答:
Your are correct. If $\langle n,q\rangle\in X\setminus P$, then $q\ne\frac1n$. Let $U=\Bbb Q\setminus\left\{\frac1n\right\}$; then $\{n\}\times U$ is an open nbhd of $\langle n,q\rangle$ disjoint from $P$, so $P$ is closed in $X$. Not only is $P$ not open: it has empty interior. Let $p=\left\langle n,\frac1n\right\rangle\in P$; $p$ has a local base of open nbhds of the form

$$\{n\}\times\left(\left(p-\frac1n,p+\frac1n\right)\cap\Bbb Q\right)$$

for $n\in\Bbb Z^+$, and clearly none of these sets is contained in $P$.

Title: Checking whether set is closed or open

Body:
Let $X=\mathbb N × \mathbb Q$ with the subspace topology of the usual topology of $\mathbb R^2$. Let $P=\{(n,\frac {1}{n}): n\in \mathbb N\}$. Then check whether $P$ is closed or open or both.

I think P is closed but not open. Since P does not have a limit point outside P it is closed. If P is open then we must have some open set U in $\mathbb R^2$ such that $P=U\cap X$. Which is not possible as $U\cap X $ will contain infinitely many points other than $(n,\frac{1}{n})$. Am I right?


问:
If $\frac{\pi}{2}<x<\pi$ and $\sqrt{\frac{1+\sin x}{1-\sin x}}=k\sec x$, then k is?
Q: If $\frac{\pi}{2}<x<\pi$ and $\sqrt{\frac{1+\sin x}{1-\sin x}}=k\sec x$, then k is?

A: From $\sqrt{\frac{1+\sin x}{1-\sin x}}=k\sec x$,

$\implies \sqrt{\frac{(1+\sin x)^2}{1-\sin^2x}}=k\sec x$

$\implies \pm\frac{(1+\sin x)}{\cos x}=k\sec x$

From inequality $\frac{\pi}{2}<x<\pi$, we know that x lies in the second quadrant, thus $\cos x$ is negative.

$\therefore -\frac{1+\sin x}{\cos s}=k\sec x$

$\implies -(1+\sin x)=k$

How can I now extract $x$ from these information? The given answer is $k=-1$, without knowing x, I don't see any way of concluding $k$ from the given data.

Thank you for any help!

答:
Original answer

$$ -1 - k = \sin(x)$$

The range of sin(x) on this domain is $(\frac{\pi}{2},\pi)$:

$$ 0 \leq \sin(x) \leq 1$$

Sub the expression:

$$ 0 \leq -1 -k \leq 1$$

Or,

$$ 1 \leq -k \leq 2$$

Or,

$$ -1 \geq k \geq -2$$

So in the whole set above, only legal value is $k=-1$ [ This is actually wrong, indeed $k \in (-1,-2)$

Note

For keeping the square root as positive,

$$ \frac{1+ \sin x}{1- \sin x} \geq 0 $$

Or,

$$ 1+ \sin x \geq 0$$

$$ \sin(x) \geq -1$$

Sub exp for sin in terms of k

$$ -1 - k \geq -1 $$

Or,

$$ 0 \geq k$$

$(-1, \infty)$ is the domain

IF you were doing the problem correctly then you would take intersection of allowed domain of square root with the domain of x which is given because at some point it may turn out funct is undefined

After thinking about it somewhere without specifying 'x' basically they are saying two functions which match for the whole domain but I don't think there exists a scaling factor 'k' for which that is possible. I made a graph of it here

ContentLicense: CC BY-SA 4.0


答:
Reduce mod $101$ and by an application of Fermat's little theorem of the form $n^{100} \equiv 1\pmod {101}$ and the symmetric nature of the polynomial, the terms cancel like so: $-8n -4n^{2}-4n^{4}-2n^{5}-n^{10}-n^{20}+8n+4n^{2}+4n^{4}+2n^{5}+n^{10}+n^{20} \equiv 0 \pmod {101}$ so that $101 | p(n)$.

Reduce $\pmod {20}$ to see that we need to show that $20|
8{{n}^{101}}+4{{n}^{202}}+4{{n}^{404}}+2{{n}^{505}}+{{n}^{1010}}+{{n}^{2020}}$.

By Ross's hint, the right side is easily seen to be divisible by $4$.

All that is left is to show $5 | 8{{n}^{101}}+4{{n}^{202}}+4{{n}^{404}}+2{{n}^{505}}+{{n}^{1010}}+{{n}^{2020}}$

By another application of Fermat's little theorem, $n^4 \equiv 1 \pmod 5$, the right side becomes $8n +4n^{2}+4+2n+n^{2}+1 \equiv 10n +5n^{2}+5 \equiv 0 \pmod 5$.

Altogether, this means that $2020 | p(n)$ for all $n$.

Title: Show that 2020 divides $p\left( n \right)$ for every integer $n$

Body:
Consider:
$$p\left( n \right)=800n+400{{n}^{2}}+400{{n}^{4}}+200{{n}^{5}}+100{{n}^{10}}+100{{n}^{20}}+8{{n}^{101}}+4{{n}^{202}}+4{{n}^{404}}+2{{n}^{505}}+{{n}^{1010}}+{{n}^{2020}}$$
Show that: $2020$ divides $p\left( n \right)$ for every positive integer $n$?

Work: the statement is indeed true for $n=1,2..100$ , of course using a computer!!


问:
Vector 2 norm and infinity norm proof
So I've already proven why $\left\lVert x\right\rVert_2\geq \left\lVert x\right\rVert_\infty$. I'm having trouble proving that $\sqrt{m}{\left\lVert x\right\rVert_\infty}\geq \left\lVert x\right\rVert_2$.

I've tried looking at the individual elements, but I'm not getting anywhere. I think I may have to use Holder's inequality, but I'm not sure if that's applicable, or how I would use it.

How should I do this?

答:
I assume you are talking about norms on $\mathbb{R}^m$. Then, for $x\in\mathbb{R}^m$,
$$
\lVert x\rVert_2^2 = \sum_{i=1}^m x_i^2 \leq m\cdot\max_{1\leq i\leq m} x_i^2 = m(\max_{1\leq i\leq m} |x_i|)^2 = m\lVert x\rVert_\infty^2
$$
so that
$$
\lVert x\rVert_2 \leq  \sqrt{m}\lVert x\rVert_\infty
$$
as claimed.</p>

ContentLicense: CC BY-SA 3.0

Title: Vector 2 norm and infinity norm proof


问:
solving an equation from trigonometry
I am trying to solve the following trigonometric equation: $$\frac{\cot\theta+\csc\theta}{\tan\theta+\sec\theta}=\cot(\pi/4+\theta/2)\cot \theta/2$$ but unfortunately, after giving a lot of effort I am unable to solve the problem, I think I am missing some obvious steps so the problem is getting messed up but I wasn't able to point out my fault, I also tried to solve it by starting from RHS but I also failed in there.

I will be very much grateful if someone helps me.

答:
Let $c=\cot\dfrac\theta2$

Use Weierstrass substitution

$$\tan\theta=\cdots=\dfrac{2c}{c^2-1}$$

$$\sin\theta=\cdots=\dfrac{2c}{c^2+1}$$

$$\cos\theta=\cdots=\dfrac{c^2-1}{c^2+1}$$

If $c\ne0,$

LHS  $=\dfrac{\dfrac{c^2-1}{2c}+\dfrac{c^2+1}{2c}}{\dfrac{2c}{c^2-1}+\dfrac{c^2+1}{c^2-1}}=\dfrac{c(c^2-1)}{(c+1)^2}=c\cdot\dfrac{c-1}{c+1}$ if $c+1\ne0$

Finally use Prove that $\cot(A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B}$

Title: solving an equation from trigonometry


问:
Mathematical induction, sum of sines

Prove that
$$\sin\theta+\sin2\theta+\sin3\theta+\cdots+\sin n\theta=\frac{\sin\frac{(n+1)\theta}2\sin\frac{n\theta}2}{\sin\frac\theta2},n\ge1,n\in\mathbb Z.$$

I have solved the base case $n = 1$ and prepared the assumption $n = k$. After I substitute the assumption into the induction step at $n = k+1$ I get lost, as I don't know how to successfully manipulate the numerator into the RHS. Any hints?

答:
BIG HINT

(Warning: I've written the entire inductive step, just cryptically)

$$\sin{0.5(n+2)}\theta\sin{0.5(n+1)\theta}-\sin{0.5(n+1)}\theta\sin{0.5n\theta}$$

$$=\sin{0.5(n+1)\theta}(\sin{0.5(n+2)}\theta-\sin{0.5n\theta})$$

$$=2\sin{0.5(n+1)\theta}\cos{\frac{0.5(n+2)\theta+0.5n\theta}{2}}\sin{\frac{0.5(n+2)\theta-0.5n\theta}{2}}$$

$$=2\sin{0.5(n+1)\theta}\cos{0.5(n+1)\theta}\sin{0.5\theta}$$

$$=\sin(n+1)\theta \sin{0.5\theta}$$

Title: Mathematical induction, sum of sines


问:
Integrating over an embedded manifold: Jacobian factor?
Let's say I want to integrate a function

$$
f(x,y),\quad x\in\Gamma_1,y\in\Gamma_2
$$ where $\Gamma_1,\Gamma_2$ are both embedded manifolds in $\Bbb{R}^3$.  The dimension of $\Gamma_1$ is 1 (a smooth curve, say), while the dimension of $\Gamma_2$ is 2 (a plane or surface).  Formally, what I want to compute is:

$$
\int_{\Gamma_1}\int_{\Gamma_2}f(x,y)dxdy
$$

If I choose (global) parametrizations of $\Gamma_1$ and $\Gamma_2$, what does the change of variables Jacobian factor look like?  The matrix of the transformation will not be square (it will be 6x3!)

I think I know the answer, based on this very useful set of lecture notes: if $\Phi:\Bbb{R}^3\rightarrow\Bbb{R}^3\times\Bbb{R}^3$ is my parametrization, the answer should be

$$
\int_{\Gamma_1}\int_{\Gamma_2}f(x,y)dxdy = \int_{\phi_1}^{\phi_2}\int_{\beta_1}^{\beta_2}\int_{\alpha_1}^{\alpha_2}f(\Phi(\alpha,\beta,\phi))\sqrt{\det (J^T_\Phi J_\Phi)}d\alpha d\beta d\phi
$$

Is this Jacobian factor correct?  Does anyone have a book reference for doing this?  The part I haven't seen before is the non-square Jacobian.  I must have somehow missed this in my calculus courses oh-so long ago.

答:
This is called the area formula. Its most general form (except for easy upgrade to manifolds) is as follows.

Theorem (Area Formula) Let $n \leq m$ and $f:\mathbb{R}^n \to \mathbb{R}^m $ be Lipschitz. Then for any Lebesgue measurable subset $A \subset \mathbb{R}^n$, we have
$$
\int_A J_f(x) \, d\mathcal{L}^n(x) = \int_{\mathbb{R}^m} N(f,A,y) \, d\mathcal{H}^n(y) \ ,
$$
where the multiplicity function $N(f,A,y)$ is the number (possibly infinite) of points from $A$ that are mapped to $y$, and
$$
J_f(X) = \sqrt{\det  \left( Df(x)^T \cdot Df(x) \right)} \, .
$$
What this Jacobian does is to see the derivative $Df(x)$ as a linear map from $\mathbb{R}^n$ to the tangent space to the manifold $f(A)$ at $f(x)$, which is a copy of $\mathbb{R}^n$, and ignore the complement to the tangent space. Now you have a linear map between equal dimensional spaces and its determinant is well-defined. It is then left to prove that this equals the one given above.

For proofs of the theorem in this form see Evans and Gariepy: Measure theory and Fine properties of functions

Many versions of the theorem appear in different contexts. Usually $f$ is assumed to be $C^1$ if one does not wish to go through Rademacher's differentiability theorem for Lipschitz maps, and for simplicity $f$ is assumed to be one-to-one, so $N(f,A,y)=1$ (i.e. it disappears).


问:
Prove a function is continuous on the union of closed sets if it is continuous on each set.
I have a function $f: (A,d) \to (B,s)$ be a function between two metric spaces. If $$A=\bigcup_{i=1}^n C_i$$ where $C_i$ are closed sets in $A$ and $f$ is continuous on each $C_i$.

Prove $f$ is continuous on $A$ and give an example where $f$ can fail to be continuous on all $A$ if instead I had a countably infinite union of closed sets $$A=\bigcup_{i=1}^{\infty }  C_i$$

So I take the open sets $O_i$ in $B$ then $f^{-1} (O_i)$ is continuous in $C_i$ so
$$f^{-1} \left( \bigcup_{i=1}^n O_i \right)$$ gives me $$\bigcup_{i=1}^n f^{-1} (O_i)$$ which is open in $A$ as union of open sets.

Is this correct? I think I am missing something.

Also I wasn't able to find an example where $f$ can fail to be continuous on all $A$ if I had a countably infinite union of closed sets $$A=\bigcup_{i=1}^{\infty } $$
Can you give an example?

答: In a metric space $X,$ every singleton (i.e. $\{x\}$ for $x\in X$) is closed, and any function with domain $\{x\}$ is continuous. Suppose $A=B=\{1/n:n\in \Bbb N\}\cup \{0\}$ with $d(u,v)=s(u,v)=|u-v|.$ Let $C_1=\{0\}$ and $C_j=\{1/(j-1)\}$ for $j>1.$ Let $f:A\to B$ be discontinuous at $0.$ E.g. $f(1/n)=0$ and $f(0)=1.$


问:
An integer valued function that generate repeating integer for n amount of times in algebraic expression
Recently I've made this function $f(x)=\frac{x+\frac{(-1)^x-1}{2}}{2}$ or $f(x)=\frac{2x+(-1)^x-1}{4}$ where $x$ is an integer. And it generate integers repeated two times, such as: $...,-3,-3,-2,-2,-1,-1,0,0,1,1,2,2,3,3,...$

So the question is: Is there any function where it could repeat itself for $n$ number of times? Like 3 times, $...,-3,-3,-3,-2,-2,-2,-1,-1,-1,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,....$

Although knowing the function is $f(x)=\frac{2x+(-1)^x-1}{4}$ , I could derive a function that repeats the integer for $2^n$ times by putting the output of the function $f(x)$ into the function $f(x)$ again, or like $f(f(x))$ and $f(f(f(x)))$ and so on.

