问:
F-weak topology and sequence convergence
In Royden's book there is:

It is easy to see that a sequence {$x_n$} in X converges to x $\in$ X with respect to the $\mathcal{F}$-weak topology iif $\lim \limits_{n \to \infty} f(x_n)=f(x)$  for all $f\in\mathcal{F}$

Here X is any set, $\mathcal{F}$ contains any function from X to R.

Logically this should be the quivalence to when $\mathcal{F}$ only contains one element. But for this one element situation, there need be an extra requirement that X is first countable for the reverse statement to hold.

Although any normed linear space as a metric space is first countable which means that the $\mathcal{F}$-weak topology (also $\mathcal{F}$'s elements are from X*) as a coarser topology than the norm-induced topology should be first countable too, which means the above quote is right for weak topology and weak-*topology, but in context of the above quote, X is only a set endowed with the $\mathcal{F}$-weak topology.

Does the book indeed lack the first countable property of the $\mathcal{F}$-weak topology there, or the first countable property can be induced from somewhere like the $\mathcal{F}$-weak topology itself (possibly from some properties of R, note that X is just a set)?

答:
For the statement above we don't need first countability. I don't know if the weak topology in general is first countable or not.

$X$ is just a set, but the $\cal{F}$-weak topology equips it with a topology. At each $x\in X$, there is a base for this topology of the form
$$\{x: |f_i(x)-f_i(y)|<\varepsilon\}$$
for some $\varepsilon>0$, $y\in X$ and some finite set $f_1,\dots, f_n$ in $\cal{F}$.

Assume $f(x_n)\to f(x_0)$ for every $f\in\cal{F}$. To prove that $x_n\to x_0$ we want to show that given an open set $U$ containing $x_0$, there exists $N$ such that if $n>N$,then $x_n\in U$. Since the above sets are a local base, there exist some $y\in X$,some $\varepsilon>0$ and some finite subset $\{f_1,\dots, f_m\}$ of $X$ such that if $|f_i(x_0)-f_i(y)|<\varepsilon$ for each $1\leq i\leq m$ then $y\in U$. By hypothesis,  $f_i(x_n)$ converges to $f_i(x_0)$, for every $1\leq i\leq m$, so we can find $N$ such that if $n>N$, then $|f_i(x_n)-f_i(x_0)|<\varepsilon$, for all $1\leq i\leq m$, so $x_n\in U$.

For the other direction -- if $x_n\to x_0$ in the $\cal{F}$-weak topology, then
for every open set $U$ containing $x_0$ there exists $N$ such that if $n>N$ then
$x_n\in U$. Given $f\in\cal{F}$, and $\varepsilon>0$, the set
$$\{x: |f(x)-f(x_0)|<\varepsilon\}$$
is an open set containing $x_0$, hence contains $x_n$ for sufficiently large $N$, which proves that $f(x_n)\to f(x_0)$.


问:
How to solve this $\lim_{n\to\infty} \frac{\sqrt{n}}{n+2}$
How to solve this?
$$\lim_{n\to\infty} \frac{\sqrt{n}}{n+2}$$

My approach is to use Squeeze theorem but no results. Like this

$$0 < \frac{\sqrt{n}}{n+2} < \frac{n}{n+2} < 1 $$

Maybe there is another solution.

答:
$0 \lt \dfrac {\sqrt{n}}{(\sqrt{n})^2+2} < \dfrac{\sqrt{n}} {(\sqrt{n})^2}= \dfrac{1}{\sqrt{n}}.$

The limit is?

ContentLicense: CC BY-SA 4.0


问:
Conditions for diagonalizability and tridiagonalizability of a given matrix

Investigate matrix $A \in \mathrm{Mat}_3 (\mathbb{R})$ for diagonalizability and tridiagonalizability, depending on parameter $\lambda$. $$A = \begin{pmatrix} 0 & 10 & 2 \\ \lambda & 2  & 1 \\ 0 & -3 & 0 \end{pmatrix} $$

I looked at the characteristic polynomial

$$p_A(t)=t^3-2t^2+t(3-10\lambda)+6\lambda$$

and tried to factorize but with no success. So with a function plotter i found out that:

for $\lambda \in[-\infty,\frac38)\ p_A(t)=0$ has only one real
solution.

for $\lambda=\frac38 \ p_A(t)=0$ has 3 real solutions ($-1,\frac32,\frac32$)

for $\lambda \in (\frac38,\infty] \ p_A(t)=0$ has 3 UNIQUE Solutions

So how can i proof these three statements?
Are there any ways to find some conditions directly from the equations $p_A(t)=0$ itself and i an missing something?

I really appreciate any help cause im stuck on this problem for 2 days now.

答:
I will write $p_\lambda$ instead of $p_A$.

To find how many roots $p_\lambda$ has, study its variations.

$$p_\lambda'(t)=3t^2-4t+(3-10\lambda)$$
We compute the discriminant : $\Delta=120\lambda-20$.

This shows that for $\lambda\le \frac{1}{6}$, $p_\lambda$ is strictly increasing therefore has only one simple real root : $A$ isn't trigonalisable.

For $\lambda>\frac{1}{6}$, $p_\lambda$ is increasing- decreasing- increasing : therefore is reached a local minimum at $$x_{\lambda,+}=\frac{4+\sqrt{120\lambda-20}}{6}=\frac{2+\sqrt{30\lambda-5}}{3}$$ and a local maximum at

$$x_{\lambda,-}=\frac{4-\sqrt{120\lambda-20}}{6}=\frac{2-\sqrt{30\lambda-5}}{3}$$

Let's study now the sign of these local extrema $p_\lambda(x_{\lambda,\pm})$. Indeed :

$p_\lambda(x_{\lambda,+}) > 0 \Longleftrightarrow p_\lambda$  has only one roots;

$p_\lambda(x_{\lambda,+}) = 0 \Longleftrightarrow p_\lambda$  has  one simple root and one double;

$p_\lambda(x_{\lambda,-})p_\lambda(x_{\lambda,+}) < 0 \Longleftrightarrow p_\lambda$  has  three distinct roots.

Let's start with the local minimum $p_\lambda(x_{\lambda,+})$.

Introduce $u=x_{\lambda,+}$. Conversely, $\lambda=\frac{(3u-2)^2+5}{30}$
$$f(u):=p_\lambda(u)=u^3-2u^2+u\left(3-10\frac{(3u-2)^2+5 }{30}\right)+6\left(\frac{(3u-2)^2+5 }{30}\right)$$

One finds $$f(u)=-2u^3+\frac{19}{5}u^2-\frac{12}{5}u+\frac{9}{5}$$

For $\lambda=\frac{3}{8}$, $u=\frac{3}{2}$ and one checks that $f(3/2)=0$.

So on can find factorize $f$. One finds  $$f(u)=(-2u+3)\left(u^2-\frac{2}{5}u+\frac{19}{5}\right)$$

Now $\Delta=(-2/5)^2-4\times 19/5<0$ : the 2nd factor is always $>0$.

This means $f$ is positive  on $]-\infty;3/2[$, zero at $3/2$ and negative on $]3/2;\infty[$.

--> Going back from $u$ to $\lambda$, the local minimum $p_\lambda(x_{\lambda,+})$ is positive on $]1/6;3/8[$, zero at $3/8$ and negative on $]3/8;\infty[$.

Study of the sign of the local maximum $p_\lambda(x_{\lambda,-})$.

Write $v=x_{\lambda,-}$ this time. Again, $\lambda=\frac{(3v-2)^2+5}{30}$.

As before, $$g(v)=p_\lambda(x_{\lambda,-})=...=(-2v+3)\left(v^2-\frac{2}{5}v+\frac{19}{5}\right)$$

So $g$ is positive on $]-\infty;3/2[$, zero at $3/2$ and negative on $]3/2;+\infty[$.

But this time, for any $\lambda> 1/6$, $v<3/2$ so the local maximum is always positive.

Let's sum everything up now :

for $1/6 < \lambda < 3/8$, we have $p_\lambda(x_{\lambda,+}) >0 $ : $p_\lambda$ has only one simples root and A isn't trigonalisable;

for $\lambda = 3/8$, we have $p_\lambda(x_{\lambda,-})>0$ and $p_\lambda(x_{\lambda,+}) = 0 $ : $p_\lambda$ has two distinct roots therefore three (one simple, one double) : $A$ is trigonalisable (but not diagonalisable because on checks that $E_{3/2}=Vect(4,1,-2)$ so $dimE_{3/2}<2$).

for $\lambda > 3/8$, we have $p_\lambda(x_{\lambda,-})p_\lambda(x_{\lambda,+}) < 0 $ : $p_\lambda$ has three distinct roots  and $A$ is diagonalisable.


问:
Nested recurrence sequence with interesting properties
This is my first post here, thanks for stopping by. The question as written below comes from the book 'A Concise Introduction To Pure Mathematics'. I've included my working (this isn't a homework problem) and points I am interested in - particularly, the thought process that you had in dealing with this problem.

Critic Ivor Smallbrain has been keeping a careful account of the number of chocolate bars he has eaten during film screenings over his career. For each positive integer $n$ he denotes by $a_n$ the total number of bars he consumed during the first n films. One evening, during a screening of the Christmas epic It's a Wonderful Proof, he notices that the sequence $a_1 , a_2 , ...$ obeys the following rules for all $ n => 1$:

$a_{n+1} > a_n $

$a_{a_n} = 3n$

Also $a_1 > 0$

(a) Find $a_1$ . (Hint: Let $x=a_1$. Then what is $a_x$?)

(b) Find $a_2 , a_3 , ... , a_8 , a_9$

(c) Find $a_{100}$

(d) Investigate the sequence $a_1 , a_2 , ...$ further.

(a) Well, we would be good to make use of the hint, so that $a_x = a_{a_1} = 3$
by the second property. Now, let us say that x is greater than or equal to four. As the sequence is strictly increasing, the terms before $a_x$ are strictly decreasing, thus we would run out of terms if this were the case. Hence x must be less than 4, so x is 1,2 or 3, remembering that the first term of the sequence is at least one.

Well, suppose that x is 1. Then we get $a_1 = a_x = a_{a_1} = 3$, an absurdity.

So, we investigate if x is 3. Then $a_3 = a_x = a_{a_1} = 3$ Yet we recall x is $a_1$ and the sequence is strictly increasing, so this is an absurdity.

(b) Knowing that $a_1 = 2$ we note $3 = a_{a_1} = a_2$ by the second property. Similarly, $6=a_{a_2}=a_3$ so $9 = a_{a_3} = a_6$. Now, we are looking for $a_5$ and $a_4$, but this is made easy by the first property; the sequence is increasing, and the third term is 6, the sixth term is 9. Hence the fourth is 7 and the fifth 8. So, to sum;

$a_1$ = 2
$a_2$ = 3
$a_3$ = 6
$a_4$ = 7
$a_5$ = 8
$a_6$ = 9

To finish, note the seventh term is obtained from the fourth, to be 12,the eighth from the fifth, to be 15, and the ninth from the sixth, to be 18.

(c) -

(d) -

Parts I am stuck on. Well, it sounds like if you could answer (d) with an explicit formula then obtaining the answer for (c) would be fine - you just plug in n = 100. However, I am struggling to find how you can 'skip' finding numbers. From the working above it appears to me that I need to find all terms before the hundredth.

I am asking for your help in not merely a method, but in illumination - please explain to me why you are doing this or that and so forth - I am hoping to learn something here, and if you can answer this, then whatever you say will be something I and all people can learn from so do not hold back - say anything.

答:
This is one of my favorite problems! One way to proceed here is by induction, and for this, since the multiplication in the recursive formula is by $3$, it may help to write integers in base $3$. Then the induction hypothesis looks like this. Let $f(n)=a_n$ for $n\ge 1$, and let $n$ in base $3$ be $n=a_{k+1} a_{k}\dots a_1a_0$ (as a string of digits), then
$$
f(1a_k\dots a_1 a_0)=2a_k\dots a_1 a_0, \quad f(2a_k\dots a_1 a_0)=1a_k\dots a_1 a_0 0.
$$
In other words,
$$
a_n=f(n)=
\begin{cases}
n+3^k, & \text{ if } 3^k\le n\le 2\cdot 3^k,\\
3n-3^{k+1}, & \text{ if } 2\cdot 3^k\le n\le 3^{k+1},
\end{cases}
$$
where $k=\left\lfloor\log_3 n\right\rfloor\ge 0$.

Notice also that
$$
n+3^k=3n-3^{k+1}=3^{k+1} \quad \text{when} \quad n=2\cdot 3^k,
$$
and
$$
n+3^{k+1}=3n-3^{k+1}=2\cdot 3^{k+1} \quad \text{when} \quad n=3^{k+1}.
$$

Since you seem to be asking for a hint rather than for a complete solution, I will leave it here and let you carry out the induction.

ContentLicense: CC BY-SA 3.0


问:
Why $v_1 \in L$ and $v_2 \in L^\perp$?
Let $L$ be a one dimensional subspace of $R^2$ (we may view $L$ as a line in the plane through the origin).

Suppose $\alpha$ is the angle from the positive x axis to $L$.

Let $v_1=(\cos \alpha, \sin \alpha)$ and $v_2=(-\sin \alpha, \cos \alpha)$.

Then $v_1 \in L$ and $v_2 \in L^\perp$.

I can't understand the last line. Why $v_1 \in L$ and $v_2 \in L^\perp$?

答:
Since $L$ is a line going through $(0,0)$ you can write
$$
  L
= \{ t \vec{d}: t \in \mathbb{R} \}
$$
for some unit vector $\vec{d} = (x,y)$. Since $\alpha$ is the angle it makes with the positive $x$-axis, drawing yourself a right triangle whose vertices are $(0,0)$, $(x,y)$ and $(x,0)$, you see that $\cos\alpha = x/1=x$ and $\sin\alpha = y/1=y$, (the fact that $\vec{d}$ is a unit vector means the hypotenuse of the triangle here is $1$).

That is to say $v_1 = (\cos\alpha,\sin\alpha) = (x,y) = \vec{d}$, so $L$ is just the span of $v_1$. This explains the first part of the statement.

