问:
Blackjack card counting, with one whole deck should the "count" end on zero?
When playing blackjack if you are card counting for a single deck, should the count always come to zero at the end of the deck? Wouldn't it depend on strategy and a corresponding betting table?

Would the optimum card counting strategy end you with a count of zero for one deck?

答:
It depends on the system of card counting.

With the Hi-Lo count (the most common), the count should sum to $0$ after every single card in the deck has been played.

```<code>2 3 4 5 6    = +1 (low cards)
7 8 9        =  0 (neutral cards)
10 J Q K A   = -1 (high cards)
</code>```

Using this chart, it is easy to see that there is an equal number of $+1$ cards and $-1$ cards.  Over the course of the deck, there will be a total of $20$ low cards and $20$ high cards (along with $12$ neutral cards).  This means that the count at the end must equal zero.

Several  of the more advanced systems, however, will not have a balance between positives and negatives, and they will then have a non-zero count at the end of the deck.

Body:
It depends on the system of card counting.

With the Hi-Lo count (the most common), the count should sum to $0$ after every single card in the deck has been played.

2 3 4 5 6    = +1 (low cards)
7 8 9        =  0 (neutral cards)
10 J Q K A   = -1 (high cards)

Using this chart, it is easy to see that there is an equal number of $+1$ cards and $-1$ cards.  Over the course of the deck, there will be a total of $20$ low cards and $20$ high cards (along with $12$ neutral cards).  This means that the count at the end must equal zero.

Several  of the more advanced systems, however, will not have a balance between positives and negatives, and they will then have a non-zero count at the end of the deck.


问:
Proof of the projection formula $(\pi^*D)\cdot D'=D\cdot (\pi_*D')$
I'm trying to prove the following (simplified) version of the projection formula

Let $S',S$ be non-singular algebraic surfaces and $\pi:S'\to S$ a birrational map between them. If $D\in \text{Div}(S)$ and $D'\in \text{Div}(S')$, then:
$$(\pi^*D)\cdot D'=D\cdot(\pi_*D')$$

By linearity, we may just consider $D, D'$ prime divisors.

Let $X\subset S'$ and $Y\subset S$ closed sets such that $\pi:S'\setminus X\to S\setminus Y$ is an isomorphism.

The claim is obviously true if $D\subset S\setminus Y$ and $D'\subset S'\setminus X$.

Now if $D'\cap X\neq \emptyset$ or $D\cap Y\neq\emptyset$, I don't know what to do.

In the particular case $\pi:S'\to S$ is the blow-up at the point $P\in S$ with $E:=\pi^{-1}(P)$, I was able to deal with some specific divsors. For example, when $D'=E$ and $D=C$ where $C$ is a curve through $P$. If $m$ is the multiplicity of $P$ in $C$, we can use the formula $\pi^*(C)=\widetilde{C}+mE$.

The result for this specific case was not so evident (in my opinion) and I still can't see how the general case should go.

答:
Nine months later, I'll post the answer I was pregnant with.

Since a birrational morphism between smooth surfaces is a composition of blowups and isomorphisms, it is enough to consider $\pi$ as a single blowup.

By linearity, it is enough to consider $D,D'$ prime divisors. We have two cases:

(1) $D'=E$. Let $\widetilde{C}$ be the stric transform of $C$. We have the formula $\pi^*C=\widetilde{C}+mE$, where $m$ is the multiplicity of $P$ in $C$. Therefore $(\pi^*C)\cdot E=\widetilde{C}\cdot E-m=0$. But since $\pi_*E=0$, we get $C\cdot (\pi_*E)=0$ and we're done.

(2) $D'\neq E$. In this case $\pi_*D'=\pi(D')$ and $D'$ is the strict transform of $\pi(D')$, therefore $\pi^*(\pi_*D)=D'+mE$. By the previous case, $(\pi^*C)\cdot E=0$, so:
\begin{align*}
(\pi^*C)\cdot D'&=(\pi^*C)\cdot(\pi^*(\pi_*D)-mE)\\
&=(\pi^*C)\cdot (\pi^*(\pi_*D))\\
&=C\cdot (\pi_*D).
\end{align*}

(Here we've used the fact that $(\pi^*D_1)\cdot (\pi^*D_2)=D_1\cdot D_2$ for any divisors $D_1,D_2$ in $S$). $_\blacksquare$

ContentLicense: CC BY-SA 4.0


问:
Can we have a set that is hereditarily equinumerous to itself, of any size?
Is it consistent to add to ZFC-Reg. the existence of a nonempty set $\mathcal H_\mathcal H$ that is hereditarily equinumerous to itself?

If that is consistent, then is it consistent that we can have $\mathcal H_\mathcal H$ being of any nonempty cardinality? Formally that is:

$\forall x (x \neq \emptyset \implies \exists y \forall z [z \in TC(\{y\}) \Rightarrow z \sim x])$

Where $``TC(x)"$ stands for "transitive closure of $x$" defined in the usual manner, and $`` z \sim x"$ stands for existence of a bijection between $z$ and $x$.

答:
Yes, and the proof is nearly identical to the argument in this answer: just start with an $M_0$ that consists of a set which is hereditarily of some size, and then formally construct a cumulative hierarchy on it.  Or, if you want to simultaneously get such hereditarily equinumerous sets of all cardinalities, take $M_0$ to be a class-sized structure which consists of such sets for all cardinalities.

More generally, that construction shows that given any structure $M_0$ in the language of set theory which satisfies extensionality, there is a class model of ZFC without regularity which contains a transitive set that is isomorphic to $M_0$.  This works even if $M_0$ is a proper class (take a direct limit of the construction over all set-sized substructures of $M_0$), as long as each element of $M_0$ has only a set of elements (this assumption is needed for Power Set to hold).


问:
Am I using this particular definition of a free variable correctly to identify a free variable?

The definition:

We use this definition for a free variable.

Suppose that $v$ is a variable and $\phi$ is a formula. We will say that $v$ is free in $\phi$ if:

$\phi$ is atomic and $v$ occurs in (is a symbol in) $\phi$, or

$\phi := (\lnot \alpha)$ and $v$ is free in $\alpha$, or

$\phi :=$ $(\alpha \lor \beta)$ and $v$ is free in at least one of $\beta$ or $\alpha$, or

$\phi :=$ $(\forall u)(\alpha)$ and $v$ is not $u$ and $v$ is free in $\alpha$.

The formula:

Now, the task is to explain, using the definition above, how $v_2$ is a free variable in the formula below:
$$\phi = (\forall v_1)(\lnot(\forall v_5) (v_2 = v_1 + v_5))$$

The answer:

We say that $v_2$ is a free variable in the $\phi$ above since first, our $\phi$ matches clause $(4)$ of the definition, where $(\forall u) = (\forall v_1)$ and $(\alpha) = (\lnot(\forall v_5) (v_2 = v_1 + v_5))$, which tells us that $v_1$ is not a free variable in this expression. We then take to looking for a free variable in $(\alpha) = (\lnot(\forall v_5) (v_2 = v_1 + v_5))$ as per the definition.

So, we define $\phi$ recursively by replacing the value of $\alpha$:

$$\phi = \lnot(\forall v_5) (v_2 = v_1 + v_5)$$

Further, we try to find a free variable in $\phi= \lnot(\forall v_5) (v_2 = v_1 + v_5)$, and realize that it matches clause $(2)$ of the definition provided to us such that $\phi = \lnot(\forall v_5) (v_2 = v_1 + v_5)$, where $\alpha = (\forall v_5) (v_2 = v_1 + v_5)$. We then take to looking for a free variable in $\alpha = (\forall v_5) (v_2 = v_1 + v_5)$ as per the definition.

So, we define $\phi$ recursively by replacing the value of $\alpha$:

$$\phi = (\forall v_5) (v_2 = v_1 + v_5)$$

Finally, when we have $\phi = (\forall v_5) (v_2 = v_1 + v_5)$, it recursively matches clause $(4)$ of the definition such that $(\forall u) (\alpha) = (\forall v_5) (v_2 = v_1 + v_5)$, which tells us that per the definition, $v_5$ not the free variable, and that it must be in $(v_2 = v_1 + v_5)$. We then take to looking for a free variable in $(v_2 = v_1 + v_5)$ as per the definition. So, we define $\phi$ recursively by replacing the value of $\alpha$:

$$\phi = (v_2 = v_1 + v_5)$$

Conclusively, we identify that within $(v_2 = v_1 + v_5)$ as $\phi$ and hitting clause $(1)$ of our definition, it is $v_2$ that occurs in $\phi$, and since $\phi$ as $(v_2 = v_1 + v_5)$ is atomic, it must be that $v_2$ is the free variable in $\phi$.

答:
Why not answer the question directly? It only asks whether $v_2$ is free, it doesn't ask you to identify all the free variables.

I would write no more than this:

$v_2$ is free in
$$ v_2 = v_1 + v_5$$
by clause (1).

Hence $v_2$ is free in
$$(\forall v_5) (v_2 = v_1 + v_5)$$
by clause (4).

Hence $v_2$ is free in
$$\phi = (\lnot(\forall v_5) (v_2 = v_1 + v_5))$$
by clause (2).

Hence $v_2$ is free in
$$\phi = (\forall v_1)((\lnot(\forall v_5) (v_2 = v_1 + v_5)))$$
by clause (4).

[I have repaired the bracketing to match the bracketing used by the rules.]


问:
A tricky probability question.
I have been asked the following question, and unfortunately I have no idea how to proceed. Here is the question:

Suppose we have 99 empty papers and we wrote numbers from 1 to 99(using each number) on one side of the papers randomly. Then we mixed all the papers randomly and started to write numbers from 1 to 99 on the other sides (empty sides) of each paper randomly. What is the probability of having the same number on both sides of at least one paper?

Thanks!

答:
Hint: Derangements.

Hint: Complement.

ContentLicense: CC BY-SA 3.0

Title: A tricky probability question.

Body:
Hint: Derangements.

Hint: Complement.


问:
Calculating $P\{\max(-x_{(1)},x_{(n)})\leq x \}$ where $X_{i}$ are i.i.d. from uniform distribution
Let $X_{i}$ be i.i.d. from uniform distribution $[-\theta,2\theta]$. I am interested in if the following is true:

$P\{\max(-x_{(1)},x_{(n)})\leq x \}= P\{-x_{(1)}\leq x , x_{(n)}\leq x \}\\ = P\{-x\leq x_{(1)} , x_{(n)}\leq x \}\\= P\{-x\leq x_{1} \leq x, ... ,-x\leq x_{(n)}\leq x \}\\= P\{-x\leq x_{1} \leq x \} ... P\{-x\leq x_{n} \leq x \}\\=(\int_{-x}^{x}\frac{1}{3\theta})^{n} $

答:
The last line is not quite correct since $x$ is not known. You will have to make cases:

$$F_{\max\{-x_{(1)},x_{(n)}\}}(x)=\begin{cases}\int_{-x}^x\frac{dy}{3\theta},&0< x\le\theta\\\int_{-\theta}^x\frac{dy}{3\theta},&\theta< x\le2\theta\\1,&2\theta< x\\0,&\text{otherwise}\end{cases}$$


问:
Dynamic system show that it has only one equilibrium point
I have the following dynamic system $A\cdot\ddot{x}+B\cdot\dot{x}+ C\cdot x=0 $ and A,B,C are positive definite matrices.
I have to show that it has only one equilibrium point and that this point is asymptotically stable. Any help will be appreciated!

UPDATE
Following the comment about treating A,B,C as scalars
$\dot{x_1}=x_2 = f_1(x_2)\\
\dot{x_2}=(-B\cdot x_2-C\cdot x_1)/A = f_2(x_1,x_2)$

$\longrightarrow \dot{x}=A(x)|_{x=0} \cdot x \\ $

$ A(x)=\begin{bmatrix}
\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\
\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \\
\end{bmatrix}=\begin{bmatrix}
0 & 1\\
-C/A&-B/A\\
\end{bmatrix}$
$det(A-\lambda\cdot I)= 0 \longrightarrow det(\begin{bmatrix}
-\lambda & 1\\
-C/A& -\lambda -B/A\\
\end{bmatrix}) =0 \longrightarrow \lambda^2+\lambda B/A+C/A=0$\

But I don't seem to get any idea on how to generalize it.

答:
I would start by calling $x=x_1$ and $\dot{x}=x_2$. Then you can rewrite the ODEs as:
\begin{equation}
\begin{matrix}
\dot{x_1}&=&&x_2\\
A\dot{x_2}&=&-Cx_1&-Bx_2
\end{matrix}
\end{equation}

or even:
\begin{equation}
\dot{\left[
\begin{matrix}x_1\\
x_2
\end{matrix}
\right]}
=
\left[
\begin{matrix}
0&I\\
-A^{-1}C&-A^{-1}B
\end{matrix}\right]
\left[\begin{matrix}
x_1\\
x_2
\end{matrix}\right]
\end{equation}
Then, the equlibria will be when $\dot{\left[\begin{matrix}x_1\\x_2\end{matrix}\right]}=\mathbf{0}$. That means, $0=x_2=\dot{x}$ and $0=-A^{-1}Cx_1=-A^{-1}Cx$ then you can apply that $A$ and $C$ are positive definite to find a value for $x$.

Now, to prove its stability we need to inspect the eigenvalues of $\left[\begin{matrix}
0&I\\
-A^{-1}C&-A^{-1}B
\end{matrix}\right]$.

$\lambda=\mathrm{zeros}(\left|\begin{matrix}
sI&-I\\
A^{-1}C&sI+A^{-1}B
\end{matrix}\right|)=\mathrm{zeros}(s^2I+A^{-1}Bs+A^{-1}C)$, using the fact that $A$, $B$ and $C$ are positive definite and using from Routh-Hurwitz criterion:

The second-degree polynomial, $P(s)=s^{2}+a_{1}s+a_{0}$ has both roots in the open left half plane (and the system with characteristic equation if and only if both coefficients satisfy $a_{i}>0$.

we can finally tell if its stable or not.


问:
How do I find the subspace of $P_{2}$ from the following subsets?
One of the following is a subspace of $P_{2}$.

A. $W= \{ax^2+bx+c\mid a+b+c=1\}$

B. $W= \{ax^2+bx+c\mid a+b+c=0\}$

C. $W= \{ax^2+bx+c\mid a \geq 0\}$

D. $W= \{ax^2+bx+c \mid b,c \geq 0\}$

So it's either $B, C$, or $D$ because the zero vector condition won't work on $A$. What I don't understand is the difference between $C$ and $D$. How does having 'a' greater than or equal zero differ from having '$b$' and '$c$' greater than or equal zero? Can someone please explain how it works? This concept is a bit new to me and I couldn't find any examples of this particular case online.

答:
Hint to answer the title question:

A subspace must be closed under multiplication by a scalar such as $−1$.


问:
Are there names for the sums of every third element from the expansion of e^x, similar to how sinh and cosh are every other element?
Without knowing if they have names, I'll call them ash, bsh, and csh. They would have the interesting property that $\frac{d\operatorname{ash}(x)}{dx} = \operatorname{bsh}\left(x\right), \frac{d\operatorname{bsh}\left(x\right)}{dx} = \operatorname{csh}\left(x\right), \frac{d\operatorname{csh}\left(x\right)}{dx} = \operatorname{ash}\left(x\right),$ and $\operatorname{ash}\left(x\right)+\operatorname{bsh}\left(x\right)+\operatorname{csh}\left(x\right) = e^x$

答:
Solving the system of equations without any boundary condition, we have
$$\color{red}{3 \,e^{\frac x 2}\operatorname{ash}(x)-\left(c_1+c_2+c_3\right) e^{\frac{3 x}2}=} $$
$$\sqrt{3}
   \left(c_2-c_3\right) \sin \left(\frac{\sqrt{3} x}{2}\right)+\left(2
   c_1-c_2-c_3\right) \cos \left(\frac{\sqrt{3} x}{2}\right)$$
$$\color{red}{3 \,e^{\frac x 2}\operatorname{bsh}(x)-\left(c_1+c_2+c_3\right) e^{\frac{3 x}2}=} $$
$$\sqrt{3} \left(c_3-c_1\right) \sin
   \left(\frac{\sqrt{3} x}{2}\right)-\left(c_1-2 c_2+c_3\right) \cos
   \left(\frac{\sqrt{3} x}{2}\right)$$
$$\color{red}{3 \,e^{\frac x 2}\operatorname{csh}(x)-\left(c_1+c_2+c_3\right) e^{\frac{3 x}2}=}$$
$$\sqrt{3} \left(c_1-c_2\right) \sin
