问:
Complement of a cyclic subgroup in a finite, abelian p-Group (Proof-Verification)
I'm trying to prove following statement, can anyone check if my proof works? In Isaacs, I.M. Algebra - A Graduate Course the proof is really long and complicated(but needs less theory), so I'm afraid I made a mistake, as the statement seems to be quite trivial with the use of the fundamental theorem of finitely generated abelian groups.

Lemma 1

In a finite, abelian p-group $G$, a cyclic subgroup $H$ of maximal possible order has a complement.

The following proof is incomplete, see my answer for a correct one.

Proof. Let $G\cong \Bbb{Z}_{p^{r_1}}\oplus\cdots \oplus \Bbb{Z}_{p^{r_s}}$ with $r_1 = \cdots =r_l \geq  ... \geq r_s$.
As $H$ is cyclic and of maximal order, $H=\Bbb{Z}_{p^{r_i}}$ for $i\in\{1,\cdots,l\}$.
Let $X$ be the subgroup consisting of all summands, but the $i$-th. Meaning $X=\bigoplus \Bbb{Z}_{p^{r_k}}$, $k \in \{1,\cdots,s\} \setminus \{i\}$.

Then $G=H \oplus X$ and $H$ has a complement in $G$.

q.e.d.

答:
Thanks to quasi I saw that my proof is incomplete. Here is now a full proof:

Lemma 1

In a finite, abelian p-group $G$, a cyclic subgroup $H$ of maximal possible order has a complement.

Proof.
If $G$ is cyclic, then $H=G$ and nothing needs to be shown. Let $G$ be not cyclic.
We make an induction over $|G|$.
If $|G|=p$, then $G$ is cyclic and we are finished.  Let $|G|>p$, and suppose, our lemma is true for all groups with order smaller then $|G|$.
Because $G$ is not cyclic, there is more than one subgroup of order $p$. Let $K \leq G$ be a subgroup with $|K|=p$ and $K \not \leq H$. $K$ has prime order, hence $H \cap K = \{e\}$ and $(HK)/K \cong H$.
For arbitrary $g \in G$ with corresonding coset $gK \in G/K$ is the following true: $ord(gK) | ord(g) \leq |H|$. Though $(HK)/K$ is a subgroup of maximal order in $G/K$. Our induction-hypothesis implies the existence of a complement $\bar{H'} \leq G/K$ of $(HK)/K$ in $G/K$. $\bar{H'}$ is the image of a group $H'$ under the canonical epimorphism $G \rightarrow G/K$ with $K \leq H \leq G$(correspondence-theorem).

Hence $G/K=((HK)/K)(H'/K)=(HKH')/K$ and finally: $$G=(HK)H'=H(KH')=HH' \; \text{with} \; H \cap H' = \{e\}.$$


问:
Why are convolutions so widely used in functional analysis
Functional Analysis has never been particularly intuitive to me, but in recent years I'm starting to see how it aids Fourier Analysis, Operator Theory and PDEs, but for the life of me I don't get the fixation on the convolution operation. I know it has some applications in engineering but it dates back to the time of Laplace. How is it used in mathematics?

One example I can think of is if you have a family of good kernels/mollifiers $D_n$, $f\in L^p$ then $D_n * f$ are smooth and converge to $f$.

答:
First of all, there's no "fixation" whatsoever. No more than there is a "fixation" with addition. Or with numbers. Or with vector spaces. Actually way way less, as convolution does not feature prominently.

In any case, convolution appears very naturally in lots of contexts. Off the top of my head, and with the caveat that I don't know/use/see convolution much:

in Probability, the distribution of the sum of two independent random variables is the convolution of the distributions;

when you multiply two polynomials, the coefficients of the products are obtained as the convolution of the coefficients;

in Group Theory, $\ell^1(G)$ is made naturally an algebra with convolution as the product. "Natural" because the canonical basis satisfies $e_{gh}=e_g*e_h$;

as mentioned by copper.hat, the Fourier transform takes convolution to pointwise product;

as mentioned in the question, convolution allows one to show that $C^\infty$ functions  with compact support are dense in $L^p(\mathbb R^n)$.

ContentLicense: CC BY-SA 4.0


问:
Zelda - Oracle of Ages tiles rooms puzzles: how can you prove if there are/aren't solutions?
In the Oracle games of Legend of Zelda, there are some puzzle rooms of this kind:

You are the green character on the bottom left. As you enter the room, all tiles (except for one, in yellow) are blue. The rules are:

You are assigned a designated starting square, somewhere in the room.

From your square, you can move to an adjacent square on your left, right, up or down (but not diagonals).

Once you leave a square, it will shift from blue to red.

You can't go back to red squares.

In the initial configuration, some squares will be 'forbidden'. In the screenshot, those are the ones with statues. You can't hop into them, but you also don't need to change their color (they can also be interpreted as 'holes' in the ground).

You solve the puzzle if you finish your path on the yellow square after changing every tile from blue to red.

If this set of rules is not properly defined, I apologize. The following gif from IGN of one of these rooms being solved may help visualize it better:

Through the game, these puzzles are you usually somewhat simple and have multiple solutions. However, the very last one, has this configuration:

As it turns out, it doesn't seems to be possible to 'solve' this room. No matter what path you try, you will eventually leave one blue tile *(I'm not claiming this as a fact, as i haven't seem a proof of this. But, considering the game, this seems to be what developers were aiming for: there is an in-game mechanic where you use an item to manually paint a single square as red - here is the intended solution:

.

The question is them: assuming you are given a rectangular room of some given size, with some set of (i) initial square, (ii) forbidden squares and (iii) ending square, is there a way (better than to try all possible paths) to be sure that no tile filling path exists? If so, is there a way to tell what would be the "best" result (in the sense of the path that leaves the smallest number of blue squares)?

Or, maybe if the general question is too broad, at least: how could the developers be sure that the last puzzle has no solution (apparently)?

答:
For the "impossible room", recolour the blue squares and yellow goal square black and white in a chessboard pattern, such that the upper-left square is white. Then

there is one more white square than black square

you start on a white square

the goal square is coloured black

If it were possible to solve the puzzle without using the Cane of Somaria, since every step goes from a white square to black and vice versa yet there is one more white square than black, you would have to end on a white square – i.e. the goal square would have to be white. But we coloured the goal square black, so we have to use the item.

When this puzzle is generalised to arbitrary graphs, this is the Hamiltonian path problem, which is NP-complete. Parity simply provides a way to quickly prove that there is no Hamiltonian path.


问:
Relations on given set
Let V={1,...,5}.
I want to specify the smallest possible relation on V that is

reflexive and symmetrical, but not transitive

symmetrical and transitive, but not reflexive

is it possible to just take for example {(1,1)} for both ?

答:
Symmetrical and transitive but not reflexive.

Smallest is the empty set.

Title: Relations on given set


问:
Show that $X$ and $Y$ are identically distributed.
Question: A hat contains $a$ slips of paper labeled with the number $1$, and $b$ slips labels with the number $2$. Two slips are drawn at random without replacement. Let $X$ be the number on the first slip drawn and $Y$ the number on the second. Show that $X$ and $Y$ are identically distributed.

Attempt: $X$ and $Y$ are identically distributed if $F_X(x)=F_Y(x)$ for $x=1,2$ (where $F_X, F_Y$ are the cumulative density functions of $X,Y$ respectively). Clearly $F_X(2)= F_Y(2) = 1$ ($X,Y$ are always $\leq 2$). To see the $x=1$ case observe

$$F_X(1) = \mathbb{P}(X \leq 1) = \mathbb{P}(X=1) = \frac{a}{a+b} $$

and also
\begin{align*}
F_Y(1) = \mathbb{P}(Y\leq 1) &= \mathbb{P}(Y=1 \  \text{given $X=1$}) + \mathbb{P}(Y=1 \ \text{given }X=2)\\
&=\frac{\mathbb{P}(Y=1, X=1)}{\mathbb{P}(X=1)} + \frac{\mathbb{P}(Y=1, X=2)}{\mathbb{P}(X=2)} \\
&= \frac{\frac{a}{a+b} \cdot \frac{a-1}{a+b-1}}{\frac{a}{a+b}}+ \frac{\frac{b}{a+b}\cdot \frac{a}{a+b-1}}{\frac{b}{a+b}} = \frac{a-1}{a+b-1}+\frac{a}{a+b-1} = \frac{2a-1}{a+b-1}
 \end{align*}

but then $F_X(1) \neq F_Y(1)$ and so $X,Y$ are $NOT$ identically distributed.  Did I do something wrong? They are supposed to be identically distributed.

答:
Since
$$
\mathsf{P}(X=1)=\frac{a}{a+b},\quad \mathsf{P}(X=2)=\frac{b}{a+b},
$$
\begin{align}
\mathsf{P}(Y=1)&=\mathsf{P}(Y=1\mid X=1)\mathsf{P}(X=1)+\mathsf{P}(Y=1\mid X=2)\mathsf{P}(X=2) \\
&=\frac{a-1}{a+b-1}\frac{a}{a+b}+\frac{b}{a+b-1}\frac{b}{a+b}=\frac{a}{a+b}.
\end{align}
and, similarly, $\mathsf{P}(Y=2)=b/(a+b)$.


问:
Find all triples of positive integers x, y and z such that x ≤ y ≤ z and $x^y + y^z = z^x$
I find one solution and that is (1,1,2), but still, I cannot get any idea on a general solution to this question. Could anyone help me with that or give me some hint?</p>

答:
It is well known that for $a>b$, if $a^b>b^a$, then we must have $(a,b)=(3,2)$ or $b=1$. In your equation, you have:
$$z^x =x^y+y^z >y^z \geqslant x^z$$

If $(z,x)=(3,2)$ then we have $2^y+y^3=9$, which clearly is not possible for positive integer $y$. Thus, we must have $x=1$. Substituting gives $y^z+1=z$. We know that $2^z>z$, so this once again forces $y=1$ and thus, $z=2$. Hence, the only solution is $(x,y,z)=(1,1,2)$.


问:
Convergence of the double series $\sum_{d|n} \mu(d) x_{n}$
Let $\sum x_n$ be a power series which behaves sufficiently nicely, for example, absolutely convergent. Can we deduce that the double series
$$
\sum_{d|n} \mu(d) x_{n}
$$
converges in the sense described below to zero? From the property of Mobius function $\mu$ that $\sum_{d|n} \mu (d)=0$, the only possible limit of this series is zero. But to show that it converges is a bit difficult, since $\mu$ does not necessarily sum nicely to zero in partial sums.

Definition of convergence:
we say that a double series $\sum a_{ij}$ converges to $l$ if for all $\epsilon>0$ there exists $N>0$ such that for all finite set $I\subseteq \mathbb N \times \mathbb N$ satisfying $\{0,1,\ldots, N\}^2 \subseteq I$, we have
$$
\left| \sum_{(i,j)\in I} a_{ij} - l \right|<\epsilon.
$$

In particular, for this question, we have
$$
a_{ij}=\mu(i) x_j \text{ if }i|j,\\
a_{ij}=0 \text{ otherwise. }
$$

Alternatively we can consider the series
$$
\sum_{d, m} \mu(d) x_{md},
$$
where $md$ is the product of $m$ and $d$.

答:
$$\sum_{n\ge 1}\sum_{d|n} \mu(d) x_{n}= x_1$$ Did you mean $$\sum_{d\ge 1}\mu(d)\sum_{m\ge 1} x_{md}$$ ? If $x_n=O( n^{-r})$ for some $r> 1$, since $\tau(n)=O( n^{(r-1)/2})$ then $$\sum_{n\ge 1}\sum_{d|n} |\mu(d) x_{n}|\le \sum_{n\ge 1}|x_{n}|\tau(n)$$ converges so we can change the order of summation to get that both series are equal.

The double series $\sum_{n,d} \mu(d) x_n$ converges independently of the order of summation (thus to $x_1$) iff it converges absolutely iff $\sum_n |x_n| 2^{\omega(n)}$ converges.

With $$\cases{x_n = k^{-1/4}-(k+1)^{-1/4} \ if \ n= \prod_{p\le k} p\\x_n=0 \ otherwise}$$ and $Lpf$ the largest prime factor we get that $$\sum_{d\ge 1}\mu(d)\sum_{m\ge 1} x_{md}=\sum_{d\ge 1} \mu(d)Lpf(d)^{-1/4}$$ which diverges (proof ?)


问:
True or False: If the function $(f + g)(x)$ is continuous, then the functions $f(x)$ and $g(x)$ are continuous
The full question comes from Advanced Calculus by Fitzpatrick, Section 3.1, Question 1a

Determine if the following is true or false and justify answer

If $f + g : \mathbb{R} \rightarrow \mathbb{R}$ is continuous, then $f : \mathbb{R} \rightarrow \mathbb{R}$ and $g : \mathbb{R} \rightarrow \mathbb{R}$ are continuous

My intuition tells me this is false, but when I work through it I'm arriving at something that looks like True

My attempt

This is of form $P \rightarrow Q$ so it will be False when we have $True \rightarrow False$.

Assume $f + g$ is continuous. Our goal is of form
$$\lim\limits_{n \to \infty}x_n = x_0 \rightarrow \lim\limits_{n \to \infty}f(x_n) = f(x_0) \quad \wedge \quad \lim\limits_{n \to \infty}x_n = x_0 \rightarrow \lim\limits_{n \to \infty}g(x_n) = g(x_0)$$

We can prove each separately. Note that each is of the form $P \rightarrow Q$ so we will assume $P$ again and show $Q$.

Let $x_0$ be arbitrary and assume $\lim\limits_{n \to \infty}x_n = x_0$.

\begin{align}
   \lim\limits_{n \to \infty}x_n = x_0 \rightarrow&& &\lim\limits_{n \to \infty}(f+g)(x_n) = (f+g)(x_0) &&\text{assumption and definition of continuity} \tag 1\\
   \lim\limits_{n \to \infty}x_n = x_0 \rightarrow&& &\lim\limits_{n \to \infty}f(x_n) + g(x_n) = f(x_0) + g(x_0) && \text{def. of sum of functions} \tag 2\\
   \lim\limits_{n \to \infty}x_n = x_0 \rightarrow&& &\lim\limits_{n \to \infty}f(x_n) + \lim\limits_{n \to \infty}g(x_n) = f(x_0) + g(x_0) && \text{sum property of limits} \tag 3
\end{align}

At this point it seems like you can get $\lim\limits_{n \to \infty}f(x_n) = f(x_0)$ and $\lim\limits_{n \to \infty}g(x_n) = g(x_0)$ which would show this all to be true, however I don't think that is a valid step.

