问:
Difficulty finding Dual
min $\left \langle a,x \right \rangle$ subject to $\left \| x \right \|^2 \le 1$

What was done:

$L(x,\lambda) = a^Tx + \lambda^T(x^Tx-1) = a^Tx + \lambda^Tx^Tx-\lambda^T$

 that is, $L(x,\lambda) = inf\{(a^T+\lambda^Tx^T)x\}-\lambda^T$

 Gradients are:

$\nabla f(x) = a$ e $\nabla g(x) = 2x$

Therefore, $\nabla L(x,\lambda) = \nabla f(x) + \lambda \nabla g(x) = 0$

$x^* = \frac{-a}{2\lambda}$

Replacing $x^*$ em $L(x, \lambda)$ we have:

$(a^T + \lambda^T(\frac{-a}{2\lambda})^T)(\frac{-a}{2\lambda}) - \lambda^T$

My doubt is, that when I do this distributivity I end up not seeing the dual.

答:
$$ is $a^Tx$ not $ax^T$

$\lambda$ is a scalar so no point in writing $\lambda^T$

When replacing $x^*$ you forget multiplying it with $a^T$

You are incorrectly replacing $x^Tx$

Title: Difficulty finding Dual

Body:
$<a,x>$ is $a^Tx$ not $ax^T$

$\lambda$ is a scalar so no point in writing $\lambda^T$

When replacing $x^*$ you forget multiplying it with $a^T$

You are incorrectly replacing $x^Tx$


问:
Definition of Norm of an Ideal
I need to prove $(23,\alpha-10)$ is a prime ideal in $Z[\alpha]/(\alpha^3-\alpha-1)$. I want to show that $Norm(23,\alpha-10)$ is a prime number, thus proving it is a prime as we are in a Dedekind domain. Can anyone help me to compute the ideal norm?

Also, in particular, I know that for a PID, $R$, $Norm(I)$, $I\in R$ is given by the generator conjugate's products. Can anyone help me to get the connection between this definition and $Norm(I)=|R/I|$?

答:
Let's label $R=\mathbb{Z}[\alpha]/\langle\alpha^3-\alpha-1\rangle$ and $I=\langle 23,\alpha-10 \rangle$.  Then we can identify (as abelian groups)
$$R\cong \mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$$
via the map $x+y\alpha+z\alpha^2\mapsto (x,y,z)$. Note that this map identifies (again as abelian groups)
$$I\cong \langle(23, 0, 0), (-10, 1, 0), (-100, 0, 1)\rangle$$
(you should check this to be sure for yourself!).  At this point the quotient (as an abelian group) on the right hand side of this identification is clear, it isomorphic to $\mathbb{Z}/23\mathbb{Z}$.

ContentLicense: CC BY-SA 4.0

Title: Definition of Norm of an Ideal


问:
In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$.

In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$. If $CD = 7$ and $[\Delta ABD] = a\sqrt{5}$ , find $a$ .

What I Tried: Here is a picture:-

Let the perpendicular bisector of $BC$ pass through $BC$ at $E$ .

Then I first noticed that $\Delta BDE \cong \Delta CDE$ from $(SAS)$ congruency.

This gives the required information in the diagram, as well as we have $BD = 7$ .

Now :- $$\Delta ABD \sim \Delta ACB$$
$$\rightarrow \frac{AD}{AB} = \frac{7}{BC} = \frac{AB}{AC}$$
So let $AD = k$ , $AB = m$ , $BE = EC = n$ . We have :-
$$\frac{k}{m} = \frac{7}{2n} = \frac{m}{(7+k)}$$

答:
Let $E=(0,0)$, $B=(-\alpha,0)$, $C=(\alpha,0)$, $D=(0,\beta)$. Then we have
$$\alpha^2+\beta^2=7^2=49,\quad\tan \angle ACB=\frac{\beta}{\alpha}\in(0,\sqrt3).$$
Let $t=\tan \angle ACB$, then
$$y_{AB}=\frac{2t}{1-t^2}(x+\alpha),\quad y_{AC}=-tx+\beta.$$
And we have
$$A=\left(-\alpha\frac{1+t^2}{3-t^2},\alpha\frac{4t}{3-t^2}\right),\quad |AB|=2\alpha\frac{1+t^2}{3-t^2}.$$
Thus, the area
\begin{align}
S_{\triangle ABD} &= \frac12|AB|\cdot|DE|\\
&=\alpha\beta\frac{1+t^2}{3-t^2}\\
&=\frac{49t}{3-t^2}\in(0,\infty).
\end{align}
If you have some other conditions such that $\frac{t}{3-t^2}=\sqrt5$, then $a=49$. If not, $a$ can be any positive real number.


问:
$A$ is a vector subspace of $ℝ^4$?

Am I right to say row space of the matrix $A$ is a vector subspace of $ℝ^4$? I think that because $\operatorname{Col}(A)$ is a vector subspace of $\mathbb{R}^4$ and $\dim \operatorname{Col}(A) = \dim \operatorname{Row}(A)$.

答:
Since the rows each have 5 entries, you can see that the row space must be a subspace of $\Bbb{R}^5.$ The column space is a subspace of $\Bbb{R}^4.$

You can generalize this by considering your matrix as a linear transformation, $A:\Bbb{R}^5 \longrightarrow \Bbb{R}^4$

Where $A: \text{domain}\longrightarrow \text{codomain},$ more generally, and so your row space will be a subspace of your domain, and your column space will be a subspace of your domain.

Title: $A$ is a vector subspace of $ℝ^4$?


问:
(practical) Codes from sphere packings
I know that good codes allow you to construct good sphere packings in Euclidean space, e.g. the binary Golay code is the key feature of the construction of the Leech lattice.

Going in the other direction, I am wondering whether sphere packings in $\mathbf{R}^n$ actually help with designing good codes (for some definition of "code" that is useful in a practical setting). I know that codes on $n$ bits are sphere packings in Hamming space, but I'm wondering specifically about Euclidean sphere packings.

I know that kissing configurations yield spherical codes (by definition). Are these actually useful in practice? Also, do dense (lattice) packings produce any practically useful codes? From searching on Google I found a small section of a paper of H. Cohn (page 6 of https://arxiv.org/pdf/1003.3053.pdf) which gives a simplified example of how one can do coding with noise using good sphere packings, but I am wondering whether this is actually done in practice.

答:
The quick answer is no, this approach to code design in not used in practice.
Take a look at the codes used in 5G or WiFi for example. Both are large scale
projects where code performance and decoder complexity are key...neither use
sphere packings in the code design</p>

Title: (practical) Codes from sphere packings


问:
Points with given curvature on cubic Bézier curve
I am looking for all points on cubic Bézier curve that have a specific curvature $\kappa_1$.

The curvature $\kappa$ is caluclated as

$$
\kappa=\frac{x'y'' - y'x''}{\sqrt{({x'}^2+{y'}^2)^3}}
$$

Solving the equation above for a curvature $\kappa_1$ would first involve getting rid of the square root, so we have

$$
\kappa_1^2=\frac{(x'y'' - y'x'')^2}{({x'}^2+{y'}^2)^3}
$$

This can be multiplied with $({x'}^2+{y'}^2)^3$, expanded, factored and moved to one side of the equation. For a cubic Bézier curve $x'$ and $y'$ have degree 2, so this will result in a polynomial of degree 12 that we have to find the roots for.

Although I am not too happy with having to solve a polynomial of degree 12, I understand what needs to be done to get there.

What I do not understand is this:

Since a cubic bezier curve can have up to five curvature extrema plus two inflection points (see this excellent answer), I assume that for non-degenerate cases there should be no more than six points with curvature $\kappa_1$ on the curve. But a polynomial of degree 12 can have up to 12 roots. What does this mean? Is my assumption wrong that there can be only six points with $\kappa_1$? Or will all roots occur twice? If so, does this mean that there is a chance to reduce our polynomial to degree 6?

答: The reason is that due to squaring, not only the solutions for $\kappa_1$, but also the solutions for $-\kappa_1$ will be found. Therefore the squared equation can have twice the number of solutions as the original equation.


问:
Draw the paths of these numbers from t=0 to t=$2\pi$ in the complex plane
Draw the paths of these numbers from t=0 to t=$2\pi$ in the complex plane:
$$e^{it}$$
$$e^{(-1+i)t}=e^{-t}e^{it}$$
$$(-1)^t = e^{t\pi i}$$

I only know the first one: unit circle since I don't see r in this case (i.e., r=1). How to interpret the other two numbers?

Can anyone tell me where I can find the graphs of these three?

答: $t$ is real.  In the second $e^{-t}$ is real and is the radius as $|e^{it}|=1$. Can you see the link between the third and the first?


问:
Counting with decimal digits
How many distinct ways are there to encode the decimal digits $0-9$ as binary sequences of length $4$? Consider only codes which represent different digits by different sequences.

Firstly, I could not understand mathematically what is asked for, whether it is string of length = $4$ which uses decimal digits, or it is something different. Moreover, what does it mean by saying "codes which represent different digits"? Aren't constructed string different with each other?

答: There are $2^4=16$ binary strings of length $4$ and there are $10$ digits so there are $16$ ways to encode $0$ and given an encoding for $0$ there are $15$ ways to encode $1$ etc. so we get the number of permutations of ten things out of sixteen things are the possible encodings.

Title: Counting with decimal digits


问:
$n$ people have participated in a party. each of whom brought an umbrella and a coat. At the end of the party...
$n$ people have participated in a party. each of whom brought an umbrella and a coat. At the end of the party, everyone picks an umbrella and a coat out of the stack and leaves.

$1$.
In how many ways these coats and umbrellas can be selected such that no one is left with his/her own umbrella or his/her own coat?

$2$.
In how many ways these coats and umbrellas can be selected such that no one is left with his/her own umbrella and his/her own coat?

$1$.

Let $A$ denote the selections such that no one left with his/her umbrella and denote by $B$ the selections that no one is left with his/her coat, then what the question wants is :

$$\left|A \cup B\right|=\left|A\right|+\left|B\right|-\left|A \cap B\right|$$
$$=D_nn!+D_nn!-(D_n)^2$$
$$=D_n(2(n!)-D_n)$$

$2$.

One may use PIE to find the answer which is :$$\sum_{k=0}^{n}\binom{n}{k}\left(n-k\right)!^{2}\left(-1\right)^{k}$$

But why we cannot say there are $D_n$ ways to distribute the umbrellas to the people and for each of these selections there are $D_n$ ways to distribute the coats to the people and so by the multiplication law $(D_n)^2$.

答:
Assuming that the author of the question uses the English language correctly, you’ve inverted their meanings. The first one is asking for the number of ways in which each person gets both the wrong coat and the wrong umbrella, which is $D_n^2$. It’s your inclusion-exclusion calculation that is wrong: you’re excluding only the ways in which the people who get the right coat also get the right umbrella, not the ways in which, for instance person $1$ is the only one to get the right coat, and person $2$ is the only one to get the right umbrella.

The second question is asking for the number of ways in which no person gets both the right coat and the right umbrella, i.e., the number of ways in which each person gets the wrong coat, the wrong umbrella, or both; this is what you calculated for the first question.


答:
There are $16$ binary strings of length $4$, ranging from $0000$ to $1111$. A code that represents each of the $10$ decimal digits $0-9$ with a different binary sequence of length $4$ can therefore be concocted in any of

$$16\times15\times14\times\cdots\times7$$

ways.  That is, you pick one of the $16$ binary strings to represent the decimal digit $0$, one of the $15$ remaining strings to represent the decimal digit $1$, one of the $14$ remaining strings for the decimal digit $2$, and so on up to $7$ choices for the decimal digit $9$. (Note, there will be six binary strings left over.)

Title: Counting with decimal digits

Body:
How many distinct ways are there to encode the decimal digits $0-9$ as binary sequences of length $4$? Consider only codes which represent different digits by different sequences.

Firstly, I could not understand mathematically what is asked for, whether it is string of length = $4$ which uses decimal digits, or it is something different. Moreover, what does it mean by saying "codes which represent different digits"? Aren't constructed string different with each other?


问:
Let $S=\{AB-BA| A,B \in M_n(K)\}$ where $K$ is a field. Prove that $S$ is closed under matrix addition.
I know there is a result that $S=$ collection of all trace $0$ matrices, and that collection forms a vector space. But I want to prove it independently i.e.

For any $A,B,C,D\in M_n(K)$ we have to find $E,F\in M_n(K)$ such that $(AB-BA)+(CD-DC)=EF-FE$.

But I don't know how to solve this. But the thing I can observe that the above equation gives rise to $n^2$ equation (equating each entries of matrices of both the sides) with $2n^2$ variables (Total number of entries of $E,F$ is $2n^2$).Can this problem be simplified if we choose special kind of $E$ say diagonal matrix.

Edit-(This is valid only for $\Bbb{R}$ or $\Bbb{C}$)
 $AB-BA=(A+aI)(B+bI)-(B+bI)(A+aI)$ for all $a,b\in\Bbb{R}$. And there is $a,b$ in $\Bbb{R}$ such that $A+aI,B+bI$ are invertible.

Hence, we can assume $A,B$ to be invertible matrices i.e. $S=\{AB-BA|A,B\in GL_n(K)\}$

答:
I have commented that fixing $E$ to have diagonal form would be too restrictive
because then $EF - FE$ would have only zeros on its main diagonal.

However, @ancientmathematician's comment
let me track down the original proof by Shoda (1936). As it turns out,
Shoda did use the idea with a diagonal $E$ after observing that similarity
transformations carry over to commutators, and in characteristic 0
a zero-trace matrix is similar to a matrix with zero main diagonal.
For the latter, Shoda referred to matrix normal forms. Actually
we do not need that much normalization, so I'll try an elementary proof here.

Lemma: Let $K$ be a field of characteristic 0,
$n$ a positive integer and $M\in\operatorname{Mat}_n(K)$
with $\operatorname{tr} M = 0$.
Then there exists $T\in\operatorname{SL}_n(K)$ such that
$T M T^{-1}$ has only zeros on its main diagonal.

Proof: For $n = 1$, $\operatorname{tr} M = 0$ implies $M = ((0))$,
and $T = I$ (identity matrix) works.

Now assume $n > 1$. We repeat the following sequence of steps,
each iteration finding a transformation $S\in\operatorname{SL}_n(K)$ such that
replacing $M$ by $S M S^{-1}$ increases the number of zero main
diagonal elements of $M$.
Multiplying all used $S$ (later ones at the left)
results in a suitable transformation $T$.

