答: For (a), if you want an explicit example: consider the sequence $$F_n(x) = \int^x_0 \frac{2}{\pi}\arctan(n(\tfrac 1 2-t))dt.$$ As $n \to \infty$, the integrand converges to the function which is $1$ on $[0,1/2)$ and $-1$ on $(1/2,1]$, so $F_n(x)$ converges to $F(x) = \frac 1 2 - \lvert x - \frac 1 2 \rvert$, which is not $C^1$, and hence cannot be written as the integral of a continuous function.

Title: Questions on a subset of $C[0,1]$

Body:
Problem: Let $$\mathcal{F} =\left\{F\in C[0,1]: F(x) = \int_0^x f(t) \, dt, \,\, f\in C[0,1],\, \, ||f||_\infty \leq 1\right\}. $$
Questions:

a) Show that $\mathcal{F}$ is not closed.

b) Show that $\overline{\mathcal{F}} = \{f \in C[0,1]:f(0) = 0 , \, \, \text{Lip}(f)\leq 1\}$ (Lip represents Lipschitz constant).

I'm not quite sure how to start with these two questions. For part (a), I was trying to construct a sequence that does not converge in $\mathcal{F}$ but this seems to be difficult. For part (b), I was given a hint to find $F_n \in \mathcal{F}$ so that $F_n(\frac{k}{2^n}) = (1-\frac{1}{n})f(\frac{k}{2^n})$ but I'm not too sure how does that help with this question.

Any suggestion or hint for these two parts? Thank you.


问:
A question about cyclic groups and their isomorphisms
If I generate some cyclic groups, $C_{8k}$ with $k\in \mathbb{N}$, can these ever be isomorphic to a direct sum of other cyclic groups?

I'm generating these groups using rotation matrices corresponding to the angles $\frac{2\pi n}{8k}, n=0,1,\ldots,k-1.$

Similarly for $C_{4\left(2k+1\right)}$.

Thank you for any help or direction.

答:
Cyclic groups can often be represented as direct sums of other cyclic groups. Here's the general result:

If $m, n$ are relatively prime then $C_n \oplus C_m \cong C_{nm}$.

If $m, n$ are not relatively prime then $C_n \oplus C_m$ is not cyclic.

The proof of these statements will both rely on the following fact: if $a, b \in G$ are elements of finite order and $ab=ba$ then $|ab| = lcm(|a|,|b|)$, where $|a|$ is the order of $a$. Recall furthermore that $lcm(n, m) = nm/gcd(n,m)$.

Now, let $m, n$ be relatively prime. Then $(1, 0) \in C_n \oplus C_m$ has order $n$ and $(0, 1)$ has order $m$. Then as stated above, $|(1, 1)| = lcm(n, m) = nm$. Since $|C_n \oplus C_m| = nm$, this element generates the group and it is therefore cyclic. This, by the way, is part of the Chinese Remainder Theorem.

Suppose now that $(a, b) \in C_n \oplus C_m$ and let $|a| = d$, $|b| = e$. Then of course, $d \mid n, e \mid m$. Now, we have $|(a, b)| = de/gcd(d,e)$. If $d < n$ or $e < m$ then certainly $de/gcd(d,e) < nm$. If $d=n, e=m$ then $|(a, b)| = nm/gcd(n,m) < nm$ as we assumed $gcd(n,m) > 1$. Hence, no element of $C_n \oplus C_m$ has order $nm$ so it is not cyclic.

ContentLicense: CC BY-SA 4.0


问:
show that $f\in C^{1}$ if $f_{n}$ is a sequence of twice differentiable functions
Given a sequence of twice differentiable functions $f_{n}: E → R$  where $E$ is an interval,
suppose that $f_{n} → f$ pointwise to $E$, that $(f′_{n} (x_{0}))$ is bounded for some $x_{0}∈E$ and that $(f′′_{n})$ is uniformly bounded on $E$. Show that $f∈C^{1}$.

I have a idea to show this, first I want to use the theorem of continuity and uniformly convergence to show that sequence $f'$ is continuos , but I need to show that $(f_{n}')\rightarrow f'$ uniformly and I was looking in theorem of differentiable and uniformly convergence to do this and but again I need to show that sequence $(f′′_{n})\rightarrow f′′$ uniformly and I don't how I do after. If you can give me some advice ,I will be grateful. Thank you.

答:
I suppose you are considering a compact interval $[a,b]$. Let $c$ be any piont in $[a,b]$.

$(f_n')$ is equi-continuous since by MVT. By Arzela-Ascoli theorem there is a subsequence $(f_{n_i})$ converging uniformly to a continuous function $g$. Now $f(x)=\lim f_{n_i}(x)=\lim f_{n_i}(c)+\lim \int_c^{x} f_{n_i}' (t) dt=f(c)+\int_c^{x} g(t)dt$. Since $g$ is continuous it follows that $f$ is continuously differentiable.


问:
Integrate ${\sec 4x}$
How do I go about doing this? I try doing it by parts, but it seems to work out wrong:

$\eqalign{
  & \int {\sec 4xdx}   \cr
  & u = \sec 4x  \cr
  & {{du} \over {dx}} = 4\sec 4x\tan 4x  \cr
  & {{dv} \over {dx}} = 1  \cr
  & v = x  \cr
  & \int {\sec 4xdx}  = x\sec 4x - \int {4x\sec 4x\tan 4xdx}   \cr
  & \int {4x\sec 4x\tan 4xdx} :  \cr
  & u = 4x  \cr
  & {{du} \over {dx}} = 4  \cr
  & {{dv} \over {dx}} = \sec 4x\tan 4x  \cr
  & v = {1 \over 4}\sec x  \cr
  & \int {4x\sec 4x\tan 4xdx}  = x\sec 4x - \int {\sec 4x} dx  \cr
  & \int {\sec 4xdx}  = x\sec 4x - \left( {x\sec 4x - \int {\sec 4xdx} } \right) \cr} $

I don't know where to go from here, everything looks like it equals 0, where have I went wrong?
Thank you!

EDIT: Is there an easier way to do this?

答:
The standard trick is $$\sec(4x)=\frac{(\sec 4x)(\sec 4x +\tan 4x )}{\sec 4x + \tan 4x}=\frac{\sec^2 4x+\sec 4x \tan 4x}{\sec 4x + \tan 4x}$$

Now take $u=\sec 4x + \tan 4x$, we have $du=4 \sec 4x \tan 4x + 4 \sec^2 4x$, so $\sec 4x dx=\frac{du}{4u}$.

ContentLicense: CC BY-SA 3.0

Title: Integrate ${\sec 4x}$


问:
Find the limit as $n \rightarrow \infty$ of $\sum_{k=1}^{n} \frac{k^3}{n^4}$
I would appreciate suggestions or hints for this homework question.

I'm asked to find the limit as $n \rightarrow \infty$ of the above series. The question is in a chapter on Riemann integration, so I assume that I need to formulate it as a Riemann sum (although I could be wrong about that).

So far, I have expanded the sum out to the partial, $S_{N} = \frac{1}{N^4} (1^3 + 2^3 + ... + (N-1)^3 + N^3)$.

But I'm not sure where to go from here.

答:
Making the problem more general
$$S_p=\sum_{k=1}^{n} \frac{k^{p-1}}{n^p}$$
Faulhaber's formula write
$$\sum_{k=1}^{n} k^{p-1}=\frac {n^p} p+\frac {n^{p-1}}2+\cdots$$ that is to say that
$$S_p \sim \frac{\frac {n^p} p+\frac {n^{p-1}}2+\cdots }{n^p}=\frac 1 p+\frac 1 {2n}+\cdots$$ and so, you have not only the limit but also how it is approached when $n\to\infty$.


问:
Simplify the Boolean expression A'BC + A'BD' + AB' + AC + ABC + ACD
I'm trying to simplify the expression: A'BC + A'BD' + AB' + AC + ABC + ACD

I got to the expression: BC + AC + A'BD' + AB'

I know this expression is equvalant to the one above because I verified using truth tables.

The simplified expression, according to online calculators is: BC + A'BD' + AB'

How do I continue with my simplification?

答:
You can finish as follows writing $1$ for "True":

\begin{eqnarray*}BC+AC+A'BD' + AB'
& = & BC + A(B+B')C +A'BD' + AB' \\
& = & (1+A)BC + AB'(C+1) + A'BD' \\
& = &  BC + AB' + A'BD'
\end{eqnarray*}


问:
To determine the cardinality of the rotation group of a cube
In Dummit and Foote's Abstract algebra a problem is given as-

Let $G$ be the group of rigid motions in $\mathbb R^3$ of a cube. Show that $|G|$=24.

For an n-gon, all the points on the n-gon can be determined by fixing the two vertex points on the n-gon.

For a 3-D solid, by fixing three adjacent vertex points on the solid all other points on the solid can be determined. This is easy to prove.

So if we label the vertices ("1","2",...., "8") of the solids and take into consideration the vertex "1". It has eight possibilities to take position, vertex "2" has 3 possibilities and the vertex "3" has 2 possibilities. So total symmetries are 8×3×2=48.

Out of 48 some are rotations and some are reflections.

How to conclude that 24 symmetries are rotations?

Like in planar n-gons, the rotation $r^k=\frac{2\pi k}{n}$, so there comes out to be n rotations. But in cube like how we define these rotations?

I have seen related post here, but some theorems are used to prove this, I have not studied these theorems yet as I am beginner to the abstract algebra.

Please clarify the doubt.

答:
Hint: If it's a rigid motion in $\Bbb{R}^3$, then it's orientation-preserving, so you don't actually have 2 possibilities for where vertex 3 goes, once you've picked vertex 1 and 2.

Edit: "Orientation-preserving" means, informally, left-handed things don't get changed to right-handed. Rotations are orientation preserving.

In the picture, I cannot rotate the cube in such a way that 1 and 2 stay put, but 3 and 4 swap places. I would need to do a reflection, which would reverse orientation. The vertices around 2 have to go 1, 3, 4 in clockwise order no matter how I rotate the cube, so if I pick locations for 1 and 2, that automatically determines where 3 and 4 end up as well.


问:
Computing the characteristic polynomial of a linear transformation
I am confused as to how to solve this problem, as the linear transformation includes an arbitrary inner product.

Let $V$ be a finite-dimensional inner product space. Fix vectors $v,w \in V$ and define $S_{v,w} (x) = \langle x,v \rangle w$. What is the characteristic polynomial of $S$?

I cannot figure out how to get started, nor do I have any intuition for what the eigenvalues of $S$ are. The vector space is arbitrary, so I cannot choose a basis. With a bit of help, I am in the process of proving that $x$ is an eigenvector of $S$ if and only if $w = ax$ for some scalar $a$ or $\langle x, v \rangle = 0$. In the latter case, $x$ has eigenvalue $0$. In the former case, $x$ has eigenvalue $\langle x, v \rangle a$. Where I am stuck, however, is the following: prove the reverse direction (that these are the only eigenvalues), finding the dimensions of the eigenspaces, and finding exactly how many eigenvectors (or eigenspaces) there are. The factors of the characteristic polynomial are $(z - \lambda_i)^{\dim V_{\lambda_i}}$ where $\dim V_{\lambda_i}$ is the dimension of the eigenspace.

答:
If $v$ or $w$ is the zero vector, then $S_{v,w}$ is the zero map whose characteristic polynomial is $x^n$.

Assume that both $v$ and $w$ are non-zero. Let $w_1 = \frac{w}{\|w\|}$ and complete $w_1$ to an orthonormal basis $w_1, w_2, \dots, w_n$. Let $v_1 = \frac{v}{\|v\|}$ and complete $v_1$ to an orthonormal basis $v_1, v_2, \dots, v_n$.

First, assume that $\langle w_1, v_1\rangle \ne 0$. Since every linear combination of $v_2, \dots, v_n$ is orthogonal to $v_1$, we see that $w_1$ is linearly independent from $v_2, \dots, v_n$. Thus, $w_1, v_2, \dots, v_n$ is a basis. Moreover,

$$
S_{v,w}(w_1) = \lambda w_1 \\
S_{v,w}(v_2) = 0 \\
\dots \\
S_{v,w}(v_n) = 0 \\
$$

where $\lambda = \langle w, v \rangle$. Thus, we have found the basis of eigenvectors of $S_{v,w}$ (and $w$ is indeed an eigenvector as you expect). In this basis, the matrix of $S_{v,w}$ is diagonal with $(\lambda, 0, \dots, 0)$ on the diagonal so its characteristic polynomial is

$$
f(x) = x^{n-1}(x - \lambda).
$$

Now, assume that $\langle w_1, v_1\rangle = 0$. Note that for every $x \in V$, $S_{v,w}(x)$ is a scalar multiple of $w_1$. Therefore, $S_{v,w}(S_{v,w}(x)) = 0$ for every $x \in V$. In other words, $S_{v,w}$ is nilpotent and so its characteristic polynomial is

$$
f(x) = x^n.
$$

Finally, note that these two cases can be written together as

$$
f(x) = x^{n-1}(x - \langle w, v \rangle).
$$


问:
Importance and Intuition of Polynomial Rings
I know that a polynomial ring $R[x]$ is the ring with elements consisting of polynomials with coefficients in $R$. However, this definition leaves me confused when I try to really understand the concept of polynomial rings, rather than just accept that definition for what it is. Where did the polynomials come from? Why are polynomial rings important? Is there a way to “construct” the polynomials in the context of Abstract Algebra, so that we may see why studying rings made up of polynomials is important? Are polynomial rings actually just sets of polynomials, or are the polynomials just a concrete way to represent some more abstract idea? What does the variable $x$ mean in polynomial rings? I apologize for all the questions, I’m just having trouble seeing the importance or intuition behind them.

答:
I will elaborate my answer on how to see the polynomial ring without the indeterminate $x,$ and on the universal property of polynomial rings. I will show you two alternative ways to understand the definition of polynomial rings. Those interpretations will allow you to see this construction without wondering about the role of indeterminate(s). Let $R$ be a commutative ring with unity $1_R.$

Monoid Ring on the Natural Numbers $R[\mathbb{N}]$:

This is the monoidal version of the group ring construction over the monoid of natural numbers. Here objects are formal finite linear combinations of natural numbers with coefficients in $R.$ For example $$r_1n_1+r_2n_2+\cdots+r_jn_j,\qquad r_1, r_2, \cdots, r_j\in R,\quad n_1, n_2,\cdots, n_j\in\mathbb{N}.$$ It is not difficult to see that this is another way to look at the polynomial ring by interpreting above element as the polynomial $$r_1x^{n_1}+r_2x^{n_2}+\cdots+r_jx^{n_j}\in R[x].$$ Furthermore, if you use $\mathbb{N}^n$ (ordered $n$-tuples of natural numbers) in this construction, you would get the  polynomial ring $R[x_1, x_2,\cdots, x_n]$ with $n$ variables.

