问:
Applicatin of Stokes' theorem to compute flux of $\nabla \times G$
I need a help on the following exercise:

Consider the following surface

$$S= \{(x,y,z): x^2+y^2+z^2 = 4, x^2+y^2 \leq 2, z>0 \}$$

Compute the flux of $\nabla \times G$ exiting from $S$, where $G(x,y,z)=(xy,zx,x^2+y^2)$

First of all, it's clear that $S$ can be parametrized with $\sigma(x,y)=(x,y,x^2+y^2)$, where $(x,y) \in B_{\sqrt{2}}(0,0)$. However, that surface integral is not so short to compute, so it's better to use Stokes' theorem and compute $$\int_{\partial S^+} G \cdot dr$$ where $\partial S$ is the boundary of the surface $S$.

The question is: what is the boundary of $S$, or better, how may it be parametrized? My guess is that it is parametrized by $r(t)=(\sqrt{2} \cos(t), \sqrt{2} \sin(t),\sqrt{2})$ with $t \in [0, 2 \pi)$. So, the integral to compute is $$\int_{\partial S} G \cdot dr = \int_0^{2 \pi} (\sin(2t),2 \cos(t),2) \cdot (-\sqrt{2} \sin(t), \sqrt{2} \cos(t),0) dt = \text{computations} = 0$$

Is everything correct?

答: the boundary of the surface $S$ is just the two dimentional ball in the xy plane which can be parametarized as $r(t)=(\sqrt{2} cos(t),\sqrt{2} sin(t),0)$ for $t\in [0,2\pi)$


问:
Determine $ \lim_{x\to 0} \frac{d}{dx} (\frac{\ln(1+x)}{x}) $
I want to determine this limit $$ \lim_{x\to 0} \frac{d}{dx} (\frac{\ln(1+x)}{x}) $$ but I can't find a way to solve it. At first I tried to put the limit inside the derivative: $$ \frac{d}{dx}(\lim_{x\to 0} \frac{\ln(1+x)}{x}) = 0 $$ which was wrong. Then I tried to solve $$\lim_{x\to 0} \frac{\frac{x}{1+x}-\ln(1+x)}{x^2}$$ but I don't know how to proceed.

答:
Since it's a $\dfrac{0}{0}$ form, you can use De l'Hôpital rule, hence you take the separate derivatives of numerator and denominator, and you get:

$$\lim_{x\to 0} \dfrac{-\frac{x}{(x+1)^2}}{2x} = \lim_{x\to 0} -\dfrac{1}{2(1+x)^2} = -\dfrac{1}{2}$$

ContentLicense: CC BY-SA 4.0


问:
Solve equation $\sin x + \cos x= i$
Could anyone help with following equation?

$$\sin(x) + \cos(x) = i$$

Solution either in C or R. First of all, sorry for dummy question. And even dummier followup. What is wrong with following reasoning?

$(\sin(x) + \cos(x))^2 = i^2$

$((\sin(x))^2 + (\cos(x))^2 + 2\sin(x)\cos(x) = -1$

$2\sin(x)\cos(x) = -2$

$\sin(2x) = -2$

Which is out of sine codomain ): ?

Really appreciate the help!

答:
To avoid spurious solutions from squaring the equation, proceed as follows. Note
$$\sin x+\cos x= \sqrt2\sin (x+\frac\pi4+2\pi n)=
i \sqrt2 \sinh[-i(x+\frac\pi4+2\pi n)]=i
$$
which leads to

$$-i(x+\frac\pi4+2\pi n)=\sinh^{-1}\frac1{\sqrt2}=\frac12\ln(2+\sqrt3)
$$
and the solutions
$$x = \frac i2\ln(2+\sqrt3)-\frac\pi4-2\pi n$$

Title: Solve equation $\sin x + \cos x= i$


问:
Geometric distribution and Conditional probability problem
I got this problem in my book.. I will directly go the last part which I could not solve.

$P(good condition) = 0.91854$ $P(bad condition)= 0.08146$

a random package check is checked by a company till there is a package in a bad condition .

what is the probability that the company will check exactly 4 packages?

2.it is known that the first 3 packages that were checked are in good condition , what is the probability that more than 8 packages will be checked(edit: the answer should be 0.654).

what I did was knowing that the last package (4th) has to be bad so it is $GGGB$ and the probability as stated for good is $P(good) = 0.91854$ and $P(bad)= 0.08146$ for bad we get $(0.91854)^3$ as in for 3 good (GGG) and one bad $0.08146$ so we multiply and get $(0.91854)^3$*$(0.08146)$= $0.063130$ which is right according to the book

from here I got that it is a geometric distribution so our function will be

$P(X=i)=(1-0.08146)^{i-1}$ *$0.08146$

$P(X=4)=(1-0.08146)^{4-1}$ *$0.08146$=$0.063130$

for the second part of the question

I didn't know how to use the Conditional probability I think that what is needed is $P(3<X≤8)$ but i got stuck.. cannot think of a way to use Geometric distribution or Conditional probability.

thanks for any tips and help much appreciated and sorry for the English mistakes it is hard to translate it , hope it is somehow understandable

答:
First note that Geometric distribution is a memoryless distribution thus the fact that you know that the first 3 packages are good changes your request in calculating

$$P(X>5)=1-P(X\leq 5)=0.91854^5\approx0.6539$$

To calculate it you can use the fact that the CDF of your rv is known...

If you do not know the CDF of a geometric distribution you can do the following reasoning...the probability to have more than 5 failures is exactly the probability of having 5 consecutive failures...after this events any event can happen....thus you probability is

$$0.91854\times0.91854\times0.91854\times0.91854\times0.91854\times 1\approx0.654$$


问:
Find the value of $\int_{0}^{\infty}\frac{x^3}{e^x-1}\ln(e^x - 1)\,dx$
I'm trying to figure out how to evaluate the following:
$$
J=\int_{0}^{\infty}\frac{x^3}{e^x-1}\ln(e^x - 1)\,dx
$$
I'm tried considering $I(s) = \int_{0}^{\infty}\frac{x^3}{(e^x-1)^s}\,dx\implies J=-I'(1)$, but I couldn't figure out what $I(s)$ was. My other idea was contour integration, but I'm not sure how to deal with the logarithm. Mathematica says that $J\approx24.307$. </p>

<p>I've asked a similar question and the answer involved $\zeta(s)$ so I suspect that this one will as well. </p>

答:
Using the change of variables $ u=e^{-x} $, we have

$$\int_{0}^{\infty}\frac{x^3}{e^x-1}\ln(e^x - 1)\,dx  = \int _{0}^{1}\!{\frac { \left( \ln  \left( u \right)  \right) ^{3}
  \ln  \left( 1-u \right)  }{u-1
}}{du}- \int _{0}^{1}\!{\frac { \left( \ln  \left( u \right)  \right)^{4}  }{u-1
}}{du}. $$

Now, just apply the technique which has been used to find the exact solution in this problem and the result will follow.

ContentLicense: CC BY-SA 3.0


问:
Characterizing open countable subsets of the Cantor set
The Cantor-Bendixson theorem implies that any closed subset of the Cantor set $\mathcal{C}$ can be described as a disjoint union of a set $\mathcal{C}_c$ that is homeomorphic to the original Cantor set, and a countable open set $\mathcal{C}_o$.

The following answer, and the referenced work by Schoenberg & Grunhage therein, implies that all noncompact open subsets of the Cantor set are homeomorphic to the Cantor set minus a point—say, $\mathcal{C}/ \{ 0\}$. But this would mean that $\mathcal{C}_o$ is homeomorphic to $\mathcal{C}/ \{ 0\}$, which would imply the Cantor set minus a point is countable, which seems strange.

Is this true, or am I missing something?

答: The "countable open set $\mathcal C_o$" is not open as a subset of the cantor set $\mathcal C$, it is open as a subset of the set that can be written as $\mathcal C_c \sqcup \mathcal C_o$. Non-empty open subsets of Cantor space $\mathcal C$ are never countable.


问:
Find $\frac{a+b}{ab}$ such that $\int_{-1/2}^{1/2} \cos x\ln\frac{1+ax}{1+bx}dx=0$
Let $f(x) = \cos(x) \ln\left(\frac{1+ax}{1+bx}\right)$ be integrable on $\left[-\frac{1}{2} , \frac{1}{2}\right]$. Let
$$\displaystyle \int_{-1/2}^{1/2}f(x)\operatorname{dx}=0$$

where $a$ and $b$ are real numbers and not equal. Find the value of $\frac{a+b}{a\cdot b}$.

At first I applied integration by parts but then it got even more complicated and I couldn't solve it. How can I solve this problem?

答:
For a start, note that when $a=-b$ we get $f(x) = -f(-x)$ since $\cos(x)$ is even and $\ln(\frac{1+ax}{1-ax}) = -\ln(\frac{1-ax}{1+ax})$. Thus, by symmetry of the integration domain, the integral evaluates to $0$. The value of $\frac{a+b}{ab}$ would therefore be zero.

Proving the opposite, i.e. that we must have $a=-b$, uses a similar idea. Note that
$$\int_{-1/2} ^{1/2} f(x) \mbox{dx} = 0 \quad \implies \int_{-1/2} ^{1/2} [f(x) + f(-x)] \mbox{dx} = 0.$$
You can easily see that the integrand on the right equals
$$ \cos(x) \ln\bigg(\frac{1-a^2x^2}{1-b^2x^2}\bigg)$$
by using that $\cos(x)$ is even and $\ln(u) + \ln(v) = \ln(uv)$. Now comes the nice part: if $a^2 \neq b^2$, then the logarithm is either always positive (when defined) or always negative, because its argument is either $<1$ or $>1$. But the cosine is always positive, so then the integral would not evaluate to zero. As such, we require $a^2 = b^2$, or $a=-b$.

Small comment: the integrability condition ensures that neither of the expressions $1-a^2x^2$, $1-b^2x^2$ can change sign on the domain.


问:
How to prove if the relation R is an equivalence relation?
Hello so I stumbled upon an exercise where I need to prove that R is an equivalence relation, normally I have no problem doing this, but with this exercise I have absolutely no clue where to begin. I think my biggest problem is understanding what's given. The exercise goes as follows:

$X$ and $Y$ are two sets. We notate $Fun(X,Y)$ is the set of all functions $f:X\rightarrow Y$. Let $R$ be the relation on $Fun(X,Y)$ given by: $(f,g) \in R$ if and only if there exist a bijective function $\sigma:Y \rightarrow Y$ with $\sigma \circ f =g $.

Prove that $R$ is an equivalence relation.

What I know

$R$ is an equivalence relation when $R$ is reflexive, symmetric and transitive (so this is what I need to prove).

$R$ is reflexive if for all $x\in X : xRx$

$R$ is symmetric if for all $x,y\in X : xRy \rightarrow yRx$

$R$ is transitive if for all $x,y,z \in X: xRy \text{ and } yRx \rightarrow xRy$

If someone could help me, it would be very kind of you :)

Thank you in advance

答:
Reflexivity

For all $f\in Fun(X,Y)$ we have that $(f,f)\in R$ since $Id \circ f=f$ with $Id:Y\to Y:y\mapsto y$ the identity operator.

Symmetric

Let $(f,g)\in R$. Then there exists a bijection $\sigma:Y\to Y$ with $\sigma\circ f=g$. Then $\sigma^{-1} \circ g=f$ and since $\sigma^{-1}$ is a bijection we have $(g,f)\in R$.

Transitivity

Let $(f,g)\in R$ and $(g,h)\in R$. Then there are bijections $\sigma ,\lambda:Y\to Y$ with  $\sigma \circ f=g$ and $\lambda \circ  g=h$. Then $\lambda \circ \sigma \circ f=h$ and since a composition of bijection is bijective it holds that $(f,h)\in R$.


问:
Let's assume 1 kilogram is equal to 2.2 pounds. But I got: 1 : 2.2 = lb : kg
Let's assume 1 kilogram is equal to 2.2 pounds.

$1 kg = 2.2 lb$

So we get:

$1/2.2 = lb/kg$

Which means:

$1 : 2.2 = lb : kg$

So if we convert 5 kg to lb and y stands for the pound number, we get:

$1 : 2.2 = ylb : 5kg$

$5kg= y*2.2*lb$

$5(2.2lb)= y*2.2*lb$

$y = 5$

It is really wierd because we all know that it should have been

$kg : lb = 1 : 2.2$

Is there a lack of thinking from me?

答:
Reconsider what it means for an object $\star$ to have a mass $5 kg$?

$$\begin{align*}
\text{mass of }\star &= 5\cdot \text{mass of something of }1kg\\
\frac{\text{mass of }\star}{\text{mass of something of }1kg} &= 5
\end{align*}$$

And what it means for 1 kilogram to be equal to 2.2 pounds?

$$\begin{align*}
\text{mass of something of }1kg &= 2.2\cdot\text{mass of something of }1lb\\
\frac{\text{mass of something of }1kg}{\text{mass of something of }1lb} &= 2.2\\
\end{align*}$$

So if you know $y\ lb$ and $5kg$ both represent the mass of object $\star$, then

$$\begin{align*}
y &= \frac{\text{mass of }\star}{\text{mass of something of }1lb}\\
&= \frac{\text{mass of }\star}{\text{mass of something of }1kg} \cdot \frac{\text{mass of something of }1kg}{\text{mass of something of }1lb}\\
&= 5 \cdot 2.2
\end{align*}$$

You can say one of the followings:

$$\begin{align*}
y:5 &= 2.2:1\\
y:5 &= kg:lb\\
y\ lb: 5kg &= 1:1
\end{align*}$$


问:
tilde notation for random variables
what does $X,Y \sim U(\{1,\dots ,n\})$ mean, where $X$ and $Y$ are independent random variables.

Does it just mean that $X$ and $Y$ are defined on $[1,n]$?

答: I suppose that means that these two random variables have a uniform distribution over integers $1, \dots, n$.

Title: tilde notation for random variables


问:
Questions on fractional Laplacian graph spectra
Both the signed ($D-A$) and unsigned ($D+A$) Laplacian are of interest in spectral graph theory, see eg Cvetkovic: Bibliography on the signless Laplacian eigenvalues: first one hundred references.

Considering the spectral function $D+\rho A$ over the interval $\rho \in [-1,1]$ rather than just the extreme values $\rho={-1,1}$, results in the curves $\lambda_i(\rho)$.

For example, the following fractional spectra are are shown as point interpolations (as opposed to curve following, as in bifurcation theory) corresponding to some random Bernoulli graphs with increasing connectivity:

These experimental results raise basic questions:

If graphs are $M$-cospectral on $\rho = {-1,1}$ are they cospectral for $\rho \in [-1,1]$?

