if $q_1,q_2$ are anistropic modulo $\pi$, then $q$ is anisotropic  to homogenous forms of degree $d$.

What will be missing is: if a homogeneous form $f$ (with the appropriate notion of nondegeneracy, which is not clear to me) of degree $d\geq 2$ has no nontrivial zeroes, then it has no nontrivial zeroes modulo $\pi$.

I don't know if it is reasonably true (i suspect it isn't).

Title: Quadrics and a theorem of Springer


答:
The Hessian can be treated as a double derivative in the same way that the gradient or Jacobian is treated as a single derivative. The catch is that in multivariable calculus, the first derivative is no longer a scalar but instead is a linear map, while the second derivative is no longer a scalar but instead is a bilinear map. (And so on: the third derivative is a multilinear map accepting three arguments).

Let $f : \mathbb{R}^n \to \mathbb{R}$ be a smooth function, then the first derivative $Df : \mathbb{R}^n \to L(\mathbb{R}^n, \mathbb{R})$ is a smooth function from $\mathbb{R}^n$ to the set $L(\mathbb{R}^n, \mathbb{R})$ of linear maps $\mathbb{R}^n \to \mathbb{R}$. This can look complicated at first, but it really just follows straight from the definition: fixing a point $x \in \mathbb{R}^n$, the derivative must satisfy
$$ f(x + \epsilon) = f(x) + ((Df)(x))(\epsilon) + o(\epsilon),$$
where $o(-)$ is a function which goes to zero fast enough (here, faster than linear).

The derivative $Df$ itself is a smooth function so we can differentiate again to get a map $D^2 f : \mathbb{R}^n \to L(\mathbb{R}^n, L(\mathbb{R}^n, \mathbb{R}))$. The codomain $L(\mathbb{R}^n, L(\mathbb{R}^n, \mathbb{R}))$ is isomorphic to the set of bilinear maps $\operatorname{Bil}(\mathbb{R}^n \times \mathbb{R}^n, \mathbb{R})$. (To see this, just think about the fact that a function which returns another function may as well be a function of two arguments - the "linearised" version is that a linear map returning a linear map may as well just be a multilinear map of two arguments). Therefore after fixing a point $x \in \mathbb{R}^n$, the second derivative $(D^2 f)(x)$ is a bilinear map $\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$, which is the same kind of object that the Hessian is: we write $((D^2 f)(x))(v_1, v_2)$ rather than $v_1^T H v_2$. We then have a kind of Taylor's formula:
$$ f(x + \epsilon) = f(x) + \frac{1}{1!} ((Df)(x))(\epsilon) + \frac{1}{2!} ((D^2 f)(x))(\epsilon, \epsilon) + o(\epsilon),$$
where $o(-)$ goes to zero faster than a quadratic.

I would contend that this is the "right" way of looking at higher derivatives from a theoretical standpoint (perhaps not a computational one), since it generalises even to smooth functions $f : \mathbb{R}^n \to \mathbb{R}^m$ immediately, and does not require a choice of basis or coordinates in order to be defined. The reason that only $Df$ (the Jacobian) and $D^2 f$ (the Hessian) in a first course is that in order to write down $D^n f$ in any other case (or even $D^2 f$ in the $f: \mathbb{R}^n \to \mathbb{R}^m$ case) requires some multilinear algebra / tensors / whatever, which is normally not worth introducing in a first course on vector calculus.

As for what the operator $(D^2 f)(x)$ does intuitively: it models the concavity of a hypersurface at a point. Symmetric bilinear forms can be orthogonally diagonalised, so there is some orthonormal basis $(v_1, \ldots, v_n)$ of $\mathbb{R}^n$ so that the sign of $((D^2 f)(x))(v_i, v_i)$ tells you whether the curve traced by $f(x + tv_i)$ is concave up or down. If the sign is zero, then more work has to be done a-la the second derivative test.

Title: What's the best way to think about the Hessian?

Body:
I've always thought about the Hessian like this:

Let $f:\mathbb R^n \to \mathbb R$ be smooth.  Let $g:\mathbb R^n \to \mathbb R^n$ such that $g(x) = \nabla f(x)$.  (I am using the convention that $\nabla f(x)$ is a column vector.)  Then the Hessian of $f$ at $x$ is, by my definition, the $n \times n$ matrix $H(x) = g'(x)$.

However, with this way of looking at the Hessian, I'm not thinking of $H(x)$ as being a quadratic form.  I'm worried that there is a "quadratic form" viewpoint of the Hessian that I am missing.  There is a rule of thumb I've heard that when a matrix (such as the Hessian) is automatically symmetric, then it's often most natural to think of it as defining a quadratic form.  I realize that you can define a quadratic form at $x_0$ by $x \mapsto \langle x, H(x_0) x \rangle$, and that this quadratic form appears in Taylor's formula.  But I still think I'm missing something, because I don't see why it is fundamentally most natural to think of the Hessian as being a quadratic form.

Is it true that there is a quadratic form viewpoint that I'm missing out on?  If so, what is it?  More generally, what do you think is the best way to think about the Hessian?


问:
Question about image and pre-image of subsets
Let $M$ and $N$ be sets and $f:M\to N$ be a map. Let $A\subset M$ and $B \subset N$. Show that
$$f(f^{-1}(B)\cap A)=B\cap f(A)$$
As $f^{-1}(B), A\subset M$ we can say that $f(f^{-1}(B)\cap A)\subset f(f^{-1}(B))\cap f(A)$. However, I am unsure as to how to proceed from here. Any help is appreciated.
Thanks!

答:
$f(x) \in B\cap f(A)$ $\Rightarrow$ $f(x) \in B \land f(x)\in f(A)$

$f(x) \in f(A)$ $\Rightarrow$ $x \in A$

$f(x) \in B$ $\Rightarrow$ $x \in f^{-1}(B)$

$x \in  f^{-1}(B)\cap A$ $\Rightarrow$ $f(x) \in f(f^{-1}(B)\cap A)$

$f(x) \in f(f^{-1}(B)\cap A)$ $\Rightarrow$ $x \in f^{-1}(B)\cap A$

$x \in f^{-1}(B)\cap A$ $\Rightarrow$ $x \in f^{-1}(B) \land A $

$x \in f^{-1}(B)$ $\Rightarrow$ $f(x) \in B$ , $x \in A$  $\Rightarrow$ $f(x) \in f(A)$

$f(x) \in B\cap f(A)$

$f(f^{-1}(B)\cap A)=B\cap f(A)$


问:
probability with combinations
A college math department consisting of 20 faculty members must choose a department head, an assistant department head, and a faculty senate representative. in how many ways can this be done?</p>

答:
So you want to find out how many combinations these three spots can be chosen.

Forgive me if I'm going to grossly oversimplify this. Let's assume you have a bag of marbles with 10 colors. You want to pick randomly 3 balls to put on your desk. To calculate how many combinations that could be and their separate chances, first do the probability of picking one color (let's say blue) which is 1/10. that leaves 9 balls, all of which now have a probability 1/9 of being chosen. You pick the second one (yellow) which leaves 8 balls, each of which have a 1/8 chance of being picked. finally you pick the last one (green) and you're done.

With 10 balls in the bag, they had 1/10 chance each. With 9 balls, each had 1/9 chance. Eight balls had 1/8 chance of being picked.

The combination is simply a multiplication of these fractions: 1/10 * 1/9 * 1/8 = 1/720

the 720 represents the total combinations of colored balls.
1/720 is the chance of achieving one of them.

So, with your example:

choosing the department head: 1/200 chance

choosing the assistant department head: 1/199 chance

choosing the faculty senate representative: 1/98 chance

200 * 199 * 198=7,880,400 combinations

1/7,880,400 is the chance of achieving one of these combinations.

postscript: the reason we reduce the denominator each step is because one person was already elected for a spot. he can't take several spots.

** Also this is assuming the numerator is always 1

Title: probability with combinations


问:
How to Prove the Intersection of two Sylow p-subgroups is non-trivial.
I am working to prove that a group of order $p^2q^2$ is solvable, where $p\neq q$ are primes. Here is my solution so far.

We can assume without loss of generality that $p>q.$ Consider the possibilities for $n_G(p^2).$ By Sylow's Theorem we have that $$n_G(p^2)\equiv 1\;\;\text{mod}\;\;p,\;\text{and}\;\;n_G(p^2)\;|\;q^2.$$
So $n_G(p^2)=1,q,$ or $q^2.$ If $n_G(p^2)=1$, then there is only one Sylow $p$-subgroup, and it is normal, call it $P.$ $P$ is a $p$-group, of order $p^2$, and so it is solvable. Also, $|G/P|=q^2$, which is a $q$-group, and so $G/P$ is solvable. So $G$ is solvable.

If $n_G(p^2)=q$, then since $n_G(p^2)\equiv 1$ mod $p$ we have that $p\;|\;q-1$. This is impossible, since $p>q.$

Now suppose that $n_G(p^2)=q^2.$ Let $P\neq Q$ be two Sylow $p$-subgroups, and let $R:=P\cap Q.$ By Lagrange's Theorem, the order of $R$ must divide $p^2q^2$. Since $P$ and $Q$ are both of order $p^2$, and they are not equal, this gives us that $|R|=1,q,pq,q^2,$ or $p$. Note that if $|R|=q,pq,$ or $q^2$, then $R$ has an element of order $q$. Thus, both $P$ and $Q$ have an element of order $q,$ which is impossible, since $q\nmid p^2=|P|=|Q|.$ So $|R|=1$ or $p.$

How can I show that $|P\cap Q|$ must equal $p$?

Since both $P$ and $Q$ have order $p^2$, there is an element of order $p$ in both $P$ and $Q$. Can I somehow show that there is only one element of order $p$ and then it will have to be in $P\cap Q$? I feel like I'm missing something obvious here.

答: Let $P,Q \in Syl_p(G)$ and $P \neq Q$. Assume that $|P \cap Q|=1$. Then $p^2 q^2 =|G| \gt |PQ|=\frac{|P| \cdot |Q|}{|P \cap Q|}=p^2 \cdot p^2$. It follows that $q \gt p$, a contradiction. Hence $|P \cap Q|=p$. (Of course $|P \cap Q| \lt |P|$, since $P$ and $Q$ are not equal).


问:
Defective laptops problem
Among $10$ laptop computers, seven are good, the rest have defects. Unaware of this, a
customer buys $5$ laptops.

a) What is the probability of exactly $2$ defective ones among them (event A)?

b) Knowing that at least $2$ purchased laptops are defective, what is the probability that exactly
$2$ are defective (event B)?

Can somebody give me some ideas,please? I can't figure out what probabilistic model to use.

Title: Defective laptops problem

Body:
a)

$$\mathbb{P}[A]=\frac{\binom{3}{2}\binom{7}{3}}{\binom{10}{5}}=\frac{5}{12}$$

b)

$$\mathbb{P}[B]=\frac{\frac{5}{12}}{1-\frac{\binom{3}{0}\binom{7}{5}}{\binom{10}{5}}-\frac{\binom{3}{1}\binom{7}{4}}{\binom{10}{5}}}=\frac{\frac{5}{12}}{1-\frac{1}{12}-\frac{5}{12}}=\frac{5}{12-1-5}=\frac{5}{6}$$


问:
Isometric completion of eventually null sequences onto $l^1$
Can we isometrically embed $(c_{00},d_1)^*$, the completion of space of eventually null sequences of real numbers with $d_1=\sum_{n=1}^{\infty}|f(n)-g(n)|$ onto the space $l^1=\{f:\mathbb{N}\to\mathbb{R}:\sum_n|f(n)|<\infty\}$?

I think the limit functional would do the job. Is it right? This is because $c_0$ is dense in $l^1$ and $c_{00}$ is dense in $c_0$. Thus, since limit functional is an isomorphic embedding of $c_{00}$ to $c_0$ and $c_0$ is dense in $l^1$, I think the result follows. Am I right? Or should we take a cauchy sequence in $c_{00}$? Any hints? Thanks beforehand.