Edit: I forget to add that I only know middle school math, if there are any formula that only use algebraic expression that would be helpful.

答:
One relatively simple function using integer $x$ inputs to get each integer repeating itself $n$ times in the output is using the floor function to get

$$f(x) = \left\lfloor \frac{x}{n} \right\rfloor \tag{1}\label{eq1A}$$

Alternatively, another function you can use is the ceiling function, with the output values shifted to the left $n - 1$ positions compared to the results of \eqref{eq1A}, to instead get

$$f(x) = \left\lceil \frac{x}{n} \right\rceil \tag{2}\label{eq2A}$$

A somewhat convoluted method is to use a piece-wise function, e.g., to get the same output as \eqref{eq1A}, use

$$f(x) =
\begin{cases}
m, & \text{if } x = mn, \; m \in \mathbb{Z} \\
m, & \text{if } x = mn + 1, \; m \in \mathbb{Z} \\
m, & \text{if } x = mn + 2, \; m \in \mathbb{Z} \\
\; \vdots & \; \; \; \; \; \; \; \; \; \; \; \; \; \vdots \\
m, & \text{if } x = mn + n - 1, \; m \in \mathbb{Z}
\end{cases}
\tag{3}\label{eq3A}$$


答:
The following function should do the job:$$f(x) = \left\lfloor \frac x3 \right\rfloor $$
where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.

(Also, see floor and ceiling functions for more information)

Note that, your function could also be written in fancy way as
$$f(x)=\frac{2x+(-1)^x-1}{4} = \left\lfloor \frac x2 \right\rfloor$$
since when $x$ is even the numerator is divisible by $4$, and when $x$ is odd the numerator gives remainder of $2$ when divided by $4$ so you subtract $2$ to make it divisible by $4$

Therefore, this function can easily be generalized by just adjusting the denominator of the floor function.

Title: An integer valued function that generate repeating integer for n amount of times in algebraic expression

Body:
Recently I've made this function $f(x)=\frac{x+\frac{(-1)^x-1}{2}}{2}$ or $f(x)=\frac{2x+(-1)^x-1}{4}$ where $x$ is an integer. And it generate integers repeated two times, such as: $...,-3,-3,-2,-2,-1,-1,0,0,1,1,2,2,3,3,...$

So the question is: Is there any function where it could repeat itself for $n$ number of times? Like 3 times, $...,-3,-3,-3,-2,-2,-2,-1,-1,-1,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,....$

Although knowing the function is $f(x)=\frac{2x+(-1)^x-1}{4}$ , I could derive a function that repeats the integer for $2^n$ times by putting the output of the function $f(x)$ into the function $f(x)$ again, or like $f(f(x))$ and $f(f(f(x)))$ and so on.

Edit: I forget to add that I only know middle school math, if there are any formula that only use algebraic expression that would be helpful.


问:
Probablistic approach
I have been reading about probabilistic approach in some problems in particular when we want to prove that something exists without explicitly constructing it. I really want to see more of this. Does anyone know a good book/paper/any kind of material where I could find different examples with this probabilistic method?

Thank you very much for your time!

答: The Probabilistic Method, by Alon and Spencer, would be the reference you're looking for.

Title: Probablistic approach


问:
Incorrect Definitions of NP
I was solving problems related to P and NP where I encountered the following problem:
Given a standard definition of NP,

if x belongs to L then there exists y such that |y| <= |x|^d and A(x, y) = 1;

if x does not belong to L then for every y with |y| <= |x|^d we have A(x, y) = 0.

what is the new class formed when we don't include the second statement?(x belongs to L)

what is the new class formed when we don't include the first statement? (x does not belong to L)

I am well versed with the definitions of P and NP but unable to figure out how to determine and prove  these new classes.

答:
If we don't include the second statement, then there is no restriction on what we do when x is not in $L$, so consider the program:

```<code>A(x,y):
  return 1
</code>```

And this shows that any language $L$ is in our new language class. A similar construction works for 2. See if you can figure it out now!

Title: Incorrect Definitions of NP

Body:
If we don't include the second statement, then there is no restriction on what we do when x is not in $L$, so consider the program:

A(x,y):
  return 1

And this shows that any language $L$ is in our new language class. A similar construction works for 2. See if you can figure it out now!


问:
Probability of winning a game of craps
The dice game craps is played as follows: The player throws 2 dice, and if the sum is 7 or 11, he/she wins. If the sum is 2, 3, or 12, he/she loses. If the sum is anything else, then he/she continues throwing until that number appears again, or throws a 7, where the game ends in a loss.

What I do know is that $P(7) = 6/36$, $P(11)=2/36$. So P(winning) in first roll is $8/36$. Furthermore, the probability of having to roll again will be $1-[P($winning 1st roll$) +P($losing 1st roll$)]=24/36$. It's what happens if the game doesn't end in the 1st roll that's got me a bit confused, since it could go on and indeterminate number of rolls. But since it can't be 7,11,2,3,or 12, it depends on if they roll a 4,5,6,8,9, or 10.

Note, this is not a homework problem, but an intriguing one I found in a different book "Probability Models, Sheldon Ross"

答:
Others have covered the same solution, but here's another method using infinite sum.

Preliminaries:
\begin{array}{|c|c|c|c|c|c|c|}
\hline
Sample \in& \{2,12\} & \{3,11\}& \{4,10\}& \{5,9\}& \{6,8\}& \{7\}\\ \hline
Probabilities&\frac{1}{36} &\frac{1}{18} &\frac{1}{12}&\frac{1}{9}&\frac{5}{36}&\frac{1}{6}\\ \hline
\end{array}

Let the RV indicating the $nth$ throw be $T_n$. Let $W$ be the event of winning
\begin{align}
\begin{split}
P(W) &= \quad P(W|T_1\in \{7,11\})P(T_1\in\{7,11\})\\&\quad+P(W|T_1\in \{4,10\})P(T_1\in\{4,10\})\\&\quad+P(W|T_1\in \{5,9\})P(T_1\in\{5,9\})\\&\quad+P(W|T_1\in \{6,8\})P(T_1\in\{6,8\})\\&\quad+P(W|T_1\in \{2,3,12\})P(T_1\in\{2,3,12\})
\end{split}\qquad\text{(Eqn.1)}
\end{align}
Calculating the individual conditional probabilities
\begin{align}
P(W^C|T_1\in \{4\})&=P(T_2=7)+P(T_2\notin\{4,7\})P(T_3=7)\\&\quad+P(T_3\notin\{4,7\},T_2\notin\{4,7\})P(T_4=7)\\&\quad+P(T_4\notin\{4,7\},T_3\notin\{4,7\},T_2\notin\{4,7\})P(T_5=7)\\&\quad+\dots\\\\
\end{align}
Since $T_i$ are independent $\forall i$
\begin{align}
P(W^C|T_1\in \{4\})&=P(T_i=7)(1+P(T_i\notin\{4,7\})+P(T_i\notin\{4,7\})^2+\dots)\\
&=\dfrac{P(T_i=7)}{1-P(T_i\notin\{4,7\})}\\
&=\dfrac{\frac{1}{6}}{1-\frac{36-9}{36}} = \dfrac{2}{3}\\
\end{align}

Taking the complement,
$P(W|T_1\in \{4\})=1-\frac{2}{3} = \frac{1}{3}$ and similarly, $P(W|T_1\in \{10\})=\frac{1}{3}$

We now calculate other quantities
\begin{equation}
P(W^C|T_1\in \{5\})=\dfrac{P(T_i=7)}{1-P(T_i\notin\{5,7\})}=\dfrac{\frac{1}{6}}{\frac{10}{36}}=\dfrac{3}{5}
\end{equation}

Then $P(W|T_1\in \{5\}) = \dfrac{2}{5}$ and $P(W|T_1\in \{9\}) = \dfrac{2}{5}$

And the same way
\begin{equation}
P(W^C|T_1\in \{6\})=\dfrac{P(T_i=7)}{1-P(T_i\notin\{6,7\})}=\dfrac{\frac{1}{6}}{\frac{11}{36}}=\dfrac{6}{11}
\end{equation}

Then $P(W|T_1\in \{6\}) = P(W|T_1\in \{8\}) = \dfrac{5}{11}$. Finally we have $P(W|T_1\in \{2,3,12\})=0$ by the rules of the game.

Substituting all this in Eqn.1 we have
\begin{align}
P(W) &= 1\times \frac{8}{36} + \frac{1}{3} \times \frac{2}{12}+ \frac{2}{5} \times \frac{2}{9}+ \frac{5}{11}\times \frac{10}{36}+ 0\times \frac{4}{36}\\
&= \frac{976}{1980}
\end{align}

Title: Probability of winning a game of craps


问:
Basic definition of Inverse Limit in sheaf theory / schemes
I read the book "Algebraic Geometry" by U. Görtz and whenever limits are involved I struggle for an understanding. The application of limits is mostly very basic, though; but I'm new to the concept of limits.

My example (page 60 in the book): Let $A$ be an integral domain. The structure sheaf $O_X$ on $X = \text{Spec}A$ is given by $O_X(D(f)) = A_f$ ($f\in A$) and for any $U\subseteq X$ by

\begin{align}
O_X(U) &= \varprojlim_{D(f)\subseteq U} O_X(D(f)) \\
&:= \{ (s_{D(f)})_{D(f)\subseteq U} \in \prod_{D(f)\subseteq U} O_X(D(f)) \mid  \text{for all } D(g) \subseteq D(f) \subseteq U: s_{D(f)\big|D(g)} = s_{D(g)}\} \\
&= \bigcap_{D(f)\subseteq U} A_f.
\end{align}

I simply don't understand the last equality: In my naive understanding the elements of the last set are "fractions" and the elements of the Inverse Limit are "families of fractions".

Any hint is appreciated.

答: My personal advice is to study a bit of category theory: it will let you understand all this stuff in a very clearer way. In fact you can easily realize that the first equality is not a definition, but a way to express a limit of an arbitrary presheaf, while the second is an isomorphism, not exactly an equality, given by the universal property defining limits. I started with Hartshorne, but without category theory as a background it's just like wandering in the dark without even a candle with you.</p>


问:
Fitting a dodecahedron inside a cube
I'm afraid I'm not fantastic at maths and am struggling with a problem.
I am a woodworker and have been asked to cut a solid dodecahedron from a 3 inch cube of wood. I am struggling to figure out what the maximum sized dodecahedron I could make is.

Is there a formula for calculating this?

答:
The most "natural" dodecahedron positioning is by placing two of its vertices on each of the $6$ faces of the cube, as follows:

(In the above image, I've bolded the edges which are tangent to the faces of the cube. Each of them passes through the center of their face.)

This is probably the way that I would try to cut a dodecahedron out of a cube, because every face in this dodecahedron is at a right angle to a face of the cube. This means that you can make easy vertical cuts to produce these faces by setting the cube flat on one of its faces.

However, it's not the largest dodecahedron possible:

This image, from Moritz Firsching's 2018 paper Computing maximal copies of polytopes contained in a polytope (PDF), depicts a maximal dodecahedron inside a cube. It has a side length of around $0.3943$ times that of the cube, compared to the original diagram's $\phi^{-2}\approx 0.382$.

Given the marginal improvement for what I assume is a much more difficult woodworking experience, I wouldn't particularly recommend it.

For an approach starting with a more cylindrical piece of wood (namely, a tree stump), see this IRL guide.

Title: Fitting a dodecahedron inside a cube


问:
Fixed point of $f:[0, 4] \to [1,3]$ such that $f'(x) \neq 1$
Let $ f:[0, 4]\to [1,3]$ be a differentiable function such that $f'(x)\neq 1$ for all $ x \in [0,4]$ Then the function $f$ has

Atmost one fixed point

Unique fixed point

No fixed point

More than one fixed point

As far I guess the function is a contraction mapping cause the range is smaller than the domain. Hence has unique fixed point. But I am unable to prove and unable to give more logical description. Is my thinking totally wrong?
Thanks in advance.

答:
First note that $f(x)-x$ is positive at $x=0$ and negative at $x=3$ and hence, by the intermediate value property it has at least a fixed point.

Next, by Darboux Theorem, $f'$ satisfies the Intermediate Value Property. Since $f'(x) \neq 1$ for all $x \in [0,4]$ you have either

$f'(x) <1$ for all $x \in [0,4]$
or

$f'(x) >1$ for all $x \in [0,4]$

Show that the second is not possible, and hence $f'(x) <1$ for all $x \in [0,4]$.

Now, you can prove that the fixed point is unique. Indeed, if $a,b$ are fixed points, apply the Mean Value Theorem on $[a,b]$.


问:
Number of ways to arrange $n$ people in a line
I came across this confusing question in Combinatorics.

Given $n \in \mathbb N$. We have $n$ people that are sitting in a row. We mark $a_n$ as the number of ways to rearrange them such that a person can stay in his seat or move one seat to the right or one seat to the left. Calculate $a_n$

This is one of the hardest combinatorics questions I have came across (I'm still mid-way through the course,) so if anyone can give me a direction I'll be grateful!

答:
HINT: It’s a little more convenient to think of this in terms of permutations: $a_n$ is the number of permutations $p_1 p_2\dots p_n$ of $[n]=\{1,\dots,n\}$ such that $|p_k-k|\le 1$ for each $k\in[n]$. Call such permutations good.

Suppose that $p_1 p_2\dots p_n$ is such a permutation. Clearly one of $p_{n-1}$ and $p_n$ must be $n$.

If $p_n=n$, then $p_1 p_2\dots p_{n-1}$ is a good permutation of $[n-1]$. Moreover, any good permutation of $[n-1]$ can be extended to a good permutation of $[n]$ with $n$ at the end. Thus, there are $a_{n-1}$ good permutations of $[n]$ with $n$ as last element.

If $p_{n-1}=n$, then $p_n$ must be $n-1$, so $p_1 p_2\dots p_{n-2}p_n$ is a good permutation of $[n-1]$ with $n-1$ as its last element. Conversely, if $q_1 q_2\dots q_{n-1}$ is a good permutation of $[n-1]$ with $q_{n-1}=n-1$, then $q_1 q_2\dots q_{n-2}nq_{n-1}$ is a good permutation of $[n]$ with $n$ as second-last element. How many good permutations of $[n-1]$ are there with $n-1$ as last element?

Alternatively, you could simply work out $a_k$ by hand for $k=0,\dots,4$, say, and see what turns up; you should recognize the numbers that you get.


答: Here is what I have done so far. I cant think of a way to manipulate the LHS into the RHS

Title: Mathematical induction, sum of sines

Body:
Prove that
$$\sin\theta+\sin2\theta+\sin3\theta+\cdots+\sin n\theta=\frac{\sin\frac{(n+1)\theta}2\sin\frac{n\theta}2}{\sin\frac\theta2},n\ge1,n\in\mathbb Z.$$

I have solved the base case $n = 1$ and prepared the assumption $n = k$. After I substitute the assumption into the induction step at $n = k+1$ I get lost, as I don't know how to successfully manipulate the numerator into the RHS. Any hints?


问:
What is the laplace transform of $(\delta(t−π/6))u(t-π/6)$
I know that  $L(f(t-a).u(t-a)) = e^{-as}.F(s)$

Does that mean the laplace transform of the above equation is $e^{-2πs}$?

Edited: sorry i meant $(\delta(t−π/6))u(t-π/6)$ not $(\delta(t−π/6))u(t-π)$ missed the 6 out when i initially asked the question.

答:
It is equal to $0$ since
$$
f(t)\delta(t-t_0)=f(t_0)\delta(t-t_0)
$$
and
$$
u({\pi\over 6}-\pi)=0
$$

Also the statement you wrote is wrong. We have
$$
L(f(t-a))=e^{-as}F(s)
$$
as long as the ROCs are treated carefully.


问:
question about the closures of a set
A subset $A$ (resp. subspace $A$) of space $  (X,\tau)$ has $\alpha$-property if  every $\tau$-open (resp.$\tau_{A}$-open) cover of $A$ has a finite subfamily whose $\tau$-closures (resp. $\tau_{A}$-closures) cover $A$.

If subspace $A$ has $\alpha$-property of topological space $(X, \tau)$, does closures $A$ have $\alpha$-property as a subspace?

答:
Notation. Let $A \subseteq X$ be any subset. For convenience, we denote the $\tau_{A}$-closure of a subset  $U$ of $A$ as $\mathrm{cl}_{A}(U)$, and the $\tau_{X}$-closure of $A$ as $\overline{A}$.

Little exercises for you.

(a) $\mathrm{cl}_{A}(U)= \overline{U} \cap A$. (Hint: Use definitions of subspace topology and closure.)

(b) $\overline{U_{1}\cup U_{2}} =\overline{U_{1}}\cup \overline{U_{2}} .$

With these preparations, we are able to prove the proposition. Suppose $A$ has $\alpha$-property. For the purpose of showing that $\overline{A}$ also has $\alpha$-property, we take any $\tau_{\overline{A}}$-open cover $\{U_{i} \cap \overline{A}\}_{i \in I}$ of $\overline{A}$. Needless to say, $\{U_{i} \cap A\}_{i \in I}$ is a $\tau_{A}$-open cover of $A$. Since $A$ has $\alpha$-property, there exists a finitely many indices $i_{1},i_{2},\dots,i_{n}$ such that $\bigcup_{k=1}^{n} \mathrm{cl}_{A}(U_{i_{k}}\cap A) = A$. According to exercises above, this means $$\bigcup_{k=1}^{n}\overline{U_{i_{k}} \cap A} \supseteq A.$$
Notice the set on the left side, as a finite union of closed subsets, is closed. We actually have $$\bigcup_{k=1}^{n}\overline{U_{i_{k}} \cap A} \supseteq \overline{A}.$$
Moreover this implies $$\bigcup_{k=1}^{n}\overline{U_{i_{k}} \cap \overline{A}} \supseteq \overline{A}.$$

The upshot is that now we obtain $\bigcup_{k=1}^{n} \mathrm{cl}_{\overline{A}}(U_{i_{k}} \cap \overline{A}) = \overline{A}$, by exercise (a) again, which means $\overline{A}$ also has $\alpha$-property.

Title: question about the closures of a set


问:
Behaviour of orthogonal matrices
I am given that A is an orthogonal matrix of order $n$, and $u, v$ are Vectors in the $R^n $ space.

I need to prove that $||u|| = ||Au||$. The first step of the solution hint I am given is that $$||Au||^2 = (Au)^T(Au)$$. Why is this so? I know that $A^{-1} = A^T$ in the definition of an orthogonal matrix, but how does this contribute to the above statement? Or is there some other property I'm missing out on?

Title: Behaviour of orthogonal matrices

Body: If $u=(u_1,u_2,\ldots,u_n)$, then\begin{align}u^\top u&=\begin{pmatrix}u_1\\u_2\\\vdots\\u_n\end{pmatrix}(u_1,u_2,\ldots,u_n)\\&=u_1^{\,2}+u_2^{\,2}+\cdots+u_n^{\,2}\\&=\|u\|^2.\end{align}


问:
Why do the French count so strangely?
Today I've heard a talk about division rules. The lecturer stated that base 12 has a lot of division rules and was therefore commonly used in trade.

English and German name their numbers like they count (with 11 and 12 as exception), but not French:

```<code>  # |   English |           German | French
-----------------------------------------------
  0 |      zero |            null  | zero
  1 |       one |            eins  | un
  2 |       two |            zwei  | deux
  3 |     three |            drei  | trois
  4 |      four |            vier  | quatre
  5 |      five |            fünf  | cinq
  6 |       six |           sechs  | six
  7 |     seven |          sieben  | sept
  8 |     eight |            acht  | huit
  9 |      nine |            neun  | neuf
 10 |       ten |            zehn  | dix
 11 |    eleven |             elf  | onze
 12 |    twelve |           zwölf  | douze
 13 | thir|teen |       drei|zehn  | treize
 14 | four|teen |       vier|zehn  | quatorze
 15 |  fif|teen |       fünf|zehn  | quinze
 16 |  six|teen |       sech|zehn  | seize
 17 |seven|teen |       sieb|zehn  | dix-sept
            18 and 19 are "regular"
 20 |    twenty |          zwanzig | vingt
 21 |twenty-one |  ein|und|zwanzig | vingt et un
 22 |twenty-two | zwei|und|zwanzig | vingt-deux
            23 - 69 are "regular"
 70 |  seven|ty |         sieb|zig | soixante-dix = 60 + 10
              ....
 80 |   eigh|ty |         acht|zig | quatre-vingts = 4*20 ?!?!
 81 |eighty-one |  ein|und|achtzig | quatre-vingt-un = 4*20 + 1
      ...
</code>```

So my question is:

Why do French count so strangely after 79?

(Are there other languages that count similar? What's the historic / mathematical reason for this system?)

Related Questions

Why is a full turn of the circle 360°?

What could be better than base 10?

答:
This is a bit of a guess but I think this has a lot to do with whether the numbering system was rationalised at any point.

If you tell someone from England that you're having trouble sleeping he'll probably suggest you count sheep. I won't attempt a weak joke about having a bedroom full of farm animals. The phrase comes from an old numbering system

```<code> 1  |  yan
 2  |  tan
 3  |  tether
 4  |  mether
 5  |  pip
 6  |  nether
 7  |  aether
 8  |  oevro
 9  |  cuevro
10  |  dick
11  |  yanadick
12  |  tanadick
13  |  tetherdick
14  |  methradick
15  |  bumfit
16  |  yanabum
17  |  tanabum
18  |  tetherbum
19  |  metherbum
20  |  jiggit
</code>```

This died out a long time ago but shepherds kept it up for longer. So to count sheep is to   If you look on Wikipedia there are loads of different ones people counted differently in different areas of the country. (It seems the one my great-granddad taught me isn't in there.)

At some point this was changed by introducing a new numbering system.  Similarly The swedes and Norwegians use a base 10 numbering system today, but they used to use the Danish system. Similarly German (I think) has undergone several standardizations to keep it as one language. It wouldn't surprise me if there used to be less rational German counting systems that died out.

If the French tried to rationalise their numbering is would have been during the revolution. But as they tried to rationalise everything, including a ten hour day and a decimal calender (Today is Carp the 25th of Flower) A lot of things didn't take off. So I would guess the numbers were one of the things that failed.

Title: Why do the French count so strangely?

Body:
This is a bit of a guess but I think this has a lot to do with whether the numbering system was rationalised at any point.

If you tell someone from England that you're having trouble sleeping he'll probably suggest you count sheep. I won't attempt a weak joke about having a bedroom full of farm animals. The phrase comes from an old numbering system

 1  |  yan
 2  |  tan
 3  |  tether
 4  |  mether
 5  |  pip
 6  |  nether
 7  |  aether
 8  |  oevro
 9  |  cuevro
10  |  dick
11  |  yanadick
12  |  tanadick
13  |  tetherdick
14  |  methradick
15  |  bumfit
16  |  yanabum
17  |  tanabum
18  |  tetherbum
19  |  metherbum
20  |  jiggit

This died out a long time ago but shepherds kept it up for longer. So to count sheep is to   If you look on Wikipedia there are loads of different ones people counted differently in different areas of the country. (It seems the one my great-granddad taught me isn't in there.)

At some point this was changed by introducing a new numbering system.  Similarly The swedes and Norwegians use a base 10 numbering system today, but they used to use the Danish system. Similarly German (I think) has undergone several standardizations to keep it as one language. It wouldn't surprise me if there used to be less rational German counting systems that died out.

If the French tried to rationalise their numbering is would have been during the revolution. But as they tried to rationalise everything, including a ten hour day and a decimal calender (Today is Carp the 25th of Flower) A lot of things didn't take off. So I would guess the numbers were one of the things that failed.


问:
How can the product of $\sqrt{x}$ be negative?
I've recently implementing my own raytracer in C.

I was reading up some paper on ray-sphere intersection from http://www.cs.unc.edu/~rademach/xroads-RT/RTarticle.html and this came up:

$d = \sqrt{r^2 - (c^2 - v^2)}$

To determine whether an intersection occurs, we compute the value of $d$. If $d >= 0$, then a valid intersection occurs. If the ray does not intersect, then d will be less than zero.

How can the value of $d$ be ever negative from  a square root?

I assume that the paper meant is:

If $r^2 - (c^2 - v^2) >= 0$, then a valid intersection occurs. If the ray does not intersect, then $r^2 - (c^2 - v^2)$ be less than zero.

But I'm not sure with this assumption as I'm still new in math.

答:
Yes indeed, they did mean what you suggest. You can see this in the pseudocode that immediately follows:

```<code>v = dot_product( EO, V )
disc = r2 - ((dot_product( EO, EO ) - v2 )
if ( disc < 0 )
    no intersection
else
    d = sqrt( disc )
    P = E + (v - d) * V
</code>```

You can convince yourself that this is correct in the following way. Look at the given diagram:

The ray doesn't intersect with the circle precisely when $b$ is greater than $r$. In this case, we see that $d^2 = r^2-b^2 \leq 0$, so that must be the right condition.

Body:
Yes indeed, they did mean what you suggest. You can see this in the pseudocode that immediately follows:

v = dot_product( EO, V )
disc = r2 - ((dot_product( EO, EO ) - v2 )
if ( disc < 0 )
    no intersection
else
    d = sqrt( disc )
    P = E + (v - d) * V

You can convince yourself that this is correct in the following way. Look at the given diagram:

The ray doesn't intersect with the circle precisely when $b$ is greater than $r$. In this case, we see that $d^2 = r^2-b^2 \leq 0$, so that must be the right condition.


问:
Is $\frac{e^z-1}{e^z+1}$ analytic?
Is the function $\frac{e^z-1}{e^z+1}$ analytic?

I tried using $\mathop{\lim}\limits_{z\to a} \frac{f(z)-f(a)}{z-a}$, but I didnt really know what to use for $a$.

答: Yes, if $G$ is the set where the denominator is nonzero, then $f(z)=\frac{e^z-1}{e^z+1}$ is analytic on $G$.  This is true because $e^z$ is analytic, constants are analytic, sums of analytic functions are analytic, and quotients of analytic functions are analytic away from zeros of the denominators (and the usual quotient rule for differentiation holds).  The set where the denominator is $0$ is $\{(2n+1)\pi i:n\in\mathbb{Z}\}$, and $f$ has a simple pole at each point in this set.</p>

ContentLicense: CC BY-SA 2.5

Title: Is $\frac{e^z-1}{e^z+1}$ analytic?


问:
A basic doubt on the definition of a Poisson random variable
What is the significance of "large city" in the definition of the following Poisson variable :

"Number of phone calls placed during a ten second interval in a large city"

I guess either $n \to \infty$ or $p$ very small is not implied by "large city".

答:
Consider the binomial random variable:
$P(X=i)=\frac{n!}{(n-i)!i!}p^i(1-p)^{n-i}$

If you let $n\rightarrow \infty$ and $p \rightarrow 0$ for fix $np$ and consider $p=\frac{\mu}{n}$

Use the following identities:

$\lim_{n \to \infty} (1-\frac{\mu}{n})^n= e^{-\mu}$ and $\frac{n!}{(n-i)!} \approx n^i$ and you arrive at the Poisson random variable:

$P(X=i)=\frac{e^{-\mu}\mu^{i}}{i!}$

Thus, to model a phenomenon with Poisson distribution its important that the population is large ($n\rightarrow \infty$) and events are rare ($p \rightarrow 0$).