Are there no intersections of $\lambda_i(\rho)$ outside $\rho \in [-1,1]$?

Does $\lim_{\lambda_i \to \infty}\frac{d\lambda_i}{d\rho} = const?$

What's the combinatorial interpretation of intersections in $\rho \in [-1,1]$?

Feel free to add your own conjectures.

答:
Let G0 be the graph with edge set
$$
  (0, 1), (0, 7), (0, 8), (1, 2), (2, 3), (2, 5), (2, 7), (2, 8), (3, 4), (3, 6), (4, 5), (5, 6), (6, 7)
$$
and let G1 be the graph with edge set
$$
  (0, 1), (0, 3), (0, 5), (0, 8), (1, 2), (1, 4), (2, 3), (2, 8), (2, 9), (3, 4), (4, 5), (4, 8), (6, 7)
$$
(Neither graph is connected and G0 has an isolated vertex.)
The Laplacian matrices of these two graphs are cospectral. As these graphs
are bipartite, their signless Laplacians are also cospectral.
However the respective degree sequences are
$$
  [0, 2, 2, 2, 3, 3, 3, 3, 3, 5],  [1, 1, 1, 2, 3, 3, 3, 4, 4, 4]
$$
and so the answer to your first question is no.</p>


问:
convergenge of Hausdorff dimension for a sequence of compact sets
Recall that

the Hausdorff distance (metric) is:

A metric in the space of subsets of a compact set $K$, defined as follows. Let $X,Y\subset K$ and let $D_{x,y}$ be the set of all numbers $\rho(x,Y)$ and $\rho(y,X)$ where $x\in X$, $y\in Y$ and $\rho$ is a metric in $K$. Then the Hausdorff metric $\operatorname{dist}(X,Y)$ is the least upper bound of the numbers in $D_{x,y}$.

the Hausdorff dimension is:
Let $(X,d)$ be a metric space. In what follows, for any subset $E\subset X$,
$\operatorname{diam}(E)$ will denote the diameter of $E$.

For any $E\subset X$, any $\delta \in ]0, \infty]$ and any $\alpha\in [0, \infty[$ we consider the outer measure
$$
\mathcal{H}^\alpha_\delta (E) :=
\inf \left\{ \sum_{i=1}^\infty (\operatorname{[diam}\, E_i)^\alpha :
E\subset \bigcup_i E_i \quad\text{and}\quad \operatorname{ diam}(E_i)< \delta\right\}.
$$
The map $\delta\mapsto \mathcal{H}^\alpha_\delta (E)$ is monotone nonincreasing and thus we can define the Hausdorff measure $\alpha$-dimensional measure of $E$ as

$$
\mathcal{H}^\alpha (E) := \lim_{\delta\downarrow 0} \mathcal{H}^\alpha_\delta (E)
$$
The map $\delta\mapsto \mathcal{H}^\alpha_\delta (E)$ is monotone nonincreasing and thus we can define the  Hausdorff α-dimensional
measure of E as
$$
\mathcal{H}^\alpha (E) := \lim_{\delta\downarrow 0} \mathcal{H}^\alpha_\delta (E).
$$

I wonder if the following is true:  If $A_n \to A$ in the Hausdorff distance, then $\operatorname{dim}_H A_n \to \operatorname{dim}_H A$, where $A$ is a compact subset of $[0,1]^2$.

答:
The conjectured result does not hold.

For $j=1,2,$ define the maps $\varphi_j : [0,1] \to [0,1]$ by
$$\varphi_1(x) = \frac{1}{3}x \qquad\text{and}\qquad \varphi_2(x) = \frac{1}{3}x + \frac{2}{3}. $$  Let $A_0 = [0,1]$, and recursively define $A_n$ by
$$ A_n = \varphi_1(A_{n-1}) \cup \varphi_2(A_{n-1}). $$
Thus, working out the first few sets in the sequence:
\begin{align}
A_0 &= [0,1] \\
A_1 &= [0,1/3] \cup [2/3,1] \\
A_2 &= [0,1/9] \cup [2/9, 1/3] \cup [2/3,7/9] \cup [8/9,1],
\end{align}
and so on.  In general, the set $A_n$ will consist of $2^n$ closed intervals, each having length $3^{-n}$.  As each $A_n$ is a finite union of compact sets, each $A_n$ is, itself, compact.

Moreover, it can be shown that $A_n \to A$ in the Hausdorff metric, where $A$ is the usual two-thirds Cantor set (a proof of this follows by noting that the map of sets defined by $\Phi(E) = \varphi_1(E) \cup \varphi_2(E)$ is a contraction mapping on the space of compact subsets of $\mathbb{R}$ with respect to the Hausdorff distance; the claimed result then follows from the Banach fixed point theorem; see Hutchinson's 1981 paper for details).

Now, note that $\operatorname{dim}_H(A_n) = 1$ for each $n$ (where $\operatorname{dim}_H$ denotes the Hausdorff dimension)—that intervals are one-dimensional is a fairly standard result.  On the other hand, $\operatorname{dim}_H(A) = \log_3(2)$—again, I would refer you to Hutchinson's 1981 paper for a proof of this result, though, again, this is a fairly standard result.  Therefore
$$ A_n \overset{d_H}{\longrightarrow} A
\qquad\text{but}\qquad
1 = \operatorname{dim}_H(A_n) \not\to \operatorname{dim}(A) = \log_3(2). $$

Note: I am working with subsets of $[0,1]$.  However, we can regard these sets as subsets of $[0,1]^2$ with no modifications.  Alternatively, we might consider a similar setup with four maps $\varphi_j : [0,1]^2 \to [0,1]^2$ defined by
\begin{align}
\varphi_1(x) &= \frac{1}{3} x, &
\varphi_2(x) &= \frac{1}{3} x + \begin{pmatrix} \frac{2}{3} \\ 0 \end{pmatrix}, \\
\varphi_3(x) &= \frac{1}{3} x + \begin{pmatrix} 0 \\ \frac{2}{3}\end{pmatrix}, &
\varphi_4(x) &= \frac{1}{3} x + \begin{pmatrix} \frac{2}{3} \\ \frac{2}{3} \end{pmatrix}. \\
\end{align}
In this case, take $A_0 = [0,1]^2$ and
$$ A_n = \bigcup_{j=1}^{4} \varphi_j(A_{n-1}). $$
The approximating sets have Hausdorff dimension $2$, while the limiting set is a two-dimensional Cantor dust, which has Hausdorff dimension $\log_3(4)$.


问:
Different ideal and Kähler differentials
Let $A$ be a Dedekind domain with fraction field $K$, $L/K$ be a finite separable extension and $B$ be the integral closure of $A$ in $L$, which is also a Dedekind domain. Assume all residue extensions are separable, e.g. if $A=\mathbb{Z}$.

Let $\delta_{L/K}$ be the different ideal, which is the inverse fractional ideal of the trace-dual of $B$. We can compute the different as the annihilator of the module of Kähler differentials of $B$ over $A$. So the zero locus of the different ideal is the set of points on which the Kähler differentials vanish.

But we also know that those primes are the primes that are ramified over some prime in $A$. And if we draw $\operatorname{Spec}(A)$ as a horizontal line (call it $x$-axis) and project down the regular curve $\operatorname{Spec}(B)$, those primes are the points with vertical on which the curve has vertical slope. "The $x$-coordinate of the tangent vector vanishes", and Dedekin's central theorem regarding different and ramification tells us moreover that it vanishes with the right order.

Is everything I have said so far correct? If so, is there any relation between the Kähler differentials and the $x$-coordinate of the "tangent vector" to the curve $B$? And if this is the case, how does one connect the definition of the different with the trace-dual and the inverse fractional ideal to this geometric interpretation?

Points where I am stuck: I don't know how to interpret the trace pairing in a geometric way, so I don't know how to interpret trace-dual ideals in a geometric way.

Extra (secondary) question: $B$ may not be generated by a single element over $A$, but $\Omega_{B/A}$ can always be generated by a single element. What is the intuition behind this? I get that if $B=A[\alpha]$ then $\Omega_{B/A}$ is generated by $d\alpha$, but if this is not the case, who is the generator and why can we expect to have a single generator?

答:
If $f:X\rightarrow Y$ is a flat morphism of schemes of finite type over a field $k$,
it follows $f$ is etale iff $\Omega^1_{X/Y}=0$. If a similar statement is valid for
for the relative situation you get something similar in your situation.

Definition. We say the morphism $f$ is unramified at $x$ iff $\mathfrak{m}_y\mathcal{O}_{X,x}=\mathfrak{m}_x$ and $\kappa(y) \subseteq \kappa(x)$ is separable. We may define $f$ to be "etale" iff it is flat and unramified at all points.

If $A:=\mathcal{O}_K, B:=\mathcal{O}_L$ with $K \subseteq L$ a finite extension of number fields it follows any residue field extension $\mathcal{O}_K/\mathfrak{p} \subseteq \mathcal{O}_L/\mathfrak{q}$ is separable. If $Y:=Spec(A), X:=Spec(B)$ and $f:X\rightarrow Y$ is the canonical map,
it follows the support $Z$ of $\Omega^1_{X/Y}$ is a closed subscheme defined by the different $\delta_{L/K}$. The map $f$ is finite and hence $Y-f(Z):=U$ is an open subscheme of $Y$. The induced map $f_U: V:=f^{-1}(U) \rightarrow U$ is flat and $\Omega^1_{V/U}=0$ by definition. Hence if a relative version of Exercise III.10.3 in Hartshorne is true, it follows $f_U$ is etale.

If $\mathfrak{p}\in U$ is a prime where the fiber $f^{-1}(\mathfrak{p})$ is unramified, we get an isomorphism

E1. $f^{-1}(\mathfrak{p})\cong Spec(B/\mathfrak{p}B)$

where $B/\mathfrak{p}B\cong B/\mathfrak{p}_1\cdots \mathfrak{p}_d$

where $\mathfrak{p}_i \neq \mathfrak{p}_j$ are distinct maximal ideals in $B$. It follows

E2 $B/\mathfrak{p}B\cong B/\mathfrak{p}_1\oplus \cdots \oplus B/\mathfrak{p}_d:=R_1\oplus \cdots \oplus R_d$

Let $B_1:=B/\mathfrak{p}B$ and $A_1:=A/\mathfrak{p}$.

We get

$\Omega^1_{B_1/A_1}\cong \Omega^1_{R_1/A_1}\oplus \cdots \oplus \Omega^1_{B_d/A_1}=0$ since the field extensions $A_1 \subseteq B_i$ are separable.

Question: "Is everything I have said so far correct? If so, is there any relation between the Kähler differentials and the x-coordinate of the "tangent vector" to the curve B? And if this is the case, how does one connect the definition of the different with the trace-dual and the inverse fractional ideal to this geometric interpretation?"

The support $Z \subseteq X$ of the module of Kahler differentials is canonically determined by the module of Kahler differentials and the image $f(Z)\subseteq Y$ is canonically a closed subscheme of $Y$ determined by an ideal $I$ in $A$. The discriminant $D(L/K)$ in $A$ is an ideal giving a characterization of primes in $A$ that ramify in $B$. Hence the two ideals $D(L/K)$ and $I$ have the same support. Maybe they are the same ideal. A problem is that if $\mathfrak{p}\in V(D(L/K))$ it follows the fiber $\pi^{-1}(\mathfrak{p})$ is "ramified", but not all closed points $\mathfrak{q}\in \pi^{-1}(\mathfrak{p})$ need to be ramified.

Ref. your other question on the "geometry" of the map $f$:

Geometric interpretation of ramification of prime ideals.


问:
Are there any exceptions to the power rule?
I solved for the derivative of $f(x)=\sqrt{e-2x}$ and ended up with $f'(x)= \frac{-1}{\sqrt{e-2x}}$ which I believe is correct.

Where I'm confused is when I apply the power rule it doesn't seem to work. The derivative of $x^n$ should be $nx^{n-1}$. When I do $\frac{1}{2}*\sqrt{e-2x}^{-1/2}$, I end up with $\frac{1}{2\sqrt{e-2x}}$ which does not seem to equal  $f'(x)= \frac{-1}{\sqrt{e-2x}}$.

Are there any exceptions to this power rule or did I mess up in any calculations?

答:
As @FormerMath points out, you missed the chain rule.
\begin{align}
& \frac{d}{dx} \sqrt{e-2x} \\
=& \frac12 (e-2x)^{-1/2} \frac{d}{dx} (e - 2x) \tag{It's '()' instead of square root} \\
=& \frac12 (e-2x)^{-1/2} (-2) \\
=& -\frac{1}{\sqrt{e-2x}}
\end{align}


Title: Determine $ \lim_{x\to 0} \frac{d}{dx} (\frac{\ln(1+x)}{x}) $

Body: I want to determine this limit $$ \lim_{x\to 0} \frac{d}{dx} (\frac{\ln(1+x)}{x}) $$ but I can't find a way to solve it. At first I tried to put the limit inside the derivative: $$ \frac{d}{dx}(\lim_{x\to 0} \frac{\ln(1+x)}{x}) = 0 $$ which was wrong. Then I tried to solve $$\lim_{x\to 0} \frac{\frac{x}{1+x}-\ln(1+x)}{x^2}$$ but I don't know how to proceed.

Body:
$$L=\lim_0\frac{\frac{x}{x+1}-\ln(1+x)}{x^2}=$$

$$\lim_0\frac{x-(1+x)\ln(1+x)}{x^2(1+x)}=$$

$$\lim_0\frac{x-(1+x)\ln(1+x)}{x^2}$$

but

$$\ln(1+x)=x-\frac{x^2}{2}(1+\epsilon(x))$$

and
$$(1+x)\ln(1+x)=$$
$$x-\frac{x^2}{2}+x^2(1+\epsilon(x))$$

thus
$$L=\lim_0\frac{-\frac{x^2}{2}+x^2\epsilon(x)}{x^2}=-\frac 12$$


问:
linear Transformation of polynomial with degrees less than or equal to 2
I would like to determine if the following map $T$ is a linear transformation:

\begin{align*}
T: P_{2} &\to P_{2}\\
A_{0} + A_{1}x + A_{2}x^{2} &\mapsto A_{0} + A_{1}(x+1) + A_{2}(x+1)^{2}
\end{align*}

My attempt at solving:

\begin{align}
T(p + q) &= p(x+1) + q(x+1)\\
&= \left[A_{0} + A_{1}(x+1) + A_{2}(x+1)^2\right] + \left[b_{0} + b_{1}(x+1) + b_{2}(x+1)^2\right]\\
&= \left(A_{0} + b_{0}\right) + \left(A_{1} + b_{1}\right)(x+1) + \left(A_{2} + b_{2}\right)(x+1)^2\\
&= T(p) + T(q)
\end{align}

Is this right so far? If not, what am I doing wrong?

