问:
How to deal with inconsistencies after doing manipulation?
Given a problem as follows.

Find the range of
$$
f(x)=\frac{4x}{\sqrt{(25+x^2)(1+x^2)}}
$$
for any real $x$.

My Attempt

\begin{gather}
y=\frac{4x}{\sqrt{(25+x^2)(1+x^2)}}\\
\sqrt{(25+x^2)(1+x^2)}=\frac{4x}{y}\\
x^4+26x^2+25 = \frac{16}{y^2}x^2\\
x^4+(26-\frac{16}{y^2})x^2+25=0
\end{gather}

As $y$ is defined for any real $x$ then $x^2$ must be non-negative real numbers.

The constraints are
\begin{gather}
D\geq 0 \quad\text{and}\quad -\frac{b}{a} \geq 0 \quad\text{and}\quad \frac{c}{a} \geq 0
\end{gather}

As you can see, the third constraint has been satisfied so only the first two constraints will be considered as follows.

First constraint:
\begin{gather}
(26-\frac{16}{y^2})^2 - 4\times 1\times 25 \geq 0\\
\frac{64(y-1)(y+1)(3y-2)(3y+2)}{y^4}\geq 0
\end{gather}

The solution is $(-\infty,-1]\cup[-2/3,0)\cup(0,2/3]\cup[1,\infty)$.

Second constraint:
\begin{gather}
-\frac{26-\frac{16}{y^2}}{1} \geq 0\\
26-\frac{16}{y^2} \leq 0 \\
\frac{(y\sqrt{26}-4)(y\sqrt{26}+4)}{y^2} \leq 0
\end{gather}

The solution is $[-4/\sqrt{26},0)\cup(0,4/\sqrt{26}]$.

By intersecting the above solutions, the range of $f(x)$ is $[-2/3,0)\cup(0,2/3]$.

Question

If we consider the $f(x)$, it will be zero if $x=0$. However the previously calculated range says it never reach zero.  How to avoid this contradictory result?

答:
When $x = 0$, we have $f(x) = 0$ from the definition of $x$, which you would write as $y = 0$.

In the second line of your "attempt", you divide by $y$, but fail to note that this is only valid (as are all subsequent steps) if $y \ne 0$ (hence $x \ne 0$).

So the answer is: keep track of your domains, especially when dividing by anything or when squaring (or applying other non-injective functions) both sides of an equality.


问:
Almost sure convergence in parameters preserves convergence in distribution
Let $X_n$, $n\in\mathbb{N}$ denote a sequence of real-valued random variables that converges in distribution to the standard normal distribution. In addition, each $X_n(c)$ is a function of a  real-valued parameter $c$. I.e. $X_n(c) \stackrel{d}{\to}\mathcal{N}(0,1)$ as $n\to\infty$.

Now, assume $c$ is unkown but the sequence $c_n$, $n\in\mathbb{N}$ converges almost sure to $c$.

What are sufficient conditions for the functions $X_n(c)$, such that $X_n(c_n) \stackrel{d}{\to}\mathcal{N}(0,1)$ as $n\to\infty$? (As stated above, at the limit point $c$ the sequence $X_n(c)$ also converges in distribution to $\mathcal{N}(0,1)$.)

This statement cannot be true in general. But, I assume it holds if $X_n$ are continuous functions of $c$. Does it?

答:
Here is another counterexample, perhaps more convincing.

Let $B_t$ be the standard Brownian motion.
Then, the following map is almost surely continuous:
$$
X_n:\mathbb{R}\rightarrow\mathbb{R},\quad c\mapsto \frac{1}{\sqrt{e^c}}B_{e^c}
$$
For all $c\in\mathbb{R}$, $X_n(x)\sim\mathcal{N}(0,1)$, by construction of the Brownian motion.
Moreover, define the following sequence:
$$
c_n:=\begin{cases}
0,&\text{if }B_1>0,\\
\ln (2),&\text{else.}
\end{cases}
$$
Trivially, this is an almost surely convergent sequence.
Note that for all $n\in\mathbb{N}$:
$$
\mathbb{P}[X_n(c_n)>0]=\mathbb{P}\left[\{B_1>0\}\cup\left\{\frac{1}{\sqrt{2}}B_{2}>0\right\}\right]
$$
$$
=\mathbb{P}[B_1>0]+\mathbb{P}\left[\left\{\frac{1}{\sqrt{2}}B_{2}>0\right\}\setminus \{B_1>0\} \right]
=\frac{1}{2}+\gamma
$$
Where $\gamma>0$.
So, $X_n(c_n)$ cannot have a normal distribution.
As $X_n(c_n)$ is identical to its limit, it does not converge to a normally distributed random variable either.

ContentLicense: CC BY-SA 4.0


问:
Why does the power method converge to an eigenvector?
I'm currently working on a way to find eigenvectors analytically. The teacher suggested that I show that

$$ x_{k+1} \gets \dfrac{Ax_k}{\|Ax_k\|}$$

where $x_{k}$ is the vector and $A$ the matrix, converges to an eigenvector.

I can't seem to think of any ideas on how to prove this. Is there any literature on this?

答:
This method is known as the power iteration, you can first check that without paying attention to $0$ division, $x_n = \frac{A^nx_0}{||A^nx_0||}$.

To see what's happening let me check this convergence when A is symmetric positive and has $d$ distinct eigenvalues and $x_0$ has a non negative component with respect to the eigenvector of largest magnitude. In this case one can introduce an orthogonal basis $(v_1,...,v_d)$ of $\mathbb{R}^d$ constituted of eigenvectors associated to $A$'s eigenvalues $0 \leq \lambda_1 < ... < \lambda_{d-1} < \lambda_d$. Then by Pythagoras's formula we have for any $x \in \mathbb{R}^d$ and $n \in \mathbb{N}$,

$$\Vert A^nx\Vert^2 = \left\Vert A^n\sum_{k=1}^d \langle v_k,x \rangle v_k\right\Vert^2 = \left\Vert\sum_{k=1}^d \langle v_k,x \rangle\lambda_k^nv_k\right\Vert^2 = \sum_{k=1}^d \langle v_k,x \rangle^2\lambda_k^{2n} $$

and thus, using that $\langle v_k,x_0 \rangle > 0$,

$$\frac{A^nx_0}{\Vert A^nx_0\Vert} = \sum_{k=1}^d \frac{\langle v_k,x_0 \rangle\lambda_k^n}{\sqrt{\sum_{k=1}^d \langle v_k,x_0 \rangle^2\lambda_k^{2n}}}v_k$$

Since $(v_1,...,v_d)$ is a basis of $\mathbb{R}^d$ we only have to check each real sequence of coordinates, if $k < d$ we have

$$\frac{\langle v_k,x_0 \rangle\lambda_k^n}{\sqrt{\sum_{k=1}^d \langle v_k,x_0 \rangle^2\lambda_k^{2n}}} = \frac{\langle v_k,x_0 \rangle(\lambda_k / \lambda_d)^n}{\sqrt{\sum_{k=1}^d \langle v_k,x_0 \rangle^2(\lambda_k / \lambda_d)^{2n}}} = \frac{o(1)}{\sqrt{\langle v_k,x_0 \rangle^2 + o(1)}} \longrightarrow 0$$

and for $k = d$,

$$\frac{\langle v_d,x_0 \rangle\lambda_d^n}{\sqrt{\sum_{k=1}^d \langle v_k,x_0 \rangle^2\lambda_k^{2n}}} = \frac{\langle v_d,x_0 \rangle}{\sqrt{\langle v_k,x_0 \rangle^2 + o(1)}} \longrightarrow 1$$

This can be adapted to the case of any number of complex eigenvalues but constants may appear.


问:
Existence of variation
Let $I[w] =\int_U L(Dw,w,x) dx$. Let $1<q<\infty$, and there exist constants $\alpha>0$,$\beta\ge0$ such that $$L(p,z,x)\ge \alpha |p|^q - \beta$$

This implies that if $I[w]$ exists,
$$I[w] \ge \alpha \|Dw\|_{L^q}^q -\beta |U|$$

Now Evans says that for $w\in W^{1,q}(U)$ that $I[w]$ is defined but possibly infinite. What is the reason for this? Is it true always that functions which are bounded below are integrable (where integral can be $\infty$)?

答:
On a set of finite measure (such as $U$ here) every measurable function that is bounded from below has a well-defined integral, which is either a real number or $+\infty$. Indeed, one can write $f=f^+-f^-$ where $f^+=\max(f,0)$ and $f^-=\max(-f,0)$. Then define $\int_U f=\int_U f^+-\int_U f^-$.  The second integral is finite, the first may be $+\infty$.

In this case, $L(Dw(x),w(x),x)$ is bounded from below by $-\beta$. One still has to check that it's measurable, which requires placing some condition on $L$ (continuity is enough, but there are more general conditions that suffice for this purpose).

ContentLicense: CC BY-SA 3.0

Title: Existence of variation


问:
Proving "A graph $G$ with chromatic number $$ has a proper coloring with a walk with $k$ vertices in $k$ different colors" by induction
I'm trying to prove "A graph $G$ with the chromatic number $$ has a proper coloring with a walk with $k$ vertices in $k$ different colors".

I tried using induction to prove this statement such as color the graph, remove edges to change it into a $(k-1)$-colorable graph, and then connect the $(k-1)$-walk with one deleted edge. From here on I'm not sure how to proceed to connect the walk, especially about how to ensure I can find such a vertex to connect.

Thanks in advance!

答:
Among all proper colorings $f:V(G)\to\{1,2,3,\dots,k\}$, choose one which maximizes the sum $\sum_{v\in V(G)}f(v)$. Then each vertex $v$ with $f(v)\lt k$ has a neighbor $w$ with $f(w)=f(v)+1$, as otherwise we could increase that sum by changing the color of $v$. Hence we can construct a path $v_1,v_2,v_3,\dots,v_k$ with $f(v_i)=i$ for each $i$.

P.S. In fact, if $\chi(G)=k$, then for any proper coloring $f:V(G)\to\{1,2,3,\dots,k\}$ there is a path $v_1,v_2,v_3,\dots,v_k$ with $f(v_i)=i$ for each $i$; see the answer to this question.


问:
Is the following inequality proof correct?
Consider the following inequality transformations:

$ |x + y|^2 = (x + y)^2 = x^2 + 2xy + y^2 \geq x^2 - 2|x| |y| + y^2 = |x|^2 - 2 |x| |y| + |y|^2 = (|x|-|y|)^2 = | |x| - |y| |^2  \Longleftrightarrow |x + y|^2 \geq | |x| - |y| |^2 \Longleftrightarrow  |x + y| \geq | |x| - |y| |$

Does this seem correct?

Edit: I made use of facts proven earlier in my course that aren't mentioned in the proof above, so my question should have rather been whether the transformations I did are correct since for being a rigorous proof I would obviously have to mention the Corollaries I made use of.

答:
It all looks correct.

It's worth noting, however, that you'll need to justify that $$xy\geq-|x||y|$$ and that $$|x+y|^2\geq \bigl||x|-|y|\bigr|^2\Longrightarrow |x+y|\geq \bigl||x|-|y|\bigr|.$$ The former is almost trivial, and I leave it to you, but the latter might take a bit of work. Personally, I'd proceed by supposing we have some $a,b\in\Bbb R$ such that $|a|^2\ge|b|^2.$ Clearly, if $a=0$ and $b=0,$ then we have $|a|\ge|b|,$ and if not, then each of the following implies the next: $$|a|^2\ge|b|^2\\|a|^2-|b|^2\ge0\\\bigl(|a|+|b|\bigr)\bigl(|a|-|b|\bigr)\ge0\\|a|-|b|\ge0\\|a|\geq|b|.$$ Do you see how to justify each step?


问:
the angle between two quaternion vectors is equal to half the angle between their corresponding 3D orientations
I have read in several sources, such as this Math Stack Exchange answer and these slides from a computer animation class (p. 44), that given two quaternions $\mathbf{q}_1, \mathbf{q}_2 \in \mathbb{R}^4$, the angle between the 4D quaternion vectors is equal to half the angle between their corresponding 3D orientations. In other words, letting $\mathbf{R}_1, \mathbf{R}_2 \in SO(3)$ denote the rotation matrices corresponding to $\mathbf{q}_1$ and $\mathbf{q}_2$ respectively, the angle between $\mathbf{q}_1$ and $\mathbf{q}_2$ is half the angle of rotation required to get from $\mathbf{R}_1$ to $\mathbf{R}_2$ (the "geodesic distance"). Note that due to the fact that the maximum angle between any 3D orientations is $\pi$, I believe this fact only holds when the angle between the 4D quaternion vectors is less than or equal to $\pi/2$.

I have done an exhaustive search and can't find any proof of why this is true. So I'm looking for proof.

my best insight/attempt:

So far, I have only been able to prove this fact for the case in which $\mathbf{q}_2$ equals the identity quaternion. Start by writing out the quaternions vectors in terms of axis-angles, where $\theta_1$ and $\mathbf{e}_1 \in \mathbb{R}^3$ denote the angle and axis of $\mathbf{q}_1$, and $\theta_2$ and $\mathbf{e}_2 \in \mathbb{R}^3$ denote the angle and axis of $\mathbf{q}_2$. We have
$$
\mathbf{q}_1 = \begin{bmatrix} \cos(\theta_1/2) \\ \sin(\theta_1/2) \mathbf{e}_1 \end{bmatrix}, ~~~
\mathbf{q}_2 = \begin{bmatrix} \cos(\theta_2/2) \\ \sin(\theta_2/2) \mathbf{e}_2 \end{bmatrix} .
$$
Taking the dot product  and using that fact that the quaternions are unit length, we can solve for the angle $\phi$ between the 4D quaternion vectors:
$$
\begin{align}
\cos \phi &= \frac{\mathbf{q}_1 \cdot \mathbf{q}_2}{\lVert \mathbf{q}_1 \rVert \lVert \mathbf{q}_2 \rVert} \\
&= \cos(\theta_1/2) \cos(\theta_2/2) + \mathbf{e}_1 \cdot \mathbf{e}_2 \sin(\theta_1/2) \sin(\theta_2/2) .
\end{align}
$$
If $\mathbf{q}_2$ is taken as the identity quaternion, we have $\theta_2 = 0$ and $\mathbf{q}_2 = [1 ~~ 0 ~~ 0 ~~ 0]^T$, and the equation above becomes
$$
\begin{align}
\cos \phi &= \cos(\theta_1/2) .
\end{align}
$$
Since we have defined $\phi$ to be non-negative, for $-\pi \leq \theta_1 \leq \pi$ we have $\phi = | \theta_1/2 |$. So this seems to offer some insight, but I'm not sure where to go from here.