   \left(\frac{\sqrt{3} x}{2}\right)-\left(c_1+c_2-2 c_3\right) \cos
   \left(\frac{\sqrt{3} x}{2}\right)$$
$$\color{blue}{\operatorname{ash}(x)+\operatorname{bsh}(x)+\operatorname{csh}(x)=\left(c_1+c_2+c_3\right) e^x}$$


问:
Construction of $\mathbb{CP}^2$ from $S^2\times S^2$
Show that $\mathbb{CP}^2$ is homeomorphic to the orbit space of $(S^2\times S^2)/\mathbb{Z}_2$. The action is given by $(x,y) \mapsto (y,x)$.

My attempt: I can see that $S^2\times S^2$ is a subset of $S^5$. I am unsuccessfully trying to restrict the action of $S^1$ on $S^5$ to $S^2\times S^2$ to get the required map.

答:
This is actually a special case of a more general statement involving the symmetric products of $S^2$. The $n$-fold symmetric product $SP_n(S^2)$ is the quotient space
$$SP_n(S^2)=(S^2)^n/\Sigma_n$$
where $(S^2)^n$ denotes the $n$-fold cartesian product, and $\Sigma_n$, the group of permutations on $n$ letters, acts by permuting the factors. The claim is that for each $n\in\mathbb{N}$ there is a homeomorphism
$$SP_n(S^2)\cong \mathbb{C}P^n.$$
Your question relates to the special case of $n=2$, but it's no more trouble to establish the general result.

To begin introduce $P_n(\mathbb{C})$ as the set of non-zero complex polynomials of degree $\leq n$. The assignment
$$P_n(\mathbb{C})\ni\sum^n_{i=1}a_iz^i\mapsto (a_0,\dots,a_n)\in\mathbb{C}^n\setminus\{0\}$$
establishes a bijection between $P_n(\mathbb{C})$ and $\mathbb{C}^n\setminus\{0\}$ which we will use to topologise $P_n(\mathbb{C})$. This yields the quotient map
$$\gamma_n:P_n(\mathbb{C})\rightarrow \mathbb{C}P^n,\qquad \sum^n_{i=1}a_iz^i\mapsto [a_0,\dots,a_n].$$
Observe that if $f,g\in P_n(\mathbb{C})$, then $\gamma_n(f)=\gamma_n(g)$ if and only if $f$ and $g$ have the same roots (counted with multiplicity).

Now identify $S^2\cong\mathbb{C}P^1\cong\mathbb{C}\cup\infty$ as the Riemann plane and define $\varphi_n:SP_n(\mathbb{C})\rightarrow P_n(\mathbb{C})$ by
$$\varphi_n\langle\xi_1,\xi_2,\dots,\xi_n\rangle=(z-\xi_1)(z-\xi_2)\cdots(z-\xi_n),$$
with the understanding to replace $(z-\infty)$ by $1$ if it appears in the product on the right-hand side. I am not claiming that $\varphi_n$ is continuous, but rather than its composite with $\gamma_n$ is a continuous map $SP_n(S^2)\rightarrow\mathbb{C}P^n$. You should convince yourself that this is true, and that $(z-\infty)\mapsto 1$ is indeed the correct replacement to make.

Now, we remarked above that the quotient $\gamma_n$ identifies two polynomials exactly when they have identical roots, so it follows that the composite $\gamma_n\varphi_n:SP_n(S^2)\rightarrow \mathbb{C}P^n$ is injective. On the other hand, to see that it is surjective, for $\xi_1,\dots,\xi_n\in\mathbb{C}\cup\{\infty\}$ we can write
$$\prod^n_{i=1}(z-\xi_i)=\sum^n_{i=0}(-1)^i\left(\sum_{1\leq j_1<j_2<\dots< j_i\leq n}\xi_{j_1}\dots\xi_{j_i}\right) z^{n-i}$$
and so check that everything in the target indeed gets hit.

Finally we have that $\gamma_n\varphi_n$ is a continuous bijection. Since $SP_n(S^2)$ is compact and $\mathbb{C}P^n$ is Hausdorff we see that this map is a homeomorphism.

Edit: Thanks to Olivier Bégassat for a correction.


问:
Starting with Klein geometry
I have read that in Klein's approach, geometry can be seen as the study of invariants under a group of transformations. We begin with a space $X$ and a subgroup $G$ of all its bijections (tough usually it would be shrink to a subgroup of the homeomorphisms or diffeomorphisms of $X$). This is called the principal group. Next step is to fix a subspace $S \subseteq X$ and look at
$$
H:=Stab_G (S) \subseteq G,
$$
the subgroup of $G$ that leaves invariant the configuration $S$. Klein's idea was that the space
$$
G/H
$$
is interesting, and this kind of spaces are called homogeneous spaces or Klein geometries. In particular, if the action is transitive and we have $S=\{x\}$ for $x\in X$,
$$
X\approx G/H
$$
that is, we recover the set of points. But  I have been told that we also recover the geometric properties. How that can be done from here?

I'd like to see it with an example. If I consider $X$ as the Euclidean space, an arbitrary point $P\in X$ and the rigid motions $E(3)$ as the principal group $G$, then $H=O(3)$ and it is clear to me that $X=G/H=\mathbb{R}^3$. But how can we deduce from here the gist of this geometry, that is, the Euclidean distance formula or the scalar product? Is there any general method that can be used with any Klein geometry to deduce its "metric" or something?

答:
I have found that the key is in the adjoint representation of $H$ over $\mathfrak{g}$. Because in this context we can look for invariant subspaces and I guess that "inside" is hidden the information for the geometry. I have seen a good explanation in Parabolic Geometries, by Andreas Cap and  Jan Slovak, page 6, but I don't fully understand yet.

I have also found useful the foreword of Differential geometry, Cartan's generalization of Klein Erlangen program, by Sharpe and Chern; although I cannot see yet how the Maurer-Cartan form let us recover the metric.

Edited

After a stop I have returned to the topic, and I have found that the whole chapter 1 of Parabolic Geometries is devoted to this issue, specifically section 1.4. When I finish reading I will edit again to write a summary, or an outline.

On the other hand, I think that the book of Sharpe is the typical book it is worth reading once you know about the topic, because you can think a lot of important stuff, but you need to have a minimum of ideas in your mind to be able to follow it.

Title: Starting with Klein geometry


问:
$\|A\|,\|B\|\leq 1$. How do I prove $\left(\begin{array}{}1&A&AB\\A^*&1&B\\B^*A^*&B^*&1\end{array}\right)$ is positive.
$A,B$ are elements of a unital C*-algebra $\mathscr A$ with $\|A\|\leq 1$ and $\|B\|\leq 1$.

$X=\left(\begin{array}{}1&A&AB\\A^*&1&B\\B^*A^*&B^*&1\end{array}\right)\in M_3(\mathscr A)=\mathscr A\otimes M_3(\mathbb C)$ .

I tried to find a matrix such that $Y^*Y=X$ and failed.

Then I tried to show the invertibility of $X+\lambda$ for $\lambda>0$. This is equivalent to the invertibility of $X$ for $\|A\|,\|B\|<1$. And failed again.

Could you please give me some hint?

Body:
Another elegant solution:

$\left(\begin{array}{}A&0&0\\1&0&0\\B^*&0&0\end{array}\right)\left(\begin{array}{}A^*&1&B\\0&0&0\\0&0&0\end{array}\right)=\left(\begin{array}{}AA^*&A&AB\\A^*&1&B\\B^*A^*&B^*&B^*B\end{array}\right)\leq \left(\begin{array}{}1&A&AB\\A^*&1&B\\B^*A^*&B^*&1\end{array}\right)$


问:
Integrals as vector spaces
I saw an integral version of the Cauchy Schwartz inequality. Could someone please explain what is the vector space, the corresponding field and the inner product space here.</p>

答:
I'm assuming that you mean the inequality
$$\left\vert\int_\Omega f\bar g\mathrm dV\right\vert^2\leq\int_\Omega\vert f\vert^2\mathrm d V\int_\Omega\vert g\vert^2\mathrm dV.$$

The space is $L^2(\Omega,\mathbb F)$, the space of square integrable functions $f:\Omega\longrightarrow\mathbb F$, where $\Omega$ is a measure space and $\mathbb F\in\{\mathbb R,\mathbb C\}$. This is a vector space over $\mathbb F$. A function is square integrable if
$$\Vert f\Vert_2^2:=\int_\Omega \vert f\vert^2\mathrm dV<\infty.$$
Such a vector space becomes an inner product space via the inner product
$$\langle f,g\rangle:=\int_\Omega f\bar g\mathrm d V,$$
where $\bar g$ is the complex conjugate of $g$. The norm induced by this inner product is exactly the norm $\Vert f\Vert_2$ from above.

As a side note: Technically speaking, what I described is $\mathcal L^2$. The space $L^2$, on which the above is a well-defined inner product, is the quotient space of $\mathcal L^2$ with respect to the relation defined as $f\sim g$ iff $f=g$ almost everywhere.

Title: Integrals as vector spaces


问:
Decomposition group and splitting polynomials
Let $f(x)$ be a polynomial over an algebraic number field $K$ with coefficients in $\mathcal{O}_K$ such that its reduction splits completely in $\mathcal{O}_K/\mathfrak{p}\mathcal{O}_K$. Let $L$ be the splitting field of $f$ over $K$ and $\mathfrak{q}$ be a prime in $\mathcal{O}_L$ such that $\mathfrak{q}|\mathfrak{p}$. Then I want to show the decomposition group of $\mathfrak{q}$ is trivial.

A polynomial splits completely by definition is that it splits as linear factors. Could someone explain how to show $D(\mathfrak{q})$ is trivial?

The motivation is this question.

Thanks.

答:
Assume $f$ is monic, and let $\mathfrak{q}|\mathfrak{p}$ where $\mathfrak{q}$ prime ideal in $L$. Then $f(X)=\prod_i{(X-a_i)}$ for some $a_i \in \mathcal{O}_L$. Reducing mod $\mathfrak{q}$, we find that $\overline{f} = \prod_i{(X-\overline{a_i})}$. As $\overline{f}$ splits in $\mathcal{O}_K/(\mathfrak{p})$, we find the $\overline{a_i}$ belong to $\mathcal{O}_K/\mathfrak{p}$, and they're distinct because $\overline{f}$ splits completely.

Now, let $\sigma$ be in the decomposition group of $\mathfrak{q}$, ie $\sigma \in Gal(L/K)$ with $\sigma(\mathfrak{q}) \subset \mathfrak{q}$. Then $\sigma$ acts on the roots of $f$, and since their reductions mod $\mathfrak{q}$ are in $\mathcal{O}_K/(\mathfrak{p})$, we know that $\sigma(a_i) = a_i$ mod $\mathfrak{q}$. But by the above, the $a_i$ mod $\mathfrak{q}$ are distinct, so $\sigma(a_i)=a_i$. $L$ is the splitting field of $f$ so is generated by $K$ and the $a_i$ so $\sigma$ is the identity and the decomposition group is trivial.


问:
Finite abelian $2$-group

If $G$ is a finite abelian group such that $o(x)=2$ for all $x \neq e$ and $|G|=2^n$ for some $n\in\mathbb N$, prove that $G \cong \mathbb{Z}_2\times\cdots\times\mathbb{Z}_2$ ($n$ factors).

Any help is appreciated. I asked this question before and was suggested to try induction but I haven't been able to write down the inductive proof properly. If anybody can help that will be really nice. Thanks!

答:
The proposition is true for $n=1$. For $n\geq 2$, take $g_1 \in G$ and let $G_1=\langle g_1\rangle$. For $i=2,\dots, n$ take $g_i \in G \text{ \ } G_{i-1}$ where $G_{i-1}=\langle g_1,\dots,g_{i-1}\rangle$. By the induction hypothesis (or directly), for $i < n, G_i \cong \mathbb{Z}_2^i$. In particular $G_{n-1} \cong \mathbb{Z}_2^{n-1}$.

By construction $g_n \in G\text{ \ }G_{n-1}$, so $\langle g_n \rangle \cap G_{n-1}=\{e\}$, they are both normal since $G$ is abelian and every element in $G$ can be written as a product of elements from $\langle g_n\rangle$ and $G_{n-1}$. (You can see this by noting that the left cosets of $G_{n-1}$ are $G_{n-1}$ and $g_nG_{n-1}$ and partition the set.) Thus $G \cong \langle g_n \rangle \times G_{n-1} \cong \mathbb{Z}_2\times\mathbb{Z}_2^{n-1}\cong\mathbb{Z}_2^n$ as required.

This is a rather brute force approach, but there's only one idea behind it i.e. "keep pulling out $\mathbb{Z}_2$ until there's nothing left". The solution is much simpler if you know about modules (a generalisation of vector spaces) or by treating it as a linear algebra problem (as egreg does below), but there's nothing wrong with this elementary approach.

Title: Finite abelian $2$-group


问:
On a correspondence between subperiodic and magnetic space groups
The Wikipedia article on space groups says

Frieze groups are magnetic 1D line groups and layer groups are magnetic wallpaper groups…

Indeed:

There are $7$ frieze groups and $7$ magnetic 1D line groups, where a magnetic group is a space group where the domains can also be time-reversed.

There are $80$ layer groups and $80$ magnetic wallpaper groups. One day, with too much free time on my hands, I worked out the exact correspondence between these two sets of groups:

(The numbers beside each layer/magnetic wallpaper group refer to their number in the International Tables for Crystallography.)

Define an $(n,m)$ space group as a (non-magnetic) space group in $n$ dimensions whose translation lattice is $m$-dimensional. Then the two equivalences mentioned at the beginning of this question may be restated as

There are as many $(2,1)$ space groups as magnetic $(1,1)$ space groups

There are as many $(3,2)$ space groups as magnetic $(2,2)$ space groups

The reason for the correspondence seems clear: time reversal corresponds to reflection in a dimension orthogonal to all other dimensions. Based on this, it seemed very reasonable for me to conjecture that

There are as many $(4,3)$ space groups as magnetic $(3,3)$ space groups

However, this is false. There are $1651$ magnetic $(3,3)$ space groups, but only $1594$ $(4,3)$ space groups. How does the discrepancy arise?

答:
B. Souvignier (2006), "The four-dimensional magnetic point and space groups", Zeitschrift für Kristallographie 221, pp. 77–82 gives the number of magnetic $(3,3)$ space groups as $1594\ (+57)$ where $57$ is the number of enantiomorphic pairs.

There are no enantiomorphic pairs among the 1D line or wallpaper groups, which explains the equality in their respective dimensions, but there are $11$ enantiomorphic pairs of 3D space groups:
$$P4_1,P4_3\ (6)\qquad P4_122,P4_322\ (8)\qquad P4_12_12,P4_32_12\ (8)$$
$$P3_1,P3_2\ (3)\qquad P3_112,P3_212\ (4)\qquad P3_121,P3_221\ (4)$$
$$P6_1,P6_5\ (4)\qquad P6_2,P6_4\ (4)\qquad P6_122,P6_522\ (6)\qquad P6_222,P6_422\ (6)$$
$$P4_132,P4_332\ (4)$$
When elaborated into magnetic groups, these $11$ pairs become $57$, explaining the discrepancy completely; the numbers in parentheses above indicate the number of pairs of magnetic groups corresponding to each pair under the BNS setting (verified by the Bilbao Crystallographic Server). An enantiomorphic pair of magnetic space groups collapses into one $(4,3)$ space group when time reversal is converted into reflection in the fourth dimension.


问:
General solution for an integral with Hermite polynomials?
I'm currently studying quantum mechanics and thus dealing with Hermite polynomials.

By looking on Wikipedia I found this identity for Hermite polynomials:

$$\int\limits_{-\infty}^{\infty} H_m(x)H_n(x)e^{-x^2} \ dx=\sqrt{\pi}2^nn!\delta_{mn}$$

While doing a problem, after a little bit of calculations, I came to the following integral:
$$\int\limits_{-\infty}^{\infty} H_m(x)\cdot x\cdot H_n(x)e^{-x^2} \ dx$$

By just looking at this integral, I figured that it has to be zero when $m-n=\pm2k, k=1,2,3...$ (cause the expression would be uneven and thus the integral would be zero).

