问:
Understanding tensor transformation under coordinate rotation formula
Given a Cartesian vector $x \in \mathbb{R}^3$, and suppose we wish to find the coordinates $(x'_1, x'_2, x'_3)$ of $x$ in a new rotated coordinate system  that shares the same origin with original coordinate system.

One can show that $x'_j = x_k C_{ki}$, where  $C_{ki}$ is the $3 \times 3$ matrix of direction cosines, and the proof is relatively easy.

Now, given a second order tensor $T_{ij}$, my textbook defines the transformation under the new rotated coordinate system as $$ T'_{mn} = C_{im}C_{jn}T_{ij}$$

Yet, the above formula is given with no proof, only a reference to Mechanics of Deformable Bodies (1964) book by Sommerfeld, which I don't have access to. I tried to prove it myself with no luck.

Note: I'm a beginner in tensor calculus, so please go easy on me.

答:
If you are learning tensor calculus with several sources of different ages and backgrounds, you will run afoul of several points of conflicts with the conventions you were exposed to.

1. Different notations developed from several backgrounds that are relatively independent and thus kept/keep their traditions for a long time.

Despite having predecessors in the 18th and 19th century, matrix and tensor notation was invented and came into wide use only in the late 1920ies. There was the systematic coordinate version of differential geometry in the Ricci calculus as the mathematical apparatus behind general relativity. Then the algebra of infinite matrices in the developing quantum theory in the Heisenberg picture. (These were later replaced by an operator calculus, and the previous matrices only appear if selected eigen-decompositions are applied.) And then several other conventions based on their own histories, related but distinct, and with not so stringent demands on generality, for applications in Euclidean geometry, the mechanics of continua etc. where it was more important to have a compact form for a restricted set of material coefficients and their application in equations.

Your source appears to fall into the latter category.

2. Some conventions are a matter of taste among conflicting requirements.

The guiding principle exposed in the first half of your question is that linear transformations be written down in the order they occur. That is, if you first apply transformation $C_1$ and then transformation $C_2$, then their combined action be represented by the matrix product $C_1C_2$. Note that this convention underlies also many computer graphics libraries, but is not the prevailing one in mathematics and physics where operators act from the left, so if they are represented as matrices the main vector objects are column vectors.

To have the matrices act properly on the coefficient vector thus needs that vector to the left, making it a row vector. But that is from interpreting it now, it is not clear that the now common identification of a $M_{ij}$ index pair as row and column index in that order was that fixed at the time of the publication of the book, or even if there was only one fixed rectangular scheme to organized these coefficients.

3. Abstract axiomatic versus concrete constructions

The next point is that in physics the point-of-view is prevailing that a tensor is the data structure with its transformation laws. From a mathematical perspective this looks the same as saying that the real numbers are the Dedekind cuts of the rationals (or some other fixed realization of the axioms).

For a mathematician a tensor is an element of a linear space that is the product (of some sorts) of other linear spaces. The transformation laws are then a consequence of the nature of the factor spaces and basis changes there. If the factor spaces are the same vector space and its dual, then these basis changes may be synchronized.

In total it means that the formula under discussion is not something to prove, but it is a defining property of $T$ as a tensor. It characterizes it as doubly covariant. That is, if ${\bf x}=x_i{\bf e}_i$, then $${\bf T}=T_{ij}({\bf e}_i\otimes {\bf e}_j)$$ acts like a "double vector" (bi-vector?), a sum of products of vectors.

To make the circle complete, if the underlying basis is changed as ${\bf e}_i=C_{im}{\bf e}'_m$, then
$$
{\bf x}=x_iC_{im}{\bf e}'_m=x'_m{\bf e}'_m
\\~\\
{\bf T}=T_{ij}(C_{im}{\bf e}'_m\otimes C_{jn}{\bf e}'_n)
=T_{ij}C_{im}C_{jn}({\bf e}'_m\otimes {\bf e}'_n)
=T'_{mn}({\bf e}'_m\otimes {\bf e}'_n)
$$
But this is truly a circle, as the nature of $T$ was divined from the given transformation law, and the last equation sequence just confirms the compatibility of the mathematical with the physical idea of a tensor.

Note that $C$ acts backwards on the basis change and forwards on the coordinate change, thus the naming of the coordinate tuple as "contra-variant" to the basis tuple.


问:
How to solve this $\lim_{n\to\infty} \frac{\sqrt{n}}{n+2}$
How to solve this?
$$\lim_{n\to\infty} \frac{\sqrt{n}}{n+2}$$

My approach is to use Squeeze theorem but no results. Like this

$$0 < \frac{\sqrt{n}}{n+2} < \frac{n}{n+2} < 1 $$

Maybe there is another solution.

答:
You must divide by the highest power of $n$, i.e. by $1$:
$$\frac{\sqrt{n}}{n+2}\cdot\frac{\frac{1}{n}}{\frac{1}{n}}=\frac{\frac{1}{\sqrt{n}}}{1+\frac{2}{n}}~\to~\frac{0}{1}=0$$

ContentLicense: CC BY-SA 4.0


问:
GIven a loop in $X$, how do I know that it is a continuous map $f: S^1 \rightarrow X$?
Very simple question, but it left me puzzled. Let $[\omega] \in \pi_1(X, x_0)$. Then, how do I show that $\omega \in [f]$ for some continuous function $f: (S^1, s_0) \rightarrow (X, x_0)$?

答:
Hint: try to loot at $S^1$ as a quotient of the topological space $[0,1]$.

Moreover, the notation $[\omega] \in \pi_1(X, x_0)$ means that $\omega$ is a representative of it's own equivalence class of continuous fonctions $\omega:[0,1]\to [X]_{x_0}$ ($[X]_{x_0}$ is the path connected class of $x_0$ in $X$) with the ratio of homotopy with respect to $\partial[0,1]$ such that:

$\omega(0)=\omega(1)=x_0$

Hence one can think about $\omega:[0,1]\to [X]_{x_0}$, $\omega(0)=\omega(1)=x_0$ as a continuous function $\omega':S^1\to [X]_{x_0}$ (try to use the universal property of quotient space).


问:
Prove that $e^{iAt}$ is unitary if $A$ is Hermitian
I would like to prove that if $A$ is Hermitian matrix, then exponential $e^{iAt}$ for $t \in \mathbb{R}$ is unitary. This means I need to show that $e^{iAt}[e^{iAt}]^\dagger=[e^{iAt}]^\dagger e^{iAt} = I$.
I have a proof (below) but I would like to verify if my reasoning is correct.

I started with spectral decomposition of $A$ and definition of a matrix exponential:
$$
[e^{iAt}]^\dagger = \Big[\sum_{j=1}^n e^{i \lambda_j t} u_j^\dagger u_j \Big]^\dagger =
\sum_{j=1}^n (e^{i \lambda_j t})^* [u_j^\dagger u_j]^\dagger = \sum_{j=1}^n e^{-i \lambda_j t} [u_j^\dagger u_j] = e^{-iAt},
$$
where $\lambda_j$ is eigenvalue of $A$ and $u_j$ is a respective eigenvector.

This is possible because firstly eigenvalues $\lambda_j \in \mathbb{R}$ thanks to hemiticity of $A$, hence complex conugate to $e^{i \lambda_j t}$ is $e^{-i \lambda_j t}$, and secondly $[u_j^\dagger u_j]^\dagger = u_j^\dagger (u_j^\dagger)^\dagger = u_j^\dagger u_j$.

So, now we have $e^{iAt}[e^{iAt}]^\dagger = e^{iAt}e^{-iAt}= e^{itA-itA}$ which is possible because $A$ apparently commutes with itself. Finnaly $e^{itA-itA} = e^O = I$ and similarly for the matrices switched.  So, the $e^{iAt}$ is unitary.

Is my proof correct?

答:
It's correct.