The second part says that, for any element $w \in L$, $v_2 \cdot w = 0$. You can check this by noting that any $w \in L$ has the form $w = tv_1$ for some $t \in \mathbb{R}$ by the first statement, and so
$$
  v_2 \cdot w
= t (-\sin\alpha, \cos\alpha) \cdot (\cos\alpha,\sin\alpha)
= t (-\sin\alpha\cos\alpha + \cos\alpha\sin\alpha)
= 0.
$$

Title: Why $v_1 \in L$ and $v_2 \in L^\perp$?


问:
How do you represent the following sentence in first-order logic?
Q) The best score in Greek is always higher than the best score in French.
AIMA book shows this solution-:
Ans=>∀s ∃x ∀y Score(x,G,s) > Score(y,F, s).

AIMA has defined-:
score(x,c,s) as  score obtained by student x in course c in semester s.

My Confusion-:
I understand ∃x Score(x,G,s) part.
But I am confused why are we saying ∀y? Shouldn't that be just ∃y as we want to compare with the best score?

答:
All we need is that all French scores are lower than the best Greek score. It then automatically follows that the best French score also obeys that. And vice versa, if the best French score is lower than the best Greek one, then the same is true for all French scores (as these are lower or equal). So implicitly the solution is using the above argument (based on transitivity of $>$) to avoid mentioning "the best score in French" at all.

It's not a literal translation of the sentence in that sense but one that is equivalent to it if $<$ is indeed transitive (and all best scores are well-defined, because the domain is finite or some such reason).

You could formulate it more literally by introducing the owners of both best scores first and then comparing them:

$$\forall s \exists x \exists x': (\forall y: \text{score}(x,G,s) \ge \text{score}(y,G,s)) \land (\forall y: \text{score}(x',F,s) \ge \text{score}(y,F,s)) \land (\text{score}(x,G,s) > \text{score}(x',F,s))$$


问:
Can we interchange conditional expectation and $L^2$-convergent series?
So I know the following holds true (I can show it by using an alternative approach, and much more complex) but I would like to keep things simpler and try to find a way to prove it directly.

An old result by Cameron & Martin states that a square integrable "Brownian functional" $X\in L^2(\Omega,\mathcal F,P)$ (where $\mathcal F$ is generated by a Brownian motion) can be decomposed as a series using "generalized Hermite polynomials" (sometimes known as Wick polynomials)

$$X(\cdot)=\sum_{\alpha\in \mathcal J} x_{\alpha}\mathcal H_{\alpha}(\cdot)$$

where $\mathcal J$ is the set of multi-indices that have finitely many non-null elements and $\mathcal H_{\alpha}$ is the Wick polynomial associated with the multi-index $\alpha$. And the convergence of the series is in the $L^2$ sense.(Notice that this is just the Wiener chaos expansion in terms of Wick polynomials)

I would like to show that

$$\mathbb E [X| \mathcal G]=\sum_{\alpha\in \mathcal J} x_{\alpha}\mathbb E [\mathcal H_{\alpha}| \mathcal G]$$

where $\mathcal G$ is some sub-sigma algebra of $\mathcal F$.

So what I want to do is to interchange an integral (wrt the conditional distribution) with an $L^2$ convergent series. I wanted to apply Fubini lemma but for that I would need (this is sufficient but may be not necessary) the series to be absolutely convergent.

Is there some simple way to show that indeed one could put the conditional expectation inside the summation ? (without recurring to second quantization operators and things like that)

答: If $Y_n \to Y$ in $L^{2}$ then $E (Y_n|\mathcal G) \to E (Y|\mathcal G)$ in $L^{2}$ for any suib-sigma filed $\mathcal G$. This follows from conditional version of Jensen's inequality: $(E (Y_n|\mathcal G) - E (Y|\mathcal G))^{2} \leq E((Y_n-Y)^{2}|\mathcal G)$. Now take expectation on both sides to finish.


问:
Calculating $\int_0^\infty ({ae^{-jw}})^t \ dt$
I know how to calculate $\int_0^\infty e^{-jwt} \ dt$ from $0$ to infinity but how do you deal with that extra $a^t$ term?

Body:
Hint:

$$a^t=e^{t\ln(a)}$$ $$$$$$$$


问:
Characteristic function of the Binomial distribution converges to that of the Poisson

Find conditions on $\lambda, n, p$, so that the characteristic function of the Binomial converges to that of the Poisson

Binomial distribution is given as $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$

Poisson distribution is given as $P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}$

So characteristic function of binomial: $\phi_X(t)=E(e^{itX})=\sum_1^n \binom{n}{k} e^{itk} p^k(1-p)^{n-k}=\sum \binom{n}{k} (e^{it}p)^k (1-p)^{n-k}=(pe^{it}+(1-p))^n$

and that of Poisson in a similar way:$\quad$$e^{\lambda(e^{it}-1)}$

Now my questions:

1) We defined Expectation with integral, why here with a sum ?

2) Can you give me a hint to solve the problem ?

答:
Answer to (1):
If $X$ is a discrete random variable then there exists a countable set $\{x_1,x_2,\ldots\}$ such that $P(X^{-1}(\Bbb{R}\setminus\{x_1,x_2,\ldots\}))=P(X\notin \{x_1,x_2,\ldots\})=0\implies P(X^{-1}(\{x_1,x_2,\ldots\}))=1$

Let $A=X^{-1}(\{x_1,x_2,\ldots\})=\bigsqcup\limits_{i=1}^\infty X^{-1}(x_i)$. Then for any event $E$, $P(E)=P(E\cap A)$.

So for any measurable map $f$,

$\int_\Omega f(X)\ dP=\int_E f(X)\ dP+\int_{E^c} f(X)\ dP=\int_E f(X)\ dP=\int_{\bigsqcup\limits_{i=1}^\infty X^{-1}(x_i)} f(X)\ dP=\sum\limits_{i=1}^\infty \int_{X^{-1}(x_i)}f(X)\ dP=\sum\limits_{i=1}^\infty f(x_i)P(X=x_i)$

Here $f(x)=e^{itx}$


问:
Differentiating the CDF of a transformed random variable
Suppose I have a random variable, $x$, distributed with CDF $G(x)$ and PDF $g(x)$ on $[0,1]$. I am trying to determine the derivative of $G(x-a)$ with respect to $x$, where $a>0$. My intuition is that I should first differentiate $G(x-a)$ with respect to $(x-a)$, giving $g(x-a)$, and then (x-a) with respect to $x$, which is $1.$ Thus I get $g(x-a)$. Is my thinking correct? Thank you.

答:
Yes correct and it is very easy to prove

$$G_X(x-a)=\int_0^{x-a}g_X(t)dt$$

$$G_X'(x-a)=g_X(x-a)\times 1=g_X(x-a)$$


问:
Boolean algebra: why does the distributive property not make intuitive mathematical sense?
I'm trying to understand the distributive law of Boolean algebra by relating it to what I know from ordinary algebra, and it seems to only work for one form of the law.

Ordinary algebra:

$x(y+z) = xy + xz$

This holds true for the first part of the distributive law.

Ordinary algebra:

$x + (yz)$

It's just that. I don't see any way of simplifying this more.

But in Boolean Algebra:

$x + (yz) = x\vee (y \wedge z) = (x\vee y) \wedge (x\vee z) = (x+y)(x+z)$

And in ordinary algebra, if you expand this, it's:

$xx+xz+xy+yz$

Could someone please help me understand this? Am I wrong in trying to relate ordinary algebra to Boolean algebra? Subquestion: if I am wrong in trying to make this connection, why?

答:
Sorry for my bad English. I'm study Boolean Algebra.
First, in ordinary algebra, it is not $xx+xz+xy+zz$ if you expand $(x+y)(x+z)$. It's $xx+xy+xz+yz$.

According to my understanding, the laws are the same for both ordinary and Boolean algebra.
In Boolean algebra, according to distributive laws, $(x+y)(x+z)$ could expand like ordinary algebra:

$(x+y)(x+z)=xx+xz+xy+yz$

But for Boolean Algebra, it's not the simplest form. Because Boolean Algebra only has two possible value, therefore:

$xx+xz+xy+yz=x+xz+xy+yz=x1+xz+xy+yz=x(1+y+z)+yz=x+yz$

Please correct me if I'm wrong.


问:
Morphism of topoi - the canonical topology
Let $T,T'$ be Grothendieck topoi and $(f^\ast,f_\ast): T' \to T$ a morphism of topoi, i.e. $f^\ast\colon T\to T'$ is left adjoint to $f_\ast\colon T\to T'$ and $f^\ast$ commutes with finite limits.

In exercise 2.N(iii) in M. Olsson's book "Algebraic Spaces and Stacks" one is asked to show that $f^\ast$ maps a covering in the canonical topology of $T$ to a covering of the canonical topology of $T'$.

My question is: How is this proven? I tried for some time now, but I did not get far. Below I give further context and a reduction that I managed to show. Maybe I made some mistake there and that is the source of my confusion.

Olsson works with pretopologies, so a covering in the canonical topology on an object $F\in\operatorname{obj}(T)$, may be given as a family of morphisms $\{F_i \to F\}_{i\in I}$, where $I$ is some index set such that the following diagram is an equalizer diagram for all $G\to F$ and $H$ in $T$:
$$
\operatorname{Hom}_T(G,H) \to \prod_{i\in I}\operatorname{Hom}_{T}(G\times_F F_i , H) \rightrightarrows \prod_{i,j\in I}\operatorname{Hom}_{T}(G\times_F F_i \times_F F_j , H).
$$
So now we want $\{f^\ast(F_i) \to f^\ast(F)\}_{i\in I}$ fulfilling the analogous criterion in $T'$, i.e. one has to show
$$
\operatorname{Hom}_{T'}(G',H') \to \prod_{i\in I}\operatorname{Hom}_{T'}(G'\times_{f^\ast(F)} f^\ast(F_i) , H') \rightrightarrows \prod_{i,j\in I}\operatorname{Hom}_{T'}(G'\times_{f^\ast(F)} f^\ast(F_i) \times_{f^\ast(F)} f^\ast(F_j) , H'),
$$
for every $G\to f^\ast(F)$ and $H$ in $T'$, is exact. If one assumes that $G'=f^\ast(G)$ for some $G\in\operatorname{obj}(T)$ and that $G'\to f^\ast(F)$ comes from some $G\to F$ adjointsness and commutativity of limits yields the desired result, but in the general case ...

If I assume that $T'=\operatorname{Shv}(\mathcal{C})$ for some site $\mathcal{C}$, I can reduce to the case that $G'$ is representable (i.e. the sheafification of $\operatorname{Hom}_{T'}(-,V)$ for some $V\in\operatorname{obj}(T')$), since given any family of morphisms
$$
\{ \varphi_i \colon G' \times_{f^\ast(F)} f^\ast(F_i) \to H' \}_{i\in I}
$$
that fulfill the equalizing condition, we can construct a $\varphi\colon G'\to H'$ via:

Taking $g\in G'(V)$ for $V\in\operatorname{obj}(\mathcal{C})$ we apply the assumption on representables to the family
$$
\{ \operatorname{Hom}_{T'}(-,V)^a \times_{f^\ast(F)} f^\ast(F_i) \to G' \times_{f^\ast(F)} f^\ast(F_i) \to H'\},
$$
where the superscript $a$ is supposed to denote sheafification and $\operatorname{Hom}_{T'}(-,V) \to G'$ is induced by $g$ via Yoneda's lemma. Thus we get a morphism $\operatorname{Hom}_{T'}(-,V)^a \to H'$ or equivalently an element $\varphi(g)\in H'(V)$. We use this to construct $\varphi\colon G' \to H'$. But is this even a step in the right direction?

It feels to me that this is some standard problem, but I could not find a solution. Any nudge in the right direction helps.

Thanks in advance and have a great day!

答:
It seems that your question is resolved as soon as you are able to identify the canonical topology on a Grothendieck topos, so let me state it explicitly.

Proposition.
The canonical topology on a Grothendieck topos has as its covering families all small jointly epimorphic sinks.

As you surmised, this is because epimorphisms in a topos are effective and stable under pullback; in other words, in a topos, epimorphism = universal effective epimorphism.

Your original question about the inverse image functor is now easy to answer: left adjoints preserve colimits, so small jointly epimorphic sinks are sent to small jointly epimorphic sinks; in other words, they preserve coverings w.r.t. the canonical topology.


问:
Laplace's eigenvalue problem
I am computing the eigenvalue of the Laplace problem on an unit square.

Consider the eigenvalue problem as follows. $$-L u(x, y) = \lambda u(x, y)$$ where $$L = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$$ and the Dirichlet boundary condition is $u = 0$ at the boundary of the unit square.

Now I am interested to calculate $N(t)$ that is, the number of eigenvalues less than or equal to $t$.

I have a formula, namely Weyl's law to calculate $N(t)$. Now Weyl's law is as follows:

Consider $\Omega$ be a bounded region in $\mathbb{R}^d$ and $\omega_{d}$ is the volume of the unit ball in $\mathbb{R}^d$, then Weyl's law gives

$${\displaystyle N(\lambda )=(2\pi )^{-d}\lambda ^{d/2}\omega _{d}\mathrm {vol} (\Omega )\mp {\frac {1}{4}}(2\pi )^{1-d}\omega _{d-1}\lambda ^{(d-1)/2}\mathrm {area} (\partial \Omega )+o(\lambda ^{(d-1)/2})}$$

My question is how to unwind the formula for $2$ dimensional case, i.e for the unit square in $\mathbb{R}^2$.

Please help me. Thanking in advanced.

答:
The notation $\mathrm{vol}(\Omega)$ and $\mathrm{area}(\partial \Omega)$ can be misleading. The term $\mathrm{vol}(\Omega)$ is the $n$-Lebesgue measure of $\Omega$ which is the area of $\Omega$ when $n=2$. Also, $\mathrm{area}(\partial \Omega)$ is the $(n-1)$-Hausdorff measure of $\partial \Omega$ which is the length of $\partial \Omega$ (i.e the perimeter of $\Omega$) when $n=2$. Hence, when $\Omega$ is the unit square in $\mathbb{R}^2$, $\mathrm{vol}(\Omega)=1$ and $\mathrm{area}(\partial \Omega)=4$.