I thought I could make the same argument by saying the integral is nonzero whene $m-n=\pm(2k+1)$. While this is true for $k=0$ (so $m-n=\pm1$), I turns out that this seems to be not true for $k>0$. For example I calculated:
$$\int\limits_{-\infty}^{\infty} H_0(x)\cdot x\cdot H_3(x)e^{-x^2} \ dx=\int\limits_{-\infty}^{\infty} e^{-x^2}x(8x^3-12x)dx=0$$

I tried to understand why this is the case by using the recursion formula of the Hermite polynomials and plugging it into the integral but didn't succeed...

I wanted to ask if there is another already existing general identity for my integral (like the one on Wikipedia). If not it would be awesome if someone could help me out here and explain why the argumentation breaks down for $k>0$... Thanks!

答: Any polynomial of degree $\leqslant n$ is a linear combination of $H_k(x)$ with $0\leqslant k\leqslant n$. In particular, there exist numbers $a_{m,k}$ such that $xH_m(x)=\sum_{k=0}^{m+1}a_{m,k}H_k(x)$. And, if $m+1<n$, then this implies $\int_{-\infty}^\infty xH_m(x)H_n(x)e^{-x^2}\,dx=0$ by the orthogonality of the Hermite polynomials. (To handle $m+1=n$, one just has to compute $a_{m,m+1}$ by comparing the coefficients of $x^{m+1}$.)


问:
Evaluating $\left.\int_0^{\pi/2}\sqrt{1+\frac1{\sqrt{1+\tan^nx}}}\text dx \middle/\int_0^{\pi/2}\sqrt{1-\frac1{\sqrt{1+\tan^nx}}}\text dx\right.$

Evaluate the integral ratio$$\dfrac{I_1}{I_2}=\dfrac{\displaystyle \int_{0}^{\frac{\pi}{2}} \sqrt{1+\frac{1}{\sqrt{1+\tan^nx}}} \mathrm{d}x}{\displaystyle \int_{0}^{\frac{\pi}{2}} \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}} \mathrm{d}x}$$

Using Wolframalpha, I found out that the ratio of the above two integrals is $\sqrt{2}+1$, that is, independent of $n$. I couldn't think of a way to solve it.

I don't think any simple substitutions would work here. Any useful hints would be appreciated.

Body:
Note

\begin{align}
\frac{I_1}{I_2}&=1+ \frac{I_1-I_2}{I_2} \\
&=1+\frac1{I_2}\int_0^{\frac\pi2} \left( \sqrt{1+\frac{1}{\sqrt{1+\tan^nx}}}-  \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}} \right)dx \\
&=1+\frac1{I_2}\int_0^{\frac\pi2}
\frac{\sqrt{\sqrt{\sin^nx+\cos^nx}+\sqrt{\cos^nx}}
 - \sqrt{\sqrt{\sin^nx+\cos^nx}-\sqrt{\cos^nx}}}{\sqrt[4]{\sin^nx+\cos^nx}}dx \\
&=1+\frac1{I_2} \int_0^{\frac\pi2}
\frac{\sqrt{2\sqrt{\sin^nx+\cos^nx}-2\sqrt{\sin^nx}}
}{\sqrt[4]{\sin^nx+\cos^nx}}dx \\
&= 1+\frac{\sqrt2}{I_2} \int_0^{\frac\pi2} \sqrt{1-\frac{1}{\sqrt{1+\cot^nx}}}dx\>\>\>\>\>\>\>\>(x\to \frac\pi2-x)\\
&=1+\frac{\sqrt2}{I_2}\int_0^{\frac\pi2} \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}}dx \\ &=1+\sqrt2 \\
\end{align}


问:
Identifying this approximation of pi
I've come across this approximation of pi, which I'm struggling to put a name on. I want to do further research on it but I can't find any evidence of it on the internet. It is an infinite sum:
$\pi = 1 + 1/2 + 1/3 + 1/4 - 1/5 + 1/6 + 1/7 + 1/8 + 1/9 - 1/10 + 1/11 + 1/12 - 1/13 + \dots $

This is the description I've been given of it:

In this formula, each fraction 1/n has a sign (±) determined by: the first two terms have
positive signs; after that, if the denominator is a prime of the form 4m − 1 (for example,
n = 3, 7, 11, . . .), the sign is positive; if the denominator is a prime of the form 4m + 1 (for
example, n = 5, 13, 17, . . .), the sign is negative; if the denominator is a composite number,
then the sign is equal to the product of the signs corresponding to its factors (for example,
n = 9 = 3 × 3, so its sign is positive (a positive times a positive), while n = 10 = 5 × 2,
so its sign is negative (a negative times a positive)).

Can anyone help me identify which approximation this is? Is it a well-known approximation?

答:
The given sum is related to the Dirichlet $L$-series associated to the quadratic character $\chi_1:(\Bbb Z/4)^\times\to\pm 1$.

en.wikipedia.org/wiki/Dirichlet_character

In a table as the one in loc. cit., Dirichlet_character#Modulus_4:
$$
\begin{array}{|r||c|c|c|c|}
\hline
n\text{ mod }4 & 0 & 1 & 2 & 3 \\\hline
\chi_1(n) & 1 & 0 & -1 & 0\\\hline
\end{array}
$$
But there are some differences.

First, hen considering the prime $2$, we have $\chi_1(2)=0$, so we have to manually insert this Euler factor.

Then when considering the signs to be implemented. The signs for the primes are "reverted". We use a minus sign for the primes congruent to $1$ modulo four, and a plus sign for the others. Let us use an other letter for this...
$$
\begin{array}{|r||c|c|c|c|}
\hline
n\text{ mod }4 & 0 & 1 & 2 & 3 \\\hline
\eta(n) & -1 & 0 & +1 & 0\\\hline
\end{array}
$$

In the cited link, the value for $L(\chi_1,s)=\beta(s)$ is referenced to be the Dirichlet beta function computed in $s$.

Putting all together, the given sum $S$ is:
$$
\begin{aligned}
S
&= \left(1+\frac 12+\frac 14+\frac 18+\dots\right)
\left(\sum_{\substack{n>0\\n\text{ odd}}}\frac{\eta(n)}n\right)
\\
&= \frac 1{1-2^{-1}}\cdot
\prod_{p\text{ odd prime}}\frac{1}{1-\eta(p)p^{-1}}
\\
&= 2\cdot
\prod_{p\text{ odd prime}}\frac{1}{1+\chi_1(p)p^{-1}}
\\
&= 2\cdot
\prod_{p\text{ odd prime}}\frac{1}{1-p^{-2}}
\Big/
\prod_{p\text{ odd prime}}\frac{1}{1-\chi_1(p)p^{-1}}
\\[2mm]
&=2\cdot \frac{\zeta(2)}{(1-2^{-2})^{-1}}\;\Big/\;L(\chi_1,s)|_{s=1}
\\
&=\frac{\pi^2}4\Big/\beta(1)
\\
&=\frac{\pi^2}4\Big/\left(1-\frac 13+\frac 15-\frac 17+\frac 19-\dots\right)
\\
&=\frac{\pi^2}4\Big/\arctan 1
\\
&=\pi\ .
\end{aligned}
$$

Title: Identifying this approximation of pi


问:
Show that two matrices are similar when they have the same characteristic polynomial and a cyclic vector
I need to show that if $A$ and $B$ have a cyclic vector and have equal characteristic polynomial, then they are similar.

Here I know that given $T: V \rightarrow V$ with $n=\operatorname{dim} V,$ a $T$ -cyclic vector $v \in V$ satisfies $B=\left(v, T(v), \ldots, T^{n-1}(v)\right)$ is an ordered basis of $V$, and $[T]_{B}=L_{\Delta_{T}}=L_{\psi_{T}}$, where $L_{f}$ is the companion matrix defined by the polynomial $f$.
But I'm not seeing how I can use this information to prove what I need.

答:
You have all of the pieces, you just need to put them together.

$A$ and $B$ have a cyclic vector and have common characteristic polynomial $f$. Let $v$ denote an $A$-cyclic vector, and let $w$ denote a $B$-cyclic vector. Let $\mathcal A = (v,Av,\dots,A^{n-1}v)$, and let $\mathcal B = (w,Bw,\dots,B^{n-1}w)$. We note that
$$
[A]_{\mathcal A} = [B]_{\mathcal B} = L_f.
$$
In other words, $A$ is similar to $L_f$ and $B$ is similar to $L_f$. Because matrix similarity is an equivalence relation, this means that $A$ is similar to $B$.


问:
$\int _0^{2\pi}\Biggl(\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}\Biggr)^2 dx$ by changing integral and summation.

Let $f(x)=\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}$. Prove that $f$ converges uniformly on $[0,2\pi]$ and evaluate $\int _0^{2\pi}f(x)^2dx$.

Since $\left|\frac{\sin nx}{2^n}\right|\le\left|\frac{1}{2^n}\right|$ and $\sum \limits _{n=1}^\infty \frac{1}{2^n}$ converges, $f(x)$ converges uniformly on $[0,2\pi]$.

After reading some related post, Changing integral and summation, and Calculate $\int_{-\pi}^\pi\bigg(\sum_{n=1}^\infty\frac{\sin(nx)}{2^n}\bigg)^2dx$. I need to prove $$\Biggl(\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}\Biggr)^2 \text{converges uniformly on }[0,2\pi].$$

Write $\Biggl(\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}\Biggr)^2=\sum_\limits{n=1}^\infty\sum_\limits{m=1}^\infty \frac{\sin nx}{2^n}\frac{\sin mx}{2^m}=\sum_\limits{k=1}^\infty c_k$, where $c_k=\sum_\limits{j=1}^k\frac{\sin jx\sin (k-j)x}{2^k}$.

So $$|c_k|\le\left|\frac{k}{2^k} \right|$$
Is $\sum_\limits{n=1}^\infty \frac{k}{2^k}$ converges?

I have another question. In the linked post, they write $$\begin{align} \int _0^{2\pi}\Biggl(\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}\Biggr)^2 dx & = \int _0^{2\pi} \sum_\limits{n=1}^\infty\sum_\limits{m=1}^\infty \frac{\sin nx}{2^n}\frac{\sin mx}{2^m} dx \\&=\sum_\limits{n=1}^\infty\sum_\limits{m=1}^\infty \int _0^{2\pi} \frac{\sin nx}{2^n}\frac{\sin mx}{2^m}dx . \end{align}$$

But I think it is a Cauchy product and so $$\begin{align} \int _0^{2\pi}\Biggl(\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}\Biggr)^2 dx &=\int _0^{2\pi}\sum_\limits{k=1}^\infty \sum_\limits{j=1}^k\frac{\sin jx\sin (k-j)x}{2^k}dx .   \end{align}$$

What is the difference between these two integrals?

答:
Yes, the series converges. You showed that $|c_k|\le k/2^k$. Note that for $|q|>1$ we have
$$\sum_{n\ge1}\frac{1}{q^n}=\frac{1}{q-1},$$
from the geometric series (which converges uniformly). Then,
$$\sum_{n\ge1}\frac{n}{q^n}=-\sum_{n\ge1}q\frac{d}{dq}q^{-n}=-q\frac{d}{dq}\left(\sum_{n\ge1}\frac1{q^n}\right)=-q\frac{d}{dq}\left(\frac1{q-1}\right)=\frac{q}{(q-1)^2}.$$ Remember, this is for any $q$ with $|q|>1$, so we can choose $q=2$ and get $$\sum_{n\ge1}|c_n|\le\sum_{n\ge1}\frac{n}{2^n}=2.$$


问:
Book Suggestions for an Introduction to Measure Theory
Couldn't find this question asked anywhere on the site, so here it is! Do you guys have any recommendations for someone being introduced to measure theory and lebesgue integrals?

A mentor has suggested a book that's in french, but unfortunately I don't know french (heck I barely know english) - so english books only please!

Thanks in advanced!

EDIT: Did not realize this had been asked here but I'm going to leave this question open to see if there are any newer books (that question was originally asked in 2011). If I am doing something wrong here, just give me a gentle shoutout in the comments and I'll be understanding and close this

答: Frank Jone's Lebesgue Measure on Euclidean space is the greatest math book that I have ever read. It gives a very slow introduction to lebesgue measure. It is fabulous; I did every homework problem for fun.</p>


问:
Expressing sides of a triangle

Express $AB$, $BC$ and $AC$ over $\sin \alpha$, $\cos \gamma$ and $CC_1$ (note that $AA_1$, $BB_1$ and $CC_1$ are heights, $\alpha =$ angle $BAC$ and $\gamma=$ angle $ACB$).
What I tried for now is expressing $\sin \alpha$ and $\cos \gamma$, and i got $\sin \alpha = \frac{CC_1}{AC} = \frac{BB_1}{AB}$ and $\cos \gamma = \frac{A_1C}{AC}=\frac{B_1C}{BC}$, but that leads me nowhere.

答:
We have that $AC=\frac{CC_1}{\sin \alpha}$, we also have that $BC=\frac{CC_1}{\sin \beta}$.
However $\sin \beta=\sin (\pi-(\alpha+\gamma))=\sin (\alpha+\gamma)=\sin \alpha \cos \gamma+\cos \alpha \sin \gamma$. Now we can express $BC$ as required (use identity $\cos^2 \alpha +\sin^2 \alpha=1$ to find $\cos \alpha$, etc.).

Finally use law of sines to express $AB$: $\frac{BC}{\sin \alpha}=\frac{AB}{\sin \gamma}$.

Title: Expressing sides of a triangle


问:
notation for the set n choose k
Lately I've found myself writing things like $ijk \in {n \choose 3}$, by which I mean that $i,j,k$ are coming from the sets enumerated by ${n \choose 3 }$. It's a slight abuse of notation, since ${ n \choose 3 }$ is usually understood to be a number and not the set it is counting.

Is there a standard notation for this set? i.e. the set of cardinality $k$ subsets of $\{1, \ldots, n\}$.

I'd be happy to learn that the answer is no and that my abuse of notation is common or clear enough.

答:
Use $A\choose k$ where $A$ is a set, like $\{1,\ldots,n\}\choose k$.
This makes sense, since

$\left\vert {\{1,\ldots,n\}\choose k}\right\vert = {n\choose k}$.

Title: notation for the set n choose k


答:
Given a set $X$, we have the notation for the set of size-$k$ subsets of $X$:

$$\binom{X}{k} := \{A~:~|A|=k,~A\subseteq X\}$$

Further, the set $\{1,2,3,\dots,n\}$ can be notated as $[n]$.  There are some authors who prefer to have this represent instead $\{0,1,2,3,\dots,n-1\}$.  In either event, $[n]$ is the prototypical $n$-element set of the first $n$ natural numbers and most scenarios where you see $[n]$ it doesn't actually matter which interpretation it was since the end result of whatever was being talked about winds up being the same.

So, a particular three-element subset $\{i,j,k\}$ of $\{1,2,3,\dots,n\}$ can be written as

$$\{i,j,k\}\in\binom{[n]}{3}$$

Title: notation for the set n choose k

Body:
Lately I've found myself writing things like $ijk \in {n \choose 3}$, by which I mean that $i,j,k$ are coming from the sets enumerated by ${n \choose 3 }$. It's a slight abuse of notation, since ${ n \choose 3 }$ is usually understood to be a number and not the set it is counting.

Is there a standard notation for this set? i.e. the set of cardinality $k$ subsets of $\{1, \ldots, n\}$.

I'd be happy to learn that the answer is no and that my abuse of notation is common or clear enough.


问:
$TS^2$ is not trivial , seeing explicitely what might fail when we try and construct a trivialization
I just did an exercise, that was around the fact that if we have an orientable $1$-dimensional vector bundle over a paracompact space then it's trivial.

It's a well know fact that $TS^2$ is not trivial do to the hairy ball theorem, and this is a $2$-dimensional vector bundle so the latter result does not apply but I was wondering what failed if I tried to construct a trivialization for this, that is , we have local trivializations of $TS^2$ and then we can glue them together using partitions of unity . What would fail for the fact that we do not have that this gives an isomorphism of vector bundles to $S^2\times \mathbb{R}^2$.  I don't know if anyone has tried to do this or knows a reference where I could look something like this up but any discussion regarding what may fail at the fiber levels, since we are using partitions of unity and I guess we will need to use formulas for the transition functions trough the tangent vector represented in different bases. Any comment is appreciated. Thanks in advance.

答:
The issue is that a convex combination of invertible linear maps may not be invertible.  Consider what your partition of unity construction would do on a point $p$ that is in two different open sets $U$ and $V$ on which you have a trivialization.  As a part of your trivialization over the open set $U$, you have chosen an isomorphism $f_U:T_pS^2\to\mathbb{R}^2$, and similarly the trivialization over $V$ gives an isomorphism $f_V:T_pS^2\to \mathbb{R}^2$.  Your partition of unity would then choose nonnegative functions $\varphi_U$ and $\varphi_V$ supported on $U$ and $V$ such that $\varphi_U(p)+\varphi_V(p)=1$, and would combine the two trivializations by taking $\varphi_U(p)f_U+\varphi_V(p)f_V:T_pS^2\to\mathbb{R}^2$.  However, $\varphi_U(p)f_U+\varphi_V(p)f_V$ might not be invertible!  For instance, it might be the case that $\varphi_U(p)=\varphi_V(p)=1/2$ and $f_V=-f_U$, and then $\varphi_U(p)f_U+\varphi_V(p)f_V$ would be $0$.

It is worthwhile to contrast what's going on here with, for instance, the construction of a Riemannian metric on any manifold using a partition of unity.  The difference there is that a Riemannian metric is positive definite, and any convex combination of positive definite bilinear forms is still positive definite.  That's how you can be sure that when you combine local Riemannian metrics using a partition of unity, the result is still positive definite (and in particular, nondegenerate).  Similarly, in the case of an orientable rank $1$ vector bundle, the orientation gives a notion of "positivity" such that a convex combination of two "positive" isomorphisms to $\mathbb{R}$ is still positive (and thus in particular, it is still an isomorphism).


问:
Continiutity of a dense subset imply continuity of the set
I would like some help proving that if $f$ and $g$ are continous on $(a,b)$ and $f(x)=g(x)$ for every $x$ in a dense subset of $(a,b)$ then $f(x)=g(x)$ for all $x$ in $(a,b)$.</p>

答:
Let $S\subset(a,b)$ be the dense subset in the question.  Take any $x\in(a,b)$, and find a sequence $\{a_n\}$ in $S$ such that $a_n\to x$.  Such a sequence exists precisely because $S$ is dense in $(a,b)$.

Now, we know that $f(a_n)=g(a_n)$ for all $n$ because they agree on $S$.  We also know that $f$ and $g$ are continuous on $(a,b)$.  By continuity, we have:

$$f(x)=\lim_{n\to\infty}f(a_n)=\lim_{n\to\infty}g(a_n)=g(x)$$


问:
Show that $\operatorname{span}(\operatorname{span}(M) \cap \operatorname{span}(N)) = \operatorname{span}(M) \cap \operatorname{span}(N)$
Can anyone guide me through how to prove that
$$\operatorname{span}(\operatorname{span}(M) \cap \operatorname{span}(N)) = \operatorname{span}(M)  \cap \operatorname{span}(N)?$$

I honestly have no idea how I should proceed here...

Cheers

答:
By definition, $$\text{Span}\big(\text{Span}(M)\cap \text{Span}(N)\big)$$ is the smallest vector space that contain $\text{Span}(M)\cap \text{Span}(N)$. Since $\text{Span}(M)\cap \text{Span}(N)$ is a vector space, then
$$\text{Span}(M)\cap \text{Span}(N)=\text{Span}\big(\text{Span}(M)\cap \text{Span}(N)\big).$$


答:
The standard way of gluing up the Riemann sphere is using gluing to copies of $\mathbb{C}$ using $z^{-1}$. Hence the way of gluing up the tangent bundle is by its derivative $\frac{-1}{z^2}$. The asymptotics of these gluing functions give a reasonably intuitive reason why the vector bundle is not trivial (even why the first chern class is $2$).

Another good exercise is to pick a Riemannian metric on $S^2$ compute $\pi_{1}$ of the $3$-manifold $$U = \{(p,v) \in TS^2 : |v| = 1 \} .$$

If the bundle where trivial then one may see that we should have $U = S^2 \times S^1$, but actually computing then fundamental group  of $U$ shows this cannot be the case.

Title: $TS^2$ is not trivial , seeing explicitely what might fail when we try and construct a trivialization

Body:
I just did an exercise, that was around the fact that if we have an orientable $1$-dimensional vector bundle over a paracompact space then it's trivial.

It's a well know fact that $TS^2$ is not trivial do to the hairy ball theorem, and this is a $2$-dimensional vector bundle so the latter result does not apply but I was wondering what failed if I tried to construct a trivialization for this, that is , we have local trivializations of $TS^2$ and then we can glue them together using partitions of unity . What would fail for the fact that we do not have that this gives an isomorphism of vector bundles to $S^2\times \mathbb{R}^2$.  I don't know if anyone has tried to do this or knows a reference where I could look something like this up but any discussion regarding what may fail at the fiber levels, since we are using partitions of unity and I guess we will need to use formulas for the transition functions trough the tangent vector represented in different bases. Any comment is appreciated. Thanks in advance.


问:
Product of correlated brownian motions
Consider that the correlation between two standard brownian motions $dB_x$  and $dB_y$ be $\rho$. And we write $\mathtt{Cor} (dB_x,dB_y)$ = $\rho$. Show that $dB_xdB_y$ = $\rho dt$  </p>

答:
I solved and one should use these rules:

.$\left(dt\right)^{n}\in$o$\left(dt\right)$

.$\mathbb{E}[f(B_t)]=f(B_t)$ if $Var[f(B_t)]=0$

.$dB_t^2=dt$

From the result that I already have i.e.

$\mathbb{E}(dBxdBy)=\rho dt$

We just have to show that $Var(dBxdBy)=0$, and apply the second rule, but

$Var(dBxdBy)=\mathbb{E}(dBxdBy)^2-\mathbb{E}^2(dBxdBy)=\mathbb{E}(dtdt)-\rho2dt^2=dt^2-\rho^2dt^2=0$

QED

Title: Product of correlated brownian motions


问:
If one stochastic process is a modification of another, then they have the same finite probability distribution.
On page 2 in Karatzas and Shreve: Brownian Motion and Stochastic Calculus it is said that a stochastic process $Y$ is a modification of $X$ if for all $t$:
$P(X_t=Y_t)=1$.

If both are stochastic processes into $\mathbb{R}^d$ it is said that they have the same finite dimensional distribution if for every $n$ and every $A \in \mathcal{B}(\mathbb{R}^{nd})$, and any real numbers $0 \le t_1<t_2< \ldots<t_n <\infty$ we have

$P((x_{t_1},x_{t_2},\ldots,x_{t_n})\in A)=P((y_{t_1},y_{t_2},\ldots,y_{t_n})\in A)$

Is it trivial that the first condition gives the second condition? How do we prove it? I think that in order to show this, you are not supposed to used the facts learned in elementary probability and statistics classes, but use product measures of some kind? Does it in some way follow from the theory that concerns the extensions of measures and product measures?

答:
Since I could't understand the second answer in this post for a while (https://math.stackexchange.com/q/1709305 (version: 2016-03-23)) I would like to add my current understanding and derivation (as I cannot add comments, I use an answer window).

For simplicity of notation, let $\mathbf{X}\equiv(X_{t_1},\dots,X_{t_n}), \mathbf{Y}\equiv(Y_{t_1},\dots,Y_{t_n})$ and some $A\in\mathcal{B}(\mathbb{R}^{nd})$.
$$
|\mathbb{P}(\{\mathbf{X}\in A\})-\mathbb{P}(\{\mathbf{Y}\in A\})|=|\mathbb{P}(\{\mathbf{X}\in A\}\cap\{\mathbf{X=\mathbf{Y}}\})+\mathbb{P}(\{\mathbf{X}\in A\}\cap\{\mathbf{X\neq\mathbf{Y}}\})-\mathbb{P}(\{\mathbf{Y}\in A\}\cap\{\mathbf{X=\mathbf{Y}}\})-\mathbb{P}(\{\mathbf{Y}\in A\}\cap\{\mathbf{X\neq\mathbf{Y}}\})|=(1)
$$
by division of set on two disjont intersections. As on set $\{\mathbf{X=\mathbf{Y}}\}$ we must have $\{\mathbf{X}\in A\}\iff\{\mathbf{Y}\in A\}$, we can write
$$
\{\mathbf{X}\in A\}\cap\{\mathbf{X=\mathbf{Y}}\}=\{\mathbf{Y}\in A\}\cap\{\mathbf{X=\mathbf{Y}}\}\implies\\\
\mathbb{P}(\{\mathbf{X}\in A\}\cap\{\mathbf{X=\mathbf{Y}}\})=\mathbb{P}(\{\mathbf{Y}\in A\}\cap\{\mathbf{X=\mathbf{Y}}\})\\\ (1)=|\mathbb{P}(\{\mathbf{X}\in A\}\cap\{\mathbf{X\neq\mathbf{Y}}\})-\mathbb{P}(\{\mathbf{Y}\in A\}\cap\{\mathbf{X\neq\mathbf{Y}}\})|\leq\mathbb{P}(\{\mathbf{X}\in A\}\cap\{\mathbf{X\neq\mathbf{Y}}\})+\mathbb{P}(\{\mathbf{Y}\in A\}\cap\{\mathbf{X\neq\mathbf{Y}}\})=(2)
$$
by triangle inequality. Both arguments of probabilities in last line are subsets of $\{\mathbf{X\neq\mathbf{Y}}\}$, so
$$
(2)\leq2\mathbb{P}(\{\mathbf{X}\neq\mathbf{Y}\})=0
$$
as a consequence, that processes are reciprocal modifications.


问:
Examples and definition of cocompact objects
An object $X$ of a locally small category $C$ that admits filtered colimits is called compact if
$$
\operatorname{Hom}_{C}(X,-)
$$
preserves filtered colimits.

Let $C$ be a locally small category that admits filtered limits. Let's define an object $X$ of $C$ to be cocompact if
$$
\operatorname{Hom}_{C}(-,X)
$$
sends cofiltered limits to filtered colimits in $Sets$.

I never heard of this definition before. Is it reasonable? If yes, what are typical examples of cocompact objects in $Sets$, $Top$, etc?

答:
I prefer to call the quoted categorical concept of compactness strong compactness, since it is stronger than the usual notion of compactness for topological spaces.

We can express the definition of strong cocompactness of an object $D$ as the requirement that any morphism $\lim_{i \in I}C_i \to D$, with $I$ a cofiltered diagram, should factor through one of the limit projections $\pi_j: \lim_{i \in I}C_i \to C_j$. We see from this reformulation that if any subset/subspace of a set/space $X$ fails to be strongly cocompact, we can therefore deduce that $X$ is not strongly compact either; we use this to help us in our search.

In $\mathbf{Set}$, consider $\prod_{n \in \mathbb{N}} 2$ as a cofiltered limit of copies of finite products of the two-element set, $2$. Given any set $X$ with two or more distinct elements, pick two such elements and call them $0$ and $1$; label the elements of $2$ the same way. Then we have a function $f: \prod_{n \in \mathbb{N}} 2 \to X$ sending a sequence to $0$ if the sequence contains a $1$ and the first $1$ in the sequence is followed by a $0$; and to $1$ otherwise. By construction, this cannot factor through the projection from $\prod_{n \in \mathbb{N}} 2$ to any finite product. Clearly any morphism from a cofiltered limit to the one-element set $1$ factors through one (indeed any) of its projection maps. However, contrary to what Zhen Lin said in the comments, $0$ is not strongly cocompact in $\mathbf{Set}$, since if we consider the codirected limit of subsets $[n,\infty) \subseteq \mathbb{N}$, whose intersection is empty, we see that this is not preserved by $\mathrm{Hom}(-,\emptyset)$.

By passing across the forgetful-indiscrete adjunction $(U \dashv I)$ between $\mathbf{Top}$ and $\mathbf{Set}$, we can use the same argument as above to deduce that any indiscrete space with more than 2 points, or indeed any non-$T_0$ space, cannot be strongly cocompact in $\mathbf{Top}$. If the subspace of $X$ on the points $0$ and $1$ is homeomorphic to the Sierpinski space, such that $0$ is open, the argument still works, since the set $f^{-1}(0)$ of sequences described above is the union of opens $$\prod_{i=0}^{n-1} \{0\}_i \times \{1\}_n \times \{0\}_{n+1} \times \prod_{i=n+2}^{\infty} 2$$
in $\prod_{n \in \mathbb{N}} 2$. This union is disjoint and there can be arbitrarily many non-trivial projection indices, so $f$ cannot factor through any of the projections to finite products of $2$.

The argument fails more generally, though, since $f$ cannot be continuous for any choice of points of $X$ when $X$ is Hausdorff, say, since the subspace consisting of the points $\{0,1\}$ is necessarily discrete in that case, and $f^{-1}(1)$ is not open in $\prod_{n \in \mathbb{N}} 2$.

Instead, consider
$$Y_0 := \{(x,y) \in \mathbb{N} \times \mathbb{N} \mid x+y \geq 0\},$$
topologized with basic open sets $U_{a,b} := \{(x,y) \mid x \geq a, \, y \geq b\}$. Then we have a directed collection of subspaces,
$$Y_n := \{(x,y) \in Y_0 \mid 2^n \text{ divides } x+y \};$$
these subspaces are all connected, but their intersection is the subspace
$$Y_{\infty} := \{(x,y) \in \mathbb{N} \times \mathbb{N} \mid x+y = 0\},$$
homeomorphic to $\mathbb{N}$ with the discrete topology. Suppose $X$ has a pair of points $\{0,1\}$ such that the subspace on these points is discrete. Then we have a continuous map $Y_{\infty} \to X$ sending $(x,y)$ to $0$ if $y \geq x$ and to $1$ otherwise. This function cannot factor through any of the $Y_n$, though, since the inverse images of $0$ and $1$ would have to non-trivially disconnect $Y_n$, which is impossible.

In summary: Since we have exhausted the possibilities for two-point subspaces, and the argument for the emptyset carries across, we have shown that in $\mathbf{Top}$, just as in $\mathbf{Set}$, the singleton space is the only strongly cocompact space.

As a final observation, if $R$ is a functor having a left adjoint $L$ which preserves cofiltered limits (if $L$ has a further left adjoint, for example) then $R$ preserves strongly cocompact objects. Indeed, letting $X$ be strongly cocompact and $I$ be a cofiltered indexing diagram, we have:
$$\mathrm{Hom}(\lim_{i \in I}Y_i,R(X)) \cong \mathrm{Hom}(L(\lim_{i \in I}Y_i),X) \cong \mathrm{Hom}(\lim_{i \in I}L(Y_i),X)$$
$$\cong \mathrm{colim}_{i \in I}\mathrm{Hom}(L(Y_i),X) \cong \mathrm{colim}_{i \in I}\mathrm{Hom}(Y_i,R(X)),$$
so even without the explicit calculations in $\mathbf{Top}$, we could have deduced that the terminal space $1$ is strongly cocompact, since it is the image of the one-element set under the indiscrete functor $\mathbf{Set} \to \mathbf{Top}$ mentioned earlier.


问:
Where are the centered trigonal lattices among the 14 Bravais lattices?
I have identified the missing entries in the Bravais lattice table for seven crystal families as the already existing entries int the table. However I'm unable to do so just for the trigonal crystal family. Can you help?

So far, I've done the following.

答:
There is no "trigonal crystal family". There is however a rhombohedral lattice system, which should take the place of "trigonal" in your table.

All centred rhombohedral lattices are equal to primitive rhombohedral lattices with smaller unit cells.


问:
Probability of generating a sequence of numbers between 1 and n
Let's say we want to generate a random number between 1 and n, n-times,then the probability that every integer between 1 and n appears once in the generated sequence of random numbers is $\frac{n!}{n^n}$. So if we were to generate a number between 1 and 10, 10-times, the probability that every number between 1 and 10 appears once in the generated sequence of numbers is $\frac{10!}{10^{10} }$, which is about 1 in 3000. So we should expect that if we do this process of generating 10 numbers 3000 times we should get a sequence of numbers where all numbers between 1 and 10 apeear once only one time. But when I tried this on my calculator(it has a button for generating integers) I got that pattern of all integers between 1 and 10 appearing without repeating after my 15th trial! So does that mean my calculator dose not actually generate random numbes?