答:
The idea is right but you are starting from  what you have to proof

Hint : $T(p+q)=T(A_0+B_0+(A_1+B_1)x+(A_2+B_2)x^2)=...=T(p)+T(q)$.

I see that you had the right idea maybe this could make you start


问:
On Closed maps between metric spaces
For any two metric spaces $(X, d_{X})$ and $(Y, d_Y)$, a map $f:X \to Y$ is said to be a closed map if whenever $F$ is closed in $X$ , then $f(F)$ is closed in $Y$ . For any subset $B$ of a metric space, $B$ is given the induced metric.The metric on $X×Y$ is given by
$d\{(x,y), (x',y')\}=\max \{d_X(x,x'), d_Y(y,y')\}  $. Which of the following are true ?

$(a)$ For any subset $A\subseteq X$ , the inclusion map $i : A \to X$ is closed map.

$(b)$ The projection map $p_1 : X×Y \to X$ given by $p_1(x,y)=x$ is closed.

$(c)$ Suppose that $f:X\to Y, g:Y\to Z$ are continous maps . If $g\circ f: X\to Z$ is closed map then $g|_{f(X)} : f(X) \to Z$ is closed map . Here $g|_{f(X)}$ means the restriction of $g$ on $f(X)$

$(d)$ If $f:X \to Y$ takes closed balls into closed sets , then $f$ is closed .

My Thinking :-

$(a)$ Let $A$ be an open subset of $X$.

Then $A$ with the induced metric is closed in itself but
$i(A)=A$ is open in $X$.

So it is false.

$(b)$ I think this is true but  can't prove.

Let $B\subseteq X×Y$ be a closed subset . Then $\overline{B}=B$

Let $y\in \{ p_1(B)\} '$( The derived set) .

Then there exist a sequence $\{p_1(x_n)\} \in  p_1(B) $ such that $\displaystyle \lim_{n\to \infty} p_1(x_n)=y$

Since $p_1$ is continous with respect to the product metric and so by sequential criterion of continuity , it follows that

$\displaystyle \lim_{n\to \infty} p_1(x_n)=p_1(\displaystyle \lim_{n\to \infty}x_n)$ (But the problem here is that $\{x_n\}$ may not be convergent !)

I can't proceed after this.

I have no idea for $(c)$ and $(d)$.

Please give hints.  Thank you.

答:
Some hints:

For b) and d), think of the projection map from $\mathbb{R}^{2}$ to $\mathbb{R}$.

For c), try to use the fact that if $f: X \to Y$ is a continuous function and $A \subset Y$ is closed, then $f^{-1}(A)$ is closed in $X$.

Title: On Closed maps between metric spaces


答:
Although @DF has already published his/her answer, I'm still writing one in response to your question.

Does it just mean that $X$ and $Y$ are defined on $[1,n]$?

No, note the difference between $\{1,\dots,n\}$ and $[1,n]$.

set
$\{1,\dots,n\}$
$[1,n]$

in words
set of integers from $1$ to $n$
closed interval from $1$ to $n$

finite set?
yes
no

Title: tilde notation for random variables

Body:
what does $X,Y \sim U(\{1,\dots ,n\})$ mean, where $X$ and $Y$ are independent random variables.

Does it just mean that $X$ and $Y$ are defined on $[1,n]$?


问:
Question about equivalent classes.

I have no idea what going on in this question, nor how to approach it.
From messing around with this question I found 14 pairs (i.e. (1,2),(2,1) is considered one pair) that I can add to the original set. But surely it cannot be right? Because then my options will be astronomically large..

答:
Recall that a partition defines an equivalence relation and vice versa.  We may look instead at the possible partitions.

Being told that it contains $(6,6)$ is irrelevant as every equivalence relation will have this per reflexivity.

We are told that $7,4,2$ must be in the same equivalence class (since (7,4) and (4,2) are required elements).  We are told that $5$ must be in a different equivalence class than $4$ (since (5,4) is a required nonelement).  We are told that $3$ must be in a different equivalence class than both $7$ and $5$ (since (7,3) and (3,5) are required nonelements).  We are told that $1$ must be in the same equivalence class as $3$ (since (1,3) is a required element).

So... we know that there are at least three equivalence classes, $\{1,3\},\{2,4,7\},\{5\}$.  The only question is then for the remaining number $6$... whether it goes into one of these already existing equivalence classes or if it forms its own equivalence class by itself.  There are merely $4$ options.

Title: Question about equivalent classes.


问:
Let $(\Bbb R,*)$ be such that $x*y = |x|y$ for all $x,y \in \Bbb R$. Show that $(\Bbb R,*)$ is not a group.
Let $(\Bbb R,*)$ be the mathematical system defined as: $x*y = |x|y$ for all $x,y \in \Bbb R$. Show that $(\Bbb R,*)$ is not a group.

Attempt:
I have shown the assosiative property.
Now, I have to find an identity element in $(\Bbb R,*)$.

Let $m \in \Bbb R$ be the identity element in $(\Bbb R,*)$. Let $y \in \Bbb R$. Then,
\begin{align*}
m*y &= y \\
|m|y &= y \\
m=1 &\vee m=-1
\end{align*}
and
\begin{align*}
y*m &= y \\
|y|m &= y \ldots (k)
\end{align*}
If $y \ge 0$, then $(k)$ becomes
$ym=y$ i.e. $m=1$.

If $y < 0$, then $(k)$ becomes
$-ym=y$ i.e. $m=-1$.

Thus, in two cases, we see that $m$ have two values. But, the identity element of a group must be unique. Hence, $(\Bbb R,*)$ has no an identity element i.e. is not a group. In particularly, $(\Bbb R,*)$ is a semigroup.

Is the above true?

答:
Note that each group $G$ has exactly one idempotent, for if $x\in G$ such that $x^2=x$, then $x=x^{-1}x^2=x^{-1}x=e$.

But in $(\Bbb R,\ast)$, we have $(-1)\ast(-1)=|-1|(-1)=-1$ and $1\ast 1=|1|(1)=1$.

Hence $(\Bbb R,\ast)$ is not a group.


答:
Note that you're hiding the core of the proof, which is simply that if $p,q$ are polynomials, we define their sum, which is also a polynomial, and $(p+q)(x)=\sum_{i=1}^n (a_i+b_i)x^i=p(x)+q(x)$.

Thus, if $T$ is your transformation, $$T((p+q)(x))=(p+q)(x+1)=p(x+1)+q(x+1)=T(p(x))+T(q(x))$$ and $$T(\lambda p(x))=(\lambda p)(x+1)=\lambda p(x+1)=\lambda T(p(x))$$

Maybe it is easier to see why the transformation is linear by seeing how it acts on the coefficients of $P$. If $P$ is $(a_0,a_1,a_2)=a_0+a_1x+a_2x^2$ then $$T(P)=(a_0+a_1+a_2,a_1+2a_2,a_2)$$ because $$\begin{align}a_0+a_1(x+1)+a_2(x+1)^2&=a_0+a_1x+a_1+a_2x^2+2a_2x+a_2\\&=(a_0+a_1+a_2)+(a_1+2a_2)+a_2x^2\end{align}$$

Title: linear Transformation of polynomial with degrees less than or equal to 2

Body:
I would like to determine if the following map $T$ is a linear transformation:

\begin{align*}
T: P_{2} &\to P_{2}\\
A_{0} + A_{1}x + A_{2}x^{2} &\mapsto A_{0} + A_{1}(x+1) + A_{2}(x+1)^{2}
\end{align*}

My attempt at solving:

\begin{align}
T(p + q) &= p(x+1) + q(x+1)\\
&= \left[A_{0} + A_{1}(x+1) + A_{2}(x+1)^2\right] + \left[b_{0} + b_{1}(x+1) + b_{2}(x+1)^2\right]\\
&= \left(A_{0} + b_{0}\right) + \left(A_{1} + b_{1}\right)(x+1) + \left(A_{2} + b_{2}\right)(x+1)^2\\
&= T(p) + T(q)
\end{align}

Is this right so far? If not, what am I doing wrong?


答:
a) Your counterexample is a good idea, but is too general. Being closed and open are not mutually disjoint properties, even for nonempty proper subsets. For instance, in $(0,1)\cup (2,3)$ with the subspace metric, $(0,1)$ is open and closed. In many examples, this will not work. For instance, take the specific inclusion $(0,1)\subseteq \mathbb R$. This is not a closed map as $(0,1)$ is closed on itself but not in $\mathbb R$

b) This is actually false. Consider $X=Y=\mathbb R$. Let $A = \{(x,y)\in \mathbb R^2 : xy=1\}$. This is the graph of $y=1/x$, so it is closed. However, $p_1[A] = \mathbb R-\{0\}$, which is not closed in $\mathbb R$.

c) Let $A\subseteq f[X]$ be closed. Then $A= f[f^{-1}[A]]$. As $f$ is continuous, $f^{-1}[A]$ is closed in $X$. Hence, $g[A]= g[f[f^{-1}[A]]]=(g \circ f)[f^{-1}[A]]$ is closed, as $g \circ f$ is closed by assumption.

d) The same counterexample as in part a) works. A closed ball in $(0,1)$ is just a closed interval $[a,b]\subseteq (0,1)$, which is certainly still closed in $\mathbb R$. However, as discussed in a) this inclusion map is not closed.

Title: On Closed maps between metric spaces

Body:
For any two metric spaces $(X, d_{X})$ and $(Y, d_Y)$, a map $f:X \to Y$ is said to be a closed map if whenever $F$ is closed in $X$ , then $f(F)$ is closed in $Y$ . For any subset $B$ of a metric space, $B$ is given the induced metric.The metric on $X×Y$ is given by
$d\{(x,y), (x',y')\}=\max \{d_X(x,x'), d_Y(y,y')\}  $. Which of the following are true ?

$(a)$ For any subset $A\subseteq X$ , the inclusion map $i : A \to X$ is closed map.

$(b)$ The projection map $p_1 : X×Y \to X$ given by $p_1(x,y)=x$ is closed.

$(c)$ Suppose that $f:X\to Y, g:Y\to Z$ are continous maps . If $g\circ f: X\to Z$ is closed map then $g|_{f(X)} : f(X) \to Z$ is closed map . Here $g|_{f(X)}$ means the restriction of $g$ on $f(X)$

$(d)$ If $f:X \to Y$ takes closed balls into closed sets , then $f$ is closed .

My Thinking :-

$(a)$ Let $A$ be an open subset of $X$.

Then $A$ with the induced metric is closed in itself but
$i(A)=A$ is open in $X$.

So it is false.

$(b)$ I think this is true but  can't prove.

Let $B\subseteq X×Y$ be a closed subset . Then $\overline{B}=B$

Let $y\in \{ p_1(B)\} '$( The derived set) .

Then there exist a sequence $\{p_1(x_n)\} \in  p_1(B) $ such that $\displaystyle \lim_{n\to \infty} p_1(x_n)=y$

Since $p_1$ is continous with respect to the product metric and so by sequential criterion of continuity , it follows that

$\displaystyle \lim_{n\to \infty} p_1(x_n)=p_1(\displaystyle \lim_{n\to \infty}x_n)$ (But the problem here is that $\{x_n\}$ may not be convergent !)

I can't proceed after this.

I have no idea for $(c)$ and $(d)$.

Please give hints.  Thank you.


问:
Determine $ \lim_{x\to 0} \frac{d}{dx} (\frac{\ln(1+x)}{x}) $
I want to determine this limit $$ \lim_{x\to 0} \frac{d}{dx} (\frac{\ln(1+x)}{x}) $$ but I can't find a way to solve it. At first I tried to put the limit inside the derivative: $$ \frac{d}{dx}(\lim_{x\to 0} \frac{\ln(1+x)}{x}) = 0 $$ which was wrong. Then I tried to solve $$\lim_{x\to 0} \frac{\frac{x}{1+x}-\ln(1+x)}{x^2}$$ but I don't know how to proceed.

答:
If you do not want to use L'hopital or series expansion, then I suggest the following :

Notice that : $$ \frac{\frac{x}{1+x}-\ln{\left(1+x\right)}}{x^{2}}=\frac{1}{1+x}\left(\frac{x-\ln{\left(1+x\right)}}{x^{2}}-\frac{\ln{\left(1+x\right)}}{x}\right) $$

You may recognize the well-known limit $ \lim\limits_{x\to 0}{\frac{x-\ln{\left(1+x\right)}}{x^{2}}} $, that, we'll be proving, equals $ \frac{1}{2} $ :

Let $ x\in\left(-1,1\right)\setminus\left\lbrace 0\right\rbrace $, note that : $$ \frac{x-\ln{\left(1+x\right)}}{x^{2}}=\int_{0}^{1}{\frac{1-y}{\left(1+xy\right)^{2}}\,\mathrm{d}y}=\frac{1}{2}-x\int_{0}^{1}{\frac{y\left(2+xy\right)}{\left(1+xy\right)^{2}}\,\mathrm{d}y} $$

Since : \begin{aligned} \left|x\int_{0}^{1}{\frac{y\left(2+xy\right)}{\left(1+xy\right)^{2}}\,\mathrm{d}y}\right|&\leq\left|x\right|\int_{0}^{1}{\frac{y\left(2+\left|x\right|y\right)}{\left(1-\left|x\right|y\right)^{2}}\,\mathrm{d}y}\\ &\leq\frac{\left|x\right|\left(2+\left|x\right|\right)}{2\left(1-\left|x\right|\right)^{2}}\underset{x\to 0}{\longrightarrow}0 \end{aligned}

Then : $$ \lim_{x\to 0}{\frac{x-\ln{\left(1+x\right)}}{x^{2}}}=\frac{1}{2} $$

Thus : \begin{aligned}\lim_{x\to 0}{\frac{\mathrm{d}}{\mathrm{d}x}\left(x\mapsto\frac{\ln{\left(1+x\right)}}{x}\right)\left(x\right)}&=\lim_{x\to 0}{\frac{1}{1+x}\left(\frac{x-\ln{\left(1+x\right)}}{x^{2}}-\frac{\ln{\left(1+x\right)}}{x}\right)}\\ &=1\times\left(\frac{1}{2}-1\right)\\&=-\frac{1}{2}\end{aligned}

$\underline{\textrm{Note :}} $ You may have noticed that I prefer to use the notation $ \frac{\mathrm{d}f}{\mathrm{d}x}\left(x\right) =f'\left(x\right) $, rather than $ \frac{\mathrm{d}f\left(x\right)}{\mathrm{d}x}=\left(f\left(x\right)\right)' $, which in my mind, doesn't stand for the derivative of $ f $ at $ x $, but stands for the derivative of the image of $ x $ by $ f $, which sould be $ 0 $.