That being said, that doesn't prove it False - I think I need a counterexample but I'm not seeing one.

Any help is appreciated to refine this.

答:
It is false. Let $f\colon \mathbb R \to \mathbb R$ be:
$$
f(x) = \begin{cases}
-1 &\text{if } x \lt 0 \\
1 & \text{otherwise}
\end{cases}
$$
and let $g\colon \mathbb R \to \mathbb R$ be:
$$
g(x) = \begin{cases}
1 &\text{if } x \lt 0 \\
-1 & \text{otherwise}
\end{cases}
$$

Then $f$ and $g$ are not continuous at $x = 0$, but their sum is the zero function, which is continuous everywhere.


答: take $f(x)$ to be constantly 0 for $x<0$ and constantly 1 for $x\geq 0$, then g to be constantly 0 for $x<0$ and constantly -1 for $x\geq 0$. The sum is the constant zero function which is continuous.

Title: True or False: If the function $(f + g)(x)$ is continuous, then the functions $f(x)$ and $g(x)$ are continuous

Body:
The full question comes from Advanced Calculus by Fitzpatrick, Section 3.1, Question 1a

Determine if the following is true or false and justify answer

If $f + g : \mathbb{R} \rightarrow \mathbb{R}$ is continuous, then $f : \mathbb{R} \rightarrow \mathbb{R}$ and $g : \mathbb{R} \rightarrow \mathbb{R}$ are continuous

My intuition tells me this is false, but when I work through it I'm arriving at something that looks like True

My attempt

This is of form $P \rightarrow Q$ so it will be False when we have $True \rightarrow False$.

Assume $f + g$ is continuous. Our goal is of form
$$\lim\limits_{n \to \infty}x_n = x_0 \rightarrow \lim\limits_{n \to \infty}f(x_n) = f(x_0) \quad \wedge \quad \lim\limits_{n \to \infty}x_n = x_0 \rightarrow \lim\limits_{n \to \infty}g(x_n) = g(x_0)$$

We can prove each separately. Note that each is of the form $P \rightarrow Q$ so we will assume $P$ again and show $Q$.

Let $x_0$ be arbitrary and assume $\lim\limits_{n \to \infty}x_n = x_0$.

\begin{align}
   \lim\limits_{n \to \infty}x_n = x_0 \rightarrow&& &\lim\limits_{n \to \infty}(f+g)(x_n) = (f+g)(x_0) &&\text{assumption and definition of continuity} \tag 1\\
   \lim\limits_{n \to \infty}x_n = x_0 \rightarrow&& &\lim\limits_{n \to \infty}f(x_n) + g(x_n) = f(x_0) + g(x_0) && \text{def. of sum of functions} \tag 2\\
   \lim\limits_{n \to \infty}x_n = x_0 \rightarrow&& &\lim\limits_{n \to \infty}f(x_n) + \lim\limits_{n \to \infty}g(x_n) = f(x_0) + g(x_0) && \text{sum property of limits} \tag 3
\end{align}

At this point it seems like you can get $\lim\limits_{n \to \infty}f(x_n) = f(x_0)$ and $\lim\limits_{n \to \infty}g(x_n) = g(x_0)$ which would show this all to be true, however I don't think that is a valid step.

That being said, that doesn't prove it False - I think I need a counterexample but I'm not seeing one.

Any help is appreciated to refine this.


问:
Probability Density of x
Suppose that I have the function $$x(t)=e^{-b(W(t))^2} \ \ \ \ \ \ \ \ (1)$$ where W(t) is just a Wiener process (i.e. a Gaussian in general). I want to know what the probability density for $x$, $P(x)$, is. I started off by just assuming that I want to measure the expectation value of an observable $f(x)$, so $$<f(x)>=\int_{W=0}^{W=t}{P(W)f(g(W))dW} \ \ ,\ \ x=g(W) $$ Then I transformed variables from $W$ to $t$ and I got $$<f(x)>=\int_{x=g(0)}^{x=g(t)}P(g^{-1}(x))f(x)(\frac{dW}{dx})dx=\int_{x=g(0)}^{x=g(t)}\frac{P(g^{-1}(x))}{g'(g^{-1}(x))}f(x)dx$$ so I just assume that the probability for x is $$P(x)=\frac{P(g^{-1}(x))}{g'(g^{-1}(x))}$$ Since $x=g(W)$, then $W=g^{-1}(x)$, but from (1) I get $$\frac{dW}{dx}=\frac{1}{2x\sqrt{blnx}}$$ and assuming that $\sigma^2=V=t$, I get $$P(x)=\frac{e^{lnx/{2bt}}2\sqrt{blnx}}{x\sqrt{2{\pi}t}}$$ Is this right? Shouldn't I get a Gaussian? Also, was I right to take the values of $x$ to be $[0,t]$ and not $(-\infty,+\infty)$ ? Thank you in advance for reading my question :)

答:
Hint:

The distribution of a (monotone) transformation $Y=g(X)$  of a random variable $X$ is given by

$$F_Y(y)=F_X(g^{-1}(y))$$
from where you can deduce that the density is given by

$$f_Y(y)=\frac{d}{dy}F_X(g^{-1}(y))\cdot |\frac{d}{dy}g^{-1}(y)|=f_X(g^{-1}(y))\cdot |\frac{d}{dy}g^{-1}(y)|.$$

In your case the random variable $X=W(t)$ is Gaussian with variance $t$, hence you know what $f_X$ is. Your transformation is not monotone, hence you have to divide the domain of $g$ into two parts ($(-\infty,0)$ and $[0,\infty)$) and then add up the two parts.

Title: Probability Density of x


问:
Number of solution of three equation in term of $x,y,z,\theta$
Find the number of all possible values of $\theta$, where $0<\theta<\pi$, for which the system of equations

$\left( {y + z} \right)\cos 3\theta  = xyz\sin 3\theta $

$x\sin 3\theta  = \frac{{2\cos 3\theta }}{y} + \frac{{2\sin 3\theta }}{z}$

$xyz\sin 3\theta  = \left( {y + 2z} \right)\cos 3\theta  + y\sin 3\theta $
have a solution $x_0,y_0,z_0$ with $y_0z_0\ne 0$ .

My approach is as follow

$\left( {y + z} \right)\cos 3\theta  = xyz\sin 3\theta  \Rightarrow \frac{{y + z}}{{xyz}} = \tan 3\theta $

$x\sin 3\theta  = \frac{{2\cos 3\theta }}{y} + \frac{{2\sin 3\theta }}{z} \Rightarrow \sin 3\theta \left( {x - \frac{2}{z}} \right) = \frac{{2\cos 3\theta }}{y} \Rightarrow \tan 3\theta  = \frac{2}{{y\left( {x - \frac{2}{z}} \right)}} = \frac{{2z}}{{y\left( {xz - 2} \right)}} = \frac{{2z}}{{\left( {xyz - 2y} \right)}}$

$xyz\sin 3\theta  = \left( {y + 2z} \right)\cos 3\theta  + y\sin 3\theta  \Rightarrow \sin 3\theta \left( {xyz - y} \right) = \cos 3\theta \left( {y + 2z} \right) \Rightarrow \tan 3\theta  = \frac{{y + 2z}}{{\left( {xyz - y} \right)}}$

$\tan 3\theta  = \frac{{y + z}}{{xyz}} = \frac{{y + 2z}}{{\left( {xyz - y} \right)}} = \frac{{2z}}{{\left( {xyz - 2y} \right)}}$

$xyz = t \Rightarrow \tan 3\theta  = \frac{{y + z}}{t} = \frac{{y + 2z}}{{t - y}} = \frac{{2z}}{{t - 2y}}$.

How do we proceed from here.

Body:
$$
\begin{cases}
x y z \sin (3 t)=y \cos (3 t)+z \cos (3 t)&(1)\\
x y z \sin (3 t)=2 y \sin (3 t)+2 z \cos (3 t)&(2)\\
x y z \sin (3 t)=y \sin (3 t)+y \cos (3 t)+2 z \cos (3 t)&(3)\\
\end{cases}
$$
$$
\begin{cases}
y \cos (3 t)+z \cos (3 t)=2 y \sin (3 t)+2 z \cos (3 t)&(1)=(2)\\
y \cos (3 t)+z \cos (3 t)=y \sin (3 t)+y \cos (3 t)+2 z \cos (3 t)&(1)=(3)\\
\end{cases}
$$

$$
\begin{cases}
z \cos (3 t)=y \cos (3 t)-2 y \sin (3 t)\\
z \cos (3 t)=-y\sin(3t)\\
\end{cases}
$$
$$y \cos (3 t)-2 y \sin (3 t)=-y\sin(3t)$$
$$\sin(3t)=\cos(3t)\to \tan(3t)=1$$
$$3t=\frac{\pi}{4}+k\pi\to t=\frac{\pi}{12}+k\frac{\pi}{3}$$
Three solutions in $[0,\pi]$
$$t_1=\frac{\pi}{12};\;t_2=\frac{5\pi}{12};\;t_3=\frac{3\pi}{4}$$


问:
How to prove this centering inequality
When X is a random variable, please prove that
$$
\|X-EX\|_{L^2}\leq\|X\|_{L^2}
$$
It might be trivial, but I'm stuck on it.Thanks in advance!

答:
$$
\mathsf{E}[X-\mathsf{E}X]^2=\mathsf{E}X^2-(\mathsf{E}X)^2\le \mathsf{E}X^2.
$$

Title: How to prove this centering inequality


问:
If a condition is true for sample, is it possible to calculate the probably that it is true for the whole population
I have spent some time Googling this, but I suspect I am using the wrong terminology, so please excuse me if this has been answered many times before.

Example: A computer system is designed to transform data records from system 1 to system 2 via a complex process that may have bugs. A human verifies the integrity of the records for a random sample of those that have been transferred.

If the human checks a random sample of records that have been transferred and finds that all these randomly chosen records are free of problems, what is the probably that this is true for all records?

答:
To handle this you may need some kind of Bayesian analysis, starting with a prior view of the distribution of the number of possible errors in the transfer of $N$ records, then combining this with the observation that $n$ selected at random without replacement do not have any errors, to get a posterior distribution for the number of remaining errors.

If $\mathbb P(X=x)$ is your prior probability of $x$ errors in total, and none are observed in the sample, then the posterior probability of the number of remaining errors also being $0$ would be $$\mathbb P(X=0 \mid \text{no errors in sample of }n)=\frac{\mathbb P(X=0) {N \choose n}}{\sum\limits_{x=0}^{N-n} \mathbb P(X=x) {N-y \choose n}}$$

Even then it might not be as helpful as you might hope, since the result is sensitive to your prior. Here are two examples of different prior distributions with the same prior mean:

Suppose your prior probability that the total number of errors is $x$ is uniform, i.e. $\frac{1}{N+1}$ for $x \in \{0,1,2,3,\ldots,N\}$. Then having observed no errors in the sample of $n$, the posterior probability of no errors at all is $\frac{n+1}{N+1}$

Suppose your prior probability that the total number of errors is $x$ is binomial with parameter $\frac12$, i.e. ${N \choose x}\frac{1}{2^{N}}$. Then having observed no errors in the sample of $n$, the posterior probability of no errors at all is $\frac{1}{2^{N-n}}$

These can be very different: suppose you transfer $10$ records and check $5$ of them as error-free. Then the first approach gives you a posterior probability of no errors at all of $\frac{6}{11} \approx 0.545$  while the second gives a posterior probability of no errors at all of $\frac{1}{2^{5}}=0.03125$


问:
Holomorphic function from upper half plane
If $f$ is a holomorphic function from the upper half-plane(without the real line) to the unit disc, and if $f(i) = 0$, how large can $|f(2i)|$ be under these conditions?

I'm not sure what theorems and techniques to use here. Schwarz's lemma or cauchy estimates? But I'm not sure how to use them here. Any suggestions? This is from a Dutch complex analysis textbook

答:
The upper half plane is conformally equivalent to the unit disc by the Cayley transform
$$\varphi:\mathbb H\longrightarrow\mathbb D,~z\mapsto\frac{z-\mathrm i}{z+\mathrm i}.$$
The Cayley transform also maps $\mathrm i\mapsto0$, so its inverse $\varphi^{-1}$ maps $0\mapsto\mathrm i$.

Now $f\circ\varphi^{-1}$ is a holomorphic map $\mathbb D\longrightarrow\mathbb D$ which maps $0$ to $f(\varphi^{-1}(0))=f(\mathrm i)=0$. Now apply the Schwarz lemma to $f\circ\varphi^{-1}(\varphi(2\mathrm i))=f\circ\varphi^{-1}(\frac{1}{3})$.

It is generally a useful trick to consider conformal maps to the unit disc, because the Schwarz lemma tells us a lot about how holomorphic functions behave on the disc. So if we have a holomorphic function $f:A\longrightarrow B$, where $A$ and $B$ are conformally equivalent to the unit disc via the conformal maps $\varphi_{A}:A\longrightarrow \mathbb D$ and $\varphi_{B}:B\longrightarrow \mathbb D$, then the function $\varphi_B\circ f\circ\varphi_A^{-1}$ is a holomorphic map from the unit disc to the unit disc. If we also manage to choose $\varphi_A$ and $\varphi_B$ such that this map $\varphi_B\circ f\circ\varphi_A^{-1}$ maps $0$ to $0$, then we can apply the Schwarz lemma to the composite function, which can yield desirable insights.


