问:
Convergence of double sum on the lattice
I am working with a commutator $T$ acting on the lattice $\ell^2(\mathbb{Z}^2;\mathbb{C})$, the function space made up by the basis elements
\begin{align}\left|\vec{x}\right>\,:\,\mathbb{Z}^2&\rightarrow \mathbb{C}\\
\vec{y}&\mapsto \delta_{\vec{x},\vec{y}}
\end{align}
I am studying a paper in which the authors claim a particular (double) sum is convergent. It is the following,
\begin{equation}
C_N\sum_{r_1\in \mathbb{Z}}\sum_{x_1\in \mathbb{Z}}\left(1+\frac{1}{2}\left(|x_1+r_1|+|x_1|\right)\right)^{-N}.\qquad (1)
\end{equation}
Here, $N$ can be any real number, to each $N$ a certain constant $C_N$ is fit. In the paper, the authors claim this sum to be convergent, but I have a hard time realising why that is the case. At first thought, for fixed $r_1'\in \mathbb{Z}$, e.g. for $N=2$, the inner sum will look very much like the sum $\sum_{x_1\in \mathbb{Z}}\frac{1}{|x_1|^2}$, which I now converge to a finite number. But if I sum over this finite number an infinite number of times, obviously it will not converge. Is it possible to pick a proper $N$ such that the double sum in (1) does indeed converge?

答:
It converges for $N>2$. For this we can use the (stupid) estimate
$$ \frac{1}{(1+\vert x_1 + r_1 \vert + \vert x_1 \vert)^N} \leq \frac{1}{(1+\vert r_1 + x_1\vert)^{N/2}} \frac{1}{(1+\vert x_1 \vert)^{N/2}}.$$
Then we have
$$ \sum_{x_1, r_1 \in \mathbb{Z}} \frac{1}{(1+\vert x_1 + r_1 \vert + \vert x_1 \vert)^N}
\leq \sum_{x_1, r_1 \in \mathbb{Z}} \frac{1}{(1+\vert r_1 + x_1\vert)^{N/2}} \frac{1}{(1+\vert x_1 \vert)^{N/2}}
= \left(\sum_{k\in\mathbb{Z}} \frac{1}{(1+\vert k \vert)^{N/2}} \right)^2.$$
For the last equality we first fixed $x_1$ and summed over $r_1$ (shifting it by $x_1$). For $N>2$ we have $N/2>1$ and so the RHS does converge and hence so does the LHS.

We can also analyse it a bit more carefully to see that it does not converge for $1<N<2$ (clearly it does not converge for $N\leq 1$, not even the series over a single variable). For this we will use the integral test, which tells us
$$ \frac{N-1}{(2+\vert x_1 \vert)^{N-1}} = \int_1^\infty \frac{1}{(1+\vert x_1\vert +r)^N} dr \leq \sum_{r\in \mathbb{N}_{\geq 1}} \frac{1}{(1+\vert x_1 \vert + r)^N} . $$
Thus, we obtain
$$ \sum_{x_1\in \mathbb{Z}} \sum_{r_1\in \mathbb{Z}} \frac{1}{(1+\vert x_1 + r_1 \vert + \vert x_1 \vert)^N}
\geq \sum_{x_1\in \mathbb{Z}} \left(  2 \sum_{r_1\in \mathbb{N}} \frac{1}{(1+\vert x_1 + r_1 \vert + \vert x_1 \vert)^N} \right)
\geq 2\sum_{x_1\in \mathbb{Z}} \frac{N-1}{(2+\vert x_1 \vert)^{N-1}}.$$
Using the integral again, one sees that the RHS blows up and so our original series does not converge.


问:
Analysis and n-adic expansions
I am trying to discern some mathematical term known as a $n$-adic expansion. For example the book I am reading says: write every $x  \in[0, 1]$ its "$4$-adic" expansion
\begin{equation}
   x = \sum\limits_{i=1}^\infty\dfrac{x_i}{4^i} = \sum\limits_{i=1}^k\dfrac{x_i} {4^i} + r_k(x)
\end{equation}
for $(x_i \in \{0, 1, 2, 3\}, 1 \leq i < \infty)$

Could someone explainwhat this $4$-adic expansion means and why it is useful? Every google result search brings up $p$-adic expansions where $p$ is a prime.

答: As explained in comments: $n$-adic is sometimes used  as a synonym of "base $n$", although this use is old-fashioned and potentially confusing. These days one is more likely to hear binary expansion than dyadic expansion. Nonetheless, an interval of $\mathbb R$ formed by all numbers with a given initial segment of binary expansion is usually called a dyadic interval.

ContentLicense: CC BY-SA 3.0

Title: Analysis and n-adic expansions


问:
Alternative expression of norm L1
Let $(X, \Sigma , \mu)$ a space with measure and $f \in L^1(X,\mu)$. Prove that
$$
||f||_{L^1(X,\mu)} = \int_{0}^{+\infty} \mu(A_r) dr $$

Where:

$$A_r = \{x \in X: |f(x)|\ge r\}$$

I tried to solve this exercise, but I couldn't. I was unable to relate the integral in X to the real integral. Can someone help me?

答:
This is an exercise in Fubini.

\begin{align}
\|f\|_{L^1(X,\mu)} &=\int_X |f(x)|\textrm{d}\mu(x)=\int_X \int_0^{|f(x)|} 1\textrm{d}r\textrm{d}\mu(x)=\int_X \int_0^{\infty} 1_{r\leq|f(x)|}\textrm{d}r\textrm{d}\mu(x)\\
&=\int_0^{\infty} \int_X 1_{r\leq |f(x)|} \textrm{d}\mu(x) \textrm{d}r=\int_0^{\infty} \mu(A_r)\textrm{d}r
\end{align}

ContentLicense: CC BY-SA 4.0

Title: Alternative expression of norm L1


问:
Solving system of equation using Macaulay2
As an algebraic curve, the Klein quartic can be viewed as a projective algebraic curve over the complex numbers $\mathbb{C}$, defined by the following quartic equation in homogeneous coordinates $[x:y:z]$ on $\mathbb{P}^2_{\mathbb{C}}$:

$$x^3 y + y^3 z + z^3 x = 0.$$

Now we want to find the eigenvectors of this curve, therefore consider the following matrix:

$$
 \begin{bmatrix}
 3x^2y+z^3 & x^3 + 3y^2z & y^3+3z^2  \\
 x & y & z  \\
 \end{bmatrix},
 $$
after computing the $2$-minors we got the following system:
\begin{equation*}
\left\{
\begin{alignedat}{3}
% R & L   &  R & L   &  R & L
3x^2y^2+z^3y & -{} &  (3y^2zx + x^4) &  = 0 \\
3z^2x^2+y^3x & -{} &  (3x^2yz + z^4)& = 0 \\
3y^2z^2+x^3z & -{} &  (3z^2xy + y^4)& = 0
\end{alignedat} \ .
\right.
\end{equation*}
I am new to Macaualy 2 and I am wondering if Macaulay2 can help me to solve the above system numerically with loading NumericalAlgebraicGeomety package, (I know that because of infinity many solution we need normalize the system on unit sphere).

答:
I can admit I got the same many solution, I was using the below code (you can try it online Macaulay2Web):

"
R=QQ[x,y,z]

f=x^3* y + y^3* z + z^3* x

x1=basis(1,R)

I=minors(2,diff(x1,f)||x1)

pI=primaryDecomposition I

loadPackage "NumericalAlgebraicGeometry"

q=x^2+y^2+z^2

I=minors(2,diff(x1,f)||x1)+ideal(q-1)

sol=solveSystem(first entries gens I)

length sol"

and you will got 26, 13 pairs on sphere. Thanks for the answering I should learn Mathematica too. I will check it precisely.


问:
Show that $4\mathbb{Z}+6\mathbb{Z}=2\mathbb{Z}$.
Show that $4\mathbb{Z}+6\mathbb{Z}=2\mathbb{Z}$.

My attempt goes something like this:

An arbitrary elt in $4\mathbb{Z}$ looks something like $4x$ for some $x \in \mathbb{Z}$.

An arbitrary elt in $6\mathbb{Z}$ looks something like $6x$ for some $x \in \mathbb{Z}$.

Then $4x+6y=2(2x+3y)$, note $(2x+3y) \in \mathbb{Z}$, which is the form of an element in $2\mathbb{Z}$.

Thus $4\mathbb{Z}+6\mathbb{Z}=2\mathbb{Z}$

Would this be correct? It seems too easy. Should I have used the fact that for ideals I and J, I+J is an ideal? Any help is appreciated.

答:
It is too easy! You've only shown one direction of the set inequality, that $4\mathbb{Z} + 6\mathbb{Z} \subseteq 2\mathbb{Z}$. Now you need to go the other direction, show that any even integer can be decomposed appropriately.

Hint: 2 = 6 - 4.


答:
You have shown that $4\mathbb{Z} +6\mathbb{Z} \subseteq 2\mathbb{Z}$. You should show that you have inclusion the other direction as well, so $2\mathbb{Z} \subseteq 4\mathbb{Z} +6\mathbb{Z}$.

I'm not sure what you mean by 'too easy'. A more complicated proof does not mean that it's better, although if it helps you understand the connections between ideas it might be helpful to you personally.

Body:
Show that $4\mathbb{Z}+6\mathbb{Z}=2\mathbb{Z}$.

My attempt goes something like this:

An arbitrary elt in $4\mathbb{Z}$ looks something like $4x$ for some $x \in \mathbb{Z}$.

An arbitrary elt in $6\mathbb{Z}$ looks something like $6x$ for some $x \in \mathbb{Z}$.

Then $4x+6y=2(2x+3y)$, note $(2x+3y) \in \mathbb{Z}$, which is the form of an element in $2\mathbb{Z}$.

Thus $4\mathbb{Z}+6\mathbb{Z}=2\mathbb{Z}$

Would this be correct? It seems too easy. Should I have used the fact that for ideals I and J, I+J is an ideal? Any help is appreciated.


问:
Extending the Supplement of Eisenstein Reciprocity
One of the supplements of Eisenstein Reciprocity states the following:

Supplement: If $m$ is an odd prime and $a$ is a rational integer relatively prime to $m$, then

$\left(\frac{1-\zeta_m}{a }\right)_m =
\left(\frac{\zeta_m}{a }\right)_m^{\frac{m+1}{2}} = (\zeta_m^{\frac{a^{m-1}-1}{m}})^{\frac{m+1}{2}} = \zeta_m^{\frac{(a^{m-1}-1)(m+1)}{2m}}.$

I would like to know if the m-th power residue symbol of $1-\zeta_m$ can be extended to prime ideals $\mathscr{P}\subseteq\mathbb{Z}[\zeta_m]$. Is it possible to evaluate

$\left(\frac{1-\zeta_m}{\mathscr{P}}\right)_m$

in terms of $N\mathscr{P}$ and perhaps the Frobenius element of $\mathscr{P}$? Part of the reason that I am curious about this is that I would like to know if

$\left(\frac{1-\zeta_m}{\mathscr{P}}\right)_m = 1$

for a set of prime ideals $\mathscr{P}\subseteq\mathbb{Z}[\zeta_m]$ of 'density' $\frac{1}{m}$. We know that this is the case with quadratic reciprocity since $\left(\frac{2}{p}\right)_2 = 1$ if and only if $p \equiv 1,-1\pmod{8}$, which consists of 'half' of the primes in $\mathbb{Z}$. Similarly, we know that for cubic reciprocity we have that $\left(\frac{3}{\pi}\right)_3 = 1$ for a primary prime $\pi = a+b\zeta_3$ if and only if $3|b$, which consists of 'one third' of the primes in $\mathbb{Z}[\zeta_3]$.

P.S. I have unfortunately not been able to find and answer to this Lemmermeyer's book, in Ireland and Rosen's book, through a google search, or even by asking my friend who specializes in higher reciprocity laws. It seems like a natural question to ask, so it must be difficult to answer.

答:
Your question is not difficult to answer, although some of the underlying proofs are technical (but have been known for over 100 years).  In what sense does your friend "specialize" in higher reciprocity laws?

You ask if the $m$th power residue symbol "can be extended" to prime ideals, but you have things backwards: the $m$th power residue symbol is first defined on prime ideals. Letting
$$
K = \mathbf Q(\zeta_m),
$$
each prime ideal $\mathfrak p$ in $\mathbf Z[\zeta_m]$ that does not divide $m$ is unramified in $K$ and $x^m - 1$ is separable in $(\mathbf Z[\zeta_m]/\mathfrak p)[x]$. For each $\alpha \in \mathbf Z[\zeta_m]$ such that $\alpha \not\equiv 0 \bmod \mathfrak p$, $\alpha^{({\rm N}(\mathfrak p)-1)/m} \bmod \mathfrak p$ is an $m$th root of unity in the residue field $\mathbf Z[\zeta_m]/\mathfrak p$, so there is a unique $m$th root of unity $\zeta$ in $K$ such that
$$
\alpha^{({\rm N}(\mathfrak p)-1)/m} \equiv \zeta \bmod \mathfrak p.
$$
This $\zeta$ is defined to be the power residue symbol $(\frac{\alpha}{\mathfrak p})_m$. When $m = 2$, so $K = \mathbf Q$, this is the Legendre symbol $(\frac{a}{p})$ when $\mathfrak p = p\mathbf Z$ for an odd prime $p$ and integer $a$ not divisible by $p$.

The power residue symbol $(\frac{\alpha}{\mathfrak p})_m$ is a Frobenius element in a Kummer extension of $K$.  Set
$$
L = K(\sqrt[m]{\alpha}).
$$
Since $K$ contains all the $m$th roots of unity, $L/K$ is abelian with a cyclic Galois group of order dividing $m$: ${\rm Gal}(L/K)$ emebds into $\mu_m$ by $\sigma \mapsto \sigma(\sqrt[m]{\alpha})/\sqrt[m]{\alpha}$.  (That ratio is independent of the choice of $m$th root of $\alpha$, since different choices have a ratio in the $m$th roots of unity, which all lie in $K$ and thus are all fixed by $\sigma$.) Since the Galois group is abelian, primes in $K$ have well-defined Frobenius elements in $L$ when they are unramified in $L$.

Claim: When $\mathfrak p \nmid (\alpha)m$ in $\mathbf Z[\zeta_m]$, the Frobenius element ${\rm Frob}_{K(\sqrt[m]{\alpha})/K}(\mathfrak p)$ is identified with $(\frac{\alpha}{\mathfrak p})_m$ under the above embedding of ${\rm Gal}(K(\sqrt[m]{\alpha})/K)$ into $\mu_m$.