问:
Given a finite number of stones. We place every stone on an integer. Prove that, given different movements, we can only make finite number of moves
Given a finite number of stones. We place every stone on an integer (a number $x$ where $x\in Z$) and maybe multiple stones on an integer.

On each move we can make one of the following movements:

Remove a stone from numbers $n-1$ and $n$ and place one stone in integer $n+1$.

Remove $2$ stones from $n$ and place them in integers $n+1$ and $n-2$.

Prove that we can only make finite number moves.

I attempted to do it as follows:

On each stone we give it the weight $x^n$ where $x^2-x-1=0$, hence $x=\frac{1+\sqrt{5}}{2}$.

Hence we have that $x^{n-1}(x^2-x-1)=0$ and $x^{n+1}+x^{n-2}-2x^n=x^{n-2}(x^3+1-2x^2)=x^{n-2}(x^2+x+1-2x^2)=x^{n-2}(x+1-x^2)=0$.

Which proves that the total weight of the stones is an invariant.

This is as far as I got. I do not know how to finish it off although I do believe that this question is logically solved like this. Could you please explain to me how to solve this question?

答:
Suppose that there is a counterexample, i.e. an infinite set of moves from some configuration with a finite number of stones, $S$ say. If $S=1$, then no moves are possible and so we can suppose that we have a counterexample with $S$ minimal.

If the sequence contained a move of Type 1 then, after that move, there would only be $S-1$ stones and, by minimality of $S$, there could only be a finite number of further moves. All moves are therefore of Type 2 and, as in the post, this means that  the total weight $W$ is unchanged for $x=\frac{1+\sqrt{5}}{2}$.  A stone on position $n$ contributes $x^n$ to $W$ and, in consequence, the integer values for any occupied position is bounded above by the log of $W$ to the base $x$.

Since we are dealing with integers, this upper bound, $m$ say, is attained. Now consider moves from a configuration with at least one stone at position $m$. No move of Type 2 from this configuration can involve stones on $m$ and so the remaining moves involve at most $S-1$ stones. By minimality of $S$ there can only be a finite number of further moves after all.


问:
Multivariate normal with diagonal variance implies its components are normally distributed.
I have a question that seems intuitive but I am not sure how to prove it. Let $X$ be a random vector with componants $X=[X_1,\ldots,X_n]$. Suppose $X\sim N(\mu,\Sigma)$ where $\Sigma = diag(\sigma_1^2,\ldots,\sigma_n^2)$ and $\mu = [\mu_1,\ldots,\mu_n]$. How can I properly show that $X_i\sim N(\mu_i,\sigma_i^2)$  (and that they are independent)?

答:
if the $\Sigma$ matrix is diagonal it means that all the covariances are zero. In a gaussian model, incorrelation and independence are equivalent thus

$$f_{X_1X_2\dots X_n}(x_1,x_2,\dots,x_n)=f_{X_1}(x_1)\cdot f_{X_2}(x_2)\cdot \dots \cdot f_{X_n}(x_n)$$


问:
does not has continuous extension of $S^1 \to S^1$ to $\Bbb{R}^2 \to S^1$
Let $S^1$ be the smooth manifold then the identity map is smooth $i:S^1 \to S^1$,prove there exist not continuous extension from $i:\Bbb{R}^2 \to S^1$

I try to construct some connected set in $\Bbb{R}^2$ to $S^1$ that fail to be connected,for example removing two point that still keeps the domain connected but not in $S^1$ is my idea correct?

答: Suppose such extension $f:\mathbb{R}^2\rightarrow S^1$, exist, $i\circ f$ is the identity, implies that $(i\circ f)_*:\pi_1(S^1)\rightarrow\pi_1(S^1)$ is the identity., we have $(i\circ f)_*=i_*\circ f_*$ and $f_*$ is constant, contradiction since $\pi_1(S^1)=\mathbb{Z}$.


问:
Find a closed form for the recurrence relation: $a_{n+2}-a_n=\sin\frac{n \pi }{2} \;\;\;a_0=a_1=1$
Find a closed form for the following recurrence relation:

$$a_{n+2}-a_n=\sin\frac{n \pi }{2} \;\;\;a_0=a_1=1\tag{$n \ge 0$}$$

This is what I've done so far:

Define a function $f(x):=\sum_{n\ge0}^{ }a_{n}x^{n}$,then:

$$\sum_{n\ge0}^{ }a_{n+2}x^{n+2}-x^{2}\sum_{n\ge0}^{ }a_{n}x^{n}=x^{2}\sum_{n\ge0}^{ }\sin\left(\frac{n\pi}{2}\right)x^{n}$$

The RHS is $\sum_{n\ge0}^{ }\sin\left(\frac{n\pi}{2}\right)x^{n}=\sum_{n\ge0}^{ }\left(-1\right)^{n}x^{2n+1}$
$$f\left(x\right)-a_{0}-a_{1}x-x^{2}f\left(x\right)=x^{3}\sum_{n\ge0}^{ }\left(-x^{2}\right)^{n}$$

Considering the initial values the equality transforms to:

$$f\left(x\right)-1-x-x^{2}f\left(x\right)=x^{3}\sum_{n\ge0}^{ }\left(-x^{2}\right)^{n}$$

\begin{align}
a_n=[x^n]f\left(x\right)
&=[x^n]\frac{2x^{3}+x^{2}+x+1}{\left(1+x^{2}\right)\left(1-x^{2}\right)}\\
&=\left[x^{n}\right]\left(\frac{5}{4}\frac{1}{1-x}-\frac{1}{4}\frac{1}{1+x}-\frac{1}{2}\frac{x}{1+x^{2}}\right)\\
&=\left[x^{n}\right]\left(\frac{5}{4}\sum_{n\ge0}^{ }x^{n}-\frac{1}{4}\sum_{n\ge0}^{ }\left(-1\right)^{n}x^{n}-\frac{1}{2}x\sum_{n\ge0}^{ }\underbrace{\binom{-1}{n}}_\text{=$(-1)^n$}x^{2n}\right)\\
&=\frac{5}{4}-\frac{1}{4}\left(-1\right)^{n}\end{align}

But the problem is that the answer is $$a_n=\frac{5}{4}-\frac{1}{4}\left(-1\right)^{n}-\frac{1}{2}\sin\frac{n\pi}{2}$$

And I don't understand why.

答:
Since a generating function answer has already been provided, I'll instead provide the direct proof. Note that

$$a_{n+2}=a_n+\sin(n\pi /2)=a_{n-2}+\sin((n-2)\pi/2)+\sin(n\pi/2).$$

But $\sin(n\pi/2-\pi)=-\sin (n\pi/2)$, so $a_{n+2}=a_{n-2}$. Hence the sequence is 4-periodic, and it suffices to know the first four entries. We have $a_0=a_1=1$ by assumption, and therefore also $$a_2=a_0+\sin(0\pi/2)=a_0=1,$$ $$a_{3}=a_1+\sin(\pi/2)=a_1+1=2.$$

We conclude that $a_{n}=2$ if $n\bmod 4=3$ and $a_n=1$ otherwise.


问:
Linear functional on $B(X,Y)$ with Compact-Open topology

Theorem(Grothendieck): Let $X$ and $Y$ be two Banach space. Let $B(X,Y)$ denotes the set of Bounded liner operator from $X$ to $Y$. Equipped $B(X,Y)$ with the standard Compact-Open topology, denote by $\tau$. Then every continuous linear functional on $(B(X,Y),\tau )$ has the following form: $$f: B(X,Y)\rightarrow \mathbb{C} : A \mapsto \sum_{n=1}^{\infty} <y_n^*, Ax_n>$$ where $\{x_n \} \subseteq X$, $\{ y_n^*\}\subseteq Y^*$ such that $\sum_{n=1}^{\infty}||x_n|| \cdot ||y_n^*||< \infty$.

It is easy to check if $f$ has above from, then $f$ continuous with $\tau$. This theorem claims there is no others. Below is my thought. It may be long, and it is not necessary to read them, if you don’t want to. Any help, hint, discussion or reference are helpful. Thanks!

My thought: Suppose $f$ continuous, then $\{A\in B(X,Y): |f(A)|<1  \}$ is a open neighborhood of zero(with $\tau$). Use the form of local base, and note that union of finite compact set is compact. There is a compact set $K\subseteq X$ and $\epsilon > 0$, such that $\{A\in B(X,Y): ||Ax||<\epsilon, \forall x\in K \} \subseteq \{A\in B(X,Y): |f(A)|< 1 \}$. By Grothendieck compactness principal, without lose of generality, we can assume $K$ is a closed convex hull of a null sequence $\{ x_n\}, i.e. $$x_n \rightarrow 0$ in X. That is: $K=\{\sum_{n=1}^{\infty} a_n x_n: \sum_{n=1}^{\infty}|a_n| \leq 1\}$. By convexity, it is clear that $Ax_n=0$ implies $A|_{K}=0$. Also note that, $L=\{A\in B(X,Y): A|_{K}=0 \}$ is a subspace of $B(X,Y)$, and f acts on $L$ is bounded, so $f(L)=0$.

Now we consider the map $\phi$ from $B(X,Y)$ to $\Pi_x Y:=\{(y_n)_{n=1}^{\infty}\in \Pi_{n=1}^{\infty}Y:||(y_n)||:=\textrm{sup}_n \frac{||y_n||}{||x_n||}< \infty  \}$, via $$\phi: A \mapsto (Ax_1,Ax_2,...,Ax_n,...).$$ It’s routine to check $\Pi_x Y$, with this “weighted-$\infty$-norm”, is Banach and $\phi$ is continuous linear map. The first paragraph shows $\textrm{ker}\phi \subseteq \textrm{ker}f$. There is a will defined linear map $g$ from $\textrm{Range}\phi$ to $\mathbb{C}$ by setting $g=f(\phi^{-1}((y_n)))$.

I tend to claim the following:

Every continuous functional on $\Pi_x Y$ has form $$(y_n)\mapsto \sum_{n=1}^{\infty}<y_n^*, y_n>, \hspace{0.2cm}\textrm{with its norm} \hspace{0.1cm} \sum_{n=1}^{\infty}||x_n||\cdot||y_n^*||<\infty.$$

$g$ is a continuous linear functional on subspace $\textrm{Range} \phi$, so use Hahn-Banach theorem, $g$ can be extension to $\Pi_x Y$.
Together with 1 and 2, the theorem established, since $f=g\circ \phi$.

I’m confident with the first claim. Because $||x_n||\rightarrow 0$, so does $||y_n||$, thus $\Pi_x Y$ is just like the “wighted-$c_0$” by replace each component $\mathbb{C}$ with $Y$(maybe just like tensor?). And the dual of $c_0$ is exactly $l^1$, so they are agree in form. However, for the second claim, I’m not so confident. Because it seems no reason for $\phi$ must be open or closed. I tried to use Arzela-Ascoli theorem on K, to prove $\textrm{Range}\phi$ is closed, then use open-mapping theorem, or something else. But all fail. I think I may miss something important.

答:
$\newcommand{\bf}[1]{\mathbb #1}$Picking up your argument from the point you say

"Now we consider the map $\phi$ from $B(X, Y)$ to $\Pi_xY$ ...",

let me instead consider
$$
  c_0(Y) = \Big\{(y_n)_{n\in {\bf N}} \in  Y^{\bf N}: \lim_{n\to \infty } y_n = 0\Big\},
  $$
equipped with the sup norm, and put
$$
  \phi:T\in B(X,Y)\mapsto \big (T(x_n)\big )_{n\in {\bf N}} \in  c_0(Y).
  $$

As you observed,  since
$$
  \phi(T)=0 \Rightarrow f(T)=0,
  $$
there exists a linear map $g$ from $\text{Ran}(\phi)$ to ${\bf C}$, such that $g\circ \phi=f$.

I claim that $g$ is continuous with norm no bigger than $2/\varepsilon $.  To see this,  let $y = (y_n)_{n\in {\bf N}} \in  \text{Ran}(\phi)$,
with $\|y\|\leq 1$.  Choose $T\in B(X,Y)$ such that $\phi(T)=y$, and notice that
$$
  \|T(x_n)\|=\|y_n\|\leq 1, \quad\forall n\in {\bf N}.
  $$
By convexity we then have that
$$
  \|T(x)\|\leq 1, \quad\forall x\in K.
  $$
Setting $A=\varepsilon T/2$, it follows that
$$
  \|A(x)\|<\varepsilon , \quad\forall x\in K,
  $$
so
$$
  1>|f(A)|=\frac{\varepsilon |f(T)|}2 = \frac{\varepsilon |g(\phi(T))|}2 = \frac{\varepsilon |g(y)|}2,
  $$
whence
$$
  |g(y)|<\frac2\varepsilon,
  $$
proving the claim.
Using Hahn-Banach we may extend $g$ to the whole of $c_0(Y)$ and, imitating the proof that $c_0^*=\ell ^1$, we may prove
that the extended $g$ has the form
$$
  g(y) = \sum_{n\in {\bf N}}\langle y_n^*, y_n\rangle ,\quad\forall y = (y_n)_{n\in {\bf N}} \in  c_0(Y),
  $$
where $y^*=(y^*_n)_{n\in {\bf N}}\in \ell ^1(Y)$.

The grand conclusion is then that
$$
  f(T) = g(\phi(T)) = \sum_{n\in {\bf N}}\langle y_n^*, T(x_n\rangle \rangle ,\quad\forall T\in B(X,Y),
  $$
while
$$
  \sum_{n\in {\bf N}}\|x_n\|\|y_n^*\| \leq      \Big (\sup_{n\in {\bf N}}\|x_n\|\Big )\Big (\sum_{n\in {\bf N}}\|y_n^*\|\Big )<\infty .
  $$


问:
How to prove if the relation R is an equivalence relation?
Hello so I stumbled upon an exercise where I need to prove that R is an equivalence relation, normally I have no problem doing this, but with this exercise I have absolutely no clue where to begin. I think my biggest problem is understanding what's given. The exercise goes as follows:

$X$ and $Y$ are two sets. We notate $Fun(X,Y)$ is the set of all functions $f:X\rightarrow Y$. Let $R$ be the relation on $Fun(X,Y)$ given by: $(f,g) \in R$ if and only if there exist a bijective function $\sigma:Y \rightarrow Y$ with $\sigma \circ f =g $.

Prove that $R$ is an equivalence relation.

What I know

$R$ is an equivalence relation when $R$ is reflexive, symmetric and transitive (so this is what I need to prove).