Here's another more elementary proof. First notice that for any $M\in\mathcal{M}_n(\mathbb{C})$,  $$\exp(M)^\dagger = \left(\sum_{k=0}^{\infty} \frac{M^k}{k!}\right)^\dagger=\sum_{k=0}^\infty \frac{(M^\dagger)^k}{k!}=\exp(M^\dagger)$$

This is correct because, $M\mapsto M^\dagger$ is ($\mathbb{R}$)-linear (therefore continuous) and because $(AB)^\dagger = B^\dagger A^\dagger$.

Now, in the case where $M=itA$ with $A$ hermitian, this implies $$\exp(itA)^\dagger=\exp((iAt)^\dagger)=\exp(-iAt)$$


答:
As you know, the simplest standard ways are,

$$\lim_{n\to\infty} \frac{\sqrt{n}}{n+2}=\lim_{n\to\infty} \frac{1}{\sqrt n+\frac{2}{\sqrt n}}=0.$$

Also, since denominator goes to infinity, we are allowed to subtract $2.$

$$\lim_{n\to\infty} \frac{\sqrt{n}}{n+2}=\lim_{n\to\infty} \frac{\sqrt{n}}{n}=\lim_{n\to\infty} \frac{1}{\sqrt n}=0.$$

Title: How to solve this $\lim_{n\to\infty} \frac{\sqrt{n}}{n+2}$

Body:
How to solve this?
$$\lim_{n\to\infty} \frac{\sqrt{n}}{n+2}$$

My approach is to use Squeeze theorem but no results. Like this

$$0 < \frac{\sqrt{n}}{n+2} < \frac{n}{n+2} < 1 $$

Maybe there is another solution.


问:
Find local isometry $f:X\rightarrow Y$,
So I'm revising for an exam in differential geometry, and have got stuck on a question regarding isometries.

We define a local isometry as a map $f:X\rightarrow Y$, (between 2 regular m-surfaces X,Y) where, for all $p\in X$ and all $u \in T_{p}X, |df_{p}(u)|=|u|$

The example in question involves finding a local isometry between 2 surfaces:

Let $X=\mathbb{R}^2${($0,0$)} and $Y$={$x$ $\in$ $\mathbb{R}^3$ $8x_{1}^2 +8x_{2}^2=x_{3}^2$ and $x_{3} >0$}.

We can use polar coordinates $q(r,\theta)=(rcos\theta, rsin\theta)$ on X to define a map $f:X\rightarrow Y$,

Now apparently, this map is
  $f(r,\theta)=(\frac{r}{3}cos\theta,\frac{r}{3}sin\theta,\frac{\sqrt{8}}{3}r)$

We then go on to prove this is well defined etc, but I really don't see where f has come from.  Has anyone got any ideas? Many thanks.

答:
Without loss of generality, note that any point in $Y$ can be written in the form $$x_1=\rho\cos \theta,x_2=\rho \sin\theta,x_3=\pm \sqrt{8x_1^2+8x_2^2}=+ \sqrt 8 \times \rho$$
just by using plane polars for $(x_1,x_2)$.

Then they set $\rho(r)=r/3$ (no idea why they make that choice, but it'll do) and their map $f$ follows by identifying $\theta$.

ContentLicense: CC BY-SA 3.0

Title: Find local isometry $f:X\rightarrow Y$,


问:
If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$
If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$

The solution goes as follows:

$a=\frac{x}{y+z}$, $b=\frac{y}{z+x}$, $c=\frac{z}{x+y}$. We can see that:

$ab+bc+ac+2abc=1$

It's enough if: $(a+b+c)^2\ge 4-14abc$

$(a+b+c)^2\ge 4(ab+bc+ab+2abc)-14abc$

From Schur it is enough if: $6abc\ge\frac{9abc}{a+b+c}$ which is true from Nesbit.

Could you please provide a more intuitive and easier approach?

答:
Well it's an attempt to delete the square root .

My only idea is to use Bernoulli's inequality and play with the coefficient $\sqrt{2}$.We have :

$$\sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}\leq \sqrt{\alpha}\Big(1+\frac{1}{2}\Big(\frac{4}{\alpha}-1-\frac{14}{\alpha}\frac{xyz}{(x+y)(y+z)(x+z)}\Big)\Big)$$

Remains to show (using the OP's substitution) when:

$$ \sqrt{\alpha}\Big(1+\frac{1}{2}\Big(\frac{4}{\alpha}-1-\frac{14}{\alpha}abc\Big)\Big)\leq a+b+c$$

Wich seems to be easier I think.

The last inequality is equivalent to :

$$\frac{y}{2}+\frac{2}{y}\leq a+b+c+\frac{7}{y}abc$$

With $y^2=\alpha$ so now a bit of algebra to get :

$$y^2+4\leq 2y(a+b+c)+14abc$$

Now putting $y=a+b+c$ we get the inequality of the OP.

The end is the same as the OP.


问:
A metric space is second-countable iff it is separable: Proposition 2.1.4 in Dudley's "Real Analysis and Probability"
I am reading Proposition 2.1.4 in Dudley's "Real Analysis and Probability (Cambridge Edition)" but I have some troubles.

That proposition and the proof are here:
A metric space $(S,d)$ is second-countable iff it is separable.

Proof: ($\longleftarrow$) Let $A$ be countable and dense in $S$. That is $\overline A=S $. Let $\mathcal U$ be the set of all balls $B(x, 1/n)$ for $x\in A$ and $n=1,2,\dots$. To show that $\mathcal U$ is a base, let $U$ be any open set and $y\in U$. Then for some $m$, $B(y, 1/m)\subset U$. Take $x\in A$ with $d(x,y)<1/(2m)$ Then $y\in B(x, 1/(2m))\subset B(y, 1/m)\subset U$, so U is the union of the elements of $\mathcal U$ that it includes, and $\mathcal U$ is a base, which is countable. Since Cartesian products of countable sets (in this proof, $A$ and $\mathbb{N}$ are so) is also countable, $\mathcal U$ is a countable base.

($\longrightarrow$) Suppose there is a countable base $\mathcal V$ for the topology, which we may assume consists of nonempty sets. By the axiom of choice, let $f$ be a function on $\mathbb{N}$ whose range contains at least one point of each set in $\mathcal V$. Then this range is dense.

I have two questions:

in the proof of ($\longleftarrow$), where did Dudley use the denseness of A?
Note that in Theorem 2.1.3 in his book says that for any $A\subset S$, $\overline A$ is the set of all $x\in S$ such that for some net $x_i \longrightarrow x$ with $x_i\in A$ for all $i$.
That is, $\overline A$ is the set of all $x\in S$ such that any neighborhood of $x$ contains all but finitely many points $x_i$ with $x_i\in A$. Do I need to use this fact?

in the proof of ($\longrightarrow$), why the range of $f$ is dense?

答:
When Dudley said 'Take $x \in A$ with $d(x,y) <1/2m$ he used the fact that $A$ is dense.

If $x \in S$ and $r>0$ then $B(x,r)$ contains a member $V$ of $\mathcal V$ with $x \in V$ (by the definition of a base). By the choice of $f$ there is at least one point $y$ in the range of $f$ which belongs to $V$. Now $d(x,y) <r$ (becasue $V \subseteq B(x,r)$). This proves that the range of $f$ is dense.


问:
Extremal of the variational problem $J[y] = \int_{0}^1 (y'(x))^2 dx$ with the conditions $y(0)=0$, $y(1) = 1$ and $\int_{0}^1 y(x)dx =0$
I want to solve the variational problem $$ J[y] = \int_{0}^1 (y'(x))^2 dx $$ with the conditions $y(0)=0, y(1)=1$ and $\int_{0}^{1}y(x)dx = 0$.

Using Euler's condition (for an extremal):
$$ \frac{\partial F}{\partial y}- \frac{\partial^2 F}{ \partial x \partial  y'}- \frac{\partial ^2 F}{\partial y \partial y'} -y'' \frac{\partial^2 F}{\partial y'^2} =0. $$

This will give: $ y'' \frac{\partial ^2 F}{\partial y}=0 $

If $y''=0$ then the solution is of the form $Ax+B$, which can not be the case here. After this point I am unable to solve the problem.