Or in another words: what is the probability that after generating 10 numbers (between 1and 10) randomly 15 times you get a sequence where all integers between 1 and 10 appear once without repeating?

答:
Paraphrasing:

Probability of seeing 10 distinct numbers is about 1 in 3000. So we should expect that if we do this process 3000 times we should get the outcome only one time.

This is not true. By your logic, since a coin flip has a $1/2$ chance of being heads, if you flip a coin twice, you should always get exactly one heads and one tails, which is not true.

In your situation, it is entirely possible (but extremely unlikely) that you get a sequence of 10 distinct numbers every single time out of the 3000 trials.

The interpretation of "the probability is about 1 in 3000" should be this: if you did this process $n$ times and computed the proportion of trials where you did see  10 distinct numbers, then this proportion will approach $1/3000$ as $n \to \infty$. (This is the law of large numbers.)

For a fixed large $n$, it will usually be close to $1/3000$, but it is still possible that it is not; things like the central limit theorem can quantify this more concretely.


答:
I recomend An introduction to measure theory  by
Terence Tao ( Fields Medalist  in 2006).  </p>

Title: Book Suggestions for an Introduction to Measure Theory

Body:
Couldn't find this question asked anywhere on the site, so here it is! Do you guys have any recommendations for someone being introduced to measure theory and lebesgue integrals?

A mentor has suggested a book that's in french, but unfortunately I don't know french (heck I barely know english) - so english books only please!

Thanks in advanced!

EDIT: Did not realize this had been asked here but I'm going to leave this question open to see if there are any newer books (that question was originally asked in 2011). If I am doing something wrong here, just give me a gentle shoutout in the comments and I'll be understanding and close this


问:
Independence of $X_{N+1}$ and $X_N$
I have recently started learning about probability and came across this question in my textbook.

Let $X_1,X_2,X_3,...$ be i.i.d discrete random variables. Let $N$ be a positive discrete random variable, not necessarily independent of $X_1,X_2,X_3,...$. Are $X_{N+1}$ and $X_N$ independent?

My first thought is that they should be independent because the entire sequence of $X$'s is i.i.d. I can prove independence by proving $P(X_{N+1} = x | X_N) = P(X_{N+1} = x)$, but I couldn't get anywhere trying this.

My second thought is that they are not independent since they both depend on the variable $N$.

I'm not sure what to do now.

答:
Let each $X_i$ be Bernoulli variables with $p = \frac12$ (i.e. fair coin flips), and let $N$ be the smallest $i$ for which $X_i = X_{i+1}$ (or $N = 0$ if it never happens).

Then $P(X_N = 0) = \frac12$, but $P(X_N = 0\mid X_{N+1} = 0) = 1$, making them dependent.

Title: Independence of $X_{N+1}$ and $X_N$


问:
Prove or disprove: $A=\left\{P^k\mid k\in\mathbb{N}\right\}$ has a finite number of elements
Let $P\in[0,1]^{n\times n}$  be a stochastic matrix, meaning for every $i\in[1,n]_{\mathbb{N}}$:

$$\sum_{j=1}^nP_{ij}=1$$

It is given that $P$ is not diagonalizable. Define the group $A$ to be a group of square $n\times n$ matrices such that:

$$A=\left\{P^k\mid k\in\mathbb{N}\right\}$$

Prove or disprove: $A$ has a finite number of elements.

I'm asking this question because it is often required to compute $P^n$ when dealing with Markov Chains. I did a little research and found out that a stochastic matrix doesn't have to be diagonalizable, what eventually led me to think - How can I compute $P^n$ if $P$ is not diagonalizable? The statement that I'm asking your help proving or disproving is somewhat related - since if it is true, then my problem is solved. It might be false, though. If it is indeed false - first, I would be glad to see why, and additionally, I'd be happy to hear your thoughts on what should I do if I am presented with a non-diagonalizable $P$ and need to compute $P^n$. Thanks!

答:
Here is a convenient counterexample: for $0 < \lambda < 1$ and $\alpha > 0$ sufficiently small (in particular $\alpha < \frac{1 - \lambda}{6}$), the matrix
$$
P = \frac{1 - \lambda}{3} \pmatrix{1\\1\\1}\pmatrix{1&1&1} +
\alpha \pmatrix{1\\1\\-2}\pmatrix{1&-1&0} + \lambda I
$$
is (doubly) stochastic with positive entries, is not diagonalizable, and has eigenvalues $1$ with multiplicity $1$ and $\lambda$ with multiplicity $2$ ($I$ denotes the identity matrix of size $3$). Because $P$ has an eigenvalue of magnitude not equal to $1$ or $0$, $\{P^k\mid k \in \Bbb N\}$ will contain infinitely many elements.

In this case, there was no reason to go about computing $P^n$ explicitly: if $P$  has an eigenvalue $\lambda$ with $|\lambda|\notin \{0,1\}$, then it necessarily holds that $\{P^k:k \in \Bbb N\}$ contains infinitely many elements. Indeed: suppose for the purpose of contradiction that there are finitely many elements. By the pigeonhole principle, there exist integers $k_1 \leq k_2$ for which $P^{k_1} = P^{k_2}$. If $v$ is the eigenvector associated with $\lambda$, then we have
$$
P^{k_1}v = P^{k_2}v \implies \lambda^{k_1}v = \lambda^{k_2}v \implies \lambda^{k_1 - k_2} = 1 \implies |\lambda|^{k_1 - k_2} = 1.
$$
However, this would imply that $|\lambda| = 1$, contradicting our premise.

For the matrix constructed above, note that the vector $v = (1,1,-2)^T$ satisfies $Pv = \lambda v$.

If you are interested in computing $P^n$ nevertheless, then we can make use of the fact that $P$ can be written in the form $P = D + N$ where $D$ is diagonalizable, $N^2 = 0$, and $DN = ND$ (this is a "Jordan-Chevalley" decomposition). In particular, we have
$$
D = \frac{1 - \lambda}{3} \pmatrix{1\\1\\1}\pmatrix{1&1&1}+ \lambda I, \\
N = \alpha \pmatrix{1\\1\\-2}\pmatrix{1&-1&0}.
$$
Verify that we have
$$
D^n = \frac{(1 - \lambda)^n}{3} \pmatrix{1\\1\\1}\pmatrix{1&1&1}+ \lambda^n I.
$$
From there, use the binomial theorem to compute
$$
P^n = (D+ N)^n = D^n + nD^{n-1}N + 0 = D^n + nD^{n-1}N\\
= D^n + n\lambda^{n-1}N.
$$

Another way to write $P$ (for ease of presentation):
$$
P = \lambda I + \pmatrix{1&1\\1&1\\1&-2} \pmatrix{(1 - \lambda)/3 & 0\\0 & \alpha} \pmatrix{1&1&1\\1&-1&0}.
$$


问:
How would I show that a function is one-to-one using symbols?
If I have a function $f: X\to Y$ and its inverse $f^{-1}: Y\to X$, how would I show that it is one-to-one using symbols?

I can not figure out if it would be correct for me to say $\forall\ y\in f^{-1}\ \exists!\ x$. We know that it is one-to-one, so there is only one x per y value. I am just not sure if this is correct.

答:
Injectivity of function $f:A\rightarrow B$ means that

$\forall a,a'\in A: f(a)=f(a')\Rightarrow a=a'$.


问:
Numerical estimation of second order derivative
Given that y" = [y(x+h)-2y(x)+y(x-h)]/h^2 (central difference) with error = O(h^2). Find  an expression for y" that has a smaller error than O(h^2).

Note: h is small (say 0 < h < 1).
Thanks in advance

答:
Thanks Mason.
So I need to do Richardson's Extrapolation which gives an error of O(h^4). Check link below:

http://home.cc.umanitoba.ca/~farhadi/Math2120/Numerical%20Differentiation.pdf


问:
Expressing decimals as a sequence of nested intervals
Let $I = [0,1]$ and $t\in[0,1]$

Define: $A_{p}(t) = \frac{p}{4}+\frac{1}{4}t,\quad$ $p\in\{0,1,2,3\}$

Let write $t$ in terms of the quaternary system

$t = 0.t_1t_2\ldots t_{n-1}t_n \ldots,\quad t_i\in\{0,1,2,3\}$

In my book there is stated that

$t = \lim_{n\to\infty} A_{t_1}A_{t_2}\ldots A_{t_n}(I)$

The thing i don't understand is the order in which the functions are evaluated. Why not this way:

$t = \lim_{n\to\infty} A_{t_n}A_{t_{n-1}}\ldots A_{t_1}(I)$ ?

答:
Think of the decimal version $A_p= \frac{p+t}{10}$.
Given a value $t=0.t_1t_2t_3...$, what this function does is add the digit $p$ to give $p+t=p.t_1t_2t_3...$, and then divide by $10$ to shift the decimal point to the left to give $A_p(t)=0.pt_1t_2t_3...$. So the function $A_p$ inserts the digit $p$ on the left end, just behind the decimal point.

So if you use these functions $A_p$ to build a decimal number, you are inserting digits on the left hand side, and when you stop the last one you inserted is the left-most digit. Therefore, to build $0.t_1t_2...t_n$ you will have to end with $t_1$ (and hence start with $t_n$).

The functions in the question do the same but in base $4$.


问:
Locally extend a morphism defined on stalks
I have the following: a locally free sheaf $F$ on a scheme $X$, an $\mathcal{O}_X$-module $G$ and, for a fixed $x\in X$, a morphisms of $\mathcal{O}_{X,x}$-modules $F_{x}\to G_{x}$. Is it true that we can extend this morphism locally, that is, to find $V$ open neighborhood of $x$ and a morphisms of $\mathcal{O}_V$-modules $F|_V\to G|_V$ inducing the original one on stalks?

My attempt really stops when trying to find this suitable $V$, because I know that, since we're working in abelian categories, for any $s^x\in F_x$ I can find an open neighborhood $U$ of $x$ and a section $s\in F(U)$ such that $s_x=s^x$, and mapping this with the morphism on stalks to, say, $t^x$, lead to consider $W\ni x$ and a section $t\in G(W)$ such that $t_x=t^x$. But it doesn't seem to be that this takes me somewhere in order to define a morphism of sheaves, not even on a small neighborhood, beacuse any of these small neighborhood that I can find depends on the element $s^x$.

(I also think that it would be easy to find such a morphism on any open $U$ if I had maps on stalks for every $x$ in $U$, but here just one morphism in given).

Any help is appreciated! Thanks in advance

答:
I write this down in full detail since others on this forum may have an interest in this question - it is a much asked question.

The question is local hence we may assume $X:=Spec(A)$, $E:=A\{e_1,..,e_n\}$ a free $A$-module of rank $n$ and $F$ any $A$-module. Let $p\in X$ be a point with corresponding prime ideal $\mathfrak{p}\in A$. Let $\phi_{\mathfrak{p}}: E_{\mathfrak{p}} \rightarrow F_{\mathfrak{p}}$ be an $A_{\mathfrak{p}}$-linear map. It follows $\phi(e_i)=x_i/t_i$ with $t_i\in A-\mathfrak{p}$ for all $i$. Let $t(i):=\prod_{k\neq i}t_k$  and let $t:=\prod_i t_i$. let $y_i:=t(i)x_i$. It follows $t(i),t\in \mathfrak{p}$ and there is an equality $y_i/t=x_i/t_i$ in $F_{\mathfrak{p}}$. The element $t$ is in $A$ and by construction it follows $\mathfrak{p}\in D(t)$. We may define an $A_t$-linear map $\phi_t:A_t\otimes E \rightarrow A_t \otimes F$ by $\phi_t(e_i):=y_i/t$. The map $\phi_t$ has the property that when we take the tensor product $A_{\mathfrak{p}}\otimes_{A_t}-$ we get back the map $\phi_{\mathfrak{p}}$. Let $\mathcal{E},\mathcal{F}$ be the sheaves correspoding to $E,F$.

Question: "Is it true that we can extend this morphism locally, that is, to find V open neighborhood of x and a morphisms of OV-modules F|V→G|V inducing the original one on stalks?"

Yes this is true. Above we have constructed a map of sheaves of $\mathcal{O}_{D(t)}$-modules $\phi_t:\mathcal{E}_{D(t)} \rightarrow \mathcal{F}_{D(t)}$ which gives back the map $\phi_{\mathfrak{p}}$ when we pass to the stalk at $\mathfrak{p}$. Here $\mathcal{E}_{D(t)}$ means we restrict the sheaf $\mathcal{E}$ to the open set $D(t)$.


问:
Does $x<y$ and $a\le b\implies x+a<y+b$? Similarly, does $x>y$ and $a\ge b\implies x+a>y+b$?
I know this is a very simple question, but I still have my doubts on this question:

Does $x<y$ and $a\le b\implies x+a<y+b$? Similarly, does $x>y$ and $a\ge b\implies x+a>y+b$? Why or why not?

Based on intuition, these seem to be true, but how would I "prove" this concretely? I'm not too sure if there are any rules that allow me to combine $<$ and $\le$(or $>$ and $\ge$) inequalities. It seems like there should be one, but I wasn't able to find any. Could someone help me on this?

答:
If you always remember this in your head: $x<y \iff x-y<0; a \le b \iff a-b \le 0$.

Then $(x+a)-(y+b) = (x-y)+(a-b) <0 \implies x+a<y+b.$

The other direction is similar.


答:
The list need not be finite. Here is an example:

$$P:=\begin{pmatrix}\tfrac{1}{4}&\tfrac{1}{4}& \tfrac{1}{2} \\
 \tfrac{1}{2} & \tfrac{1}{4} & \tfrac{1}{4} \\
 \tfrac{1}{2} & \tfrac{1}{2} & 0\end{pmatrix}$$

It is not diagonalizable, and the coefficients of $P^n$ do not repeat (the denominator of $P_{11}$ grows as $4^n$).

Title: Prove or disprove: $A=\left\{P^k\mid k\in\mathbb{N}\right\}$ has a finite number of elements

Body:
Let $P\in[0,1]^{n\times n}$  be a stochastic matrix, meaning for every $i\in[1,n]_{\mathbb{N}}$:

$$\sum_{j=1}^nP_{ij}=1$$

It is given that $P$ is not diagonalizable. Define the group $A$ to be a group of square $n\times n$ matrices such that:

$$A=\left\{P^k\mid k\in\mathbb{N}\right\}$$

Prove or disprove: $A$ has a finite number of elements.

I'm asking this question because it is often required to compute $P^n$ when dealing with Markov Chains. I did a little research and found out that a stochastic matrix doesn't have to be diagonalizable, what eventually led me to think - How can I compute $P^n$ if $P$ is not diagonalizable? The statement that I'm asking your help proving or disproving is somewhat related - since if it is true, then my problem is solved. It might be false, though. If it is indeed false - first, I would be glad to see why, and additionally, I'd be happy to hear your thoughts on what should I do if I am presented with a non-diagonalizable $P$ and need to compute $P^n$. Thanks!


问:
Binomial identity involving Harmonic numbers
Using numeric experiments, I have found the following identity
$$
k! H_k+\sum _{j=1}^k \frac{\sum _{i=j}^k (-1)^i \binom{k}{i} \left((j-i)^k-(-i-n+1)^k\right)}{j+n-1}=0
$$
where $H_k = \sum_{i=1}^k 1/i$. This seems to hold for all $k \ge 1$ and $n \ge 1$.

It's fairly easy to check this for any fix $k$. But I am not able to prove it for general $k$. Induction also does not seem to work well in this case.

Update, you can check this with the following Mathematica code

