问:
Conditional expectation on $p \in [1,+ \infty]$
If $X \in L_{p}(\Omega,\mathcal{F},P)$ for some $p \in [1,+\infty]$ how can I show that for the conditional expectation $Y = \mathbb{E}[X|\mathscr{p}]$ we have $$Y \in L_{p}(\Omega,\mathcal{F},P)$$

答:
To make the notations clearer, I'll assume that $Y$ is the conditional expectation of $X$ given a sub-$\sigma$-algebra $\mathcal{G}$. Therefore we have $Y := \mathbb{E}[X|\mathcal{G}]$.

Now, if $p \in [1,+\infty[$, then the map $x \mapsto |x|^p$ is convex, and we can apply Jensen's Inequality for conditional expectation, therefore :

$$|Y|^p \le \mathbb{E}\left[|X|^p\,|\mathcal{G}\right] \tag{1} $$

Taking expectations of both sides in (1), we get

$$ \mathbb{E}\left[|Y|^p\right] \le \mathbb{E}\left[\mathbb{E}\left[|X|^p\,|\mathcal{G}\right]\right] =\mathbb{E}\left[|X|^p\right] $$

Hence, if $X \in L_{p}(\Omega,\mathcal{F},\mathbb{P})$, then $\mathbb{E}[X|\mathcal{G}] \in L_{p}(\Omega,\mathcal{F},\mathbb{P})$ too.


问:
Prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions over $\mathbb{R}$
The problem goes as follows:

Using elementary methods prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions for $x \in \mathbb{R}$.

I first came across the problem in a school book targeted towards students just exposed to factorisation and it intrigued me how one could prove this without Sturm's Theorem. Maybe one could try to factorise into squares, but I can get no result from that. Thanks in advance.

答:
We may observe that our polynomial has the form

$$ (x^2+x+1)\left(x^3 - \frac{5}{2}\right)^2 + \frac{1}{4} (3x^2+3x-1) $$
The first term is always non-negative (and positive for $x \neq \sqrt[3]{\frac{5}{2}}$, that case we can check separately), and the second one is non-negative outside interval $(-2,1)$, so we just need to check the values inside this interval. But $x^2+x+1 = (x+ \frac{1}{2})^2) + \frac{3}{4} \geqslant \frac{3}{4}$, and $(x^3-\frac{5}{2})^2$ has on this interval values not less than $\frac{9}{4}$. It suffices to show that minimum of $\frac{1}{4}(3x^2+3x-1)$ is greater than $\frac{27}{4}$, which is easy to check with elementary methods.

ContentLicense: CC BY-SA 4.0


问:
Let $G$ be a group in which $aba = b \; \forall a,b \in G$. Prove that $G$ is Abelian.
I was wondering if someone could tell me if my method is correct. I feel like the substation $b=e$ is a stretch and that there is a better way to do it.

We know that $G$ is Abelian $\iff (ab)^{-1} = a^{-1}b^{-1}$. So I know that we will have to manipulate $aba = b$ into $(ab)^{-1} = a^{-1}b^{-1}$.

But this is where I get stuck.
$$aba = b \implies ab = ba^{-1} \implies (ab)^{-1} = (ba^{-1})^{-1} \implies (ab)^{-1}=ab^{-1}$$
This is close, but not what I want. One thing that I did notice is that if $b=e$, then $aba=b \implies a^2 = e \implies a=a^{-1}$. Since $a^{-1}$ is a unique element we know that $a=a^{-1}$ regardless of what $b$ is. So if we make that substitution into out previous equation we get
$$(ab)^{-1} = a^{-1}b^{-1}$$
Therefore $G$ is Abelian.

答:
Choose

$a = b; \tag 1$

then

$aba = b \tag 2$

becomes

$a^3 = a, \tag 3$

which implies

$a^2 = e, \tag 4$

$e$ being the identity element of $G$; thus,

$a = a^{-1}, \tag 5$

whence (2) becomes

$aba^{-1} = b, \; \forall a, b \in G, \tag 6$

or

$ab = ba, \tag 7$

and we see that $G$ is abelian. $OE\Delta$.


问:
Nash equilibrium and common knowledge
If NE is a CK? It seems that yes since given all information about payoffs/strategies players can derive NE based on the procedures similar to that of in the common knowledge, but I'm not sure. </p>

答:
If the full payoff matrix for everyone was part of the common knowledge, then of course everyone will be able to observe that a given position is a Nash equilibrium, by observing that no-one will be able to improve their position by changing individually, and this calculation would also be part of the common knowledge (provided it was common knowledge that everyone could perform this calculation). So in this case, it would be common knowledge whether a given position was a Nash equilibrium.

But if the payoff matrix was not part of the common knowledge, for example, if people only know their own payoffs, then it could be that they find themselves in a Nash equilibrium, in the sense that no-one can improve their payoff by changing only their own value, but not knowing sufficiently much of the full payoff matrix, they wouldn't know whether someone else could improve their position, and in this imperfect information situation being in a Nash equilibrium wouldn't be common knowledge.

Even if everyone knows the full payoff matrix, but doesn't know that everyone knows it (or perhaps they know everyone else knows it, but don't know that everyone else knows that everyone knows it), then it could be that everyone knows the current position is a Nash equilibrium, but it still might not be part of the common knowledge, because perhaps not everyone knows that everyone knows it.

ContentLicense: CC BY-SA 3.0

Title: Nash equilibrium and common knowledge


问:
Characterising the affine image of a polar convex body
Let $S\subseteq \mathbb R^n$. Its polar is defined by $$S^\circ  = \{z\in\mathbb R^n:z^Tx\le 1,\forall x\in S\} $$

Now, let's say we are interested in some affine image of $S$ (and its polar). Let $T = AS + b = \{Ax+b:x\in S\}$ be the affine image of interest. Then $$T^\circ  = \{z\in\mathbb R^n:z^Tx\le 1,\forall x\in T\}$$

Is it possible to nicely characterize $T^\circ$ in terms of $S^\circ$ somehow? I know that $$x\in S \Longleftrightarrow Ax+b\in T$$ which gives me $T^\circ  = \{z\in\mathbb R^n:z^Tx\le 1,\forall x\in T\} =\{z\in\mathbb R^n:z^T(Ax+b)\le 1,\forall x\in S\}$. Is this correct so far?

I'm not able to work with this expression further and obtain a nice expression for $T^\circ$ in terms of $S^\circ$. Intuitively I feel that $T^\circ = AS^\circ + b$ could be a good guess, but I'm not sure. Any ideas?

Thanks a lot!

答:
Your derivations are correct. You can take a few more steps if $b = 0$:
\begin{align}
\{z\in\mathbb R^n:z^T(Ax+b)\le 1,\forall x\in S\}
& =\{z\in\mathbb R^n: (A^Tz)^Tx\le 1,\forall x\in S\} \\
& =\{z\in\mathbb R^n: y^Tx\le 1,\forall x\in S, y=A^Tz\}
\end{align}
which is the inverse image of $S^\circ$ under $A^T$.

For $b\neq 0$ there is no simple transformation. For example, the polar of $[-1,1]$ is $[-1,1]$ while the polar of $[0,2]$ (corresponding to $A=b=1$) is $(-\infty,0.5]$.


问:
Probability that two throws with three dice each will show the same configuration if the dice are indistinguishable
The question was asked couple of times here. First this question and second this question.

But unfortunately, I couldn't really understand that logic.

Since the dice are indistinguishable, there are only 41 ways of getting different results in the first throw of three dice. These are

All distinct (20)

(1,2,3), (1,2,4), (1,2,5), (1,2,6), (1,3,4), (1,3,5), (1,3,6), (1,4,5), (1,4,6), (1,5,6), (2,3,4), (2,3,5), (2,3,6), (2,4,5), (2,4,6), (2,5,6), (3,4,5), (3,4,6), (3,5,6), (4,5,6)

Two repetition (30)

(1,2,2), (1,3,3), (1,4,4), (1,5,5), (1,6,6), (2,1,1), (2,3,3), (2,4,4), (2,5,5), (2,6,6), (3,1,1), (3,2,2), (3,4,4), (3,5,5), (3,6,6), (4,1,1), (4,2,2), (4,3,3), (4,5,5), (4,6,6), (5,1,1), (5,2,2), (5,3,3), (5,4,4), (5,6,6), (6,1,1), (6,2,2), (6,3,3), (6,4,4), (6,5,5)

Three repetition (6)

(1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6)

Thus we got total of 56 distinct ways of arranging the first throw of the dice. Now the second throw should exactly match the first throw. For each of the first throw, this can happen only way. So the probability has to be $\frac{56}{6^6}$.

What am I doing wrong ?

答:
The $20$ all distinct "configurations" correspond to $20 \times 3! = 120$ actually different throws.

The $30$ from your second group correspond to $30 \times 3 = 90$ actually different throws.

(check $120 + 90 + 6 = 216 = 6^3$)

Call the event that their the same configuration in the end $S$, and call the groups I,II,III. If the first throw is from I we have $\frac{6}{216}$ chance to repeat the configuration. If from II, that is $\frac{3}{216}$, and from III, only $\frac{1}{216}$ so that (conditioning on the event that the first throw is in the respective group):

$$P(S) = P(S|I)P(I)+ P(S|II)P(II) + P(S|III)P(III)=
  \frac{6}{216}\frac{120}{216} + \frac{3}{216}\frac{90}{216} + \frac{1}{216}\frac{6}{216} \simeq .0213$$


问:
Prove $\sup\{r\geq 0: \exists M\in\mathbb{R}[\forall n\in \mathbb{N} [\, |r^n a_n|<M\ ]] \}= \sup \{ r\geq 0 :\sum_n |r^n a_n| \in \mathbb{R} \}$
Given a complex power series $\sum_n a_n z^n$ for $z,a_n\in\mathbb{C}$, let
$$A:=\{r\geq 0:  \exists M\in\mathbb{R}[\forall n\in \mathbb{N} [\, |r^n a_n|<M\ ]] \}$$
$$B := \{ r\geq 0 :\sum_n |r^n a_n| \in \mathbb{R} \}$$

Prove that $\sup A = \sup B$.

Now my colleagues and I are becoming crazy with this one. We know that $B\subset A$, which implies that $\sup B \leq \sup A$, so I am trying to prove that $\sup B < \sup A$ leads to a contradiction, but after many trials, I can't find a reason to imply the conclusion, although I am not able either to find a contre-example to $\sup A = \sup B$, for every time I try I get simply that the radius of convergence of the failed counter-example series is actually the $\max A$, which still confirms the expected result.

On the other hand,

$$\begin{split}
&r>\sup B\ &\Rightarrow l :=\limsup_{n\to\infty} |a_n r^n|\neq 0\\
&l \in \mathbb{R} &\Rightarrow \exists M\in\mathbb{R}[\forall n\in\mathbb{N}[\ |a_n r^n| < M\ ]]\\
&&\Rightarrow l \in A\\
&&\Rightarrow \sup A \neq \sup B
\end{split}$$

which is a reasoning that I find correct, but I guess it is not, right? I am completely puzzled, and I've been thinking in circles for hours. Please, is $\sup A = \sup B$? Give a proof, or at least lead me to it. Thank you in advance.

答:
Clearly $B\subseteq A$ and hence $\sup B\leq\sup A$. We go to show
that $\sup A\leq\sup B$. Prove by contradiction. Suppose the contrary
that $\sup A>\sup B$. Choose $r_{1}$ such that $\sup B<r_{1}<\sup A$.
Since $r_{1}$ is not an upper bound of $A$, there exists $r\in A$
such that $r_{1}<r$. Choose $M>0$ such that $|r^{n}a_{n}|<M$ for
all $n$. Observe that
\begin{eqnarray*}
 &  & \sum_{n=0}^{\infty}\left|r_{1}^{n}a_{n}\right|\\
 & = & \sum_{n=0}^{\infty}\left|\left(\frac{r_{1}}{r}\right)^{n}r^{n}a_{n}\right|\\
 & \leq & \sum_{n=0}^{\infty}M\left(\frac{r_{1}}{r}\right)^n\\
 & < & \infty
\end{eqnarray*}
because $0<\frac{r_{1}}{r}<1$. This shows that $r_{1}\in B$, which
is a contradiction.


问:
Solving for theta
I am trying to solve some a math problem in a 3D software project.

I know that:

$$\cos(\theta)\sin(75°)a+\cos(75°)b+\sin(\theta)\sin(75°)c=0$$

How do I write an equation to solve for $\theta$? Could someone go through all the steps please? Because I have to solve many similar problems.

I already know that there may be zero, one, or (usually) two solutions.

Thanks.

[edit]

By the way, what would the solutions for $\theta$ be in this case if a = 4, b = 5, and c = 6? (Just so I can work it out on my own.)

答:
Your question suggests that $a$, $b$ and $c$ are known constants. So are the $\sin$ and $\cos$ of $75^\circ$. So you want to solve an equation of the form
$$
r\cos \theta + s \sin \theta + t = 0
$$
for $\theta$ in terms of $r$, $s$ and $t$.

Write
$$
r \cos \theta = r\sqrt{1-\sin^2 \theta} =
- s \sin \theta - t.
$$

Square both sides of that equation and solve the resulting quadratic equation for $\sin \theta$.