To calculate $\omega_1$ remember that the unit ball in $\mathbb{R}$ is just the interval $(-1,1)$ which has length 2, so $\omega_1=2$. Similarly, the unit ball in $\mathbb{R}^2$ is just the region enclosed by a circle of radius 1, so $\omega_1=\pi$. Plugging all of this into Weyl's Law gives \begin{align*}
N(\lambda) &= (2 \pi)^{-2}\cdot \pi  \lambda \mp \frac 1 4 (2 \pi)^{-1} \cdot 2 \cdot 4 \lambda^{\frac 1 2} + o (\lambda^{\frac 1 2 }) \\
&= \frac \lambda {4 \pi} \mp \frac{\lambda^{\frac 1 2 }}{\pi}+ o (\lambda^{\frac 1 2 })
\end{align*} as $\lambda \to \infty$.

Title: Laplace's eigenvalue problem


问:
Homotopic to wich identity function
I'm having some problem understanding when a function is homotopic to the identity function.
Let's say that i have the function $\Delta:X\to X\times X$ such as $\Delta(x)=(x,x)$ and I want to show that it's homotopic to the identity function, but i can't understand which identity function I'm talking about.

From the definition I have two function $f,g:X\to Y$ are homotopic to each other if $\exists H:X\times[0,1]\to Y$ such as $H(x,0)=f(x)$ and $H(x,1)=g(x)$, so $\Delta$ can't be homotopic to $id_X$ but neither to $id_Y$.

But at the same times I can find two function $\Delta:X\to X\times X$ and $i:X\times X\to X$ to show that $X\simeq X\times X$.
Am I making any mistakes? Is there a way to show that $\Delta$ is homotopic to one of those identity function?

Can I show that $id_X\simeq id_{X\times X}$?

答: No, a continuous map $X\to Y$ can't be homotopic to "the identity" when $X\neq Y$. Maps $X\to Y$ can only be homotopic to other maps $X\to Y$. However, when you have maps $f\colon X\to Y$ and $g\colon Y\to X$ you obtain $f\circ g\colon Y\to Y$ and $g\circ f\colon X\to X$ which can be homotopic to $\operatorname{id}_Y$ and $\operatorname{id}_X$, respectively. In this case $f$ (and $g$) is called a homotopy equivalence between $X$ and $Y$ and we write $X\simeq Y$.

Title: Homotopic to wich identity function


问:
$X$ exponential random variable - probability
We know that $X$ is an exponential random variable with parameter $\gamma$.

I need to calculate $\mathbb{E}[X|X>1]$.

How I do it?

答:
The conditonal density is

$$f_{X|X>1}=e^{\theta}\theta e^{-\theta x}=\theta e^{\theta(1-x)}$$

To calculate the conditional expectation you get

$$E(X|X>1)=\int_1^{\infty}x\theta e^{\theta(1-x)}dx=1+\frac{1}{\theta}$$

But you can calculate your expectation quickly using memoryless property

$$E(X|X>1)=1+E(X)=1+\frac{1}{\theta}$$


问:
Why does ${x}^{x^{x^{x^{\,.^{\,.^{\,.}}}}}}$ bifurcate below $\sim0.065$?
When you calculate what ${x}^{x^{x^{x\cdots }}}$ converges to between $0$ and $1$, before approximately $0.065$ the graph bifurcates.

Why does this happen and is there a reason for it happens at that number?

答: See Projective Fractals for an explanation of why the bifurcation is at $e^{e^{2 \pi i x-{e^{2 \pi i x}}}}$ where $x=-1$.


问:
Volume of irregular solid
I need to calculate volume of irregular solid which is having fix $200 \times 200$ width and breadth but all four points varies in depth.

I have table which gives depth at each point.

How to calculate volume of such solid?

Hi, I am giving here my main problem definition.
I have a grid with size  $200 \times 200$  and the depth at each point is given in array.
For  $2 \times 4$ grid, below is the depth level.

And i need to find volume of such solid.

537.52,    707.52,    742.52,    719.52,    654.52

631.52,    783.52,    795.52,    764.52,    727.52

597.52,    868.52,    846.52,    793.52,    707.52

In Image i have tried to plot first grid cell.

Any help/pointer/suggestion would be real help.

I found some post related to this. But not sure which will give best result.
http://tutorial.math.lamar.edu/Classes/CalcI/MoreVolume.aspx (Example 2) and link 2 answers.yahoo.com/question/index?qid=20080804220134AA8skGw

Please comment your views.
Thanks a lot in advance.

答:
It appears that @Blue and @MvG have landed on the same formula which I believe is correct. I came up with the same formula using calculus. It can be described by "Area of the base times the average height". The calculus is done by extruding a right trapezoid in the XY plane along the Z axis, where the heights of the sides are described by $y_0$ and $y_1$

$$\begin{align}y_0 &= a(z) = \Biggl(\frac{{y_a}_f - {y_a}_i}{L}\Biggr)z + {y_a}_i\\\\y_1 &= b(z) = \Biggl(\frac{{y_b}_f - {y_b}_i}{L}\Biggr)z + {y_b}_i\\\\A &= \frac{W}{2}(y_0 + y_1)\\\\&=\frac{W}{2}[a(z) + b(z)]\\\\V &= \int^L_0A(dz)\\\\&=\frac{W}{2}\int^L_0[a(z) + b(z)](dz)\\\\&NOTE: \,\,\,\, a(0) = b(0) = 0\\\\&=\frac{W}{2}\Biggl[\frac{({y_a}_f - {y_a}_i)z^2}{2L} + {y_a}_iz + \frac{({y_b}_f - {y_b}_i)z^2}{2L} + {y_b}_iz\Biggr]\\\\&= \frac{W}{2}\Biggl(\frac{z}{2L}\Biggr)\Biggl[({y_a}_f - {y_a}_i)z + 2L{y_a}_i + ({y_b}_f - {y_b}_i)z + 2L{y_b}_i\Biggr]\\\\&Evaluate\,\,\,at\,\,\,z = L\\\\&= \frac{WL^2}{4L}({y_a}_f - {y_a}_i + 2{y_a}_i + {y_b}_f - {y_b}_i + 2{y_b}_i)\\\\&= \frac{WL}{4}({y_a}_f + {y_a}_i + {y_b}_f + {y_b}_i)\\\\&=\frac{WL}{4}\sum h \\\\&\Rightarrow WL\frac{\sum h}{4}\\\\&= Area\,\,\,of\,\,\,base\,\,\,(WL)\,\,\,times\,\,\,average\,\,\,height\,\,\,\Biggl(\frac{\sum h}{4}\Biggr)\end{align}$$

Right trapezoid math figures

Title: Volume of irregular solid


问:
Pigeonhole question involving choosing $6$ distinct integers

Prove that if $6$ distinct integers from one to ten are chosen, then
there will be at least two of them which sum to eleven.

I have identified the five only possible pirs: $(10,1),(9,2),...,(6,5)$ but I am having trouble expressing how this is guaranteed if six digits are chosen. I'm not sure if this helps, but I know for the first digit chosen, there is only one way to get the desired outcome (sum to eleven). Since there are 6 digits chosen, $\lceil 6/5 \rceil=2$? So At least two will produce a sum of eleven? Sounds very flawed to me.

答:
The thing that sounds very flaw to you is, in fact, correct.

Put the number $1$ to $10$ in five boxes, as you have done: $(10, 1), (9, 2), \dots, (6, 5)$.

Now you want to choose $6$ numbers from the five boxes. By pigeonhole principle, there must be two numbers that are chosen from the same box.

These two numbers will sum up to $11$.


问:
Decompose the group $U_{60}$ as direct product of cyclic groups
Decompose the following group as a direct product of cyclic groups: $U_{60}$.

Here is what i've done so far:

$60 = 2\cdot2\cdot5\cdot3$.

Therefore the answers are

$U_{60} \cong C_2\times C_2\times C_3\times C_5$ or $U_{60}\cong C_4\times C_3\times C_5$.

Sorry about the format.

答:
Notice first that $U_{60}$ is not the same as $C_{60}$. $U_{60}$ is the group consisting of those numbers $1\le k<60$  that satisfy $gcd(k,60)=1$, and the group operation is multiplication modulo $60$. So the size of $U_{60}$ is considerably smaller than $60$. Here is what you need to do:

Find out how many elements $U_{60}$ does have (hint: count them! There are not too many so you can (and should!) just find all of them. Then you'll have all the elements in the group in front of you (general remark: there is a formula for the number of elements in $U_n$, look up Euler's totient function).

Now, notice that $U_{60}$ is abelian (just like any other $U_n$). Now you know how many elements it has so decompose that number into prime factors and proceed to use the fundamental theorem of finite abelian groups.


问:
How to find the inverse function for $y = x^5 +2x + 1$
I'm using a self study pre-calculus book and have to invert function $y = x^5 + 2x + 1$. I know I have to swap the variables and put everything on one side of the equation such as $y^5 + 2y -x + 1 = 0$ but from here I don't know what to do.  All of the examples I can find on the internet either use calculus which I don't know or show examples with the degree of $2$ rather than $5$ and so they use the quadratic equation to solve for $y$.  So I am stuck.

答: You either misread the book, misunderstood the bigger picture or it's playing a cruel joke on you because there's no closed form inverse of this function in terms of elementary functions.  The form that exists and can be found using a CAS (in terms of the generalized hypergeometric function) certainly could not be found by a student at that level.</p>


问:
Size of a compact set in space from its shadows
Say we have a finite number of disjoint balls in the space forming the set $A.$ We have some bound $R^2\pi$ for the areas of all its projections to planes. Is it true that the size of the set $A$ in space is bounded by $\frac{4}{3}R^3\pi$?

答: Ok, I got the following by myself. I only need to assume that the set $A$ is bounded and closed in space, that is $$A \subset C$$ for some ball $C > 0.$ Now, via scaling let $$c^34/3\pi = \mu(A) = \int_{B(0,1)} 1_{A}\mu$$ and $A \subset B(0,1).$ So $1_A$ is the indicator function of $A$. Let $e$ be an unit vector. So we need to show a bound $$\mu_{2}(A|e^{\bot}) \geq c^2\pi$$ for one of the projections, but I only get a bound $$\mu_{2}(A|e^{\bot}) \geq c^3\frac{4}{3}.$$ Now, we assume some uniform density $\gamma$ on $B(0,1)$. This is a scaled Lebesgue-measure in space so it projects to measure $\gamma^{2}\mu$ to a plane. We write $$c^34/3\pi =\mu(A) = \int_{B(0,1)}\gamma\mu$$ and we now should have $$\gamma^{2/3}\mu_{2}(B(0,1)|e^{\bot}) = \mu_{2}(B(0,\gamma^{1/3})|e^{\bot}) = c^{2}\pi.$$ So that $c^3 = \gamma.$ It was not easy to me to show that any kind of projection bound holds. But if we have $$\mu_{2}(A|e^{\bot}) < \frac{4c^3}{3},$$ then there exist some $\gamma_1 < \gamma = c^3$ s.t $$\mu_2(A|e^{\bot}) = \gamma_1\pi < \frac{4c^3}{3}.$$ It's clear that we have $A \cap H^e \subset (A \cap (A|e),$ where $H^e$ is a subspace orthogonal to $e.$ But now we have $A \subset \bigcup(A \cap H^e)$ and for some $H^e$ that $$\mu_2(A \cap H^e) = \gamma_1\pi.$$ Let $\delta > 0$ be small. Then we have for   $(A \cap H^{e}_{\delta})$ that $$\mu(A) \leq  \mu(A \cap \bigcup H^e_{\delta}) \leq \sum_{i=1}^m \mu(A \cap H^{e_i}_{\delta}) \leq \sum_{i=1}^{m} \pi\gamma_1\delta < \gamma\frac{4}{3}\pi = \mu(A),$$ which is a contradiction. Above I used compactness of $A$ in order to get a finite sum, where we should have $m \leq \frac{\pi}{\delta}.$ Now, $m \leq \frac{\pi}{\delta}$  is the right bound because $$A \subset B(0,1) \subset \bigcup_{i=1}^{\frac{\pi}{\delta}}B_2(0,1) \cap H^{e_i}.$$ So we have $$\mu_{2}(A|e^{\bot}) \geq \frac{4c^3}{3}\,$$ for some projection $(A|e^{\bot})$. This does not imply my original question. But it's clear that my proof for analogical bound works in every dimension. In other words I proved that small projections imply small volume. But what happens if the largest projections equal the largest sections? Then we would have a that a small sections imply small volume. My proof works for sections also. (EDIT) However, if we have an annulus $S_2$ of thickness $r$ in the  unit ball, then I got that$$\frac{4}{3}\pi(1-(1-r)^3) = 4\pi\int_{1-r}^1 r^2 dr = \int_S = \frac{4}{3}\gamma\pi.$$ Now, an annulus has a slice of size $2\pi\int_{1-r}^1 r dr = \pi*r(2-r).$ Which means that there is a maximum slice of the annulus smaller than $\gamma^{2/3}\pi.$ So this is a counter example.


问:
The antiderivative of $\frac{1}{x-ib}$
I know if my domain $D_{\mathbb{R}}\subset\mathbb{R}$ and $b$ is a real constant, then:

$$\int_{D_\mathbb{R}}\dfrac{1}{x-ib}\mathrm{dx}=\ln(|x-ib|)$$

But if my domain $D_{\mathbb{C}}\subset\mathbb{C}$:

$$\int_{D_\mathbb{C}}\frac{1}{x-ib}\mathrm{dx}=\log(x-ib)=\ln|x-ib|+i\mathrm{arg}(z)$$

This feels a bit counterintuitive to me, but is it correct?

答:
In the complex case, even for real domain, do not use $|\cdot|$. That is incorrect.

Correct is: [some branch of] $\log(x-ib)$.

Example

Computed numerically, we have to $10$ decimals
$$
\int_0^1 \frac{1}{x-i}\;dx \approx 0.3465735903 + i\; 0.7853981634
$$
But
$$
\log\big(|x-i|\big)\Big|_{x=0}^{x=1} = \log\big(|1-i|\big) - \log\big(|-i|\big)
= \log(\sqrt2) - \log(1)  \approx 0.3465735903
$$
incorrect.

On the other hand,
$$
\log(1-i) - \log(-i) = \big(\log\big(\sqrt2\big) - \frac{i\pi}{4}\big) - \big(\frac{-i\pi}{2}\big) = \log\big(\sqrt{2}\big) + \frac{i\pi}{4} \approx 0.3465735903 + i\;0.7853981635
$$
correct

I used a branch of $\log$ that is continuous on the line segment from $1-i$ to $-i$.