问:
Dividing a square into $n$ equal size parts with minimal fence
Let's divide a square into $n$ parts with equal area. How to do this using minimum amount of fence? (I assume the dividing should be done with line segments(?) so the meaning of fence length is sum of their lengths).

Here is an example with $n=5$. I don't know if it is minimal though:

答:
A minimal configuration will satisfy the two-dimensional version of Plateau’s laws: each fence will be straight line or circular arc, and on either end, it will meet two others at 120° angles, or the boundary at a 90° angle.

Your configuration (left) has fences that don’t meet at 120°, so it can’t be minimal.  Its length is $2 + \frac{4}{\sqrt{5}} - \frac{4}{\sqrt{10}} \approx 2.52394$.  By rounding out the inside square enough to make the fences meet at 120° (right), we decrease the length to $2 + 2 \sqrt{\frac{\pi + 3 - 3\sqrt{3}}{15}} \approx 2.50211$.


问:
Galois group of an octic over $\mathbb{Q}(i)$
Let $K=\mathbb{Q}(i)$ and $L=K(\beta)$ where $\beta$ is a root of $f(x)=27x^8+72x^4-16$. This problem comes from Silverman and Tate; $\beta$ is the $y$-coordinate of one of the 3-torsion points of $y^2=x^3+x$. Let
\begin{equation*}
\beta = \sqrt[4]{\frac{8\sqrt{3}-12}{9}},
\end{equation*}
and if we set
\begin{equation*}
\alpha = \sqrt{\frac{2\sqrt{3}-3}{3}} = \frac{\beta^2\sqrt{3}}{2},\quad
\alpha' = -\frac{i}{\alpha\sqrt{3}},\quad \beta' = \sqrt[4]{\frac{-8\sqrt{3}-12}{9}},
\end{equation*}
then the roots of $f(x)$ are $\pm\beta, \pm i\beta, \pm\beta', \pm i\beta'$, and the 3-torsion points of $y^2=x^3+x$ are (together with the point at infinity)
\begin{equation*}
(\alpha, \pm\beta),\ (-\alpha, \pm i\beta),\ (\alpha', \pm \beta'),\ (-\alpha', \pm i\beta').
\end{equation*}
I am trying to understand the Galois group of $L/K$ ($L$ is Galois over $K$). My first attempt was to show that the group is cyclic, with generator $\beta\mapsto \beta'$. However, $\beta\mapsto\beta'$ also maps $\alpha$ to $-\alpha'$, and $(-\alpha',\beta')$ is not a 3-torsion point of the curve. Thus this cannot be an automorphism of $L/K$, although I cannot see why it isn't. (Of note is that Sage seems to think that this is an automorphism of $L/K$.) The map $\beta\mapsto i\beta$ preserves the 3-torsion points, but has only order four.

So: 1) what is going on with $\beta\mapsto\beta'$? Is it an $L/K$-automorphism? If so, I must have done something wrong with my calculation of the image of $\alpha$ under that automorphism. And 2) if it is not, what is a second generator of the Galois group other than $\beta\mapsto i\beta$?

I'd appreciate not getting back a complete solution, as I'm really trying to understand this group in preparation for doing another exercise.

答:
Assuming that $\beta$ is real and positive, and $\beta'$ has argument $\pi/4$, then I don't see the problem. If $\sigma$ is an automorphism with the property $\sigma(\beta)=\beta'$ then:

$\sigma(\beta^4)=\beta'^4$ implying that $\sigma(\sqrt3)=-\sqrt3$, and hence

$\sigma(\alpha)=-\sqrt3\beta'^2/2$, a complex number with argument $-\pi/2$, agreeing with the argument of $\alpha'$.

This suggests to me that $\sigma(\alpha)=\alpha'$, just as you would expect from the list of 3-torsion points. I think this settles your first question, and seems to make the second question moot, but the structure of the relevant Galois groups is not clear to me yet.


问:
if $|f''(x)|\le |f'(x)|,\forall x\in R$ show that either $f'(x)>0$ or $f'(x)<0,\forall x\in R$
Edit:let $f:R\to R$ be twice differenriable on $R$, such that
$$|f''(x)|\le |f'(x)|,\forall  x\in R$$
show that either:
$f'(x)>0$
or $f'(x)<0,\forall x\in R$

since
$$0\ge (f''(x))^2-(f'(x))^2=(f''(x)-f'(x))(f''(x)+f'(x))$$
so
$$f"(x)+f'(x)\ge 0,\mbox{and} f''(x)-f'(x)\le 0$$
or
$$f"(x)+f'(x)\le 0,\mbox{and} f''(x)-f'(x)\ge 0$$

答: let $f(x)=-2x$. One has  $f'(x) = -2$ and $f''(x) = 0$ for all $x$. We see that $|f''(x)|\le |f'(x)|$ for all $x$ but $f'(x) = -2<0$ for all $x$.


问:
Is there a difference between 'inconsistent', 'contrary', and 'contradictory'
Is there a difference between 'inconsistent' 'contrary' and 'contradictory'? As far as I understand, two statements are inconsistent when they can not both be true; two statements are contradictory when they can not both bear the same truth value.

Does 'contrary' have a unique meaning?

答:
There is already an accepted answer that describes the issue pretty well.  To add to it, however, contrary and contradictory are used with distinct meanings in Aristotelian logic.  The square of opposition shows four types of relations between sentences.  These are:

contrary;

subcontrary;

contradictory; and

subalternation.

Therein, the contrary of the sentence "All S are P" is "No S are P", while its contradictory is "Some S is not P."  The contradictory is logically equivalent to the negation, but the contrary is not.  The linked Wikipedia article covers all the cases in much more detail.  You're probably not likely to encounter these terms in casual use, or even in formal logic today unless someone is making a point of speaking about Aristotelian logic.


问:
Limit of ln summation
I am trying to find the limit $$ \lim_{n\to\infty}\frac{1}{n}\sum_{i=\left\lceil \sqrt{n}\right\rceil }^{n}\ln\left(\frac{n}{i}\right) $$ I can't  find a good way to handle it. I tried bounding it from below and from above, but the bounds I got to are not tight enough.

答:
$$
\sum_{i=[\sqrt{n}]}^n \ln (n/i)\ge \sum_{i=[\sqrt{n}]}^n \ln (\sqrt{n})=\frac{1}{2}(n-[\sqrt{n}]+1)\ln n>\frac{n}{4}\ln n
$$
the last inequality, for $n\ge 4$.

Hence
$$
\frac{1}{n}\sum_{i=[\sqrt{n}]}^n \ln (n/i)\ge \frac{1}{4}\ln n\to\infty.
$$

Title: Limit of ln summation


问:
Where can I find old articles by Paul Erdős?
I am looking for a few old articles by Paul Erdős. I have tried to google the names and have looked for them on other sources like arvix without finding them. I am wondering if anyone knows how I can find these articles. To be more specific I am looking for the 6 first references in the following article : http://maths.nju.edu.cn/~zwsun/Cref.pdf. They are:

$[\mathrm{E} 52]$
P. Erdős, On a problem concerning congruence systems (Hungarian), Mat. Lapok $\mathbf{3}(1952), 122-128 .$ MR $17,14 \mathrm{~d}$

$[\mathrm{E} 62]$
P. Erdős, Remarks on number theory IV: Extremal problems in number theory $I$, Mat. Lapok $13(1962), 228$ - $255 .$ MR $33 \# 4020 .$

$[\mathrm{E} 73]$
P. Erdős, Problems and results on combinatorial number theory, in: A Survey of Combinatorial Theory (J. N. Srivastava et al., eds.), North-Holland Publ. Comp., Amsterdam, $1973,$ Ch. $12,117-138 .$

$[\mathrm{E} 77]$
P. Erdős, Problems and results on combinatorial number theory III, in:
Number Theory Day (M. B. Nathanson, ed.), Lecture Notes in Math. 626 , Springer, New York, $1977,43-72 .

$ [\mathrm{EG}80]$  P. Erdős and R. L. Graham, Old and New Problems and Results in Combinatorial Number Theory, (Monogr.; Vol. 28), L'Enseignement Math. Geneve, 1980. MR 82j:10001.

$[\mathrm{E} 81]$
P. Erdős, Problems and results in number theory, in: Recent Progress in Analytic Number Theory (H. Halberstam and C. Hooley, eds.), Vol. 1, Academic Press, London, $1981,1-14$.

Any information will help!

答:
I would never share an illegal platform, where the unscrupulous might find every paper they've hypothetically ever looked for (not that I would be speaking from experience, how dare you suggest such a thing!)

Instead I will drop a wikipedia link which describes such a website.

Sci-hub wikipedia article

I hope this helps ;)


问:
Proving $\left|\sum_{r=1}^{n}\vec{a_{r}}\right|^2=\sum_{r=1}^{n}\left|\vec{a_{r}}\right|^2+2\sum_{r\lt j}\vec{a_{r}}\cdot\vec{a_{j}}$
How to prove the following result. I can interpret the equality for $n=2$ boiling down to the cosine rule in disguise.

$$\left|\sum_{r=1}^{n}\vec{a_{r}}\right|^2=\sum_{r=1}^{n}\left|\vec{a_{r}}\right|^2+2\sum_{r\lt j}\vec{a_{r}}\cdot \vec{a_{j}}$$

And so a proof I came up with was to keep using it like cosine rule and treating the term inside the modulus as sum of two vectors, which I think is just similar to induction applied backwards. Apart from these two proofs, is there a more elegant way to prove it? Thanks.

答:
The trick here is to recognize that $$ \left| \sum_{r=1}^n \overrightarrow{a_r} \right|^2 = \left(\sum_{r=1}^n \overrightarrow{a_r}\right)\cdot\left(\sum_{r=1}^n \overrightarrow{a_r}\right).$$
To make the result really obvious we'll get rid of the sigma notation.
That is,
$$ \left| \sum_{r=1}^n \overrightarrow{a_r} \right|^2 = (\overrightarrow{a_1}+\overrightarrow{a_2}+\dots+\overrightarrow{a_n} )\cdot(\overrightarrow{a_1}+\overrightarrow{a_2}+\dots+\overrightarrow{a_n} ). $$ Expanding the RHS as normal because the dot product is distributive leads to,
$$ \left| \sum_{r=1}^n \overrightarrow{a_r} \right|^2 = \overrightarrow{a_1}\cdot \overrightarrow{a_1} + \overrightarrow{a_1}\cdot \overrightarrow{a_2} + \dots + \overrightarrow{a_1}\cdot \overrightarrow{a_n}+\overrightarrow{a_2}\overrightarrow{a_1}+\dots +\overrightarrow{a_n}\cdot \overrightarrow{a_n}. $$ Recall that the dot product is commutative and that $\overrightarrow{a_r}\cdot \overrightarrow{a_r}=\left|\overrightarrow{a_r}\right|^2$.

By simplifying, we have $$ \left| \sum_{r=1}^n \overrightarrow{a_r} \right|^2 = \left|\overrightarrow{a_1}\right|^2 + 2\overrightarrow{a_1}\cdot \overrightarrow{a_2} + \left|\overrightarrow{a_2}\right|^2 + 2\overrightarrow{a_2}\cdot \overrightarrow{a_3} + \dots + \left|\overrightarrow{a_n}\right|^2 + 2\overrightarrow{a}_{n-1}\cdot \overrightarrow{a_n}.$$
Therefore,
\begin{align*} \left| \sum_{r=1}^n \overrightarrow{a_r} \right|^2 &= \left( \left|\overrightarrow{a_1}\right|^2 + \dots + \left|\overrightarrow{a_n}\right|^2 \right) + 2\left( \overrightarrow{a_1}\cdot \overrightarrow{a_2} + \dots + \overrightarrow{a}_{n-1}\cdot \overrightarrow{a_n}  \right) \\ &= \sum_{r=1}^n \left|\overrightarrow{a_r}\right|^2 + 2\sum_{r<j}\overrightarrow{a}_{j}\cdot \overrightarrow{a_r}\end{align*}
for all $2\leq j \leq n$ where $r\leq j-1$.


问:
Solving a problem on differentiation
$f(x)$ is differentiable function and $G\times{f(x)}$ satisfies the relation $$G\times{f(x)}= \lim_{h\to 0}\frac{({f(x+h)}){^m}-{({f(x)}}{^m})}{h}$$ where m is a definite natural number.

I have been asked to find the relation: $$G\times (f(x)+g(x))$$ the provided answer is: $$G\times f(x)[\frac{f(x)+{g(x)}^{
(m-1)}}{{f(x)}^{(m-1)}}]+G\times g(x)[\frac{f(x)+{g(x)}^{(m-1)}}{{g(x)}^{m-1}}]$$I have no idea about how to solve this kind of problem, I ask for your kind  help regarding this problem, I will be also grateful if someone teach me the general theme of this problem or some perquisites required to solve this types of sums.

Title: Solving a problem on differentiation

Body:
Well, $G*f(x)$ is just the derivative of $f(x)^m$

So $G*f(x) = mf(x)^{m-1}f'(x)$ call this eqn(1)

So $G*h(x) = mh(x)^{m-1}h'(x)$

So just substitute in $h(x)=f(x)+g(x)$

So

$$G*(f(x)+g(x)) $$
$$= m(f(x)+g(x))^{m-1}(f'(x)+g'(x))$$
$$= (mf'(x)+mg'(x))(f(x)+g(x))^{m-1}$$

From eqn (1), we get

$mf'(x) = (G*f(x))f(x)^{1-m}$

$mg'(x) = (G*f(x))g(x)^{1-m}$

plutting these back into our main equations

$$=[(G*f(x))f(x)^{1-m}+(G*g(x))g(x)^{1-m}](f(x)+g(x))^{m-1}$$

$$=(G*f(x))\left[\frac{f(x)+g(x)}{f(x)}\right]^{m-1}+(G*g(x))\left[\frac{f(x)+g(x)}{g(x)}\right]^{m-1}$$


问:
Integrate $\int sin(x)^6 $ using integration by parts?
I know how to solve it using the double angle formula for $cos(2x)$. I also know how to integrate $sin(x)^2$ by parts like this:

$$\begin{align}\int \sin(x)\sin(x)dx &= −\cos(x)\sin(x)-\int(-\cos(x))\cos(x)dx
\\&= -\cos(x)sin(x)+\int(1-\sin(x)^2)dx
\\&= -\cos(x)\sin(x)+x-\int \sin(x)^2dx
\\2\int \sin(x)^2dx &= -\cos(x)sin(x)+x
\\\int \sin(x)^2dx &= -\frac 12 \cos(x)\sin(x)+\frac 12 x \end{align}$$

Is it possible to integrate $sin(x)^6$ by parts?

