问:
Matrix multiplication and determinant
Let $A$ and $B$ be two matrices. The rows of matrix $AB$ can be seen as the linear combinations of the rows of matrix $B$. Thus, if the determinant of a matrix is invariant to row operations, shouldn't $\det(AB) = \det(B)$?

I know it's $\det(AB) = \det(A) \det(B)$. Where is the fallacy in my argument? Thanks in advance.

答:
$\det(B)$ is a measurement of the "size" of (the rows of) $B$ in a certain sense. You are correct that $AB$ is a linear combination of rows of $B$, but linear combinations do not preserver size. For a slightly different example, consider the vectors $\{(1, 0), (0,1)\}$. They each have length $1$, but we certainly don't expect all linear combinations of them to have length $1$! For another example, $||\frac{1}{x^2+1}||_2=\frac{\pi}{2}$ and $||\sin(x)/x||_2=\sqrt{\pi}$. But I can make $||a\frac{1}{x^2+1} + b\frac{\sin(x)}{x}||_2$ as large as I like by picking appropriate $a$ and $b$.

Typically, there are special types of linear transformations that do preserve size. In the case of vectors in $\mathbb{R}^k$, these are rotations. For the case of matrices, they are precisely multiplication by matrices of determinant $1$.

Title: Matrix multiplication and determinant


问:
Help Prove the equality of sets. I know very little math analysis
$$ (B\setminus A)\cup(C\setminus  A) = (B\cup C)\setminus  A$$

I found a very similar example but it was De Morgan's law. I would like to be explained and shown how this example is solved

答:
Intuitively, this just says that if an element is either in $B$ but not $A$ or in $C$ but not $A$, then either way it must be in $B$ or in $C$ and cannot be in $A$.  Conversely if it is in $B$ or $C$, but not in $A$ then either it must be in $B$ but not $A$ or $C$ but not $A$.

Formally you can prove this by considering the $8$ cases for inclusion in $A,B,C$ and showing that in each case an element is in the LHS in precisely the same cases as when it is in the RHS.

I'll get you started with one row of the table:

$$
\begin{array}{ccc|ccc|cc}
A&B&C&B\backslash A&C \backslash A & B\cup C& (B\backslash A)\cup(C \backslash A) &(B\cup C)\backslash A\\\hline
0&1&0&1&0&1&1&1
\end{array}
$$

ContentLicense: CC BY-SA 4.0


问:
Prove that the sequence $\{q_n\}_{n \geq 1}$ diverges to $\infty.$

Let $\{x_n \}_{n \geq 1}$ be a sequence of rational numbers converging to $\pi.$ Let for all $n \geq 1,$ $x_n = \frac {p_n} {q_n},$ where $p_n,q_n \in \Bbb Z$ with $\text {gcd}\ (p_n,q_n) = 1.$ Prove that the sequence $\{q_n \}_{n \geq 1}$ diverges to $\infty.$

How do I prove that? Any help will be highly appreciated.

Thanks in advance.

答: If the sequence $\{q_{n}\}$ is bounded then clearly $\{p_{n}\}$ must also be bounded as otherwise $\{x_{n}\}$ does not converge. Then $\{x_{n}\}$ must be bounded, and most importantly, finite. Since $\pi$ is irrational, the closest we can get to $\pi$ is obtained by taking the $x_{k}$ (which exists in $\{x_{n}\})$ such that $\inf\{|{x_{n}}-\pi|\}$ is satisfied by $n=k$. Since $x_{k} \neq \pi$, we see we cannot get arbitarily close to $\pi$ using a finite set of fractions. Take $\epsilon < \inf\{|{x_{n}}-\pi|\}$ and we see that $\{x_{n}\}$ does not converge to $\pi$, a contradiction.


问:
A normal random variable independent of each component of a multivariate normal random vector.
Suppose that $\mathbf{Y}\sim N_3(0,\,\sigma^2\mathbf{I}_3)$ and that $Y_0$ is $N(0,\,\sigma_0^2)$, independently of the $Y_i$'s.
My question is that does $(\mathbf{Y}, Y_0)$ also have multivariate normal distribution? By using moment generating function, it suffices that $\mathbf Y$ and $Y_0$ are independent. But is this the case?

答:
I think not: there can exist random variables which are individually Gaussian, and pairwise independent, but not jointly Gaussian.  Let $X_0,X_1,X_2,X_3$ be iid $N(0,1)$.  Now let $$(Y_0,Y_1,Y_2,Y_3)=(|X_0|\operatorname{sgn}(X_1 X_2 X_3), X_1, X_2,X_3),$$ where $\operatorname{sgn} (X_1 X_2 X_3)=\pm1$ according to the sign of  $X_1X_2X_3$. One can check that $Y_0$ is Gaussian, and independent of each of the other individual $Y_i$. Why is $Y_0$ Gaussian?  Well, its absolute value is distributed as the absolute value of an $N(0,1)$ rv, and that its sign is uniformly distributed on $\pm1$.  Why is it independent of (say) $Y_1$, that is, of $X_1$?  Because $X_2$ and $X_3$ obliterate any control $X_1$ might hold over $\operatorname{sgn} (X_1 X_2 X_3).$  And finally, why are the $Y_i$ not jointly Gaussian?  Well, their product is $\ge0$ with probability $1$.

This is similar to the example of $T=RS$ when $R$ and $S$ are iid uniform $\pm1$ rvs, three rvs which are pairwise independent but not jointly independent.


问:
Can AB be anything other than the Identity matrix?
Say we have two square, non-singular matrices A and B, which are not equal to each other. And such that both AB and BA is defined. So will the only solutions to the equation AB = BA be that either one of them is the Identity matrix or A is the inverse of B or B is the inverse of A. But my teacher had said that the last two are true only when AB = BA = I (Identity matrix). But how can AB = BA be equal to something other than the Identity matrix ? Is it possible that : $$AB = BA \neq I$$

答: any set of matrix that are simultaneously diagonalizable (meaning they have the same eigenvectors) verify your equality</p>


问:
Explicit realization of Eilenberg-MacLane spaces and TOP/PL

What is the explicit realization of following Eilenberg-MacLane spaces? There are some examples I know but not for the higher

$${\displaystyle K(\mathbb {Z}  ,1)}=S^1$$
$${\displaystyle K(\mathbb {Z}  ,2)}=\mathbb {CP}^\infty$$
$${\displaystyle K(\mathbb {Z}  ,3)}=B\mathbb {CP}^\infty=?$$

$${\displaystyle K(\mathbb {Z} /2,1)}=\mathbb {RP}^\infty.$$
$${\displaystyle K(\mathbb {Z} /2,2)}=B\mathbb {RP}^\infty=?$$
$${\displaystyle K(\mathbb {Z} /2,3)}=B^2\mathbb {RP}^\infty=?$$

Why does it mean to state there a weak equivalence ${\displaystyle TOP/PL\sim K(\mathbb {Z} /2,3)}$ of TOP/PL with an Eilenberg–MacLane space?

答:
For your first question, there are many reasonable constructions of these. One way is via linearization of spheres (see the section at https://ncatlab.org/nlab/show/Eilenberg-Mac+Lane+space). Another way to think about it is through the Dold-Kan correspondence. In this lens, an Eilenberg-MacLane space is just the space associated to the chain complex with that group concentrated in that degree.

I can say more about the second question:

Recall that $Top$ is the colimit of the $Top(n)$ which is the homeomorphism group of $\mathbb{R}^n$. Similarly, $PL$ is the colimit of the $PL(n)$ which is the PL homeomorphism group of $\mathbb{R}^n$. I will first go over the argument of Kirby and Siebenmann, and then I will explain how it relates to Wikipedia's statement.

Now usually it is the case that when we prove a stabilization result, it is easier to get at the stabilized and then use the result to get at the unstable object. Here Kirby and Siebenmann go the other way around. They prove the result that the inclusion $Top(m)/PL(m) \rightarrow Top/PL$ is an isomorphism on the first $m$ homotopy groups if $m>4$ and will use this to get at the homotopy groups of the stable object. So with this result is suffices to prove that actually if $m>4$ that $Top(m)/PL(M)$ is a $K(\mathbb{Z}/2,3)$.

This is done by using a very great result of Morlet that $PL_\partial(D^m) \simeq \Omega^{m+1}(Top_m/PL_m)$. The space on the left is contractible by what is called the Alexander trick, so we get that the homotopy groups above $m$ are trivial. This result holds for $m \neq 4$. So if we take $m=5$ and manually check that the homotopy groups below $6$ are what we want, then we are done. This is their theorem 5.3 and is rather long. My understanding is that they compare triangulations (see the latter half of my answer) and "homotopy triangulations". The latter are accessible through surgery type arguments.

So using these three results: stability, triviality of high homotopy groups, and having the correct lower homotopy groups, we see that $Top/PL$ is a $K(\mathbb{Z}/2,3)$. Now how does this relate to what wikipedia says?

Well the reason we care about $Top$ and $PL$ so much is that their classifying spaces classify stable Top and PL $\mathbb{R}^n$ bundles. These are like vector bundles, but there is no vector space structure on the fibers. From the fiber sequence $Top/PL \rightarrow BPL \rightarrow BTop$ we can deduce that the homotopy groups of $Top/PL$ are important in asking when we can lift a Top bundle to a PL bundle and to what extent this is unique.

But in fact these spaces are even more important than just for these reasons. The question of whether a manifold can be given a PL structure, and if this is unique, is actually equivalent (if we are not in dimension 4) to whether or not the stable tangent bundle of the manifold (this is not a vector bundle if we are not smooth) can be given a PL structure. This might be called the fundamental theorem of triangulation theory.

By obstruction theory, we know that the obstructions to lifting a map along a fibration lie in the cohomology of the domain with coefficients in the homotopy groups of the fiber, shifted by 1. In our case we are using the fibration $Top/PL \rightarrow BPL \rightarrow BTop$, if it were the case that every homotopy group of $Top/PL$ was realized as an obstruction to lifting a map $[M, BTop]$ to a map $[M,BPL]$ (I am not sure how one could argue this without Kirby and Seibenmann's result), for some topological manifold M, then the fact that the Kirby-Siebanmann obstruction is the only obstruction and lies in $H^4 (M;\mathbb{Z}/2)$ would imply that $Top/PL$ is a $K(\mathbb{Z}/2,3)$. But to be clear, this is not how the result was proven, and I believe the Kirby-Siebanmann invariant is defined with the knowledge that the fiber is an Eilenberg-MacLane space.


问:
Proof about $k$-partite graphs in which all maximal cliques are size $k$ and a way to remove edges so that all maximal cliques remain size $k$
I have written this graph-theoretic proof about certain symmetric $k$-partite graphs, and a way to filter their edges so that all maximal cliques in the graph remain size $k$. The reason why I am not confident the proof is right is because it is a strong implication that $P=NP$ by way of $SAT$'s reduction to the clique problem. I can't find any errors in the proof, but the implication seems too strong for the proof to be correct so I'd appreciate any help pointing out a problem I may have missed. I've tried to use standard terminology as much as possible, but anything I've created is bold and italicized, so please let me know if there is a better term or definition out there

Definition: Given some graph $G$, $V(G)$ is its set of vertices and $E(G)$ is its set of edges. A vertex is an element $v \in V(G)$ and an edge is an element $vw \in E(G)$ with adjacent endpoints $v, w \in V(G)$. Two graphs $G$ and $H$ are called isomorphic if there is a bijection $f: V(G) \rightarrow V(H)$ such that any $vw \in E(G)$ implies $f(v)f(w) \in E(H)$, and are called non-isomorphic otherwise

Definition: A $k$-clique $\Delta$k is a graph on $k$ vertices where all vertices are pairwise adjacent. A clique which is not a proper subset of any other is called maximal

Definition: A $k$-partite graph $G$ is a graph whose vertex set has been partitioned into $k$ independent sets. An independent set is a set of vertices which are all pairwise non-adjacent. If $G$ is complete, then vertices are pairwise adjacent between all partitions. The restoration of $G$ is the set of edges needed to make $G$ complete under this definition. The complete $k$-partite graph formed by adding the restoration to $E(G)$ is defined as $*(G)$

Definition: The clique complex of a graph $G$ is the abstract simplicial complex whose faces are the vertex set of each clique in $G$. A face which is not a proper subset of any other face is called a facet. The dimension of a face $X$ is defined as $|X|-1$. The dimension of the clique complex is equal to the dimension of the largest face. A clique complex is said to be pure if each facet is the same dimension

Theorem 1: If $G$ is a $k$-partite graph with no edges and $\Gamma$k is some subset of all the possible $\Delta$k in $G$, then adding the edge set of each $\Delta$k$ \in \Gamma$k to $G$ will result in a graph which has a pure clique complex of dimension $k-1$

Proof: Assume that $\Delta$m is some maximal clique in $G$ where $m<k$. We know that each vertex in $V(\Delta$m$)$ belongs to at least one $\Delta$k$ \in \Gamma$k. However, since the vertex set can't be a subset due to the clique being maximal, they can't all belong to the same $\Delta$k. If we have some vertex $v \in V(\Delta$m$)$ and $v \notin V(\Delta$k$)$, we know that $v \in V(\Delta$k*$)$ for some $\Delta$k*$ \in \Gamma$k which is non-isomorphic to $\Delta$k. Since we know that $\{vw~|~w \in V(\Delta$k$)\} \subset E(\Delta$k*$)$, the definition of a clique tells us that those vertices $w \in V(\Delta$k$)$ which are connected to $v$ are all in $V(\Delta$k*$)$ as well. This means that $V(\Delta$m$) \subset V(\Delta$k*$)$, therefore it is not a maximal clique and we have arrived at a contradiction. Since no edges will be added between any of the vertices within a partition, we will have no cliques with more than $k$ vertices due to the Pigeon Hole Principle. Therefore, all maximal cliques must have exactly $k$ vertices and $G$ has a pure clique complex of dimension $k-1$

Definition: Given a graph $G$ and a subset of its edges $T \subset E(G)$, a subgraph on $T$ is a new graph whose edges are the set $T$