Title: Solving for theta


问:
Finding limit of sequence quite hard
Im trying to find this limit as n goes to infinity. I raised it to 1/n power bit I cant get it to a  form so I can substitute n with infinity . Form answer book I know its 0, bit I just cant see how to deal with this.

$$\lim_{ n\rightarrow\infty} (-1)^n\frac{\sqrt{1+n}}{n}$$

答:
Note that $\lim_{n\rightarrow \infty} a_nb_n = \lim_{n\rightarrow\infty}a_n\lim_{n\rightarrow\infty}b_n$ if $a_n, b_n$ are convergent.
Consider $$(-1)^n\frac{\sqrt{1+n}}{n} = \frac{(-1)^n}{\sqrt{n}}\frac{\sqrt{1+n}}{\sqrt{n}}.$$ Put $a_n = \frac{(-1)^n}{\sqrt{n}}$ and $b_n = \frac{\sqrt{1+n}}{\sqrt{n}}.$ It is easy to show that both $a_n$ and $b_n$ are convergent with $\lim_{n\rightarrow\infty}a_n = 0$, and $\lim_{n\rightarrow\infty}b_n = 1$. Using the aforementioned result, $$\lim_{n\rightarrow\infty}(-1)^n\frac{\sqrt{1+n}}{n} = 0.$$

Title: Finding limit of sequence quite hard


Title: Prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions over $\mathbb{R}$

Body:
The problem goes as follows:

Using elementary methods prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions for $x \in \mathbb{R}$.

I first came across the problem in a school book targeted towards students just exposed to factorisation and it intrigued me how one could prove this without Sturm's Theorem. Maybe one could try to factorise into squares, but I can get no result from that. Thanks in advance.

Body:
Define the real polynomial
$$ p(x)\!:=\!x^8\! + \!x^7\! + \!x^6\! - \!5x^5\!-
 \!5x^4\!- \!5x^3\! + \!7x^2\! + \!7x\!+\!7. \tag{1} $$
By observation, the polynomial easily factors as the product
$$ p(x) = (x^2+x+1)(x^6-5x^3+7) = p_1(x)p_2(x^3). \tag{2} $$
If we can prove for all real $\,x\,$ that $\,p(x)>1\,$ then
also $\,p(x)-1>0\,$ and hence $\,p(x)-1,\,$ the polynomial
in question, has no real roots.

Note that
$$ p_1(x) := x^2 + x + 1 = (x+1/2)^2+(3/4) \ge 3/4
 \tag{3} $$
and
$$ p_2(x) := x^2-5x+7 = (x-5/2)^2+(3/4) \ge 3/4.
 \tag{4} $$
Note that $\,p_1(x)\le 1\,$ iff $\,-1\le x\le 0\,$ and
$\,p_2(x^3)\le 1\,$ iff $\,2\le x^3\le3\,$ which are disjoint intervals. Now prove that when $\,p_1(x)<1\,$ that $\,p_2(x^3)>4/3\,$ and also when $\,p_2(x^3)<1\,$ that $\,p_1(x)>4/3.\,$


问:
How to evaluate the following integral along the principal branch of Logz
$\int_{-1}^1 \log{z} dz$, where the principal branch of $\log z$ is the negative real axis, and hence $Arg(z) \in (-\pi,\pi]$. \

This is what I did.

\begin{align*}
\int_{-1}^1 \log{z} dz &= z\log z - z|_{-1}^1 \\
&= \log(1) - 1 - (-\log(-1) + 1) \\
&= \log(1) + \log(-1) - 2 \\
&= ln|1| + iArg(1) + \ln|-1| + iArg(-1) - 2 \\
&= i\pi - 2
\end{align*}

I used FTC, but I'm not sure if I'm allowed to since the line segment from $-1$ to $1$ goes through the branch cut. So I'm pretty sure my computation does not make theoretical sense. So where have I went wrong?

答:
Fundamental Theorem of Calculus works for a complex function $f$, when there is a function $F$ such that its complex derivative exists and equal to $f$ along the integration path. For the complex derivative to exist, $F$ should be holomorphic in some open set, that includes the path of integration.

You can change the definition of  $\log$ by taking $\operatorname{Arg}(z)\in(-\pi+\varepsilon, \pi+\varepsilon]$ for some small $\varepsilon>0$. Then $F(z) = z\log z - z$ will be holomorphic in open sets

$$
\begin{align}
&U_1 = \{z \in \mathbb{C} \;| \; \operatorname{Arg}(z) \in (\pi-\varepsilon, \pi+\varepsilon) \text{ and } z\ne 0\}, \\
&U_2 = \{z \in \mathbb{C} \;| \; \operatorname{Arg}(z) \in (-\varepsilon, \varepsilon) \text{ and } z\ne 0\},
\end{align}
$$
and its complex derivative will equal to $f(z) = \log z$ within this sets.

You can then calculate
$$
\int_{-1}^{0} f(z)dz = -1+i\pi \quad \text{and} \quad \int_{0}^{1} f(z)dz = -1
$$
without any worries.


问:
Showing the smoothness of the function with two variables
In the last step of the proof, I have to show that $f:(x,y)\mapsto\sqrt{1-x^2-y^2}$ is a smooth mapping, where $x,y\in\mathbb{R}$. From direct computation with the usage of induction, I know that this map is $C^{\infty}$ if we view this map with respect to $x$ and $y$, and view $y$ and $x$ as a constant, respectively. But to show that $f_{xyxyxyyxyy}$ is continuous, the only thing I can think of is clairaut theorem, which I am uncertain I can use here. Getting my hands dirty for this function seems also like an impossible mission. What should I do? Is there any general theorem regarding this?

Edit: $x,y$ is in the disc, not the entire real.

答: Your function is defined for $x^2+y^2\leq 1$, which represents the unit disc (with boundary) centered at the origin of $\mathbb R^2$. Now note that $f = g\circ h$, where $h(x,y) = 1-x^2-y^2$ and $g\colon \mathbb R_{\geq 0} \to \mathbb R$ is defined by $g(x) = \sqrt{x}$. Showing that $h$ is smooth is easy, because it is polynomial. The map $g$ is smooth in the open domain $(0,+\infty)$, whereas for $x \to 0^+$ its first derivative explodes to $+\infty$. So your $f$ is a composition of smooth functions in the interior of your disc, and is thus smooth (see e.g. here). There is no way to prove smoothness on a bigger subset of $\mathbb R^2$ including your disc, because such a set would contain points on the boundary of the disc where the gradient of $f$ is not defined.


答: The answer of @lmaosome does the work.  Also notice that any time you see an alternating sequence as $(-1)^n$ then the only way for $(-1)^nb_n$ to converge, it is that $b_n\to 0$. So in your case you should check that $b_n=\frac{\sqrt{n+1}}{n}\to 0$. Another way to see why this happens, $$\frac{\sqrt{n+1}}{n}=\frac{\sqrt{n+1}}{n}\cdot \frac{\sqrt{n+1}}{\sqrt{n+1}} =\frac{1}{\sqrt{n+1}}+\frac{1}{n\sqrt{n+1}}$$ since the denominators $n\sqrt{n+1}\to\infty$ and $\sqrt{n+1}\to\infty$ you get $b_n\to0$. So using this you get $(-1)^nb_n\to 0$ which in case of convergence is the only limit possible.

Title: Finding limit of sequence quite hard

Body:
Im trying to find this limit as n goes to infinity. I raised it to 1/n power bit I cant get it to a  form so I can substitute n with infinity . Form answer book I know its 0, bit I just cant see how to deal with this.

$$\lim_{ n\rightarrow\infty} (-1)^n\frac{\sqrt{1+n}}{n}$$


问:
p-adic numbers vs real numbers
Could anyone give a concrete example of a p-adic number that is not a "real number"? that is, do we create "new numbers" (non real numbers) by completing Q with a non Archimedean norm? If so, what are they?</p>

答:
Could anyone give a concrete example of a p-adic number that is not a "real number"?

1. If one is very very formal, one could say that no $p$-adic number is a real number. The sets $\mathbb Q_p$ and $\mathbb R$ can be thought of as disjoint.

However, we usually don't want to be that formal. Every field of characteristic $0$ contains a unique copy of $\mathbb Q$, the rational numbers. Hence we can, and usually do, say that both $\mathbb Q_p$ and $\mathbb R$ "contain" $\mathbb Q$. What that means in practice is that if you have any rational number, i.e. a fraction $m/n$ with $m,n \in \mathbb Z, n \neq 0$, you should be easily able to express it as a real number as well as as a $p$-adic number. For example $3/4$ "is" the real number $0.75$, and it also "is" the $5$-adic number $...111112.$ Likewise, $1/7$ "is" the real number $0.1428571428571428571....$ and it also "is" the $7$-adic number $...00.1$ (here, I expand a $p$-adic number in base $p$ to the left, and a real number in base $10$ (we're all so used to the decimal system) to the right). Note that for real numbers one could choose any other base and still the rational numbers would be the ones whose expansions are eventually periodic. There is a crucial fact here that any given rational number is uniquely determined as element of $\mathbb R$ as well as as element of $\mathbb Q_p$, i.e. e.g. $0.75$ "is" the rational number $3/4$ in the sense that no other real number solves $4x-3=0$.

2. That being said, usually questions like this get answered with examples like "$\sqrt{1-p^3} \in \mathbb Q_p$, but $\notin \mathbb R$" and "$\sqrt{p} \in \mathbb R$, but $\notin \mathbb Q_p$". That's fair and will make a beginner careful (and hopefully eager to learn how to come up with examples like these), but I think it is a bit informal, and the imprecision can cause serious trouble later. I will explain this in point 3. For now, I claim these examples are much better phrased as follows: "The equation $x^2-1+p^3=0$ has a solution in $\mathbb Q_p$, but none in $\mathbb R$; hence there is no field embedding $\mathbb Q_p \hookrightarrow \mathbb R$. And the equation $x^2-p=0$ has a solution in $\mathbb R$, but none in $\mathbb Q_p$; hence there is no field embedding $\mathbb R \hookrightarrow \mathbb Q_p$. (Note that for all this, we tacitly presuppose that $\mathbb Q$ is contained in both the $p$-adics and the reals as per 1.)

3. Now I will expand Mark Bennet's comment. I find this absolutely crucial because I remember I needed some time to get this right, and I have seen many students who don't get it. (Also, compare https://math.stackexchange.com/a/3260194/96384) Every beginner will want to say:

"But I learned that, for example, $\sqrt 2 \in \mathbb Q_7$. And everyone knows since high school that $\sqrt 2 \in \mathbb R$. So is not $\sqrt 2 \in \mathbb Q_7 \cap \mathbb R$ ?"

No, it is not. (Or: Only in a sense that is subtly but distinctly different from how a beginner would understand this. And the trouble with the usual phrasing of the examples in 2. is that it hides exactly this subtlety.) In a nutshell, the issue is that the symbol $\sqrt 2$ means one thing in the statement "$\sqrt 2 \in \mathbb R$", but another thing in the statement "$\sqrt 2 \in \mathbb Q_7$".

There are two elements in $\mathbb R$ whose square is $2$. As real numbers, they can be expressed to arbitrary precision, and in decimal expansion they start $x_1=1.41423562...$ and $x_2=-1.41423562...$ (without ever getting periodic), respectively.

There are two elements in $\mathbb Q_7$ whose square is $2$. As $7$-adic numbers, they can be expressed to arbitrary precision, and their $7$-adic expansions start $y_1=...16213.$ and $y_2=...50454.$ (without ever getting periodic), respectively.

We do have $x_1=-x_2$ and $y_1=-y_2$. (This is a purely algebraic fact, as the negative of a square root is also a square root. For example, there are also two elements in $\mathbb F_{17}$ which are square roots of $\bar 2$, namely, $z_1=\bar 6$ and $z_2=\bar{11}$, and they are also each others negatives, $z_1 = -z_2$.)

We can look at the subfields $K:= \mathbb Q(x_1) = \mathbb Q(x_2) \subset \mathbb R$ as well as $E:= \mathbb Q(y_1) = \mathbb Q(y_2) \subset \mathbb Q_7$.

We can also define purely algebraically, without knowing anything about real or $p$-adic numbers, the field $F:= \mathbb Q[X]/(X^2-2)$.

We can then see that the fields $K, E, F$ are all isomorphic. We desperately want to just call all of them $\mathbb Q(\sqrt 2)$. Well, as long as we promise to only do algebra, of course we can do that.

(By "doing only algebra", I mean things like $(1+\sqrt2)^2 = 3+2\sqrt2$, or $(1+\sqrt2)^{-1} = \sqrt2-1$, which are true just by the property that the symbol "$\sqrt2$" signifies a number whose square is $2$. The symbol can stand for each of $x_1,x_2,y_1,y_2$. (This should look somewhat surprising when a beginner thinks about it. Yes, if $\sqrt2$ would mean $\color{red}{-}1.4142...$, these two equations, and many others, still hold.)

But as soon as we want to invoke anything else (like an "order" in the field, talk about somethig being "positive" vs. "negative", say some numbers are big or small, close to each other, ...; i.e. metric or topologies), we should absolutely not do that.

And even if we do only do algebra, we should notice that: Because although the fields $K, E, F$ are all isomorphic, these isomorphisms are not quite canonic. Rather, for each pair of them there are two isomorphisms.

In particular there are two isomorphisms $K \simeq E$. One of them sends $x_1$ to $y_1$, and the other sends $x_1$ to $y_2$. Now we are used to giving the real number starting $1.4142...$, the one I called $x_1$ above, the name and notation "$\sqrt 2$". As mathematicians we have to unlearn this: There is a purely conventional choice here, namely, we just picked the positive (and this notion is already beyond algebra, a property of $\mathbb R$) solution of $x^2-2=0$ as "the" square root. As long as we only do algebra, this is arbitrary, and we could have chosen the other one as well. Everything which distinguishes $x_1$ from $x_2$ is a property of them as real numbers, not just algebra. And everything which distinguishes $y_1$ from $y_2$ is a property of them as $7$-adic numbers, not just algebra.

$x_1$ is positive, $x_2$ is negative. But neither $y_1$ nor $y_2$ are positive or negative. Which one should we match to $x_1$?

$x_1$ is closer to $1$ than $x_2$ is (as real numbers). The powers of $x_1-1$ get smaller and smaller, the powers of $x_2-1$ do not (in $\mathbb R$).