答:
If you are asking  if there exists an isometry from $(c_{00},d_1)$ onto $\ell^{1}$ the answer is NO. The first space is not complete and the second one is complete.

Answer for the edited version: the completion of $(c_{00},d_1)$  is exactly $\ell^{1}$.


问:
Bounded Variation is Bounded

Let $f:[a,b]\to \mathbb{R}$ and let $D=\{x_o,x_1,...,x_n\}$ be a division of $[a,b]$.
  We say that $f$ is of bounded variation on $[a,b]$ if
  $\displaystyle \sup_{D\in \mathscr{D}} \sum_{i=1}^n |f(x_i)-f(x_{i-1})|<\infty$, where $\mathscr{D}$ is a collection of divisions of $[a,b]$.

My question is: Does Bounded Variation implies that a function is Bounded?
Thanks.

答: Yes, because for arbitrary $x$ you can easily find $D$ showing that $|f(x)-f(a)|\le $ the given bound.</p>

Title: Bounded Variation is Bounded


问:
How can one describe the closure of the image of a morphism of schemes?
Let $(f,f^{\#}):(X,\mathcal{O}_{X})\to(Y,\mathcal{O}_{Y})$ be a morphism of schemes. Is it possible to describe $\overline{f(X)}$ concretely? Note that if $X$ and $Y$ are affine, then $\overline{f(X)}=V(\ker f^{\#}(Y))$. Also more generally, if $U\subseteq Y$ is affine open, then for the induced morphism $f_U:f^{-1}U\to U$ I am able to prove $\operatorname{cl}_U(\operatorname{im} f_U)=V(\ker f^{\#}(U))$, at least provided that $X$ is Noetherian. But then we only have $\operatorname{cl}_U(\operatorname{im} f_U)=U\cap\overline{U\cap\operatorname{im}f}$, and it would be more practical to have a description of $U\cap \overline{\operatorname{im}f}$ instead. Can something general, yet useful be said? Perhaps under some mild conditions? Thanks in advance!

答:
If $f$ is quasi-compact or $X$ is reduced, then the underlying set of the scheme-theoretic image is indeed the closure of the set-theoretic image. One cannot remove both of these assumptions.

Vakil's book does a good job with these items in section 8.3, and this post follows that exposition. Briefly, what you would want to do is given an affine open subset $\operatorname{Spec} A\subset Y$, cover $f^{-1}(\operatorname{Spec} A)$ with affine opens $\operatorname{Spec} B_i\subset X$ and declare the scheme-theoretic image to be the subscheme cut out by the kernel of $A\to\prod B_i$. In order to glue these together correctly, you need to show that $\ker(A\to \prod B_i)_g = \ker(A_g\to \prod (B_i)_g)$ for any $g\in A$.

In order for this to be true, given an $a/g^n\in \ker(A_g\to \prod (B_i)_g)$, one needs to find an $m>0$ so that $g^ma\in B_i$ is zero for all $i$. If there are finitely many $B_i$ (the case when $f$ is quasi-compact), you're good - just take the maximum. Alternatively, in the case that $X$ is reduced, $m=1$ works: if $g^na=0$ in $B$, then $ga$ is a nilpotent, so $ga=0$.

To see that there are examples of bad behavior when $X$ is nonreduced and $f$ is not quasi-compact, see a recent example here, copied for posterity:

Let $X=\coprod_{n\geq 0} \operatorname{Spec} k[x]/(x^n)$, let $Y=\operatorname{Spec} k[x]$, and define $f:X\to Y$ by the obvious map $\operatorname{Spec} k[x]/(x^n) \to \operatorname{Spec} k[x]$ on each component. Then the set-theoretic image of $f$ is just $(x)$, but the scheme-theoretic image is all of $Y$: the statement about factorization means we're looking to find the kernel $I$ of the map $k[x]\to \prod_{n\geq 0} k[x]/(x^n)$, and then the closed subscheme which is the scheme-theoretic image is $\operatorname{Spec} k[x]/I$. It is straightforward to see that $I$ must be zero: if a polynomial in $x$ is zero modulo every $x^n$, it must be zero.


问:
Proof that harmonic functions $u,v$ satisfy this condition.
Question: Suppose $u,v$ are harmonic and satisfy the Cauchy-Riemann equations in $\mathbb{R}^2$. Show that $f = u + iv$ satisfies $f'(x) = u_x(x,0) - iu_y(x,0)$ for real $x$.

My attempts: I considered the fact that $\displaystyle \frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} = 0$ and from there came to the conclusion that $\displaystyle \frac{\partial ^2 u}{\partial x^2} = -\frac{\partial ^2 u}{\partial y^2}$ and so $\displaystyle \frac{\partial u}{\partial x} = -\frac{\partial u}{\partial y} + C$. I'm not entirely sure where to go from here. Is this on the right track? Any help or guidance would be greatly appreciated!

答:
We have $u,v \in C^2( \mathbb R^2,\mathbb R)$, hence $u$ and $v$ are real differentiable, thus $f$ is real differentiable. Since $u,v$ satisfy the Cauchy-Riemann equations , $f$ is holomorphic on $ \mathbb C.$ Hence

$$f'(z)=u_x(x,y)+iv_x(x,y)$$

for all $z=x+iy \in \mathbb C.$ From $v_x=-u_y$ we get

$$f'(x) = u_x(x,0) - iu_y(x,0)$$

for $x \in \mathbb R.$


问:
How does the rewriting of the following two equations work?
I am failing to understand the proof of coming to the steady-state formula in queueing theory. This is probably due to the fact that I may have forgotten (and cannot find it back) some of the algebra from my school days. The step I am missing is the transition from the first to the second statement below:

$$(\lambda + \mu)\frac{\lambda}{\mu}P_0 = \lambda P_0 + \mu P2$$

$$\frac{\lambda^2}{\mu}P0 + \lambda P_0 = \lambda P_0 + \mu P2$$

My main question is how does $$(\lambda + \mu)\frac{\lambda}{\mu}P_0$$ become
$$\frac{\lambda^2}{\mu}P_0 + \lambda P_0$$
and what did happen to: $$\mu$$

Can anybody help me out?

答:
This is rather simple actually, all you need to do is expand (distribute the elements).

Expanding this:
$$
(\lambda + \mu)\frac{\lambda}{\mu}P0
$$

We have this:
$$
\frac{\lambda^2}{\mu}P0+\frac{\mu\lambda}{\mu}P0
$$

And the final result after simplification is :
$$
\frac{\lambda^2}{\mu}P0 + \lambda P0
$$

To answer your question about what happened to $\mu$. When you simplify: $\frac{\mu\lambda}{\mu}$ it becomes $1\lambda$, which is the same as just $\lambda$.


问:
How to understand whether a given ideal is a prime ideal in a Dedekind domain
$\DeclareMathOperator{\Norm}{Norm}$I have started Algebraic Number Theory, and I have a basic doubt about how to understand whether a given ideal is prime or not in a Dedekind domain. What I thought is, if ideal $I\subset J$, then if $\Norm(I)\mid \Norm(J)$, then by showing $\Norm(P)$ to be prime, we can prove ideal $P$ as prime. But I am confused whether, $I\subset J\rightarrow \Norm(I)\mid \Norm(J)$ is true or not.

答:
I assume you are working with a Dedekind domain that is the integral closure of some other Dedekind domain in a finite field extension, and the norm is with respect to that field extension.

By definition $N(I)$ is the ideal generated by norms of elements of $I$. Thus $I \subset J$ implies $N(I) \subset N(J)$, in which case $I = J$ iff $N(I) = N(J)$. (This last fact requires properties of Dedekind domains.)

Hence $J \mid I$ implies $N(J) \mid N(I)$, and $N(I)$ prime implies $I$ prime.

(But the converse, $I$ prime implies $N(I)$ prime, is false in general.)


问:
Proof: $f: A \to B$, if $f$ is bijection then $\forall a\in B \exists ! b\in A (f(b) = a) $
I musti proof the following:

" let $f: A \to B$, if $f$ is injection then $\forall a\in B \exists ! b\in A (f(b) = a)$"

proof: by contraddiction, therefore the negation of $\forall a\in B \exists ! b\in A (f(b) = a) $ is $\forall a \in B \exists c \in A (f(c) = a$ and $ c \neq b)$, but by hypothesis $f$ is bijection, injection and surjection, so $c \neq b$ then $f(c) \neq f(b)$, but $f(b) = a $ and $f(c) = a $ then $a \neq a$.. a absurd!!

Is correct?

Thanks in advance!

答:
First of all: Why “$a ∈ B$” and “b ∈ A”? Why? And why “$\triangleq$”?

Secondly, your statement is wrong:
It basically says that every injective function is bijective.
But there are easy counter-examples:

Set $A = [0..∞)$, $B = ℝ$ and let $f$ be $[0..∞) → ℝ, x ↦ x$.
Then $f$ is certainly injective but for $a = -1 ∈ B$ there is no $b ∈ A$ such that $f(b) = a$, or much simpler:

$\{1\} → \{1,2\}, \, 1 ↦ 1$ or even simpler:

the empty map $∅ → B$ for any nonempty set $B$.

And thirdly, the negation of $∀a ∈ B\, ∃! b ∈ A:\, (f(b) = a)$ is really:
$$∃a ∈ B\, \big(∀b ∈ A:\, f(b) ≠ a \big) ∨ \big( ∃ b_1, b_2 ∈ A:\, (f(b_1) = a ∧ f(b_2) = a ∧ b_1 ≠ b_2)\big)$$
And in English: There is an $a ∈ B$ which is either not hit by $f$ or hit by two different $b_1$, $b_2$ in $A$.

Also, your argument is hard to follow and does not seem very logical.
But my guess is that you mixed up some notions.


问:
Group presentation, central subgroup
In a proof that I am reading there is the following statement.
For a group $G$ with presentation
$$ G= \langle \gamma_1 ,\gamma _2,\gamma _3 ,c \mid \gamma _1^pc^{-1} =\gamma_2^qc^{-1}= \gamma _3^r c^{-1}=\gamma_1\gamma _2\gamma _3 c^{-1} =1 \rangle $$
the element $c$ generates a central subgroup $C$, with quotient $G/C$ isomorpic to the group
$$ \langle \gamma_1 ,\gamma _2,\gamma _3  \mid \gamma _1^p =\gamma_2^q= \gamma _3^r =\gamma_1\gamma _2\gamma _3  =1 \rangle  . $$
Can someone maybe give me a hint why that is true?

答:
Here is a general fact: Let $G=\langle S\rangle$. Then $g\in Z(G)$ if and only if $gx=xg$ for all $x\in S$.

That is, an element is contained in the center of a group $G$ if and only if the element commutes with every element of a generating set for $G$.

In your set-up, $G=\langle\gamma_1, \gamma_2, \gamma_3\rangle$, while the element $c$ is a power of each of generator $\gamma_i$, and so commutes with each $\gamma_i$, so is central by the above fact. For example, $\gamma_1c=\gamma_1\gamma_1^p=\gamma_1^p\gamma_1=c\gamma_1$.

Therefore, $G/\langle c\rangle$ makes sense and we obtain the presentation by adding the relator $c=1$ to get:
\begin{align*}
&\langle \gamma_1 ,\gamma _2,\gamma _3 ,c \mid \gamma _1^pc^{-1} =\gamma_2^qc^{-1}= \gamma _3^r c^{-1}=\gamma_1\gamma _2\gamma _3 c^{-1} =1, c=1 \rangle\\
&\cong
\langle \gamma_1 ,\gamma _2,\gamma _3 ,c \mid \gamma _1^p=\gamma_2^q= \gamma _3^r=\gamma_1\gamma _2\gamma _3=1, c=1 \rangle&\text{simply using $c=1$}\\
&\cong
\langle \gamma_1 ,\gamma _2,\gamma _3 \mid \gamma _1^p=\gamma_2^q= \gamma _3^r=\gamma_1\gamma _2\gamma _3=1 \rangle
\end{align*}
as required. (In the last step we removed the generator $c$ via a Tietze transformation.)