Can you please help me understand what am I doing wrong. Any help is highly appreciated.

答:
Consider the lagrangian

$$
\int_0^1(\dot y)^2dt +\lambda\int_0^1 y dt
$$

then from Euler-Lagrange

$$
2\ddot y -\lambda = 0
$$

and solving with boundary conditions

$$
y = \frac 14(4t-\lambda t+\lambda t^2)
$$

The solution should satisfy the restriction so

$$
\int_0^1 ydt = \frac 12-\frac{\lambda}{24}=0\Rightarrow \lambda = 12
$$

and the solution is

$$
y = 3t^2-2 t
$$


Body:
$$\lim_{n\to\infty}\frac{\sqrt{n}}{n+2}$$
$$=\frac{1}{\lim_\limits{n\to\infty}{\frac{n+2}{\sqrt{n}}}}$$
$$=\frac{1}{\lim_\limits{n\to\infty}{\sqrt{n}}+\frac{2}{\sqrt{n}}}$$
$$=\frac{1}{\infty+\frac{2}{\infty}}$$
$$=\frac{1}{\infty+0}$$
$$=\frac{1}{\infty}$$
$$=0$$


问:
What is larger -- the set of all positive even numbers, or the set of all positive integers?
We will call the set of all positive even numbers E and the set of all positive integers N.

At first glance, it seems obvious that E is smaller than N, because for E is basically N with half of its terms taken out.  The size of E is the size of N divided by two.

You could see this as, for every item in E, two items in N could be matched (the item x and x-1).  This implies that N is twice as large as E

On second glance though, it seems less obvious.  Each item in N could be mapped with one item in E (the item x*2).

Which is larger, then?  Or are they both equal in size?  Why?

(My background in Set theory is quite extremely scant)

答:
Mathematics is the art of clever forgetting.  The first mathematical breakthrough, numbers, came about when people realized that if you just forgot about whether it was 5 cows + 3 cows, or 5 rocks + 3 rocks or whatever you always got 8.  Numbers are what you get when you look at collections of objects and forget what kind of object they are.

When you say "as sets" you mean you're forgetting a lot of information, in particular you don't care about what the names of the elements in that set are or what properties those elements have.  As sets the positive numbers and the positive even numbers are "the same" (that is are in bijection) because you can take 1,2,3,...  and just rename 1 to 2, and 2 to 4, and n to 2n, and you've just renamed all the elements and got the even numbers!

However, if you want to remember more about these sets, for example that they're not arbitrary sets they're both subsets of the natural numbers, then they become distinguishable.  Depending on how you want to measure "size of a subset of the natural numbers" they might be different sizes.  For example, a common way to measure "size of a subset of the natural numbers" is by its "density."  That is you look at the first N numbers and calculate what fraction of them are in your set, and then take the limit as N goes to infinity (warning for sufficiently complicated sets this limit might not exist).  So for your two examples, one has density 1 and the other has density 1/2, which is one way to make precise the intuition that the former is bigger as a subset of the natural numbers (though not as a set) than the latter.

ContentLicense: CC BY-SA 2.5


问:
How to calculate the expected distribution of results using random numbers?
See this SO thread: Calculating which item is next in a percentage distribution.

To summarize,

you calculate

```<code>max ((rand1 * 10), (rand2 * 20), (rand3 * 30), (rand4 * 40))
</code>```

for four random numbers in a range (say 1-100).

If the max is

```<code>rand1 * 10, you choose 'A'
rand2 * 20, you choose 'B'
rand3 * 30, you choose 'C'
rand4 * 40, you choose 'D'
</code>```

What is the expected distribution of choosing A, B, C and D?

How to derive it mathematically?

(Using some C code, the percentages come out to be 1, 10, 31 and 56 respectively)

答:
The probabilities are given by:

$p_i = \int_0^1 \prod_{j \in {1,2,3,4}, j\ne i} min(\frac{ix}{j},1)  dx, i = 1, 2, 3, 4$

each factor in the integrand computes the probability density that the i-th random variable is larger than the j-th one.

The exact result is

[1/96, 31/288 91/288 163/288]


答: Hint: $\ \rm mod\,\ 2n\!:\ 24\equiv 77\,\Rightarrow\, mod\,\ 2\!:\ 24\equiv 77\,\Rightarrow\, 0\equiv 1$</p>

Title: Modulo question about equality

Body:
True or false?

$$24 \equiv 77 \mod 16 $$

$1.$ $77/16 = 4.8125 $

$2.$ $4.8125 - 4 = .8125$

$3.$ $0.8125 \times 16 = 13$

$4.$ $24 != 13$

So the answer is false? Am I right?


问:
Definition of a field in maths and physics
Is the field as defined in analysis ( i.e a set with multiplication and addition defined on it such that they obey certain field axioms like closedness and associativity .... ) related by anyway to the " field" in physics ( e.g electric field, force field …).

Additionally, if they are not equivalent, you would do me great justice if you can refer me to some rigorous definition of "field" in physics. Thank You

答:
I don't  think that there is any connection; it is just a coincidence that a mathematician chose field for one concept and a physicist chose it for another. Such terms, especially in mathematics, are often completely arbitrary.  A mathematical field does not reasonable a place where you might play football or graze a horse, a ring in algebra does not look much like a thing that you might put on your finger, etc.

One way to see the arbitrary nature is to look at the corresponding terms in other languages.  A neat way to do this is to find the concept in English Wikipedia and then switch the language.  So, let's find the German for the two uses of field.

Here is the English article for field in the mathematical sense: Field (mathematics) and here is the German: Körper (Algebra). Now, let's find the day to day meaning of Körper (using Google Translate): Body.  So, again an arbitrary word but a quite different one.  This explains why $K$ is a popular symbol for a field in mathematics.

Let's do the same trick for the physics sense: Field (physics) and Feld (Physik).  This time, the meaning is close to the English.  So, a German speaker would be unlikely to ask a question like yours.

Playing the same trick for French gives a similar result: Corps (body) for the mathematical object and Champ (field) for the mathematical one.

Let's get more exotic.  Chinese for the mathematical object is 域 (Domain) and the physics object is 场 (field).

So, in my sample of four languages, I find that the physics term is quite consistent but the mathematics one varies.  Only in English (*) do they coincide.

(*) In my small sample, I expect that they match in some other languages.

I could not resist checking a few more languages.  Spanish and Portuguese follow French with body for the mathematical object but Italian follows English with field for both.  Danish follows German but with the less easily guessed word: Legeme.  I could not find any that did not use a close equivalent for the physics object.


问:
The antiderivative of $\frac{1}{x-ib}$
I know if my domain $D_{\mathbb{R}}\subset\mathbb{R}$ and $b$ is a real constant, then:

$$\int_{D_\mathbb{R}}\dfrac{1}{x-ib}\mathrm{dx}=\ln(|x-ib|)$$

But if my domain $D_{\mathbb{C}}\subset\mathbb{C}$:

$$\int_{D_\mathbb{C}}\frac{1}{x-ib}\mathrm{dx}=\log(x-ib)=\ln|x-ib|+i\mathrm{arg}(z)$$

This feels a bit counterintuitive to me, but is it correct?

答:
For the first one, since $D\subset\mathbb R$, we know that $D$ is an interval such as $(a,b)$ with $a<b$ since $D$ is connected.

This means the first integral becomes
$$
\int_{D_\mathbb{R}}\dfrac{1}{x-ib}\mathrm{dx}=\ln(|x-ib|)\Bigg|_{x=a}^b.
$$

For the integral over $\mathbb C$: Look up the definition of the complex logarithm function first (on Wikipedia or any standard textbook). You'll find that there is no unique logarithm function.