问:
If $\space\lim _{x\to a}f\left(x\right) = L\space$ for every$\space a\in R,$ then $\space f\left(x\right) = L\space$ for every$\space x \in R$.
Let $\space f:\mathbb R\rightarrow \mathbb R.\space$ If $\space\lim _{x\to a}f\left(x\right) = L\space$ for every$\space a\in R,$ then $\space f\left(x\right) = L\space$ for every$\space x \in R$.

prove or disprove.

I believe it's true.
I assumed my contradiction that $\space f\left(x\right) \ne L\space$ , so it's either $\space f\left(x\right) > L\space$ or $\space f\left(x\right) < L\space$

so that means that $|\space f\left(x\right) - L\space| > 0\space$ and i can choose $\epsilon > 0\space$ to be $|\space f\left(x\right) - L\space|\space$ and get a contrediction.

but I don't know if this direction is even possible. I would love to hear your thoughts.

thanks!

答:
Let $f(x) = \begin{cases} 1,& x = 0 \\ 0, & \text{otherwise}\end{cases}$.

Then we have $\lim_{x \to 0} f(x) = 0$ and for $a \neq 0$ we have $\lim_{x \to a} f(x) = 0$.


问:
Help solving the initial value problem dy/dt = 2t/(y-t)
I am trying to solve the following initial value problem for this differential equation:

$$\frac{dy}{dt} = \frac{2t}{y-t} , \hspace{20px} y(1) = 0$$

I have tried the substitution $v=\frac{y}{t}$ which transforms the ODE into a separable equation, but I don't think it was the best approach to solve this problem. Can you help me figuring out what to do?

Thanks!

答:
$$\frac{dy}{dt} = \frac{2t}{y-t} , \hspace{20px} y(1) = 0$$
Substitute $y=vt$:
$$v't+v = \frac{2}{v-1}
$$
$$v't = \frac{2-v^2+v}{v-1}
$$
This differential equation is separable.


答:
Other answers have provided you with the counterexamples you need, but as to where you went wrong, it's in your third item, the one you labelled "sum property of limits", where you seem to have used

$$ \lim (f+g) = \lim f + \lim g,$$

which is true SO LONG AS we know that $\lim f$ and $\lim g$ exist. Which you don't.


问:
Prove that $\sum_i {n\choose i}{2n\choose n-i}={3n\choose n}$.
I want to calculate Example 9 on page 130 of Herbert S. Wilf's book generatingfunctionology (2. edition)link to the pdf of the book via the Snake Oil method. I followed the suggestion given by the author to temporarily replace $2n$ by $m$ and $n-i$ by $r-i$ and I got the following generating function
\begin{align}
F(x)&=\sum_{n\geq 0}x^n\sum_i{n\choose i}{m\choose r-i}\nonumber\\
 &=\sum_i{m\choose r-i}\sum_{n\geq 0}{n\choose i}x^n\nonumber\\
 &=\sum_i{m\choose r-i}\frac{x^i}{(1-x)^{i+1}}\nonumber\\
 &=\frac{x^r}{(1-x)^{r+1}}\sum_i{m\choose r-i}\left(\frac{1-x}{x}\right)^{r-i}\nonumber\\
 &=\frac{x^r}{(1-x)^{r+1}}\left(1+\frac{1-x}{x}\right)^{m}\nonumber
\end{align}
The 3rd equality is obtained by the identity  $(4.3.1)$ on page 120 of the book and the last equality is obtained by the binomial theorem. The Snake Oil method suggest to find the coefficients of this generating function because this is the answer we are looking for, but I don't know how to do this.

Note that I used the customary conventions about binomial coefficients and the ranges of summation variables given in the book. These are: first that the binomial coefficient ${x\choose m}$ vanishes if $m < 0$ or if $x$ is a nonnegative integer that is smaller than $m$. Second, a summation variable whose range is not
otherwise explicitly restricted is understood to be summed from $-\infty$ to $\infty$.

答:
We have
\begin{align*}
\frac{x^r}{(1-x)^{r+1}}\left(1+\frac{1-x}{x}\right)^{m}\nonumber
&=
\frac{x^r}{(1-x)^{r+1}}\left(\frac{1}{x}\right)^{m}\nonumber\\
&=
\frac{x^{r-m}}{(1-x)^{r+1}}\\
&=
x^{r-m} \sum_{k=0}^{+\infty} \binom{k+r}{r}x^k\\
&= \sum_{k=0}^{+\infty} \binom{k+2n}{n}x^{k}.
\end{align*}
Since $m=2n$ and $r=n$ and by letting $k=n$, we get
$$
\sum_i{n\choose i}{2n\choose n-i}=\binom{3n}{n}.
$$


问:
A finite ring without zero divisors is a division ring
Question: Let R be a finite ring without zero divisors and |R| > 1. Then show that R is a division ring.

I used the following logic, but I'm not sure if this is correct.

Let |R| = n, a finite number.
$Let 0 \not = a \in R.$

Therefore, $a^n = a => a(a^{n-1} - 1) = 0 => a^{n-1} = 1 \,\,\, (\text{Since} \,\, a \not = 0)$

Therefore, $a a^{n-2} = 1$

Therefore $a^{n-2}$ is the inverse of a for all a.
Therefore R is a division ring.

Does this work?

答:
Why do you think that $a^n=a$? This is not obvious at all. It will end up being true, because if it is a division ring then the nonzero elements will form a group under multiplication, and then it follows from Lagrange's theorem. But you have to prove it first.

First of all, by induction it follows that if $a\ne 0$ then $a^n\ne 0$ for all $n\in\mathbb{N}$, this is because there are no zero divisors. Now, it is true that there are some $j>i$ such that $a^j=a^i$, because the set $\{a,a^2,a^3,a^4,...\}$ must be finite. Then $0=a^j-a^i=a^i(a^{j-i}-1)$. Since $a^i\ne 0$ it follows that $a^{j-i}-1=0$, again because there are no zero divisors. So now $a^{j-i-1}$ is an inverse of $a$.


问:
Proof Regarding the Upper Bound of the Product $\alpha(G) \omega(G)$ in a Graph
I came across this question in my textbook, but I am having trouble solving it.

Prove that $\alpha(G) \cdot \omega(G) \leq \frac{n(n+2)}{4}$ when $n$ is even, and $\alpha(G) \cdot \omega(G) \leq \frac{(n+1)^2}{4}$ when $n$ is odd. Here $\alpha(G)$ is the independence number and $\omega(G)$ is the clique number in a graph with $n$ vertices.

I tried to prove this by induction, but got stuck. I would appreciate any help on this!

答: Note that $\alpha(G) = \omega(\overline{G})$. The problem boils down to maximizing the product $\omega(G)\omega(\overline{G})$. You will need to show that $\omega(G) + \omega(\overline{G}) \le n+1$. Maximizing the sum implies maximizing the product, so assuming equality of the sum leads to $$\omega(G)\omega(\overline{G}) = \omega(G)(n+1 - \omega(G)) = (n+1)\omega(G) - (\omega(G))^2.$$ This is a quadratic (in $\omega(G)$) with a maximum, so find the vertex (of the quadratic function). This is where the product is maximized. What remains is to think about what the product looks like when $n$ is even or odd.


问:
Example of a curve which is C1 continuous but not G1 continuous.
I'm reading about parametric and geometric continuity using  these UC Berkeley Lecture notes.

In the section "Graph of the curve", this is written:

While the conditions for parametric continuity seem stronger than
geometric continuity, they are not. There are C1 curves that are not
G1.

Can anyone give an example for such a curve?

Found a similar question here but the example given is of a curve which is G1 continuous but not C1 continuous.

These Cornell lectures seemingly give an example of such a curve but it is not very clear. (Page 6 of the pdf)

答:
Consider the parameterized curve $\gamma(t)=(t^3,t^2)$. This parameterization is clearly $C^1$, because each component is (infinitely!) differentiable.

However, the graph of the curve looks like this:

As you can see, the left and right tangent vectors at the origin (where $t=0$) do not match, so the curve is not $G^1$.

This is possible because $\gamma'(0)=\left<0,0\right>$. That is, the parameterization comes gradually to a stop at $t=0$, which allows it to smoothly traverse a geometrically unsmooth curve.


Title: Prove that $\sum_i {n\choose i}{2n\choose n-i}={3n\choose n}$.

Body:
I want to calculate Example 9 on page 130 of Herbert S. Wilf's book generatingfunctionology (2. edition)link to the pdf of the book via the Snake Oil method. I followed the suggestion given by the author to temporarily replace $2n$ by $m$ and $n-i$ by $r-i$ and I got the following generating function
\begin{align}
F(x)&=\sum_{n\geq 0}x^n\sum_i{n\choose i}{m\choose r-i}\nonumber\\
 &=\sum_i{m\choose r-i}\sum_{n\geq 0}{n\choose i}x^n\nonumber\\
 &=\sum_i{m\choose r-i}\frac{x^i}{(1-x)^{i+1}}\nonumber\\
 &=\frac{x^r}{(1-x)^{r+1}}\sum_i{m\choose r-i}\left(\frac{1-x}{x}\right)^{r-i}\nonumber\\
 &=\frac{x^r}{(1-x)^{r+1}}\left(1+\frac{1-x}{x}\right)^{m}\nonumber
\end{align}
The 3rd equality is obtained by the identity  $(4.3.1)$ on page 120 of the book and the last equality is obtained by the binomial theorem. The Snake Oil method suggest to find the coefficients of this generating function because this is the answer we are looking for, but I don't know how to do this.

Note that I used the customary conventions about binomial coefficients and the ranges of summation variables given in the book. These are: first that the binomial coefficient ${x\choose m}$ vanishes if $m < 0$ or if $x$ is a nonnegative integer that is smaller than $m$. Second, a summation variable whose range is not
otherwise explicitly restricted is understood to be summed from $-\infty$ to $\infty$.

Body:
\begin{eqnarray*}
\sum_{i=0}^{n} \binom{n}{i} \binom{2n}{n-i} &=& [x^n] \sum_{i=0}^{n} \binom{n}{i} x^{i} (1+x)^{2n} \\
&=& [x^n] (1+x)^n (1+x)^{2n} = [x^n]: (1+x)^{3n} = \binom{3n}{n}.\\
\end{eqnarray*}


问:
Divisibility rules and congruences
Sorry if the question is old but I wasn't able to figure out the answer yet.
I know that there are a lot of divisibility rules, ie: sum of digits, alternate plus and minus digits, etc... but how can someone derive those rules for any number $n$ let's say. I know it could be done using congruences, but how ?

Thank you !

ContentLicense: CC BY-SA 3.0

Title: Divisibility rules and congruences

Body:
Bill Dubuque's answer gave me the motivation to calculate the $\text{modulus-}7$ residue 'on a computer'.

In $\text{base-}2$

$\quad 43211 =1010100011001011_2$

Computing the $\text{mod-}7$ residue of $43211$:

Initialize the algorithm input, $1010100011001011 \sim [1]010100011001011$.

$\quad [1]010100011001011; \; [2]10100011001011; \; [5]0100011001011; \; [3]100011001011;$

$\quad [0]00011001011; [0]11001011; \; [1]1001011; \; [3]001011; \; [6]01011; \; [5]1011;$

$\quad [4]011; \; [1]11; \; [3]1; \; [0]$

So $43211 \equiv 0 \pmod7$.


问:
$\int_{-\infty}^{\infty}\!e^{- \pi (x+iy)^2}\,dx = 1$ for all $y$.
Can anyone provide a proof of why $\int _{-\infty} ^ {\infty} e^{-\pi (x+iy)^2} dx$ equals 1, for all y ? $x$ and $y$ are real numbers.

EDIT: We already know this for y=0.

Thank you.

答:
By the Cauchy integral theorem,  $\int_{\partial R}f(z)\,dz=0$ where $R$ is the rectangle with vertices $\pm A$, $\pm A+iy$, and $A$ can be any positive number.

For $z$ lying on either vertical side of $R$ we have $\mathrm{Re}\,(z^2)\ge A^2-y^2$. (Why?)

When $A\ge 2|y|$, the function $e^{-\pi z^2}$ does not exceed $e^{- \pi A^2/2}$ on either vertical side of $R$. (Why?)

The contribution of the vertical sides of $R$ to the integral  $\int_{\partial R}f(z)\,dz$ tends to zero as $A\to\infty$. (Why?)

$\lim_{A\to\infty}\int_{-A}^A e^{-\pi (x+iy)^2}\,dx = \lim_{A\to\infty}\int_{-A}^A e^{-\pi x^2}\,dx$.

I assumed you already know that the integral converges. If this is not known, then either replace endpoints $\pm A$ with $A,B$ where $A<0<B$, or argue convergence separately, using the comparison test.


问:
can we say that $\lim\limits_{x\to0} \sqrt{-x^2}=0$ even if the function domain is a singleton $\{0\}$
In Real analysis, the strict definition of a limit, we take a sequence of point that converging to the point a "limit point". which is not verified here in this case we have only one element in the function domain :

$$\lim_{x\to 0} \sqrt{-x^2}$$

but by replacing $0$ in the function we get $0$

so my question is : in a formal way can we say that

$$\lim_{x\to 0} \sqrt{-x^2}=0$$

答:
We can say so with perfectly good conscience. Ignoring all the notational baggage, the function in question is
$$f:\{0\}\longrightarrow\mathbb R,~0\mapsto0.$$
Here, $\{0\}$ is a metric space with the only possible metric on a singleton set: $d(0,0)=0$. With this metric, the set $\{0\}$ gains the discrete topology, with respect to which every function to any other metric space (or even topological space) is continuous. It's not a particularly interesting topology, but it's a perfectly fine one.

Another argument why it's perfectly fine to allow such a function to be continuous: Let $f:X\longrightarrow Y$ be a constant map where $X,Y$ are metric spaces (or even topological ones). Then $f$ is continuous. This is a very intuitive statement which is also easy to prove in the context of topology. Your function is a constant function between metric spaces, so we'd have to make an exception for this theorem, which in my opinion would be way worse than calling your function continuous.


问:
Limit of integral with gamma function
By some relations of Whittaker functions and a comment I read in a paper I came up with the following identity
$$\lim_{s\to 0+} \frac{1}{\Gamma(s)} \int_{0}^\infty e^{-vt} t^{s-1} (1+t)^{s+1}dt =1,$$
which should hold for all positive $v$ (the $0+$ under the limit shall indicate that I approch $s=0$ with positive values of $s$ since that are the values for which the expression is defined). Is there a direct way to prove it?