Proof: When $\mathfrak p \nmid (\alpha)m$, $\mathfrak p$ is unramified in $L$.  Letting $\mathfrak P$ be a prime lying over $\mathfrak p$ in $\mathcal O_L$, by definition
$$
{\rm Frob}_{L/K}(\mathfrak p)(\beta) \equiv \beta^{{\rm N}(\mathfrak p)} \bmod \mathfrak P
$$
for all $\beta$ in $\mathcal O_L$.  Taking $\beta = \sqrt[m]{\alpha}$, we get
$$
{\rm Frob}_{L/K}(\mathfrak p)(\sqrt[m]{\alpha}) \equiv \sqrt[m]{\alpha}^{{\rm N}(\mathfrak p)} \bmod \mathfrak P.
$$
Since
$$
\sqrt[m]{\alpha}^{{\rm N}(\mathfrak p)} =
\sqrt[m]{\alpha}^{({\rm N}(\mathfrak p)-1)}\sqrt[m]{\alpha} =
(\sqrt[m]{\alpha}^m)^{({\rm N}(\mathfrak p)-1)/m}\sqrt[m]{\alpha}
=
\alpha^{({\rm N}(\mathfrak p)-1)/m}\sqrt[m]{\alpha},
$$
we have
$$
{\rm Frob}_{L/K}(\mathfrak p)(\sqrt[m]{\alpha}) \equiv \alpha^{({\rm N}(\mathfrak p)-1)/m}\sqrt[m]{\alpha} \bmod \mathfrak P.
$$
By the definition of the power-residue symbol,
$\alpha^{({\rm N}(\mathfrak p)-1)/m} \equiv (\frac{\alpha}{\mathfrak p})_m \bmod \mathfrak p$.  A congruence modulo $\mathfrak p$ is also true modulo $\mathfrak P$, so
$$
{\rm Frob}_{L/K}(\mathfrak p)(\sqrt[m]{\alpha}) \equiv
\left(\frac{\alpha}{\mathfrak p}\right)_m\sqrt[m]{\alpha} \bmod \mathfrak P.
$$

Under the embedding of ${\rm Gal}(L/K)$ into $\mu_m$,
the Frobenius element $\sigma := {\rm Frob}_{L/K}(\mathfrak p)$ is mapped to the ratio $\sigma(\sqrt[m]{\alpha})/\sqrt[m]{\alpha}$. Call this $m$th root of unity $\zeta$, so $\sigma(\sqrt[m]{\alpha}) = \zeta\sqrt[m]{\alpha}$.  Therefore
$$
\zeta\sqrt[m]{\alpha} \equiv \left(\frac{\alpha}{\mathfrak p}\right)_m\sqrt[m]{\alpha} \bmod \mathfrak P.
$$
Since $\mathfrak p \nmid (\alpha)$, also $\mathfrak P \nmid (\sqrt[m]{\alpha})$ in $\mathcal O_L$, so we can divide both sides of the congruence by $\sqrt[m]{\alpha}$:
$$
\zeta \equiv \left(\frac{\alpha}{\mathfrak p}\right)_m \bmod \mathfrak P.
$$
Since $\mathcal O_L/\mathfrak P$ is a field of characteristic not dividing $m$, $m$th roots of unity that are congruent modulo $\mathfrak P$ must be equal, so
$$
\zeta = \left(\frac{\alpha}{\mathfrak p}\right)_m.
$$
In other words, for each prime ideal $\mathfrak p$ in $\mathbf Z[\zeta_m]$ not dividing $(\alpha)m$, the standard embedding of ${\rm Gal}(K(\sqrt[m]{\alpha})/K)$ into $\mu_m$ maps the Frobenius element of $\mathfrak p$ to the power residue symbol $(\frac{\alpha}{\mathfrak p})_m$. That finishes the proof of the claim.

The Chebotarev density theorem, in the case of an abelian Galois extension of degree $n$, says each element of the Galois group is a Frobenius element for a set of prime ideals in the base field with (natural) density $1/n$.  In the case of the abelian extension $K(\sqrt[m]{\alpha})/K$, where $K = \mathbf Q(\zeta_m)$ and we embed the Galois group into $\mu_m$ in order to identity (all but finitely many) Frobenius elements with power residue symbols $(\frac{\alpha}{\mathfrak p})_m$, we can say that for each $m$th root of unity $\zeta$ that is in the image of the embedding of ${\rm Gal}(K(\sqrt[m]{\alpha})/K)$ into $\mu_m$, the set of primes $\mathfrak p$ in $\mathbf Z[\zeta_m]$ such that $(\frac{\alpha}{\mathfrak p})_m = \zeta$ has natural density $1/[K(\sqrt[m]{\alpha}):K]$.

In particular, if $x^m - \alpha$ is irreducible over $\mathbf Q(\zeta_m)$ then for each $m$th root of unity $\zeta$, the set of primes $\mathfrak p$ in $\mathbf Z[\zeta_m]$ not dividing $(\alpha)m$ such that $(\frac{\alpha}{\mathfrak p})_m = \zeta$ has natural density $1/m$.

You are asking specifically about primes $\mathfrak p$ in $\mathbf Z[\zeta_m]$ such that $(\frac{\alpha}{\mathfrak p})_m = 1$.  That means $\mathfrak p$ splits completely in $\mathbf Q(\zeta_m,\sqrt[m]{\alpha})$ (an unramified prime splits completely in a Galois extension if and only if its Frobenius conjugacy class is trivial). The Chebotarev density theorem implies that the set of primes splitting completely in a Galois extension of degree $n$ has density $1/n$, and in your case $1/n = 1/m$ if and only if $x^m - \alpha$ is irreducible over $\mathbf Q(\zeta_m)$.  So you need to determine whether $x^m - (1 - \zeta_m)$ is irreducible in $\mathbf Q(\zeta_m)[x]$, where the parentheses around the constant term just collect terms together; they do not indicate a principal ideal.

The Chebotarev density theorem is almost 100 years old (proved by Chebotarev in the 1920s), but to compute the density of primes with a given Frobenius element in an abelian Galois group we can use instead a weaker result than Chebotarev's density theorem called the Frobenius density theorem (proved in 1880). This result of Frobenius is about the density of primes with Frobenius element lying in something possibly larger than a conjugacy class of a Galois group when the group is not abelian, but for  abelian Galois groups, the Frobenius and Chebotarev density theorems become the same statement. Since $K(\sqrt[m]{\alpha})/K$ is an abelian Galois extension, the fact that the density of $\mathfrak p$ such that $(\frac{\alpha}{\mathfrak p})_m$ has a particular value (like $1$) is $1/[K(\sqrt[m]{\alpha}):K]$ could be considered a consequence of the Frobenius density theorem.


问:
Show the unilateral forward shift is essentially normal
Let $H$ be a complex, infinite dimensional, separable complex Hilbert space with orthonormal basis $(e_n)_n$.

Let $T$ be the unilateral forward shift with $Te_n= e_{n+1}$.

We say that $T \in \mathcal{B(H)}$ is essentially normal if the image $\pi(T)$ of $T$ under the canonical quotient map $\pi: \mathcal{B(H)} \rightarrow {\mathcal{B(H)}}/{\mathcal{K(H)}}$ in a sense that $\pi(T)\pi(T^*)= \pi(T^*)\pi(T)$

Show that $T$ is essentially normal.

What I've tried:

Since the quotient map is linear, I want to show that

$\pi(T)\pi(T^*) - \pi(T^*) \pi(T)=\pi(TT^*) - \pi(T^*T)= \pi(TT^*- T^*T)=0$

But not I'm confused because isn't that equivalent to showing that $T$ is normal?
i.e. $TT^* - T^*T = 0$.

Also, I'm not sure how to find $TT^* - T^*T$.

Any help will be appreciated!

Thank you in advance!

答: Hints: since $\langle T^*e_n,e_m\rangle=\delta_{n(m+1)}$ show that $\langle T^*e_n,x\rangle=\langle e_{n-1},x\rangle$ and therefore that $T^*e_n=e_{n-1}$. Then, $T^*T-TT^*=I-TT^*.$ which you can show by direct calculation, to be projection on the rank one subspace spanned by $e_0.$


问:
Why would a 2 degree polynomial with 2 variables on intersecting with a line give either 0, 1 or 2 points?
For clarification, the 2nd degree polynomial with two variables is $f(x,y)$, and the intersecting line is also a (first degree) polynomial in $x$ and $y$.

Why would the 2nd degree polynomial with 2 variables intersect with a line in either 0, 1 or 2 points?

I have no clue on how even to begin this, so any basic idea/hint would be helpful. I have tried a couple of straightforward easy examples, but I haven't been able to generalize it.

答:
There is an exception to what the problem describes.  Let $f(x,y)$ be a second degree polynomial and $g(x,y)$ a first degree polynomial.   If $g$ divides $f$ as polynomials, then all (infinitely many) solutions to $g(x, y)= 0$ also solve $f(x,y)=0$.

Otherwise we can show that the equations $f( x,y)=0$ and $g(x,y)=0$ have at most two solutions in common.  To begin note that since $g(x,y)$ is first degree,  at least one of the variables $x,y$ will have a nonzero coefficient.   Assume for the sake of argument that $y$ has a nonzero coefficient (otherwise we exchange the roles of $x$ and $y$ in the steps that follow).

Now rewrite the equation $g(x,y)=0$ as giving $y=mx+b$.

Substitute that for $y$ in the equation $f(x,y)=0$.  The result is an equation only in $x$ of degree (at most) two.

The point to be careful of is that because $g$ doesn't divide $f$, the result cannot simplify to $0=0$.  In other words there will not be more than two solutions for $x$ and corresponding solutions for $y=mx+b$.


问:
Find the limit of the sequence $(x_n)$, where $x_1=1$ and $x_{n+1}=(2+x_n)^{1/2}$
I have been given this hint in the problem: In order to determine the limit, it may be helpful to use the fact that if $y_n \rightarrow y$ and $y_n > 0$ for every $n$,then $\sqrt{y_n} \rightarrow \sqrt{y}$. You can use this without proving it. You can also use without proof the fact that if $0\le a \le b,$ then $\sqrt{a} \le \sqrt{b}$.

答: You have: $|x_n - x_{n-1}| = \left|\sqrt{x_{n-1}+2}-\sqrt{x_{n-2}+2}\right|=\dfrac{|x_{n-1} - x_{n-2}|}{\sqrt{x_{n-1}+2}+\sqrt{x_{n-2}+2}}\le \dfrac{|x_{n-1}-x_{n-2}|}{2\sqrt{2}}$. This shows it's a Cauchy sequence, hence converges to $L$ that also satisfies: $L = \sqrt{L+2} \implies L = 2$.


答: You can show more generally that $a\mathbb{Z} +b\mathbb{Z} =\text{gcd} (a,b) \mathbb{Z} $.

Title: Show that $4\mathbb{Z}+6\mathbb{Z}=2\mathbb{Z}$.


问:
Showing the Fibonacci inequality $f_1^{f_1}f_2^{f_2}f_3^{f_3}\cdots f_n^{f_n}\leq f_1!f_2!f_3!\cdots f_n!\;e^{({f_{n+2}-n-1)}}$ without induction.
Today I saw the following 2 beautiful inequalities on a facebook page.

$$1^12^23^34^4\cdots n^n\leq 1!2!3!4!\cdots n!e^{\frac{n(n-1)}{2}}$$

$$f_{1}^{f_{1}}f_{2}^{f_{2}}f_{3}^{f_{3}}\cdots f_{n}^{f_{n}}\leq f_{1}!f_{2}!f_{3}!\cdots f_{n}!e^{({f_{n+2}-n-1)}}$$

Here $f_{n}$ denotes the fibonacci numbers $f_{1}=f_{2}=1$ and $f_{n+2}=f_{n+1}+f_{n}$ for $n\in N$

Here is How I proved the first one.

Method 1:-

$$1^12^23^34^4...n^n\leq 1!2!3!4!...n!e^{\frac{n(n-1)}{2}}$$

$$\implies\bigg(\frac{1^1}{1!}\bigg)\bigg(\frac{2^2}{2!}\bigg)\bigg(\frac{3^3}{3!}\bigg)...\bigg(\frac{n^n}{n!}\bigg)\leq e^{\frac{n(n-1)}{2}}$$

$\implies \displaystyle\prod_{x=1}^{n} \frac{x^x}{x!}\leq
e^{\frac{n(n-1)}{2}}$

Now taking logarithm on both sides of inequality.

$\implies \displaystyle\sum_{x=1}^{n} \ln\bigg(\frac{x^x}{x!}\bigg)\leq
{\frac{n(n-1)}{2}}$

$\implies \displaystyle\sum_{x=1}^{n} \ln\bigg(\frac{x^x}{x!}\bigg)\leq
{\frac{n(n+1)}{2}}-n$

$\implies \displaystyle\sum_{x=1}^{n} \ln\bigg(\frac{x^x}{x!}\bigg)\leq
\bigg(\sum_{x=1}^{n} x \bigg)-n$

$\implies \displaystyle \ \sum_{x=1}^{n} x -\displaystyle\sum_{x=1}^{n} \ln\bigg(\frac{x^x}{x!}\bigg)\geq n$

$\implies \displaystyle \ \sum_{x=1}^{n} x + \ln\bigg(\frac{x!}{x^x}\bigg)\geq n$

$$\implies 1+1.30+1.49+1.63+1.74+1.82+1.90+1.97+2.02+...\geq n$$

$\implies 1+(1+0.30)+(1+0.49)+(1+0.63)+(1+0.74)+...\geq  {\smash[b]{1+\! \underbrace{1+1+\cdots1\,}_\text{$n$ times}}}$

This proves our first inequality.

Method 2:-

Using $\frac{n^n}{n!}\leq e^{n-1}$ for $n\in N$.

$\frac{1^1}{1!}\leq e^{0}$

$\frac{2^2}{2!}\leq e^{1}$

$\frac{3^3}{3!}\leq e^{2}$

$\vdots\\$

$\frac{n^n}{n!}\leq e^{n-1}$

Now multiply all above inequalities we get,

$\implies \displaystyle\prod_{x=1}^{n} \frac{x^x}{x!}\leq
e^{\frac{n(n-1)}{2}}$

Using the same approach for $2^{nd}$ inequality we get,

$$\sum_{x=1}^{n} \ln\bigg(\frac{{f_{x}}^{f_{x}}}{f_{x}!}\bigg)\leq f_{n+2}-n-1$$

Now I got stuck on this step.