NOTE.  When you ask Maple or Mathematica for $\int \frac{dx}{x}$ they give you $\log (x)$ and not $\log\big(|x|\big)$.  This is (one of) the reasons.  The "answer" $\log\big(|x|\big)$ is just wrong in many cases.

Title: The antiderivative of $\frac{1}{x-ib}$


问:
Is the $\Bbb{R}^2\setminus\{P_1,...,P_n\}$ plane homeomorphic to $\Bbb{R}^2\setminus\{Q_1,...,Q_n\}$?
Let $K=\{P_1,...,P_n\}\in\Bbb{R}^2$ and let $\Bbb{R}^2\setminus K$ be the plane with $n$ holes. Show that for any $K':=\{Q_1,...,Q_n\}$, $\Bbb{R}^2\setminus K \cong \Bbb{R}^2\setminus K'$.

As I understood it, I may need to find an $f:X \to Y$ that is a homeomorphism. At first I thought of an $f$ that sends each $(x,y)\in\Bbb{R}^2\setminus(K\cup K')$ to $(x,y)$, and, assuming WLOG that $K\cap K'=\emptyset$, $f(Q_i)=P_i$. But I see this doesn't make a continuous map.

Then I thought I should take $f(x,y)=(x,y)+(x_{P_1}-x_{Q_1},y_{P_1}-y_{Q_1})$ and try to show that it is surjective by setting $(x_i,y_i)=(x_{P_i}-x_{P_1}+x_{Q_1},x_{P_i}-y_{P_1}+y_{Q_1})$ for $i\neq1$ and that way achieve $f(x_i,y_i)=P_i$. But what if $(x_i,y_i)=P_j$ where $f$ is not defined? I am confused. Maybe I didn't get it at all. I could use some help here.

答:
As John Samples comments, it suffices to find a homeomorphisms $h : \mathbb R^2 \to \mathbb R^2$ such that $h(P_i) = Q_i$. This is even a stronger result than $h(K) = K'$ which would also show that  $\Bbb{R}^2\setminus K \cong \Bbb{R}^2\setminus K'$.

Note that it suffices to consider the special points $Q_i = (i,0) \in X = \mathbb R \times \{0\} $.

We can do it by induction. As the base case we can take $n = 0$ (nothing to prove). But also for $n =1$ it is obvious: Take a translation shifting $P_1$ to $Q_1$.

Now assume that we have $h : \mathbb R^2 \to \mathbb R^2$ such that $h(P_i) = Q_i$ for $i \le n$. In other words, we may assume w.l.o.g. that $P_i = Q_i$ for $i \le n$. For $(a, b) \in \mathbb R^2$ and $r > 0$ let us define $$\psi_{a,b,r} : \mathbb R^2 \to \mathbb R^2, \psi_{a, b,r}(x,y) =  \begin{cases} (x,y) & \lvert x - a  \rvert \ge r \\ (x,y + \frac{(\lvert x - a \rvert - r)b }{r}) & \lvert x - a  \rvert \le r \end{cases}$$
This is well-defined continuous map which shifts $(a,b)$ to $(a,0)$ and $(a,0)$ to $(a,-b)$ and moreover keeps all points $(x,0) \in X$ with $\lvert x - a \rvert \ge r$ fixed. It is a homeomorphism whose inverse is $\psi_{a,-b,r}$ (since $\psi_{a,b,r} \circ \psi_{a,-b,r} = id$ and $\psi_{a,-b,r} \circ \psi_{a,b,r} = id$).

$P_{n+1} \notin X$.

Write $P_{n+1} = (a, b)$ with $b \ne 0$. There is a unique linear map $\phi : \mathbb R^2 \to \mathbb R^2$ such that $\phi(1,0) = (1,0)$ and $\phi(a,b) = (n+1,b)$ since $(1,0)$ and $(a,b)$ form a basis of $\mathbb R^2$. Clearly $\phi$ is a linear isomorphism, hence a homeomorphism. It keeps $X$ fixed. Then $h_{n+1} = \psi_{n+1,b,1} \circ \phi$ keeps  the $P_i$ with $i \le n$ fixed and maps $P_{n+1}$ to $Q_{n+1}$.

$P_{n+1} \in X$.

Write $P_{n+1} = (a, 0)$. We have $P_{n+1} \notin \{P_1,\dots,P_n\}$, hence there exists $r > 0$ such that for $\lvert x - a  \rvert \le r$ we have $(x,0) \notin \{P_1,\dots,P_n\}$. Hence $\psi_{a,1,r}$ keeps the $P_i$ with $i \le n$ fixed and shifts $P_{n+1}$ to $(a,-1) \notin X$. Now proceed as in 1.

By the way, this proof shows that $h$ can always be chosen to be orientation preserving.


问:
Is the space of continuous functions on $\mathbb{R}$ separable/compact with respect to the topology of uniform convergence in compact sets?
Consider the space of continuous functions $C(\mathbb{R}, \mathbb{R})$, equipped with the metric
$$ d(f,g) := \sum_{n=1}^\infty \frac{1}{2^n} \frac{\sup_{x \in [-n,n]} |f(x)-g(x)|}{1 + \sup_{x \in [-n,n]} |f(x)-g(x)|} . $$

The topology induced by this metric should be the topology of uniform convergence in compact sets.

This space is complete.

Is it separable?

Is it compact?

EDIT: It's not a Banach space, since the metric $d$ does not induce a norm.

答:
It is not a Banach space, as $d$ is no norm.
It is a locally convex completely metrisable TVS, also called a Fréchet space in this context.

It is certainly not compact as the constant functions from a non-compact closed copy of $\Bbb R$.

It is separable: can use (rational) polynomials to approximate continuous functions uniformly on some $[-N,N]$ for $N$ large enough..


问:
$\mathbb{C} \setminus \{z \in \mathbb{C} \mid \operatorname{Re}(z)=0\}$ is not simply connected.
Let $X := \mathbb{C} \setminus \{z \in \mathbb{C} \mid \operatorname{Re}(z)=0\}$ be the space in question.

It suffices to show that $X$ is not path-connected. For this, I reckon that $-1+0\mathrm{i}$ and $1+0\mathrm{i}$ cannot be connected with a continuous path within $X$ (since such a path would have to cross the imaginary axis, which is not in $X$). But how do I write this down rigorously?

答: If $\gamma\colon[a,b]\longrightarrow\Bbb C$ is a path such that $\gamma(a)=-1$ and that $\gamma(b)=1$, let $f=\operatorname{Re}\circ\gamma$. Then $f(a)=-1$ and $f(b)=1$ and therefore, by the intermediate value theorem, $f(t)=0$, for some $t\in(-1,1)$. But then $\gamma(t)\notin X$.


答: $X$ is not even connected: $U = \{z \in \Bbb C; \text{Re}(z) >0\}$ is open and $V = \{z \in \Bbb C; \text{Re}(z) < 0\}$ too, and the form a disjoint cover of $X$.

Title: $\mathbb{C} \setminus \{z \in \mathbb{C} \mid \operatorname{Re}(z)=0\}$ is not simply connected.

Body:
Let $X := \mathbb{C} \setminus \{z \in \mathbb{C} \mid \operatorname{Re}(z)=0\}$ be the space in question.

It suffices to show that $X$ is not path-connected. For this, I reckon that $-1+0\mathrm{i}$ and $1+0\mathrm{i}$ cannot be connected with a continuous path within $X$ (since such a path would have to cross the imaginary axis, which is not in $X$). But how do I write this down rigorously?


答:
It is easier to show that $X$ is not connected directly.
For this note that the open sets $\{z\in\mathbb{C}|\Re(z)>0\}$ and $\{z\in\mathbb{C}|\Re(z)<0\}$ are open in $X$ and cover the complete set $X$. Therefore $X$ is not connected.


问:
Expectation value of 2^X.
Three players throw a dice and sum the values. Then 2 is raised to the obtained degree.
What's the expected value of the obtained random variable? Enter the value:

For example, let:

On three dice we get 1, 2 and 5 respectively. Then the sum is 8. Hence:

$x_{(1,2,5)} = 2^8$

we need expectation value of:

$E(X) = \sum_{i=1}^6\sum_{j=1}^6\sum_{k=1}^6 x_{(i,j,k)} * p_{i,j,k}$

Title: Expectation value of 2^X.

Body:
Note that we can write:

$$E[X] = \frac{1}{6^3}\sum_{i=1}^6\sum_{j=1}^6\sum_{k=1}^6 2^{i+j+k}$$

$$E[X] = \frac{1}{6^3}\sum_{i=1}^62^i\sum_{j=1}^62^j\sum_{k=1}^6 2^{k}$$

If $f(n) = \sum_{k=1}^n 2^k$ then $f(n) = 2^{n+1} - 2$, proof left as an exercise to the reader (hint: induction).

$$E[X] = \frac{f(6)^3}{6^3}= {\left(\frac{2^7 - 2}{6}\right)}^3 = 21^3 = 9261$$


问:
Show that The integral is $(\Bbb E[X_i^2-\tfrac13])^2=0$
Show that The integral is $(\Bbb E[X_i^2-\tfrac13])^2=0$

Why does this hold?

答: The integral is $(\Bbb E[X_i^2-\tfrac13])^2=0$, since famously $X_i\sim U(0,\,1)\implies\Bbb E[X_i^2]=\tfrac13$.


问:
Reward Probability on Biased Coin
I tried so hard to figure out this question, but I couldn't. I thought ((1/7)^3)^2

I would appreciate your help.

```<code>You and your friend play a game with a significantly biased coin where the
probability of a head is
1/7. The coin will be flipped 3 times. Let's model the total number of heads
with X (you can use a known random variable for X) and your reward points
after the game will be R = X^2.
</code>```

Then, your expected reward after the game R will be?

答:
simply you can use a binomial $B(3;1/7)$ and you have to calculate

$$E(X^2)=V(X)+E^2(X)=3\times \frac{1}{7}\times\frac{6}{7}+\left(\frac{3}{7}\right)^2$$

Title: Reward Probability on Biased Coin


问:
Way of collapsing set of rules into few universal rules - mathematical Occam's razor
I was thinking about learning rules of a system by just looking at it. Let's say you know nothing about chess and you watch inifinite number of games - you watched all possible transitions of states. If you are dumb(or a neural network) you can write really big set of rules, where each state can evolve into all possible next states and call it rules. It's pretty easy to guess that you would end up with billions of billions of rules - this goes very much against Occam's razor.

Let's simplify the problem then. Let's say we have $x$ by $x$ chessboard and just one tower - we watch infinite moves of this tower. Now we can derive rules of this system, let's say again we are dumb and come up with separate rules that say that from any $x*x$ possible positions we can move to any other $2(x-1)$ positions, but these are all SEPARATE rules. How could we "collapse" this enormous set of rules into something more comprehensible and UNIVERSAL?

I would like to say in advance: I am really sorry for not using a lot of mathematical notation, this problem was a bit more philosophical in my mind and I am not advanced enough in this field of mathematics to write something that makes sense.

答:
A 'mathematical Occam's razor' is the minimum description length principle.

Roughly speaking it states that the best model for a set of observations is the number of bits needed to represent that model, plus the number of bits needed to represent the corrections to the model for observations it gets wrong. The model that minimizes this sum is the best model for a set of observations.

Note that this (just like Occam's razor) does not tell you how to find this model. But it can, given two models, tell you which of the two is better.

As for finding these models, well that's an entire field: machine learning. Tons upon tons of methods exist (linear models, decision trees, (convolutional) neural networks, (non-linear) SVMs, rule-based models, etc).


答:
You can solve this by a direct computation. For $i\neq j$ \begin{align*}
\int_{(0,1)^n} \bigg (x_i^2-\frac 1 3 \bigg ) \bigg (x_j^2-\frac 1 3 \bigg )dx &= \int_0^1 \cdots \int_0^1 \bigg (x_i^2-\frac 1 3 \bigg ) \bigg (x_j^2-\frac 1 3 \bigg ) dx_1 \cdots dx_n \\
&= \bigg ( \int_0^1 \bigg ( t^2 - \frac 1 3 \bigg) d t \bigg )^2 \\
&= 0.
\end{align*} As an aside we can also compute the case $i = j$: \begin{align*}
\int_{(0,1)^n} \bigg (x_i^2-\frac 1 3 \bigg ) \bigg (x_j^2-\frac 1 3 \bigg ) &= \int_{(0,1)^n} \bigg (x_i^2-\frac 1 3 \bigg )^2 dx_i \\&= \frac 4 {45}.
\end{align*}

Body:
Show that The integral is $(\Bbb E[X_i^2-\tfrac13])^2=0$

Why does this hold?


答:
The expectation is:
$$
\sum_{k=0}^3 k^2\binom3k\left(\frac17\right)^k\left(\frac67\right)^{3-k}.
$$

Title: Reward Probability on Biased Coin

Body:
I tried so hard to figure out this question, but I couldn't. I thought ((1/7)^3)^2

I would appreciate your help.

```<code>You and your friend play a game with a significantly biased coin where the
probability of a head is
1/7. The coin will be flipped 3 times. Let's model the total number of heads
with X (you can use a known random variable for X) and your reward points
after the game will be R = X^2.
</code>```

Then, your expected reward after the game R will be?


问:
Is the polynomial $x^{105} - 9$ reducible over $\mathbb{Z}$?

Is the polynomial $x^{105} - 9$ reducible over $\mathbb{Z}$?

This exercise I received on a test, and I didn't resolve it. I would be curious in any demonstration with explanations. Thanks!

答:
For binomials there is a classical irreducibility test (below). It implies that $\,x^{105}-c\,$ is irreducible over a field $\,F\,$ if $\,c\,$ is not a third, fifth, or seventh power in $\,F,$ since $\,105 = 3\cdot 5\cdot 7.$

Theorem $\ $ Suppose $\:c\in F\:$ a field,  and $\:0 < n\in\mathbb Z.$

$\quad x^n - c\ $ is irreducible over $\:F \iff c \not\in F^p\:$ for all primes $\,p\mid n\:$ and $\ c\not\in -4F^4$ when $\: 4\mid n$.

Proofs can be found in many Field Theory textbooks, e.g. see Lang's Algebra, or see Karpilovsky, Topics in Field Theory, Theorem 8.1.6.