$R$ is reflexive if for all $x\in X : xRx$

$R$ is symmetric if for all $x,y\in X : xRy \rightarrow yRx$

$R$ is transitive if for all $x,y,z \in X: xRy \text{ and } yRx \rightarrow xRy$

If someone could help me, it would be very kind of you :)

Thank you in advance

答:
Although you already have a fine answer, I'm posting this because I think a picture makes it easier to understand.

The top left diagram shows the definition.  If we follow the arrows across and down, first applying $f$ and then $\sigma$, we get the same thing as we would if we just took the shortcut via $g$.

The top right diagram shows the situation for reflexivity, when we have $f=g$.  Clearly, $\sigma=\operatorname{id}$ is required.

The middle diagram is for symmetry.  We are given $\sigma$ and we must find a $\tau$ so that if we first go diagonally by $g$ and then up by $\tau$, we get the same thing as if we went straight across by $f$.  It's easy to guess $\tau=\sigma^{-1}$ from the diagram, isn't it?

The third picture is for transitivity.  We are given $\sigma$ and $\tau$ such that $g=\sigma\circ f$ and $h=\tau\circ g$ and we must find $\rho$ such that $h=\rho\circ f.$  It's clear from the picture that $\rho=\tau\circ\sigma$ is what we need.


问:
Double Integral of absolute value
How should I calculate the integral of $\int _0^1\:\:\int _0^1\:\left|x-y\right|dxdy$?

I understand that I need to split in two parts: $x<y$ and $x\ge y$

But I don't understand how to calculate it.

答:
Your integral can be understood as an integral which its integrand is a piecewise function with respect to the region of integration.

$$\int _0^1\int _0^1f(x,y)\,dxdy$$

$$f(x,y) = |x-y| = \begin{cases}+(x-y), \: x\geq y \\ \\ -(x-y), \: x<y \end{cases}$$

So you have to split the region of integration in separate regions until you end up with "normal functions" for every one of them.

In your case, Yalikesifulei's answer and tommik's answer do the trick because $f(x,y)$ is fixed on $[+(x-y)]$ in the first region and fixed on $[-(x-y)]$ in the second one.

$$\int \int_D f(x,y)\,dxdy = \int \int_{D_1} f(x,y)\,dxdy + \int \int_{D_2} f(x,y)\,dxdy$$

$$D = D_1 \cup D_2$$
$$\left\{ (x,y) : 0\leq x\leq 1, 0 \leq y \leq 1\right\} = \left\{ (x, y) : 0\leq x \leq 1, 0 \leq y < x\right\} \cup  \left\{ (x, y) : 0\leq x \leq 1, x\leq y \leq 1\right\} $$

Title: Double Integral of absolute value


问:
Does the set of all formal Laurent series form an ordered field?
Let $R$ be an ordered field. Consider the set $R[[x]]:=\{\sum_\limits{n \in \mathbb{Z}}a_nx^n:a_n \in R\}$

Addition:
$\sum_\limits{n \in \mathbb{Z}}a_nx^n+\sum_\limits{n \in \mathbb{Z}}b_nx^n:=
\sum_\limits{n \in \mathbb{Z}}(a_n+b_n)x^n$

Multiplication:
$\sum_\limits{n \in \mathbb{Z}}a_nx^n \times \sum_\limits{n \in \mathbb{Z}}b_nx^n:=
\sum_\limits{n \in \mathbb{Z}}(a_nb_n)x^n$

Does these operations on $R[[x]]$ makes it a field?

Have I correctly defined the multiplication?

Is the notation $R[[x]]$, i.e does the notation $R[[x]]$ have some other meaning?

If yes, is there any standard notation for this set?

What order relation can be defined on $R[[x]]$ in order to turn it into an ordered field?

My motive is to show that $R[[x]]$ is Cauchy Complete but is not Order Complete(i.e lub property)

Edit: As said by @ the multiplication defined above doesn't even make it an integral domain.

So I am changing the set to $R[[x]]:=\{\sum_\limits{k=-n}a_kx^k: a_k \in R, n \in \mathbb{N} \}$  and multiplication as defined in the answers.

答:
To define an order, it suffices to say which elements are $\geq 0$, since then we can define $a\leq b$ if and only if $(b-a)\geq 0$. (And then one needs to check that this order actually gives us an ordered field.)

One way to define the order is this: a formal Laurent series in $R((x))$, where $R$ is an ordered field, is $\geq 0$ if and only if it is $0$ or $c_n \geq 0$ in $R$, where $n$ is the least power of $x$ with nonzero coefficient $c_n$.

This ordering makes lower powers of $x$ "more significant", so you can think of it as an ordering of $R((x))$ in which $x$ is positive and infinitesimal.


问:
Proof's involving Zorn's Lemma
my assignment is to prove a theorem using Zorn's Lemma.

What are some good theorems that Zorn's Lemma is applicable to (on the easy/moderate level).

答: Here's one: Let $F$ be a field, $L_1, L_2$ algebraic extensions of $F$ with $L_2$ algebraically closed. Then any homorphism  $\sigma: F\rightarrow L_2$ has an extension  $\tau:L_1\rightarrow L_2$ such that the restriction of $\tau$ on $F$ is $\sigma$.

Title: Proof's involving Zorn's Lemma


问:
Further explanation needed :Finding all $z$ such that the modulus of $f(z)=e^{(z+1)/(z-1)}$ is equal to/at most $1$
I was solving a previous exam paper and there I got stuck on the following problem:

Let $f(z)=e^{\frac{z+1}{z-1}}$. Then find all $z \in \Bbb C$ for which

$|f(z)|=1$,

$|f(z)|\le 1$.

Any idea about how to tackle this problem will be highly appreciated. Thanks for your time.

答:
Let $ z = a + bi, (a, b) \in \mathbb{R}^2 $. Hence $$\begin{align*} \left|e^{\frac{z+1}{z-1}}\right| &= \left|e^{\frac{(a + 1 + bi)(a - 1 - bi)}{(a - 1 + bi)(a - 1 - bi)}} \right| \\&= \left|\ e^{\frac{a^2 - 1 + b^2 -2bi}{(a - 1)^2 + b^2}}\ \ \right| \\ &= \left|\ e^{ \frac{a^2 -1 + b^2}{(a - 1)^2 + b^2}}\ \ \ \ \ \right|\cdot\left|e^{-\frac{2bi}{(a-1)^2 +b^2}}\right|\\&=\ \ e ^{\frac{a^2 - 1 + b^2}{(a - 1)^2 + b^2)}}\end{align*}$$
Hence, the modulus is $ 1 $ when $ a^2 + b^2 = 1 $, which is a unit circle. Note that the point $ (1, 0) $ should be excluded because $ f(1) $ is not defined.

Also, the modulus is less than $1 $ on the entirety of the disk that is contained by that circle, i.e. $ a^2 + b^2 < 1 $.


问:
Kähler differential and higher derivations (geometric interpretation of diagonal here)
I am studing Kähler differentials and I tried to understand the geometric motivation behind these settings. What I do not understand is the role which plays the diagonal in all these theory. The cotangent sheaf is later defined in terms of the diagonal map. Why is this geometrically interesting? I tried to write a short introduction to Kähler differential to make the geometric nature more available, but I do not now if it make sense. Here it is:

Differential $ 1 $-forms are linear transformations $ \omega_{p}:T_p X\mapsto K $  assigning an element in $ K $ to a tangent vector of the tangent space $ T_p X $ of a point $ p\in X $ in some differential manifold $ X $. Differential $ 1 $-forms can be viewed as \textit{infinitesimal} direction vectors $ \triangle p $. This means in physical terms, that the scalar $ \omega_p(\triangle t)\in K $ with $ \triangle t $ a tangent vector represents the \textit{work} required to move from $ x_i $ to $ x_{i+1} $ with $ p\in (x_i,x_{i+1}) $ along some curve. In other words, differential forms are cotangent vectors over some field $ K $, which give information about the work which is locally required to move along some curve. However, they can be generalized and captured by sheaf theory. For this attempt, we observe first that $ \triangle p $ is related to Taylor expansions. Indeed, let $ f $ be a smooth function, that is $ f\in C^{\infty}(\mathbb{C}) $, on a differential manifold $ X$ and let $\mathfrak{J}$ be the ideal of smooth functions vanishing at the point $p\in X$. The zero order part of the Taylor series of a smooth function $f$ is the value of $f$ at the point $ p $, let us say, $ f(p)=c $, so that $ f-c\in\mathfrak{J}$. Now the first order derivatives of $f-c$ correspond to the first order terms in the Taylor series and these are given by the image of $f$ in $\mathfrak{J}/\mathfrak{J}^2$. Let us denote this map by $ d(f) $ with $ d: \mathcal{O}_X\rightarrow \mathfrak{J}/\mathfrak{J}^2 $ and where $ \mathcal{O}_X $ denotes the ring of smooth functions on $ X $. Moreover, if $ f $ is constant, this means a fixed vector, then $ d(f)=0 $. Another important input is that $ \triangle p$ is required to be non zero, as there is no direction available for the zero vector. But $ \triangle p=0 $ is satisfied if and only if the two endpoints of the tangent vector $ \triangle p $ are choosen to be the same, which happens if and only if the point $ p\in X $ correspond to an element on the diagonal of $ X\times X $. So we demand that we just consider elements in $ X\times X $ vanishing on the diagonal (or in the complement of the diagonal).

Summarizing up, my two questions are the following:

1) which geoemtric interpretation have the diagonal in these context?

2) Higher derivations seemed to me a generalization of Kähler differentials, but what is their motivation or geometric nature (analogy to differential geometry), since I cannot see any connection between Kähler differentials and higher derivations ?

答:
This is an old question, but the question may have interest for others on this site.

Question 1. "Which geometric interpretation have the diagonal in these context?"

Answer. If $\phi:A \rightarrow B$ is a map of commutative rings and $I:=ker(m) \subseteq B\otimes_A B$ is the ideal of the diagonal, we get a closed immersion

D1. $\Delta:X:=Spec(B) \rightarrow Spec(B\otimes_A B):=X\times_S X.$

The sheafication $\Omega:=\Omega^1_{X/S}$ of the module of Kahler differentials $\Omega^1_{B/A}$ is a sheaf of $\mathcal{O}_{X\times_S X}$-modules with support on $\Delta(X)$. The two projection maps $p,q:X\times_S X\rightarrow X$ induce a left and right $\mathcal{O}_X$-module structure on $\Omega$, hence $\Omega$ is a sheaf of left and right $\mathcal{O}_X$-modules with the property that $a(\omega b)=(a\omega)b$ for all local setions $a,b,\omega$. The following formula holds:

D2. $a\omega = \omega a.$

hence the left and right structure coincide. There is an isomorphism of $B$-modules

D3.  $Hom_B(\Omega^1_{B/A},B) \cong Der_A(B)$

where $Der_A(B)$ is the set of $A$-linear maps $\delta:B\rightarrow B$ with $\delta(ab)=a\delta(b)+\delta(a)b$. At the level of sheaves you get an ismorphism

D4. $\Omega^* \cong \Theta_{X/S}$

hence the dual of the module of Kahler differentials is the "tangent bundle" $\Theta_{X/S}$.

Let $k\subseteq A$ be a map of commutative unital rings. A "higher derivation of $A/k$" is constructed as follows: Let $B:=A[t]/(t^{n+1})$
and let $D:A \rightarrow B$ be a map of $k$-algebras. It follows

D5. $D(a)=\sum_{i\geq 0}D_i(A)t^i$

where $D_i:A \rightarrow A$ are $k$-linear maps and
$D_k(ab)=\sum_{i+j=k}D_i(a)D_j(b)$. The map

D6. $D_1: A\rightarrow A$

is a derivation, but the other maps $D_i$ are not. See Matsumura's book "Commutative ring theory" for some applications to the study of local rings.

Question: "Higher derivations seemed to me a generalization of Kähler differentials, but what is their motivation or geometric nature (analogy to differential geometry), since I cannot see any connection between Kähler differentials and higher derivations?"

Answer: If $k$ is a field of characteristic zero it follows for any derivation $\delta \in Der_k(A)$ we get a higher derivation $D^n(\delta)$

$$D^n(\delta)\left(\sum_{i=0}^n a_it^i\right):=\sum_{i\geq 0}\frac{1}{i!}\delta^i(a_i)t^i$$

where $\delta^i:=\delta \circ \cdots \circ \delta.$ This construction and it's relation to smoothness is studied in Matsumura's book Chapter 27.

They also go under the name "Hasse-Schmidt derivations". In some cases (over a field of charateristic zero) they are related to the notion "flat connection".
You may find some results on this on the arXiv preprint server.


问:
Quick clarifying question regarding eigenvalues
Suppose a $3 \times 3$-matrix $A$ has real entries. Consider the 2 cases over real and complex numbers. Over the real numbers the eigenvalue 3 is the only one. over the complex numbers the eigenvalues $3, i, -i$ appear. Is the eigenspace and associated eigenvector(s) for the eigenvalue 3 the same exact ones or are they different since the matrix associated with the eigenvalues are over different numbers?

答:
Let $v$ denote a non-zero real vector for which $Av = 3v$. Technically speaking, the eigenspace associated with $\lambda = 3$ for the real matrix $A$ (or the associated linear map over $\Bbb R^3$) is the set of real multiples of $v$. That is, the eigenspace is
$$
\{kv : k \in \Bbb R\}.
$$
The eigenspace associated with $\lambda = 3$ for $A$ considered as a complex matrix (or for the associated linear map over $\Bbb C^3$) is the set of complex multiples of $v$. That is, the eigenspace is
$$
\{kv : k \in \Bbb C\}.
$$


问:
Prove that the Galois group of $x^5+x^4-4x^3-3x^2+3x+1$ is cyclic of order $5$
I have been asked to prove that the Galois group of $x^5+x^4-4x^3-3x^2+3x+1$ (presumably over $\mathbb{Q}$ is cyclic of order $5$.  Unfortunately, I have no idea where to start.

The hint says to show that this is the minimal polynomial of $\zeta_{11} + \zeta_{11}^{-1}$.  Which I guess I could do by plugging it into the polynomial, but there is most likely a better way to do it.  Anyway, even if I could establish that, I wouldn't know any of the other roots so it's not like I could find any of the automorphisms.  I guess it would kind of help because it would show that the Galois group had order divisible by 5.  Except a lot of them are.  I think I am completely missing the point of the hint.

The section was about how you could reduce polynomials mod $p$, and the Galois group would have to contain a permutation with cycle type $(n_1 n_2 ... n_k)$, where $n_1$, ..., $n_k$ are the degrees of the irreducible factors (mod $p$).  But this is not very relevant because since the group is cyclic of prime order, all the permutations would be 5-cycles.  And lots of other subgroups of $S_5$ have 5-cycles too.

So at this point I am kind of stuck.  Can you give me a hint as to how to solve the problem?  Thanks!