Title: Group presentation, central subgroup


问:
Is the probability density for c = np.random.random() and d = 30 * np.random.random() same?
import numpy as np

c = np.random.random()

Then we know c is uniformly distributed over the half-open interval [0,1) and the probability density function (pdf) of c is given by,
$$
p(c) =
\begin{cases}
\frac{1}{1-0}~\text{ for}~ 0 \leq c < 1\\
0,~\text{otherwise}
\end{cases}
$$
Now, if we generate d as follows:

scalar = 30

d = scalar * np.random.random()

then what is the pdf of d i.e. p(d)?

答:
d also has a uniform distribution, but now it's on the range $[0,30)$. So
$$
p(d) =
\begin{cases}
\frac{1}{30-0}~\text{ for}~ 0 \leq d < 30\\
0,~\text{otherwise}
\end{cases}
$$


问:
Limit $\lim_{z\to i\pi} \left( z-i\pi \right)\frac{e^z}{1+e^z}$
I'm about to finding a residue at $z\to i\pi$, but I'm stuck in finding the limit:

$$\lim_{z\to i\pi} \left( z-i\pi \right)\frac{e^z}{1+e^z}$$

What are the steps to eliminate the $z-i\pi$ part and $1+e^z$ part?

答: Hint: Note that that limit is equal to$$\frac{\displaystyle\lim_{z\to i\pi}e^z}{\displaystyle\lim_{z\to i\pi}\frac{e^z-e^{i\pi}}{z-i\pi}}.$$


问:
Convergence of $\sum_{k=1}^\infty \frac {(-1)^{k-1}}{k^{2+(-1)^k}}$
So I first I tried to use Comparison Test with $\frac {(-1)^{k-1}}{k^{2+(-1)^k}} \le \frac {(-1)^{k-1}}{k}$ but I think Comparison Test only works with series with positive terms. Then I tried Alternating Series Test, but the sequence $\frac {1}{k^{2+(-1)^k}}$ is not decreasing for small values of $k$ and I don't think it will be decreasing for large $k$ either. How can I proceed with this problem?

答: The sum of the non-negative terms diverges and the sum of the negative terms converges. Therefore the series diverges to $\infty$: namely, if you call $a_k=\frac{(-1)^{k-1}}{k^{2+(-1)^k}}$ and $L=\sum_{k=1}^\infty a_{2k-1}$, then eventually $$\sum_{k=1}^n a_k\ge L+\varepsilon_n+\sum_{1\le 2k\le n} a_{2k}$$ with $\varepsilon_n\to 0$. A slight variation works for the general case where the subsequence of all the negative terms ended up being a generic $\left\{a_{n_k}\right\}_{k\in\Bbb N}$.


问:
How to represent the floor function using mathematical notation?
I'm curious as to how the floor function can be defined using mathematical notation.

What I mean by this, is, instead of a word-based explanation (i.e. "The closest integer that is not greater than x"), I'm curious to see the mathematical equivalent of the definition, if that is even possible.

For example, a word-based explanation of the factorial function would be "multiply the number by all integers below it", whereas a mathematical equivalent of the definition would be $n! = \prod^n_{i=1}i$.

So, in summary: how can I show what $\lfloor x \rfloor$ means without words?

答:
I suspect that this question can be better articulated as: how can we compute the floor of a given number using real number field operations, rather than by exploiting the printed notation, which separates the real and fractional part, making nearby integers instantly identifiable.

How about as Fourier series? If we subtract the function $y = x$ from $\lfloor x\rfloor$ then we get a periodic sawtooth wave. Hey look, that Wikipedia page even mentions this relationship.

So if you obtain the Fourier series (a sum of sinusoidal functions of various frequencies and amplitudes) for the sawtooth wave $\lfloor x\rfloor - x$ and add $x$, then you have an approximation for $\lfloor x\rfloor$ which is as good to whatever accuracy you care to apply in evaluating the series. Obtaining the Fourier series for $\lfloor x\rfloor - x$  is a basically a matter of scaling the function to fit: finding the parameters to substitute into the generic equation to produce the right amplitude, frequency and offset.

But another answer is that the printed notation which approximates a number is computable. When a computer evaluates floor(x), it's all math. The printed (or electronically stored) notation for a rational number which approximates a real number is a mathematical object.


答:
Let $f(z):=e^z$. Then

$$\frac{e^z+1}{z-i \pi}=\frac{f(z)-f( i \pi)}{z-i \pi} \to f'( i \pi)$$

as $z \to  i \pi.$

Can you take it from here ?

Title: Limit $\lim_{z\to i\pi} \left( z-i\pi \right)\frac{e^z}{1+e^z}$

Body:
I'm about to finding a residue at $z\to i\pi$, but I'm stuck in finding the limit:

$$\lim_{z\to i\pi} \left( z-i\pi \right)\frac{e^z}{1+e^z}$$

What are the steps to eliminate the $z-i\pi$ part and $1+e^z$ part?


答:
Hint : $$\frac{1+e^z}{z-i\pi} = \frac{e^z - e^{i\pi}}{z- i\pi}$$

Taking the limit should make appear a derivative.


答:
Hint

$$y= \left( z-i\pi \right)\frac{e^z}{1+e^z}$$ Let  $z=x+i \pi$ to make
$$y=\frac{e^x x}{e^x-1}$$ since $e^{i \pi}=-1$. Now, use composition of Taylor series around $x=0$.

When done, replace $x$ by $z-i \pi$


问:
Difference between optimisation on manifolds and Lagrange multipliers
I have few reference I'm currently reading through but I still don't quite get the difference between optimising a function over a manifold and simply use constrained optimisation.

Do the algorithms end up being simpler maybe in the manifold cases?

Taking as example the line direction (i.e. gradient descent). In the constrained case I would add all the Lagrange multipliers and check the KKT conditions hold, while on a manifold I would just implement the retraction operator (I'm over simplifying I know...)

But apart from this difference I don't really see an advantage of one over the other.

Is there a computational advantage maybe? I know that a "large class" of constrained optimisation problems can be rephrased as optimisation over manifold but why would I want to do it? I'm pretty sure there's some subtlety I'm missing but so far I wouldn't really bother learning manifolds/riemannian manifolds just for a different framework.

答:
The optimization on the manifold is a special class, where constraints must satisfy the property of the manifold. No matter how complex the objective function is, we can treat it as a unconstrainted optimization problem when optimizing on the manifold with the Euclid gradient replaced by the manifold gradient.

However, the points that satisfy KKT conditions may not be checked easily when the objective function is very complex.

In my opinion, we should choose a different method according to the concrete problem.


问:
Can a transpose of a 2d vector geometrically be interpreted as 90º rotation?
See the (octave) code below.

```<code>pkg load geometry;

v1 = [3 4 0];
v2 = [2 5 0];
vr = rotateVector(v2, deg2rad(90));
vr = [vr(1) vr(2) 0];

dp = dot(v1, v2);
cp = cross(v1, vr);

disp('Dot p.: '), disp(dp);
disp('Parallelogram surface area: '), disp(cp);
</code>```

This is the output:

```<code>Dot p.:
    26
Parallelogram surface area:
    0   -0   26
</code>```

Here the cross product represents the surface area of the parallelogram between v1 and the rotated vector v2 by 90 degrees. (3d vectors are used to be able to use the cross product, but the Z in X, Y and Z of the vectors is 0, hence be ignored).

The cross product (or technically the length of this resulting vector) equals the dot product.

This cannot be a coincidence.

So can I conclude  that the transpose in the dot product v1Tv2 means: rotate vector by 90 degrees, in case of 2d vectors.

(If not, how is this reasoning invalid?)

答:
The answer is no.

Let's consider an example. You have a vector with coordinate x=0, y=2. It means the vector lies on the y axis and has the length of 2 unit. You can present it as a row matrix, namely (0, 2).
Transposing the vector means rewriting it as a column matrix, but coordinates remain unchanged (x=0 and y=2).


问:
What are the prerequisites to Jech's Set theory text?
I'm looking for a book to self-study axiomatic set theory, and heard this was a classic.  What are the main prerequisites for this text? My knowledge of set theory isn't too great. Probably the only time I came across nontrivial set theory was when I read the proof that every nonzero ring has a maximal ideal (Zorn's Lemma).</p>

答:
Reading Jech generally requires no prerequisites, but it is good to be familiar with the basics of logic and very naive set theory before reading it.

The first part of the book, which is the introductory part, covers a lot of basic topics in modern set theory, and most of the proofs are relatively complete. However Jech often skips small and minor steps, leaving them as exercises. If you are not used to this sort of writing it may be hard to read the book in full, on the other hand it may give you the push to fill in the details.

My best advice is to find someone to guide you through the book, so you could get an additional structure and help when reading it. You can ask questions here, but I'm not sure if that would be completely enough.

If you prefer to start with a different reading instead there are the books in Arthur's post as well as:

Kunen's book, Set Theory (which has a 2011 edition).

Halbeisen's book, Combinatorial Set Theory: With a Gentle Introduction to Forcing (Springer Monographs in Mathematics) (which is available freely on the author's website as for 2013).

Levy's book, Basic Set Theory.

Enderton's book, Elements of Set Theory.

I did not read the last two books, but I did hear a lot of good things about them from people in whom I have great trust on the matter. Note that each book is aimed to cover slightly different material in a slightly different approach. It all depends on what level of set theory you are aiming to have at the end of the book.


问:
Bott Tu Exercise 6.14, integration along the fiber
Suppose $\pi:E\to M$ is an oriented $C^\infty$ vector bundle of rank $n$. We denote by $\Omega_{cv}^k(E)$ the set of all differential $k$-forms $\omega$ on $E$, such that for each compact $K\subset M$, $\pi^{-1}(K)\cap \text{supp}(\omega)$ is compact. In particular, the support of the restriction $\omega|_F$ to each fiber is compact.  Assume $\{(U_\alpha,\phi_\alpha)\}$ is an oriented trivialization for $E$. On $\pi^{-1}(U_\alpha)$, such a form $\omega$ is uniquely expreesed as a sum of the forms either of type $(\pi^*\phi)f(x,t_1,\dots,t_n)dt_{i_1}\cdots dt_{i_r}$ with $r<n$, or $(\pi^*\phi)f(x,t_1,\dots,t_n)dt_1\cdots dt_n$. (Here $x_1,\dots,x_n$ are coordinate functions on $U_\alpha$ and $t_1,\dots,t_n$ are fiber coordinates on $\pi^{-1}(U_\alpha)$ given by $\phi_\alpha$. We define a map $\Omega_{cv}^*(E)\to \Omega^*(M)$ by sending the forms of first type to zero, and the forms of second type to $\phi \int_{\Bbb R^n} f(x,t_1,\dots,t_n)dt_1 \dots dt_n$.

Exercise 6.14 askes to show that this map is well-defined. Suppose $U_\alpha \cap U_\beta $ is nonempty. Then on $\pi^{-1}(U_\alpha \cap U_\beta)$, a form of second type can be expressed as
$$(\pi^*\phi)f(x,t_1,\dots,t_n)dt_1\cdots dt_n=(\pi^* \tau)g(y,u_1,\dots,u_n) du_1\cdots du_n.$$
Then I have to show that
$$\phi \int_{\Bbb R^n} f(x,t_1,\dots,t_n)dt=\tau \int_{\Bbb R^n} g(y,u_1,\dots,u_n)du,$$
but I got stuck. Any hints?

答:
Assume that $U_a$ and $U_b$ are relatively compact and $\varphi$, $\psi$ its charts on $M$. By a linearity argument, you  can assume that

$$
\phi = h_1 dx_{i_1} \wedge \cdots dx_{i_k}
$$

and the same for $\tau$ with coefficient $h_2$. Without loss of generality assume $k = m = \dim M$. Let $T$ and $S$ the coordinates for $t$ and $s$ in $\mathbb{R}^n$. Take the adaptated chats in $E$ $(U_a \times \mathbb{R}^n,\varphi \times T)$ and $(U_b \times \mathbb{R}^n,\psi \times S)$.