Definition: Given a graph $G$ and a subset of its vertices $S \subset V(G)$, an induced subgraph on $S$ is a subgraph of $G$ formed by the set of all edges $T \subset E(G)$ whose endpoints are both in $S$

Definition: Given a $k$-partite graph $G$, if for every $\Delta$m in $G$ where $1<m<k$, there exists a vertex $v$ such that the induced subgraph of $G$ on $V(\Delta$m$) \cup \{v\}$ forms a $\Delta$m+1, then $G$ is said to have the clique extension property

Theorem 2: If $G$ is a $k$-partite graph with a pure clique complex of dimension $k-1$, then $G$ has the clique extension property

Proof: If we have a $\Delta$m in $G$, where $1<m<k$, then by the definition of a pure clique complex, $\Delta$m is not a maximal clique as its vertex set does not form a face of dimension $k-1$. Since $\Delta$m is not a maximal clique, it must be a subset of a larger one. Therefore, there must be some vertex $v \notin V(\Delta$m$)$ such that the induced subgraph of $G$ on

$V(\Delta$m$) \cup \{v\}$ forms a $\Delta$m+1

Definition: Given a graph $G$ and its clique complex, as well as two edges $vw, xy \in E(G)$, the edge $vw$ is said to cover $xy$ if every facet containing $\{x, y\}$ also contains $\{v, w\}$. The coverage of an edge is the set of all edges which $vw$ covers

Definition: Given a graph $G$, the neighborhood of a vertex $v \in V(G)$, or $N(v)$, is the set of all vertices adjacent to $v$. The neighborhood of a clique, or $N(\Delta$m$)$, is the intersection of the neighborhoods of each vertex in $V(\Delta$m$)$. The use of square brackets means that the vertices used to define the neighborhood are included in the neighborhood or $N[v] = \{v\} \cup N(v)$

Definition: Given a $k$-partite graph $G$ with a pure clique complex of dimension $k-1$, the tangential of some clique, or $\tau(\Delta$m$)$, is defined as the set of edges in the induced subgraph of $G$ on $N[\Delta$m$]$

Theorem 3: If $G$ is a $k$-partite graph with a pure clique complex of dimension $k-1$ and $vw \in E(G)$, then $\tau(vw)$ is precisely the union of the edge set of each $\Delta$k which contains $vw$

Proof: From the definition of a clique, $V(\Delta$k$) \subset N[vw]$ if it contains some edge $vw$. Following this, the set of potential edges of any such maximal clique can be partitioned into three sets, $\{vw\}$, $\{vx, wx~|~x \in N(vw)\}$,

and $\{xy~|~x, y \in N(vw)\}$. The first partition is the edge which we are using to define the tangential. This is where we get the term tangential from, as all of the maximal cliques represented by the edge set will be "tangent" to this edge. The second partition is the set of all edges connecting $vw$ to any one of its neighbors $x$. We know every edge in this set is a part of some $\Delta$3 formed by the edges $\{vw, vx, wx\}$. Since $G$ has the clique extension property according to theorem 2, each $\Delta$3 can be extended to some $\Delta$$k$, meaning every edge between $vw$ and its neighbors will be in a maximal clique. The third partition is the set of edges which connect any two vertices $x, y \in N(vw)$. If $xy$ exists, we know that $\{vx, wx, vy, wy\} \subset E(G)$ from our definition of the neighborhood of a clique. Therefore, we also know that we can create a $\Delta$$4$ such that $E(\Delta$4$) = \{vw, vx, wx, vy, wy, xy\}$. If $\Delta$$4$ is not already maximal, we can again use theorem 2 to guarantee the clique can be extended to some $\Delta$$k$. Therefore, if $xy$ exists, it will be in a maximal clique. Taken together, the edge set of all maximal cliques which contain $vw$ is precisely all of the edges with both endpoints in $N[vw]$, or the induced subgraph of $G$ on $N[vw]$

Theorem 4: If $G$ is a $k$-partite graph with a pure clique complex of dimension $k-1$, for two edges $vw, xy \in E(G)$, if $vw$ covers $xy$ then $\tau(xy) \subset \tau(vw)$

Proof: According to theorem 3, the set of edges in $\tau(vw)$ are each part of some maximal clique containing $vw$. Therefore, for two edges $vw, xy \in E(G)$, if $\tau(xy) \subset \tau(vw)$, there are no maximal cliques that $xy$ is a part of that $vx$ isn't also a part of and $vw$ covers $xy$ according to our definition. We can also say $\tau(xy)$ represents the edge set of maximal cliques which happen to be tangent to both $vw$ and $xy$

Definition: Given a $k$-partite graph $G$ with a pure clique complex of dimension $k-1$ and an edge $vw \in E(G)$, a facet reduction, or $Red(G, vw)$ is the subgraph $H$ formed from $G$ by removing the edge $vw$ and each edge in its coverage. A facet reduction is also known as reducing $G$ by $vw$

Theorem 5: If $G$ is a $k$-partite graph with a pure clique complex of dimension $k-1$ and $H=Red(G, vw)$, then $H$ is either an independent set or it is a $k$-partite graph with a pure clique complex of dimension $k-1$ containing all of the $\Delta$k in $G$ which do not contain $vw$

Proof: Removing the edge $vw$ removes all $\Delta$k which contain $vw$. If we have some edge $xy$ covered by $vw$, then $xy$ can't be in any remaining $\Delta$k. Therefore, for any maximal clique $\Delta$m which contains $xy$, $m < k$. However, by removing $vw$ and its coverage, any remaining edge is contained in some $\Delta$k which does not contain $vw$. Alternatively, $H$ is formed from the edge set of all $\Delta$k in $G$ which did contain $vw$. Therefore, according to theorem 1, if $E(H) \neq \emptyset$, then $H$ has a pure clique complex of dimension $k-1$. If there are no $\Delta$k without $vw$, then $H$ is an independent set as no edges will remain

Theorem 6: If $G$ is a $k$-partite graph and $R$ is its restoration, then there is some induced $\Delta$k in $G$ if and only if $*(G)$ is not reduced to an independent set after being subjected to the facet reduction with each edge in $R$

Proof: If any $\Delta$k exists in $G$, then $E(\Delta$k$) \subset E(G)$, and as such, cannot contain any edge in $R$. Since $*(G)$ is $k$-partite and formed from the set of all possible $\Delta$k in $G$, it has a pure clique complex of dimension $k-1$ according to

theorem 1. Using theorem 5, reducing $*(G)$ by any $vw \in R$ will either produce another $k$-partite graph with a pure clique complex of dimension $k-1$ or it will produce an independent set. In the first case, we know that the new graph has no $\Delta$k which contain $vw$, but it does contain some other $\Delta$k. To find it, start with any $\Delta$2, add a vertex from its intersection to find a larger clique, then repeat until you have $k$ vertices. Since the new graph still has the clique extension property, we'll always end up with some $\Delta$k with the added property that it does not contain $vw$. Since the new graph also satisfies the conditions needed to perform a facet reduction, we can reduce it again by any remaining edges in $R$. If we reduce the graph by each of the edges in the restoration and it still has a non-empty edge set, then we know that $G$ contains some $\Delta$k which is able to be constructed from $E(G)$. Otherwise, at some point we reduce the graph to an independent set and we can determine that there are no $\Delta$k which can be formed using the edge set of $G$

答:
So the errors in the proof are theorem 1 and theorem 5 and 6

Theorem 1 can be shown to be wrong through a counterexample. Take the 4-partite graph where the 4 blocks in the vertex partition are [1, 2, 3], [4, 5, 6], [7, 8, 9], and [10, 11, 12]. If we add the edge set for the $\Delta_4$ with vertices {1, 4, 8, 10}, {1, 5, 7, 11}, and {2, 4, 7, 12}, we get a maximal $\Delta_3$ with vertex set {1, 4, 7}

Theorem 5 is wrong because now it's not guaranteed that the subgraph we create will have a pure clique complex and theorem 6 is wrong because it relies on theorem 5

If we do have a $k$-partite graph with a pure clique complex of dimension $k-1$, theorems 2, 3, and 4 still stand. If we only use two cliques from the above counterexample, we have an example of such a graph


问:
If $S= \overline {abc}_j$ and $S = \overline{cba}_{j+1}$ with $a, c \neq 0,$ find $S$.
If $j\ge 3$ is a positive integer, and $S= \overline {abc}_j$ and $S = \overline{cba}_{j+1}$ with $a, c \neq 0,$ find all possible $S$.

My progress:
I set up the equation
$$aj^2+bj+c = c(j+1)^2+b(j+1) + a,$$
and simplified it into
$$a(j^2-1) = c(j^2+2j)+b.$$
But I'm not sure how to proceed from here. Any help?

EDIT: It seems as though they only exist for odd $j$ and are of the form $a=\lceil j/2\rceil, b = c =a-1,$ in base $j$. How can I prove these are the only ones? Or, are they even the only ones?

答:
Instead you should write it as a quadratic equation as follows:

$$aj^2+bj+c=c(j+1)^2+b(j+1)+a=cj^2+2cj+c+bj+b+a$$
$$(a-c)j^2-2cj-(b+a)=0$$

Suppose $j^*$ is a positive integer solution to the equation above. By considering the expansion of $(mj+n)(j-j^*)$ (with $m,n \ge 0$), we see that $j^* \mid (b+a)$.

However we must have $0 \le a,b < j^*$. This forces $a+b = j^*$. We now have

$$(a-c){j^*}^2-2cj^*-j^*=j^*((a-c)j^*-2c-1)=0$$

This implies $j^* = \dfrac {2c+1}{a-c}$. Suppose $a-c > 1$. $a-c \ne 2$ since $j^*$ is an integer.

This gives $a-c \ge 3$, and we will have $j^* \le \dfrac {2c+1}3\le \dfrac {2c+c}3=c$, which is a contradiction.

Hence $a-c = 1$ and $a+b = j^* = 2c+1$. (Here it is more obvious that $j^*$ is odd.)

Solving the equations in terms of $j^*$, we have

$$a = \frac {j^*+1}2, b = c = a-1 = \frac {j^*-1}2$$

and finally we check the result for any odd $j^* \ge 3$.


问:
Improper Integral $\int_0^1\frac{dx}{x^p}$
Is this integral convergent only for $p<1$?

$$\int_0^1\frac{dx}{x^p}$$

答:
Yes.
\begin{align}
\int \dfrac{dx}{x^p} = \begin{cases} \dfrac{x^{1-p}}{1-p} + \text{constant} & \text{ for } p \neq 1\\ \log(x) & \text{ for }p=1\end{cases}
\end{align}
From the above, note that for $p\geq1$ as $x \to 0^+$, we have $x^{1-p} \to \infty$ and for $p<1$ as $x \to 0^+$, we have $x^{1-p} \to 0$.
Hence,
\begin{align}
\int_0^1 \dfrac{dx}{x^p} = \begin{cases} \dfrac{1}{1-p} & \text{ for } p < 1\\ \text{does not exist} & \text{ for }p\geq1\end{cases}
\end{align}</p>

ContentLicense: CC BY-SA 3.0


问:
Co-primes in Gaussian integers
Say $A=\mathbb{Z}[i]$ the Gaussian integers with multiplication and addition of complex numbers. Now we define the Norm $N$ on $\mathbb{Z}[i]$ by: $\forall a=x+yi\in\mathbb{Z}[i], N(a)=x^2+y^2$.
Now suppose $x=a+bi, y=c+di, m=N(x), n=N(y)$, where $gcd(a,b)=gcd(c,d)=gcd(m,n)=1$. I have to prove - for $xy = (ac-bd)+(ad+bc)i, gcd(ac-bd, ad+bc)=1$.
I tried assuming by contradiction that there's a prime number $p$ that divides both $ac-bd, ad+bc$ - therefore $p$ divides both $(a-b)(c+d)$ and $(a+b)(c-d)$ but wasn't able to come up with a contradiction.
Are there any $gcd$ arithmetic rules I can apply here? any hint would help.

答:
If $p$ is a prime integer such that $p|xy$ then either (case 1) $p$ is a prime Gaussian integer, in which case $p|x$ or $p|y$ contradicting $\gcd(a,b)=1$ or $\gcd(c,d)=1$, or (case 2) $p=\alpha\bar{\alpha}$ where $\alpha$ is a prime Gaussian integer.

Again in case 2 we cannot have $p|x$ or $p|y$ as we cannot have $\gcd(a,b)=1$ or $\gcd(c,d)=1$.  Therefore we must have $\alpha|x$ and $\bar{\alpha}|y$ (or vice versa).  But then $$p=\alpha\bar{\alpha}|x\bar{x}=N(x),\qquad p=\alpha\bar{\alpha}|y\bar{y}=N(y),$$
contradicting $\gcd(n,m)=1$.

Title: Co-primes in Gaussian integers


问:
How to say that one set is the smallest containing another set?
Imagine that we have a collection of sets $\cal C$, let $S\notin \cal C$ and $A \in \cal C$.

In algebra, topology, etc... we often see statements like " ----- is the smallest set containing -----", but I'm always not sure how to formalize these kinds of statements. How can I write that $A$ is the smallest set in $\cal C$ containing the set $S$? How can we measure the "length" of these sets in order to say that one is smaller than the other?

答: In this use we order sets by inclusion, so you would say that $A$ is in $\mathcal C$, contains $S$, and all other sets in $\mathcal C$ that contain $S$ also contain $A$.


答:
The statement usually means the property $$A\in\mathcal C\land A\supseteq S\land (\forall B\in \mathcal C,(A\subseteq B\lor S\nsubseteq B))$$
id est that $A\in\mathcal C$ is a set containing $S$ and, for all other sets $B\in\mathcal C$ which contain $S$, $A\subseteq B$.

Personally, I would not use the word "smallest" when I'm referring to cardinality. I would say "of least cardinality" (or "least size", or "least measure", or "least greatest lower bund of the sine of an odd integer contained in it"...).

Title: How to say that one set is the smallest containing another set?

Body:
Imagine that we have a collection of sets $\cal C$, let $S\notin \cal C$ and $A \in \cal C$.