On the other hand, $y_1$ and $y_2$ as $7$-adic numbers both have the same distance from $1$. The powers of $y_1-1$ and $y_2-1$ get neither smaller nor bigger, but remain of the same size. However, $y_1$ is close to $3$, and $y_2$ is close to $4$. So which one of $y_1$ and $y_2$ do you want to call $\sqrt 2$ and which one $-\sqrt 2$, the one that is close to $3$ or the one that is close to $4$?

Or: The polynomial $p(X)= X^2 - 8X + 14$ has a pair of solutions in each of the fields $K, E, F$. We could just call each of those pairs $4 \pm \sqrt 2$. Then we would only do algebra and be fine (and keep the ambiguity $\pm$). But if somebody tried to nail us down, well which of the solutions is ... bigger (further away from $0$ in absolute value)? Hm. If $\pm \sqrt 2$ mean $x_1, x_2$, then the solution $4+x_1$ is bigger than $4+x_2= 4-x_1$ as real numbers (has bigger real absolute value).

But if $\pm \sqrt 2$ mean $y_1, y_2$, then the solution $4+y_2$ is bigger than $4+y_1= 4-y_2$ (has bigger $7$-adic absolute value).

So you see there is a lot of ambiguity and traps here as soon as we leave the field $\mathbb Q$. And here I only wanted to add a square root, which have the nice extra feature that they always come in pairs and we are tempted to just forget about the $\pm$ ambiguity! Now what will happen if we want to identify solutions of cubic polynomials in $\mathbb R$ with ones in $\mathbb Q_p$? For example, the equation $X^3-X+\frac{7}{20}=0$ has three different solutions $X_1,X_2,X_3$ in $\mathbb R$ (graph them with calculus: one of them is "close to $-1$, the other two are "to the left and right of $1/2$"). It also has three different solutions in $\mathbb Q_7$: One of them is close to $1$, one is close to $0$, and one is close to $-1$. If you want to identify them (and that is what you want to do when you say that $X_1, X_2, X_3$ are "also" $7$-adic numbers): Which one is which? And why?

And now imagine higher degree polynomials ... and maybe several at once, or various layers of field extensions ...

Upshot: If one wants to (and for purely algebraic reasons one might want to) identify certain subfields of $\mathbb R$ and of $\mathbb Q_p$ (or, for that matter, of some $\mathbb Q_p$ and a different $\mathbb Q_\ell$), one can do that. But one has to be very careful then to not invoke properties of them which belong to one of them and not to the other.

(Added:) Also, e.g. here is a recent question where the underlying problem is the false belief that a symbol string like "$\,\mathbb Q(\sqrt[3]{2})$" would well-define a subfield of $\mathbb Q_p$, or, for that matter, e.g. "$\,\mathbb Q_5(\sqrt[3]{2})$" would well-define a field extension of $\mathbb Q_5$. These root symbols and the way we are taught to just accept and use them in high school, as if they are well-defined in themselves, will forever be obstacles on beginners' paths to algebra and number theory.

4. If one believes in the axiom of choice, then there are embeddings $\mathbb Q_p \hookrightarrow \mathbb C$ (without axiom of choice, there are none; with choice, there are many, many more than anyone can imagine, and none of them is explicit). See e.g. Is there an explicit embedding from the various fields of p-adic numbers $\mathbb{Q}_p$ into $\mathbb{C}$?.

I want to say: Unfortunately. Because this might re-enforce the false belief that I just tried to argue against in number 3.

For example, in your response to Lubin's comment about $i:= \sqrt{-1} \in \mathbb Q_5$, you say something to the effect of "oh but $i \in \mathbb C$, I know that". Well. As long as "$i$" is just a placeholder for any number whose square is $-1$, yes you can just use that. For example you have $(1+i)^2 = 2i$, or $i^{-1} = -i$, or $(4-i)(3+i)=13+i$ for all such numbers $i$: it doesn't matter if you mean one of the complex or one of the $5$-adic roots of the equation $x^2+1=0$ with $i$ here. But I bet you imagine the complex numbers $\pm i$ one unit away from $0$, perpendicular to the real number line? That's right in the complex numbers as Gaussian plane. But those two numbers we also call $\pm  i$ in $\mathbb Q_5$: One of them is close to $2$, and the other is close to $3$. Actually, the first is close to $7$, and the other close to $18$ (yes, the natural numbers $7$ and $18$!). Actually, the first is pretty close to $57$, the other is very close to $68$ ... Actually, I can find natural numbers that are closer to them than any given $\epsilon$. They are not far away from $\mathbb Q$ at all (nothing in $\mathbb Q_5$ is)!

Also, you know of course that $\mathbb C$ has an automorphism, complex conjugation, which flips $i$ and $-i$. But $\mathbb Q_5$ has no automorphism which flips those two numbers we call $i$ and $-i$ in it. (Indeed, $\mathbb Q_5$ has no non-trivial automorphism at all.)

5. One more example: Take the polynomial $x^3-6 =0$.

This polynomial has three roots in $\mathbb C$. I assume you are familiar with writing them conventionally as $$\sqrt[3]{6}, \qquad e^{\frac{2\pi i}{3}}\cdot \sqrt[3]{6}, \qquad e^{\frac{4\pi i}{3}}\cdot \sqrt[3]{6}.$$ You see that one of them is real, while the other two are truly complex and conjugates of each other.

Well, the polynomial also has three roots in $\mathbb Q_7$. One of them is $$\text{close to } 24, \text{ another one close to } 40, \text{ and a third one close to } 34.$$

If you want to write down an embedding $\mathbb Q_7 \hookrightarrow \mathbb C$, you have to decide: Which one of those three numbers in $\mathbb Q_7$ do you want to send to the real number $1.817 ...$ we call $\sqrt[3]{6}$?

The answer, which is actually a problem, is that you could send any of them there. If you like the number $24$ better than $40$, go for it and identify that third root of $6$ with the real number $\sqrt[3]{6}$. The axiom of choice says, if I throw more and more polynomials (and then also a bunch of transcendentals) at you, you will always be able to identify them with a bunch of complex numbers in a compatible way. However, your neighbour can also say the spell "axiom of choice", but in her compatible choice $\mathbb Q_7 \hookrightarrow \mathbb C$, she identified  a different root (say, the one close to $34$) with the real number $\sqrt[3]{6}$. She might also have matched a different number to $\sqrt{17e}+4.51\pi$ than we did; there's many numbers; we might never even know. Of course for some purposes, one might not care about those. Sometimes one gets away with that. Sometimes, not.

6. (Added in response to comment by Lubin:) Above I mentioned a (presumably) transcendental number at the end. If I were to expand on that, I would actually continue my theme and say that with the transcendental numbers, almost more so than with algebraic numbers, the problem is not so much that "no $p$-adic number can be identified with $\pi$" but (with axiom of choice) "literally every $p$-adic number $t$ which is transcendental over $\mathbb Q$ [of which there are ... plenty] could be identified with $\pi$", in the sense that by axiom of choice there are field isomorphisms $\mathbb C \simeq \mathbb C_p$ sending $\pi$ to $t$.

What a beginner might not get is that for the goal of calling some element of $\mathbb C_p$ by the name "$p$-adic $\pi$", it would be totally useless to say the spell "choice" and claim there is an iso as above, by which one would have merely fixed one's favourite transcendental element $t$ of $\mathbb C_p$ and given it a fancy name.

Rather, one would have to investigate what "roles" the classical $\pi$ "plays" in the real numbers (e.g.: area and / or half the circumference of the unit circle; first positive zero of the sine function; four times the value of arctan at $1$; certain rational multiple of $\zeta$-values; certain multiple of the period of the complex logarithm; ...), and then investigate if the $p$-adic analogues of any of these properties single out a certain $p$-adic number. I understand that's exactly what Professor Lubin is doing in his answer to Can $\pi$ be defined in a p-adic context?, and it turns out that many of those properties either do not have a meaningful analogue in $p$-adics to begin with, or the numbers singled out by them "over there" are something trivial (like $0$ or $1$) and certainly should not be called $p$-adic $\pi$'s.

That does not exclude that some more involved "roles" that the classical $\pi$ (or a related number like $2\pi i$, or another classical special number like $e$, $\gamma$ [?], ...) "plays" in the archimedean world might have well-defined analogues in the $p$-adic world which are played by well-defined unique elements of some $p$-adic field or ring. I do find it fascinating that e.g. in Fontaine's $B_{dR}$ (an overring of $\mathbb C_p$) there is an element which in a certain sense "plays the role" of the complex period $2\pi i$.

Title: p-adic numbers vs real numbers


问:
Sum of a number of terms of $\left\lceil \frac{n}{2} \right\rceil $
The sum of integers from $1$ to $n$ is the following if we ignore the first half of the terms on the left side below:

$ 1+2+...+n\ge \left\lceil \frac{n}{2} \right \rceil + (\left\lceil \frac{n}{2} \right \rceil +1) + ... +n$

$\left\lceil \frac{n}{2} \right \rceil + (\left\lceil \frac{n}{2} \right \rceil +1) + ... +n \ge \left\lceil \frac{n}{2} \right \rceil +...+\left\lceil \frac{n}{2} \right \rceil$

$\left\lceil \frac{n}{2} \right \rceil +...+\left\lceil \frac{n}{2} \right \rceil = (n-\left\lceil \frac{n}{2} \right \rceil+1)*\left\lceil \frac{n}{2} \right \rceil$

$(n-\left\lceil \frac{n}{2} \right \rceil+1)\left\lceil \frac{n}{2} \right \rceil \ge (\frac{n}{2})(\frac{n}{2})$

Could you please explain how we get:

$\left\lceil \frac{n}{2} \right \rceil +...+\left\lceil \frac{n}{2} \right \rceil = (n-\left\lceil \frac{n}{2} \right \rceil+1)*\left\lceil \frac{n}{2} \right \rceil$

$(n-\left\lceil \frac{n}{2} \right \rceil+1)\left\lceil \frac{n}{2} \right \rceil \ge (\frac{n}{2})(\frac{n}{2})$

答:
$$1+2+\dots+n\ge
\left\lceil \frac n2 \right \rceil
+ \left(\left\lceil \frac n2 \right \rceil +1 \right) + \dots + n$$

Either $n = 2k$ or $n = 2k-1$ for some positive integer, $k$.

In either case, $\left\lceil \dfrac n2 \right\rceil = k \,$ and

\begin{align}
   \left\lceil \frac n2 \right \rceil
+ \left(\left\lceil \frac n2 \right \rceil +1 \right) + \dots + n
&= k +(k+1) + \dots + n \\
&\le (1 + 2 + \dots +k) + (k +(k+1) + \dots + n)
\end{align}


问:
subrings and ideals in $\mathbb{Z}_4[i]$

Find, with proof, all the subrings and ideals of $\mathbb{Z}_4[i].$

The subrings of $\mathbb{Z}_4[i]$ that I could find are:

$\{0\}, \{0,2\}, \{0,2i\}, \{0,2+2i\}, \{0,1,2,3\}, \{0,2,2i, 2+2i\}, \{0,2,1+i, 3+i, 2i, 2+2i, 1+3i, 3+3i\}, \mathbb{Z}_4[i]$.

The ideals of $\mathbb{Z}_4[i]$ that I found are:
$\{0\}, \langle 2 + 2i\rangle = \{0, 2+2i\}, \langle 2\rangle = \{0,2,2i, 2+2i\},\langle 1\rangle =\mathbb{Z}_4[i],$ but I'm not sure if this is all of them.

I think I'm missing some ideals and subrings; I was wondering if there's some sort of algorithm-like approach I could use to find these subrings and ideals more easily? If so, then could this algorithmic approach be used to find all the subrings and ideals of $\mathbb{Z}_6[i]$ or $\mathbb{Z}_3[i]$? For $\mathbb{Z}_3[i],$ some subrings would be $\{0\}, \{0, 1, 2\}, \mathbb{Z}_3[i],\cdots$ and some ideals would be $\{0\}, \mathbb{Z}_3[i]$ (in fact $\mathbb{Z}_3[i]$ is a field so these are the only ideals).

答:
Factorize $n=\prod_j a_j^{e_j}$ in $\Bbb{Z}[i]$ where the $a_j$ are irreducibles, the ideals of $\Bbb{Z}[i]/(n)$ are $(\prod_j a_j^{d_j}),0\le d_j\le e_j$.

$4 = (1+i)^4$.

Enumerating the subrings seems harder.


答:
I came up with a weird proof, but the idea behind is kind of simple: the fact is that I expect that for $x$ outside an interval the terms $x^8+x^7+x^6$ will be greater than the negative terms, and in this range the remaining terms will be positive. On the other side, for small $x$ the -5 will paly his role and make this negative.

I don't know if this actually happens, but this led me to factorize in this way:

$$ (x^3-5)(x^5+x^4+x^3) + 7x^2+7x+6 = 0$$

Now, continuing with algebraic manipulation:

$$ (x^3-5)x^3(x^2+x+1) + 7(x^2+x+1) = 1 $$
$$ (x^6 - 5x^3 + 7)(x^2+x+1) = 1 $$

The nice thing here is that we have two big terms and they have to multiplicate to 1. Weird. Let us calculate the minima of the two terms. Imposing derivatives equal to zero, we see that the minima of the first is reached in $ x^3 =(15/6)$, which yields

$$ \frac{225}{36} - \frac{450}{36} + 7 = - \frac{225}{36} + \frac{252}{36} = \frac{27}{36} = \frac{3}{4} $$

The second reaches its minimus at $x= -1/2$, which yields $ 1/4 -1/2+1 = 3/4$.
The strategy now is the following: it is enough to show that when one term is $\le 1$, the other is $> 4/3$: in this way the product will be always $>1$.