答:
Of course, you can. To integrate $\sin^{6}(x)$, use method how to integrate $\sin^{2}(x)$ again and again.

$\int \sin^{6}(x) dx$

$=\int \sin(x) \sin^{5}(x)dx$

$=-\cos(x) \sin^{5}(x)-\int (-\cos(x)) (5 \sin^{4}(x) \cos(x)) dx$

$=-\cos(x) \sin^{5}(x)+5 \int \sin^{4}(x) \cos^{2}(x) dx$

$=-\cos(x)\sin^{5}(x)+5 \int \sin^{4}(x)dx-5 \int \sin^{6}(x) dx$

From this integration, you get $\int\sin^{6}(x)dx=-\dfrac{1}{6}\cos(x)\sin^{5}(x)+\dfrac{5}{6} \int \sin^{4}(x)dx$.

Repeating this for 3 times, you finally get desired integral result.

Or simply using the fact that $\sin^{4}(x)=\left(\dfrac{1-\cos(2 x)}{2} \right)^{2}=\dfrac{\cos^{2}(2 x)-2 \cos(2 x)+1}{4}$

$=\dfrac{\dfrac{\cos(4 x)+1}{2}-2 \cos(2 x)+1}{4}=\dfrac{\cos(4 x)-4 \cos(2 x)+3}{8}$ gives result quicker than repeating iteration of integrating by parts


问:
How many n satisfy this condition?
How many $n$ (natural number) satisfies this condition:
$$2^{n} + 3^{n} = \beta \cdot7, \beta  \in  N$$
In another words, how many $n$ is possible in such a way that this number is divisible by seven?
I am really poor in the theory of numbers/figures, so I would appreciate a tip.

答:
Since the OP does not seem to know FLT, I would suggest the following:

Prove by induction that $2^{3+6k}+3^{3+6k}$ is divisible by $7$ for all $k\geq 0$.

Title: How many n satisfy this condition?


问:
Can we prove $ℝ^n ∖ \{\text{straight line}\}$ and $ℝ^n ∖ \text{\{an axis\}}$ homeomorphic by proving their complements homeomorphic?
I'm trying to prove that $\mathbb R^n \setminus \{\text{straight line}\}$ is homeomorphic to $\mathbb R^n\setminus\text{\{an axis\}}$.

Their complements, a straight line and an axis, are isomorphic, since the bijective map we find is just a translation from one line to the other. So I thought it could be a good idea to prove that $G\cong H \Leftrightarrow G^c\cong H^c$, but I'm not sure that this is generally true.

If it is true, which map could we take to prove it? And if it is false, how should I say that $\mathbb R^n \setminus \{\text{straight line}\} \cong \mathbb R^n \setminus \{\text{axis}\}?$ Or at the very least, that $\mathbb R^n \setminus \{\text{straight line}\} \simeq \mathbb R^n \setminus \{\text{axis}\}$?.

答: In $\mathbb R$, $(0,1)$ is homeomorphic to $(1,\infty)$ (via $x \to \frac 1 x)$ but the complement of the first one is not connected whereas the complement of the second interval is connected.


答:
The statement about complements that you claim is not true. First of all, not all topological spaces embed into $\mathbb R^n$. Even for those that do, we can find counterexamples. For instance, in $\mathbb R^2$ the subsets $A = \{(x, 0) : x \in (0, 1)\}$ and $B = \{(x, 0) : x \in \mathbb R\}$ are homeomorphic. However, $\mathbb R^2 - A$ is connected but $\mathbb R^2 - B$ is disconnected. Thus, they are not homeomorphic, so the proof method you had in mind will not work.

Here's a proof for the statement at hand. Let $L = \{u + tv : t \in \mathbb R\}$ for some $u,v \in \mathbb R^n$ and $v \neq 0$ be a line in $\mathbb R^n$ and $A = \{(t, 0, \dots, 0) : t \in \mathbb R\}$ be an axis. We define a homeomorphism $f: \mathbb R^n \longrightarrow \mathbb R^n$ such that $f[A] = L$. Indeed, extend $\{v\}$ to a basis $v=v_1, v_2, \dots, v_n$. We define $f(x_1, \dots, x_n) = u + \sum x_i v_i$. This is clearly continuous with a continuous inverse (they are both affine linear isomorphisms). Furthermore, $f(t, 0,\dots, 0) = u + tv \in L$ so $f[A] = L$ and $f$ restricts to a homeomorphism $f: \mathbb R^n - A \longrightarrow \mathbb R^n - L$.

Title: Can we prove $ℝ^n ∖ \{\text{straight line}\}$ and $ℝ^n ∖ \text{\{an axis\}}$ homeomorphic by proving their complements homeomorphic?

Body:
I'm trying to prove that $\mathbb R^n \setminus \{\text{straight line}\}$ is homeomorphic to $\mathbb R^n\setminus\text{\{an axis\}}$.

Their complements, a straight line and an axis, are isomorphic, since the bijective map we find is just a translation from one line to the other. So I thought it could be a good idea to prove that $G\cong H \Leftrightarrow G^c\cong H^c$, but I'm not sure that this is generally true.

If it is true, which map could we take to prove it? And if it is false, how should I say that $\mathbb R^n \setminus \{\text{straight line}\} \cong \mathbb R^n \setminus \{\text{axis}\}?$ Or at the very least, that $\mathbb R^n \setminus \{\text{straight line}\} \simeq \mathbb R^n \setminus \{\text{axis}\}$?.


问:
How many permutations of the letters A to H if the three letters ABC must appear together but not necessarily consecutively?
How many permutations of the letters A to H exist if the three letters ABC must appear together but not necessarily consecutively?

Pretty confused on how to start...
Can I group ABC as one letter? Then there will be 6 "letters", making it 6!, and then multiplied by 3! because there are 3! ways in which ABC can be assigned.

答:
$A,B,C,D,E,F,G,H$

$ABC$ always appear together. Consider this as a single letter X.

$X,D,E,F,G,H$

Number of permutation=$6!$=$720$

The number of permutation in X is also $3!$.

Therefore total permutations=$6! \times 3!$=$720 \times 6$=$4320$


问:
Finite ring of sets
I have some questions about finite rings of sets and I'll be very grateful for any help.

Let E be some fixed non-empty set.
Suppose we are given some finite ring of subsets of set E, i.e. some non-empty system $S \subset E$ such that

$\forall A,B \in 2^E ~~ A \vartriangle B \in S$

$\forall A,B \in 2^E ~~ A \cap B \in S$

1) Does it necessarily have a unity? (or in more abstract form: is there necessarily a unity in the finite commutative ring in which multiplication is idempotent?)

2) Suppose we're given some finite system of subsets of E. Is there any algorithm for building the minimal ring of sets which contains this system? I know that from n sets using union, intersection and set difference we mat build at most $~2^{2^n}$ different sets so this ring must be finite.

3) If we know that S is a ring with unity (and so it's a boolean ring) how can we build an isomorphism from our ring of sets to some ring $ B^n = (\{0,1\}^n, +, \cdot)$ where

$0 + 0 = 1 + 1 = 0$

$0 + 1 = 1 + 0 = 1$

$0 \cdot 1 = 0 \cdot 0 = 1 \cdot 0 = 0$

$1 \cdot 1 = 1$

Thanks in advance!

答:
Some elementary help seeing why it has unity:

Notice that $(A\vartriangle B)\vartriangle(A\cap B)=A\cup B$, so the ring actually contains finite unions of its members. Since the ring is finite, $E:=\cup S\in S$. Therefore, $E$ contains all members of $S$, and so $E\cap A=A$ for all $A$.

This also gives a partial start on item 2. Suppose that $G$ is a collection of sets that you want to generate a ring of sets with. Notice that both operations only produce sets which have elements in the two sets you started with. So, any set you generate will not escape $\cup G$. Then the generated ring is contained inside the powerset $\mathcal{P}(\cup G)$.

Ring theoretically, the collection of finite sums of finite products of generators produces the ring.

Title: Finite ring of sets


问:
How to find the angle formed by an isosceles triangle next to another one?
The problem is as follows:

Using the figure from below: Find the unknown angle indicated as $x$.
Assume $AD=BC$ and $BD=DC$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&12^{\circ}\\
2.&10^{\circ}\\
3.&15^{\circ}\\
4.&16^{\circ}\\
4.&14^{\circ}\\
\end{array}$

What I attempted to do here was to add the angles in the isosceles which adds up to $4x$, this can be added to the $3x$ triangle but that's where I'm stuck.

In other words the only thing which I could spot was:

$\angle BDA = 2x+2x$

How exactly can it be used congruency to solve this problem?. Can someone help me here?. The intended approach is relying in euclidean geometry postulates, but I don't know exactly which sort of congruency of triangles identity should be used.

Please include a drawing in your answer because this part is difficult for me to spot with accuracy. Can you please use an explanation step-by-step.

答:
Draw a line $DF$ such that $\angle ADF = 3x$. Draw lines $AE$ and $DE$ such that $\angle DAE = \angle ADE = 2x$.

So $\triangle AED \cong \triangle BDC$ (by A-A-S, as $BC = AD$)

So, $AE = DE = BD = DC$

Now $AF = DF$ so $\triangle AFE \cong \triangle DFE$ (by S-A-S).

Similarly, $\triangle DFB \cong \triangle DFE$

So, we have $\angle AFE = \angle EFD = \angle DFB = 60^0$

So $\angle FAD = \angle FDA = 3x = 30^0$

So, $x = 10^0$.


问:
How to use differentiation/integration to solve this problem
The question

So my lecturers gave our class a challange to solve this question a few weeks ago, however he has foegotten about it and I am really keen to know how to solve this quesion as a similar patter may come out for our mid sems.
Here is the question:

You work in a company that designs packaging for food products. Your team is tasked to design open top
storage bins with square bases. As part of the design, your team has to determine the dimensions of the
storage bin that can hold the largest volume given that 192 sq centimeters of material are used to fabricate each bin.

I can easily solve this question by using trial and error or using 3D modeling software to model and solve it. However, my teacher insisted to use calculus to solve this question. I remember him telling us to use maxima and minima to solve this question. Does anyone know how to go around doing this question? Any help is appreciated.

答:
The way I see it:

The constraints of the problem are as follows -
i) s^2 + 4hs = 192 (s being base width/length & h being the bin hight)
ii) Vol = s^2 * h

Now using (i) u can easily replace h in (ii).
you will get:

Vol = s^2 * ( 192 - s^2 ) / 4s

Calculus only comes in at this point. Vol is a function of s, and most likely some sort of curve. What you would want to do is find the maxima of this curve as that would give u Vol-max, which is your required answer.

f(s) = Vol(s)
Find f'(s) and f''(s) and solve f'(s) = 0 st. f''(s) < 0.

The value of s would be your required answer. Use that s to find h.


问:
Behaviour of orthogonal matrices
I am given that A is an orthogonal matrix of order $n$, and $u, v$ are Vectors in the $R^n $ space.

I need to prove that $||u|| = ||Au||$. The first step of the solution hint I am given is that $$||Au||^2 = (Au)^T(Au)$$. Why is this so? I know that $A^{-1} = A^T$ in the definition of an orthogonal matrix, but how does this contribute to the above statement? Or is there some other property I'm missing out on?

答:
Assume that $A$ is orthogonal, i.e. that $A^T A = I$ (observe that this is equivalent to $A^{-1} = A^T$). We consider

$$ || A x ||^2 = (Ax)^T (Ax) = x^T A^T A x = x^T x = || x || ^2, $$

which shows that $|| A x || = || x ||$.

Comment: On advise from another user, I posted a more thorough version of my earlier comment as this answer.

Title: Behaviour of orthogonal matrices


问:
Prove that gcd exists and it is unique
How can I show using division algorithm on $\mathbb{N}$ that there is a gcd for every pair of number $a,b \in \mathbb{N}$ and this gcd is unique?</p>

答:
If $d$ is the greates common divisor of $a$ and $b$, then by def.  $d|a$ and $d|b$ and if $c$ is any other common divisor then $c|d$. Now suppose on the contrary. Suppose $a$ and $b$ have another greatest common divisor $d'$ then $d'|d$ and $d|d'$ i.e $d=d'h_1$ and $d'=dh_2$  where $h1, h2 \in \mathbb{N}$

$d'=d'h_1h_2$ which means
$h_1h_2=1$. This is possible only when $h_1=h_2=1$.
So $d=d'$.
Hence proved that the gcd of $a$ and $b$ is unique.

Title: Prove that gcd exists and it is unique


问:
Uniqueness of the solution of the functional equation $ \frac{\phi(x) + \phi(y)}{2} = \phi\left(\frac{2xy}{x+y}\right) $
TL;DR. I'm trying to understand why the parameter $\beta$ in the Gibbs measure is the inverse of the temperature $\frac{1}{T}$ in thermal dynamical context.

In the space of smooth bijections (diffeomorphisms) from $(0,\infty)$ to $(0,\infty)$, the function

$$ x \mapsto \frac{1}{x}$$

satisfies the functional equation

$$ \frac{\phi(x) + \phi(y)}{2} = \phi(\frac{2xy}{x+y}).$$

Indeed,

$$ \frac{\frac{1}{x} + \frac{1}{y}}{2} = \frac{x+y}{2xy}.$$

Question

Is this the only solution?