In algebra, topology, etc... we often see statements like " ----- is the smallest set containing -----", but I'm always not sure how to formalize these kinds of statements. How can I write that $A$ is the smallest set in $\cal C$ containing the set $S$? How can we measure the "length" of these sets in order to say that one is smaller than the other?


问:
What powers of $p$, where $p$ is prime, divide $\binom {p^{\alpha} m}{p^{\alpha}}$?
My question is about a short combinatorial discussion in Herstein's Topics in Algebra, on page 92, preceding one of his three proofs of Sylow's Theorem. I will cite the important part:

If $p^r | m$ but $p^{r+1} \nmid m$, consider $$\binom {p^\alpha m}{p^{\alpha}} = \frac{(p^{\alpha}m)!}{(p^{\alpha})!(p^{\alpha}m-p^{\alpha})!} = \frac {p^{\alpha}m(p^{\alpha}m-1) \dots (p^{\alpha}m-i) \dots (p^{\alpha}m-p^{\alpha} + 1)}
{p^{\alpha}(p^{\alpha}-1) \dots (p^{\alpha}-i) \dots (p^{\alpha}-p^{\alpha} + 1)}.$$
  The question is, What power of p divides  $\binom {p^{\alpha} m}{p^{\alpha}}$? Looking at this number, written out as we have written it out, one can see that except for the term $m$ in the numerator, the powers of $p$ dividing $(p^{\alpha}m-i$$)$ is the same as that dividing $p^{\alpha}-i$, so all powers of $p$ cancel out except the power which divides $m$. Thus $p^r | \binom {p^{\alpha} m}{p^{\alpha}}$  but $p^{r+1} \nmid \binom {p^{\alpha} m}{p^{\alpha}}$.

My understanding is as follows. $\binom {p^\alpha m}{p^{\alpha}} = m *\frac{\binom {p^\alpha m}{p^{\alpha}}}{m}$, consider the unique factorization of both factors. The right will not contain any nonzero powers of $p$, and so the power of $p$ dividing our binomial is the largest power in the unique factorization of $m$. But there are two gaps in my understanding: why is the second factor neccesarily an integer, and why does the second factor not contain powers of $p$? Herstein explains this second question, but I don't get his argument.

答:
Observe that $\displaystyle\binom{p^\alpha m}{p^\alpha} = m \displaystyle\binom{p^\alpha m - 1}{p^\alpha - 1}$, so the second factor is an integer.

Now the claim is that the factors of $p$ in $\displaystyle\binom{p^\alpha m - 1}{p^\alpha - 1}$ all cancel out perfectly from the numerator and denominator. This will follow if we show that, for all $i$ with $1 \leq i \leq p^\alpha - 1$, a power of $p$ (say $p^k$) divides $p^\alpha m - i$ if and only if it divides $p^\alpha - i$.

Proposition: Let $1 \leq i \leq p^\alpha - 1$. Let $k$ be a positive integer. (i) If $k > \alpha$, then $p^k \nmid p^\alpha - i$ and $p^k \nmid p^\alpha m - i$. (ii) If $1 \leq k \leq \alpha$, then $p^k \mid p^\alpha - i$ if and only if $p^k \mid p^\alpha m - i$.

Proof: (i) Since $k > \alpha$, $p^k > p^\alpha > p^\alpha - i$ so $p^k \nmid p^\alpha - i$. And $p^k \nmid p^\alpha m - i$, otherwise if say $p^\alpha m - i = p^k l$, then $i = p^\alpha m - p^kl = p^\alpha (m - p^{k - \alpha}l)$ (note $k - \alpha > 0$) so $p^\alpha \mid i$, a contradiction since $i \leq p^\alpha - 1 < p^\alpha$.

(ii) Suppose $p^\alpha - i = p^kl$ for some $l$, so $i = p^\alpha - p^kl = p^k(p^{\alpha - k} - l)$ (note $\alpha - k \geq 0$) so $p^k \mid i$. Combining this with $p^k \mid p^\alpha m$, we get $p^k \mid p^\alpha m - i$. Conversely suppose $p^\alpha m - i = p^k l$ for some $l$. So $i = p^\alpha m - p^k l= p^k(p^{\alpha - k}m - l)$ (note $\alpha - k \geq 0$), so $p^k \mid i$. Combine this with $p^k \mid p^\alpha$, we get $p^k \mid p^\alpha - i$. $\square$


问:
If I have a certain relation R, how do I find R^2 and R^3?
Suppose I have A = {a,b,c,d} and R = {(a,b),(b,c),(c,b),(c,d)}. How do I find R^2 and R^3?</p>

答:
At this point, you should be familiar with functions.  If you know that $f(a)=b$ and you know that $f(b)=c$ you should know that $f^2(a)=f(f(a))=f(b)=c$.  Composition of relations is just like composition of functions except that rather than always having one "output" for each input you could potentially have had many outputs and you need to keep track of each one.

You could picture this graphically as well... drawing a directed edge from one element to another if the first element is related to the second element.  Think of it like a board game like snakes and ladders.  The original relation, $R$, is the collection of all ways you can move from one point to another in a single turn.  $R^2$ is the collection of all ways you can move from one point to another in exactly two turns.  $R^n$ is the ways of moving from one point to another in exactly $n$ turns and so on...

So, you could go from $a$ to $c$ in two turns by taking the two individual steps $a\mapsto b\mapsto c$.  You could go from $b$ back to $b$ in two turns as well taking the individual steps $b\mapsto c\mapsto b$, or from $b$ to $d$ in two turns taking the steps $b\mapsto c\mapsto d$.  See if you can find any other paths you could take over two turns and then see if you can find all ways you could go somewhere in three turns.

If you are well organized and can spot things like this quickly and easily, just going through by inspection can be just fine and may be the most efficient way to approach the problem.  If the problem is much larger, convoluted, or if you just want a more formal generic approach that doesn't rely on inspection, you may create an "adjacency matrix" $A$ for your relation.  You may then calculate the adjacency matrix for $R^n$ by simply computing the matrix $A^n$, keeping all zero entries as zero and replacing all nonzero entries with $1$, making such replacements in inbetween steps during calculations as well if desired.  (Not replacing the nonzero entries can be of interest though as those numbers can tell you "how many different ways you could get from one point to another in that number of steps)

Here, we have the adjacency matrix (a matrix where the $i$'th row $j$'th column entry is $1$ if the $i$'th element is related to the $j$'th element in our set according to our relation):

$$A=\begin{bmatrix}0&1&0&0\\0&0&1&0\\0&1&0&1\\0&0&0&0\end{bmatrix}$$

We can then calculate $A^2$ and $A^3$ and interpret as the adjacency matrices of $R^2$ and $R^3$ respectively.

 $R^2 = \{(a,c),(b,b),(b,d),(c,c)\}$, $R^3 = \{(a,b),(a,d),(b,c),(c,b),(c,d)\}$


问:
Show that monotone function f: I->R is measuable on an interval
Show that a monotone function f: I -> R is borel-measureable.

I know, that this question has been asked and answered multiple times here, but only for f: R->R und not for f: I->R.
The first case could be proofed by proofing that {x $\in$ R: f(x) < a} $\in$ B(R) (in case of decreasing) for any a $\in$ R.

First of all, why is that enough? In my opinion, I have to proof that f^(-1) (A) $\in$ B(R) for all A $\in$ J with J as set of any kind of intervals in R (since it creates B(R) = $\sigma$(J) )

Secondly, how can I consider the interval I in that proof?
I think, there is a huge difference between proofing f:R->R and f: I->R as measurable functions.

答:
I am not quite satisfied with that answer: Yes, I is an interval and our measure spaces are:

(I, B(I) = $\sigma$ ($O_I$) )  and (R, B(R) = $\sigma$($O_R$) = $\sigma$ (J)) with $O_I$ and $O_R$ as topologies.

It holds: B(I) = $\sigma$ ($O_I$) $\subseteq$ $\sigma$(J) = B(R). Any kind of interval C $\in$ J is automatically in B(R)=$\sigma$(J), since J $\subseteq$ $\sigma$(J). But I can't quite formally see, why C should be in B(I) = $\sigma$ ($O_I$).

Edit: $C$ = f^(-1) ($A$) = f^(-1) ($\inf$,a]


问:
Can two fields that have monomorphism to each other not be isomorphic?
More precisely, let $F_1, F_2$ be fields, and we have none zero homomorphisms $f: F_1 \to F_2$ and $g: F_2 \to F_1$ (actually both are monomorphism since $F_1, F_2$ are fields), is there a possibility that $F_1$ is not isomorphic to $F_2$?

答:
So apparently, it is possible for this to fail, though I don't profess to understand the counterexamples given in the comments. You are asking if fields satisfy the Cantor–Schröder–Bernstein, and it appears that they do not.

Edit: I take that back, I think I understand one example, if not its construction. There is a natural inclusion $\mathbb{C} \hookrightarrow \mathbb{C}(x)$ and a (complicated) inclusion $\mathbb{C}(x) \hookrightarrow F$ where $F \cong \mathbb{C}$, where $\mathbb{C}(x)$ is the field of rational functions over $\mathbb{C}$. However, the complex field is algebraically closed whereas $\mathbb{C}(x)$ is (apparently) not, so they cannot be isomorphic.


问:
Consider the symmetric RW on a graph constructed from two complete graphs such a way that the two complete graphs have 2 common vertices
Consider the symmetric RW on a graph constructed from the two complete graphs $K_{m}$ and $K_{n}$ in such a way that the two complete graphs have exactly 2 common vertices. S is one of the common vertices. Let A and B be "Blackholes" or endpoints on $K_{m}$ and $K_{n}$ respectively, and let $m = 10^{6}$ and $n = 2 * 10^{6}$ (i.e., million and two-million). Find the average travel time to any blackhole. Figure for reference where m=4, n=5

答:
The standard way to solve a hitting time problem is to define a variable $t_v$ for every vertex $v$ which denotes the expected hitting time for a random walk starting at $v$. You'd set $t_A = t_B = 0$; for all other $v$, $t_v$ would be $1$ more than the average value of all its neighbors.

Here, you don't want to do that, because you'd have around three million variables.

Fortunately, we can simplify the process: the average travel time is the same for any vertex (other than $A$) in the first clique but not the second; the same for any vertex (other than $B$) in the second clique but not the first; and the same for either of the vertices in both cliques.

So let's just define three variables: $t_1$ for the expected hitting time starting in the first clique, $t_2$ for the expected hitting time starting in the second clique, and $t_{12}$ for the expected hitting time starting in the overlap. We can write down a linear equation for each one of them: for example, for $t_1$ we get the equation
$$
    t_1 = 1 + \frac1{m-1} \cdot t_A + \frac{2}{m-1} \cdot t_{12} + \frac{m-4}{m-1} \cdot t_1 \implies \frac{3}{m-1} t_1 - \frac{2}{m-1} t_{12} = 1.
$$
Solve all three equations and you get the three hitting times you want; I think in your case, you're starting in the middle, so you want to know $t_{12}$.


问:
What happens when t tends to infinity in the heat equation?
I have read that for the solution $u$ of the heat equation:

$$u_t(t,x) - \frac{1}{2} \nabla^2 u(t,x) =0, \\(t,x) \in (0, \infty)\times\mathbb{R}^n$$
$$u(0,x)=f(x), x \in \mathbb{R}^n$$

with $f \in L^1(\mathbb{R}^n), n \in \mathbb{N}$

When time tends to infinity the u function tends to zero, but I don't know how to prove it, could someone help me?

答:
The unique solution (assuming the usual reasonable growth conditions) is given by convolution with a rescaled version of the heat kernel:
$$
u(t,x) = \int_{\mathbb{R}^n} \frac{e^{-|x-y|^2/(2t)}}{(2\pi t)^{n/2}}f(y) dy.
$$
Since $f \in L^1(\mathbb{R}^n)$ we can simply bound
$$
|u(t,x)| \le \int_{\mathbb{R}^n} \frac{1}{(2\pi t)^{n/2}} |f(y)| dy = \frac{1}{(2\pi t)^{n/2}} \Vert f \Vert_{L^1},
$$
and from this it's clear that $\lim_{t\to\infty} \sup_{x \in \mathbb{R}^n} |u(t,x)| =0$, i.e. the solution decays uniformly to zero as $t \to \infty$.


问:
Spectrum of integral operator $A$: $(A f)t = \int_0^t f$ by definition of $Sp$
Exercise 7.15 from Ciprian Foias, Michael Jolly, "Differential Equations in Banach Spaces" (with some edits).

Let $X = C([0, 1], \mathbb C)$ (a set of continuous functions from $[0, 1]$ to $\mathbb C$ with the uniform norm) and $A \in B(X)$ (bounded linear operator from $X$ to $X$) defined by

$$(A f)(t) = \int_0^t f$$

Determine the spectrum of $A$ (i.e. the set of $\lambda$ s.t. $A - \lambda I$ is not invertible).

I can show that $\|A^n\|^{1/n} \to 0$, and therefore by Theorem 7.14 from the book $\max \{|\lambda| \colon \lambda \in Sp(A)\} = 0$. I.e. $Sp(A) = \{0\}$.

Question: is there a way to show that by the definition of spectrum? Currently, it's not clear to me at all why $A$ is not invertible, and why $A + \lambda I$ is invertible for any $\lambda \ne 0$. Also, is there a way to guess the answer by just looking at the operator?

Motivation: While I can solve this particular problem, I'm interested in a general method of how a spectrum can be found.

I found this answer: https://math.stackexchange.com/a/199730/743044, but again it doesn't show how to do this by definition.

答:
Why is $A$ not invertible?

Note that for any element $g$ of the image of $A$ (i.e. $g = Af$ for some $f$), we have $g(0) = 0$. From this, it is clear that $A$ is not onto and therefore not invertible.

Why is $A - \lambda I$ invertible for all $\lambda \neq 0$?