P.S. We are allowed to use the computer program SAGE for this assignment, but not the command that calculates the Galois group.  Also, I tried to use SAGE to calculate the roots but it gave me numerical answers.

答:
Suppose you have shown that the minimal polynomial of $\zeta_{11}+\zeta_{11}^{-1}$ is the given polynomial.

What does that mean $[\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1}):\mathbb{Q}]$ is?

For any $n$, we know that $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is Galois, and $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong(\mathbb{Z}/n\mathbb{Z})^\times$ (or, if this has not come up in class, you should prove it separately to make your argument complete). In particular, $$\text{Gal}(\mathbb{Q}(\zeta_{11})/\mathbb{Q})\cong(\mathbb{Z}/11\mathbb{Z})^\times\cong \mathbb{Z}/10\mathbb{Z},$$ the cyclic group of order 10.

Note that $\mathbb{Q}(\zeta_{11})\supset\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1})$.

The cyclic group of order $n$ has a single subgroup of size $d$, for each $d\mid n$. Therefore it also has a single quotient group of size $d$, for each $d\mid n$.

We know that $[\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1}):\mathbb{Q}]=|\text{Gal}(\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1})/\mathbb{Q})|$.

What does that mean $\text{Gal}(\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1})/\mathbb{Q})\cong\text{Gal}(\mathbb{Q}(\zeta_{11})/\mathbb{Q})\bigg/\text{Gal}(\mathbb{Q}(\zeta_{11})/\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1}))$ is?


问:
A basic doubt on axiom of foundation of Zermelo-Fraenkel set theory
I have read in a book that the "axiom of foundation prevents anomalies such as a
set being an element of itself".

Now, axiom of foundation says that there exist an element in every set which is disjoint with the set.

Now if $S$ be a set then according to the above axiom $S=\{S\}$ can't be a set. But, $S=\{\{1\},S\}$ satisfies axiom of foundation and contains $S$.

So, is the statement mentioned in the book correct ?

答:
You seem to think that you can find a set $S$ such that $S=\{\{1\},S\}$.

But can you really? Go on, write down this set explicitly. Truth is...you won't be able to.

Put it in another way: assume ab absurdo, that you can find such a set $S$. So $S \in S$.
Then the set $\{S\}$ violates the axiom of regularity since both $\{S\}$ and $S$ contain $S$, so  $\{S\} \cap S = \{S\}$ and thus non empty.


问:
Does there exist a function $f(x)$ such that $f'(x)$ exists for all $x \in A$ but $f'(x)$ is not continuous on $A$?
I don't think this can be the case, since the derivative is defined as a limit

$$f'(x) = \lim_{a\to x} \frac{f(x)-f(a)}{x-a}$$

And a function that equals its limit is by definition continuous but this feels like a non-rigorous proof so I do not want to depend on it.

If a function does exist such that $\forall \, x \in A \,\, \exists \,y : f'(x) = y $, a good answer would give an example of such a function, or if such a function does not exist, a good answer would give pointers as to how I could prove its non existence

答:
There do exist functions for which the derivative is defined for all $x \in \mathbb{R}$ but not continuous. An example of such a function is

$$f(x) = \begin{cases}x^2 \sin \left(\frac{1}{x}\right) &\text{if }x\neq0\\
0 &\text{if }x=0
\end{cases}$$
whose derivative is
$$f'(x) = \begin{cases}2x \sin \left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) &\text{if }x\neq0\\
0 &\text{if }x=0
\end{cases}$$

$f'(0)$ exists and is equal to zero, however $\lim_{x\to0} f'(x)$ does not exist due to the term $\cos\left(\frac{1}{x}\right)$ whose limit is undefined as $x\to0$.

Therefore, $f(x)$ is an example of a function that is differentiable everywhere but whose derivative is not continuous.

I found this function in this question, however that question did not pop up in my initial search$^1$, so, I decided to use the "Answer you own question" feature so that future people with the same question may find the answer more easily.

Probably because it is much more specific.


问:
Inverse of $(I+QAQ^{T})$ where $Q$ is the eigenvector matrix, $A$-eigenvalue matrix
I would like to find $(I+QAQ^{T})^{-1}$ where $Q$ is the eigenvector matrix, $A$-eigenvalue matrix. I saw somewhere that it is equal to $Q(I+A)^{-1}Q^{T}$ is this so? I don't understand why should one be able to take the eigenvectors matrices out of the brackets of the inverse sign?

答:
When you say that "$Q$ is the eigenvector matrix", I assume that in this context $Q$ is an orthogonal matrix (i.e. $Q^T = Q^{-1}$).

With that said, note that
\begin{align}
(I + QAQ^T)^{-1} & = (I + QAQ^{-1})^{-1}
\\ & = (QIQ^{-1} + QAQ^{-1})^{-1}
\\ & = (Q[I + A]Q^{-1})^{-1}
\\ & = (Q^{-1})^{-1} (I + A)^{-1} Q^{-1}
\\ & = Q(I + A)^{-1} Q^{-1} = Q(I + A)^{-1}Q^T.
\end{align}
The more general phenomenon is that for any "analytic function" $f(x)$ (for example, a polynomial $f(x)$, $f(x) = 1/(1 + x)$, $f(x) = e^x$), if $f(A)$ is defined "in the natural way" (via the power series of $f$ or a functional calculus), then it necessarily holds that
$$
f(PAP^{-1}) = Pf(A)P^{-1}
$$
for any invertible matrix $P$.


问:
Equation of Partial derivatives
I am currently learning multivariable calculus and the following was done in my lecture to prove one of the properties of Jacobian

The property:

($JJ'=1$), where $J$ is $\frac{\partial (x,y)}{\partial(u,v)}$ and $J'$ is $\frac{\partial(u,v)}{\partial (x,y)}$

The equation in the proof:

$\partial u = \frac{\partial u}{\partial x} \partial x +\frac{\partial u}{\partial y} {\partial y}$

Then the professoor proceeded to divide by $\partial u$ on both sides so you end up with $\frac{\partial u}{\partial u}$ on the left hand side which is essential 1. He then did this for $v$ also.

My question: $\partial u\ , \partial x\ , \partial y$ on their own makes no sense. And since partial derivative is defined for suppose u with respect to some other variable, dividing or multiplying by $\partial (some \ variable)$ should be wrong. What is the correct theory ?

答: There's nothing too deep going on here. The point is that if $f : \mathbb{R}^n \to \mathbb{R}^n$ and $g : \mathbb{R}^n \to \mathbb{R}^n$ satisfy $f \circ g = I$, then $Df|_{g(x)}Dg|_{x} = DI|_{x} = I$. All we used was the chain rule and the fact that the Jacobian of the identity transformation is the identity matrix.

Title: Equation of Partial derivatives


问:
Show that a point inside an acute triangle is the orthocenter
Point $M$ lies inside an acute triangle $\triangle ABC$ such that the circumscribed circles of $\triangle ABM,\triangle BCM$ and $\triangle ABC$ have equal radii. Show that $M$ is the orthocenter.

I have been trying to solve the problem for an hour using the law of sines because we have just studied it, but I can't come up with anything. What about a solution using vectors?

答:
These definitions are from this book: 'An introduction to the modern geometry of the triangle and the circle'; Nathan Altshiller Court.You can see these and their proof in there.

1-Definition:  the midpoints of segments connecting the orthocenter and vertices of a triangle is called the Euler points of the triangle.

2-Theorem: In any triangle the mid points of sides , the foot of altitudes and Euler points are cyclic (on a circle). This is called nine point circle of a triangle.

3-Triangles (AMB), (BMC), (AMC) and(ABC) are called group of orthocenter's triangles.

4-The radius of circumscribed circles of the group of orthocenter triangle are equal to the diameter of nine point circle of triangle.

Now your statement say the radius of circumscribed circles of ABM, AMC and BMC are equal, therefore M must be orthcenter of ABC.


问:
Find $\frac{a+b}{ab}$ such that $\int_{-1/2}^{1/2} \cos x\ln\frac{1+ax}{1+bx}dx=0$
Let $f(x) = \cos(x) \ln\left(\frac{1+ax}{1+bx}\right)$ be integrable on $\left[-\frac{1}{2} , \frac{1}{2}\right]$. Let
$$\displaystyle \int_{-1/2}^{1/2}f(x)\operatorname{dx}=0$$

where $a$ and $b$ are real numbers and not equal. Find the value of $\frac{a+b}{a\cdot b}$.

At first I applied integration by parts but then it got even more complicated and I couldn't solve it. How can I solve this problem?

答:
Note that the log function in the integrand can be decomposed as a sum of odd and even ones
\begin{align}
\ln\frac{1+ax}{1+bx}& =\frac12\ln\frac{1+ax}{1-ax}-\frac12\ln\frac{1+bx }{1-bx} + \frac12 \ln\frac{1-a^2x^2}{1-b^2x^2} \\
& =\tanh^{-1}(ax)-\tanh^{-1}(bx) + \frac12 \ln\frac{1-a^2x^2}{1-b^2x^2}
\end{align}

where the odd functions $\tanh^{-1}()$ do not contribute to the integral and the even function has to vanish for the integral to vanish, i.e.
$$\ln\frac{1-a^2x^2}{1-b^2x^2} =0$$
which leads to $a=-b$, hence $\frac{a+b}{ab }=0$.


问:
Argument of complex number: what am I doing wrong?

Find $\arg z$ where
$$z=-\frac{i\omega-w_0}{i\omega+w_0}$$
where $\omega$ and $\omega_0$ are positive numbers.

My attempt was this:

$$\arg z=\pi +\arctan\left(\frac{\omega}{-\omega_0}\right)+\pi-\arctan\left(\frac{\omega}{\omega_0}\right)$$$$=2\pi -\arctan\left(\frac{\omega}{\omega_0}\right)-\arctan\left(\frac{\omega}{\omega_0}\right)$$$$=2\pi-2\arctan\left(\frac{\omega}{\omega_0}\right)$$

However, the correct answer seems to be:
$$\arg z =\pi-2\arctan\left(\frac{\omega}{\omega_0}\right)$$

What am I doing wrong?

答:
You seem to have misunderstood concepts.
$$\arg(z_1\pm z_2)\neq \arg z_1\pm \arg z_2$$
The proper and intended way would be to write $z$ in the form $a+bi$. This can be done as follows:
$$z=\frac{\omega_0-i\omega}{\omega_0+i\omega}=\frac{(\omega_0-i\omega)(\omega_0-i\omega)}{(\omega_0+i\omega)(\omega_0-i\omega)}$$
$$z=\frac{\omega_0^2-\omega^2-2\omega_0\omega i}{\omega_0^2+\omega^2}$$
$$z=\frac{\omega_0^2-\omega^2}{\omega_0^2+\omega^2}+\frac{-2\omega_0\omega}{\omega_0^2+\omega^2}i$$
And with this, we can easily calculate $\arg z$.

Hope this helps. Ask anything if not clear :)


问:
linear Transformation of polynomial with degrees less than or equal to 2
I would like to determine if the following map $T$ is a linear transformation:

\begin{align*}
T: P_{2} &\to P_{2}\\
A_{0} + A_{1}x + A_{2}x^{2} &\mapsto A_{0} + A_{1}(x+1) + A_{2}(x+1)^{2}
\end{align*}

My attempt at solving:

\begin{align}
T(p + q) &= p(x+1) + q(x+1)\\
&= \left[A_{0} + A_{1}(x+1) + A_{2}(x+1)^2\right] + \left[b_{0} + b_{1}(x+1) + b_{2}(x+1)^2\right]\\
&= \left(A_{0} + b_{0}\right) + \left(A_{1} + b_{1}\right)(x+1) + \left(A_{2} + b_{2}\right)(x+1)^2\\
&= T(p) + T(q)
\end{align}

Is this right so far? If not, what am I doing wrong?

答:
Why is $T(p+q)=p(x+1)+q(x+1)$? That's basically what you're trying to show.

You could try to just expand $T(p)$ and see that $T$ can be easily expressed as a matrix in base $1,x,x^2$.


问:
Mean value theorem for multivariable integrals (With slightly less demands)
Define: A set $ E \subset \mathbb{R}^n $ is admissible if it is bounded in $ \mathbb{R}^n $ and its boundary is a set of measure zero (in the sense of Lebesgue).

Now let $ f $ be an integrable functio over an admissble and connected set $ E $, such that $ f $ is also continuous. Prove that there exists $ \zeta\in E $ such that  $ \intop_{E}f\left(x\right)dx=f\left(\zeta\right)\mu\left(E\right) $.

Now if I know that the set $ E $ is a closed set then it would be a compact set and I know how to prove it given this information. But as I see it, $ E $ is not necessarily a closed set. So what can do I in this case?

Thanks in advacnce.

答:
Suppose that $|E|>0$, otherwise there is nothing to prove.
If $f$ is continuous on $E$ then
$$
\inf f(E) \leq \frac{1}{|E|} \int_Ef(x) dx \le \sup f(E).
$$
Since $E$ is connected, $f(E)$ is an interval and, therefore, it contains any point in between $\inf f(E)$ and $\sup f(E)$.


问:
More Questions about Two 2-sphere with points identified.
Suppose I have two 2-spheres $S_1$ and $S_2$ and $p_1, q_1\in S_1$ and $p_2, q_2\in S_2$. Then I identify $p_1$ to $q_1$ and $p_2$ to $q_2$. Let X denote this space.

Here is my previous question: Two $S^2$ with points identified.

Jonathan commented that the possible CW structure is 2 0-cell, $p_1$ on $S_1$ and $p_2$ on $S_2$, two lines $a, b$ and 2-cells via $aa^{-1}$ and $bb^{-1}$.

I want to study this space further by calculating $H_n$ and $H^n$ using two different methods respectively as follows.

My questions are: for all $n\geq 0$,

How can we compute $H_n(X)$ by using Mayer-Vietoris sequence and Cellular homology.

How can we compute $H^n(X)$ by using cellular cohomology and universal coefficient theorem.

Universal Coefficient Theorem for Cohomology: Hatcher page 195

I know

$H_2(X^2, X^1)=\mathbb{Z}\oplus\mathbb{Z},$

$H_1(X^1, X^0)=\mathbb{Z},$

$H_0(X^0)=\mathbb{Z}$.

答:
As I pointed out in the comment, we may compute the (co)homology for each component first.

Let $M$ denote one component of your space, then it looks like this (I tried my best...):

Since $M$ is connected, $H_0(M)\cong\Bbb{Z}$. For $k>2$, $H_k(M)\cong 0$.

Take $U\approx B^2$ be an open ball on the attached disk, and $V$ be a nhbd of $M-\overline{B^2}\simeq S^1$ that deformation retracts onto it. So, $U\cap V\approx S^1$.