Title: The antiderivative of $\frac{1}{x-ib}$


问:
$B_{\mathcal{l_2}}$ convex in $\mathcal{l_2}$
In the space $\mathcal{l_2}=\{x=(x_n)_{n=1}^\infty:\displaystyle{\sum_{n=1}^\infty}x_n^2<\infty\}$, I am trying to prove that $B_{\mathcal{l_2}}=\{x=(x_n)_{n=1}^\infty:\displaystyle{\sum_{n=1}^\infty}x_n^2\leq1\}$ is convex.

To do so, I am trying to prove that if $x,y\in B_{\mathcal{l_2}}$, that is $\sum_{n=1}^\infty x_n^2\leq1$ and $\sum_{n=1}^\infty y_n^2\leq1$, and $\lambda\in [0,1]$ , then $(1-\lambda)x+\lambda y\in B_{\mathcal{l_2}}$. I need to prove that $\sum_{n=1}^\infty ((1-\lambda)x_n+\lambda y_n)^2\leq1 $. What I have done is this:

$\sum_{n=1}^\infty ((1-\lambda)x_n+\lambda y_n)^2\leq1 = (1-\lambda)^2\sum_{n=1}^\infty x_n^2 + \lambda^2 \sum_{n=1}^\infty y_n^2+2(1-\lambda)\lambda \sum_{n=1}^\infty x_ny_n \leq (1-\lambda)^21 + \lambda^2 1+(1-\lambda)\lambda (\sum_{n=1}^\infty x_n^2 +\sum_{n=1}^\infty y_n^2) \leq (1-\lambda)^2 + \lambda^2 +2(1-\lambda)\lambda = (1-\lambda+\lambda)^2=1$,

where I have used that $x_ny_n \leq \frac{1}{2}(x_n^2+y_n^2)$.

However I am not sure this proof is correct, because there is case that worries me: could it be possible that $(1-\lambda)<0$ and $\sum_{n=1}^\infty x_ny_n$ is $-\infty$ or a negative number that makes that the sum of everything is greater than 1?

It would be very helpful if anyone could point out how to prove this correctly.

Thanks in advance.

答: $\sum x_n^{2} \leq 1$ if and only if $\|(x_n)\| \leq 1$ where $\|(x_n)\|=\sqrt {\sum x_n^{2}}$. In any normed linear space $\{x: \|x|| \leq 1\}$ is a convex set since $\|\lambda x+(1-\lambda)y\| \leq \lambda \|x\|+(1-\lambda) \|y\|$.


问:
What is the explicit expression of a plane wave in the frequency domain?
A plane wave in the time domain can be written (using notation for an electric field):
$$\boldsymbol{E}(\boldsymbol{r},t)=\boldsymbol{E}_0 e^{i(\boldsymbol{k} \boldsymbol{r}-\omega t)}$$
what is the corresponding expression for a plane wave, $\boldsymbol{E}(\boldsymbol{r}, \omega)$, in the frequency domain?

答:
The key for all these formulas is
$$\tag{1}
\int_{\mathbb R^d} e^{ix\cdot y}\,\frac{dy}{(2\pi)^{\frac d 2}} = \delta(x), $$
where $\delta $ is the $d$-dimensional Dirac distribution. This question is probably concerned with $d=1$ only, but there is no extra difficulty in doing the general case.

Of course, (1) is to be suitably interpreted; it is not an absolutely convergent integral. My favorite reference for this is the book "Quantum field theory: a tourist guide for mathematicians", by Gerry Folland. (There, he states that (1) is to the mathematical physicist as $\int_{-\infty}^\infty e^{-x^2}\, dx =\sqrt{\pi}$ is to the mathematician. I love that quote).

To finally answer the question, we apply (1) to obtain
$$
\int_{\mathbb R} E_0e^{ik\cdot x -\omega t} e^{-i ts}\, ds = (2\pi)^\frac 1 2 E_0e^{i k\cdot x}\delta(t+\omega).$$
This is the Fourier transform of the given function, the "plane wave", in the time variable.


问:
Pigeonhole problem finding the minimum
There are 20 meetings. each neeting has 8 attendees. However, no pair of attendees appear more than once. What is the minimum number of people? To be more specific once a specific pair of people appear in a meeting, this pair will not appear again in any of the other meetings.

My soluton
8C2=28
So there are at leat 28×20=560 distinct pairs
Minimum number of people for there to be 560 distinct pairs= 34
But textbook ans is 35???

答:
As you calculated, the 20 meetings generate 560 pairs of people, and each pair can only appear once in all meetings.
Imagine that there was a solution with only 34 people, resulting in 561 possible pairs of people. Then apart from one pair $(p_1,p_2)$, each pair should appear exactly once. This means that there certainly is one person (for example $p_3$) who had meetings with every of the 33 other persons.

However, the meetings of $p_3$ should now partition all 33 people in groups of 7, but clearly this is impossible since 7 doesn't divide 33.

Of course, this reasoning does not imply that a schedule could effectively be made with 35 people.

Title: Pigeonhole problem finding the minimum


问:
Asymptotic expansion of a function
If you take this function:

$f(x) = \frac {e^{-x^2}}{1-erf(x)}$

and ask software like Wolfram Alpha to give you a Taylor expansion at infinity, it will output this:

$f(x) \approx \sqrt \pi x + \frac {\sqrt \pi} {2 x} - \frac {\sqrt \pi} {2 x^3} + \frac {5 \sqrt \pi} {4 x^5} + O((\frac 1 x)^6)$

This is very nice, and in some systems it works even better than the function itself when one wants to calculate $f(x)$ for $x > 5$.

However, I would like to understand how such result is obtained, or more precisely, how one could do it without recurring to CAS.

I looked up 'Taylor expansion at infinity', and found several posts and resources describing the $x = \frac 1 y$ substitution trick to be able to do an expansion at $0$.

In most cases, however, the functions being handled could be manipulated into the product of some rational term not defined at $0$ (e.g. $\frac 1 y$ itself) and a function $h(y)$ that was defined and differentiable at $0$, so one could expand $h(y)$, multiply it by the left-out factor and do the back-substitution $y = \frac 1 x$.

I would say this is not possible with my $f(x)$, is it?

$h(y) = \frac {e^{-{\frac 1 {y^2}}}}{1-erf(\frac 1 y )}$

Where do I go from here? What do I factor out?

In this post it is suggested to differentiate the function (after the $x = \frac 1 y$ substitution), yielding an expression that can then be 'Taylored' to a polynomial and integrated to get the Taylor expansion of the original function.

But again, not only the expression one manipulates is rational, but also defined and differentiable at $0$...

This still does not work in my case, I think.

$h'(y) = \frac {2 e^{- \frac 1 {y^2}}} {y^3 (1-erf(\frac 1 y))} - \frac {2 e^{- \frac 2 {y^2}}} {\sqrt \pi y^2 (1-erf(\frac 1 y))^2}$

I really do not see how this expression takes me any further in solving the problem.

I also considered expanding separately the numerator and denominator of my $f(x)$.

It turns out that the numerator cannot be expanded at infinity (that's what Wolfram Alpha says), and for $erf(x)$, Wolfram Alpha yields this:

$erf(x) \approx e^{- x^2} (- \frac {1} {\sqrt \pi x} + \frac {1} {2 \sqrt \pi x^3} - \frac {3} {4 \sqrt \pi x^5} + O((\frac 1 x)^7) + 1$

Obviously I can see that if I use this expression, my $f(x)$ becomes:

$f(x) \approx \frac {e^{- x^2}} {1 - [e^{- x^2} (- \frac {1} {\sqrt \pi x} + \frac {1} {2 \sqrt \pi x^3} - \frac {3} {4 \sqrt \pi x^5} + O((\frac 1 x)^7) + 1]} = \frac {1} {\frac {1} {\sqrt \pi x} - \frac {1} {2 \sqrt \pi x^3} + \frac {3} {4 \sqrt \pi x^5} + O((\frac 1 x)^7}$

which is yet another expression yielding a pretty close approximation of my function for $x > 5$.