We want to find an inverse for $(A - \lambda I)$. Correspondingly, we want a solution to $(A - \lambda I) f = g$ for every $g \in C([0,1],\Bbb C)$. First, consider the case where $g$ is differentiable. We have
$$
- \lambda f(t) + \int_0^t f(x)\,dx = g(t) \iff
- \lambda f'(t) + f(t) = g'(t) \quad \text{and} \quad f(0) = -(1/\lambda) g(0).
$$
We can solve this differential equation as follows:
$$
- \lambda f'(t) + f(t) = g'(t)\\
[e^{-\lambda t} f(t)]' = e^{-\lambda t}g'(t)\\
e^{-\lambda t}f(t) = e^{-\lambda \cdot 0}f(0) + \int_0^x e^{-\lambda x}g'(x) \,dx =
- \frac 1{\lambda}g(0) + \int_0^x e^{-\lambda x}g'(x) \,dx\\
f(t) = -\frac 1{\lambda}e^{\lambda t} g(0) + e^{\lambda t} \int_0^x e^{-\lambda x}g'(x) \,dx.
$$
So, let $B$ be the map defined by the Riemann Stieltjes integral
$$
(Bg)(t) = -\frac 1{\lambda}e^{\lambda t} g(0) + e^{\lambda t} \int_{x = 0}^{t} e^{-\lambda x}dg(x).
$$
Note that $B$ is a bounded (i.e. continuous) linear map. With the help of the above work, we can see that for all differentiable $g$, we have $(A - \lambda I)Bg = g$ and $B(A - \lambda I)g = g$. Because the differentiable functions form a dense subspace of $C([0,1],\Bbb C)$, we can conclude by continuity that $(A - \lambda I)Bg = g$ and $B(A - \lambda I)g = g$ hold for all elements $g$ of $C([0,1],\Bbb C)$.

So, $A - \lambda I$ is invertible with inverse $B$.


问:
Minimizing a quantity is equivalent to minimize the squared quantity...?
On the formulation of the Support Vector Machine problem, at some point, after the scaling of my constraints I arrive at the following formulation:

$$ \max_{\textbf{w},w_0} \|w\|^{-1}$$

subject to constraints $t_i (\textbf{w}^T\phi(x_i) +w_0) \ge 1$. Of course searching for the $max$ of the above quantity is the same as searching for the $\min_{\textbf{w},w_0} \|w\|$. But then, I read that it is equivalent again to minimzing $\|w\|^2$. I'm pretty new to optimization theory... so it is always true that searching for the minimum of something is equivalent to searching for the minimum of the same quantity squared? And what about the maximum? Thanks.

答: $\min\limits_{x} f(x)$ is equivalent to $\min\limits_{x}g(f(x))$ in the sense that their minimizers are the same if $g:\mathbb{R}\to\mathbb{R}$ is a strictly increasing function and $f:\mathbb{R}\to\mathbb{R}$ is convex.


问:
How to approach the negative root of the series $\sum\limits_{k=1}^{\infty} \sigma_1(k) x^k$
Let's define
$$
f(x) = \sum\limits_{k=1}^{\infty} \sigma_1(k) x^k, \quad \text{where } \sigma_1(k) = \sum\limits_{d|k}d
$$

Is there a way to approach the root $x_0$ of this series, where $-1 < x_0 < 0$ ? I tried to squeeze it and using geometric series, but its too far away.

Body:
I truncated the sum at $n$ terms and applied three iterations of Newton's Method, starting at $x_0=-0.5$. Here are the results:
$$
\left(
\begin{array}{cc}
 1 & 0 \\
 2 & -0.333384146341463 \\
 3 & 0.0961538461538462 \\
 4 & -0.397485670060330 \\
 5 & -0.500000000000000 \\
 6 & -0.407129807891880 \\
 7 & -0.420277540201558 \\
 8 & -0.409817509129324 \\
 9 & -0.413179097089750 \\
 10 & -0.411220988091159 \\
 11 & -0.411739696601906 \\
 12 & -0.411245203711637 \\
 13 & -0.411344738057402 \\
 14 & -0.411275353266776 \\
 15 & -0.411303377636828 \\
 16 & -0.411288790446980 \\
 17 & -0.411292174341874 \\
 18 & -0.411289268468018 \\
 19 & -0.411289851523367 \\
 20 & -0.411289380413744 \\
 21 & -0.411289515069802 \\
 22 & -0.411289460359527 \\
 23 & -0.411289472717242 \\
 24 & -0.411289463450467 \\
 25 & -0.411289464495063 \\
 26 & -0.411289464563588 \\
 27 & -0.411289464215141 \\
 28 & -0.411289464649207 \\
 29 & -0.411289464488793 \\
 30 & -0.411289464727583 \\
\end{array}
\right)
$$ Based on this, I would say with high certainty that the root satisfies $x_0\approx -0.41128946$.


问:
Convolution of cosine with exponential
As part of an exercise, I'm trying to find the output of a cosine wave entering a low-pass filter by using a convolution integral.

The impulse response of the filter is $h(t) = \frac{1}{RC}\exp\left({-\frac{t}{RC}}\right)$

The solution should look as follows (if I'm not mistaken):

ouput = $\int_0^t\cos(wt)h(t-\tau) d\tau$

I was told that the output should be a solution of the form $A\cos(wt)$, but when I try doing this integral I get some fairly messy expressions... Could someone help me out with figuring out where I'm going wrong?

Thanks!

p.s. sorry for the lack of "math font", I'm fairly new to this site and can't quite figure it out....

答:
The convolution is set up incorrectly. The independent variable of the cosine should be a tau, not a t. Furthermore the lower limit should be negative infinity. If this integral is performed, you will see that the answer is smaller than the input cosine by a factor 1/sqrt(1+(wRC)^2), and phase-shifted by the phase -atan(wRC/sqrt(1+(wRC)^2)).

Sincere apologies for my lack of equation formatting, but I see that this is the top Google result for "sinusoid convolved with decaying exponential," and currently a student might get the wrong result.

Title: Convolution of cosine with exponential


问:
Linearly Independent Elements in the Néron–Severi group
I have a question regarding to the proof given in lemma 1.9 (Bend and Break II) pg 11. given in the book ‘Birational Geometry of Algebraic Varieties’ by János Kollár and Shigefumi Mori.

In the second paragraph we assume that $\tilde{G}: \tilde{S} \rightarrow X$ is a morphism. By fixing an ample divisor $H$ on $X$, and setting $C_0, C_\infty \subset S$ to be sections extending $\{0\} \times D, \{\infty\} \times D$ we have $(\tilde{G}^*H)>0$ and $C_0 \cdot \tilde{G}^*H = C_\infty \cdot \tilde{G}^*H =0$. By the Hodge index theorem, we have $C_0^2 , C_\infty^2 <0$.

The author went on to claim that $\{\tilde{G}^*H, C_0, C_\infty\}$ must be an linearly independent set in the Néron–Severi group of $S$. How do we see this? Given a linear combination $a\tilde{G}^*H + b C_0 + bC_\infty =0$ we easily see that $a=0$. So we are left with $bC_0 + cC_\infty =0$. How do we use the fact that $C_0^2, C_\infty^2 <0$ to show that $b=c=0$ necessarily?

Edit: I think I have figured it out. We have $b(C_0 - C_\infty)=-(b+c) C_\infty$. Then taking product, $-(b+c) C_\infty^2 = b (C_0^2 - C_0C_\infty)$.

We may assume $b \ge 0$ and so the RHS is $<0$. This forces $b+c \le 0$. Now, by the Hodge Index theorem applied to $C_0-C_\infty$, we have $(C_0 - C_\infty)^2 \le 0$.

So $0 \ge b(C_0-C_\infty)^2= -(b+c)(C_0C_\infty -C_\infty^2)$. So $-(b+c) \le 0$. So $b+c \ge 0$. All in all, $b=-c$ and so we have $C_0 = C_\infty$, which cannot be since they are distinct.


问:
Integrating $\int \sin^3{x} \cos^2{2x}\,dx$
Is there a trick to this integral?
$$\int \sin^3{x} \cos^2{2x}\,dx$$

I've tried to solve this by integration by parts and by expanding $\cos^2{2x}$ but these seem to make it more complicated.

Is there something I'm missing like a clever substitution or using the trigonometric identities?

答:
With $c:=\cos x$, the integral becomes polynomial,

$$I=-\int (1-c^2)(2c^2-1)^2dc.$$

Then

$$\int(4c^6-8c^4+5c^2-1)\,dc$$ is immediate.


问:
How to lift the Frobenius map on $W(k)[[t]]$?
Let $k$ be a perfect field of characteristic $p$ with ring of Witt vectors $W(k)$.

Define the Frobenius endomorphism $F(t)=t^p$ on $W(k)/pW(k)$.

Now define the ring $W(k)[[t]]$ be extending the ring $W(k)$.

How to lift the Frobenius $F$ to the map $F(t)=t^p+p^2t$ on the ring $W(k)[[t]]$ ?

Is the lift again a Frobenius map ?

Here is a concept of $p$-derivation. If $\delta_p$ be the $p$-derivation map on $W(k)$ then the lift of the Frobenius $F(t)=t^p$ $\left(\text{mod} \  pW(k) \right) $ is given by $$F(t):=t^p+p \delta(t).$$
This is bijective if $W(k)$ is $p$-torsion free.

Any hints please.

答:
Remark: the Frobenius on $W(k)$ is not given by $F(w) = w^p.$ Since $p\neq 0$ in $W(k)$, this won't be a ring morphism!

To check if your map $F : W(k)[\![t]\!]\to W(k)[\![t]\!]$ is a lift of Frobenius, where $F$ is the  Frobenius on $W(k)$ and $F(t) = t^p + p^2t,$ you need to check if it is the Frobenius modulo $p.$ However, this is clear: $F(t)\equiv t^p\pmod{p},$ and you already know that the Frobenius lift on $W(k)$ satisfies the desired property.


问:
Is $M= \{(x,y) \in \mathbb{R^2} : x \in \mathbb{N} \lor y \in \mathbb{N}\} $ polygonally connected?
I'm trying to prove that $M= \{(x,y) \in \mathbb{R^2} : x \in \mathbb{N} \lor y \in \mathbb{N}\} $ is polygonally connected.

I know that polygonally connected implies arc-connected so if I prove that for an $x, y \in M$  there is no arc $\varphi$ joining x and y that verifies $\varphi(t) \in A$ $ \forall t \in [a,b]$ then it won't be polygonally connected, but I'm not quite sure how I could prove it or if I'm taking the wrong path.

Any suggestions?

答:
Let $(x,y)$ and $(x',y')$ be points in $M$ and let $g(x)=\max(0,\lfloor x\rfloor)$. Note that $g(n)=n$ for all $n\in\mathbb N$, $g(x)\in\mathbb N$ for all $x\in\mathbb R$ and hence the following polygonal chain is completely contained in $M$:
$$
(x,y) \to (g(x), y) \to (g(x), g(y)) \to(g(x'),g(y)) \\\to (g(x'), g(y')) \to (g(x'), y') \to (x',y').
$$


问:
Why are this all the eigenvalues/eigenvectors of this transformation?
Consider the linear transformation: $$ f(x)\longmapsto \int_{-x}^x f(t)dt,$$ in the linear space (over $\mathbb{R}$) that is span by $ \langle 1,\cos(x),sin(x),\dots,\cos(mx),\sin(mx)\rangle$ for $m\in \mathbb{N}$. My professor in his solution claims that the only eigenvalue of this transformation is $0$ and is the same eigenvalue for all eigenvectors and these have the form $\sin(kx)$ for some $k\in \mathbb{N}$.

Is not hard to calculate the integrals from all elements of the set given and see that is true that $0$ is an eigenvalue and indeed some eigenvectors have this form, but I don't understand why these are the ONLY eignevalues/vectors. Can you give me some help here?

答:
First, suppose $f \in \mathrm{span}\left(1,\cos,\sin,\ldots,x\mapsto\cos(nx),x\mapsto \sin(nx)\ldots \right)$ and write $u(f) : x \mapsto \int_{-x}^x f$. Then $u(f)$ is an odd function because:
\begin{align}
\forall x\in \mathbb{R},~u(f)(-x) &= \int_{x}^{-x} f \\
&= - \int_{-x}^x f\\
&= -u(f)(x)
\end{align}

Then, if $\lambda\neq 0$ is an engeinvalue with eigenvector $f$ of $u$, $u(f) = \lambda f$, and $f$ is forced to be an odd function. Thus, $f$ is an odd function in $\mathrm{span}\left(1,\cos,\sin,\ldots,x\mapsto\cos(nx),x\mapsto \sin(nx)\ldots \right)$. There exists $n_1,\ldots, n_k$ integers and $\lambda_1,\ldots,\lambda_k$ reals so that
\begin{align}
\forall x \in \mathbb{R},~f(x)= \sum_{j=1}^k \lambda_j\sin\left(n_j x \right)
\end{align}
But every odd function in the above vector space is an eigenvector with eigenvalue $0$! This is an easy computation. So the only eigenvalue for $u$ is $0$.


问:
Random Variable Transformation With Dirac delta function
Consider  $f_X(x) = 1/2+x$ where $0 <x\leq 1$. We want to find the pdf of $Y=X^{-1}$ by utilizing the following transformation.

$$f_Y(y) = \int_0^1 f_X(x) \delta (y - g(x)) dx$$

Where $u =g(X) = Y$.

From my understanding we can represent $x$ as $x=g^{-1}(g(x))$ and then we can write $dx$ as $dx = \frac{d(g(x))}{dx}dx \implies dx=-x^2du$. Since $x \in (0,1]$, we can conclude that $y \in [1,\infty)$. So we can rewrite our expression as:

$$f_Y(y) = \int_1^\infty f_X(g^{-1}(g(x))) \delta (y - g(x)) (-\frac{1}{y^2})du$$
However, after this point I don't really know how to continue. Yes I can factor out a negative sign, but i just don't see what my next step is. Any ideas?