答:
I assume by "angle of rotation required to get from $R_1$ to $R_2$" you mean simply the by angle which $R_2{R_1}^{-1}$ rotates around its axis. But this is the same as the angle to rotate the identity frame to $R_2{R_1}^{-1}$, because both transitions between frames are achieved by the same rotation, $R_2{R_1}^{-1}$!

And similarly, the angle between quaternions $q_1$ and $q_2$ is the same as the angle between $1$ and $q_2{q_1}^{-1}$, because multiplying by a unit quaternion is a 4D isometry. But then, the quaternion $q_2{q_1}^{-1}$ corresponds to the rotation $R_2{R_1}^{-1}$... so if you say

I have only been able to prove this fact for the case in which $q_2$ equals the identity

then you're already done.


问:
A theorem on book Functional Analysis by Brézis
Theorem 2.10, on page 37

Let $E$ be a Banach space. Assume that G and L are two closed linear sub spaces such that $G+L$ is closed. Then there exists a constant $C \geqslant 0$ such that every $z \in G+L$ admits a decomposition of the form $z=x+y$ with $x \in G, y \in L, \vert\vert x \vert\vert \leqslant C \vert\vert z \vert\vert$ and $\vert\vert y \vert\vert \leqslant C \vert\vert z \vert\vert$.

The proof uses the open mapping theorem, consider the map:
$$T:G \times L \to G+L$$
The norm of $G \times L$ defined as $\vert\vert [x,y] \vert\vert = \vert\vert x \vert\vert + \vert\vert y \vert\vert $.
The theorem guarantees that the image of $\{[x,y] : \vert\vert x \vert\vert + \vert\vert y \vert\vert\}$ conclude $\{z : \vert\vert z \vert\vert < c\}$. By homogeneity,
$$z = x+y,\qquad \vert\vert x \vert\vert + \vert\vert y \vert\vert \leqslant 1/c \cdot \vert\vert z \vert\vert$$

I would like to ask the final step. Why $\vert\vert x \vert\vert + \vert\vert y \vert\vert \leqslant 1/c \cdot \vert\vert z \vert\vert$ ? I think if $z$ is quite small, it cannot prove this result.

答:
Open Mapping Theorem tells you that there exists $\delta >0$ such that for any $z \in G+L$ with $\|z\| <\delta $ we ca n write $z=x+y$ with $\|x\|+\|y\|<1$.
Now take any $z \neq 0$ in $G+L$. Let $z'=\frac  {\delta z} {2 \|z\|}$. Then $\|z'\|<\delta$. So $z'=x+y$ with $x \in G, y \in L$ and $\|x\|+\|y\|<1$. Now let $x'=\frac {2\|z||} {\delta}x$ and $y'=\frac {2\|z||} {\delta}y$. Then $z=x'+y'$ and $\|x'\|+\|y'\| \leq \frac 1 c \|z\|$ where $c=\frac {\delta} 2$.


问:
$pW_1(t) + \sqrt{1-p^2}W_2(t)$ Wiener process
Let's define two independent wiener processes: $W_1, W_2$ and take $p \in (0,1)$.

I want to prove that $W(t) = pW_1(t)+\sqrt{1-p^2}W_2(t)$

I want to check four statements:

(i) $W(0) = 0$

(ii) W has independent increasments

(iii) W has Gaussian incresments ($W_t - W_s \sim \mathbb{N}(0, t-s))$

(iv) W has continuous paths: $W_{t}$ is continuous in $t$

My work so far

(i) W(0) = $0 \cdot p + \sqrt{1-p^2}\cdot 0  = 0$ $\\$

(iii) Let's fix $t \ge s$

$W_t(t) - W_s(t) = pW_1(t) + \sqrt{1-p^2}W_2(t) - pW_1(s) - \sqrt{1 - p^2}W_2(s) = $

$=p(W_1(t) - W_1(s)) + \sqrt{1-p^2}(W_2(t) - W_2(s))$

First expression has distrubution $\mathbb{N}(0, p^2(t-s))$ and second one $\mathbb{N(0, \sqrt{1-p^2}(t-s))}$ . So the sum of those expressions is distributed $\mathbb{N}(0, t-s)$.

(iv) $W_t$ is continous because it's a sum of continous functions

I have two questions: Is my justification correct ? And also - ow can I prove second condition i.e. how to prove that $W$ hasindependent increasments ?

答:
Fix $0 \leq t_0 < t_1<t_2<...<t_n$, and consider the increments $W(t_i) - W(t_{i-1})$ for $i=1,2,...,n$. We need to show that these are independent.

Note that $W(t_i) - W(t_{i-1}) = p(W_1(t_i) - W_1(t_{i-1})) + \sqrt{1-p^2}(W_2(t_i) - W_2(t_{i-1}))$.

We use a theorem called the disjoint blocks theorem for independent random variables.

Let $X_1,...,X_n$ be independent random variables. Let $S_1 \cup S_2 \cup ... \cup S_m = \{1,2,...,n\}$ be a disjoint decomposition of $1,2,...,n$. Let $f_i: \mathbb R^{|S_i|} \to \mathbb R$ be measurable functions. Then $f_i((X_i : i \in S))$ are independent random variables.

For example, if $X_1,X_2,X_3,X_4$ are independent, then using the disjoint blocks theorem we get that $e^{X_4}+\cos X_2$ and $X_3X_1^2 + \sin X_3$ are independent (see what the $S_i$ and $f_i$ are in this case).

Now, we have the random variables $W_j(t_i) - W_j(t_{i-1})$ for $i=1,2,...,n$ and $j=1,2$. By independence of increments for $W_1$ and $W_2$, and by independence of $W_1$ from $W_2$, these random variables are independent.

Now form the blocks $S_i = \{W_1(t_i) - W_1(t_{i-1}), W_2(t_i) - W_{2}(t_{i-1})\}$ which are a disjoint decomposition of those random variables. Then the functions $f_i$ are the same i.e. $f_i(x,y) = px + \sqrt{1-p^2}y$. Applying the disjoint blocks theorem tells you instantly that the increments of $W(t)$ are independent.

Hence, we are done.

ADDED FOR FURTHER READING

Note that $p^2 + (\sqrt{1-p^2})^2 = 1$ and both $p,\sqrt{1-p^2}$ are non-negative. $W$ can be seen as the first component of an orthogonal transformation (a rotation) of the two-dimensional Brownian motion $(W_1(t),W_2(t))$.

That is, $(p,\sqrt{1-p^2})$ is the first row of an orthogonal matrix $O$, such that $W(t) = [O(W_1(t),W_2(t))]_1$ where $[]_1$ means the first column.

It is well known that combining independent Brownian motions as components provides a Brownian motion of higher dimensions.

Orthogonal transformations preserve the standard Brownian motion

and that components of Brownian motion are lower-dimension Brownian motions.

Combining these facts, $W$ is a Brownian motion.


问:
Gaussian estimate
I am trying to find some $a\gt 0$ such that $$\int_{-2}^{2}\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}dt\leq e^{-a}$$ Any good method to do so?
I'm looking for an analytical argument rather than a numerical solution.
Edit: there was a typo in the integral.

答:
with standard procedure you can see that

$$\int_{-\infty}^2 \phi(x)dx\approx 0.97725$$

thus you can find an approximated value for $a$, that is

$$a\approx0.023$$

and it is valid with $=$ instead of $\leq$

Title: Gaussian estimate


问:
Is $f,g$ increasing on $[0,\infty)$
$f(x)=\sin x-x+\frac{x^3}{3!}$, $g(x)= \cos x-1+\frac{x^2}{2!}$ which of the following are correct statements?

$1$. $f(x)\ge 0\forall x>0$

$2$. $g$ is an increasing function on $[0,\infty)$

$3.$  $g$ is an decreasing function on $[0,\infty)$

$4$. $f$ is an decreasing function on $[0,\infty)$

$1$ True as $f'(x)=g(x)>0\forall x>0$ as $x\in(\frac{n\pi}{2},n\pi),n\in\mathbb{N}$   we have $\cos x\in (0,-1)$.

$2$ is true as By Taylor's Expansion of $\cos x$ near $0$ we get $\cos x=1-\frac{x^2}{2}+O(x^3)$ so clearly $\cos x\ge1-\frac{x^2}{2}$

$3$ is False as $2$ Is true.

$4$. is false as $1$ is true.

Are my logics are correct?

答: Since $g''(x)=1-\cos x\ge 0$, the function $g'$ is increasing. But $g'(0)=0$. So, $g'$ is nonnegative on $[0,\infty)$, and $g$ is increasing there. Finally, use $f'=g$ to answer the questions about $f$.   </p>

Title: Is $f,g$ increasing on $[0,\infty)$


问:
Probability and Counting Question: 8 People Choosing 3 Rooms under Conditions
If there are eight people and three rooms, and each person chooses a room with equal probability (1/3), what is the probability that there is at least 1 person in each room?
I have tried solving the problem by using
P(at least 1) = 1 - P(none)
But in this case, there are three cases:

One room is empty

Two rooms are empty

Thus I decided that the solution wasn't neat.
If possible, is there a neat solution that computes the probability of having at least 1 person per room? It would be great if I have steps and a solution.
Thank you!

答:
I'll try to give a structural way to think about the situation. Consider first the trinomial expansion:
$$
P(x,y,z):=
\left(\frac 13(x+y+z)\right)^8
=\left(\frac {x+y+z}3\right)^8
=\frac 1{3^8}
\sum_{\substack{0\le i,j,k\le 8\\i+j+k=8}}
\binom 8{i\ j\ k}x^iy^jz^k\ .
$$
Here the trinomial coefficients involved are given by the formula:
$$
\binom 8{i\ j\ k}=\frac{8!}{i!\; j!\;k!}\ .
$$
The combinatorial relation to the given people and rooms is as follows.
We let the first person enter the corridor with the three rooms $x,y,z$ in it. Her choice of the room determines the term with the same label from the first $(x+y+z)$ factor in the product $P(x,y,z)$. Now the second person enters the corridor, her choice of a room corresponds to a choice of a term in the second factor $(x+y+z)$. And so on. Our normalization factor $\frac 13$ makes $P(1,1,1)=1$, corresponding to a total probability one. So $\frac 1{3^8}
\binom 8{i\ j\ k}$
is the probability that finally there are $i$ persons in room $x$, $j$ persons in room $y$ and $k$ persons in room $z$.

Now let us write explicitly:

The probability that the room $x$ remains empty is the sum of the coefficients in $P(x,y,z)$ where $x$ appears at power $0$, i.e. $P(0,1,1)$. Plugging in the zero instead of $x$ makes a zero evaluation for all $x^iy^jz^k$ with $i>0$. Similarly...

The probability that the room $y$ remains empty is $P(1,0,1)$,

The probability that the room $z$ remains empty is $P(1,1,0)$,

The probability that the rooms $y,z$ remain empty is $P(1,0,0)$,

The probability that the rooms $z,x$ remain empty is $P(0,1,0)$,

The probability that the rooms $x,y$ remain empty is $P(0,0,1)$,

The probability that all rooms $x,y,z$ remain empty is $P(0,0,0)=0$.

By the inclusion-exclusion principle, the probability to have at least one empty room is:
$$
\begin{aligned}
p&=
P(0,1,1)+P(1,0,1)+P(0,1,1)\\
&\qquad-P(1,0,0)-P(0,1,0)-P(0,0,1)\\
&\qquad\qquad+P(0,0,0)
\\
&=
3\cdot \left(\frac 23\right)^8
-
3\cdot \left(\frac 13\right)^8\ .
\\
&\qquad\text{So the wanted probability is }
\\
1-p &=
1 -3\cdot \left(\frac 23\right)^8
+
3\cdot \left(\frac 13\right)^8
=\frac{644}{729}\approx 0.883401920438957475994513031\dots\ .
\end{aligned}
$$


问:
Variety of Nilpotent Matrices
Let $k$ be an algebraically closed field and view $M_n(k)$ as $\mathbb{A}^{n^2}$.  $A\in M_n(k)$ is nilpotent if and only if $A^n=0$.  Since the equation $A^n=0$ is given by $n^2$ polynomial equations, it defines a variety in $\mathbb{A}^{n^2}$.  What is the dimension of this variety?  Is it irreducible?  If not, what are its irreducible components?</p>

答:
To prove irreducibility, note that any two regular nilpotent $n\times n$ matrices (i.e. nilpotent matrices whose minimal poly. is $X^n$) are conjugate (by putting them into Jordan normal form), and that any non-regular matrix can be written as a "limit" of regular ones (put it in Jordan form, than put an entry $t$ in any of the entries immediately above the diagonal that are zero, and let $t \to 0$).

From this we see that (a) the locus of regular nilpotent elements is irreducible (it is the image of $GL(n)$ under the conjugation action on any particular regular nilpotent), and (b) the whole nilpotent cone (another common synonym for the word nullcone in Jim's answer) is the irreducible, being the closure of the regular nilpotent elements.

A fairly simple computation shows that the centralizer of a regular nilpotent is $n$-dimensional (it is just the span of $I, A, \ldots, A^{n-1}$), and hence the image of a regular nilpotent under the conjugation action of $GL(n)$ is $n^2 - n$ dimensional.  Again considering (a) and (b) above, we see that the nilpotent cone is $n(n-1)$-dimensional.

Note also that the equation $char(A) = 0$ gives $n$ equations cutting out the nilpotent cone, so it is a complete intersection, and in particular Cohen--Macaulay (although it is not smooth, unless $n = 1$).  A little more argument builds on this to show it is regular in codimension one (i.e. its singular locus lies in codimension two), and hence that it is normal (by Serre's criterion for normality).