Case 1. $x^6-5x^3+7 \le 1$. Substituting $t=x^3$, this is equivalent to
$$ (t-3)(t-2) \le 0$$
that is $2 \le t \le 3$. Substituing back $t=x^3$ we get $\sqrt[3]{2} \le x \le \sqrt[3]{3}$. Since the derivative of $x^2+x+1$ is positive for $x \ge 1$, in the above interval it will yield at least
$$ \sqrt[3]{4} +\sqrt[3]{2} +1 \ge 3 > 4/3 $$

Case 2. $x^2+x+1 \le 1$. This is equivalent to $x(x+1) \le 0$, that is $-1 \le x \le 0$. Using again $t=x^3$, we see that $(t-3)(t-2)+1$ is decreasing in the above interval. Thus the minima is attained at $t=0$, for which it is $7 > 4/3$.


问:
All subspaces of $2\times 2$diagonal matrix
I would like to know all subspaces of a $2\times 2$
diagonal matrix. This is a problem from a book, Introduction to Linear Algebra by Dr. Strang. In the problem, I would like to find all subspaces of a $2\times 2$ diagonal matrix.

The answer says that

"The subspaces of D are D itself, the zero matrix by itself, and all
the “one-dimensional” subspaces that contain all multiples of one
fixed matrix."

the fixed matrix is \begin{bmatrix} d1 & 0\\ 0 & d2 \end{bmatrix}

I understand that the subspaces contain D itself and the zero matrix. However, what does it mean that “one-dimensional” subspaces that contain all multiples of one
fixed matrix" ? The related question is posted in here, but I do not really understand it.

答:
A subspace must have all the properties of a vector space. In particular, summing two elements must yield a new element in the same subspace, and multiplying an element by a scalar will also result in an element of the subspace. What does that mean for this problem? First we need to find out the dimensionality of $D$, which is $2$. So a subspace can have dimensionality $2$, $D$ itself, dimensionality $0$, the zero matrix, or dimensionality $1$. This implies that you have a basis of the subspace with exactly one element. What element is that? It can be any element of $D\ne 0$. Let's call this basis vector the matrix $M$, where
$$M=\begin{pmatrix}a&0\\0&b\end{pmatrix}$$ Then the set $S$ of all matrices of the form $\lambda M, \lambda\in\mathbb R$ is a subspace of $D$. You have from definition that $\lambda_1M\in S$. If $\lambda_1M,\lambda_2M\in S$, then $$\lambda_1M+\lambda_2M=(\lambda_1+\lambda_2)M\in S$$
If $M=0$, then $\lambda M=0$, so the subspace has only one element.


问:
Proving theorem about basis for some topology
Definition:Let $\tau$ be a topology on a set $X$, and let $\mathcal{B} \subset \tau$. Then $\mathcal{B}$ is a basis for the topology $\tau$ if and only if every open set in $\tau$ is the union of elements of $\mathcal{B}$. If $B \in \mathcal{B}$, we say $B$ is a basis element. Note that $B$ is an element of the basis of $\mathcal{B}$, but a subset of the set $X$.

Theorem(used in proof):

Let $(X,\tau)$ be a topological space, and let $\mathcal{B}$ be a collection of subsets of $X$. Then $\mathcal{B}$ is a basis for $\tau$ if and only if

$(1) \mathcal{B} \subset \tau$ and

$(2)$ for each set $U$ in $\tau$ and point $p$ in $U$ there is a set $V$ in $\mathcal{B}$ such that $p \in V \subset U$.

Theorem(trying to prove):Suppose $X$ is a set and $\mathcal{B}$ is a collection of subsets of $X$. Then $\mathcal{B}$ is a basis for some topology if and only if

$(1)$ each point of $X$ is in some element of $\mathcal{B}$ and

$(2)$ if $U$ and $V$ are sets in $\mathcal{B}$ and $p$ is a point in $U \cap V$, there is a set $W$ in $\mathcal{B}$ such that $p \in W \subset (U \cap V)$.

I am confident with the first part however the other direction of the biconditional has been confusing me.

($\Rightarrow$)

Since $\mathcal{B}$ is a basis, $X$ can be expressed as the union of elements of $\mathcal{B}$, so each point in $X$ is in some basis element. If $U,V \in\mathcal{B}$ since $U,V$ are open, $U \cap V$ is also open.Let $x \in U \cap V$. Since $U \cap V$ is open there exists a basis element $W$ such that $x \in W \subset U \cap V$.

($\Leftarrow$)

I have been struggling deciding what I need to show for this direction. Do I need to show for every point in every open set in $\tau$, there is a subset containing the point, contained in the open set, or do I need to show every element of $\tau$ can be expressed as a union of elements satisfying $(1)$ and $(2)$. Thinking about this, I realize it is wrong to already claim there is a set $\tau$, being a topology, without having proven it.

I will try and do the latter. Since each $x \in X$ is in a set satisfying $(1)$ and $(2)$, $X$ is the union of these sets. $\varnothing$ is in the union of these sets vacuously. I am having difficulty deciding what to do, to prove the second half of this theorem. Any hints will be appreciated.

答:
The topology we define from $\mathcal{B}$ is $\tau$, the set of all unions of subfamilies of $\mathcal{B}$.

(1) says that $X \in \tau$ and $\emptyset$ follows as we can take the empty subfamily.

$\tau$ is closed under unions by construction (a union of unions is a union...)

If $U$ and $V$ are in $\tau$, we pick for each $x \in U \cap V$ some $B_1 \in \mathcal{B}$ such that $x \in B_1 \subseteq U$ ($B_1$ is from the subfamily than makes up $U$)
and some $B_2 \in \mathcal{B}$ such that $x \in B_2 \subseteq V$. (ditto for $V$).

Then $(2)$ tells us we can pick $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq B_1 \cap B_2$.

But having done this for all such $x \in U \cap V$ we can easily see that $\bigcup \{B_x\mid x \in U \cap V\} = U \cap V$ (each $x \in U \cap V$ is in its "own" $B_x$, and hence in the union, for one inclusion and all $B_x \subseteq U \cap V$, hence so is the union, for the other), so we can also write $U \cap V$ as a union of a subfamily of $\mathcal{B}$, and so we're done checking $\tau$ is a topology.


问:
How do I know when I can do something to both sides in an inequality?
This question is basically derived from one that I asked previously but with a different focus, that's why I chose not to just edit the question. Please let me know if I should've. The previous question: How do I know when I can multiply by something on an inequality?

I ask about a general rule about when to do something to both sides in an inequality, because saying "whenever the thing you are multiplying is negative, flip it" isn't complete as we can for example, log on both sides, sqrt both sides, raise to 5 both sides, etc. I have also been told that this has something to do with being monotonically increasing or not? Not sure, but it somewhat rings a bell on this topic. I'd love to be given some insight, I know I can't multiply by negative numbers, or if I want to, I have to separate into cases. But I'd love to be given some insight of everything I've mentioned on this post

答:
You can use intuition to know whether an inequality manipulation is valid. For example, it is easy to see that adding $2$ to both sides of $a \gt b$ makes $a+2 \gt b+2$ true. However, if you multiply both sides by $-1$, $-a  \gt -b$ is FALSE. This is because negative numbers with big magnitudes, like $-7$ are smaller than negative numbers with smaller magnitudes, like $-2$, thus if you negate both sides of an inequality, the side with the bigger magnitude side becomes smaller.

However, you can also use rigorous math, as well, to see when an inequality manipuation is valid. If a function $f(x)$ is increasing for all $x$, then $f(a) > f(b)$ if $a \gt b$, which you can see intuitively. So, if we want to see if adding $2$ to both sides of an inequality preserves the inequality, we can let $f(x) = x+2$. From the graph of this, we can tell $f(x) = x+2$ is increasing, thus if $a \gt b$, then $f(a) = a+2 \gt f(b) = b+2$ or $a+2 \gt b+2$. However, a function like $f(x) = -x$ is decreasing for all $x$, which you can tell from the graph, thus $f(a) < f(b)$ if $a \gt b$. Thus, if we have $a \gt b$, we have $f(a) = -a < f(b) = -b$ or $-a \lt -b$.

See if you can apply similar logic on inequality manipulations like taking the reciprocal of both sides, or the absolute value.


问:
True or false statements

Two of the following statements are true and one is false

a) For all rational numbers $q$, there exists an integer $n$ so that $q+n=271$.

b) For all integers $n$, there exists a rational number $q$ so that $q+n=271$.

c) There exists an integer $n$ so that $271-n$ is even.

I believe a and c are true and b is false

答:
Hints:

For any integers $x$ and $y$, their difference $x-y$ is an integer.

Every integer is a rational number.

$\frac{1}{2}$ is a rational number.

Title: True or false statements


问:
Is this definition for primeness known?
A natural number $p$ is prime if and only if it has exactly two solution sets $\{x,y\}$ to the equation $$\frac1p = \frac1x + \frac1y, \text{ or equivalently, }p=\frac{xy}{x+y}.$$

Those solution sets are $\{p,p(p+1)\}$ and $\{2p\}$ (where $x=y=2p$).

Is this a known fact about prime numbers? If so, could somebody provide a reference?

A proof is given below.

Proof: $(\implies)$ Suppose $p$ is prime. Then if $$p=\frac{xy}{x+y},$$ we can express $x=p^ak$ and $y=p^bl$ for $k$ and $l$ coprime to $p$. Without loss of generality, suppose $0\leq a\leq b$ and $1\leq b$. Then we have $$p=p^b\frac{kl}{k+lp^{b-a}}.$$

Suppose that $b>a$. We then have that $p\not\mid k+lp^{b-a}$, and therefore $b=1$ and $a=0$. Therefore $kl=k+lp$, which implies $k(l-1)=lp$. Thus, $p\mid l-1$.

Consider any natural number $m\mid k$. Then $m\mid k(l-1)=lp$. Since $p\not\mid k$, we have $p\not\mid m$. Thus, $m\mid l$.

Similarly, for any natural $n\mid l$, we have $n\mid l(k-p)=k$.

So $l$ and $k$ have the exact same divisors, and thus $l=k$. Thus,

$$k^2=k+kp=k(p+1),$$ meaning $k=p+1$. So $x=p+1$, which implies $y=p(p+1)$. This is the first solution set.

Next, consider the case in which $b=a$. Then $$p=p^b\frac{kl}{k+l},$$ so $$p^{1-b}=\frac{kl}{k+l}.$$

Since $k$ and $l$ are coprime to $p$, it follows $b>0$. We also have $$\frac{1}{p^{b-1}}=\frac1k+\frac1l,$$ implying that $$\frac{1}{p^{a+b-1}}=\frac{1}{p^{2b-1}}=\frac1x+\frac1y,$$ so $2b-1=1$ and thus $a=b=1$. Finally, that implies $$p=p\frac{kl}{k+l},$$ so $kl=k+l$ and therefore $k=l=2$. This gives the solution $x=y=2p$. So if $p$ is prime then it has two solution sets.

$(\Longleftarrow)$ If $p$ is not prime, then $p=qr$ where $q$ is prime and $r\neq 1$. Then $$\frac1p=\frac1{p+r}+\frac1{q(p+r)}$$ gives us another solution set.

答:
This is a known fact, and here follows a simple proof.

Let $p$ be a prime. Suppose $x \ge y \ge 1$ are integers such that
$$\frac{1}{p} = \frac{1}{x}+ \frac{1}{y}$$
Clearly this equation implies that $x,y >p$. Multiplying everything by $xyp$ we get
$$xy-px-py=0$$
$$xy-px-py+p^2=p^2$$
$$(x-p)(y-p)=p^2$$
$(x-p)$ and $(y-p)$ are two positive integers whose product is $p^2$: there are only two possibilities:

$x-p=p^2$ and $y-p=1$: this gives us the solution $(x;y)=(p^2+p;p+1)$

$x-p=p$ and $y-p=p$: this gives us the solution $(x;y)=(2p;2p)$

Title: Is this definition for primeness known?


问:
Deriving derivatives with differentials
What does it mean to derive the rule $d(x^3)=3x^2dx$ using differentials? Are differentials derivatives? In that case what is the question?

答:
There's a variety of ways to derive differential rules.  I'm not sure exactly what you are looking for, but a fairly straightforward way is to find the difference between a function at $x = x$ and $x = x + dx$, and simplify using certain rules.

So, for instance,
$$
d(x^3) = (x + dx)^3 - (x)^3 \\
d(x^3) = x^3 + 3x^2\, dx + 3x\,dx^2 + dx^3 - x^3 \\
d(x^3) = 3x^2\, dx + 3x\,dx^2 + dx^3 \\
d(x^3) \simeq 3x^2\,dx
$$

To understand the last simplification you can think of  multiplying decimals.  If I multiply $0.001$ by $0.001$, the result will be even smaller, $0.000001$.  If you think about a differential as essentially having an infinite number of zeroes after the decimal, multiplying two differentials together gives a value that is essentially infinitely tinier than either differential alone.  Therefore, they are essentially ignorable.

A derivative is simply a taking a differential of both sides of an equation and solving for some ratio of differentials.

I'm basing the process here on the Calculus Made Easy method.  However, if that's the book you are using, it is normally done by simply replacing every instance of $x$ with $dx$, $y$ with $dy$, etc.  So, you would transform the equation:

$$ y = x^3 $$
into
$$ (y + dy) = (x + dx)^3  \\
y + dy = x^3 + 3x^2\, dx + 3x\,dx^2 + dx^3 \\
dy = 3x^2\, dx + 3x\,dx^2 + dx^3 \\
dy \simeq 3x^2\, dx \\
\frac{dy}{dx} \simeq 3x^2
$$
The step between line 2 and line 3 is because, since $y = x^3$, we can subtract $y$ from the left and $x^3$ on the right and it will be the same thing.

Title: Deriving derivatives with differentials


问:
Hartogs Operation
I know that this question has already appeared, but I would like to share my attempt to solve it as well.

Let H be the formula operation of Hartogs . Prove that  ∀A(A < A + H(A)).

Proof:
With  H(A) we denote the smallest ordinal α, such that H(A)=μα[¬α≤A].
So I will use that if A is isomorphic (0) to some ordinal β, then β<α<μα.
Hence H(A)+A is equivalent to β+μα>β.
However, (0) I think that only holds if A is well ordered.

Any help is welcome.