But again, I have no idea how that expression was obtained, or in fact why there is a $e^{- x^2}$ factor in it, as I assumed all Taylor expansions (or Laurent etc.) could only have powers of $x$.

Could anybody please tell me where I am going wrong / point me to posts or resources explaining how to do this without recurring to CAS?

Thanks!

答:
$$f(x)=\frac{e^{-x^2}}{1-\text{erf}(x)}\implies \frac 1{f(x)}=e^{x^2} (1-\text{erf}(x))$$
The asymptotics of $\text{erf}(x)$ being
$$\text{erf}(x)=1+e^{-x^2} \left(-\frac{1}{\sqrt{\pi } x}+\frac{1}{2 \sqrt{\pi } x^3}-\frac{3}{4
   \sqrt{\pi } x^5}+O\left(\frac{1}{x^7}\right)\right)$$
$$\frac 1{f(x)}=\frac{1}{\sqrt{\pi } x}-\frac{1}{2 \sqrt{\pi } x^3}+\frac{3}{4 \sqrt{\pi }
   x^5}+O\left(\frac{1}{x^7}\right)$$ Now, long division
$$f(x)=\sqrt{\pi } x+\frac{\sqrt{\pi }}{2 x}-\frac{\sqrt{\pi }}{2
   x^3}+O\left(\frac{1}{x^5}\right)$$

Title: Asymptotic expansion of a function


问:
2nd derivative test fail
I trying to solve this problem in Advanced Calc by Buck, sec 3.6 problem 9:

Let $f(x,y)=(y-x^{2})(y-2x^{2})$.
Show that the origin is a critical point for $f$ which is a saddle
point, although on any line through the origin, $f$ has a local minimum
at $(0,0)$.

Solution:

We have
\begin{align*}
f(x,y) & =(y-x^{2})(y-2x^{2})\\
 & =y^{2}-3x^{2}y+2x^{4}
\end{align*}

so that
\begin{align*}
f_{1}(x,y) & =-6xy+8x^{3} &  & (1)\\
f_{2}(x,y) & =2y-3x^{2} &  & (2)
\end{align*}
set (1) and (2) equal to zero and solve:

from (2)
$2y-3x^{2}=0\,\Longrightarrow y=\frac{3}{2}x^{2}$

in (1)
$-6xy+8x^{3}=0\,\Longrightarrow-9x^{3}+8x^{3}=0\,\Longrightarrow x=0,\Longrightarrow y=0$
hence (0,0) is a critical point of $f$.

(0,0) is a saddle point:
By Theorem 19 page 157: If the determinant of the hessian (2nd order
partial derivatives matrix) is negative at $p_{0}$, then $p_{0}$is
a saddle point.

hence
$H=\left[\begin{array}{cc}
f_{11} & f_{12}\\
f_{21} & f_{22}
\end{array}\right]=\left[\begin{array}{cc}
-6y+24x^{2} & -6x\\
-6x & 2
\end{array}\right]$

$\Delta=-12y+48x^{2}-36x^{2}=12(x^{2}-y)$

at (0,0) $\Delta=0$, therefore the second derivative test is inconclusive.

However since $f(x,y)=(y-x^{2})(y-2x^{2})$, then if $x^{2}<y<2x^{2}$
we have $f(x,y)<0$, also if ($y<x^{2}$ or $y>2x^{2}$) we have $f(x,y)>0$,
and we have $f(x,y)=0$ if $x=y=0$. Therefore the critical point
(0,0) is a saddle point.

I don't feel confident about the proof of (0,0) is a saddle point. Also I am not sure how to do the last part (any line through the origin, $f$ has a local minimum
at $(0,0)$) any help appreciated.

答:
This reply is intended to illustrate the situation presented by this function.  As you found the Hessian determinant function to be $ \ \mathbf{H_f} \ = \ 12·(x^2-y) \ , $ the value of the determinant is not only zero at the origin, but at all points on the parabola $ \ y \ = \ x^2 \ . $  Some authors refer to a critical point with zero determinant as "degenerate"; these seem often to occur where the critical point lies on a line or curve where $ \ \mathbf{H_f} \ = \ 0 \   $ everywhere.  So we must look at properties of the function (and sometimes the first partial derivatives as well) to understand better what is occurring at that critical point, which you did.

These graphs may help to give an idea of what the surface for $  \ f(x,y) \ = \ (y-x^{2})·(y-2x^{2})\ $ is like in the neighborhood of the origin.  You found the first partial derivatives to be zero for

$$ f_{x}(x,y) \  = \ 8x^3 - 6xy \ = \ 2x·(4x^2-3y) \ = \ 0 \ \ , $$
which gives the union of the parabola $ \ y \ = \ \frac43 x^2 \ $ and the $ \ y-$ axis [marked in blue in the first graph], and

$$ f_{y}(x,y) \  = \ 2y-3x^{2} \ = \ 0 \ \ , $$

yielding the parabola $ \ y \ = \ \frac32 x^2 \   $ [in green].  The intersection of all three curves is the single critical point $ \ ( 0 , 0 ) \ , $ as you found.

We find that the first partials have a complicated behavior in the vicinity of the point.  For $ \ x > 0 \ , $ both are positive "above" $ \ y \ = \ \frac32 x^2 \ , $  both negative "below" $ \ y \ = \ \frac43 x^2 \ , $ with the sign of $ \ f_y \ $ changing from positive to negative in the "sliver" between the two parabolas.  To the "left" of the $ \ y-$ axis [$ x < 0 $ ] , however, the sign of $ \ f_x \ $ is reversed from what it is in the "right" half-plane.  This is not at all what one expects near a local extremum.

This second graph of $  \ f(x,y) \ $ is pertinent to the proof you gave for the origin being a saddle point:  the function is positive in the region marked in green, which is everywhere except the parabolic "sliver" in red, where the function is negative.  A local extremum would appear surrounded by a region of only one color, so we see that the origin is a saddle point.

In regard to the demonstration provided by Przemysław Scherwentke, we observe that the function has been contrived so that very close to the origin, no straight line through the origin will be found in the region in red.  Along each line, the intersection of the function surface with a plane $ \ y = mx \ $ is a quartic curve $ \ z = 2x^4 - 3mx^3 + m^2x^2 \ $ , which has a local minimum at $ \ x = 0 \ $ for all values of $ \ m \ $ .

Title: 2nd derivative test fail


问:
Show that $G$ has four inequivalent 1-dimensional complex representations
I am looking at the exercise below in relation to representation theory

We define a group structure on $\lbrace \pm 1, \pm a, \pm b , \pm c \rbrace$ through the relations $a^2 =b^2 =c^2 =abc = 1 $ and denote the resulting group by G.

Show that $G$ has four inequivalent  1-dimensional complex
representations, and find those (Hint: $\lbrace \pm 1 \rbrace $ is a
normal subgroup of G and $G / \lbrace \pm 1 \rbrace  \simeq
\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z} $.)

Define $\pi: G \to GL_2(\mathbb{C})$ by $\pi (\pm 1 )= \pm I_2 ,
    \pi(\pm a)=\pm \begin{pmatrix} i & 0 \\ 0 & -i  \end{pmatrix}, \pi
    (\pm b)=\pm \begin{pmatrix} 0 & 1 \\
    -1 & 0  \end{pmatrix}, \pi(\pm c)= \pm \begin{pmatrix} 0 & i \\ i & 0  \end{pmatrix} $. Prove that $\pi$ is a well-defined
irreducible
representation of $G$.

Use (1) and (2) to find the character table of G.

I have shown part 2. by using the inner product of the character, but I am having some difficulties in part 1. (and therefore also in part 3). In part 1, I am not quite sure how I am to use a normal subgroup in order to show the wanted. Additionally, I am having some difficulties in deciding what the representations explicitly are.