问:
Spectral Projection and GNS in Commutative Case
For $N>4$, let $A=C([0,N])$ and $1_A=\mathbf{1}_{[0,N]}$ (the identity on $A$). Consider $\operatorname{id}:=\operatorname{id}_{[0,N]}=(t\mapsto t)$ (not the identity on $A$ but $f(t)=t$) and the functional $\varphi:=\operatorname{ev}_{N-2}$, $f\mapsto f(N-2)$. I am interested in the spectral projection $\mathbf{1}_{\{N-2\}}(\operatorname{id})$ and the GNS representation $(\mathsf{H}_\varphi,\pi_\varphi,\xi_\varphi)$. I guess I'll start with the GNS representation as that seems more straightforward.

The nullspace of $\varphi$, $N_\varphi$ consists of those functions that vanish at $N-2$. We quotient to get the (one-dimensional?) space of functions $A/N_\varphi$ which agree at $N-2$. We have an inner product:
$$\langle f+N_\varphi,g+N_\varphi\rangle=\overline{f(N-2)}g(N-2).$$
We take the (trivial?) Hilbert space completion $\overline{A/N_\varphi}=\mathsf{H}_\varphi$. The representation $\pi_\varphi:A\rightarrow B(\mathsf{H}_\varphi)$ is:
$$\pi_\varphi(f)(g+N_\varphi)=fg+N_\varphi.$$
The cyclic vector is $\xi_\varphi=1_A+N_\varphi$. The importance of this vector is that it is an eigenvector of eigenvalue $N-2$ for $\pi_\varphi(\operatorname{id})$ as:
$$\pi_\varphi(\operatorname{id})(1_A+N_\varphi)=\operatorname{id}+N_\varphi=(N-2)(1_A+N_\varphi),$$
as $\operatorname{id}(N-2)=N-2=(N-2)\varphi(1_A)=\varphi((N-2)1_A)$. So in this representation $N-2$ is an eigenvalue.

Now onto $\mathbf{1}_{N-2}(\operatorname{id})$. I am wondering is it possible to describe this concretely. When I try and play around with $L^\infty([0,N])$ and $L^2([0,N])$ and multipliers and such I cannot get it to work, and the best I think I can do is something like:
$$\mathbf{1}_{(N-2-\varepsilon,N-2+\varepsilon)}(\operatorname{id})(f)(t)=\begin{cases}f(t)& \text{ if }|t-(N-2)|<\varepsilon \\ 0 & \text{otherwise}\end{cases}$$
I don't think it should be zero though in the generality I am interested in. I think I should be looking at the enveloping von Neumann algebra $\pi_U(A)''$, and then my understanding is that $\mathbf{1}_{\{N-2\}}(\operatorname{id})$ is the projection onto the kernel of $\operatorname{id}-(N-2)1_A$, and if I have the big $\mathsf{H}_U$ then I at least have $\xi_\varphi\in\mathsf{H}_U$ and so we should have something like:
$$\mathbf{1}_{\{N-2\}}(\pi_U(\operatorname{id}))\xi_{\varphi}=\xi_\varphi,\qquad (*),$$
which might be written as $\mathbf{1}_{\{N-2\}}(\operatorname{id})\xi_{\varphi}=\xi_\varphi$, and so  $\mathbf{1}_{\{N-2\}}(\operatorname{id})\neq 0$.

My questions I guess are:

Does this all check out? In particular do I have to be more careful with the quotient map in (*)

If $\mathbf{1}_{\{N-2\}}(\operatorname{id})$ is indeed non-zero in $\pi_U(A)''\subset B(\mathsf{H}_U)$, is there any hope of describing it concretely?

What I am hoping to do eventually is condition states $\omega$ (on what algebra now I am not sure, $A$ or normal extensions to $\pi_U(A)''$) by $p_{N-2}:=\mathbf{1}_{\{N-2\}}(\operatorname{id})$:
$$\omega\mapsto \frac{\omega(p_{N-2}\cdot p_{N-2})}{\omega(p_{N-2})},$$

so that, for example, we could condition $(\operatorname{ev}_{N-3}+\operatorname{ev}_{N-2})/2$ to $\varphi=\operatorname{ev}_{N-2}$.

答:
Let $A=C_0(X)$ where $X$ is a Hausdorff space. Then the pure states are the evaluation maps, which are also the characters. As such $A/N_\varphi$ is one-dimensional for every pure state $\varphi$.

So while your GNS construction works it is pretty much overkill, in particular every non-zero element of $A/N_\varphi$ is a cyclic vector and $\pi(f)$ is always proportional to the identity (ie has an eigenvalue). The spectral projection of the point $\{N-2\}$ on the representation $A/N_{ev_{N-2}}$ is given by $\Bbb1$, the identity operator.

If you want to find the spectral projection in the enveloping von Neumann algebra it goes like this:

One construction of $A''$ is as the bi-commutant of the image of $A$ in the representation

$$\bigoplus_{\varphi} A/N_{\varphi}$$
where $\varphi$ sums over all pure states / irreducible representations. In this case these are the characters so you are summing over
$$\bigoplus_{x\in X} \Bbb C = \ell^2(X)$$
where the action of $f\in C_0(X)$ one the summand corresponding to $x\in X$ is given by multiplication with $f(x)$. The bi-commutant of this image is $\ell^\infty(X)$.

The spectral projection of a set $S\subseteq X$ in $A''$ is then given by pointwise multiplication with the characterstic function $\chi_S$ of $S$ on $\ell^2(X)$.


问:
radical and radical ideal
I am trying to get used to the terminology of commutative algebra. Let $R$ be a commutative ring with unity. Then I know radical $\sqrt{I} := \{ a\in R| a^n \in I, \textrm{for some $n \in \mathbb{Z}$} \}$ and an ideal $I$ is called a radical ideal if $I=\sqrt{I}$.

Since $\sqrt{I}$ itself is a ideal containing $I$, I am confused with the terminology a "radical ideal".  Radical and Radical ideal are different.  Am I right?

It seems this radical ideals are important tools for Hilbert's many theorem thus important to commutative algebras and algebraic geometry. So there seems to reason the mathematician defined "radical ideals".

I know some property of radical and radical ideals for prime ideal. For example, for prime ideal $P$, $\sqrt{P} =P$. so Every prime ideal is a radical ideal.

答:
I'm not entirely sure what confuses you, but the point is that you can define "radical ideals" in two equivalent ways:

an ideal $J$ is a radical ideal if $\sqrt{J}=J$

an ideal $J$ is a radical ideal if $J = \sqrt{I}$ for some ideal $I$ (in other words $J$ is the radical of some ideal).

This is the same thing because $\sqrt{\sqrt{I}}=\sqrt{I}$. So the first definition obviously implies the second one, and the converse is true because if $J=\sqrt{I}$ then $\sqrt{J} = \sqrt{\sqrt{I}} = \sqrt{I}=J$.

Title: radical and radical ideal


问:
Topologically equivalent metrics but not strongly equivalent in $\mathrm{Lip}_M(\mathbb{R})=\{f:[0,1]\rightarrow\mathbb{R}:|f(y)-f(x)|\leq M·|y-x|\}$
Let's consider this set $\mathrm{Lip}_{M}(\mathbb{R})=\{f: [0,1] \rightarrow \mathbb{R} : |f(y)-f(x)|\leq M\cdot|y-x| \}$ (i.e Lipschitz functions in $[0,1]$). How can I prove that $(\mathrm{Lip}_{M}(\mathbb{R}), d_{\infty})$ and $(\mathrm{Lip}_{M}(\mathbb{R}), d_{1})$ are topologically equivalent (this means that if a sequence converges for $d_{\infty}$ then it converges for $d_{1}$ and viceversa or that a subset $A$ is open for $d_{\infty}$  if and only if it is open for $d_{1}$) but that they are not strongly equivalent (i.e there is no constant C such that $d_{\infty}(f,g)\leq C \cdot d_{1}(f,g)\ \ \forall f, g \in \mathrm{Lip}_{M}(\mathbb{R})$)?

Here is my proof for the latter. By definition,

$$d_{\infty}(f,g)=\sup_{x \in [0,1]} |f(x)-g(x)|$$

$$d_{1}(f,g)=\int_{0}^{1} |f(x)-g(x)|\,dx$$

Clearly,

$$
\begin{multline}d_{1}(f,g)=\int_{0}^{1} |f(x)-g(x)|\,dx \leq \int_{0}^{1} \sup_{x \in [0,1]}|f(x)-g(x)|\,dx \\
= \sup_{x \in [0,1]}|f(x)-g(x)| \cdot (1-0) = d_{\infty}(f,g).
\end{multline}
$$

To show that these metrics are not strongly equivalent, let's consider the sequence of functions $f_{n}(x) = (1/n) \cdot \cos(n \pi x/2)$, each of which is Lipschitz in $[0,1]$.
So, $d_{\infty}(f_{n},0) = 1/n$, but
$$d_{1}(f_{n},0) = \int_{0}^{1} \left|\frac{1}{n}\cdot\cos\left(nx \frac{\pi}{2}\right)\right|\,dx = \frac{1}{n} \cdot \sin\left(nx \frac{\pi}{2}\right)\Big|_{0}^{1} \cdot \frac{2}{n\pi}=(-1)^n\ \frac{2}{n^2\pi}$$
So if there is a constant $C$ such that $d_{\infty}(f,g)\leq C \cdot d_{1}(f,g)\ \ \forall f, g \in \mathrm{Lip}_{M}(\mathbb{R})$, we should have that, for all $n \in \mathbb{N}$, $$\frac{1}{n} \leq C \cdot (-1)^n\ \frac{2}{n^2\pi} \quad\text{and hence}\quad n\leq C\cdot \frac{2}{\pi}$$
which is absurd. Is this correct?

But I don't know how to show that this metrics are topologically equivalent.

答:
Your proof of the second statement ($d_1$ and $d_\infty$ are not strongly equivalent) is a solid attempt, but it's not quite there: some details slipped through.

For instance,
$$\sin\left(nx \frac{\pi}{2}\right) \Big|_{x=0}^1 = \sin\left(n\cdot 1 \cdot \frac{\pi}{2}\right) - \sin\left(n\cdot 0 \cdot \frac{\pi}{2}\right) = \sin\left(n\frac{\pi}{2}\right)$$
which is not equal to $(-1)^n$—rather, it cycles between $1$, $0$, $-1$ and $0$ starting at $n = 1$. Also note that if that were true, then the inequality would yield $\frac{1}{n} \le -C \frac{2}{n^2\pi}$ for odd $n$, which is impossible since $n$ and $C$ are positive.  (By the way, you seem to have made the $(-1)^n$ factor magically disappear in the following inequality ;-) .)

But more importantly, you can't ignore the absolute value sign in the integral:
$$\int_{0}^{1} \left|\frac{1}{n}\cdot\cos\left(nx \frac{\pi}{2}\right)\right|\,dx$$
is not the same as
$$\int_{0}^{1} \frac{1}{n}\cdot\cos\left(nx \frac{\pi}{2}\right)\,dx = \frac{2}{n^2\pi} \cdot \sin\left(nx \frac{\pi}{2}\right)\Big|_{0}^{1} $$
because $\cos(nx\frac{\pi}{2})$ changes sign $\lfloor\frac{n}{2}\rfloor$ times in between 0 and 1.

A way we can compute this integral is to exploit the periodicity and symmetry of $\cos$: the area under the curve of $|\cos(x \frac{\pi}{2})|$ between $0$ and $n$ is made up of a number of "bumps" of the same area, so we can simply compute the area of one of these half-bumps and count the number of half-bumps on the interval. For example, for $n=6$, there are $6$ copies of the area labelled $A$:

In fact, for any $n \in \mathbb{N}$, there are exactly $n$ half-bumps. This is because at every multiple of $\frac{\pi}{2}$, $\cos$ has either a maximum, a minimum, changes sign from positive to negative, or from negative to positive, and these are precisely the points that limit our half-bumps; for any $n$, these multiples are at $k = 0, 1, 2, \ldots, n$, so there are $n + 1$ of them, meaning there are $(n + 1) - 1 = n$ half-bumps.

Also notice that the number of half bumps in $|\cos(nx \frac{\pi}{2})|$ between $0$ and $1$ is the same as the number of half bumps in $|\cos(x \frac{\pi}{2})|$ between $0$ and $n$.

Finally, the area of one of these half-bumps (for $|\cos(nx\frac{\pi}{2})|$ now, not $|\cos(x\frac{\pi}{2})|$) is
$$\int_{0}^{\frac{1}{n}} \cos\left(nx \frac{\pi}{2}\right)\,dx = \frac{1}{n} \int_{0}^{1}  \cos\left(x \frac{\pi}{2}\right)\,dx = \frac{2}{n\pi} \cdot \sin\left(x \frac{\pi}{2}\right)\Big|_{0}^{1} = \frac{2}{n\pi}$$

Therefore, we have
$$\int_{0}^{1} \left|\frac{1}{n}\cdot\cos\left(nx \frac{\pi}{2}\right)\right|\,dx = \frac{1}{n} \int_{0}^{1} \left|\cos\left(nx \frac{\pi}{2}\right)\right|\,dx = \frac{1}{n} \cdot n \cdot \frac{2}{n\pi} = \frac{2}{n\pi}$$

Now, if we apply the inequality, we don't get anything contradictory:

$$\frac{1}{n} \le C \frac{2}{n\pi} \quad\rightsquigarrow\quad C \ge \frac{\pi}{2}$$

Note that this doesn't mean that there is such a $C$, just that this is not sufficient proof to show that there isn't.


问:
Are the conjugacy classes normal subgroups in $S_n$?
I am working on proving that if $C$ is a conjugacy class in $S_n$ then $C^g =C$ for all $g \in S_n$.

I went over similar questions which only prove that a normal subgroup is contained in union of conjugacy classes, but not claiming a conjugacy class is normal subgroup.

答:
Let $G$ be any group, $a\in G$ and $\operatorname{cl}(a):=\{g'ag'^{-1}, g'\in G\}$. Then:

\begin{alignat}{1}
g\operatorname{cl}(a)g^{-1} &= \{gg'ag'^{-1}g^{-1}, g'\in G\} \\
&= \{(gg')a(gg')^{-1}, g'\in G\} \\
\tag 1
\end{alignat}

Since, for every $g\in G$, the map $g'\mapsto gg'$ is onto $G$, from $(1)$ follows: $g\operatorname{cl}(a)g^{-1}=\{g''ag''^{-1}, g''\in G\}=\operatorname{cl}(a)$, for every $g\in G$.