If $M$ has only zeros on its main diagonal, $S = I$ works, and we are done.

Otherwise, $M$ must have at least two nonzero main diagonal elements
$m_{ii}, m_{jj}$ with $1\leq i < j\leq n$
because if there were only one, it would equal the trace,
which is zero, contradiction.

Furthermore, not all nonzero main diagonal elements can be equal because
otherwise the zero-trace requirement would imply that their number is
a multiple of the field characteristic, but that's zero, contradiction.

Therefore we can choose $i,j$ such that $0\neq m_{ii}\neq m_{jj}\neq 0$.
We now define a transformation $S\in\operatorname{SL}_n(K)$ that differs from
the identity matrix only in its principal submatrix with rows/columns $i,j$:

$$\begin{pmatrix}s_{ii} & s_{ij}\\s_{ji} & s_{jj}\end{pmatrix}
= \begin{cases}
    \begin{pmatrix}1 & 0\\\frac{m_{ii}}{m_{ij}} & 1\end{pmatrix}
    & \text{if $m_{ij}\neq 0$, else}
\\  \begin{pmatrix}1 & -\frac{m_{ii}}{m_{ji}}\\0 & 1\end{pmatrix}
    & \text{if $m_{ji}\neq 0$, else}
\\  \begin{pmatrix}1 & \frac{m_{ii}}{m_{jj}-m_{ii}}
    \\ 1 & \frac{m_{jj}}{m_{jj}-m_{ii}}\end{pmatrix}
    & \text{if $m_{ij} = 0 = m_{ji}$}
\end{cases}$$

Then $M' = S M S^{-1}$ has the same main diagonal elements as $M$
except that $m_{ii}' = 0$ and $m_{jj}' = m_{ii} + m_{jj}$, preserving the
trace.
Off-diagonal elements with row or column index in $\{i,j\}$ may have changed
as well, but that does not matter.

Armed with the above Lemma, we can now assume $M$ to have an all-zero main
diagonal.
Fix $E\in\operatorname{Mat}_n(K)$ to be a diagonal matrix with pairwise distinct diagonal elements. Characteristic 0 ensures their existence.
To obtain $F$, let

$$f_{ij} = \begin{cases}
\frac{m_{ij}}{e_{ii}-e_{jj}}    & \text{for $i\neq j$}
\\  \text{(arbitrary)}              & \text{for $i = j$}
\end{cases}$$

Then $M = EF - FE$.

References

K. Shoda (1936): Einige Sätze über Matrizen. Japanese J. Math. 13,
pp. 361--365.

M. Rosset, S. Rosset (2000): Elements of trace zero that are not commutators.
Communications in Algebra 28:6, 3059-3071,
DOI: 10.1080/00927870008827009.


问:
Dimension of Subspaces Involving Linear Inequalities
In my introductory linear algebra course, one of our assignment questions is to determine whether the set of all vectors in $\mathbb{R}^2$ such that for every$\begin{bmatrix}x \cr y\end{bmatrix}$, $x\le2y$, is a subspace or not. By my thinking, this should not be a subspace because it is neither a point, a line, or all of $\mathbb{R}^2$. I've tried to prove this using the definitions of a subspace (for every $\bar{v}\in S$, $\bar{w}\in S$, and scalar $c$, $\bar{v}+\bar{w}\in S$, $c\bar{v}\in S$, and $\bar{0}\in S$), but I don't really know how to do that with inequalities. Can I represent S as the span of a set of vectors? I really just don't know what to do.

答: It is not a subspace because for instance, $\left[\begin{smallmatrix}0\\1\end{smallmatrix}\right]$ belongs to it, but $-\left[\begin{smallmatrix}0\\1\end{smallmatrix}\right]\left(=\left[\begin{smallmatrix}0\\-1\end{smallmatrix}\right]\right)$ doesn't.


问:
Compute cartesian components of the curl
Given a vector $v$, the curl of $v$ is defined as the unique vector field with the property  $$(\nabla v - \nabla v^T) a = (\text{curl } v) \times a$$
for every vector $a$. (See pag. 32 of Gurtin's book)

I want to find its cartesian components (pg. 33 of the link above): so I try to compute the l.h.s., since I know that $$(\nabla v)_{ij} = \frac{\partial v_i}{\partial x_j}$$ and $$(\nabla v^T)_{ij} = \frac{\partial v_j}{\partial x_i}$$

Therefore, since $a = a_k e_k$ (I'm writing $a$ w.r.t the basis ${\{ e_k\}}_k$) I obtain

$$\Bigl( \frac{\partial v_i}{\partial x_j} - \frac{\partial v_j}{\partial x_i} \Bigr)(e_i \otimes e_j)a_ke_k = \Bigl( \frac{\partial v_i}{\partial x_j} - \frac{\partial v_j}{\partial x_i} \Bigr)a_je_i  $$

I'm using the Einstein notation for repeated indices, but I don't know how to obtain the result. The last term seems to be in the right path, but I can't continue to get the classical formulas for the cartesian components of the curl

答: Write $w:=\nabla\times v$. Pardon my slight relabelling of indices. Taking components of your equation in blue,$$\epsilon_{ijk}\epsilon_{jlm}\tfrac{\partial v_m}{\partial x_l}a_k=\color{blue}{(\tfrac{\partial v_i}{\partial x_k}-\tfrac{\partial v_k}{\partial x_i})a_k=\epsilon_{ijk}w_ja_k}.$$So $w_j=\epsilon_{jlm}\tfrac{\partial v_m}{\partial x_l}$.


问:
Showing that $C^{-1}B(C^{-1}B+(AC)^{-1})^{-1}$ has real eigenvalues
Suppose $\Sigma$ is a real-valued covariance matrix partitioned into equal-sized blocks as follows
$$
\left(\begin{matrix}
A&C\\
C^T&B
\end{matrix}\right)
=
\Sigma
$$

The following matrices are not symmetric yet seem to always have real eigenvalues, can anyone see a way to prove this?

$$D_1=C^{-1}B(C^{-1}B+(AC)^{-1})^{-1}$$
$$D_2=(BC^{-T}+CA^{-1})^{-1}CA^{-1}$$

答:
Note that $D_1$ has real eigenvalues iff $D_1^{-1}$ has real eigenvalues. We find that
$$
D_1^{-1} = (C^{-1}B + (AC)^{-1})B^{-1}C =
C^{-1}BB^{-1}C + C^{-1}A^{-1}B^{-1}C\\
= I + C^{-1}A^{-1}B^{-1}C.
$$
So, it suffices to show that $D_1^{-1} - I = C^{-1}A^{-1}B^{-1}C$ always has real eigenvalues. We note that this matrix is similar to
$$
C(D_1^{-1} - I)C^{-1} = C[C^{-1}A^{-1}B^{-1}C]C^{-1} = A^{-1}B^{-1},
$$
which is in turn similar to
$$
A^{1/2}A^{-1}B^{-1}A^{-1/2} = A^{-1/2}B^{-1}A^{-1/2}.
$$
So, we see that $D_1^{-1} - I$ is similar to $A^{-1/2}B^{-1}A^{-1/2}$. $B$ is a principal submatrix of a positive definite matrix and is therefore positive definite, which means that $B^{-1}$ is positive definite, which means that $A^{-1/2}B^{-1}A^{-1/2} = [A^{-1/2}]^TB^{-1}A^{-1/2}$ is positive definite, which means that $A^{-1/2}B^{-1}A^{-1/2}$ has real eigenvalues, which means that $D_1^{-1} - I$ has real eigenvalues.

A similar line of reasoning can be applied to $D_2$.


问:
Let $x$ be a real number such that $|x|<1$. Determine $\lim_{n \to \infty} \prod_{i=1}^{n} \left(1+x^{2^{i}}\right)$
I've written out the first few terms of the series up to n = 3 and I've figured out that they always take the form $1+x^2+x^4+...+x^{2k}$. How can I figure out what the limit will be as $n$ goes to infinity?

答:
Well if the form you found is the right one and x<1 then I would say that limit of $x^k=0$(it's a geometric progession with a common ratio inferior to one) when n goes to infinity. then it means you'll get  $\frac{1}{1-x^2}$

I hope it'll help


问:
Cancellation law for addition in Fischer's Intermediate Real Analysis
Taken from Problem 2.6 in Intermediate Real Analysis by Emanuel Fischer

Prove: (a) $x+z<y+z$ implies $x<y$, and (b) $x+z=y+z$ implies $x=y$.

The thing is, the author has not introduced the symbol $0$ nor the additive inverse yet (till the next page). How would you answer to this problem based on the previous axioms the author stated (p. 4-6)?

答:
a) Suppose that $x > y$. By Axiom $0_4$ this would implies $x+z>y+z$ a contradiction with Axiom $0'_2$. If $x=y$ then we would have $x+z=y+z$ [1], again a contradiction. From Axiom $0'_2$, $x<y$.

b) Suppose $x < y$. By Axiom $0_4$ this would implies $x+z<y+z$ a contradiction with Axiom $0'_2$. Using the same argument $x>y$ yields a contradiction. Hence by Axiom $0'_2$, $x=y$.

[1] This holds because the operation $+:\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is well defined. In fact, given $(x,y) \in \mathbb{R} \times \mathbb{R}$, from Axiom $A_1$, there is an unique $r \in \mathbb{R}$ such that $r=+(x,y)=x+y$. You can find discussions about it in these links Algebra: What allows us to do the same thing to both sides of an equation?
Why do we have to do the same things to both sides of an equation?


问:
Proving that the set of limit points of a set is closed
From Rudin's Principles of Mathematical Analysis (Chapter 2, Exercise 6)

Let $E'$ be the set of all limit points of a set $E$. Prove that $E'$ is closed.

I think I got it but my argument is a bit hand wavy:

If $x$ is a limit point of $E'$, then every neighborhood of $x$ contains some $y\in E'$, and every neighborhood of $y$ contains some $z\in E$. Therefore every neighborhood of $x$ contains some $z\in E$, and so $x$ is a limit point of $E$. Then $x\in E'$, so $E'$ is closed.

The thing that's bugging me is the leap from one neighborhood to another. Is this formally correct?

答:
Let $X$ be a $T_1$ space and $A \subseteq X$. Notice that $A'$ is closed $\iff \overline{A'} = A' \iff A'' \subseteq A' $ since $\overline{A'} = A' \cup A'' $. Hence, we show that $A''\subseteq A'$.

Let $x \in A''$. Thus, for every open nbhd $U_x$ of $x$, one has $U_x \cap (A' \setminus \{x\}) \neq \emptyset $.

Pick $y \in U_x \cap (A' \setminus \{x\})$. As $y \in U_x$ and $U_x$ is an open set, there exists an open nbhd $V_y$ of $y$ entirely contained in $U_x$. Moreover, as $X$ is $T_1$, singletons are closed, so $\{x\}$ is a closed subset of $X$. Therefore, $V_y \setminus \{x\} = V_y \cap \{x\}^C$ is the intersection of open subsets of $X$ and so, is also open in $X$, contains $y$, and is entirely contained in $U_x$. (This is the subtlety missed by OP and accepted answer)

Now, as $y \in A'$, then for every open nbhd $U_y$ of $y$, one has $U_y \cap (A \setminus \{y\}) \neq \emptyset $. In particular, this property also applies to the nbhd $V_y \setminus \{x\} $ of $y$.

Picking now $z \in (V_y \setminus \{x\} ) \cap (A \setminus \{y\}) $, we have that $z \in V_y \setminus \{x\} \subset U_x$, where $z \neq x$. Therefore, for every open nbhd $U_x$ of $x$, one has $U_x \cap (A \setminus \{x\}) \neq \emptyset$.

Hence, $x \in A'$.

ContentLicense: CC BY-SA 3.0


问:
Chromatic number of $G$, the graph on $n$ vertices obtained from $K_n$ by removing $n$ edges forming an $n$-cycle.
Let $G$ be the graph on $n$ vertices $(n \geq 4)$ obtained from $K_n$ by removing $n$ edges
forming an $n$-cycle. Determine the chromatic number of $G$.

For any even $n$ I know that removing the $n$-cycle edges leaves a graph with a $\frac{n}{2}$ clique, so the chromatic number $\chi(G) \geq \frac{n}{2}$. Similarly, for odd $n$, removing the $n$-cycle edges leaves a graph with another (odd) $n$-cycle, so $\chi(G) \geq 3$. I'm not exactly how to proceed from here. Any help would be appreciated!

答:
You should check that for $n>3$, the graph $G$ (commonly denoted as $\overline{C_n}$, the complement of the cycle graph) does not have an independent set of size $3$. Equivalently, for $n>3$, $C_n$ does not contain a triangle.

This means that every color can be used on at most two vertices, and therefore $\chi(\overline{C_n}) \ge \lceil \frac n2 \rceil$. To show that this bound is tight, and conclude that $\chi(\overline{C_n}) = \lceil \frac n2 \rceil$, you should find a coloring that uses this many colors.

Again, it might be easier to think about working with the cycle graph $C_n$, where we want the complement of a proper coloring. That's a vertex coloring where the vertices of each color form a clique. In this case, for each color, there is either only one vertex of that color, or two adjacent vertices.


问:
Existence of a neutral element in set with an associative internal law
Let $E$ be a set with an associative internal law $\star$ (i. e. such that $\star : (x, y) \in E^2 \mapsto x \star y \in E$ and for all $(x, y, z) \in E^3, (x \star y) \star z = x \star (y \star z)$).

We assume that there exists $a \in E$ such that:
$$
\forall y \in E, \exists x \in E : y = a \star x \star a
$$

I have to show that $E$ admits a neutral element $e$, which means I have to show:
$$
\exists e \in E, \forall x \in E : x \star e = e \star x = x
$$

When I look for a candidate value of $e$, I don't find a way to solve this. Any help is welcome.

答:
How can you try to find the answer?

If this were a group, then the condition holds (just take $x=a^{-1}ya^{-1}$). In that case, setting $y=a$ will give $a=axa$, which tells you that $ax=xa=e$. So perhaps one should try to see if that holds in this case: if we pick $x$ such that $a=axa$, will $ax=xa$ be the identity element?

Indeed it will.

We know that there exists $x$ such that $a=axa$.