Ring of $R$-sequences of finite support $R^{(\mathbb{N})}$:

Another way to look at the linear combination $r_1n_1+r_2n_2+\cdots+r_jn_j$ is as a function $$f:\mathbb{N}\to R$$ whose values are $f(n_i)=r_i$ for all $i=1, 2, \cdots, j$ and zero everywhere else. Since the domain of this function is non-negative integers, this is nothing but a sequence. Also, the sequence is zero for all but finitely many terms, therefore we say it is finitely supported. On the other hand if you are given a such sequence we can easily create a unique polynomial (and a linear combinaion). For example: Say $R=\mathbb{Z}/3\mathbb{Z},$ then the finitely supported sequence $(1, 1, 1, 0, 0, 0, \cdots)$ represents the polynomial $1+x+x^2=(2+x)^2$ and vise versa. Note that even this factorization carries all the information about the associated sequence.

Above two equivalent (ring isomorphic) constructions shows us that powers of indeterminate $x$ are nothing but placeholders. We can identify $x$ with the linear combination $1_R1$ or with the sequence $(0, 1_R, 0, 0, \cdots).$ Also, if we accept any of them as the polynomial ring, say the first construction, then we have an injective ring homomorphism (embedding) $$\iota : R \to R[\mathbb{N}]$$ given by $r\mapsto r0$ (or in the other construction $r\mapsto (r, 0, 0, 0, \cdots)\in R^{(\mathbb{N})}$). This map is universal in the sence that any other such map factor through $\iota$ while sending $x$ to a priscribe value. This is the universal property of polynomial ring (of a single variable) over $R$:

For every commutative unital ring $S$ that has an injective ring homomorphism $R\xrightarrow{\phi}S$ and for any chosen $a\in S,$ there is a unique extension (evaluation homomorphism) $R[\mathbb{N}]\xrightarrow{\hat\phi_a}S$ such that $\hat\phi_a\circ\iota=\phi$ and $\hat\phi_a(x)=a.$
$$\require{AMScd}
\require{cancel}
\def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1}
   \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}}
\begin{CD}
&& R[\mathbb{N}]\\
& \diaguparrow{\iota} @VV\hat\phi_a V \\
R @>>\phi> S
\end{CD}$$

One can easily formulate similar universal property for polynomial ring with any number of variables.


问:
Continuous functions and open sets/covers
I'm taking a course in elementary algebraic geometry, but I seem to be lacking a topological background. The following result is often used:

Let $X,Y$ be topological Spaces, and $U_i$ for $i \in I$ an open cover for $X$ (the convention here is that $\bigcup_{i \in I} U_i = X$). Then, a function $f: X \to Y$ is continuous if and only if $f|_{U_i}: U_i \to Y$ is continuous for $\forall i$ where $U_i$ is endowed with the subspace topology.

How does one prove this? I've not seen this lemma before. (Apparently it's called 'local property of continuity'?)

答:
One direction is trivial. For the other direction, suppose that $f\upharpoonright U_i$ is continuous for each $i\in I$. Let $V$ be an open set in $Y$; to show that $f$ is continuous, we need to show that $f^{-1}[V]$ is open in $X$.

For each $i\in I$ let $W_i=(f\upharpoonright U_i)^{-1}[V]$; by hypothesis $W_i$ is open in $U_i$, and since $U_i$ is open in $X$, this means that $W_i$ is open in $X$. And clearly

$$f^{-1}[V]=\bigcup_{i\in I}W_i\,,$$

so $f^{-1}[V]$ is open in $X$.


问:
How to find the numbers efficiently given their sum and product?
Problem Statement: This is a rather simple problem to describe. You will be given three numbers $S,P$ and $k$. Your task is to find if there are integers $n_1,n_2,\dots,n_k$ such that $n_1+n_2+\dots+n_k=S$, and $n_1 n_2\dots n_k=P$.

If such integers exist, print them out. If no such sequence of integers exist, then print "NO".

For example if $S=11,P=48$, and $k=3$ then $(3,4,4)$ is a solution. On the other hand, if $S=11,P=100$ and $k=3$, there is no solution and you should print "NO".

Given that, $1\leq k \leq 4$, $1 \leq S \leq 1000$ and $1\leq P \leq 1000$.

My Approach: $n_1,n_2,\dots,n_k$ have to satisfy two conditions: $n_1+n_2+\dots+n_k=S$, and $n_1 n_2\dots n_k=P$.

The second condition implies that $n_i$ is a divisor of $P$ for all valid $i$, hence we can narrow down the search using this fact**. So I factored the number $P$ and stored its divisors. Next I implemented a recursive solution to fill each variable and solve the sub problem with one less variable to fill each time. I did this only for $k-1$ variables as the $k$-th variable is fixed automatically by the first condition.

** I came to this conclusion because I found that any number less than or equal to $1000$ has at most $32$ divisors (for reference https://www.geeksforgeeks.org/querying-maximum-number-divisors-number-given-range/), which means a total of $64$ (taking both the negative as well as positive divisors as the problem only says integers and not positive integers) resulting in a total of $64^4 = 16777216$ possibilities which can easily be searched through in time.

I did get an "AC" (i.e. the solution passed all the test cases (using only the positive numbers, though nothing like "only positive integers" was mentioned)).

But I just wanted to know that can we do better, and find the solution more efficiently, because this approach will take up a lot of time with more number of variables. Is there any other way to solve this problem in that case?

答:
I'm not sure I understand exactly what you did, but limiting the possibilities to the factors of $P$ must be the most important step.  The biggest any single summand can be is $S-(k-1)$ which may allow one to substantially cut down on the possibilities.  In the example, $S=11, P=100, k=3$ this observation immediately limits the possible $n_i$ to $1,2,4,5$.

You also may be able to cut the process short.  When you find out that you can't do it if $n_3=5$, you can note that the largest product you can make is $4^3=64<100$ and so it's impossible.

No doubt there are other observations one can make, but you probably reach the point of diminishing returns quickly.


问:
Integral of $\cot^2 x$?
How do you find $\int \cot^2 x \, dx$? Please keep this at a calc AB level. Thanks! </p>

答:
We use the identity: $$\cot x = \pm \sqrt{\csc^2 x - 1} \implies \cot^2 x = \csc^2 x - 1$$

So we can rewrite the integral as follows:

$$\int \cot^2 x \,dx = \int \left(\csc^2x - 1\right)\, dx$$

$$ \int \left(\csc^2x - 1\right) dx \; = \; \int \csc^2 x \, dx\; -\; \int \,dx  \;\;= \;\;-\cot x - x + \text{constant}$$

Recall, $$\dfrac{d}{dx}\left(\cot x\right) = - \csc^2 x$$ hence $$\int \csc^2 x \,dx = -\cot x + C$$

Title: Integral of $\cot^2 x$?


问:
Decay of fractional derivative of Schwartz function
Let $\phi$ be a Schwartz function and let $\alpha>0$. I want to analyze the decay as $x\rightarrow\infty$ of: $$\int_\mathbb{R}e^{2\pi i x\xi}|\xi|^\alpha\phi(\xi)\,d\xi$$Heuristically, for $\xi\ll x$, we have one oscillation from $\xi=0$ to $\xi=1/x$ at which point the amplitude is of order $|x|^{-\alpha}$ so the answer should be $|x|^{-\alpha-1}$. However, I have had trouble making this intuition precise. I have tried smoothly partitioning into $|\xi|\ll x$ and $|\xi|\gg |x|$ and separately doing a dyadic decomposition at level $|x|^{-1}$, but the details have not worked out. I am interested a precise solution as well as any heuristic ways of thinking about this sort of integral. If it would help, I can share what I've tried in more detail.

答:
Your intuition seems spot on in terms of how to think about this, so I don't really have anything to add in terms of the heuristics. I'm not sure what went awry when you tried out the dyadic decomposition, but here's how to make a fairly precise argument of it to establish that there's decay like $|x|^{-\alpha - 1}$ as $x \to \infty$.

Step 1: The dyadic decomposition

First, we can smoothly partition to $|\xi| \gtrsim 1$ and then also to $|\xi| \approx 2^{-k}$ for $k \geq 1$. The first part ($|\xi| \gtrsim 1$) more straightforwardly has rapid decay in $x$ since this avoids the lack of $C^{\infty}$ smoothness at the origin, i.e. we're taking the inverse Fourier transform of a straight Schwartz function.

We're then left to analyze the dyadic decomposition
$$
\sum_{k = 1}^{\infty}
\int_{\mathbb{R}} e^{2 \pi i x \xi}
|\xi|^{\alpha} \eta(2^k \xi) \phi(\xi) \, d\xi,
$$
where $\eta$ is the smooth function being used to produce the dyadic decomposition.

Step 2: Setting up integration by parts

As is often the case in this type of set-up, integration by parts is our friend. We use the fact that
$$
e^{2 \pi i x \xi}
= (2 \pi i x)^{-1} D_{\xi} (e^{2 \pi i x \xi}),
$$
with $D_{\xi}$ denoting differentiation with respect to $\xi$. So then we have
$$
\int_{\mathbb{R}} e^{2 \pi i x \xi}
|\xi|^{\alpha} \eta(2^k \xi) \phi(\xi) \, d\xi
= (-2 \pi i x)^{-\ell} \int_{\mathbb{R}} e^{2 \pi i x \xi}
D_{\xi}^{\ell}
\bigl[
|\xi|^{\alpha} \eta(2^k \xi) \phi(\xi)
\bigr] \, d\xi
$$
for any integer $\ell \geq 0$.

The derivative here can essentially be written as
$$
\sum_{m_1 + m_2 + m_3 = \ell}
C_{m_1, m_2, m_3, \alpha}
|\xi|^{\alpha - m_1}
\cdot 2^{m_2 k} \eta^{(m_2)}(2^k \xi)
\cdot \phi^{(m_3)}(\xi),
$$
I think modulo signum terms that come from differentiating $|\xi|$.

Step 3: Estimating the integrand

From here, the idea is to use "crude" size estimates: the derivatives of $\eta$ and $\phi$ can be bounded by additional constants (i.e. independent of $\xi$, though with dependence on $m_2$ and $m_3$), while for $\xi \approx 2^{-k}$ we have $|\xi|^{\alpha - m_1} \lesssim 2^{k (m_1 - \alpha)}$. So in all, the derivative term can be bounded by
$$
\sum_{m_1 + m_2 + m_3 = \ell}
C_{m_1, m_2, m_3, \alpha}
2^{k (m_1 - \alpha)} \cdot 2^{m_2 k}
\leq C_{\ell, \alpha} 2^{k (\ell - \alpha)},
$$
with constants different than those that appeared above.

Therefore we can bound one integral term in the dyadic decomposition as follows:
\begin{align*}
\left|
(-2 \pi i x)^{-\ell} \int_{\mathbb{R}} e^{2 \pi i x \xi}
D_{\xi}^{\ell}
[ |\xi|^{\alpha} \eta{2^k \xi} \phi(\xi) ] \, d\xi
\right|
&\leq C_{\ell, \alpha} |x|^{-\ell}
\int_{|\xi| \approx 2^{-k}} 2^{k (\ell - \alpha)} \, d\xi \\
&= C_{\ell, \alpha} |x|^{-\ell} 2^{k (\ell - \alpha - 1)}.
\end{align*}
Noting that this bound holds for any $\ell \geq 0$, we can see that if we don't actually use integration by parts ($\ell = 0$) we get a bound on the order of $2^{-k (\alpha + 1)}$, and that integration by parts provides an improvement if $|x| \gtrsim 2^{k}$.

Step 4: Re-assembling the dyadic pieces

So for a fixed $x$, we want to use integration by parts (with $\ell > \alpha + 1$) on the dyadic terms where $2^k \leq |x|$, but no integration by parts ($\ell = 0$) on the terms where $2^k > |x|$. This allows us to estimate as follows:
\begin{align*}
\left|
\sum_{k = 1}^{\infty}
\int_{\mathbb{R}} e^{2 \pi i x \xi}
|\xi|^{\alpha} \eta(2^k \xi) \phi(\xi) \, d\xi
\right|
&\leq \sum_{2^k \leq |x|}
C_{\ell, \alpha} |x|^{-\ell} 2^{k (\ell - \alpha - 1)}
+ \sum_{2^k > |x|} C' 2^{-k (\alpha + 1)} \\
&\leq C_{\ell, \alpha} |x|^{-\ell} \cdot |x|^{\ell - \alpha - 1}
+ C_{\alpha} |x|^{-\alpha - 1} \\
&= C_{\alpha} |x|^{-\alpha - 1}.
\end{align*}
At the end we've dropped the constant's dependence on $\ell$ since we can really think of $\ell$ depending on $\alpha$ (for example by taking $\ell$ to be the be the smallest integer greater than $\alpha + 1$).

This gives the claimed decay of the original (inverse) Fourier transform as $x \to \infty$. Of course, if $\phi$ vanishes to some degree at the origin we can get a correspondingly higher degree of decay at $\infty$, but in general $|x|^{-\alpha - 1}$ is the best we can do.


问:
Probability of getting one item from $n$ items after $x$ tries, such that in each try a picked unwanted item is removed.
Suppose there's items $A,B,C,D,E$ where the desired pick will be to pick $C$. At first try, I randomly picked $B$, hence now the pick-pool will be $A,C,D,E$ and second try I picked $E$ and now it's $A,C,D$ and so on.

So if there's $n$-items and $x$-tries ($x \leq n$). What is the probability that I get the desired item out of these random items after $x$-tries. If we have $x=1$, then the probability is $1/n$ and if $x=n$, then it's $1$

答: This is just the probability that if you pick $x$ items out of $n$, the desired object will be among those you picked, so the probability of success is $$\frac xn$$


问:
If $\mathbb f$ is analytic and bounded on the unit disc with zeros $a_n$ then $\sum_{n=1}^\infty \left(1-\lvert a_n\rvert\right) \lt \infty$
I'm going over old exam problems and I got stuck on this one.
Suppose that $\mathbb{f}\colon \mathbb{D} \to \mathbb{C}$ is analytic and bounded. Let $\{a_n\}_{n=1}^\infty$ be
the non-zero zeros of $\mathbb{f}$ in $\mathbb{D}$ counted according to multiplicity. Show that
$$ \sum_{n=1}^\infty \left( 1 - \left|a_n \right|\right)\lt\infty $$
I can understand that $\left|a_n \right|$ goes to $1$ since zeros are isolated, but it doesn't help showing the series is convergent. Any help will be appreciated!</p>

答:
This is known as Blaschke's condition and is in fact also true for functions in the so called Nevanlinna class. The simplest way to prove this is using Jensen's formula.