```<code>Table[k!*HarmonicNumber[k] +
       Sum[((-1)^(1 + i + k)*(-1 + i + n)^k*Binomial[k, i] +
              Eulerian[k, i])*HarmonicNumber[-1 + i + n], {i, 0,
     k}] ==
     0, {k, 1, 6}, {n, 1, 10}]
</code>```

答:
This is not a whole solution but I think it can be salvaged. If we apply the well known identity $a^k - b^k = (a-b) \cdot \sum\limits_{i=1}^{k-1} a^{k-1-i} b^i $ for $(a,b) = ((j-i),(-i-n+1))$  then the conjecture above boils down to proving that:

\begin{eqnarray}
rhs_\eta := \sum\limits_{j=1}^k \sum\limits_{i=j}^k \sum\limits_{\xi=\eta}^{k-1}
(-1)^i \binom{k}{i} \binom{\xi}{\eta} (j-i)^{k-1-\xi} (1-i)^{\xi-\eta} = -k! H_k \cdot \delta_{\eta,0}
\end{eqnarray}
for $\eta=0,1,2,\cdots$. This comes from the fact that the sum in question is now a polynomial in the variable $n$ whose all coefficients except for the zeroth term must vanish.

Let us take a look at the coefficient at the zeroth power of $n$ (meaning $\eta=0$). We have:

\begin{eqnarray}
rhs_0 &=& -k + \sum\limits_{j=2}^k \sum\limits_{i=j}^k (-1)^i \binom{k}{i} \left( \frac{-(1-i)^k + (-i+j)^k}{j-1}\right) \\
&=& -k + \sum\limits_{i=1}^k (-1)^{i-1-k} (i-1)^k \binom{k}{i} \left( H_{i-1} - \psi^{(0)}_{i+1} + \psi^{(0)}_{k+1} \right) \\
&=& -k + \sum\limits_{i=0}^{k-1} (-1)^{i+k}  i^k \binom{k}{i+1} \left(-\frac{1}{i+1} + H_k \right) \\
&=& -k+ \sum\limits_{i=0}^{k-1} (-1)^{i+k}  i^k \binom{k}{i+1} \left(-\frac{1}{i+1} \right) + (1-k!) H_k \\
&=& -k+ (k - H_k) + (1-k!) H_k
\end{eqnarray}

In the first line from the top we carried out a sum over $\xi$ since  it is a geometric series.  In the second line from the top we split the double sum into two sums and in the first sum we did the sum over $j$  whereas in the second sum we substituted for $(i-j) \rightarrow i $ and then we did the sum over $j$ as well. In the third line we substituted $(i-1) \rightarrow i $ and then we used the the relationship between the polygamma function and harmonic numbers. In the last two lines we did the remaining single sums by doing the usual tricks $i^k = \left. d_t^k \exp(i t) \right|_{t=0} $ and then using the binomial expansion to resum.