Let $r\in E$; then there exists $z$ such that $r=aza$. Hence $(ax)r = ax(aza) = (axa)za = aza = r$. And $r(xa) = (aza)xa = az(axa) = aza = r$.

Thus, $ax$ is a left unit, and $xa$ is a right unit. But then
$$\begin{align*}
ax &= (ax)(xa) &&\text{(because }xa\text{ is a right unit)}\\
&= xa &&\text{(because }ax\text{ is a left unit)}
\end{align*}$$
so $ax=xa$ is the two-sided unit.


问:
Distributional Fourier Transform of $f(x)=\frac{e^{ix\cdot \xi_0}}{\vert x-x_0\vert}$ where $x\in\mathbb R^2$ and $\xi_0,x_0\in\mathbb R^2$ fixed
Problem: Find the Fourier transform of $f:\mathbb R^2\to\mathbb C$ given by
$$f(x)=\frac{e^{ix\cdot \xi_0}}{\vert x-x_0\vert}\quad\text{where }\xi_0,x_0\in\mathbb R^2\text{ are fixed.}$$

My Attempt: Note that $f\notin L^1(\mathbb R^2)$ and also $f\notin L^2(\mathbb R^2)$. Therefore, we compute the Fourier transform of $f$ in the distributional sense since $f\in L^1_{\operatorname{loc}}(\mathbb R^2)$. Let $\psi\in\mathcal C_c^\infty(\mathbb R^2)$. Then
\begin{align*}
\widehat{f}(\psi)&=f(\widehat{\psi})=\int_{\mathbb R^2}f(x)\widehat\psi(x)\,dx\\
&=\frac{1}{2\pi}\int_{\mathbb R^2}\frac{e^{ix\cdot \xi_0}}{\vert x-x_0\vert}\int_{\mathbb R^2}\psi(\xi)e^{-ix\cdot\xi}\,d\xi\,\,dx\\
&=\frac{1}{2\pi}\int_{\mathbb R^2}\psi(\xi)\int_{\mathbb R^2}\frac{e^{ix\cdot(\xi_0-\xi)}}{\vert x-x_0\vert}\,dx\,d\xi.
\end{align*}
Therefore, it suffices to evaluate the integral
$$I=\frac{1}{2\pi}\int_{\mathbb R^2}\frac{e^{ix\cdot(\xi_0-\xi)}}{\vert x-x_0\vert}\,dx.$$
The one thing that is confusing me about the problem is the multidimensionality, if on the other hand we were in $\mathbb R$, then some manipulations with the Heaviside function and the delta distribution would most likely do the trick. Also, the integral above is not absolutely convergent and the integrand is not nonnegative, whence we cannot even apply polar coordinates, as stated in Folland's book.

Therefore, my question is: how does one deal with the integral above, or is there a different and more efficient way to go about this problem?

Thank you for your time and appreciate any feedback.

答:
Here's an evaluation that should provide some hints as to what the correct answer is.

Consider the Fourier integral

$$F_{\epsilon}(\xi)=\int_{\mathbb{R}^2}d^2x~\frac{e^{-i(\xi-\xi_0)x}}{|x-x_0|}e^{-\epsilon |x-x_0|}$$

where $\epsilon$ is a positive number (can be readily generalized to a complex number with $\text{Re}( \epsilon)>0$). This regulated Fourier transform does converge absolutely, in  contrast to it's unregulated counterpart. This allows for an easy evaluation of the integral as follows. It suffices to consider the simpler integral

$$I(p)=\int_{\mathbb{R}^2}d^2x~\frac{e^{-ipx}}{|x|}e^{-\epsilon |x|}$$

since of course

$$F_{\epsilon}(\xi)=e^{-i(\xi-\xi_0)x_0}I(\xi-\xi_0)$$
One can evaluate $I$ in spherical coordinates in various ways. Here's one of them
$$I(p)=\int_{0}^{2\pi}d\theta\int_{0}^{\infty}dr~e^{-\epsilon r-i|p|r\cos\theta}=\int_{0}^{2\pi}\frac{1}{\epsilon+i|p|\cos\theta}d\theta=\frac{2\pi}{\sqrt{|p|^2+\epsilon^2}}$$
where the last integral can be easily performed with a standard contour integration trick. It would appear that one can easily take the limit $\epsilon\to 0^+$ to get that the requested FT is given by
$$\hat{f}(\xi)=\frac{e^{-i(\xi-\xi_0)x_0}}{|\xi-\xi_0|}$$
and that the limit is quite regular, despite the fact that actually if it wasn't for the cutoff parameter we wouldn't have been able to evaluate the double integral in the order we did. One may worry that this procedure is misleading, since the integral diverges if evaluated with $\epsilon=0$ directly step by step as above. However it turns out that the cutoff and distributional considerations are not needed if the right order of integration is employed since

$$I_0(p)=\int_{0}^{\infty}dr\int_{0}^{2\pi}d\theta~e^{-i|p|r\cos\theta}=2\pi\int_0^{\infty}J_0(|p|r)dr=\frac{2\pi}{|p|}$$
where the last step can be derived from the Laplace transform of a Bessel function
$$\int_{0}^{\infty}J_0(px)e^{-sx}dx=\frac{1}{\sqrt{p^2+s^2}}$$
In summary, this integral exists and is well defined but poses computational difficulties due to the lack of absolute covergence, which also precludes the free use of Fubini's theorem. If the integral exists, evaluating with a regulator is enough to calculate it's value if it makes the evaluation easier, but eventually one has to prove its existence somehow. In this example there is a way to do that without resorting to distributional results. Let us note, however, that this integral can also be shown to have the value above by using the following distributional identity(calculation is left to the reader)
$$\int_{0}^{\infty}dx~e^{-ipx\cos\theta}=\pi\delta(|p|\cos\theta)-i ~PV\left(\frac{1}{p\cos\theta}\right)$$


答:
Hint: Multiply by $ ( 1 - x^2)$.

There's a telescoping series going on.

Title: Let $x$ be a real number such that $|x|<1$. Determine $\lim_{n \to \infty} \prod_{i=1}^{n} \left(1+x^{2^{i}}\right)$

Body: I've written out the first few terms of the series up to n = 3 and I've figured out that they always take the form $1+x^2+x^4+...+x^{2k}$. How can I figure out what the limit will be as $n$ goes to infinity?


答:
If you have not figured it out, you can use a telescoping product by multiplying the product by $\frac{1-x^2}{1-x^2}$ and repeatedly using difference of squares. Then you get $$\prod_{k=1}^{n}{(1+x^{2^k})} = \frac{1-x^{2^{n+1}}}{1-x^2}=\frac{1}{1-x^2}-\frac{x^{2^{n+1}}}{1-x^2}.$$
As $n\to\infty,$ the first term stays constant, the denominator of the second term stay constant and the numerator of the second term goes to $0$ because $|x|<1.$ Thus, the answer is $\frac{1}{1-x^2}.$


问:
Show that $\exp: \mathfrak{sl}(n,\mathbb R)\to \operatorname{SL}(n,\mathbb R)$ is not surjective
It is well known that for $n=2$, this holds. The polar decomposition provides the topology of $\operatorname{SL}(n,\mathbb R)$ as the product of symmetric matrices and orthogonal matrices, which can be written as the product of exponentials of skew symmetric and symmetric traceless matrices. However I could not find  out the proof that  $\exp: \mathfrak{sl}(n,\mathbb R)\to\operatorname{SL}(n,\mathbb R)$ is not surjective for $n\geq 3$.</p>

答:
I remember learning somewhere that every matrix in the image of the exponential from $\mathfrak{sl}(n,\Bbb R)$ is either diagonalizable or is unipotent. So, any defective element which is not unipotent would be missed.

Defective matrices with determinant 1 can be built by installing a defective block. For example:

$$
\begin{bmatrix}-1&1\\0&-1\end{bmatrix},\begin{bmatrix}-1&1&0\\0&-1&0\\0&0&1\end{bmatrix},\begin{bmatrix}-1&1&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}\text,{etc.}
$$


问:
Simple algebra clarification needed for Gaussian entropy formula
Norwich (2003) gives the following equality based on a substitution as shown in the image.

I don't get it though because $\frac{1}{2}$ appears in two places in the first formula. Does the substitution written occur at both places? Even if I do this, I don't know how to distribute the multiplication of "$\ln e$"

Could someone show the steps in-between involving the substitution, and how to arrive at the final equation shown.

答:
We have
$$
\frac12\ln(2\pi\sigma^2) + \frac12 = \frac12\ln(2\pi\sigma^2) + \frac12\ln e\\
= \frac12\left(\ln(2\pi\sigma^2) + \ln e\right)\\
= \frac12\ln(2\pi e\sigma^2)
$$
It is the ${}+\frac12$ on the right side that gets an extra factor of $\ln e$. Then we use distribution, and finally elementary logarithm rules to put the two logarithms together into one.


问:
Definition of Norm of an Ideal
I need to prove $(23,\alpha-10)$ is a prime ideal in $Z[\alpha]/(\alpha^3-\alpha-1)$. I want to show that $Norm(23,\alpha-10)$ is a prime number, thus proving it is a prime as we are in a Dedekind domain. Can anyone help me to compute the ideal norm?

Also, in particular, I know that for a PID, $R$, $Norm(I)$, $I\in R$ is given by the generator conjugate's products. Can anyone help me to get the connection between this definition and $Norm(I)=|R/I|$?

答:
Since ${\rm disc}(x^3-x-1) = -23$ is squarefree, the ring of integers of the field $\mathbf Q(\alpha)$ is $\mathbf Z[\alpha]$ (do you know why?), so the way an ideal $(p) = p\mathbf Z[\alpha]$ decomposes in $\mathbf Z[\alpha]$ matches how $x^3 - x - 1$ decomposes in $(\mathbf Z/(p))[x]$.   Is this a result you've heard of already?  Since $x^3 - x - 1 \equiv (x -3)(x - 10)^2 \bmod 23$, we have the prime ideal factorization $(23) = (23, \alpha-3)(23,\alpha-10)^2$ where both prime ideals $(23, \alpha-3)$ and $(23,\alpha-10)$ have norm $23^1$ since both irreducible factors of $x^3 - x - 1 \bmod 23$ have degree $1$.

More generally, the Dedekind--Kummer theorem says that in every number field $K$ with ring of integers $\mathcal O_K$, if $\alpha$ is an algebraic integer  generating $K$ over $\mathbf Q$ then (even if $\mathcal O_K \not= \mathbf Z[\alpha]$) for each prime $p$ not dividing $[\mathcal O_K:\mathbf Z[\alpha]]$, the way $(p) = p\mathcal O_K$ decomposes into prime ideals in $\mathcal O_K$ resembles how $f(x)$ decomposes in $(\mathbf Z/(p))[x]$ where $f(x)$ is the minimal polynomial of $\alpha$ over $\mathbf Q$ (necessarily $f(x)$ is monic in $\mathbf Z[x]$): if
$$
f(x) \equiv \pi_1(x)^{e_1} \cdots \pi_g(x)^{e_g} \bmod p
$$
with irreducible $\pi_i(x)$ in $(\mathbf Z/(p))[x]$ then
$$
p\mathbf Z[\alpha] = \mathfrak p_1^{e_1} \cdots \mathfrak p_g^{e_g}
$$
where $\mathfrak p_i = (p,\pi_i(\alpha))$ and the norm of $\mathfrak p_i$ is $p^{\deg \pi_i}$. (Although $\pi_i(x)$ is defined as a polynomial in $(\mathbf Z/(p))[x]$, the ideal $(p,\pi_i(\alpha))$ is well-defined because the ambiguity of the lift of $\pi_i(x)$ to $\mathbf Z[x]$ is washed out in the ideal $(p,\pi_i(\alpha))$ since the ideal contains $p$.) See here for further details.

Your description of the norm of a principal ideal as a product of conjugates of a generators is not the right way to think about the norm as a foundational concept. You should define the field norm ${\rm N}_{K/\mathbf Q}(\alpha)$ of an element $\alpha$ in a number field $K$ as $\det(m_\alpha)$ where $m_\alpha \colon K \to K$ is the multiplication-by-$\alpha$ map viewed as a $\mathbf Q$-linear map.  That this definition for ${\rm N}_{K/\mathbf Q}(\alpha)$ agrees with the "product of conjugates" viewpoint is here (see Theorem 5.9 and Corollary 5.15 to relate minimal and characteristic polynomials of an element in a finite extension field). For every nonzero ideal $\mathfrak a$ in $\mathcal O_K$, the ideal norm ${\rm N}(\mathfrak a)$ of $\mathfrak a$ is defined to to be the index $[\mathcal O_K:\mathfrak a]$. The compatibility between these two concepts of norm (of an element and an ideal) occurs in the case where  $\mathfrak a = (\alpha)$ is a principal ideal for some nonzero $\alpha \in \cal O_K$, where the compatibility is ${\rm N}((\alpha)) = |{\rm N}_{K/\mathbf Q}(\alpha)|$. That formula is a special case of Theorem 5.15 here by taking $M = \mathcal O_K$, $M' = \alpha\mathcal O_K$, and letting $(c_{ij})$ be the matrix expressing a $\mathbf Z$-basis of $\alpha\mathcal O_K$ in terms of a $\mathbf Z$-basis of $\mathcal O_K$.  Moreover, that also gives an approach to computing the norm
of an arbitrary nonzero ideal $\mathfrak a$ in $\mathcal O_K$: ${\rm N}(\mathfrak a) = |\det(c_{ij})|$ where $(c_{ij})$ is the matrix describing a $\mathbf Z$-basis of $\mathfrak a$ in terms of a $\mathbf Z$-basis of $\mathcal O_K$.
A completely different method of computing ${\rm N}(\mathfrak a)$ by using a field norm for $K(T)/\mathbf Q(T)$, due to Kronecker, is described in the top MO answer here.

Title: Definition of Norm of an Ideal


问:
Does $\overline{A}$ be dense-in-itselft implies A is dense-in-itselft?
Let ($X, τ$) ve a topological space and $A\subseteq X$ we say that $A$ is dense-in-itselft if $A \subseteq A'$ where $A'$ denotes the set of acumulation points of A.
If $\overline{A}$ is dense-in-itselft then A is dense-in-itselft?
I think yes but I have no idea of how to proof it or any counterexample

答:
Partial answer: Yes, provided $(X,\tau)$ is $T_1$.