We have $\pi^*\phi = h_1 \circ \pi \pi^*(dx) = h_1dx$ in these coordinates. So

$$
\pi^* \phi f(x,t)dt = h_1(x)f(x,t)dx\wedge dt
$$
On the other hand, using the change of variables for diff. forms
$$
\pi^* \tau gds = (h_2 \circ \psi \circ \varphi^{-1})(g \circ \psi \circ \varphi^{-1}) \cdot Jac(\psi \circ \varphi^{-1}) Jac(S \circ T^{-1}) dx \wedge dt = h_1 fdx\wedge dt
$$
We deduce that $$h_1f = (h_2 \circ \psi \circ \varphi^{-1})(g \circ \psi \circ \varphi^{-1}) \cdot Jac(\psi \circ \varphi^{-1}) Jac(S \circ T^{-1})$$

The last expression can be integrated into $(U_c = U_a \cap U_b) \times \mathbb{R}^n$ and using Fubini and the change of variable for integrals and the previous definitions
$$
\int_{U_c \times \mathbb{R}^n}\phi f(x,t)dx\wedge dt = \int_{U_c \times \mathbb{R}^n}\tau g(y,s)dy \wedge ds.
$$
Finally, by a parametrical argument and using $U_a$ and $U_b$ be compact

$$
\phi \int_{\mathbb{R}^n} f(x,t)dt = \tau \int_{\mathbb{R}^n}g(y,s)ds.
$$

You must be a little more cautious for $k < n$ because the change of variable for differential forms is a bit different.  If I have some typo please fix it.


问:
Maximal solution and its domain of $x'(t) = x(t)^{x(t)}$.
How can I proof that the maximal solution to

\begin{align}
x'(t) = x(t)^{x(t)}, \quad x(0) = x_0,
\end{align}

where $x_0 > 0$ and $t \geq 0$, is not (globally) defined on $\mathbb{R}_{0}^{+}$?

I am given the hint that it might help to first look at $x_0 > 1$. Unfortunately, that does not help. I am close to giving up on this one so any help is appreciated.

答:
Perhaps a little more direct, we know that $x^x=e^{x\ln x}$ has a minimum of $e^{-e^{-1}}>0.6922>\frac12$ at $x=e^{-1}$. So we know that the solution is growing at least with the rate $\frac12$. So we can be sure that $x(t)\ge 2$ for $t\ge 4$, assuming the domain of the maximal solution $x$ extends far enough to contain $t=4$.

After that, $x'\ge x^2$ so that
$$
x(t)\ge\frac{x(4)}{1-x(4)(t-4)}
$$
forces a divergence to infinity shortly after that point (as $x(4)\ge 2$).


问:
Finding fourth corner of parallelogram

If three comers of a parallelogram are (1, 1), (4,2), and (1,3), what
are all three of the possible fourth comers?

I am solving this question from linear algebra textbook. In my opinion there will be only one possible fourth point. But answer is three possible points. How is this possible?

答:
This is sort of a combinatorics question. Given three points from a parallelogram, how many of the edges connecting the three points will be sides of the parallelogram?

You probably drew two of the sides but which two? Well, it depends on the parallelogram...

Title: Finding fourth corner of parallelogram


问:
Equilibrium of simple SIS model
I have a simple SIS model with a non-constant population size due to disease-related mortality:

\begin{aligned}
  \frac{dS}{dt} &= a I - b S I\\\\
  \frac{dI}{dt} &= b S I - a I - c I
\end{aligned}

where $S$ and $I$ are the susceptible and infected classes, $a$ is the rate of transition back to the susceptible class after being infected, $b$ is the per-capita transmission rate, and $c$ is the disease-related mortality for infecteds.

Finding the equilibria of this model is stumping me for some reason. First, I factor the second equation to get:

\begin{equation}
  I ( bS - a - c ) = 0
\end{equation}

so $\hat{I} = 0$ or $\hat{S} = \frac{a+c}{b}$ must be accord with $\dot{I} = 0$. The first means that:

\begin{equation}
  \dot{S} = a(0) - b S (0) = 0
\end{equation}

and so $\hat{S}$ must be its initial condition, $S_0$.

However, there is another equilibrium where the disease spreads and dies out, meaning $\hat{S} < S_0$, which I assumed is equal to $\hat{S} = \frac{a+c}{b}$.

However, using simultation in $R$, this equilibrium isn't correct - $\frac{a+c}{b}$ is still off by a fair amount.

Can anyone see where I'm going wrong? Thanks!

答:
For $I=0$ you can have any value of $S$ to get an equilibrium position, the equilibrium equations do not impose further conditions on $S$. If you add the stability at $I=0$, then you have to add the condition $bS−a−c<0$. Outside that region the values for $I$ will increase.

If you divide both equations, you find that
$$
\frac{dS}{dI}=\frac{a-bS}{bS-a-c}
$$
is a separable equation, so that
$$
I+S-\frac{c}{b}\ln|bS-a|
$$
is a conserved quantity. That is, for the limit with $\hat I=0$ you get the equation
$$
\hat S-\frac{c}{b}\ln|b\hat S-a|=I_0+S_0-\frac{c}{b}\ln|bS_0-a|
\\~\\
e^{b\hat S-a}(b\hat S-a)=e^{b(I_0+S_0)-a}(bS_0-a)
\\~\\
\hat S=\frac1b\left(a+W_0(e^{b(I_0+S_0)-a}(bS_0-a))\right)
$$
for the limit value $\hat S$. The sign of $bS-a$ has to stay constant, $W_0$ is the Lambert-W function.

This only works for the simplest of models, more complex models will usually not give rise to conserved quantities/constants of motion.

Title: Equilibrium of simple SIS model


问:
Check the continuity and differentiability of $f(x)= \sin^{-1}(\cos x)$ at $x=0$

Check the continuity and differentiability of  $f(x)= \sin^{-1}(\cos x)$  at $x=0$

This is how I tried to solve the problem:

$$f(x)= \sin^{-1}(\cos x)=???$$

$$\lim_{x \to 0^+}f(x)=\lim_{x \to 0^-}f(x)=f(0)=???$$
Therefore the function is continuous at $x=0$
$$\lim_{x \to 0^+}\frac{f(x)-f(0)}{x-0}=???$$and
$$\lim_{x \to 0^-}\frac{f(x)-f(0)}{-x}=???$$

答:
Both $\arcsin$ (I prefer this notation to $\sin^{-1}$) and $\cos$ are continuous functions, so it's also their composition, wherever it's defined; since $-1\le\cos x\le 1$ for all real $x$, your function $f$ is continuous on the real line.

Let's compute its derivative:

$$
f'(x)=\arcsin'(\cos x) \cos'x=
-\frac{1}{\sqrt{1-\cos^2 x}}\sin x=
-\frac{1}{\lvert\sin x\rvert}\sin x
$$

whenever this expression is defined. So your function is differentiable on all $x$ that are not integer multiples of $\pi$. What about differentiability in $0$? It's the same as in the other points where the above expression is undefined.

We have
$$
\lim_{x\to 0^-}f'(x)=1,\quad \lim_{x\to 0^+}f'(x)=-1
$$
so, by l'Hôpital's theorem the function $f$ is not differentiable in $0$.

What can you deduce from this? Let's limit ourselves to the interval $(0,\pi)$. Here your function has the same derivative as $x\mapsto-x$, so there is $c$ such that $f(x)=-x+c$:
$$
f(\pi/2)=\arcsin\cos\frac{\pi}{2}=\arcsin0=0=-\frac{\pi}{2}+c
$$
so that you can surely write
$$
f(x)=\frac{\pi}{2}-x \quad \text{for }x\in(0,\pi).
$$

For $x\in(-pi,0)$ the situation is similar: your function has the same derivative as $x\mapsto x$, so you can write $f(x)=x+d$:
$$
f(-\pi/2)=\arcsin\cos\left(-\frac{\pi}{2}\right)=\arcsin0=0=-\frac{\pi}{2}+d
$$
so that $d=\pi/2$. Therefore
$$
f(x)=\frac{\pi}{2}+x \quad \text{for }x\in(-\pi,0).
$$

You can go on and write an explicit expression for $f$, but observing that it is periodic with period $\pi$ makes this unnecessary. In any case, be very careful, because $\sqrt{1-\cos^2x}=\sin x$ is, in general false; the correct identity is

$$\sqrt{1-\cos^2x}=\lvert\sin x\rvert$$


问:
Intuition on exp function
So in my calc1 class we introduced $e$ and $\exp(x)$ by searching for a function with the property that it is equal to its derivative. We get the well known infinite sum, all good.
With the Cauchy product and the Binomial theorem we proof that $\exp(x+y)=\exp(x)\exp(y)$ is correct.
What I don't understand is why the function that is equal to its derivative coincidentally satisfies $\exp(x+y)=\exp(x)\exp(y)$ and I'm looking for a non-formal explanation. Thanks in advance, and I apologize if I didn't find any obvious resources on that question on the internet.

答: Fix $y\in\Bbb R$ and consider the functions $f,g\colon\Bbb R\longrightarrow\Bbb R$ defined by$$f(x)=\exp(x+y)\quad\text{and}\quad g(x)=\exp(x)\exp(y).$$Then $f'=f$ and $g'=g$. So$$\left(\frac fg\right)'=\frac{gf'-fg'}{g^2}=0.$$Therefore, $\frac fg$ is constant. But $f(0)=g(0)\bigl(=\exp(y)\bigr)$, and so $f=g$.

Title: Intuition on exp function


问:
Definition of vector bundles - what is meant by a structure of a real vector space?
I am currently reading John Lee's book about Differential geometry and I don't fully understand his definition of vector bundles. Unfortunately, Wikipedia didn't help.

A vector bundle of rank $k$ is a continuous map $\pi\colon E\to M$ with the following property (among others):

For each $p\in M$, $\pi^{-1}(p)$ is endowed with the structure
of a $k$-dimensional real vector space.

On page 13 of "Introduction to
Smooth Manifolds", he defines a smooth structure to be a smooth atlas, so I thought that a structure might be an atlas. But even is if this is true, I don't see what he means by the addition "of a $k$-dimensional real vector space".

答:
It's structure in the general mathematical sense of "adding something" to a set to get a richer object:

To endow a set $G$ with a group structure is to pair it with an operation $*$ such that $(G,*)$ is a group;

To endow a set $V$ with a vector space structure is to pair it with operations $+, \cdot$ satisfying the vector space axioms;

To endow a set $M$ with a smooth structure is to pair it with a smooth atlas.

In this case, what is meant is that part of the definition of a vector bundle is the choice of a $k$-dimensional vector space structure on each fiber $\pi^{-1}(p),$ i.e. $+$ and $\cdot$ satisfying the vector space axioms are defined on each $\pi^{-1}(p).$


问:
Double for loop doesn't work
I have 100000 iid normal variables X1, . . . , X100000 from N(0,1) and for i ∈ {2, . . . , 100000} and want to plot the graph of the function \

$$ R_i = \frac{\max{X_j}}{\sqrt{2 \log i}} $$

where  j=1,...,i. And Repeat the above experiment 10 times and plot the respective trajectories of Ri
on the same graph. I tried this code

```<code>sample<- rnorm(100000,mean=0,sd=1)
i=2:100000
r=0
for (i in 2:100000){
  for (j in 1:i){
    r[i]=(max(sample[j]))/(sqrt(2*log(i)))
  }
}
plot(r)
</code>```

But the results are not logical. I think the for loop doesn't work properly in max(X_j) but I don't know where. Also how to repeat it 10 times, could you please help? Thank you!

答:
This is more of a programming question as it is not quite about the law of the iterated logarithm.

R has a cummax function so you could do something like

```<code>set.seed(2020)
r <- function(samplesize){
  s <- rnorm(samplesize, mean=0, sd=1)
  cummax(s) / sqrt(2*log(1:samplesize))
  }
simulations <- replicate(10, r(10^5))
matplot(simulations, type="l")
</code>```

to get something like

Title: Double for loop doesn't work

Body:
This is more of a programming question as it is not quite about the law of the iterated logarithm.