Assume $f \in H^\infty$. You may as well assume that $f(0) \neq 0$ and that $f$ has infinitely many zeros. Let $n(r)$ be the number of zeros in the disc $D_r$. Fix any integer $k$ and choose $r < 1$ so large that $n(r) > k$. By Jensen's formula,
$$
|f(0)| \prod_{n=1}^{n(r)} \frac{r}{|a_n|} = \exp\left( \frac{1}{2\pi} \int_0^{2\pi} \log|f(re^{i\theta})|\,d\theta \right).
$$

Hence (if $|f(z)| < M$ on $D$):
$$
|f(0)| \prod_{n=1}^{k} \frac{r}{|a_n|} \le |M|.
$$

In other words
$$
\prod_{n=1}^{k} |a_n| \ge \frac{|f(0)|}{|M|} r^k
$$
for every $k$. Letting $r \to 1$ and $k\to\infty$ it follows that
$$
\prod_{n=1}^{\infty} |a_n| \ge \frac{|f(0)|}{|M|} > 0,
$$
which implies that $\sum_{n=1}^\infty (1-|a_n|) < \infty$.


问:
Given $dy/dt = 6y/39 -(y^2)/2002$ and the point $(0,14)$, what is the original equation in "$y=\dots$" form?
This is how I tried to solve it:
$$\frac{dy}{dt}=\frac{6y}{39}-\frac{y^2}{2002}$$
I multiplied and divided:
$$\frac{dy}{\frac{6y}{39}-\frac{y^2}{2002}}=dt$$
I simplified:
$$\frac{dy}{\frac{2y}{13}-\frac{y^2}{2002}}=dt$$
I found a common denominator:
$$\frac{dy}{\frac{308y-y^2}{2002}}=dt$$
I divided the fractions:
$$\frac{2002dy}{308y-y^2}=dt$$
I completed the square:
$$\frac{2002dy}{308y-y^2-154^2+154^2}=dt$$
$$\frac{2002dy}{-(y^2-308y+154^2)+154^2}=dt$$
$$\frac{2002dy}{-(y-154)^2+154^2}=dt$$
Now I am temporarily looking at only the left side of the equation to find the integral:
$$\int\frac{2002}{-(y-154)^2+154^2}dy$$
I substituted u:
$$2002\int\frac{1}{(-1)(u)^2+154^2}du$$
At this point I was very confused so I tried to use arctan when I probably wasn't allowed to. I used the following formula:
$$\int\frac{b}{b^2+a^2x^2}dx=\arctan(\frac{ax}{b})+c$$
My integral:
$$\frac{2002}{154}\int\frac{154}{(-1)(u)^2+154^2}du$$
$$\frac{2002}{154}\arctan(\frac{iu}{154})$$
$$13\arctan(\frac{i(y-154)}{154})$$
The integral of $1dt$ is $t$ so I got this equation:
$$13\arctan(\frac{i(y-154)}{154})=t+c$$
I then solved for y:
$$\frac{i(y-154)}{154}=\tan(\frac{t+c}{13})$$
$$y=\frac{154\tan(\frac{t+c}{13})}{i}+154$$
I then solved for c by substituting the initial point (0,14) into the equation:
$$14=\frac{154\tan(\frac{0+c}{13})}{i}+154$$
$$\frac{-140i}{154}=\tan(\frac{c}{13})$$
$$\arctan(\frac{-10i}{11})=\frac{c}{13}$$
$$c=13\arctan(\frac{-10i}{11})$$
And I substituted the found $c$ value into the equation:
$$y=\frac{154\tan(\frac{t+13\arctan(\frac{-10i}{11})}{13})}{i}+154$$
Is this technically correct or did do I something wrong?
If I did do something wrong, what did I do wrong?
What would be an alternative method of solving this?

Body:
After getting some feedback I tried to resolve this going back to right after I substituted u:
$$2002\int\frac{1}{154^2-u^2}du$$
I used the common integral:
$$\int\frac{1}{a^2-x^2}dx=\frac{1}{2a}ln(\frac{a+x}{a-x})$$
And I got:
$$2002(\frac{1}{308}ln(\frac{154+u}{154-u}))=t+c$$
$$ln(\frac{154+u}{154-u})=\frac{2(t+c)}{13}$$
$$\frac{154+u}{154-u}=e^{\frac{2(t+c)}{13}}$$
$$\frac{154+u}{154-u}=e^{\frac{2(t)}{13}}*e^{\frac{2(c)}{13}}$$
$10^{\frac{2(c)}{13}}$ can be given the name "A" because it is a constant:
$$\frac{154+y-154}{154-(y-154)}=e^{\frac{2(t)}{13}}A$$
$$\frac{y}{308-y}=10^{\frac{2(t)}{13}}A$$
Multiplying the denominator and distributing:
$$y=(308)e^{\frac{2(t)}{13}}A-(y)e^{\frac{2(t)}{13}}A$$
$$y+(y)e^{\frac{2(t)}{13}}A=(308)e^{\frac{2(t)}{13}}A$$
$$y=\frac{(308)e^{\frac{2(t)}{13}}A}{1+e^{\frac{2(t)}{13}}A}$$
Then to solve for A I am going back a little to where there was only one A:
$$\frac{y}{308-y}=e^{\frac{2(t)}{13}}A$$
I plugged in the point (0,14):
$$\frac{14}{308-14}=e^{\frac{2(0)}{13}}A$$
$$A=\frac{1}{21}$$
I plugged in A to the "y =" equation
$$y=\frac{(308)e^{\frac{2(t)}{13}}\frac{1}{21}}{1+e^{\frac{2(t)}{13}}\frac{1}{21}}$$
$$y=\frac{(44)e^{\frac{2(t)}{13}}}{3(1+e^{\frac{2(t)}{13}}\frac{1}{21})}$$
$$y=\frac{(44)e^{\frac{2(t)}{13}}}{\frac{21+e^{\frac{2(t)}{13}}}{7}}$$
I divided the fractions:
$$y=\frac{(308)e^{\frac{2(t)}{13}}}{21+e^{\frac{2(t)}{13}}}$$


问:
Determinant and inverse matrix calculation of a special matrix
Is there any smart way of calculating the determinant of this kind matrix?

\begin{pmatrix}
1 & 2 & 3 & \cdots &  n \\
2 & 1 & 2 & \cdots & n-1 \\
3 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & \cdots & 1 \end{pmatrix}

I encountered this in a problem for the case $n=4$. I need to find the inverse matrix.

I doubt the idea of this problem is to calculate all the cofactors and then the inverse matrix in the usual way.

I see the pattern here and some recursive relations... but I am not sure if this helps for calculating the determinant of the existing matrix.

答:
Treat $n$ as a parameter, let

$M$ be your $n \times n$ matrix.

$\eta$ be the $n\times n$ matrix with $1$ on the lower diagonal and $0$ otherwise.

$\bar{\eta}$ be the transpose of $\eta$.

$u$ be the $n \times 1$ column vector with all entries $1$.

$e$ be the $n \times 1$ column vector with $1$ on first row and $0$
otherwise.

$f$ be the $n \times 1$ column vector with $1$ on last row and $0$

There are several algebraic facts we need.

$\eta$ and $\bar{\eta}$ are nil-potent. More precisley, $\eta^n = \bar{\eta}^n = 0$.

$\eta\bar{\eta} = I - ee^T$ and $\bar{\eta}\eta = I -ff^T$.

$uu^T = (I-\eta)^{-1} + \bar{\eta}(I-\bar{\eta})^{-1} = (I-\eta)^{-1}\eta + (I-\bar{\eta})^{-1}$.

$(I - \eta)u  = e$, $(1 - \bar{\eta}) u = f$ and $(I - \bar{\eta}) e = e$.

If one look at the lower triangular part of $M$, it equals to
$$I + 2\eta + 3\eta^2 + \cdots + n\eta^{n-1} = (I - \eta)^{-2}$$
Similar things happens to the upper triangular part of $M$. This leads to:

$$
\begin{align}M
&= (I - \eta)^{-2} + (I-\bar{\eta})^{-2} - I\\
&= (I-\eta)^{-1}(uu^T - \bar{\eta}(1-\bar{\eta})^{-1})
 + (uu^T - (1-\eta)^{-1}\eta)(1 - \bar{\eta})^{-1} - I\\
&= (I-\eta)^{-1}((ue^T - \bar{\eta})  + (eu^T - \eta))(1-\bar{\eta})^{-1} - I
\end{align}
$$
As a result, following product of matrix has a relatively simple structure:

$$\begin{align}
N \stackrel{def}{=} (1-\eta)M(1-\bar{\eta})
&= (eu^T - \eta) + (ue^T - \bar{\eta}) - (I - \eta)(I - \bar{\eta})\\
&= ee^T + eu^T + ue^T - 2I
\end{align}$$

Let $v = \frac{u - e}{\sqrt{n-1}}$. It is easy $e, v$ are orthogonal unit vectors. Extend $\{ e, v \}$ to an orthonormal basis. In the new basis, $N$ is block-diagonalized and has the form
$$N_{new} = \begin{bmatrix}
1 & \sqrt{n-1} & 0 & 0 & \ldots & 0\\
\sqrt{n-1} & -2 & 0 & 0 & \ldots & 0\\
0 & 0 & -2 & 0 & \ldots & 0\\
0 & 0 & 0  & -2 & \ldots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & 0 & \ldots & -2
\end{bmatrix}$$
Since $\det(1-\eta) = \det(1-\bar{\eta}) = 1$, we find

$$\det(M) = \det(N) = (-2-(n-1))(-2)^{n-2} = -(n+1)(-2)^{n-2}$$

In the new basis, inverting $N$ comes down to inverting a $2 \times 2$ matrix. In terms of $e, v$, we have

$$\begin{align}
N^{-1} &= \frac{1}{n+1}(2 ee^T + \sqrt{n-1}(ve^T + ev^T) - vv^T) - \frac12(I - ee^T - vv^T)\\
&= \frac{ue^T + eu^T}{n+1}
+ (\frac12-\frac{1}{n+1})vv^T - \frac12(I- ee^T)\\
&= \frac{ue^T + eu^T}{n+1} + \frac{(u-e)(u-e)^T}{2(n+1)}
 - \frac12(I - ee^T)\\
&= \frac{(u+e)(u+e^T)}{2(n+1)} - \frac12(I- ee^T)
\end{align}
$$
This leads to
$$\begin{align}M^{-1}
&= (1-\bar{\eta})N^{-1}(1-\eta)\\
&= \frac{(e+f)(e+f)^T}{2(n+1)} - \frac12((I-\bar{\eta})(I-\eta)-ee^T)\\
&= \frac{(e+f)(e+f)^T}{2(n+1)} - \frac12(I - \eta - \bar{\eta} + \bar{\eta}\eta - ee^T)\\
&= \underbrace{-I + \frac12(\eta + \bar{\eta})}_J + \underbrace{\frac{(n+2)(ee^T+ff^T) + (ef^T+fe^T)}{2(n+1)}}_K
\end{align}
$$
This implies $M^{-1}$ is very close to a tri-diagonal matrix

$$J = -I + \frac12(\eta + \bar{\eta})
\quad\text{ with }\quad J_{ij} = \begin{cases}
-1, & i = j\\
\frac12, & |i - j| = 1\\
0, & \text{ otherwise }
\end{cases}$$

The difference $K$ is non-zero only at following $4$ entries:

$$K_{11} = K_{nn} = \frac{n+2}{2(n+1)},\quad K_{n1} = K_{1n} = \frac{1}{2(n+1)}$$

As an example, following is what $M^{-1}$ looks like at $n = 7$.
$$\begin{bmatrix}-\frac{7}{16} & \frac{1}{2} & 0 & 0 & 0 & 0 & \frac{1}{16}\cr \frac{1}{2} & -1 & \frac{1}{2} & 0 & 0 & 0 & 0\cr 0 & \frac{1}{2} & -1 & \frac{1}{2} & 0 & 0 & 0\cr 0 & 0 & \frac{1}{2} & -1 & \frac{1}{2} & 0 & 0\cr 0 & 0 & 0 & \frac{1}{2} & -1 & \frac{1}{2} & 0\cr 0 & 0 & 0 & 0 & \frac{1}{2} & -1 & \frac{1}{2}\cr \frac{1}{16} & 0 & 0 & 0 & 0 & \frac{1}{2} & -\frac{7}{16}\end{bmatrix}$$

One can see that
$M^{-1}_{17} = M^{-1}_{71} = \frac{1}{2(7+1)} = \frac{1}{16}$ and
$M^{-1}_{11} = M^{-1}_{77} = \frac{7+2}{2(7+1)} - 1 = -\frac{7}{16}$. Matching what we have computed before.


问:
Showing $\prod_{n\geq 1} (1+q^{2n}) = 1 + \sum_{n\geq 1} \frac{q^{n(n+1)}}{\prod_{i=1}^n (1-q^{2i})}$
I want to show

\begin{align}
\prod_{n\geq 1} (1+q^{2n}) = 1 + \sum_{n\geq 1} \frac{q^{n(n+1)}}{\prod_{i=1}^n (1-q^{2i})}
\end{align}

I know one proof via self-conjugation of partition functions with Young Tableaux.  But it seems not natural for me. [In the process of the proof it appears Durfee square, etc]

Is there any other (simple?) proof for this equality?

答:
Substituting $\sqrt{q}$ for $q$, we get
$$
\prod_{n=1}^{\infty}(1+q^n)=\sum_{k=0}^{\infty}{\frac{q^{\binom{k+1}{2}}}{\prod_{i=1}^{k}{(1-q^i)}}}.
$$

The left-hand side is the generating function for partitions with distinct parts (a part of each size $n$ occurs $0$ or $1$ times).

On the other hand, if a partition with distinct parts has $k$ parts, then the $i$th smallest part is of size at least $i$. Given a partition $0<\lambda_1<\lambda_2<\dots<\lambda_k$, subtract $i$ from the size of the $i$th smallest part to get a partition with parts $0\le\lambda_1-1\le\lambda_2-2\le\dots\le\lambda_k-k$ with $\le k$ parts (after exluding the $0$s), whose conjugate is a partition with the largest part $\le k$. The "staircase" we subtracted has $1+2+\dots+k=\binom{k+1}{2}$ cells. That yields the summand on the right-hand side:
$$
q^{\binom{k+1}{2}}\prod_{i=1}^{k}{\frac{1}{1-q^i}}.
$$

I think this is about as simple as it gets. The Durfee square need not be involved, as you can see.