Fix $x\in A$ and $U\in\tau$ with $x\in U$. By assumption, $x\in \overline A\subset (\overline  A)'$, so there is some $y\in \overline A$ with $y\in U$ and $y\neq x$. By assumption, there is some $V\in \tau$ with $y\in V$ and $x\not \in  V$. Now $y\in \overline A$ and $y\in V\cap U$, so there is some $z\in A$ with $z\in V\cap U$. Since $x\not \in V$, we can infer that $z\neq x$.

But then every open neighbourhood of $x$ contains  a point of $A$ different from $x$. Therefore, $x\in A'$.


问:
Find finite sum $ \sum_{k = 1}^{n} {k2^k}$
How do you find the finite sum  of $ \sum_{k = 1}^{n} {k2^k}$ .
Any hint?

No integration and differentiation.

答:
Hint:
Generalise the problem to computing, for any $x\ne 1$,
$$\sum_{k=1}^nk\,x^k=x\sum_{k=1}^nk\,x^{k-1}=x\Bigl(\sum_{k=1}^nx^k\Bigr)'.$$


问:
"Topological disjoint union" vs "disjoint union"
I want to understand the definition of topological disjoint union. Let $X$ be a topological space such that we can write $X=\bigsqcup X_i$. We say that $X$ is the "topological disjoint union" of the $X_i$ if each $X_i$ is an open subset of $X$, but if the $X_i$ are not necesserily open subsets of $X$ we say that $X$ is a the "disjoint union" of the $X_i$ only as sets (without the topological adjective) which means that $X =\cup X_i$ is the union of the $X_i$ such that all the $X_i$ are pairwise disjoint as sets. Is my understanding correct? thank you for your help!

答:
A topological space is a pair $(X,T)$ where $T$ is a topology on the set $X,$ but it is common to call it "the space $X$" when a particular $T$ is assumed.

If $(X_i,T_i)$ is a topological space for each $i\in I$ then the topological sum (topological disjoint union) is $(X, T)$ where $X=\cup_{i\in I} X_i\times \{i\}$ and $T$ is the topology on $X$ generated by the base (basis) $\cup_{i\in I}\{t\times \{i\}: t\in T_i\}.$


答:
I wrote this solution before noticing that Arturo published his own solution which is more elegant. I'm still sharing my solution with a bit of the reasoning for choosing a candidate because it's a bit different.

First when I tried to solve it, I wasn't what will be a good candidate of $e$, just like you mentioned. I thought $a$ might be a good candidate. But then when I tried to look at the group $\mathbb{Z}$, I noticed that $a=3$ satisfy the condition, so I figured $a$ is a bad candidate. I noticed that plugging $y=a^n$, gives $a^n = axa$ for some $x$, and if we want $x$ to be the identity, then using $n=2$ is our best choice (then we'll have $a^2 = axa = aea = aa$).

And now the dirty work:

$a^2 = aba$

Then, for some $y = axa$ we have:

$yba = ax(aba) = axa^2 = ya$

$aby = (aba)xa = a^2xa = ay$

And now for some $y = axa$ and some $z$ we have:

$ybz = (axa)bz = ax(abz) = ax(az) = (axa)z = yz$

Now, $b$ also have to satisfy the condition so for some $c$ we have $b=aca$. So now:

$bca = (aca)ca = ac(aca) = acb$

And then for some $y=axa$ using the previous 2 results we get:

$by = b(axa) = (aca)axa = (ac)aaxa = (abc)aaxa = a(bca)axa = a(acb)axa = aa(cba)xa = aa(ca)xa = a(aca)xa = abxa = (abx)a = (ax)a = axa = y$

and pretty much the same for $yb$.

Title: Existence of a neutral element in set with an associative internal law

Body:
Let $E$ be a set with an associative internal law $\star$ (i. e. such that $\star : (x, y) \in E^2 \mapsto x \star y \in E$ and for all $(x, y, z) \in E^3, (x \star y) \star z = x \star (y \star z)$).

We assume that there exists $a \in E$ such that:
$$
\forall y \in E, \exists x \in E : y = a \star x \star a
$$

I have to show that $E$ admits a neutral element $e$, which means I have to show:
$$
\exists e \in E, \forall x \in E : x \star e = e \star x = x
$$

When I look for a candidate value of $e$, I don't find a way to solve this. Any help is welcome.


答:
No, consider $X=\{1,2\}$ with $\tau=\{\emptyset,X\}$. Then, $\{1\}'=\{2\}$ and $\overline{\{1\}}=X=X'$.

Your claim is true for T1 spaces.
If $X$ is T1 and $\overline{A}'\subseteq\overline A$, let $a\in A$ and let $U\ni a$ be open. Since $a\in\overline A$, there is some $b\in \overline A\cap U$ such that $b\ne a$. Now, consider and open set $V$ suc that $b\in V$ and $a\notin V$. Since $b\in\overline A$ and $b\in U\cap V$, there is some $a'\in A\cap U\cap V$. Therefore $a'\in A\cap U\setminus\{a\}$.

Title: Does $\overline{A}$ be dense-in-itselft implies A is dense-in-itselft?

Body:
Let ($X, τ$) ve a topological space and $A\subseteq X$ we say that $A$ is dense-in-itselft if $A \subseteq A'$ where $A'$ denotes the set of acumulation points of A.
If $\overline{A}$ is dense-in-itselft then A is dense-in-itselft?
I think yes but I have no idea of how to proof it or any counterexample


问:
Explicit proof that $\mathbb{P}_B^n\cong\mathbb{P}_A^n\times_{\operatorname{Spec}A}\operatorname{Spec} B$
On page 103 of Hartshore, just before the definition of projective morphisms, he states that if $\varphi:A\to B$ is a ring morphism and $(f,f^{\#}):\operatorname{Spec}B\to \operatorname{Spec}A$ is the induced morphism on spectra, then we have $\mathbb{P}_B^n\cong \mathbb{P}_A^n\times_{\operatorname{Spec}A}\operatorname{Spec} B$. He gives no explanation, and at least for a beginner like me, this doesn't seem so obvious. If we denote by $\iota_A:A\hookrightarrow A[x_0,...,x_n]$ resp. $\iota_B:B\hookrightarrow B[x_0,...,x_n]$ the natural inclusions and by $\hat\varphi:A[x_0,...,x_n]\to B[x_0,...,x_n]$ the induced morphism such that $\hat\varphi\circ\iota_A=\iota_B\circ\varphi$, then I think these should induce the implicit structure morphisms $(\hat f,\hat{f}^{\#}):\mathbb{P}_B^n\to\mathbb{P}_A^n$, $(p_A,p_A^{\#}):\mathbb{P}_A^n\to\operatorname{Spec} A$ resp. $(p_B,p_B^{\#}):\mathbb{P}_B^n\to\operatorname{Spec} B$. I know of this post, however I would like to find a more explicit proof. That is, I would like to construct from morphisms $X\to\mathbb{P}_A^n$ and $X\to \operatorname{Spec}$ the product morphism $X\to\mathbb{P}_B^n$. To construct such a morphism, one should be able, given prime ideals $\mathfrak{p}\in\mathbb{P}_A^n$ and $\mathfrak{b}\in\operatorname{Spec} B$ such that $\iota_A^{-1}(\mathfrak{p})=\varphi^{-1}(\mathfrak{b})$, a prime ideal $\mathfrak{q}\in\mathbb{P}_B^n$ such that $\iota_B^{-1}(\mathfrak{q})=\mathfrak{b}$ and $\hat\varphi^{-1}(\mathfrak{q})=\mathfrak{p}$. But as prime ideals don't behave well under extension, I'm not sure how to accomplish this. So how to proceed?

答:
Suppose we have a pair of morphisms $\newcommand{\Spec}{\operatorname{Spec}}\newcommand{\PP}{\mathbb{P}}\newcommand{\AA}{\mathbb{A}} X\to \PP^n_A$ and $X\to\Spec B$ which are equal after composing with the maps $\PP^n_A\to\Spec A$ and $\Spec B\to \Spec A$. Covering $\PP^n_A$ with the standard affine opens $\AA^n_A\to\PP^n_A$, we can form a cover of $X$ by open subschemes $U_i$ which all map in to a single standard affine open. Refining this cover to a cover by affine open subsets, we can assume that $X$ is covered by a collection of open affine subschemes which each land inside one of the standard affine opens.

For any affine scheme $U$ with maps $U\to\AA^n_A$ and $U\to \Spec B$ which agree after composing with the maps $\AA^n_A\to\Spec A$ and $\Spec B\to \Spec A$, we have a unique induced morphism $U\to \AA^n_A\times_A\Spec B$. As fiber products of affine schemes are given by the spectrum of the tensor products of their coordinate rings, we have that $\AA^n_A\times_A\Spec B = \Spec (A[x_1,\cdots,x_n]\otimes_A B)=\Spec B[x_1,\cdots,x_n]=\AA^n_B$. Composing with the inclusion $\AA^n_B\to \PP^n_B$, we see that we obtain a compatible collection of maps to $\PP^n_B$ from the $U_i\subset X$ to $\PP^n_B$. By the uniqueness of gluing, this implies we have a unique map $X\to\PP^n_B$ and thus $\PP^n_B=\PP^n_A\times_A\Spec B$.

Alternatively, one can use the direct construction of the fiber product to prove this. $\PP^n_A\times_A \Spec B$ is glued together from the fiber products $\AA^n_A\times_A\Spec B=\AA^n_B$ as $\AA^n_A$ runs over the standard affine opens of $\PP^n_A$. The gluing data for this scheme is the same as the gluing data for $\PP^n_B$, so by uniqueness of gluing we have that the schemes are isomorphic.


问:
"Explicitization" of Implicit Euler
Consider the ODE $\dot y=f(t,y)$.

Explicit Euler says: $$y_{n+1}=y_n+hf(t_n,y_n)\tag{1}$$

Implicit Euler says: $$y_{n+1}=y_n+hf(t_{n+1},y_{n+1})\tag{2}$$

The above schemes are well-known. In a way similar to what is done in Heun's method, we could make Explicit the Implicit Euler scheme as follows:

Plug (1) in the right handside of (2) to get: $$y_{n+1}=y_n+hf(t_{n+1},y_n+hf(t_n,y_n))=0\tag{3}$$

We could also plug (2) (possibly multiple times) into the right handside of (2) to get $$y_{n+1}=y_n+hf(t_{n+1},y_n+hf(t_{n+1},y_{n+1}))=0\tag{4}$$
or $$y_{n+1}=y_n+hf(t_{n+1},y_n+hf(t_{n+1},y_n+hf(t_{n+1},y_{n+1})))=0\tag{5}$$

We could finally make (5) explicit by plugging (1) into its right handside, as follows:
$$y_{n+1}=y_n+hf(t_{n+1},y_n+hf(t_{n+1},y_n+hf(t_{n+1},y_n+hf(t_n,y_n))))=0\tag{6}$$
I have never heard of the schemes above. Maybe they bear no specific good numerical features. Are there known results on this?

答: Repeated iterations do not offer improved convergence. Note that any solution to $(2)$ is also a solution to $(4)$ and $(5)$. Provided that the iterations converge, this is just a fixed-point estimate of $(2)$ and is thus no more accurate than $(2)$. This approach is mentioned on Wikipedia.

Title: "Explicitization" of Implicit Euler


问:
Examples of systems conforming the Lorenz Attractor
Might sound like a trivial question but would you please show me some examples of real systems conforming the Lorenz Attractor?

It can be any kind of system, just a little list. It can be a system arising from Economical, Social, Biological or Physical matters.

答: The chaotic waterwheel! It was literately build to physically realise the Lorenz equations (see DIY version and Harvard version). A nice discussion about it can be found in the book by Strogatz.</p>


问:
Let $a$ and $b$ be integers such that $a|b$ and $b|a$. Show that $a ∈ \{b,−b\}$.
Let $a$ and $b$ be integers such that $a|b$ and $b|a$. Show that $a ∈ \{b,−b\}$.

My attempt

Since  $a\mid b$ and $b\mid a$, we have that $a\neq 0$ and $b\neq 0.$

I think I can use this result that appears in the book I am studying... If  $a\mid b$ and $b\neq 0$, then $a\leq |b|$.  Thus, in this case $a\leq |b|$ and $b\leq|a|$. The problem is that I don't see how to relate these two inequalities to get what I want.

答:
Another approach

If $a\mid b$, then  $b= n a$ for some integer $ n$.

Analogously, since $b\mid a$, $a= m b$ for some integer $m$.

So, $a= m b= m n a$, hence $mn=1$. But $m$ and $n$ are integers, so $m=n=1$ or $m=n=-1$.


问:
Number of non-identity elements of order $2$ in this set. (Related to abelian group of order $34.$)
This question was asked in a bachelor exam for which I am preparing and I was unable to solve it.

Let $G$ be an abelian group of order $34$ and $S=\{ g \in G \mid g=g^{-1}\}$ . Then what is the  number of non- identity elements in $S$?

I used sylow theorem: There is $1$ sylow subgroup of order $17$ and $17$ sylow subgroups of order $2$ but $17$ is not the answer ( I'm not even close!) .

What is wrong in my approach? Can you please tell?

答: By the classification theorem for finite abelian groups, $G$ is cyclic of order 34. A finite cyclic group of order $n$ has exactly one subgroup of order $d$ for any $d\mid n$, so there is exactly one subgroup of order two. Since any element of order two generates a subgroup of order two, there is precisely one such element. So there are two elements with $g=g^{-1}$: the identity and the unique element of order two.


答:
What does $a\mid b$ mean literally?

We have that $a\mid b$ so there is an integer, $k$, so that $b=ak$ and $b\mid a$ so there is an integer, $m$, so that $a = mb$ so $a = m(ak) = a(km)$.

If $a\ne 0$ then $km = 1$ and the only divisors of $1$ are $1$ and $-1$ so either $m=1$ and $a = 1\cdot b = b$.  Or $m=-1$ so $a = (-1)b = -b$.  So $a \in \{b,-b\}$.

But what if $a = 0$. (I don't see why you can claim $a\ne 0; b\ne 0$). We then $b =ak = b\cdot 0 = 0 = a$ and $a = b$.  So $a \in \{b,-b\}$.

So either way $a\in \{b,-b\}$.

Body:
Let $a$ and $b$ be integers such that $a|b$ and $b|a$. Show that $a ∈ \{b,−b\}$.