Attempts and motivation

I've used some techniques.. like investigating limits, finding special values, or differentiating $x\phi(x)$, etc. This question comes from statistical mechanics. It will help me understand, after accepting that the Gibbs measure

$$ \mu(s) \sim \exp(-\beta s) $$

is natural, why the parameter $\beta$ introduced from Lagrange multiplier method naturally corresponds to the inverse of the temperature $\frac{1}{T}$ in thermal dynamical context.

答:
Hints for finding $\phi$: Differentiate w.r.t. $x$ to get $\frac  1 2 \phi'(x)=\phi '(\frac {2xy} {x+y}) \frac {2y^{2}} {(x+y)^{2}}$. Now put $x=1$ to get $ \phi '(\frac {2y} {1+y})$. Put $t=\frac {2y} {1+y}$ and you will get $\phi'(t)$ for every $t \in (0,1)$.

Treating $\phi'(1)$ as any given constant $c$, then you'll have

$$t^2 \phi'(t) = c.$$

By mean value theorem, $\phi(t) = c/t$ are the only solutions.


问:
Series manipulation doubt
Say that I have a series whose $n$th term is given by $a(n)$ and this converges to a limit $l$ as $n$ tends to infinity.
We know that there exists some N for which $|a(n)-l|<|l|/3$. How can I manipulate this to prove that $|a(n)|\geq2 |l|/3$ for all $n>N$?

答:
This follows from the triangle inequality $|x|+|y|\geq |x+y|$.

Now suppose $|a(n)|<2|l|/3$ and $|l-a(n)|<|l|/3$. By the triangle inequality we get $$|l|=|(l-a(n))+a(n)|\leq |l-a(n)|+|a(n)|<|l|/3+2|l|/3<|l|,$$
a contradiction.

Note that your initial statement that such an $N$ exists assumes that $l\neq 0$. (Your second statement is trivially true if $l=0$, however.)

Title: Series manipulation doubt


答:
I hope this answer may serve as a complement to GreginGre's. We would like to show his answer does give all desired $n$'s.

Fermat's Little Theorem (Special case):  $2^{6k} \equiv 3^{6k} \equiv 1 \: \text{(mod 7)}$ for any natural number $k$.

Proof. By direct calculation, $2^6 \equiv 64 \equiv 1  \: \text{(mod 7)}$. So for any natural number $k$, we have $2^{6k} \equiv (2^{6})^{k} \equiv 1^{k} \equiv 1  \: \text{(mod 7)}$. One  can prove similarly for $3$.

FLT basically shows the periodicity of $2^n + 3^n  \text{(mod 7)}$ (as a congruence class modulo $7$). Since the period is $6$, it suffices to check for $n = 0,1,\dots,5$. After some computation, one finds $3$ is the only candidate.

Title: How many n satisfy this condition?

Body:
How many $n$ (natural number) satisfies this condition:
$$2^{n} + 3^{n} = \beta \cdot7, \beta  \in  N$$
In another words, how many $n$ is possible in such a way that this number is divisible by seven?
I am really poor in the theory of numbers/figures, so I would appreciate a tip.


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

答:
I studied set theory by myself too, so my experience might be relevant to you. I was aware of Jech's book when I became interested to read up on the subject, but I choose not to use it for self study. I thought its overwhelming depth would take away the fascination and curiosity I had. It must be a good reference textbook by looking at its contents and structure, but I would only recommend it for self study if you have someone to guide you through it, as Asaf has advised already.

The two books I ended up with and really loved are Naive Set Theory by Paul Halmos, and Axiomatic Set Theory by Patrick Suppes. Do not be mislead by the title of Halmos' book, it is not that naive. It is rather axiomatic, he is just not using  ZFC explicitly.

Both authors have a very good narrative style. They put things in historical context considering the developments of set theory, and emphasise the importance of different proofs in a very mind-opening way. The exercises are very instructive too, giving you the playground to try what you have just read.

I remember how much I was stunned for months by the powerful construct for families in Halmos' book. Suppes' family construct was a bit different, significantly less powerful I would say. By the time I got to the cardinal and ordinal numbers, $\varepsilon_0$, and friends  I was a totally different person. Anyhow, both authors have a fascinating style, and they complemented each other very well for my purpose, i.e. to blow my mind.

Someone mentioned above that a prerequisite to set theory could be a good introductory textbook on logic. Suppes in fact has one, that is how I got to know his style even before reading his set theory book. I was not disappointed with either.


问:
Understanding vector calculation in support vectors
I was going through Andrew Ng's lecture notes for Machine Learning course at Stanford. I have small possibly silly doubt.

On page 48, 49 of the pdf, it defines geometric margin ($\gamma^{(i)}$) between a training point point $A=x^{(i)}$ and a decision boundary ($w^Tx+b=0$) in th context of SVMs:

It says following:

How can we find the value of $γ^{(i)}$? Well, $w/\Vert w\Vert$ is a unit-length vector pointing in the same direction as $w$. Since $A$ represents $x(i)$, we therefore find that the point $B$ is given by $x^{(i)}\gamma{(i)}.w/\Vert w\Vert$. But this point lies on the decision boundary, and all points $x$ on the decision boundary satisfy the equation $w^Tx+b=0$. Hence, $$w^T\left(x^{(i)}-\gamma^{(i)}\frac{w}{\Vert w\Vert}\right)+b=0$$.
Solving for $\gamma^{(i)}$ yields, $$\gamma^{(i)}= \frac{w^Tx^{(i)}+b}{\Vert w \Vert}$$

My doubt is, I am getting $\gamma^{(i)}= w^Tx^{(i)}+b$:

$$w^Tx^{(i)}-w^T\gamma^{(i)}\frac{w}{\Vert w\Vert}+b=0$$
$$w^Tx^{(i)}-\gamma^{(i)}+b=0 \quad\quad(\because w^Tw=\Vert w\Vert)$$
$$\therefore\gamma^{(i)}= w^Tx^{(i)}+b$$

Am I plain wrong here? Or something more is going on here?

答:
The error is where you have $w^Tw = ||w||$. This should be $w^Tw = ||w||^2$ because $||w|| = \sqrt{w^Tw}$. Using the correction we have
$$w^T\bigg(x^{(i)} - \gamma^{(i)}\frac{w}{||w||}\bigg) + b = 0$$
$$w^Tx^{(i)} - \gamma^{(i)}\frac{||w||^2}{||w||} + b = 0$$
$$w^Tx^{(i)} - \gamma^{(i)}||w|| + b = 0$$
This leads to
$$\gamma^{(i)} = \frac{w^Tx^{(i)} + b}{||w||}$$


问:
Identify bifurcation and sketch diagram
I am given the equation

$$\dot x = rx + \frac {x}{1+x^2} $$

I believe I've found fixed points at $x=0$ and $x= \sqrt \frac{-1-r}{r}$ and I think its stable for $-1<r<0$ but I'm not sure how to sketch this.  I looked online but couldn't find any tools to do this.  I also need to determine if it is a subcritical pitchfork or not. Thanks in advance.

答:
You can easily determine the sign structure of the vector field on the one-dimensional phase space. The right side factors as
$$
\dot x=x\left(r+\frac1{1+x^2}\right).
$$

For $r\ge0$ there are no roots from the second factor, and it is always positive. The resulting sign structure is $(-,+)$ for increasing $x$ values, $x=0$ is an unstable equilibrium.

For $r<0$ the vector field points inwards if $|x|$ is large enough

for $-1<r<0$ the second factor has a pair of opposite roots, starting close to infinity for $r$ close to zero and meeting each other at $r=-1$ with a vertical tangent at $x=0$. The sign structure along the $x$ axis is $(+,-,+,-)$, meaning the outer equilibrium points are stable, the inner one at $x=0$ remains unstable.

for $r<-1$ the second factor is strictly negative, there is only one equilibrium at $x=0$. The sign structure is $(+,-)$, making this a stable equilibrium.

Title: Identify bifurcation and sketch diagram


问:
Bayes theorem and three shooters
Three shooters $A,B,C$ aim at a target. Each of them shoots one. The accuracy of the $A$ shooter is equal to $0.6$, $B$ is $0.7$, $C$ is $0.8$. What's the probability that $A$ hit the target if:

the target was hit once,

the target was hit twice,

the target was hit at least once,

the target was hit at least twice.

It's obvious that
$$ P(A)=0.6$$
$$P(B)=0.7$$
$$ P(C)=0.6$$
where $P(\cdot)$ is the probability that the shooter hit the target.

For 1) I have:
$$P(A|T_{1})=\frac{P(T_{1}|A)\cdot P(A)}{P(T_{1}|A)\cdot P(A) + P(T_{1}|B)\cdot P(B) + P(T_{1}|C)\cdot P(C)} $$
where $P(T_{1}|A)=0.6*(1-0.7)*(1-0.8)$ and respectively for the other ones.

For 2)
$$P(A|T_{2})=\frac{P(T_{2}|A\cap B)\cdot P(A \cap B) + P(T_{2}|A\cap C)\cdot P(A \cap C)}{...}$$
Now, I'm kind of lost. What I wanted to obtain is the probability that A hit the target and the target was hit twice - so either B or C can aim with A. But what's $P(A \cap B)$? Maybe I want $P(A \cup B)$?

For 3)
My first thought was to make this with the opposite event. So the target wasn't hit at all. But I can't imagine how to appy Bayes theorem to this case. I mean if the target wasn't hit then the probability that A hit the target is 0. Something doesn't feel right. For 4) I have a similar problem.

答:
First question:

$$\mathbb{P}[A|\text{1 hit}]=\frac{\mathbb{P}[A]\mathbb{P}[\overline{B}]\mathbb{P}[\overline{C}]}{\mathbb{P}[A]\mathbb{P}[\overline{B}]\mathbb{P}[\overline{C}]+\mathbb{P}[\overline{A}]\mathbb{P}[B]\mathbb{P}[\overline{C}]+\mathbb{P}[\overline{A}]\mathbb{P}[\overline{B}]\mathbb{P}[C]}=\frac{36}{36+56+96}=\frac{9}{47}$$

Second question:

Consider that if the target was hit twice the following cases are possible

$$AB,AC,BC$$

Thus the solution is very similar to 1)

$$\mathbb{P}[A|\text{2 hits}]=\frac{0.6\times0.7\times0.2+0.6\times0.3\times0.8}{0.6\times0.7\times0.2+0.6\times0.3\times0.8+0.4\times0.7\times0.8}=\frac{228}{452}=\frac{57}{113}$$

Third question:

$$\mathbb{P}[A|\text{hits} \geq 1]=\frac{36+228+336}{1000-24}=\frac{75}{122}$$

Fourth question:

It is up to you...

To speed up the calculation of 3. I used the numerators previously calculated, which are the favourable cases with 1 and 2 hits, respectively.

I used integer numbers because I do not like decimals...

Title: Bayes theorem and three shooters


问:
Wrong theorem - The number 1 is the largest integer
I have a theorem and its proof. I am trying to find the flaw in this wrong theorem.

Theorem : The number 1 is the largest integer.

Proof -> Suppose the conclusion is false. Then let $ n> 1$ be the largest integer. Multiplying both sides of this inequality by $n$, yields $n^2 > n$. This is contradiction, because $n^2$ is another integer larger than $n$. Thus the theorem is proved.

Now, using the quantifiers, I think theorem will be expressed as follows.

$$ \forall \,x \in \mathbb{Z}\left[ \, x < 1 \, \right] $$

And negation of this will be as follows.

$$ \exists \,x \in \mathbb{Z}\left[ \, x \geqslant 1 \, \right] $$

So, we have $ x \in \mathbb{Z} $ and $ x \geqslant 1$. So, I think that the negation which they have done is wrong. Is my intuition correct ?

I have seen few other posts talking about a relation question. But I could not follow their arguments. Hence the new question.

答: Yes, when you do the negation, what you get is "1 is not the largest integer". Then, the wrong part comes when it says "Then let n>1 be the largest integer", the problem is that that $n$ may not exist.


答: The negation of "$1$ is the largest integer" should be "there exists some larger integer $n$ than $1$", in lieu of "there exists some largest integer $n$ larger than $1$". So you actually had assumed more than the negation of the statement, which is not allowed.

Title: Wrong theorem - The number 1 is the largest integer

Body:
I have a theorem and its proof. I am trying to find the flaw in this wrong theorem.

Theorem : The number 1 is the largest integer.

Proof -> Suppose the conclusion is false. Then let $ n> 1$ be the largest integer. Multiplying both sides of this inequality by $n$, yields $n^2 > n$. This is contradiction, because $n^2$ is another integer larger than $n$. Thus the theorem is proved.

Now, using the quantifiers, I think theorem will be expressed as follows.

$$ \forall \,x \in \mathbb{Z}\left[ \, x < 1 \, \right] $$

And negation of this will be as follows.

$$ \exists \,x \in \mathbb{Z}\left[ \, x \geqslant 1 \, \right] $$

So, we have $ x \in \mathbb{Z} $ and $ x \geqslant 1$. So, I think that the negation which they have done is wrong. Is my intuition correct ?

I have seen few other posts talking about a relation question. But I could not follow their arguments. Hence the new question.


答: The claim here is$$(\forall n\in\Bbb Z):n\leqslant1.$$Denying this is to assert that$$(\exists n\in\Bbb Z):n>1.$$And the error lies in assuming that there is a largest integer.


问:
If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$
This is the $y=2^k$ case of this question.

Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?

Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that these two divisors are $2^k-1$ and $2^k+1$?

答:
I can reduce the problem to an infinite family of generalized Pell equations, which explains why the problem is hard. Maybe someone who is familiar with this corner of number theory can finish it off?