答:
Presumably the context is without the axiom of choice, otherwise you'd only be asked to prove that $A<H(A)$. So you definitely cannot use the assumption that $A$ can be well-ordered.

The concept of Hartogs numbers is usually studied in the context of cardinals, so $<$ here means that there is an injection (but that part should be really obvious), but there is no bijection. This requires you to simply observe the defining property of $H(A)$.

Title: Hartogs Operation


答:
You have a family $\mathscr{B}$ of subsets of $X$ such that each $x\in X$ belongs to at least one $B\in\mathscr{B}$, and whenever $p\in U\cap V$, where $U,V\in\mathscr{B}$, there is a $W\in\mathscr{B}$ such that $p\in W\subseteq U\cap V$. You want to prove that $\mathscr{B}$ is a base for a topology $\tau$ on $X$. Since this is the case if and only if

$$\tau=\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{B}\right\}\,,\tag{1}$$

the natural approach is to define $\tau$ by $(1)$ and prove that $\tau$ is a topology on $X$. You’ve already observed that $\varnothing,X\in\tau$, so it remains to be shown that $U\cap V\in\tau$ whenever $U,V\in\tau$, and that $\bigcup\mathscr{U}\in\tau$ whenever $\mathscr{U}\subseteq\tau$.

It’s helpful to prove a small lemma first: if $B_0,B_1\in\mathscr{B}$, then there is a $\mathscr{U}\subseteq\mathscr{B}$ such that $B_0\cap B_1=\bigcup\mathscr{U}$ (and hence $B_0\cap B_1\in\tau$). I’ll leave this to you: it follows immediately from the second condition on $\mathscr{B}$.

Suppose that $U,V\in\tau$. Then there are $\mathscr{U},\mathscr{V}\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{U}$ and $V=\bigcup\mathscr{V}$. Then

$$\begin{align*}
U\cap V&=\left(\bigcup\mathscr{U}\right)\cap\bigcup\mathscr{V}\\
&=\bigcup_{B\in\mathscr{U}}\left(B\cap\bigcup\mathscr{V}\right)\\
&=\bigcup_{B\in\mathscr{U}}\bigcup_{C\in\mathscr{V}}(B\cap C)\,.\tag{2}
\end{align*}$$

By the lemma each of the sets $B\cap C$ in $(2)$ is the union of members of $\mathscr{B}$, so $U\cap V$ is a union of members of $\mathscr{B}$ and is therefore in $\tau$, as desired.

The proof that $\tau$ is closed under arbitrary unions is essentially the same as the last step of the previous argument.

Title: Proving theorem about basis for some topology

Body:
Definition:Let $\tau$ be a topology on a set $X$, and let $\mathcal{B} \subset \tau$. Then $\mathcal{B}$ is a basis for the topology $\tau$ if and only if every open set in $\tau$ is the union of elements of $\mathcal{B}$. If $B \in \mathcal{B}$, we say $B$ is a basis element. Note that $B$ is an element of the basis of $\mathcal{B}$, but a subset of the set $X$.

Theorem(used in proof):

Let $(X,\tau)$ be a topological space, and let $\mathcal{B}$ be a collection of subsets of $X$. Then $\mathcal{B}$ is a basis for $\tau$ if and only if

$(1) \mathcal{B} \subset \tau$ and

$(2)$ for each set $U$ in $\tau$ and point $p$ in $U$ there is a set $V$ in $\mathcal{B}$ such that $p \in V \subset U$.

Theorem(trying to prove):Suppose $X$ is a set and $\mathcal{B}$ is a collection of subsets of $X$. Then $\mathcal{B}$ is a basis for some topology if and only if

$(1)$ each point of $X$ is in some element of $\mathcal{B}$ and

$(2)$ if $U$ and $V$ are sets in $\mathcal{B}$ and $p$ is a point in $U \cap V$, there is a set $W$ in $\mathcal{B}$ such that $p \in W \subset (U \cap V)$.

I am confident with the first part however the other direction of the biconditional has been confusing me.

($\Rightarrow$)

Since $\mathcal{B}$ is a basis, $X$ can be expressed as the union of elements of $\mathcal{B}$, so each point in $X$ is in some basis element. If $U,V \in\mathcal{B}$ since $U,V$ are open, $U \cap V$ is also open.Let $x \in U \cap V$. Since $U \cap V$ is open there exists a basis element $W$ such that $x \in W \subset U \cap V$.

($\Leftarrow$)

I have been struggling deciding what I need to show for this direction. Do I need to show for every point in every open set in $\tau$, there is a subset containing the point, contained in the open set, or do I need to show every element of $\tau$ can be expressed as a union of elements satisfying $(1)$ and $(2)$. Thinking about this, I realize it is wrong to already claim there is a set $\tau$, being a topology, without having proven it.

I will try and do the latter. Since each $x \in X$ is in a set satisfying $(1)$ and $(2)$, $X$ is the union of these sets. $\varnothing$ is in the union of these sets vacuously. I am having difficulty deciding what to do, to prove the second half of this theorem. Any hints will be appreciated.


问:
Analytic expression for $\mathbb E_X[(X^\top u)^p (X^\top v)^p]$, where $u$ and $v$ are fixed vectors in $\mathbb R^d$ and $X$ is uniform on sphere
Let $u$ and $v$ be fixed vectors in $\mathbb R^d$. Let $X$ be a random vector uniformly distributed on the unit-sphere in $\mathbb R^d$, and let $f_p$ be a real polynomial of degree $p \ge 1$ (for simplicity, we might simply consider $f_p(t) \equiv t^p$).

Question. What is an analytic expression for the correlatetion $c(u,v) := \mathbb E_X[f(X^\top u)f(X^\top v)]$ ?

Special case $p=1$

In this case, a simple computation gives

$$
c(u,v) = \mathbb E[X^\top uv^\top X] = \mbox{trace}(\mbox{cov}(X)uv^\top) = \mbox{trace}((1/d) I_d uv^\top) = \frac{u^\top v}{d}
$$

Is it too crazy to conjecture that in general, $c(u,v) = K_{d,p} \cdot (u^\top v)^p$, for some constant $K_{d,p}$ which only depends on $d$ and $p$ ?

答:
Here's an alternate solution I came up with. So, let $Z$ be a random vector in $\mathbb R^d$ with iid components distributed according to $N(0,1)$. Set $R := \|X\|$ and note that

$X$ and $Z/R$ have the same distribution.

$R$ and $Z$ are independent.

Let $A := (uv^\top + vu^\top)/2$, a psd matrix, and note that $E[(X^\top u)^p(X^\top v)^p] = E[(X^\top A X)^p]$. One computes

$$
E[(Z^\top A Z)^p] = E[R^{2p}]\cdot E[(X^\top AX)] = E[R^{2p}]\cdot E[(X^\top u)^p(X^\top v)^p]. \tag{1}
$$
Now, $E[R^{2p}] = E[(\|Z\|^2)^p] = 2^p \cdot \dfrac{\Gamma(p + d/2)}{\Gamma(d/2)}$, the $p$th moment of a chi-squared random variable with $d$ degrees of freedom It remains to compute. Combining with (1) then gives

$$
E((X^\top u)^p(X^\top v)^p] = \frac{\Gamma(d/2)}{2^p\Gamma(p+d/2)}\cdot\sigma_p(A).\tag{2}
$$
It remains to compute $\alpha_p(A) := E[(Z^\top A Z)^p]$.

Now, since as early as this paper https://www.janmagnus.nl/papers/JRM003.pdf, it is known that the $\sigma_p(A)$ is a polynomial in the first $p$ cumulants $\kappa_1,\ldots,\kappa_p$ of $N(0,1)$, and in fact for the first $2$ values of $p$, evaluates to

$$
\begin{split}
\sigma_1(A) &= \mbox{tr} A = u^\top v\\
\sigma_2(A) &= (\mbox{tr} A)^2 + 2\mbox{tr} A^2 = 2(u^\top v)^2 + \|u\|^2 \|v\|^2\\
&\;\;\vdots
\end{split}
$$

In particular, using (2) gives $E[(X^\top u)(X^\top v)] = \dfrac{u^\top v}{d}$ (which was already computed directly) and
$$
E[(X^\top u)^2(X^\top v)^2] = \dfrac{2(u^\top v)^2 + \|u\|^2 \|v\|^2}{4(d/2 + 1)(d/2)} = \dfrac{2(u^\top v)^2 + \|u\|^2\|v\|^2}{d(d+2)}.
$$


问:
Derive Cauchy-Bunyakovsky by taking expected value
In my notes, it is said that taking expectation on both sides of this inequality $$|\frac{XY}{\sqrt{\mathbb{E}X^2\mathbb{E}Y^2}}|\le\frac{1}{2}\left(\frac{X^2}{\mathbb{E}X^2}+\frac{Y^2}{\mathbb{E}Y^2}\right)$$ can lead to the Cauchy-Bunyakovxky (Schwarz) inequality $$\mathbb{E}|XY|\le\sqrt{\mathbb{E}X^2\mathbb{E}Y^2}$$ I am not really good at taking expected values, may anyone guide me how to go about it?

Note: I am familiar with the linearity and monotonicity of expected values, what I am unsure about is the derivation that leads to the inequality, especially when dealing with double expectation.

Thanks.

答:
Hint. All you need to know here is that expected value is linear, i.e. for arbitrary random variables $X,Y$ and all $a,b\in \mathbb{R}$ $$\mathbb{E}(aX+bY)=a\mathbb{E}X +b\mathbb{E}Y.$$
Is this a fact you are familiar with?

Monotonicity is also important. Once you have $X\le Y$, then you also know that $\mathbb{E}X\le\mathbb{E}Y.$ This is exactly the case you are dealing with. For now, you have that one random variable (namely $\frac{\vert XY\vert}{\sqrt{\mathbb{E}X^2\mathbb{E}Y^2}}$) is not greater than the other. If you take the expected value of both sides then the inequality will remain true. It will be essentially the Cauchy-Schwarz you are looking for.


问:
A Euclidean Geometry Construction
I have constructed the following configuration, as shown in the image below.

Click for image

$ABC$ is a triangle with the circumcircle and incircle drawn, and point $O$ is the incenter. Furthermore, point $D$ is the point of contact of the incircle with $BC$. $E$ is a point on the circumcircle such that $\angle OEA=90^\circ$.

I can see various properties in this setup, however I am struggling to prove them. How can I prove that line $OE$ is parallel to line $BC$, or equivalently that $\angle EOD=90^\circ$, or even that points $E, A, B, D$ are concyclic.

Any help with this would be appreciated.

答:
I don´t know if you are trying to troll but anyway. Here is a similar configuration that proves your first two conjectures to be false

The third one is obviously false because the circumference through $E,A,B$, which is the excircle of the triangle, doesn´t pass through $D$.

Title: A Euclidean Geometry Construction


问:
Show that $x^2+4x+17$ is $\mathcal{O}(x^3)$ but that $x^3$ is not $\mathcal{O}(x^2+4x+17)$
This proof for Big-O analysis of the the function mentioned in question is done in two parts:

Part I $x^2+4x+17$ is $\mathcal{O}(x^3)$: In this part we can set $k=10$ and we can still get that $x^2+4x+17$ is $\mathcal{O}(x^3)$, but I found based on an answer that the value is $k=17$. We can see that:

$$x^2+4x+17\le x^2 + x^2 + x^2$$
$$ = 3x^2$$

Question 1: We can conclude that $C=3$ and $k=17$. But I am not sure why $k=17$ and not $k=10$ for example?

Part II $x^3$ is $\mathcal{O}(x^2+4x+17)$: In this part we can set $k=17$, and then proceed the proof by contradiction to arrive that:

$$\mid x^3\mid \le C\mid 3x^2\mid$$
$$$$

Question 2: in the proof it's mentioned that we have a contradiction because $\mid x^3\mid \gt \mid 3C x^2\mid$ when $x>3C$. I think taht $\mid x^3\mid$ is always greater than $\mid 3C x^2\mid$ not only $x>3C$, so I don't understand his conclusion here.

答:
The definition of Big-O states that $f(x) \in \mathcal{O}(g(x))$ if and only if there exist constants $C, k \in \mathbb{Z}^{+}$ such that $|f(x)| \leq C \cdot |g(x)|$ for all $x \geq k$.

The solutions may have picked $k = 17$ for convenience, but $k = 10$ seems to also check out. Note that the quantifier in the definition is existential. It does not exclude other valid pairs for $(C, k)$.

Edit: For the second part, you need to rule out all $(C, k)$ pairs, not just pairs where $k = 17$. Try instead the following.

Suppose that $(C, k)$ satisfies that $|x^{3}| \leq C|x^{2} + 4x + 17|$ for all $x \geq k$. We exhibit a value of $x \geq k$ for which this inequality fails to hold. From the previous part, we showed that $x^{2} + 4x + 17 \leq 3x^{2}$ for $x \geq 17$. So to achieve our contradiction, it suffices to show that $|x^{3}| \geq 3C \cdot |x^{2}|$ for some $x \geq k$. Solving for $x$, we obtain that $|x^{3}| \leq C|x^{2} + 4x + 17|$ fails to hold whenever $x > \max\{3C, k, 17\}$.


问:
Rings with non-zero intersection of all non-zero ideals
Let $R$ be a ring such that $\bigcap_{I\neq 0} I \neq 0$, ie. has a non-zero intersection of all non-zero ideals. This is equivalent to ask for the existence of an element $a\in R$ which is a multiple of all non-zero elements of the ring.

What can be said of such rings? Can they be classified? Right now the only examples I can think of are $\Bbb Z/(p^k)$ and fields. I also know that if $R$ is an integral domain with this property then it must be a field. Are there other examples?

Thank you in advance.

答:
This is equivalent to being a subdirectly irreducible ring. (See Lam’s First course in noncommutative rings p 192, for example.

Such rings are not expressable as a subdirect product of two nonzero rings.