问:
$\sigma_x^2 \leq \frac{R_x^2}{4}$
Can anyone please help me to prove this - $\sigma_x^2 \leq \frac{R_x^2}{4}$ ? Here $\sigma_x$  is the variance of the given data set and $R_x$  is the range of the given data set.

答:
Also known as the Popoviciu inequality, a proof sketch is as follows : if $X$ is the random variable generating the data set (i.e. whose samples are being taken) with mean $\mu$ and variance $\sigma^2$, then for any $t$ we have : $$
\mbox{Var}(X) = E[(X-\mu)^2] = E[(X-t)^2] - (t - \mu)^2
$$

which can be checked by expansion. This implies that $\mbox{Var}(X) \leq E[(X-t)^2]$ for all $t$. Let $X$ have range $[m,M]$ and put $t = \frac{m+M}{2}$. Try to now bound the RHS of the inequality above by $\frac{(M-m)^2}{4}$ using elementary inequalities.

Title: $\sigma_x^2 \leq \frac{R_x^2}{4}$


答:
First, assume $f(0) \ne 0$. Let $r < 1$ and $n(r)$ be the number of zeros of $f$ inside $\overline{D}(0, r)$. Let $k$ be a positive integer so that $k < n(r)$. By Jensen's formula:
$$
\left|f(0)\right| \prod_{n=1}^{n(r)}\frac{r}{\left|\alpha_n\right|} = \exp \left\{\frac{1}{2\pi}\int_{-\pi}^\pi \log\left|f(re^{i\theta})\right|\right\}
$$

Since $f$ is bounded (say by $M > 0$), we have:
$$
\prod_{n=1}^{k}\left|\alpha_n\right| \ge \frac{\left|f(0)\right|r^k}{M}
$$

Now, by letting $r \to 1$, we get:
$$
\prod_{n=1}^{\infty}\left|\alpha_n\right| \ge \frac{\left|f(0)\right|}{M} > 0
$$

Which implies that:
$$
\sum_{n=1}^\infty \left(1 - \left|\alpha_n\right|\right) < \infty
$$

Finally, if $f(0) = 0$. Put $f(z) = z^m g(z)$ where $g(0) \ne 0$ and apply the reasoning above to $g$. Since $g$ and $f$ have the same zeros apart from $z = 0$, the result follows for $f$.

Title: If $\mathbb f$ is analytic and bounded on the unit disc with zeros $a_n$ then $\sum_{n=1}^\infty \left(1-\lvert a_n\rvert\right) \lt \infty$

Body:
I'm going over old exam problems and I got stuck on this one.
Suppose that $\mathbb{f}\colon \mathbb{D} \to \mathbb{C}$ is analytic and bounded. Let $\{a_n\}_{n=1}^\infty$ be
the non-zero zeros of $\mathbb{f}$ in $\mathbb{D}$ counted according to multiplicity. Show that
$$ \sum_{n=1}^\infty \left( 1 - \left|a_n \right|\right)\lt\infty $$
I can understand that $\left|a_n \right|$ goes to $1$ since zeros are isolated, but it doesn't help showing the series is convergent. Any help will be appreciated!</p>


问:
If $A$ is a proper subset of $B$ then $|A| < |B|$
Why is this statement false? If $A$ is a proper subset of $B$ then $\forall x \in A, x \in B$ but since $A \ne B$ then $\exists x \in B, x\notin A$

This should mean that $|B| > |A|$, right?

答: $|A|\le|B|$, and that corresponds to the fact that $A$ injects into $B$.


答:
It is first important to realize that if this is true for the case $n = 2$, then it will be true for all cases. This is because for all datasets $X$ taking $X_1, X_n$ as a new dataset will only increase the LHS and keep the RHS constant. Now the problem is simple algebra.

$$Var(\{X_1, X_2\}) = \frac{(X_1 - \frac{X_1 + X_n}{2})^2 + (X_2 - \frac{X_1 + X_n}{2})^2}{2} = \frac{(X_1 - X_n)^2 + (X_n - X_1)^2}{8} = \frac{R_x^2 + R_x^2}{8} = \frac{R^2_x}{4}$$

Title: $\sigma_x^2 \leq \frac{R_x^2}{4}$

Body: Can anyone please help me to prove this - $\sigma_x^2 \leq \frac{R_x^2}{4}$ ? Here $\sigma_x$  is the variance of the given data set and $R_x$  is the range of the given data set.


问:
finitely generated projective module over local ring is free
I have a question about an argument from a proof in Hideyuki Matsumura's "Commutative Ring Theory" on page 9, Theorem 2.5:

Let $(A,\mathfrak{m})$ be a local ring; then a finitely generated projective module $M$ over $A$ is free.

The proof works as follows:

Choose a minimal $A$-basis $\omega_1,...,\omega_n$ of
$M$. Take into account that "minimal" means that there cannot exist another system of generators $b_1,..., b_m \in M$ with $m < n$ and $M = \sum_{i=1} ^m A b_i$.

Define a surjective map $\varphi:F \to M$ from the free module $F = Ae_1 \oplus \cdots \oplus Ae_n$ to $M$ by $\varphi(\sum a_i e_i) = \sum a_i\omega_i$. If we set $K = \operatorname{Ker}(\varphi)$ then, from the minimal basis property

$$\sum a_i \omega_i =0 \Rightarrow a_i \in \mathfrak{m} \text{ for all } i. $$

Thus $K \subset \mathfrak{m}F$.  Because $M$ is projective, there exists $\psi: M \to F$ such that
$F = \psi(M)\oplus K$, and it follows that $K = \mathfrak{m}K$. (???)

On the other hand, $K$ is a quotient of $F$, therefore finite over $A$, so that $K = 0$ by Nakayama and $F = M$.

Question: why the fact that $F = \psi(M)\oplus K$ implies that $K = \mathfrak{m}K$?

$K \subset \mathfrak{m}F$,
then we can deduce  $K＝K∩mF$＝$K∩（m \psi(M)\oplus mK）$
But why the last is equal to $mK$?

Thank you for your kind help.

答: Let $k=A/\mathfrak{m}$ be the residue field of $A$. Tensor the exact sequence $G \rightarrow F\rightarrow M\rightarrow 0$ with $k$ (where $G$ is a free module minimally projecting on $K$) to get $F/\mathfrak{m}F\simeq M/\mathfrak{m}M$ ($K\subseteq \mathfrak{m}F$). Now tensor the equality $F=\psi(M)\oplus K$ with $k$ and count vector space dimension.


问:
Large deviations Exercise (Durret 2.7.6)
I'm studying probability theory and doing the exercises in Durret v5.

Let $X_1,\ldots, X_n$ iid with $EX_1 = 0$. Show that if $\epsilon,a>0$, then
then  $$\liminf_{n\to\infty }\frac{P(S_n \geq na)}{nP(X_1 \geq
n(a+\epsilon))}\geq 1$$

Hint: Let $F_n =\{ X_i \geq n(a+\epsilon)
> \text{for exactly one } i\leq n\}$.

Edit: I tried using the inequality $P(S_n \geq na) \geq P(S_{n-1}\geq -n\epsilon)P(X_n \geq n(a+\epsilon))$, but i don't know what to do next.

答:
We have
$$
\frac{P(S_n\geq na)}{nP(X_1\geq n(a+\epsilon))}\geq \frac{P(S_{n-1}\geq -n\epsilon \vert X_j<n(a+\epsilon) \text{ for all }j<n)P(F_n)}{nP(X_1\geq n(a+\epsilon))}. \ \ (1)
$$

Note that the conditional probability should be used due to lack of independence of "All $X_j <n(a+\epsilon)$" and "$S_{n-1}>-n\epsilon$".

Let $\alpha = P(X_1\geq n(a+\epsilon))$. Then $P(F_n)=n\alpha(1-\alpha)^{n-1}$, which gives (1) is at least
$$
P(S_{n-1}\geq -n\epsilon\vert X_j<n(a+\epsilon) \text{ for all }j<n)(1-\alpha)^{n-1}.
$$
By Markov inequality, we have
$$
\alpha=P(X_1\geq n(a+\epsilon))\leq \frac{E(|X_1| I_{X_1\geq n(a+\epsilon)})}{n(a+\epsilon)}, \ \mathrm{and}
$$
$$
1-\alpha\geq 1-\frac{E(|X_1| I_{X_1\geq n(a+\epsilon)})}{n(a+\epsilon)}.
$$
Thus,
$$
1\geq (1-\alpha)^{n-1} \geq \left(1-\frac{E(|X_1| I_{X_1\geq n(a+\epsilon)})}{n(a+\epsilon)}\right)^{n-1} \rightarrow 1 \ \mathrm{as} \ n\rightarrow\infty.
$$
By Weak Law of Large Numbers (Textbook Theorems 2.2.12 and 2.2.14), $P(S_{n-1}\geq -n\epsilon\vert X_j<n(a+\epsilon) \text{ for all }j<n)\rightarrow 1$ as $n\rightarrow\infty$. The result follows.


问:
If f is onto, is the inverse one-to-one
If $f$ is onto, is the inverse $f^{-1}$ one-to-one?

答:
Not necessarily. If $f$ is onto, then the inverse $f^{-1}$ may or may not exist.

If the inverse $f^{-1}$ does exist, then $f$ is onto and one-to-one and $f^{-1}$ is onto and one-to-one. (Thus, if we have the inverse there is no need to separately assume that $f$ is onto.)

Conversely, if $f$ is onto and one-to-one, then $f^{-1}$ exists.

Title: If f is onto, is the inverse one-to-one


问:
Show $f(x) =x-\dfrac{1}{2}(1+\log(x)) $ for $p\geq 1$ and $x\in(0,+\infty) $ is in $L^p$, and find value of $p$
How to show following function lies in $L^p$ space?
Actually, I have to find condition on $p$ so that this function belongs to $L^p$ space:

$$f(x) =x-\dfrac{1}{2}(1+\log(x)),\ \text{ for }\ p\geq 1\ \text{ and }\ x\in(0,+\infty) $$

I am not able to integrate
$$\int_{0}^{\infty}|f(x)|^p dx =\int_{0}^{\infty}\left| x-\dfrac{1}{2}(1+\log(x))\right|^p dx $$

and show that it is finite. Show that the following function $f \in L_p$ if and only if $p=2$

答: The function $f(x)=x-\frac{1}{2}(x-\log(x))$ is not in $L^p((0;\infty))$ for $p\geq 1$. Note that $f'(x)=\frac{1}{2}(1+1/x)>1/2$, thus our function is not bounded and we have $f(x) \geq f(1)=1/2$ for all $x\in (1;\infty)$. Hence, $f\notin L^p((0;\infty))$ for any $p\geq 1$.


答:
$$\int \cot^2(x) \, dx = \int (\csc^2(x) - 1) \, dx = \int \csc^2(x) \, dx - x +\text{const}$$
\begin{align}
\int \csc^2(x) \, dx & = \int \dfrac{dx}{\sin^2(x)}\\
& = \int \dfrac{\cos^2(x) + \sin^2(x)}{\sin^2(x)} \, dx\\
& = \int \dfrac{\cos(x) \, d (\sin(x)) - \sin(x) \, d(\cos(x))}{\sin^2(x)}\\
& = -\int d\left(\dfrac{\cos(x)}{\sin(x)}\right) = -\cot(x)
\end{align}
Hence,
$$\int \cot^2(x) \, dx = \int (\csc^2(x) - 1) \, dx = -\cot(x) - x +\text{const}$$</p>

Title: Integral of $\cot^2 x$?

Body: How do you find $\int \cot^2 x \, dx$? Please keep this at a calc AB level. Thanks! </p>


问:
Integral of $\exp(-\frac{c_1}{x}-\frac{c_2}{x^2})$ on positive real line
Is it possible to find
$$
\int_{0}^{K}
\exp\left(-\,{c_{1} \over x} - {c_{2} \over x^{2}}\right)
\,\mathrm{d}x
$$
or at least a good approximation to it, where $c_{1}$ and $c_{2}$ are positive and $K$ is a very large number $?$.

I tried to consider the change of variables
$-u = -c_{1}/x - c_{2}/x^{2}$ but to no avail.

Note: Approximation can include known functions such as Erf, or Gamma, if necessary.