Let $y=2^k$. Then $y^2-x^2\bigm|y^2-1$. In other words,
$$y^2-1=n(y^2-x^2)$$
for some $n\geq1$.
Rearranging terms gives
$$nx^2-(n-1)y^2=1.$$
It suffices to show that this equation has no solutions for $y$ even and $n\geq2$.
Equivalently, it suffices to show that the equation
$$nx^2-4(n-1)y^2=1$$
has no solutions for $n\geq2$.

For each $n\geq2$, this is a generalized Pell equation.

I plugged this generalized Pell equation into this solver for all $n\leq30$, and in each case there are no solutions.


问:
Separable C* algebra has a cyclic representation that is isometric.
I'm studying Conway's Functional Analysis.

In chapter 8.5, we learn about GNS representation. Theorem 8.5.17 states that

$\mathcal{A}$ is separable then, we can select representation $(\pi, \mathcal{H})$ that $\mathcal{H}$ is separable.

And next line, every separable C*-algebra has a cyclic representation that is isometric.

I want to prove this, and Conway suggests that $f_n$ is countable weak* dense subset of $S_\mathcal{A}$(which is state space of $\mathcal{A}$), then if we set $f = \sum 2^{-n}f_n$ then $\pi_f$ is an isometry.($\pi_f$ is from GNS construction)

Since GNS Construction suggest that if we make $\pi_f$, then it is cyclic representation, so I need to prove that $\pi_f$ is isometry. But I cannot get any way to get $\pi_f$'s norm.

if $e_f$ is cyclic vector of $\pi_f$,
$$\|p_f(a) \|^2 \geq \|\pi_f(a) e_f\|^2 = \langle  \pi_f(a^* a) e_f, e_f\rangle = f(a^* a) = \sum 2^{-n}f_n(a^* a)$$

and we know that $\|a\| = \sup\{f(a) : f \in S_\mathcal{A} \}$ for positive element $a$ but I don't know that $\sum 2^{-n}f_n(a^* a) \geq  \|a^* a\| - \epsilon$. How can i deal with this problem?

答:
So $f$ is a state. Lets first show that it is faithful, meaning that $f(a^*a)>0$ whenever $a\neq0$:

Suppose $a\in A$ and $a\neq0$, then there is a state $\varphi$ with $\varphi(a^*a)>0$. Since $f_n$ is weak* dense you have a sub-sequence with $f_{n_k}\to \varphi$ in the weak* topology, but this topology is just so that you can recover $f_{n_k}(a^*a)\to\varphi(a^*a)$, hence there is some $f_n$ with $f_n(a^*a)>0$. In particular $f(a^*a)>0$.

Now lets show a general statement. Namely that if $\pi_f$ is the GNS representation onto a faithful state then $\pi$ is an isometry. For this statement we use the non-trivial statement that injective C* morphisms are isometries. So all we check is that $\pi$ doesn't have a kernel:

Suppose that $\pi(a) =0$, then $\pi(a^*a)=\pi(a)^*\pi(a)=0$. It follows that for the ground-state $|f\rangle$ you get $\langle f,\pi(a^*a)f\rangle = f(a^*a) =0$, implying that $a=0$ by faithfulness of $f$.


答:
Note that the solution $(x,5,2,y,t)=(1,5,2,9,8,8)$ is not found by your computer.

By the divisibility tests for $\overline{x52yzt}$, we conclude that $(x+2+z)-5+y+t)=11a$

Since we already have $(x+2+z)+(5+y+t)=11m$, then we get

$$\Big((x+2+z)-(5+y+t)\Big) \pm \Big((x+2+z)+(5+y+t)\Big)=11(a \pm m)$$

Thus both $x+2+z$ and $5+y+t$ are divisible by $11$

Thus all we need to do is to determine the number of solutions to the following equations

$$x+z=9$$ or $$x+z=20$$

and

$$y+t=6 $$ or $$y+t=17$$

Since $x \ne 0$ then $(x,z)$ has $9$ possibilites, while $(y,t)$ has $9$ possibilites. That gives you a total of $81$ solutions.

Title: How would you find number of solutions without solving this$N=\overline{x52yzt}=11k$, such that $x+5+2+y+z+t=11m$?

Body:
How would you find number of solutions without solving this problem:

Find N with following condition:

$N=\overline{x52yzt}=11k$, such that $x+5+2+y+z+t=11m$?

I saw this problem in a book and it gives 8 solutions, but I found 56 solutions by computer which includes all solutions in the book. The method used in the book needed to solve numerous system of Diophantine equations. I have two questions:

1-How would you find number of solutions without solving it.

2- Is there a simpler method to solve this kind of problems without computer?

Solutions by computer:

(x, 5, 2, y, z, t)=(1, 5, 2, 0, 8, 6) (1, 5, 2, 1, 8, 5) (1, 5, 2, 2, 8, 4) (1, 5, 2, 3, 8, 3) (1, 5, 2, 4, 8, 2) (1, 5, 2, 5, 8, 1) (1, 5, 2, 6, 8, 0) (2, 5, 2, 0, 7, 6) (2, 5, 2, 1, 7, 5) (2, 5, 2, 2, 7, 4) (2, 5, 2, 3, 7, 3) (2, 5, 2, 4, 7, 2) (2, 5, 2, 5, 7, 1) (2, 5, 2, 6, 7, 0) (3, 5, 2, 0, 6, 6) (3, 5, 2, 1, 6, 5) (3, 5, 2, 2, 6, 4) (3, 5, 2, 3, 6, 3) (3, 5, 2, 4, 6, 2) (3, 5, 2, 5, 6, 1) (3, 5, 2, 6, 6, 0) (4, 5, 2, 0, 5, 6) (4, 5, 2, 1, 5, 5) (4, 5, 2, 2, 5, 4) (4, 5, 2, 3, 5, 3) (4, 5, 2, 4, 5, 2) (4, 5, 2, 5, 5, 1) (4, 5, 2, 6, 5, 0) (5, 5, 2, 0, 4, 6) (5, 5, 2, 1, 4, 5) (5, 5, 2, 2, 4, 4) (5, 5, 2, 3, 4, 3) (5, 5, 2, 4, 4, 2) (5, 5, 2, 5, 4, 1) (5, 5, 2, 6, 4, 0) (6, 5, 2, 0, 3, 6) (6, 5, 2, 1, 3, 5) (6, 5, 2, 2, 3, 4) (6, 5, 2, 3, 3, 3) (6, 5, 2, 4, 3, 2) (6, 5, 2, 5, 3, 1) (6, 5, 2, 6, 3, 0) (7, 5, 2, 0, 2, 6) (7, 5, 2, 1, 2, 5) (7, 5, 2, 2, 2, 4) (7, 5, 2, 3, 2, 3) (7, 5, 2, 4, 2, 2) (7, 5, 2, 5, 2, 1) (7, 5, 2, 6, 2, 0) (8, 5, 2, 0, 1, 6) (8, 5, 2, 1, 1, 5) (8, 5, 2, 2, 1, 4) (8, 5, 2, 3, 1, 3) (8, 5, 2, 4, 1, 2) (8, 5, 2, 5, 1, 1) (8, 5, 2, 6, 1, 0)

Solutions found in book:

$152086, 152680, 852016, 352869, 352968, 652839, 652938$


问:
Ruled Surfaces and Rational Fibrations
I have a few doubts about Ruled Surfaces.

Edit:

We're working on ground field $\Bbb{C}$

The term ruled stands here for birationally ruled

Let $S$ be an algebraic smooth surface and $C$ a smooth projective curve.

If there is a morphism $p:S\rightarrow C$ with each fiber birational to $\Bbb{P}^1$ let's say that $S$ is a rational fibration. By the Noether-Enriques theorem each rational fibration is a ruled surface.

Claim: If $S\neq\Bbb{P}^2$ then $S$ is a rational fibration if and only if $S$ is ruled.

Assume this is true. Then given a ruled surface $S\neq\Bbb{P}^2$ we consider $S$ as a rational fibration and there can be at most a finite number of fibers which are birational but not isomorphic to $\Bbb{P}^1$. So they are singular irreducible curves (right?). Hence by a finite sequence of blow-ups we get a geometrically ruled surface, i.e. a rational fibration with each fiber isomorphic to the line. Is this argument correct?

Do you think that claim is true?

Edit: Since Asal's answer suggests the claim is false. Let me re-question: suppose $S\neq\Bbb{P}^2$ is birationally ruled. Can we show there exists a morphism $p:S\rightarrow C$ whose generic fiber is isomorphic to $\Bbb{P}^1$ ?

答:
Let me try to answer the modified question. (I assume we are working over an algebraically closed field of characteristic zero.)

The answer to your question depends crucially on what is meant by ruled. (I think this was a source of confusion in the comments.) There are two different defintions I know of, neither one standard:

$S$ is geometrically ruled (what Hartshorne calls ruled) if $p: S \rightarrow C$ has every fibre isomorphic to $\mathbf{P}^1$;

$S$ is birationally ruled (what some other people called ruled) if $S$ is birational to a product $C \times \mathbf{P}^1$.

Now Noether--Enriques tells you that a rational fibration is birationally ruled (at least once you know there are some smooth fibres, which you get from Bertini's second theorem) but you might worry that there are some singular fibres that stop it from being geometrically ruled. However, this can't actually happen, for the following reason.  By your definition of rational fibration, every fibre of $p$ is an irreducible reduced rational curve, but any such curve which is not isomorphic to $\mathbf{P}^1$ must be singular, hence have arithmetic genus greater than 0; however, this cannot happen for a flat family such as $p: S \rightarrow C$. (See Hartshorne for justifications of all these assertions.)

So a rational fibration is the same thing as a geometrically ruled surface. Just to be sure that these are really different from the birationally ruled surfaces, let's give an example: take $p: S \rightarrow C$ a geometrically ruled surface, and let $\pi: S' \rightarrow S$ be the blowup of any point on $S$. Then $p \circ \pi: S' \rightarrow C$ is a birationally ruled surface, but one of its fibres is a reducible curve, so it is not geometrically ruled.

Actually, this last example also highlights a problem with your idea in the final paragraph: blowing up points unfortunately can't turn a fibration with singular fibres into one with smooth fibres, because it introduces extra components into some of the fibres.

I hope this helps.

Edit: Finally I think I can answer the intended question. Let $S$ be a birationally ruled surface, not isomorphic to $\mathbf{P}^2$. Contract $(-1)$-curves on $S$ until we reach a minimal surface $S_m$. So we have a morphism $p: S \rightarrow S_m$. Now according to Enriques' classification, there are three possibilities for $S_m$:

It is a geometrically ruled surface $\pi: S_m \rightarrow C$ over a curve of genus $>0$;

It is a Hirzebruch surace $\pi: S_m \rightarrow \mathbf{P}^1$;

It is $\mathbf{P}^2$.

The first two types are geometrically ruled, so $\pi \circ p$ gives a morphism to a curve with generic fibre $\mathbf{P}^1$. (We get $S$ from $S_m$ by blowing up a finite number of points, so that doesn't affect the generic fibre.)

The only problem is if we arrive at $\mathbf{P}^2$. But in that case we just stop contracting curves at the penultimate step, to get a morphism $p: S \rightarrow \Sigma_1$ where $\Sigma_1$ is the blowup of $\mathbf{P}^2$ in one point. Now $\Sigma_1$ is also a geometrically ruled surface, so we can argue as in the other cases.

And if that doesn't answer the question, I quit! :)

Title: Ruled Surfaces and Rational Fibrations


问:
Limit of ln summation
I am trying to find the limit $$ \lim_{n\to\infty}\frac{1}{n}\sum_{i=\left\lceil \sqrt{n}\right\rceil }^{n}\ln\left(\frac{n}{i}\right) $$ I can't  find a good way to handle it. I tried bounding it from below and from above, but the bounds I got to are not tight enough.

答:
If you know about Pochhammer symbols
$$\sum_{i=a}^n\log\left(\dfrac{n}{i}\right)=\log \left(\frac{n^{n-a+1}}{a (a+1)_{n-a}}\right)$$
$$\dfrac{1}{n}\sum_{i=\left\lceil \sqrt{n}\right\rceil}^n\log\left(\dfrac{n}{i}\right)=\frac 1n\log\left(\frac{\Gamma \left(\left\lceil
   \sqrt{n}\right\rceil \right)}{\Gamma (n)}\,n^{n-\left\lceil \sqrt{n}\right\rceil }\right) $$

Now, use Stirling approximation.

Title: Limit of ln summation


问:
Find $W_1,W_2\neq \mathbb R^2$ so that $U+W_1=U+W_2=\mathbb R^2$
in the vector space $ \mathbb R_{col}^{2} $, there's a vector subspace $U=\mathbb R  w$, and $$ w=\left[\begin{array}{c}
1\\
-1
\end{array}\right] $$ find two different vector subspace $W_1,W_2\neq \mathbb R_{col}^{2}$ so that:
$$ U+W_1=U+W_2=\mathbb R_{col}^{2}$$

Body:
Take
$$W_1=\{(x,0): x\in \mathbb{R}\} \\
 W_2=\{(0,y):y\in \mathbb{R}\} \\
U=\{(x,y): x,y\in \mathbb{R}\textrm{ and }y=x\}$$.


问:
Find Tangents to a Rotated Translated Ellipse at 2 points through which a straight line passes.
This image (view in this link) is a problem I can't seem to complete (for a personal project).

I have a 2D array of points (or pixels) that represent an ellipse and I'm trying to to find tangents to the ellipse at 2 points on the ellipse (through which a straight line passes not necessarily parallel to the horizontal). See picture for an understanding of the problem.

I can't seem to figure this out because of the rotation. I imagine it would involve a change in coordinate systems perhaps (I may be wrong).

答:
Comment only. Is this the problem statement?

From a fully defined inclined ellipse two points (say D,A) are selected on it. Equations of two tangents from them are required.


问:
Find the limit of $\dfrac{\phi(n)}{n}$ where $\phi$ is a permutation of $\Bbb N$
Find the limit of $\dfrac{\phi(n)}{n}$ where $\phi$ is a permutation of $\Bbb N$.