问:
Transitivity of a square of a relation

Transitivity

$\def\p{\mathrel p}$If $\p$ is a relation on a set $A$, define $\p^2$ by $a \mathrel{\p^2} b$ if and only if there exists $c$ with $a \p c$ and $c \p b$.

If $p$ is reflexive/symmetric/transitive does $p^2$ have the same properties?

So i understand and have proven the reflexive and symmetric properties (see also this question: Reflexivity, Transitivity, Symmertry of the square of an relation), however is this correct for transitivty.

Suppose $\p$ is transitive

Let $a,b,c$ $\in$ A

WTS - a $\p^2$ b, b  $\p^2$ c, therefore a $\p^2$ c

So if p is transitive that means :

$a \p b$

$b \p c$

$a \p c$

For  b  $\p^2$ c we need $a \p c$ and $c \p c$ for it to be related

So since p is transitive we can say that $a \p c$  and that $c \p c$ since its reflexive, so then we have our condition satisfied so therefor we can say that if

a  $\p^2$ b and b  $\p^2$ c

Then a  $\p^2$ c since condition $a \p c$, $c \p c$ is satisfied because p is transitive

qed

am i on the right track here?

答:
HINT: You’ve correctly identified what you want to show, but your argument isn’t really on the right track. Your assumptions are that $a\mathbin{p^2}b$ and $b\mathbin{p^2}c$, and you want to show that $a\mathbin{p^2}c$.

Now $a\mathbin{p^2}b$ means that there is some $x\in A$ such that $a\mathbin{p}x$ and $x\mathbin{p}b$. Do not use $c$ where I have $x$: $c$ already refers to a specific element of $A$, and this $x$ may be a completely different element. Similarly, $b\mathbin{p^2}c$ means that there is some $y\in A$ such that $b\mathbin{p}y$ and $y\mathbin{p}c$. That is, we have the following chain of relationships:

$$a\mathbin{p}x\mathbin{p}b\mathbin{p}y\mathbin{p}c\;.\tag{1}$$

In order to show that $a\mathbin{p^2}c$, you must show that there is some $z\in A$ such that $a\mathbin{p}z$ and $z\mathbin{p}c$. Is there any way to get that from $(1)$?

Yes, and here’s where we use the fact that $p$ is transitive. Remember that $a\mathbin{p}x$ and $x\mathbin{p}b$; since $p$ is transitive, this means that $a\mathbin{p}b$. Thus, we can replace the chain $(1)$ by

$$a\mathbin{p}b\mathbin{p}y\mathbin{p}c\tag{2}\;.$$

Can you take one more step to find a $z\in A$ such that $a\mathbin{p}z$ and $z\mathbin{p}c$? (Note that this $z$ may be one of the elements of $A$ to which we’ve already given names.)

Title: Transitivity of a square of a relation


问:
Maximum number of independent constants of motion in involution
Consider a holonomic system. Let $M$ be the phase space $T^{\ast}Q$ where $Q$ is the configuration space with $\dim(Q)=n$. Two functions $f,g: M \to \mathbb{R}$ are in involution if they Poisson-commute ($\{f,g\}=0$).  I was told that the maximum number of (functionally) independent constanst of motion in involution is $n$, but I can't prove that (or even see why). If you know about this stuff and you're able to prove this fact, thank you so much.

My idea is that it's probably related to the fact that the symplectic form $\omega$ is not-degenerate, but I couldn't get anything from this.

答:
You can start noticing that $\omega(X_f,X_g)=\{f,g\}$, where $X_h$ is the Hamiltonian vector field of $h:T^*Q\rightarrow \mathbb{R}$. At this point assume to have $f_1,...,f_n:Q\rightarrow \mathbb{R}$ which are  pairwise in involution functionally independent functions, i.e. $df_i\wedge df_j\neq 0$, $\{f_i,f_j\}=0$ for any $i\neq j$.

Assume now that there exists another function $\bar{f}: Q\rightarrow \mathbb{R}$ which is in involution with all the other $n$s and does not depend on them.

You then get that $\omega(X_{\bar{f}},X_{f_i})=0$ for any $i=1,...,n$. However this gives an immediate contradiction. Indeed, let us take an arbitrary point $q\in Q$. The tangent vectors $X_{f_1}(q),...,X_{f_n}(q)\in T_qQ$ form a basis of this linear space. Therefore, any $v\in T_q Q$ can be written as $$v = \sum_{i=1}^n \lambda_i X_{f_i}(q)$$ and hence, by bilinearity of the symplectic form, you get that $$ \omega(X_{\bar{f}},v) = \sum_{i=1}^n\lambda_i\omega(X_{\bar{f}},X_{f_i})=0.$$

This is a contradiction since the symplectic form is non-degerate by construction, which means if $\forall v\in T_qQ$ $\omega(X,v)=0$, then necessarily $X=0$. However, we assumed $\bar{f}$ to be functionally independent from the other functions, which is not the case when $\bar{f}=0$. We can then conclude the maximum number of independent functions in involution on $Q$ is $n$.

The same can be said in terms of constants of motion, for example when you talk about Liouville-Arnold's integrability theorem you assume to have the maximum number of functionally independent first integrals of the Hamiltonian vector fields you are considering.

I am sure this is not the shortest proof of this fact, however, at this moment this seemed more clear to me than other possibilities.


问:
What are some examples of incomplete nonextendible manifolds?
A connected Riemannian manifold $(M,g)$ is said to be extendible if it is isometric
to a proper open subset of a larger connected Riemannian manifold (this is the definition given in Lee's Introduction to Riemannian manifolds). I was able to show that a complete Riemannian manifold is nonextendible. I would like to find an example of incomplete Riemannian manifold which is not extendible. According to Wikipedia such manifolds exist, but no reference is given.

What are some (relatively simple) examples of connected incomplete nonextendible Riemannian manifolds ?

I tried looking at the case of the real numbers $\mathbf{R}$, without great success. More specifically I tried to find conditions on a metric on $\mathbf{R}$ to be extendible. Suppose that there is a metric $g$ on $\mathbf{R}$ and an isometric embedding $\varphi:(\mathbf{R},g)\hookrightarrow (M,g_0)$ with $M$ a connected 1-manifold and with $\varphi(\mathbf{R})\subsetneq M$. For example we can look at the case $M=\mathbf{R}$ and $\varphi(\mathbf{R})=(-\infty,0)$ with $\varphi$ increasing. Writte $g_0 = f(x)dx^2,~f>0$. Then, $\varphi$ bieng an isometry, we get

$$g = \underbrace{f\circ\varphi(x)\varphi^\prime(x)^2}_{h(x):=}dx^2.$$

Now $\varphi$ is strictly increasing and $\lim_{x\to+\infty} \varphi(x)=0$ so there is a sequence $(x_n)_n\in\mathbf R^\mathbf N$ converging to $+\infty$ such that

$$\lim_{n\to\infty}\varphi^\prime(x_n)=0.$$

Since $f$ in bounded near $0$, we get
$$\lim_{n\to\infty}h(x_n)=0.$$

Therefore if we want to construct nonextendible metric $g=h(x)dx^2$ on $\mathbf R$ "by hand", we might want to assume $h$ is bounded bellow by a positive constant $\varepsilon>0$ (so that there is no sequence $(x_n)_n\in\mathbf R^\mathbf N$ with $\lim_{n\to\infty} \vert x_n\vert = +\infty$ and $\lim_{n\to\infty}h(x_n)=0$). However I am pretty sure that these metrics are all complete (even if I don't have an actual proof of this fact), so I don't think I am gonna find an example in this way.

In dimension two, the only incomplete manifolds I can think of are punctured surfaces, but these are all extendible.

Any kind of help will be greatly appreciated.

答:
Although this has been answered a while ago, I cannot resist describing one of my favorite examples, which is related to the comment of @TedShifrin.

Start with the ordinary Euclidean metric on $\mathbb R^2 - \{(0,0)\}$ and then lift it to the universal cover. The resulting space is isometric to $(0,\infty) \times (-\infty,+\infty)$ with coordinates $r \in (0,\infty)$, $\theta \in (-\infty,+\infty)$, equipped with the metric $dr^2 + r^2 d\theta^2$. A nonconvergent sequence $(r_n,\theta_n)$ in this space is Cauchy if and only if $r_n \to 0$: indeed the infimal path length (aka distance) from $(r_1,\theta_1)$ to $(r_2,\theta_2)$ is $\le r_1 + r_2$. For similar reasons, all nonconvergent Cauchy sequences are equivalent. Thus the completion is obtained by adding one point.


答:
One dimensional subspaces that contain multiples of one fixed matrix are multiples of $\left(\begin{matrix}d1&0\\0&0\end{matrix}\right), \left(\begin{matrix}0&0\\0&d2\end{matrix}\right), \text { or }\left(\begin{matrix}d1&0\\0&d2\end{matrix}\right), d1,d2\ne 0$. It is clear that the vector space $D$ has dimension 2. That is to say, each diagonal matrix in that vector space has 2 degrees of freedom. You can get from any diagonal matrix d to another diagonal matrix d' in at most two moves if you add/subtract a multiple of two arbitrarily chosen linearly independent diagonal matrices.

Now what happens when we restrict the subspace to be a multiple of one fixed matrix- Any matrix in the resulting subspace has only one degree of freedom because you can get to another matrix in the subspace by adding a multiple of a single matrix. Visually, perhaps the matrices all lie on one line. Any diagonal matrix from the original vector space had 2 degrees of freedom though, so this second degree of freedom is lost when you restrict to only a multiple of one fixed matrix.

Looking at the other post, you've got subspaces of dimension 2 (the entire D), dimension 1 (given n, the set of all d such that $\langle n, d\rangle_F=0$ (this reduces the degree of freedom on any element d in the sub space by 1 because it is no longer allowed to vary in the direction spanned by n)), and dimension 0 $\langle n_1, d\rangle_F=0 \&\langle n_2, d\rangle_F=0$, $n_1,n_2$ linearly independent (this reduces the degree of freedom of any element in the sub space by 2 because d is not allowed to vary in the directions of n_1 and n_2; but since there were only two degrees of freedom to begin with, the resulting dimension is 0).

Title: All subspaces of $2\times 2$diagonal matrix

Body:
I would like to know all subspaces of a $2\times 2$
diagonal matrix. This is a problem from a book, Introduction to Linear Algebra by Dr. Strang. In the problem, I would like to find all subspaces of a $2\times 2$ diagonal matrix.

The answer says that

"The subspaces of D are D itself, the zero matrix by itself, and all
the “one-dimensional” subspaces that contain all multiples of one
fixed matrix."

the fixed matrix is \begin{bmatrix} d1 & 0\\ 0 & d2 \end{bmatrix}

I understand that the subspaces contain D itself and the zero matrix. However, what does it mean that “one-dimensional” subspaces that contain all multiples of one
fixed matrix" ? The related question is posted in here, but I do not really understand it.


问:
Strong convergence of a sequence in $B(X,Y)$, where $X$ is Banach
Let $X, Y$ be normed spaces, where $X$ is Banach.

Let $(T_n)_n$ be a sequence in $B(X,Y)$ such that for each $x\in X$, $T_n(x)$ converges in norm to some $T(x)\in Y$.

Show $T \in B(X,Y)$.

I showed that $T$ is linear. Now I have to show that it's bounded.

I know that there exists some $N>0$ such that for all $n>N$, $\| T_n(x) - T(x)\| < \epsilon$.

Also, by the principle of uniform boundedness, we know that for each $n\in \mathbb{N}$, we can find a constant $M_n$ such that $\| T_n(x)\| \leq M_n \| x\|$.

Now I'm not sure how to finish this.

Thanks in advance!

答: Unifrom Boundeness Principle tells you that there exists one constant $M$ (independent of $n$) such that $\|T_nx\| \leq M\|x||$ for all $x$ and all $n$. Letting $n \to \infty$ in this we get $\|Tx\| \leq M\|x||$ for all $x$.


问:
Sum of a number of terms of $\left\lceil \frac{n}{2} \right\rceil $
The sum of integers from $1$ to $n$ is the following if we ignore the first half of the terms on the left side below:

$ 1+2+...+n\ge \left\lceil \frac{n}{2} \right \rceil + (\left\lceil \frac{n}{2} \right \rceil +1) + ... +n$

$\left\lceil \frac{n}{2} \right \rceil + (\left\lceil \frac{n}{2} \right \rceil +1) + ... +n \ge \left\lceil \frac{n}{2} \right \rceil +...+\left\lceil \frac{n}{2} \right \rceil$

$\left\lceil \frac{n}{2} \right \rceil +...+\left\lceil \frac{n}{2} \right \rceil = (n-\left\lceil \frac{n}{2} \right \rceil+1)*\left\lceil \frac{n}{2} \right \rceil$

$(n-\left\lceil \frac{n}{2} \right \rceil+1)\left\lceil \frac{n}{2} \right \rceil \ge (\frac{n}{2})(\frac{n}{2})$

Could you please explain how we get:

$\left\lceil \frac{n}{2} \right \rceil +...+\left\lceil \frac{n}{2} \right \rceil = (n-\left\lceil \frac{n}{2} \right \rceil+1)*\left\lceil \frac{n}{2} \right \rceil$

$(n-\left\lceil \frac{n}{2} \right \rceil+1)\left\lceil \frac{n}{2} \right \rceil \ge (\frac{n}{2})(\frac{n}{2})$

答:
The sum is just a arithmetic series:

$\begin{align*}
   \sum_{0 \le k \le m} k
     &= \frac{m (m + 1)}{2} \\
   \sum_{\lceil n / 2 \rceil \le k \le n} k
     &= (n - \lceil n / 2 \rceil + 1) \cdot \lceil n / 2 \rceil
          + \sum_{0 \le k \le n - \lceil n / 2 \rceil} k \\
     &= (n - \lceil n / 2 \rceil + 1) \cdot \lceil n / 2 \rceil
          + \frac{(n - \lceil n / 2 \rceil) (n - \lceil n / 2 \rceil + 1)}{2}
\end{align*}$


问:
What am I doing wrong with my probability formula?
The question is

I have a class of 10 students. The class consists of 4 statistics majors, 3 philosophy majors, and 3 sociology majors.