问:
Derivative Integral Problem
Preface: I am sorry if there is a blatently obvious error or problem with my logic, I am only just learning about the fundamental theorem of Calculus and have been stumped by this example problem and why it has the solution that it does.

I will first show my understanding and what I originally thought was the solution, the problem is:
$$\frac{d}{dx}\int _{a}^{f(x)} g(t)dt = ?$$
My solution:
$$ \text{let } h(x) = \int_{a}^{f(x)}  g(t)dt$$
By the chain rule:
$$ \frac{d}{dx}h(f(x) = h'(f(x)\cdot{}f'(x)$$
$$ h'(x) = d/df \int_a^{f(x)}g(t)dt $$
By the fundamental theorem of Calculus, the derivative and the integral should cancel correct?
$$\frac{d}{df} \int_a^{f(x)}g(t)dt = g(t) $$
Therefore:
$$\frac{d}{dx}\int _{a}^{f(x)} g(t)dt = g(t)\cdot{}f'(x)$$

This however is not the correct answer, I have been told it is really this:

$$\frac{d}{dx}\int _{a}^{f(x)} g(t)dt = g(f(x))\cdot{}f'(x)$$

Again, I apologize if there is some obvious error, Any and all help is appreciated thanks.

答: Thank you to the comments, my idea of the fundamental theorem of calculus was flawed, I have realized I wrote a few things down wrong which carried out through my work, I will fix this now and post this answer to resolve the question. (I can only resolve my own answer in 2 days unfortunately)</p>

Title: Derivative Integral Problem


问:
What does this mean: Every element $\pmod{2^n}$ that is $1 \pmod 4$ is a power of $5$?
For example, consider $17$. Then $17 \pmod{2^n}$ for any $n$ is always $\equiv 1 \pmod 4$, but $17$ isn't a power of $5$.

Just in case I have misunderstood anything, please see this video where the claim is made.

答:
It means: for any power of $2$, say $2^n$, and any residue $0<k<2^n$ such that $k\equiv 1\pmod 4$, there is a value $r\geq 0$ satisfying
$$5^r\equiv k\pmod{2^n}.$$
Clearly these are all the values of $k$ (mod $2^n$) for which such a statement can be true, since $5^r\equiv 1\pmod 4$ for any $r$.

For example, with $k=17$ and, say, $n=5$ we get $5^4=625\equiv 17\pmod {32}$, and a (different) value of $r$ can be found which works for any given larger power of $2$. (If we were to take $n=6$ the first value which works is $r=12$, since $244140625\equiv 17 \pmod{64}$.)


问:
Morphism of topoi - the canonical topology
Let $T,T'$ be Grothendieck topoi and $(f^\ast,f_\ast): T' \to T$ a morphism of topoi, i.e. $f^\ast\colon T\to T'$ is left adjoint to $f_\ast\colon T\to T'$ and $f^\ast$ commutes with finite limits.

In exercise 2.N(iii) in M. Olsson's book "Algebraic Spaces and Stacks" one is asked to show that $f^\ast$ maps a covering in the canonical topology of $T$ to a covering of the canonical topology of $T'$.

My question is: How is this proven? I tried for some time now, but I did not get far. Below I give further context and a reduction that I managed to show. Maybe I made some mistake there and that is the source of my confusion.

Olsson works with pretopologies, so a covering in the canonical topology on an object $F\in\operatorname{obj}(T)$, may be given as a family of morphisms $\{F_i \to F\}_{i\in I}$, where $I$ is some index set such that the following diagram is an equalizer diagram for all $G\to F$ and $H$ in $T$:
$$
\operatorname{Hom}_T(G,H) \to \prod_{i\in I}\operatorname{Hom}_{T}(G\times_F F_i , H) \rightrightarrows \prod_{i,j\in I}\operatorname{Hom}_{T}(G\times_F F_i \times_F F_j , H).
$$
So now we want $\{f^\ast(F_i) \to f^\ast(F)\}_{i\in I}$ fulfilling the analogous criterion in $T'$, i.e. one has to show
$$
\operatorname{Hom}_{T'}(G',H') \to \prod_{i\in I}\operatorname{Hom}_{T'}(G'\times_{f^\ast(F)} f^\ast(F_i) , H') \rightrightarrows \prod_{i,j\in I}\operatorname{Hom}_{T'}(G'\times_{f^\ast(F)} f^\ast(F_i) \times_{f^\ast(F)} f^\ast(F_j) , H'),
$$
for every $G\to f^\ast(F)$ and $H$ in $T'$, is exact. If one assumes that $G'=f^\ast(G)$ for some $G\in\operatorname{obj}(T)$ and that $G'\to f^\ast(F)$ comes from some $G\to F$ adjointsness and commutativity of limits yields the desired result, but in the general case ...

If I assume that $T'=\operatorname{Shv}(\mathcal{C})$ for some site $\mathcal{C}$, I can reduce to the case that $G'$ is representable (i.e. the sheafification of $\operatorname{Hom}_{T'}(-,V)$ for some $V\in\operatorname{obj}(T')$), since given any family of morphisms
$$
\{ \varphi_i \colon G' \times_{f^\ast(F)} f^\ast(F_i) \to H' \}_{i\in I}
$$
that fulfill the equalizing condition, we can construct a $\varphi\colon G'\to H'$ via:

Taking $g\in G'(V)$ for $V\in\operatorname{obj}(\mathcal{C})$ we apply the assumption on representables to the family
$$
\{ \operatorname{Hom}_{T'}(-,V)^a \times_{f^\ast(F)} f^\ast(F_i) \to G' \times_{f^\ast(F)} f^\ast(F_i) \to H'\},
$$
where the superscript $a$ is supposed to denote sheafification and $\operatorname{Hom}_{T'}(-,V) \to G'$ is induced by $g$ via Yoneda's lemma. Thus we get a morphism $\operatorname{Hom}_{T'}(-,V)^a \to H'$ or equivalently an element $\varphi(g)\in H'(V)$. We use this to construct $\varphi\colon G' \to H'$. But is this even a step in the right direction?

It feels to me that this is some standard problem, but I could not find a solution. Any nudge in the right direction helps.

Thanks in advance and have a great day!

答:
This is basically what @ZhenLin explained above, I just want to summarize it to close this question.

(EDIT: @ZhenLin posted his answer right before mine, so I just leave this up to not let the effort go to waste.)

The definition for a covering in the canonical topology that was given in the question is general and applies to any category with fiber products. In the case that the underlying category is a topos $T$, i.e. the category of sheaves for some site $\mathcal{C}$ one finds, that for every $F\in\operatorname{obj}(T)$ a covering of $F$ is already given by a family $\{F_i \to F\}_{i\in I}$ which is jointly epimorphic. Since small coproducts are available in $T$ this is equivalent to $\bigsqcup_{i\in I} F_i \to F$ being an epimorphism.

Now since epimorphisms are preserved by any left adjoint, we arrive at the claim of the exercise.

I just want to fix a proof for the transition from the definition of a covering in the canonical topology as in the question to the characterization above, because   that is what kept me from seeing this in the first place.

So given an object $F\in\operatorname{obj}(T)$ and a jointly epimorphic family $\{F_i \to F\}_{i\in I}$ in $T$, we show that
$$
\operatorname{Hom}_T(G,H) \to \prod_{i\in I} \operatorname{Hom}(G\times_F F_i,H) \rightrightarrows \prod_{i,j\in I} \operatorname{Hom}(G\times_F F_i \times_F F_j , H)
$$
is exact, for every $G\to F$ and $H$ in $T$.

Lemma 1: Every epimorphism $\varphi\colon K\to L$ in $T$ is an effective epimorphism, i.e.
$$
K \times_L K \rightrightarrows K \xrightarrow{\varphi} L
$$
is exact, where the two arrows on the left are the canonical projections.

Proof: In order to show the claim, we have to check that for every $Z\in\operatorname{obj}(T)$ the following sequence is exact:
$$
\operatorname{Hom}_T(L,Z) \hookrightarrow \operatorname{Hom}_T(K,Z) \rightrightarrows \operatorname{Hom}(K\times_L K,Z).
$$
As $K\to L$ is an epimorphisms the $\operatorname{Hom}_T(L,Z) \to \operatorname{Hom}_T(K,Z)$ is indeed a monomorphism. And any $L\to Z$ fits into a diagram
$$
\require{AMScd}
\begin{CD}
K \times_L K @>\operatorname{pr}_2>> K \\
@V\operatorname{pr}_1VV @VV{\varphi}V \\
K @>>{\varphi}> L @>>> Z.
\end{CD}
$$
So let $\psi\colon K\to Z$ be a morphism such that $\psi\circ \operatorname{pr}_1 = \psi\circ\operatorname{pr}_2$ holds. Now we construct a $\psi'\colon L\to Z$ such that $\psi = \psi' \circ \varphi$. We therefore use that $T$ is the category of sheaves over the site $\mathcal{C}$.

Take $C \in \operatorname{obj}(\mathcal{C})$ and $f\in L(C)$. Since $\varphi$ is an epimorphism there is a covering $\{C_i \to C\}_{i\in I}$ in $\mathcal{C}$ and $g_i\in K(C_i)$ such that $\varphi(g_i) = f_{\restriction C_i}$. From $\psi\circ \operatorname{pr}_1 = \psi\circ\operatorname{pr}_2$ we can deduce that whenever $\varphi(g_1) = \varphi(g_2)$ for some $g_1,g_2\in K(C')$ with $C'\in\operatorname{obj}(\mathcal{C})$ holds, $\psi(g_1)=\psi(g_2)$ follows. Since we have
$$
\varphi(g_{i\restriction C_i \times_C C_j}) = f_{\restriction C_i\times_C C_j} = \varphi(g_{j\restriction C_i \times_C C_j})
$$
we get
$$
\psi(g_{i\restriction C_i \times_C C_j}) = \psi(g_{j\restriction C_i \times_C C_j})
$$
for all $i,j \in I$. Since $Z$ is a sheaf, we can glue the $\psi(g_i)$ to obtain an element $\psi'(f)\in Z(C)$. Modulo checking naturality and well-definedness this gives the desired morphism $\psi'\colon L\to Z$.

Lemma 2: Epimorphisms in $T$ are stable under pullbacks.

Proof: Let $\varphi\colon K\to L$ be an epimorphism and $\xi\colon M\to L$ some morphism in $T$. We want to check that $\varphi'\colon K\times_L M \to M$ is an epimorphism, so fix $\lambda_1,\lambda_2\colon M\to M'$ with $\lambda_1 \circ \varphi' = \lambda_2 \circ \varphi'$.

Let there a $C\in\operatorname{obj}(\mathcal{C})$ and some $f\in M(C)$. We set $f':=\xi(f)$. Since $\varphi$ is an epimorphism of sheaves there is a covering $\{C_i\to C\}_{i\in I}$ in $\mathcal{C}$ and $g_i\in K(C_i)$ such that $\varphi(g_i) =f'_{\restriction C_i}$ for all $i\in I$. Now we may view $(g_i,f_{\restriction C_i})$ as an element of $(K\times_L M)(C_i)$, whence $\lambda_1(f_{\restriction C_i}) = \lambda_2(f_{\restriction C_i})$ for all $i\in I$. Since $M'$ is a sheaf $\lambda_1(f)=\lambda_2(f)$ follows and thus $\varphi'$ is an epimorphism.

Now we can conclude: As $\{F_i\to F\}_{i\in I}$ is jointly epimorphic, we have that $\bigsqcup_{i\in I} F_i \to F$ is an epimorphism, by Lemma 1 and Lemma 2 this morphism is even a universally effective epimorphism. But this is just the definition of a covering of $F$ in the canonical topology given in the question.


问:
How do I compute the limit of this product without delving into epsilon theorems?
So, basically, those few lasts I've been struggling to compute the limit of a product without having to delve into epsilon stuff, that is:

$$\lim \limits_{n \to \infty} \prod_{i = 2}^n (2 - \sqrt[i]2)$$

What I've observed that it does not work with ratio test (it is inconclusive), Stolz-Cesaro cannot work and so do not most of the classical theorems, which made me think it works with Squeeze Theorem, as I just have to find two majorants.

That made me think of a problem I saw earlier this year, which was $$\lim \limits_{n \to \infty} \prod_{i = 2}^n (2 - \sqrt[i]e)$$ which was solvable by saying that $(1+ \frac{1}{n})^n < e < (1 + \frac{1}{n-1})^n (1) $, or something along those lines. I've tried raising it to the power $ \ln{2} $, but it is not solvable afterwards as we get something ugly. Can someone give me a hint or something? Thank you.

P.S.: I've thought of using Taylor Series as well, but I'm not sure how that would work

答:
First, notice that : $\sqrt[i]{2}=\exp\left(\frac{\ln(2)}{i}\right)=1+\frac{\ln(2)}{i}+o(\frac{1}{i})$ because $\frac{\ln(2)}{i}\to_{i\to\infty} 0$.

Therefore $2-\sqrt[i]{2}=1-\frac{\ln(2)}{i}+o(\frac{1}{i})$ and $\ln(2-\sqrt[i]{2})=\ln(1-\frac{\ln(2)}{i}+o(\frac{1}{i}))\sim_{i\to\infty} -\frac{\ln(2)}{i}$.

This implies $\sum_{i=2}^{n}\ln(2-\sqrt[i]{2}) \to_{n\to\infty}-\infty$ and passing to the exponential, $\prod_{i=2}^{n}(2-\sqrt[i]{2}) \to 0$.


问:
Are there any famous/notable mathematicians who have their own YouTube channel?
Are there any famous/notable mathematicians who have their own YouTube channel?

I found this amazing  video regarding the book.
The YouTube channel name is The Math Sorcerer,
but I don't know what the background of The Math Sorcerer is. From the analysis of the video I think he is a famous/notable mathematician.

I want to know more about YouTube channels of notable mathematicians.

答:
Field medalist and Abel Prize winner Timothy Gowers

You tube channel link


问:
Transitive Action of a Finite Group
If $G$ is a finite group acts transitively on a set $\mathbb{X}$. Does this imply that the set $\mathbb{X}$ is finite?