My attempt

Since  $a\mid b$ and $b\mid a$, we have that $a\neq 0$ and $b\neq 0.$

I think I can use this result that appears in the book I am studying... If  $a\mid b$ and $b\neq 0$, then $a\leq |b|$.  Thus, in this case $a\leq |b|$ and $b\leq|a|$. The problem is that I don't see how to relate these two inequalities to get what I want.


问:
Using a sequence to prove a subspace has no interior points
I am working on this problem:

Prove explicitly that $L^2[0, 1]$ as vector subspace of $L^1[0, 1]$ has no interior points

Suggestion: It is enough to show that for each element of $L^2[0, 1]$ it is possible to construct a sequence of elements belonging to $L^1[0; 1] \setminus L^2[0, 1]$ which converges in $L^1[0, 1] $ to that element

Can someone please explain the logic of the suggestion? I don't follow it. I know points in the closure of a set can be characterized as limit of sequence, but for interior points all I know is the definition that says: "x is an interior point of a set S if there exist an open ball contained in that set". I was trying to  prove that the suggestio follows from this definiton but not luck
I already have the solution of the problem, but I don't understant it, because I don't understand the suggestion that they have implemented.
Can someone shed some some light?

答: Suppose $f$ is an interior point of $L^{2}$ in $L^{1}$. Then there exists an  open ball $B(f,r)$ contained in $L^{2}$. Now suppose we can find a sequence $(f_n)$ in $L^{1}\setminus L^{2}$ converging to $f$. Then $f_n \in B(f,r)$ for $n$ sufficiently large. But this is a contradiction because $f_n \in B(f,r) \subseteq L^{2}$ and $f_n \notin L^{2}$.


问:
The geometric interpretation
In the course of mathematical analysis, there was one problem that i excited to know more about it:
What is the geometric interpretation of
$$ \int_a^b f(x)\,d(\alpha(x)) $$
and $\alpha(x)$ is function in $[a,b]$</p>

答:
$$d\alpha(x)=\frac{d\alpha}{dx}dx$$

So the integral can be rewritten as follows:

$$\int_a^b{f(x)d\alpha(x)}=\int_a^bf(x)\frac{d\alpha}{dx}dx$$

So a geometric interpretation is, it's the integral of $f(x)$ multiplied by the derivative of $\alpha$. However, this calculation is only valid if $\alpha$ is both continuous and differentiable on $[a,b)$.

Title: The geometric interpretation


问:
Diagonal is closed in $(X, \tau_1) \times (X, \tau_2)$ (both are Hausdorff)
From Willard's General Topology:

Let $\tau_1$ and $\tau_2$ be Hausdorff topologies on the same set $X$. Let $\tau_3 = \tau_1 \cap \tau_2$.
If $(X, \tau_3)$ is Hausdorff, then the diagonal is closed in $(X, \tau_1) \times (X, \tau_2)$.

I know that since $\tau_1, \tau_2,$ and $\tau_3$ are all Hausdorff, that the diagonal $\Delta = \{(x,x) : x \in X\}$ is closed in each of the following product topologies $(X, \tau_1) \times (X, \tau_1)$, $(X, \tau_2) \times (X, \tau_2)$, and $(X, \tau_3) \times (X, \tau_3)$.

Without thinking much about the question, it seems intuitive that yes, $\Delta$ will also be closed in $(X, \tau_1) \times (X, \tau_2)$. I've tried to proceed by contradiction below, but I can't figure out how to finish it off. I would appreciate any help finishing it off.

I tried to proceed by contradiction:

Supposing that $\Delta$ is not closed in $(X, \tau_1) \times (X, \tau_2)$, then $\exists y \in X, y \neq x$, such that $(x,y) \in \overline{\Delta}$ (the closure of $\Delta$ in $\tau_1 \times \tau_2$). Then $\forall U \times V$ basic nbhoods of $(x,y)$, we must have that $(U\times V) \cap \Delta \neq \emptyset$. So $\exists z \in X$, such that $(z,z) \in U \times V$.

Also, since $U \times V$ is a basic nbhood of $(x,y)$ in $(X, \tau_1) \times (X, \tau_2)$, then $U$ is a basic nbhood of $x$ in $(X, \tau_1)$ and $V$ is a basic nbhood of $y$ in $(X, \tau_2)$. But then, since $(z,z) \in U \times V$, we must have $z \in U$ and $z \in V$. So $U, V$ are not disjoint. But $U$ was an arbitrary basic nbhood of $x$ in $(X, \tau_1)$ and $V$ was an arbitrary basic nbhood of $y$ in $(X, \tau_2)$, so there cannot exist any open disjoint sets $A \in \tau_1$, $B \in \tau_2$ such that $x \in A$ and $y \in B$...

I'm guessing this contradicts the Hausdorff-ness of $(X, \tau_3)$, but I'm having trouble putting it together.

Thanks.

答:
You are making your life overly complicated.

Take a point $(x,y)$ off the diagonal. We know there is an open set $U$ of $(X,\tau_3)$ and an open set $V$ of $(X,\tau_3)$ such that $x\in U$, $y\in V$, and $U\times V$ does not intersect the diagonal; i.e., $U\cap V=\varnothing$. This because $(X,\tau_3)$ is Hausdorff, so the diagonal in $X\times X$, with the $\tau_3\times\tau_3$ topology, is closed.

But $\tau_3=\tau_1\cap\tau_2$. Thus, $U\in \tau_1$ and $V\in \tau_2$. That means that $U\times V$ is open in $\tau_1\times\tau_2$, contains $(x,y)$, and does not intersect the diagonal.


问:
Composition factors of two groups are the same up to permutation
Show that the composition factors of the composition series of $S_3$ and $\mathbb{Z}_6$ have the same composition factors up to permutation. I was given this problem and was just asked to find a composition series for each of the two groups, and then was told to show that the two groups have the same composition factors up to permutation. I found the composition factors but did not know how I should proceed to show this. What is the way to do this?

答:
Observe, for $\mathbb Z_6,$ we can have two composition series:
$$\mathbb Z_6 \trianglerighteq \mathbb Z_6 / \langle 3 \rangle \cong \mathbb Z_2 \trianglerighteq \{e\}$$
$$\mathbb Z_6 \trianglerighteq \mathbb Z_6 / \langle 2 \rangle \cong \mathbb Z_3 \trianglerighteq \{e\}$$
By Jordan-Holder program, we know these to have equivalent composition factors. In particular,
$$\{ \mathbb Z_2, \mathbb Z_3 \}$$
which are cyclic groups of prime order and therefore simple. Similarly, considering $S_6,$ we see
$$S_3 \trianglerighteq A_3 \cong \mathbb Z_3 \trianglerighteq \{e\}$$

which has the composition factors
$$\{ S_3 / A_3 \cong \mathbb Z_2, \mathbb Z_3 \}$$

Therefore, the composition series for both groups are equivalent under their composition factors.


问:
Does $\bigtriangleup u =0$ imply $u=0$?
If $u \in H^2(\Omega) \cap H_0^1(\Omega)$ such that $\bigtriangleup u =0 $.

Is it necessarily that $u=0$?

Thank you!

答:
$$
           \int_{\Omega}u\Delta u \;dv = \int_{\Omega}\nabla\cdot(u\nabla u)-|\nabla u|^2 dv \\
    = \int_{\partial\Omega}u\frac{du}{dn}ds-\int_{\Omega}|\nabla u|^2dv \\
    = -\int_{\Omega}|\nabla u|^2dv
$$
Therefore, if $\Delta u=0$, with $u\in H^2\cap H^1_0$, then $\int_{\Omega}|\nabla u|^2dx=0$, which forces $u$ to be constant, and that constant must be $0$ in order for $u\in H_0^1(\Omega)$ to hold.

Title: Does $\bigtriangleup u =0$ imply $u=0$?


问:
True or False: Let $n ≥ 4$. There is an $n\times n$ matrix A such that $A^4 = 0$ and $\text{rank}(A+I_n) < 4$.
I don't know how to judge this and I am a beginner at linear algebra so I need an answer which explains all necessary terms.</p>

答:
If $\text{rank}(A + I_n) < 4$, then by the rank-nullity theorem it's nullity is at least $1$, i.e. there exists some non-trivial vector $v \neq 0$ such that $(A + I_n) v = 0 \implies A v = -v$. In other words, $v$ would be an eigenvector of $A$ with eigenvalue $-1$. But then
$$A^4 v = A(A(A(Av))) = v \neq 0$$
But why would this lead to a contradiction? (Hint: We haven't yet used the fact that $A^4 = 0$.)


问:
Which values of $\alpha$ allows $p_k$ to be a probability mass function?
I am in front of this problem :

The question is : for which values of alpha is it a probability mass function ?

I suppose that I should use the property that :

$$ \sum\limits_{i} p_{i} = 1 $$

I don't know where to start or how to address this problem...

答:
Note that for any positive integer $n$, $$\sum_{k=1}^n k = \frac{n(n+1)}2. $$ This can be shown by induction; it is trivially true for $n=1$, and assuming it is true for some $n\geqslant 1$, then
\begin{align}
\sum_{k=1}^{n+1} k &= \sum_{k=1}^n k + (n+1)\\
&= \frac{n(n+1)}2 + (n+1)\\
&= \frac{n(n+1) +2(n+1)}2\\
&= \frac{(n+1)(n+2)}2,
\end{align}
hence showing it is true for $n+1$. Now from $$\sum_{k=1}^n \mathbb P(X=k) = \sum_{k=1}^n \alpha k = 1  $$
we immediately find that
$$
\alpha = \frac2{n(n+1)}.
$$


问:
Solving recurrence relation, $a_n=6a_{n-1} - 5a_{n-2} + 1$
I'm trying to solve this recurrence relation:

$$
a_n = \begin{cases}
    0 & \mbox{for } n = 0 \\
    5 & \mbox{for } n = 1 \\
    6a_{n-1} - 5a_{n-2} + 1 & \mbox{for } n > 1
    \end{cases}
$$

I calculated generator function as:
$$
A = \frac{31x - 24x^2}{1 - 6x + 5x^2} + \frac{x^3}{(1-x)(1-6x+5x^2)} =
\frac{31x - 24x^2}{(x-1)(x-5)} + \frac{x^3}{(1-x)(x-1)(x-5)}
$$

(I'm not sure if that's right)

and its partial fractions decomposition looks like:
$$
A = \left(\frac{-7}{4} \cdot \frac{1}{x-1} - \frac{445}{4} \cdot \frac{1}{x-5}\right) +
\left( \frac{39}{16} \cdot \frac{1}{x-5} + \frac{3}{4} \cdot \frac{1}{(x-1)^2} - \frac{375}{16} \cdot \frac{1}{x-5} \right)
$$
(again - I'm not sure if it's ok)

I'm stuck here... From solutions I know that I should get:
$$
a_n = \frac{-21}{16} - \frac{1}{4}n + \frac{21}{16}5^n
$$

but I have no idea how it's solved... I hope somebody can help me (I spend more than 3h trying to solve this myself...)

答:
I did not check your work, so I’ll outline what you need to do to finish. You have something of the form

$$A(x)=\frac{a}{1-x}+\frac{b}{5-x}+\frac{c}{(1-x)^2}=\frac{a}{1-x}+\frac{b/5}{1-\frac{x}5}+\frac{c}{(1-x)^2}\;.$$

Expand these three terms into power series:

$$\begin{align*}
A(x)&=a\sum_{n\ge 0}x^n+\frac{b}5\sum_{n\ge 0}\left(\frac{x}5\right)^n+c\sum_{n\ge 0}(n+1)x^n\\\\
&=\sum_{n\ge 0}\left(a+\frac{b}{5^{n+1}}+c(n+1)\right)x^n\;.
\end{align*}$$

Now you can read off the coefficient of $x^n$.

Added: I’ve now had a chance to check your work, and it appears to be a bit off. I’ll use my preferred approach, which begins by assuming that $a_n=0$ for all $n<0$. Then the recurrence can be written

$$a_n=6a_{n-1}-5a_{n-2}+1-[n=0]+4[n=1]\;,$$

for all $n\ge 0$, where the last two terms contain Iverson brackets and are added to make the recurrence give the correct values to $a_0$ and $a_1$. Now multiply through by $x^n$ and sum over $n\ge 0$:

$$\begin{align*}
\sum_{n\ge 0}a_nx^n&=\sum_{n\ge 0}\Big(6a_{n-1}-5a_{n-2}+1-[n=0]+4[n=1]\Big)x^n\\\\
&=6x\sum_{n\ge 0}a_{n-1}x^{n-1}-5x^2\sum_{n\ge 0}a_{n-2}x^{n-2}+\sum_{n\ge 0}x^n-1+4x\;.
\end{align*}$$

Since $A(x)=\sum_{n\ge 0}a_nx^n=\sum_{n\ge 0}a_{n-1}x^{n-1}=\sum_{n\ge 0}a_{n-2}x^{n-2}$ (remember the blanket assumption that $a_n=0$ for $n<0$), we have

$$A(x)=6xA(x)-5x^2A(x)+\frac1{1-x}-1+4x\;,$$

and hence

$$A(x)=\frac1{(1-x)^2(5-x)}+\frac{4x-1}{(1-x)(5-x)}\;.$$

I’ll leave the partial fraction decomposition to you, at least for now.


问:
Vector Analysis- Curves in Space

I have attached both the problem and approach to the solution. But I am confused how to proceed. Kindly help in providing the simplified version of the solution.

PS- I tried pasting the problem and solution from a word file, but I don't understand why the editor is not accepting my text. That's why I have attached the images.

答:
I'm not at all sure that the following remarks are in any real sense a simplification of the argument given in our OP Sanket Garud's question and answer, but perhaps I can expand upon the essential points and thus provide some clarification.

