问:
$\frac{a^{2}-1}{b+1}+\frac{b^{2}-1}{a+1}$ an integer $\Rightarrow \frac{a^{2}-1}{b+1}$ and $\frac{b^{2}-1}{a+1}$ are integers.
If $\frac{a^{2}-1}{b+1}+\frac{b^{2}-1}{a+1}$ is an integer, prove that also $\frac{a^{2}-1}{b+1}$ and $\frac{b^{2}-1}{a+1}$ are integers.

By doing the math, I get $\frac{(a-1)(a+1)^{2}+(b-1)(b+1)^{2}}{(a+1)(b+1)}$ is an integer which means that $(a+1)(b+1)$ divides $(a-1)(a+1)^{2}+(b-1)(b+1)^{2}$.

but I don't know how to continue.

By the way, this is not homework or anything. I just found it in a book.

答:
The problem states that, for some natural numbers $a$ and $b$, there's an integer $i$ where

$$\frac{a^2 - 1}{b + 1} + \frac{b^2 - 1}{a + 1} = i \tag{1}\label{eq1A}$$

If the first term is an integer, say $\frac{a^2 - 1}{b + 1} = j$, then $\frac{b^2 - 1}{a + 1} = i - j$ is also an integer. Doing the same thing with the second term instead shows that if either term is an integer, the other one is as well.

Assume neither left side term in \eqref{eq1A} is an integer. As you did, combining the terms using a common denominator gives

$$\frac{(a - 1)(a + 1)^{2} + (b - 1)(b + 1)^{2}}{(a + 1)(b + 1)} = i  \tag{2}\label{eq2A}$$

Since $a + 1 \mid (a - 1)(a + 1)^{2}$, this means $a + 1 \mid (b - 1)(b + 1)^{2} = (b + 1)(b^2 - 1)$. However, since it's assumed $a + 1 \not\mid b^2 - 1$, this means there's a prime $p$ where, using the $p$-adic order function, we have

$$\nu_{p}(a + 1) \gt \nu_{p}(b^2 - 1) \tag{3}\label{eq3A}$$

but

$$\nu_{p}(a + 1) \le \nu_{p}((b + 1)(b^2 - 1)) \tag{4}\label{eq4A}$$

This means $p \mid b + 1$. Define

$$\nu_{p}(a + 1) = k_1 \tag{5}\label{eq5A}$$

$$\nu_{p}(b + 1) = k_2 \tag{6}\label{eq6A}$$

If $p \neq 2$, then $p \not\mid b - 1$ and $p \not\mid a - 1$. Thus, \eqref{eq3A} gives

$$k_1 \gt k_2 \tag{7}\label{eq7A}$$

Since $\nu_{p}((a - 1)(a + 1)^2) = 2k_1$ and $\nu_{p}((b - 1)(b + 1)^2) = 2k_2$, this means with the numerator of \eqref{eq2A},

$$\nu_{p}((a - 1)(a + 1)^{2} + (b - 1)(b + 1)^{2}) = 2k_2 \tag{8}\label{eq8A}$$

However, with the denominator,

$$\nu_{p}((a + 1)(b + 1)) = k_1 + k_2 \gt 2k_2 \tag{9}\label{eq9A}$$

Thus, the fraction can't be an integer. Since it is an integer, this means the assumption $p \neq 2$ must be incorrect, i.e., $p = 2$. Thus, $a$ and $b$ are odd, with $p \mid a - 1$ and $p \mid b - 1$. Define

$$\nu_{p}(a - 1) = m_1 \tag{10}\label{eq10A}$$

$$\nu_{p}(b - 1) = m_2 \tag{11}\label{eq11A}$$

Using \eqref{eq6A} and \eqref{eq11A} in \eqref{eq3A} gives

$$k_1 \gt k_2 + m_2 \tag{12}\label{eq12A}$$

Note the same arguments as above could have been used to check $b + 1$ not dividing into $a^2 - 1$, with it also reaching the conclusion the prime involved must be $2$. Thus, this would give \eqref{eq12A} but with the $1$ and $2$ indices switched around, i.e.,

$$k_2 \gt k_1 + m_1 \tag{13}\label{eq13A}$$

Combining \eqref{eq12A} and \eqref{eq13A} gives

$$k_1 \gt k_2 + m_2 \gt (k_1 + m_1) + m_2 \implies 0 \gt m_1 + m_2 \tag{14}\label{eq14A}$$

This is not possible since $m_1$ and $m_2$ must be positive integers. Thus, the original assumption of the two left terms in \eqref{eq1A} not being integers must be incorrect, i.e., they are both integers.


问:
Pontryagin dual for non abelian $\widehat G := \operatorname{Hom}(G, T) $?
For a locally compact abelian group  $G$ , the Pontryagin dual  is the group  $\widehat G$  of continuous  group homomorphisms  from $G$  to the circle group  $T$ . That is,
:

$$\widehat G := \operatorname{Hom}(G, T). $$

The Pontryagin dual $\widehat G$ is usually endowed with the  topology  given by uniform convergence on  compact sets (that is, the topology induced by the  compact-open topology  on the space of all continuous functions from $G$ to $T$).

For example,

$$\widehat {\mathbb Z} = T,\  \widehat {\mathbb R} = \mathbb R,\ \widehat T = \mathbb Z.$$

questions

do we have a relation that
$$ \operatorname{Hom}(G_1, G_2)= \operatorname{Hom}(G_2, G_1)?$$
for any groups? I think the answer is no. If no, can you give the reasonings behind why and for counter examples? (for abelian groups examples? for non-abelian groups examples?)

How do we define $\widehat G$ for non-abelian $G?$ Is that $\widehat SU(2)=SO(3)$ and $\widehat SO(3)=SU(2)$?

答:
For abelian groups $G_1$  and $G_2$, the map
$$
  \varphi  \in  \text{Hom}(G_1, G_2) \mapsto  \hat \varphi \in \text{Hom}(\hat G_2, \hat G_1),
  $$
given by
$$
  \hat \varphi (\gamma ) = \gamma \circ \varphi , \quad \forall \gamma \in  \hat G_2,
  $$
is a bijection.  Now, for every finite abelian group $G$ one has that  $\hat G$ is isomorphic to $G$.  So, if $G_1$ and $G_2$ are
finite abelian, one gets a bijection
$$
  \text{Hom}(G_1, G_2) \buildrel \widehat{} \over \longrightarrow \text{Hom}(\hat G_2, \hat G_1) = \text{Hom}(G_2, G_1).
  $$

Besides finite abelian groups, things fail miserably.

Without finiteness: Notice that $\text{Hom}(\mathbb T, \mathbb Z_2)$ has only one element,  namely the trivial homomorphism,  but in
$\text{Hom}(\mathbb Z_2, \mathbb T)$ one may find the trivial homomorphism in addition to the map sending the generator to
$-1$.

Without commutativity: Since the alternating group $A_n$ is simple for $n\geq 5$, one has that
$\text{Hom}(A_n, \mathbb Z_2)$ only has  the trivial homomorphism, while
$\text{Hom}(\mathbb Z_2, A_n)$ has as many elements as order-two even permutations.

Regarding the dual of non-abelian groups,  one usually defines $\hat G$ as the set of equivalence classes of irreducible unitary
representations on Hilbert spaces.  Here $\hat G$ does not have a group  structure  and the theory is simultaneously  a lot more
sophisticated and a lot less powerful than Pontryagin duality.

If $G$ is abelian, Schur's Lemma implies that every unitary irreducible representation
$$
  π:G→U(H)
  $$
is one-dimensional, that is,  $H$ is a one-dimensional Hilbert space.   In this case $U(H)=\mathbb T$, so $π$ is
actually a  homomorphism from $G$ to $\mathbb T$.  In other words,  the dual  reduces to the Pontryagin dual.

However,
for lots of non-abelian groups, e.g. $A_n$,  there are no nontrivial
one-dimensional representations, so
$
  \text{Hom}(G, \mathbb T)
  $
may be empty!

ContentLicense: CC BY-SA 4.0


问:
What is a subinterval?
I got a specific question, and I think I'm forgetting some of my terminology, specifically what a subinterval is. Here is the question

3 engineers work on a cluster which has a daily limit of 2 compute ’credits’. When the total daily amount of compute credits of the three engineers exceeds that daily limit, the company is charged a one thousand dollar penalty. Assume each engineer’s daily cluster usage in compute credits is distributed between 0 and 1 and that the chance the usage belongs in some subinterval of [0, 1] is equal to the length of the subinterval. Assume also that the daily usages of the engineers across the year are independent. Estimate the total cost over 100 days.

I changed the question around some because I want to solve it myself but needed to provide context.