答:
Are you aware of the filtering/sifting property of Dirac Delta function? The filtering property says that $\forall\epsilon,\gamma>0$ the value of$$\int_{a-\epsilon}^{a+\gamma} h(x)\delta(a-x)dx=\lim_{x\to a}h(a)$$
Take $g(x)=1/x$ as $m$, then you have$$f_Y(y)=\int_1^\infty\frac1{m^2}f_X(1/m)\delta(y-m)~dm$$

For $y<1,$ $\delta(y-m)=0$ so $f_Y(y)=0,y<1$. For $y>1$, we have $1/y\in(0,1)$ and so$$f_Y(y)=\lim_{m\to y}\frac{f_X(1/m)}{m^2}=\lim_{1/m=k\to1/y}k^2f_X(k)=\frac{f_X(1/y)}{y^2}$$since $k^2f_X(k)$ is continuous in $(0,1)$. Finally you get$$f_Y(y)=\begin{cases}0,&y\le1\\\frac1{y^2}\left(\frac12+\frac1y\right),&y>1\end{cases}$$


答:
Let $T$ denote the transformation. Note that the eigenspace associated with the eigenvalue $0$ consists of the set of all odd functions within the domain. Now, show that $Tf$ is itself an odd function for all $f$. It follows that for all functions $f$ from the domain, we have $T(T(f)) = 0$.

Now, suppose that $\lambda$ is an eigenvalue of $T$ and that $f \neq 0$ is an associated eigenvector. We have $T(T(f)) = 0 \implies \lambda^2 f = 0$. This implies that $\lambda^2 = 0$, which means that $\lambda = 0$. So, the only possible eigenvalue of $T$ is $0$.

Title: Why are this all the eigenvalues/eigenvectors of this transformation?

Body:
Consider the linear transformation: $$ f(x)\longmapsto \int_{-x}^x f(t)dt,$$ in the linear space (over $\mathbb{R}$) that is span by $ \langle 1,\cos(x),sin(x),\dots,\cos(mx),\sin(mx)\rangle$ for $m\in \mathbb{N}$. My professor in his solution claims that the only eigenvalue of this transformation is $0$ and is the same eigenvalue for all eigenvectors and these have the form $\sin(kx)$ for some $k\in \mathbb{N}$.

Is not hard to calculate the integrals from all elements of the set given and see that is true that $0$ is an eigenvalue and indeed some eigenvectors have this form, but I don't understand why these are the ONLY eignevalues/vectors. Can you give me some help here?


问:
generalization of fibration
A Hurewicz fibration $E\rightarrow B$ is a map so that for all maps $X\rightarrow E$ and $X\times I\rightarrow B$ making the obvious square commute, there is a lifting $X\times I\rightarrow E$ so that everything commutes.  This is known as the homotopy extension property for $X$.  A Serre fibration is a map that satisfies the homotopy extension property for cubes.

My question is: first, given a Hurewicz/Serre fibration, for what pairs $A\rightarrow X$ mapping into $E\rightarrow B$ can we find a map $X\rightarrow E$ making the diagram commute?  For example, this is true for $X\rightarrow X\times I$ for Hurewicz fibrations; I would like to know a complete characterization for such pairs.

Also, is there a name for maps $E\rightarrow B$ satisfying this sort of lifting property for a specific pair $A\rightarrow X$?  I would guess it would be named something like "homotopy lifting property relative to $A\rightarrow X$" but I have failed to find something like this in my searches.

In addition, I would be interested in the dual questions for cofibrations.  These questions feel simple enough that there should be references for them, but I haven't found them.

答:
The fibrations in any model catetgory are characterised by their lifting property with respect to a particular class of maps. In detail:

A morphism $p:E\rightarrow B$ in a model category $\mathcal{M}$ is a fibration if and only if it has the right lifting property with respect to all morphisms $i:A\xrightarrow\simeq X$ which are both cofibrations and weak equivalences.

Below I will call any morphism which is both a cofibration and a (weak) equivalence an acyclic cofibration. Similarly for fibrations.

Now, let $Top$ denote either the category of all topological spaces or, say, all compactly-generated weak Hausdorff spaces. Then $Top$ supports two relevant model structures:

The first has homotopy equivalences as weak equivalences, Hurewicz fibrations as fibrations, and closed cofibrations as cofibrations. Call this the Strom model structre.

The second has weak homotopy equivalences as weak equivalences, Serre fibrations as fibrations, and the cofibrations are the maps which have the left lifting property with respect to all acyclic Serre fibrations. Call this the Serre model structure.

So this answers your first question. Working with the Strom model structure we see that the Hurewicz fibration are exactly the maps which have the right lifting property with respect to all maps $i:A\xrightarrow\simeq X$ which are both homotopy equivalences and closed cofibrations. Note that a map which is both a cofibration and a homotopy equivalence is a strong deformation retract, although the converse is not true. Note also that any cofibration between weak Hausdorff spaces is closed, although there do exist non-closed cofibrations (which are even homotopy equivalences) in the larger category.

The acyclic cofibrations in the Serre model structure have a more awkward description. They are maps which are retracts of transfinite compositions of spaces constructed as pushouts of sums of inclusions of the form $D^n\hookrightarrow D^n\times I$. One of the benefits of Serre model is that it is cofibrantly generated, and you'll find a slightly more accurate description of these maps by looking up this term.

Turning to your second question. In general it's said that a map $p:E\rightarrow B$ has the homotopy lifting property (HLP) with respect to a space $X$ if for any map $f:X\rightarrow E$ and any homotopy $H_t:X\times I\rightarrow B$ with $H_0=pf$, there exists a homotopy $G:X\times I\rightarrow E$ with $G_0=f$ and $pG_t=H_t$.

Then the Hurewicz fibrations are those maps which have the homotopy lifting property with respect to all spaces, and the Serre fibrations are those maps which have the homotopy lifting property with respect to all discs $D^n$, $n\geq0$. Note that for any $X$ the inclusion $in_0:X\hookrightarrow X\times I$ is an acyclic cofibration in both model structures.

Useful examples abound: A map has the HLP with respect to the one-point space if and only if it has path-lifing. A locally trivial map has the HLP with respect to all paracompact $T_2$ spaces. A map has the HLP with respect to all CW complexes if and only if it is a Serre fibration.

The last thing to adress is that yes, all of this is dualisable. This is really the power of realising the model category framework: the existence of the Strom and Serre model structures really should be celebrated facts. There is an equally valid description of the cofibrations, acyclic cofibrations and acyclic fibrations in any model category.

An accessible reference for model categories would have to be Mark Hovey's book Model Categories. If you are interested in seeing the full statement with respect to the Strom model structure on topological spaces, then let me direct you to Jeff Strom's book Modern Classical Homotopy Theory $\S$5.6.4.

Title: generalization of fibration


问:
How to simplify nested cubic radicals $\sqrt[3]{a+b\sqrt c}$
While trying to answer this question, I got stuck showing that

$$\sqrt[3]{26+15\sqrt{3}}=2+\sqrt{3}$$

The identity is easy to show if you already know the $2+\sqrt{3}$ part; just cube the thing. If you don't know this, however, I am unsure how one would proceed.

That got me thinking ...

If you have some quadratic surd $a+b\sqrt{c}$, where $a$, $b$, and $c$ are integers, and $c$ is not a perfect square, how do you find out if that surd is the cube of some other surd, i.e. how to simplify nested cubic radicals of the form
$$\sqrt[3]{a+b\sqrt c}$$

Body:
Apply the known denesting formula
$$\sqrt[3]{a+b \sqrt c}=\frac12\sqrt[3]{3bt-a}\left(1+\frac1t \sqrt c\right)$$
where $t$ satisfies $t^3-\frac{3a}bt^2+3c t-\frac{ac}b=0$. Take the example $\sqrt[3]{26+15\sqrt{3}}$ in question and solve
$$t^3-\frac{26}5t^2+9t-\frac{26}5=\frac15(t-2)(5t^2-16t+13)=0$$
which yields $t=2$ and the denestation
$$\sqrt[3]{26+15\sqrt{3}}=2+\sqrt{3}$$

Other examples are listed below, along with their resolvent equations
\begin{align}
\sqrt[3]{7+5\sqrt{2}}= 1+\sqrt{2}&\>\>\>\>\>\>\>
t^3-\frac{21}5t^2+6t-\frac{14}5 =0,\>\>\>t=1\\
\sqrt[3]{90-34\sqrt{7}}= 3-\sqrt{7}&\>\>\>\>\>\>\>
t^3+\frac{135}{17} t^2+21 t+\frac{315}{17}=0 ,\>\>\>t=3 \\
\sqrt[3]{\frac{99}2+\frac{59}2\sqrt{\frac52}}= 3+\sqrt{\frac52}&\>\>\>\>\>\>\>
t^3-\frac{297}{59} t^2+\frac{15}2 t-\frac{495}{118}=0 ,\>\>\>t=3 \\
\sqrt[3]{25+10\sqrt{5}}= \frac52+\frac12\sqrt{5}&\>\>\>\>\>\>\>
t^3-\frac{15}{2} t^2+15 t-\frac{25}{2}=0 ,\>\>\>t=5\\
\sqrt[3]{70-22\sqrt{7}}= \sqrt[3]{49}\left(1-\frac{\sqrt{7}}7\right)&\>\>\>\>\>\>\>
t^3+\frac{105}{11} t^2+21 t+\frac{245}{11}=0 ,\>\>\>t=7\\
\end{align}


问:
Explicit general formula for $n!$ permutation matrices for column vector $ \pi = (1 \cdots n)^{T} $
I am looking for an explicit general formula of $ n!$ permutation matrices for the initial column vector $\pi = \begin{pmatrix} 1,  2  \cdots \ n \end{pmatrix}^{T}$.

The first permutation vector is $\pi^{I} = \pi$ and the first permutation matrix is the identity matrix:

$
\pi^{I}= P^{I}_{\pi} \pi
$

$
P^{I}_{\pi} = \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 1 \end{pmatrix}
$

The second permutation vector is $\pi^{II} = \begin{pmatrix} 2,  1 \ \cdots \ n \end{pmatrix}^{T}$ and the corresponding permutation matrix is $P^{II}_{\pi}$ as follows:

$
\pi^{II}= P^{II}_{\pi} \pi
$

$
P^{II}_{\pi} = \begin{pmatrix} 0 & 1 & \cdots & 0 \\ 1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 1 \end{pmatrix}
$

I would like to get the expression of the $K^{th}$ permutation matrix $P^{K}_{\pi}$ for the $K^{th}$ permutation, $1 \leq K \leq n!$

Any idea?

Thanks

答: Permutations can be represented by the corresponding permutations of the columns of the identity matrix. i.e. $$\begin{bmatrix} 1&0&0\\0&0&1\\0&1&0\end{bmatrix}$$ represents the permutation $(1,3,2)$, since we swapped columns one and two, then we get the permutation that maps $2\mapsto 3$ and $3\mapsto 2$.


问:
Find the integrating factor of the differential equation $(y\log y)dx = (\log y-x)dy$
I am familiar with linear differential equations of the form
$$\frac{dy}{dx} + P(x)y = Q(x)$$

But this equation doesn’t seem to be of that form, or any other ones I know. It could be converted into the linear form, but I don’t know how. How should I simplify?

答: Hint: $$\frac{dx}{dy}+\frac x{y\log y}=\frac1y$$is first order linear ODE in $x$.


问:
Restricting the codomain of a smooth map to submanifold is smooth
Let $F:N\to M$ be a smooth map from a smooth manifold $N$ to a smooth manifold $M$ (without boundary) and $S$ an immersed submanifold of $M$ such that $F(N)\subset S$. If $F$ is continuous as a map from $N$ to $S$, then $F:N\to S$ is smooth.

This is theorem $5.29$ in John Lee's introduction to smooth manifolds $2$nd edition and in that same book he wrote the following proof (I will not give the full proof but only upto the part where I don't understand):

"$Proof$: Let $p\in N$ and $q=F(p)\in S$. Then as $S$ is an immersed submanifold, it is locally embedded. Therefore there is a neighborhood $V$ of $q$ such that $V$ is embedded in $M$, that is the inclusion $i|_V:V\hookrightarrow M$ is a smooth embedding. This implies $V$ satisfies the local slice criterion, that is there exist a smooth chart $(W, \psi)$ for $M$ that is a slice chart for $V$ centerd at $q$. The fact that $(W,\psi)$ is a slice chart means that $(V_0, \widetilde{\psi})$ is a smooth chart of $V$, where $V_0 = W\cap V$ and $\widetilde{\psi}= \pi\circ \psi|_{V_0}$, with $\pi: \mathbb{R}^n\to \mathbb{R}^k$ the projection onto the first $k=\dim S$ coordinates."

I understand everything upto this line. The next line below is where my confusion started.

"Since $V_0 = \left( i|_V\right)^{-1}(W)$ is open in $V$, it is open in $S$ in its given topology, and so $(V_0, \widetilde{\psi})$ is also a smooth chart for $S$."

I know that $V_0$ is open in $S$ and I know that $(V_0, \widetilde{\psi})$ will be a (continuous) chart but I dont understand the claim that $(V_0,\widetilde{\psi})$ is a smooth chart for S. How do we make sure that $(V_0, \widetilde{\psi})$ belongs to the smooth structure of $S$ by merely knowing $V_0$ is open? Is there a relationship between the smooth structure of $V$ and the smooth structure of $S$. Please help me I'm so confuse.

答:
Personally I do not like the concept of an immersed submanifold as presented by Lee and other authors. The situation is this:

We have an injective immersion $j : P \to M$. This induces a unique topology and a unique smooth structure on the image set $S = j(P)$ such that $\bar j :  P \stackrel{j}{\to} S^*$ becomes a diffeomorphism, where $S^*$ denotes the set $S$ endowed with the appropriate topology and smooth structure. This $S^*$ is called an immersed submanifold of $M$.

The problem here is that $j$ is in general no homeomorphism from $P$ onto its image $j(P)$ with the subspace topology inherited from $M$. Thus, although we call $S^*$ an immersed submanifold, it is in general not even a topological subspace of $M$. In my opinion this is somewhat confusing, but of course it is a matter of taste.

However, even if $S^*$ should not be a topological subspace of $M$ globally, it is always locally one. This means that each $q \in S^*$ has an open neigborhood $V$ in $S^*$ which is an emdedded submanifold of $M$. Note that $V$ is open in $S^*$, but in general not open in $S$ with the subspace topology inherited from $M$.