Given a unit speed curve $\gamma(s)$ in $\Bbb R^3$, where $s$ is the arc-length, its unit tangent vector $T(s)$ is defined via

$T(s) = \dot \gamma(s); \tag 1$

the unit normal vector $N(s)$ and curvature $\kappa(s)$ are defined by the first Frenet-Serret equation

$\dot T = \kappa N, \; \vert N \vert = 1, \; \kappa(s) > 0; \tag 2$

we also have the unit binormal vector

$B(s) = T(s) \times N(s); \tag 3$

with the hypothesis $\tau(s) = 0$, the third Frenet-Serret equation yields

$\dot B(s) = -\tau(s) N(s) = -(0)N(s) = 0; \tag 4$

that is, $B(s)$ is constant along $\gamma(s)$; we thus adopt the shorthand

$B = B(s); \tag{4.5}$

in accord with the definition (3) we have

$B \cdot N(s) = T(s) \times N(s) \cdot N(s) = 0 \tag 5$

and

$B \cdot T(s) = T(s) \times N(s) \cdot N(s) = 0. \tag 6$

Now everything which has been written in the above is either a well-known and accepted fact of elementary differential geometry or has been covered by our OP in his question/answer.  We next deviate from the presented work of Sanket Garud and observe that (1) implies

$\gamma(s) - \gamma(s_0) = \displaystyle \int_{s_0}^s T(u) \; du, \tag 7$

whence

$B \cdot (\gamma(s) - \gamma(s_0)) = B \cdot \displaystyle \int_{s_0}^s T(u) \; du = \int_{s_0}^s B \cdot T(u) \; du = \int_{s_0}^s 0 \; du = 0, \tag 8$

which shows that $\gamma(s) - \gamma(s_0)$ lies in a plane normal to $B$; thus

$\gamma(s) = (\gamma(s) - \gamma(s_0)) + \gamma(s_0) \tag{8.5}$

lies in the plane normal to $B$ and containing $\gamma(s_0)$; furthermore, in accord with (5) and (6), both $T$ and $N$ lie in, are tangent to, this plane at every $\gamma(s)$; thus the vector

$C(s) = \gamma(s) + \dfrac{1}{\kappa}N(s) \tag 9$

lies in this plane as well, since $N(s)$ does; we have

$\dot C(s) = \dot \gamma(s) + \dfrac{1}{\kappa} \dot N(s) = \dot \gamma(s) + \dfrac{1}{\kappa} \dot N(s)$
$= T(s) + \dfrac{1}{\kappa} (-\kappa T(s) + \tau B(s)) = T(s) + \dfrac{1}{\kappa}(-\kappa T(s)) = T(s) - T(s) = 0; \tag{10}$

thus,

$C(s) = C \tag{11}$

is constant, and (9) becomes

$\gamma(s) = C - \dfrac{1}{\kappa}N(s); \tag{12}$

thus,

$\vert \gamma(s) - C \vert = \left \vert \dfrac{1}{\kappa} N(s) \right \vert = \dfrac{1}{\kappa} \vert N(s) \vert = \dfrac{1}{\kappa}, \tag{13}$

which shows that $\gamma(s)$ lies in the circle of radius $1/\kappa$ centered at $C$ in the plane normal to $B$ containing $\gamma(0)$.  We note that if $\gamma(s)$ is not restricted to lie in this plane, then (13) still binds but now tells us that
$\gamma(s)$ lies in the sphere with the same radius and center.  Since both the circle and sphere described by (13) are of the same radius $1/\kappa$, the circle is in fact a great circle in this sphere.

Title: Vector Analysis- Curves in Space


问:
Does $\overline{A}$ be dense-in-itselft implies A is dense-in-itselft?
Let ($X, τ$) ve a topological space and $A\subseteq X$ we say that $A$ is dense-in-itselft if $A \subseteq A'$ where $A'$ denotes the set of acumulation points of A.
If $\overline{A}$ is dense-in-itselft then A is dense-in-itselft?
I think yes but I have no idea of how to proof it or any counterexample

答:
The restiction to $T_1$ spaces is in general necessary.

Let $S=\{0,1\}$ with the topology $\{\varnothing,\{0\},S\}$, let $[0,1]$ have its usual topology, let $Y=[0,1]\times S$ have the product topology, and let $X=Y\setminus\{\langle0,1\rangle\}$. Let

$$A=\{\langle 0,0\rangle\}\cup\big((0,1]\times\{1\}\big)\,;$$

then $\langle 0,0\rangle$ is an isolated point of $A$, so $A$ is not dense-in-itself, but $\operatorname{cl}A=X$, which has no isolated points and is therefore dense-in-itself.


问:
Very basic question about integrals and substitution
I am a bit frustrated since I have done that a few years ago but know I suddenly forgot and it's such a "strange" question that I don't even have the keywords to search for it. My questions is: how exactly do we treat the argument of a function like $f(x-x_0)$ when it comes to the $dx$ in the integral, when we perform integration on such a thing?

My questions is related to this proof:

Proof: Let $t'=t±t_0$, i.e., $t=t'\mp t_o$, we have

$$\begin{align*}
\mathcal{F}[x(t±t_0)]\quad &: \quad \int^\infty_{-\infty}x(t±{t_0})e^{-j \omega t}dt'=\int^\infty_{-\infty}x(t')e^{-j \omega(t'\mp t_0)}dt\\
&:e^{±j\omega t_0}\int^\infty_{-\infty}x(t')e^{-j\omega t'}dt'=X(j \omega)e^{±j \omega t_0}
\end{align*}$$

''Naively'' one would consider that since we write the first integral (the integral of $x(t \pm t_0)$, one could be tempted to write $d(t \pm t_0)$ instead of the $dt$ for the infinitesimal in the integral. However, I am not sure if that even as a meaning and therefore why should that not be done/what is the justification?

Source: http://fourier.eng.hmc.edu/e101/lectures/handout3/node2.html

P.S.
Maybe, following this idea, one could develop $d(t \pm t_0)$ into $dt \pm dt_0$ and consider that since $t$ is the variable and $t_0$ is a constant... that $dt_0 =0$, but I don't know if that makes any sense at all.

答:
The basis of substitution is that $d(\phi(x))=\phi'(x) dx$, which gives $$\int  f(\phi(x))\phi'(x) dx =\int f(\phi(x))d(\phi(x))$$

Supposing that $\phi(x)=x+\alpha$, we have $d(\phi(x))=dx$, that is: $$\int_\Bbb R f(x+\alpha) dx =\int_\Bbb R f(x+\alpha) d(x+\alpha)$$
Naturally, you must account for limits of integration, but since yours are infinite, they will stay the same.

Essentially what you are doing is translating the function by $\pm t_0$. Integrating across the whole real space will still give the same answer at the end.


问:
genus$=2$ implies Hyperelliptic.
In the book of Rick Miranda (Algebraic Curves and Riemann Surfaces), in Proposition 1.10 of Chapter VII (p. 198), the claim is that every compact Riemann Surfaces of genus $2$ is hyperelliptic. The proof is:

Let $K_X$ the canonical divisor has degree $2g-2=2$, and by Riemann-Roch Theorem we have $\dim L(K_X)=2$, therefore we may suppose that $K_X>0$, take $f\in L(K_X)$ nonconstant so when you look as a map to $\overline{\mathbb{C}}$ has degree $=2$ so it's Hyperelliptic.

It's a good proof but a just don't understand why we can suppose $K>0$.

I'm tried write $K_X=P-N$ where $P,N\in \operatorname{Div}(X)$ are non-negative divisors and use Riemann-Roch but I failed, thanks for any comments.

答:
Recall that
$$
\DeclareMathOperator{\div}{div}
L(D) = \{f \in \mathbb{C}(X) : D + \div(f) \geq 0\} \cup \{0\}
$$
where $D$ is a divisor on $X$. Since $\ell(K_X) = 2 > 0$, then there is a nonzero function $f \in L(K_X)$, i.e., a function $f \in \mathbb{C}(X)$ such that $K_X + \div(f) \geq 0$. Since $K_X$ is only defined up to linear equivalence we can replace $K_X$ by $K_X' := K_X + \div(f)$: we have $K_X' \geq 0$ by the definition of $L(K_X)$, and as pointed out in the comments, $K_X' \neq 0$ since $\ell(0) = 1$ and $\ell(K_X') = \ell(K_X) = 2$.

Title: genus$=2$ implies Hyperelliptic.


问:
Isomorphism $\left(\mathbb{C}^{n}\setminus\{0\}\right)/\mathbb{Z}$ with $S^{1} \times S^{2n-1}$
I have to prove that there is this isomorphism:
$$\frac{\mathbb{C}^{n}{\setminus\{0\}}}{ \mathbb{Z}} \simeq S^{1} \times S^{2n-1},$$
where there is this equivalence relation in the left side:
$(w_1, \dots, w_n) \sim (z_1, \dots, z_n)$ if exists $\lambda \in \mathbb{C}$ with $|\lambda| \ge 1$ and a fixed $d \in \mathbb{N}$ such that
$$(w_1, \dots, w_n) \sim (\lambda^d z_1, \dots,\lambda^d z_n).$$</p>

答:
Consider the map $$\phi:z\in\mathbb C^n\setminus\{0\}\mapsto\Bigl(\frac{z}{|z|},\log|z|\Bigr)\in S^{2n-1}\times\mathbb R.$$

Show that it is an homeomorphism.

Next, define on the codomain of $\phi$ an equivalence relation $\equiv$ such that $\phi$ induces an homeomorphism from $(\mathbb C^n\setminus\{0\})/\sim$ to $(S^{2n-1}\times\mathbb R)/\equiv$.

Figure out what $(S^{2n-1}\times\mathbb R)/\equiv$ is.


答:
I'm not sure if this is your point, but see: put $t'=t+t_0$, then $dt'/dt=1$.

That is: $t=t'-t_0$ and $dt'=dt$.
Hence
$$\int_{-\infty}^\infty x(t+t_0)e^{-j\omega t}dt=\int_{-\infty}^\infty x(t')e^{-j\omega (t'-t_0)}dt'$$

Title: Very basic question about integrals and substitution

Body:
I am a bit frustrated since I have done that a few years ago but know I suddenly forgot and it's such a "strange" question that I don't even have the keywords to search for it. My questions is: how exactly do we treat the argument of a function like $f(x-x_0)$ when it comes to the $dx$ in the integral, when we perform integration on such a thing?

My questions is related to this proof:

Proof: Let $t'=t±t_0$, i.e., $t=t'\mp t_o$, we have

$$\begin{align*}
\mathcal{F}[x(t±t_0)]\quad &: \quad \int^\infty_{-\infty}x(t±{t_0})e^{-j \omega t}dt'=\int^\infty_{-\infty}x(t')e^{-j \omega(t'\mp t_0)}dt\\
&:e^{±j\omega t_0}\int^\infty_{-\infty}x(t')e^{-j\omega t'}dt'=X(j \omega)e^{±j \omega t_0}
\end{align*}$$

''Naively'' one would consider that since we write the first integral (the integral of $x(t \pm t_0)$, one could be tempted to write $d(t \pm t_0)$ instead of the $dt$ for the infinitesimal in the integral. However, I am not sure if that even as a meaning and therefore why should that not be done/what is the justification?

Source: http://fourier.eng.hmc.edu/e101/lectures/handout3/node2.html

P.S.
Maybe, following this idea, one could develop $d(t \pm t_0)$ into $dt \pm dt_0$ and consider that since $t$ is the variable and $t_0$ is a constant... that $dt_0 =0$, but I don't know if that makes any sense at all.


问:
Calculus: Why can't a we define 1 as a critical point when we have $h(x) = \frac{4x^2}{x^2-1}$?
I'm trying to understand an explanation in Khan Academy on the topic "Finding the critical point".

I understand that we need to get the derivative down to either zero or undefined to find that critical point, but I don't get why we can't use use 1 in this equation's derivative. $h(x) = \frac{4x^2}{x^2-1}$

The derivative is $\frac{-8x}{(x^2-1)^2}$

Is it because the graph itself already has undefined at 1?

答:
A critical point is a point in the domain of the function where the derivative is either undefined or zero. As you notice, $-1$ and $1$ are not in the domain of the function, so they don't meet the criteria.

We would say that $x=-1$ and $x=1$ are asymptotes of the function.


问:
Given  = 〈1, −4〉   = −2 + 6 find w such that...
Given  = 〈1, −4〉   = −2 + 6, I need to find a vector W such that $$-4u - 1v +w = 0$$
My intuition tells me to just put the vectors $u$ and $v$ in the equation and isolate for $w$, is this the correct way to go about this problem?

答: Isolate for w  $$ w = 4u + v$$ and substitute vectors into equation.


问:
Probability of an RNG?
For fun (relatively speaking), I'd like to do a little research on probabilities related to a pseudo-random number generator.

Specifically, how would I get started figuring out "what is the probability that a prng will generate the same random number twice?"

This is part of an investigation into a larger issue, but I'm not sure where to begin (or if my question is even relevant).

Edited to add:

After reading the comments and answers I see that my question isn't clear. As an example: the rng outputs a 5 What is the probability that another 5 will be the next output? How many "rolls" of the RNG must I make before I see another 5?

If we assume a good rng has uniform distribution, then is it 1 out of (2^32) (assuming 32bit integers)?

For a larger example, lets say I use the RNG to generate numbers that are used as an index into an array of characters. The array is 25 numbers long, so the output of the RNG is mod 25 I will generate a sequence of 5 characters.

How would I determine the probability of generating the same 5 character sequence?

If we assume that we have a uniform distribution of output numbers from the RNG, then the RNG's output shouldn't factor into the probability of generating the same character sequence.

However, is my assumption valid?

答:
Others have answered your question well, so this answer will be directed more to cryptographic security of pseudorandom random number generators.

The only totally secure encryption scheme is the one-time pad, which requires pre-sharing random keys of arbitrary length. In practice this is rarely done, as pre-sharing is not computationally feasible or potentially not possible at all. However, if both parties can use a cryptographically secure pseudorandom number generator that somehow started at the same point, then they could create a stream cipher that would be a fairly good approximation to the one-time pad. Note that the deterministic nature of the PRNG makes it possible for it to be used in this way. If it were nondeterministic the receiver would be unable to recreate the keystream, and thus would be unable to decrypt and recover the message.

How do we make sure that a PRNG is "sufficiently random" to be secure? It is not enough for the probability of each outcome to be equiprobable. We need it to have forward secrecy, meaning that if someone reads an arbitrary length of the output they will have no advantage in predicting the next outcome. The most common way of measuring this is to use Claude Shannon's definition of entropy. The basic definition of entropy is
$$J = - \sum_{i=1}^k \, p(x_i) \cdot \log_2 p(x_i).$$
where the $x_i$ are the potential outcomes and $p(x_i)$ is the probability of that outcome occurring. If the PRNG is uniformly distributed, then we have $J = \log_2 k$. But we desire to test for more than just the randomness of each individual output, so we look at nth order entropy:
$$J_n = \frac{-1}{n} \sum_{i=1}^k \, p(x_i) \cdot \log_2 p(x_i)$$
where the $x_i$ are now a sequence of $n$ outputs concatenated together. With a PRNG, many of the outputs will never happen, so many of the probabilities will be $0$. Then finally, we can define the absolute minimum entropy
$$E = \min_{1 \leq n \leq \infty} J_n$$
We should note that for a PRNG, which is periodic, the absolute minimum entropy must be $0$ (why?). But, by looking at approximations to the absolute minimum entropy using large values of $n$, we can get a better idea of the probabilistic nature of the PRNG. This is something you could potentially compute for a RNG of smaller period.