There is one sentence here that I don't quite understand

```<code>Assume each engineer’s daily cluster usage in compute credits is distributed between 0 and 1
and that the chance the usage belongs in some subinterval of [0, 1] is equal to the length of the subinterval
</code>```

What is this sentence telling me? For the 1st part, do I assume normal distribution?
What does it the 2nd part mean, about the chance of the usage belonging in the subinterval? If the length is 1, then wouldn't the chance be 100%?

The sense I get is that each engineer uses somewhere between 0-1 usage every day and that usage is normally distributed, so on average they would use 1.5 credits a day. Is my thinking correct here?

答:
$J$ is a subinterval of $I$ if $J$ and $I$ are intervals and $J\subset I$.

So, taking the subinterval $[a,b]$ of $[0,1]$ (for $0\le a\le b\le 1$) that  sentence says that if $0\le a\le b\le 1$ then $P(a\le X\le b)=b-a$; in other words, $X$ is uniformly distributed on $[0,1]$.

Title: What is a subinterval?


问:
How to calculate the following probability from density function?
Suppose that the variable $X$ takes values on the interval $[0,2]$.

Its Probability density function here is:

$f(x) =c(2-x)^3.$

Question : How can I calculate the following possibility:

$P(1 < X < 4) $  ?

After integrated the density I got that:

$\int^2_0 f(x) =[−c(x−2)^4/4]_0^2 = 4c  =1$

From that I got:

$ c = 0.25$

So the Cumulative distribution function should be : $F(x) = -0.25  * (x-2)^4/4$

Now I know that $P(1 < X < 4)  = F(4) - F(1)$, but I cannot get the given solution ($0.06$)

I get $F(4) =-1$

and $F(1) = -1/16$

Any help would be appreciated!

答:
A distribution function cannot take negative values. You did not integrate properly. $F(x)=-c\frac {(2-t)^{4}} 4|_0^{x}=4c-c\frac {(2-x)^{4}} 4$. Since $X$ takes values in $[0,2]$ we have $X <4$ with probability $1$. Hence $P(1<X<4)=1-F(1)=\frac 1 {16}$.

$0.06$ is only approximate value for this probability.


问:
Confusion about rank, matrix, linear transformations
i have very recently started studying linear algebra, and I have the following doubt. I know that the image of a linear transformation T is span{T(v1),T(v2),...T(vn)}. Now, the rank of a linear transformation is the dimension of that span. I understand that it's equivalent to say that [1] : the rank of the matrix associated with that transformation is the dimension of the span of the columns of the matrix. What I don't understand is why I usually find on textbooks that [2] : the rank of a matrix is the number of linearly independent columns of the matrix. How can we go from [1] to [2] ?

答:
Let's say the columns of an $m\times n$ matrix $A$ are given as $\xi_1,\dots, \xi_n \in \Bbb{F}^m$. The span of its columns is the vector subspace $S=\text{span}\{\xi_1,\dots,\xi_n\}=\left\{\sum_{i=1}^nc_i\xi_i\,| \, c_1,\dots, c_n\in \Bbb{F}\right\}$.
The dimension of a vector space is by definition the size of one (hence any) basis. Now, what is a basis? A basis is a set of vectors which are linearly independent and which span the vector space.

In finite dimensions, if you have a spanning set (like $\{\xi_1,\dots, \xi_n\}$) then some subset of it must be linearly independent and it has the same span (this should be one of the first few theorems proven in any linear algebra book). In other words, some subset $\beta\subseteq \{\xi_1,\dots, \xi_n\}$ is actually a basis for $S$. So, $\dim S = |\beta|$ is the number of linearly independent columns of the matrix.


答:
One explanation is that this is a consequence of the dimension theorem, which states that any basis for the image of the transformation $T$ will have the same number of vectors.  With that in mind, let $S = \{T(v_1),\dots,T(v_n)\}$. I claim that the following is true as a consequence of the dimension theorem:

A linearly independent subset $S' = \{T(v_{i_1}),\dots,T(v_{i_k})\} \subsetneq S$ can be extended to a larger linearly independent of $S$ if and only if we have $\operatorname{span}(S) \subsetneq \operatorname{span}(S')$.

(Actually, this weaker statement is not too difficult to prove without using the dimension theorem).

With that in mind, let $d = \dim \operatorname{span}(S)$. There will necessarily exist a linearly independent set $S' \subseteq S$ with $d$ elements (why?). In fact, this set $S'$ of $d$ elements must form a basis for the image of $T$, i.e. $\operatorname{span}(S)$ (why?). Moreover, any set with more than $d$ elements cannot be a basis for the image of $T$ (why?).

So, it is indeed the case that the dimension $d$ of the image of $T$ is equal to the maximal size among the linearly independent subsets of $S$.

Title: Confusion about rank, matrix, linear transformations

Body: i have very recently started studying linear algebra, and I have the following doubt. I know that the image of a linear transformation T is span{T(v1),T(v2),...T(vn)}. Now, the rank of a linear transformation is the dimension of that span. I understand that it's equivalent to say that [1] : the rank of the matrix associated with that transformation is the dimension of the span of the columns of the matrix. What I don't understand is why I usually find on textbooks that [2] : the rank of a matrix is the number of linearly independent columns of the matrix. How can we go from [1] to [2] ?


问:
Property of Hom-functor
How to prove $$\operatorname{Hom}_{R}(A,\operatorname{Hom}_{\mathbb{Z}}(R,B))\cong \operatorname{Hom}_{\mathbb{Z}}(A,B)$$ where $R$ is a commutative ring, $A$ an $R$-module and $B$ an abelian group?

Is the following proof valid?

By adjunction of hom and tensor, we know, $\operatorname{Hom}_{R}(A,\operatorname{Hom}_{R}(M,B))\cong \operatorname{Hom}_{R}(A\otimes_{R}M,B)$.
Replacing $M$ by $\mathbb{Z}$, $A\otimes_{R}\mathbb{Z}\cong A$ (can we consider $\mathbb{Z}$ as a $R$-module?) and using $\operatorname{Hom}_{R}(\mathbb{Z},B)\cong \operatorname{Hom}_{\mathbb{Z}}(R,B)$ (is this valid?) we have the above result.

答:
Find explicit isomorphisms. I assume that $R$ is unitary.

If $f:A\to B$ is a $\Bbb Z$-morphism, then let $g(a):=r\mapsto f(ra)$.

If $g:A\to \hom_{\Bbb Z}(R,B)$ is an $R$-morphism, then $f(a):=g(a)\,(1)$.

Check that these defined mappings $g$ and $f$ are as stated, and that $f\mapsto g$ and $g\mapsto f$ inverses to each other.

ContentLicense: CC BY-SA 3.0

Title: Property of Hom-functor


问:
Is $U = V$ in the SVD of symmetric matrices?
Consider the SVD of matrix $A$:

$$A = U \Sigma V^\top$$

If $A$ is a symmetric, real matrix, is there a guarantee that $U = V$?

There is a similar question here that also posits $A$ is positive semi-definite. But I'm wondering whether $U$ would be equal to $V$ if $A$ is symmetric?

答:
No: if $\Sigma$ is diagonal with non-negative entries, then $U \Sigma U^T$ will necessarily be positive semidefinite. Indeed, we note that for any column vector $x$, we have
$$
x^\top(U\Sigma U^\top)x = (U^\top x)^\top \Sigma (U^\top x) \geq 0.
$$


问:
Teams in a Volleyball Tournament ;Prove that in the entire tournament, there is a team that has lost to no more than $4$ of the remaining $109$ teams.
$110$ teams participate in a volleyball tournament. Every team has played every other team exactly once (there are no ties in volleyball). Turns out that in any set of $55$ teams, there is one which has lost to no more than $4$ of the remaining $54$ teams. Prove that in the entire tournament, there is a team that has lost to no more than $4$ of the remaining $109$ teams.

I couldn't start, but I did take examples. It's Russia 2015 11.3.

答:
Suppose that the result is false, and let $m>55$ be minimal such that there is a set $T=\{t_1,\ldots,t_m\}$ of $m$ teams such that each team in $T$ has lost to at least $5$ teams in $T$. Let $k\in[m]=\{1,\ldots,m\}$, and let

$$L_k=\big\{i\in[m]\setminus\{k\}:t_i\text{ lost to at most }4\text{ teams in }T\setminus\{t_k\}\big\}\,.$$

The minimality of $m$ implies that $L_k\ne\varnothing$, and we let $f(k)=\min L_k$. Since $t_{f(k)}$ lost to at least $5$ teams in $T$ and at most $4$ teams in $T\setminus\{t_k\}$, it must be the case that $t_{f(k)}$ lost to $t_k$ and to exactly $4$ other teams in $T$.

Let $i\in[m]$, and suppose that $i=f(k)$; then $t_i$ lost to $t_k$. Moreover, $t_i$ lost to exactly $4$ other teams in $T$, so $t_i$ lost to just $5$ teams, and there are therefore at most five $k\in[m]$ such that $f(k)=i$. Let $T_0=\{t_{f(k)}:k\in[m]\}$; we’ve just shown that each team in $T_0$ lost exactly $5$ matches to teams in $T$. This means that the map $f$ is at most $5$-to-$1$, so $m=\big|[m]\big|\le 5|T_0|$, and

$$|T_0|\ge\frac{m}5\ge\left\lceil\frac{56}5\right\rceil=12\,.$$

Now consider just the $\binom{|T_0|}2$ matches between teams in $T_0$. Each of those matches resulted in a loss for some team, and there are $|T_0|$ teams, so the mean number of losses per team is

$$\frac1{|T_0|}\binom{|T_0|}2=\frac{|T_0|-1}2\ge\frac{11}2>5\,,$$

meaning that at least one team in $T_0$ lost more than $5$ matches to teams in $T$. This contradiction shows that no such $m$ exists, i.e., that the result is in fact true.


答:
In this question, $A$ is symmetric and therefore must be square.

However, in the general case, if $A$ is not a square matrix, it can be converted into a square matrix $AA^T$ or $A^TA$ and an eigendecomposition can be done on that to get the SVD components. The $V$ are eigenvectors of $A^TA$ and the $U$ are eigenvectors of $AA^T$. The singular values in $\Sigma$ are the square roots of the non-zero eigenvalues of $A$. This (the square root of the eigenvalues) is where the positive semi-definite condition for the SVD to be equivalent to the eigendecomposition comes from. See The SVD and the EVD in http://www.math.kent.edu/~reichel/courses/intr.num.comp.1/fall11/lecture7/svd.pdf for a more detailed explanation.

[Edited to incorporate Joppy's comment.]

Title: Is $U = V$ in the SVD of symmetric matrices?

Body:
Consider the SVD of matrix $A$:

$$A = U \Sigma V^\top$$

If $A$ is a symmetric, real matrix, is there a guarantee that $U = V$?

There is a similar question here that also posits $A$ is positive semi-definite. But I'm wondering whether $U$ would be equal to $V$ if $A$ is symmetric?


问:
Under what condition does $-(x-1)^3+x(a^2+b^2+c^2)-(a^2+b^2+c^2)=0$ have a positive solution $x$?

Consider the equation $$-(x-1)^3+x(a^2+b^2+c^2)-(a^2+b^2+c^2)=0$$ where $a,b,c$ are real numbers. Through what condition does $x$ have positive real solution?

Answer: $a^2+b^2+c^2<1$

How to prove this?

Thank you.

答:
False, perhaps typo in the query.

Let $k = a^2 + b^2 + c^2.$

The query factors into $(x-1)^3 - k(x-1) = 0.$

Counter example:

$x = 1 + \sqrt{3}, 1 = a = b = c \implies k = 3.$


问:
Truncating $p$-series
Let $p \in \mathbb{R}$. Series of the form $$S_p = \sum_{k \in \mathbb{N^*}} \frac{1}{k^p}$$ converge if and only if $p > 1$. Let us define $$S_{p, n} \triangleq \sum_{k = 1}^{n} \frac{1}{k^p},$$ the truncation of the $p$-series at its $n$-th term. (It is evident that $S_{p, n} \space \xrightarrow{n\to\infty} \space S_p$.) Is there a closed expression for $S_{p, n}$, at least for some values of $p$ (integers must be easier)?</p>

答: In Maple notation, for $p \ge 2$ integer, $$S_{p,n} = \zeta(p) + \frac{(-1)^{p-1}}{(p-1)!} \Psi(p-1,n+1),$$ where $\psi(m,\space·)$ is the $m$-th polygamma function.</p>

Title: Truncating $p$-series


问:
Questions about projective modules.
Let $P$ be a projective module and $M$ a submodule of $P$. We know that $M$ is also a projective module. Can we conclude that $P=M\oplus N$ for some module $N$? Thank you very much.</p>

答: This even fails when $P,M$ are free (advice: always check simple special cases before asking!), even when they are of rank $1$. Take any nontrivial principle ideal of an integral domain.</p>

Title: Questions about projective modules.


问:
Let $n \ge 3$. For $[a] \in (\Bbb Z /{2^n} \Bbb Z)^\times$ the relation $x^2 \equiv a \pmod {2^n}$ has either $0$ or $4$ residue solutions.

Supply a proof for the title question.

My work

I became interested in this after seeing lab bhattacharjee's answer to

$\quad$Find out all solutions of the congruence $x^2 \equiv 9 \mod 256$.

I've convinced myself that I can prove this using induction and elementary group theory, but it would be interesting to see any rigorous arguments.

If nobody supplies a proof, I will give one in a week or so.

答:
Using the fact that every nonempty subset of natural numbers has a least element, I can prove (minimal criminal technique) that

$\tag 1 x^2 \equiv 1 \pmod{2^n}$

has exactly $4$ solutions.

An alternate method is to attempt to 'position' a fifth solution to the known solutions;

see Bill Dubuque's proof.

Suppose that a solution $[b] \in (\Bbb Z /{2^n} \Bbb Z)^\times$ exists for

$\tag 2 x^2 \equiv a \pmod{2^n}$

Let $Q = \{1, 2^{n-1}-1, 2^{n-1}+1, 2^{n}-1\}$ be the $4$ solutions to $\text{(1)}$. Using elementary group theory we can show that the integers

$\quad b, (2^{n-1}-1)b, (2^{n-1}+1)b, (2^{n}-1)b$

represent $4$ distinct solutions to $\text{(2)}$.

If $c$ is any solution to $\text{(2)}$ then

$\quad \large c b^{-1} \in Q$

and so $\large (c b^{-1})b$ has already been accounted for.