答:
If you are interested in large values of $K$, an asymptotic expansion is possible. Using integration by parts on the equivalent integral

$$I(c_1,c_2;K)=\int_{1/K}^{\infty}\exp[-c_1 t-c_2 t^2]\frac{dt}{t^2}$$

one can isolate the divergent terms as follows

$$I=(K-c_1\log K)e^{-c_1/K-c_2/K^2}+2c_2\int_{1/K}^{\infty}dt\exp[-c_1 t-c_2 t^2]-c_1^2\int_{1/K}^{\infty}dt\log t\exp[-c_1 t-c_2 t^2]-2c_1c_2\int_{1/K}^{\infty}t\log t\exp[-c_1 t-c_2 t^2]dt$$

This is an exact statement so far and all the integrals on the RHS possess a finite limit as $K\to \infty$. The first of the three integrals is easily obtainable:

$$\int_{1/K}^\infty dt\exp[-c_1 t-c_2 t^2]=\frac{e^{c_1^2/4c_2}}{\sqrt{c_2}}\text{erfc}\left(\frac{\sqrt{c_2}}{K}+\frac{c_1}{2\sqrt{c_2}}\right)$$

As a matter of fact, Mathematica states that all of these integrals can be evaluated in terms of derivatives of hypergeometric functions with respect to one of their arguments in the limit $K\to\infty$. To wit,

$$8c_2^{3/2}\int_{0}^{\infty}t\log t\exp[-c_1 t-c_2 t^2]dt=\sqrt{\pi } c_1 e^{\frac{c_1^2}{4 c_2}} \left(_1F_1^{(1,0,0)}\left(0,\frac{3}{2},-\frac{c_1^2}{4 c_2}\right)+(\log c_2+\gamma ) \text{erfc}\left(\frac{c_1}{2 \sqrt{c_2}}\right)-2+\log (4)\right)-2 \sqrt{c_2} \left(- ~_{1}F_1^{(1,0,0)}\left(1,\frac{1}{2},\frac{c_1^2}{4 c_2}\right)+\log c_2+\gamma \right)$$

and

$$-4c_2\int_{0}^{\infty}\log t\exp[-c_1 t-c_2 t^2]dt=\sqrt{\pi }\sqrt{c_2} e^{\frac{c_1^2}{4 c_2}} \left(_1F_1^{(1,0,0)}\left(0,\frac{1}{2},-\frac{\text{c1}^2}{4 c_2}\right)+(\log c_2+\gamma) \text{erfc}\left(\frac{c_1}{2 \sqrt{c_2}}\right)+\log 4\right)+c_1 ~_{1}F_1^{(1,0,0)}\left(1,\frac{3}{2},\frac{c_1^2}{4 c_2}\right)$$

where $_1F_1^{(1,0,0)}(a,b,z)=\frac{\partial}{\partial a}~_1F_1(a,b,z)$. Assembling these terms one obtains the first few terms of the asymptotic series for the result
$$I(c_1,c_2)\sim (K-c_1\log K)e^{-c_1/K-c_2/K^2}+2\sqrt{c_2}e^{c_1^2/4c_2}\text{erfc}\left(\frac{\sqrt{c_2}}{K}+\frac{c_1}{2\sqrt{c_2}}\right)-c_1^2\int_{0}^{\infty}dt\log t\exp[-c_1 t-c_2 t^2]-2c_1c_2\int_{0}^{\infty}t\log t\exp[-c_1 t-c_2 t^2]dt+\mathcal{O}\left(\frac{\log K}{K}\right)$$


问:
Can someone please check the solutions of the given equation?
I have a system of $8-$equations for $\{x,y,z,t,w,p,r,s\}$ and I want to solve this system.
$$ 4 \pi  (w+y)+x+y=w+4 \pi  (x+z)+z,\\4 \pi  r+r+s=4 \pi  (s+w-z)+w+z,\\4 \pi  (p+s)+r+s=p+4 \pi  (r+t)+t,\\4 \pi  p+p+t=4 \pi  (t+x-y)+x+y,\\p+e^{8 i \pi ^2} (t-x)=y,\\p+y=e^{8 i \pi ^2} (t+x),\\r+s=w+z,\\r+z=s+w $$

I use Mathematica (the code is given below) and the solution is as follows
$$x=0,\quad y=0,\quad t=0,\quad w=-\frac{(-1-4 \pi ) z}{4 \pi -1},\quad s=z,\quad r=-\frac{(-1-4 \pi ) z}{4 \pi -1},\quad p=0$$

From the nature of my problem, I expect that at least one of the variables $x$ and $y$ to be nonzero. My question is can I trust this result? Or if I use another method and manually, will I get a different result with one of $x$ and $y$ nonzero?

```<code>eqs := {x + y + 4 π (w + y) == w + z + 4 π (x + z),
   r + 4 π r + s == w + 4 π (s + w - z) + z,
   r + s + 4 π (p + s) == p + t + 4 π (r + t),
   p + 4 p π + t == x + 4 π (t + x - y) + y,
   p + E^(8 I π^2) (t - x) == y,
   p + y == E^(8 I π^2) (t + x),
   r + s == w + z,
   r + z == s + w};

Solve[eqs, {x, y, z, t, w, s, r, p}]
<span class="math-container">```
```

答:
This being a linear system, I have problems to  conceive that another solution could exist. In any manner, notice that the results express as functions of $z$ which means that there are somewhere two identical equations.

Just for the "fun" of it, I solved the system with pen and paper using successive eliminations.

Adding $(7)$ to $(8)$, we immediately have $r=w$ and each of them gives $s=z$ (this is where we have the redundancy). Now $(2)$ is satisfied.

Adding $(5)$ to $(6)$, $p=e^{8 i \pi ^2} x$.

Reusing $(6)$, $y=e^{8 i \pi ^2} t$. At this point are left $(1)$, $(3)$ and $(4)$ for $t$, $w$, $x$ and $z$.

From $(4)$, $t=\frac{4 \pi  +1}{1-4 \pi }x$ and remain $(1)$ and $(3)$ for $w$, $x$ and $z$.

Adding $(1)$ to $(3)$ gives $x=0$ and $w=-\frac{4 \pi  +1}{4 \pi -1}z$.

Finally $z$ stays undefined.

By the end
$$x=y=t=p=0 \qquad s=z \qquad r=w=-\frac{4 \pi  +1}{4 \pi -1}z$$ which is your result.


问:
Must compact sets be contained in the topology?
Can a set be compact but not be in the topology itself? For example, if $S$ is compact in a topological space $X$, must $S$ be in the topology $X$? Or does $S$ merely have to be covered by (finitely many) sets that are in the topology of $X$?

答: Usually, compact sets do not belong to the topology, since usually they are not open. For instance, the only compact subset of $\Bbb R$ (with its usual topology) which belongs to the topology is the empty set.


答:
The sets in the topology are the open sets. Compact sets can be open, but in general they are not. For example, every finite subset of $\Bbb R$ is compact, as are the middle-thirds Cantor set and the set $\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$, and the only open set in the lot of them is $\varnothing$.

Every subset of every space is covered by an open set in the space, because the space itself is an open set. What makes a set compact is not that it can be covered by finitely many open sets: it’s that it has the property that every open cover of it has a finite subcover.

Title: Must compact sets be contained in the topology?

Body: Can a set be compact but not be in the topology itself? For example, if $S$ is compact in a topological space $X$, must $S$ be in the topology $X$? Or does $S$ merely have to be covered by (finitely many) sets that are in the topology of $X$?


答:
Given a function $f:A\to B$, a section of $f$ (id est, a function $g:B\to A$ such that $f\circ g=id_B$) is always injective, because it's the right factor in an injective composition. A section exists only if $f$ is surjective (because $f$ is, in turn, the left factor in a surjective composition). Under the axiom of choice, the latter is an if and only if.

Existence of an inverse, id est of a section $g:B\to A$ such that $g\circ f=id_A$, requires more. Specifically, there is one if and only if $f$ is bijective. This holds regardless of the axiom of choice.

It should be pointed out that the discussion on the existence of an inverse, as interesting as it might be, is something of an idle talk in this instance, because I would evaluate your assertion as "true" in the case where $f$ has no inverse.

Title: If f is onto, is the inverse one-to-one

Body: If $f$ is onto, is the inverse $f^{-1}$ one-to-one?


答:
If all the derivatives $\phi^{(k)}(0)=0$ then $|\xi|^a\phi$ is Schwartz thus so is its Fourier transform.

Otherwise let $\phi^{(k)}(0)$ be the first non-vanishing derivative, wlog assume that $a\in (0,1)$ and $l\in 0,1$ then $$h=sign(\xi)^l |\xi|^a\phi-\frac{\phi^{(k)}(0)}{k!} \xi^k sign(\xi)^l |\xi|^a e^{-|\xi|} =O(|\xi|^{a+k+1})$$ is $k+2+\lfloor a\rfloor$-times differentiable in $L^1+\Bbb{C}\delta$ so that for $|x|\ge 1$ $$\mathcal{F}^{-1}(sign(\xi)^l|xi|^a\phi)$$ $$=\mathcal{F}^{-1}(\frac{\phi^{(k)}(0)}{k!} \xi^k sign(\xi)^l |\xi|^a e^{-|\xi|})+\mathcal{F}^{-1}(h)$$ $$=\frac{\phi^{(k)}(0)}{k!}  ((1-2i\pi x)^{-a-k-1}+(-1)^{k+l} (1+2i\pi x)^{-a-k-1}) \Gamma(a+k+1) + (-2i\pi x)^{-k-2-\lfloor a\rfloor}\mathcal{F}^{-1}(h^{(k+2)})$$
$$ = C x^{-a-k-1}+O(x^{-k-2-\lfloor a\rfloor})$$

Title: Decay of fractional derivative of Schwartz function

Body: Let $\phi$ be a Schwartz function and let $\alpha>0$. I want to analyze the decay as $x\rightarrow\infty$ of: $$\int_\mathbb{R}e^{2\pi i x\xi}|\xi|^\alpha\phi(\xi)\,d\xi$$Heuristically, for $\xi\ll x$, we have one oscillation from $\xi=0$ to $\xi=1/x$ at which point the amplitude is of order $|x|^{-\alpha}$ so the answer should be $|x|^{-\alpha-1}$. However, I have had trouble making this intuition precise. I have tried smoothly partitioning into $|\xi|\ll x$ and $|\xi|\gg |x|$ and separately doing a dyadic decomposition at level $|x|^{-1}$, but the details have not worked out. I am interested a precise solution as well as any heuristic ways of thinking about this sort of integral. If it would help, I can share what I've tried in more detail.


问:
Is it possible to have an infinitely repeating whole number
So I would like to know whether I could have a repeating whole number, such as 9999... repeating or if that's a mathematical fallacy that breaks things. I don't really want to know if it would be natural or not (I assume not), I'm just wondering if its possible to have a number fitting my specifications. Also I know that repeating decimals are a thing, I'm wondering if the same concept can apply to whole numbers.</p>

答:
Just observe that any natural number $m$ with $n+1$ number of digits can be uniquely written as $$m = \sum_{k= 0}^{n } a_k 10^k$$ where each $0 \leq a_k \leq 9$ and $a_n \ne 0$. So it is clear that $m > 10^n$.

Now if the number has infinite digits then the series $\sum_{k=0}^{\infty} a_k 10^k$ should diverge since $10^n \to \infty$ as $n \to \infty$. So there is no possibility of natural number whose number of digits are infinite.

If $m$ is a negative integer then consider $-m$ and apply the above argument


答:
Proof:

Let $\min_{1\leq i \leq n} x_i = a$ and $\max_{1\leq i \leq n} x_i = b$

Then $R_x = b - a$

$\sigma_x^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2$

Now $\sum_{i=1}^n (x_i - c)^2$ is least when $c = \bar{x}$

Hence $\sum_{i=1}^n (x_i - \bar{x})^2 \leq \sum_{i=1}^n (x_i - \frac{a+b}{2})^2 = \sum_1 (x_i - \frac{a+b}{2})^2 + \sum_2 (x_i - \frac{a+b}{2})^2$

where $\sum_1$ includes the values of $x$ less than or equal to $\frac{a+b}{2}$ and $\sum_2$ includes values of $x$ greater than $\frac{a+b}{2}$

Or, $\sum_{i=1}^n (x_i - \bar{x})^2 \leq \sum_1 (a - \frac{a+b}{2})^2 + \sum_2 (b - \frac{a+b}{2})^2 = \sum_1 \frac{R_x^2}{4} + \sum_2 \frac{R_x^2}{4} = n \frac{R_x^2}{4}$

Or, $\sigma_x^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 \leq \frac{R_x^2}{4}$

Title: $\sigma_x^2 \leq \frac{R_x^2}{4}$


问:
Maximum number of stable states of a Markov matrix
I wondered if the maximum number of stable states (solutions) for a sequence modeled by a Markov matrix is 1 (which means, either no stable solution or one stable solution), or it could be any number.

In other words, might there exist a Markov matrix with two or more stable states? Is there any boundary on the number of possible states?

Note: I mean by Markov matrix a matrix with all nonnegative entries and columns adding up to 1.

答:
Any non-negative matrix gives rise to a directed graph with edges from vertices $i$ to $j$ whenever $a_{ij}>0$. Such a directed graph decomposes into a number of strongly connected components, and these components can be ordered from source components to intermediate components to sink components. Here is an example:

$\qquad\qquad\qquad$

The vertices 1 and 2 form source components, then {3,4,6,7} form an intermediate component, and 5 and 8 form sink components.

For an example with $n$ stable states, take one source vertex connected to $n$ sink vertices. Its matrix for $n=5$ is $$\begin{pmatrix}1&0&0&0&0&0\\1&1&0&0&0&0\\1&0&1&0&0&0\\1&0&0&1&0&0\\1&0&0&0&1&0\\1&0&0&0&0&1\end{pmatrix}\to\begin{pmatrix}\tfrac{1}{6}&0&0&0&0&0\\\tfrac{1}{6}&1&0&0&0&0\\\tfrac{1}{6}&0&1&0&0&0\\\tfrac{1}{6}&0&0&1&0&0\\\tfrac{1}{6}&0&0&0&1&0\\\tfrac{1}{6}&0&0&0&0&1\end{pmatrix}$$ Each vector $e_2,\ldots,e_6$ is a stable state. Of course you can pick any connected directed graph: there will be as many (independent) stable states as there are sink components.


问:
Degree of a Cartier Divisor under pullback
This is question 7.2.3 in Liu's book Algebraic Geometry and Arithmetic Curves and I have been trying with this for some time now.

Let $f:X \rightarrow Y$ be a morphism of Noetherian schemes, and suppose that X and Y are integral and that f is finite surjective. We will let $Div(Y)$ resp. $Div(X)$ stand for Cartier Divisors on Y resp. X.

Associated to a Cartier divisor $D \in Div(Y)$ we have a notion of multiplicity at a point $y \in Y$, defined as $$\text{mult}_y D = \text{length } \mathcal{O}_{Y,y}/D_y \mathcal{O}_{Y,y}.$$

I am trying to show that for a Cartier Divisor $D \in Div(Y)$, we have that $$\text{mult}_x f^\ast(D) = \text{length } \mathcal{O}_{X,x}/(\mathcal{m}_y\mathcal{O}_{X,x}) \cdot \text{mult}_y(D)$$
where $m_y$ is the maximal ideal of $\mathcal{O}_{Y,y}$.

Now, it is quite easy to do this if f is a flat morphism, since we can simply use 50.13 of this document http://stacks.math.columbia.edu/download/algebra.pdf (Stacks, Algebra). But I don't see how to do it in the general case, and I have been trying for some time, so any hint would be nice.

Thank you!

答:
I just e-mailed Qing Liu about this exercise and got a nice reply (as always), and I will share it in case other people try this exercise and get stuck. Please excuse that I refer to numbers in the book

"7.2.3 a) is incorrect for the case f finite surjective, see EGA IV 5.6.11 . Maybe it can be corrected by assuming further that Y is normal.

In b) one should keep the hypothesis that Y is normal. "

I think I know how to do this in this case and will post a solution then, by editing this answer.

Update:

OK - here is how I think one can do it in the case Y is normal. Let us note that the property is local on Y, so that we can assume Y to be a normal integral, local noetherian ring of dimension 1. As such, Y is a Dedekind scheme. Now, we know that for a non-constant morphism $f: X \rightarrow Y$ with X integral and Y Dedekind , f is flat (see Liu, Corollary 4.3.10) . Hence, it reduces to the flat case and this is an easy exercise in commutative algebra.

And I forgot to add, if I am wrong, please tell me!