R has a cummax function so you could do something like

set.seed(2020)
r <- function(samplesize){
  s <- rnorm(samplesize, mean=0, sd=1)
  cummax(s) / sqrt(2*log(1:samplesize))
  }
simulations <- replicate(10, r(10^5))
matplot(simulations, type="l")

to get something like


答:
By the additive property, if you shift the argument, you get a function proportional to the original:

$$f(x+1)=f(x)f(1).$$

Hence the slope of the chord is also proportional:

$$\frac{\Delta f}{\Delta x}=f(x+1)-f(x)=(f(1)-1)f(x)=cf(x).$$

The same goes for the slope of a tangent.

If we consider $f(\lambda x)$ instead of $f(x)$, which is a stretched version of $f$, the slopes are modified accordingly, and it will be possible to find a $\lambda$ such that $c=1$.

On this plot, the function $3^x$ (blue), the slope of chords for unit intervals (green) and the slope of tangents (magenta). All proportional.

Title: Intuition on exp function

Body:
So in my calc1 class we introduced $e$ and $\exp(x)$ by searching for a function with the property that it is equal to its derivative. We get the well known infinite sum, all good.
With the Cauchy product and the Binomial theorem we proof that $\exp(x+y)=\exp(x)\exp(y)$ is correct.
What I don't understand is why the function that is equal to its derivative coincidentally satisfies $\exp(x+y)=\exp(x)\exp(y)$ and I'm looking for a non-formal explanation. Thanks in advance, and I apologize if I didn't find any obvious resources on that question on the internet.


问:
d-metric on the space of types.
Reading this paper http://math.univ-lyon1.fr/~begnac/articles/mtfms.pdf by Itaï Ben Yaacov, Alexander Berenstein, C. Ward Henson, and Alexander Usvyatsov I have encountered the following problem. In section 8, subsection 'The d-metric on types', they define the distance between two types $p$, $q\in S_n(T_A)$ to be the infimum of the distance between their realisations. More precisely:
$$ d(p,q)=\inf\{ d(a,b) : \mathcal{M}_A\models p[a] ,  \mathcal{M}_A\models q[b]\}.$$
I have problems proving the triangular inequality, I have tried to use compactness theorem and also tried to think this metric as a pseudometric induced in the quotient space in the usual way. However, I am not able to end the proof.

答:
Claim: The infimum above is always attained at a pair of points $a,b\in M^n$ such that $\mathcal{M}_A\models p[a]$, $\mathcal{M}_A\models q[b]$.

Proof: Since the set of conditions $p(x)\cup q(y)\cup \{ d(x,y)\leq d(p,q)+\frac{1}{n}: n\geq 1 \}$ is finitely satisfiable, the saturation of $\mathcal{M}$ implies that there exists an element satisfying the whole set. Hence, if $d(p,q)=0$, they have a common realization, so $p=q$.

Lemma: The distance $d$ defined above, defines a metric in $S_n(T_A)$.

Proof: The only property of a metric that is nontrivial is the triangular inequality. To prove it, we first show that given $a'\in M^n$ such that $\mathcal{M}_A\models p[a']$, there exists $b'\in M^n$, $\mathcal{M}_A\models q[b']$, such that $d(a',b')=d(p,q).$ Suppose that $d(p,q)$ is attained in a pair of points $a,b\in M^n$. As $a$ and $a'$ both realize the same type $p$ over $A$, there exists an elementary map
\begin{equation*}
  h:A\cup\{a\}\to A\cup\{a'\}.
  \end{equation*}

Now, we use that $\kappa$-saturation implies $\kappa$-homogeneity to extend this elementary map to an elementary map
\begin{equation*}
  h':A\cup\{a,b\}\to A\cup\{a',b'\}\subseteq {M}.
  \end{equation*}
Hence $\mathcal{M}\models q[b']$ and $d(a,b)=d(a',b')=d(p,q)$.

Now, let $p,q,r\in S_n(T_A)$. Suppose that $d(p,r)=d(a,c_1)$ and $d(r,q)=d(c_2,b)$. As, $c_1$ and $c_2$ realize the same type $r$ over $A$, by the discussion above, there exist $b'$, $\mathcal{M}\models q[b']$, such that $d(c_1,b')=d(c_2,b)$. Hence, $$d(p,r)+d(r,q)=d(a,c_1)+d(c_1,b')\geq d(a,b')\geq d(p,q),$$
Where the middle inequality is just the tiangular inequality for the metric on $M^n$.

Title: d-metric on the space of types.


问:
Finding the area between two curves and the x-axis
The two curves $y = 3\sqrt{4-x}$ and $y = 3\sqrt{x-2}$ are given. The task is to find the area enclosed by the curves and the x-axis.

As far as I see, the root of the first function is $x = 4$ and the root of the second function is $x = 2$. Their intersection is at $x = 3$.

My question is now the following: How do I have to chose the intervals of integration such that I get the area mentioned?

Simply integrating $\int_{2}^4 \left( 3\sqrt{4-x} - 3\sqrt{x-2} \right) dx$ seems to be wrong, isn't it?

Thanks.

答:
Notice that $3 \sqrt{4 - x}$ is only defined for $x \le 4$, where it's values are non-negative.  Similarly, $3 \sqrt{x - 2}$ is only defined for $x \ge 2$, where it's values are non-negative.

As you've correctly observed, the two curves intersect at $x = 3$, where $y = 3$ as well.  The two curves, together with the $x$-axis (which is yet another curve, namely $y = 0$) form the boundary of a region.

Sketch the picture.  (You ought to know what the graph of $y = \sqrt{x}$ looks like.  Hint: it's a parabola.  Now, your curves are obtained by some transformations.)

Now, the area is calculated as:
$$
\begin{align}
A &= \int_2^4 \min\{ 3 \sqrt{x - 2}, 3 \sqrt{4 - x} \}\;dx \\
&= \int_2^3 3 \sqrt{x - 2}\;dx + \int_3^4 3 \sqrt{4 - x}\;dx.
\end{align}
$$


问:
Doubt in proving homomorphism from $S_n$
My question is :

Let us assume that we have to find the number of homomorphisms from $S_n \to D_{2n}$ when $n > 3$

How  to prove  that the elements of the form $xyx^{-1}y^{-1}$ always belong to the kernel.

So  all the even number of $2$-cycles can be written in the form $xyx^{-1}y^{-1}$ and they belong to the kernel. [which is what I know as $A_n$ is the only normal subgroup of $ S_n$ except when $n \ne 4$,but I can't use this result as I will have to prove this using sylows theorem which we haven't been taught so can't be used in exam ]

答:
I really like @ancientmathematician's argument above but if you want here's another argument.

Suppose $N$ is the kernel of your homomorphism, then $N$ has index dividing $2n$, so all $2n$th powers are in $N$, in particular, those elements whose order is coprime to $2n$ are in $N$. Now choose odd number $k\leq n$ coprime to $2n$ (the only time you can't do this is when $n=3$ which as the others mentioned is a counterexample, otherwise choose $k=n-1$ or $n-2$ works) then all $k$-cycles by the above are in $N$.

All $k$-cycles for $k$ odd generate $A_n$ so $A_n\subset N$. If you don't yet know this see here or here And it's clear that all commutators are in $A_n$.


问:
A puzzling point in proof of Eisenstein Criterion for irreducible polynomials on Integral Domain
Theorem (Eisenstein Criterion).
Let $P$ be a prime ideal of the integral domain
$R$ and let $f(x) = x^n +a_{n-1}x^n+\cdots+a_1 x +a_0$ be a polynomial in $R [x]$ (here $n \geq 1$).
Suppose $a_{n-1},\dots, a_1, a_0$ are all elements of $P$ and suppose $a_0$ is not an element of $P^2$. Then $f(x)$ is irreducible in $R[x]$.

Proof. Suppose $f(x)$ were reducible, say $f(x) = a(x)b(x)$ in $R[x]$, where $a(x)$
and $b(x)$ are nonconstant polynomials. Reducing this equation modulo $P$ and using
the assumptions on the coefficients of $f(x)$ we obtain the equation $x^n = \overline{a(x)b(x)} $ in $(R/ P)[x]$, where the bar denotes the polynomials with coefficients reduced mod $P$.
Since $P$ is a prime ideal, $R/P$ is an integral domain, and it follows that both $\overline{a(x)}$ and
$\overline{b(x)}$ have $0$ constant term, i.e., the constant terms of both $a (x)$ and $b(x)$ are elements
of $P$ . But then the constant term $a_0$ of $f(x)$ as the product of these two would be an
element of $P^2$, a contradiction. $\Box$

This proof is from Abstract Algebra by Dummit & Foote. The puzzling point is bold-italic in the proof. Why it is not possible that only one constant of $\overline{a(x)}, \overline{b(x)}$ is $0$ and the other is not? I don't think it violates the rule of integral domain in this case.

Could any one give me some idea ? Thank you in advance!

答:
That proof of Eisenstein's criterion is clearer using unique factorization of prime products.

$\rm\color{#0a0}{Assume}$ that $f$ is reducible $\,f = g\,h,\,$ $\,i = \deg g,\ j = \deg h,\ \color{#c00}{i,j\ge 1}$, wlog $\,g,h\,$ monic.

This factorization maps to $\,x^n = \bar g\,\bar h\,$ in $\,\bar R := R/P,\,$  a domain, so $\,x\,$ is prime in $\bar R[x]$.

Since it is a prime power: $\,x^n = x^k\, x^{n-k}\,$ are the only possible monic factorizations.

Thus $\,\bar g = x^i,\ \bar h = x^j,\,$ so $\,\color{#c00}{i,j\ge 1}\Rightarrow\,\bar g(0)\!=\!0\!=\!\bar h(0),\,$ i.e. $\,g(0),h(0)\in P$.

Therefore $\,f(0) = g(0)h(0)\in P^2,\,$ contra $\rm\color{#0a0}{hypothesis},\,$ hence $f$ is irreducible.

Alternative direct inductive proof that $\,\bar g(0)\!=\!0\!=\!\bar h(0),\,$ i.e. $\,x\mid \bar g,\bar h.\,$ If not, wlog $\, x\nmid \bar g,\,$ then $\,x^n = \bar g\,\bar h\,$ $\Rightarrow\,x\mid \bar g\,\bar h,\,$ so $\,x\nmid \bar g\Rightarrow x\mid \bar h,\,$ by $\,x\,$ prime. Repeating this shows all $\,n\,$ factors of $\,x\,$ must divide into $\,\bar h,\,$ so $\,n = \deg \bar h = \deg h,\,$ so $\,\color{#c00}i = \deg g \color{#c00}{= 0},\,$ contra hypothesis.


问:
Possible applications for this semi-linear first order PDE.
I am cross-posting this question https://physics.stackexchange.com/questions/590549/possible-applications-for-this-semi-linear-first-order-pde

First of all I should say that I work in probability and my knowledge about PDEs is quite small, so this question could make little sense, please let me know if something is not well stated.

While dealing with approximations methods for SDEs a I've noticed a particular connection between an SDE and a deterministic PDE of the form:

$$\partial_t u+\sigma(t)\partial_x u=b(t,x,u)$$
where $b$ is a Lipschitz continuous function in $u$.

I've tried searching online for application of this kind of equation but unfortunately I wasn't able to find anything concrete.
In the book by  Moussiaux, Zaitsev and Polyanin "Handbook of first order PDEs" they discuss methods for solving this kind of equations but they don't provide examples of applications.

I suspect this could be connected somehow to the transport equations, but I am not entirely sure.
Do you know of some references for applications of this particular equation?