答:
Integrating by parts with $u=e^{-vt}(1+t)^{s+1}$ and $v=t^s/s$, we find that

$$\begin{align}
\int_0^\infty e^{-vt}t^{s-1}(1+t)^{s+1}\,dt&=\left.\left(\frac{t^s}{s}e^{-vt}(1+t)^{s+1}\right)\right|_0^\infty -\frac1s \int_0^\infty t^s\frac{d}{dt}\left(e^{-vt}(1+t)^{s+1}\right)\,dt\\\\
&=-\frac1s \int_0^\infty t^s\frac{d}{dt}\left(e^{-vt}(1+t)^{s+1}\right)\,dt
\end{align}$$

Inasmuch as $\Gamma(s)\sim \frac1s$ as $s\to 0^+$, the limit of interest is

$$\begin{align}
\lim_{s\to 0^+}\frac1{\Gamma(s)}\int_0^\infty e^{-vt}t^{s-1}(1+t)^{s+1}\,dt&=-\lim_{s\to 0^+}\int_0^\infty t^s\frac{d}{dt}\left(e^{-vt}(1+t)^{s+1}\right)\,dt\\\\
&=-\lim_{s\to 0^+}\int_0^\infty \frac{d}{dt}\left(e^{-vt}(1+t)^{s+1}\right)\,dt\\\\
&=-\lim_{s\to 0^+}\left.\left((e^{-vt}(1+t)^{s+1}\right)\right|_0^\infty\\\\
&=1
\end{align}$$

as was to be shown!

Title: Limit of integral with gamma function


问:
Evaluating limits involving absolute values: $\lim_{x\to\pm6} \frac{2x+12}{|x+6|}$
I have two questions regarding limits involving absolute values. How do I evaluate the following:

$\displaystyle\lim_{x\to -6} \frac{2x+12}{|x+6|}$

$ \displaystyle \lim_{x\to 6} \frac{2x+12}{|x+6|}$.

To handle the first problem, I considered two cases: $x\gt -6$ and $x\lt -6$ and thus get $2$ and $-2$ respectively for the right and left hand limits.

However, I am bit confused regarding the second problem. How do I handle that?

答:
For the second problem, when $x$ approaches $6$, $x+6$ will be positive — the absolute values are unecessary, they're just here for show.

(more generally, if $a\neq 0$ and $x_n\operatorname*{\to}_{n\to\infty} a$, then there exists $N \geq 0$ such that $\forall n\geq N$ $x$ and $a$ have same sign)


问:
Problem with proving Brownian Motion
Hey i want to prove that $Y_t= \frac{1}{\sqrt{2}}M_t $ is a B.M, while
$$
M_t = X_t-X_0 + \int_{0}^{t} X_s ds
$$
and $$
X_t=e^{-t} B_{e^{2t}}, $$
where $(B_t)_{t \geq 0}$ is a standart B.M..

In order to show this i wanted to show that $<Y>_t=t$.

We have that $dM_t=dX_t+X_tdt.   $ Hence $$ d<Y>_t=dY_t * dY_t = \frac{1}{2}dM_t*dM_t = \frac{1}{2} (dX_t+X_tdt)(dX_t+X_tdt)=(1).$$
Since $dX_t= d(e^{-t}B_{e^{2t}}) = -e^{-t} B_{e^{2t}} dt + e^{-t} dB_{e^{2t}}, $

we get $$ (1)= \frac{1}{2} e^{-2t}dB_{e^{2t}} dB_{e^{2t}} = \frac{1}{2} e^{-2t}e^{2t} dt = \frac{1}{2} dt.  $$
Hence $<Y>_t \neq t.$ So $Y_t$ is not a Brownian Motion? So what am i missing?
Furthermore i am quiet sure that $Y_t$ is not a martingale.
Any help would be appreciate.

答:
Hey i tried this again and got this:

$<Y>_t=t$, since i forgot a factor $2$. (From the answer from @Hoil Lee)
Hence it remains to show that $Y_t$ is a martingale. We have
$$ \mathbb{E}[M_t-M_s \vert \mathcal{F}_s]= \mathbb{E}[X_t-X_s + \int_s^t X_u du \vert \mathcal{F}_s]= \mathbb{E}[X_t-X_s \vert \mathcal{F}_s]+ \mathbb{E}[\int_s^t X_u -X_s+X_u \vert \mathcal{F}_s]
= \mathbb{E}[X_t-(t-s-1)X_s \vert \mathcal{F}_s] + \int_s^t \mathbb{E}[ X_u-X_s \vert \mathcal{F}_s]=(1)$$
Since the fact from @Jan Stuller, we have that $ X_t \sim e^-t \frac{\sqrt{e^{2t}}}{\sqrt{t}}B_t $ and we know that $B_t-B_s \amalg \mathcal{F}_s $, we can conclude that $$
(1)=\mathbb{E}[X_t-(t-s-1)X_s] + \int_s^t \mathbb{E}[ X_u-X_s]=0,$$
which proofs that $Y_t$ is a Martingale and hence a Brownian Motion.

Title: Problem with proving Brownian Motion


问:
How do I solve this exercise of double integral $\iint_{D}\frac{1}{\sqrt{x^{2}+y^{2}}}dA$?
I have the double integral $\iint_{D}\frac{1}{\sqrt{x^{2}+y^{2}}}dA$ where $D$ is the domain given by:

$D = \left \{ (x, y ) \in \mathbb R^{2} : x^{2}+y^{2} \le 2y\right \} $

I tried this in polar coordinates: $$\left ( r cos \Theta \right )^{2} + \left ( r sin \Theta - 1 \right )^{2} = 1 $$

$$r = 2sin\Theta $$ $$\int_{-\Pi /4}^{\Pi /4}\int_{0}^{2sin\Theta } \frac{1}{r} rdrd\Theta$$

Am I the right way? How to resolve then?

答: Draw a diagram: the correct range for $\theta$ is from $\pi/4$ to $3\pi/4$, giving$$\int_{\pi/4}^{3\pi/4}\int_0^{2\sin\theta}drd\theta=\int_{\pi/4}^{3\pi/4}2\sin\theta d\theta=[-2\cos\theta]_{\pi/4}^{3\pi/4}=2\sqrt{2}.$$


问:
Quadratic and arithmetic mean inequality
I want to prove by induction that

$$\left(\tfrac{1}{n}\sum_{k=1}^{n} x_k\right)^2\leq\tfrac{1}{n}\sum_{k=1}^{n} x_k^2$$

But I have no idea how. For n=1 it is trivial that
$$\left(\tfrac{1}{1}\sum_{k=1}^{1} x_k\right)^2\leq\tfrac{1}{1}\sum_{k=1}^{1} x_k^2=x_k^2$$

But if i want to conclude n+1 from n I get

$$\left(\tfrac{1}{n+1}\sum_{k=1}^{n+1} x_k\right)^2\leq\tfrac{1}{n+1}\sum_{k=1}^{n+1} x_k^2$$

And now I cannot seperate $\tfrac{1}{n+1}$ in order to substitute the rest with the induction hypothesis.

Maybe one needs to use a different version of induction? Thanks for any help.

答:
Hint:

you can resolve the $1/n+1$ hurdle by considering weighted averages of previous $n$ points and new added one.

Then consider that $f(x)=x^2$ is convex


问:
Second countable space $\implies [x_n\to a, f(x_n)\to L \implies \lim\limits_{x\to a}f(x)=L]$
Let $f:S\to Y$ where $S\subset X$. I wanted to know if $\lim\limits_{x\to a}f(x) = L$ (where $a$ is a limit point of $A$) was equivalent to $\lim\limits_{n\to \infty}f(x_n) = L$ for all sequences $x_n\to a$. I think that the first implies the second. Indeed I have to show that if I fix an open $V\subset Y$ arbitrary then $\exists N$ such that $f(x_n)\in V, \forall n>N$. Now form the definition of $\lim\limits_{x\to a}f(x) = L$, there exists an open $U$ such that $a\in U\subset S$ and $f(U)\subset V$. But then since $x_n\to a$ and $U$ contains $a$, there exists a natural $N$ such that $n>N\implies x_n\in U$. So for $n>N f(x_n)\in f(U) \subset V$ and we are done.

Now for the converse, I read from the internet that the space needs to be second countable, so with a countable basis. I don't really understand how the second countability plays a role here.

答:
Actually, we only need $S$ to be first countable. Suppose $S$ is first countable, and let $(U_n)_{n=1}^\infty$ be a local basis of the point $a$. We would like it to be monotone, so let's define $W_n=\cap_{k=1}^n U_k$ for every $n$.

Now we can show that the second condition implies the first. Indeed, suppose $\lim_{x\to a} f(x)\ne L$. Then there must be some neighborhood $y\in V\subseteq Y$ such that for every neighborhood $U$ of $a$, the image $f(U)$ is not contained in $V$. In particular, for every $n\in\mathbb{N}$ there is a point $x_n\in W_n$ such that $f(x_n)\notin V$. Note that we have $x_n\to a$. Indeed, let $U$ be a neighborhood of $a$. Then there is some $n$ such that $U_{n_0}\subseteq U$. And then for every $n>n_0$:

$x_n\in W_n=\cap_{k=1}^n U_k\subseteq U_{n_0}\subseteq U$

So indeed we eventually have $x_n\in U$. However, $f(x_n)$ obviously can't converge to $L$, so we have a contradiction.

This direction is not true without first countability. For example, define the co-countable topology on $X=\mathbb{R}$. A sequence converges in this topology if and only if it is eventually constant. So in that case $x_n\to a$ always implies $f(x_n)\to f(a)$, for any function $f$. If that would imply the first condition of your statement then it would mean every function $f:X\to Y$ is continuous. But this is not true, for example take $Y=X$ with the discrete topology. The identity map $f:X\to Y$ is not continuous.


答: If $x\rightarrow 6$, eventually it has to be greater than $-6$ and so, we will have that $|x+6| = x+6$. You can then just do regular division and you'll get that the answer is 2.</p>

Title: Evaluating limits involving absolute values: $\lim_{x\to\pm6} \frac{2x+12}{|x+6|}$

Body:
I have two questions regarding limits involving absolute values. How do I evaluate the following:

$\displaystyle\lim_{x\to -6} \frac{2x+12}{|x+6|}$

$ \displaystyle \lim_{x\to 6} \frac{2x+12}{|x+6|}$.

To handle the first problem, I considered two cases: $x\gt -6$ and $x\lt -6$ and thus get $2$ and $-2$ respectively for the right and left hand limits.

However, I am bit confused regarding the second problem. How do I handle that?


问:
Why is the vector $v = i+j+4k$ perpendicular to the plane $x+y+4z=0$?
Why is the vector $v = i+j+4k$ perpendicular to the plane given by $x+y+4z=0$ ?

I understand that given an equation defining a plane, you can just read off the coefficients for each variable to get a normal vector. It's been about 60 years since I learned this and would appreciate a refresher... thank you and God Bless!

答:
If you have two vectors $a_1i+a_2j+a_3k$ and $b_1i+b_2j+b_3k$ in $\Bbb R^3$, then the angle $\theta$ between them is such that$$\cos\theta=\frac{a_1b_1+a_2b_2+a_3b_3}{\sqrt{a_1^{\,2}+a_2^{\,2}+a_3^{\,2}}\sqrt{b_1^{\,2}+b_2^{\,2}+b_3^{\,2}}}.$$

Now, take any element $xi+yj+zk$ from that plane. Then the angle $\theta$ between $xi+yj+zk$ and $i+j+4k$ is such that$$\cos\theta=\frac{x+y+4z}{\sqrt{x^2+y^2+z^2}\sqrt{6}}=0.$$So, $\theta=90^\circ$.


问:
Maximum area of circles inscribed in sectors of the unit circle

We divide the unit circle into equal sectors and inscribe circles within each sector.
Assuming we make smaller and smaller sectors, what does the total area of the inscribed circles converges to?

I think I have solved it, but since I'm not a mathematician, I wonder if this is correct.

I began by asking: What is the total area of all the circles in a single sector?

We will start with the unit circle as the outer perimeter, with $\theta$ being half of the subtended angle, and $r_0$ being the radius of the first ring of circles, $r_1$ being the radius of the second ring of circles, and so on.

Since the unit circle has a radius of 1, we can immediately deduce the following:
$$r_{0} = ({\sin{\theta} \over 1+\sin{\theta}})$$
and we also know that:
$$r_{n+1} = r_{n} ({1-\sin{\theta} \over 1+\sin{\theta}})$$
that means that:
$$r_{1} = ({\sin{\theta} \over 1+\sin{\theta}}) ({1-\sin{\theta} \over 1+\sin{\theta}}) = {(\sin{\theta})({1-\sin{\theta})} \over (1+\sin{\theta})^2}$$
and:
$$r_{2} = {(\sin{\theta})({1-\sin{\theta})^2} \over (1+\sin{\theta})^3}$$
so generally we can say that:
$$r_{n} = {(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}}$$
the area of a single circle in a sector is:
$$A_{n} = \pi r_{n}^2 = \pi ({(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}})^2$$
so the area of a single sector of circles will be:
$$A_{s} = \pi \sum_{n=0}^{\infty} ({(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}})^2 $$
and since we're dealing with two or more sectors, we can say that:
$$ 0 < \theta \leq {\pi \over 2} $$
so this means the sum of areas of all the circles inscribed in a sector converges to:
$$A_{s} = \pi \sum_{n=0}^{\infty} ({(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}})^2 = {\pi \sin{\theta} \over 4}$$

I then continued with: What is the total area of all the circles in all the sectors?

Assuming we have $k$ sectors, then $\theta = {\pi \over k}$ and our sum would be:
$$A_{k} = {k A_{s}} = {\pi k \sin{\pi \over k} \over 4}$$
and if we take the limit of $k$, we will get:
$$\lim_{k \to \infty} {\pi k \sin{\pi \over k} \over 4} = {\pi^2 \over 4} \tag*{$\blacksquare$}$$

Is this valid?