This completes the proof.


问:
Sum of divisors in terms of their prime factors
The sum of divisors of a number is given by the following expressions:

$$\sigma(n)= \prod_{i=1}^l\left(1+p_i+{p_i}^2+··· +{p_i}^{a_i}\right)$$
$$\sigma(n)=\prod_{i=1}^l\left(\frac{{p_i}^{a_i+1}-1}{p_i-1}\right)$$

for

$$n=\prod_{i=1}^l{p_i}^{a_i}$$

What I'm having trouble understanding is how the first expression can be turned into the second, and while I'm sure the answer should be quite simple I can't see it. Could someone explain it to me?

答:
Using the formula for a geometric series (initial term $1$, ratio $p_i$),

$1+p_i+{p_i}^2+··· +{p_i}^{a_i} =  \dfrac{{p_i}^{a_i+1}-1}{p_i-1}. $

You could verify that

$(p_i-1)(p_i^{a_i}+p_i^{a_{i-1}}\cdots+p_i+1)$

$=p_i^{a_i+1}-p_i^{a_i}+p_i^{a_i}-p_i^{a_{i-1}}+\cdots+p_i^2-p_i+p_i-1$

$=p_i^{a_i+1}-1.$


问:
Residue of a complex function with essential singularity
Compute the residue of the function:
$$
\frac{e^{1/z}}{1-z}
$$

when $z_0 = 0$

I don't know how to proceed with this exercise. I can't find the Laurent Series of the function around $z_0 = 0$. Could someone give me a hint?

Thanks in advance

答:
You have:

$$\begin{split} e^{1/z} &= \sum_{n=0}^\infty \frac{1}{n!}\ z^{-n} \\ \frac{1}{1 - z} &= \sum_{n=0}^\infty z^n \end{split}$$

hence you can write:

$$\begin{split} e^{1/z} &= \sum_{n=-\infty}^\infty a_n\ z^n \\ \frac{1}{1 - z} &= \sum_{n=-\infty}^\infty b_n z^n \end{split}$$

where:

$$\begin{split} a_n &= \begin{cases} \frac{1}{(-n)!} &\text{, if } n\leq 0 \\ 0 &\text{, if } n>0 \end{cases} \\ b_n &= \begin{cases} 0 &\text{, se } n<0 \\ 1 &\text{, if } n \geq 0 \end{cases} \end{split}\; .$$

Therefore the Laurent series of your function is given by:

$$\frac{e^{1/z}}{1 - z} = \left( \sum_{n=-\infty}^\infty a_n\ z^n \right) \cdot \left( \sum_{n=-\infty}^\infty b_n z^n \right) = \sum_{n=-\infty}^\infty \underbrace{\left( \sum_{h+k=n} a_hb_k \right)}_{=:c_n}\ z^n\; .$$

Now, you're interested only in $c_{-1} = \operatorname{Res}(f;0)$ and this coefficient evaluates into:

$$\begin{split} c_{-1} &= \sum_{h+k=-1} a_hb_k \\ &= \sum_{k=0}^\infty a_{-1-k} b_k \\ &= \sum_{k=0}^\infty \frac{1}{(k+1)!} \\ &\stackrel{i=k+1}{=} \sum_{i=1}^\infty \frac{1}{i!} \\ &= e-1\; .\end{split}$$


问:
Equivalent Definitions of the Socle of a Module

Wikipedia gives the following definitions of the socle of an $R$-module $M$:
$$\text{Soc}(M)=\sum \left\{S:S\subseteq M\text{ is simple}\right\}:=S_1$$
and
$$\text{Soc}(M)=\bigcap\left\{ E:E\subseteq M\text{ is essential}\right\}:=S_2$$

I'm trying to show that these are equivalent.

I can show that $S_1\subseteq S_2$:

Suppose that $S\subseteq M$ is simple. If $x\in S$ is non-zero then $Rx=S$. For any essential $E\subseteq M$ we have $Rx\cap E\neq0$, and so $Rx\cap E=Rx$ by simplicity. Then $Rx\subseteq E$, so $x\in E$.

However I'm struggling to show the converse. Here is what I have tried so far:

Suppose that $e\in E$ for every essential $E\subseteq M$. I need to show that $e$ can be written as a sum of elements in simple submodules, so I thought I'd try to show that $Re$ is simple.

If not, then we have some $0\subsetneq N\subsetneq Re$, so there exists some $r\in R$ such that $re\notin N$. If $e\in E$ for every essential $E\subseteq M$, then $re$ does also.

Then it would be enough to show that $N\subseteq M$ is essential for a contradiction. Since
$$N\subsetneq Re\subseteq E\subseteq M$$
it would then be enough to show that $N\subsetneq Re$ and $Re\subseteq E$ are essential extensions. Unfortunately I can't seem to prove either, and so I'm beginning to doubt that this is the right approach.

Any help would be much appreciated.

答:
This proof follows Proposition 8.8 in these notes.

Let $N$ be any submodule of $S_2$. By Zorn's Lemma, we can find a module $N'\subseteq M$ which is maximal with respect to the property that $N\cap N'=0$.

Then $N\oplus N'\subseteq M$ is essential, since if $L\cap(N\oplus N')=0$ then $N'\oplus L$ would contradict the maximality of $N'$.

This proves that $N\subseteq S_2\subseteq N\oplus N'$, since $S_2$ is the intersection of all essential submodules of $M$.

Then
$$S_2=S_2\cap(N\oplus N')=N\oplus(S_2\cap N')$$
so any submodule of $S_2$ is a direct summand.

By the proof linked to here, this shows that $S_2$ is semisimple, and so is the direct sum of its simple submodules.

Then $S_2\subseteq S_1$ and we are done.


问:
$E/L$, $L/K$ normal then $E/K$ normal
I have to decide if the following statement is true:

If $E/L$, $L/K$ are normal, then $E/K$ is normal

I have though of this proof but I don't know if it is correct, could someone help me?

Let $p(x) \in K[x]$ and let $\alpha \in E$ be a root of $p(x)$. If $\alpha \in L$ then all the other roots of $p(x)$ are in $L$ (because $L/K$ is normal) and therefore they are also in $E$. Thus $E/K$ is normal. If $\alpha \notin L$ then we can also see $p(x) \in K[x]$ as a polynomial in $L[x]$. If $p(x) \in L[x]$ has a root $\alpha \in E$ then all its other roots are also in $E$ because $E/L$ is normal. Therefore $E/K$ is normal

答:
The statement is false: Consider $K = \mathbb{Q}$, $L = \mathbb{Q}[\sqrt{2}]$, and $E = \mathbb{Q}[\sqrt[4]{2}]$.

Indeed, in this case we have $E/L$ and $L/K$ are degree 2, and thus automatically Galois extensions. However, $E/K$ is not normal, as $x^4 - 2$ has a root in $E$, but does not split completely.

Title: $E/L$, $L/K$ normal then $E/K$ normal


问:
Dividing under ultraproduct
Let $\{M_i\}_{i<\omega}$ be a family of structures and let $\phi(x)$ be a formula which $k$-divides in $M_i$ for each $i<\omega$. Let $M^*$ be an ultraproduct of $M_i$'s.

Question. Does $\phi(x)$ k-divide (or just divide) in $M^*$? In other words; is dividing preserved under ultraproduct?

答:
We need to be a little bit careful about the parameters here - recall that a formula without parameters trivially does not divide (unless it is inconsistent)!

So I'll assume we're given a formula $\varphi(x,y)$ and a tuple $b_i\in M_i$ such that $\varphi(x,b_i)$ $k$-divides (over $\varnothing$) in $M_i$ for all $i<\omega$. Then, writing $b^* = [b_i]_{i<\omega}$ for the equivalence class of $(b_i)_{i<\omega}$ in the ultraproduct, we have that $\varphi(x,b^*)$ $k$-divides (over $\varnothing$) in $M^*$.

This follows pretty easily from Łoś's Theorem. The $k$-dividing is witnessed by, in each $M_i$, a sequence $(b_{i,j})_{j<\omega}$ such that:

$\text{tp}(b_{i,j}) = \text{tp}(b_i)$ for all $j<\omega$. This is witnessed by $M_i\models \psi(b_{i,j})\leftrightarrow \psi(b_i)$ for all $j<\omega$ and all formulas $\psi(y)$.

$\{\varphi(x,b_{i,j})\mid j<\omega\}$ is $k$-inconsistent. This is witnessed by $M_i\models \lnot  \exists x\, (\varphi(x,b_{i,j_1})\land \dots \varphi(x,b_{i,j_k}))$ for all $j_1<\dots<j_k<\omega$.

Now define $b^*_j = [b_{i,j}]_{i<\omega}$. By Łoś's Theorem, we have:

$M^*\models \psi(b^*_j)\leftrightarrow \psi(b^*)$ for all  $j<\omega$ and all formulas $\psi(y)$.  So $\text{tp}(b^*_j) = \text{tp}(b^*)$ fro all $j<\omega$.

$M^*\models \lnot  \exists x\, (\varphi(x,b^*_{j_1})\land \dots \varphi(x,b^*_{j_k}))$ for all $j_1<\dots<j_k<\omega$. So $\{\varphi(x,b^*_j)\mid j<\omega\}$ is $k$-inconsistent.

Thus $(b_j^*)_{j<\omega}$ witnesses $k$-dividing of $\varphi(x,b^*)$ in $M^*$.

Note that is was important that the formula $\varphi$ and the number $k$ was uniform across the $M_i$! It is not true in general that if $\text{tp}(a_i)$ divides (over $\varnothing$) in each $M_i$, then $\text{tp}(a^*)$ divides (over $\varnothing$) in $M^*$, where $a^* = [a_i]_{i\in \omega}$. To lift dividing to the ultraproduct, we need to know that the same formula witnesses $k$-dividing for the same $k$ in all $M_i$ (or at least in a large set of the $M_i$, with respect to the ultrafilter).

Title: Dividing under ultraproduct


问:
Infinite product of $\sqrt{2}$
I am struggling to show the following relation
$$
\prod_{k=1}^\infty \left(1 - \frac{(-1)^k}{(2k-1)}\right) = \sqrt 2.
$$
I have tried to compute the sum
$$
\sum_{k=1}^\infty \log \left(1 - \frac{(-1)^k}{2k-1}\right),
$$
by using the expansion for $\log(1+x)$, however, I was not able to evaluate the double sum. Furthermore, I tried to square and reorder (although it should not be possible), but haven't quite got the right track.

Could someone give me a hint for this problem?

答:
It follows from the Euler/Weierstrass factorisation for sine: for $x \in \mathbb{R}$

$$\frac{\sin{x}}{x} = \prod_{j \ge 1} \left(1-\frac{x^2}{\pi^2 j^2}\right)$$

Separating the odd even parts we can rewrite your product as:

$$\begin{aligned} \mathcal{P} & = \prod_{k \ge 1 } \left(1 - \frac{(-1)^k}{2k-1}\right)  \\& = \frac{1}{2}\prod_{k \ge 0} \bigg(1-\frac{1}{4k-1}\bigg)\bigg(1+\frac{1}{4k+1}\bigg) \\& = \frac{1}{2} \prod_{k \ge 0} \frac{16k^2-4}{16k^2-1} \\& = 2 \prod_{k \ge 1} \frac{16k^2-4}{16k^2-1} \\& = 2 \prod_{k \ge 1}\frac{\bigg(1-(\dfrac{1}{2})^2/k^2\bigg)}{\bigg(1-(\dfrac{1}{4})^2/k^2\bigg)} \\& =  2 \cdot \frac{\sin{\frac{1}{2} \pi}}{\frac{1}{2}\pi} \bigg/\frac{\sin{\frac{1}{4} \pi}}{\frac{1}{4}\pi} \\& = 2 \cdot \frac{1}{\sqrt{2}} \\& = \sqrt{2}. \end{aligned}$$

Title: Infinite product of $\sqrt{2}$


问:
Norm of composition of linear maps
I am trying to prove that for normed vector spaces $U$, $V$, and $W$ with $S : U \rightarrow V$ and $T : V \rightarrow W$ linear maps. We have  $||T \circ  S|| \leq || T|| \cdot ||S||$.

I have that $\|T\|=\sup \{\|T f\|: f \in V \text { and }\|f\| \leq 1\}$ and $\|T \circ S \|=\sup \{\|T(Sf)\|: f \in V \text { and }\|f\| \leq 1\}$. But don't really know where to go from here. Thanks in advance :)

答:
Hint:
$$
\|T \circ S\| = \|S\| \cdot \sup\left\{\left\|T \left(\frac{Sx}{\|S\|} \right)\right\|: f \in V \text{ and } \|f\| = 1\right\}.
$$

Title: Norm of composition of linear maps


问:
Abel criterion proof
I want to prove the following statement:

Let be $\sum\limits_{k=1}^{\infty} a_k$ a convergent series and $\left(b_k\right)_{n\in\mathbb{N}}$ a monotone and bounded sequence. Then $\sum\limits_{k=1}^{\infty} a_kb_k$ is also convergent.

I know there already exist a few questions on this problem, however they mostly have additional assumptions (i.e. $\left(b_k\right)_{k\in\mathbb{N}}$ with $b_k\geq 0$ for all $k$).

My approach:

We define $A_n:=\sum\limits_{k=1}^{n} a_k$. As $A_n$ is convergent there exists a bound $A$ such that $|A_n|\leq A$ for all $n$. We know that $\left(b_k\right)_{k\in\mathbb{N}}$ is convergent and hence the sequence $\left(A_kb_k\right)_{k\in\mathbb{N}}$ is also convergent (product of two convergent sequences). Let be $n_1$ and $n_2$ two indices such that for all $n,m$ with $n>m>n_1$ it holds $|A_nb_n-A_mb_m|<\frac{\epsilon}{2}$ and for all $n,m$ with $n>m>n_2$ it holds $|b_n-b_m|<\frac{\epsilon}{2A}$. Now we define $n_0:=\max\{n_1,n_2\}$. With this in mind we apply Abel's lemma (summation by parts) and it follows for all $n>m>n_0$:

$$
|\sum\limits_{k=m+1}^{n} a_kb_k|=|A_nb_n-A_mb_m+\sum\limits_{k=m}^{n-1} A_k(b_k-b_{k+1})|\leq |A_nb_n-A_mb_m|+\sum\limits_{k=m}^{n-1} |A_k(b_k-b_{k+1})| \cdots
$$
If $\left(b_k\right)_{k\in\mathbb{N}}$ is monotonically decreasing it follows:
$$
\cdots<\frac{\epsilon}{2}+ \sum\limits_{k=m}^{n-1} |A_k|(b_k-b_{k+1})\leq \frac{\epsilon}{2}+ \sum\limits_{k=m}^{n-1} A(b_k-b_{k+1})=\frac{\epsilon}{2}+A (b_m-b_n)<\frac{\epsilon}{2}+\frac{\epsilon A}{2A}=\epsilon.
$$

If $\left(b_k\right)_{k\in\mathbb{N}}$ is monotonically increasing it follows:
$$
\cdots<\frac{\epsilon}{2}+ \sum\limits_{k=m}^{n-1} |A_k|(b_{k+1}-b_k)\leq \frac{\epsilon}{2}+ \sum\limits_{k=m}^{n-1} A(b_{k+1}-b_k)=\frac{\epsilon}{2}+A (b_n-b_m)<\frac{\epsilon}{2}+\frac{\epsilon A}{2A}=\epsilon.
$$
So in both cases $\sum\limits_{k=1}^{\infty} a_kb_k$ satisfies the Cauchy criterion and hence is convergent.

Is this correct or is there a more elegant/faster approach?

答:
For another approach, we can show that the sequence of partial sums converges without using the Cauchy criterion.  We have existence of the limits $\lim_{n \to \infty}A_n = \lim_{n \to \infty}\sum_{k=1}^na_k =A$ and $\lim_{n\to \infty}b_n = b .$

Summing by parts, we get

$$S_n =\sum_{k=1}^n a_kb_k = a_1b_1+\sum_{k=2}^n (A_k - A_{k-1})b_k = a_1b_1+\sum_{k=2}^{n} A_k b_k-  \sum_{k=2}^{n} A_{k-1} b_k \\ = \sum_{k=1}^{n} A_k b_k-  \sum_{k=1}^{n-1} A_{k} b_{k+1} = A_nb_{n+1} + \sum_{k=1}^{n} A_k (b_k - b_{k+1})$$

The series $\sum(b_k - b_{k-1}) $ converges since $\sum_{k=1}^n (b_k - b_{k+1}) = b_1 - b_{n+1} \to b_1 - b$ as $n \to \infty$.  Since $(A_k)$ is a bounded sequence and the terms $(b_k- b_{k+1})$ are all of the same sign,   it follows that $\sum A_k(b_k - b_{k+1})$ is convergent.

Therefore, the series $\sum a_kb_k$ converges since

$$\sum_{k=1}^\infty a_kb_k = \lim_{n \to \infty}A_nb_{n+1} + \lim_{n \to \infty}\sum_{k=1}^{n} A_k (b_k - b_{k+1}) = Ab + \sum_{k=1}^\infty A_k(b_k - b_{k+1})$$

Title: Abel criterion proof


问:
Derivative of integral with varying domain? Fundamental theorem of calculus?
Say I want to find the $n^{th}$ derivative of
$$\int_{V_x} F{(x_1,...,x_k)}dV$$
where $V_x$ is some $k$-dimensional volume dependent on $x$. In the $1$-dimensional case this is the fundamental theorem of calculus for $n=1$ and we can take higher derivatives after applying the fundamental theorem. To be concrete, say $V_x$ is the cube $[0,x]^k$. What can I do in general? Is this Stokes theorem?