答:
If $G$ acts transitively on $X$, then, given $x\in X$, for every $y\in X$ there is some $g\in G$ such that $y=g\cdot x$. This is precisely the condition for a surjection $f_x\colon G\longrightarrow X$ to exist. Therefore the cardinality of $G$ is at least equal to the one of $X$.

(Incidentally, since by group action's axioms distinct $x$'s induce distinct $f_x$'s, there are $|X|$ of such surjections.)

Title: Transitive Action of a Finite Group


问:
Find the norm of operator $S:C[0,1]\rightarrow C[0,1]$, $(Sx)(z)=z\int_{0}^{1}x(t)dt$.
I'm suspecting that $||S||=1$. Beacause $$ ||S(x)||_{\infty}=\sup_{z\in [0,1]}\left|z\int_{0}^{1}x(t)dt\right| =\sup_{z\in [0,1]}|z|\cdot \left|\int_{0}^{1}x(t)dt\right|\leq ||x||_{\infty}.$$
Hence, $||S||\leq 1$. Now, if we show that $||S||\geq 1$, then $||S||=1$. So I imagine that it remains to show that $||S||\geq 1$.

答: You are right. Pick $ x(t) = 1$ $\forall t \in [0,1]$ and you get the complete result.


问:
What exactly is a Dimension of a Matrix?
As far as I know , Dimension is the number of elements in the basis of a matrix . Basis deals with linearly independent vectors. So for instance , if we have an nxn matrix and we reduce the matrix to it's row echelon form , the basis comprises of the linearly independent rows .  So as I understand it , dimension of a matrix ≤ order of the matrix. (Due to the fact that ,no. of linearly independent rows can never be more than the rows of the matrix itself)

Recently I came across this answer, which proposed that a dimension of a 2x2 matrix can be 4.

https://math.stackexchange.com/a/320688/708144

Can someone help me comprehend what a dimension actually is?

答:
A single matrix does not have any dimension, as this notion is reserved for vector spaces. In general a single matrix does not constitute a vector space.

The vector space of matrices of a fixed size $m\times n$ does have a dimension, namely $mn$, as it is generated under addition by the matrices with precisely one nonzero entry 1.

A matrix itself induces two vector spaces: its kernel and its image. The dimension of the image is determined by the number of linear independent columns of the matrix, which is called the rank of the matrix.


答:
There is no such thing as a basis for a matrix. A basis is a maximal linearly independent set in a vector space. For example the set
$$\{\pmatrix{1&0\\0&0},\pmatrix{0&1\\0&0},\pmatrix{0&0\\1&0},\pmatrix{0&0\\0&1}\}$$
in the space of $2\times 2$-matrices with real entries $\mathcal{M}_2(\mathbb{R})$, which is thus of dimension $4$. There is a rank of a matrix, which is defined as the dimension of the space generated by  its columns .

Title: What exactly is a Dimension of a Matrix?

Body:
As far as I know , Dimension is the number of elements in the basis of a matrix . Basis deals with linearly independent vectors. So for instance , if we have an nxn matrix and we reduce the matrix to it's row echelon form , the basis comprises of the linearly independent rows .  So as I understand it , dimension of a matrix ≤ order of the matrix. (Due to the fact that ,no. of linearly independent rows can never be more than the rows of the matrix itself)

Recently I came across this answer, which proposed that a dimension of a 2x2 matrix can be 4.

https://math.stackexchange.com/a/320688/708144

Can someone help me comprehend what a dimension actually is?


问:
Does this ODE question have closed form solution?
These days, I am struggling with following ODE problem when I build up my research model:

$1/2f''(x)+a(b - x) f'(x) -(c+ e^{A+Bx})f(x)=0$
where f(x) is a smooth function, and $a,b,c, A,B$ are all constants.
How to get the closed form of f(x)?

I tried the Laplace transform to work on it, say $F(s) = L(f(x)) $, but because of $e^{A+Bx}$, there will be a term $F(s-B)$ in the transformed equation. How to deal with this term?

I also tried the power series method, but got some very complicate coefficients, which stops me going further.

I think the term $e^{A+Bx}$ is the difficult part.

Could anyone here tell me how to deal with this kind of problem? Does the solution exit? I tried several ODE books but cannot find similar examples. Or could any one can suggest some relevant books?

Thank you very much.

Body:
Maple doesn't find a closed-form solution, so I suspect there isn't any (other than the trivial $f=0$).  Here are the first few terms of some series solutions: with $f(0)=1$, $f'(0)=0$,
$f(x) = 1+ \left( c+{{\rm e}^{A}} \right) {x}^{2}+ \left( -2/3\,abc-2/3\,ab{{\rm e}^{A}}+1/3\,{{\rm e}^{A}}B \right) {x}
^{3}$

$~~~~~~~~~~~+(~1/6\,{c}^{2}+1/3\,c{{\rm e}^{A}}+1/6\, \left( {{\rm e}^{A
}} \right) ^{2}+1/3\,{a}^{2}{b}^{2}c+1/3\,{a}^{2}{b}^{2}{{\rm e}^{A}}-
1/6\,ab{{\rm e}^{A}}B+1/3\,ac+1/3\,a{{\rm e}^{A}}$

$~~~~~~~~~~~+1/12\,{{\rm e}^{A}}{
B}^{2}~) {x}^{4}+ ( 2/15\,{{\rm e}^{A}}Bc+2/15\, ( {
{\rm e}^{A}} ) ^{2}B+{\frac {1}{60}}\,{{\rm e}^{A}}{B}^{3}-2/15
\,ab{c}^{2}-{\frac {4}{15}}\,cab{{\rm e}^{A}}$

$~~~~~~~~~~~-2/15\,ab ( {{\rm e}
^{A}} ) ^{2}-2/15\,{a}^{3}{b}^{3}c-2/15\,{a}^{3}{b}^{3}{{\rm e}^
{A}}+1/15\,{a}^{2}{b}^{2}{{\rm e}^{A}}B-1/3\,{a}^{2}bc-1/3\,{a}^{2}b{
{\rm e}^{A}}$

$~~~~~~~~~~~-1/30\,ab{{\rm e}^{A}}{B}^{2}+1/10\,a{{\rm e}^{A}}B
 ) {x}^{5}+O( {x}^{6}) $

With $f(0)=0$, $f'(0)=1$:

$f(x) = x-ab{x}^{2}+ \left( 2/3\,{a}^{2}{b}^{2}+1/3\,a+1/3\,c+1/3\,{{\rm e}^{A}} \right) {x}^{3}$

$+\left( -1/3\,abc-1/3\,ab{
{\rm e}^{A}}-1/3\,{a}^{3}{b}^{3}-1/2\,{a}^{2}b+1/6\,{{\rm e}^{A}}B
 \right) {x}^{4}$

$+(~1/5\,{a}^{2}{b}^{2}c+1/5\,{a}^{2}{b}^{2}{
{\rm e}^{A}}+2/15\,{a}^{4}{b}^{4}+2/5\,{a}^{3}{b}^{2}-1/6\,ab{{\rm e}^
{A}}B+1/10\,{a}^{2}+2/15\,ac+2/15\,a{{\rm e}^{A}}$

$~~~~+1/30\,{c}^{2}+1/15\,
c{{\rm e}^{A}}+1/30\, ({{\rm e}^{A}}) ^{2}+1/20\,{{\rm e}
^{A}}{B}^{2}) {x}^{5}+O(x^{6})~) $


问:
How to find the inverse function for $y = x^5 +2x + 1$
I'm using a self study pre-calculus book and have to invert function $y = x^5 + 2x + 1$. I know I have to swap the variables and put everything on one side of the equation such as $y^5 + 2y -x + 1 = 0$ but from here I don't know what to do.  All of the examples I can find on the internet either use calculus which I don't know or show examples with the degree of $2$ rather than $5$ and so they use the quadratic equation to solve for $y$.  So I am stuck.

答:
I'm thinking you could solve this by making use of the geometric interpretation of the inverse of a function, namely that its graph is symmetric with respect to a line at $45^{\circ}$.

In principle, I think you could compute this symmetric function by rotating each point such that it lands at the corresponding position. Take the image below as an example. Given the graph of the $y=x^5+2x+1$ function, you can take a random point and notice that you could rotate it clockwise by $2\alpha$ to get to the point corresponding to the inverse graph. This could be done, at least I think it could, at a pre-calculus level, but you kind of have to derive the formula for the rotation and then generalize it for all the points, except the one where the graph intersects the line at $45^{\circ}$.

One way of doing it, but not at a pre-calculus level is by using the rotation matrix

$$R(-2\alpha) = \left(\matrix{\cos(2\alpha) & \sin(2\alpha) \\ -\sin(2\alpha) & \cos(2\alpha)}\right)$$
which can be applied as follows

$$\left(\matrix{\cos(2\alpha) & \sin(2\alpha) \\ -\sin(2\alpha) & \cos(2\alpha)}\right) \cdot \left(\matrix{x \\ x^5+2x+1}\right)=\left(\matrix{x' \\ y'}\right)$$
with $\alpha = \pi/4-\arctan \left(\frac{x}{x^5+2x+1}\right)$.

If I did not make some errors in my reasoning, this should at least give you the coordinates for the inverse and then you could in principle consider the job done.

However, I repeat, I am not sure you are supposed to have to derive the formula for the rotation, even more so to use it as I have done above using a vector-matrix approach. Still, one could argue that it is just geometry and some trigonometry so ... pre-calculus?


问:
Prove that R and P(R) does not have the same cardinality.
How can I show such a thing in an elegant, valid way? I know it shouldn't be hard. Well it does have to be plausible and mathematically logic, but I guess I am not expected to rediscover the great, massive proof founded by Cantor. If I am to use what he says, what's the point then? I mean, how can I prove this elegantly as an "Introduction to Set Theory" mere student? I'd appreciate your help.</p>

答: If $R$ = $∅$, $℘$($R$) = { {$∅$} }. Suppose that $R$ ≠ ∅. Then, there is a one-to-one mapping  $ƒ$, $R$ onto ℘($R$). We now consider the set of elements of $R$ which are not mapped to the subset of R under $ƒ$. Let $B$ = {$r$ $∈$ $R$ ; $r$ $∉$ $ƒ$($r$)}. We will try to arrive at some contradiction. Suppose $x$ $∈$ $B$. $⇒$ $x$ $∈$ {$r$ $∈$ $R$ ; $r$ $∉$ $ƒ$($r$)} $⇒$ $x$ $∉$ $ƒ$($x$) $=$ $B$. Next, suppose $x$ $∉$ $B$. Then $x$ $∉$ $ƒ$($x$) and so $x$ $∈$ {$r$ $∈$ $R$ ; $r$ $∉$ $ƒ$($r$)} $=$ $B$, a contradiction.


问:
Minimum ticket price to break even
A football team plays in a stadium that has a seating capacity of $30,000$ spectators. With the ticket price set at $\$79$, average attendance has been $21,692$. A market survey indicates that for each $\$4$ the ticket price is lowered, the average attendance increases by $1000$. Assume that the stadium has a fixed cost of $50,000$ and a variable cost of $2$ per ticket sold.

a) Determine the ticket price(s) that the football stadium breaks even.

b) Algebraically determine the price that maximizes profit for the football stadium.

c) What does the ticket price have to be to ensure that the profit is $\$1,000,000$?

I'm having trouble creating equations for this word problem. I'm not sure where I would start to make solve the 3 questions.

答:
Let the ticket price be $p$ and the number of people be $n$. $n=21692-250p$, so the profit is $p(21692-250p)-50000$ where the $50000$ is the stadium price. Expanding gives $-250^2+21692p-50000$

a) This is just asking to solve $-250p^2+21692p-50000=0$. Using the quadratic formula you get $p=\frac{21692\pm\sqrt{420542864}}{500}$. Rounding to the nearest cent gives $p=\$2.37,\$84.40$.

b) The min/max of a quadratic equation is $-\frac{b}{2a}$, so it is $p=\frac{5423}{125}$. Rounding to the nearest cent gives $p=\$43.38$.

c) The maximum is at $p=\frac{5423}{125}$, so plugging it in gives $\frac{52567858}{125}$, or $\$420542.86$ rounded to the nearest cent. This is less than $\$1000000$, so the answer is never.

Title: Minimum ticket price to break even


答:
When $m\equiv (1(\bmod 4)\bmod 2^n$, it means we can write:

$$m=k2^n+ 4t+1$$

k can be $k=2^n$ ant t can be  $t=\frac{\sum {n\choose i}4^i\cdot 1^{n-i}}4$such that:

$$m=(2^n)^2 + \sum {n\choose i}4^i\cdot 1^{n-i}+1=(4+1)^n=5^n$$

This shows the condition, that is why numbers such as 17 or 33 . . . can not be condidates.

Title: What does this mean: Every element $\pmod{2^n}$ that is $1 \pmod 4$ is a power of $5$?

Body:
For example, consider $17$. Then $17 \pmod{2^n}$ for any $n$ is always $\equiv 1 \pmod 4$, but $17$ isn't a power of $5$.

Just in case I have misunderstood anything, please see this video where the claim is made.


问:
Making a matrix positive semidefinite
I have a symmetric matrix $A$. Is there a way to have a positive semidefinite matrix $B$, so that solving a problem $X$ with $A$ or with $B$ is similar?

I thought about doing $\alpha \cdot I + A$, so that alpha is big enough. Does it work in all cases? ie find alpha such that $\alpha \cdot I + A$ is positive semidefinite?

Thank you in advance

答: If your matrix is real, then there is a basis in which it is diagonal. Just use $\alpha>\min_{\lambda\in\text{Sp}(A)}\lambda$ where $\text{Sp}$ denotes the spectrum of $A$ and you will have a semi definite positive matrix. However, this does not mean that the solution of your problem with this new matrix if "close" to the real solution. It depends on what is your problem at hand, and what you mean by "close" (a metric could be used to quantify this).

Title: Making a matrix positive semidefinite


问:
Question about increasing sequences

We've started going over sequences and I'm not sure how to prove that the sequence converges in the extended sense.. I thought about splitting it into cases, one in which the sequence is bounded and one where it's not but without any luck so far. :(

答:
If the sequence is bounded above by $M \in \Bbb R$, then $s:=\sup\{a_n\mid n=1,2,3\ldots\}$ exists as a real number. Then show that $\lim_{n \to \infty} a_n = s$, using the definition of supremum and the increasingness.