In order to complete the proof one needs to show that $rhs_\eta =0 $ for all integer $\eta >0 $ . This should not be hard to do.

```<code>In[530]:= n =.;

Table[Sum[(-1)^(i + xi - eta) Binomial[k, i] Binomial[xi, eta] If[
       i == j && k - 1 == xi, 1, (j - i)^(k - xi - 1)] If[
       i == 1 && xi == eta, 1, (i - 1)^(xi - eta)], {j, 1, k}, {i, j,
      k}, {xi, eta, k - 1}] (-n)^eta, {k, 1, 5}, {eta, 0, k - 1}] //
  Expand // MatrixForm

Table[-k +
  Sum[(-1)^i Binomial[k, i] ((-(1 - i)^k + (-i + j)^k)/(-1 + j)) , {j,
     2, k}, {i, j, k}], {k, 1, 5}]

Table[-k +
  Sum[(-1)^(i - 1 + k) ((i - 1)^k)  Binomial[k,
     i] (HarmonicNumber[i - 1] - PolyGamma[0, 1 + i] +
      PolyGamma[0, 1 + k]), {i, 1, k}], {k, 1, 5}]

Table[-k +
  Sum[(-1)^(i + k) ((i)^k)  Binomial[k,
     i + 1] (-1/(i + 1) + HarmonicNumber[k]), {i, 0, k - 1}], {k, 1,
  5}]
Table[-k +
  Sum[(-1)^(i + k) ((i)^k)  Binomial[k, i + 1] (-1/(i + 1)), {i, 0,
    k - 1}] + (1 - k!) HarmonicNumber[k], {k, 1, 5}]
Table[-k + (k - HarmonicNumber[k]) + (1 - k!) HarmonicNumber[k], {k,
  1, 5}]
</code>```

Body:
This is not a whole solution but I think it can be salvaged. If we apply the well known identity $a^k - b^k = (a-b) \cdot \sum\limits_{i=1}^{k-1} a^{k-1-i} b^i $ for $(a,b) = ((j-i),(-i-n+1))$  then the conjecture above boils down to proving that:

\begin{eqnarray}
rhs_\eta := \sum\limits_{j=1}^k \sum\limits_{i=j}^k \sum\limits_{\xi=\eta}^{k-1}
(-1)^i \binom{k}{i} \binom{\xi}{\eta} (j-i)^{k-1-\xi} (1-i)^{\xi-\eta} = -k! H_k \cdot \delta_{\eta,0}
\end{eqnarray}
for $\eta=0,1,2,\cdots$. This comes from the fact that the sum in question is now a polynomial in the variable $n$ whose all coefficients except for the zeroth term must vanish.

Let us take a look at the coefficient at the zeroth power of $n$ (meaning $\eta=0$). We have:

\begin{eqnarray}
rhs_0 &=& -k + \sum\limits_{j=2}^k \sum\limits_{i=j}^k (-1)^i \binom{k}{i} \left( \frac{-(1-i)^k + (-i+j)^k}{j-1}\right) \\
&=& -k + \sum\limits_{i=1}^k (-1)^{i-1-k} (i-1)^k \binom{k}{i} \left( H_{i-1} - \psi^{(0)}_{i+1} + \psi^{(0)}_{k+1} \right) \\
&=& -k + \sum\limits_{i=0}^{k-1} (-1)^{i+k}  i^k \binom{k}{i+1} \left(-\frac{1}{i+1} + H_k \right) \\
&=& -k+ \sum\limits_{i=0}^{k-1} (-1)^{i+k}  i^k \binom{k}{i+1} \left(-\frac{1}{i+1} \right) + (1-k!) H_k \\
&=& -k+ (k - H_k) + (1-k!) H_k
\end{eqnarray}

In the first line from the top we carried out a sum over $\xi$ since  it is a geometric series.  In the second line from the top we split the double sum into two sums and in the first sum we did the sum over $j$  whereas in the second sum we substituted for $(i-j) \rightarrow i $ and then we did the sum over $j$ as well. In the third line we substituted $(i-1) \rightarrow i $ and then we used the the relationship between the polygamma function and harmonic numbers. In the last two lines we did the remaining single sums by doing the usual tricks $i^k = \left. d_t^k \exp(i t) \right|_{t=0} $ and then using the binomial expansion to resum.

In order to complete the proof one needs to show that $rhs_\eta =0 $ for all integer $\eta >0 $ . This should not be hard to do.

In[530]:= n =.;

Table[Sum[(-1)^(i + xi - eta) Binomial[k, i] Binomial[xi, eta] If[
       i == j && k - 1 == xi, 1, (j - i)^(k - xi - 1)] If[
       i == 1 && xi == eta, 1, (i - 1)^(xi - eta)], {j, 1, k}, {i, j,
      k}, {xi, eta, k - 1}] (-n)^eta, {k, 1, 5}, {eta, 0, k - 1}] //
  Expand // MatrixForm

Table[-k +
  Sum[(-1)^i Binomial[k, i] ((-(1 - i)^k + (-i + j)^k)/(-1 + j)) , {j,
     2, k}, {i, j, k}], {k, 1, 5}]

Table[-k +
  Sum[(-1)^(i - 1 + k) ((i - 1)^k)  Binomial[k,
     i] (HarmonicNumber[i - 1] - PolyGamma[0, 1 + i] +
      PolyGamma[0, 1 + k]), {i, 1, k}], {k, 1, 5}]

Table[-k +
  Sum[(-1)^(i + k) ((i)^k)  Binomial[k,
     i + 1] (-1/(i + 1) + HarmonicNumber[k]), {i, 0, k - 1}], {k, 1,
  5}]
Table[-k +
  Sum[(-1)^(i + k) ((i)^k)  Binomial[k, i + 1] (-1/(i + 1)), {i, 0,
    k - 1}] + (1 - k!) HarmonicNumber[k], {k, 1, 5}]
Table[-k + (k - HarmonicNumber[k]) + (1 - k!) HarmonicNumber[k], {k,
  1, 5}]


问:
Derived category and so on
I am looking for an introductive reference to the theory of derived categories. Especially I need to start from the very beginning and I need to know how to use this in examples which comes from algebraic geometry. I don't want a too rigorous approach, made of a lot of definition and propositions but instead I would like to find an introduction which gives the main ideas and many examples.
Thank you</p>

答:
I like "Sheaves in Topology" by Dimca a lot.

I also second the suggestion of Gelfand/Manin. They actually have two books for some reason, "Homological Algebra" and "Methods of Homological Algebra", which are quite similar but have slightly different focus/applications. Both of them are worthwhile, and I think either one of them could be useful to you.

Title: Derived category and so on


问:
How to find the confidence interval for a prediction using a multiple linear regression model?
I have a dataset of 100 observations and 5 numerical variables ($x_1, x_2, ..., x_{5}$) which I used to generate a linear model which to predict some value $y$. I plugged in some values into the model and got a result for $\hat{y}$.

I would like to calculate a 95% confidence interval for this value.

Using Python I get the following model summary:

```<code>Adj. R-squared: 0.892
F-statistic:    27.63
Prop(F-stat):   2.45e-18

        coef   std err   t      P>|t|   [0.025    0.975]
const  -12.71  3.713    -3.424  0.001   -20.156  -5.268
x1       0.29  0.255     1.134  0.262   -0.222   -0.800
x2       0.17  0.031     5.283  0.000    0.103    0.228
x3       0.44  0.089     4.970  0.000    0.262    0.617
x4      -0.47  0.194    -2.414  0.019   -0.856   -0.079
x5       0.36  0.226     1.587  0.118   -0.094    0.811
</code>```

I am aware that in order to calculate a 95% confidence interval for a simple linear regression the formula is $\hat{\beta} \pm t_{0.95, f}\: \cdot \: \hat{\sigma}_{B_0}$ where $f$ is the degree of freedom and $\hat{\sigma}_{B_0}$ is the std err. But how do I calculate this for multiple linear regression?

答:
You cannot compute it from the output you have available in the screenshot you pasted because that only provides the standard errors of the estimated coefficients while ignoring any covariance between them. In other words the variance of the predicted value cannot be simply computed combining the variances of the estimated coefficients because these are not independent.

In a Bayesian setting, what you are looking for is called posterior predictive distribution.

Assuming an improper prior $p(\beta, \sigma^2) \propto \frac1{\sigma^2}$ on the parameters and that your new test points are conditionally independent of your training data, given $\beta$ and $\sigma^2$, then you can find the distribution of new test points as described in [1]. In a nutshell, if you try to predict more than one new test point, the vector of these will follow a multivariate t distribution (since they will all depend on the estimated regression coefficients, the predicted test points will not be independent). If you are only interested in a single test point $\tilde y$, then the maths simplifies and following holds:

$$\frac{\tilde y - \tilde X \hat\beta}{\sqrt{h}} \sim t_{(T-K)}$$

where $T$ is the number of training examples, $K$ is the number of coefficients, $\tilde X$ is the row vector of regressors for the new point, and $\hat\beta$ and $h$ are as follows:
\begin{align}
\hat\beta &= (X' X)^{-1} X'y \\
s^2 &= \frac{1}{T-K} \, (y-X\hat\beta)'(y-X\hat\beta)\\
h &= \frac{1}{s^2 \, [1 + \tilde X(X' X)^{-1}\tilde X']},
\end{align}
where $y$ is the $T\times 1$ vector of observations and $X$ is the $T \times K$ design matrix corresponding to your training data.

[1] Zellner, A.; Chetty, V. K., Prediction and decision problems in regression models from the Bayesian point of view, J. Am. Stat. Assoc. 60, 608-616 (1965). JSTOR.


问:
Writing an equation in self-adjoint form
I'm trying to solve a S-L problem and I need to write this equation in self adjoint form. How do I go about doing this?

My equation is:
$X''(x)+ 2X(x) +\lambda X(x)=0$

答:
The self-adjoint form is
$$
(p(x)X'(x))'+q(x)X(x)=0.
$$
In its expanded form
$$
p(x)X''(x)+p'(x)X'(x)+q(x)X(x)=0
$$
this has to be a multiple of the given equation. That is, $(p(x),p'(x),q(x))$ has to be proportional to $(1,2,λ)$. From $p'=2p$ one gets $p(x)=e^{2x}$. $q=λp$ gives the second coefficient function.


问:
Probability of getting a white ball.
I have the following tasks:

I have two boxes:

Box 1: 4 white balls and 3 black balls

Box 2: 2 white balls and 1 black ball.

Two random balls are picked from box 1 and transferred to box 2.
Then two random balls are picked from box 2 and transferred to box 1.

Then we pick a random ball from box 2. What is the probability the ball to be white?

So my aproach was to calculate all possible cases.

Case 1: Two white balls are transferred (box 1->box2) then two white balls are transfered(box2->box1). Now what is the probability of picking a white ball.

Case 2: ...

and so on. And then sum the cases.
The problem is that this is not an easy tasks. There are 9 cases and the calculations are heavy.

Is there an easier way for solving this problem? Any tips?

答:
Note: I think this response should be in the comments rather than as an answer, but I don't have enough reputation, so... yeah.

The second step (i.e. 2 balls transferred from Box 2 to Box 1) doesn't actually affect the probability.

Think about the case where you have 5 balls, say 3 are white and 2 are black, and you first remove 2 balls randomly then pick one ball out of the remaining 3. You can do some math on it, but what is really happening is just you picking one ball out of the 5 possible balls from the bag. 4 balls are left unpicked anyway, no matter how many balls you first remove.

After disregarding the second transfer, you can split the 3 cases of the first transfer (2W, 1W1B, 2B) and add up the probabilities to get the final answer.

Title: Probability of getting a white ball.


问:
Generating function for distributing $k$ indistinguishable elements into $n$ distinguishable bins such that each bin has even number of elements.
Generating function for distributing $k$ indistinguishable elements into $n$ distinguishable bins such that each bin has even number of elements.

So for the bins to have exactly even number of elements we assume that $k$ is even and we distribute them into pairs so there are $x = k/2$ indistinguishable elements to insert in $n$ bins. How do I generate the function for the series?

答:
We have $k$ balls and $n$ bins where $k$ is even.

$\sum \limits_1^n b_i = k \,, \, b_i$ is number of balls in each bin and it is even.

If the restriction of even was not there, your polynomial for each factor (box) would be $(1 + x + x^2 + .. + x^k)$.

Given the restriction, we only pick even terms.

So the polynomial for each factor would be $(1 + x^2 + x^4 + .. + x^k) \,$ (considering empty box as even)

For $n$ factors, the generating function would be

$g(x) = (1 + x^2 + x^4 + .. + x^k)^n \,$. You have to take the coefficient of term $x^k$.

If you need to take this further,

$g(x) = (1+x^2+x^4+\dots +x^k)^n = \left( \frac{1-x^{k+2}}{1-x^2} \right)^n$

$ = (1-x^{k+2})^n \times (1-x^2)^{-n}$

$= \sum (-1)^i \binom{n}{i} x^{(k+2)i} \times \sum \binom{n + j - 1}{j} x^{2j}$

and take the coefficient of term $x^k$ which means

$(k+2)i + 2j = k \implies i = 0, j = \frac{k}{2} = m \, $ (say).

So we get ${n + m - 1 \choose m}$

Or you can simply use the infinite geometric series -

$g(x) = (1 + x^2 + x^4 + .. + ...)^n \, = \frac{1}{(1+x^2)^{n}} = (1+x^2)^{-n}$

$ = \sum \limits_{i=0}^{\infty} {n + i - 1 \choose i} x^{2i}$.

So $, \,i = \frac{k}{2} = m \,$ will be the $x^k$ term and coefficient is
${n + m - 1 \choose m}$.


问:
Noetherian $R$-algebra corresponds to a coherent sheaf of rings on $\operatorname{Spec}(R)$
Let $R$ be a ring and $A$ a Noetherian $R$-Algebra. Let $\newcommand{\m}{\mathcal} \m{A} = \tilde{A}$ be the corresponding $\m{O}_X$-Module, where $(X, \m{O}_X) = \operatorname{Spec}(R)$. I would like to show that $\m{A}$ then is a coherent $\m{A}$-Module on X.

Is the following attempt at a proof correct?

We first need that $\m{A}$ is of finite type, which is trivial. The nontrivial part is to prove that for every short exact sequence of $\m{A}$-Modules
$$
0 \to \m{K} \to \m{A}^n \to \m{A} \to 0
$$
the kernel $\m{K}$ is an $\m{A}$-Module of finite type.

Since $\m{A}$ and $\m{A}^n$ are quasi-coherent $\m{O}_X$-Modules, $\m{K} = \tilde{K}$ is quasi-coherent, with $K = \Gamma(X, \m{K})$. $K$ is an $\Gamma(X, \m{A})$-Module and by the equivalence of categories between $\mathsf{qCoh}(X)$ and $R$-$\mathsf{Mod}$,
$$
0 \to K \to A^n \to A \to 0
$$
is an exact sequence (in $R$-$\mathsf{Mod}$, and hence also in $A$-$\mathsf{Mod}$). Since $A$ is Noetherian, it follows that $K$ is a $A$-Module of finite type, i.e. there is a surjection $A^m \to K \to 0$. Applying $\tilde{}$ gives an exact sequence
$$
\m{A}^m \to \m{K} \to 0
$$
(again, at first this is only exact in $\m{O}_X$-$\mathsf{Mod}$). $\square$

Background: Proposition 16.1.8 in EGA IV (part 4), which (I think) uses this fact.

(Grothendieck writes that this follows from the exactness of inverse images and $X = Y \times_{Y \times_S Y} (X \times_S Y)$, however I don't see how this is helpful at all. Alternatively - my french is not very good - these facts could be explaining why $\operatorname{gr}_I^\bullet A$ is Noetherian for $A$ Noetherian. If you can make sense of any of this please drop me a comment.)