答:
Do a change of variable to eliminate the dependence on $x$ in the integral bounds and note that $x$ inside the integral is a dummy variable. Namely, $$\int_{V_x} F(x)\,{\rm d}x = \int_{[0,x]^k} F(y)\,{\rm d}y$$Now write $g\colon [0,1]^k \to [0,x]^k$ as $y = g(z) = xz$. Then $Dg(z) = x{\rm Id}_{\Bbb R^k}$, so $\det Dg(z) = x^{k}$. So ${\rm d}y = x^k\,{\rm d}z$, and so $$\int_{[0,x]^k} F(y)\,{\rm d}y =\int_{g([0,1]^k)} F(y)\,{\rm d}y =\int_{[0,1]^k} F(xz)\, x^k\,{\rm d}z = x^k\int_{[0,1]^k} F(xz)\,{\rm d}z.$$Hence $$\frac{{\rm d}}{{\rm d}x} \left(x^k\int_{[0,1]^k}F(xz)\,{\rm d}z\right) = kx^{k-1}\int_{[0,1]^k} F(xz)\,{\rm d}z + x^k \int_{[0,1]^k} \langle \nabla F(xz), z\rangle\,{\rm d}z.$$To bring the domain back to what it was in the beginning, we undo the change of variable, to get: $$\frac{{\rm d}}{{\rm d}x}\left(\int_{[0,x]^k}F(y)\,{\rm d}y\right) = \frac{k}{x} \int_{[0,x]^k}F(y)\,{\rm d}y + \frac{1}{x}\int_{[0,x]^k} \langle \nabla F(y), y\rangle\,{\rm d}y.$$
Note that it makes perfect sense for this formula to become singular for $x=0$, as this is the case where your domain of integration $[0,x]^k$ degenerates to a point.


问:
Use the Handshake Lemma to determine the number of edges in GK_n
In chess, a knight's move consists of two spaces either vertically or horizontally, followed by one space in the perpendicular direction. In this way, every knight's move results in an L shaped displacement from the original position. To model all of the legal knight's moves on an n x n chessboard, we can create a graph where each vertex represents a square on the board, and two vertices are adjacent if an only if a knight can legally move between the corresponding squares.
Use the handshake lemma to determine the number of edges in GK_n

Is GK_n always, sometimes or never Eulerian

Does GK_n always, sometimes or never contain an Euler trail

By use of the Handshake Lemma edges are twice the amount of degree sum so if you had a graph GK_4 with 16 vertices, it would have degree sum 48 and 24 edges. I'm not sure how to use this information to determine the number of edges in GK_n. The abstractness is what confuses me.

答: HINT: You need to know the number of odd vertices. Squares that are at least two rows and two columns away from the edges of the board have degree $8$: they're all even. What about squares along the edges of the board, or one row or column in? Be a little careful: they're not all the same.</p>


Title: Infinite product of $\sqrt{2}$

Body:
I am struggling to show the following relation
$$
\prod_{k=1}^\infty \left(1 - \frac{(-1)^k}{(2k-1)}\right) = \sqrt 2.
$$
I have tried to compute the sum
$$
\sum_{k=1}^\infty \log \left(1 - \frac{(-1)^k}{2k-1}\right),
$$
by using the expansion for $\log(1+x)$, however, I was not able to evaluate the double sum. Furthermore, I tried to square and reorder (although it should not be possible), but haven't quite got the right track.

Could someone give me a hint for this problem?

Body:
$$\begin{align}\frac{2}{1}\frac{2}{3} &= 4^1\frac{2!^3}{4!1!^2}\\\frac{2}{1}\frac{2}{3}\frac{6}{5}\frac{6}{7} &= 4^2\frac{4!^3}{8!2!^2}\\\frac{2}{1}\frac{2}{3}\frac{6}{5}\frac{6}{7}\frac{10}{9}\frac{10}{11} &= 4^3\frac{6!^3}{12!3!^2}\\\ldots\end{align}$$
Use Stirlings approximation $n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ to establish the limit (the $e$'s and $\pi$'s will factor out).


问:
Tips for a recursive function, proof by induction
Suppose the following

$$ f(0)= 0$$

$$ f(1)=2$$

$$ f(n)= 4f(n-1)-3f(n-2)$$

I want to prove that

$$f(n)= 3^n -1$$

by induction.

I started by trying this

$$f(n+1)=4f(n)-3f(n-1)$$

Which gives me this:

$$ f(n+1)= 4(3^n-1)-3f(n-1)$$

However I do not understand how the second part

$$3f(n-1)%$$

of the term is solvable. Can someone give me some pointers on how to solve this?

答:
HINT

You are probably used to the inductive scheme where you go from $P(n)$ to $P(n+1)$, where $P(n)$ is some claim about the number $n$.

That is, once you show the base $P(0)$, and you have also shown that $P(n)\to P(n+1)$, you have shown $P(n)$ for all $n$:

You have shown $P(0)$

But given $P(0)$ and $P(n)\to P(n+1)$, this means $P(1)$

But given $P(1)$ and $P(n)\to P(n+1)$, this means $P(2)$

But given $P(2)$ and $P(n)\to P(n+1)$, this means $P(3)$

Etc.

OK, so that's a very 'typical' form of induction.... where each 'domino stone' knocks over the next.

However, in this case, each domino stone needs not just the very previous domino stone to fall over, but the previous two ... you can immediately see this by the fact that the recursive function refers to the previous two terms .. and that the sequence starts with two base cases.

Thus, what you need to show is that $P(n-1)$ and given that $P(n)$, you can show $P(n+1)$

Also, there are now two base cases. That is, also show that $P(0)$ and that $P(1)$

With that, the claim $P(n)$ should hold for any $n$, because:

You have shown $P(0)$ and $P(1)$.

But given $P(0)$ and $P(1)$ and $(P(n-1) \land P(n)) \to P(n+1)$, this means $P(2)$

But given $P(1)$ and $P(2)$ and $(P(n-1) \land P(n)) \to P(n+1)$, this means $P(3)$

But given $P(2)$ and $P(3)$ and $(P(n-1) \land P(n)) \to P(n+1)$, this means $P(4)$

Etc.

So, applied to your problem, you need to show that $f(n+1)=3^{n+1}-1$ if you assume that:

$f(n)=3^{n}-1$
and
$f(n-1)=3^{n-1}-1$

And, for the base, you need to show that $f(n)=3^{n}-1$ for $n=0$ and $n=1$


问:
Why should a transverse intersection give intersection multiplicity one?
My setting is that of Hartshorne chapter I ("classical varieties" over an algebraically closed field, which are irreducible [not schemes]).

Background

I've been working on problem 7.7(a) and I've hit a bit of a wall with part (a):

Problem 7.7. Let $Y$ be a variety of dimension $r$ and degree $d>1$ in $\Bbb P^n$. Let $P\in Y$ be a nonsingular point. Define $X$ to be the closure of the union of all lines $PQ$, where $Q\in Y$, $Q\neq P$.

(a) Show that $X$ is a variety of dimension $r+1$.

Clearly $X$ contains $Y$, so $X$ must be of dimension at least $r$. On the other hand, we can define a map $f$ from $(Y\setminus\{P\})\times\Bbb P^1$ to $X$ by sending $(Q,[a:b])\mapsto aQ+bP$ which surjects on to the collection of points in $X$ which are on a line $PQ$ with $Q\in Y$ and $Q\neq P$. Since $(Y\setminus\{P\})\times\Bbb P^1$ is irreducible, this shows that $X$ is irreducible, and the map is dominant, so we get an extension of function fields $k(X)\to K(Y\times\Bbb P^1)$. This shows that $\dim X \leq r+1$. So all I have to do finish the problem is to show that $X$ contains one point which isn't in $Y$. This is turning out to be harder than I expected!

Here's how I'd want to solve this problem: Pick a hyperplane $H$ through $P$ which is transverse to $Y$ (we can do this because $r<n$ by the degree assumption plus the fact that $P$ is a smooth point). Then the algebraic set $Y\cap H$ is smooth at $P$ (I can prove this via the Jacobian criteria), so there's a unique irreducible component $Z\subset Y\cap H$ through $P$. If I could prove that the intersection multiplicity along $Z$ is equal to one, I'd win: either $Y\cap H$ is reducible and I can connect $P$ with a point $Q$ on some other irreducible component and find a point in $X\setminus Y$ on the line $PQ$, or $Y\cap H=Z$, a variety of dimension $r-1$ and degree $d$, and eventually I can get it down to the case of a curve and I understand what to do in that case.

Question

Suppose $Y\subset \Bbb P^n$ is a variety of dimension $r$ and degree $d>1$, and $P$ a smooth point on $Y$. Suppose $H$ is a hyperplane through $P$ so that $Y\cap H$ is smooth at $P$ and $Y\cap H$ is irreducible. Why should $i(Y,H;Y\cap H)=1$?

My thoughts

The problem is that I feel very unsure about why or how I should be able to get this result. I'm supposed to compute the length of $(S/(I_Y+h))_\mathfrak{p}$ over the local ring $S_\mathfrak{p}$, where $\mathfrak{p}$ is the homogeneous prime ideal corresponding to the irreducible subvariety $Y\cap H$ and $h$ is a generator for the ideal of $H$. I can show that this the same as what you'd want in the affine case - that is, if we work in some standard affine patch $U$ containing $P$, I can show that the intersection multiplicity is the same as the length of $(k[Y\cap U]/(\widetilde{h}))_{\mathfrak{p}'}$ over $k[Y\cap U]_{\mathfrak{p}'}$ where $\mathfrak{p}'$ is the ideal corresponding to $Y\cap H\cap U$, but I don't see how I'm supposed to conclude this is $1$ if I know smoothness at $P$.

Edit: I've made a little progress - I can show that $f$ is in the maximal ideal of $P$ in $Y$  but not it's square, so $f$ is in the ideal of $Y\cap H$ but not it's square. I still don't see how this gives that the intersection multiplicity is one - I would need to know something along the lines of $\mathcal{O}_{Y,Y\cap H}$ having principal maximal ideal, which would mean it's a DVR? Still pretty stuck, though.

Or, if you think I'm barking up the wrong tree, is there some other way I'm supposed to finish problem 7.7(a)?

答: In the language of the post, the local ring of $Y\cap H\cap U$ at $P$ is $\mathcal{O}_{Y,P}/\sqrt{(f)}=k[x_1,\cdots,x_n]_{(x_1,\cdots,x_n)}/\sqrt{(I_{Y\cap U}+(f))}_{(x_1,\cdots,x_n)}$, which is the reduction of $k[x_1,\cdots,x_n]_{(x_1,\cdots,x_n)}/(I_{Y\cap U}+(f))_{(x_1,\cdots,x_n)}$. But Hartshorne's theorem I.5.1 says that this last ring is actually a regular local ring by the Jacobian criterion and therefore reduced. So this is an equality, and $\mathcal{O}_{Y\cap U,P}/\sqrt{(f)}=\mathcal{O}_{Y\cap U,P}/(f)$. By the transitivity of localization, we have that $\mathcal{O}_{Y\cap U,Z\cap U}=(\mathcal{O}_{Y\cap U,P})_{I_{Z\cap U,P}}$, so the multiplicity of $k[U]/(I_{Y\cap U}+(f))$ is one. Since this is equal to the multiplicity of the homogeneous version, we have that the intersection multiplicity is one.


问:
If the order of a finite abelian group is not divisible by a square, show that the group must be cyclic.
If the order of a finite abelian group is square free, show that the group is cyclic.

This is a question from "basic abstract algebra" by bhattacharya

答: $|G| = p_1 \dots p_n$, a product of distinct primes. Use Cauchy's theorem to get $g_i \in G$ of order $p_i$ for $1 \leq i \leq n$. Then, use induction and the assumption that $G$ is abelian to prove $g_1\ldots g_i$ has order $p_1 \ldots p_i$.</p>


问:
Why does calculators have differing limits reaching $\infty$?
Discovering that $\infty$ can be varied, it might make sense that very big numbers will limit theirselves to become $\infty$.

I went to use my phone for experimenting this. Once I reach $1e309$, it becomes $\infty$.

I went to see if my computer will limit as the same value as in my phone. Turns out $1e10000$ is the limit compared to my phone's $1e309$.

If you're confused what limit I'm talking about, it is the first quantity that turns into $\infty$. Also, I used a Windows for this experiment.

So why is it that these limits vary? Is it the same way when it comes to regular calculators (actual ones, not digital app)?

答:
As a programmer, I think it's safe to say that memory is the reasoning for this.

Your phone has limited memory which is why $1e309$ would be considered $\infty$ by your calculator, and it differs comparing from the computer's bigger memory making $1e10,000$ into $\infty$ instead..

Or rather calculators themselves say that "a value that's this big will be considered an infinite digit". It can vary in any way.

They are programmed for many values, but they cannot fully withstand all. Which is why they limit their bigger digits as $\infty$.


问:
Modified NIM game : which player has a winning strategy
A game consists of two players and one pile with n beans. In each move, a player can take P^k beans.( P is a prime number and k is an integer. P and k are not fixed. ) The winner is the player who does the last move.

Determine which player has a winning strategy based on n.

I tried using the binary table and the xor function often explained in the standard game where you can take any amount of beans in each move but I can't quite figure out how to apply that solution properly here.

This question came up during my discrete math class during the Game Theory presentation and we are to search for the answer. I would appreciate some help.

答: Hint. Try working up from $n=1$. That's a winning position for the first player, who can take $2^0 = 1$ . What about $n=2,3,4,5,6,\ldots$? ($6$ is the first interesting case.)


问:
Is this Stochastic Process bounded from above?
I'm dealing with a problem in my thesis that involves proving the boundedness of a Stochastic Process. This question is related to another one I did recently, but the results found in the other one sadly didn't help me sufficiently.

Let $\text{d}X_t= f(t,X_t) \text{d}t +\text{d}W_t$ be a stochastic process defined for the times $t\ge0,$ such that $X_0=0$ and $f(t,x)<a<0$ for all $t\ge0,x>k$ for a certain constant $k$.

Prove that $$\mathbb{P}[\exists C>0: X_t<C \ \forall t\ge 0]=1.$$