How can we prove the $2^{nd}$ inequality. Moreover can we show without numerically calculating(as I do above) that $\displaystyle \ \sum_{x=1}^{n} x + \ln\bigg(\frac{x!}{x^x}\bigg)\geq n$

答:
Are known sums of the Fibonacci numbers
$$F_{n+2}-1 =\sum\limits_{k=1}^n F_n\tag1$$
and the Stirling's inequality
$$m!\ge\sqrt{2\pi m}\, m^m\,e^{-m},\quad m\in \mathbb N.\tag2$$
Then, taking in account $(2),$
\begin{align}
&1!\,e^{1-1} = 1 = 1^1,\\[4pt]
&2!\,e^{2-1} \ge \sqrt{4\pi}\, 2^2 e^{-2+2-1} \ge 2^2,\dots,\\[4pt]
&m!\,e^{m-1} \ge \sqrt{2\pi m}\, m^m e^{-m+m-1} \ge m^m,\dots,\\[4pt]
\end{align}
$$m!\,e^{m-1} \ge m^m,\quad m\in\mathbb N.\tag3$$

Finally,
$$e^{F_{n+2}-1-n}\prod\limits_{k=1}^n F_k! = \prod\limits_{k=1}^n F_k!e^{F_k-1}
\ge \prod\limits_{k=1}^n F_k^{F_k}.$$

The first inequality allows the same approach.


问:
How many distinct orbits there are of $S_4$ when $S_4$ acts on itself by conjugation?
The question is, how many distinct orbits there are of $S_4$ when $S_4$ acts on itself by conjugation?

I was thinking the number would be the same as the cycle types of $S_4$. However, the orbits must be distinct, so here is where I get slightly confused.

答:
As discussed in the comments, two permutations are conjugates iff they have the same cycle type. Thus, it comes down to the number of cycle types in $S_4$. This is the same as the number of integer partitions of $4$. We have:

$1 + 1 + 1 + 1$

$2 + 1 + 1$

$2 + 2$

$3 + 1$

$4$

How do we know we've got all? Well, count the number of permutations in each of the cycle type listed above. It is not too tough to see that they are (in the same order) given as:

$1$

$6$

$3$

$8$

$6$

Since $1 + 6 + 3 + 8 + 6 = 24 = 4! = |S_4|$, we know that we've covered all.


问:
How do I evaluate the limit $\;\lim\limits_{x\to 0}\dfrac{\sin3x}{x\cos2x}$?
$$\lim_{x\to 0}\dfrac{\sin3x}{x\cos2x}$$

I'm having trouble doing this problem: the farthest I've gotten is just using a limit law for division and then moving the constant $x$ in the denominator in front. I also thought about leaving $\lim_{x\to 0}\sin3x/1$ and then somehow introducing $3x$ so it turns into a special limit.

Otherwise, I'm just stuck. Hmm, the original problem does resemble tangent, not sure if that has to do with anything?

答:
We want to evaluate
$$\lim\limits_{x\to 0}\dfrac{\sin3x}{x\cos2x}.$$
Set $p=3x$, so $x=p/3$. Note that as $x\to0, p\to0.$
$$3\lim_{p\to 0}\dfrac{\sin p}{p\cos\tfrac{2p}{3}}=3\lim_{p\to 0}\dfrac{\sin p}{p}\lim_{p\to 0}\frac{1}{\cos\tfrac{2p}{3}}=3.$$
Here is a nice geometric proof for
$$\lim_{p\to 0}\dfrac{\sin p}{p}=1.$$


答:
hint

To be sure the sequence is well-defined, Prove by induction that

$$(\forall n\ge 1)\;\;  x_n\ge 0$$

then use

$$|x_{n+1}-2|=\frac{|x_n-2|}{\sqrt{2+x_n}+2}$$
$$\le \frac 12|x_n-2|$$
$$\le \frac{1}{2^n}|x_1-2|$$

Title: Find the limit of the sequence $(x_n)$, where $x_1=1$ and $x_{n+1}=(2+x_n)^{1/2}$

Body: I have been given this hint in the problem: In order to determine the limit, it may be helpful to use the fact that if $y_n \rightarrow y$ and $y_n > 0$ for every $n$,then $\sqrt{y_n} \rightarrow \sqrt{y}$. You can use this without proving it. You can also use without proof the fact that if $0\le a \le b,$ then $\sqrt{a} \le \sqrt{b}$.


问:
Image of closed unit ball under a compact operator
Let $X,Y$ be Banach spaces and $A\in\mathcal L(X,Y)$ . The task is to prove the following:

$A$ is compact if and only if the image of the closed unit ball in $X$ is compact in $Y$.

I have proven this when $X$ is a reflexive space.

Proof. Let $X$ be a reflexive space, $\bar B$ the closed unit ball in $X$, and $A$ a compact operator. Let further $y_n=Ax_n$ be a sequence in $A(\bar B)$.

In reflexive spaces $\bar B$ is weakly compact, so there exists a subsequence $x_{n_j} \to x$ weakly.

Because $A$ is compact, $Ax_{n_j}\to Ax$ strongly.

On the other side, $A(\bar B)$ is relatively compact, so there exists $z_k=Ax_{n_{j_k}}$ that converges strongly to $y\in Y$.

But $z_k\to Ax$ strongly. So by unicity of the limit $y=Ax$ and the image is compact.

It's easily proved that if the image is compact, the operator is also compact.

But I don't know what to do in case of nonreflexive spaces. Is there any counterexample or proof in such case?

答:
It is false in general that the image of the closed unit ball under a compact operator is closed (and hence compact). Here is an easy example:

Consider $X = C[0,1]$ with the uniform norm, and the compact operator $A \in B(X)$ defined by the formula:

$\displaystyle\qquad Af(x) = \int_0^x f(t)\,dt$.

Compactness of $A$ is easily proven using Arzelà–Ascoli. Our operator $A$ produces an anti-derivative of any input given to it, and the image of the closed unit ball of $X$ under $A$ is the set

$\displaystyle\qquad \{f \in C^1[0,1] \mathrel: f(0)=0,\ \lVert f'\rVert \leq 1\}$

which certainly is not a closed subset of $X$.


问:
Find a holomorphic injection $f$
Find a holomorphic injection $f$ such that $$f\Big(\{ z\in \mathbb C: |\text{Re}z|<1\}\Big)=\{z\in \mathbb C: \text{Re}z>0, \text{Im}z>0\}$$
and$$f(0)=1+i.$$ My try: $$z\mapsto i\cdot z\mapsto z+i$$

However, I don't know how to proceed further.

答:
Via $z\mapsto iz +i$ you map the strip $S_1:=\{|\Re z|<1\}$ to the strip
$S_2:=\{0<\Im z<2\}$. By multiplication with $\frac \pi 4$ as image you get
$S_3:=\{0<\Im z<\frac \pi 2\}$. Now apply the exponential map to obtain
$Q:=\{\Re z>0, \Im z >0\}$. The composition of these maps is $g(z)= \exp\left ( \frac{\pi} 4i (z+1)\right)$ which sends $0$ to $\frac{\sqrt 2} 2(1+i)$.

Finally multiplying by $\sqrt 2$ you keep $Q$ as image  of the map but you map $0$ to $1+i$:
the final map is $$f(z)=\sqrt{2} \exp\left ( \frac{\pi} 4i (z+1)\right)\,.$$

Note that this map is actually injective, since the "height" of $S_3$ is $\frac \pi 2 <2\pi$, hence the exponential map restricted to it is injective.

Title: Find a holomorphic injection $f$


问:
Derivations of a ring
I am trying to show that Der$(\mathbb{Q}) = \{ 0 \}$, where Der($R)$, $R$ a commutative ring, is the set of endomorphisms $\varphi: R \rightarrow R$ such that $\varphi(x \cdot y) = y \cdot \varphi(x) + x \cdot \varphi(y)$ for all $x,y \in R$.

Now clearly $\{ 0 \} \subseteq$ Der$(\mathbb{Q})$. My problem is showing the containment in the other direction. If I let $\varphi$ be any endomorphism of $\mathbb{Q}$ such that $\varphi \in$  Der$(\mathbb{Q}) $ and I took elements $x$ and $y$ in $\mathbb{Q}$, why is it that $\varphi(x \cdot y) = y \cdot \varphi(x) + x \cdot \varphi(y) = 0$. Here $0$ refers to the $0$ mapping.

答:
Derivations are usually defined with a notion of "scalars," e.g. $k$-derivations of a $k$-algebra $R$.

At the very least, the scalars of any derivation will include $\mathbb{Z}$ (or the image of $\mathbb{Z}$ in $R$, the "prime subring" integer multiples of 1), i.e. $\phi(\{n\cdot1 : n\in\mathbb{Z}\})=0$.  In the case at hand, this extends to the prime subfield $\mathbb{Q}$
$$
0=\phi(p)=\phi(pq/q)=q\phi(p/q)+\phi(q)p/q \Longrightarrow \phi(p/q)=0.
$$

Title: Derivations of a ring


问:
Probability Mass Function Understanding What To Look For!
Problem:
For which value of $\alpha$ is the function $p_i=\frac{(\alpha+1)(i-\alpha)+2}{120}$ of $\{1,2,\cdots,10\}$ a p.m.f?

My understanding is that $p_i=1$ and that $i=\{1,2,\cdots,10\}$, but trying to solve for $\alpha$ seems overly complicated.

I tried using a formula simplifier on another website and got the following to solve for $\alpha$:
$$\alpha=-\frac{1-i+ \sqrt{-472+2i}}{2},-\frac{1-i- \sqrt{-472+2i}}{2}$$

But then of course it broke my calculator because the value inside the $sqrt{}$ is negative. So then I tried understanding how to fix the negative $sqrt{}$ and it took me to the next step of trying to understand if this is a radical expression and only when the $i$ is replaced with a number can it be distilled down. However, even then it creates an $i$ out front of the $sqrt$ and I don't know if that $i$ is the same as the $i$ in my equation. Example: $2i\sqrt{118}$.

So at this point, I am sufficiently lost, but my thoughts are that what I'm solving for is to see which of the numbers $i=\{1,2,\cdots,10\}$ is the correct number. What constitutes the right answer? Is it if $\alpha$ is a whole number?

Can someone help me understand the whole question/problem?  I fear I am in the weeds on this one.

Thank you

答:
Hints:

$i$ here is a index, not $\sqrt{-1}$

You want $\sum\limits_{i=1}^{10}\frac{(\alpha+1)(i-\alpha)+2}{120}=1$ as this is a sum of all the probabilities

When finding the sum, $i$ should not appear in the result as it is a bound variable

$$\sum\limits_{i=1}^{10}\frac{(\alpha+1)(i-\alpha)+2}{120} \\ = \frac{2-\alpha^2-\alpha}{120}\sum\limits_{i=1}^{10}1 + \frac{\alpha+1}{120} \sum\limits_{i=1}^{10}i \\ = \frac{2-\alpha^2-\alpha}{120}\times 10 + \frac{\alpha+1}{120} \times 55 $$

Simplifying this and setting this equal to $1$ will give you a quadratic equation with two real roots for $\alpha$.

You should check each potential solution to ensure they give probabilities for $p_i$: in particular the probabilities should be non-negative.  One works and the other does not.


答: Note that $\lim_{x\to0}\frac1{\cos(2x)}=1$. On the other hand$$\lim_{x\to0}\frac{\sin(3x)}x=3\lim_{x\to0}\frac{\sin(3x)}{3x}=3.$$Therefore,$$\lim_{x\to0}\frac{\sin(3x)}{x\cos(2x)}=3\cdot 1=3.$$

Title: How do I evaluate the limit $\;\lim\limits_{x\to 0}\dfrac{\sin3x}{x\cos2x}$?

Body:
$$\lim_{x\to 0}\dfrac{\sin3x}{x\cos2x}$$

I'm having trouble doing this problem: the farthest I've gotten is just using a limit law for division and then moving the constant $x$ in the denominator in front. I also thought about leaving $\lim_{x\to 0}\sin3x/1$ and then somehow introducing $3x$ so it turns into a special limit.

Otherwise, I'm just stuck. Hmm, the original problem does resemble tangent, not sure if that has to do with anything?


问:
Why do we use Complex numbers and not other systems?
I just started studying Complex analysis and have in fact just switched field to mathematics recently and so please forgive me if this is question seems trivial for a mathematics student to ask.

Question: Why do we use Complex numbers instead of another algebraic field or number system? i.e. It is "natural" to have the hierarchy $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$ and not others.

Attempt: I have often been told that $\mathbb{C}$ is algebraically closed (every polynomial with complex coefficients have a root in $\mathbb{C}$) and that it contains $\mathbb{R}$ as a isomorphic subfield.

But should there be other fields or number systems with the same properties as well? Did I miss anything "extra" about $\mathbb{C}$ that makes it special?

答:
There are split-complex numbers and dual numbers, and they are used in certain fields. But the both are "worse" than $\mathbb{C}$: they are not algebraically closed and not fields (have zero divisors).

Still, they have many properties of numbers, allowing full-fledged calculus on them.


问:
Show that the Scattered Line (also called the Michael Line) is $T_4$
I just finished reading through the section about normal spaces (resp. $T_4$-spaces) from Willard General Topology text, and came across the following problem:

Show that the Scattered Line (also called the Michael Line) is $T_4$.

I'm not sure how to prove this, let alone get started on the problem. Any help would be much appreciated.

Note: The Scattered Line (also called the Michael Line) is given as follows: Let $\mathbb{R}$ be the real number line. Let $\mathbb{P}$ be the set of all irrational numbers. Let $\mathbb{Q}=\mathbb{R}-\mathbb{P}$, the set of all rational numbers. Let $\tau$ be the usual topology of the real line $\mathbb{R}$. The following is a base that defines a topology on $\mathbb{R}$.

$$\mathcal{B}=\tau \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}$$

The real line with the topology generated by $\mathcal{B}$ is called the Scattered Line (also called the Michael Line).

答:
This can be viewed (as Brian M. Scott also does) from a more general point of view:

A $T_1$ space is called monotonically normal iff there is a so-called $\mu$-operator that assigns an open subset $\mu(x,O)$ for every pair $x,O$ where $x \in O$ and $O$ is open in $X$ and so that

$x \in \mu(x,O) \subseteq O$ and

Whenever we have $\mu(x,U) \cap \mu(y,V) \neq \emptyset$ we can conclude that $x \in V$ or $y \in U$.