In practice, a massive amount of tests are done on a potential CSPRNG, including Knuth's algorithms in The Art of Computer Programming Vol.2, the DIEHARD tests (mentioned by J.M.), the Crypt-XS package, standards from the Handbook of Applied Cryptography, and the powerful NIST Statistical Test Suite. The latter contains several of the tests from each of the others, in addition to many of its own. To elaborate on some of the subtler points mentioned throughout the answer, the process of getting a CSPRNG to "start at the same place" is done by seeding. A seed is a relatively short input to a PRNG that, in a very loose sense, gives it a starting position. The seed should ideally be truly random, which in practice is best approximated by hardware random number generators, and can be shared via some encryption scheme between the two parties due to its small length. And finally, randomness can be deterministically increased from insufficient sources of entropy. While the built-in random functions in most programming languages are not sufficiently random, several outputs can be concatenated together and hashed with a cryptographic hash function such as one of the SHA family. This increases the first order entropy, and if concatenations of different (random) length are used, then higher order entropy also increases. This forms the basic idea of randomness extractors.

Title: Probability of an RNG?


答:
Let's write your recurrence relation for $n$ and $n+1$:

$a_{n}-6a_{n-1}+5a_{n-2}-1=0$

$a_{n+1}-6a_{n }+5a_{n-1}-1=0$

Now we subtract one from another:
$a_{n+1}-7a_{n }+11a_{n-1}-5a_{n-2}=0$ (relation 2)

Then, from Theorem on wiki we build a characteristic polynomial
$x^3-7x^2+11x-5 $ with roots $1,1,5$. Hence, by that theorem, the solution of our recurrence relation (2) is $c_1 1^n+ c_2n1^n+c_3 5^n$. All you have to do now is to find those constants $c_i$ from initial conditions and from the fact that it should satisfy your initial recurrence relation.

Title: Solving recurrence relation, $a_n=6a_{n-1} - 5a_{n-2} + 1$

Body:
I'm trying to solve this recurrence relation:

$$
a_n = \begin{cases}
    0 & \mbox{for } n = 0 \\
    5 & \mbox{for } n = 1 \\
    6a_{n-1} - 5a_{n-2} + 1 & \mbox{for } n > 1
    \end{cases}
$$

I calculated generator function as:
$$
A = \frac{31x - 24x^2}{1 - 6x + 5x^2} + \frac{x^3}{(1-x)(1-6x+5x^2)} =
\frac{31x - 24x^2}{(x-1)(x-5)} + \frac{x^3}{(1-x)(x-1)(x-5)}
$$

(I'm not sure if that's right)

and its partial fractions decomposition looks like:
$$
A = \left(\frac{-7}{4} \cdot \frac{1}{x-1} - \frac{445}{4} \cdot \frac{1}{x-5}\right) +
\left( \frac{39}{16} \cdot \frac{1}{x-5} + \frac{3}{4} \cdot \frac{1}{(x-1)^2} - \frac{375}{16} \cdot \frac{1}{x-5} \right)
$$
(again - I'm not sure if it's ok)

I'm stuck here... From solutions I know that I should get:
$$
a_n = \frac{-21}{16} - \frac{1}{4}n + \frac{21}{16}5^n
$$

but I have no idea how it's solved... I hope somebody can help me (I spend more than 3h trying to solve this myself...)


问:
Proving the Inf of a given set
Given $A=\{(n-5)/4n : n \in \mathbb{N}\}$

I want to show the $\inf(A)=-1.$ I've claimed this and shown that $(n-5)/4n \ge -1$. I now usually use archimedian property to say that there exists an $n>5/(1-4y)$ in order to prove that $y$ can't be the $\inf(A)$ as $y>(n-5)/4n$. However, solving this gives me $y<(n-5)/4n$ instead and working backwards from the correct answer gives me $n<5/(1-4y)$ which isn't shown by the archimedian property? Can anyone help

答:
If $A\subset \Bbb R$ and $A$ has a least member $\min A$  then $\min A  =\inf A .$

Because $\min A$ is a  lower bound for $A$  and no $y>\min A$  can be a lower bound for $A,$

(... because if $y>\min A$ then $y$ is greater than some member of $A,$ that is, $y>\min A\in A$...)

so $\min A$ must be the largest lower bound for $A$, i.e. $\min A=\inf A .$

For the set in your Q:

$-1 \in A$ because if $n=1$ then $-1=(n-5)/4n \in A.$

If $n\in \Bbb N$ then $(n-5)/4n\ge -1$ because for any $n\in\Bbb N$ we have $$(n-5)/4n\ge-1\iff 4n\cdot (n-5)/4n\ge 4n\cdot (-1)\iff n-5\ge -4n \iff 5n\ge 5.$$

So $\min A=-1 .$

Title: Proving the Inf of a given set


问:
Does the prisoner dilemma have a non-pure strategy Nash equilibrium?
I would like to know whether this Prisoner Dilemma has a non-pure strategy Nash Equilibrium:

here is the link.

答: A pure strategy is considered a special case of a mixed strategy, so it technically has a mixed strategy of playing defect with probability one and cooperate with probability zero. However there is no non-pure mixed strategy Nash equilibrium in a prisoner's dilemma because both players have strictly dominant strategies. Any possible mix with a positive weight on cooperating would be strictly dominated by a pure strategy of defecting.</p>


问:
Does the following alternating series converge or diverge?
I have the following series that I have to check for convergence or divergence:

$$
\sum\limits_{n=1}^{\infty} \frac{\sin (n+1/2)\pi}{1 + \sqrt{n}}
$$

I know that it is an alternating series therefore I have to check for two conditions to be satisfied in order for it to be convergent; the limit has to equal 0 as n approaches infinity and that for series $a_n$ $a_n < a_{n+1}$. I am able to prove that the limit is approaches zero but could someone help me prove that for the series above that:
$ a_n < a_{n+1}$

答:
$$
\sum\limits_{n=1}^{\infty} \frac{\sin (n+1/2)\pi}{1 + \sqrt{n}}=\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt n}
$$
so this is an alternating series. Generic term converges to $0$ $$\underset{n\to \infty }{\text{lim}}\frac{1}{\sqrt{n}+1}=0$$
Therefore the series  converges and
$$\sum _{n=0}^{\infty } \frac{(-1)^n}{\sqrt{n}+1}\approx 0.721717$$


问:
If $f(x)=x^2+ax+b$ has integer roots, and $f(x+\frac1x)=f(x)+f(\frac1x)$, then find $a^2+b^2$

Let $f(x) = x^2 + ax + b$. If for all non-zero real $x$
$$f\left(x + \frac1x\right) = f(x) + f\left(\frac1x\right)$$
and the roots of $f(x) = 0$ are integers, what is the value of $a^2 + b^2$?

Here is the photo of the question if you are finding difficult to look at it here

EDIT: The answer must be in two-digits

EDIT: What I have done so far is that I added temporary values to the equation
$$f(x) = 0 = x^2 + ax + b$$ and tried to solve it in the form of $a + b$ to which then I squared and solved and got,
$$a^2 + b^2 = x^2 - 2ab$$

I don't really know what I am doing is correct, so help me out please!

答:
As described in Servaes' answer, deduce that $b=2$ by comparing coefficients and also that the roots belong to the set $\{\pm 1, \pm 2 \}$ by the use of Rational Root Theorem.

At this point you don't have to solve any equations. You can use Vieta's formulas on the sums and products of roots to deduce that the product of the roots must equal $b=2$, so they must be of the same sign, that is either of the pairs $1,2$ or $-1,-2$. Then deduce the sum of the roots must be equal to $a$, which gives $a = \pm 3$. Then $a^2 + b^2 = 13$, as before.


问:
Bertrand's Paradox and uniform distribution
This is a problem from Hans-Otto Georgii's textbook.

Recall the situation of Bertrand’s paradox, and let X be the distance of the random chord to the centre of the circle. Find the distribution density of X if

(a) the midpoint of the chord is uniformly distributed on the disk $|x|<r$,

(b) the angle under which the chord is seen from the centre of the circle is uniformly distributed on the interval $\Omega_2$ = $[0,\pi]$.

The solutions are given as:
a. the distribution density of X on [0,r] is $\rho_1(x)$ = $2x/r^2$.

b. $\rho_2(x) = 2/(\pi*\sqrt(r^2 - x^2))$, x $\in [0,r]$.

edit:
the overall approach is to derive cdf then differentiate.

a. cdf = $\pi x^2/ \pi r^2$; pdf = $2x/r^2$ .

my thoughts for b:

The probability formula for continuous uniform distribution in my textbook:
$U_\Omega(A) = \lambda^n(A) / \lambda^n(\Omega)$. So I tried expressing cdf like some probability:

$\Omega = [0,\pi]$, so $\lambda^n(\Omega) = \pi$. A is the range of angle POQ when the distance is in $[0,x]$. PQ is the chord.

A = $2arccos(x/r)$ for x in $[0,x]$.

cdf = $2arccos(x/r) / \pi$; pdf = $- 2 / \pi \sqrt(r^2 - x^2)$.

However there is an extra negative sign here. could anybody help correct my workings?

答:
Although the question is somewhat older, let's derive the solution to both parts in more detail than provided in the book (and comments).
In the end, the exercise consists of basic trigonometry and calculus together with some probability fundamentals.

Solution to (a).

Consider the following image created in Tikz which shows some circle $Z$ and a (uniformly drawn) random point within the circle.

The distance of the circle center to that point is $x$. The set of points $A$ that have a distance $\leq x$ from the circle center is $A := \{y \in Z: \Vert y \Vert \leq x\}$ and is hatched blue. In other words, all points of $A$ lie in a smaller circle with radius $x$ around the circle center of circle $Z$. Since $\vert A \vert = \pi x^2$ and $\vert Z\vert = \pi r^2$, we have that

\begin{equation}
F_X(x) = P(X \leq x) = \frac{\vert A \vert}{\vert Z\vert} = \frac{\pi x^2}{\pi r^2} = \Big(\frac{x}{r}\Big)^2
\end{equation}

as you already edited into your question.

But then

\begin{equation}
\frac{\text{d}}{\text{d}x} F_X = \frac{2x}{r^2}
\end{equation}

Solution to (b).

Consider the following image, also created in Tikz.

Let $r$ be the radius and $X$ the distance of the circle center to the midpoint $M_1$ of the chord.
Based on the image and basic trigonometry, it is clear that $\frac{\alpha_1}{2} = \arccos(\frac{X}{r})$. Conversely, it is $X = \cos\Big(\frac{\alpha_1}{2}\Big)r$.

The probability for a distance $X \leq x$ with $x \in [0, r]$ is hence

\begin{equation}
 F_X(x) = P(X \leq x) = P(\cos\Big(\frac{\alpha_1}{2}\Big)r \leq x) = P(\alpha_1 \geq 2\arccos(\frac{x}{r})) = 1 - P(\alpha_1 < 2\arccos(\frac{x}{r})) = 1 - \frac{2}{\pi}\arccos\Big(\frac{x}{r}\Big)
\end{equation}

and the probability density therefore

\begin{equation}
 f_X(x) = \frac{\text{d}}{\text{d}x} F_X(x) = \frac{2}{\pi}\frac{1}{\sqrt{1 - \Big(\frac{x}{r}\Big)^2}}\frac{1}{r} = \frac{2}{\pi}\frac{1}{\sqrt{r^2 - x^2}}
\end{equation}

since $r > 0$.


问:
Prove that $d(A,B)=d(x,y).$
Problem: Define the distance between two subsets $A$ and $B$ of a metric space $X$ by
\begin{equation*}
d(A,B)=\inf\{d(a,b):a\in A,b\in B\}.
\end{equation*}
Prove that if $A$ and $B$ are compact, then there exist $x\in A$ and $y\in B$ such that
\begin{equation*}
d(x,y)=d(A,B).
\end{equation*}
** I know there's something missing in this proof because I can't guarentee that my $x\in A, y\in B$ will cover all the possible point in the set. We're currently using Baby Rudin as our textbook and haven't finished the chapter on series and sequences yet. I would appreciate any feedback.

My work:
Given $A$ and $B$ are compact, choose a sequence $\{a_n\}_n \in A, \{b_n\}_n\in B.$ Then by thrm 3.6(a), there exists a subsequence $a_{n_\ell}\to x\in A, b_{n_\ell}\to y\in B.$ This means for all $\varepsilon >0,$ there exists a $N\in \mathbb{N}$ s.t. $d(a_{n_\ell}, x) <\varepsilon$ for $\ell\geq N$. Similarly for all $\varepsilon >0,$ there exists $M\in \mathbb{N}$ s.t.  $d(b_{n_\ell}, y)<y$ for $\ell \geq M.$ Then choosing $\ell\geq \max\{N,M\},$ $$d(a_{n_\ell}, b_{n_\ell})\leq d(a_{n_\ell}, x)+d(b_{n_\ell},y)+d(x,y)$$
$$d(x,y)\leq d(a_{n_\ell}, x)+d(b_{n_\ell},y)+d(a_{n_\ell}, b_{n_\ell})$$
So together we have $|d(a_{n_\ell}, b_{n_\ell})-d(x,y)|\leq d(a_{n_\ell}, x)+d(b_{n_\ell},y)<2\varepsilon$ thus $\lim_{\ell\to \infty} d(a_{n_\ell}, b_{n_\ell})=d(x,y).$

Let $D:=\{d(a, b) \mid a\in A, b\in B\}$; need to show $\inf\{D\}=d(x,y).$ We have $d(x,y)\in D.$

By thrm 3.7 we have that all the limits of subsequences in a metric space form a closed set. i.e.  \begin{align*}
    \lim_{\ell \to \infty}d(a_{n_\ell},b_{n_\ell})&=d(x_1,y_1)\\
    \lim_{k \to \infty}d(a_{n_k},b_{n_k})&=d(x_2,y_2)\\
    &\vdots
\end{align*}
Then all the $d(x_i, y_i)$ form a closed set where $x_i\in A, y_i\in B$. All of theses $\{d(x_i,y_i)\mid i\in\mathbb{N} \} \subset \mathbb{R}$ and in particular, they form a closed subset of $\mathbb{R}$ so there is a minimum  element we call $d(x,y)$.
\begin{align*}
    d(A,B)&=\inf\{d(a, b) \mid a\in A, b\in B\}\\
    &=\inf\left\{\lim_{\ell \to \infty} d(a_{n_\ell},b_{n_\ell})=d(x,y) \mid  a_{n_\ell} \in A, y, b_{n_\ell} \in B\right\}.
\end{align*}
The infimum of a closes sets is an element that is in the set because every set contains all its limit points so $\inf\{D\}\in D.$

Reference:
Thrm 3.6(a) If $\{p_n\}$ is a sequence in a compact metric space $X$, then some subsequence of $\{p_n\}$ converges to a point of $X$.