答:
The ratio of the total area of the disks over the outer circle is the same as the ratio of the area of a single disk over the tight annular sector that bounds it.

If there are $k$ circles of radius $r$ in the outer ring, the subtended angle is $\theta=\dfrac{2\pi}k$.

The area of a sector is
$$\theta(1-(1-2r)^2)$$

and the requested ratio

$$\frac{kr}{4(1+2r)}.$$


问:
Prove this square matrix has no square root
Prove the matrix with filled with 0s everywhere except 1s just above the main diagonal has no square root.
The matrix looks like this:
enter image description here</p>

答:
This is quite visual. The main trick is that $M^2 $ and $M$ always commute, as associativity
$$ M M^2 =  M (M M) = (M M) M = M^2 M $$
Therefore, assuming $M^2 = J$ where $J$ is the matrix given, we know that $JM = MJ$

I will display intermediate steps for dimension $4.$

$$
J =
\left(
\begin{array}{cccc}
0&1&0&0 \\
0&0&1&0 \\
0&0&0&1 \\
0&0&0&0 \\
\end{array}
\right)
$$
$$
M =
\left(
\begin{array}{cccc}
a&b&c&d \\
e&f&g&h \\
i&j&k&l \\
m&n&o&p \\
\end{array}
\right)
$$

$$
JM =
\left(
\begin{array}{cccc}
e&f&g&h \\
i&j&k&l \\
m&n&o&p \\
0&0&0&0 \\
\end{array}
\right)
$$

$$
MJ =
\left(
\begin{array}{cccc}
0&a&b&c \\
0&e&f&g \\
0&i&j&k \\
0&m&n&o \\
\end{array}
\right)
$$

Equating the two matrices gives
$$  e=i=m=n=o= 0$$
which then leads to
$$  j=e=0  $$
Next $$ a=f=k=p$$
which are permitted nonzero for now.
Then $$ b=g=l$$
and finally $$ c = h$$
$$
M =
\left(
\begin{array}{cccc}
a&b&c&d \\
0&a&b&c \\
0&0&a&b \\
0&0&0&a \\
\end{array}
\right)
$$
That is as far as commutativity takes us. Now we explicitly demand $M^2 = J$
The $(1,1)$ position in $M^2$ is calculated as $a^2$ (because of all the zero entries). Therefore, $M^2 = J$ tells us that

$$
M =
\left(
\begin{array}{cccc}
0&b&c&d \\
0&0&b&c \\
0&0&0&b \\
0&0&0&0 \\
\end{array}
\right)
$$

is strictly upper triangular.

However, we now see that the $(1,2) $ position in $M^2$ is equal to $0,$ which contradicts the fact that the entry of $J$ in the $(1,2)$ position is set to $1.$

The idea of commuting with a nilpotent Jordan block was in my mind as I had left a comment at

commutes with a nilpotent matrix and invertible

The theorem on commuting with Jordan blocks is this: if a square matrix $M$ has just a single Jordan block for each eigenvalue, the only possible matrices that commute with it are polynomials in $M,$  that is
$$ aI + b M + c M^2 + \cdots + z M^{n-1}  $$


问:
Application of Intermediate Value Theorem in a proof

Let $p(x)_{x \in \mathbb R} = x^n + a_{n - 1}x^{n  - 1} + \ldots + a_1x + a_0$ be a polynomial. Then all the zeroes of $p$ are in $(-M, M)$ where $M = 1 + \max\{|a_{n - 1}|,\ldots, |a_1|, |a_0|\}$.

Proof:

Let $r(x)_{x \ne 0} = \frac{p(x)}{x^n} - 1$. Then for $|x| > 1, \ $ $|r(x)| < \frac{M}{|x| - 1}$ meaning $|x| \ge M$ implies $|r(x)| < 1$. Now we have $p(x) = x^n(1 + r(x))$ for $|x| \ge M$ where $1 + r(x) > 0$ since $|r(x)| < 1.$ It follows $p(x)$ and $x^n$ have same sign for $|x| \ge M$ meaning $\color{red}{\text{ any zero of $p$ must be in $(-M, M)$}}.$

How does the $\color{red}{\text{part in red}}$ above follow? If $|x| < M$, then $x \in (-M, M)$ and  $p(x), x^n$ will have opposite signs. How does that connect to $p(-M) \le 0 \le p(M)$ (or $p(-M) \ge 0 \ge p(M))$ granted they are using the Intermediate Value Theorem?

答:
I think the proof has a small bug that is easily corrected. Also, I don't think the intermediate value theorem is needed.

The only real change is that $M$ is replaced by $M-1$ below.

Suppose $|x| \ge M$.

You have
\begin{eqnarray}
|{p(x) \over x^n} -1| &\le& {|a_{n-1}| \over |x|}+ \cdots + {|a_{0}| \over |x|^n} \\
&\le& \color{red}{(M-1)} ({1 \over |x|}+ \cdots + {1 \over |x|^n} )\\
&<& (M-1) ( {1 \over 1-{1 \over |x|}} -1) \\
&=& {M-1 \over |x|-1} \\
&\le& 1
\end{eqnarray}
In particular you have
$-1 < {p(x) \over x^n} -1 < 1$ or
$0 < {p(x) \over x^n} < 2$
and so $p(x) \neq 0$.


答:
Let us consider the plane $P = \{(x,y,z)\in\mathbb{R}^{3} \mid x + y + 4z = 0\}$.

Clearly, $O = (0,0,0)\in P$, $A = (1,-1,0)\in P$ and $B = (0,-4,1)\in P$.

Let us denote $w = A - O = (1,-1,0)$ and $z = B - O = (0,-4,1)$.

Thus the lines
\begin{align*}
\begin{cases}
L_{1} = \{(x,y,z)\in\mathbb{R}^{3} \mid (x,y,z) = O + tw\}\\\\
L_{2} = \{(x,y,z)\in\mathbb{R}^{3} \mid (x,y,z) = O + tz\}
\end{cases}
\end{align*}
are contained in $P$, are distinct and are not parallel.

Moreover, we do also have that
\begin{align*}
\begin{cases}
\langle v,w\rangle = \langle(1,1,4),(1,-1,0)\rangle = 1 - 1 + 0 = 0\\\\
\langle v,z\rangle = \langle(1,1,4),(0,-4,1)\rangle = 0 - 4 + 4 + 0 = 0
\end{cases}
\end{align*}

Consequently, since $v\perp L_{1}$ and $v\perp L_{2}$, we conclude that $v\perp P$, and we are done.

Title: Why is the vector $v = i+j+4k$ perpendicular to the plane $x+y+4z=0$?

Body:
Why is the vector $v = i+j+4k$ perpendicular to the plane given by $x+y+4z=0$ ?

I understand that given an equation defining a plane, you can just read off the coefficients for each variable to get a normal vector. It's been about 60 years since I learned this and would appreciate a refresher... thank you and God Bless!


答:
For any

$0 \ne a \in R \tag 1$

consider the map

$\theta_a: R \to R \tag 2$

given by

$\theta_a(r) = ar, \; \forall r \in R; \tag 3$

if

$\theta_a(r) = \theta_a(s), \tag 4$

then

$ar = as, \tag 5$

that is,

$a(r - s) = 0; \tag 6$

since $R$ has no zero divisors, by virtue of (1) we conclude that

$r - s = 0, \tag 7$

whence

$r = s, \tag 8$

and we see that $\theta_a$ is injective; $R$ being finite, we also have $\theta_a$ is surjective, and therefore

$\exists b \in R, \; ab = \theta_a(b) = 1, \tag 9$

and $b$ is a right inverse of $a$; now

$ab = 1 \Longrightarrow (ba)b = b(ab) = b \cdot 1 = b, \tag{10}$

from which

$(ba -1)b = (ba)b - b = 0, \tag{11}$

and once more invoking the hypothesis that $R$ has no zero divisors (and observing that (9) implies $b \ne 0$) we may write

$ba - 1 = 0 \Longrightarrow ba = 1 \tag{12}$

as well; we see that $b$ is also a left inverse of $a$; (9) and (12) together imply that

$b = a^{-1}; \tag{13}$

since every non-zero element $a$ of $R$ has a (two-sided) inverse, $R$ is indeed a division ring.  $OE\Delta.$

Title: A finite ring without zero divisors is a division ring

Body:
Question: Let R be a finite ring without zero divisors and |R| > 1. Then show that R is a division ring.

I used the following logic, but I'm not sure if this is correct.

Let |R| = n, a finite number.
$Let 0 \not = a \in R.$

Therefore, $a^n = a => a(a^{n-1} - 1) = 0 => a^{n-1} = 1 \,\,\, (\text{Since} \,\, a \not = 0)$

Therefore, $a a^{n-2} = 1$

Therefore $a^{n-2}$ is the inverse of a for all a.
Therefore R is a division ring.

Does this work?


问:
Sketch complex function without any software

I was asked for a school project to sketch this and we were advised to do it without sny software, because we must be able to sketch something like that in our exams.But I am generally interested in the question. I tried to manipulate it with euler's identity to bring into a more familiar form but I can 't see how i can sketch something like that.
I meant , to bring into the form $a+jb$ in order to have the real - part graph and and the arg(X) part for the angle but it did not work. Any help appreciated.

$X(\Omega) = (\frac{1}{1 - a^{-1} e^{-j\Omega}})^3 = (\frac{e^{j\Omega}}{e^{-j\Omega} - a^{-1} })^3 = (\frac{cos(\Omega) + j sin(\Omega)}{cos(\Omega) + j sin(\Omega) - a^{-1} })^3$

update* : $X(\Omega)=(\frac{cos(\Omega) + j sin(\Omega) - a^{-1} }{cos(\Omega) + j sin(\Omega)})^{-3} = (1 - a^{-1}cos(\Omega)+ja^{-1}sin(\Omega))^{-3}$

I wonder:

Can I say that ? $Re{X(\Omega)} = ((1 - a^{-1}cos(\Omega))^{-3}$ and

$Im{{X(\Omega)}} = (a^{-1}sin(\Omega))^{-3}$ ?

and $|X(\omega)| = \sqrt{Re^2{X(\omega)} + Im^2{X(\Omega)} }$

and $Arg(X) = tan^{-1}(\frac{Im{X}}{Re{X}})$

答:
There are two possible ways to proceed.

I)
$$
\left\{ \matrix{
  1 - {1 \over a}e^{\, - j\,\Omega }  = 1 - {1 \over a}\cos \Omega  + j{1 \over a}\sin \Omega  =  \hfill \cr
   = {1 \over a}\sqrt {\left( {a - \cos \Omega } \right)^{\,2}  + \sin ^{\,2} \Omega } \;e^{\,j\,\Phi }  =  \hfill \cr
   = {1 \over a}\sqrt {1 + a^{\,2}  - 2a\cos \Omega } \;e^{\,j\,\Phi }  = R\,e^{\,j\,\Phi }  \hfill \cr
  \Phi  = \arctan _{\,4} \left( {{{\sin \Omega } \over {a - \cos \Omega }}} \right) \hfill \cr}  \right.
\quad  \Rightarrow \quad \left( {1 - {1 \over a}e^{\, - j\,\Omega } } \right)^{\, - 3}
  = R^{\, - 3} e^{\, - j\,3\,\,\Phi }
$$
where $ \arctan _{\,4}$ denotes the arctan over the four quadrants.

II)
$$
\eqalign{
  & {1 \over {1 - {1 \over a}e^{\, - j\,\Omega } }}
 = {1 \over {1 - {1 \over a}\cos \Omega  + j{1 \over a}\sin \Omega }} =   \cr
  &  = {{1 - {1 \over a}\cos \Omega  - j{1 \over a}\sin \Omega } \over {R^{\,2} }} \cr}
$$

from which
$$
\eqalign{
  & \left( {{1 \over {1 - {1 \over a}e^{\, - j\,\Omega } }}} \right)^{\,3}  =   \cr
  &  = {1 \over {R^{\,6} }}\left( \matrix{
  \left( {1 - {1 \over a}\cos \Omega } \right)^{\,3}
 - j3{1 \over a}\left( {1 - {1 \over a}\cos \Omega } \right)^{\,2} \sin \Omega  +  \hfill \cr
   - {3 \over {a^{\,2} }}\left( {1 - {1 \over a}\cos \Omega } \right)\sin ^{\,2} \Omega
 + j{1 \over {a^{\,3} }}\sin ^{\,3} \Omega  \hfill \cr}  \right) =   \cr
  &  = {1 \over {a^{\,3} R^{\,6} }}\left( \matrix{
  \left( {a - \cos \Omega } \right)^{\,3}  - 3\left( {a - \cos \Omega } \right)\sin ^{\,2} \Omega  +  \hfill \cr
   + j\left( {\sin ^{\,3} \Omega
  - j3\left( {a - \cos \Omega } \right)^{\,2} \sin \Omega } \right) \hfill \cr}  \right) =   \cr
  &  = {1 \over {a^{\,3} R^{\,6} }}\left( \matrix{
  \left( {a - \cos \Omega } \right)\left( {\left( {a - \cos \Omega } \right)^{\,2}
  - 3\sin ^{\,2} \Omega } \right) +  \hfill \cr
   + j\sin \Omega \left( {\sin ^{\,2} \Omega
  - 3\left( {a - \cos \Omega } \right)^{\,2} } \right) \hfill \cr}  \right) \cr}
$$


问:
homogenous equation substitution mistake
My teacher gave me this equation to solve which is a homogeneous equation. In this type of equation, we usually use t substitutions. But in this one I cannot find what should I substitute t with? Any help?

$$ 2xdy+(x^2y^4+1)ydx=0$$

Any help would be appreciated)

Body:
$$2xdy+(x^2y^4+1)ydx=0$$
Multiply by $y$:
$$2yxdy+(x^2y^4+1)y^2dx=0$$
$$(xdy^2+y^2dx)+x^2y^6dx=0$$
$$d(xy^2)+x^2y^6dx=0$$
$$\dfrac {d(xy^2)}{(xy^2)^3}+\dfrac {dx}x=0$$
Integrate:
$$\dfrac {1}{(xy^2)^2}-2\ln |x|=C$$


问:
Find the equation of the hyperbola given foci and the minor axis
first time posting and using the site. I have a quick problem that I need some help with. I need to find the equation of a hyperbola given the foci and the length of the minor axis.