This looks rather innocent but is in fact quite powerful: it's a hereditary property: if $X$ is monotonically normal and $A \subseteq X$ then so is $A$ in its subspace topology: if $O \cap A$ is an open subset of $A$ (so $O$ is open in $X$) we define $\mu_A(x, O \cap A) = \mu(x,O) \cap A$, when $\mu$ is the operator on $X$.

Fact

If $X$ is monotonically normal, then $X$ is $T_5$ (aka hereditarily normal /completely normal).

Proof: as being monotonically normal is hereditary, it suffices to show $X$ is normal (the $T_1$ is part of the definition). So let $A$ and $B$ be disjoint closed sets of $X$.

Define $$U = \bigcup \{\mu(a, X\setminus B)\mid a \in A\}, V = \bigcup \{\mu(b, X\setminus A)\mid b \in B\}$$

and note that $A \subseteq U$, $B \subseteq V$ and these sets are open. They are also disjoint, for if there not, there must be some $a_0 \in A$ and some $b_0 \in B$ such that $\mu(a_0, X\setminus B) \cap \mu(b_0, X\setminus A) \neq \emptyset$ and by property 2 we'd then have $a_0 \in x \setminus A$ or $b_0 \in X\setminus B$, both of which are absurd. So that completes the proof.

Useful fact:

It suffices to define $\mu(x,O)$ only for $O$ from some given fixed base $\mathcal{B}$ for $X$ and all $x \in O$, having the right properties. Then defining $\mu(x, \bigcup_{i \in I} O_i) = \bigcup \{\mu(x,O_i)\mid x \in O_i, i \in I\}$, where all $O_i \in \mathcal{B}$, defines a $\mu$-operator for all open subsets.

The proof is an easy exercise in the definitions.

Some examples: Any metric space is monotonically normal, we can define $\mu$ on a base by $\mu(x, B(x,r)) = B(x, \frac{r}{2})$, and the triangle inequality takes care of defining property 2.

The Sorgenfrey line is monotonically normal: define $\mu(x, [a,b)) = [x,b)$ on base elements, and checking property 2 is a simple matter of case checking.

So we then have "for free" that metric spaces are $T_5$ and so is the Sorgenfrey line. In fact all ordered spaces are monotonically normal too, this has a longer proof that I won't give here now.

But back to the Michael Line, or Scattered line: this is a special case of a more general construction: if $(X, \mathcal{T}_X)$ is a space and $A \subseteq X$ a subspace of $X$, then $X_A$ is the space with point set $X$ and topology generated by $\mathcal{T}_X \cup \mathscr{P}(X\setminus A)$; in effect: $A$ keeps its original subspace topology and all points outside of $A$ become isolated points.

The Michael line is just $\Bbb R_{\Bbb Q}$, where $\Bbb R$ has its usual topology.

Proposition

If $X$ is monotonically normal and $A \subseteq X$ then $X_A$ is also monotonically normal

Proof: $T_1$ is trivial, as we have a finer topology.
Let $\mu$ be the operator on $X$. Then a $\mu$ operator on the generating set of $X_A$ is defined by $\mu'(x,O)=\mu(x,O)$ if $O \in \mathcal{T}_X$ and $\mu'(x,B)=\{x\}$ if $B \subseteq X\setminus A$. This works, as is quite easy to check.

Corollary

The Michael line is monotonically normal and hence hereditarily normal.

Many spaces with good high separation properties have it for the reason that they're monotonically normal. No longer a very fashionable notion/property, but I quite like it.

Hope this helps someone.


问:
What is sum of reciprocal?
Is ${1\over 1}+{1\over 2}+{1\over 3}+\cdots +{1\over n}$ computable? I couldn't find any formulas to find the result. The explanation would be very helpful. Thanks before.</p>

答: I think what you are looking for is the harmonic number.</p>

Title: What is sum of reciprocal?


答:
We can use also use power series for this limit. Using the Maclaurin series expansion for $\sin$ and $\cos$ we see that
$$\lim_{x\to 0}\frac{\sin3x}{x\cos2x}=\lim_{x\to 0}\frac{3x-\frac{(3x)^3}{3!}+\cdots}{x(1-\frac{(2x)^2}{2!}+\cdots)}=\lim_{x\to 0}\frac{3-\frac{3^3x^2}{3!}+\cdots}{1-\frac{(2x)^2}{2!}+\cdots}=3$$

I hope that was helpful. If you have any questions please don't hesitate to ask :)


问:
How does the definition of outer measure guarantee the existence of sequence of intervals $I_n$ such that $|A|>\sum_n|I_n|-\epsilon.$
I am trying to understand the solution by @Matematleta to this question. This is the response:

Hint: by definition of outer measure, there is a sequence of intervals $I_n$ such that $A\subseteq \bigcup_n I_n$ and $|A|>\sum_n|I_n|-\epsilon.$ Consider $\{I_n\cap (-t,t)\}_n$ and $\{\mathbb R\setminus I_n\cap (-t,t)\}_n$.

My question is this: How exactly does the definition of outer measure guarantee the existence of a sequence of intervals $I_n$ such that $A\subseteq \bigcup_n I_n$ and $|A|>\sum_n|I_n|-\epsilon$? Thanks.

答: Hint: Think of what infimum means in terms of epsilon, try to draw a picture(also maybe move the epsilon to the other side, it might help you see thinks clearly)</p>


问:
Integral challenge
The challenge is: if $f:\mathbb{R}\to\mathbb{R}$ is a continuous function, then $$\lim_{t\to0+}\frac{1}{\pi}\int_{-1}^1\frac{tf(x)}{t^2+(x-a)^2}\,\mathrm{d}x=f(a)$$ for every $a\in[-1,1]$

My attemps:

Under certainly conditions, i put the limit symbol under the integral sign, but the limit would be $0$. So it is a bad idea.

Therefore, i think use Weierstrass theorem for $f$ in $[-1,1]$, but i can't solve the problem in the case of $f(x)=x^n$.

What do you recommendme?

答:
I consider the case $a=0$.

Note that $$\lim_{t \to 0 } \int_{-1}^1 \frac{tf(0)}{t^2+x^2} dx = f(0) \lim_{t \to 0} \left( \arctan(1/t) - \arctan(-1/t) \right) = \pi f(0).$$

This essentially reduces the problem to the case where $f(0)=0$.

If $f(0)=0$, by continuity for every $\epsilon > 0$ there exists a $\delta > 0$ such that $|f(x)| \le \epsilon$ for $|x| \le \delta$. It then follows that $$\lim_{t \to 0} \int_{|x| > \delta} \frac{tf(x)}{t^2+x^2} dx = \int_{|x| > \delta} \lim_{t \to0}  \frac{tf(x)}{t^2+x^2} dx = \int_{|x|>\delta} 0 \, dx = 0.$$
On the other hand, we have
$$
\left| \lim_{t \to 0} \int_{-\delta}^\delta \frac{tf(x)}{t^2+x^2} dx \right|  \le \lim_{t \to 0} \int_{-\delta}^\delta \left| \frac{tf(x)}{t^2+x^2} \right| dx \le 2\epsilon \lim_{t \to 0} \int_{0}^\delta  \frac{t}{t^2+x^2}  dx = \lim_{t \to 0} 2\epsilon \arctan(\delta/t)  = 2\pi \epsilon.
$$
It follows that
$$\left| \lim_{t \to 0} \int_{-1}^1 \frac{tf(x)}{t^2+x^2} dx \right| \le 2\pi \epsilon$$
for all $\epsilon > 0$, so $\lim_{t \to 0} \int_{-1}^1 \frac{tf(x)}{t^2+x^2} dx = 0$.

Title: Integral challenge


答:
Yes. The result is:
$$S=\Psi(n+1)+\gamma$$
where:$$S=\sum_{k=1}^n \frac{1}{k}$$
For the explanation see: http://en.wikipedia.org/wiki/Harmonic_number</p>

Title: What is sum of reciprocal?

Body: Is ${1\over 1}+{1\over 2}+{1\over 3}+\cdots +{1\over n}$ computable? I couldn't find any formulas to find the result. The explanation would be very helpful. Thanks before.</p>


问:
100 Gambles of Roulette
I am stuck on the following problem:
You go to Las Vegas with $\$1000$ and play roulette $100$ times by betting $\$10$ on red each time.
Compute the probability of losing more than $\$100$. Hint: Each bet you have chance 18/38
of winning $\$10$ and chance 20/38 of losing $\$10$.

So essentially this is recurring 100 times.  Now each time you do it you have an 18/38 chance of winning and 20/38 chance of losing.  You have a greater chance of losing than winning, so based on speculation you need to lose 56/100 in order to lose more than $\$100$, to counterbalance your winnings . How would I set up an equation to solve for this?

Thank you!

答:
This is a classical Bernoulli experiment:
You already mentioned that we need to loose at least $56$ times to loose more than $100\$$. Let $X$ describe the number of times you loose. We want to compute $P(X > 56) = 1 - P(X \leq 56)$.  We also have:
$$P(X \leq 56) = \sum\limits_{k = 0}^{\ 56}{100 \choose k}\left(\frac{18}{38}\right)^{100-k} \left(\frac{20}{38}\right)^{k}\space\space\approx 0.7805$$  which is the probability that we loose at most 56 times. Thus we finally obtain $1-0.7805 = 0.2195 = 21.95\%$ as our end result.

Title: 100 Gambles of Roulette


问:
What are the uses of split-complex numbers?
The set of Complex numbers can be used in lots of domains like geometry, vectorial calculations, solving equation with no real solution etc. But what are the uses of split-complex number that can't be done with complex numbers? I think you could do the same works in geometry or vectorial calculation in a "split-complex" plane but what advantages gives us to know that j is a solution of the equation $x^2=1$?

What I've thought so far is that using complex numbers and split-complex numbers together, we can have numbers of the form $a+bi+cj$ so everything that can be done in the complex-plan could be extended to 3 dimensional space by adding a split-complex part.

答:
What I've thought so far is that using complex numbers and split-complex numbers together

You will get exactly tessarines, a commutative associative algebra. But it is 4-dimensional: $ij$ is a separate unit, similar to complex unit. $i^2=-1$, $j^2=1$, $(ij)^2=-1$.

So, your number will look $a+bi+cj+dij$ or $(a+bi)+j(c+di)$ or $(a+bj)+i(c+dj)$.


答:
The class equation will give this information: the number of terms is the number of classes.

The class equation for $S_4$ is $1+6+3+8+6$.

Thus we have $5$ orbits.

Title: How many distinct orbits there are of $S_4$ when $S_4$ acts on itself by conjugation?

Body:
The question is, how many distinct orbits there are of $S_4$ when $S_4$ acts on itself by conjugation?

I was thinking the number would be the same as the cycle types of $S_4$. However, the orbits must be distinct, so here is where I get slightly confused.


问:
A book of set theory, exercise.
Let $\{A_i\}_{i∈I}$ and $\{B_j \}_{j∈J}$ be indexed families of classes. Suppose that $∀i ∈ I, ∃j ∈ J$ such that $B_j ⊆ A_i$
. Prove
that $$\bigcap_{j\in J}B_j\subseteq \bigcap_{i\in I}A_i$$

For this demo, I'm going to take a $ x \in \bigcap_ {j \in J} B_j $. So for all $ j \in J $ we have $ x \in B_j $. But how can I use my hypothesis to guarantee that my $ B_j \subseteq A_i $ for all $ i \in I $, and thus conclude.

答:
Let $x \in \bigcap_{j\in J} B_{j}$, then $x \in B_{j} \quad \forall \, j\in J$

And we have that for each $i\in I$ there exists some $j_{i}\in J $ such that $B_{j_{i}} \subset A_{i}$

Then for each $i \in I $ we have that $x \in B_{j_{i}} \subset A_{i} \implies x\in A_{i} \quad \forall \, i \in I$

Thus $x \in \bigcap_{i \in I} A_{i}$

Title: A book of set theory, exercise.


问:
Radius of circle tangent to six tangent parabolas

Six congruent copies of the parabola $y = x^2$ are arranged in the plane so that each vertex is tangent to a circle, and each parabola is tangent to its two neighbors. Find the radius of the circle.

I'm honestly not sure how to start this problem. The vertices of the parabolas are spaced $60^\circ$ apart on the circumference of the circle, and the points where two parabolas are tangent are also spaced $60^\circ$ apart, $30^\circ$ from a vertex. But I'm not sure how this is useful.

I also know that the focus and directrix of each parabola are the same distance from the vertex, and the directrixes (is that the plural?) intersect $60^\circ$ apart, $30^\circ$ from a vertex. But again, I don't know how useful that is.

Normally on problems like these I'd try to draw a radius and form some right triangles but I really don't see how I can do that here.

TIA!

答:
Geometrically, consider a parabola with focus $F$, vertex $V$, and directrix through $D$ (so that $\overline{TF}\cong\overline{TT'}$, where $T'$ is the foot of the perpendicular from $T$ to the directrix). We require $T$ make $\angle TKF=30^\circ$, where $K$ is the center of the ostensible circle.

The Reflection Property tells us that $\overline{TK}$ bisects $\angle FTT'$, and we deduce that $\triangle FTT'$ is equilateral, and thus also is $\triangle FKT'$. Consequently, the directrix bisects $\overline{FK}$. Since $V$ bisects $\overline{FD}$, we conclude that $|KV| = 3|FV|$.

For the specific case of the parabola defined by $y=1x^2=\dfrac{1}{4\,|FV|}x^2$, we have $|FV|=1/4$, so that $|KV|=3/4$.

Inspired by @JeanMarie's generalization, here's a revised argument:

For arbitrary $\theta:=\angle FKT$, and vertex-to-focus distance $a$, a little angle-chasing and the Reflection Property ultimately yield $\triangle FVM\sim\triangle MVK$, so that
$$|KV|=a\cot^2\theta$$
In the context of fitting $n$ tangent $y=x^2$ parabolas together about $K$, we have $\theta=\frac{\pi}{n}$ and $a=\frac14$, so that this agrees with @JeanMarie's result. $\square$


问:
Show that if $X$ has a countable dense subset, every collection of disjoint open sets in $X$ is countable.
Show that if $X$ has a countable dense subset, every collection of disjoint open sets in $X$ is countable.