Title: Variety of Nilpotent Matrices


问:
Why are abelian groups of interest? What is their usefulness?
I am reading about Abelian groups

So apparently it is a set, with an associative binary operation, and identity element, an inverse operation and the binary operation must also be symmetric.

But it is not clear to me how they are useful. Trying to find why they are important it seems they arise as "additive structures" in various systems but this is too abstract for me.

Could someone give some less formal/more practical application or usages showing what is important about abelian groups (layman's terms basically)?

答:
The same question could really be asked about any abstract algebraic structure: why do we study fields or rings or groups or semigroups, etc.?  For that matter, why do we study abstract algebra at all?

As other answers have noted, algebraic structures are studied because they are useful abstractions.  For example, the reason why we study groups is because the definition of what counts as a group is:

broad enough that a lot of things in mathematics satisfy it, but

specific enough that we can prove quite a few useful theorems using only the group axioms, which means that these theorems automatically apply to everything that satisfies those axioms.

For example, the integers (and rationals and reals and even complex numbers) under addition, positive rational (and real) numbers under multiplication and invertible $n \times n$ matrices under matrix multiplication all satisfy the definition of a group.  This means that any theorems that we prove about groups in general apply to all of those systems, as different as they may seem on the surface.

And we study abelian groups because, as useful as groups are, it turns out that there are also quite a few useful theorems that we cannot prove just from the group axioms unless we also assume commutativity.  Adding the commutativity axiom thus allows us to prove more useful results — the tradeoff, of course, being that those results won't apply to groups that are not commutative, such as the group of $n \times n$ invertible matrices mentioned above.

This is a common situation in abstract algebra (and in many other areas of "abstract" mathematics): the more specific we make our abstractions, the more results we can prove, but the fewer things those results will apply to.  The trick here is to find abstractions that happen to hit a sweet spot where there are just enough axioms to prove useful results, but no unnecessary ones that we could remove and still prove essentially the same results.  Of course, in practice, there are many such "sweet spots" of varying specificity.  Experience has shown that groups are certainly one of them, while abelian groups are another.


问:
smooth vector fields over the 2-sphere is not a free module
I keep seeing that the hairy ball theorem implies that smooth vector fields over $S^2$ are not a free module is implied by the hairy ball theorem; I don't understand how. How do I fill in the gaps?

The hairy ball theorem tells us that we cannot have a smooth nonvanishing vector field over $S^2$. For contradiction, assume that $\mathfrak X(S^2)$ is a free module over the ring $C^\infty (S^2)$. Thus, $\mathfrak X(S^2) \simeq \oplus_i C^\infty(S^2)$, since a free module is isomorphic to a direct sum of copies the ring.

I am unsure how to continue. I have some ideas:

First, see that we need at least two copies of $C^\infty(S^2)$, because the sphere locally looks like $\mathbb R^2$.

If we have exactly two basis vector fields $V_1, V_2$, then these must both vanish at points $p_1$, $p_2$ by the hairy ball theorem. Consider the neighbourhood of $p_1$: since $v_1$ vanishes, we have only one vector field $V_2$ which is not sufficient to locally span $\mathbb R^2$.

If we have three vector fields, we can have $V_{1, 2, 3}$ vanish at distinct $p_{1 2, 3}$ thereby leaving two vector fields even when one of them vanishes. However, now we can pick a point where none of $V_{1, 2, 3}$ vanish. (Why does such a point exist?). At this point, locally, we have three degrees of freedom $V_{1, 2, 3}$, but the vector space looks like $\mathbb R^2$ so they cannot be linearly independent.

Unfortunately, as is obvious from the above, I have no idea how to make this rigorous. I would appreciate whether this proof is repairable, and if now, how does one actually prove that smooth vector fields over $S^2$ are not  free module?

答:
Recall the following two basic facts:

Given two vectors bundles $E,F$ over $M$, any homomorphism $f \colon \Gamma(E) \rightarrow \Gamma(F)$ of $C^{\infty}(M)$-modules is induced by a unique smooth homomorphism of vector bundles $F \colon E \rightarrow F$ (i.e a "smoothly varing" family of linear maps $F_p \colon E_p \rightarrow F_p$). This is sometimes called the "tensor characterization lemma" when the bundles involved are tensor bundles. For a proof, see for example Lemma 10.29 in Lee's Introduction to Smooth Manifolds. If $f$ is an isomorphism of $C^{\infty}(M)$-modules then $F$ is also an isomorphism of smooth vector bundles.

Given a vector bundle $E$ over $M$, the module $\Gamma(E)$ is finitely generated as a $C^{\infty}(M)$-module. For a proof, see here.

Now, assume that $\Gamma(TS^2)$ is free so that $\Gamma(TS^2) \cong \oplus_{i} C^{\infty}(S^2)$. By $(2)$, the right hand side must be a finite direct sum $\oplus_{i=1}^n C^{\infty}(S^2) \cong \Gamma(\underline{\mathbb{R}^n})$ where $\underline{\mathbb{R}^n}$ is the trivial bundle $S^2 \times \mathbb{R}^n$. Now, by $(1)$, we have an vector bundle isomorphism $F \colon \underline{\mathbb{R}^n} \rightarrow TS^2$ (which in particular implies that if such an $F$ exist, $n = 2$). Choose any nonwhere zero section of $\sigma$ of $\underline{\mathbb{R}^n}$ and consider the section $F_{*}(\sigma)$ of $TS^2$ given by $F_{*}(p) = F_p(\sigma(p))$. Since $F_{p} \colon \mathbb{R}^n \rightarrow T_p S^2$ is a linear isomorphism, $F_{*}(\sigma)$ is a nonwhere zero section of $TS^2$, i.e, a non-vanishing vector field on $S^2$, a contradiction.


问:
Integrating state space equations using Runge-Kutta 4?
I have a state-vector which is of size 12x1, an input-vector $u$ which is of size 4x1, and A and B matrices which are of sizes 12x12 and 12x4 respectively, such that $\dot{x}=Ax+Bu$. Using an initial state, my code uses forward Euler integration to integrate $\dot{x}$ over intervals of $dt$ and add it to the previous state to get the next state.

Now I'd like to make this more accurate using Runge-Kutta 4 integration. I watched some videos about RK4 for this task, so my knowledge about it might be a bit incorrect, but my understanding is that you find the step that you will add to the previous state by doing
$$
\frac{k_1 + 2k_2 + 2k_3 + k_4}{6}
$$

How do I find the various $k$ coefficients? I found some pointers online but none of them work for a case like mine. All of the cases I've found use time as an input, but I don't think I can do that here. As I said, I'm pretty new to RK4 integration, so any kind of help would be greatly appreciated!

答:
If your equation is time-independent, simply omit time from the method.

From definition: $k_1 = f(x_k, t_k) = f(x_k, u_k) = A x_k + B u(t_k)$.

$$ k_i := f\Bigr( x_k + \Delta t \sum_{j=1}^{i-1} a_{i,j} k_j , u(t_k + c_i \cdot \Delta t)\Bigl)$$

There are many RK methods with 4 stages, but I am guessing you are using the "main one" (explicit one). The sum of $k$'s you have there is already multiplied with the $b$ coefficients. The coefficients for $a$ and $c$ are already inserted below.

Then
$$k_2 = A(x_k + \Delta t \frac{1}{2} k_1 ) +  B\cdot u(t_k + \frac{1}{2}\Delta t)$$
$$k_3 = A(x_k + \Delta t \frac{1}{2} k_2 ) +  B\cdot u(t_k + \frac{1}{2}\Delta t)$$
$$k_4 = A(x_k + \Delta t  k_3 ) +  B\cdot u(t_k + \Delta t)$$

The $u$ is just part of your function, I assume that this is a specified input.

To be completely clear:

$$
x_{k+1} = x_k + \Delta t \frac{k_1 + 2k_2 + 2k_3 + k_4}{6}
$$


问:
Determinant of a sum of square matrices
Let $$A=\begin{bmatrix}0&1&0&\cdots 0\\
0&0&1&\cdots 0\\
\vdots\\
0&0&0&\cdots 1 \\
1&1&1&\cdots1
 \end{bmatrix}_{n\times n}$$

i.e. it has ones above the main diagonal except for the last row and the last row has all ones.

I am trying to find $\det(A+A^2+\cdots+A^t)$ for $t\leq n $ $\tag{1}$

I have checked that for a few $n$, $\det(A)=\det(A^2)=\cdots=\pm 1$. But I am not sure how to prove that. Also determinant of sum of matrices is not equal to sum of determinant of those matrices, so I'm not sure how to find $(1)$? Any ideas?

**EDIT:**I realize that it is hard to find exact value of $(1)$ and so an upper bound for $(1)$ is also useful to me.

答:
The characteristic polynomial of $A$ is $$\lambda^n-\lambda^{n-1}-\lambda^{n-2}-\ldots-1=0$$.  This has one root $\lambda_1\approx2$, and all the other roots are inside the unit circle.

$$det(I-A^t)=\prod (1-\lambda_i^t)\approx (\lambda_1^t)\lt2^t$$

Title: Determinant of a sum of square matrices


问:
Prove $(p\to q)\land(p\to\neg q)\iff\neg p$
How can I prove $(p\to q)\land(p\to\neg q)\iff\neg p$?

答:
$(p\to q)\land(p\to\neg q)\iff \\
(\lnot p \lor q)\land (\lnot p \lor \lnot q) \iff\\
(\lnot p)\lor(q \land \lnot q) \iff \\
(\lnot p)\lor FALSE \iff \lnot p$


问:
Artin Algebra, 10.3.7
$R$ will denote a commutative unitary ring.

(3.4) Proposition. Substitution Principle: Let $\varphi: R \to R'$ be
a ring homomoprhism. Then given elements $\alpha_1,$...$,\alpha_n \in
R'$ there is a unique homomorphism $\phi: R[x_1,...,x_n] \to R'$ from
the polynomial ring in $n$ variables to $R'$ which agrees with
$\varphi$ on constant polynomials and which sends $x_i \to \alpha_i$
for $i = 1,$...$,n$.

The author then says that the Substitution Principle is an efficient way to prove that various constructions of polynomial rings are equivalent, like $R[x,y] \cong R[x][y]$.

(3.7) Corollary. There is a unique isomorphism $R[x,y] \to R[x][y]$
which is the identity on $R$ and which sends the variables to
themselves.

Proof. Consider the inclusion map $\varphi : R \to R[x][y]$. The
Substitution Principle tells us that there is a unique homomorphism
$\phi: R[x,y] \to R[x][y]$ which extends this map and sends the
variables $x,y$ wherever we like. So we can send the variables to
themselves. (...)

This sounds like non-sense. The substitution principle as stated sends the variables to elements of a ring, not to "variables". My first thought is that maybe we could extend the codomain ring with transcendental elements to act like the variables, but then it'd have to be the image of some element of $R$ which also doesn't seem to make sense. And this section is before ring extensions anyway. Is this proof incorrect or am i missing something?

答:
There's nothing wrong with the author's proof.

The "variables" in question are just the elements $x,y$ in the respective rings.

The rings $R[x,y]$ and $R[x][y]$ both have elements named $x,y$.

When the author says

"send the variables to themselves"

the intended meaning is

"send the elements named $x,y$ in the ring $R[x,y]$ to the correspondingly named elements in the ring $R[x][y]$"

the possibility of which follows immediately from the Substitution Principle.

Finally, to show that the homomorphism is an isomorphism, simply verify that the homomorphism is surjective and has kernel $0$.

Title: Artin Algebra, 10.3.7


问:
Factorization of a unimodular matrix
Let $S = \big(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \big)$ be a matrix with determinant $ad - bc  =1$ (this is the case I'm interested in, but I don't think it is essential).
Then the matrix
$$ T = \left( \begin{matrix} a^2 & 2ac & c^2 \\ ab & ad+bc & cd \\ b^2 & 2bd & d^2  \end{matrix} \right) $$
occurs in the theory of binary quadratic forms and has determinant $\det T = (\det S)^3$.

My question: Can $T$ be written as a product of three matrices whose determinant is obviously $ad-bc$, such as, for example, the following?
$$  \left( \begin{matrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & 1  \end{matrix} \right) $$
For what it's worth, the map $S \mapsto T$ induces a homomorphism SL$_2({\mathbb Z}) \longrightarrow $ SL$_3({\mathbb Z})$. Preferably, this should be a consequence of a factorization of $T$ as the one I'd like to have.  Relevant references are also welcome.

答:
We can do the first step and write
$$
T=\begin{pmatrix} a & b & 0 \cr c & d & 0 \cr 0 & 0 & 1\end{pmatrix}\cdot
\begin{pmatrix} \frac{a(ad-b^2)}{ad-bc} & \frac{2acd-abd-b^2c}{ad-bc} & \frac{cd(c-b)}{ad-bc} \cr \frac{a^2(b-c)}{ad-bc} & \frac{a(ad+bc-2c^2)}{ad-bc} & \frac{c(ad-c^2)}{ad-bc} \cr
b^2 & 2bd & d^2 \end{pmatrix}
$$
The matrix $T$ also "arises" in the description of Lie algebra automorphisms of $\mathfrak{sl}_2(K)$ with $S\in GL_2(K)$. But I don't see how this helps.

Title: Factorization of a unimodular matrix


问:
Question about a statement involving system of linear equations and rank of matrix.
the statement is:

let $A\in \mathbb{R}^{m\times n}$,   if for every $c\in \mathbb{R}^m$ there exists a solution for $Ax=c$, then $\operatorname{rank}(A)=m$.

Now I can understand that this is a true statement, since if $\operatorname{rank}(A)<m$ we can get no solution for the equation.

What I wanted to ask if the given information was for every $c\in \mathbb{R}^m$ there exists a unique solution, could I infer that the matrix is square? ($m = n$).

This is coming from the idea that if $Ax=c\:$ has a unique solution, then $\operatorname{rank}(A) = n$, so if it has a solution for every $c$ that means $\operatorname{rank}(A) = m $ and so $m=n$.

I would appreciate any feedback and would love to know if there's more interesting stuff to infer from the given information.