答:
Your equation is quasilinear, but sometimes it can be connected with a semilinear equation
$$ v_s +v_x = c(s,x,v). $$
Specifically, define $\Sigma(t) = \int_0^t \sigma (\tau)d\tau$.
If $\Sigma$ is an invertible function, for example if $\sigma$
is increasing, then the change of variables
$$ s = \Sigma(t), \qquad u(x,t) = v(x,s) $$
gives
$$ u_t+\sigma u_x = \sigma(v_s+v_x) = b(\Sigma^{-1}(s),x,v) $$
so dividing by $\sigma$ you have $v_x+v_x = c(s,x,v)$ with
$$c(s,x,v) = \frac{b(\Sigma^{-1}(s),x,v)}{\Sigma'(\Sigma^{-1}(s))}$$

Edit: for an example application, suppose $w(x,t)$ is the
depth of a layer of particles that are being transported to the right
by some process, or perhaps the density of animals at longitude $x$ during
an eastward migration. Consider a conservation law of the form
$$
 \frac{d}{dt}\int_a^{a+h} w(x,t)dx
 = kw(a,t)-kw(a+h,t)
 +\int_a^{a+h}c(w,x,t)dx
$$
where some function $c(w,x,t)$ represents a source of particles/length, or
a birth rate of animals, and $kw$ is the flux of particles from left
to right, $k$ being a fraction of the local density that are able
to migrate past.

Thus the conservation law says that the time rate of change of the population
in the interval $[a,a+h]$ is given by flux in (imigration), minus
flux out (emigration), plus
the birth rate.

Divide by $h$ and take the limit as $h$ tends to zero; you get by the
Fundamental Theorem of Calculus that
$$
 w_t = -kw_x+c(w,x,t).
$$


问:
Seating Problem Formula Derivation
I am trying to solve this problem, but I am just confused on how to approach it.

Problem Statement: A class should consist of N students. Inside the
class the students have to be seated in a row of N chairs. In the
class of N students, only fewer than B boys can sit together and fewer
than G girls can be seated together. In how many different ways can a
class of N students be formed? (Assume that you have more than N boys
and N girls available).

This is not a homework question, but a question from one of the exercises that I am trying to solve from a book (self-learning). I am able to solve questions where the variable are known (this is from advance section where variables are unknown and we should derive formulas). It would be grateful if someone could explain how to approach such problems (or at least point me to some resource for such problems). Also I would be grateful if someone could tell me how to solve this particular problem as well.

Thank You.

Denise

答:
This solution is taking the assumption that it has to be in algebric form rather than be able to give an actual absolute value.

So taking first condition means that for the problem you can assume that all the students are in one long line (rather than a typical classroom)

The second condition means that you can only have limited number of the same gender students next to each other (eg you can't have a class of just boys or girls)

The last statement just means that there are essentially unlimited 'resources' so any combination of boys and girls is possible.

So using the last statement you can work out the absolute maximum number of combinations will be 2^N as each place has 2 possibilties as each thing is independent of the other hence using num of possibilites to the power of number of spaces.

As said the second condition means you can only have a limited amount of boys and girls next to eachother. Meaning that it is a divisor of the maximum of 2^N.

I sincerely apologies but this is as much as I think I can give without risking making something up creating a false answer and leading you down wrong path but I believe what I have given you is a good way to help breakdown the question to help understand constriants and is most likely a good way for you to attempt any similar questions.

Title: Seating Problem Formula Derivation


问:
why $(\varphi \circ \varphi^{-1})(I) = I$ implies that $\varphi^{-1}(I) \subset I$?
Let $R$ be a ring and $I \subset R$ a two-sided ideal, with quotient homomorphism $\pi : R \rightarrow R/I.$ Let $\operatorname{End_{I}(R)}$ be the set of $\varphi \in \operatorname{End(R)}$ such that $\varphi(I) \subset I,$ and let $\operatorname{Aut_{I}(R)} = \operatorname{End_{I}(R) \cap Aut(R)}.$

$(a)$ Given $\varphi \in \operatorname{End_{I}(R)},$ show that there exists $\bar{\varphi} \in \operatorname{End(R/I)}$ such that $\bar{\varphi} \pi = \pi \varphi.$

$(b)$ Given $\varphi, \psi \in \operatorname{End_{I}(R)},$ show that $\overline{\varphi \psi} = \bar{\varphi} \bar{\psi}.$

$(c)$ Given $\varphi \in \operatorname{Aut_{I}(R)},$ show that $\varphi(I) = I$ and $\bar{\varphi} \in \operatorname{Aut(R/I)}.$\ Hint: Use part $(b).$

My question is:

Here is the solution:

1- I do not understand why $(\varphi \circ \varphi^{-1})(I) = I$ implies that $\varphi^{-1}(I) \subset I, $ I can not see the details of this proof, could anyone explain that for me please?

2- I do not understand the definition of $\operatorname{Aut(R/I)}$ used in showing that $\bar{varphi} \in \operatorname{Aut(R/I)},$ Could anyone explain that for me please? I know that  $\operatorname{Aut(R/I)}$ is the set of all isomorphisms from $R/I$ to $R/I$ but how is that definition related to the existence of inverse for $\bar{varphi},$ could anyone explains that for me please?

答:
You wrote that we have $\varphi^{-1} \in \operatorname{Aut}_I(R)$. By definition, this means that $\varphi^{-1}(I)\subset I$, because this is meant by the subindex $I$: "
Let $\operatorname{End_{I}(R)}$ be the set of $\psi \in \operatorname{End(R)}$ such that $\psi(I) \subset I.$ "
So take $\psi=\varphi^{-1}$. So the claim from the title follows.


问:
Separation of variables for Aronsson's equation in $\mathbb{R}^2$
I am working on the following problem

I don't seem to get it done. I've tried with $u(x_1,x_2)=v(x_1)w(x_2)$ and got the following
$$w^3(v')^2v''+2wv(w')^2(v')^2+v^3(w')^2w''=0.$$
But this doesn't help since no matter what I do, I get terms involving both $v$ and $w$.

Then I found this post: Separation of Variables for PDE

I tried that change of variables in that post and after a lengthy calculations I still did not get rid off the mixed terms. Any help on this one? The problem is from Evans book on PDEs from chapter 4.

答:
$$w^3(v')^2v''+2wv(w')^2(v')^2+v^3(w')^2w''=0.$$
$$\frac{(v')^2v''}{v^3}+2\left(\frac{v'}{v}\right)^2\left(\frac{w'}{w}\right)^2+\frac{(w')^2w''}{w^3}=0 $$
Among several possibilities : Let $\frac{v'}{v}=\lambda=$constant.
$$v(x_1)=c_1e^{\lambda x_1}\qquad\lambda=\text{constant}$$
$$\lambda^4+2\lambda^2\left(\frac{w'}{w}\right)^2+\frac{(w')^2w''}{w^3}=0 $$
Solving for $w(x_2)$ would lead to many solutions. More simply, among those solutions :
$$w(x_2)=c_2e^{\mu x_2}\qquad\mu=\text{constant}$$
with condition
$$\lambda^4+2\lambda^2\mu^2+\mu^4=0 $$
$$\mu=\pm i\:\lambda$$
$$u(x_1,x_2)=v(x_1)w(x_2)=c\:e^{\lambda(x_1\pm i\:x_2)}$$
The constants $c$ and $\lambda$ can be complex. Real solutions with sinusoidal functions can be derived instead of exponential of complex variable.


问:
Solving systems of linear equations non directly
Consider the system of linear equations

$$\mathbf A \mathbf x = \mathbf b,$$

where $\mathbf A \in \mathbb R^{n \times n}$ is invertible and $\mathbf b \in \mathbb R^n$. Let $\mathbf D \in \mathbb R^{n \times n}$ be a diagonal and invertible.

We can solve the system $\mathbf A \mathbf x = \mathbf b$ directly. However, I want to show that if we solve it such a way that we calculate $\mathbf D \mathbf A \mathbf D$ and $\mathbf D\mathbf b$, solve the system $\mathbf D \mathbf A \mathbf D \mathbf y = \mathbf D\mathbf b$, and calculate $\mathbf x = \mathbf D \mathbf y$. Then this  $\mathbf x$ will be the same  $\mathbf x $ as solving it directly.

Using Algorithm GEPP (Gaussian elimination with partial pivoting), how would you go about showing that the vector $\mathbf x$ calculated in Approach 2 indeed solves $\mathbf A \mathbf x = \mathbf b$.

Thank you in advance.

答:
if we solve your second equation, we get :

$$y=\left(DAD \right)^{-1}Db=D^{-1}A^{-1}D^{-1}Db=D^{-1}A^{-1}b $$

so you get :

$$Dy=A^{-1}b=x $$ by your first definition


问:
On the limit $\lim \limits_{x \rightarrow +\infty} \left( e^{\sqrt{x+2}} + e^{\sqrt{x-2}} - 2 e^{\sqrt{x}} \right)$
Evaluate the limit

$$\ell = \lim_{x \rightarrow +\infty} \left( e^{\sqrt{x+2}} + e^{\sqrt{x-2}} - 2 e^{\sqrt{x}} \right)$$

without using differential calculus.

I'm interested in evaluating the above limit using only pure limit theory without using MVT , Taylor etc. The limit is equal to $+\infty$ which is easy to extract using Taylor for example. I'm looking for  a way to evaluate it avoiding the big guns.

答:
Let $f(t)=(e^t-1-t)/t^2$; then $e^\sqrt{x+2}+e^\sqrt{x-2}-2e^\sqrt{x}=e^\sqrt{x}g(x)$, where
\begin{align*}
g(x)&=g_0(x)+g_-(x)+g_+(x),
\\g_\pm(x)&=h_\pm^2(x)f\big(h_\pm(x)\big),
\\h_\pm(x)&=\sqrt{x\pm 2}-\sqrt{x}=\frac{\pm 2}{\sqrt{x}+\sqrt{x\pm 2}},
\\g_0(x)&=h_+(x)+h_-(x)
\\&=\frac{2}{\sqrt{x}+\sqrt{x+2}}-\frac{2}{\sqrt{x}+\sqrt{x-2}}
\\&=\frac{2(\sqrt{x-2}-\sqrt{x+2})}{(\sqrt{x}+\sqrt{x+2})(\sqrt{x}+\sqrt{x-2})}
\\&=-\frac{8}{(\sqrt{x}+\sqrt{x+2})(\sqrt{x}+\sqrt{x-2})(\sqrt{x+2}+\sqrt{x-2})}.
\end{align*}
Trivially $\lim\limits_{x\to+\infty}xh_\pm^2(x)=1$ and $\lim\limits_{x\to+\infty}xg_0(x)=0$. Now if we know that $\lim\limits_{t\to 0}f(t)=1/2$, we obtain $\lim\limits_{x\to+\infty}xg(x)=1$, and finally $\lim\limits_{x\to+\infty}e^\sqrt{x}/x=+\infty$ implies that the given limit is $+\infty$.


问:
Limits and double integral?
Determine the double integral $\int \int_{R} ye^y dxdy$, where $R$ is $1\leq x-y \leq 2,\ 1 \leq xy \leq 2$

I'm having trouble to grasp how i should think to get the limits.

$x\ne y$ becaue $1\leq xy \leq 2$.

$x$ must be atleast 1 greater then $y$ because $\ 1 \leq x-y \leq 2$.

This give me $\begin{matrix}
x &y   \\
2 & 1  \\
\end{matrix}$

Can i thefore say that the limits are $0 \leq y \leq 1, 0\leq x\leq 2y$ ?

答:
We have that $R$ corresponds to the strip between the lines

$y=x-2$

$y=x-1$

and the hyperbolas

$y=2/x$

$y=1/x$

One way to proceed is to find the intersection points for these region and then integrate accordingly for each part.

Title: Limits and double integral?


问:
Spectrum of adjoint of linear operator
The following result is confusing me; I have not been able to find a clear proof that I understand.