Consider the non-trivial part of the reduced MV sequence,

$$0\to H_2(M)\overset{\partial_*}{\longrightarrow}H_1(S^1)\overset{(i_*,-j_*)}{\longrightarrow}H_1(S^1)\overset{k_*+l_*}{\twoheadrightarrow} H_1(M)\to 0$$

We just need to know $(i_*,-j_*)$. It's actually a trivial map because if the generator $a\in H_1(U\cap V)$ is represented by $\alpha$, then $i_*:H_1(U\cap V)\to H_1(U), \alpha\mapsto 0$, since $U\approx B^2$. $-j_*:H_1(U\cap V)\to H_1(V), \alpha\mapsto -(\beta-\beta)=0$ for the generator $\beta$ of $H_1(V)$ because $\alpha\sim(\beta-\beta)$ (homologous). Therefore, both $(k_*+l_*)$ and $\partial_*$ are isomorphisms. For the disjoint union $X$, you just take the direct sum.

For cellular homology, I think I can provide a hint with the cellular chain complex
$$0\overset{}{\to}\Bbb{Z}\overset{\partial}{\to}\Bbb{Z}\overset{0}{\to}\Bbb{Z}\to0$$
where the map $\partial$ is related to the attaching map (which you already have)

For cohomology using UCT, note that the short exact sequence splits, which means (Try yourself before click on the spoiler)

 $$H^k(X)\cong \operatorname{Hom}(H_k(X);\Bbb{Z})\oplus\operatorname{Ext}(H_{k-1}(X);\Bbb{Z})$$ For a finite CW complex such that its homology is finitely generated,  $$H^k(X)\cong \operatorname{Hom}(H_k(X);\Bbb{Z})\oplus T_{k-1}(X)$$ $\operatorname{Ext}$ reflects the torsion (in our case, it's torsion-free).

It is worth mentioning that $M$ is the deformation retraction of $S^2$ with an arc attaching its north and south pole, which is homotopic equivalent to $S^2\vee S^1$. It would be clearer if we use this to compute its (co)homology.


问:
compute $\prod_{n=2}^{N} \frac{n(n+1) +1}{ n(n-1)+1}$ .
I am not able to compute this infinite product and so I am asking for help here.

Compute the product $\prod_{n=2}^{N} \frac{n(n+1) +1}{ n(n-1)+1}$ .

I tried by factoring $\frac{n(n+1) +1}{ n(n-1)+1}$ into  $(1+ \frac{2n}{n^2-n+1}) $ but this doesn't seems to help and I don't  have any other ideas.

Please help.

答:
That is a telescopic product, for $ N\geq 2 $, we have : \begin{aligned} \prod_{n=2}^{N}{\frac{n^{2}+n+1}{n^{2}-n+1}}=\prod_{n=2}^{N}{\frac{\left(n+1\right)^{2}-\left(n+1\right)+1}{n^{2}-n+1}}&=\frac{\left(N+1\right)^{2}-\left(N+1\right)+1}{2^{2}-2+1}\\ &= \frac{N^{2}+N+1}{3}\end{aligned}

A telescopic product, has the form of $ \prod\limits_{k=p}^{q}{\frac{a_{k+1}}{a_{k}}}=\frac{a_{q+1}}{a_{p}} $, were $ p,q\in\mathbb{N} $, such that $ q\geq p $, and $ \left(a_{k}\right)\in\left(\mathbb{R}^{*}\right)^{\mathbb{N}} $.


问:
Applicatin of Stokes' theorem to compute flux of $\nabla \times G$
I need a help on the following exercise:

Consider the following surface

$$S= \{(x,y,z): x^2+y^2+z^2 = 4, x^2+y^2 \leq 2, z>0 \}$$

Compute the flux of $\nabla \times G$ exiting from $S$, where $G(x,y,z)=(xy,zx,x^2+y^2)$

First of all, it's clear that $S$ can be parametrized with $\sigma(x,y)=(x,y,x^2+y^2)$, where $(x,y) \in B_{\sqrt{2}}(0,0)$. However, that surface integral is not so short to compute, so it's better to use Stokes' theorem and compute $$\int_{\partial S^+} G \cdot dr$$ where $\partial S$ is the boundary of the surface $S$.

The question is: what is the boundary of $S$, or better, how may it be parametrized? My guess is that it is parametrized by $r(t)=(\sqrt{2} \cos(t), \sqrt{2} \sin(t),\sqrt{2})$ with $t \in [0, 2 \pi)$. So, the integral to compute is $$\int_{\partial S} G \cdot dr = \int_0^{2 \pi} (\sin(2t),2 \cos(t),2) \cdot (-\sqrt{2} \sin(t), \sqrt{2} \cos(t),0) dt = \text{computations} = 0$$

Is everything correct?

答:
Please note your parametrization is correct. Also note that the surface is above $z \geq \sqrt2$ as $x^2 + y^2 \leq 2, z \gt 0$ so based on your integral, the orientation of the boundary and the surface also matches. But your integral is not zero.

$\displaystyle \int_{C} \vec{G} \cdot d\vec{r} = \int_0^{2 \pi} (\sin(2t),2 \cos(t),2) \cdot (-\sqrt{2} \sin(t), \sqrt{2} \cos(t),0) \, dt$

$\displaystyle = 2\sqrt2\int_0^{2\pi} (\cos^2t- \sin^2t \cos t) \, dt = 2 \sqrt2 \, \pi$

By the way the double integral is not very complicated either.

$\nabla \times \vec{G} = (2y-x, -2x, z-x)$. Unit normal vector to the surface is $(\frac{x}{2}, \frac{y}{2}, \frac{z}{2})$

In spherical coordinates $x = 2 \cos \theta \sin \phi, y = 2 \sin \theta \sin \phi, z = 2 \cos \phi$

So your integral becomes,

$8 \displaystyle \int_0^{2\pi} \int_{0}^{\pi/4} (cos^2\phi - cos^2 \theta \sin^2\phi - 2 \cos \theta \sin\phi \cos\phi) sin\phi \, d\phi \, d\theta = 2 \sqrt2 \, \pi$


问:
Lecture notes of complex analysis
I just want to go through the topics of complex analysis, in order to get an overview of the subject. Does there exists any good lecture notes for complex analysis which covers all the topics quickly? Like Milne's notes in Algebraic number theory. I am currently referring Tao's lecture notes on complex analysis, but at many points he has referred to Wikepedia so everything is not covered there...</p>

答:
Try Donald Sarason's Complex Function Theory.!

In preface:

These are the notes for a one-semester introductory course in the theory of
functions of a complex variable. The aim of the notes is to help students
of mathematics and related sciences acquire a basic understanding of the
subject, as a preparation for pursuing it at a higher level or for employing
it in other areas. The approach is standard and somewhat old-fashioned.

As these notes are a rather bare-bones introduction to a vast subject, the
student or instructor who uses them may well wish to supplement them with
other references

Title: Lecture notes of complex analysis


问:
Unable to reconstruct a new polynomial to find a given value
Suppose $f(x)= x^3+2x^2+3x+3$ and has roots $a , b ,c$.
Then find the value of
$\left(\frac{a}{a+1}\right)^{3}+\left(\frac{b}{b+1}\right)^{3}+\left(\frac{c}{c+1}\right)^{3}$.

My Approach :
I constructed a new polynomial $g(x) = f\left(\frac{x^{\frac{1}{3}}}{1-x^{\frac{1}{3}}}\right)$ and then used the Vieta's formula for the sum of roots taken one at a time to solve the sum.

But then I realised that I won't be able to do so as $g(x)$ is not a polynomial anymore.
Can anyone help me please.

Body:
We shall make use of the following well-known identity:
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz).$$
Now, using Vieta's Formula, $a+b+c=-2, ab+ac=bc=3$, and $abc=-3$.
Thus, by direct expansion, we have that
$(a+1)(b+1)(c+1)=-1.$

\begin{align}
\dfrac{a}{a+1}+\dfrac{b}{b+1} + \dfrac{c}{c+1} &= 3- \left(\dfrac{1}{a+1} +
\dfrac{1}{b+1} + \dfrac{1}{c+1} \right) \\
& = 3- \dfrac{(b+1)(c+1) + (a+1)(c+1) + (b+1)(a+1)}{-1} \\
& = 3   + bc+b+c+1 + ac+ a + c+1  + ab+ a +b+1 \\
& = 5.
\end{align}

$$\dfrac{a}{a+1} \cdot \dfrac{b}{b+1} \cdot \dfrac{c}{c+1} = \dfrac{abc}{-1}=3.$$

\begin{align}
\dfrac{a}{a+1}\cdot \dfrac{b}{b+1} + \dfrac{a}{a+1}\cdot \dfrac{c}{c+1} + \dfrac{b}{b+1}\cdot \dfrac{c}{c+1} & = \dfrac{3(c+1)}{c} + \dfrac{3(b+1)}{b} + \dfrac{3(a+1)}{a} \\
& = 3 \left(3+ \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a} \right) \\
& = 3 \left(3 + \dfrac{ab+bc+ca}{abc}\right) \\
& = 6.
\end{align}

Let $A=\dfrac{a}{a+1}, B=\dfrac{b}{b+1} , C=\dfrac{c}{c+1}$. Then,
\begin{align}
A^3 + B^3 + C^3 & = (A+B+C)(A^2+B^2+C^2-AB-BC-CA)+3ABC \\
& = 5 \left((A+B+C)^2-2(AB+BC+CA)-AB-BC-CA \right) + 3ABC& \\
& = 5 (25 -3(6))+3(3) \\
& = \mathbf{44}.
\end{align}


问:
Inverse of a function with square root
I'm trying to change order of integration of the following double-integral:
$$\int^2_1dx\int^{y=\sqrt{2x-x^2}}_{y=x-2}f(x,y)\,dy$$
I drew both curves and came to the conclusion that it's the same as writing $$\int^1_0dy\int^{x=?}_{x=y+2}f(x,y)\,dx$$where $?$ is the inverse function of $\sqrt{2x-x^2}$, which I'm not able to find.

How does one find the inverse of such function, or is there another way to do change of order here?

答:
Let $g:[1,2]\to\mathbb{R}:x\mapsto \sqrt{2x-x^2}$. We can calculate the inverse of $g$ by inverting the equation $y=\sqrt{2x-x^2}$ as follows. Squaring both sides yield $x^2-2x+y^2=0$. This is a second degree polynomial with discriminant $4-4y^
2$ which is greater than zero since $y\in g([1,2])=[0,1]$. Hence we have $x= 1+\sqrt{1-y^2}$ thus $g^{-1}:g([1,2])\to [1,2]: y\mapsto 1+ \sqrt{1-y^2}$.

Title: Inverse of a function with square root


问:
Decompose $P$ into the direct sum of irreducible representations.
Note: I need help with part (c).

Consider the representation $P: S_3 \rightarrow GL_3$ where $P_{\sigma}$ is the permutation matrix associated to $\sigma$.

a) Determine the character $\chi_P : S_3 \rightarrow \mathbb{C}$

b) Find all the irreducible representations of $S_3$.

c) Decompose $P$ into the direct sum of irreducible representations.  That is, find a single matrix $Q$ so that $Q^{-1}P_{\sigma}Q$ is block diagonal where the blocks along the diagonal are either $T_{\sigma}$, $\Sigma_{\sigma}$ or $A_{\sigma}$

My Attempt

Here is my overall progress for the problem:

I let $e$ to be the identity permutation, $x = (1 \ 2 \ 3)$ and $y = (1 \ 2)$

Then, I let $P_x = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$, $P_y = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $P_e = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

The conjugacy classes are as followed:

$\{e\}$

$\{y, xy, yx\}$

$\{x, x^2\}$

The character table shows that:

$$\chi (\{e\}) = 3$$
$$\chi (\{x, x^2\}) = 0$$
$$\chi (\{y, xy, yx\}) = 1$$

By the theorem I have applied, I have found three representations, which is congruent to the number of conjugacy classes.  They are:

trivial representation $T$

sign representation $\Sigma$

two-dimensional representation $A$, presented as the symmetries of equilateral triangle

The sum of their dimensions corresponds to the theorem I applied:

$$d_1^2 + d_2^2 + d_3^2 = |S_3| = 6$$

The only possibility for equality to hold is $d_1 = d_2 = 1$ and $d_3 = 2$.

Another character table shows that:

$\chi_T (\{e\}) = 1$

$\chi_T (\{x, x^2\}) = 1$

$\chi_T (\{y, yx, xy\}) = 1$

$\chi_A (\{e\}) = 2$

$\chi_A (\{x, x^2\}) = -1$

$\chi_A (\{y, xy, yx\}) = 0$

$\chi_{\Sigma} (\{e\}) = 1$

$\chi_{\Sigma} (\{x, x^2\}) = 1$

$\chi_{\Sigma} (\{y, yx, xy\}) = -1$

Now, I am stuck in determining what is the matrix $Q$ for $Q^{-1}P_{\sigma}Q$.

I know that I need to do "change of basis" and work out the vectors and stuff like this, but I can't seem to find the thorough approach.

EDIT:

Here is what I currently have:

For the trivial representation, I have the vector $(1 , 1 , 1)$ spanning the invariant subspace.

For the two-dimensional representation, I need to find two vectors $v$ and $w$ such that:

$P_x v = -v/2 + \sqrt{3}w/2$

$P_x w = -v/2 - \sqrt{3}w/2$

$P_y v = v$

$P_y w = -w$

I found the $Q$ matrix, which is:

$$\begin{bmatrix} 1 & 1 & -\frac{(1 + \sqrt{3})}{2} \\ 1 & 1 & \frac{(1 + \sqrt{3})}{2} \\ 1 & -2 & 0 \end{bmatrix}$$

But it is wrong.

Any advices or comments you have?

答:
The permutation representation of $S_3$ on $\mathbb{C}^3$ (though everything works over the rationals in exactly the same way) decomposes as the direct sum of the trivial representation and the two dimensional reflection representation, which you have called $A$.

More explicitly, the subspace spanned by the vector $v=(1,1,1)^t$ is a copy of the trivial representation, with $S_3$-stable complement spanned by the vectors $\alpha_1=(1,-1,0)^t$ and $\alpha_2=(0,1,-1)^t$, isomorphic to $A$. The matrix $Q$ can therefore be taken to have these three vectors as its columns.

In case these vectors $\alpha_1$ and $\alpha_2$ seemed to come out of the blue, here is where they actually came from: the usual Hermitian inner product on $\mathbb{C}^3$ is $S_3$-invariant, so the orthogonal complement of $v=(1,1,1)^t$ is an $S_3$-stable complement to its span. But this orthogonal complement consists exactly of the vectors whose coordinates sum to $0$, and the $\alpha$'s are just what I consider the most obvious basis of this space.