答:
Yes, it is true. Note that the rank of the matrix $A$ is also the dimension of the subspace generated by the columns of the matrix, and if evety vector in $\mathbf R^m$ is attained, this dimension is $m$.

On another hand $\operatorname{rank}A\le \min(m,n)$, and  if the solution is unique; thr rank-nullity formula shows thatt the rank is the codimension of the kernel of the associated linear map, i.e.
ˆ$$\dim(\ker A)=\operatorname{rank}A=n\iff 0+m=n.$$


问:
What if we define the area of the unit square to be any other number (say 2)?
When formalizing the notion of area, among other postulates the area of a unit square can be defined to be $1$. Using this we can prove that the area of square of length $x$ has area $x^2$. Here a link for such a proof.

What if we define it to be something other than 1? Say 2, the formula for area of a square will then becomes $2x^2$. Using which we can derive the formulas for areas of other shapes.

Can something go wrong? Is the choice of $1$ as the multiplicative factor arbitrary?

Note : By Unit Square I mean a square of side length 1.

答:
Without that axiom, we cannot deduce the area of anything (with nonzero area). We can prove statements like "this polygon has $3.5$ times the area of that other polygon" or "the area of this region is the difference between the areas of those other two regions", but we are limited to relative statements, not absolute ones.

As a result, it is consistent with the other axioms for any square to have area $1$. So we pick one square, call it the unit square, and decide that it will have area $1$. (We could instead pick a square and decide it will have area $2$; we probably shouldn't call it a "unit" square, though.)

Another problem this axiom solves is that it tells us that not everything has area $0$. (That's otherwise consistent with the other axioms for area, in most formulations.)


问:
Can two sets $A,A'\subset\mathbb{R}^n$ have the same measure but different expectation?
I have three 'simple' questions regarding measure theory. For every measurable set $A\subset\mathbb{R}^n$, we define the avarage value of $A$ (denoted as $\mathbb{E}[A]$) by
\begin{gather}
\mathbb{E}[A]=\frac{1}{\mu(A)}\int_A \vec{x} \mathrm{d}\mu(\vec{x})
\end{gather}
where  $\mu$ is any measure defined on the Borel $\sigma$-algebra of $\mathbb{R}^n$.

My three questions are as follows:

Can two sets $A,A'\subset\mathbb{R}^n$ such that $A\neq A'$ have the same measure (i.e., $\mu(A)=\mu(A’)$) but different average values (i.e., $\mathbb{E}[A]\neq\mathbb{E}[A’]$)?

When will two sets $A,A'\subset\mathbb{R}^n$ such that $A\neq A'$ have the same avarage value (i.e., $\mathbb{E}[A]=\mathbb{E}[A’] $)?

Will any set $A\subset\mathbb{R}^n$ with measure $\mu(A)=\infty$ have avarage value $\mathbb{E}[A]=\infty$?

My attempt to reasoning about these questions is the following.

For question 1, consider the Real line $\mathbb{R}$ and sets $A=[1,2]$ and $A'=[3,4]$. Since the notion of measure in the Real line boils down to the concept of length, the measure $\mu$ of each set is just $\mu(A)=2-1=1$ and $\mu(A')=4-3=1$. However, the average value of each set is just $\mathbb{E}[g_A]=1.5$ and $\mathbb{E}[g_{A'}]=3.5$. Therefore, for the case in which $n=1$, it is possible to construct two sets $A,A'\subset\mathbb{R}$ with the same measure but different average values. Hence, I'm inclined to believe that the answer to question 1 is 'yes', but I do not know if this argument generalises beyond $n=1$.

For question 2, I do not have a clear argument, but I guess that two sets $A,A'\subset\mathbb{R}^n$ will have the same average value when their measure is zero (i.e., $\mu(A)=\mu(A')=0$), when their measure is infinite (i.e., $\mu(A)=\mu(A')=\infty$) or when they are not disjoint and the only sets that receive strictly positive measure belong to the intersection $A\cap A'$. Am I correct?

For question 3, I do not have an argument... My intuition just tells me that the answer is 'yes'.

As you can probably tell from my questions, I don't quite understand what does it mean for a set to have a certain mesure. Any explanation will be very welcome.

答:
Let us start presenting some "intuition". A measure $\mu$ can be thought as way to determine the sizes of sets.  The average value a set $A$ with respect to $\mu$ is the center of $A$ if we measure sizes using $\mu$.

Now for your questions:

Your argument is correct and valid in $\Bbb R^n$. Just an example in $\Bbb R^2$:
$A= [1,2]\times [0,1]$  and $A'= [3,4]\times [0,1]$.  Let us use Lebesgue measure in $\Bbb R^2$. Then $\mu(A)=\mu(A') = 1$, but for the average values, we have: $\mathbb{E}[A] =(1.5,0.5)\neq  (3.5,0.5)  =\mathbb{E}[A'] $.

The are many situation were distinct sets can have the same average value. For instance:

a. $A = [-1, 1]$  and $A'=[-100, 100]$. Let us use Lebesgue measure in $\Bbb R$. They are distinct sets with different "sizes" but the same average value, that is $0$.
Note that there are subsets of $A'\setminus A$ that have strictly positive measure.

b. $A = [-1, 1]$ and and $A'=[-100, -2] \cup [2,100]$. Let us use Lebesgue measure in $\Bbb R$. They are disjoint sets with different "sizes" but the same average value, that is $0$.

If $\mu(A) = \infty$ then, in general, the average point is not defined. The formula
\begin{gather}
\mathbb{E}[A]=\frac{1}{\mu(A)}\int_A \vec{x} \mathrm{d}\mu(\vec{x})
\end{gather}
will produced an undetermined value similar to $\frac{\infty}{\infty}$ in the case of $\Bbb R$.

Remark 1: There are some cases where $\mu(A)=\infty$ and the average value is defined. For instance, consider $\Bbb R$ and let $\mu$ be a measure defined by $\mu(E)=\infty$ if $0 \in E$ and  $\mu(E)=0$ if $0 \notin E$. Let $A =[-1,1]$. In such case, $\mu(A) = \infty$  and
$$ \int_A x \mathrm{d}\mu(x) = 0$$
So
\begin{gather}
\mathbb{E}[A]=\frac{1}{\mu(A)}\int_A x \mathrm{d}\mu(x) = 0
\end{gather}

Remark 2: Using the formula
\begin{gather}
\mathbb{E}[A]=\frac{1}{\mu(A)}\int_A \vec{x} \mathrm{d}\mu(\vec{x})
\end{gather}
It is easy to see that, if $\mu(A)=\infty$ and the average value is defined, then necessarily, the average value is $0$.

Remark 3:
There are some ways to extend the notion of average value to broader classes of sets of infinite measure, but which way is suitable depends on the context (for instance, using limits of  integral, in a certain way, we could say that, for the Lebesgue measure, the average value of $\Bbb R$ is $0$).


问:
Questions about solving equations.
Should a backward or bi-directional arrow be used to link every statement in the process of solving an equation?

It is often difficult to find what implies a statement than what the statement implies. I guess solving simple elementary equations is not that hard is because most operations are reversible.

答:
The resolution of an equation is asking for a necessary and sufficient condition, or if you prefer, the explicit enumeration of the solutions.

You can present the steps the way you like, but the conclusion will be like the example below:

$$x^2-3x+2=0\iff x=1\lor x=2.$$

Sometimes you are asked to find some solution. Then you establish a sufficient condition.

$$x=1\implies x^2-3x+2=0.$$

Questions that require a necessary condition seem less frequent. For instance

$$x^2-3x+2=0\implies x\in[1,2].$$

If in the process of resolution you introduce alien solutions, you need to eliminate them after the fact.

E.g.

$$\sqrt x=2\to(\sqrt x)^2=4\to x=4\to x=\pm 2$$

but $\sqrt{-2}$ does not exist.

Title: Questions about solving equations.


问:
Debugging an analysis text
I'm having some problems understanding the following paragraph, which I read in a analysis script (hopefully I haven't made any translation errors):

"A map $f:U \rightarrow Y$, where $U$ is open and $X,Y$ are Banach spaces, is continuous at $x' \in U$ if $$f(x')=\lim_{x\rightarrow x'} f(x)=\lim_{h\rightarrow 0} f(x'+h),$$
  where $h=x-x'$. We can decompose $h$ in a "polar" fashion in $h=ts$, where $\left\Vert h \right\Vert \geq 0$ and $s=\frac{1}{\left\Vert h \right\Vert } h$. Then
  $f(x')=\lim\limits_{h\rightarrow 0} f(x'+h)$ iff $f(x'+ts)\rightarrow f(x')$ for $t \rightarrow 0^+$ uniformly with respect to $\left\Vert s \right \Vert = 1$.
  No matter from which direction $s$ with $\left\Vert s \right\Vert=1$ we approach $x'$, the value of the function has to converge to $f(x')$ with a to all $s$ common "minimal speed"  ".

What I don't understand is this:

1) What does in means to decompose anything in a "polar" fashion ?

2) I thought only sequences of function can converge uniformly...and what does it mean, if something converges uniformly with respect to another thing ?

3) What does the author mean with "common "minimal speed""

Thanks in advance.

答:
I guess in (1) it should be $t=\|h\|$, so the analogy would be with the polar decomposition of a complex number $z=ts$ where $t=|z|$ and $s=z/|z|$.

In (2), the uniform convergence probably means that $\sup\limits_{\|s\|=1} \|f(x'+ts)-f(x')\|\to 0$ as $t\to 0_+$.

I suppose the minimal speed thing is meant to be a way of understanding the uniformity condition in (2). One way to keep convergence but violate this uniform convergence would be if you could let $h$ approach $0$ along different paths $\gamma_1,\gamma_2,\dots,$ say with $\|\gamma_i(t)\|=\|\gamma_j(t)\|$ for every $i,j$, but so that the rate of convergence of $f(x'+\gamma_i(t))$ to $f(x')$ gets slower and slower as $i$ increases. So by saying that there should be a lower bound on such rates of convergence (a "common minimal speed", I think) you'd avoid that issue.

Title: Debugging an analysis text


问:
Cardinality and Measurability
We can show that $\mathbb{R}$ and $\mathbb{R}^2$ or ($\mathbb{R}^n$) have same cardinality using the following one-to-one and onto mapping:

Say x = (0.123456789....)

Then f(x) = {(0.13579...),(0.2468..)}

My question is can we claim that the Borel sigma algebra in $\mathbb{R}$ has a correspondent sigma algebra (if it is, is it THE Borel sigma algebra in $\mathbb{R}^2$?) by mapping each Borel set D to $\mathbb{R}^2$ by f(D)? Why or why not could you give the argument??

Similarly (I am more certain of this)
When we have N^n and N using this mapping could we claim that the sigma algebra in N has a correspondent sigma algebra in N^n? Does the statement hold for N^infinity and N as well, if N^infinity and N have the same cardinality ( i am not sure about my last statement)?

Thank you!

答:
The fact that there are $1-1$ mapping between two sets says almost nothing about the measure of the two sets. Consider a silly example of the Cantor set, you can show relatively easily that it has cardinality $c$ while having zero measure. </p>

<p>The $f$ you defined from $\mathbb{R}^{1}\rightarrow \mathbb{R}^{2}$ is measurable (because it is continuous), and hence $f(D)$ does generate a $\sigma$-algebra in $\mathbb{D}^{2}$. But this $\sigma$-algebra is "twisted" in some sense. For example, the image of $(0.4,0.5)$ in the plane includes points "outside" of $(0.4,0),(0.5,0)$ like $(0.4,0.9)$ from $0.49$ or $(0.49,0.95)$ from $(0.4995)$. It is intuitively not the right choice for us to define a $\sigma$-algebra on the plane, where the open sets arise from open balls or open retangles. </p>

Title: Cardinality and Measurability

Body:
The fact that there are $1-1$ mapping between two sets says almost nothing about the measure of the two sets. Consider a silly example of the Cantor set, you can show relatively easily that it has cardinality $c$ while having zero measure.

The $f$ you defined from $\mathbb{R}^{1}\rightarrow \mathbb{R}^{2}$ is measurable (because it is continuous), and hence $f(D)$ does generate a $\sigma$-algebra in $\mathbb{D}^{2}$. But this $\sigma$-algebra is "twisted" in some sense. For example, the image of $(0.4,0.5)$ in the plane includes points "outside" of $(0.4,0),(0.5,0)$ like $(0.4,0.9)$ from $0.49$ or $(0.49,0.95)$ from $(0.4995)$. It is intuitively not the right choice for us to define a $\sigma$-algebra on the plane, where the open sets arise from open balls or open retangles.


答:
Since $\;p\to q\;$ is TRUE if and only if $p$ is FALSE or $q$ is TRUE and since $\;p\to\neg q\;$ is TRUE if and only if $p$ is FALSE or $\neg q$ is TRUE, then

$(p\to q)\land(p\to\neg q)\iff(\lnot p\lor q)\land(\lnot p\lor\lnot q)$

By applying the distributive property of OR with respect to AND, we get that

$(\lnot p\lor q)\land(\lnot p\lor\lnot q)\iff
(\lnot p)\lor(q\color{blue}{\pmb{\land}}\lnot q)$

But $\;q\color{blue}{\pmb{\land}}\lnot q\;$ is always FALSE, consequently

$(\lnot p)\lor(q\color{blue}{\pmb{\land}}\lnot q)\iff(\lnot p)\lor\color{blue}{FALSE}$

Finally,

$(\lnot p)\lor\color{blue}{FALSE}\iff \lnot p$

Title: Prove $(p\to q)\land(p\to\neg q)\iff\neg p$

Body: How can I prove $(p\to q)\land(p\to\neg q)\iff\neg p$?


问:
Is the intersection of any subset of $X$ and a $\lambda$-measurable set $\lambda$-measurable?
Suppose $A\subset X$ is $\lambda$-measurable. Is $B\cap A$ $\lambda$-measurable for any $B\subset X$? I have tried on playing with set theories but still cannot come up with a close solution. All I have to do is just show that for any $C\subset X$, $\lambda(C)\geq\lambda(C\cap(A\cap B))+\lambda(C\cap(A\cap B)^c)$. How should I proceed? Thanks!