The foci coordinates are as follows:

F(-5, 4) and F'(3, -2)

The length of the minor axis is 2√11.

Any help is very much appreciated.

答:
I would recommend first applying a linear transformation that rotates the axis to become horizontal.  If you plot the foci, you'll find that we need to rotate the axis counterclockwise an angle $\theta$, where $\cos\theta=4/5$.  Here is the corresponding rotation matrix:

$$\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}=\begin{pmatrix}\frac{4}{5}&-\frac{3}{5}\\\frac{3}{5}&\frac{4}{5}\end{pmatrix}$$

Under this rotation, the foci become $(-32/5,1/5)$ and $(18/5,1/5)$.  Then, using the standard form of a hyperbola, we have the following equation:

$$\frac{(x+\frac{7}{5})^2}{14}-\frac{(y-\frac{1}{5})^2}{11}=1$$

Finally, we rotate our hyperbola back to its original orientation.  The final result is:

$$\frac{(\frac{4x}{5}-\frac{3y}{5}+\frac{7}{5})^2}{14}-\frac{(\frac{3x}{5}+\frac{4y}{5}-\frac{1}{5})^2}{11}=1$$

or $2x^2+28x-24xy-5y^2-14y=133$.


问:
Upper bound of this fraction?
Do we have an upper bound for $\frac{ax+by+cz}{a_{2} x +b_{2} y +c_2 z}$. What about when $x,y,z<1$ ?

I don't know how to go about it? any tips?

EDIT: Someone kindly presented a counterexample. But are there nice restrictions to put so that it is upper bounded by some parameter. Like can it somehow be upper bounded by say $\frac{a}{a_2}+\frac{b}{b_2}+\frac{c}{c_2}$. (This looks familiar and I've seen it somewhere :) )
Thanks!

答:
Numerator and denominator represent two planes , when each is put equal to $d_1, \, d_2$ .

So either they are parallel and there are no common points, or they are coincident, and the ratio is constant everywhere,  or they have a common intersection line where the ratio is $d_1 / d_2$.

Then if you increase $d_1$ the upper plane will move .. and the intersection line will move ..

Title: Upper bound of this fraction?


问:
simple inequality for vector norm
For vectors x,y $\in \mathbb{R}^n_+$ with entries ordered decreasingly: if
$$
\sum_{i=1}^kx_i \leq \sum_{i=1}^ky_i \ \forall 1\leq k\leq n
$$
then $\|x\|_p\leq\|y\|_p$.

From https://math.stackexchange.com/a/2819688/780264

答:
This result is a consequence of Corollary II.3.4 of the same reference in the linked (Bhatia's Matrix Analysis). Here is an adaptation of the proof given in the text.

I will use $x \prec_w y$ to mean that when $x,y$ are reindexed in decreasing order, $\sum_{i=1}^kx_i \leq \sum_{i=1}^ky_i$ for $1\leq k\leq n$. $x \prec y$ denotes the same with the additional condition that $\sum_{i=1}^kx_i = \sum_{i=1}^ky_i$. I will use $x \leq y$ to mean that $x_i \leq y_i$ for all $i$. We will first prove the following:

Claim: If $x,y \in \Bbb R_+^n$ satisfy $x\prec y$, then $x^p \prec_w y^p$ (where $(x_1,\dots,x_n)^p = (x_1^p,\dots,x_n^p)$).

We need the following facts:

The function $f(x) = x^p$ is convex over $\Bbb R_+$ for $p \geq 1$

If $x \leq y$, then $\|x\|_p \leq \|y\|_p$

If $x \prec y$, then there exists a doubly stochastic matrix $A$ (i.e. the row and column sums of $A$ are all $1$) for which $x = Ay$ (cf. Bhatia Theorem II.1.10)

The Birkhoff von-Neumann theorem: such a matrix $A$ can be written in the form $\sum_j P_j$ for permutation matrices $P_j$ and positive $t_j$ with $\sum_j t_j = 1$.

With that, the proof proceeds as follows. We have
$$
x^p = [Ay]^p = \left[\sum_j t_j P_j y \right]^p \leq
\sum_j t_j (P_j y)^p
= \sum_j t_j P_j(y)^p = A y^p
$$
So, we have $x^p \leq Ay^p$ and $Ay^p \prec y^p$. It follows that $x^p \prec_w y^p$.

From the fact that $x^p \prec_w y^p$, it follows that the sum of the entries of $x^p$ is less than that of $y^p$, from which it follows that $\|x\|_p \leq \|y\|_p$.

It now remains to be shown that $x \prec_w y \implies \|x\|_p \leq \|y\|_p$. To that end, it suffices to note that if $x \prec_w y$, there exists a vector $u$ (ordered decreasingly) for which
$$
x \leq u \prec y,
$$
(cf. Bhatia Theorem II.2.8, a consequence of fact 3). It follows that $\|x\|_p \leq \|u\|_p \leq \|y\|_p$ .

Title: simple inequality for vector norm


问:
A possible domain for the Log function
When I choose some usual branch of the function(like $\mathbb{C}$ minus $0$ and the negative real axis), I know that $log(z) = log|z| + $$i$ $arg(z)$.

If I define the domain of the log, as a different but simply conected region like A, does the formula that I stated before to calculate log keeps making sense?

Thanks in advance

答:
Yes, as long as, for each $z\in A$, you take the appropriate argument of $z$. That is, there is a function $\operatorname{arg}_A\colon A\longrightarrow\Bbb R$ such that:

For each $z\in A$, $\operatorname{arg}_A(z)$ is an argument of $z$.

The function $\operatorname{arg}_A$ is continuous.

For each $z\in A$, $\log z=\log|z|+\operatorname{arg}_A(z)$.

Note that then, in the set $A$ from your picture (which intersects $(0,\infty)$ in two distinct intervals), the positive reals $z$ closer to $0$ will give a certain value of $\operatorname{arg}_A(z)$, whereas those from the segment further away from $0$ will have another value of $\operatorname{arg}_A(z)$.

Title: A possible domain for the Log function


问:
Transformation of Uniform Distribution using $Y = X^2$
I know this has already been answered somewhere but I can't seem to find where I am going wrong.

$X$ has a uniform distribution on $[-1,1]$

Let $Y = X^2$

P.d.f of X:

$f_X(x) = \frac{1}{2}$ for $-1 \le x \le 1$

Then using $f_Y(y) = f_X(g^{-1}(  y))$ $|\frac{d}{dy}[g^{-1}(  y)]|$

My new range of $y$ is $0 \le y \le 1$

Then $|\frac{d}{dy}[g^{-1}(  y)]|$ = $\frac{1}{2\sqrt{y}}$

Subbing in I get that there is a an extra factor of  one half stopping the p.d.f summing to 1 over the interval. Where have I gone wrong?

答:
You can use the definition of CDF

$$F_Y(y)=P(Y\leq y)=P(-\sqrt{y}\leq X\leq \sqrt{y})=F_X(\sqrt{y})-F_X(-\sqrt{y})=$$

$$=\frac{\sqrt{y}+1}{2}-\frac{-\sqrt{y}+1}{2}=\sqrt{y}$$

Thus derivating you get your density

$$f_Y(y)=\frac{1}{2\sqrt{y}}\mathbb{1}_{(0;1]}(y)$$


问:
Basic question about the definition of an integral on a measure space
Let $(X,\mathcal{B},\mu)$ be a measure space.

$\bf{\text{Definition:}}$  For a non-negative measurable function $f$ on $X$, $E\in \mathcal{B}$,

$$\int_{E}f d\mu := \text{inf}\int_{E}\varphi d\mu$$

where the infimum is taken over all simple functions $\varphi$ such that $0\leq \varphi \leq f$.

$\bf{\text{My question:}}$

When I am using the definition to prove something for such a function $f$, I want to take a sequence of simple functions $\varphi_{n}$ with $0\leq \varphi_{n}\leq f$ such that

$\int_{E}fd\mu = \lim\limits_{n\to\infty}\int_{E}\varphi_{n}d\mu$.

Is it possible to choose the sequence so that $\varphi_{n}(x)\to f(x)$ almost everywhere in $E$?

答:
As Christopher A. Wong said, infimum should be supremum.

Yes, for example, $\varphi_n(x) = \min(2^n, 2^{-n} \lfloor 2^nf(x) \rfloor)$ is a sequence of simple functions that converges to $f$ at every point where $f(x)$ is finite. I used $2^n$ instead of $n$ to make the sequence increasing: $\varphi_n\le \varphi_{n+1}$.


问:
Pool table and Poincaré recurrence theorem
Consider a square pool table $1\times1$. We hit a ball, and observe what happens. There is no friction and this system obeys law of reflections. I am intrested where the ball is going to hit.

Now let's take a point $x_0$, where the ball hit the wall for the first time. Consider a segment of wall $I$. $|I| = \epsilon$ and it has $x_0$ as the middle point.

Is this true, that the ball will be hitting the interval $I$ infinitely often? How to (dis)prove this?

I have an intuition, that i should use Poincaré recurrence theorem.

The function here is $T(p_1,\alpha_1) = (p_2,\alpha_2)$.

$p_1$-is the previous point where the ball hit, $p_2$ is the next point.

$\alpha_1$ - is an angle between the wall where the previous point lies and the dircetion to the next point, we define $\alpha_2$ similarly

But the question is how to prove that this function preserves Lebesgue measure.

Regards

答:
This is true, but you don't need Poincare recurrence theorem.

As a hint, try to prove that either the trajectory of the ball is periodic (in this case, it will hit exactly $x_0$ an infinity of times), either it is dense in the square (in this case, it will not hit $x_0$ again, but it will hit $I$ an infinity of times).


问:
Need help with Recursive sequence
Given the following recursive sequence $$a_n=1\\ a_{(n+1)}=\sqrt{\frac{1}{n^2}+a_n^2}$$
I've computed the first couple of values to be $1, 1.414, 1.5, 1.5365, 1.5567, 1.5695$. It's clearly increasing, however I'm having trouble trying to find the limit (if it exists that is). I assume we would have to show some upper bound? Even if we assumed the limit existed then we get the following $$L^2=\frac{1}{n^2}+L^2$$ which doesn't really yield any solution. Anything I'm doing wrong?

Title: Need help with Recursive sequence

Body:
Note that you have: $a_{n+1}^2 = a_n^2 + \dfrac{1}{n^2}$. Thus,
\begin{align*}
a_{n+1}^2 - a_n^2 & = \dfrac{1}{n^2} \\
\implies \sum_{r = 1}^n (a_{r+1}^2 - a_r^2) & = \sum_{r = 1}^{n} \dfrac{1}{r^2} \\
\implies a_{n+1}^2 - a_1^2 & = \sum_{r = 1}^{n} \dfrac{1}{r^2} \\
\implies \lim_{n \to \infty} a_{n+1}^2  & = 1 + \lim_{n \to \infty}  \sum_{r = 1}^{n} \dfrac{1}{r^2} = \frac{\pi^2}{6} + 1.
\end{align*}

Thus, $\lim_{n \to \infty} a_n = \sqrt{\dfrac{\pi^2}{6} + 1}$.


问:
Use of definition of limit to prove $\lim_{x→2} \dfrac{x}{x^2-2}=1$
I know that by definition I have to prove that $$\lim_{x→2}\dfrac{x}{x^2−2}=1⟺∀ϵ>0,∃δ>0,\, 0<|x−2|<δ⟹\left|\frac{x}{x^2−2}−1\right|<ϵ.$$

I have:
$\left|\dfrac{-x^2+x+2}{x^2-2}\right|= \left|\dfrac{-(x^2-x-2)}{x^2-2}\right|= \left|\dfrac{-(x^2-x+1-3)}{x^2-2}\right|=$
$\left|\dfrac{-(x-1)^2+3}{x^2-2}\right|$.

So, I don't know how to continue.

答: Note that $-(x^2-x-2)=-(x-2)(x+1)$ and that therefore$$\frac x{x^2-2}-1=-(x-2)\frac{x+1}{x^2-2}.$$Besides, if $|x-2|<\frac12$, then $\frac32<x<\frac52$ and therefore $\frac14<x^2-2<\frac{17}4$; in particular, $|x^2-2|>\frac14$. On the other hand, $\frac52<x+1<\frac72$; in particular, $|x+1|<\frac72$. So,\begin{align}\left|-(x-2)\frac{x+1}{x^2-2}\right|&=|x-2|\frac{|x+1|}{|x^2-2|}\\&<|x-2|\frac{7/2}{1/4}\\&=14|x-2|.\end{align}Therefore, given $\varepsilon>0$, take $\delta=\min\left\{\frac\varepsilon{14},\frac12\right\}$, and then$$|x-2|<\delta\implies\left|\frac x{x^2-2}-1\right|<\varepsilon.$$


问:
Requirements of Lebesgue measure theorems
We are looking at the Lebesgue measure on $\mathbb{R}$ and ${(f_n)_{n\in \mathbb {N}}\colon \mathbb {R} \to \overline {\mathbb {R}}}$ with $f_n(x) = 1+\sum _{k=1}^n |x|^k$.

Does $(f_n)_{n\in \mathbb {N}}$
fulfill the requirements of

a. the dominated convergence theorem

b. Fatou's theorem

c. the monotone convergence theorem

?

I think that the answer to b is yes and to c no, is that correct? I'm not really sure about a.

答: For (b)&(c) the conditions are fulfilled, albeit trivially, since all of the integrals are infinite. None of these functions are integrable on $\mathbb{R}$, nor are they dominated by such, so the requirements of (a) are not fulfilled.


问:
Solving $2^y = 615+x^2$ for $y$
How can I transform the following equation:

$$2^y = 615+x^2$$

to something of the form

$$y = \cdots$$

SideQuest:

What would the values of $x$ and $y$ that make the equality True, and how would you get them?

Thanks!