I choose three students at random, without replacement. (Order doesn’t matter.)

Let event A be "All three students I choose are statistics majors."

Let event B be "At least one student I choose is a philosophy major."

For P(A) I did $$4/10 \cdot 3/9 \cdot 2/8 = .033$$

For P(B) I thought it would be $$(7!\cdot 3!)/(10!) = .833$$ But that is not correct. I am a little confused on where I am going wrong, I thought it would be number of events that satisfy B/number of equally likely events some help in the right direction is all im asking for, thank you in advance.

答:
So, to use your approach: for the first, you have indeed
$$\Pr[A] = \frac{\binom{4}{3}}{\binom{10}{3}} = \frac{1}{30}
$$
since you have $\binom{4}{3}$ events where the 3 students you choose belong to the 4 statistics majors, and a total of $\binom{10}{3}$ events (choose $3$ students out of 10).

For the second, $\Pr[B]$, this is not the fastest way, but you can do the same. The number of events where at least one student is a philosophy major is
$$
\binom{3}{1}\binom{7}{2} + \binom{3}{2}\binom{7}{1}  + \binom{3}{3}\binom{7}{0} = 85
$$
since you choose either exactly 1, exactly 2, or exactly 3 of your 3 picks to be philosophy majors. In the denominator, of course, you'll still have $\binom{10}{3}$ events.

A faster way is to note that the complement of $B$, which has probability $1-\Pr[B]$, is "none of the students I chose is a philosophy major," and the number of events satisfying this is more simply
$
\binom{7}{3}
$. Computing the probability of the complement is much easier.


问:
How to denote a sequence in terms of a set and a function?
Let $s=(a,b,c)$ be a sequence. This sequence can be expressed as a tuple $(S,f)$, where $S=\{a,b,c\}$, $f:\{1,2,3\} \rightarrow S$, and $f$ is a bijective function. In fact, $f(1)=a$, $f(2)=b$, and $f(3)=c$.

My question is, what is the appropriate notation for denoting that $(a,b,c)$ and $(S,f)$ are representations of the same object?

Is it correct to use the equal sign? $(S,f)=(a,b,c)$?

Or should I use the equivalence sign? $(S,f)\equiv (a,b,c)$?

Or is there a more suitable sign?

I appreciate any help you can provide.

答:
There is no "the" appropriate notation. The use of mathematical notations depends very much on context and the conventions of targeted mathematical audience.

There are two ways to denote a (real) "sequence" of length $n$, which are essentially the same:

an element of $n$-fold Cartesian product $\mathbb{R}^n$,

an element of the set $\mathbb{R}^{\{1,2,\cdots,n\}}$, i.e., a function $f:\{1,2,\cdots,n\}\to\mathbb{R}$

Similarly, if you are talking about a set $S=\{a,b,c\}$, the triple $(a,b,c)$ can be thought as

an element in $S^3$,

an element of the set $S^{\{1,2,3\}}$, i.e., a function $f:\{1,2,3\}\to S$


问:
In how many permutations of $\lbrace a,b,c,d,e \rbrace$ a comes before e and e come after c?
In how many permutations of $\lbrace a,b,c,d,e \rbrace$ $a$ comes before $e$ and $e$ come after $c$?

I know that the answer is $\frac{5!}{3}$ but I don't understand why?

My approach is:

In half of the permutations, $a$ comes before $e$, and then we should determine the order of $e$ and $c$.  Then $c$ may be between $a$ and $e$ or $c$ comes after $e$ or $c$ comes before $a$ and $\frac{2}{3}$ of possible situations are just as we wanted. so $\frac{2}{3} \times \frac{1}{2} = \frac{1}{3}$ of all permutations is the answer.

Is my approach correct? and is there another way to solve this? Thanks a lot!

答: It is correct! Another way is just choose the $3$ spots in which you are placing $a,e,c$ in $\binom{5}{3}\cdot 2$ ways because you pick the $3$ spots and then you can have either "ac" or "ca", and then multiply by $2!$ to know what to do with $b,d$ so $$\binom{5}{3}2!\cdot 2=\frac{5!}{3!}\cdot 2=\frac{5!}{3}.$$


问:
Question Regarding a Multilinear Map
In my math textbook, I found the following problem.

Given an $n$-dimensional vector space $V$ over a field $\mathbb{K},$ consider a nonzero map $f : V \times V \times \cdots \times V \rightarrow \mathbb{K}$ that is multilinear map for all integers $k \geq 2,$ where $k$ is the number of copies of $V$ in the product.

 a.) What values can $k$ take?

 b.) Prove that $f$ is not $\mathbb K$-linear.

What I'm generally confused about is what the problems ask.

If I pick a basis $B = \{v_1, \dots, v_n\}$ for each copy of $V,$ then I get $f(v_{1_{N_1}}, \dots, v_{k_{N_k}}) = C_{N_1 N_2 \cdots N_k}$ is a constant for one of the $n^k$ possible $k$-tuples of basis vectors, where $N_1, N_2, \dots, N_k \in \{1, 2, \dots ,n\}.$

So, as for a.), can't $k$ just take any value in $\mathbb{N}$?

As for b.), this is a multilinear map, which means it's linear in each argument separately, so is it not linear by virtue of having more than one argument?

答:
a) Yes, $k$ can be any positive integer.

On the other hand, if $f$ is also assumed to be alternating (gives $0$ whenever two of its arguments are equal), then $k$ must be at most $n=\dim V$.

b) We have $f(2v_1,2v_2,\dots)=2^kf(v_1,v_2,\dots)$
whereas if $f$ were linear, we should get $f(2v_1,2v_2,\dots)=2f(v_1,v_2,\dots)$.

At least for the case when characteristic of $\Bbb K$ is $0$, this proves that $k$ must be $1$ (or $f$ must be constant $0$).

Title: Question Regarding a Multilinear Map


问:
Visual proof for $\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$
I've found visualizations of simple mathematical concepts to be a really useful tool in building intuition for more complex mathematical concepts. For example, this visualization of $(a + b)(c + d) = ac + ad + bc + bd$ can be used to visualize the calculus product rule.

That being said, does anyone know of a visual way to show the following equation?
$$
\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}
$$

答:
Maybe something like this:

See also this one: https://www.geogebra.org/m/DV6Ehjnx#material/aEsvBzN2


问:
Geometry problem prove $\angle{ADB}=3\angle{BAC}$
Acute $\triangle{ABC}$, point $D$ is on side $AC$ so that $AD=BC$, $CF$ bisects $\angle{ACB}$. Point $E$ on side $AB$ so that $DE \parallel CF$ and $AE=CD$. Prove: $\angle{ADB}=3\angle{BAC}$

答: In quadrilateral ABCD, we have AD = BC and                ∠DAB = ∠CBA.                In ΔABD and ΔBAC,                AD = BC[Given]                AB = BA[Common]                ∠DAB = ∠CBA[Given]                ∴ Using SAS criteria, we have ΔABD ≌ ΔBAC          (ii) ∵ ΔABD ≌ ΔBAC                ∴ Their corresponding parts are equal.                ⇒ BD = AC          (ii) Since ΔABD ≌ ΔBAC                ∴ Their corresponding parts are equal.                ⇒ ∠ABD = ∠BAC. Hence Proved</p>


答:
Obviously the easiest approach is $1-P(\text {no philosophy majors})=1-\frac 7 {10} \frac 6 9\frac 5 8$

If you were to do $\frac{\text{number of groups with at least one philosophy major}}{\text{number of groups}}$, you'd have to calculate the number of groups all of which have at least one philosophy major. Notice that this is equivalent to $\frac{\text{number of groups}-\text{number of groups without a philosophy major}}{\text{number of groups}}=\frac{{10 \choose 3}-{7 \choose 3}}{{10 \choose 3}}$. Notice that working with groups is the same as working with probabilities, as this simplifies to $1-\frac{{7 \choose 3}}{{10 \choose 3}}=1-\frac{7 *6 *5}{10*9*8}$

Title: What am I doing wrong with my probability formula?

Body:
The question is

I have a class of 10 students. The class consists of 4 statistics majors, 3 philosophy majors, and 3 sociology majors.

I choose three students at random, without replacement. (Order doesn’t matter.)

Let event A be "All three students I choose are statistics majors."

Let event B be "At least one student I choose is a philosophy major."

For P(A) I did $$4/10 \cdot 3/9 \cdot 2/8 = .033$$

For P(B) I thought it would be $$(7!\cdot 3!)/(10!) = .833$$ But that is not correct. I am a little confused on where I am going wrong, I thought it would be number of events that satisfy B/number of equally likely events some help in the right direction is all im asking for, thank you in advance.


问:
bijective function verification

Let $A$ and $B$ be sets. Suppose that $|A|\leq |B|$ and $|B| \leq |A|.$ Then $|A|= |B|.$

Below is an almost complete proof.

Choose injective functions $f:A\to B$ and $g :B\to A$. Since the functions $f :A\to f(A), g : B\to g(B),$ and $f : g(B) \to f(g(B))$ are bijective we have $|A| = |f(A)| $ and $|B| = |g(B)| = |f(g(B))| .$ Let $X = f(g(B)), Y=f(A), Z=B.$ Then $|X|=|Z|$ and $X\subseteq Y\subseteq Z.$ We need to show $|Y| = |Z|.$

The function $h = f\circ g : Z\to X$ is a bijection. Define for $n\in\mathbb{N}, Z_0 = Z,Z_n := h(Z_{n-1})$ and $Y_0 = Y, Y_n = h(Y_{n-1}).$ Since $Y_0 = Y, Z_0 = Z, Z_1 = h(Z_0) = h(Z)=X,$ so $Z_1 \subseteq Y_0\subseteq Z_0.$ Assume inductively that for some $1\leq n\in\mathbb{N}, Z_n\subseteq Y_{n-1}\subseteq Z_{n-1}.$ Then $Z_{n+1} = h(Z_{n})\subseteq h(Y_{n-1})\subseteq h(Z_{n-1}),$ since $h$ is injective, so $Z_{n+1}\subseteq Y_n\subseteq Z_n$. Now let $U_n = Z_n\backslash Y_n, U= \cup_{n=0}^{\infty} U_n$ and $V = Z\backslash U.$ Define $H : Z\to Y$ by $H(x) =\begin{cases}h(x),&\text{ if $x\in U$}\\
x,&\text{ if $x\in V$}.\end{cases}$

My question is, why is $H$ bijective? Clearly, it's injective because if $H(x)=H(y) \in Y,$ then since $U$ and $V$ are disjoint, $x,y \in V$ or $x,y\in U$ and in both cases it's easily verified that $x = y.$ Now to show $H$ is surjective, I was thinking of showing that $h(U) = Y\backslash V$ because clearly $H(V) = V$ and $H(U\cup V) = H(U)\cup H(V) = h(U)\cup V$. Also, $h(Z_n\backslash Y_n) = h(Z_n)\backslash h(Y_n).$ Indeed, suppose $x \in h(Z_n\backslash Y_n)\Rightarrow x = h(a), a \in Z_n\backslash Y_n.$ Then if $x = h(y)$ for some $y \in Y_n,$ we have $a = y = h^{-1}(x),$ which is a contradiction as $y\in Y_n$ while $a$ is not in $Y_n.$ sO $h(Z_n\backslash Y_n)\subseteq h(Z_n)\backslash h(Y_n).$ Now if $x = h(b)\in h(Z_n)\backslash h(Y_n),$ then $b\not \in Y_n$ because otherwise $h(b)\in h(Y_n),$ a contradiction. Thus, since $b\in Z_n, h(b)\in h(Z_n\backslash Y_n).$ The above two justifications imply that $h(U)=h(\cup_{n\geq 0} Z_n\backslash Y_n) = \cup_{n\geq 0} h(Z_n)\backslash h(Y_n) = \cup_{n\geq 0}Z_{n+1}\backslash Y_{n+1},$ but I'm not sure how to show that this equals $Y\backslash V$; the sets $Z_n\backslash Y_n$ may not be disjoint. Am I doing something wrong?