In my opinion a better formulation of Lee's Theorem 5.29 would be this:

Let $f :N \to P$ be a continuous function from a smooth manifold $N$ to a smooth manifold $P$  and let $j : P \to M$ be an immersion (which is not required to be injective). If $F = j \circ f$ is smooth, then $f$ is smooth.

Lee's theorem is a corollary. We are given an immersed submanifold $S^*$ of $M$ and a smooth $F : N \to M$ such that $F(N) \subset S^*$ and $\bar F : N \stackrel{F}{\to} S^*$ is continuous. This $S^*$ is associated to an injective immersion $j : P \to M$ such that $\bar j : P \stackrel{j}{\to} S^*$ is a diffeomorphism. The function $f = (\bar j)^{-1} \circ \bar F : N \to P$ is continuous and $j \circ f = F$ is smooth, thus $f$ is smooth. Hence $\bar F = \bar j \circ f$ is smooth.

Our above theorem can be proved exactly as Lee's Theorem 5.29. Let us now see why $(V_0, \widetilde{\psi})$ is a smooth chart for $S^*$.

Since $V$ is open in $S^*$ and $V$ is an embedded submanifold of $M$, it suffices to show that $(V_0, \widetilde{\psi})$ is a smooth chart for $V$. But this follows from a theorem in Lee's book which says that each subspace $E \subset M$ which is an embedded submanifold has a unique smooth structure such that the inclusion map $E \to M$ is a smooth embedding. A smooth atlas on $E$ generating this smooth structure is given precisely by the collection of all $(W_0 = W \cap E, \pi \circ \psi \mid_{W_0})$ where the $(W,\psi)$ are smooth charts for $M$ which are slice charts for $E$ centered at the points $q \in E$.

Perhaps it is somewhat confusing that the smooth structure on $V$ is primarily induced by $j_V : j^{-1}(V) \stackrel{j}{\to} V$. However, since we know that $V$ is an embedded submanifold of $M$, this smooth structure agrees with the above smooth structure induced by slice charts.


问:
Confused about Serge Lang's proof of hermitian operator has real eigenvalue.
On Serge Lang's linear algebra page 225:

Theorem 5.1: Let $A:V\to V$ be a hermitian operator. Then every eigenvalue of $A$ is real.

Proof: Let $v$ be an eigenvector with an eigenvalue $\lambda$. By Theorem 2.4 of Chapter VII we know that $<Av,v>$ is real. Since $Av=\lambda v$, we find $<Av,v>=\lambda <v,v>$. But $<v,v>$ is real $>0$ by assumption. Hence $\lambda$ is real, thus proving the theorem.

Theorem 2.4 Let $V$ be a finite dimensional vector space over the complex numbers. Let $A$ be an operator such that $<Av,v>=0$ for all $v\in V$. Then $A =o$.

I don't understand why Theorem 2.4 can prove that $<Av,v>$ is real, and what assumption makes $<v,v>$ real.

(There're some proofs of the title much easier to understand, for example https://proofwiki.org/wiki/Hermitian_Matrix_has_Real_Eigenvalues, but I just can't understand the specific proof method which Lang use.)

答: if $v \neq 0$ is an eigenvector, then $||v||^{2} = \langle v | v \rangle >0$ (from the definition of an inner product), and $\lambda \langle v | v\rangle = \langle \lambda v | v\rangle =  \langle Av | v \rangle = \langle v | Av \rangle = \langle v | \lambda v\rangle = \lambda ^{*} \langle v | v \rangle  $, and since $\langle v | v\rangle \neq 0 $ we get $\lambda = \lambda^{*}$ so $\lambda$ is real.


问:
What is suboptimality gap in reinforcement learning?
I was reading some research papers on Reinforcement Learning Theory, and I constantly encountered a term called the suboptimality gap. As I searched the internet, I couldn't find any information about this term. So, I wonder whether anyone here knows what this means?

答:
In Non-Asymptotic Gap-Dependent Regret Bounds for Tabular MDPs,  suboptimality gap associate with action $a$ at state $x$ is defined to be

$$gap_\infty(x,a)=V^{\pi^*}(x)-Q^{\pi^*}(x,a),$$

It is the difference in the value of a particular action from a particular state as compared to the optimal move.

Similar term has been used in bandit problem as well.


问:
Showing that $R = \{ \pm(e_i-e_j) : 1 \leq i < j \leq l+1\}$ is a root system
Let $e_i$ be the standard unit basis vector in $\mathbb{R}^{l+1}$

I'm trying to show that the $R = \{ \pm(e_i-e_j) : 1 \leq i < j \leq l+1\}$ is a root system for $E = spanR = \{ \Sigma \alpha_i e_i : \Sigma \alpha_i = 0\}$

I'm struggling to show that given $\alpha \in R$, we have that the reflection about the hyperplane normal to $\alpha$ permutes the elements of $R$. The reflection operator of $\alpha$ is denoted $s_\alpha$.

We have that $s_\alpha(\gamma) = \gamma - 2\frac{(\gamma,\alpha)}{(\alpha,\alpha)}\alpha$.

My calculations have not been working out. Can somebody help me with this? Thank you

答:
Let $v_{ij} = e_i - e_j$. We find that
$$
s_{v_{ij}}\left(\sum_k \alpha_k e_k \right) = \alpha_j e_i + \alpha_i e_j + \sum_{k \notin\{i,j\}} \alpha_k e_k.
$$
From there, I think it should be clear that $s_{v_{ij}}$ indeed permutes the elements of $R$.

With that established, we find the following: if $\{i,j\},\{p,q\}$ are disjoint,
$$
s_{v_{ij}} v_{pq} = v_{pq} \in R.
$$
If $p = i$ and $q \notin \{i,j\}$, we have
$$
s_{v_{ij}} v_{pq} = s_{v_{ij}} v_{iq} = v_{jq} \in R.
$$
Since $s_{v_{ij}} = s_{v_{ji}}$ and $v_{pq} = -v_{qp}$, all cases where there is one common index are similar to this one. Finally, if $\{i,j\} = \{p,q\}$, we find that $s_{v_{ij}}v_{pq} = -v_{pq} \in R$.

In each case, $s_{v_{ij}}$ takes elements of $R$ to $R$. Because $s_{v_{ij}}$ is invertible, the map induced on $R$ must be bijective.


问:
Find all $C^1$ functions from the reals to the reals satisfying $f'(a-x)=f(x)$ for some $a>0$.
My try : it is easy to see that $f$ is twice differentiable so by differentiations we get :
$f''(x)=-f'(a-x)=f(x)$ and so $f''(x)+f(x)=0$ and the solution must be of the form $a\cos(x)+b
\sin(x)$ for some constants $a,b$.

Now plugging this in the original equation i get some ugly equations (well, actually not that ugly, but it seems kinda strange) on $a$ that i have to solve and i am unnsure of whether this is the right thing to do or not.
Is it the right approach ? If not not, could you show me what i've done wrong ?
By the way, if anyone could post good resources to learn basic strategies for solving differential equations of order 1 and 2 at beginner's level i would be extremely grateful to them.

Body:
$f(x)=c_1e^{ix}+c_2e^{-ix}$

$ic_1e^{i(a-x)}-ic_2e^{-i(a-x)}=c_1e^{ix}+c_2e^{-ix}$

$ic_1e^{ia}-c_2=0\\c_1+ic_2e^{-ia}=0$

Thus $f(x)=ce^{ix}+ice^{ia}e^{ix}$


问:
A representation system of $\Gamma_0(Nl) \subseteq \Gamma_0(N)$
For an integer $n$, let
$$ \Gamma_0(n) = \left\{ \left( \begin{matrix}a & b \\ c & d \end{matrix} \right) \in \operatorname{SL}_2 (\mathbb{Z})\bigg | c \equiv 0 \mod n\right\}.$$

Let $N, l$ be different primes.
Then what is a representation system of $\Gamma_0(Nl) \subseteq \Gamma_0(N)$?

It seems that $\beta_i = \left( \begin{matrix}1 & 0 \\ Ni & 1 \end{matrix} \right)$ for $i = 0, \dots l-1$,
and $\beta_\infty = \left( \begin{matrix}a & b \\ N & l \end{matrix} \right)$,
where $al -Nb = 1$,
represent it.

But I can't show.

答:
Let $B(l)$ be the image of $\Gamma_0(l)$ in $SL_2(\Bbb{Z}/l\Bbb{Z})$. Take some $r_j,s_i$ such that $$SL_2(\Bbb{Z}/l\Bbb{Z})=\bigcup_j r_j B(l)=\bigcup_i B(l)s_i\qquad (disjoint\ union)$$

The reduction $\bmod l$ is an homomorphism
$$f:\Gamma_0(N)\to SL_2(\Bbb{Z}/l\Bbb{Z})$$ Since $N,l$ are coprime then it is surjective. The kernel is contained in $\Gamma_0(Nl)$, and the image of $\Gamma_0(Nl)$ is $B(l)$. This implies that

$$\Gamma_0(N)/\Gamma_0(Nl)\cong f(\Gamma_0(N))/f(\Gamma_0(Nl))=SL_2(\Bbb{Z}/l\Bbb{Z})/B(l)$$
and similarly for the quotient on the left.

Take some representatives $R_j,S_i\in \Gamma_0(N),f(R_j)=r_j,f(S_i)=s_i$. Then $$\Gamma_0(N)=\bigcup_j R_j \Gamma_0(Nl)= \bigcup_i \Gamma_0(Nl) S_i\qquad (disjoint\ union)$$
Finally, show that $\{r_j\}$ can be taken to be $ \pmatrix{1&0\\*&1}\cup \pmatrix{0&-1\\1&0}$ equivalently $\pmatrix{1&0\\*&1}\cup \pmatrix{0&b\\N&0}$ where $-Nb\equiv 1\bmod l$.

What about the $s_i$ ?


问:
Improper Integral $\int_0^1\frac{dx}{x^p}$
Is this integral convergent only for $p<1$?

$$\int_0^1\frac{dx}{x^p}$$

答:
If $p\le 0$, there is no problem of any kind: the function is continuous in the interval $[0,1]$, so the integral exists.

But if $p\gt 0$, there is a potential problem, since the function blows up as $x$ approaches $0$ from the right. So to find out whether the improper integral converges or not,  for any $\epsilon \gt 0$, we let
$$I_\epsilon=\int_\epsilon^1 \frac{1}{x^p}\,dx,$$
and examine the behaviour of $I_\epsilon$ as $\epsilon$ approaches $0$ from the right.

If $\lim_{\epsilon\to 0+} I_\epsilon$ exists, then the improper integral converges, otherwise it diverges. (Note that $\lim_{\epsilon\to 0^+}I_\epsilon$ means the limit as $\epsilon$ approaches $0$ from the right, that is, through positive values.)
If $p\ne 1$, then
$$I_\epsilon=\left. \frac{1}{1-p}x^{-p+1}\right|_\epsilon^1=\frac{1}{1-p}(1-\epsilon^{1-p}),$$
while if $p=1$ then
$$I_\epsilon =-\ln(\epsilon).$$
First deal with the case $p=1$. As $\epsilon$ approaches $0$ from the right, $\ln(\epsilon)\to-\infty$, so $\lim_{\epsilon\to 0^+}I_\epsilon$ does not exist.
If $p\gt 1$, then $1-p$ is negative, and again $I_\epsilon$ blows up as $\epsilon$ approaches $0$ from the right.

Finally, if $p\lt 1$, then $1-p$ is positive, so $\epsilon^{1-p}$ approaches $0$ as $\epsilon$ approaches $0$ from the right, so the improper integral converges.


问:
Basic question (clarification) on tensor notation of a matrix: Up-Down or Down-Down?
Easy to muddle through while reading along, but a recurrent source of frustration for me (Wikipedia reference):

$A_{ij}$ would seem to suggest that we are writing $A_{ij}\;e^i\otimes e^j$. But this makes no sense, since a matrix would "eat" one vector and one covector.

On the other hand $A^{i}{}_{j}$ would make more sense interpreted as an object $A^{i}{}_{j} \;e_i \otimes e^j$ that would take in one vector and one covector. In the case of a vector in Euclidean coordinates:

$$\begin{align}
 A^{i}{}_{j}v^j &= u^i\\[2ex]
\text{or}\\[2ex]
\left( A^{i}{}_{j} \;e_i \otimes e^j \right) v^j e_j &= u^i e_i
\end{align}$$

Likewise, the matrix product can be expressed as

$$\begin{align}
 A^{i}{}_{j} B^{j}{}_k &= C^{i}{}_k\\[2ex]
\text{or}\\[2ex]
A^{i}{}_{j} \;e_i \otimes e^j \, B^{j}{}_k \; \;e_j \otimes e^k &= C^{i}{}_k \;e_i \otimes e^k
\end{align}$$

So are $A_{ij}$ and $A^{i}{}_{j}$ at all compatible and equivalent, or is $A_{ij}$ misleading when one considers the basis vectors?

Context:

I am confused by this passage in page 9 of Introduction to Tensor Analysis by H.D. Block:

If

$$ {\bf{\bar e_j}} = a^i{}_j {\bf{ e_i}}$$

is the $j$-th entry in a vector in the new basis, and the transformation matrix is

$$A=\begin{bmatrix}
a^1{}_1 & a^1{}_2 & \cdots & a^1{}_n\\
a^2{}_1 & a^2{}_2 & \cdots & a^2{}_n\\
\vdots & \vdots  & \ddots & \vdots\\
a^n{}_1 & a^n{}_2 & \cdots & a^n{}_n\\
 \end{bmatrix}$$

I can't see the notation in

$$ {\bf{\bar e_*}} = A^\top {\bf{ e_*}}$$

答: ${A^i}_j$ is what you must write if you are using Einstein notation. $A_{ij}$ is only what normal mortals use.


问:
Does uniform convergence of $X_n$ imply $L^p$ convergence of $f(X_n)$?
Suppose I have $X_n \sim F_n$, $X \sim F$ such that $\Vert F_n - F \Vert_\infty \rightarrow 0$, does this imply that $f(X_n) \xrightarrow{L^p} f(X)$ for any continuous $f$?

I know that, since we are dealing with probability measures/distribution functions, $X_n \xrightarrow{L^p} X$ holds.