答:
For sidequest : hint

Assuming that you want a positive integer solution in $x,y$.

$2^y$ can not be congruent to $0$ mod 3.

$x^2$ can not be congruent to $2$ mod 3.

$615$ congruent to $0$ mod 3.

Therefore, $2^y$ and $x^2$ must both be congruent to $1$ mod 3.

Therefore, $y$ must be an even.

This greatly reduces the problem, because now,

$2^y - x^2$ can be readily factored.

Title: Solving $2^y = 615+x^2$ for $y$


问:
Function to create a "stepped," diagonal line
First off, sorry if this is a basic question or one that has been asked before. I really don't know how to phrase it, so it's a hard question to google.

I'm looking for a function that will generate a line similar to the one below

```<code>       __/
    __/
 __/
/
</code>```

I'm pretty good at math, but for some reason this seems to be stumping me as it seems like it should be really simple.

In case it helps, I am planning on using it to drive an animation, so that it moves, pauses, moves, pauses, etc. using the current time (zero through infinity) as the input.

I am using an "Absolute," system (IE: if I were to jump to frame 35, the math needs to be able to calculate frame 35 without knowing the frames before it), so I can't do anything like if (floor(sin(time)) + 1 > 0) { add 1 }

答:
John Wayland's answer was good, and worked on the project I needed, but after playing around on a different project I came up with a more "tweakable," function that does the same thing, but allows changing of parameters.

D is the duration of one cycle, T is the amount of time the transition should take, and A is the amount of change between cycles.

$$cycle(x) = A(max\{\frac{(x \mod D) -D}{T} + 1, 0\} + \lfloor \frac{x}{D} \rfloor)$$

Breakdown

Create saw wave with intended duration
$$(x \mod D)$$
clamp it so that only 1 unit is greater than 0
$$max\{(x \mod D) - D + 1, 0\}$$
set duration of transition
$$max\{\frac{(x \mod D) -D}{T} + 1, 0\}$$
create stepped line, where each step is the duration of one cycle
$$\lfloor\frac{x}{D}\rfloor$$
Add the two together so that each step of the previous functions "lifts," each cycle up by 1 unit
$$max\{\frac{(x \mod D) -D}{T} + 1, 0\} + \lfloor \frac{x}{D} \rfloor$$
Finally multiply the whole thing by A to set the amount of desired change between cycles.

As a bonus, you can also wrap the max function in a smoothstep function and get a very nice smooth graph for animation

$$S(x)=x^2(3-2x)$$
$$cycle(x) = A(S(max\{\frac{(x \mod D) -D}{T} + 1, 0\}) + \lfloor \frac{x}{D} \rfloor)$$


问:
Under what conditions is a zero divisor element $a$ in commutative ring $R$ nilpotent?
Suppose that $R$ is a commutative ring with identity $1$

Let $a\in R$ with $ab=0$ for some $b\ne0$.

Under what conditions $a$ must be also nilpotent?

答:
This is technically an answer, though I don't know if it's particularly useful...

Claim: Let $a$ and $b$ be elements of a commutative unital ring such $R$ that $ab = 0$. Then $a$ is nilpotent if and only if every minimal prime containing $b$ also contains $a$.

Proof: If $a$ is nilpotent, then it is contained in all (minimal) prime ideals of $R$. Conversely, suppose that every minimal prime ideal containing $b$ also contains $a$. It is enough to show that $a$ is a member of every minimal prime of $R$ (for then $a$ is a member of every prime of $R$, and is therefore in the nilradical of $R$).  So let $P$ be a minimal prime of $R$. If $b \in P$ then by hypothesis $a \in P$ also. On the other hand, if $b \notin P$ then $ab = 0 \in P$ with $P$ prime implies that $a \in P$.  QED

In particular, if $b \in R$ is a zero divisor that is not a member of any minimal prime, then every element of $R$ that annihilates $b$ is nilpotent. (For if $ab = 0$, then it's vacuously true that every minimal prime containing $b$ also contains $a$.)

For a specific example of this situation, take the ring $R = k[x,y]/(x^2,xy)$. This has unique minimal prime ideal $(x)$, with $y \notin (x)$. So every element of $R$ that annihilates $y$ (for instance, $x$) must be nilpotent. (In fact, the annihilator of $y$ is quite easily seen to be $(x)$. So it's not a terribly interesting example.)

(You may also wish to see the following question on MathOverflow for vaguely related information:
https://mathoverflow.net/questions/20826.)


问:
Trying to make sense of a proof about metric spaces.
Theorem. Let $(X,d)$ be a metric space, $K\subseteq X$ a compact subset, and $E\subseteq X$ a closed subset, such that $K\cap E=\emptyset$. Show that there is some $D>0$ such that $d(x,y)\geq D$ for all $x\in K$ and $y\in E$.

I will break the proof down into segments and place questions where I am not understanding things.

Proof. Suppose otherwise. Then for $D>0$ we claim $d(x,y)\leq D$ for some $x\in K$ and some $y\in E$. Then there are points $x_n \in K$ and $y_n\in E$ such that $d(x_n,y_n)\leq1/n$.

I am not entirely sure why we've chosen $x_n$ and $y_n$, is it having to do with the compactness of $K$ and the fact that $E$ is closed?

Now, define some set $S=\{x_n|n\in\mathbb{N}\}$ such that $S$ is the set of all $x_n's$. We will consider the two cases where either $S$ is infinite or $S$ is finite.

Case 1: $S$ is finite. This means there is a finite collection of points $\{z_1,z_2,z_3,...,z_M\}$ such that for all $i \in \mathbb{N}$, there exists $j\leq M$ such that $x_i=z_j$. Now define the set $N_j\subseteq\mathbb{N}$ with $i\leq j\leq M$ such that $N_j=\{i\in\mathbb{N}|x_i=z_j\}$. By assumption, this means that $\mathbb{N}$ is the union of the $N_j's$. But since there are only finitely many $N_j's$ at least one of them must be infinite.

Why must at least one be infinite? Because $\mathbb{N}$ is an infinite set and the union of the $N_j's$?

So pick $j$ such that $N_j$ is infinite. We claim that $z_j$ is a limit point of $E$. So we must have $z_j \in E$, a contradiction to the assumption.

The proof does not go into detail here and I'm left trying to figure out how I can prove this claim. Any suggestions?

I am pretty sure I understand the infinite case. There are not any gaps like the finite case. Your pointers and suggestions would be helpful!

答:
We choose $x_n$ and $y_n$ that way because that's convenient for what we will do next.

Yes, that is correct.

There a strictly increasing sequence $(n_k)_{k\in\Bbb N}$ of natural numbers such that $(\forall k\in\Bbb N):x_{n_k}=z_j$. But then $d(z_j,y_{n_k})=d(x_{n_k},y_{n_k})\leqslant\frac1{n_k}$. Besides, $\lim_{k\to\infty}\frac1{n_k}=0$. So, $\lim_{k\to\infty}d(x_{n_k},y_{n_k})=0$; in other words, $\lim_{k\to\infty}y_{n_k}=z_j$. So, and since reach $y_n$ belongs to $E$, $z_j$ is a limit point of $E$.


答: If $K$ and $E$ are not separated by a minimum distance, then you can find a sequence of pairs of points which are getting arbitrarily close. Since the $x_n$ are in $K$, they have an accumulation point. You should look at that accumulation point for your contradiction.

Title: Trying to make sense of a proof about metric spaces.

Body:
Theorem. Let $(X,d)$ be a metric space, $K\subseteq X$ a compact subset, and $E\subseteq X$ a closed subset, such that $K\cap E=\emptyset$. Show that there is some $D>0$ such that $d(x,y)\geq D$ for all $x\in K$ and $y\in E$.

I will break the proof down into segments and place questions where I am not understanding things.

Proof. Suppose otherwise. Then for $D>0$ we claim $d(x,y)\leq D$ for some $x\in K$ and some $y\in E$. Then there are points $x_n \in K$ and $y_n\in E$ such that $d(x_n,y_n)\leq1/n$.

I am not entirely sure why we've chosen $x_n$ and $y_n$, is it having to do with the compactness of $K$ and the fact that $E$ is closed?

Now, define some set $S=\{x_n|n\in\mathbb{N}\}$ such that $S$ is the set of all $x_n's$. We will consider the two cases where either $S$ is infinite or $S$ is finite.

Case 1: $S$ is finite. This means there is a finite collection of points $\{z_1,z_2,z_3,...,z_M\}$ such that for all $i \in \mathbb{N}$, there exists $j\leq M$ such that $x_i=z_j$. Now define the set $N_j\subseteq\mathbb{N}$ with $i\leq j\leq M$ such that $N_j=\{i\in\mathbb{N}|x_i=z_j\}$. By assumption, this means that $\mathbb{N}$ is the union of the $N_j's$. But since there are only finitely many $N_j's$ at least one of them must be infinite.

Why must at least one be infinite? Because $\mathbb{N}$ is an infinite set and the union of the $N_j's$?

So pick $j$ such that $N_j$ is infinite. We claim that $z_j$ is a limit point of $E$. So we must have $z_j \in E$, a contradiction to the assumption.

The proof does not go into detail here and I'm left trying to figure out how I can prove this claim. Any suggestions?

I am pretty sure I understand the infinite case. There are not any gaps like the finite case. Your pointers and suggestions would be helpful!


问:
Method of moments estimator of $\lambda$ for $f(x|\lambda)=\lambda e^{-\lambda(x-1)}$
I’m trying to understand what the method of moments is, exactly.  I haven’t seen a perfectly clear (to me) statement of this, so here’s my current belief about what it is:  You want to estimate some parameter as some formula using the $r$th central sample moments.  Sometimes this is trivial because the parameter you’re estimating just is one of the central moments.  But in a case where it’s not trivial because the parameter is not exactly one of the central moments, you may need to find a way to relate the parameter to a formula using central moments.

So for example suppose we want the method of moments estimate of $\lambda$ for $f(x|\lambda)=\lambda e^{-\lambda(x-1)}$ when $x >1$ and $f(x|\lambda)=0$ for $x\leq 1$.  This is close to an exponential but not quite, so we can’t rely on the mean being the parameter.  We compute the mean in order to see if we can relate it to the parameter:

$$E[X] = \int_1^\infty x\lambda e^{-\lambda(x-1)} \ dx $$
$$   = \lambda\left(x\cdot \frac{e^{-\lambda(x-1)}}{-\lambda} \Bigg|_1^\infty - \int_1^\infty \frac{e^{-\lambda(x-1)}}{-\lambda}\ dx \right) $$
$$ = \lambda \left(\frac{1}{-\lambda}-\frac{1}{-\lambda}\int_1^\infty e^{-\lambda(x-1)}\ dx\right) $$
$$ =\lambda \left(-\frac{1}{\lambda}+1\right) $$
$$ = \lambda-1 $$

So in this case, we can estimate $\hat\mu_k \approx E[X] = \lambda -1$ and therefore $\lambda \approx \hat\mu_k+1$?

答:
That's right. Basically the method of moments says, solve the equations $E[X \mid \vec{\lambda}]=\overline{X},E[X^2 \mid \vec{\lambda}]=\overline{X^2},\dots,E[X^n \mid \vec{\lambda}]=\overline{X^n}$ for $\vec{\lambda}$ and call that result your estimator.

These equations will usually be independent, so generally there will be as many of them as there are parameters, assuming $X$ is itself scalar.


问:
Compositions of $n$ with largest part at most $m$
This is a problem from Stanley's Enumerative Combinatorics that I'm failing at a bit (lot):

Let $\bar{c}(m,n)$ denote the number of compositions of $n$ with largest part at most $m$.  Show that $$\sum_{n\geq 0}\bar{c}(m,n)x^n={{1-x}\over{1-2x+x^{m+1}}}$$

Some definitions:  A composition of $n$ is an $ordered$ list of positive integers that equals $n$.  If $\{a_1,...,a_k\}$ is one such composition, we say that the composition has $k$ $parts$.

I know it's pretty traditional to list a "what you've done so far" but I'm really about as blindly stuck as can be.

答:
Hazy idea: I'd try to derive a relation $(\dagger)$ between for instance $\bar{c}(m, n)$ and $\bar{c}(m, n-1)$ (or even with multiple terms $\bar{c}(m, n-k)$), or something similar. The goal would then be to find an EDP for $f\colon x \to \sum_{n=0}^\infty \bar{c}(m, n)x^n$ by differentiating term by terms and using $(\dagger)$.
Then, olving the EDP for $f$, one could get a closed-form expression of it — hopefully the one you're asked to prove.

Of course, the main step is to find (via combinatorics?) the said relation.


问:
What are some examples of "visually beautiful" math texts?
Are there good examples of math textbooks that seamlessly combines art, cartoon drawing or computer graphics with the texts? I know that there are a few books on the market, which are not text-book per se, that are geared towards general audience, where there are plenty of artistic figures naturally generated by the phenomena under study, usually things of geometric nature, Escher tiling, or things like chaos.

Wolfram's a New Kind of Science comes to mind. What are some other examples of "visually beautifu" or "artistic" math textbooks or texts.

答:
A Topological Picturebook by George K. Francis

is full of enlightening pictures of topological structures. Many of them are little masterpieces of drawing art and support this way a better understanding of the theme.

Sometimes it might also be interesting to see behind the curtain.

In A mathematical typography a grandmaster shares with us some ideas about this subject. Here we can get a glimpse about the extraordinary creativity and the deep knowledge which is necessary to produce aesthetically pleasing mathematical texts and pictures.


问:
Every non-zero element of order of a number field belongs to finitely many prime ideals

Let $K$ be a number field and  $\mathcal{O}$ be an order of $K$. If $0\neq x \in \mathcal{O}$ then is it true that $x$ belongs to only a finite number of prime ideals of $\mathcal{O}$?

Since $\mathcal{O}$ is noetherian every ideal has a primary decomposition. Let
\begin{equation}
(x)=\mathfrak{q}_1\cap\mathfrak{q}_2 \cap \cdots \cap \mathfrak{q}_n
\end{equation}
be the primary decomposition of the ideal generated by $x$, where $\mathfrak{q}_i$ are primary ideals. Suppose $(x) \subset \mathfrak{p}$ for some prime ideal $\mathfrak{p}$. From the above equation we have,
\begin{align*}
\mathfrak{q}_1\cap\mathfrak{q}_2 \cap \cdots \cap \mathfrak{q}_n \subset \mathfrak{p}
\end{align*}
Taking radical on both sides and noting that the radical of a prime ideal is itself and the radical of a primary ideal $\mathfrak{q}_i$ is prime (call it $\mathfrak{p}_i$) We get the following equation.