问:
Let $X_n$ a sequence such that $x_n \leq 7$ and $x_n < x_{n+1}$
Let $X_n$ a sequence such that $x_n \leq 7$ and $x_n < x_{n+1}$ show that $x_n$ converges without using Theorem of the Convergence of Monotonic Sequences

the problem is I don't understand how the sequence is defined, any hint of what should I do to prove it?

答:
The proof can be written with the following outline. The questions in parentheses are questions that should be answered in order to make the proof complete.

Let $S = \{x_n : n \in \Bbb N\}$, and let $\alpha$ denote the supremum $\alpha = \sup(S)$ (why does the supremum exist/why is it finite?) We will show that $x_n$ converges to $\alpha$.

Let $\epsilon > 0$. Take $m \in \Bbb N$ to be such that $x_m > \alpha - \epsilon$ (how do we know that such an $m$ exists?). For any $n \geq m$, we see that $|x_n - \alpha| \leq |x_m - \alpha| < \epsilon$ (how?). Thus, $x_n$ converges to $\alpha$.


答:
$$ y=\sqrt{2x-x^2} $$

is equivalent to

$$ x^2 - 2x = -y^2 $$

$$ x^2-2x+1 =1-y^2 $$

$$ (x-1)^2=1-y^2 $$

and therefore

$$ x = 1 + \sqrt{1-y^2} $$

I took into account the fact that here, $0 \le y \le 1$ and $1 \le x \le 2$

Title: Inverse of a function with square root

Body:
I'm trying to change order of integration of the following double-integral:
$$\int^2_1dx\int^{y=\sqrt{2x-x^2}}_{y=x-2}f(x,y)\,dy$$
I drew both curves and came to the conclusion that it's the same as writing $$\int^1_0dy\int^{x=?}_{x=y+2}f(x,y)\,dx$$where $?$ is the inverse function of $\sqrt{2x-x^2}$, which I'm not able to find.

How does one find the inverse of such function, or is there another way to do change of order here?


问:
A little help integrating this torus?

Let $\mathbf{F}\colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be given by

  $$\mathbf{F}(x,y,z)=(x,y,z).$$

  Evaluate $$\iint\limits_S \mathbf{F}\cdot dS$$ where $S$ is the surface of the torus given by

  $$\begin{align*}
x&=(R+\cos(\phi))\cdot \cos(\theta)\\
y&=(R+\cos(\phi))\cdot \sin(\theta)\\
z&=\sin(\phi)
\end{align*}$$
  and$$0\leq{\theta}\leq{2\pi},\qquad 0\leq{\theta}\leq{2\pi}.$$

  Assume $S$ is oriented outward using the outward unit normal.

My Take

Ok, so I know I have to start by taking the cross product of the partial derivatives of theta and phi, but when it says using the outward unit normal, which vector has to be positive in order for it to be the outward normal? Is it the $i$, $j$ or $k$ vector? How do I know which vector it is? What will my next step be after I find the cross product?

答:
I don't know if you wanna compute the surface integral the hard way, which is, computing the surface outward normal pointwisely and then integrating. For me this problem looks rather like a standard "applying the divergence theorem" exercise.

Denote the interior of the torus as $T$, and $\partial T = S$, then by Divergence theorem:
$$
\iint_S \mathbf{F}\cdot d\mathbf{S} = \iiint_T \nabla \cdot \mathbf{F} \,dV  = 3 |T|,
$$
where $|T|$ is the volume of $T$. Now we compute the volume using the parametrization you gave: based on the parametrization, if we assume $R>1$, then the torus has $R$ as the distance from the center of the tube to its center, the radius of the tube is 1. Thus the interior's parametrization is:
$$
\begin{aligned}
&x =  (R + r \cos \phi) \cos{\theta}
\\
&y=  (R + r \cos \phi) \sin{\theta}
\\
&z=  r \sin \phi
\end{aligned}
$$
for $(r,\phi,\theta)\in [0,1]\times[0,2\pi]\times[0,2\pi]$. Compute the Jacobian:
$$
\left|\frac{\partial (x,y,z) }{\partial (r,\phi,\theta)} \right|= r(R+r\cos\phi).
$$
Hence the volume $|T|$ is:
$$
|T| = \int^1_0 \int^{2\pi}_0 \int^{2\pi}_0 r(R+r\cos\phi) d\phi d\theta dr = 2\pi^2 R,
$$
and
$$
\iint_S \mathbf{F}\cdot d\mathbf{S} = 6\pi^2 R.
$$
Wikipedia's entry of Torus uses Pappus's centroid theorem to compute the volume, which is much simpler, but I guess we are supposed to use the integral to compute for we are given a parametrization here.

Title: A little help integrating this torus?


问:
Associated Legendre Polynomials from the Schrodinger Equation for the Hydrogen Atom
EDIT: See below for answer, I found an alternative approach which is better in my opinion than trying to find a power series solution. I still don't fully understand why $m>l$ i.e. $P_l^m=0$ is not an acceptable solution. Intuitively I guess this results in the wave function being equal to zero and maybe we have to be able to normalise the wave function.

How can I study the convergence of the following series:

$$ P_l^m=(1-x^2)^{m/2}[c_0+c_1x+\sum_{k=2}^{\infty}c_kx^k]$$

where $$c_{k+2}=\frac{((m+k+1)(m+k)-A)}{(k+2)(k+1)} c_k$$

In particular, how is it possible to show that the series only converges for $A=l(l+1)$ where $l$ is a nonnegative integer, $-l\leqslant m\leqslant l$?

This series which comes up when solving the Schrodinger equation for the hydrogen atom represents an associated Legendre polynomial (according to this page).

Note that it is possible to show the series is equivalent to the following;

$$P_l^m= (1-x^2)^{m/2}[c_0 + \sum_{k=1}^{\infty} c_0x^{2k}\prod_{i=1}^{k}\frac{(m+2i-1)(m+2i-2)-A}{2i(2i-1)}] + (1-x^2)^{m/2}[c_1x + \sum_{k=1}^{\infty}c_1x^{2k+1}\prod_{j=1}^{\lfloor{(2k+1)/2}\rfloor}\frac{(m+2j)(m+2j-1)-A}{2j(2j+1)}]$$

which is equivalent to;

$$(1-x^2)^{m/2}(c_0\sum_{k=0}^{\infty} \frac{c_{2k}}{c_0}x^{2k}+c_1\sum_{k=0}^{\infty}\frac{c_{2k+1}}{c_1}x^{2k+1})$$

Alternatively rather than examining the convergence of the series it might be easier to show that the series is equivalent to the following definition of Legendre polynomials:

$$P_l^m = (-1)^m(1-x^2)^{m/2}\frac{d^m}{dx^m}P_l(x)$$

where

$$P_l(x)=\frac{1}{2^l l!}\frac{d^l}{dx^l}[(x^2-1)^l]$$

Instead of proving the values of $l$ and $m$ for which the series converges it might be easier to show that it is equivalent to this definition from which it is quite intuitive that $l$ is a nonnegative integer since it is the order of a derivative and for the same reason $m$ must be an integer. Taking a derivative of order $l+m$ in the definition of $P_{l,m}$ implies that $-l$ is a lower bound on $m$. If $m>l$ then $P_{l,m}=0$ although I'm not sure why this is not possible.

In order to solve the polar differential equation I made the substitution $y(x)=(1-x^2)^{m/2}f(x)$ which gives another differential equation in terms of f(x) which can be solved this using a power series giving the result above.

Notice that;

$$c_k = \frac{((m+k-1)(m+k-2)-A)}{k(k-1)} c_{k-2}, k=2,3,4,5,...$$

It's possible to get a product formula rather than a recursive formula for the coefficients.

$$c_k= c_0\prod_{i=1}^{k/2}\frac{(m+2i-1)(m+2i-2)-A}{2i(2i-1)}, \quad  \text{k even}$$

$$c_k= c_1\prod_{i=1}^{\lfloor{k/2}\rfloor}\frac{(m+2i)(m+2i-1)-A}{2i(2i+1)}, \quad \text{k odd}$$

then a series of all the even terms is $c_0 + \sum_{k=1}^{\infty} c_{2k}x^{2k}$ and a series with all the odd terms is $c_1x + \sum_{k=1}^{\infty} c_{2k+1} x^{2k+1}$ where we use the appropriate even/odd formula for $c_{2k}$ and $c_{2k+1}$.

There is some discussion here although I didn't find it very helpful, if someone can provide a more in depth explanation that would be very much appreciated.

答:
I found a better approach as described here.

When solving the polar equation you get an equation $$(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+(l(l+1)-\frac{m^2}{1-x^2})y=0$$

If you make the substitution $y=(1-x^2)^{m/2}f(x)$ you get a new differential equation;

$$(1-x^2)\frac{d^2f}{dx^2}-2x(m+1)\frac{df}{dx}+(l(l+1)-m(m+1))f(x)=0$$

If you set $m=0$ you get the regular Legendre equation for which the ordinary Legendre polynomial $P_l$ is a solution. If you differentiate the equation above you get again the regular Legendre equation except $f$ is replaced by $f'$ and $m$ is replaced by $m+1$ that means that since $P_l$ solves the equation for m=0, $P_l'$ solves the equation for m=1. You can continue differentiating and the same holds for a given m the solution of the above equation is the mth derivative of the Legendre polynomial. You could prove that it holds for all $m$ by induction. Then since we made a substitution $y=(1-x^2)^{m/2}f(x)$ the solution to the original equation is $$P_l^m=(1-x^2)^{m/2}\frac{d^m}{dx^m}P_l$$ which is called an associated Legendre polynomial. You can deduce from the azimuth equation that $m$ must be an integer as described here, I found this easiest if you express the solution to the azimuth equation in terms of sines and cosines rather than exponentials.

If you set $m=0$ you get that the solution is the Legendre polynomial $P_l$ which involves a derivative of order $l$ therefore $l$ must be a nonnegative integer since it is the order of a derivative. Then if you expand the definition of the associated Legendre polynomial you have a derivative of order $l+m$ which means that $m$ must be bounded below by $-l$. We have deduced that $l$ is a nonnegative integer and that $m$ is an integer bounded below by $-l$.

How can we deduce that $m\le l$? If $m>l$ then $P_l^m=0$. Maybe this is not allowed because we have to be able to normalise the wave function. Is this an allowable solution?


问:
A generic terminology for left and right parts - i.e. M and N - of a general lambda application term MN?
As pointed out in the comments, the rhs and lhs are both terms, so "rhs term" and "lhs term" works. There may be a more evocative, and equally concise term.

Presumably (?) the only important role of the lhs is as a "substitution scope" for the rhs term. So the rhs term of an application might be the "argument" of the application and the lhs it's "scope". Note: lhs term need not generally be an abstraction (it might be a variable).

A subset of application terms - those with lhs an abstraction - are so called $\beta-redex$'s.

答: In a $\lambda$-calculus, an expression of the form $M\,N$ is called an application, where $M$ is the applicand and $N$ is the argument. These terms appear in most texts on the $\lambda$-calculus (though "applicand" is not as common as the other two). See, for instance, the Wikipedia article on combinatory logic.


问:
Prove that an isometry of a compact metric space is not topologically mixing.
Let $(X,d)$ be a compact metric space and let $f:X\to X$ be an isometry. We need to show that $f$ is not topologically mixing. Recall that if $Y$ is a topological space and a map $g:Y \to Y$ is continuous then $g$ is topologically mixing if for every pair of non-empty open sets $U, V \subseteq X$ there exists $n_0 \in \mathbb{N}$ such that $f^{-n}(U) \cap V \neq \emptyset \, \forall n \geq n_0.$

My attempt: Fix an arbitrary $x \in X.$ Suppose $f$ was topologically mixing and applying the mixing condition to $U= V = B(x, \frac{1}{k})$ we can find a strictly increasing sequence $(n_{k}) \in \mathbb{N}$ such that $f^{-n}(B(x, \frac{1}{k}))\cap B(x, \frac{1}{k}) \neq \emptyset \, \forall n \geq n_k$

$\Rightarrow \, \, \exists \, (x_k) \in B(x, \frac{1}{k})$ such that $d(f^{n}(x_{k}), x) < \dfrac{1}{k}$ for $n\geq n_{k}$

By the fact that an isometry of a compact metric space must be a homeomorphism and that $f$ is an isometry we can say:

$\Rightarrow d(x_k, x)< \dfrac{1}{k}$ and $d(x_k, f^{-n}(x))< \dfrac{1}{k}$ for $n \geq n_{k}$

$\Rightarrow d(x, f^{-n}(x))< \dfrac{2}{k}$ for $n \geq n_{k}$ (by the triangle inequality)

$\Rightarrow d(x, f^{n}(x))< \dfrac{2}{k}$ for $n \geq n_{k}$

Then taking limits as $k \to \infty$, we note that $n_{k} \to \infty$ and hence $n \to \infty.$

$\Rightarrow f^{n}(x) \to x$ so by continuity, $f^{n+1}(x)\to f(x)$ and we must therefore have $f(x)=x.$

But $x \in X$ was arbitrary. So $f$ must be the identity, but it's immediately clear that the identity cannot be mixing (take two disjoint non-empty open subsets of $X.$) and we get a contradiction.

Therefore no such $f$ can be mixing.

My question: Is this proof correct? If so, what might be a simpler proof perhaps avoiding use of the fact that such an $f$ is in fact invertible?

答:
Although it has been a long time, I think it's useful to provide an answer.

First of all, we need one more assumption that $X$ contains at least two points. Let $x \neq y \in X$. Suppose $d(x, y) = 4\delta$ then choose open balls $B(x, \delta)$ and $B(y, \delta)$ with $B(x, \delta) \cap B(y,\delta) = \emptyset$. Then consider the open ball $U$ with radius $\frac{\delta}{2}$. If $f$ is topologically mixing then we have there exist $n_x, n_y$ such that $f^n(U) \cap B(x, \delta) \neq \emptyset$ and $f^n(U) \cap B(y, \delta) \neq \emptyset$ for all $n \geq \max\{n_1, n_2\}$. However, as $f$ is an isometry, then we have \begin{align} d(x,y) &\leq d(x, f^n(z_1)) + d(f^n(z_1), f^n(z_2)) + d(f^n(z_2), y) \\
&= d(x, f^n(z_1)) + d(z_1, z_2) + d(f^n(z_2), y) \\
&\leq \delta + \delta + \delta = 3\delta \end{align} where $f^n(z_1) \in B(x, \delta)$ and $f^n(z_2) \in B(y, \delta)$.
Then it contradicts to the fact that $d(x, y) = 4\delta$. Therefore we have $f$ is not topologically mixing.