I know that, for example, similar results hold for convergence in probability or weak convergence (using unif convergence implies $\xrightarrow{p}$, and then using CMT), but I was wondering if I can say something about $L^p$ convergence.

答:
No, you can't say anything about that convergence.

The reasoning behind that is the different nature of these two types of convergence.

While the condition that you give provides a lot of info on the convergence in law, it gives nothing on the spatial relation between $(X_n;n \ge 0)$.

A good thought counterexample is when $(X_n; n \ge 0)$ are independent and identically distributed.


问:
Two specific irrational numbers summing to rational number that is not zero?
There already exist pages that show how two irrational numbers can sum to a rational number. However, is there an actual example of this? What are some of irrational numbers $x$ and $y$ such that $x + y$ is rational and $x ≠ -y$ (for an example $\sqrt{2}$ and -$\sqrt{2}$ doesn't count)?

If there is an example of an irrational number $y$ for every irrational number $x$ that will sum to a rational number, does this property apply to all irrational numbers?

答:
For any $x \in \mathbb R$ and any $q \in \mathbb Q$, since $\mathbb R$ contains $\mathbb Q$, $y = q-x$ will satisfy $y+x\in \mathbb Q$.

For any algebraic number, it is possible to find the minimal polynomial over $\mathbb Q$ and thus the algebraic conjugates of the number, a pile of numbers that, along with $x$, both add up to a rational number and multiply together to a rational number!

Bad news though: the algebraic conjugate of $\sqrt{2}$ specifically is $-\sqrt{2}$.


答:
You have four maps, all equivalent under index-raising/lowering:

$A \doteq A^{\rm operator}: V \to V$. This is described by $Ae_j = A^i_je_i$

$A^{(1,1)}\colon V^* \times V \to \Bbb R$, given by $A^{(1,1)}(f,v) = f(Av)$. This is described by $A^{(1,1)} = A^i_j e_i \otimes e^j$.

$A^{(0,2)}\colon V \times V \to \Bbb R$, given by $A^{(0,2)}(v,w) = \langle Av,w\rangle$. This is $A^{(0,2)} = A_{ij}e^i\otimes e^j$.

$A^{(2,0)} \colon V^*\times V^* \to \Bbb R$, given by $A^{(2,0)}(f,g) = A^{(0,2)}(f^\sharp, g^\sharp)$, where $\sharp\colon V^* \to V$ is induced by the metric. This is $A^{(2,0)} = A^{ij} e_i\otimes e_j$.

Title: Basic question (clarification) on tensor notation of a matrix: Up-Down or Down-Down?

Body:
Easy to muddle through while reading along, but a recurrent source of frustration for me (Wikipedia reference):

$A_{ij}$ would seem to suggest that we are writing $A_{ij}\;e^i\otimes e^j$. But this makes no sense, since a matrix would "eat" one vector and one covector.

On the other hand $A^{i}{}_{j}$ would make more sense interpreted as an object $A^{i}{}_{j} \;e_i \otimes e^j$ that would take in one vector and one covector. In the case of a vector in Euclidean coordinates:

$$\begin{align}
 A^{i}{}_{j}v^j &= u^i\\[2ex]
\text{or}\\[2ex]
\left( A^{i}{}_{j} \;e_i \otimes e^j \right) v^j e_j &= u^i e_i
\end{align}$$

Likewise, the matrix product can be expressed as

$$\begin{align}
 A^{i}{}_{j} B^{j}{}_k &= C^{i}{}_k\\[2ex]
\text{or}\\[2ex]
A^{i}{}_{j} \;e_i \otimes e^j \, B^{j}{}_k \; \;e_j \otimes e^k &= C^{i}{}_k \;e_i \otimes e^k
\end{align}$$

So are $A_{ij}$ and $A^{i}{}_{j}$ at all compatible and equivalent, or is $A_{ij}$ misleading when one considers the basis vectors?

Context:

I am confused by this passage in page 9 of Introduction to Tensor Analysis by H.D. Block:

If

$$ {\bf{\bar e_j}} = a^i{}_j {\bf{ e_i}}$$

is the $j$-th entry in a vector in the new basis, and the transformation matrix is

$$A=\begin{bmatrix}
a^1{}_1 & a^1{}_2 & \cdots & a^1{}_n\\
a^2{}_1 & a^2{}_2 & \cdots & a^2{}_n\\
\vdots & \vdots  & \ddots & \vdots\\
a^n{}_1 & a^n{}_2 & \cdots & a^n{}_n\\
 \end{bmatrix}$$

I can't see the notation in

$$ {\bf{\bar e_*}} = A^\top {\bf{ e_*}}$$


问:
Why is $(k L - y) \mod L = L - y \mod L$?
Why is $(k L - y) \mod L = (L - y) \mod L$? What's the rule that allows the constant $k$ to be ignored?

答: In your notation $a\pmod b = c\pmod b$ means that $a-c \pmod p =0 \pmod p$ this is just a consequence of the division algorithm. By doing that, notice that the $-y$ cancel and you end up having a $(k-1)\cdot L \pmod L,$ what is the remainder by dividing a number by itself? hint: $L = 1\cdot L +0.$


问:
Why I am not getting all solutions for $2\cos^2x+3\sin x=0$?
The question is Find the principle solution for

$$2\cos^2x+3\sin x=0$$

so, $\sin x=\frac{-1}{2}$ and $\alpha=-\frac{\pi}{6}$.

So plugging it in the formula

$$2n\pi+\alpha$$

I am only getting $\frac{11\pi}{6}$ .

My teacher said this form covers all solutions but I am getting only one of the two, Is he wrong?

答: The general solution of $\sin x=\sin p$ is $x=n\pi+(-1)^np,n\in\Bbb Z$. Plugging $p=-\pi/6$ gives two principal solutions $2\pi-\pi/6$ and $\pi+\pi/6$.


问:
Is there a way to prove that the results of quadratic equation are always odd?
Given the following quadratic equation:

$$f(x) = x^2-3x-1$$

For $x, 1\leq{x}\leq{100}, x \equiv 1\mod 2$

Is this true for all $x \in{\mathbb{Z}}$? If so, is there a way to prove it?

Body:
If $x = 2k + 1$, then

$$x^2 - 3k - 1 = (2k + 1)^2 - 3(2k + 1) - 1 = 2k(2k - 1) -3 \equiv 1 \pmod{2},$$

so the result is always odd.


问:
$A^3+A=0$ We need to show $\mathrm{rank}(A)=2$

Let $A\ne 0$ be a $3\times 3$ matrix with real entries such that $A^3+A=0$. We need to show $\mathrm{rank}(A)=2$.

$\det A(A^2+I)=0\Rightarrow\det A=0\Rightarrow \mathrm{rank}(A)<3$,  Suppose $\mathrm{rank}(A)=1$, Then I showed one matrix with rank $1$ which do not satisfies the given relation, is my answer is ok? Thank you for help and discussion

答: By the Cayley Hamilton theorem, the minimal polynomial of a square matrix divides its characteristic polynomial. Therefore, since $A \neq 0$ by assumption, $(x^2 + 1) | f_A(x)$ which means that $A$ has at least the eigenvalues $\pm i$. Since these are distinct eigenvalues, the corresponding eigenvectors are linearly independent. Since they lie in the column space of $A$, the rank of $A$ is at least $2$. As you mentioned, a determinant argument can be used to show that it is less than $3$.</p>


答:
We have that

$$2 \cos^2x+3\sin x=0 \iff 2 \sin^2x-3\sin x-2=0 \implies \sin x = -\frac12$$

which has two families of solutions

$$x=-\frac \pi 6 +2k\pi \quad \lor \quad x=\frac 7 6 \pi +2k\pi$$

Title: Why I am not getting all solutions for $2\cos^2x+3\sin x=0$?

Body:
The question is Find the principle solution for

$$2\cos^2x+3\sin x=0$$

so, $\sin x=\frac{-1}{2}$ and $\alpha=-\frac{\pi}{6}$.

So plugging it in the formula

$$2n\pi+\alpha$$

I am only getting $\frac{11\pi}{6}$ .

My teacher said this form covers all solutions but I am getting only one of the two, Is he wrong?


问:
By using primitive roots how does one solve $x^2 \equiv -1 \pmod p$ for $x$, given prime $p$.
By using primitive roots how does one solve $x^2 \equiv -1 \pmod p$ for $x$, given prime $p$ (when solvable).

In particular I want to solve for very large $p$ efficiently.

答:
This uses Euler's Criterion

it is not necessary to find a primitive root. First we find a quadratic nonresidue $\pmod p.$  That is, an $a$ such that Legendre symbol
$$ ( a,p) = -1  $$
Stan Wagon points out that the search for a nonresidue is quick. This is in a column in February 1990 called The Euclidean Algorithm Strikes Again.

Then $$  a^{\frac{p-1}{2}} \equiv -1 \pmod p $$
while $\frac{p-1}{2}$ is even. Thus
$$ \left(  a^{\frac{p-1}{4}} \right)^2 \equiv -1 \pmod p $$


问:
How is $g(12)(34)g^{-1} = (g(1) g(2))(g(3)g(4))$
Firstly I think (please confirm) the notation $g(x)$ (for $x \in \{1,2,3,4\}$) caters to the position $y \in \{1,2,3,4\}$ such that $x \to y$.

Is it true that $g((12)) = (g(1) g(2))$? I can not wrap around my head if this is even defined. If so, an explanation would be really helpful.

How is $g(12)(34)g^{-1} = (g(1) g(2))(g(3)g(4))$ where $g \in S_4$?

I am fine if hints or heuristics are provided so I can think how it works. A definition for $(1)$ would enable me to think $(2)$.

答:
The definitions you may be missing: On the set $A=\{1,2,3,4\}$:

$g$ is any permutation (i.e. any bijection $A\to A$). $g^{-1}$ is its inverse.

$(1 2)$ is the permutation that swaps $1$ with $2$ (i.e. maps $1$ to $2$, $2$ to $1$, $3$ to itself and $4$ to itself). Similarly, $(3 4)$ is the permutation that swaps $3$ with $4$. Similarly again, $(g(1) g(2))$ is a permutation that swaps $g(1)$ with $g(2)$ - whatever those two elements of $A$ are. I am leaving to you to guess what is $(g(3) g(4))$.

Multiplying permutations means composing them into new permutations by applying the composed permutations one after another, right-to-left. This means: if $g, h$ are permutations, then $gh$ is a permutation such that $gh(x)=g(h(x))$.

Does this help? Please try to take it from there, and the rest of the solution is hidden in the spoiler below.

 Let's see how $g(1 2)(3 4)g^{-1}$ works. What does it map the element $g(1)\in A$ into? Remember we apply the maps right-to-left, so, first, $g^{-1}$ acts on $g(1)$, mapping it into $1$. Then, $(3 4)$ maps $1$ into itself. Then, $(1 2)$ maps $1$ into $2$. Finally, $g$ maps $2$ into $g(2)$. Can you finish this now by proving that $g(2)$ also maps to $g(1)$, that $g(3)$ maps to $g(4)$ and $g(4)$ maps to $g(3)$? In other words, the whole permutation swaps $g(1)$ with $g(2)$ and $g(3)$ with $g(4)$, which is the same thing that $(g(1) g(2))(g(3) g(4))$ does.


问:
To prove $EA = FB$ or that $CQ'$ is radical axis
Given disjoint circles $c_1 = \odot(P,PA), c_2 = \odot(O,OB)$ such that $B$ and $A$ are in the same half-plane wrt $OP$ and that $PA \parallel OB \perp OP$.

Line $CDQ$ is the perpendicular bisector of $AB$, $D \in AB,Q \in OP$.

Point $Q'$ is the reflection of $Q$ wrt to $M$, the midpoint of $OP$.

Point $C$ on line perpendicular to $OP$ in $Q'$.

$E = CA \cap c_1, F = CB \cap c_2$.

I want to prove $EA = FB$.

We know that $DC = DQ$ because $DM$ is midsegment of trapezium $OBAP$.

答:
Instead of starting with $MQ'=MQ$ as a given condition, and trying to prove the equality of tangents, in order to show that $CQ'$ is the radical axis, this approach starts from the equality of tangents $LH$, $LJ$ and proves $MQ'=MQ$, and hence that $CQ'$ is the radical axis.

$DQ$ is perpendicular bisector of $AB$, $M$ is midpoint of $OP$, as in the posted question.

$BA$ and $OP$ extended meet at $G$.  $HJ$, tangent to the circles at $H$, $J$, is known to concur with $OP$ and $BA$ at $G$.

From $L$, the midpoint of $HJ$, drop $LQ'$ perpendicular to $OP$, and extend it upward to meet $QD$ at $C$.

Join $HO$, $DM$, $ML$, $PJ$.

We prove$$QM=Q'M$$

Since $M$, $L$ bisect $OP$, $HJ$, and $OH\perp HJ$, then $$ML\perp LJ$$

Therefore$$\triangle LMQ'\sim\triangle GML\sim\triangle GOH$$and$$\triangle DQM\sim\triangle GQD\sim\triangle GBO$$But$$OH=OB$$therefore$$QM=Q'M$$

And $LH$, $LJ$ are equal tangents. Therefore $L$ lies on the radical axis.

Repeating the same construction and argument in the lower halfplane, we get a second point $L'$ on $LQ'$ extended. Therefore $CQ'$ is the radical axis.


问:
Probability that in a group of 25 people there are at least 10 unique birth months
I am trying to find the probability that in a group of 25 people there are at least 10 unique birth months using the principle of inclusion exclusion.

Principle of inclusion exclusion

$\bigcup_{i=1}^{n} A_{i}\left|=\sum_{i=1}^{n}\right| A_{i}\left|-\sum_{1 \leq i<j \leq n}\right| A_{i} \cap A_{j}\left|+\sum_{1 \leq i<j<k \leq n}\right| A_{i} \cap A_{j} \cap A_{k}\left|-\cdots+(-1)^{n-1}\right| A_{1} \cap \cdots \cap A_{n} \mid$

I approached the problem as follows.

Let $M_1,... M_{12}$ represent the months January, ..., December.