In other words, I wish to show that if the function $f$ is negative when $X_t$ goes above a certain value $k$, then the stochastic process is bounded from above, because the drift part is going to be negative.

I know that $Y_t=at+W_t$ is bounded from above if $a<0$, but what I can't manage to deal with is the fact that we cannot control the moments when $X_t$ goes above $k$: every time it goes above $k$ it might reach a higher peak since the bound on $Y_t$ is not $\omega$-wise uniform, and the sequence of those peaks might be unbounded.

Here's a simulation of the SDE $\text{d}Y_t=-Y_t\text{d}t+\text{d}W_t$, which seems to fit these hypotheses, as the time taken is quite long.

EDIT: my SDE, in particular, is $$\text{d}X_t=X_t\frac{D(X_t)-S}{D(X_t)+S}\text{d}t+X_t\sigma\text{d}t,$$ with $D$ being $D(x)=D_0\exp(-\alpha x),$ for some $D_0>S>0$, and $X_0>0$. This SDE is equivalent to $$\text{d}Y_t=\left(\frac{D(\exp(Y_t))-S}{D(\exp(Y_t))+S}-\frac{\sigma^2}2\right)\text{d}t+\sigma\text{d}t,$$ thanks to the substitution $Y_t=\log(X_t)$: this second equation is the one I was referring to. Actually, $S$ is not a constant but a bounded function, which breaks the Markovianity, but I think that if the result is not true for constant $S$ the it's not true also for bounded $S$.

Any help would be immensely appreciated.

答:
This might not be good news for you.

I don't think your setting will work generally. The reason -why your simulation seems to be confirming your hunch - might be due to your limited simulated time interval.

The main reason based on which I came to that conclusion is the Markov nature of your SDE. ( if there is no $t$ in f)

Now, let us together examine your setting to see if I have made any mistake.

Some assumptions I made on $f$:

(i)  $f$ is time stationary, that is $ f(t,x)=f(x)$

(ii) $f(0)= 0$

(iii)$f(x) \ge 0$ if $x \le 0$ and $ f(x) \le 0$ in the other case.

(iv) $f$ is bounded under by a value $b<0$, that is $f(x) \ge b$

Let $\tau_1, \gamma_M$ denote:

$ \tau_1 := \inf \{ t \ge 1: X_t=0\}$ , the first time after $1$ that $X$ revisits $0$.

$ \gamma_M:= \inf \{ t \ge 0: X_t= m\}$ for some $M>0$

So the desired conclusion is that: $$\lim_{M \rightarrow +\infty} \mathbb{P}\left( \gamma_M < \infty \right) =0$$

What I will show is that: It seems to me, under these assumptions :
$$  \mathbb{P}\left( \gamma_M < \infty \right) =1 \space \forall M>0$$

Heuristically, under my third assumptions (iii), $f$ acts as a repulsive force that drags $X$ back to $0$, and I don't think it is hard to prove that:
$$ \mathbb{P}( \tau_1 < \infty)=1 $$
Then, we have:
$$\mathbb{P}\left( \gamma_M< \infty \right) =\mathbb{P}\left( \gamma_M< \tau_1 \right)+\mathbb{P}\left( \tau_1 \le \gamma_M<\infty \right)$$
$$ \underbrace{=}_{ \text{Strong Markov's property}} \mathbb{P}\left( \gamma_M< \tau_1 \right)+\mathbb{P}\left( \tau_1 \le \gamma_M \right)\mathbb{P}\left(  \gamma_M <\infty \right)$$
Which is equivalent to:
$$\left[ \mathbb{P}\left( \gamma_M< \infty \right) -1 \right]\mathbb{P}\left( \gamma_M< \tau_1 \right)=0$$
$$\Leftrightarrow \mathbb{P}\left( \gamma_M< \infty \right)=1$$
Because the fourth assumption gives us clear a reason for  which $\mathbb{P}\left( \gamma_M< \tau_1 \right) >0$. Indeed, we have:
$$\mathbb{P}\left( \gamma_M< \tau_1 \right) \ge \mathbb{P}\left( \gamma_M< 1 \right)  \ge \mathbb{P}\left( X_1 > M \right)  \ge \mathbb{P}\left( W_1+b > M \right) >0$$

**QED **

*Discuss *: So there are more than just only the control over the negativity of $f$ in order for your result to happen.


问:
Under what condition does $-(x-1)^3+x(a^2+b^2+c^2)-(a^2+b^2+c^2)=0$ have a positive solution $x$?

Consider the equation $$-(x-1)^3+x(a^2+b^2+c^2)-(a^2+b^2+c^2)=0$$ where $a,b,c$ are real numbers. Through what condition does $x$ have positive real solution?

Answer: $a^2+b^2+c^2<1$

How to prove this?

Thank you.

Body:
$$-(x-1)^3+x(a^2+b^2+c^2)-(a^2+b^2+c^2)=0$$

$$-\frac{(x-1)^3}{(a^2+b^2+c^2)}+(x-1)=0$$

$$-(x-1) \left[ \frac{(x-1)^2}{(a^2+b^2+c^2)} - 1 \right]=0$$

Focusing in

$$ \frac{(x-1)^2}{(a^2+b^2+c^2)} - 1 = 0 $$

$$ (x-1)^2 = (a^2+b^2+c^2) $$

$$ (x-1) = \pm\sqrt{a^2+b^2+c^2} $$

$$ x = 1\pm\sqrt{a^2+b^2+c^2} $$

$ 0 < a^2+b^2+c^2 < 1 $ guarantees that both solutions are real and positive.


答:
Short version: The max for your phone is (slightly less than) $2^{2^{10}}$ because your phone is using double precision floating points to represent numbers.

Computers typically represent numbers as strings of bits of some fixed length - so might say, "a number is represented by some string of $32$ bits" giving rise to $2^{32}$ possible values (at most) - of which there is necessarily some largest and smallest value that may be represented and where the exact behavior that happens when you exceed that varies. Largely, they may represent numbers in two forms: integral types (representing integers) and floating point types (trying to represent real numbers to some accuracy).

The simpler case is that integral types do the obvious thing and say, "I represent an integer in binary" - so, if we had $8$ bits to work with $01001111$ would represent $135$ and we'd could represent all integers $0$ to $255$ this way - and, if we tried to exceed this range, we'd wrap around as in modular arithmetic (mod $256$). Sometimes, to represent negative numbers, we'd use something called two's complement, which essentially says that the most significant place value is negated - so $10000000$ represents $-128$ and $10010100$ represents $-128+16+4=-108$ and $11111111$ represents $-1$. One can do this with any number of bits (or even a flexible number of bits), although most commonly these representations are used for $8$ or $16$ or $32$ or $64$ bits.

General purpose calculators tend to use a floating point representation of a number - essentially, they break the bit string up into pieces, an extremely common format for doing this with $32$ bits is as follows:

We use the first bit to represent the sign of the number.

The next $8$ bits represent an exponent $e$ ranging from $1$ to $254$.*

The next $23$ bits represent some number $n$ between $0$ and $2^{23}-1$.

Note that this representation contains two integral representations - one for an exponent and one for a "significand". The number represented by this format is $2^{e - 127}\cdot (1 + 2^{-23}\cdot e)$. This represents an odd set of numbers - note that there are about equally many numbers between $0$ and $1$ as there are exceeding $1$ - essentially, it represents writing down binary numbers to some precision, assuming the most significant place value is neither too large nor too small. The largest number representable is about $2^{127}\cdot 2 = 2^{2^7}==3.4082\times 10^{38}$. Floating points usually, if they would exceed this range, become some sort of representation of infinity - i.e. the computer no longer tracks the number in any meaningful way.

There are also longer floating point formats in common use - notably a double precision floating point uses 64 bits, using 11 bits for the exponent, and can represent numbers below $2^{2^{10}}=1.797\times 10^{308}$. This is almost certainly what is going on with your phone's calculator.

There are also larger formats for floating point numbers, which can represent an even greater range - your computer is likely using some larger representation and it may be the case that it is programmed to not expose the full range of its representation to the user, but rather to softly max out at some round number (e.g. to fail intentionally because of software rather than because of the limits of the floating point format). There are also some systems that use decimal representations of numbers, which would lead to round bounds, although I believe that's fairly rare these days. It's also possible to program systems that will compute to arbitrary precision - though, with the drawbacks to performance and memory usage that this entails. This is very common in computer algebra systems and other more heavy duty mathematical applications.

(*One might note that $8$ bits can represent $256$ values and I've only prescribed $254$ exponents - the remaining two values are used for special floating point values such as infinity)

Title: Why does calculators have differing limits reaching $\infty$?

Body:
Discovering that $\infty$ can be varied, it might make sense that very big numbers will limit theirselves to become $\infty$.

I went to use my phone for experimenting this. Once I reach $1e309$, it becomes $\infty$.

I went to see if my computer will limit as the same value as in my phone. Turns out $1e10000$ is the limit compared to my phone's $1e309$.

If you're confused what limit I'm talking about, it is the first quantity that turns into $\infty$. Also, I used a Windows for this experiment.

So why is it that these limits vary? Is it the same way when it comes to regular calculators (actual ones, not digital app)?


问:
What is the equivalent of a diagonal in a non-square matrix or array?
I have a non-square matrix $M$, that looks something like this:

$M=\left[
\begin{array}
& a & b & c \\
d & e & f \\
g & h & i \\
j & k & l \\
\end{array}\right]$

I would like to refer the cells $M_{ij}$ where $i=j$, like $\{a,e,i\}$ in this case. If this was a square matrix, that would be the diagonal. Is that still the right word, or is there something more fitting that avoids confusion?

答: Yes, those entries: $m_{ij}$ in $M$, with $i = j$, constitute what can technically be called the "main diagonal" of the rectangular matrix, though the diagonal of such a matrix is not necessarily as "useful" as it is in a square matrix. See, e.g., the Main Diagonal entry in Wikipedia.</p>


问:
How to imply $T^*=-T \Rightarrow T^2+I=0$
I have a question:

$$
T:V \to V
$$
Linear transformation in inner product space.

Given:

$$
T^* = -T
$$

Prove that $\forall a \in R \to I-aT$ Invertible.

I saw that i had a mistake, please close the question or delete it. Sorry and thanks for the help in seeing my error.

答: $(I-aT)(I+aT)=I-a^2T=(1+a^2)I$. This implies that the inverse of $I-aT$ is ${1\over{1+a^2}}(I+aT)$.


问:
Let $\{Y_n\} =\{X_n\}^{1/n}$. Show that there is a constant $c$ such that $\{Y_n\}$ converges to $ c$ almost surely, and determine that $c$.
Especially for part b:
image for the problem</p>

答:
Man, consider this new sequence:
$$S_n= \ln(X_n)$$
and,
$$T_n= S_n-S_{n-1} \forall n \ge 0$$
Under your assumptions,

$T_n\underbrace{=}_{def.} \ln\left( \frac{X_n}{X_{n-1}}\right)$ is independent of $T_{n-1},T_{n-2},...,T_1$

$S_n= T_n+T_{n-1}+...+T_1$

And your term of interest $X_n^{1/n}$ is in fact $$\ln(X_n^{1/n})=\dfrac{T_n+T_{n-1}+...+T_1}{n}$$

Do you recall something now?


问:
If $f_n$ converges uniformly to $f$, then $\int f_n\ d\mu \to \int f\ d\mu$ as $ n \to \infty $
Let ($\Omega, \mathcal{F}, \mu)$, with $\mu (\Omega) <\infty$ and suppose  $f_n , n \geq 1$ is a sequence of integrable functions that converges uniformly on $\Omega$ to $f$. Given that $f$ is integrable, show that  $$\int f_n\ d\mu  \to  \int f\ d\mu \quad\text{ as }\quad n \to \infty $$

I tried :

$\left|\int_\Omega f_n \, d\mu - \int_\Omega f\, d\mu\right| $

: (By linearity of the integral)

$= |\int_\Omega (f_n - f)\, d\mu|\leq \int_\Omega |f_n - f|\, d\mu$

How do we prove the inequality above?

I tried the following but  I am not sure if it's correct or sufficient mathematical proof.

Since $f_n$ integrable, so  $\int_\Omega f_n \ d\mu$ is finite which means it exists.

$f$ integrable, so$\int_\Omega f\ d\mu$ is finite,   which means it also exists.

Thus, since they both exist  $(f_n - f)$ exists   $= |\int_\Omega (f_n - f)\, d\mu|$ is finite.

Hence from basic property of integral since  $= |\int_\Omega (f_n - f)\, d\mu|$ exists

Finally  how do we apply the Uniform convergence on      $  \int_\Omega |f_n - f|\, d\mu$ to conclude that

$\int_\Omega f\ d\mu  \to  \int_\Omega f\ d\mu$ as $ n \to \infty $ ?

答:
Since $f_n$ converges uniformly on $\Omega$ to $f$, for $\forall \varepsilon>0$, there is $N>0$ which is independent of $x\in\Omega$ such that
$$ |f_n(x)-f(x)<\frac{\varepsilon}{\mu(\Omega)}, \forall n\ge N, x\in\Omega. $$
So for above $\varepsilon>0$ and $N>0$, when $n\ge N$,
$$\bigg|\int_\Omega (f_n - f)\, d\mu\bigg|\le\int_\Omega |f_n-f|\ d\mu\le\int_{\Omega}\frac{\varepsilon}{\mu(\Omega)}d\mu=\varepsilon $$
which implies
$$ \lim_{ n \to \infty}\int_\Omega f_n \, d\mu=\int_\Omega f_n \, d\mu. $$


问:
Finding $\mathbb E(X\mid Y)$ where $P(X > t) =e^{−t}$ with $t >0$ and $Y=\min(X,t)$

Find $\mathbb E(X\mid Y)$ where $X$ is a random variable such that $P(X >
 t)  =e^{−t}$ with $t >0$ and $Y=\min(X,t)$.

I have not done a problem before where we condition on the minimum or maximum value. I thought about breaking it up into cases as follows:

$$\mathbb E(X\mid X>t)\mathbb P(X>t)+\mathbb E(X\mid X<t)\mathbb P(X<t)$$

but this gives $\mathbb E(X)$ rather than $\mathbb E(X\mid Y)$. I tried thinking of this conceptually but what makes this problem tricky is that (if I'm understanding the problem correctly) we are not told whether $X<t$ or $X>t$ but rather we're just given the smaller of the two values. If we are (unknowingly) given the information that $Y=\min\{X,t\}=t$ with probability $e^{-t}$ then by the memoryless property, $\mathbb E(S\mid t)=t+1$ and if we are (unknowingly) given the information that $Y=\min\{X,t\}=X$ with probability $1-e^{-t}$ then $\mathbb E(X\mid X=x)=x$ so that