答:
If that were so, then all subsets would necessarily be $\lambda$-measurable, since $X,$ itself, is $\lambda$-measurable. Simply take $A=X$ to see why.

Thus, it really depends on whether there are any subsets of $X$ that are not $\lambda$-measurable.


问:
$\prod_{n=1}^N (1+f_n(x))$ converges uniformly to a function on $A$ as $N\to\infty$.
I am reading "Calculus vol.2" (in Japanese) by Shizuo Miyajima.

There is the following theorem in this book:

Theorem 8.2

Let $f_1, f_2, \dots$ be functions on a set $A$.

Let $M_1, M_2, \dots$ be real numbers such that $|f_n(x)|\leq M_n$ holds for any $x\in A$ and any $n\in\{1,2,\dots\}$, and $\sum_{n=1}^\infty M_n$ converges.

Then, $\prod_{n=1}^N (1+f_n(x))$ converges uniformly to a function on $A$ as $N\to\infty$.

And the author's proof of this theorem is the following:

Since $\sum_{n=1}^\infty M_n$ converges, so $\lim_{n\to\infty} M_n = 0$.

So, there exists $n_0$ such that $M_n < \frac{1}{2}$ for $n > n_0$.

We can assume that $M_n < \frac{1}{2}$ for all $n$.

Since $\frac{d}{dx} \log(1+x) = \frac{1}{1+x}$, by the mean value theorem, $|\log(1+x)|\leq2|x|$ for $|x| <\frac{1}{2}$.

So, $|\log(1+f_n(x))|\leq2 M_n$.

So, by Weierstrass test, $\sum_{n=1}^\infty \log(1+f_n(x))$ converges uniformly to a bounded function on $A$.

So, $\prod_{n=1}^N (1+f_n(x)) = \exp(\sum_{n=1}^N \log(1+f_n(x)))$ also converges uniformly on $A$.

My questions are here:

Does $\sum_{n=1}^\infty \log(1+f_n(x))$ really converges uniformly to a bounded function on $A$?

Does $\prod_{n=1}^N (1+f_n(x)) = \exp(\sum_{n=1}^N \log(1+f_n(x)))$ really converges uniformly on $A$?

答:
Let $\phi(x) = \sum_{n=1}^\infty \log(1+f_n(x))$.

$\lvert \phi(x) \rvert \le \sum_{n=1}^\infty \lvert \log(1+f_n(x)) \rvert \le \sum_{n=1}^\infty 2M_n = 2 \sum_{n=1}^\infty M_n =: R$. Thus $\phi$ is bounded.

Note that also for each $N$ we have $\lvert \sum_{n=1}^N \log(1+f_n(x)) \rvert \le \sum_{n=1}^\infty \lvert \log(1+f_n(x)) \rvert \le R$.

Clearly  $\prod_{n=1}^N (1+f_n(x)) = \exp(\sum_{n=1}^N \log(1+f_n(x)))$ converges pointwise to $\psi(x) = \exp(\phi(x))$.  Moreover $\exp$ is uniformly continuous on $[-R,R]$ since this interval is compact.

Now let $\epsilon > 0$. There exists $\delta > 0$ such that $\lvert \exp(a) - \exp(b) \rvert < \epsilon$ for $a,b \in [-R,R]$ with $\lvert a - b \rvert < \delta$. There exits $N_0$ such that $\lvert \phi(x) - \sum_{n=1}^N \log(1+f_n(x))\rvert < \delta$ for $N \ge N_0$ and all $x \in A$. We therefore get for $N \ge N_0$ and all $x \in A$
$$\left\lvert \psi(x) - \prod_{n=1}^N (1+f_n(x)) \right\rvert = \left\lvert \exp(\phi(x)) - \exp(\sum_{n=1}^N \log(1+f_n(x))) \right\rvert < \epsilon .$$


问:
A problem related to localization
Let $A$ be integrally closed (not necessary be Dedekind), $K=\operatorname{Frac}(A),L/K$ Galois, $B$ is the integral closure of $A$ in $L,p$ is a maximal ideal in $A$. We know that the galois group $G$ acts transitively on maximal ideals in $B$ lying over $p$. If we weaken $p$ maximal to prime, let’s consider localization. Then we know that $G$ acts transitively on prime ideals in $B_{p}$ lying over $A_{p}$’s unique maximal ideal $m_p$. But now does $G$ acts transitively on prime ideal in $B$ lying over $p$?

答:
Thanks for Kneey Lau's illuminating proof, and I try to explain the original proof more detailedlly.

First, supposing that $p$ is maximal in A and $\beta_1$, $\beta_2$ are different prime ideals in B lying over p ( using $\beta|p$ as a symbol to express that), if for any $\sigma$ in G we have $\sigma(\beta_1)\not=\beta_2$, using CRT we can find an element x suffices $$x=0\ (module\  \beta_1) $$ $$x=1\ (module\  \sigma(\beta_2)$$ thus $$\sigma(x)=1\ (module\  \beta_2$$
thus $N_{L/K}(x)=\prod\sigma(x)\in A\cap\beta_1\ but\notin\beta_2$, causing a contridiction.

Second, for $p$ is a prime ideal, considering localize A at p, let $m_p$ be the unique maximal ideal in $A_p$, and in the former case, (omitting some verification here), we can let $q_1$, $q_2$ be maximal ideal in $B_p$ and $B_p|m_p$, let $\beta_i=q_i\cap B$.

Then the problem is already solved  , the prolem I missed before is that $\sigma(\beta_1) \subseteq q_2\cap B=\beta_2$ can conclude $\sigma(\beta_1)=\beta_2$

Title: A problem related to localization


问:
An autohomeomorphism on a normed linear space.
Let $X$ be a normed linear space. Let $x_{0}\in B_{1}^{(0)}=\{x\in X : \|x\|<1\}$ and define

$$
\begin{array}{lcccl}
h & : & X & \longrightarrow & X\\
 & & x & \longrightarrow &  h(x)=\left\{\begin{array}{ccc}
x+(1-\|x\|)x_{0} & if & \|x\|<1 \\\\
x & if & \|x\|\geq 1
\end{array}
\right.
\end{array}
$$
I have managed to show that for all $x,y\in X$, we have that $(1-\|x_{0}\|)\|x-y\|\leq \|h(x)-h(y)\|\leq   (1+\|x_{0}\|)\|x-y\|$. Then $h$ is injective, my question is how can I show that $h$ is surjective? In this case, by the inequalities above, $h$ is an autohomeomorphism on $X$.
Thanks

答:
Clearly $h$ is surjective on $X \setminus B_1^{(0)}$ so let $y \in B_1^{(0)}$ be fixed. We wish to find $x \in B_1^{(0)}$ such that $h(x) = y$.

Consider the continuous function $f : [0,1] \to \Bbb{R}$ given by
$$f(t) = \|y-tx_0\|+t-1.$$
We have
$$f(0) = \|y\|-1 < 0, \quad f(1) = \|y-x_0\| \ge 0$$
so by the intermediate value theorem there is some $t \in (0,1]$ such that $f(t) = 0$.

Define $$x := y-tx_0.$$
Then $$0 = f(t) = \|x\|+t-1$$ so $t = 1-\|x\|$ and hence
$$x = y - (1-\|x\|)x_0.$$
Also $\|x\| = 1-t \in [0,1)$ so $x \in B_1^{(0)}$. Furthermore
$$h(x) = x+(1-\|x\|)x_0 = y - (1-\|x\|)x_0 + (1-\|x\|)x_0 = y.$$


问:
How do I evaluate $\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx$?
I need to calculate the following definite integral:

$$\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx.$$

The only thing that I've found is:

$$\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx = \int_{1/3}^3 \frac{\arctan \frac{1}{x}}{x^2 - x + 1} \; dx,$$

but it doesn't seem useful.

答:
What you have found is actually very useful. Let $I$ be your integral. Note that for $x>0$ one has that
$$\arctan(x)+\arctan(1/x)=\pi/2,$$
so we have the relation
$$2I=\int_{1/3}^3 \frac{\pi/2}{x^2-x+1}~dx,$$
which can be computed by completing the square on the denominator.


问:
What is the asymptotic upper bound of a variable in the functional equation $f(x)=\left\lceil\frac{f(x+1)}{\lceil\log_2(f(x+1))\rceil}\right\rceil$?
We are given a recursive function $ f ( x ) = \left\lceil \frac { f ( x + 1 ) } { \lceil \log_2 ( f ( x + 1 ) ) \rceil } \right\rceil $. We know that $ f ( 1 ) = 2 $ and $ f ( a ) = n $. What is the asymptotic upper bound of $ a $ expressed in terms of $ n $?

答:
$ \require {begingroup} \begingroup
\def \Z {\mathbb Z ^ +}
\def \Zs {\mathbb Z ^ + \setminus \{ 1 \}}
\def \C {\mathcal C}
\def \t {\quad \longrightarrow \quad} $
Let $ \Z $ be the set of positive integers and $ h : \Zs \to \Zs $ be the function defined with $ h ( y ) = \left\lceil \frac y { \left\lceil \log _ 2 y \right\rceil } \right\rceil $ for all $ y \in \Zs $. Here's one reading of your question:

Let $ \C $ be the family of all functions $ f : \Z \to \Zs $ such that $ f ( 1 ) = 2 $ and $ f ( x ) = h \big( f ( x + 1 ) \big) $ for all $ x \in \Z $. For $ n \in \Zs $, let $ A _ n $ be the set of all $ a \in \Z $ for which there is some $ f \in \C $ with $ f ( a ) = n $. What is the least upper bound of $ A _ n $?

Reading the question like this, the answer is that there is no upper bound (assuming $ A _ n \ne \varnothing $, which we will later show to be true). Consider any $ f \in \C $ with $ f ( a ) = n $ and define $ g : \Z \to \Zs $ with
$$ g ( x ) = \begin {cases} 2 & 1 \le x \le m \\ f ( x - m + 1 ) & x > m \end {cases} $$
Then it's easy to see that $ g \in \C $ and $ g ( a + m - 1 ) = n $. As $ a + m - 1 $ can be as large as you want, $ A _ n $ is unbounded.

But the above only happens because $ h ( 2 ) \nless 2 $. $ 2 $ is the only element of $ \Zs $ with this property, and if we only consider those $ f \in C $ that $ f ( 2 ) \ne 2 $, this won't happen. I think that is what you intended, so I'll continue to follow this road. Note that as $ h ( y ) < y $ for all $ y \in \Zs $ with $ y \ne 2 $, any $ f \in \C $ with $ f ( 2 ) \ne 2 $ will be strictly increasing. In particular, for any $ n \in \Zs $ there will be at most one $ a \in \Z $ with $ f ( a ) = n $. For $ n \in \Zs $, let $ B _ n $ be the set of all $ a \in \Z $ for which there is some $ f \in \C $ with $ f ( 2 ) \ne 2 $ and $ f ( a ) = n $. Opposite to what happened to $ A _ n $, $ B _ n $ will have no more than a single element, since for any $ f \in \C $ and any $ x \in \Z $, $ f ( x ) $ is uniquely determined by $ f ( x + 1 ) $ through applying $ h $; this means that if $ f ( a ) = n $, you apply $ h $ repeatedly to $ n $, $ h ( n ) $ and so on, until you reach $ 2 $. This process always ends for those $ f $ with $ f ( 2 ) \ne 2 $, as they are strictly increasing, and when reaching $ 2 $ we know that that is $ f ( 1 ) $ and not $ f ( x ) $ for any other $ x $, and thus $ a $ will be the number of times you applied $ h $ plus one. The following lemma makes this precise, together with showing that starting with any $ n \in \Zs $ is always possible.

Lemma: Let $ n \in \Zs $.

For any $ f , g \in \C $ with $ f ( 2 ) \ne 2 \ne g ( 2 ) $, if there exist $ a , b \in \Z $ such that $ f ( a ) = n = g ( b ) $, then $ a = b $, and for all $ x \in \Z $ with $ x \le a $, $ f ( x ) = g ( x ) $.

There is some strictly increasing $ f \in \C $ and some $ a \in \Z $ such that $ f ( a ) = n $.

$ B _ n $ is a singleton, and its only element is $ a ( n ) $, where $ a : \Zs \to \Z $ is recursively defined with $ a ( 2 ) = 1 $ and $ a ( y ) = a \big( h ( y ) \big) + 1 $ for $ y > 2 $.

Proof:

Without loss of generality, assume that $ a \le b $. Use induction on $ x $ to prove $ f ( a - x ) = g ( b - x ) $ for all $ x < a $, noting that we have $ f ( a ) = g ( b ) $ and $ f , g \in \C $. This will also show that $ g ( b - a + 1 ) = f ( 1 ) = 2 = g ( 1 ) $, which by the fact that $ g $ is strictly increasing, gives $ a = b $.

Since $ h ( y ) < y $ for any $ y \in \Zs $ with $ y \ne 2 $, there is some smallest $ a \in \Z $ such that $ h ^ { a - 1 } ( n ) = 2 $, where $ h ^ m $ denotes $ m $-th iteration of $ h $. Let $ f ( x ) = h ^ { a - x } ( n ) $ for $ 1 \le x \le a $, and continue assigning values of $ f ( x ) $ for $ x > a $ by using the fact that $ h $ is surjective. The fact that constructed this way, $ f $ will become strictly increasing, again comes from $ h ( y ) < y $ for $ 2 \ne y \in \Zs $.

By part 1, $ B _ n $ has at most one element, and by part 2, it has at least one element, and hence it is a singleton. The fact that $ a ( n ) $ is its only element comes from the construction in the previous part.

Now that we have all these, we can see that the function $ a $ defined above can be considered as some sort of "inverse" to any $ f \in \C $, which helps with making sense of what you've asked. So, we're finally ready to reformulate the question in a precise manner, and start solving the main problem.

Define $ a : \Zs \to \Z $ recursively with $ a ( 2 ) = 1 $ and $ a ( n ) = a \big( h ( n ) \big) + 1 $ for $ n > 2 $. Find asymptotic bounds for $ a $ in terms of elementary functions.