Let $M_k^{c}$ be the event that no one is born in $M_k$. Then $P(M_k)^{c} = (11/12)^{25}$. Similarly, the probability that no one is born in any of x months is $((12-x)/12)^{25}$.

Using the principle of inclusion exclusion,

$$P(M_{1}^{c} U M_{2}^{c} U ... U M_{10}^{c}) = {10 \choose 1}(11/12)^{25} - {10 \choose 2}(10/12)^{25} +{10 \choose 3}(9/12)^{25} - {10 \choose 4}(8/12)^{25} + ... -{10 \choose 10}(2/12)^{25}$$

The probability that at least one person is born in each of the months $M_{1}$ through $M_{10}$ is 1 - $P(M_{1}^{c} U M_{2}^{c} U ... U M_{10}^{c})$.

I wrote the following Matlab code to evaluate this expression. The output is approximately 0.25.

```<code>fprintf("%.60f\n",findp2(25))
function output = findp2(p)
    %the output is the probability that there are at least 10 unique
    %birthmonth among p people
pc = 0;
for k = 1:10
pc = pc + (-1)^(k-1)*nchoosek(10,k)*((12-k)/12)^(p);
end
format long
output = 1-pc;
end
</code>```

Because this is the probability for 10 specific months, I multiplied this by ${12 \choose 2}$ to obtain 16.5 as the probability that there are at least 10 unique birth months among 25 people.

Can someone explain what I am doing wrong?

答:
"Can someone explain what I am doing wrong?"  You correctly applied inclusion-exclusion for the first part when talking about the probability at least one of the months Jan-Oct were missing.  You then completely ignored inclusion-exclusion when talking about the probability there was at least some group of ten months who all appeared, having just multiplied by $\binom{12}{2}$.  You should have then applied inclusion-exclusion on this as well, subtracting $12$ times the probability that some eleven months all appeared and then adding the probability that all twelve months appeared.

As for a more streamlined calculation (but requires stronger machinery), consider using Stirling Numbers of the Second Kind.  There are ${{25}\brace{10}}\cdot 12\frac{10}{~}$ ways to break $25$ people into $10$ nonempty groups and assign a month to each of those groups.  Similarly for $11$ months it would have been ${{25}\brace{11}}\cdot 12\frac{11}{~}$ and all $12$ months it would have been ${{25}\brace{12}}\cdot 12!$.  Adding these and then dividing the result by $12^{25}$ completes the problem, giving:

$$\dfrac{{{25}\brace{10}}\cdot 12\frac{10}{~}+{{25}\brace{11}}\cdot 12\frac{11}{~}+{{25}\brace{12}}\cdot 12!}{12^{25}}\approx 0.886841\dots$$

Notation notes:  ${n\brace k}$ represents the Stirling Number of the Second Kind denoting the number of ways $n$ distinct objects can be partitioned into $k$ unlabeled nonempty parts.  $n\frac{k}{~}$ is another way of writing the falling factorial $\frac{n!}{(n-k)!}$ and denotes the number of injective functions from a set with $k$ elements to a set with $n$ elements.


问:
Determining if a set is a $\mathfrak{A}$-big
The Problem

My professor gave us some homework exercises and there is one my colleague and I could not solve.
First the definitions:

Let $A$ be a non-empty set and $R \subseteq A \times A$ a relation. Elements of the relation $R$ will be denoted as $xRy$, the shorthand notation for $\langle x,y\rangle \in R$. Let $\mathfrak{A}$ denote the structure of $(A,R)$. Let $V$ be a subset of $A$. We call $V$ $\mathfrak{A}-$small if and only if $\exists_{a \in A} [V = \{b \in A \mid bRa\}]$. $V$ is $\mathfrak{A}-$big iff $V$ is not $\mathfrak{A}-$small.

The exercise showed us some sets $V$ and most of them were determined by us to be $\mathfrak{A}-$big. However, this last problem is giving the headaches

Determine if $$V = \{a\in A \mid \neg\exists_{f:\mathbb{N} \to A}[\forall_{n \in \mathbb{N}}[f(n+1)Rf(n)] \land f(0) = a]\}$$
is $\mathfrak{A}-$big or not. Prove the result.

Attempts

First of all, we believe that $V$ is $\mathfrak{A}-$big.
We tried to get to a contradiction when assuming that $V$ is $\mathfrak{A}-$small. There must therefore be an $a \in A$ such that $V = \{b\in A \mid bRa\}$.

We then made a case distinction: $a \in V$ and $a\notin V$.
We managed to determine that if $a\notin V$ then a contradiction occurs and therefore $a \in V$.
If you assume that $a \notin V$ then there is a function $f$ for which $f(0) = a$ and $\forall_n[ f(n+1)Rf(n)]$. It must hold then that $f(1)Ra$ and thus $f(1) \in V$. But we can construct a function $g:\mathbb{N} \to A$ as $n \mapsto f(n+1)$. It can now be seen that $g(0) = f(1)$ and $g(n+1)Rg(n)$ (This last one because $f(n+2)Rf(n+1)$. Thus $f(1)$ cannot be in $V$ so $a$ must be in $V$.

We tried something similar for the assumption that $a \in V$ but we soon realized that this method wouldn't work because we now cannot determine an element which is not $a$ to be in $V$.
Even if we assume more statements (such as $\forall_n [f(n+1)Rf(n)]$), we ended up in a dead end because we need more assumptions we cannot rectify.

Can someone give us a hint into the right direction?

Thank you in advance!

答: If $a\in V$, then $aRa$ by your choice of $a$.  Now consider the function $:\mathbb N\to A$ that is constant with value $a$.


问:
Can two fields that have monomorphism to each other not be isomorphic?
More precisely, let $F_1, F_2$ be fields, and we have none zero homomorphisms $f: F_1 \to F_2$ and $g: F_2 \to F_1$ (actually both are monomorphism since $F_1, F_2$ are fields), is there a possibility that $F_1$ is not isomorphic to $F_2$?

答:
To elaborate on the example given by perpetuallyconfused, indeed $\mathbb C$ and $\mathbb C(x)$ provide a counterexample.

Some additional elements.

$\mathbb C$ is algebraically closed: this is well known. $\mathbb C(x)$ is not. In particular the polynomial $p(t) = t^2-x \in \mathbb C(x)[t]$ can't have a root $\frac{r(x)}{s(x)} \in \mathbb C(x)$. If it was the case, you would have $r^2(x)=x s^2(x)$ with the contradiction that the left polynomial of the equality is having an even degree and the right one an odd one. Therefore $\mathbb C$ and $\mathbb C(x)$ are not isomomorphic.

Also, the identity is an obvious embedding $\mathbb C \hookrightarrow \mathbb C(x)$.

Regarding an embedding $\mathbb C(x) \hookrightarrow \mathbb C$, you have to know that Two algebraically closed fields are isomorphic if and only if they have the same transcendence degree over their prime fields (proof provided in the link). And also that the cardinality of the algebraic closure of an infinite field $F$ has the cardinality of $F$. As the cardinality of $\mathbb C(x)$ is the one of $\mathbb C$, the algebraic closure $\overline{\mathbb C(x)}$ of $\mathbb C(x)$ is isomorphic to $\mathbb C$ and therefore you can embed $\mathbb C(x)$ into $\mathbb C$.


问:
How to approach the negative root of the series $\sum\limits_{k=1}^{\infty} \sigma_1(k) x^k$
Let's define
$$
f(x) = \sum\limits_{k=1}^{\infty} \sigma_1(k) x^k, \quad \text{where } \sigma_1(k) = \sum\limits_{d|k}d
$$

Is there a way to approach the root $x_0$ of this series, where $-1 < x_0 < 0$ ? I tried to squeeze it and using geometric series, but its too far away.

Body:
With the help of Mathematica I defined
$$f(x)=\sum _{k=1}^{1000} \sigma _1(k) x^k$$
then  I applied the Newton's method starting from $x_0=-0.411248$

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

$$
\begin{array}{r|l}
n   &x_n\\
\hline
0 &-0.411248\\
1 &-0.4112484791789996\\
2 &-0.4112484791779548\\
3 &-0.4112484791779548\\
4 &-0.4112484791779547\\
5 &-0.4112484791779548\\
\end{array}
$$

bonus

Define Pochhammer symbol
$$Poch(q)=\prod _{k=0}^{\infty } \left(1-q^{k+1}\right)$$
The root of $Poch'(q)$ in the interval $(-1,0)$ is the same as the root of $\sum _{k=1}^{\infty} \sigma _1(k) x^k$ in the same interval.


问:
How many different roots does $p(x) = x^{12}+2x^{6}+1 \in \mathbb{F}_3[x]$ have?
I want to find out how many different roots does $p(x) = x^{12}+2x^{6}+1 \in \mathbb{F}_3[x]$ have (not necessarily in $\mathbb{F}_3$, but in an body that has a subgroup isomorphic to $\mathbb{F}_3$), where $\mathbb{F}_3$ is the finite body with $3$ elements ($\cong \mathbb{Z}_3$).

$p'(x)=12x^{11}+12x^{5} \equiv 0$ so $p(x)$ has roots of multiplicity $\geq 2$. I actually know that the roots with multuplicity $\geq 2$ are the zeroes of g.c.d$(p(x), p'(x))$ = g.c.d$(p(x), 0)=p(x)$. Therefore $p(x)$ only has roots with multiplicity $\geq 2$.

$p(x)$ doesn't have any roots in $\mathbb{F}_3$ so I have to build a body where $p(x)$ has at least a root. For example
$$\mathbb{F}_3[T]/\langle T^{12}+2T^6+1\rangle$$
I know that here $\overline{T}$ is a root. Using Frobenius' homomosphism I get that $\overline{T}^3, \overline{T}^9, \overline{T}^{27} = 2\overline{T}^9$ are roots. If I use Frobenius once again, I go back to $\overline{T}^9$. Does this mean I've found all the roots? How can I be sure there aren't any more roots?

答:
HINT:
In $\mathbb{F}_3[x]$
$$p(x) = x^{12}+2x^{6}+1 = (x^{6} + 1)^2$$
How many roots do you expect to find in a splitting field, counting multiplicity?


问:
Proving whether or not a certain recursively defined sequence is a Cauchy sequence or not
Question:

Let $\{x_{n}\}$ be a recursively defined sequence defined as:
$$x_{1} = \frac{1}{2}$$
$$x_{n+1} = \frac{1-x_{n}}{4}$$

I want to show first that $|x_{n+1}-x_{n}| \le \frac{1}{4}|x_{n}-x_{n-1}|$. Using this, I have to prove that the sequence is Cauchy and then find its limit. Here is what I have done:

Part 1:

$$|x_{n+1}-x_{n}| = \left|\frac{1-x_{n}}{4} - \frac{1-x_{n-1}}{4}\right| \\
=\left|\frac{-x_{n}}{4} + \frac{x_{n-1}}{4}\right|
= \frac{1}{4}|-x_{n}+x_{n+1}|$$

But, I do not know how to go from this equality to the inequality that the question has. How may I manipulate the equation so that the $=$ changes to a $\le$?

Part 2:

Now as for the second subpart where I have to show that the sequence is Cauchy, I am not sure where to start, so I started with the definition of Cauchy sequence, which is as follows:

A sequence $\{a_{n}\}$ is called a Cauchy sequence if $\forall \epsilon>0 \exists      N \in \mathbb{N}$ s.t. if $m,n \ge N$, then
$$|a_{m}-a_{n}| < \epsilon$$
But the problem I face is that if I try to use this definition and try to apply it here, I am getting that:
$$x_{m}-x_{n} = \left|\frac{-x_{m-1}}{4}+\frac{x_{n-1}}{4}\right|$$
But after this, how can I use my previous result to prove that this is always bounded by $\epsilon$?

Could someone help me?

答:
Let $f : [0,1] \rightarrow [0,1]$ be the function defined for all $x \in [0,1]$ by $$f(x)=\frac{1-x}{4}$$

First, $f$ is well-defined because it is decreasing over $[0,1]$, and $f(0), f(1) \in [0,1]$. Because $x_1 = 1/2 \in [0,1]$, you deduce by an easy induction that the sequence $x_{n+1} = f(x_{n-1})$ is well-defined and takes all its values in $[0,1]$.

For any integer $n \geq 1$, the MVT applied to the function $f$ between $x_n$ and $x_{n-1}$ gives $$|x_{n+1}-x_n|=|f(x_n)-f(x_{n-1})| = \frac{1}{4}|x_n -x_{n-1}| \quad \quad (*)$$

(because $f' \equiv -1/4$ over $[0,1]$). This is the first part.

Now, for every integers $q>p$, you can see by an easy induction (using the equality $(*)$), that

$$|x_q-x_p|= \frac{1}{4^{p-1}}|x_{q+1-p}-x_1| \leq \frac{1}{4^{p-1}}$$

This proves that the sequence is Cauchy, and you are done.


问:
Efficient Evaluation of the Expression
I have stumbled upon this problem and couldn't really get to an efficient approach :

Problem:

We have an array $P$ having $N$ elements denoted by $P_i$. Our task is to find the value of the below expression.

$$X=\sum_{j=1}^k \prod_{i=1}^N(P_i-j)$$
where $k$ is the minimum element in the array. We have to return the value of X modulo $10^9+7$.

Constraints:

$1 \le P_i \lt 10^9+7$
$ 1 \le N \le 10^5 $

The only solution that I have been able to come up with is a brute force approach of complexity $O(N\times K)$ where $K$ is the minimum element in the array. I think maybe something like FFT or generating functions could be used but I couldn't understand how.

Any sort of help is appreciated!!

答:
Let $ f(x) = \prod_{i=1}^N (-1)( x - P_i) \sum_{i=0}^N a_i x^i$.

We want to calculate $ \sum_{x=1}^k f(x),$ which is a lot easier to do termwise with a change of variable, namely:

$$ = \sum_{x=1}^k \sum_{i=0}^N a_i x^i = \sum_{i=0}^N \left( a_i \sum_{x=1}^k x^i\right), $$

This is already quite a big simplification.

Note that $ \sum_{x=1}^k x^i$ could be further simplified via Faulbaher's formula.

Title: Efficient Evaluation of the Expression