Attempt:Let $U=\{U_i\}$ be a collection of disjoint open sets in $X$ that is countable. Let $A$ be a countable dense subset in $X$. Let $U_i \in U$. Then $U_i \cap A \neq \varnothing$, since $\overline A=X$. Since $U_i \cap U_j =\varnothing$ for each $i \neq j$, $U_i,U_j$ contain different elements of $A$.So there is a one to one correspondence between elements in $U$ and subsets of $A$. So $U$ is countable, since the result holds for any collection of disjoint open sets.

Comment:I am struggling to think of why each element of $U$ being associated with a corresponding subset of $A$ means $U$ is countable. Would it be correct that the intersection of each $U_i$ and $A$ partition the set $A$. Are there any issues with my attempt at a proof for this?Also I now see that it is wrong to say the collection of intersections of elements in $U$ with $A$ must contain the whole set $A$, since the result must hold for any collection of disjoint open sets.

答:
Not a 1-1 correspondence really, but an injection: let $D$ be dense, and $\mathcal{U}$ be a family of pairwise disjoint non-empty open sets.

Using AC we pick for each $U \in \mathcal{U}$ some $f(U) \in D\cap U$ which is possible as $D$ is dense.

Then if $U \neq U' \in \mathcal{U}$ we know that $U \cap U' = \emptyset$ so $f(U) \neq f(U')$ and so $f$ is 1-1 from $\mathcal{U}$ to $D$. So $|\mathcal{U}| \le |D|$

If $X$ is separable, we can pick a fixed countable $D$ and then for any such pdof (pairwise disjoint (non-empty) open family) $\mathcal{U}$ we get by the previous that $|\mathcal{U}| \le \aleph_0$ and we're done. QED: A separable space is ccc.


问:
In search of a literature or theory that deals with $\int_{[a,b]}f(x)dx \geq \int_{g([a,b])}f(x)dx$
I stumble upon the following problem. Given an increasing function $g$ is there a function $f$ (not identically null) s.t.

$$\int_{[a,b]}f(x)dx \geq \int_{g([a,b])}f(x)dx$$

And, if there is, what can we say about $f$. The problem could be reinstated in terms of pushforward, i.e. find a signed measure $\mu$ s.t. $\mu([a,b]) \geq g_{*}(\mu)([a,b])$. There are also other analogous ways to approach it (in terms of differential equation the problem is a composition of function). However, none of this seems to shed light upon the problem. I tried to find a theory of some sort that could help me with it, but so far have not been successful.

$\textbf{Edit:}$ Considering avs answer the following idea emerged. Let $g$ be a strictly increasing continuouosly differentiable function. Let $A = \{x : g'(x) < 1\}$ and $B = \{x : g'(x) > 1\}$. Then $g$ is a contraction in every open interval in $A$ (actually, a subcontraction) and is an expansion (in the sense that its inverse is a contraction) in every open interval in $B$. Thus let $f$ be s.t. $f \leq 0$ in A, $f \geq 0$ in $B$ and $f = 0$ in $(A \cup B)^c$. Then $f$ must be a solution.

答:
A simple experiment gives a less than encouraging result: let
$$
g(x) = 2x, \quad a = 0, \quad b > 0.
$$
It follows that
$$
g([a, b]) = g([0, b]) = [0, 2b].
$$

Now imagine a measure $\mu$ on the (Borel $\sigma$-algebra of) $\mathbb{R}$ satisfies the required inequality.  Then we would need to have
$$
\mu([0, b]) \geq \mu ( g ([0, b]) ) = \mu([0, 2b]) = \mu([0, b]) + \mu(]b, 2b]),
$$
which would imply that
$$
\mu\left(]b, 2b]\right) = 0 \mbox{ for all positive }b.
$$
Thus, the existence of a measure $\mu$ with the desired property (a, sort of, "monotonicity w.r.t. $g$") depends a lot on what $g$ is.

If $g$ is a simple contraction, e.g., $g(x) = x / 2$, then it is not hard to show that $\mu$ can be taken to be the Lebesgue measure on $\mathbb{R}$.

It's also not that hard to construct a $\mu$ that will fail the desired property for the latter $g$; e.g., let $\mu$ have the density function
$$
f(x) = {C \over 1 + x^2}, \quad \mbox{where $C$ is large enough}.
$$
Basically, $f(x)$ assigns large measures to subsets near $0$, and small ones far away from $0$, but $g$ takes subsets that are far from $0$ and brings them closer to $0$, thus increasing their measure.

If you are looking for the field of mathematics that studies this, I'd try to look at measures as linear functionals.  The general concept of monotonicity of functionals (which are generally not in a Euclidean field, hence devoid of a "natural" linear ordering) is usually developed by using the concept of a cone.


答:
Since you want to prove that every family of pairwise disjoint non-empty open subsets of $X$ is countable, you certainly can’t start by assuming that you have a countable family of open sets. Even before you know just where you’re going, the natural way to start is something like this:

Let $D$ be a countable dense subset of $X$, and let $\mathscr{U}$ be a family of pairwise disjoint non-empty open subsets of $X$; we want to prove that $\mathscr{U}$ is countable.

Your next step, however, was in the right direction, though it can be phrased rather better:

Since $D$ is dense in $X$, and each $U\in\mathscr{U}$ is non-empty, for each $U\in\mathscr{U}$ there is an $x_U\in U\cap D$.

Now you have in effect a function from $\mathscr{U}$ to $D$.

Let $\varphi:\mathscr{U}\to D:U\mapsto x_U$. If $U,V\in\mathscr{U}$, and $U\ne V$, then $U\cap V=\varnothing$, so $x_U\ne x_V$. Thus, $\varphi$ is injective. It follows that $|\mathscr{U}|\le|D|\le\aleph_0$, i.e., that $\mathscr{U}$ is countable.

To answer your last question (while sticking with my notation), $\{U\cap D:U\in\mathscr{U}\}$ is not a partition of $D$: it is a pairwise disjoint collection of non-empty subsets of $D$, but in general its union is not $D$.

Body:
Show that if $X$ has a countable dense subset, every collection of disjoint open sets in $X$ is countable.

Attempt:Let $U=\{U_i\}$ be a collection of disjoint open sets in $X$ that is countable. Let $A$ be a countable dense subset in $X$. Let $U_i \in U$. Then $U_i \cap A \neq \varnothing$, since $\overline A=X$. Since $U_i \cap U_j =\varnothing$ for each $i \neq j$, $U_i,U_j$ contain different elements of $A$.So there is a one to one correspondence between elements in $U$ and subsets of $A$. So $U$ is countable, since the result holds for any collection of disjoint open sets.

Comment:I am struggling to think of why each element of $U$ being associated with a corresponding subset of $A$ means $U$ is countable. Would it be correct that the intersection of each $U_i$ and $A$ partition the set $A$. Are there any issues with my attempt at a proof for this?Also I now see that it is wrong to say the collection of intersections of elements in $U$ with $A$ must contain the whole set $A$, since the result must hold for any collection of disjoint open sets.


问:
Closed form of minimum value of A
I want the minimum value of A. (closed form)

$$A = \int_{0}^{p} \sin(x) dx + \int_{p}^{\frac{\pi}{2}} (\sin(x)-\sin(p))dx$$
$$0 \leq p \leq \frac{\pi}{2}$$

So...
$$A = p \cdot \sin(p) - \frac{\pi}{2}\sin(p) + 1$$

$$\frac{d}{dp} A = \sin(p) + p \cdot \cos(p) - \frac{\pi}{2}\cos(p) = 0$$

$$\tan(p) + p - \frac{\pi}{2} = 0$$

I stopped here.
I know that:

$$A = 1 - \tan(p) \cdot \sin(p)$$

$$\sin(p) = \cos(\tan(p))$$
$$\cos(p) = \sin(\tan(p))$$
$$\tan(p) = \cot(\tan(p))$$
$$\csc(p) = \sec(\tan(p))$$
$$\sec(p) = \csc(\tan(p))$$
$$\cot(p) = \tan(\tan(p))$$

$$p \approx 0.71046$$
$$A \approx 0.4389$$

EDIT:

$$\tan(p) + p - \frac{\pi}{2} = 0$$

$$\tan(p) = \frac{\pi}{2} - p$$

$$-cot\left(p-\frac{\pi}{2}\right) = \frac{\pi}{2} - p$$

$$cot\left(p-\frac{\pi}{2}\right) = p - \frac{\pi}{2}$$

$$B:=p-\frac{\pi}{2}$$

$$cot(B) = B$$

"This kind of equations which mix polynomial and trigonometric terms do not, in general, show solutions which have a closed form expression..." as said by Claude Leibovici here.

答:
$$A = \int\limits_{0}^{p} \sin(x)\,\mathrm{d}x + \int\limits_{p}^{\pi/2} (\sin(x)-\sin(p))\,\mathrm{d}x=\int\limits_{0}^{\pi/2} \sin(x)\,\mathrm{d}x-\sin p\int\limits_{p}^{\pi/2} \,\mathrm{d}x.$$
Here the first integral is equal to $1$. The second term simpliefies to $\sin p(\pi/2-p)$.
So
$$A=1-\sin p\left(\frac{\pi}{2}-p\right).$$
Now take the derivative of this function. We will get
$$A'=-\frac{\pi}{2}\cos p+\sin p+p\cos p.$$
Set $A'=0$ and solve for $p$ and pick out the $p$-value that was in your original domain. I believe the solution has not a trivial closed form.

Title: Closed form of minimum value of A


问:
Can linear programs in canonical form be infeasible?
Is one of the stipulations of putting a linear program in canonical form that it must have at least one feasible solution? Also, if a canonical linear program can be infeasible, does that mean any linear program can be converted to canonical form?</p>

答:
Programs in canonical form can be infeasible. For example, the following LP is infeasible:

$$
\begin{aligned}
\mbox{maximize } & x_1 + x_2 \\
\mbox{s.t. }     & x_1 - x_2 = -1 \\
                 & x_1 + x_2 = -4 \\
                 & x_1, x_2 \geq 0
\end{aligned}
$$

Finally, any LP can be brought into canonical form. Inequalities can be turned into equalities by introducing slack variables, and "free" variables can be rewritten as the difference of two nonnegative variables.


问:
Fill in a $5 \times 5$ square to make equal products in rows and columns
Let's say you fill in a $5 \times 5$ square with $1, 2,\dots, 25$. Is there a way to fill it so that the product of the first row is equal to the product of the first column, the product of the second row is equal to the product of the second column, etc?

What about generalization for a $n \times n$ square filled with $1, 2,\ldots, n^2$?

Edit: Here are some updates on progress. For $n=3$, it is a possibility:

$$\begin{matrix} 5 & 1 & 6\\
                 2 & 7 & 4\\
                 3 & 8 & 9\end{matrix}$$

I'm not sure how to do $n=4, 5$. I know that if there are $>n$ prime number between $n^2/2$ and $n^2$ then $n$ doesn't work. I do not know how to proceed.

答:
Okay.... $\prod_{k=1}^{25} = 2^{24}3^{10}5^6c7^3*11*13*17*19*23$

The odd prime powers must be distributed along the diagonals.  For example if $11$ is the $k$th row then the product of the $k$th row is divisible by $11$ so the product of the $k$th column is divisible by $11$ but $11$ is the only cell divisible by $11$ so $11$ must appear in the $k$th row of the $k$th column.

So the $5$ cells in the diagonals must be $11,13,17,19,23$.  But where will put the $7, 14,$ and $21$?.  If we put $7$ in the $n$th row and $m$th column, $ n \ne m$ , then the $m$th row and the $n$th column will both need multiples of $7$.  BUt we only have three such multiples total.

If we put a multiple of $7$ in the $m$th row and $n$ column we have one extra unclaimed one.  An if we put it in the $m$th row $j$ column $j\ne m, j\ne n$ the we need a fourth one int the $j$th row.

So not possible.

And the for and


问:
Maximal ideals in polynomial rings over a field
Let $K$ be an algebraically closed field and let $k$ be a subfield of $K$ such that the field extension $K \mid k$ is algebraic. Let $B$ be the polynomial ring $K [x_1, \ldots, x_n]$ and let $A$ be the subring $k [x_1, \ldots, x_n]$. It is clear that any $k$-subalgebra of $K$ is a field, so it follows that the intersection of $A$ and any maximal ideal of $B$ is a maximal ideal of $A$. Moreover, every maximal ideal of $A$ arises in this way.

Question. What are the maximal ideals of $A$, described in terms of the maximal ideals of $B$?

It is well-known that the maximal ideals of $B$ are all of the form $(x_1 - a_1, \ldots, x_n - a_n)$ for some $n$-tuple $(a_1, \ldots, a_n)$ of elements of $K$. Thus, for each maximal ideal $\mathfrak{m}$ of $A$, we may define the set
$$P(\mathfrak{m}) = \{ (a_1, \ldots, a_n) \in K^n : (x_1 - a_1, \ldots, x_n - a_n) \cap A = \mathfrak{m} \}$$
and this set is finite (because it is a 0-dimensional closed subset of $\mathbb{A}^n_K$). Consider $G = \mathrm{Aut}(K \mid k)$. Of course, $G$ acts on $B$, and every element of $A$ is invariant under the action of $G$, so it follows that $P (\mathfrak{m})$ is closed under the action of $G$ in $K^n$ as well, i.e. $P (\mathfrak{m})$ is the union of $G$-orbits.

Conjecture. $G$ acts transitively on $P (\mathfrak{m})$.

This is certainly true if $k$ is a perfect field and $n = 1$ – after all, in that case, $\mathfrak{m}$ is a principal ideal generated by a polynomial in one variable irreducible over $k$. In fact this is true even without the assumption that $k$ is perfect: we know that the residue field $\kappa (\mathfrak{m}) = A / \mathfrak{m}$ is a finite field extension of $k$ and that the right $G$-set of $k$-algebra embeddings $\kappa (\mathfrak{m}) \to K$ is isomorphic to the right $G$-set of zeros in $K$ for the generator of $\mathfrak{m}$; but the homomorphism extension property of $\kappa (\mathfrak{m}) \to K$ implies that any two embeddings can be connected by an isomorphism of $K$, so $G$ acts transitively on the set of $k$-algebra embeddings $\kappa (\mathfrak{m}) \to K$.