答:
The cyclic group $\mathbb{Z}/n\mathbb{Z}$ has 1D representations $x\mapsto\zeta^x$ where $\zeta=\exp(2\pi i/n)$ is a primitive $n$th root of unity. In general, a finite abelian group has only 1D irreducible representations. If $A$ is abelian and $K$ the kernel of a representation $\rho:A\to GL_1(\mathbb{C})$, then the image is a cyclic subgroup of $S^1$ (roots of unity), and by the first isomorphism theorem, $K/A$ must be cyclic.

Given a normal subgroup $N\trianglelefteq G$ and a representation $\rho:G/N\to GL(U)$ of its quotient, there is also a projection map $\pi:G\to G/N$, and the composition $(\rho\circ\pi):G\to GL(U)$ is a rep of $G$.

In particular, for $V=\mathbb{Z}_2\times\mathbb{Z}_2$, there are three (proper nontrivial) subgroups $W$, each isomorphic to $\mathbb{Z}_2$ generated by a nontrivial element of $V$. Each of the quotients $V/W$ has a unique nontrivial representation, taking the coset $W$ to $+1$ and the other coset of $W$ to $-1$. The corresponding representation of $V$ produces $\pm1$ depending on if an element is in $W$ or not (this only works because $W$ is index $2$).

Similarly, 1D reps of $Q_8/C_2=V$ become 1D reps of $Q_8$. The three aforementioned subgroups of $V$ become the three subgroups $H=\langle\mathbf{i}\rangle,\langle\mathbf{j}\rangle,\langle\mathbf{k}\rangle$ of $Q_8$, each index $2$. Then we also have the three 1D reps of $Q_8$ defined as producing $\pm1$ (within $GL_1(\mathbb{C})$) based on whether the element of $Q_8$ is in $H$ or not. (And of course the trivial rep is 1D.)


问:
Relationships between $2$ mutually tangent circles in a semi-circle
In the attempt to answer this question, I was brought to another one, connected to a certain family of circles.

Consider the semicircle made of the upper half of the unit circle and the horizontal diameter with endpoints $A(-1,0)$ and $B(1,0)$.

Let us now consider the family $(F)$ of circles inscribed into this semicircle as can be seen on Fig. 1.

Fig. 1: Certain circles of family $(F).$

They can be described in a parametric way as circles

$$\text{with center} \ (x_c=-\cos a, y_c=\frac12 \sin^2a) \ \text{and radius} \ y_c, \ a \in (0,\pi) \tag{1}$$

or equivalently as circles $(C_T)$

$$\text{with center} \ \left(x_c=\frac{T^2-1}{(1+T)^2}, y_c=\frac{2T}{(1+T)^2}\right) \ \text{and radius} \ y_c, \ T \in (0,+\infty) \tag{2}$$

(Connection between the two parameterizations: $T=t^2$ where $t:=\tan \tfrac{a}{2}$).

The locus of these centers is parabola $(P)$ with cartesian equation:

$$y=\tfrac12(1-x^2),\tag{3}$$

represented on Fig. 2 as the light blue curve.

Each circle $(C_T) \in F$ is associated with a unique circle $(C_{T'}) \in F$ which is tangent to $(C)$ as depicted on Fig. 2.

Fig. 2: Two circles of family $(F)$ are mutually tangent if their parameters $T$ and $T'$ comply with relationship (4).

Let $I$ be the point of tangency of $(C_T)$ and $(C_{T'})$.

Question: how can we describe the locus of $I$?

(sketched as the magenta curve on Fig. 2.)

My work: I have obtained through tedious calculations an unexpected linear relationship between parameters $T$ and $T'$ which is:

$$T=(3-2 \sqrt{2})T'\tag{4}$$

Using (4), I have tried to obtain the locus of $I$ as the envelope of chords defined by the line segment joining the centers of circles $C_T,C_{T'}$ using homographies but to no avail.

Edit: Nice solution using inversion (a little different from the solution given in a comment by Blue).

'Refer to Fig. 3). Consider inversion with pole $S(0,-1)$ and power $2$ meaning that $M$ is exchanged with $M'$ iff $S,M,M'$ are aligned and $SM.SM'=2$. Its circle of inversion is  the black circle (center $S$, radius $\sqrt{2}$). It is easy to see that any circle of family (F) is globally invariant. This is in particular the case for the two blue circles. As the image by inversion of two tangent curves in a certain point $I$ is two tangent curves with contact point $I'$ where $M'$ is the image of $M$, point $I$ is invariant, establishing that $I$ belongs to the circle of inversion.$\square$

As a consequence, tangency points $T_1$ and $U_1$, $T_2$ and $U_2$, are exchanged, meaning that $S,T_1,U_1$ and $S,T_2,U_2$ are aligned.

Fig. 3.

Remark: Two properties of circles of family (F) deserving to be known:

http://www.gogeometry.com/problem/p311_circle_inscribed_semicircle_chord.htm

http://www.gogeometry.com/problem/p310_semicircle_circle_tangent_angle.htm

答:
There's something more fundamental going on here.

Consider any circle $\bigcirc P$ tangent to the semicircle $ACB$ at $T$ and the diameter at $T'$. (My figure is rotated.) Let $\overline{CC'}$ be the perpendicular diameter with $C$ the midpoint of the semicircle.

Thales' Theorem guarantees that right angle $\angle CTC'$ subtends parallel diameters of $\bigcirc P$ and $\bigcirc O$; thus, $C'$, $T$, $T'$ are collinear, and $\triangle CTC'\sim\triangle T'OC'$, so that
$$|C'T||C'T'|=|C'C||C'O|=2r^2$$
That is, the power of $C'$ with respect to $\bigcirc P$ is independent of $P$. Consequently,

the tangent segments from $C'$ to any circle is $\sqrt{2r^2}=r\sqrt{2}$; and

$C'$ lies on the radical axis of any pair of circles.

In the specific case of a pair of inscribed circles that are tangent to each other at a point $I$, the mutual tangent line through $I$ is their radical axis, necessarily meeting $C'$. Moreover, $\overline{C'I}$, being a tangent segment to each circle, has length $r \sqrt{2}$. We conclude that $I$ lies on the circle of radius $r\sqrt{2}$ about $C'$. $\square$


问:
Questions about Proof of Theorem 2.7, Rudin's RCA
I know several questions related to this already exist on this site, but I didn't find those particularly helpful. In some places, I wish to clarify my line of thought, and in others, I need some help.

$U$ is open, so $C = U^c$ is closed. $\overline G, \overline{W_p}$ are closed by definition. So, $C\cap \overline G\cap \overline{W_p}$ is also closed. We know that $\overline G$ is compact, and that $C \cap \overline G\cap \overline{W_p} \subset \overline G$. Can we use this to conclude that $C \cap \overline G\cap \overline{W_p}$ is compact, since a closed subset of a compact set is also compact?

I understand that $$V = G\cap W_{p_1}\cap\ldots\cap W_{p_n}$$ is open, since it is a finite intersection of open sets. I must show that $\overline V$ is compact. How does the author conclude $$\overline V \subset \overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$$
I think this is how: $\overline V$ is the smallest closed set containing $V$, in that every other closed set containing $V$ must be a superset of $\overline V$. $V \subset \overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$ is verified easily, since $A\subset \overline A$ for every set $A$, and inclusion is preserved under intersections. Is that right?

How does $\overline V \subset \overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$ help us conclude that $\overline V$ is compact? Here's what I'm thinking: $\overline G$ is compact, and $\overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}} \subset \overline G$. Moreover, $\overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$ is closed, since it is an intersection of closed sets. A closed subset of a compact set is also compact, so $\overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$ is compact. We can apply a similar argument to $\overline V$ and $\overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$ to conclude that $\overline V$ is compact - correct?

Not a technical doubt as such, but could you provide some intuition as to why the author chose to prove the theorem in the way he did? In particular, what may have motivated bringing $W_p$'s etc. in the picture?

Proof attached for reference:

Thanks a lot!