You can check that $ a ( n ) \in \Theta \left( \frac { \log _ 2 n } { \log _ 2 \log _ 2 n } \right) $. In fact, the explicit bounds
$$ 1 + \left\lceil \frac { \log _ 2 n - 1 } { \log _ 2 \left\lceil \log _ 2 n \right\rceil } \right\rceil \le a ( n ) \le 2 + \left\lfloor \frac { \log _ 2 n - 1 } { \log _ 2 \log _ 2 n - 1 } \right\rfloor $$
can be given for $ n \ge 5 $. To see that how one can get these bounds, let's take a look at the sequence
$$ n \t h ( n ) \t h \big( h ( n ) \big) \t h \Big( h \big( h ( n ) \big) \Big) \t \dots $$
starting from $ n $ and ending in $ 2 $. Remember that $ a ( n ) $ is equal to the length of the above sequence (i.e. the number of steps needed for reaching $ 2 $) plus one. Instead of the actual sequence
$$ n \t \left\lceil \frac n { \left\lceil \log _ 2 n \right\rceil } \right\rceil \t \left\lceil \frac { \left\lceil \frac n { \left\lceil \log _ 2 n \right\rceil } \right\rceil } { \left\lceil \log _ 2 \left\lceil \frac n { \left\lceil \log _ 2 n \right\rceil } \right\rceil \right\rceil } \right\rceil \t \left\lceil \frac { \left\lceil \frac { \left\lceil \frac n { \left\lceil \log _ 2 n \right\rceil } \right\rceil } { \left\lceil \log _ 2 \left\lceil \frac n { \left\lceil \log _ 2 n \right\rceil } \right\rceil \right\rceil } \right\rceil } { \left\lceil \log _ 2 \left\lceil \frac { \left\lceil \frac n { \left\lceil \log _ 2 n \right\rceil } \right\rceil } { \left\lceil \log _ 2 \left\lceil \frac n { \left\lceil \log _ 2 n \right\rceil } \right\rceil \right\rceil } \right\rceil \right\rceil } \right\rceil \t \dots $$
let's take a look at another one which has a simpler look. Since $ h ( y ) \ge \frac y { \left\lceil \log _ 2 y \right\rceil } $ and $ h ( y ) \le y $ for $ y > 2 $, we can see that the sequence
$$ n \t \frac n { \left\lceil \log _ 2 n \right\rceil } \t \frac n { \left\lceil \log _ 2 n \right\rceil ^ 2 } \t \frac n { \left\lceil \log _ 2 n \right\rceil ^ 3 } \t \dots $$
consists of terms no greater than their corresponding terms in the original sequence. This means that this sequence may reach a value less than or equal to $ 2 $ in fewer steps. If $ b $ is the mentioned number of steps, it means that we have $ \frac n { \left\lceil \log _ 2 n \right\rceil ^ b } \le 2 $. By taking logarithms and rearranging, we get $ b \ge \frac { \log _ 2 n - 1 } { \log _ 2 \left\lceil \log _ 2 n \right\rceil } $, and since $ a ( n ) \ge b + 1 $, we get the desired lower bound for $ a ( n ) $. Similarly, and this time using $ h ( y ) \le \left\lceil \frac y { \log _ 2 y } \right\rceil $ and $ h ( y ) \le \frac y 2 $ for $ y > 3 $, we can see that the sequence
$$ n \t \left\lceil \frac n { \log _ 2 n } \right\rceil \t \left\lceil \frac { 2 n } { ( \log _ 2 n ) ^ 2 } \right\rceil \t \left\lceil \frac { 2 ^ 2 n } { ( \log _ 2 n ) ^ 3 } \right\rceil \t \dots $$
consists of terms which are no less than corresponding terms in the original sequence. Thus $ b $, the number of steps needed for reaching a value less than or equal to $ 2 $ is more than that of the original sequence. In case that $ b \ge 1 $ (i.e. at least one step is needed, which happens for $ n \ge 5 $) $ b - 1 $ steps are not enough, which gives $ \frac { 2 ^ { b - 2 } n } { ( \log _ 2 n ) ^ { b - 1 } } + 1 \ge \left\lceil \frac { 2 ^ { b - 2 } n } { ( \log _ 2 n ) ^ { b - 1 } } \right\rceil \ge 3 $. Taking logarithms and rearranging, we get $ b \le 1 + \frac { \log _ 2 n - 1 } { \log _ 2 \log _ 2 n - 1 } $, which by $ a ( n ) \le b + 1 $, gives the desired upper bound for $ a ( n ) $.

Finally, note that using the bounds, by squeezing we have
$$ \lim _ { n \to \infty } \frac { a ( n ) \log _ 2 \log _ 2 n } { \log _ 2 n } = 1 $$
or in other terms
$$ a ( n ) \sim \frac { \log _ 2 n } { \log _ 2 \log _ 2 n } $$
which is even better than mere $ a ( n ) \in \Theta \left( \frac { \log _ 2 n } { \log _ 2 \log _ 2 n } \right) $.
$ \endgroup $


问:
Infinite linear ordered set which doesn't include infinte well-ordered subsets
Need an example of such set.

Thank you for your time

答: What about the negative integers, usual order? Verification that this has no infinite well-ordered subset should not be difficult. </p>


问:
Writing down homeomorphisms in a formula
Hey I am dealing with this question since some months.
How rigorously do I have to prove that two "obviously homeomorphic" spaces are really homeomorphic? Example: Cup homeomorphic to a donut... Is it even possible to write down a homeomorphism in this example?</p>

答:
ZPlaya: The comments suggest that this homeomorphism comes by magic. But the way the proof that `an orientable surface with $\chi(\Sigma) = 2-2g$ is diffeomorphic to the standard $\Sigma_g$' is by constructing an explicit homeomorphism.

There are many ways this is done. Almost all of them start by `triangulating the surface' which is a difficult process but still done explicitly if you chase back all relevant results. This decomposes the surface into triangles. In your case, imagine that your coffee mug, instead of being smooth, was made of hammered gold with triangle-like faces. This gives a combinatorial sense of what's going on with your coffee mug.

We then identify the building-blocks of the surface. In your case, the handle is a copy of $S^1 \times [0,1]$ (the cylinder, or the annulus), and you need to show that the "coffee-mug without a handle" is too. This is harder, but the coffee mug has a portion starting at the lip and going inside homeomorphic to $D^2$, and a portion starting at the lip and going to the bottom, homeomorphic to $D^2$ minus two smaller discs (think: face, where I have removed where the eyes would go; or think pair-of-pants).

Then I observe that pasting these together --- the disc-bits, I mean --- I get what looks like the 2-sphere with two discs removed. This is a theorem you prove when studying quotient topologies, that 'two discs pasted together along their boundary is homeomorphic to a sphere'. Then you are almost done --- you now need to prove that the sphere with two discs removed, with $S^1 \times [0,1]$ (a `handle') glued on to those pieces, is homeomorphic to a torus.

I am being inexplicit here if only because making all of this explicit --- with formulas --- would require massive amounts of piecewise data on each piece I've described above. It is possible, but topologists are content to know the existence of a formula for a continuous function rather than write it down, because the formula rarely provides insight into what is going on. But we do prove that a homeomorphism exists, by proving that every surface can be chopped up into building-blocks, and then proving that the building-blocks can only be pasted together in relatively simple ways.

A very careful proof of this will require every single piece of knowledge you learn in a topology course, so it is not suitable for an answer here. :)


问:
Math behind this SQL problem
I have the following 'sorted by row' lists (2nd column), in which every row produces an output (3rd column, and 4th column). This output has been found without using formulas and it represents a sequence of sql database language statements which have been verified and are correct. Just for completeness, these statements are related to evaluating all aggregations from a sequential list of values (2nd column) in order to not evaluate already evaluated groups. These aggregations can be evaluated in SQL with a simple statement, like for example:

group by cube(F1,F2,F3,F4)

which groups by all 2 raised to 4 combinations without repetitions of 4 elements taken 0,1,2,3,4 at a time, and more exactly:

F1,F2,F3,F4, (F1,F2), (F1,F3), (F1,F4), (F2,F3), (F2,F4), (F3,F4), (F1,F2,F3), (F1,F2,F4), (F1,F3,F4), (F2,F3,F4), (F1,F2,F3,F4), ()

where () means grouping by no effective combination (just evaluates the sql aggregation function).

If we consider one example (i.e. 4th example), we have that these 6 statements (which evaluate already evaluated groups, generating duplicates that are removed at the end of evaluation using UNION statement and so are not good from a performance point of view):

```<code>SELECT F1,F5, COUNT(*) FROM TAB GROUP BY CUBE(F1,F5)
UNION SELECT F1,F6, COUNT(*) FROM TAB GROUP BY CUBE(F1,F6)
UNION SELECT F1,F7, COUNT(*) FROM TAB GROUP BY CUBE(F1,F7)
UNION SELECT F4,F5, COUNT(*) FROM TAB GROUP BY CUBE(F4,F5)
UNION SELECT F4,F6, COUNT(*) FROM TAB GROUP BY CUBE(F4,F6)
UNION SELECT F4,F7, COUNT(*) FROM TAB GROUP BY CUBE(F4,F7)
</code>```

are equivalent to the following ones (which not suffer from the performance problems, duplicates and multiple evaluation), represented in the 3rd and 4th columns:

```<code>SELECT F1,F5, COUNT(*) FROM TAB GROUP BY CUBE(F1,F5)
UNION ALL SELECT F1,F6, COUNT(*) FROM TAB GROUP BY GROUPING SETS ((F6)), GROUPING SETS (CUBE(F1))
UNION ALL SELECT F1,F7, COUNT(*) FROM TAB GROUP BY GROUPING SETS ((F7)), GROUPING SETS (CUBE(F1))
UNION ALL SELECT F4,F5, COUNT(*) FROM TAB GROUP BY GROUPING SETS ((F4)), GROUPING SETS (CUBE(F5))
UNION ALL SELECT F4,F6, COUNT(*) FROM TAB GROUP BY GROUPING SETS ((F4,F6))
UNION ALL SELECT F4,F7, COUNT(*) FROM TAB GROUP BY GROUPING SETS ((F4,F7))
</code>```

Starting from the following examples, I would like to find a math algorithm to get the 3rd column values (or alternatively the 4th column values), given the ordered list of the 2nd column.

The examples are:

Ordered list from cartesian product
3rd column
4th column

1st example:
1: F3,F7,F8
2: F4,F5,F6
F3,F4

F3,F4

F3,F5
F5
F3

F3,F6
F6
F3

F7,F4
F7
F4

F7,F5
F7,F5

F7,F6
F7,F6

F8,F4
F8
F4

F8,F5
F8,F5

F8,F6
F8,F6

Ordered list from cartesian product
3rd column
4th column

2nd example:
1: F3,F4
2: F5,F6
3: F7,F8
F3,F5,F7

F3,F5,F7

F3,F5,F8
F8
F3,F5

F3,F6,F7
F6
F3,F7

F3,F6,F8
F6,F8
F3

F4,F5,F7
F4
F5,F7

F4,F5,F8
F4,F8
F5

F4,F6,F7
F4,F6
F7

F4,F6,F8
F4,F6,F8

Ordered list from cartesian product
3rd column
4th column

3rd example:
1: F1,F2
2: F3,F4
3: F5,F6
4: F7,F8
F1,F3,F5,F7

F1,F3,F5,F7

F1,F3,F5,F8
F8
F1,F3,F5

F1,F3,F6,F7
F6
F1,F3,F7

F1,F3,F6,F8
F6,F8
F1,F3

F1,F4,F5,F7
F4
F1,F5,F7

F1,F4,F5,F8
F4,F8
F1,F5

F1,F4,F6,F7
F4,F6
F1,F7

F1,F4,F6,F8
F4,F6,F8
F1

F2,F3,F5,F7
F2
F3,F5,F7

F2,F3,F5,F8
F2,F8
F3,F5

F2,F3,F6,F7
F2,F6
F3,F7

F2,F3,F6,F8
F2,F6,F8
F3

F2,F4,F5,F7
F2,F4
F5,F7

F2,F4,F5,F8
F2,F4,F8
F5

F2,F4,F6,F7
F2,F4,F6
F7

F2,F4,F6,F8
F2,F4,F6,F8

Ordered list from cartesian product
3rd column
4th column

4th example:
1: F1,F4
2: F5,F6,F7
F1,F5

F1,F5

F1,F6
F6
F1

F1,F7
F7
F1

F4,F5
F4
F5

F4,F6
F4,F6

F4,F7
F4,F7

Ordered list from cartesian product
3rd column
4th column

5th example:
1: F1,F4
2: F5,F6,F7
3: F8,F9
F1,F5,F8

F1,F5,F8

F1,F5,F9
F9
F1,F5

F1,F6,F8
F6
F1,F8

F1,F6,F9
F6,F9
F1

F1,F7,F8
F7
F1,F8

F1,F7,F9
F7,F9
F1

F4,F5,F8
F4
F5,F8

F4,F5,F9
F4,F9
F5

F4,F6,F8
F4,F6
F8

F4,F6,F9
F4,F6,F9

F4,F7,F8
F4,F7
F8

F4,F7,F9
F4,F7,F9

Ordered list from cartesian product
3rd column
4th column

6th example:
1: F1,F5
2: F2,F6,F9
3: F3,F7
4: F4,F8,F10
F1,F2,F3,F4

F1,F2,F3,F4

F1,F2,F3,F8
F8
F1,F2,F3

F1,F2,F3,F10
F10
F1,F2,F3

F1,F2,F7,F4
F7
F1,F2,F4

F1,F2,F7,F8
F7,F8
F1,F2

F1,F2,F7,F10
F7,F10
F1,F2

F1,F6,F3,F4
F6
F1,F3,F4

F1,F6,F3,F8
F6,F8
F1,F3

F1,F6,F3,F10
F6,F10
F1,F3

F1,F6,F7,F4
F6,F7
F1,F4

F1,F6,F7,F8
F6,F7,F8
F1

F1,F6,F7,F10
F6,F7,F10
F1

F1,F9,F3,F4
F9
F1,F3,F4

F1,F9,F3,F8
F9,F8
F1,F3

F1,F9,F3,F10
F9,F10
F1,F3

F1,F9,F7,F4
F9,F7
F1,F4

F1,F9,F7,F8
F9,F7,F8
F1

F1,F9,F7,F10
F9,F7,F10
F1

F5,F2,F3,F4
F5
F2,F3,F4

F5,F2,F3,F8
F5,F8
F2,F3

F5,F2,F3,F10
F5,F10
F2,F3

F5,F2,F7,F4
F5,F7
F2,F4

F5,F2,F7,F8
F5,F7,F8
F2

F5,F2,F7,F10
F5,F7,F10
F2

F5,F6,F3,F4
F5,F6
F3,F4

F5,F6,F3,F8
F5,F6,F8
F3

F5,F6,F3,F10
F5,F6,F10
F3

F5,F6,F7,F4
F5,F6,F7
F4

F5,F6,F7,F8
F5,F6,F7,F8

F5,F6,F7,F10
F5,F6,F7,F10

F5,F9,F3,F4
F5,F9
F3,F4

F5,F9,F3,F8
F5,F9,F8
F3

F5,F9,F3,F10
F5,F9,F10
F3

F5,F9,F7,F4
F5,F9,F7
F4

F5,F9,F7,F8
F5,F9,F7,F8

F5,F9,F7,F10
F5,F9,F7,F10

P.S. I posted a reduced version of this question on stackoverflow (with no answers), but I think this is a math problem.