What about the general case where $n > 1$? I had always taken for granted that the indicated conjecture was true, but I have not been able to find a proof in the usual places. By descent theory, one can show that
$$\operatorname{Spec} B \otimes_k B \rightrightarrows \operatorname{Spec} B \to \operatorname{Spec} A$$
is a coequaliser diagram in the category of schemes, so $\operatorname{Spec} A$ is a scheme-theoretic quotient of $\operatorname{Spec} B$ by an internal equivalence relation, but it is not clear to me what this internal equivalence relation is when $n > 0$. (For $n = 0$ and $k$ perfect, something quite remarkable happens: $\operatorname{Spec} K \otimes_k K$ has a canonical continuous free transitive action of $\mathrm{Gal}(K \mid k)$!)

答:
It is well-known that the set of maximal ideals of $k[x_1,\dotsc,x_n]$ corresponds to $K^n / \mathrm{Aut}(K/k)$. In fact, every maximal ideal $\mathfrak{m}$ has a residue field which is finite over $k$ (Corollary of Noether normalization), hence has an embedding into $K$, so $\mathfrak{m} = \{f \in k[x_1,\dotsc,x_n] : f(a)=0\}$ for some $a \in K^n$. When $a,b$ are conjugate under the action of $\mathrm{Aut}(K/k)$, the maximal ideals clearly coincide. Conversely, if $\{f : f(a)=0\} = \{f : f(b)=0\}$, then $k(a_1,\dotsc,a_n) \cong k(b_1,\dotsc,b_n)$ via $a_i \mapsto b_i$, and this extends to an automorphism of the algebraic closure $K$. Hence $a$ and $b$ are conjugate.

Hence, $\mathrm{Spm}(k[x_1,\dotsc,x_n]) \cong \mathrm{Spm}(K[x_1,\dotsc,x_n]) / \mathrm{Aut}(K/k)$ as sets.

More generally, if $R  \subseteq R'$ is an integral extension of normal domains such that $Q(R')/Q(R)$ is a normal field extension, then $\mathrm{Spm}(R) \cong \mathrm{Spm}(R') / \mathrm{Aut}(R'/R)$ and the same with $\mathrm{Spec}$ (Cohen-Seidenberg).


问:
Intersection of subgroups of a finite abelian group
Let $G$ be a finite abelian group and let $g \in G$ be a non-trivial element. I want to show that the intersection over all subgroups $G'$ of $G$ such that $G/G'$ is cyclic  and $g \not \in G'$  is trivial.

My attempt: By the Fundamental theorem of abelian groups, $G \cong \mathbb Z/n_1 \mathbb Z \times \cdots \times  \mathbb Z/n_m \mathbb Z$. We prove by induction on the number of cyclic factors of $G$.

For the base case ($m=1$), $G$ is cyclic. Then if we let $G'=\{e\}$,  $G/G'$ is cyclic and $g \not \in G'$. Thus, the proposition is true in this case.

Now for the general case, write  $G = \mathbb Z/n_1 \mathbb Z \times \cdots \times  \mathbb Z/n_m \mathbb Z$ as before and let $g=(g_1, \ldots, g_m)$. Since $g$ is non-trivial, at least of its components is not zero. WLOG, assume $g_1 \neq 0$. Then $h:=(g_1, \ldots, g_{m-1})$ is a non-zero element of $H:=\mathbb Z/n_1 \mathbb Z \times \cdots \times  \mathbb Z/n_{m-1} \mathbb Z$. By the inudction hypothesis, the intersection over all subgroups $H'$ of $H$ such that $H/H'$ is cyclic  and $h \not \in H'$  is trivial.

I am not sure how to proceed from here and I will be very grateful for any advice on this attempt or some ideas for an alternative attempt.

答:
I think the key is the $k=2$ case. Once you figure out how to do the $n=2$ case, the general case follows the same idea. So let me do the $k=2$ case explicitly.

Let $G=C_{n_1}\times C_{n_2}$, with $n_1|n_2$. Let $x_1$ and $x_2$ be the corresponding generators, and let $g=(g_1,g_2)$ be a nontrivial element of $G$.

If $g_1\neq e$ and $g_2\neq e$, then we can take $H_1=C_{n_1}\times\{e\}$ and $H_2=\{e\}\times C_{n_2}$; each of those excludes $g$, $G/H_i$ is cyclic, and $H_1\cap H_2=\{e\}$, so we are fine.

If either $g_1$ or $g_2$ are trivial, let $y=(x_1,x_2^{n_2/n_1})$, and let $H=\langle y \rangle$. Note that $H\cap H_1=H\cap H_2 = \{e\}$, because both components have order $n_1$. This also means that $g\notin H$. Finally, $(e,x_2)H$ has order $n_2$ in $G/H$ (because $H$ does not contain any elements of $\{e\}\times C_{n_2}$ that are nontrivial), and $G/H$ has order $n_2$, so $G/H$ is cyclic. Now, either $H_1$ or $H_2$ do not contain $g$, so whichever one does not contain $g$, together with $H$, show the intersection you want is trivial. This finishes the $k=2$ case.

Note: To better mimic the general case, we could do the following instead: if $g_2$ is trivial, we replace $x_1$ and $x_2$ with $y$ and $x_2$ to get new "coordinates" for $G$: $G\cong H\times H_2$, but in these coordinates, both components of $g$ are nontrivial, so we are reduced to the first case. And if $g_1$ is trivial, then we take $H_1$ and $H$ as our subgroups.

So, now let $G=C_{n_1}\times\cdots\times C_{n_k}\times C_{n_{k+1}}$ with $n_1|\cdots|n_k|n_{k+1}$, $x_i$ a fixed generator for $C_{n_i}$, and let $g=(g_1,\ldots,g_{k+1})\neq e_G$. Let $H=C_{n_1}\times\cdots\times C_{n_k}$, so $G=H\times C_{n_{k+1}}$, and write $u=(g_1,\ldots,g_k)$. We identify $g$ with $(u,g_{k+1})$ and $H$ with the corresponding subgroup of $G$ when necessary (I'm going to play fast and loose with the internal vs. external direct product, but being more careful is going to be annoying; I hope that's okay).  We assume by induction that if $u\neq e_H$, then we can find subgroups of $H$ that do not include $u$, with cyclic quotient, with trivial intersection.

If $u\neq e$ and $g_{k+1}\neq e$, then let $K_1,\ldots,K_r$ be subgroups of $H$ with $u\notin K_i$, $H/K_i$ cyclic, and $K_1\cap\cdots\cap K_r$ trivial. Then we have that $K_i\times C_{n_{k+1}}$ does not contain $g$, and $G/(K_i\times C_{n_{k+1}})\cong H/K_i$ is cyclic. In addition, $g\notin H\times\{e\}$, and $G/(H\times\{e\})\cong C_{n_{k+1}}$ is cyclic. Now we notice that $H\times\{e\}\cap (K_1\times C_{n_{k+1}})\cap\cdots\cap (K_r\times C_{n_{k+1}})$ is trivial, so we are done.

If either $u=e$ or $g_{k+1}=e$, then for $i=1,\ldots,k$, let $y_i = x_{i}x_{n_{k+1}}^{n_{k+1}/n_i}$, and let $H_i = \langle x_1,\ldots,x_{i-1},y_i,x_{i+1},\ldots,x_k\rangle$.

Note that an element of $H_i$, viewed as a $(k+1)$-tuple, has trivial $i$th coordinate if and only if it has trivial $(k+1)$st coordinate. In particular, $H\cap H_1\cap\cdots\cap H_k$ is trivial. And $G/H_i$ has order $n_{k+1}$, and as in the $k=2$ case, the element $x_{k+1}H_{i}$ has order $n_{k+1}$, so $G/H_i$ is cyclic.

If $u\neq e$ and $g_{k+1}=e$, then say the $i$th component of $u$ is nontrivial. Then replacing $H$ with $H_i$ we have that $G\cong H_i\times\langle x_{k+1}\rangle$, but now expressing $g$ in terms of $x_1,\ldots,x_{i-1},y_i,x_{i+1},\ldots,x_{k+1}$ will have both the $H_i$ and the $x_{k+1}$ component nontrivial, so we are reduced to the first case and we are done.

And if $u=e$ and $g_{k+1}\neq e$, then each of $H$, $H_1,\ldots,H_k$ are subgroups that do not contain $g$, with $G/H$ and $G/H_j$ cyclic, and $H\cap H_1\cap\cdots\cap H_k$ trivial, so we are done.


问:
Is non-increasing the same as monotone decreasing?
My understanding of a non-increasing sequence is that for every $n$, $a_{n} \geq a_{n+1}$.

This appears to be the same definition as a sequence being monotone decreasing, except I don't understand why we would have two names for the same thing. I'm wondering if there's a subtle detail I'm missing in one of the definitions that is contributing to my confusion.

答:
There is two different systems of nomenclature.

$a_{n+1} \le a_n\quad$ ... $\quad a_{n+1} < a_n$

correspond, respectively, to

nonincreasing ... decreasing

or to

decreasing ... strictly decreasing

You must consult the book you are reading to find out which is used there.


答:
Some writers distinguish

monotone decreasing : $a_n \geq a_{n+1}$, and

strictly monotone decreasing : $a_n > a_{n+1}$.

Some writers use "monotone decreasing" to mean "strictly monotone decreasing" and others use it to mean non-strictly monotone decreasing, as recited above.

Some writers use the forms "strictly increasing", "increasing", "decreasing", and "strictly decreasing" in parallel with the above.  At least one writer I've read used "monotone" to mean strict and increasing/decreasing to mean non-strict.

I have, at times, wondered if any two humans can agree on which of these terms to use and what their exact meanings are.  Consequently, I have to check any author's definitions to make sure I know what they mean when they use these terms.

Title: Is non-increasing the same as monotone decreasing?

Body:
My understanding of a non-increasing sequence is that for every $n$, $a_{n} \geq a_{n+1}$.

This appears to be the same definition as a sequence being monotone decreasing, except I don't understand why we would have two names for the same thing. I'm wondering if there's a subtle detail I'm missing in one of the definitions that is contributing to my confusion.


问:
A closed form expression for the finite-difference Euler's method
Consider the ODE

$$f(u) \equiv u' = au+b, \>\>\> u(t_0 = 0)\equiv u^0 = u_0$$

For a timestep $k$, the forward-difference approximator for $u'$ gives the Euler update as

$$
u^{n+1} = u^n + kf(u^n) = u^n + k(au^n + b)
$$

I am trying to come up with a closed form expression for $u^n$, that is, one that involves only the constants $a, b, u_0, k$.

I've tried to do this by first writing $u^{n}$ as a cumulative sum

$$
u^n = u^n + k(au^n + b) = u_0 + \sum_{i=0}^{n-1}k(au^i+b)
$$

I then tried to expand the sum on the right and see if there was any easily identifiable pattern to write more generally, or if it would perhaps resemble an expansion of some other known function.After some ugly algebra, this is something like

$$
\begin{align}
u^n &= u_0 + k(au_0 + b) + k(au_1 + b) + k(au_2 + b) + ...\\
&= u_0 + k(au_0 + b) + (k^2a + k)(au_0 + b) + (k^3a^2 + 2k^2a + k)(au_0 + b) + ...\\
&= u_0 + (k^3a^2 + 3k^2a + 3k + ...)(au_0+b)
\end{align}
$$

Nothing jumps out to me, and I'm not sure this approach is promising. Any ideas?

I've massaged it some more and found that it can be written as

$$u^n = (1+ka)^nu_0 + C_n$$

Where C is a function of $a,b,k,n$, which I've yet to deduce...

答:
I did not solve your specific problem, but your problem can be written as a non-homogenous linear recurrence equation with constant coefficients. C. f. https://en.wikipedia.org/wiki/Recurrence_relation#Solving_non-homogeneous_linear_recurrence_relations_with_constant_coefficients .

$$u^{n+1} = (1 + ka) u^n + kb$$


问:
Reducing the Exponent of an $\ell^p$ Sequence / Counterexample
Assume $x=(x_k)_{k\in\mathbb N}\in\ell^p$ for some $p\in(0,\infty)$, i.e., $\sum\limits_{k\in\mathbb N}|x_k|^p<\infty $. Then, is there always some $q\in(0,p)$ (possibly depending on $x$), such that $x\in\ell^q$?

I doubt it, but couldn't find a counter example. Any suggestions (or proofs that such a $q$ always exists) are welcome.

I am especially interested in the case $p=1$, so any insight on that side would also help.

答:
Consider $x_k := \frac{1}{k \log(k)^2}$. By the Cauchy condensation test $\sum_{k = 2}^\infty \lvert x_k \rvert^p$ for $p >0$ converges iff
$$
\sum_{k = 2}^\infty 2^k\lvert x_{2^k} \rvert^p = \sum_{k = 2}^\infty \frac{1}{2^{k(p-1)}k^{2p}\log(2)^{2p}}
$$
converges. Note that for $p=1$ this series clearly converges. For $p \in (0, 1)$ we have $k(p-1)<0$ and therefore
$$
\frac{1}{2^{k(p-1)}k^{2p}\log(2)^{2p}} \rightarrow \infty
$$
as $k \rightarrow \infty$. Therefore no convergence in this case by the divergence test.


问:
normal bundle of level set
Let $M$ be a Riemannian manifold and $S \subset M$ a regular level set of a smooth function $f:M\rightarrow \mathbb{R}^k$. How can I show that the normal bundle of $S$ is trivial?

If $k=1$ then $\text{grad}f$ is a global frame for $NS$ but I am not sure how to show it formally for the general case. Thanks!

答:
I'd like to defend the point of view that this triviality result for the normal bundle has actually very little to do with the riemannian structure of $M$.

As user 8268 points out, $f_{\ast y}: T_{\ast y}M\to T_{\ast f(y)}N$ is surjective for each $y\in M$ and this is translated into the exact sequence of vector bundles on $S$
  $$0\to T(S) \to TM_{|S}\to f^\ast T\mathbb (R^k)_{|S} \to 0$$
Now the crucial point is that the genuine normal  bundle of $S$ in $M$ is by definition the quotient bundle $N_{S/M}=T(M)_{|S}/TS$.
This genuine normal bundle is here isomorphic to $ f^\ast T\mathbb (R^k)_{|S}$, hence trivial as requested.

And what has this to do with the riemannian structure? It is just that we can then, if we so wish, replace the genuine normal bundle, which is a quotient of $T(M)_{|S}$ by the orthogonal complement $T(N)^\perp$ of $T(N)$ inside $T(M)_{|S}$, which is a subbundle of $T(M)_{|S}$, isomorphic to $N_{S/M}$. The success of this vision of the normal bundle in the riemannian case is that everybody prefers subspaces to quotient spaces, but we have to keep in mind that what I have called the genuine normal bundle is completely canonical whereas the more concrete isomorphic riemannian normal bundle varies with the riemannian structure we impose on $M$.