A permutation of $\Bbb N$  is a bijection from $\Bbb N$ to $\Bbb N$, based on pure intuition it seems like the limit should be $1$ however I can't come up with a proof.

An already established result is that  $(\phi(n))_n$ diverges to infinity.

答:
Here is one without limit: Partition the natural numbers at the powers of 2. Let the permutation reverse each partition. The fraction $\frac{\phi(n)}n$ will bounce between $2$ and $\frac12$ infinitely many times.

If you pick partitions that grow in size faster than exponential (say you partition at each factorial), then the fraction goes arbitrarily close to $0$ and arbitrarily large as $n$ grows.


问:
Derivation of Divergence of a Vector Field Formula
Assume a vector field $\boldsymbol{V}=M(x,y)\boldsymbol{i}+N(x,y)\boldsymbol{j}$
The divergence is then:
$$\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}$$

My book(Thomas's calculus) explains this using a square in the x-y plane whose area goes to zero which is similar to this answer How would one arrive at the formulas for divergence and curl?
Following the same argument, why is the change in the x For N (i.e $N(x+\Delta x,y))$) as we move along the x-axis not included, similarly for the change in y for M.

i.e why are these not included in the derivation of the divergence formula?

$$(-N(x,y)+N(x+\Delta x,y))\Delta x$$
and
$$(-M(x,y)+M(x,y+\Delta y))\Delta y$$

答:
I'd like to offer an alternative explanation as to why the divergence is equal to the sum of the partial derivatives of $M(x,y)$ and $N(x,y)$ in the $x$ and $y$ directions respectively. This will be done alongside with a pictorial representation of the same arbitrarily "small" rectangle depicted below.

Intuitively (and also formally provable), the net flux can be split into four parts:
$$\text{net flux} = \text{right side flux}+\text{left side flux}+\text{top side flux}+\text{bottom side flux}$$

This is because the boundary of the rectangle $\text{Rect}$ (denoted $\partial \text{Rect}$) consists of all the points on the "outline" of this rectangle.

More formally,

$$\text{net flux}=\oint_{\partial \text{Rect}} \vec{V}\cdot \hat{n}\;\mathbb{d}s=\int_{R} \vec{V}\cdot \hat{n}\;\mathbb{d}s+\int_{L} \vec{V}\cdot \hat{n}\;\mathbb{d}s+\int_{T} \vec{V}\cdot \hat{n}\;\mathbb{d}s+\int_{B} \vec{V}\cdot \hat{n}\;\mathbb{d}s$$

Note that this "decomposition" of $\partial \text{Rect}$ makes easy work of using $\vec n$ on each subset $R,L,T$ and $B$ (assuming $\vec n$ points outwards from $\text{Rect}$).

Following the coordinate system provided, every line integral is expressible as a definite integral over the intervals representing each side.

$$F_R=\int_{R} \vec{V}\cdot \hat{n}\;\mathbb{d}s=\int_{y}^{y+\Delta y} \left(M(x+\Delta x,Y)\hat
 i+N(x+\Delta x,Y)\hat j\right)\cdot \hat i\;\mathbb{d}Y=\int_{y}^{y+\Delta y}M(x+\Delta x,Y)\;\mathbb{d}Y$$
Similarly for the left side L (but now noting that $\hat n=-\hat i$)
$$\int_{L} \vec{V}\cdot \vec{n}\;\mathbb{d}s=-\int_{y}^{y+\Delta y}M(x,Y)\;\mathbb{d}Y$$

Likewise for the top and bottom sides $T$ and $B$,
$$\int_{T} \vec{V}\cdot \hat{n}\;\mathbb{d}s=\int_{x}^{x+\Delta x} \left(M(X,y+\Delta y)\hat
 i+N(X,y+\Delta y)\hat j\right)\cdot \hat j\;\mathbb{d}X=\int_{x}^{x+\Delta x}N(X,y+\Delta y)\;\mathbb{d}X$$
Similarly for the bottom side B (but now noting that $\hat n=-\hat j$)
$$\int_{B} \vec{V}\cdot \hat{n}\;\mathbb{d}s=-\int_{x}^{x+\Delta x}N(X,y)\;\mathbb{d}X$$

Therefore, we have that the net flux on $\partial R$ is
$$\text{net flux}=\int_{y}^{y+\Delta y}M(x+\Delta x,Y)\;\mathbb{d}Y-\int_{y}^{y+\Delta y}M(x,Y)\;\mathbb{d}Y+\\\int_{x}^{x+\Delta x}N(X,y+\Delta y)\;\mathbb{d}X-\int_{x}^{x+\Delta x}N(X,y)\;\mathbb{d}X\tag{1}$$

Relationship to the book's approximation

Note that at this point it can be seen that the (approximation of the) flux on the right side R used in the book is the first term of the Taylor expansion of the exact flux, $F_R(\Delta y)$ (can be calculated using the Leibniz integral rule). Written using Big-O notation, the flux on the right side R is

$$\int_{y}^{y+\Delta y}M(x+\Delta x,Y)\;\mathbb{d}Y=M(x+\Delta x,y)\Delta y+\mathcal{O}(\Delta y^2)\tag{2} \quad \quad \text{as } \Delta y \to 0$$

At this point we could say that they are approximately equal to one another. However,  there is no fine line for a small enough $\Delta y$ besides one's judgement or application:
$$\int_{y}^{y+\Delta y}M(x+\Delta x,Y)\;\mathbb{d}Y\approx M(x+\Delta x,y)\Delta y$$

Below is the corresponding picture:

Note that they would be equal if $M(x+\Delta x,y)$ was constant for all $y\in [y,y+\Delta y]$, but that is not necessarily the case.

Relationship between this rectangle construction and the divergence of V at a point (x,y)

Combining integrals with "like" bounds in (1) you obtain:
$$\oint_{\partial \text{Rect}} \vec{V}\cdot \hat{n}\;\mathbb{d}s=\int_{y}^{y+\Delta y}M(x+\Delta x,Y)-M(x,Y)\;\mathbb{d}Y+\int_{x}^{x+\Delta x}N(X,y+\Delta y)-N(X,y)\;\mathbb{d}X$$

Applying the definition of the divergence for this rectangle construction we have:

$$\text{div}\;\vec V\;\Biggr|_{(x,y)}=\lim_{(\Delta x,\Delta y)\to \vec 0}\frac{1}{\Delta x\Delta y}\oint_{\partial \text{Rect}} \vec{V}\cdot \hat{n}\;\mathbb{d}s$$

Substituting in the expression for the net flux for the rectangle:

$$\small{\begin{align}\text{div}\;\vec V\;\Biggr|_{(x,y)}&=\lim_{(\Delta x,\Delta y)\to \vec 0}\frac{1}{\Delta x\Delta y}\left(\int_{y}^{y+\Delta y}M(x+\Delta x,Y)-M(x,Y)\;\mathbb{d}Y+\int_{x}^{x+\Delta x}N(X,y+\Delta y)-N(X,y)\;\mathbb{d}X\right)\\&=\underbrace{\lim_{(\Delta x,\Delta y)\to \vec 0}\frac{1}{\Delta x\Delta y}\left(\int_{y}^{y+\Delta y}M(x+\Delta x,Y)-M(x,Y)\;\mathbb{d}Y\right)}_{A}\\&+\underbrace{\lim_{(\Delta x,\Delta y)\to \vec 0}\frac{1}{\Delta x\Delta y}\left(\int_{x}^{x+\Delta x}N(X,y+\Delta y)-N(X,y)\;\mathbb{d}X\right)}_B\end{align}}$$

Using iterated limits, term $A$ can be evaluated:
$$\begin{align}A&=\lim_{(\Delta x,\Delta y)\to \vec 0}\frac{1}{\Delta x\Delta y}\left(\int_{y}^{y+\Delta y}M(x+\Delta x,Y)-M(x,Y)\;\mathbb{d}Y\right)\\&=\lim_{\Delta x\to 0}\lim_{\Delta y\to 0}\frac{1}{\Delta x\Delta y}\left(\int_{y}^{y+\Delta y}M(x+\Delta x,Y)-M(x,Y)\;\mathbb{d}Y\right)\\&=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\lim_{\Delta y\to 0}\frac{1}{\Delta y}\left(\int_{y}^{y+\Delta y}M(x+\Delta x,Y)-M(x,Y)\;\mathbb{d}Y\right)\\&=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left(M(x+\Delta x,y)-M(x,y)\right)\end{align}$$

Going from line 3 to 4 can be achieved in at least two ways I'm aware of. Either by using the Taylor expansion as discussed in (2) and taking the limit (all higher order terms disappear), or by realising that:

$$\lim_{h\to 0}\frac{1}{h}\int_{a}^{a+h}f(x)\;\mathbb{d}x=\lim_{h\to 0}\left(\text{average value of } f \text{ on } [a,a+h]\right)=f(a)$$
when $f(x)$ is continuous. In our case the inside of the integral is assumed continuous.

Notice that this is precisely the definition of the partial derivative of $M$ at $(x,y)$,
$$\therefore A=\frac{\partial M}{\partial x}$$

Applying the same process for term $B$, but in the opposite order for iterated limits, you obtain that

$$B=\frac{\partial N}{\partial y}$$

Therefore, the result is obtained:
$$\text{div} \vec V=\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}\quad \blacksquare$$

This is a result that holds for all $(x,y)$ in the domain of $\vec V(x,y)$.

Remark on transitioning from integrals over boundary (line integral) to definite integrals with bounds:

This is a standard result that takes into consideration the paramaterisation of the curve. Assuming the curve R (or L) is paramaterised by arclength, then the transition from a line integral to definite integral (with bounds) is straightforward (simply add the bounds and change the differential to $\mathbb{d}Y$ as done here).


问:
Centralizer of a $p$-element modulo the $p'$-core and conjugacy class sizes in quotient groups

Does $[ G : C_G(x) ] = [ G/K : C_{G/K}(x) ] [ K : C_K(x) ]$ hold for all finite groups $G$ and $p$-elements $x$, where $K = O_{p'}(G)$ is the largest normal subgroup of $G$ with order coprime to $p$?

A related statement is true:  Let $G = H \ltimes K$ be a semidirect product of $H$ with the (arbitrary) normal subgroup $K$, and let $x$ in $H$.  Suppose $hk$ centralizes $x$.  Then considering the group mod $K$, one gets that $h$ centralizes $x$.  Hence $k$ centralizes $x$, and $C_G(x) = C_H(x) \cdot C_K(x)$.  Importantly, we have the equality:

$$[ G : C_G(x) ] = [ G/K : C_{G/K}(x) ] [ K : C_K(x) ] .$$

If $G$ splits over $K = O_{p'}(G)$ so that $G = H \ltimes K$, then every $p$-element is conjugate to some element of $H$, and the previous implies the conjugacy class sizes behave as I expect.  However, $G$ often does not split over $K$ and I am not sure what happens then.

So suppose $G$ is a finite group with normal subgroup $K$.  One still has that every $h$ centralizing $x$ gives rise to an $hK$ in $C_{G/K}(x)$ and of course $C_G(x)$ is a union of cosets of $C_K(x)$.  However, it is not clear that all elements $hK$  of $C_{G/K}(x)$ give rise to elements $h$ in $C_G(x)$.  In other words, I can only prove that:

$$[ G : C_G(x) ] \geq [ G/K : C_{G/K}(x) ] [ K : C_K(x) ] .$$

I checked for counterexamples (small groups up to order $1000$ except $768$, primitive groups up to degree $500$ except $343$, and perfect groups up to order $10^6$ available in GAP), but found no counterexamples to equality.  I am not sure if the coprime hypothesis is truly relevant, but it is the case of interest and I could not think of any other way to say "$x$ in $H$" when $G$ did not split.  With absolutely no hypotheses on $x$ or $K$, then $G$ dihedral of order 8 with $K = Z(G)$ and $x$ a non-central involution gives a counterexample, but this is not at all similar to the case I am interested in.

答:
If $H$ is a p-subgroup of $G$, then it is true that:
$$[G:C_G(H)]=[G/K:C_{G/K}(H)]\ [K:C_K(H)].$$

This follows from two facts:

$C_K(H) = C_G(H)\cap K$; this is clear.

$ C_{G/K}(H)=C_G(H)K/K$.  To prove this, note first that $N_{G/K}(H)=N_G(H)K/K$.  This is equivalent to $N_G(HK)=N_G(H)K$, and this follows from the Frattini argument applied to $HK$ in $N_G(HK)$. Now the centralizer case follows because $N_G(H)$ is mapped to $N_{G/K}(H)$, $H$ is mapped to an isomorphic copy of itself, and the quotient map respects the action of $N_G(H)$ on $H$ (so, in particular, respects the kernel of this action, $C_G(H)$).


问:
Write the expressoin in terms of $\log x$ and $\log y \log(\frac{x^3}{10y})$
What is the answer for this? Write the expression in terms of $\log x$ and $\log y$ $$\log\left(\dfrac{x^3}{10y}\right)$$

This is what I got out of the equation so far. the alternate form assuming $x$ and $y$ are positive $$3\log(x)-\log(10y)$$ or maybe $$\log\left(\dfrac{x^3}{y}\right)-\log(5)=\log(2)$$?

I would appreciate a solution or an input thanks so much.

答:
As you have your answer I will simply give some help for future.

There are 3 basic(1,2 and 3) formula regarding logarithm:

1.$\ log_am+\ log_an=\ log_a(mn)$

2.$\ log_am-\ log_an=\ log_a(\dfrac {m}{n})$

3.$\ log_am^n=\ n\times\ log_a(m)$

4.$\ log_aa=1$

5.$\ log_ab\times \ log_na=\log_nb$

6.$\dfrac {\ log_am}{\ log_an}=\ log_nm$

and the basic which useful for solving log equation

$\log_{a}{b}=n\Longleftrightarrow a^n=b$

These can solve many problems regarding logarithm.