Edit: Just to treat the problem as a math one, we can consider it like finding the correct use of the moltiplication of 2 math operators in the 3rd and 4th column, which meaning can be explained by the following table:

3rd column:
operator
GROUPING SETS(())
4rd column:
operator
GROUPING SETS (CUBE())
Equivalent: 2^3  = 8 tuples
(3 = number of elements of the CUBE
operator in the 4rd column).
Note that 3rd column elements
are always present

F1,F2
F4,F5,F6
(F1,F2),(F1,F2,F4), (F1,F2,F5), (F1,F2,F6), (F1,F2,F4,F5), (F1,F2,F4,F6), (F1,F2,F5,F6), (F1,F2,F4,F5,F6)

So that, considering another example (i.e the 2nd example mentioned above), we have:

Produced sets (note several
incremental duplicates)
Using operators in
3rd and 4th column
Produced sets (note there
are no duplicates).
Order in a set is not
important, (a,b)=(b,a)

CUBE(F3,F5,F7)
F3,F5,F7,(F3,F5),(F3,F7),(F5,F7),(F3,F5,F7),(),
GROUPING SETS(CUBE(F3,F5,F7))
F3,F5,F7,(F3,F5),(F3,F7),(F5,F7),(F3,F5,F7),(),

CUBE(F3,F5,F8)
F3,F5,F8,(F3,F5),(F3,F8),(F5,F8),(F3,F5,F8),(),
GROUPING SETS(F8), GROUPING SETS(CUBE(F3,F5))
F8,(F3,F8),(F5,F8),(F3,F5,F8)

CUBE(F3,F6,F7)
F3,F6,F7,(F3,F6),(F3,F7),(F6,F7),(F3,F6,F7),(),
GROUPING SETS(F6), GROUPING SETS(CUBE(F3,F7))
F6,(F3,F6),(F6,F7),(F3,F6,F7)

CUBE(F3,F6,F8)
F3,F6,F8,(F3,F6),(F3,F8),(F6,F8),(F3,F6,F8),(),
GROUPING SETS(F6,F8), GROUPING SETS(CUBE(F3))
(F6,F8),(F3,F6,F8)

CUBE(F4,F5,F7)
F4,F5,F7,(F4,F5),(F4,F7),(F5,F7),(F4,F5,F7),(),
GROUPING SETS(F4), GROUPING SETS(CUBE(F5,F7))
F4,(F4,F5),(F4,F7),(F4,F5,F7)

CUBE(F4,F5,F8)
F4,F5,F8,(F4,F5),(F4,F8),(F5,F8),(F4,F5,F8),(),
GROUPING SETS(F4,F8), GROUPING SETS(CUBE(F5))
(F4,F8),(F4,F5,F8)

CUBE(F4,F6,F7)
F4,F6,F7,(F4,F6),(F4,F7),(F6,F7),(F4,F6,F7),(),
GROUPING SETS(F4,F6), GROUPING SETS(CUBE(F7))
(F4,F6),(F4,F6,F7)

CUBE(F4,F6,F8)
F4,F6,F8,(F4,F6),(F4,F8),(F6,F8),(F4,F6,F8),(),
GROUPING SETS(F4,F6,F8)
(F4,F6,F8)

At the end we have 27 distinct sets and we produce the same distinct sets incrementing row by row.

答: Sometimes solutions are easier than expected, luckily no algorithms needed. The 3rd column is the result of the 2nd column without the elements in the first row of the 4th column. The 4th column is the remaining elements of the 2nd one. It's incredible the time I spent to this. I'm sorry for wasting your time.

Title: Math behind this SQL problem


问:
How to explicitly find a natural number N that satisfies the statement in the definition of the limit?
For example, in this question:

$\lim _{x\to \infty \:}\left(\frac{x^2}{x^2+2x}\right)=1$

For every epsilon > 0, I am asked to find explicitly a natural number N which
satisfies the statement in the definition of the limit.

What does find explicitly $N$ means? Do I just prove it as usual?

$|\frac{x^2}{x^2+2x} -1|< Ɛ$

Continue with this I got:

$n+2 > \frac{2}{Ɛ}$

and then I take:

$ N = 2[\frac{1}{Ɛ}] + 1$

then since $n >= N$:

$n+2 >= 2[\frac{1}{Ɛ}] + 1 + 2$

therefore, $N = 2[\frac{1}{Ɛ}] + 3$, did I just found my $N$ or is it asking for something numerical? I am confused?

答:
Per back and forth with OP, this answer is intended to eliminate any confusion around the problem.  The answer will (mostly) use analysis already done by the OP.

For example, in this question:

$\lim _{x\to \infty \:}\left(\frac{x^2}{x^2+2x}\right)=1$

For every epsilon > 0, I am asked to find explicitly a natural number N which
satisfies the statement in the definition of the limit.

For all positive integers $n$, let $a_n$ be defined as
$\left(\frac{n^2}{n^2+2n}\right).$

There are two ways that the problem might be interpreted.  The narrow way is that given $\epsilon > 0$, you are being asked to find $N \in \mathbb{Z^+}$ such that $|a_N - 1| < \epsilon.$

The alternative intrepretation, which I will adopt, and which might not have been intended is that you are to find $N \in \mathbb{Z^+}$ such that for every element $a_n$ in the subsequence
$\{a_{N}, a_{(N+1)}, a_{(N+2)}, \cdots \}$ you have that

$|a_n - 1| < \epsilon.$

The reason that I have chosen the 2nd interpretation, which is moderately more onerous, is that this interpretation represents the standard definition of a sequence converging to a limit.

That is, normally, you show that a sequence $\langle a_n\rangle$ converges to a limit $L$ by showing that for any $\epsilon > 0$ there exists a positive integer $N$, such that for all positive integers $n \geq N$ (or $n > N$), $|a_n - L| < \epsilon.$

Hijacking the analysis already done by the OP:

it is clear that each element $a_n$ will be between $0$ and $1$.

Therefore, the first objective is to find an $N$ such that

$$1 - a_N < \epsilon \implies \tag{Constraint-1}$$

$1 - \frac{N^2}{N^2 + 2N} < \epsilon \implies$

$\frac{N^2 + 2N - N^2}{N^2 + 2N} < \epsilon \implies$

$\frac{2N}{N^2 + 2N} < \epsilon \implies$

$\frac{2}{N+2} < \epsilon \implies$

$\frac{N+2}{2} > \frac{1}{\epsilon} \implies$

$$(N+2) > \frac{2}{\epsilon}.\tag{Constraint-2}$$

What you actually have is that if Constraint-1 is satisfied for a specific positive integer $N$, then Constraint-2 is also satisfied.  However, what you want is the reverse.

Notice which way the $\implies$ symbols are directed.  However, if you examine the actual analysis step-by-step, you will see that each step has a (hidden) bidirectional implication.  That is a given step is true $\iff$ the next step is true.

This means that if Constraint-2 is satisfied for a specific positive integer $N$, then Constraint-1 will also be satisfied.

It further means that if $n$ is any positive integer that satisfies Constraint-2, then $n$ will also satisfy Constraint-1.

Therefore, for a given fixed $\epsilon > 0$, specify $N$ by Constraint-2.  This implies that $a_N$ satisfies Constraint-1.

But there is more.

Suppose that $n$ is any positive integer such that $n > N$.

Then, $(n+2) > (N+2).$

Therefore, $n$ also satisfies Constraint-2.

Therefore, $a_n$ also satisfies Constraint-1.

Thus, Constraint-2 provides the specification for $N$, based on $\epsilon$ such that each element in

$\{a_N, a_{(N+1)}, a_{(N+2)}, \cdots \}$ satisfies Constraint-1.


问:
A statement in the book "Differential and Integral Calculus, 6th ed" (Love and Rainville)
In page 285 of "Differential and Integral Calculus, 6th ed." by Love and Rainville, the example used in 'Substitution suggested by the problem' about plane areas is this:

Find the area of the ellipse $(x = a \cos{\varphi}$, $y = b \sin{\varphi})$.

Then, the solution is given by substitution.

At once,
\begin{gather*}A = 4\int_{0}^{a} y\,dx\end{gather*}.
When $x = 0$, $\varphi = \frac{1}{2}\pi$, and when $x =  a$, $\varphi = 0$. Therefore
\begin{align*}A &\;=\; 4\int_{\frac{1}{2}\pi}^{0}(b\sin\varphi)(-a\sin\varphi\,d\varphi) \\ &\;=\;4ab\int^{\frac{1}{2}\pi}_{0}\sin^{2}\varphi\,d\varphi \\ &\;=\; 4ab \cdot \frac{1}{2}\cdot \frac{\pi}{2} \\ &\;=\; \pi ab.\end{align*}

It is then followed by these sentences:

The transformations suggested in this section are intuitively reasonable. Rigorous justification of them belongs to a course in advanced calculus.

Because of this, I had two questions in mind.

1. Is the "them" the given solution? [Answered by @user170231]

Why does it need a rigorous justification? Isn't the given solution enough?

答:
"Rigorous justifications" likely means nothing given that it has it doesn't seem to have a justified context within the content of the book, unless they have left a footnote on the context they wish to express; either this, or they are leaving out elliptic integrals which may be in a further chapter and which are not necessary for the computation of the area of an ellipse. That said, I can't tell you any more about their statement beyond reasonable speculation. As observed, if

$$\left(\dfrac{x}{a}\right)^2+\left(\dfrac{y}{b}\right)^2=1,$$ then
$$(x,y)=(a\cos\theta,b\sin\theta)$$ produces the correct identity
$$\cos^2\theta+\sin^2\theta=1,$$
and so the transformation is justified. Moreover, it should be noted that upon substituting $$(x,y)=(a\cos\theta,b\sin\theta)$$ that the parametrization is still circular with respect to the angle, and so the bounds of integration with respect to the angle $\theta$ ranging from $0$ to $\dfrac{\pi}{2}$ produce a quarter of the area of the ellipse. The boundaries of integration can also be derived from the inverse sine function in the sense that

$$[\arcsin(0),\arcsin(1)]=[0,\dfrac{\pi}{2}]$$

when the inverse sine is restricted to the region on the plane where $x\in [-1,1], y\in [-\frac{\pi}{2},\frac{\pi}{2}]$.

It seems that the wording of the sentences "The transformations suggested in this section are intuitively reasonable. Rigorous justification of them belongs to a course in advanced calculus." are not faithful to the material they present, on the basis that there is not much further rigor to discuss here. If there is a greater context they would share in some other book or by some other mathematicians, then let them state it as such. But, as it stands (and according to their wording), math is not objective in the sense they imply; any further rigor comes from time-elaborated research-level discoveries around the context of the subject which can then be used to further re-enforce the truth of the result. The only requirements for contributing to the body of mathematics is that your discovery is new, interesting, and true by appropriately rigorous proof. For example, the logical proof that $1+1=2$ is very long, but inappropriate for an elementary school student, because the proof itself is actually the interesting result.

That said, if they have some other context in mind, then I would gladly like to learn it. Nevertheless, I am upset they decided to make such a remark. Mainly, however, I hope to reinforce your confidence in the subject and to not rely on other's remarks as considerations simply because it seems as if they demand it.

I hope this answer itself is appropriate to the question.


问:
The true nature of division in light of Euclid's understanding of multiplication
I'm not a math major, but a philosophy major that likes to know that he knows what he's talking about. This may seem like a super stupid question, but here I go.

So Euclid made a lot of sense when he gave the example of the nature of multiplication. For example.
"2 x 3" is really 2 added to itself 3 times, and vice versa, and this is different from 2 + 3, which is five, two units added to three.

But, what about division?

It seems to be the opposite of multiplication like a mirror image from 3rd grade, but I'm starting to doubt that after digging deeper.

For example, this site gave the following diagram for division:

Based on this image, for 6/3, division seems to be the counting of groups, not the inverse of multiplication according to Euclid.

Can someone explain the true nature of division in light of the euclid example I gave?

答:
One really should not currently consider Euclid's as an authoritative treatment of mathematics (2300 or so years since his "Elements" were written). Even his axiomatic development of geometry was faulty, as it was realized in 19th century. Still, one can use "Elements" when teaching axiomatic geometry these days.

It is worse with arithmetic/algebra. Euclid's notion of "numbers" is very limited. (Only natural numbers were considered as "proper" numbers and, even 1 was not uniformly regarded as a number.) Explaining division as "counting in groups" if you want to explain the concept to an 8 year old child is fine (as long as you limit the discussion to the case when the outcome of the division is a natural number). But once you learn about negative integers, rational numbers, real numbers, complex numbers, you realize that "counting in groups" looses any meaning. If you follow the axiomatic treatment of, say, real numbers, you realize that the only meaningful definition of division is

$$
\frac{a}{b}= a\cdot b^{-1},
$$
(for $b\ne 0$) where $b^{-1}$ is defined as the unique real number such that $b\cdot b^{-1}=1$.  (The existence of $b^{-1}$ is one of the axioms.) You can also say that
$$
\frac{a}{b}=c \iff a= bc,
$$
hence, division appears as the inverse to the multiplication function when $a$ is chosen in advance and $b, c$ are the variables.