Title: normal bundle of level set


答:
There is no specific difference. "Non-increasing" and "decreasing" can be used interchangeably as long as "strictly decreasing" (i.e. only decreases, as opposed to staying constant at some points) is separately defined.

However, from a language point of view, it would make more sense to describe a sequence that stays constant over a wider range as "non-increasing" and one that decreases almost everywhere as "decreasing".


答: Maybe related to your question about the canonical form (or standard form in some optimization literature) having at least one feasible solution: a standard form LP, when it is feasible, must have at least one basic feasible solution. In other words, the polyhedron associated with the constraints must have an extreme point, when nonempty. This is one of the reasons that an LP is transformed to standard form for computation, like using the Simplex method.

Title: Can linear programs in canonical form be infeasible?

Body: Is one of the stipulations of putting a linear program in canonical form that it must have at least one feasible solution? Also, if a canonical linear program can be infeasible, does that mean any linear program can be converted to canonical form?</p>


问:
The number of ways an even integer can be written as the sum of two primes
We all know the famous Goldbach Conjecture, namely that every even intger $>2$ is the sum of two primes. I was recently playing around with this conjecture and found out that there are MANY ways of writing an even integer as the sum of two primes.

Definition: Let $A(n)$ be the number of ways $n$ can be written as the sum of two primes, where $n$=$0$(mod $2$).

Then we have that Goldbach's conjecture is equivalent to that $A(n)$ is at least $1$.

If it isn't true, please (if possible) provide a counterexample, or some sort of triviality that comes with this statement.

So if you graph this for large enough values, one observes that $A(n)$$\geq$$\sqrt n$. But $\sqrt n$$\geq1$. So would we then have Goldbach's conjecture? In other words, is the following statement true?

Conjecture(?): If one proves $A(n)$$\geq$$\sqrt n$, this would imply $A(n)$$\geq$$1$, which is Goldbach's conjecture.

答:
According to your definition of $A(n)$, Goldbach conjecture would be equivalent to
saying that every even number $n >2$ verifies
$$A(n) \geq 1.$$
In order to prove Goldbach conjecture, it would be sufficient to prove that for any even number larger than $2$, $A(n) \geq \sqrt{n}$ (this is because $\sqrt{n} \geq 1$ when $n$ is a positive integer).


问:
Show that $P = Q^2$

Suppose $P$ is a positive semi-definite $n\times n$ matrix. Show that there exists a unique positive semi-definite matrix $Q$ such that $P = Q^2$.

In class we've been going over singular value decompositions of the form:

$A=USV^\dagger$

and we are recommended to utilize this to prove that statement. However I am having trouble understanding how two completely generalized matrixes can be related to the equation $P=Q^2$ without any more information.

Can anyone provide suggestions?

答: The singular value decomposition of a positive semidefinite matrix has a particularly simple form: $Q=USU^*$. Writing $S$ as the sum of $n$ matrices with at most one non-zero diagonal entry, we arrive at a geometric interpretation of this decomposition: $Q=\sum_{j=1}^n c_j Q_j$ where $c_j\ge 0$ and $Q_j$ are orthogonal projections on mutually orthogonal lines. Squaring $Q$, we find $Q^2=\sum_{j=1}^n c_j^2 Q_j$ because $Q_iQ_j=0$ when $i\ne j$. Therefore, for $Q^2$ to be equal to $P = \sum_{j=1}^n b_j P_j$, we must have $P_j=Q_j$ up to relabeling, and $c_j=\sqrt{b_j}$.</p>

Title: Show that $P = Q^2$


问:
Can this be a proof of Euler's formula?
Let $z=\cos\theta+i\sin\theta$.

We know $z\overline{z}=1$.

$$\overline{z}=\cos\theta-i\sin\theta=\cos(-\theta)+i\sin(-\theta)$$

Suppose $z=e^{i\theta}$. If this is true, this satisfies $z\overline{z}=1$.

$$\overline{z}=\cos(-\theta)+i\sin(-\theta)=e^{-i\theta}$$

$$e^{i\theta}e^{-i\theta}=e^{i\theta-i\theta}=e^0$$

So $z=e^{i\theta}$

$$\boxed{e^{i\theta}=\cos\theta+i\sin\theta}$$

答: See https://en.wikipedia.org/wiki/Begging_the_question. You cannot a priori suppose to be right, then try to not find a contradiction.

Title: Can this be a proof of Euler's formula?


问:
What is the size of the largest subset with a pairwise hamming distance of 3
Consider all binary strings of length $n$. Is there any known bounds on the size of the maximum subset such that the pairwise hamming distance between any two elements is at least 3.

答:
Edit: I found a much better solution, so I have overhauled this answer.

You already have the upper bound $$A_2(n,3)\le \frac{2^{n}}{n+1}.$$ which is just the sphere bound. For a lower bound, you can use the Hamming code  of length $n$, which proves
$$
A_2(n,3)\ge 2^{n-\lfloor \log_2(n)\rfloor-1}\ge \frac{2^{n-2}}{n}
$$
Therefore, this is within a factor of $4$ of the optimal code.

In general, finding the optimal code is an open problem. There is a table of best known bounds for binary codes maintained here.


问:
Is it true that the domain of this function is finite?
(This is a subproblem that came up as I was proving that $\{e: \phi_e(x) \text{ has infinite domain}\}$ is $\Pi_2$-complete.)

Let $A$ be a set of natural numbers and suppose $$n\in A\iff \forall y_1\exists y_2 T(n,y_1,y_2)$$

(where $T$ is a (computable) relation).

Consider this  partial function on pairs of natural numbers:

$$(n,x) \mapsto 1\text{ if }\forall y_1\leq x\exists y_2 T(n,y_1,y_2)\\(n,x)\text{ is undefined otherwise} $$

If $n\in A$, then the above function as a function of $x$ is total. Is it true that if $n\notin A$, then this function (again as a function of $x$) has finite domain? I think so, and I think this is a trivial result but I can't wrap my head around proving it in a formal way.

Here's what I think informally: if $n\notin A$, then the first condition in the definition may be true for small $x$, but starting from some $x_0$, that wouldn't be true. Is this indeed the case? I can't connect this with the negation of $x\in A$ i.e. with $\exists y_1\forall y_2 \neg  T(n,y_2,y_3) $.

Edit: I modified the original function. Does the new function has a finite domain as a function of $x$ when $x\notin A$? If not, I was wondering if there's a way to modify this function further so that (1) it remains computable, (2) if $n\in A$, the domain is still infinite, (3) if $n\notin A$, the domain is finite?

答:
Let $V$ denote the partial function from the question. Suppose $n\notin A$. Then $\exists x_0 \forall y_2 \neg T(n, x_0, y_2)$. Then the domain of $V(n,-)$ is a subset of $\{x\in N: x < x_0 \}$ (hence finite). Here's why nothing outside this finite set can lie in the domain. If $x \ge x_0$, then the condition $$\forall y_1 < x \exists y_2 T(n,y_1,y_2)$$

or equivalently $$\forall y_1\exists y_2 (\neg (y_1 \le x) \lor T(n,y_1,y_2))$$
is violated. Indeed, its negation is $$\exists y_1\forall y_2 ((y_1 \le x)\land \neg T(n,y_1,y_2))$$
and this negation is true (take $y_1=x_0$).


问:
maximum edges of a bipartite graph
Show that if a graph G = (V,E) is bipartite, then it has at most n^2/4 edges if n is even, and at most (n^2 − 1)/4 edges if n is odd. Here n = |V| is the number of vertices in G.

I want to prove this using induction. The basis step would be to just inspect the graphs with 1 and 2 vertices and see that the theorem holds. I don't know where to go from here, any help would be appreciated.

答:
First show that the conclusion is equivalent to the assertion that $G$ has at most $\left\lfloor\frac{n^2}4\right\rfloor$ edges.

Suppose that the result is true for bipartite graphs with fewer than $n$ vertices, and let $G$ be a bipartite graph with $n$ vertices. Let $U$ and $V$ be the parts of $G$, where $|U|\le|V|$; then $|U|\le\left\lfloor\frac{n}2\right\rfloor$. Remove one vertex from $V$, along with all of the edges incident at that vertex; the resulting graph $G'$ is bipartite on $n-1$ vertices, so by the induction hypothesis it has at most $\left\lfloor\frac{(n-1)^2}4\right\rfloor$ edges. We removed at most $\left\lfloor\frac{n}2\right\rfloor$ edges from $G$ to get $G'$, so $G$ has at most

$$\left\lfloor\frac{(n-1)^2}4\right\rfloor+\left\lfloor\frac{n}2\right\rfloor$$

edges. To complete the induction step, show that

$$\left\lfloor\frac{(n-1)^2}4\right\rfloor+\left\lfloor\frac{n}2\right\rfloor\le\left\lfloor\frac{n^2}4\right\rfloor\,.$$

If you’re interested, you can also show that the upper bound of $\left\lfloor\frac{n^2}4\right\rfloor$ edges is achieved in the complete bipartite graph $K_{\left\lfloor\frac{n}2\right\rfloor,\left\lceil\frac{n}2\right\rceil}$.

Title: maximum edges of a bipartite graph


问:
Number of elements of prime order $p$ in $\mathrm{SL}_2(\mathbb{F}_q)$
Let $p,q$ be prime numbers.
It is known that there is one element of order $2$ in $\mathrm{SL}_2(\mathbb{F}_q)$.

How many elements of order $p>2$ are there in $\mathrm{SL}_2(\mathbb{F}_q)$?

答:
I'll answer to a more general question: if $q$ if an odd prime power, how many elements of order $n$ does $\newcommand\Fq{\Bbb F_q}\newcommand\SL{\mathrm{SL}_2(\Fq)}\SL$ have, for $n>2$? I will do so using only linear algebra. I'll put $p=\operatorname{char}(\Fq)$, which conflicts with the use of $p$ in the question, but will be needed only in the case where $n=p$.

Since $n>2$ excludes multiples of the identity, the minimal polynomial of candidates $A\in\SL$ will be equal to their characteristic polynomial$~\chi_A$, which is of the form $\chi_A=X^2-2a+1$ with $a\in\Fq$. One has $\chi_A=(X-a)^2$ if $a\in\{1,-1\}$ and otherwise $\chi_A$ is square-free; in the latter case $\chi_A$ will split over$~\Fq$ if its quarter-discriminant $a^2-1$ is a square in$~\Fq$, and is irreducible (but split over $\Bbb F_{q^2}$) otherwise.

Since $\chi_A$ is the minimal polynomial, $A$ is not diagonalisable over$~\overline\Fq$ in the case $\chi_A=(X-a)^2$ with $a\in\{1,-1\}$. Therefore in this case $A$ is conjugate to a matrix of the form
$$
  \begin{pmatrix}a&x\\0&a\end{pmatrix}
  \qquad\text{with $a\in\{1,-1\}$ and $x\in\Fq^\times$},
$$
which has order $p$ if $a=1$ and order $2p$ if $a=-1$. Moreover since there is only one $p$-th root of unity in characteristic$~p$, every element$~A$ of order $p~$or$~2p$ must have minimal polynomial $(X-1){}^2$ respectively $(X+1){}^2$. The subgroups conjugate to the upper triangular matrices are characterised by the $1$-dimensional subspace of which they are the stabiliser, so the number of such subgroups is $q+1$, and each one contains $q-1$ elements each of orders$~p$ and$~2p$, for a total of $q^2-1$ elements in$~\SL$ of each of the orders$~p$ and$~2p$.

If $a^2-1$ is a square in$~\Fq$, then $A$ is diagonalisable, with eigenvalues $a\pm\sqrt{a^2-1}$ (which are mutually inverse elements of $\Fq^\times\setminus\{1,-1\}$), and has order dividing $|\Fq^\times|=q-1$. Since $\Fq^\times$ is cyclic, every $n\mid q-1$ is the order of $\phi(n)$ different elements in $\Fq^\times$, and so there are $\phi(n)/2$ different pairs of eigenvalues giving rise to elements of such orders$~n>2$. For each pair one can choose a decomposition into eigenspaces in $(q+1)q$ different ways, for a total of $\frac{\phi(n)}2q(q+1)$ elements in$~\SL$ of order $n>2$ dividing$~q-1$.

Finally if $a^2-1$ is not a square in$~\Fq$, then $\chi_A$ is irreducible but splits over $\Bbb F_{q^2}$; moreover its eigenvalues $\lambda,\lambda^q$ have product$~1$, so $\lambda$ is a $q+1$-st root of unity in $\Bbb F_{q^2}$. Also every $n>2$ dividing $q+1$ is the order of $\phi(n)$ distinct such roots, with again $\phi(n)/2$ distinct characteristic polynomials giving rise to elements of order$~n$. For matrices$~A$ with such $\chi_A$, the first canonical basis vector $e_1$ together with $A\cdot e_1$ form a basis (as there are no eigenvectors), on which $A$ is given by the companion matrix for$~\chi_A$. Fixing $n$ there are $\phi(n)/2$ choices for$~\chi_A$ and independently $q^2-q$ choices for$~A\cdot e_1$, for a total of $\frac{\phi(n)}2q(q-1)$ elements in$~\SL$ of order $n>2$ dividing$~q+1$. In summary:

for $n=1,2$ there is one element each of order$~n$,

for $n=p,2p$ there are $q^2-1$ elements each of order$~n$,

for $n>2$ dividing $q-1$ there are $\frac{\phi(n)}2q(q+1)$ elements of order$~n$,

for $n>2$ dividing $q+1$ there are $\frac{\phi(n)}2q(q-1)$ elements of order$~n$,

these are al the orders that occur in$~\SL$.

One checks easily that this accounts for all $q(q^2-1)$ elements of$~\SL$.


问:
Existence of small extensions
Let $\mathsf{CNL}$ be the category of complete noetherian local rings with (fixed) residue field $\mathbb{F}$. The morphisms are local homomorphisms fixing the residue field. The full subcategory of artinian objects (i,e, artinian complete local rings residue field $\mathbb{F}$) is denoted by $\mathsf{Ar}$.

A homomorphism $f: A \rightarrow B$ in $\mathsf{Ar}$ is called a small extension or small homomorphism, if $f$ is surjective and its kernel $\ker f = (t)$ is a principal ideal, which is annihilated by $\mathfrak{m}_A$, the maximal ideal of $A$. (i.e. $\mathfrak{m}_A (t) = (0)$.)