答:
I'll go through the proof, adding explainers:

So $K \subseteq U$ is given, where $K$ is compact and $U$ is open. The space we're in is locally compact and Hausdorff.

The first and easiest case is when $U=X$ itself: every $x \in K$ has an open neighbourhood $U_x$ with compact closure, by definition of local compactness.
$\{U_x: x \in K\}$ is an open cover of the compact $K$ so finitely many also cover $K$, say $\{U_{x_1}, \ldots, U_{x_n}\}$ so $K \subseteq \bigcup_{i=1}^n U_{x_i}$. We can use $V = \bigcup_{i=1}^n U_{x_i}$, which is open and whose closure is $\bigcup_{i=1}^n \overline{U_{x_i}}$, a finite union of compact sets, hence compact (this is not mentioned in the text before but follows easily from the definition of compactness). So this $V$ is then as required.

We don't need Hausdorffness in this case. This is the proof as sketched in the first paragraph. He keeps one thing from this case: we always know that $K$ has an open neighbourhood $G$ with compact closure (in fact the $V = \bigcup_{i=1}^n U_{x_i}$ will do). He's going to use that in the second case too.

So we can assume we're in the case $U \neq X$ so that $C= X\setminus U$ is non-empty.

For each $p \in U$ we have $p \notin K$ so that 2.5 applies which says

Suppose $X$ is a Hausdorff space and $K \subset X$ and $K$ compact and $p \notin K$. Then there exist open sets $U$ and $W$ such that $p \in U$, $K \subset W$ and $U \cap W= \emptyset$.

So this extends Hausdorffness from two points we can separate by open sets to a compact set and a point we can separate, which is a handy improvement.

So for $p \in U$ we have open (unnamed) $U_p$ containing $p$ and some $W_p$ containing $K$ such that $W_p \cap U_p = \emptyset$. Then $U_p$ is a witness to the fact that $p \notin \overline{W_p}$ and we just use that fact and forget $U_p$ itself.

So in all we indeed have $K \subseteq W_p$ and $p \notin \overline{W_p}$ and $W_p$ open.

Every set $C \cap \overline{W_p} \cap \overline{G} \subseteq \overline{G}$ is thus a closed subset of $\overline{G}$ which is compact by 2.4 as $\overline{G}$ is compact. That covers your 1.

The intersection $\bigcap \{ C \cap \overline{W_p} \cap \overline{G} \mid p \in C\}$ is empty: if $x$ were in it, it would be a point of $C$ so one of the $p \in C$ but then it cannot be in $C \cap \overline{W_p} \cap \overline{G}$ for that $p$ (as $p \notin \overline{W_p}$ for each $p$ by construction).

So Rudin then applies 2.6 to conclude we have $p_1, \ldots p_n$ in $C$ so that

$$C \cap \overline{G} \cap \overline{W_{p_1}} \cap \ldots \cap \overline{W_{p_n}} = \emptyset\tag{1}$$

Then $$V= G \cap W_{p_1} \cap \ldots \cap W_{p_n}$$

is open and contains $K$, as a finite intersection of open sets that all contain $K$.

It's also clear that

$$\overline{V} \subset \overline{G} \cap \overline{W_{p_1}} \cap \ldots \cap \overline{W_{p_n}} $$

as the right hand side is closed and contains $V$. But more is true: the right hand intersection is compact as it is still a closed subset of the compact $\overline{G}$ so 2.4 applies still. And anotehr application of 2.4 then concludes that $\overline{V}$ is also compact.

So you're correct about 2 and 3 too.

Finally note that $(1)$ can be rewritten as

$$\overline{G} \cap \overline{W_{p_1}} \cap \ldots \cap \overline{W_{p_n}} \subset X\setminus C = U$$

which allows us to indeed say $\overline{V} \subseteq U$, as we needed.

Rudin didn't need to set it up this way, using the complement of $U$ and the intersection of compact sets being empty, but it works with minimal theorems needed to prove as "lemma's": 2.4 to 2.6 suffice, and are needed later anyway. The "intuition" could be: we can do it for $K$ and one point outside $U$ with $W_p$ but we use the compactness to extend it from one point to the whole of $U^\complement$, using the one compact $\overline{G}$ to embed everything inside a compact context so we can do this (so all closures of $W_p$ become compact in the intersection with $\overline{G}$).

In a general topology book, maybe he would have shown this fact for a point and an open neighbourhood $U$ first, and then apply a compactness argument to extend to $K$, just as in the first case. All ways lead to the same results though.


问:
I want to know if the Laurent series for $(e ^ z/ z) + e ^ {1 / z}$ is resolved correctly
I tried to solve it this way but I want to check if it is correct

$f(z)=\frac{e^z}{z}+e^\frac{1}{z}=\frac{1}{z}\sum_{n=0}^\infty{\frac{z^n}{n!}}+\sum_{n=0}^\infty{\frac{1}{n!z^n}}$

$=\sum_{n=0}^\infty{\frac{z^{n-1}}{n!}+\frac{1}{n!z^n}}=\sum_{n=0}^\infty{\frac{z^{2n-1}+1}{n!z^n}}$

$=1+\frac{1}{z}+\sum_{n=1}^\infty{\frac{z^{2n-1}+1}{n!z^n}}$

答:
No, that is not correct. A Laurent series centered at $0$ is an expression of the form $\sum_{n=-\infty}^\infty a_nz^n$ and what you got is not of that form.

Your computations are correct though. You should be able to deduce from them that$$(\forall z\in\Bbb C\setminus\{0\}):\frac{e^z}z+e^{1/z}=\sum_{n=-\infty}^{-2}\frac1{(-n)!z^n}+\frac2z+2+\sum_{n=1}^\infty\frac{z^n}{(n-1)!}.$$


问:
Example of a space that is Semi-locally Simply connected that is neither Simply connected nor Locally simply connected.
I know that Semi-locally Simply connectedness (SLSC) is strictly weaker than either one of Simply connectedness (SC) and Locally Simply Connectedness (LSC), with the latter two not necessarily implying each other ($S^1$ is LSC and not SC, the Cone of the Hawaiian earring is SC but not LSC).

I wanted to understand the SLSC property better, but the most revealing example that is given is the Hawaiian earrings, which is NOT SLSC, and so neither SC nor LSC. I can't seem to find an example of a space that is SLSC but neither SC nor LSC.

The examples I mentioned above are one of SC or LSC but not the other, so they are good examples for spaces that are SLSC and not SC or LSC, but I can't seem to put the ideas together to construct a space that is only SLSC.

Hope I'm not missing something obvious, Any examples are appreciated.

答:
Take the wedge (one-point union) $S^1 \vee CH$ of the circle $S^1$ and the cone $CH$ of the Hawaiian earring.

More generally, take any LSC but not SC space X and any SC but not LSC space $Y$ and form their wedge.


问:
Way of writing this direct proof
In Richard Hammack's Book of Proof, he writes:

For the text I highlighted, wouldn't it be more accurate to say:

Then $x=2a+1$ for all $a\in \mathbb{Z}$

答: No. No number is equal to $2a+1$ for all $a\in\Bbb Z$. How could that be, since $2\times1+1\ne2\times2+1$?

Title: Way of writing this direct proof


答:
No. To say that $x = 2a + 1$ for all $a \in \Bbb{Z}$ would be to say that $x = 2a + 1$ is simultaneously true for every possible integer you substitute into $a$. For example, it would be true that $x = 2(1) + 1$, and it would also be true that $x = 2(100) + 1$. But, together, this implies
$$201 = 2(100) + 1 = x = 2(1) + 1 = 3,$$
a blatant contradiction!

Title: Way of writing this direct proof

Body:
In Richard Hammack's Book of Proof, he writes:

For the text I highlighted, wouldn't it be more accurate to say:

Then $x=2a+1$ for all $a\in \mathbb{Z}$


问:
A metric space is second-countable iff it is separable: Proposition 2.1.4 in Dudley's "Real Analysis and Probability"
I am reading Proposition 2.1.4 in Dudley's "Real Analysis and Probability (Cambridge Edition)" but I have some troubles.