Suppose $T$ is a bounded linear operator over a Hilbert space $\mathcal H$. Then $\sigma(T^*)=\{\overline\lambda:\lambda\in\sigma(T)\}$.

What I've done so far: I know that if $\lambda\in\sigma(T)$ then the map $T-\lambda I$ is not invertible. However, I don't know how to use this definition to conclude that $\sigma(T^*)=\{\overline\lambda:\lambda\in\sigma(T)\}$. If I know that $T-\lambda I$ is not invertible, then it could be the case that $T-\lambda I$ is not injective, or the case that $T-\lambda I$ is not surjective.

From linear algebra there is a theorem that states that, on a finite-dimensional vector space, a number $\lambda$ is not a eigenvalue for $T$ if and only if $\overline\lambda$ is not an eigenvalue for $T^*$, because if $T-\lambda I$ is invertible, then
$$
(T-\lambda I)^{-1}(T-\lambda I)=(T-\lambda I)(T-\lambda I)^{-1}=I,
$$
which gives
$$
(T^*-\overline\lambda I)((T-\lambda I)^{-1})^*=((T-\lambda I)^{-1})^*(T^*-\overline\lambda I)=I.
$$
This means $\lambda$ is an eigenvalue for $T$ if and only if $\overline\lambda$ is an eigenvalue for $T^*$.

But since we are dealing with $\sigma(T)$ over a not-necessarily finite dimensional Hilbert space, I don't believe this proof really applies. Any help on this result would be greatly appreciated!

答:
Suppose $T-\lambda I$ has a bounded inverse. Call the inverse $S$. Then $S(T-\lambda I)=(T-\lambda I)S=I$. This implies that $(T-\lambda I)^{*}S^{*}=S^{*}(T-\lambda I)^{*}=I$ or $(T^{*}-\overline {\lambda }I)^{*}S^{*}=S^{*}(T^{*}-\overline {\lambda} I)^{*}=I$  and $S^{*}$ is a bounded operator. Thus  $T^{*}-\overline {\lambda} I% R$ has a bounded inverse. Similarly you can prove then converse also. Hence $T-\lambda I$ and $T^{*}-\overline {\lambda }I$ have the same resolvent sets which implies that they have the same spectrum.

[I have used the fact that $(AB)^{*}=B^{*}A^{*}$].

Title: Spectrum of adjoint of linear operator


问:
To construct a set with a limit point.
I learned how to construct a Cantor Set, and I am asked to do the following.

"Construct a bounded set with exactly 3 limit points."

Since the Cantor set contains infinitely many points, I don't think something like it will not work.
But this is the only thing that I have learned from the book that tells me anything about constructing a set that has a limit point.

I am also considering the interval $[0,1]$ and constructing a set so that the limit points are $\{0,1/2,1\}$.

If possible, I would like to see more than one simple examples because I am new to analysis and I have no teacher. It's very tough.

答:
Take a sequence $\{ a_n \}_{n \in \mathbb{N}}$ converging to $0$, another sequence
$\{ b_n \}_{n \in \mathbb{N}}$ converging to $\frac{1}{2}$, and another sequence
$\{ c_n \}_{n \in \mathbb{N}}$ converging to $1$. Then just take the set
$$
   S
:= \{ a_n,b_n,c_n; n \in \mathbb{N} \}
$$
to be their union. By definition $0,\frac{1}{2}$ and $1$ will be limit points. For example, you can let
$a_n := \frac{1}{n}$, $b_n := \frac{1}{2} - \frac{1}{2^n}$, $c_n := 1 - \frac{1}{n!}$. This choice also conforms to your wish that $S \subseteq [0,1]$.</p>

Title: To construct a set with a limit point.


Title: Limits and double integral?

Body:
Determine the double integral $\int \int_{R} ye^y dxdy$, where $R$ is $1\leq x-y \leq 2,\ 1 \leq xy \leq 2$

I'm having trouble to grasp how i should think to get the limits.

$x\ne y$ becaue $1\leq xy \leq 2$.

$x$ must be atleast 1 greater then $y$ because $\ 1 \leq x-y \leq 2$.

This give me $\begin{matrix}
x &y   \\
2 & 1  \\
\end{matrix}$

Can i thefore say that the limits are $0 \leq y \leq 1, 0\leq x\leq 2y$ ?

Body:
You have to integrate over two regions. Intersection points are

$(\frac{\sqrt5 + 1} {2}, \frac{\sqrt5 - 1} {2}), (2, 1), (\sqrt3 + 1, \sqrt3 - 1), (\sqrt 2 +1, \sqrt2 - 1)$

$(\frac{-\sqrt5 + 1} {2}, \frac{-\sqrt5 - 1} {2}), (-1, -2), (-\sqrt3 + 1, -\sqrt3 - 1), (-\sqrt 2 +1, -\sqrt2 - 1)$


问:
Lambert's related proof of the irrationality of $\pi$
It appears in the book of number theory, and I don't understand the proof actually. There is a corollary (from a theorem) that states that

Corollary If $x\in\mathbb{Q}, x\not = \dfrac{\pi}{2}+k\pi,\exists k\in\mathbb Z$. Then $\tan x\not\in\mathbb{Q}$.

The proof that $\pi$ is irrational (in the book) is that they assumed $\pi\in\mathbb Q$ and so is $\dfrac{\pi}{4}$ but $\tan\left(\dfrac{\pi}{4}\right)=1\in\mathbb{Q}$, which contradicts that $\tan (\pi/4)\not\in\mathbb Q$ from the corollary. I don't understand why we didn't use $\pi$ itself instead of $\pi/4$ as it satisfies the condition and yields the contradiction as well. I wonder that I miss some points in the case, so that the book has to use $\pi/4$?

答:
For Lambert's proof see for example here, where he used $\tan(\frac{\pi}{4})=1$. As noted in this text, we could also have used $\tan(\pi)=0$ for the proof, but Lambert didn't. For a detailed account of this proof see also this post:

Lambert's Original Proof that $\pi$ is irrational.

Remark: It is funny, how many "difficulties" arise with this proof by Lambert, see for example

Proving $\pi$ irrational: help with Lambert's proof. "Circularity"?


问:
Determining whether a metric space is Compact or not
If $X$ is a compact metric space and $f$ is a continuous function from $X$ to real numbers, then $f$ is uniformly continous.

My question is is there any metric space such that every continous function is uniformly continous? If there exist such metric space can we conclude that $X$ is compact?

答:
As already observed there are non-compact spaces like $\mathbb N$ where every continuous function is uniformly continuous. I will answer the question raised in the comments about the case of connected metric spaces.

Suppose every real continuous function on a connected
metric space $X$ is uniformly continuous. We claim that $X$ is compact. In fact connectedness can be replaced by the weaker condition that $X$ has no isolated points.

Suppose $X$ has a sequence $\{x_{n}\}$ with no convergent subsequence.
Choose positive numbers $\delta _{n},n=1,2,...$ such that $B(x_{n},\delta
_{n})$ does not contain any $x_{m}$ with $m\neq n$ and $\delta _{n}<\frac{1}{%
n}$ $\forall n.$ Since $x_{n}$ is not an isolated point for any $n$  we can find distinct points $x_{n,m},m=1,2,...$
in $B(x_{n},\delta _{n})\backslash \{x_{n}\}$ converging to $x_{n}$ as $%
m\rightarrow \infty $ (for $n=1,2,...$). We may suppose $d(x_{n,m},x_{n})<%
\frac{1}{n}$ $\forall n,m.$ Let $A=\{x_{n,n}:n\geq N\}$ and $B=\{x_{n}:n\geq
1\}$. Claim: $A$ is closed. Suppose $x_{n_{j},n_{j}}\rightarrow y$. Then $%
d(y,x_{n_{j}})\leq
d(y,x_{n_{j},n_{j}})+d(x_{n_{j},n_{j}},x_{n_{j}})<d(y,x_{n_{j},2m_{j}})+%
\delta _{n_{j}}\rightarrow 0$. Thus $x_{n_{j}}\rightarrow y.$ Since $%
\{x_{n}\}$ has no convergent subsequence it follows that $\{n_{j}\}$ is
eventually constant and hence $y=\underset{j}{\lim }$ $x_{n_{j},n_{j}}$
belongs to $A.$ Clearly $B$ is also closed. Further, $A\cap B=\emptyset $.
Let $f:X\rightarrow \lbrack 0,1]$ be a continuous function such that $%
f(A)=\{0\}$ and$\ f(B)=\{1\}$. By hypothesis $f$ is uniformly continuous.
Note that $d(A,B)\leq d(x_{n},x_{n,n})<\frac{1}{n}$ $\forall n.$ Thus $%
d(A,B)=0$ and, by uniform continuity $d(f(A),f(B))=0.$ This contradicts the fact
that $f(A)=\{0\}$ and$\ f(B)=\{1\}$. This contradiction shows that $X$ is
compact.


问:
How to find basis for an image of a subset under a transformation?
I was watching the khan acadamey videos on linear algebra.

Sal shows that the transformation $T$ of a subspace $S$ is another subspace $T(S)$. Since $S$ and $T(S)$ can have different dimensions, how do I find the basis for $T(S)$ if know the matrix $\mathbf A$ such that $T(\mathbf x) = \mathbf A\mathbf x$ and I know the basis for $S$?

For example if $S$ is a line in $\mathbb R^2$, and $T:\mathbb R^2 \rightarrow \mathbb R^3$; does the transformation of the basis of $S$ form a basis of $T(S)$? Is it possible $T(S)$ be a plane in $\mathbb R^3$ or must it be a line or a point?

答:
Denote $X_S$ to be a basis of $S$. You aren't guaranteed $T(X_S)$ to be a basis of $T(S)$, just by a counting argument: Suppose $T: U \to V$ is a linear transformation, and dim $U = 3$ but dim $V = 2$, then $T(X_U)$ can have at most 2 linearly dependent elements. If dim $U = 2$, dim $V = 3$, there is no way that $T(X_U)$ can span $V$. </p>

<p>Now, since $T$ is always (just from the definition) onto its image $T(S)$, we will know that $T(X_S)$ is a basis if $T$ is injective i.e. preserves linear independence of vectors. This happens when $A$ has a trivial null space, if you follow the definitions through. </p>

Body:
Denote $X_S$ to be a basis of $S$. You aren't guaranteed $T(X_S)$ to be a basis of $T(S)$, just by a counting argument: Suppose $T: U \to V$ is a linear transformation, and dim $U = 3$ but dim $V = 2$, then $T(X_U)$ can have at most 2 linearly dependent elements. If dim $U = 2$, dim $V = 3$, there is no way that $T(X_U)$ can span $V$.

Now, since $T$ is always (just from the definition) onto its image $T(S)$, we will know that $T(X_S)$ is a basis if $T$ is injective i.e. preserves linear independence of vectors. This happens when $A$ has a trivial null space, if you follow the definitions through.


问:
Finding fourth corner of parallelogram

If three comers of a parallelogram are (1, 1), (4,2), and (1,3), what
are all three of the possible fourth comers?

I am solving this question from linear algebra textbook. In my opinion there will be only one possible fourth point. But answer is three possible points. How is this possible?

答:
Let $A(1,1),B(4,2),C(1,3)$. There are thee possibilities for point D to make a parallelogram with A,B,C.

1-ADCB (D opposite AC): Let  D(h,k), then slope equations are $m_{CD}=m_{AB}$ and $m_{BC}=m_{AD}$, we get $3k-h=6, h+3k=4 \implies k=2,h=-2 \implies  D(-2,2).$

2-ABDC (D opposite BC): $m_{DC}=m_{AB}, m_{DB}=m_{AD}$, we get $h=4, 3k-h=3 \implies D(4,4)$

3- ADBC (D opposite AB): $m_{AD}=m_{BC}, m_{AC}=m_{DB}$ we get $h+3k=4, h=4 \implies D(4,0)$.