Thrm 3.7 The subsequential limits of a sequence $\{p_n\}$ in a metric space $X$ from a closed subset of $X$.

答:
I'll be honest, I didn't read all of what you wrote, but you're overcomplicating it for yourself.

For every $n > 0$, there are $x_n \in A, y_n \in B$ such that $d(x_n, y_n) < d(A, B) + 1/n$.

Let $\{x_n\}$ and $\{y_n\}$ be the associated sequences in $A, B$. So by compactness both have convergent subsequences, and suppose that those convergent subsequences converge to some $x, y$ respectively. For convenience let $\{x_n\}$ and $\{y_n\}$ now refer to the subsequences that converge to $x,y$
.
Note that for any $\epsilon > 0$, there is some $N$ such that for $n \geq N$ for $d(x, x_n), d(y,y_n) < \epsilon /2$, which means thanks to the triangle inequality that $d(x, y) \leq d(x, x_n) + d(x_n, y_n) + d(y, y_n) \leq \epsilon + d(x_n, y_n)$.

Therefore for every $\epsilon > 0$ there is $N$ such that for all $n \geq N$, $d(x,y) \leq d(x_n, y_n) + \epsilon/2$. We also know by construction of the sequences that there is some $M$ such that for all $n \geq M$, $d(x_n, y_n) < d(A, B) + \epsilon/2$. So for $n \geq max (M, N)$, $d(x, y) \leq d(x_n, y_n) + \epsilon/2 \leq d(A, B) + \epsilon$.

So then $d(x, y) \leq d(A, B) + \epsilon$, and $\epsilon > 0$ was arbitrary. So $d(x, y) \leq d(A, B) \implies d(x,y) = d(A, B)$.

Title: Prove that $d(A,B)=d(x,y).$


问:
Section of blowup map of schemes
Let $X$ be a scheme. Let $\mathcal{I} \subseteq \mathcal{O}_X$ be a quasi-coherent sheaf of ideals on $X$, and let $Z \subseteq X$ be the closed subscheme corresponding to $\mathcal{I}$. Let $X' := \operatorname{Bl}_Z(X)$ be the blowup, and let $b\colon X' \to X$ be the blowup morphism. Suppose there exists a section of $b$, that is, a morphism $s\colon X \to X'$ such that $b \circ s = \operatorname{id}_X$. Is it necessarily true that $b$ is an isomorphism?

(This is a generalization of this question that was asked in the comments; I decided to make this a separate question since it goes beyond the scope of the original question. I initially thought I only had a partial answer, but in the process of writing it up, I realized it works in full generality, so I've provided an answer myself.)

答:
Yes, if the blowup morphism $b$ has a section, then $b$ is an isomorphism.

By Stacks tag 01O2, the blowup morphism $b$ is separated. By Stacks tag 01KT, the section $s$ is a closed immersion. Since $s$ is a closed immersion, the image $s(X)$ is closed. By Stacks tag 02OS, the blowup morphism $b$ restricts to an isomorphism on the open subscheme $b^{-1}(X \setminus Z)$, so $b^{-1}(X \setminus Z) \subseteq s(X)$. Since $s(X)$ is closed, we in fact have $\overline{b^{-1}(X \setminus Z)} \subseteq s(X)$.

In particular, $s(X)$ contains all irreducible components of $X'$ that have nonempty intersection with $b^{-1}(X \setminus Z)$, that is, all irreducible components of $X'$ that are not contained in $b^{-1}(Z)$. By Stacks tag 0BFM, no irreducible components of $X'$ are contained in $b^{-1}(Z)$, so $s(X) = X'$. That is, $s$ is surjective. In particular, $s$ is a homeomorphism with continuous inverse $b$.

By Stacks tag 02OS, the exceptional divisor $E = b^{-1}(Z)$ is an effective Cartier divisor on $X'$. We show that $Z = s^{-1}(E)$ is an effective Cartier divisor on $X$. This is a local question.

By Stacks tag 0804 and the definition of effective Cartier divisor, we reduce to the following local situation: we have a commutative ring $A$, an ideal $I \subseteq A$, an element $a \in I$, and a surjective map $\varphi\colon A[\frac{I}{a}] \to A$ such that the composition $A \overset{i}{\to} A[\frac{I}{a}] \overset{\varphi}{\to} A$ is the identity, where $A[\frac{I}{a}]$ denotes the affine blowup algebra and $i(x) = x/a^0$ for all $x \in A$. We want to show that, for each non-zerodivisor $f \in A[\frac{I}{a}]$, the element $\varphi(f)$ is also a non-zerodivisor.

Write $f = x/a^n$ with $x \in I^n$. Let $z \in A$ such that $\varphi(f) z = 0$. Let $z' = y/a^m \in A[\frac{I}{a}]$ such that $\varphi(z') = z$ and $y \in I^m$. Then
$$xy = \varphi(i(x) i(y)) = \varphi(a^{m+n} fz') = \varphi(a^{m+n}) \varphi(f) \varphi(z') = a^{m+n} \varphi(f) z = 0,$$
so $fz' = (xy)/a^{m+n} = 0$. Since $f$ is a non-zerodivisor, this implies $z' = 0$, so $z = 0$ and $\varphi(f)$ is also a non-zerodivisor.

Thus $Z$ is an effective Cartier divisor on $X$. By the universal property of blowing up, $b$ is an isomorphism.

Title: Section of blowup map of schemes


问:
Symmetry of higher order mixed partial derivatives under weaker assumptions
I have asked this question on Mathoverflow, but it did not receive much attention there.

Suppose $U$ is an open subset of $\mathbb{R}^n$, and $f:U\to \mathbb{R}$. When $f$ is $C^2$ we know that the mixed partial derivatives are symmetric, i.e.
$\partial_i\partial_jf= \partial_j\partial_if.$
But as it is famous  the continuity of the 2nd order partial derivatives is not necessary for this to happen. For example if $\partial_if$, $\partial_jf$ exist on $U$ and they are both differentiable (in the sense of Fréchet) at some point $a\in U$ then
$$\partial_i\partial_jf(a)= \partial_j\partial_if(a).$$

Now for the 3rd order partial derivatives we can obtain the symmetry if we assume that the 1st order partial derivatives of $f$ are differentiable on $U$ and its 2nd order partial derivatives are differentiable at $a$. Let me explain the proof for the particular case
$$\partial_3\partial_2\partial_1f(a)= \partial_2\partial_1 \partial_3f(a).\tag{$\star$}$$
First as $\partial_1 f$ has 1st order partial derivatives in $U$ and they are differentiable at $a$ we have
$$\partial_3\partial_2\partial_1f(a)= \partial_2\partial_3 \partial_1f(a).\tag{1}$$
Then since the 1st order partial derivatives of $f$ are differentiable in $U$ we have $\partial_3\partial_1f(x)= \partial_1\partial_3 f(x)$ for all $x\in U$. Hence we can differentiate to obtain
$$\partial_2\partial_3\partial_1f(a)= \partial_2\partial_1 \partial_3f(a).\tag{2}$$
By combining (1) and (2) we get ($\star$).

As you can see the full force of differentiability of the 1st order partial derivatives of $f$ on all of $U$ is only used for the equality of the 3rd order partial derivatives appeared in (2). So my question is

Question: Can we prove the symmetry of 3rd order mixed partial derivatives of $f$ at $a$ by merely assuming that the 1st and 2nd order partial derivatives of $f$ exist on $U$ and they are all differentiable at $a$? If not, can you provide a counterexample? Finally, if the answer is positive, can we generalize it to higher order mixed partial derivatives?

答: For the accurate statement and rigorous proof see Theorem 3 of the following paper https://www.mdpi.com/2227-7390/8/11/1946/htm</p>


问:
Given $\frac{f(2^n)}{f(2^{n+1})} = 11/18 + n/18$, how do you find $f(x)$ in terms of $x$?
If I have a semi-recursive equation for a function $f$, namely,

$f(2^n)/f(2^{n+1}) = 11/18 + n/18$ for all real values of n,

How do I convert this (without a programmable computer) to a non-recursive formula for $f(x)$ in terms of $x$?

Assume a value of $f(1) = H$ for some constant $H$.

Edit: It would be best to get a closed formula for $f(x)$ that does not include factorials, because I will eventually differentiate a function that includes this one. Also, factorials do not allow me to include negative numbers or nonintegers in the domain.

答:
Let $g(x)=f(2^x)$. So $$\frac{g(n)}{g(n+1)}=\frac{11+n}{18}$$ In other words, $$g(n+1)=\frac{18}{11+n}g(n)$$
Let $g(x)=18^xk(x)$. So
$$k(n+1)=\frac{1}{11+n}k(n)$$
Let $k(x)=\frac{1}{(10+x)!}m(x)$, so
$$m(n+1)=m(n)$$

Let $m(n)$ be constant: $m(n)=m$.

So $k(n)=\frac{m}{(10+n)!}$.

So $g(n)=\frac{18^nm}{(10+n)!}$.

So $f(n)=\frac{18^{\log_2(n)}m}{(10+\log_2(n))!}$. aThe factorial applied to general real numbers should be interpreted as the Gamma function, shifted by $1$: $$f(n)=\frac{18^{\log_2(n)}m}{\Gamma(11+\log_2(n))}$$
For $f(1)=H$, you need $m=H\cdot\Gamma(11)$.


问:
Expanding squared L2 norm of difference of two vectors and differentiating
I'm confused on how to expand and differentiate a squared $L^2$-norm when more than one vector and a matrix is involved. For example:

$$\nabla_x(\| z-Zx\|^2_2)$$

where $z \in \mathbb{R}^n$, $Z \in \mathbb{R}^{n\times n}$  and $x = [x_1,x_2,\ldots,x_n]^T \in \mathbb{R}^n$.

For the case where $\nabla_x(\cdot)$ is just $\mid\mid x\mid \mid^2_2$, it's fairly easy in that you apply the definition of the $L^2$-norm, square, and differentiate which would yield $2x$. What properties can be used here to make this problem solvable?

EDIT: The solution set claims that the expression above equates to:

$2(Z^TZx-Z^Tz)$

My original question still stands

答:
$\def\p{\partial}$
For ease of typing, define the vector
$$w=Zx-z$$
Write the function in terms of this new vector, then calculate its gradient.
$$\eqalign{
 \lambda &= \|w\|^2 = w^Tw \\
d\lambda &= 2w^Tdw = 2w^TZ\,dx = (2Z^Tw)^Tdx \\
\frac{\p\lambda}{\p x} &= 2Z^Tw = 2(Z^TZx - Z^Tz) \\
}$$
Exactly as claimed by the solution set. Yay!


问:
Changing bounds of integration question; spot the mistake.
Suppose we wanted to compute $$\int \frac{1}{x^2 + x + 1} ~ dx.$$ One such way is to fix $c$ and let $x$ vary as bounds, thus we have that $$\int \frac{1}{x^2 + x + 1} ~ dx = \int_c^x \frac{1}{t^2 + t + 1} ~ dt = \int_c^x \frac{1}{(t + 1/2)^2 + 3/4} ~ dt.$$ We will now shift the integral by $1/2,$ so we have $$\int_c^x \frac{1}{(t + 1/2)^2 + 3/4} ~ dt = \int_{c+1/2}^{x + 1/2} \frac{1}{t^2 + 3/4} ~ dt.$$ The expansion theorem for integrals tells us that $$\int_{a}^b f(x + k) ~ dx = \frac{1}{c} \int_{ac + k}^{bc + k} f(x/c) ~ dx.$$ Using the aforementioned theorem, we have that $$\int_{c+1/2}^{x + 1/2} \frac{1}{t^2 + 3/4} ~ dt = \frac{\sqrt{3}}{2} \int_{\frac{2}{\sqrt{3}}c+1/2}^{\frac{2}{\sqrt{3}}x + 1/2} \frac{1}{(3/4)t^2 + 3/4} ~ dt.$$ Factoring, we have $$\frac{\sqrt{3}}{2} \int_{\frac{2}{\sqrt{3}}c+1/2}^{\frac{2}{\sqrt{3}}x + 1/2} \frac{1}{(3/4)t^2 + 3/4} ~ dt = \frac{\sqrt{3}}{2}\cdot\frac{4}{3} \int_{\frac{2}{\sqrt{3}}c+1/2}^{\frac{2}{\sqrt{3}}x + 1/2} \frac{1}{t^2 + 1} ~ dt.$$ This of course evaluates to $$\frac{2\sqrt{3}}{3} \arctan\left( \frac{2}{\sqrt{3}}x + \frac{1}{2} \right) + \text{constant}.$$ But then notice, taking the derivative of the expression before will not give us $1/(x^2 + x + 1).$ But, taking the derivative of $$\frac{2\sqrt{3}}{3} \arctan\left( \frac{2}{\sqrt{3}}x + \frac{1}{\sqrt{3}} \right) + \text{constant}$$ does give us $1/(x^2 + x + 1).$ Where is the mistake?

答:
$$\int_{a}^b f(x + k) ~ dx = \frac{1}{c} \int_{ac + k}^{bc + k} f(x/c) ~ dx.$$ is wrong. It should be:
$$\int_{a}^b f(x + k) ~ dx = \frac{1}{c} \int_{a(c + k)}^{b(c + k)} f(x/c) ~ dx.$$