答:
Not true. Consider $f : \mathbb{Z} \to \mathbb{R}$ where $f(x) = x$.

$f(x)$ is one-to-one since $f(x) = f(y)$ implies $x=y$, but $f^{-1}(x)$ does not exist since some elements like $\frac{1}{2}$ don't have an inverse in $\mathbb{Z}$.

Title: If f is onto, is the inverse one-to-one


问:
Can I write e(mathematical constant) in fraction?
Proof of e can be in fraction

e is 2.718281828.....
so as you can see, there is a pattern. So, I break it down into 2.7 + 0.01828 + 0.0000001828 .....

and then I use sum of infinity to get the fraction

Well, in mathematics theory , e cannot be in fraction. But I tried to use sum of infinity and proof it that it can be in fraction. Am I correct?

Body:
$e$ can be represented in a continued fraction. See the List of representations of $e$.

$e=2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{4+\cfrac{1}{1+\cfrac{1}{1+\ddots\,}}}}}}}=1+\cfrac{2}{1+\cfrac{1}{6+\cfrac{1}{10+\cfrac{1}{14+\cfrac{1}{18+\cfrac{1}{22+\cfrac{1}{26+\ddots\,}}}}}}}$

$e=2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{2}{3+\cfrac{3}{4+\cfrac{4}{5+\ddots}}}}}= 2+\cfrac{2}{2+\cfrac{3}{3+\cfrac{4}{4+\cfrac{5}{5+\cfrac{6}{6+\ddots\,}}}}}$

$e=2+\cfrac{1}{1+\cfrac{2}{5+\cfrac{1}{10+\cfrac{1}{14+\cfrac{1}{18+\ddots\,}}}}} =1+\cfrac{2}{1+\cfrac{1}{6+\cfrac{1}{10+\cfrac{1}{14+\cfrac{1}{18+\ddots\,}}}}}$


问:
Representation of a transposition in group theory
While reading Dummit and Foote's Abstract Algebra, he briefly mentions that any transposition $(i \, j)$ can be written as $$(i \, j) = \lambda(1 \, 2)\lambda,$$ where $\lambda$ is the permutation that 'interchanges $1$ and $i$, interchanges $2$ and $j$, and leaves all other numbers fixed (if $i = 1$ or $j = 2$, then $\lambda$ fixes $i$ or $j$ respectively)'. I don't understand what it is that $\lambda$ does, and hence cannot prove the above expression.

So does $\lambda$ take $1$ to $i$ and $i$ to $1$ only if $i$ is not already $1$? What does the condition in the bracket mean? Could someone explain it with an example?

答:
That basically means $\lambda=(1i)(2j)$.

For example, if $i=4$ and $j=5$, then $\lambda=(14)(25)$ and notice $$(45)=(14)(25)(12)(14)(25).$$

But if, for example, $i=1$ and $j=5$, then $\lambda=(25)$ and notice $$(15)=(25)(12)(25).$$

The parenthetical comment simply addresses the fact that we wouldn't write $\lambda=(11)(25)$ in this example.


Title: What is inverse of a matrix whose diagonal elements are all zero.

Body:
I am a science researcher and I got a problem to find the generic inverse of the following matrix:

$$
A_n = \left(\begin{array}{ccc}
        0 & a_2 & a_3 & ... & a_{n-1} & a_n \\
      a_1 &   0 & a_3 & ... & a_{n-1} & a_n \\
      a_1 & a_2 &   0 & ... & a_{n-1} & a_n \\
      ... & ... & ... & ... & ...     & ... \\
      ... & ... & ... & ... & ...     & ... \\
      a_1 & a_2 & a_3 & ... &       0 & a_n \\
      a_1 & a_2 & a_3 & ... & a_{n-1} &   0 \\
      \end{array}\right)
$$

I figured out that for n=2,3

$$
A_2^{-1} = \left(\begin{array}{ccc}
        0 & a_2^{-1}  \\
      a_1^{-1} &   0  \\
      \end{array}\right)
$$

$$
A_3^{-1} = \frac{1}{2} \left(\begin{array}{ccc}
        -a_1^{-1} &  a_1^{-1} &  a_1^{-1} \\
         a_2^{-1} & -a_2^{-1} &  a_2^{-1} \\
         a_3^{-1} &  a_3^{-1} & -a_3^{-1} \\
      \end{array}\right)
$$

but can we extend it to a general case?
Can anyone help?? Thanks!

[later]

It looks like it is

$$
m_{ij} =  -\frac{n-2}{n-1} a_{i}^{-1}\ (i=j)
$$
$$
       = \frac{1}{n-1} a_{i}^{-1}(else)
$$

Body:
Your pattern extends to any $n$. You can see this by multiplying out:

$$\begin{pmatrix} -(n-2)a_1^{-1} & a_1^{-1} & ... & a_1^{-1} \\
      a_2^{-1} &   -(n-2)a_2^{-1} & ...  & a_2^{-1} \\
      ... & ... & ...   & ... \\
      ... & ... & ...   & ... \\
      a_n^{-1} & a_n^{-1} & ... &  -(n-2)a_n^{-1} \end{pmatrix}
\begin{pmatrix} 0 & a_2 & a_3 & ... & a_{n-1} & a_n \\
      a_1 &   0 & a_3 & ... & a_{n-1} & a_n \\
      ... & ... & ... & ... & ...     & ... \\
      ... & ... & ... & ... & ...     & ... \\
      a_1 & a_2 & a_3 & ... & a_{n-1} &   0 \end{pmatrix}=(n-1)I$$

For $i=j$, there are $(n-1)$ $a_i^{-1}$ in the $i$th row each multiplied by $a_i$ in the $j$th ($=i$) column to give $n-1$.

For $i\ne j$, one of $a_i^{-1}$ is multiplied by $0$; so in fact there are $(n-2)a_i^{-1}a_j$, canceled out by a single $(n-2)a_i^{-1}a_j$.


问:
Limit $\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}$
I want to compute $$\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}.$$

Is it OK how I want to do?

$$\sin{\sqrt{x+1}}-\sin{\sqrt{x}}=2\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}=2\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}$$

I'm not surem but I want to say that $$|\cos \frac{\sqrt{x+1}-\sqrt{x}}{2}| \leq 1$$ $$\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}} \to 0 \mbox{ when } x \to \infty$$
So the limit is $0$ ?

I thinks is not because  if I say that $ \displaystyle|\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}| \leq 1$ I will obtain that another limit than $0$.

Thanks :)

答:
Assuming you have the Mean Value Theorem at your dispoeal, that's the easiest way to show that the limit equals $0$: with $g(x) = \sin\sqrt x$, we have
$$
\sin\sqrt{x+1}-\sin\sqrt x = g(x+1)-g(x) = ((x+1)-x)g'(\xi) = \frac{\cos\sqrt\xi}{2\sqrt\xi}
$$
for some $x\le\xi\le x+1$. In particular,
$$
|\sin\sqrt{x+1}-\sin\sqrt x| = \frac{|\cos\sqrt\xi|}{|2\sqrt\xi|} \le \frac1{2\sqrt x},
$$
and so $\lim_{x\to\infty} (\sin\sqrt{x+1}-\sin\sqrt x) = 0$ by the Squeeze Theorem.</p>


答:
Notice that
$$
\left| {f(x)} \right| =f(x) \ge \frac{x}{2}
$$
for all $x>0$. Hence,
$$
\int_0^{ + \infty } {\left| {f(x)} \right|^p dx}  \ge \frac{1}{{2^p }}\int_0^{ + \infty } {x^p dx}  = \frac{1}{{2^p (p + 1)}}\mathop {\lim }\limits_{w \to  + \infty } w^{p + 1}  =  + \infty
$$
for any $p\geq 1$. Thus, $f$ is not in $L^p((0,+\infty))$ for any $p\geq 1$.

Title: Show $f(x) =x-\dfrac{1}{2}(1+\log(x)) $ for $p\geq 1$ and $x\in(0,+\infty) $ is in $L^p$, and find value of $p$

Body:
How to show following function lies in $L^p$ space?
Actually, I have to find condition on $p$ so that this function belongs to $L^p$ space:

$$f(x) =x-\dfrac{1}{2}(1+\log(x)),\ \text{ for }\ p\geq 1\ \text{ and }\ x\in(0,+\infty) $$

I am not able to integrate
$$\int_{0}^{\infty}|f(x)|^p dx =\int_{0}^{\infty}\left| x-\dfrac{1}{2}(1+\log(x))\right|^p dx $$

and show that it is finite. Show that the following function $f \in L_p$ if and only if $p=2$


问:
Proof Verification: $T$ is injective iff $\ker(T) = \{0_V\}$
I was given this question in my introductory linear algebra class:

Given a linear transformation $T: V \rightarrow \tilde{V}$, Prove that $T$ is injective iff $\ker(T) = \{0_V\}$.

I came up with the following proof:

Let us assume to the contradictory that $T$ is not injective even though $\ker(T) =\{0_V\}$. Then there exist $u, v \in V$ such that $T(u) = T(v)$ for $u \neq v$. We write $u = v +x$ for some $x \in V$, $x \neq 0$. Thus, $$T(u) = T(u+x)$$
$$T(u) = T(u) + T(x)$$
$$T(u) + (-T(u)) = T(u) + T(x) + (-T(u))$$
$$0_{\tilde{V}} = T(x) + 0_{\tilde{V}}$$
$$0_{\tilde{V}} = T(x)$$

Since $\ker(T) = \{0_V\}$, it follows that $x = 0_V$. However this is a contradiction which implies that $u = v$.

Hence our assumption is incorrect and $T$ must be injective.

To prove that $T$ cannot be injective if $\ker(T) \neq \{0_V\}$, we take $k \in V$ such that $T(k) = 0_{\tilde{V}}$ and $k \neq 0_V$, then repeat a similar procedure till we get a contradiction.

Is this proof correct? Is there anything I am missing?

答:
Here is a simple approach, suppose $KerT ={0}$ and suppose $T(u) = T(v)$ then it implies that $T(u-v) = 0$. Since kernel consists only $0$ vector we must have $u=v$. So $T$ is injective.

Suppose $T$ is injective then consider $u \in KerT$ and we know that $T(u) = 0 = T(0)$, since $T$ is injective we have $u=0$ and consequently $KerT$ consists only $0$ vector.


问:
Factoring a quadratic mod $p$
I have $p=1 \mod 3$.
I am trying to show that $x^2-x+1=(x-a)^2 \mod p$ for some $a$.
So far I have that $x^2-x+1=x^2-2ax+a^2 \mod p$.  Which mean that $2a=1 \mod p$ and $a^2=1 \mod p$

The first identity tells us that $a=\frac{p+1}{2}$ Using this in the second identity we have $(\frac{p+1}{2})^2=1\mod p$. So we get $\frac{p^2+2p+1}{4}=1\mod p$. But now we run into trouble because this implies $1=4\mod p$ or in particular $p=3$. Have I done something wrong here, or am I mistaken that its possible to have $p=1\mod 3$.

for added context, $x^2-x+1$ is the minimal polynomial of $\frac{1+\sqrt{-3}}{2}$.

答:
You need an element $a$ of order $3$ then $$x^2 -x+1=(x+a)(x+a^{-1})\in \Bbb{F}_p[x]$$

$x^{p-1}-1$ (of degree $p-1$) has $p-1$ distinct roots in $\Bbb{F}_p$ thus $$x^{p-1}-1=\prod_{n=1}^{p-1}(x-n)\in \Bbb{F}_p[x]$$
Take any root $b$ of $x^{p-1}-1$ which is not a root of $x^{(p-1)/3}-1$, the order of $b$ divides $p-1$ but not $(p-1)/3$ thus it is $3m$ so that $a=b^m$ works.

Title: Factoring a quadratic mod $p$


问:
Prove a planar graph with an arbitrary extra edge can be embedded on a torus
I want to prove the any two points x,y can be placed on the exterior face the same time. But i have no idea.

If you know, please tell me. Thanks.

答:
As the original graph is planar, you can embed it into a plane.
Let $uv$ be the extra edge.  Pick a face $f_u$ incident with $u$ and a face $f_v$ incident with $v$. Then you can add a handle to the plane such that one end is in $f_u$ and the other in $f_v$. Using this handle, you can draw an arc from $u$, into $f_u$, along the handle, into $f_v$, to $v$. Finally, add a point to infinity to turn your plane-with-handle into a sphere-with-handle, aka torus.


问:
Does linearization interfere with analysis of chaotic behavior?
If one has a nonlinear 3+ system of differential equations, then it is linearized, is the linearization capable of accurately portraying the chaotic behavior in some manner?</p>

答: No, the solutions to linear (ordinary) differential equation systems are known. They are not chaotic (also see this question on Physics SE). Non-linearity is a precondition for chaos in a finite dimensional system.</p>


问:
How to tell if a directed graph is acyclic from the adjacency matrix?
Suppose you have an adjacency matrix $A$ for a directed graph $G=\{V,E\}$, so $A_{ij} = 1$ if $V_i\rightarrow V_j \in E$, and $A_{ij}=0$ otherwise. Many properties of the graph can be derived from this adjacency matrix. For instance, $(A^n)_{ij}$ tells you the number of paths of length $n$ going from $V_i$ to $V_j$, and for an undirected graph the number of connected components is equal to the number of eigenvalues of $A$ equal to zero.

Is there a nice linear-algebraic quantity that can tell you whether or not this graph has cycles?

My gut says that $A^* = I + A + A^2 + A^3 + \ldots = (I-A)^{-1}$ being finite (that is, $I-A$ being nonsingular) is a necessary and sufficient condition, because it means there are a finite number of paths of arbitrary length between any two vertices. Is this correct?

答: It's necessary but not sufficient. $G$ has no directed cycles iff $A$ is nilpotent. This is a stronger condition than $A$ not having an eigenvalue of $1$ (most graphs don't have eigenvalue $1$). </p>


问:
Subgradient of $(|x_1| + |x_2|)^2$
Let $f(x) = (|x_1| + |x_2|)^2$ for $x \in \mathbb{R}^2$

How do we find the subgradient $\partial f(x)$?

I tried to express $f(x)$ as $x_1^2 + x_2^2 + 2|x_1||x_2|$. I know how to handle the cases where neither of $x_1$ nor $x_2$ are 0,  but I'm not sure how to handle the cases where $x_1$ or $x_2$ are $0$.