答:
If $R$ is a commutative ring and $A$ a commutative $R$-algebra, it seems you may do the following: Let $X:=Spec(R)$. You want to define a sheaf of commutative $\mathcal{O}_X$-algebras on $X$. Let $\mathcal{O}:=\mathcal{O}_X$. For any open set $U \subseteq X$ you define $\mathcal{O}(U)$ as the limit over $R_f$ as $f\in A$ and $D(f)\subseteq U$. You may define $\mathcal{A}(U)$ as follows. View $A$ as a left $R$-module and define $\mathcal{A}(U)$ as the limit of $R_f\otimes_R A$ for $f \in R$ with $D(f)\subseteq U$. It follows $\mathcal{A}$ is a sheaf of $\mathcal{O}$-modules. Since tensor product commute with limits, it follows there is a canonical isomorphism $\mathcal{A}(U)\cong \mathcal{O}(U)\otimes_R A$. The $\mathcal{O}(U)$-module $\mathcal{A}(U)$ is canonically a commutative ring since it is the tensor product of commutative rings and all restriction maps are maps of rings, and it follows you have defined a sheaf of commutative $\mathcal{O}$-algebras $\mathcal{A}$ with the property that $\mathcal{A}(X)=A$.

Conversely given any quasi coherent sheaf of $\mathcal{O}$-algebras $\mathcal{A}$ it follows $\mathcal{A}(X):=A$ is a commutative $R$-algebra, whose sheafification gives back $\mathcal{A}$.

It seems: There is an equivalence of categories between the category of commutative $R$-algebras and maps of $R$-algebras and the category of sheaves of commutative $\mathcal{O}$-algebras and morphism of sheaves of commutative $\mathcal{O}$-algebras.

In the non-commutative case there are problems due to the fact that localization does not behave well for non-commutative rings. There is a class of rings where we can localize: rings of differential operators.

Example 0. The module of derivations. Let $\phi: A \rightarrow B$ be a map of commutative rings and let $T\subseteq A, S\subseteq B$ be  multiplicative subsets with $\phi(T) \subseteq S$. Let $\Omega^1_{B/A}$ be the module of Kahler differentials of $B/A$. There is a canonical isomorphism

$S^{-1}\Omega^1_{B/A}\cong \Omega^1_{S^{-1}B/T^{-1}A}$.

Hence when we dualize we get an isomorphism

$S^{-1}Der_A(B) \cong Der_{T^{-1}A}(S^{-1}B)$.

Hence the module of derivations localize to give a sheaf of $\mathcal{O}_Y$-modules on $Y:=Spec(B)$ relative to the canonical map $\pi: Y \rightarrow S:=Spec(A)$. Note that the localized module $S^{-1}Der_A(B)$ is canonically a $T^{-1}A$-Lie algebra. This localize and it follows the tangent sheaf $\Theta_{Y/S}$ (which is the sheafification of the module of derivations) is a sheaf of $\mathcal{O}_S$-Lie algebras. More precisely: it is a sheaf of $\mathcal{O}_S$-Lie-Rinehart algebras/Lie algebroids. This is a much studied subject.

Example 1: Jet bundles/principal parts. More generally if $\mathcal{P}^k_{B/A}:=B\otimes_A B/I^{k+1}$ is the $k$'th module of principal parts/$k$'th jet bundle of $B/A$, it follows

$S^{-1}\mathcal{P}^k_{B/A} \cong \mathcal{P}^k_{S^{-1}B/T^{-1}A}$.

When dualizing we get an isomorphism

$S^{-1}Diff^k_A(B) \cong Diff^k_{T^{-1}A}(S^{-1}B)$

where $Diff^k_A(B)$ is the module of $k$'th differential operators of $B/A$.
This gives a lozalization of the ring of differential operators $\mathcal{D}_A(B)$: There is an isomorphism

$S^{-1}\mathcal{D}_A(B) \cong \mathcal{D}_{T^{-1}A}(S^{-1}B)$.

Hence if we define for an open set $U \subseteq Y$ (with $\pi(U) \subseteq V$ where $V\subseteq S$ is an open set) $\mathcal{D}_{Y/S}^k(U)$ as the limit of $B_f \otimes_B Diff^k_A(B)$ for $D(f) \subseteq U$, it follows

$\mathcal{D}_{Y/S}^k(U)\cong Diff^k_{\mathcal{O}_S(V)}(\mathcal{O}_Y(U))$

Similarly we get

$\mathcal{D}_{Y/S}(U)\cong Diff_{\mathcal{O}_S(V)}(\mathcal{O}_Y(U))$.
It follows $(\mathcal{D}, \mathcal{D}^k)$ becomes a sheaf of filtered associative rings with the property that the associated graded ring is commutative. The structure sheaf $\mathcal{O}_Y$ is not in the center of $\mathcal{D}$.

Example 2: If $A=k$ is a field and $B:=k[x]$ is the polynomial ring in one variable it follows $\mathcal{D}_{B/A}$ is the Weyl algebra $W(1):=k[x,\partial]$ in one variable. Here $\partial$ is partial derivative wrto the $x$-variable. It follows $B$ is not in the center of $W(1)$ in general.

Example 3. Generalized ring of differential operstors. In general if $(U, U_i)$ is an $\mathbb{N}$-filtered associative ring  with $U_0:=B$ and $A$ in the center $c(U)$ such that the associated graded ring is commutative, we may localize. The filtered associative ring $(U,U_i)$ gives rise to a filtered sheaf $(\mathcal{U}, \mathcal{U}_i)$ of associative rings on $Y$ with $\mathcal{O}_S$ in the center, whose associative graded sheaf of rings is a sheaf of commutative $\mathcal{O}_Y$-algebras. Hence this type of "generalized ring of differential operators" localize well and give rise to sheaves of associative rings on $Y$. Many of the non-commutative rings appearing in mathematics/mathematical physics are of this type.

Example 4. We may view "non-commutative algebraic geometry" as the study of the following pairs: A non-commutative space is a pair $((Y,\mathcal{O}_Y), (\mathcal{U}, \mathcal{U}_i))$ where $(Y,\mathcal{O}_Y)$ is a scheme and $(\mathcal{U}, \mathcal{U}_i)$ is a sheaf of generalized differential operators in the above sense. There is a subject called "D-Lie algebras" where this notion is systematized. Such D-Lie algebras have many properties similar to Lie-Rinehart algebras/Lie algebroids/D-modules. We may introduce notions such as enveloping algebras, connections, Ext and Tor groups, characteristic varieties etc, and many of the results valid for Lie algebroids may be generalized to this new situation.

Example 5. Characteristic $p>0$. If $k$ is a field of characteristic $p$ and if $A:=k[X]$, it follows the ring $R:=k[x^p]\subseteq A$ is in the center of the ring $\mathcal{D}_{A/k}$. Hence in characteristic $p>0$ it follows the ring of differential operators has a non-trivial center.


问:
Find closures in functional analysis
I have many exercises about finding the closures of different sets.

For example: Find the closure in $C[-1/2,1]$ of the vector space generated by the functions $ [e^{kx^2};k\in \mathbb N ]$

I would like to know the idea of the solution of this kind of exercises.

Thank you.

答:
I don't have a general recipe.

Here you can note that all your functions are even, so linear combinations and limits will also be even. If you focus on the $[0,1]$ part, you can show that your functions separate points (take logarithms, note that numbers are positive). Since the span is an algebra, by Stone Weierstrass you get all of $C[0,1]$. So the closure is
$$
\{t\longmapsto f(|t|):\ f\in C[0,1]\}.
$$

Title: Find closures in functional analysis


问:
Bernoulli trials and permutations
While solving questions using the Binomial probability theorem, the result is the probability of r successes among n trials, given that there are only two outcomes of probability p and q where p represent success(occurrence of desired event) and q represents failure(nonoccurrence of desired event), yes?

My doubt is this; does this theorem take into account all the different possibilities? For example, if I were applying this theorem to the situation of a family having three children:

If p is the probability of success of the birth of one boy, p=1/2 and q=1/2. Can we obtain the probability of 1 boy and 2 girls by

$³C1$x$(1/2)²$x$1/2=3/8?$ Does this account for all the permutations, viz. BBG, BGB and GBB? I am confused because we have used nCr here and not nPr.

Please share your insights and help me to understand this. Much thanks in advance :) Regards.

答:
Here when we calculate P(2 Girls and 1 boy) using Combination nCr, as we donot care about the order in which the children are born.
If we had used nPr, we are actually considering all possible orders(BGG, GGB, GBG..), which is not needed, so while using nCr we are actually considering combination of 1 Boy and 2 Girls.

P(A and B) = P(A) * P(B|A)
P(2 Girls and 1 Boy) = P(2 Girls) * P(1 boy| 2 Girls were born)

P(2 Girls) = Number of ways in which 2 girls are born out of 3 children(in any order) * P(Girl being born)^r * P(Boy being born)^(n-r)

P(2 Girls) = 3C2 * 0.5^2 * 0.5^1 =  3 * 0.25 * 0.5 = 0.375

P(1 boy | 2 Girls were born) = 1, as the total children are 3 and 2 are girls

P(2 Girls and 1 Boy) = P(2 Girls) * P(1 boy | 2 Girls were born)= 0.375 *1 = 0.375

Title: Bernoulli trials and permutations


问:
Movie Earnings and Relative Ticket Sizes
Given that movie A made 308 million and a 1 by 3 cm ticket represented A and movie B made 609 million, assuming that prices have been adjusted for inflation and ticket B is the same shape as ticket A, what would the size of ticket B be? </p>

<p>I said if the area of A was 3 cm which represented 308 million than ticket B would just need to be about 1.97 $cm^2$ larger i.e. have an area of about 5.91. I got this because since its a graphical representation then ticket B should be proportional to ticket A. Not really sure if it's right so if anyone would check it out and let me know, thanks! </p>

答: Making the areas proportional, as you've suggested, is the most reasonable way to represent the data; however I've also seen in some places scaling just one axis proportionally.  If you did this, you would make the long side $3\times \frac{609}{308}=5.93$cm, and the other side $1\times \frac{609}{309}=1.98$cm.</p>


问:
$X^2 + Y^2 \equiv 0 \pmod p$ implies that $-1$ is a square in $\mathbb{Z}/p\mathbb{Z}$
I know that similar questions have been asked several times in the past but I want a precision on a part of the proof.

Let $n = x^2 + y^2$ and $p$ be a prime factor of $n$. Let $d = \text{gcd}(x,y)$ where $x=dX$ and $y=dY$ with $(X, Y)$ being coprime integers.

I have shown that $X^2 + Y^2 \equiv 0 \pmod p$.

Now I want to show that $\bar{-1}$ is a square in $\mathbb{Z}/p\mathbb{Z}$.

Here is a sketch of my proof:

\begin{align*}
    X^2 + Y^2 &\equiv 0 \pmod p \\
    \iff X^2 &\equiv - Y^2 \pmod p \\
    \iff \Biggl(\frac{X}{Y}\Biggr)^2 &\equiv -1 \pmod p
\end{align*}

where $Y^{-2}$ exists since we know that for $p$ a prime, $\mathbb{Z}/p\mathbb{Z}$ is a field.

My question is: how do we justify that $Y \not\equiv 0 \pmod p$ ? (or alternatively $X \not\equiv 0 \pmod p$). It has to do with $X$ and $Y$ being coprime integers I guess.

答:
If $Y\equiv0\pmod p$, then $X^2\equiv-Y^2\equiv0\pmod p$,

so $p|X^2$, so $p|X$,

so $p|X$ and $p|Y$,

so $X$ and $Y$ are not co-prime, as you surmised.


问:
A problem in geometry(tangents) to show that $CD= 2AB \cdot (AC^2−AB^2)/(AC^2+AB^2)$
Suppose that $O$ is the circumcentre of Triangle $BCD$. Draw the tangent to the circumcircle of $BCD$ at $B$. Extend $CO$ meet this tangent at $A$. Show that

$$CD= 2AB \,\frac{AC^2−AB^2}{AC^2+AB^2}$$

What I thought was to apply Cevas theorem or Menelaus theorem. But things are not working out.I need some little help. I can't show what I have done as absolutely other than construction I could not do anything.
So I would be obliged if anyone can help.

答:
Comment: We have:

$AC^2=AB^2+AC^2-2AB\cdot AC cos(\angle BAC)$

$CD^2=AC^2+AD^2-2 AC\cdot AD cos(\angle CAD)$

Now applying Simson and Steward theorems may help you to find the relation.


问:
Determine the characteristic function of $X=\sum_{n=1}^{N} Y_{n}$, with $N \sim P(\lambda)$ and $Y_{i}$ i.i.d random variables
I need to determine the characteristic function of $X$,  $E(X)$ and $\operatorname{Var}(X)$, where $X=\sum_{n=1}^{N} Y_{n}$, with $N \sim P(\lambda)$ and $Y_{i}$ i.i.d random variables such as: $P(Y_{i}=1)=1/2$, $P(Y_{i}=2)=1 / 3$ and $P(Y_{i}=3)=1 / 6$.

This seems rather straightforward, but, for some reason, I can't seem to easily determine the characteristic function of $X$.

答:
Suggestion:

Define $S_n=\sum^n_{k=1}Y_k$ so that  $X=S_N$. Then

$\mathbb{E}[e^{itX}]=\mathbb{E}\big[\sum^\infty_{n=0}\mathbb{1}(N=n)e^{itS_N}\big]=\sum^\infty_{n=0}\mathbb{E}\big[\mathbb{1}(N=n)e^{itS_n}\big]$

Under the assumptions that $N$ is independent from $(Y_n:n\in\mathbb{N})$ and $(Y_n:n\in\mathbb{N})$ are i.i.d.

$$
\mathbb{E}\big[\mathbb{1}(N=n)e^{itS_n}\big]=\mathbb{P}[N=n]\Big(\mathbb{E}[e^{itY_1}]\Big)^n
$$

I hope you can continue from this.


问:
Area of paraboloid on top of a triangle

Calculate the area of a paraboloid $z=y^2$ that lies on top of the triangle whose tips are located at the origin and $(0,1)$ & $(1,1)$.

The situation looks as follows:

However, I'm not entirely sure how I can integrate this to find the area of the "patch" that the $xy$-triangle bounds. Should I define equations for the different lines that the points on the triangle defines or what? Any hints are appreciated.

Title: Area of paraboloid on top of a triangle

Body:
Evaluate the area as follows
\begin{align}
Area&= \int_S \sqrt{1+(z_x’)^2 +(z_y’)^2}\>dxdy\\
&= \int_0^1 \int_0^y \sqrt{1+4y^2}\>dxdy\\
&= \int_0^1 \sqrt{1+4y^2}y\>dy\\
 &=\frac14  \int_0^2 \sqrt{1+t^2}t\>dt\>\>\>\>\>\>(t=2y)\\
&=\frac1{12}(5\sqrt5-1)
\end{align}