Title: Finding fourth corner of parallelogram


问:
Why is the condition $\|Z\| < 1$ equivalent to $I - ZZ^{\top} > 0$?
As the title says, for a matrix $Z \in \mathbb{R}^{p \times q}$, the condition $\begin{Vmatrix}Z\end{Vmatrix} < 1$ equivalent to $I - ZZ^{\top} > 0$. How can I show the equivalence?

Attempt:

$\begin{Vmatrix}Z\end{Vmatrix} = \sup_{|x| = 1} \begin{Vmatrix}Zx\end{Vmatrix}$
$\implies$ $\begin{Vmatrix}Zx\end{Vmatrix} < 1$ $\implies$ $(Zx)^{\top}(Zx) < 1$ $\implies$ $x^{\top}Z^{\top}Zx < 1$.

Multiplying by the identity matrix on both sides,

$\implies$ $(x^{\top}Z^{\top}Zx)I  0 $.

答:
You are almost there.
\begin{align}
I-ZZ^T\succ0
&\Leftrightarrow x^T(I-ZZ^T)x>0 \text{ for every unit vector } x\\
&\Leftrightarrow 1-\|Zx\|^2>0 \text{ for every unit vector } x\\
&\Leftrightarrow \|Zx\|<1 \text{ for every unit vector } x\tag{1}\\
&\Leftrightarrow \|Z\|<1\tag{2}.
\end{align}
In $(1)\Rightarrow(2)$, we have used the fact that the value of $f:x\mapsto\|Zx\|$ attains maximum on the unit sphere because the unit sphere is compact and $f$ is continuous.


问:
Approximating Definite Integrals as Sums
I have a function $\Omega(x,y)$ that I want to integrate over $[x_1,x_2]$ and $[y_1,y_2]$.
$$I = \frac{1}{2\pi}\int_{x_1}^{x_2}\int_{y_1}^{y_2} \Omega(x,y)dydx$$
here $\Omega(x,y)$ is actually Berry curvature (I think its whole expression is not relevant here, if it is, I can edit). The limits are $x: [-2\pi/3,4\pi/3]$, $y: [-2\pi/\sqrt{3},2\pi/\sqrt{3}]$. This integration involves matrices and eigenvalues/vectors of a matrix. I want to evaluate it using the numerical sum approximation. The research paper that I am following claims that the value of the integration (using Gaussian meshes with 2500 points) is $-1$.

I am using the approximation as
$$I = \sum_{x=x_1}^{x_2}\sum_{y=y_1}^{y_2} \frac{\Omega(x,y)}{2\pi}\Delta x\Delta y$$
with $\Delta x = (x_2-x_1)/N$, $\Delta y = (y_2-y_1)/N$ and $N$ is the number of points taken for approximation. I put all this in MATLAB, I get answer $-0.6778$ with $N = 500$. I increased $N=5000$ but still the answer is $-0.6779$.

I wonder what exactly am I doing wrong here? How can I increase the accuracy? Is there any limitation on the approximation that I have used above? Is there any other approximation for numerical integrations?

Note: Using the same code and settings I get the correct result ($-1$) if I use MATLAB's integral2() function. which means that there is no bug in my code for $\Omega(x,y)$.

答:
You can use a better numerical quadrature,like the trapezoidal rule. In general, you would be considering approximations of the form $\int_a^b f(x) dx \approx \sum_{i=1}^N \omega_i f(x_i)$. Assuming that you are using similar integration points on both $x,y$ directions,

$$
\frac{1}{2 \pi} \int_{x_1}^{x_2} \int_{y_1}^{y_2} \Omega(x,y) dy dx\approx \frac{1}{2 \pi} \sum_{i,j = 1}^N \omega_i \omega_j \Omega(x_i,y_j)
$$

For the trapezoidal rule, you have that $\omega_1=\omega_N = \frac{h}{2}$ and $\omega_i = h, i=2, \cdots, N-1$. (here $h = \Delta x = \Delta y$)


问:
Finding the area between two curves and the x-axis
The two curves $y = 3\sqrt{4-x}$ and $y = 3\sqrt{x-2}$ are given. The task is to find the area enclosed by the curves and the x-axis.

As far as I see, the root of the first function is $x = 4$ and the root of the second function is $x = 2$. Their intersection is at $x = 3$.

My question is now the following: How do I have to chose the intervals of integration such that I get the area mentioned?

Simply integrating $\int_{2}^4 \left( 3\sqrt{4-x} - 3\sqrt{x-2} \right) dx$ seems to be wrong, isn't it?

Thanks.

答: As you can see here, you must integrate $$\int_2^33\sqrt{x-2}\;\mathrm{d}x + \int_3^4\sqrt{4-x}\;\mathrm{d}x$$</p>


问:
Conway's surreal numbers and the Collatz iteration as a game?
Let us define a game based on the Collatz function $C(n) = n/2$ if $n$ is even, otherwise $=3n+1$.

Each number $n$ represents a game played by left $L$ and right $R$:

$$n = \{L_n | R_n \}$$

The rules of the game are:

If $n$ is even, the left player can play the game:

$$n = \{n/2 | \emptyset \}$$

If $n \neq 1$ is odd, the right player can play the game:

$$n = \{\emptyset| 3n+1 \}$$

If $n=1$, the game is over as no player can make a move:

$$n = \{\emptyset| \emptyset \}$$

Assuming the Collatz conjecture, this game starting with an arbitrary number $n$, has always an end.

So these games define surreal numbers and we could add and multiply, negate those games, which I think would be fun.

Q: How do I find the numeric value of these surreal numbers? Is there an algorithm to do this?

Example:

```<code>x7  =  (frozenset(), frozenset({x22}))
x22  =  (frozenset({x11}), frozenset())
x11  =  (frozenset(), frozenset({x34}))
x34  =  (frozenset({x17}), frozenset())
x17  =  (frozenset(), frozenset({x52}))
x52  =  (frozenset({x26}), frozenset())
x26  =  (frozenset({x13}), frozenset())
x13  =  (frozenset(), frozenset({x40}))
x40  =  (frozenset({x20}), frozenset())
x20  =  (frozenset({x10}), frozenset())
x10  =  (frozenset({x5}), frozenset())
x5  =  (frozenset(), frozenset({x16}))
x16  =  (frozenset({x8}), frozenset())
x8  =  (frozenset({x4}), frozenset())
x4  =  (frozenset({x2}), frozenset())
x2  =  (frozenset({x1}), frozenset())
x1  =  (frozenset(), frozenset())
</code>```

sorted by value:

```<code>[1, 5, 13, 17, 11, 7, 9, 2, 10, 26, 34, 22, 14, 4, 20, 52, 28, 8, 40, 16]
</code>```

Example Sage-Math-Script.

答:
Notation Issue

In Combinatorial Game Theory, integers like $4$ each denote a particular game or game value, like $\{3\mid\,\}$ (note that we don't write $\varnothing$ in this notation).

As nombre pointed out in the comments, the written equations like "$n=\{\frac{n}{2}\mid\varnothing\}$" are rarely/never true under standard notation for combinatorial games.

If you don't intend to be referencing the standard meanings of $n$, $3n+1$, and $\frac{n}{2}$ in this notation, you should either have a gigantic disclaimer that the usual notation does not apply, or just use something else for the games you would like to describe.

I'll use $g(n)$ where you have $n$, etc. So we have $g(1)=\{\,\mid\,\}$, $g(n)=\{g(n/2)\mid\,\}$ for even $n$, and $g(n)=\{\,\mid g(3n+1)\}$ for odd $n>1$. Technically, this is only a valid definition for all $n$ if the Collatz conjecture is true.

Example Numeric Values

Let's start building a table and see if we see any patterns. $g(1)=\{\,\mid\,\}=0$. $g(2)=\{g(1)\mid\,\}=\{0\mid\,\}=1$. $$\begin{align}g(3)&=\{\,\mid g(10)\}\\&=\{\,\mid \{g(5)\mid\,\}\}\\&=\{\,\mid \{\{\,\mid g(16)\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{g(8)\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{g(4)\mid\,\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\{g(2)\mid\,\}\mid\,\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\{1\mid\,\}\mid\,\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{2\mid\,\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{3\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid 4\}\mid\,\}\}\\&=\{\,\mid \{0\mid\,\}\}\\&=\{\,\mid 1\}\\&=0\end{align}$$ $g(4)=2$, $g(5)=0$, $g(6)=\{g(3)\mid\,\}=\{0\mid\,\}=1$.
$$\begin{align}g(7)&=\{\,\mid g(22)\}\\&=\{\,\mid \{g(11)\mid\,\}\}\\&=\{\,\mid \{\{\,\mid g(34)\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{g(17)\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid g(52)\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{g(26)\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{g(13)\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{\{\,\mid g(40)\}\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{\{\,\mid \{g(20)\mid\,\}\}\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{\{\,\mid \{\{g(10)\mid\,\}\mid\,\}\}\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{\{\,\mid \{\{1\mid\,\}\mid\,\}\}\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{\{\,\mid \{2\mid\,\}\}\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{\{\,\mid 3\}\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{0\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{1\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid 2\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{0\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid 1\}\mid\,\}\}\\&=\{\,\mid \{0\mid\,\}\}\\&=\{\,\mid 1\}\\&=0\end{align}$$

Claim

In general, the numeric value of $g(n)$ appears to be the highest exponent $m$ such that $2^m$ divides $n$.

Proof

Let's assume, for induction, that the claim is true of all the values of $g$ that arise during the calculation of $g(n)$. Note that $g(1)=0$. If $n$ is even, then $g(n)=\{g(n/2)\mid\,\}=g(n/2)+1$, which matches the highest power of $2$ for $n$. If $n$ is odd and greater than $1$, then $g(n)=\{\,\mid g(3n+1)\}$. Since $3n+1$ is even, $g(3n+1)\ge1$, so that $g(n)=0$, as desired.


问:
$\frac{p_{n+1}}{p_n}\leq\frac{3}{2}$ for $n\geq 5$
Consider $p_k$ the $k^{th}$ prime number.

It is well known that:

$\forall \epsilon>0$, there exists $n_0\in\mathbb{N}$ such that $\forall n\in\mathbb{N}$, $n\geq n_0$ we have $\frac{p_{n+1}}{p_n}<\epsilon+1$

For example, case $\epsilon\geq 1$ is Bertrand's postulate and in this case $n_0=1$.

I am looking for an elementary (or as elementary as it can get) proof for the following:

$\frac{p_{n+1}}{p_n}<\frac{3}{2}$ for $n\geq 5$ $\big($i.e. $\epsilon=\frac{1}{2}$ and we want to prove $n_0=5\big)$

Thank you.

答:
M. El Bachraoui, Primes in the interval $[2n, 3n]$, Int. J. Contemp. Math. Sci., Vol. 1, 2006, no. 13, 617 - 621, proves

Theorem 1.3. For any positive integer $n > 1$ there is a prime number between $2n$ and $3n$.

The proof is too long to give here, but it looks to be similar to, and about as elementary as, the Erdos proof of Bertrand's Postulate.


问:
Prove that for any integer $n > 3$ there exist positive integers $a_1 > \dots > a_n$ such that $1 = \frac{1}{a_1} + \dots + \frac{1}{a_n}$
I have tried to approach this problem in a variety of ways, like triyng to construct a solution, looking at telescoping series, geometrically... But I have been unable to solve it. I would appreciate any hint. Thanks.</p>

答:
Since OP has solved the question, here is another solution to the question (besides using Sylvester's sequence). Observe for $n\ge3$:

$$1 - \frac1{2^{n-2}}=\frac12+\frac14+\dots+\frac1{2^{n-2}}$$

and

$$\frac1{2^{n-2}} = \frac 1{3 \times 2^{n-1}}+\frac1{3\times 2^{n-2}}$$

and thus

$$1 = \frac12+\frac14+\dots+\frac1{2^{n-2}}+\frac 1{3 \times 2^{n-1}}+\frac1{3\times 2^{n-2}}$$

which consists of exactly $n$ terms.