The same definition of division applies to more general algebraic structures called fields. There are more general algebraic structures where division is defined, called division rings, but one then has to distinguish between $ab^{-1}$ and $b^{-1}a$.


答:
The Borel sigma-algebra of $\mathbb{R}^2$ is defined as that generated by the subsets of $\mathbb{R}^2$ that are open in the product topology.  It is almost, but not quite, correct to say that this is equal to the collection of pointwise images of Borel subsets of $\mathbb{R}$ under your a correspondence given by splitting and interleaving decimal expansions.

The problem, as Ittay pointed out in a comment, is that this correspondence is not quite a bijection between $\mathbb{R}$ and $\mathbb{R}^2$.  Nevertheless, by changing the argument a bit, one can still show that $\mathbb{R}$ and $\mathbb{R}^2$ are Borel-isomorphic.  One way to do this is to get a Borel injection from $\mathbb{R}^2$ to $\mathbb{R}$ and then use the Borel version of the Cantor–Bernstein–Schroeder Theorem.

Indeed, a Borel injection $f: \mathbb{R}^2 \to \mathbb{R}$ can be defined as follows: Given $(x,y) \in \mathbb{R}^2$, let $x_0 x_1 x_2 \ldots$ and $y_0y_1y_2 \ldots$ be the unique decimal expansions of $x$ and $y$ respectively with the property that the digits are not eventually all 9's from some point on.
Then let $f(x,y)$ be the real number whose decimal expansion is $x_0y_0x_1y_1 \ldots.$  (For simplicity of notation I am ignoring the issue of where the decimal point goes, but this is easy to deal with.)
Unfortunately the range of this function is missing uncountably many real numbers (such as those whose even digits are eventually all 9's and whose odd digits are not) and the easiest way that I see to deal with this issue is to use the Cantor–Bernstein–Schroeder Theorem, although it may be overkill.

The sets $\mathbb{R}$ and $\mathbb{R}^2$ with these Borel structures (and indeed $\mathbb{R}^n$ for any positive integer $n$) are examples of
standard Borel spaces, which are the objects one gets from Polish topological spaces by forgetting the topology and remembering only the Borel structure. The Borel isomorphism we established above proves a special case of a theorem of Kuratowski, which says that every uncountable Polish space is Borel-isomorphic to $\mathbb{R}$, so up to isomorphism there is only one uncountable standard Borel space.

The sets $\mathbb{N}$ and $\mathbb{N}^n$ with the discrete Borel structure (every subset is a union of countable many singletons and is therefore Borel in the usual topology) are Borel-isomorphic to one another.  They are examples of the countably infinite standard Borel space, which is also unique up to isomorphism.

The set $\mathbb{N}^\infty$ (also known as $\mathbb{N}^\mathbb{N}$) of integer sequences with the Borel algebra generated by the product topology is an uncountable standard Borel space—it is Borel-isomorphic to $\mathbb{R}$ rather than to $\mathbb{N}$.

Title: Cardinality and Measurability

Body:
We can show that $\mathbb{R}$ and $\mathbb{R}^2$ or ($\mathbb{R}^n$) have same cardinality using the following one-to-one and onto mapping:

Say x = (0.123456789....)

Then f(x) = {(0.13579...),(0.2468..)}

My question is can we claim that the Borel sigma algebra in $\mathbb{R}$ has a correspondent sigma algebra (if it is, is it THE Borel sigma algebra in $\mathbb{R}^2$?) by mapping each Borel set D to $\mathbb{R}^2$ by f(D)? Why or why not could you give the argument??

Similarly (I am more certain of this)
When we have N^n and N using this mapping could we claim that the sigma algebra in N has a correspondent sigma algebra in N^n? Does the statement hold for N^infinity and N as well, if N^infinity and N have the same cardinality ( i am not sure about my last statement)?

Thank you!


问:
$S \neq R^n$ or $\emptyset \implies \partial S \neq \emptyset$
I want to show that $S \neq \mathbb R^n$ and $S \neq \emptyset \implies \partial S \neq \emptyset$ using the following property : Let $S \subset \mathbb R^n$ a compact set and $T \subset \mathbb R^n$ a closed set, both are not empty et $S \cap T = \emptyset$. Then $\exists a \in S$ and $b \in T$ such that $$inf(\lvert\lvert x-y \rvert \rvert : x \in S, y \in T) = \lvert\lvert a-b \rvert \rvert > 0$$
I hesitate about the way to do this, I don't know what I need to take to be my compact set and closed set and the next steps are also not very clear.

(I already know the easy proof with $\partial S = \emptyset \implies \overline{S}=\mathring{S}=S \implies S = \mathbb R^n$ or $S = \emptyset $, I want to use the given property to show the proposition.)

答:
Let $S \ne \emptyset, \Bbb{R}^n$ be a set such that $\partial S = \emptyset$. Then there exists a closed ball $K$ large enough such that $\overline{S} \cap K$ and $\overline{S^c} \cap K$ are both nonempty. Hence they are two disjoint nonempty compact sets so there exist
$$a \in \overline{S} \cap K, \quad b \in \overline{S^c} \cap K$$
which minimize the distance between $\overline{S} \cap K$ and $\overline{S^c} \cap K$.

Consider the element $\frac{a+b}2$. It is certainly in $K$ since $K$ is convex. We have
$$\left\|a-\frac{a+b}2\right\| = \left\|b-\frac{a+b}2\right\| = \frac12\|a-b\|$$
so

$\frac{a+b}2$ is closer to $a$ than $b$ is and hence it cannot be in $\overline{S^c} \cap K$,

$\frac{a+b}2$ is closer to $b$ than $a$ is and hence it cannot be in $\overline{S} \cap K$.

This is a contradiction since clearly
$$K = (\overline{S} \cap K) \cup (\overline{S^c} \cap K).$$


问:
What is the difference between relations xRy in Z and Z+ such that xy=1?
I am trying to analyze the relation on both cases in integers and positive integers, and here are what I got so far:

Both are symmetric and transitive. Because xy = yx = 1 on both cases, and  if xy = 1, yz = 1, then xz = 1 also work on both cases, but I am not sure whether they're reflexive and anti-symmetric, please give me a hand, thank you

答:
As correctly noted by the other answer, the relations in question happen to be $\{(-1,-1),(1,1)\}$ and $\{(1,1)\}$ if it were taken over $\Bbb Z$ and $\Bbb Z^+$ respectively.

These relations are indeed transitive (as any choice of $x,y,z$ (repetitions allowed) such that $x\sim y$ and $y\sim z$ result in $x\sim z$ as well) and symmetric (as any choice of $x,y$ such that $x\sim y$ results in $y\sim x$ as well), easily verified since the only examples of times where we have $x\sim y$ or similar are when $x,y$ (and $z$) are all equal to $1$ or are all equal to $-1$.

The relations are both not reflexive.  This is because there exists at least one example of an element who is not related to itself.  For instance, $2$ is not related to itself since $2\times 2\neq 1$.

The relations are both not irreflexive.  To have been irreflexive would mean that every element is not related to itself, but this is not the case since $1$ is related to itself in both cases since $1\times 1=1$ is true.

The relations are both antisymmetric.  This is because every case of $x$ being related to $y$ we have that $y$ is not related to $x$ or that $y=x$.  It is indeed the case here since the only examples of times we have one thing related to another it was because they were both $1$ or were both $-1$.

I encourage you to check out my answer to Can a relation be both symmetric and antisymmetric where I give graphical interpretations of symmetry and antisymmetry described using "single-sided arrows," "doublesided arrows," and "loops."  This relation is a quintessential example of one which is both symmetric and antisymmetric at the same time, the graph for it consisting only of loops.


问:
Integrating $\sqrt\frac{x}{{x+a}}$?
How can I integrate this expression?
$$\sqrt\frac{x}{{x+a}}$$

I substituted $x=a\tan^2t$. After substituting, I reach
$$\int \sec x\tan^2x$$
Is this way correct?

Thanks in advance.

答:
EDIT: Natasha J, here is your new answer!
Let $$t=\dfrac{x}{x+a}, \ \dfrac{at}{(1-t)}=x$$

$$\int \sqrt{\dfrac{x}{x+a}}dx=a\int \sqrt{t}\dfrac{1(1-t)-(-1)t}{(1-t)^2}dt$$
$$=a\int\dfrac{\sqrt{t}}{(1-t)^2}dt$$
by uv-substitution
$$=\dfrac{a\sqrt{t}}{1-t}-\dfrac{a}{2}\int \dfrac{1}{(1-t)\sqrt{t}}dt$$ where the left side integral is, with $t=u^2$ $$-\dfrac{a}{2}\int \dfrac{1}{1-u}+\dfrac{1}{1+u}du=-\dfrac{a}{2}\left(\ln|1+u|-\ln|1-u|\right)+C$$

which gives

$$\boxed{\int\sqrt{\dfrac{x}{x+a}}dx=\sqrt{x+a}\sqrt{x}-\dfrac{a}{2}\ln\dfrac{|\sqrt{x+a}+\sqrt{x}|}{|\sqrt{x+a}-\sqrt{x}|}+C}$$ Also,

$$0=\lim_{a\to 0}\dfrac{a}{2}\ln\dfrac{|\sqrt{x+a}+\sqrt{x}|}{|\sqrt{x+a}-\sqrt{x}|}$$

It was simpler than it seemed. Using similar steps,

$$\int \sqrt[n]{\dfrac{ax+b}{cx+d}}dx$$ can be known.

Title: Integrating $\sqrt\frac{x}{{x+a}}$?


问:
Lagrange Interpolation Basic Proof
I have been given this question, and I am unsure what the intuition behind answering it would be, nor where to start:

Let $x_1 < x_2 < \dots < x_n$. Show that, for some given function $f$, the polynomial
$$P(x) = \sum_{k=1}^n f(x_k) L_k(x), \quad\text{where}\quad L_k(x) = \prod_{i=1\\i\neq k}^n \dfrac{x-x_i}{x_k-x_i}$$, satisfies $$P(x) = f(x) \quad \text{for} \quad x=x_1, x_2, \ldots, x_n.$$

How would one approach this question, and what does it intuitively show by answering it?

答:
You will find that $L_k(x_j)=\delta_{jk}$ with the Kronecker delta. The zeros originate with zeros of the linear factors in the defining product of $L_k(x)$, while the ones result from cancellation in the quotient factors of the product.

So in the end you compute $P(x_j)=\sum_k\delta_{kj}f(x_j)=f(x_j)$.

Title: Lagrange Interpolation Basic Proof


问:
Show $\bmod m\!:\ ar\equiv b,\ br\equiv a\Rightarrow a^2\equiv b^2$
Suppose that $ar\equiv b$ (mod $m$) and $br\equiv a$ (mod $m$) for some $r$. Show that $a^2\equiv b^2$ (mod $m$).

I've tried quite a few different things and I'm unable to get rid of terms with $r$ in it (if that is even necessary) to show that $a^2=b^2=mz$ for some integer $z$.

I know that $ar-b=mx$ and $br-a=my$ for integers $x,y$. I tried using different mod properties like sum, difference, and product to see if it helped, but didn't have any luck. I ended up with $(a^2-b^2)(r^2-1)=m^2xz=m(mxy)$, but this is was no help and just made things more complicated. Any hints are greatly appreciated. I know there is something obvious that I haven't considered and my mind is just stuck in a certain mode of thinking and I can't escape it at the moment.

答: $$a^2\equiv abr\equiv bar\equiv b^2\pmod m.$$


问:
What's wrong in this Dedekind cut inverse?
I have been trying to understand why the additive inverse of a dedekind cut $\alpha$ has to be
$$\beta=\{p \in \mathbb{Q} : \exists r>0 \text{ with} -p-r \notin \alpha\}$$
For that, I decided to consider a simple example,

Let $\alpha=\{p \in \mathbb{Q} : p<2\}$

Then shouldn't the inverse be just $\beta=\{q \in \mathbb{Q}:q<-2\}$?

Point out if there is something wrong in my inverse. Also please tell how can i build from my example towards the general inverse of a dedekind cut.

答:
Not every Dedekind cut can be constructed in the form

$$ \alpha=\{p \in \mathbb{Q} : p < \theta\} $$

where $\theta$ is a known constant.
You might be able to do such a construction for $\theta \in \mathbb Q,$
but to construct the Dedekind cut of an irrational number this way
would be circular. (Where do you get $\theta$ from?)

The construction
$$\{p \in \mathbb{Q} : \exists r>0 \text{ with} -p-r \notin \alpha\}$$
works for any Dedekind cut $\alpha,$ not just for the ones you can construct by writing $ \alpha=\{p \in \mathbb{Q} : p < \theta\}. $


问:
Derivative function $\frac 1f$ in the sense of distributions
$\newcommand\interval{\mathopen]0,T\mathclose[}$Let $f \in W^{1,\infty }(\interval)$ and $f(x)>1$ a.e. $x\in \interval$.  Then $g = \frac 1f \in W^{1,\infty}(\interval)$. (g  is Lipschitz in $\interval$.)

I want to show it again by use derivative of g.

Is it reasonable to say that  $g'= \frac{-f '} {f^2}$ in $D'$?. If yes, I need a way to prove it

答:
For $\phi \in C^\infty_c,\phi \ge 0, \int\phi=1$.

$F_n=F\ast n \phi(nx)$ is $C^\infty$ and it converges to $F$ locally uniformly.

Since $F>1$ then $\frac{1}{F_n}$ converges to $\frac1{F}$ uniformly.

And $\frac{-F_n'}{F_n^2}$ converges to $\frac{-F'}{F^2}$ in $D'$.

ie. $\frac{-F'}{F^2}$ is the distributional derivative of $\frac1{F}$.

In $D'$: $\frac{-F'}{F^2}=\frac{-f'}{f^2},\frac1{F}=\frac1f$.