Given such a small extension, we can obtain an exact sequence
$$
0 \rightarrow tA \rightarrow A \xrightarrow{f} B \rightarrow 0.
$$

My Question 1: How to prove (or disprove) the following claim: For ANY object $A$ in $\mathsf{Ar}$, there exists a small extension $f: A \rightarrow A_0$ for some $A_0 \in \mathsf{Ar}$ with $\ell(A) > \ell(A_0)$, where $\ell(R)$ is the length of the artinian ring $R$.

Motivation: The claim is used in proofs by induction in many contexts, such as the Appendix A on Page 106 of this one : Galois Deformations and Hecke Curves by B. Mazur.

My question 2: Does every surjective homomorphisms in $\mathsf{Ar}$ factors as the composition of small homomorphisms? I think this is related to the construction of Question 1 but not sure about this.

Thank you all for anwering or commenting this post and sorry for any mistakes!

答:
For question 1: the only restriction seems to be that $A \neq \mathbb{F}$. $A$ is artinian, so, if $\mathfrak{m}$ is its maximal ideal, there is a minimal integer $p \geq 2$ such that $\mathfrak{m}^p=\{0\}$. Take some nonzero $t \in \mathfrak{m}^{p-1}$ and define $A_0=A/tA$.

Question 2 asserts that any local surjection $s: A \rightarrow B$ of complete Noetherian Artinian local rings with residue field $F$ is a product of small extensions.

So consider such a surjection $s: A \rightarrow B$. Let’s show that if $s$ isn’t an isomorphism, then it factors through a small extension $A \rightarrow C$. Indeed, let $I$ be the kernel of $s$, and let $t \in I$ be nonzero (else it’s an empty product). There is $p \geq 1$ minimal such that $t\mathfrak{m}^p=\{0\}$, so choose $q \in t\mathfrak{m}^{p-1}$ nonzero and take $C=A/(q)$.

Now, repeating this with $C \rightarrow B$, we know that there is a sequence of small extensions $A \rightarrow C_1 \rightarrow C_2 \rightarrow C_3 \rightarrow \ldots \rightarrow C_n$ that is a factor of $s$. In particular (I hope I’m not using it correctly) if $B$ wasn’t reached at some point, $\ell(A) \geq n+\ell(B)$ and we get a contradiction if $n$ is too large. So the statement should be true.

Title: Existence of small extensions


问:
Vector Application Question
I have been working on a few problems related to vector applications and I can never seem to get the angle right when it comes to my final result. For example, I have been working on this question:

A small aircraft, on a heading of $225^\circ$ is cruising $150$ km/h. It is encountering a wind blowing from a bearing of $315^\circ$ at $35$ km/h. Determine the aircraft's velocity.

My answer was approximately $154$ km/h on a bearing of $103^\circ$
I got this by drawing a vector diagram to provide a visual. Then as the wind vector approaches the aircraft vector, it ended making a 90° angle. So I just used the Pythagorean theorem to solve for the resultant vector. Then I used the sine law to determine the angle and got a 77°. After this, is where I think I am getting mixed up. I extended the wind vector so it would end up in the second quadrant and subtracted the angle I got with 180° where I ultimately got 103°
The bearing is actually supposed to be $211.9° but I can’t seem to understand why that may be the case.

If anyone has any suggestions I would really appreciate it

答: Bearing angles are measured clockwise with respect to the positive $y-$axis. The velocity of the airplane before encountering the wind is the vector $$v=\Big<150\cos\Big(\frac{5\pi}{4}\Big),150\sin\Big(\frac{5\pi}{4}\Big)\Big>$$ The velocity of the wind (which blows from a bearing of $315^{\circ})$ is the vector $$w=\Big<35\cos\Big(\frac{7\pi}{4}\Big),35\sin\Big(\frac{7\pi}{4}\Big)\Big>$$ The resultant velocity becomes $$v+w=\Big<-\frac{115 \sqrt{2}}{2},-\frac{185 \sqrt{2}}{2}\Big>$$ which has magnitude $\|v+w\|\approx 154 \text{ km/hr}$ and makes an angle of $\arctan\Big(\frac{37}{23}\Big)+\pi$ radians measured counterclockwise from the positive $x-$ axis. This angle corresponds to a bearing of $\approx 211.9^{\circ}$.

Title: Vector Application Question


问:
Axiom of choice and Euclid's first proposition
Euclid's first proposition constructs an equilateral triangle given a segment $AB$ via two circles using the segment as radiuses, using $C$ as one of the intersection points of the circles.

I understand that there assumptions made which are not part of the original set of axioms. E.g. that the circles intersect (i.e. such a $C$ exists). Additional axioms would be required e.g. involving some continuity axiom for intersection issue.

But when it turn out that the circles intersect twice, by what criteria is one of the two points chosen? Does this require something similar/equivalent to the axiom of choice or not?

答:
No, the axiom of choice is not needed to provide an element of a single finite set.

What the axiom of choice says is that the product of a set of nonempty sets is itself nonempty.


问:
Companion Lecture Notes to Atiyah-MacDonald?
Is there a set of lecture notes that follow Atiyah-MacDonald and expand on the dense passages, point out typos and so forth?</p>

答:
(Sorry, this post should really be a comment to the answer by Oliver E. Anderson, but I don't have enough reputation to be able to comment yet. Please someone re-post it as a comment instead of me. If you notify me, I'll delete this answer.)

The latest revision of the textbook by Altman and Kleiman can be found at this MO answer. Be sure to get at least the 2017 version or newer, which was a substantial expansion.


问:
What is an invariant mean on $L^\infty(\mathbb{R})$?
Consider the von Neumann algebra $M:=L^\infty(\mathbb{R})$, which consists of (classes) of essentially bounded measurable functions $\mathbb{R}\to \mathbb{C}$. Here $\mathbb{R}$ has the classical Lebesgue measure.

In a paper I'm reading, one talks about an invariant mean on $M$. This is supposed to be some state $m:M \to \mathbb{C}$, but what exactly does it mean that $m$ is invariant?

I'm aware of a notion of invariance when we have a comultiplication $L^\infty(\mathbb{R})$, so I was thinking to consider a comultiplication
$$\Delta: L^\infty(\mathbb{R})\to L^\infty(\mathbb{R}) \otimes L^\infty(\mathbb{R}) \cong L^\infty(\mathbb{R}\times \mathbb{R}):f \mapsto ((s,t) \mapsto f(s+t))$$
and then require that $m (\omega \otimes \iota)\Delta = m\omega(1) = m(\iota \otimes \omega)\Delta$. Is this the correct notion of invariant mean in this context? Does such a mean always exist?

答:
Despite my earlier comment that the existence of an invariant mean on $L^\infty (\mathbb R)$ is a non-trivial result, let me
give a reasonably simple proof of this fact.

A Fölner sequence is, by definition, a sequence $\{F_n\}_{n\in {\mathbb N}}$, where each $F_n$ is a measurable subset of $\mathbb R$,
with finite positive measure, and
such that
$$
  \lim_{n\to \infty }  \frac{|(F_n + s)\mathop{\Delta } F_n|}{|F_n|} = 0,
  $$
for every $s$ in ${\mathbb R}$.

Here I am using "$|\cdot|$" to denote Lebesgue measure, and by "$\Delta $" I mean the symmetric difference of sets, namely
$$
  A\mathop{\Delta }B = (A\setminus B)\cup  (B\setminus A).
  $$

The sets in a Folner sequence should be  thought of as being asymptotically invariant under translation.

As an example, notice that  the sets $F_n=(-n,n)$ form a Folner sequence.

Given any Folner  sequence $\{F_n\}_{n\in {\mathbb N}}$, consider for each $n$,  the linear functional $\varphi _n$ on  $L^\infty (\mathbb R)$
defined by
$$
  \varphi _n(f) = \frac 1 {|F_n|}\int_{F_n}f(t)\,  dt.
  $$

It is easy to see that each $\varphi _n$ is a positive linear functional with norm 1 (aka a state).

Given any real number $s$, let
$$
  \tau _s:L^\infty (\mathbb R)\to L^\infty (\mathbb R)
  $$
be the translation operator  defined by
$$
  \tau _s(f)|_t = f(t-s), \quad \forall f \in  L^\infty (\mathbb R).
  $$

For  $f$ in $L^\infty (\mathbb R)$ notice that
$$
  \varphi _n(\tau _s(f)) -   \varphi _n(f) =
  \frac 1 {|F_n|}\Big(\int_{F_n}f(t-s)\,  dt - \int_{F_n}f(t)\,  dt\Big) = $$$$ =
  \frac 1 {|F_n|}\Big(\int_{F_n+s}f(t)\,  dt - \int_{F_n}f(t)\,  dt\Big) = $$$$ =
  \frac 1 {|F_n|}\Big(\int_{(F_n+s)\setminus F_n}f(t)\,  dt - \int_{F_n\setminus (F_n+s)}f(t)\,  dt\Big),
  $$
from where it easily follows that
$$
  |\varphi _n(\tau _s(f)) -   \varphi _n(f)| \leq  \frac{|(F_n + s)\mathop{\Delta } F_n|}{|F_n|}\Vert f\Vert .
  \tag 1
  $$

Thanks to  Alaoglu's Theorem,  the set of all states is compact for the weak$^*$ topology,  so there is some state $m$
which is a cluster point
of the set $\{\varphi _n:n\in {\mathbb N}\}$.  By  (1) one then concludes that $m$ satisfies
$$
  m(\tau _s(f)) = m(f),
  $$
which is to say that $m$ is an invariant state.


问:
Necessary and sufficient condition on a set for a continuous function $f$ to be bounded by the subsets
Question: Find  a  necessary  and  sufficient  condition  on  the  set $A \subset \mathbb{R}$ such that $\forall B \subset A, f(B)$ is bounded, where $f$ is any continuous function $f:A \rightarrow \mathbb{R}$.

My attempt: I think the answer should be $A$ must be closed in $\mathbb{R}$. Suppose A is not closed, then we want to show that there exist a function $f:A \rightarrow \mathbb{R}$ and a subset $B$ such that $f(B)$ is unbounded. Since $A$ is not closed, there exist $a_1, a_2, ... , a_n \in A$ such that $a_n \rightarrow a$ and $a \notin A$. Consider the function $f(x)=\frac{1}{x-a}$ which is continuous on $A$. We can find an $\epsilon > 0$ such that $B_a(\epsilon) \cap A$ is a subset of $A$. Then $f(B_a(\epsilon) \cap A)$ is clearly unbounded. However, I don't know how to prove the other direction ($A$ is closed $\implies \forall B \subset A, f(B)$ is bounded). Any help would be greatly appreciated.

答: Your argument that this fails for non closed $A$ is correct, but your conjecture that 'closed' is a sufficient condition is wrong. Take $A= [a, \infty]$ and as $f= id_R$ the identity. Now take $B=A$. I think the correct answer is that the  necessary and sufficient condition you are looking for are compact sets. In $\mathbb{R}$ these are exactly the closed & bounded sets and compact sets have nice feature that the images of comact sets by continuous maps are also compact, therefore also closed & bounded. What does it say about images of all $B \subset A$?


问:
decomposition of order preseving injective map $\phi:[n-k] \to [n]$
Let $d^j:[n-1]\to [n]$ be the order preserving injection that skip $\{j\}$  in the range,where $[n] = \{0,...,n\}$.

Prove the following two fact:

Let $\phi:[n-k] \to [n]$ is  the injective order preserving map,then it has unique composition of the form :$\phi = d^{j_k}...d^{j_1}$ where $0\le j_1<...<j_k\le n$

$d^id^j = d^{j+1}d^i$ with $i\le j$.

I have no idea how to prove these theorems formally,I can check these two facts holds,but seems a bit messy to prove it

答:
$d^j$ is explicitly given by
$$d^j(r) =  \begin{cases} r & r < j \\ r+1 & r \ge j  \end{cases}$$

No. 2. :

We have $j \ge i$.

For $r \ge j$ we get $r+1 \ge i$, $r+1 \ge j+1$ and $r \ge i$ and therefore
$$d^i d^j(r) = d^i(r+1) = r+2 = d^{j+1}(r+1) = d^{j+1} d^i(r)$$
For $j > r \ge i$ we get $r+1 < j+1$ and therefore
$$d^i d^j(r) = d^i(r) = r+1 = d^{j+1}(r+1) = d^{j+1} d^i(r)$$
For $i > r$ we get $j+1 > r$ and therefore
$$d^i d^j(r) = d^i(r) = r = d^{j+1}(r) = d^{j+1} d^i(r)$$

No. 1. :

We do it by induction. For technical reasons let us consider order preserving injections $\phi : [n] \to [n+k]$. The base case $k = 1$ is obvious because there is a unique $j_1 \in [n+1]$ which is not in the image of $\phi$. Hence $\phi = d^{j_1}$.For the step $k \mapsto k+1$ pick any element $s_{k+1} \in [n+k+1] \setminus \phi([n])$. Define $$\phi' : [n] \to [n+k], \phi'(i) = \begin{cases} \phi(i) & \phi(i) < s_{k+1} \\ \phi(i)-1 & \phi(i) > s_{k+1} \end{cases}$$
This is an order preserving injection such that $\phi = d^{s_{k+1}}  \phi'$. Write $\phi'  = d^{s_k} \ldots d^{s_1}$ with $0 \le s_1 < \ldots < s_k \le n+k$. Thus $\phi = d^{s_{k+1}} d^{s_k} \ldots d^{s_1}$. If $s_k < s_{k+1}$ we set $j_r = s_r$ and are done. If $s_k \ge s_{k+1}$, we use 2. to get $\phi = d^{s_k+1} d^{s_{k+1}} d^{s_{k-1}} \ldots d^{s_1}$ where $s_{k+1} < s_k+1$. If $s_{k-1} \ge s_{k+1}$ we repeat this procedure and shift $d^{s_{k+1}}$ one position to the right. Continue this procedure until $s_r < s_{k+1}$ for some $r$ or until $d^{s_{k+1}}$ has been shifted to the last position. Then $\phi = d^{s_k+1} \ldots d^{s_{r+1}+1} d^{s_{k+1}} d^{s_r} \ldots d^{s_1}$ or $\phi = d^{s_k+1} \ldots d^{s_1+1} d^{s_{k+1}}$. By construction the sequence of upper induces is strictly increasing from right to left.