If the sequence is unbounded, for every $M\in \Bbb R$, there must be some $n$ such that $a_n > M$ but then the tail of the sequence (from this $n$ onwards) is in $(M, +\infty]$ by increasingness, so $\lim_{n \to \infty} a_n = +\infty$ by definition.

Title: Question about increasing sequences


问:
finding the second partial derivative of a function that depends on y and a function of y
I am interested in finding the second partial derivative of a function that depends on $y$ and a function of $y$
$$\frac{\partial^2f(x(y), y)}{\partial x \ \partial y}$$

My approach (see this question) is to solve this directly using the chain rule for partial derivatives, resulting in
$$\require{cancel} \frac{\partial}{\partial x }(\frac{\partial f}{\partial x} \frac{\partial x}{\partial y} + \frac{\partial f}{\partial y} \xcancel{\frac{\partial y}{\partial y}}) = \frac{\partial}{\partial x }(\frac{\partial f}{\partial x} \frac{\partial x}{\partial y}) + \frac{\partial^2 f}{\partial x \partial y}$$

However this seems to be incorrect as it just results in the original expression plus some potentially non zero term.

My question is how to calculate second order partial derivatives w.r.t. y and a function of y?

答: The map $g : y \mapsto f(x(y),y)$ is only having one independent variable. Namely $y$. Hence  $$\frac{\partial^2f(x(y), y)}{\partial x \ \partial y}$$ doesn't have any meaning.


问:
Relative estimation of Bernoulli parameter -- reference request
Suppose I can make i.i.d. samples from the Bernoulli distribution with bias $p$ and want to find approximation $\hat{p} \in [(1-\epsilon)p,(1+\epsilon)p]$ with constant probability. I suppose that this should be possible to do in $O(\frac{1}{\epsilon^2 p})$ samples. Is this so? If so, are there any papers about this? I have found many papers which looked at maximum likelihood estimates or a Bayesian setting. I am interested in a reference, rather than in a solution as it must exist and I want to reference it in a paper.

答:
I was finally able to find a reference, though it took me way too long. This paper [1] shows an algorithm that does exactly this, see Theorem 5 in that paper. The idea does not come from this paper, see the references they give.

[1] Sequential sampling techniques for algorithmic learning theory, Osamu Watanabe, https://core.ac.uk/download/pdf/82665979.pdf


问:
What are some open research problems in Stochastic Processes?
I was wondering, what are some of the open problems in the domain of Stochastic Processes. By Stochastic Processes.

Any examples or recent papers or similar would be appreciated.

The motivation for this question is that I was studying stochastics from a higher level (i mean, brownian motion and martingales and stuff; beyond the undergrad markov chains and memoryless properties) and was wondering what are the questions that still lie unanswered in this field?

答:
Various academics have lists on their website

Richard Weber, University of Cambridge (operations research)

David Aldous, University of California, Berkeley with updates from Thomas M. Liggett

Krzysztof Burdzy, University of Washington

Hermann Thorisson, University of Iceland

Other resources that might be of interest

The journal Queueing Systems published a Special Issue on Open Problems in 2011. James Cruise maintains an ungated copy of the papers along with research progress on the highlighted problems.

John Kingman published a 2009 paper titled The first Erlang century—and the next where open problems and future reserch were discussed

Jewgeni H. Dshalalow edited a book in 1985 titled Advances in queueing: theory, methods, and open problems

Lyons, Russell, Robin Pemantle, and Yuval Peres. Unsolved problems concerning random walks on trees Classical and modern branching processes. Springer New York, 1997. 223-237.

Questions tagged open problem and probability on mathoverflow

Chapter 23 Open Problems from Levin, Peres, and Wilmer's book Markov Chains and Mixing Times


问:
Does Cross Product Makes Rank-1 Tensor??
The problem itself is very simple. I have to prove that the cross product of two vectors, $C=A \times B$, satisfies vector transformation and thus rank-1 tensor. In other words, I have to prove that

$C'_i = \lambda_{ij} C_j$

where

$C_j=\varepsilon_{jmn}A_mB_n$

and

$C_i'=\varepsilon_{ikl}A_k'B_l'=\varepsilon_{ikl}(\lambda_{km}A_m)(\lambda_{ln}B_n)$

I can clearly see that this statement is true when I draw the diagram of each vector $C, C', A, B, A', B'$, but failed to prove it with algebra. I've tried putting 1, 2, 3 into each index, and failed with miserably long lines. Also tried to alter $i, k, l$ into $j,m,n$ as follows. I think this one is the most probable trial.

$\varepsilon_{ikl}\lambda_{km}\lambda_{ln}=
\varepsilon_{jmn}\lambda_{mm}\lambda_{nn}+\varepsilon_{jnm}\lambda_{nm}\lambda_{mn}+
\varepsilon_{mjn}\lambda_{jm}\lambda_{nn}+\varepsilon_{mnj}\lambda_{nm}\lambda_{jn}+
\varepsilon_{nmj}\lambda_{mm}\lambda_{jn}+\varepsilon_{njm}\lambda_{jm}\lambda_{mn}
$

which simplifies to

$\varepsilon_{ikl}\lambda_{km}\lambda_{ln}=\varepsilon_{jmn}
(\lambda_{mm}\lambda_{nn}-\lambda_{nm}\lambda_{mn}
-\lambda_{jm}\lambda_{nn}+\lambda_{nm}\lambda_{jn}
-\lambda_{mm}\lambda_{jn}+\lambda_{jm}\lambda_{mn})$

I expected this to be something like $\varepsilon_{jmn}(\lambda_{jj}+\lambda_{mj}+\lambda_{nj})$so that

$C_i'=\varepsilon_{ikl}A_k'B_l'=\varepsilon_{jmn}(\lambda_{jj}+\lambda_{mj}+\lambda_{nj})A_mB_n=
\varepsilon_{jmn}\lambda_{ij}A_mB_n$ But I always fail to convert $\lambda_{km}\lambda_{ln}$ into $\lambda_{ij}$.

Can somebody help me?

答:
In a coordinate free formulation, you have the cross product defined as
$$
\forall v\in\Bbb R^3:\langle a\times b,v\rangle=\det(a,b,v)
$$
Applying a coordinate transformation $\Lambda$ then has the effect that
$$
⟨Λa×Λb,Λv⟩=\det(Λa,Λb,Λv)=\det(Λ)·\det(a,b,v)=\det(Λ)·⟨a×b,v⟩
$$
Consequently
$$
\det(Λ)(a×b)=Λ^T(Λa×Λb)
$$
This only simplifies to the usual formula if $Λ$ is a special orthogonal transformation, $\det(Λ)=1$ and $Λ^T=Λ^{-1}$. So if you want to reconstruct that with the explicit coordinates, you have to use these relations somewhere.

Note that $\det(Λ)Λ^{-T}$ is the co-factor matrix of $Λ$.


问:
Bound of functional $g(0) - 7g(\frac12)$
Let's consider functional

$$f\colon (C[0,1], \Vert\cdot \Vert_\infty) \ni g\rightarrow g(0) - 7g(\frac12)$$

I'm trying to judge whether functional $f$ is bounded or not, however I'm dealing with problems of interpretation of condition to consider.

We want to check whether there is some ball that $f$ will be inside it.

But I'm not sure how exactly this ball will look in this space

In my opinion it should look like following:

$$B(C[0,1], r) = \{h \in C[0,1] : \lVert f-h \rVert < r\}$$

But I'm not sure if it's true. Could you please tell me if I'm thinking correctly, or maybe this ball looks differently in this space?

答:
For $g \in V = (\mathcal C([0,1], \mathbb R)$ you have

$$\vert f(g) \vert = \left\vert g(0) - 7 g\left(\frac{1}{2}\right) \right\vert \le \Vert g \Vert_\infty + 7 \Vert g \Vert_\infty = 8 \Vert g \Vert_\infty.$$

This proves that $f$ is bounded: for $\Vert g \Vert_\infty \le 1$, you have $\vert f(g) \vert \le 8$. As $f$ is linear, you immediately get for $g,h \in V$

$$\vert f(h) - f(g) \vert \le 8 \Vert h - g \Vert_\infty.$$

Even more, the norm of $f$ is equal to $8$ as for the map $h(x)=\vert 4x - 2 \vert -1$ we have $\Vert h \Vert_\infty = 1$ and $f(h)=8$.


答:
Let $g \in C([0,1])$ be arbitrary and let $\epsilon>0$ be given. Then we have:
$$
|f(g)-f(h)|=|g(0)-7g(0.5)+h(0)-7h(0.5)| \leq |g(0)-h(0)|+|7g(0.5)-7h(0.5)| \leq \\
||g-h||_{\infty}+7||g-h||_{\infty}=8||g-h||_{\infty}<^! \epsilon
$$
So if we choose $||g-h||_{\infty} < \frac{\epsilon}{8}=\delta$, we have checked the epsilon-delta definition of continuity using open balls. You then have
$$
f(B_{\delta}(g)) \subset B_{\epsilon}(f(g))
$$
A short note on your notation for balls:
$$
B_r(f)=\{g \in C([0,1]) | \; ||f-g||_{\infty}<r   \}
$$
You can also use $B_r(f)=B(f,r)$ as a notation. You usually just mention the centre and the radius, not the space you are taking this ball in. However, you can or even have to denote it somewhere in your text/question/whatever.

Title: Bound of functional $g(0) - 7g(\frac12)$

Body:
Let's consider functional

$$f\colon (C[0,1], \Vert\cdot \Vert_\infty) \ni g\rightarrow g(0) - 7g(\frac12)$$

I'm trying to judge whether functional $f$ is bounded or not, however I'm dealing with problems of interpretation of condition to consider.

We want to check whether there is some ball that $f$ will be inside it.

But I'm not sure how exactly this ball will look in this space

In my opinion it should look like following:

$$B(C[0,1], r) = \{h \in C[0,1] : \lVert f-h \rVert < r\}$$

But I'm not sure if it's true. Could you please tell me if I'm thinking correctly, or maybe this ball looks differently in this space?


问:
Zero set of ideal of an algebraic set is the algebraic set itself
An algebraic set defined by $S \subseteq k[x_1,...,x_n]$ is given by

$$X = \{a \in \mathbb{A}^n(k) | f(a) = 0, \forall f\in S\}$$

So this would the set of points that make polynomials in $S$ vanish.

The ideal of a set $X \subset \mathbb{A}^n(k)$ is given by:

$$I(X) = \{ f \in k[x_1,...,x_n] | f(a) = 0, \forall a \in X \}$$

So this is the set of polynomials that vanish on points in $X$.

Lastly, the zero set of $I(X)$ is given by:

$$V(I(X)) = \{a \in \mathbb{A}^n(k) | f(a) = 0, \forall f \in I(X)\}$$

which is the set of points that make polynomials in $I(X)$ vanish.

Now I am trying to prove that if $X$ is an algebraic set then $V(I(X)) = X$, which is an exercise that I have taken from a book I'm reading ("Algbraic Geometry: A Problem Solving Approach"). My intuition tells me that this would be the case if I could prove that $S = I(X)$. I can see that $S \subseteq I(X)$ since if a polynomial $f$ is in $S$ then of course we have that $f(a) = 0, \forall a \in X$. Then it would naturally follow that $X = V(S) = V(I(X))$, however I am struggling to see why $I(X) \subseteq S$ since $I(X)$ might contain a polynomial $g$ s.t. $g(a) = 0$ but $g \notin S$.

答:
Note that for every ideal $J\subseteq k[x_1,...,x_n]$ we have $J\subseteq I(V(J))$, and for every subset $X\subseteq \mathbb{A^n}$ we have $X\subseteq V(I(X))$. This is obvious from the definitions.

Now, suppose $J\subseteq k[x_1,...,x_n]$ is an ideal. We'll prove that $V(J)=V(I(V(J)))$. First of all, we already know that $J\subseteq I(V(J))$, and hence $V(I(V(J)))\subseteq V(J)$, as taking $V$ inverts the order. For the reverse inclusion, let $X=V(J)$. Then:

$V(J)=X\subseteq V(I(X))=V(I(V(J)))$

So indeed we have $V(J)=V(I(V(J)))$, for every ideal $J\subseteq k[x_1,...,x_n]$.

Finally, in your case we have $X=V(S)=V(J)$ where $J$ is the ideal generated by $S$. Thus:

$X=V(J)=V(I(V(J)))=V(I(X))$


问:
Factorising a spacial polynomial
I have to factorise $2(x^2 + x^3 + x^4 + x^5 + x^6 + x^7)$. It seems that there a very simple expression to which simply this polynomial.

The only thing I found so far is $2 x^2 (1 + x + x^2 + x^3 + x^4 + x^5)$. Can you explain to me in  details how to factorise this even more? If so, is there a general rule for such similar polynomial?

答:
$x^6 -1 = (x-1)(1+x+x^2+x^3+x^4+x^5)$

so all complex $6$-th roots of unity except $1$ are roots of $(1+x+x^2+x^3+x^4+x^5)$, group them by conjugate pair and you have a factorisation over $\Bbb R$.

Title: Factorising a spacial polynomial


Title: Factorising a spacial polynomial

Body:
I have to factorise $2(x^2 + x^3 + x^4 + x^5 + x^6 + x^7)$. It seems that there a very simple expression to which simply this polynomial.

The only thing I found so far is $2 x^2 (1 + x + x^2 + x^3 + x^4 + x^5)$. Can you explain to me in  details how to factorise this even more? If so, is there a general rule for such similar polynomial?

Body:
$ 2x^2(1+x+x^2+x^3+x^4+x^5) $

$ 2x^2(1(1+x)+x^2(1+x)+x^4(1+x)) $

$ 2x^2(1+x^2+x^4)(1+x) $

$ 2x^2(x^2+x+1)(x^2-x+1)(1+x) $

Hope this help you

Edit: Here's how $ (1+x^2+x^4) $ can be factorized to $ (x^2+x+1)(x^2-x+1) $

$ x^4+x^2+1 $

$ =(x^2)^2+2(x^2)(1)+1^2−x^2$

$ =(x^2+1)^2−x^2$

$ =(x^2+1−x)(x^2+1+x)$