答:
The sets $Z_n\setminus Y_n$ are pairwise disjoint: if $n<m$, we have

$$Y_m\subseteq Z_m\subseteq Y_n\subseteq Z_n\,,$$

so

$$\begin{align*}
(Z_m\setminus Y_m)\cap(Z_n\setminus Y_n)&\subseteq Z_m\cap(Z_n\setminus Y_n)\\
&\subseteq Y_n\cap(Z_n\setminus Y_n)\\
&=\varnothing\,.
\end{align*}$$

The crude diagram of $B$ shown below may be helpful: $H$ simply pushes each $U_n$ down to $U_{n+1}$, thereby translating $U$ downward one block, and leaves the blocks of $V$ alone, more or less as shown on the right.

```<code>      -------------
      |    U_0    |
      -------------        -------------
      |  Y_0\Z_1  |        |  Y_0\Z_1  |
      -------------        -------------
      |    U_1    |        |   H[U_0]  |
      -------------        -------------
      |  Y_1\Z_2  |        |  Y_1\Z_2  |
      -------------        -------------
      |           |        |   H[U_1]  |
      |     |     |        -------------
      |     V     |        |     |     |
      V           V        V     V     V
</code>```

Title: bijective function verification

Body:
The sets $Z_n\setminus Y_n$ are pairwise disjoint: if $n<m$, we have

$$Y_m\subseteq Z_m\subseteq Y_n\subseteq Z_n\,,$$

so

$$\begin{align*}
(Z_m\setminus Y_m)\cap(Z_n\setminus Y_n)&\subseteq Z_m\cap(Z_n\setminus Y_n)\\
&\subseteq Y_n\cap(Z_n\setminus Y_n)\\
&=\varnothing\,.
\end{align*}$$

The crude diagram of $B$ shown below may be helpful: $H$ simply pushes each $U_n$ down to $U_{n+1}$, thereby translating $U$ downward one block, and leaves the blocks of $V$ alone, more or less as shown on the right.

      -------------
      |    U_0    |
      -------------        -------------
      |  Y_0\Z_1  |        |  Y_0\Z_1  |
      -------------        -------------
      |    U_1    |        |   H[U_0]  |
      -------------        -------------
      |  Y_1\Z_2  |        |  Y_1\Z_2  |
      -------------        -------------
      |           |        |   H[U_1]  |
      |     |     |        -------------
      |     V     |        |     |     |
      V           V        V     V     V


问:
Integral of the absolute value of an unknown function
I was just wondering if there is a general way (method/formula) to antidifferentiate the absolute value of a function. As in, is there any way to find $\int|f(x)|dx $ given that I already know what $\int f(x)dx$ is? I am referring to the indefinite integral.

答:
$$\int|f()| = \frac{|f(x)|}{()}\cdot ∫f(x)dx + c$$

Proof:

$$\frac{d}{dx}\left(\frac{|f(x)|}{f(x)}\cdot\int f(x)dx\right)$$

$$= \frac{d}{dx}(\text{sgn}(f(x)))\cdot\int f() + \frac{d}{dx}\left(\int f()\right)\cdot \frac{|f()|}{f()}$$ as $$\text{sgn}(f(x))=\frac{|f(x)|}{f(x)}$$

$$= 0 + f(x)\cdot \frac{|f()|}{f()}$$ as $$\frac{d}{dx}(\text{sgn}(f(x)))=0$$ for all $f(x) \neq 0$

$$= |f(x)|$$

Therefore, $\frac{|f(x)|}{f(x)}\cdot\int f(x)dx$ is a solution to $\int|f(x)|dx$ for all $f(x) \neq 0$ so the general solution must be $\frac{|f(x)|}{()}\cdot\int f(x)dx + c$ for all $f(x) \neq 0$


问:
For what real values of $a$ does $\sum_{n=1}^{\infty}{(\frac{an}{n +1})^n}$ converge?

For what real values of $a$ does the series
$$\sum_{n=1}^{\infty}\left(\frac{an}{n
+1}\right)^n$$
converge?

At first, I wanted to do the root test, but the problem is that $a$ could be negative.  If I take it out of the fraction to get $$\left(\frac{a}{n+1}\right)^n$$ it doesn't meet the criteria of either the Dirichlet or the Abel tests.

答:
The question asks about real values of $a$, but, in order to make this answer more general (so that other versions of this question may be linked to this question as duplicates), I will assume that $\alpha \in \mathbb{C}$.  An expedient approach to the problem is to consider various tests of convergence.  Personally, I like to start with the simple ones, and work my way up.  To my mind, the simplest test of convergence is the ratio test:

Theorem (Ratio Test):  Let $(a_n)_{n=1}^{\infty}$ be a sequence of complex numbers, define
$$L = \lim_{n\to \infty} \left| \frac{a_{n+1}}{a_n} \right|, $$
and let $S$ denote the infinite series $\sum_{n=1}^{\infty} a_n. $

If $L < 1$, then the series $S$ is absolutely convergent (and, therefore, convergent).

If $L > 1$, then the series $S$ is divergent.

If $L = 1$, then the ratio test is inconclusive—the series $S$ may be divergent, conditionally convergent, or absolutely convergent.

In the case of the series in this question, $a_n = \left( \frac{an}{n+1} \right)^n$.  Thus
$$ L
= \lim_{n\to \infty} \left| \frac{ a^{n+1}(n+1)^{n+1} / (n+2)^{n+1} }{ a^n n^n / (n+1)^n } \right|
= \lim_{n\to\infty} \left| \frac{a(n+1)^{2n+1} }{ n^n(n+2)^{n+1} } \right|. \tag{1}$$
It can actually be shown that this limit is $|a|$, but the argument is a little tedious.  If we allow ourselves the use of big-$O$ notation,
$$ \frac{(n+1)^{2n+1}}{n^n(n+2)^{n+1}}
= \frac{n^{2n+1} + O(n^{2n})}{n^n(n^{n+1} + O(n^n))}
= \frac{n^{2n+1} + O(n^{2n}) }{n^{2n+1} + O(n^{2n})}
= \frac{n^{2n+1}}{n^{2n+1} + O(n^{2n})} + \frac{O(n^{2n}) }{n^{2n+1} + O(n^{2n})}. $$
As $n$ goes to infinity, the first term goes to $1$, and the second term vanishes.  Thus
$$ L
= \lim_{n\to\infty} \left| \frac{a(n+1)^{2n+1} }{ n^n(n+2)^{n+1} } \right|
= |a|.$$
That being said, this looks like the kind of problem which is often given to students in a first year calculus course (or, perhaps, a third year complex analysis course), which means that big-$O$ notation may not feel very rigorous.  A more convincing argument might be to compute the limit via binomial expansion of the terms in the fraction, but this is going to be quite tedious, as well.  Doable, but tedious.

Personally, I probably would have looked at the limit in (1) and decided to move on.  The next test I usually choose to apply is the root test.

Theorem (Root Test): Let $(a_n)_{n=1}^{\infty}$ be a sequence of complex numbers, define
$$ L = \lim_{n\to \infty} \sqrt[n]{|a_n|^{n}}, $$
and let $S$ denote the series $\sum_{n=1}^{\infty} a_n$.

If $L < 1$, then the series $S$ is absolutely convergent (and, therefore, convergent).

If $L > 1$, then the series $S$ is divergent.

If $L = 1$, then the ratio test is inconclusive—the series $S$ may be divergent, conditionally convergent, or absolutely convergent.

With $a_n = \left( \frac{an}{n+1} \right)^n$, the limit of interest is
$$
L = \lim_{n\to\infty} \left| \left( \frac{a n}{n+1} \right)^n \right|^{1/n}
= |a| \lim_{n\to\infty} \left| \frac{n}{n+1} \right|
= |a|. $$
This is, I think, a much more straight-forward computation, but it gives us the same result as the ratio test; namely, the series of interest converges absolutely when $|a| < 1$, and diverges when $|a| > 1$.

On the other hand, when $|a| = 1$, both tests are inconclusive.  In general, there is not really a good way of checking for convergence on the boundary of the disc of convergence.  Indeed, there are examples of series which converge for some points on this boundary, and diverge for others (as a simple example, consider the harmonic series, as compared with the alternating harmonic series).  In this case, however, we can do a little better.  Observe that if $|a| = 1$, then
$$ \lim_{n\to \infty} \left|  \frac{an}{n+1} \right|^{n}
= \lim_{n\to\infty} \left[ |a|^n \left( 1 + \frac{1}{n} \right)^{-n} \right]
= \frac{1}{\mathrm{e}}.
$$
In other words, we know that when $|a| = 1$, the general term of the series fails to converge to $0$, which means that the series fails to converge (by the divergence test).  Therefore, we may conclude:

If $|a| < 1$, then the series converges absolutely.

If $|a| \ge 1$, then the series diverges.


问:
What is wrong with my derivation of the exponential form of $\sinh(x)$
I know the exponential form of $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$. I wish to derive the exponential form of $\sinh(x)=\dfrac{e^x-e^{-x}}{2}$, and I am stuck.

I try to replace $\sin(x)$ with $\sin(i^4x)$, then multiply $i=\sqrt{-1}\implies i^4=1$ with $\sin(i^4x)$ and I have

$$i\sin(i^4x)=i\dfrac{e^{i^4x}-e^{-i^4x}}{2i}=\dfrac{e^{x}-e^{-x}}{2}\tag{1}$$

But this is a nonsensical derivation, because it implies that $i\sin(x)=\sinh(x).$ The correct identity must be $i\sin(x)=\sinh(ix)$

I think my result is clearly wrong. What is wrong with it? How should one derive the exponential form of $\sinh(x)$

答:
You know the correct answer $i \sin(x)=\sinh (ix)$, but you want to get to the RHS via the LHS, so let $y=ix$, and the above becomes

$$i\sin(-iy)=\sinh(y)$$

Now calculate the LHS using the definition of $\sin$ (first take out the minus in the $\sin$)

$$-i\frac{e^{i(iy)}-e^{-i(iy)}}{2i}=\frac{e^y-e^{-y}}{2}=\sinh (y)$$

To answer “what’s wrong with your answer”: replacing 1 with $i^4$ is pointless —- you should see why we should substitute $x=-iy$ from the above.


问:
existence of certain continuous function

Let $(X,d)$ be a metric space and $\emptyset \neq A\subseteq X.$ Define for $x\in X,\mathrm{dist}(x,A) := \inf\{d(x,a) : a\in A\}.$

Show that there is a sequence $U_1, U_2,\cdots$ of open subsets in $(X,d)$ for which $\overline{A} = \cap_{n=1}^{\infty} U_n.$

Let $\emptyset\neq K\subset X$ be a compact set with $K\cap \overline{A} = \emptyset.$ Show that there is a continuous function $g : X\to [0,1]$ so that $g(x) = 0$ for $x\in A$ and $g(y) = 1$ for $y \in K.$

For the first part, define for each $n\in\mathbb{N}, U_n := \{x \in X : \mathrm{dist}(x,A) < \frac{1}n\}.$  We claim that each $U_n$ is open. Fix $n\in\mathbb{N}.$ Let $y \in U_n.$ Then $\mathrm{dist}(y, A) =\inf\{d(x,a) : a \in A\} < \frac{1}n,$ so $\exists a \in A$ so that $d(y,a) < \frac{1}n.$ Consider the open ball $B(y, \epsilon), \epsilon := \frac{1}{n} -d(y,a) > 0.$ Then for any $x \in B(y, \epsilon), d(x,a)\leq d(x,y) + d(y,a) < \epsilon + d(y,a) = \frac{1}n$ and hence $\mathrm{dist}(x,A) \leq d(x,a) < \frac{1}n,$ so that $B(y,\epsilon)\subseteq U_n.$

Now we claim that $\cap_{n\in \mathbb{N}}U_n = \overline{A}.$ Let $x \in \cap_{n\in \mathbb{N}} U_n.$ Then for all $n\in\mathbb{N}, \mathrm{dist}(x,A) < \frac{1}n$ and since $\mathrm{dist}(y,A)\geq 0$ for all $y \in X, $ this implies $ \mathrm{dist}(x,A) = 0.$ One can then show using the definition of infimum that $x\in \overline{A}.$ Thus, $\cap_{n\in\mathbb{N}} U_n \subseteq \overline{A}.$ Now let $b\in \overline{A}.$ Then $\mathrm{dist}(b,A) = 0$ so $b \in \cap_{n\in\mathbb{N}}U_n.$ Thus, $\overline{A} = \cap_{n\in\mathbb{N}} U_n.$

Is this correct?

For the second part, I should make use of the properties of compact sets (e.g. being closed and bounded) and the fact that $K$ and $\overline{A}$ are disjoint. I know the function $f : X\to \mathbb{R}, f(x) = \mathrm{dist}(x,A)$ for all $x$ is continuous. However, I'm not sure how to define the function $g.$

答:
The first part is correct. For the second part, you can define

$g(x)=\min\left(1,\frac{d(A,x)}{d(K,x)}\right)$.

You can check the continuity of this function in the closed sets $\{x\in X;\frac{d(A,x)}{d(K,x)}\leq1\}$ and in $\{x\in X;\frac{d(A,x)}{d(K,x)}\geq1\}$ separately (it´s easy in both cases). And then by the glueing lemma $g$ is continuous.

In fact, it´s enough for $K$ to be closed and disjoint with $A$ for this function to be continuous.


问:
Are absolute extrema only in continuous functions?
The Extreme Value Theorem says that if $f(x)$ is continuous on the interval $[a,b]$ then there are two numbers, $a≤c$ and $d≤b$, so that $f(c)$ is an absolute maximum for the function and $f(d)$ is an absolute minimum for the function.

So, if we have a continuous function on $[a,b]$ we're guaranteed to have both absolute maximum and absolute minimum, but functions that aren't continuous can still have either an absolute min or max?

For example, the function $f(x)=\frac{1}{x^2}$ on $[-1,1]$ isn't continuous at $x=0$ since the function is approaching infinity, so this function doesn't have an absolute maximum.
Another example: suppose a graph is on a closed interval and there is a jump discontinuity at a point $x=c$, and this point is the absolute minimum.

The extreme value theorem requires continuity in order for absolute extrema to exist, so why can there be extrema where the function isn't continuous?

答:
First of all, $f(x) = x^{-2}$ isn't well defined on the domain $[-1,1]$, specifically where $x = 0$, so you can't really say that it is discontinuous. But if you assign it any value at $x = 0$, so $f(0) := c$ for $c \in \mathbb{R}$, it is discontinuous.
Then, as shown by yourself, continuity isn't needed to find a function on an interval $[a,b]$ having absolute extrema. This is reflected by the extreme value theorem, because it only guarantees extreme values for a continuous, real function on an interval $[a,b]$, but does not say that a function, having such extreme values, necessarily is continuous.</p>


问:
MLE of difference is difference of MLEs
Suppose a distribution has two parameters $\alpha$ and $\beta$, and let the maximum likelihood estimators for these parameters be $\hat{\alpha}$ and $\hat \beta$.

Is the maximum likelihood estimator for $\alpha-\beta$ necessarily equal to $\hat{\alpha}-\hat{\beta}?$

We cannot assume the independence of $\hat \alpha$ and $\hat \beta$ since they are both functions of the same data set.

答: Yes, the MLE of $\alpha - \beta$ is $\hat{\alpha} - \hat{\beta}$. It is the invariance property of the MLE, namely,  the MLE of $g(\theta)$ is $g(\hat{\theta})$. The only requirement is measurability of the $g$. You can find the proof here.

Title: MLE of difference is difference of MLEs