答:
It's pretty straight forward. The subgradient of $|x|$ when $x\neq 0$ is $\text{sgn{x}}$, and $[-1,1]$ when $x=0$. Therefore $$ \partial\frac{1}{2}(|x_1|+|x_2|)^2 = \begin{cases}
&x_i +\text{sgn}(x_i)|x_j|,  &x_i\neq 0 \\
&[-1, 1]|x_j|, &x_i = 0
\end{cases}  $$

Additional comment: I calculated this after unpacking the quadratic term, but notice that the chain rule would also work in this case. For the general case, let $h=f\circ g$. Assume $f$ is a nondecreasing convex function and $g$ is a convex function. Then for a point $x$, assuming that $f$ is differentiable at $g(x)$, the subdifferential $\partial h(x)=f'(g(x))\partial g(x)$. In your case, we may define $f(t)=\max\{t,0\}^2$ to get a nondecreasing convex function.

Title: Subgradient of $(|x_1| + |x_2|)^2$


答:
Your proof is almost correct! The one weird part is

To prove that $T$ cannot be injective if $\ker(T) \neq \{0_V\}$, we take $k \in V$ such that $T(k) = 0_{\tilde{V}}$ and $k \neq 0_V$, then repeat a similar procedure till we get a contradiction.

Remember how we prove implications like "$T$ cannot be injective if $\ker(T) \neq \{0\}$": we assume that $\ker(T) \neq \{0\}$, and we try to deduce that $T$ is not injective. So indeed, the correct thing to do is pick $k \in V$ such that $T(k) = 0_{\widetilde{V}}$ and $k \neq 0_V$. Now we have two elements of $V$ ($k$ and $0_V$) which are not equal but have the same image under $T$, and therefore $T$ is not injective. It's unclear what you mean by "repeat a similar procedure till we get a contradiction" -- for example, this proof did not use contradiction.

There's also one typo / mixup: if you wanted to get $T(u) = T(u+x)$, you should have defined $v = u + x$, not $u = v + x$.

If you don't mind, I'll also make a couple small notes on style:

The first half of your proof (showing that $T$ is injective if $\ker(T) = 0$) is secretly a direct proof in disguise! The direct version of this proof is a bit shorter and cleaner, so it might be worthwhile to think about.

Don't say "write $u = v + x$ for some $x \in V \setminus \{0\}$", because this misleads the reader. We actually know exactly what $x$ is! Instead, say something like "let $x = u - v$ so that $u = v + x$, and note that $x \neq 0_V$".

Title: Proof Verification: $T$ is injective iff $\ker(T) = \{0_V\}$

Body:
I was given this question in my introductory linear algebra class:

Given a linear transformation $T: V \rightarrow \tilde{V}$, Prove that $T$ is injective iff $\ker(T) = \{0_V\}$.

I came up with the following proof:

Let us assume to the contradictory that $T$ is not injective even though $\ker(T) =\{0_V\}$. Then there exist $u, v \in V$ such that $T(u) = T(v)$ for $u \neq v$. We write $u = v +x$ for some $x \in V$, $x \neq 0$. Thus, $$T(u) = T(u+x)$$
$$T(u) = T(u) + T(x)$$
$$T(u) + (-T(u)) = T(u) + T(x) + (-T(u))$$
$$0_{\tilde{V}} = T(x) + 0_{\tilde{V}}$$
$$0_{\tilde{V}} = T(x)$$

Since $\ker(T) = \{0_V\}$, it follows that $x = 0_V$. However this is a contradiction which implies that $u = v$.

Hence our assumption is incorrect and $T$ must be injective.

To prove that $T$ cannot be injective if $\ker(T) \neq \{0_V\}$, we take $k \in V$ such that $T(k) = 0_{\tilde{V}}$ and $k \neq 0_V$, then repeat a similar procedure till we get a contradiction.

Is this proof correct? Is there anything I am missing?


问:
Normal subgroup of $A_4$
Prove that $\{id, (12)(34), (13)(24), (14)(23)\}$ is a normal subgroup of $A_4$.
I see that it's possible to prove this with the following theorem: $$N \trianglelefteq G \text{ if } N\leq G \text{ and } \forall n\in N, g\in G: g^{-1}ng \in N.$$  But is there an easier way to do this? This way isn't hard but it takes quite some time to work through.

答:
It is a clear that the set is a subgroup. This requires minimal calculations. What you could do next is to calculate the conjugacy class of $(12)(34)$ in $A_4$. Almost hands-down (conjugation preserves cycle structures!!) you will see that this is $Cl_{A_4}((12)(34))=\{(12)(34),(13)(24),(14)(23)\}$. Hence your subgroup is the disjoint union of two conjugacy classes: $\{(1)\} \cup \{(12)(34),(13)(24),(14)(23)\}$ whence normal. Note that the subgroup is isomorphic to $V_4$, Klein's $4$-group. It is also equal to the commutator subgroup $[A_4,A_4]$ of $A_4$, which yields another way of showing normality.

So what I used is the fact that if $N$ is a subgroup of $G$, then $N \unlhd G$ if and only if $N=\bigcup_{n \in N}Cl_G(n)$, which you might try to prove yourself. Note that in general, unions of $G$-conjugacy classes form a normal set (meaning: closed under conjugation), but do not need to form a subgroup!

Title: Normal subgroup of $A_4$


问:
discrete math proposition
Let x be any one of the four students in a college.
Tamang attended only football games.
Raj attended both football and basketball games.
Venkat attended only basketball games.
Ram attended neither of the games.
Let $F(x)$ be “x attended football games” and $B(x)$ be “$x$ attended basketball games”,
where $x$ denotes one of the above students. Rewrite the following statements in
simple English sentence and use above information to determine its truth values.
(a) $ꓱ x$ such that $F(x) ꓥ B(x)$.
(b) $ꓯ x, F(x) ꓦ B(x)$
(c) $ꓱ x$ such that $¬ (F(x) ꓦ B(x))$

答:
a-)There exists at least a student among Tamang, Raj,Venkat,Ram such that he attend both of the games.

b-)All students among Tamang, Raj,Venkat,Ram attended either football game or basketball game.

c-)There exists at least a student among Tamang, Raj,Venkat,Ram such that he did not attend any of the games.

a-) This part is TRUE because Raj attended both of the games.

b-)This part is FALSE because Ram does not attend any of them.

c-)This part is TRUE because Ram does not attend any of them

Title: discrete math proposition


问:
Could two distinct prime number $p,q$ both be the quadratic residue of another prime $t$ with certain constraints?
Problem: let $p,q$ be two distinct odd prime numbers with
$$p \equiv 1 \pmod 8;\quad  q \equiv 3 \pmod 4 ; \quad  \left(\frac{p}{q}\right) = 1.$$

I wonder if $\left(\frac{p}{t}\right)$ and $\left(\frac{q}{t}\right)$ could both be $1$ or not ($\forall\ t$ is a prime where $t \neq p,q$).

I tried to apply the Law of Quadratic Reciprocity but still, I cannot see a clear connection between $\left(\frac{t}{q}\right)$ and $\left(\frac{t}{p}\right)$. Is there anyone could help me out? Thanks.

答: Yes, they can both be $1$. For example $p=41$, $q=31$, $t=5$ or $t=83$.


问:
Beginner's question about representation weights in Lie algebras and their relation to eigenvalues
I am hitting some rough patches in self-learning the representation theory in Lie theory, and specifically, the weight space decomposition. I understand this is not surprising, given the complexity of the topic, but I will not give up based on intimidating nomenclature or definitions.

So I'm asking for an English explanation of what the weights, weight spaces and weight space decomposition are in general based on the particular case below, and explained in here:

In the representation of ${SU}(2)$ restricted to diagonal matrices $\begin{bmatrix}e^{i\theta}&0\\0&e^{-i\theta} \end{bmatrix}$

$$R\left( \begin{bmatrix}e^{i\theta}&0\\0&e^{-i\theta} \end{bmatrix}\right)=\small \begin{bmatrix}e^{im_1\theta}&&&&\\&e^{-im_2\theta}\\&&\ddots\\&&&e^{im_n\theta} \end{bmatrix}$$

with the integers $m_1,m_2,\dots,m_n$ called weights and $V_i=e^{im_1\theta}$ called the weight spaces, so that $V=V_1\oplus V_2\oplus \cdots \oplus V_n$ is the weight space decomposition in the representation $R\vert_{\left\{\begin{bmatrix}e^{i\theta}&0\\0&e^{-i\theta} \end{bmatrix}\right\}}: \begin{bmatrix}e^{i\theta}&0\\0&e^{-i\theta} \end{bmatrix}\to GL(V).$

In the standard representation $R:{SU(2)}\to GL(2,\mathbb C)$ restricted to the diagonal matrices the

weight spaces are the lines spanned by $\begin{bmatrix}1\\0 \end{bmatrix}$ and $\begin{bmatrix}0\\1 \end{bmatrix}$ and

weights are $1$ and $-1.$

My confusion arises in that in trying to follow up the idea behind this construct I "see" the "weights" as the eigenvalues, in which case, the weights would be $e^{i\theta}$ and $e^{-i\theta}$. This, together with the eigenvectors, which nicely coincide with the "weight spaces" above would provide a definition I can follow.

But instead of $e^{i\theta}$ and $e^{-i\theta}$ we are using just $1$ and $-1,$ which it (naively) feels a bit ad hoc, because we won't always have a diagonalized matrix with exponentials and integers.

Further, is the assumption that $\begin{bmatrix}1\\0 \end{bmatrix}$ and $\begin{bmatrix}0\\1 \end{bmatrix}$ are eigevectors correct, or are they just standard basis vectors of $V$ that coincide in this case?

I presume that the issue is that we not always will go with the standard representation (leaving the matrices as are), but then what is the systematic way of get these weights? Or is it done on a case-by-case basis? What would be other examples?

"...a weight of a representation is a generalization of the notion of an eigenvalue, and the corresponding eigenspace is called a weight space."

In the real Lie algebra ${\frak su}(2)$ has basis given by $\begin{bmatrix}0 & i\\ i&0\end{bmatrix}, \begin{bmatrix}0 & -1\\ 1&0\end{bmatrix},\begin{bmatrix}i & 0\\ 0&-i\end{bmatrix},$ while its complexification of ${\frak su}(2,\mathbb C)$ has basis $H=\begin{bmatrix}1 & 0\\ 0&-1\end{bmatrix}, X=\begin{bmatrix}0 & 1\\ 0&0\end{bmatrix},Y=\begin{bmatrix}0 & 0\\ 1&0\end{bmatrix},$ satisfying the commutator relations $[H,X]=2X, [H,Y]=-2Y, [X,Y]=H.$

The eigenvalues of $H$ are the weights of the representation and if $v$ is an eigenvector of $H$ with eigenvalue $\alpha,$

$$\begin{align}\small
H\cdot (X\cdot v) &= (HX)v= (2X+XH)v=2Xv +XHv=2Xv+X\alpha v= (2+\alpha)X\cdot v\\ \small
H\cdot (Y\cdot v) &=  (-2+\alpha)Y\cdot v
\end{align}$$

Initially, and because of the difficulty of handwriting Gothic symbols, I didn't realize the lecture is dealing with the group (not the algebra). I should have realized even intuitively that there is an exponential function there.

答:
The answer consists of two parts: a short 'birds-eye' overview (sections 1-3) and more details and actual definitions to accompany them (sections 4- 6). I once was planning on writing a section 7 but decided I didn't have the time.

ACTUAL ANSWER:

There are two things that come together here:

1. Representation theory of abelian groups is much easier than that of non-abelian groups

In fact it is not really different from the theory of eigenvectors and eigenvalues of linear algebra. You just have to be a bit careful in keeping track of the precise relation between the two.

I will edit in later how this plays out in the example of your post, where the Abelian group in question is a subgroup of $SU(2)$ isomorphic to $U(1)$.

2. If you have a representation $R$ of a group $G$ on a space $V$ and $A$ is a subgroup of $G$, then $R$ is also a representation of $A$.

This is trivial if you think about it, but I mention it explicitly because on the surface it might be tempting to think that smaller groups also have smaller representations.

3. How do these two thing work together?

You start with some big non-abelian group $G$, in your example $SU(2)$. Inside is sitting a smaller abelian subgroup $T$, in your example the diagonal matrices. Now when you have an $n$-dimensional representation $R$ of $G$, perhaps irreducible, on a space $V$ then we can first view it as a representation of $T$. $T$ being abelian, everything is much easier and we can understand what this representation looks like. In particular it need no longer be irreducible and we can decompose it as a direct sum of easier to understand subrepresentations (just like in my answer to your earlier question): the weight spaces.

Now this structure of $V$ as a direct sum of smaller spaces $V_i$ is not necessarily a decomposition into subrepresentations of $G$ (unless $G$ and $T$ were equal from the start) but it might still be useful and simplify computations, especially those involving the Lie algebra.

For instance in your $\mathfrak{su}(2)$ example: the original basis can also serve as a basis of the complexification, but we prefer the other one ($\{X, Y, H\}$) for practical purposes, because it works really well together with the weight space decomposition of representations. As I said I will add in details later when I find the time.

4. Representations of abelian groups and eigenvalues of matrices

Let's start with your example. We have the group $T = \left\{ \begin{pmatrix} e^{i \theta} & 0 \\ 0 & e^{- i \theta} \end{pmatrix} \colon \theta \in \mathbb{R} \right\}$ consisting of all diagonal matrices in the bigger group $G = SU(2)$. We'll ignore $G$ for now and look at the representation $R \colon \, T \to GL(\mathbb{C}^{4})$ given by $R(\begin{pmatrix} e^{i \theta} & 0 \\ 0 & e^{- i \theta} \end{pmatrix}) = \begin{pmatrix}
e^{i m_1 \theta} & 0  & 0 & 0 \\
0 & e^{i m_2 \theta} & 0 & 0 \\
0 & 0 & e^{ i m_3 \theta} & 0 \\
0 & 0 & 0 & e^{ i m_4 \theta}
\end{pmatrix}$ from your post.

Let's focus on two different elements of $T$: $g = \begin{pmatrix} e^{i \theta} & 0 \\ 0 & e^{- i \theta} \end{pmatrix}$ and $h = \begin{pmatrix}  e^{i \phi} & 0 \\ 0 & e^{- i \phi} \end{pmatrix} $.

(Of course purely by notation alone we might have $g = h$ but for educational purposes we asssume that $g \neq h$. Things like $h = g^3$ are no problem.)

Now as you already noticed, the vector $\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$ is an eigenvector for the linear operator $R(g)$ with eigenvalue $e^{i m_1 \theta}$. Similarly $\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}$ is an eigenvector for $R(g)$ with eigenvalue $e^{i m_2 \theta}$ etc. Also, when it happens that $m_1 = m_2$ we have even more eigenvectors: in that case every vector $\begin{pmatrix} a \\ b \\ 0 \\ 0 \end{pmatrix}$ is an eigenvector for $R(g)$ with eigenvalue $e^{i m_1 \theta} = e^{i m_2 \theta}$.