\begin{align*}
\mathfrak{p}_1\cap\mathfrak{p}_2 \cap \cdots \cap \mathfrak{p}_n \subset \mathfrak{p}
\end{align*}
By prime avoidance lemma the above equation implies $\mathfrak{p}_i \subset \mathfrak{p}$ for some $i$. Since in an order every prime ideal is maximal $\mathfrak{p}_i = \mathfrak{p}$.

Is the above proof correct? Is there an alternate proof which does not require one to know primary decomposition in Noetherian rings but uses properties of Dedekind domains and order alone (like the index of the ideal in $\mathcal{O}$ in finite and that $\mathcal{O}$ is a finitely generated $\mathbb{Z}$ module)?

答:
Your proof looks fine to me, but I agree that it feels like overkill to use primary decomposition.

Here's an alternate proof that doesn't use primary decomposition: Let $R$ be a Noetherian commutative ring, and let $x \in R$ be arbitrary. Prime ideals containing $x$ correspond to prime ideals of $R/(x)$, which is also Noetherian. The prime spectrum of a Noetherian ring is a Noetherian space (that is, it satisfies the descending chain condition on closed subsets), which has only finitely many irreducible components, and the irreducible components correspond to minimal prime ideals. So $R/(x)$ has only finitely many minimal prime ideals.

If $R$ is a one-dimensional integral domain and $x$ is nonzero, then $R/(x)$ is zero-dimensional, so all prime ideals of $R/(x)$ are minimal, and we get the result you want. (There's no need to use more properties of Dedekind domains than this.)


问:
Proving $C^{A \times B} \cong (C^B)^A$ without using the "same cardinality" argument
I'd like to prove that there is a bijection between the two sets $C^{A \times B}$ (the set of all functions from $A \times B$  to $C$) and $(C^B)^A$ (the set of all functions from $A$ to the set of all functions from $B$ to $C$) by showing one, without considering that this is immediate because they have the same cardinality.

Could anyone give me a hint of how to construct such a bijection?

答: Map a function $f:A\times B\to C$ into a function $\phi:A\to C^B$ by setting, for all $a\in A$, that $\phi(a)$ maps $B$ to $C$ this way: $\phi(a)(b)=f(a,b)$ for all $b\in B$. That is all.


问:
The halting problem solved if there is an effective positive test for first-order entailment?
given any Turing machine m and any input n, we can construct a finite set of sentences ∆ and a corresponding sentence H, halting sentence, such that ∆ ⊨ H  iff  TMm eventually halts on input n. Thus if there were an effective method for determining that ∆  ⊭  H then we could solve the halting problem and thereby refute the Church-Turing Thesis.
If we know that there is an effective positive test for first-order entailment, can we solve the halting problem by considering the negation of the halting sentence, viz. ∆ ⊨ ¬H?</p>

答: There is an effective positive test for first order entailment: enumerate proofs until you find one. And no, this doesn’t solve the halting problem. There is no guarantee that we can prove that a non-halting machine doesn’t halt. (In other words, no guarantee that $\Delta\models \lnot H$ in the first place.)


问:
Calculate the following infinite sum in a closed form $\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$?
Is it possible to calculate the following infinite sum in a closed form? If yes, please point me to the right direction.
$$\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$$</p>

答:
We can have the following integral representation

$$ \sum_{n=1}^\infty(n\ \text{arccot}\ n-1)=\int _{0}^{\infty }\!{\frac {x\cos \left( x \right) -\sin \left( x
 \right) }{{x}^{2} \left( {{\rm e}^{x}}-1 \right) }}{dx} \sim - 0.4152145872, $$

which agrees with Wolfram.


答:
The bijection $\Phi : C^{A \times B} \to (C^A)^B$ is given by

$$\Phi(f)(b)(a) = f(a,b).$$

Now prove this works!

Title: Proving $C^{A \times B} \cong (C^B)^A$ without using the "same cardinality" argument

Body:
I'd like to prove that there is a bijection between the two sets $C^{A \times B}$ (the set of all functions from $A \times B$  to $C$) and $(C^B)^A$ (the set of all functions from $A$ to the set of all functions from $B$ to $C$) by showing one, without considering that this is immediate because they have the same cardinality.

Could anyone give me a hint of how to construct such a bijection?


答:
I'm not sure what you mean by “entailment”.  But I think the answer to your questions is essentially ‘yes’. Say we're given a turing machine $M$ and a string $s$.  Suppose we could construct a set $\Delta_{M,s}$ of first-order formulas that expresses the statement “Machine $M$ halts on input $s$”. That is, $$\Delta_{M,s} \vDash H(M, s). $$

We can in fact construct such a set of formulas; that's not the problem.  Doing this is straightforward!  Then if we could prove the formulas, we would know that the machine halted.

The problem is, the formulas don't help, because we there is no way to find out (in general) whether $\Delta_{M,s}$ is provable, or not.

There is a procedure that will tell us if $\Delta_{M,s}$ is provable: enumerate every possible proof, and check each one to see if the thing it proves is $\Delta_{M,s}$.  But if $\Delta_{M,s}$ is not provable, this procedure fails to halt.  So it's no better in general than just running the machine and waiting to see if it halts.

Is there a procedure that is guaranteed to tell us if $\Delta_{M,s}$ is not provable?  In particular cases there is; just try to prove the negation of $\Delta_{M,s}$, and if you succeed, then $\Delta_{M,s}$ is false and therefore not provable.  Then we would know that $M$ won't halt.  But this doesn't work in general because we may not be able to prove either $\Delta_{M,s}$ or its negation.

And how do we know that such cases exist?  Easy: if either $S$ or $\lnot S$ was provable for every $S$, we would be able to solve the halting problem, and we know we can't.

I'm not sure if this will help or just cause more confusion, but this comment I wrote elsewhere might give you something to think about:

It might be instructive to consider the simpler relation $HN(m,i,n)$ which holds if machine $m$, given input $i$, halts after exactly $n$ steps. Clearly this relation is decidable: Simulate $m$ for at most $n$ steps, or until it halts. If it halts on the $n$th step, the answer is yes; if it halts too soon, or not by the $n$th step, the answer is no. But the decidability of $HN$ doesn't help you decide if $m$ halts, because that question is not $HN(m,i,n)$ for any fixed $n$, but rather $∃n.HN(m,i,n)$…

If you want to think about what your $\Delta$ actually looks like, thinking about $HN(m,i,n)$ might be a good place to start. And you should consider the possibility that  you might be able to prove $\lnot HN(m,i,0), \lnot HN(m,i,1), \ldots$ for every $n$, but still not be able to prove what you need, which is $\forall n.\lnot HN(m,i,n)$.

Title: The halting problem solved if there is an effective positive test for first-order entailment?

Body:
given any Turing machine m and any input n, we can construct a finite set of sentences ∆ and a corresponding sentence H, halting sentence, such that ∆ ⊨ H  iff  TMm eventually halts on input n. Thus if there were an effective method for determining that ∆  ⊭  H then we could solve the halting problem and thereby refute the Church-Turing Thesis.
If we know that there is an effective positive test for first-order entailment, can we solve the halting problem by considering the negation of the halting sentence, viz. ∆ ⊨ ¬H?</p>


问:
Calculating $\sum_{k=0}^{n} {n\choose k} \sqrt{k}$
I am trying to calculate the following expression but not sure how this can be solved yet.
$$\sum_{k=0}^{n} {n\choose k} \sqrt{k}$$
The wiki page (https://en.wikipedia.org/wiki/Binomial_coefficient) contains a similar expression with $\sqrt{k}$ replaced by $k$ or $k^2$. If the above expression can be solved, I also wonder if we can solve it for any $k^j$ with any real number $j>0$?

答: For asymptotics, one might call for Watson's lemma. Using $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$ for $n,k>0$, our sum is $$S_n=n\sum_{k=1}^n\binom{n-1}{k-1}\frac{1}{\sqrt{k}}=\frac{n}{\sqrt\pi}\sum_{k=1}^n\binom{n-1}{k-1}\int_0^\infty e^{-kx}\frac{dx}{\sqrt{x}}=\frac{n}{\sqrt\pi}\int_0^\infty(1+e^{-x})^{n-1}\frac{e^{-x}}{\sqrt{x}}\,dx$$ and, substituting $1+e^{-x}=2e^{-t}$, we get $$S_n=\frac{2^n n}{\sqrt{2\pi}}\int_0^{\log2}t^{-1/2}e^{-nt}\phi(t)\,dt,\quad\phi(t)=\left(\frac{-2t}{\log(2e^{-t}-1)}\right)^{1/2}.$$ Now the expansion $\phi(t)=1-\dfrac{t}{4}-\dfrac{5t^2}{32}-\dfrac{47t^3}{384}-\dfrac{237t^4}{2048}-\ldots$ gives, by the lemma, $$S_n\asymp 2^n\sqrt\frac{n}{2}\left(1-\frac{1}{8}n^{-1}-\frac{15}{128}n^{-2}-\frac{235}{1536}n^{-3}-\frac{24885}{32768}n^{-4}-\ldots\right)\quad(n\to\infty)$$


问:
Formal relationship between rules of inference and the material conditional
I am not $100\%$ clear as to what constitutes the difference between a rule of inference and the material conditional, at least in classical logic. I am using the truth-functional definition of the material conditional, commonly visualised through its truth table, but I'm not entirely sure what the formal definition of a rule of inference is. The wikipedia article defines it to be a particular kind of logical form, which seems to be a term from philosophical logic that I'm not familiar with, but reading that article didn't really answer my question. It pertains more to the mathematical side of things, and I am specifically interested in the interplay between the concepts on the syntactic and semantic level. As far as I can tell, any rule of inference can be 'captured' by a corresponding material conditional: if we take modus ponens as a well-known example, what is the difference between $$(a\land (a\to b))\to b$$ and $${a\to b,\text{ } a \over b}?$$ On a functional level, both statements seem to be expressing the same thing. What determines the need to use two separate terms and notations, and what, if anything, separates them?

答:
Relation between semantic inference an material conditional

For classical logic, the deduction theorem holds:

$$A_1, \ldots, A_n \vDash B\ \Longleftrightarrow\ \vDash (A_1 \land \ldots \land A_n) \to B$$

i.e. the inference from premises $A_1, \ldots, A_n$ to conclusion $B$ is valid if and only if the material conditional $(A_1 \land \ldots \land A_n) \to B$ is a tautology.

Both are model-theoretic notions: $A_1, \ldots, A_n \vDash B$ means that the conclusion $B$ is true in every model in which the premises $A_1, \ldots, A_n$ are true; $\vDash B$ is a special case of that without the restriction on models of the premises.

Relation between formal proof/inference rule and semantic inference

An inference rule in a formal proof system is a rewrite rule: $\dfrac{a \to b \quad a}{b}$ says "If you have the string of symbols $a \to b$, and another string $a$ on on the same level, then you are allowed to draw a line beneath them and write the symbol $b$ below it". A formal proof system is "syntactic" in the sense that it only operates by manipulations of strings of symbols, without making implicit reference to "semantic" (actually more precisely, model-theoretic/truth conditional) notions such as models and truth values.

Of course, these manipulations of strings of symbols are not just random:

A useful proof system will be sound w.r.t. the semantics, meaning that if there is a formal proof ($\vdash$) of some conclusion from a number of premises, then the inference is semantically valid ($\vDash$) in the sense that every model of the premises is also a model of the conclusion -- i.o.w., soundness guarantees that the system doesn't prove nonsense: $$A_1, \ldots, A_n \vdash B \Longrightarrow A_1, \ldots, A_n \vDash B$$

In addition, many proof systems are semantically complete, meaning that the system makes it possible to find a formal proof ($\vdash$) for every semantic inference ($\vDash$): $$A_1, \ldots, A_n \vDash B \Longrightarrow A_1, \ldots, A_n \vdash B$$

Soundness and completness are useful properties which hold for many proof systems, but they are not self-evident; by itself, $\vdash$ is indpependent of $\vDash$, and one has two prove that the two actually match.

A proof system as a whole can only be sound if all its inference rules are sound, i.e. each step in a formal proof consitutes a semantically valid inference; that is, we have

$$\dfrac{A_1 \quad \ldots \quad A_n}{B} \Longrightarrow A_1, \ldots, A_n \vDash B$$

The converse direction does not hold directly, because not any semantically valid inference has a single inference rule corresponding to it; but any inference will have a derivation with several inference rules plugged together corresponding to it ($\vdash$).

Relation between formal prof and material conditional

Combining the soundness + completeness theorem with the deduction theorem, we have that

$$\dfrac{A_1 \quad \ldots \quad A_n}{B}\ \Longrightarrow\ A_1, \ldots, A_n \vDash B\ \Longleftrightarrow\ \vDash (A_1, \ldots, A_n) \to B$$

so also

$$\dfrac{A_1 \quad \ldots \quad A_n}{B}\ \Longrightarrow\ \vDash (A_1, \ldots, A_n) \to B$$

-- every formal inference rule is a tautological material implication --

and more generally,

$$ A_1, \ldots, A_n \vdash B\ \Longleftrightarrow\ A_1, \ldots, A_n \vDash B\ \Longleftrightarrow\ \vDash (A_1, \ldots, A_n) \to B$$

so in particular

$$ A_1, \ldots, A_n \vdash B\ \Longleftrightarrow\ \vDash (A_1, \ldots, A_n) \to B$$

-- every formal proof consisting of inference rules plugged together is a tautological material implication, and every tautological material implication has a formal proof consisting of inference rules plugged together.

For reasons why formal proof systems are at all needed, see Why should we care about syntactic proofs if we can show semantically that statements are true?.