$$\mathbb E(X\mid Y)=(t+1)e^{-t}+x\left(1-e^{-t}\right)$$

but here I'm basically doing the same thing as above so I'm not sure how to correctly think about this type of problem. Any help would be appreciated!

答: First observe that $Y=\min\{X,t\}$ is supported on $(0,t]$ and that $E(X|Y=y)$ is necessarily a function of $y$. On one hand, if $y\in (0,t)$, we have $\{Y=y\}=\{X=y\}$ giving us $$E(X|Y=y)=E(X|X=y)=y$$ On the other hand, if $y=t$, then $\{Y=t\}=\{X\geq t\}$ so in this case $$E(X|Y=y)=E(X|X \geq t)$$ The RHS evaluates to $$\int_{t}^{\infty}\frac{xf_X(x)}{P(X\geq t)}dx=t+1$$ Here we're using the fact that $X\sim \text{Exponential}(1)$.


问:
How to formally show that $f(z)$ is analytic at $z=0$?
Let $z$ be a complex number. Let $$f(z)=\dfrac{1}{\frac{1}{z}+\ln(\frac{1}{z})}.$$ How to formally show that $f(z)$ is analytic at $z=0$?
I know that for small $z$ we have $$\left|\tfrac{1}{z}\right|>\left|\ln(\tfrac{1}{z})\right|$$ and that implies $|f(0)|=0.$
Are there multiple ways to handle this ?</p>

答:
To me, "analytic" means "locally represented by its Taylor series". With this interpretation $f$ is not analytic. Indeed, suppose $f(z)=z^r\sum_{n=0}^\infty c_n z^n $ in a neighborhood of $0$, where $c_0\ne 0$. Then
$$\frac{1}{z}+\ln \frac{1}{z} = z^{-r} \sum_{n=0}^\infty b_n z^n $$ in some (possibly smaller) neighborhood of $0$. It follows that $\ln \frac{1}{z}$ has a pole or a removable singularity at $0$. If this is not evidently absurd already, apply the same to $\ln z=-\ln \frac{1}{z}$ and conclude that the logarithm is a rational function. </p>


问:
Perfect image of locally finite closed coverings.
Let  $A_i, i \in I$ is a locally finite closed covering of $X$ and $p:X\rightarrow Y$ be a perfect map(closed continuous surjective map such that $p^{-1}\{y\}$ is compact for each $y \in Y$). Then is it true that the image of $A_i, i \in I$ is a locally finite closed covering of $Y$?
How can I prove it?

答:
HINT: Let $\mathscr{A}=\{A_i:i\in I\}$, and let $K$ be a compact subset of $X$. Each $x\in K$ has an open nbhd $U_x$ that meets only finitely many members of $\mathscr{A}$. Since $K$ is compact, there is a finite $F\subseteq X$ such that $K\subseteq\bigcup_{x\in F}U_x$. Let $U=\bigcup_{x\in F}U_x$; $U$ is an open nbhd of $K$ that meets only finitely many members of $\mathscr{A}$.

Now let $y\in Y$, let $K=f^{-1}[\{y\}]$, and let $J=\{i\in I:U\cap A_i\ne\varnothing\}$; $J$ is finite. Let $C=\bigcup_{i\in I\setminus J}A_i$; $C$ is closed — why? — so $f[C]$ is closed in $Y$. Now consider the open set $V=Y\setminus f[C]$.


问:
Solve $f''+\lambda f'=g$ with Fourier series.
Suppose that $f\in\mathcal{C}^2$ and $g\in\mathcal{C}^1$. Furthermore, they are such that $g,g'\in\mathcal{PC}(2\pi)$ and $f,f',f''\in\mathcal{PC}(2\pi)$ where $\mathcal{PC}(2\pi)$ is the set of continuous by parts and $2\pi$-periodic functions. Suppose also that $f$ and $g$ satisfy the differential equation

$$f''+\lambda f=g$$

with $\lambda\neq n^2, n=1,2,3,\cdots$. Find the Fourier series of $f$ in terms of the Fourier coefficients of $g$ and prove that it converges everywhere.

This is what I have done so far.

Let's define

$f=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos(nt)+b_n\sin(nt)$

$g=\frac{c_0}{2}+\sum_{n=1}^\infty c_n\cos(nt)+d_n\sin(nt)$

Differentiating term by term we obtain

$f''=\sum_{n=1}^\infty -n^2a_n\cos(nt)-n^2b_n\sin(nt)$

Thus, the differential equation becomes

$\frac{\lambda a_0}{2}+\sum_{n=1}^\infty (\lambda-n^2)a_n\cos(nt)+(\lambda-n^2)b_n\sin(nt)=\frac{c_0}{2}+\sum_{n=1}^\infty c_n\cos(nt)+d_n\sin(nt)$

By uniqueness of the Fourier coefficients we conclude that $a_n=\frac{c_n}{\lambda-n^2}$ and $b_n=\frac{d_k}{\lambda-n^2}$.

Is this correct? What I am not sure is if it is valid to differentiate term by term as I did. How can I justify that?

答: You can differentiate term by term because $f \in C^2$, so the Fourier series of both $f$ and $f'$ are uniformly convergent on the circle.


问:
Derivative of Quadratic form using chain rule
I want to differentiate
$$
\frac{\partial}{\partial b} \left( [Z^{\top} (y - X b)]^{\top} W [Z^{\top}(y - X b)] \right)
$$

Note that $b : K \times 1$, $y:G \times 1$, $X:G \times K$,$Z : G \times L$,$W:L \times L$.

I know that $\frac{\partial}{\partial b} A^{\top} x A = (A^{\top} + A)x$, but I can't apply that to chain rule.

How to solve that?

(This equation is needed to solve GMM in econometrics.)

答:
Let $B := ZWZ^T$, then it should be obvious that
$$
\frac{\partial}{\partial b} \left( [Z^{\top} (y - X b)]^{\top} W [Z^{\top}(y - X b)] \right)
=\frac{\partial}{\partial b}  (y-Xb)^T B (y-Xb)
$$
Then, expand to get:
$$
\frac{\partial}{\partial b} [y^TBy - y^T BX b - b^T X B y + b^T X^T B X b
]$$
then, differentiate each term individually with respect to $b$ using standard rules as is listed on the wiki page.


问:
Inverse power of a sequence
If $x_n$ is a sequence of real such that ${x_n}^3$ has a convergent sequence ${x_{k_n}}^3$ then can we say that $x_{k_n}$ is a convergent sequence of the original sequence ?

I was thinking that inverse cube will always have a real root so this might be true.

Any insight. Thank you

答:
This is trivially true. $\sqrt[3]{x} : \mathbb{R} \to \mathbb{R}$ is a continuous function; so it commutes with limits. So if $x_{k(n)}^3 \to L$ (say), then
$$
\lim_{n \to \infty} x_{k(n)} = \lim_{n \to \infty}\sqrt[3]{x_{k(n)}^3} = \sqrt[3]{\lim_{n \to \infty}x_{k(n)}^3} = \sqrt[3]{L}
$$

Title: Inverse power of a sequence


问:
Use the Handshake Lemma to determine the number of edges in GK_n
In chess, a knight's move consists of two spaces either vertically or horizontally, followed by one space in the perpendicular direction. In this way, every knight's move results in an L shaped displacement from the original position. To model all of the legal knight's moves on an n x n chessboard, we can create a graph where each vertex represents a square on the board, and two vertices are adjacent if an only if a knight can legally move between the corresponding squares.
Use the handshake lemma to determine the number of edges in GK_n

Is GK_n always, sometimes or never Eulerian

Does GK_n always, sometimes or never contain an Euler trail

By use of the Handshake Lemma edges are twice the amount of degree sum so if you had a graph GK_4 with 16 vertices, it would have degree sum 48 and 24 edges. I'm not sure how to use this information to determine the number of edges in GK_n. The abstractness is what confuses me.

答:
Instead of the handshake Lemma, you can use the more general principle of Double Counting.

We can split up the knight's moves into those of the form $ (2, 1), (2, -1), (1, 2), (1, -2) $. There are clearly $ (n-1) (n-2)$ of each move, and hence a total of $ 4 (n-1)(n-2)$ edges in the graph.


问:
Write the solution as a Fourier Series. $u_{tt}=u_{xx},~ u(t,0)=u(t,\pi)=0,~u(0,x)=1, u_t(0,x)=0$
I'm running into a roadblock on this IBV problem

$$u_{tt}=u_{xx},~ u(t,0)=u(t,\pi)=0,~u(0,x)=1, u_t(0,x)=0$$

Assuming separability:

$$u(t,x)=w(t)v(x) \implies  w^{\prime\prime}(t)=\lambda w(t)~~\mbox{and}~~v^{\prime\prime}(x)=\lambda v(x)$$

If $\lambda=-\omega^2<0$ then
\begin{align}
w(t) &= \sin(\omega t) \ \mbox{or} \ w(t) = \cos(\omega t) \\ v(x) &= \sin(\omega x) \ \mbox{or} \ v(x) = \cos(\omega x)
\end{align}

If $\lambda=0$ then
\begin{align}
w(t) &= 1 \ \mbox{or} \ w(t) = t \\
v(x) &= 1 \ \mbox{or} \ v(x) = x
\end{align}

If $\lambda=\omega^2$ then
\begin{align}
w(t) &= e^{-\omega t} \ \mbox{or} \ w(t) = e^{\omega t} \\
v(x) &= e^{-\omega x} \ \mbox{or} \ v(x) = e^{\omega x}
\end{align}

By initial boundary condition $u(t,0) = u(t,\pi) = 0$, we have $v(0) = v(\pi) = 0 \implies v(x) = \sin(\omega x)$.

Here's where I encounter my roadblock. By the IC $u(0,x)=1$, we have $w(0)=\frac{1}{v(x)}$ (which is not a possible solution for $w(t)$. Am I to conclude in this scenario that the function $u(t,x)$ is not separable and I must seek another strategy for finding the solution?

Further Analysis:
Based on $v(x)=\sin(\omega x)$ in conjunction with $u(0,x)=1\implies \sin(\omega x)=1$ at $w(0)=1$.  Thus, $\omega=n\pi$ and $w_n(t)=\cos(n\pi t)$ and therefore $$u_n(t,x)=\cos(n\pi t)\sin(n\pi x)~~~n=1,2,3,...$$

I think I may have solved my own problem.  If I'm making an error here, please correct me.

This brings me to a new stumbling block.  Translating this solution into a Fourier Series solution:

Logically (in my mind at least) $u(t,x)=\sum_{n=1}^{\infty}u_n(t,x)=\sum_{n=1}^{\infty}b_n \cos{(nt)}\sin(nx)$

Solving for $b_n$ is a little tricky for me here.  I'm using v(x) to find the $b_n$ values in the following integral.  I don't see how to put the result back into the Fourier Sum.

$$b_n=\frac{2}{\pi}\int_0^\pi{\sin(\omega x)\sin(kx)}=2~~~\forall ~\omega=k$$

End Further Analysis

This problem is from Chapter 4.2 of "Introduction to Partial Differential Equations" by Peter Olver.  The section is on Separation of Variables, hence the approach I am trying to use.

答:
With $v(x)=A\cos(\omega x)+B\sin(\omega x)$ and
$u(t,0)=u(t,\pi)=0\implies v(0)=v(\pi)=0\implies A=0~$ and $v(x)=B\sin(\omega x)$ and $$v_n(x)=B_n\sin(nx)~~\mbox{with}~~~\lambda=-n^2$$

Similarly, $$w_n(t)=C_n\cos(nt)+D_n\sin(nt)$$
$$u=\sum_{n=1}^{\infty}[C_n\cos(nt)\sin(nx)+D_n\sin(nt)\sin(nx)]$$
IC $u(0,x)=1\implies u=\sum_{n=1}^{\infty}[C_n\cos(0)\sin(nx)+D_n\sin(0)\sin(nx)]=\sum_{n=1}^{\infty}C_n\sin(nx)=1$
$$C_m=\frac{2}{\pi}\int_0^\pi{1\cdot\sin(mx)dx}= \frac{2}{\pi}\left[\frac{-1}{m}cos(m\pi)+\frac{1}{m}cos(0)\right]=\frac{4}{m\pi}$$

$w^\prime=\frac{-C}{\omega}\sin(\omega t)+\frac{D}{\omega}\cos(\omega t)$ and

$u_{t_n}=\left[\frac{-C_n}{n}\sin(nt)+\frac{D_n}{n}\cos(nt)\right]\sin(nx)$.  With IC $u_t(0,x)=0$ $$\implies u_t(0,x)=\sum_{n=1}^\infty\left[\frac{-C_n}{n}\sin(0)+\frac{D_n}{n}\cos(0)\right]\sin(nx)=\sum_{n=1}^\infty\frac{D_n}{n}\sin(nx)=0$$ $$\implies D_n=0$$
Thus, $$u(t,x)=\frac{4}{\pi}\sum_{n=1}^\infty\frac{\cos((2n-1)t)\sin((2n-1)x)}{(2n-1)}$$


问:
Relations between elements and eigenvalues of symmetric matrix
I encountered this problem while reading Titu's Linear Algebra book and have no clues how to start to solve this. Any helps is appreciated.

Let $A \in M_n(\mathbb{R})$ be a symmetric matrix. Let $\lambda_1, \dots , \lambda_n$ be the eigenvalues (counted with multiplicities) of $A$. Prove that
\begin{align*}
\sum_{i,j=1}^n a^2_{ij} = \sum_{i=1}^n\lambda^2_i.
\end{align*}

答:
Let's denote $\langle A, B \rangle = tr(A^t B)$, this is called the Frobenius inner product. And by elementary computation, we see that $\langle A, A \rangle = \sum a_{ij}^2$. Now, as $A$ is symmetric and real we can apply the spectral theorem to write $A = PDP^t$ where $P^t=P^{-1}$ and $D$ is diagonal. Of course, the entries on the diagonal of $D$ are precisely the eigenvalues of $A$. Now, we compute as follows:

$$
\begin{align*}
\sum a_{ij}^2 &= \langle A, A \rangle\\
&= tr(A^tA)\\
&= tr((PDP^t)^t (PDP^t))\\
&= tr(P D^t P^t P D P^t)\\
&= tr(PD^2 P^t)\\
&= tr(D^2)\\
&= \sum \lambda_i^2.
\end{align*}
$$


问:
expected value of random variable defined in terms of uniformly distributed variables

For each $1\leq i\leq n, $ let $U_i$ be a random variable with the $U(0,1)$ distribution, and assume the $U_i$'s are independent. Define $N = \min\{ n : (\sum_{i=1}^n U_i \geq k)\}, k\in\mathbb{R}^+.$ What is the expected value of $N$? If $k\to \infty,$ what would the expected value approximately be?

I know the definition of expected value for continuous variables: $E[g(X)] := \int_{\mathrm{Range}(X)} g(x)f(x)dx.$ Here I think it makes sense to consider $N$ to be the minimum possible sum of the $U_i$'s that exceeds $k$. So it seems difficult to find a concrete formula for $N,$ but this might help find the expected value. Also, $N$ can vary quite a bit; many of the $U_i$'s may be $0.$ Maybe the range $(0,1)$ has some significance; changing it may result in a different answer, or maybe there is some easy generalization to the case where each $U_i\sim U(a,b)$?

答:
(Well, at first, I didn't want to send this reply. But with the fear that no one would give you another answer, I'll give your question a start).

If you have learned about Markov chain, $N$ is just a stopping time.

But I think you still did not touch anything related to that subject.

So in this case, you may notice that $S_n= \sum_{i=1}^n U_i$ is strictly increasing wrt to $n$.

Hence, for all $n \in \mathbb{N}^*$, we have:

$\mathbb{P}( N > n ) =  \mathbb{P}( S_n < k )=\mathbb{P}\left( \frac{S_n-\mathbb{E}(S_n)}{\sqrt{n}} < \frac{k-n/2}{\sqrt{n}} \right) =(*)$

and, $\mathbb{E}(N)= \sum_{n \ge 0}\mathbb{P}( N > n )$

The term $(*)$ can be controlled by some approximations on the tail of a normal variable. But this is somehow messy so I don't go any further.


问:
Is there an elegant way to find the equation of a plane given its line and angle of intersection with another plane?
This is a problem I give sometimes to my calculus students:

Let $\mathcal{Q}$ be the plane defined by $x+2y+3z=6.$ Find a plane $\mathcal{P}$ that meets $\mathcal{Q}$ at an angle of $\theta = \frac{\pi}{3}$ such that $\mathcal{P},\mathcal{Q}$ intersect along the line $\langle 3,0,1 \rangle+t \langle -3,1,\frac{1}{3} \rangle.$ (Hint: Write $\mathcal{P}$ as $ax+by+cz=6$, and use the conditions to solve for $a,b,c.$)

The intended solution: We know that $\mathcal{P}$ contains the point $\langle 3,0,1\rangle$ and also $\langle 3,0,1 \rangle+t \langle -3,1,\frac{1}{3} \rangle$ for any $t$, in particular for $t=1.$ Then we have:
$$
\langle 3,0,1\rangle\cdot \langle a,b,c\rangle = 6,\qquad \langle 0,1,4/3\rangle\cdot \langle a,b,c\rangle = 6
$$
We also know the angle of intersection:
$$
\frac{\langle a,b,c\rangle \cdot \langle 1,2,3 \rangle}{\sqrt{a^2+b^2+c^2}\;\sqrt{14}}=\cos(\pi/3)=\frac{1}{2}
$$Putting these together, we are solving:
$$
\begin{cases}
3a+c=6\\
b+\frac{4}{3}c = 6\\
\frac{a+2b+3c}{\sqrt{14(a^2+b^2+c^2)}}=\frac{1}{2}
\end{cases}
$$Solving the first two equations for $a$ gives $c=6-3a$, $b=4a-2.$ Plugging this into the third equation gives:
$$
56=a^2+(4a-2)^2+(6-3a)^2
$$
This is just a quadratic equation in $a.$ Solving it produces
$$
\{a,b,c\}=\left\{1\mp\frac{\sqrt{273}}{13},2\mp\frac{4\sqrt{273}}{13},3\pm\frac{3\sqrt{273}}{13} \right\}
$$

Not terrible but not entirely straightforward. I'm wondering if there is a simpler way to reach this conclusion, maybe by exploiting the normal vectors of the planes?

答:
a direction within the plane $Q$  that is orthogonal to the line is
$(1,2,3)$ crossed with $(-9,3,1);$ this is $7(-1,-4,3).$  Let us name $u= (-1,-4,3).$
Length of $u$ is $\sqrt{26},$   so make a unit vector
$$  v = \frac{1}{\sqrt{26}}(-1,-4,3) $$
The unit normal to $Q$ is
$$  w = \frac{1}{\sqrt{14}} (1,2,3)  $$
Next let $$  N = v \cos \theta + w \sin \theta  $$
be a unit vector at angle $\theta$ from $v.$  You want $\theta = \frac{\pi}{3}$


问:
Why are numbers of the form $\frac{k}{3^n}$ for $0<k<3^n$ non recurrent under times $3$ map?
Why are numbers of the form $\frac{k}{3^n}$ for $0<k<3^n$ non recurrent under times $3$ map?

I am sure there is an easy answer to this question but I just cannot see it.

If you do not know, $x$ is a recurrent point if there is a subsequence $n_k$ s.t $f^{n_k}(x)\to x$.
In this case $f(x)=3x$ and we are on $S^1$

答:
Given a natural number $k$, $\frac{k}{3^n}$ will map to $\frac{k}{3^{n-1}}$. So, after $n$ iterations of the map, this takes us to an integer, hence $0$ modulo $1$, which is a fixed point of this map (the only other such being $\frac12$).

To sketch out an explicit example:

$$\frac{13}{27}\to \frac{13}9=\frac49\to\frac43=\frac13\to1=0\to0\to0\to\ldots$$

Note how the denominator decreases each time.


问:
Asymptotic formula for $\sum_{n\leq x}\mu(n)[x/n]^2$ and the Totient summatory function $\sum_{n\leq x} \phi(n)$
I would like to show (for $x \ge 2$) that $$\sum_{n \le x}\mu(n)\left[\frac{x}{n}\right]^2 = \frac{x^2}{\zeta(2)} + O(x \log(x)).$$

I already have the identity $$\sum_{n \le x}\mu(n)\left[\frac{x}{n}\right] = 1$$ but the method to show this does not extend to my case.

I don't know how to approach this problem so any advice would be very welcome!

Since $$\left[\frac{x}{n}\right]^2 = \left(\frac{x}{n}\right)^2 + O(\frac{x}{n})$$ we have $$\sum_{n \le x}\mu(n)\left[\frac{x}{n}\right]^2 = \sum_{n \le x} \mu(n) \left(\frac{x}{n}\right)^2 + O(\sum_{n \le x} \mu(n) \frac{x}{n})$$ and the value estimate is $\frac{1}{\zeta(2)}$ so it remains to show that the error estimate is $O(x \log(x))$.

答:
So you have
\begin{align*}
\sum\limits_{n \leq x} \mu(n)\biggl[\frac{x}{n}\biggr]^{2} &= \sum\limits_{n \leq x} \mu(n)\cdot\biggl(\frac{x}{n}-\biggl\{\frac{x}{n}\biggr\}\biggr)^{2} \\ &= x^{2}\sum\limits_{n \leq x }\frac{\mu(n)}{n^{2}} -2x \cdot\mathcal{O}(\log{x}) + \mathcal{O}(x)
\end{align*}

To find the first sum note that:
\begin{align*}x^{2} \sum\limits_{n \leq x } \frac{\mu(n)}{n^{2}} &= x^{2}\sum\limits_{n=1}^{\infty} \frac{\mu(n)}{n^{2}} - x^{2}\sum\limits_{n > x} \frac{\mu(n)}{n^{2}} \\ &= \frac{x^{2}}{\zeta(2)} + \mathcal{O}\biggl(x^{2}\sum\limits_{n >x}\frac{1}{n^{2}}\biggr) \\ &= \frac{x^{2}}{\zeta(2)} + \mathcal{O}\Biggr(x^{2}\Biggl[\sum\limits_{n=1}^{\infty} \frac{1}{n^{2}}-\sum\limits_{n \leq x } \frac{1}{n^{2}}\Biggr]\Biggr) \\ &= \frac{x^2}{\zeta(2)} + \mathcal{O}\biggl(\frac{x^{2}}{\zeta(2)} + \sum\limits_{n \leq x} \frac{1}{n^{2}}\biggr)
\end{align*}

and finally note that for $s > 0$ $$\sum\limits_{n \leq x} \frac{1}{n^s} = \frac{x^{1-s}}{1-s} + \zeta(s) + \mathcal{O}(x^{-s})$$


问:
Are the axioms of set theory "special" in any way?
Sometimes I see statements such as

Regarding what's required for a complete foundation, it's at least 99.9% correct to say set theory is enough.

I don't understand what that is supposed to mean, as this implies "we don't need more (of something)". More of what? Math obviously needs more axioms, e.g. the axioms of vector spaces.

The ZFC axioms postulate the existence of $\varnothing$ and state other things. On the other hand, the axioms of vector spaces postulate the existence of $0$ and state other things. Both sets of axioms are just sets of sentences in logic.

Aside from arguably being more widely used, what, if anything, makes the axioms of set theory special?

答:
You're misunderstanding the role of the $\mathsf{ZFC}$ axioms. Their goal is to basically be a "once-and-for-all" framework for doing mathematics; rather than describe any particular mathematical object necessarily, the idea is that they should describe the whole universe of mathematics. This is a universe which, among other things, contains vector spaces - so one of the things the $\mathsf{ZFC}$ axioms should be able to do is prove things about vector spaces. By contrast this just isn't something the vector space axioms are intended to do: we don't e.g. ask "Can the vector space axioms prove Fermat's last theorem?"

The $\mathsf{ZFC}$ axioms therefore need to be considered along with general implementation strategies - ideas for how a priori non-set-theoretic mathematical concepts can be implemented in set theory. These strategies - or rather strategy really, there's never much of a trick to it, it's just really tedious - mean that we can express all of mathematics in the language of sets alone in a very real sense.

Having fixed a way of expressing mathematical statements in the pure language of set theory, we now want to actually prove/disprove them. The success of $\mathsf{ZFC}$ is measured in large part by its ability to prove things about other mathematical objects, implemented according to the previously-mentioned strategies, which we can prove in "naive mathematics" already. This is what the quote is referring to: while there are indeed mathematical questions which the $\mathsf{ZFC}$ axioms do not resolve, they are few and far between (outside of mathematical logic anyways :P).


问:
Abel criterion proof
I want to prove the following statement:

Let be $\sum\limits_{k=1}^{\infty} a_k$ a convergent series and $\left(b_k\right)_{n\in\mathbb{N}}$ a monotone and bounded sequence. Then $\sum\limits_{k=1}^{\infty} a_kb_k$ is also convergent.

I know there already exist a few questions on this problem, however they mostly have additional assumptions (i.e. $\left(b_k\right)_{k\in\mathbb{N}}$ with $b_k\geq 0$ for all $k$).

My approach:

We define $A_n:=\sum\limits_{k=1}^{n} a_k$. As $A_n$ is convergent there exists a bound $A$ such that $|A_n|\leq A$ for all $n$. We know that $\left(b_k\right)_{k\in\mathbb{N}}$ is convergent and hence the sequence $\left(A_kb_k\right)_{k\in\mathbb{N}}$ is also convergent (product of two convergent sequences). Let be $n_1$ and $n_2$ two indices such that for all $n,m$ with $n>m>n_1$ it holds $|A_nb_n-A_mb_m|<\frac{\epsilon}{2}$ and for all $n,m$ with $n>m>n_2$ it holds $|b_n-b_m|<\frac{\epsilon}{2A}$. Now we define $n_0:=\max\{n_1,n_2\}$. With this in mind we apply Abel's lemma (summation by parts) and it follows for all $n>m>n_0$:

$$
|\sum\limits_{k=m+1}^{n} a_kb_k|=|A_nb_n-A_mb_m+\sum\limits_{k=m}^{n-1} A_k(b_k-b_{k+1})|\leq |A_nb_n-A_mb_m|+\sum\limits_{k=m}^{n-1} |A_k(b_k-b_{k+1})| \cdots
$$
If $\left(b_k\right)_{k\in\mathbb{N}}$ is monotonically decreasing it follows:
$$
\cdots<\frac{\epsilon}{2}+ \sum\limits_{k=m}^{n-1} |A_k|(b_k-b_{k+1})\leq \frac{\epsilon}{2}+ \sum\limits_{k=m}^{n-1} A(b_k-b_{k+1})=\frac{\epsilon}{2}+A (b_m-b_n)<\frac{\epsilon}{2}+\frac{\epsilon A}{2A}=\epsilon.
$$

If $\left(b_k\right)_{k\in\mathbb{N}}$ is monotonically increasing it follows:
$$
\cdots<\frac{\epsilon}{2}+ \sum\limits_{k=m}^{n-1} |A_k|(b_{k+1}-b_k)\leq \frac{\epsilon}{2}+ \sum\limits_{k=m}^{n-1} A(b_{k+1}-b_k)=\frac{\epsilon}{2}+A (b_n-b_m)<\frac{\epsilon}{2}+\frac{\epsilon A}{2A}=\epsilon.
$$
So in both cases $\sum\limits_{k=1}^{\infty} a_kb_k$ satisfies the Cauchy criterion and hence is convergent.

Is this correct or is there a more elegant/faster approach?

答:
It follows from Dirichlet's criterion. Indeed, suppose $b_k \le b_{k+1}$ and $\lim b_k = b$.

Then

\begin{align}
\sum_{k=1}^n a_k b_k
&= \sum_{k=1}^n a_k b - a_k (b - b_k) \\
&= b\sum_{k=1}^n a_k - \sum_{k=1}^n a_k (b - b_k).
\end{align}

The first sum converges as it is a hypothesis and the second sum's convergence follows from the aforementioned theorem.

But this is not an improvement over your argument since the usual proof of Dirichlet's criterion is done through summation by parts.

Title: Abel criterion proof


问:
why $\nu = \mu_1 - \mu_2$ is $\sigma $ additive
Let $(X,\mathcal{M},\mu_i)$ with $(i= 1,2)$ are two difference measure space.

Now if at least one of them is finite we need to prove $\nu = \mu_1-\mu_2$ is sigma-additive.

We can show this by countable partition that $\mu_1(\bigcup E_i) - \mu_2(\bigcup E_i) = \sum\mu_1(E_i) - \sum\mu_2(E_2)$

then comes the problem why we can rearange the infinite sequence to $\sum(\mu_1-\mu_2)(E_i)$?  Since the sequence does not absolutely convergence?

答:
In general, if you have two non-negative sequences of numbers $\{a_n\},\{b_n\}$, and we have $\sum_{n=1}^{\infty}a_n=\infty$, and $b:=\sum_{n=1}^{\infty}b_n$ is finite, then we can subtract their sums as usual, and you don't have to appeal to any rearrangement theorems. This is because the partial sums satisfy
\begin{align}
\sum_{i=1}^n (a_i-b_i)\geq \left(\sum_{i=1}^na_i\right) - b
\end{align}
So, if we take the limit on both sides as $n\to \infty$, we see that $\sum_{i=1}^{\infty}(a_i-b_i)=\infty$ as well. (The case for finite limits follows simply because of limit rules for sequences).


问:
Let $X_1,X_2,\dots$ be i.i.d. with cdf $F(x)$ with $\lim_{x\to\infty}x^\alpha\left[1-F(x)\right]=b$.
Problem: Let $X_1,X_2,\dots$ be i.i.d. random variables with distribution function $F(x)$. Denote the maximum of the first $n$ elements by $M_n$. Show that if
$$\lim_{x\to\infty}x^\alpha\left[1-F(x)\right]=b$$
with fixed positive constants $\alpha,b$ then $n^{-1/\alpha}M_n$ converges in distribution and identify the limiting distribution.

What I have so far: Put $Y_n=n^{-1/\alpha}M_n$. Then we have
\begin{align*}
F_{Y_n}(x)
&=P\left(\max_{1\leq k\leq n}X_k\leq n^{1/\alpha}x\right)\\
&=[F(xn^{1/\alpha})]^n\\
&=[1-P(X_1>xn^{1/\alpha})]^n\\
&=\left[1-\frac{x^\alpha n}{x^\alpha n}P(X_1>xn^{1/\alpha})\right]^n.
\end{align*}
Now I need to evaluate the limit above. It seems to me that this limit evaluates to $e^{-b/x^\alpha}$, but upon graphing this function, I note that it is not a valid CDF. Therefore, this cannot be the right answer.

Could anyone help with a hint on how to evaluate the limit above rigorously?

Thank you for your time and appreciate any feedback.

答:
To be honest, you have finished all the necessary work( except a typo in the end).

And $F(x)=e^{-b/x^{\alpha}}$ is obviously an acceptable CDF( for $\alpha>0$).

Note that: $F(x)=0$ when $x \le 0$


问:
Classify the odd primes $q$ such that a NEGATIVE number is a quadratic residue $\mod{q}$
Suppose we are given $y < -1$. I wish to classify all primes $q$ such that $y$ is a quadratic residue $\pmod{q}$, i.e. such that there exists a number $x$ satisfying $$y \equiv x^2 \pmod{q}.$$

How would I go about doing this? I'm aware that if instead $y>1$ then I can write $y$ as a product of primes and use the law of quadratic reciprocity. However, this doesn't seem to be applicable if $y<-1$, as the Legendre/Jacobi symbol requires a positive number in the denominator, and since we do not know $q$, it seems (although I may be wrong) we cannot find a meaningful positive representative modulo $q$.

答:
You can take out the $-1$ using the multiplicativity of the Legendre symbol, and use the first supplementary law:
$$\left(\frac{-a}{q}\right)=\left(\frac{-1}{q}\right)\left(\frac{a}{q}\right)=(-1)^{(q-1)/2}\left(\frac{a}{q}\right)$$</p>


问:
Suppose that X is a random variable. E(X) = µ and Var(X) = σ^2 and let c be a constant. What is E[(X − c)^2]=?
I want to know how it can be solved? one solution i did is:

VarX=Var(X−c)
VarX=E(X−c)2−(E(X−c))2 E(X−c)2=VarX+(E(X−c))2 E(X−c)2=σ2+(μ−c)2

is this answer correct?

答:
This has been addressed in multiple other questions before.  Please search the site.

That said, it is straightforward to do the decomposition of squares:

$$(X-c)^2 = (X-\mu+\mu-c)^2 = (X-\mu)^2 + 2(\mu-c)(X-\mu) + (\mu-c)^2,$$ therefore $$\operatorname{E}[(X-c)^2] = \operatorname{E}[(X-\mu)^2] + 2(\mu-c)\operatorname{E}[X-\mu] + (\mu-c)^2$$ and because $\operatorname{E}[X] = \mu$, we immediately obtain $$\operatorname{E}[(X-c)^2] = \sigma^2 + (\mu-c)^2.$$

Your work is poorly formatted and incomprehensible.


问:
Question on finding more than one group homomorphisms from one group to another
Question:

How many group homomorphisms from $(\mathbb Z/3, +)$ to $S_4$ are there?

My attempts:

For the group of $\mathbb Z/3$ with addition as the operation, I know that it refers to the group of modulo 3.

And for $S_4$, I think it refers to the symmetry group of the all the permutations of 4 elements, which is a group consisting of 24 elements.

And I know that for the group $(\mathbb Z/3, +)$ to be a homomorphism to $S_4$, I need to check if $f(ab) = f(a)f(b)$, for f being the mapping from group $(\mathbb Z/3, +)$ to $S_4$ and $a, b $ are any elements in $(\mathbb Z/3, +). $

My Question 1:
As I am not sure on the operation in $S_4$, I am quite stuck at the checking here.

My Question 2:
If $f(ab) = f(a)f(b)$ is true, it stands for there is only ONE homomorphism exist from $(\mathbb Z/3, +)$ to $S_4$, right? But I think the question gave hints that there is more than ONE homomorphism from $(\mathbb Z/3, +)$ to $S_4$. So how to find out the other group homomorphisms from $(\mathbb Z/3, +)$ to $S_4$?

Please assist, thank you so much!

答:
The group operation for $S_4$ is composition of functions. Elements of $S_4$ are just bijective functions $\{1,2,3,4\} \longrightarrow \{1,2,3,4\}$, so this is sensible.

The homomorphism property does not imply that there is only one homomorphism from $\mathbb Z/3$ to $S_4$. In fact, there are many. Here's how I would go about finding them.

So the elements of $\mathbb Z/3$ are the equivalence classes $\{0, 1, 2\}$. Now, in this group we of course have $1+1 = 2$ and $1+1+1 = 0$. So supposing we had a group homomorphism $f: \mathbb Z/3 \longrightarrow S_4$ then by the definition of a group homomorphism, $f(2) = f(1+1)=f(1)^2$ and $f(0) = f(1+1+1)=f(1)^3$. In other words, any group homomorphism is completely determined by what $f(1)$ is, as the values of $f(2)$ and $f(0)$ are implied by the value of $f(1)$.

So we've determined that to define a group homomorphism $f: \mathbb Z/3 \longrightarrow S_4$ we need only specify the value of $f(1)$, and the rest follows. So does that mean that there are 24 such group homomorphisms - one per element of $S_4$? No! For instance, let $f(1) = (12)$ (here I am using cycle notation). Then by what we determined above, $f(2)=(12)^2=id$ and $f(0)=(12)^3=(12)\neq id$. And there's our problem, every group homomorphism has to send the identity element of the domain to the identity element of the codomain (I'll leave this as an exercise). What this tells us is that our map $f$ must satisfy $id = f(0) = f(1+1+1)=f(1)^3$. Furthermore, it turns out that for every $\sigma \in S_4$ such that $\sigma^3 = id$, $f(1) = \sigma$ defines a group homomorphism from $\mathbb Z/3 \longrightarrow S_4$. There is more abstract and general reasoning you can apply to prove that, but these groups are pretty small so you could instead find all $\sigma \in S_4$ and check manually that $f(1) = \sigma$ yields a well defined group homomorphism. I'll give you a hint: there are exactly 9 such elements. Thus, there are 9 such group homomorphisms.

So this method I'm using is a special case of something more general, which was hinted in the comments. I first said that where 1 is sent determines where the rest of the group is sent. This means that $1$ "generates" $\mathbb Z/3$. Another examples is $\mathbb Z/3 \times \mathbb Z/3$, which is generated by the set $\{(1, 0), (0, 1)\}$. I then noted that not every choice of where 1, the generator, is sent is valid. This is because $1 \in \mathbb Z/3$ satisfies the "relation" $1+1+1=0$. Thus, whatever it is mapped to must satisfy the same relation on the other side: $f(1)*f(1)*f(1)=id$.

I can't go into much more detail than that without knowing your level of knowledge, but to learn more about this very powerful general method, you would want to learn about free groups and quotients.


问:
Need help with: given $\forall x.\forall y.p(x,y)$, use the Fitch System to prove $\forall x.\forall y.p(y,x)$.
as per the title I need help with this question:

Given $\forall x.\forall y.p(x,y)$, use the Fitch System to prove $\forall x.\forall y.p(y,x)$.

I have got to the point where I have used universal elimination to get $p(a,b)$ with $a$ and $b$ being my dummy variables but I do not know how to use universal introduction to introduce the $x$ and $y$ back in.

Using this Fitch system
http://intrologic.stanford.edu/coursera/exercise.php?exercise=exercise_08_05

答:
The trick with Stanford's implementation is that universal elimination substitutes the bound variable for the input, while universal introduction just binds the input variable.  So you have to go through a little rigmarole involving a third variable to swap.

Here's the last few lines to give you the idea.

$$\rm{AX:AY:p(X,Y)\\\vdots\\AX:p(Z,X)\hspace{10ex}\textit{Try to get here}\\AZ:AX:p(Z,X)\hspace{5ex}\textsf{Universal Introduction (input Z)}\\AX:p(Y,X)\hspace{10ex}\textsf{Universal Elimination (input Y)}\\AY:AX:p(Y,X)\hspace{4.5ex}\textsf{Universal Introduction (input Y)}}$$


问:
Finding the expectation of two absolutely continuous variables that doesn't have a joint distribution
I'm having yet another confusion with my probability homework and I would love some help.

3.2.5: $X$ follows Uniform$[3,7]$, $Y$ follows Exponent$(9)$. Find $\mathop{{}\mathbb{E}}(-5X -6Y).$

3.2.6: $X$ follows Uniform$[-12,-9]$, $Y$ follows N$(-8,9)$. Find $\mathop{{}\mathbb{E}}(11X+14Y+3).$

For the life of me, I don't understand how 3.2.5 $\neq 5E(X) + -6E(Y)$, and how 3.2.6 $\neq 11E(X)+14E(Y)+3$. I must be missing some theory here.  Is there a theorem that could help me out?

答: The linearity of expectation does not require the random variables to be independent.  For any random variables $X$, $Y$ with finite expectation, and for any scalars $a, b, c$, we have $$\operatorname{E}[aX + bY + c] = a \operatorname{E}[X] + b \operatorname{E}[Y] + c.$$


问:
Inner product justification with an example
i would need an explanation with an example on why is the inner product defined as it is and not like a dot product:

\begin{align}
\text{inner pr.:}& & &\boxed{a\cdot b = a_1\overline{b_1} + a_2\overline{b_2}+...a_n\overline{b_n}}\\
\text{dot pr.:}& & &\boxed{a\cdot b = a_1b_1 + a_2b_2+...a_nb_n}
\end{align}

Here it says:

For vectors with complex entries, using the given definition of the
  dot product would lead to quite different properties. For instance the
  dot product of a vector with itself would be an arbitrary complex
  number, and could be zero without the vector being the zero vector.

Can anyone provide me an example on this and show me how the definition for inner product above solves this?

答:
Hint: With $u = (1+i, 0)$, what is the dot product $u\cdot u$?

Hint: With $v = (1, i)$, what is the dot product $v \cdot v$?

How do we classify all 2-d vectors such that $v \cdot v = 0$?


问:
How to draw a diagram for this right triangle problem?
Recently I came across this problem:

Mathlete was 100 yards from the foot of a vertical rock climbing all and could spot Slowpoke at a 60°angle of elevation still on his way up the wall.  Speedy reached the top of the wall and spotted his car on the ground at a 60° angle of depression.  The car was 425 yards from the foot of the wall.  How many more yards up the wall did Slowpoke need to climb?(Assume that Mathlete, Slowpoke, Speedy, and the care are all on the same vertical plane perpendicular to the wall.)

I'm having a really hard time drawing the diagram for this.  For one, I'm not sure which angle they mean when they say elevation and depression, and the last part saying they are all on the same vertical plane perpendicular to the wall is really confusing me.  I thought that Mathlete was on the ground and Speedy and Slowpoke are on the wall?  I'd really appreciate it if someone could either provide me with a diagram or walk me through how to get one.  Thank you!

答:
Angles of elevation and depression are always measured from the horizontal. Letting $C,M,F,S,P$ represent the car, Mathlete, the foot of the wall, Speedy and Slowpoke respectively, we get

The wall itself is $FP$. The "same perpendicular plane" condition prevents the problem from becoming 3D (and therefore introducing new variables that would have to be resolved).


问:
On the correspondence between normal Lie subgroup and ideal of Lie algebras.
I have proved the following:

Proposition. Let $G$ be a Lie group and let $H$ be an embedded Lie subgroup of $G$. Let $\mathfrak{g}$, respectively $\mathfrak{h}$, be the Lie algebra of $G$, respectively of $H$.

$1.$ If $H$ is normal in $G$, then $\mathfrak{h}$ is an ideal of $\mathfrak{g}$.

Assume that both $G$ and $H$ are connected.

$2.$ If $\mathfrak{h}$ is an ideal of $\mathfrak{g}$, then $H$ is normal in $G$.

In the proof of $2$, I actually established without using the connectedness of $H$ and $G$ that for all $X\in\mathfrak{g}$ and $Y\in\mathfrak{h}$, one has $\exp(X)\exp(Y)\exp(X)^{-1}$, which is only a local information. This is why I suspect that the result does not hold anymore droping the connectedness assumption on either $H$ or $G$.

If $G$ is connected and $H$ is any non-central discrete subgroup of $G$, then we have a counterexample of $2$ when $H$ is disconnected. For an explicit counterexample, let $G=\textrm{GL}(n,\mathbb{C})$ and $H$ be the subgroup generated by a non-central matrix with finite order, e.g. a diagonal matrix with distinct $n$th roots of the unity on the diagonal.

Now I am looping in vain for the following setup:

$G$ disconnected, $H$ connected, $\mathfrak{h}$ is an ideal of $\mathfrak{g}$ and $H$ is not normal in $G$.

I have already investigated a lot of real matrix Lie groups.

Any enlightenment will be greatly appreciated!

答:
The multiplication given by Tsemo is not associative.
Here is an example given by my course instructor Prof. Rui Loja Fernandes.

Let $G=\mathbb{R}^3 \ltimes S_{3}$, where $S_{3}$ acts on $\mathbb{R}^3$ by $\sigma(a_{1},a_{2},a_{3})=(a_{\sigma(1)},a_{\sigma(2)},a_{\sigma(3)})$. The connected component $G^{0} $of $G$ containing the identity is $\mathbb{R}^3\times \{I\}$, which is abelian, so the Lie algebra $\mathfrak{g}$ is abelian.

Let $H=\{(x,0,0)\mid x\in \mathbb{R}\} \times \{I\}$ be a conncted Lie subgroup of $G$. The Lie algebra of $H$ is an ideal of $\mathfrak{g}$ since $\mathfrak{g}$ is abelian.

But $H$ is not a normal subgroup of $G$.

$$((0,-1,0),(1,3))\star ((1,0,0),I)\star((0,1,0),(1,3))=((,0,1),I)\notin H$$


问:
Proving expectations for different functions
Currently studying probability from Sheldon Ross' A first course in probability. Came across the following question:

Let N be a nonnegative integer-valued random variable. For non-negative values $a_{j}, j
\ge 1$, show that:

$\sum_{j=1}^{\infty}(a_{1} + ..... + a_{j})P(N=j)$ = $\sum_{i=1}^{\infty}a_{i}P(N\ge i)$

I was able to do this just by expanding the LHS. But I have been beating my head up over the following subparts.

a. E[N] = $\sum_{i=1}^{\infty}P(N \ge 1)$

If I set all the $a_{i}$ = 1, but the issue is how do I show the LHS to be equal to E[N].

b. E[N(N+1)] = $2\sum_{i=1}^{\infty}iP(N \ge i)$.

I do not even know how to get started here even though I know that we can make use of the fact that the given random variable has some form of arithmetic summation formula .

But full answers will be greatly appreciated.

答: If you set $a_j=1$ for every $j\in \mathbb{N}$ then $$a_1+ \dots + a_j=j$$ making the LHS $E(N)$. Now if you set $a_j=j$ for every $j\in \mathbb{N}$ then $$a_1+\dots + a_j=\frac{j(j+1)}{2} $$ which makes the LHS $E\Big(\frac{N(N+1)}{2}\Big)$


答: This is just a rough sketch, but I hope it helps! I haven't labeled it so you can try to figure out some things on your own!

Title: How to draw a diagram for this right triangle problem?

Body:
Recently I came across this problem:

Mathlete was 100 yards from the foot of a vertical rock climbing all and could spot Slowpoke at a 60°angle of elevation still on his way up the wall.  Speedy reached the top of the wall and spotted his car on the ground at a 60° angle of depression.  The car was 425 yards from the foot of the wall.  How many more yards up the wall did Slowpoke need to climb?(Assume that Mathlete, Slowpoke, Speedy, and the care are all on the same vertical plane perpendicular to the wall.)

I'm having a really hard time drawing the diagram for this.  For one, I'm not sure which angle they mean when they say elevation and depression, and the last part saying they are all on the same vertical plane perpendicular to the wall is really confusing me.  I thought that Mathlete was on the ground and Speedy and Slowpoke are on the wall?  I'd really appreciate it if someone could either provide me with a diagram or walk me through how to get one.  Thank you!