Now what I wanted to point out to you is what happens if instead of computing the eigenvalues and eigenvectors of the operator $R(g)$ we do the same for the operator $R(h)$. Please do it yourself if you haven't already.

What we see is:

The eigenvectors (or more generally: eigenspaces) are the same, but the eigenvalues are different.

Of course the first part is the miracle here: a priori we expect everything to be different from one group element to the next. However it is also good to be aware of the different eigenvalues as it indicates that not everything is exactly as easy as we would hope.

So what is going on here? There is a famous theorem in linear algebra stating:

Theorem Let $S$ be a set of commuting linear operators on the same space $V$. Then there is a decomposition $V = V_1 \oplus \ldots \oplus V_n$ such that each $V_n$ is a generalized eigenspace for every linear operator $L$ from the set $S$. (Though possible at different eigenvalues for different $L$)

Since $T$ is commutative, all the linear operators $R(g), R(h)$ etc commute and so we can take $R(T)$ to be the set $S$ from the theorem. In fact in the example we see that we are even a bit better off and are dealing with actual eigenspaces rather than generalized ones, this is a result of our $T$ being compact in addition to abelian. I will come back to that issue later.

So in short: what are the weight spaces? They are the joint eigenspaces for the operators $R(T)$.

We also see that answering the question: what are the corresponding weights for each weight space? is slightly more tricky since each $R(g)$ can act by a different eigenvalue on a given weight space. I'll write a separate chapter about that.

5. What are the weights?

So how to think about the eigenvalues? If we fix an eigenvector ('weight vector') $v$ then apparently there is some map $\chi \colon T \to \mathbb{C}$ that assigns to every group element $g$ the eigenvalue by which the operator $R(g)$ acts on the eigenvector $v$. Now let's see if we can understand this map.

It cannot be completely random since $R$ is still a representation. If $R(g)$ acts by eigenvalue $a$ and $R(h)$ by eigenvalue $b$ then $R(g)R(h)$ acts by eigenvalue $ab$ by the definition of eigenvalue. But since $R$ is a representation this means that group element $gh$ also has to act by eigenvalue $ab$. In other words $\chi(gh) = \chi(g)\chi(h)$.

Since every group element has an inverse this last equation implies that $\chi$ cannot take the value 0 and so takes values in $\mathbb{C}^* = GL(1, \mathbb{C})$, and more importantly the last equation states that the map $\chi: T \to GL(1, \mathbb{C})$ is a group homomorphism. Now we have a word for group homomorphisms into $GL(1, \mathbb{C})$: one-dimensional representations!

So $\chi$ is a one-dimensional representation of $T$ (one-dimensional representations of groups are sometimes called characters, that is why I chose the name $\chi$) and perhaps this is not so surprising: it is just the representation $R$ where each $R(g)$ is restricted to the one-dimensional subspace $\mathbb{C}v$.

But... whenever we have a group representation we also have a Lie algebra representation. To every character $\chi$ obtained in the way above there is a corresponding Lie algebra representation $\chi' \colon \mathfrak{t} \to \mathfrak{gl}(1, \mathbb{C}) = \mathbb{C}$, where $\mathfrak{t}$ is the Lie algebra of $T$.

This Lie algebra representation is the weight for the representation $R$ of the group $T$ corresponding to the weight space $\mathbb{C}v$.

6. Some notational issues (including: why are weights for $SU(2)$ integers?)

One-dimenisonal Lie algebra representations of abelian groups are only ever called weights if they appear in the context discussed in chapter 2, so if the representation $R$ of $T$ that we started with was the restriction to $T$ of a representation $R$ of a bigger, non-abelian group $G$.

When there is talk about 'a weight of $G$ for the abelian subgroup $T$' without mentioning the specific representation, it means that it is the weight of some finite dimensional representation of $G$.

If also the group $T$ is omitted from mentioning and people just talk about 'a weight of $G$' (or a weight of a specific representation of $G$, but without mentioning which Abelian subgroup $T$ we are working with) then the tacit assumption is that $T$ is a maximal abelian subgroup.

The term 'One-dimensional Lie algebra representation' is incredibly pretentious in case the Lie algebra $\mathfrak{t}$ of which we take the representation is abelian: in that case the Lie bracket sends every 2 elements to zero and hence $\mathfrak{t}$ contains no more information than its underlying vector space. As a result the more common term for 'one-dimensional Lie algebra representation of $\mathfrak{t}$' is 'element of the linear dual $\mathfrak{t}^*$ of $\mathfrak{t}$'. Hence every weight (in the sense of the 3rd bullet point) of $G$ is an element of $\mathfrak{t}^*$. The converse is not true: the set of all weights (of all finite dimensional reps of $G$ for a fixed $T$) is an infinite but still only discrete and hence tiny subset of the vectorspace $\mathfrak{t}^*$.

In case $T$ and hence $\mathfrak{t}$ and hence $\mathfrak{t}^*$ are one dimensional (as is the case for $G = SU(2)$) we can identify $\mathfrak{t}^*$ with $\mathbb{C}$ and hence look at elements of $\mathfrak{t}^*$ (including all weights) as numbers.

Why in this case these numbers are moreover integers can be seen when going back up one level to groups instead of Lie algebras: each $\chi$ ('the exponent of the weight') maps the circle group $T$ into $\mathbb{C}^*$ by definition, but in practice more specifically into the unit circle inside $\mathbb{C}^*$. (Because of compactness.) So... each character $\chi$ (corresponding to a weight) is a group homomorphism from the circle group to itself and it is easy to see that the only maps of that type are all of the form $g \mapsto g^n$ for some fixed integer $n$.


答:
As $p\not\equiv 3\pmod 4$, we have $\left(\frac pq\right)=\left(\frac qp\right)$ and $\left(\frac pt\right)=\left(\frac tp\right)$. If we additionally demand $t\not\equiv 3\pmod 4$, we will also have $\left(\frac qt\right)=\left(\frac tq\right)$. So in order to have $\left(\frac tp\right)=\left(\frac tq\right)=1$, we impose the following modular restrictions on $t$:

$t\equiv 1\pmod 4$

$t\equiv a\pmod p$ for some quadratic residue $a\bmod p$

$t\equiv b\pmod q$ for some quadratic residue $b\bmod q$

Of course, we can simply take $a=b=1$, and as $4,p,q$ are coprime, we only need

$t\equiv 1\pmod{4pq}$.

Such primes $t$ exist by Dirichlet.

Note that ultimately we only used that $p,q$ are distinct odd primes. And even distinctness is irrelevant.

Title: Could two distinct prime number $p,q$ both be the quadratic residue of another prime $t$ with certain constraints?

Body:
Problem: let $p,q$ be two distinct odd prime numbers with
$$p \equiv 1 \pmod 8;\quad  q \equiv 3 \pmod 4 ; \quad  \left(\frac{p}{q}\right) = 1.$$

I wonder if $\left(\frac{p}{t}\right)$ and $\left(\frac{q}{t}\right)$ could both be $1$ or not ($\forall\ t$ is a prime where $t \neq p,q$).

I tried to apply the Law of Quadratic Reciprocity but still, I cannot see a clear connection between $\left(\frac{t}{q}\right)$ and $\left(\frac{t}{p}\right)$. Is there anyone could help me out? Thanks.


问:
Completeness of a certain Lipschitz function space (w/o measure theory)
I found this problem in an old analysis 2 (before I learned about anything measure theory related) exam of mine. We had to prove that
$$
X := \left \lbrace f: [0, 1]^2 \rightarrow \mathbb{R}: \lvert f(x) - f(y) \rvert \leq \lVert x - y \rVert_2, ~\forall x, y \in [0, 1]^2 \right \rbrace
$$
is complete with respect to the metric $d: X \times X \rightarrow \mathbb{R}$ is given by
$$
d(f, g) := \int_{[0, 1]^2} \lvert f(x) - g(x) \rvert \lVert x \rVert_2^2~\mathrm{d}x.
$$

I really have trouble tackling this without any measure theorey results. Even with measure theory it seems rather hard. I would start by choosing a $d$-Cauchy-sequence $(f_k)_{k \in \mathbb{N}} \subseteq X$   and trying to prove that a subsequence $(f_{k_m})_{m \in \mathbb{N}}$ with $d(f_{k_m}, f_{k_\ell})$ very, very small  converges. I would love argue with converging sums of $\lvert f_{k_m}(x) - f_{k_\ell}(x) \rvert$ like one does in measure theory...

(I would prefer not using Fatou's Lemma, the fact that $L^p$ convergent series converge pointwise a.e. in a subsequence etc...)

答:
Sequences in $X$ are equi-continuous. It is sufficient to show that a Cauchy sequence in $X$ is bounded in the $\max$-norm.

Let $x_0\in(0,1)^2$. Then
$$
|f(x_0)-g(x_0)| \le |f(x)-g(x)| + 2 \|x-x_0\|_2 \le \frac12 |f(x)-g(x)|^2 + 2\sqrt2+\frac12
$$
for all $x$ (using $2|a| \le a^2+1$). Now we can integrate with respect to $x$, but only on a subset, where $\|x\|_2$ is bounded away from zero. Then
$$
|f(x_0)-g(x_0)| \le \frac12 \int_{x\in (1/2,1)} |f(x)-g(x)|^2 dx + C
\le \int_{x\in (1/2,1)} |f(x)-g(x)|^2\cdot \|x\|_2^2 dx + C\le d(f,g)+C
$$
with some constant independent of $x_0$.
The right-hand side is independent of $x_0$, so we can take the maximum over $x_0$. This proves that bounded sequences in $X$ are bounded with respect to the $\max$-norm. Arzela-Ascoli theorem gives convergence, passing to the pointwise limit in the inequality defining $X$ shows that the limit is in $X$.  In, addition convergence in the metric $d$ can be proven.


问:
Probability of getting one item from $n$ items after $x$ tries, such that in each try a picked unwanted item is removed.
Suppose there's items $A,B,C,D,E$ where the desired pick will be to pick $C$. At first try, I randomly picked $B$, hence now the pick-pool will be $A,C,D,E$ and second try I picked $E$ and now it's $A,C,D$ and so on.

So if there's $n$-items and $x$-tries ($x \leq n$). What is the probability that I get the desired item out of these random items after $x$-tries. If we have $x=1$, then the probability is $1/n$ and if $x=n$, then it's $1$

答:
Thought experiment.

You have $n$ items, one of them with the label X.

Randomly assign each of the items a number from $\{1,2,\cdots,n\},$ without replacement.

The assigned number represents on which turn the item will be picked.

Given any two numbers $i,j$ such that $i,j \in \{1,2,\cdots, n\}$ and $i \neq j$, is the item labeled X more likely to be assigned the number $i$ rather than the number $j$?

Is the item labeled X more likely to be assigned the number $j$ rather than the number $i$?


问:
Lebesgue integral, $f(x) = |x|^\alpha$ for $x \neq 0$ and $f(0)=0$
Looking at $f(x) = |x|^\alpha$ for $x \neq 0$ and $f(0)=0$. Which of these following statements is true?

a. A $\alpha$ exists such that $f \in \mathcal {L}(\mathbb {R})$.

b. A $\alpha$ exists such that $f \in \mathcal {L}([-1,1])$.

c. A $\alpha$ exists such that $f \in \mathcal {L}(\mathbb {R} \setminus [-1,1])$.

I think that a is true, that b is wrong and that c is true; is that correct?

答:
a) is false and b) and c) are true.

$\int_{-1}^{1}|x|^{\alpha} dx<\infty$ iff $\alpha >-1$

$\int_{|x|>1} |x|^{\alpha} dx <\infty$ iff $\alpha  <-1$

Since these two conditions cannot both be satisfied it follows that $f$ can never be integrable on $\mathbb R$.


问:
Finding ordered pair of polynomial $(f(x),g(x))$ with coefficient $\mathbb{Z}_{p}$ for a prime number $p$
In process of solving a problem, I had to find some ordered pair of polynomial $(f(x),g(x))$ with coefficient $\mathbb{Z}_{p}$, $p$ is prime number. (i.e $f(x),g(x) \in \mathbb{Z}_{p}[x])$
such that
$\frac{f(x+r)}{g(x+r)}=\frac{f(x)}{g(x)}$ for all $r \in \mathbb{Z}_{p}$.

I find out that $\mathbb{Z}_{p} \simeq \{(r,1) \mid r\in \mathbb{Z}_{p} \}$ is one of them, but can't find out another.
i thought about $f(x^{p}), g(x^{p})$ because $(x+r)^p=x^{p}+r^p$ but $r^p$ doesn't disappear when $1 \leq r <p $ by Fermat's theorem.
is there something else? and if there's no more, how can i prove it?

P.s.: $x$ is transcendental over $\mathbb{Z}_{p}$, so all the members of $\left\{\frac{f(x)}{g(x)} \ \middle\vert\ f(x),g(x) \in \mathbb{Z_p} \right\}$ is simple extension of $x$ over $\mathbb{Z_p}$.

答:
You are looking for the fixed points of the group of automorphisms $G$ of the function field $L=\Bbb{Z}_p(x)$, where $G$ consists of the automorphisms
$$
\sigma_r:\frac{f(x)}{g(x)}\mapsto\frac{f(x+r)}{g(x+r)}
$$
and $r$ ranges over $\Bbb{Z}_p$. We have $\sigma_{r_1}\circ\sigma_{r_2}=\sigma_{r_1+r_2}$, so as a group it is isomorphic to the additive group of $\Bbb{Z}_p$. In particular, it is cyclic of order $p$.

By general Galois theory it follows that the fixed field of $G$ is a subfield $K=\operatorname{Inv}(G)$ of $L$ such that $[L:K]=p$. We need to identify $K$.

Your idea of using the Frobenius is good. All that is missing is to recall Little Fermat, stating that $r^p=r$ for all $r\in\Bbb{Z}_p$. It follows that $u=x^p-x\in K$. Simply because
$$
\sigma_r(x^p-x)=(x+r)^p-(x+r)=x^p+r^p-x-r=x^p-x.
$$
It follows that the purely transcendental extension $F=\Bbb{Z}_p(u)\subseteq K$. I claim that we actually have equality here, $K=F$. This follows from the observation that the relation
$$
x^p-x=u
$$
implies that $x$ is algebraic of degree $\le p$ over $F$. This, in turn, implies that $[L:F]\le p$. Together with the Galois theoretic fact the claim follows.