That proposition and the proof are here:
A metric space $(S,d)$ is second-countable iff it is separable.

Proof: ($\longleftarrow$) Let $A$ be countable and dense in $S$. That is $\overline A=S $. Let $\mathcal U$ be the set of all balls $B(x, 1/n)$ for $x\in A$ and $n=1,2,\dots$. To show that $\mathcal U$ is a base, let $U$ be any open set and $y\in U$. Then for some $m$, $B(y, 1/m)\subset U$. Take $x\in A$ with $d(x,y)<1/(2m)$ Then $y\in B(x, 1/(2m))\subset B(y, 1/m)\subset U$, so U is the union of the elements of $\mathcal U$ that it includes, and $\mathcal U$ is a base, which is countable. Since Cartesian products of countable sets (in this proof, $A$ and $\mathbb{N}$ are so) is also countable, $\mathcal U$ is a countable base.

($\longrightarrow$) Suppose there is a countable base $\mathcal V$ for the topology, which we may assume consists of nonempty sets. By the axiom of choice, let $f$ be a function on $\mathbb{N}$ whose range contains at least one point of each set in $\mathcal V$. Then this range is dense.

I have two questions:

in the proof of ($\longleftarrow$), where did Dudley use the denseness of A?
Note that in Theorem 2.1.3 in his book says that for any $A\subset S$, $\overline A$ is the set of all $x\in S$ such that for some net $x_i \longrightarrow x$ with $x_i\in A$ for all $i$.
That is, $\overline A$ is the set of all $x\in S$ such that any neighborhood of $x$ contains all but finitely many points $x_i$ with $x_i\in A$. Do I need to use this fact?

in the proof of ($\longrightarrow$), why the range of $f$ is dense?

答:
Both have the same reason: $A$ is dense in $X$ iff for all open balls $B(x,r), x\in X, r>0$ we have $A \cap B(x,r) \neq \emptyset$.

He uses this in $\longrightarrow$: if $B(x,r)$ is some ball, it contains some $V \in \mathcal{V}$ and then $f(V) \in V $ intersects that ball.

In $\longleftarrow$ he says: "pick $x \in A$ with $d(x,y) < \frac{1}{2m}$", which is picking a member of $A \cap B(y, \frac{1}{2m})$, so it applies the denseness of $A$ (intersecting all balls).


问:
eHarmony combinatoric question, probability that I should get at least 1 compatible match.
Ok.. (as I type this with a smirk on my face) - in all seriousness I am trying to figure out, given 29 degrees of compatibility and 40 million members if I should be getting at least 1 match a day.  There are of course a lot of variables, so I am trying to simplify things.  I also see this as similar to the "birthday problem" in which one wants to see, out of n people, the probability of not having any birthday's in common.

In this case however, we don't care if others get a match, only that I do (muahahaha!), so I am guessing this is, instead of being in line with $_{n}C_{k}$ this is $_{n}C_{1}$ Where I am the $1$!

This is a bit more advanced in that we are looking at the "percentage of compatibility".  So I am going start by taking the 29 degrees of freedom and looking at things in a binary way i.e. either a match is compatible on a degree of freedom (same answer to a question) or they are not. Now by my calculations there are 29 questions, so as with light switches, there are $2^{29} = 536870912$ ways to answer the questions.  So if half the eH members are women, then there is a 20,000,000/536,870,912 = 3.7% chance that I would answer exactly as a potential match.

It seems there should or could be much more to it in that say I went about the computation from the point of view that the probability of not answering the questions the same.  So I think I would have something such as $$(1-2^{28}/2^{29})(1-2^{27}/2^{29})(1-2^{26}/2^{29})\space...\space(1-2^{0}/2^{29})$$

which seems to reduce to: $$((2^{28})(3 \cdot 2^{27})(7 \cdot 2^{26}) \space ... \space(2^{2}(2^{27}-1))(2^{1}(2^{28}-1)) \cdot (2^{29}-1))/(2^{29})^{29}$$  after multiplying the numerator by  $(2^{29})^{29}$.

I am not sure what this reduces too - hopefully more matches than I am currently getting..

I wonder though if I am on the right track?  I however wonder about the 20,000,000.  If I were to take the possible "state spaces" or the $2^{29}$ possible choices for the 29 degrees of freedom (treated as binary yes/no), as it were the possible 365 days in a year birthday state space, then I come up with:

$$\frac{2^{29}!}{(2^{29})^{20,000,000}\cdot (2^{29} - 20,000,000)!}$$

which just seems insane.

Any thoughts (or dating advice ha!)

Thanks,

Brian

答: The probability of at least 1 match is $$1- \mathsf{\text{probability of no matches}} = 1-(1-p)^n$$ where $p = \frac{1}{2^{29}}$ and $n=20000000$. Assuming the choices are equaly likely, and the candidates choose independently. Approximately 3.66%.</p>


问:
Which of the ZFC Axioms do Classes Fail to Satisfy?
In Set Theory , Thomas Jech states

Classes

Although we work in ZFC which, unlike alternative axiomatic
set theories, has only one type of object, namely sets, we
introduce the informal notion of a class. We do this for practical
reasons: It is easier to manipulate classes than formulas.

I don't understand the difference between them. My question is:
Which of the ZFC axioms should a class satisfy, and which axioms are not necessary?

In Jech's text, the ZFC axioms are stated as follows:

1.1. Axiom of Extensionality. If $X$ and $Y$ have the same
elements, then $X = Y$.

1.2. Axiom of Pairing. For any $a$ and $b$ there exists a set
$\{a,b\}$ that contains exactly $a$ and $b$.

1.3. Axiom Schema of Separation. If $P$ is a property (with
parameter $p$), then for any $X$ and $p$ there exists a set
$Y = \{u \in X : P(u, p)\}$ that contains all those $u \in X$ that
have property $P$.

1.4. Axiom of Union. For any $X$ there exists a set
$Y = \bigcup X$, the union of all elements of X.

1.5. Axiom of Power Set. For any $X$ there exists a set
$Y = P(X)$, the set of all subsets of $X$.

1.6. Axiom of Infinity. There exists an infinite set.

1.7. Axiom Schema of Replacement. If a class $F$ is a
function, then for any $X$ there exists a set
$Y = F(X) = \{F(x) : x \in X\}$.

1.8. Axiom of Regularity. Every nonempty set has an
$\in$-minimal element.

1.9. Axiom of Choice. Every family of nonempty sets has a
choice function.

The theory with axioms 1.1–1.8 is the Zermelo-Fraenkel
axiomatic set theory ZF; ZFC denotes the theory ZF with the Axiom
of Choice.

Definition of class, also from Jech's book:

If $\varphi(x,p_1,...,p_n)$ is a formula, we call
$C = \{x : \varphi(x, p_1, \ldots, p_n)\}$ a class. Members of the
class $C$ are all those sets $x$ that satisfy
$\varphi(x, p_1, \ldots, p_n)$:
\begin{equation*}
    x \in C \quad\text{if and only if}\quad
    \varphi(x,p_1, \ldots, p_n).
\end{equation*}

Note here the definition is by formula, not by axioms, this is where I'm lost.

There are similar posts on MSE, but I'm still not sure;
in What is the difference between a class and a set? , Sylvain's answer states

The powerset axiom is the main exception, the one axiom that cannot
be properly justified.

OK, so Axiom of Powerset is out. How about others?

HellCat's answer says

the axioms of Set Theory apply only to sets and not to classes.

Also in Difference between a set and a class, Asaf says:

Sets are elements of the model of set theory, and they have to
satisfy the axioms, e.g. the axiom of power set. Classes are
collections of elements from a model of set theory, but they don't
have to correspond to any element in the model, and they don't have
to obey the axioms.

Does this mean all the axioms above are not applicable to classes? What are the properties of classes restricted to?

答:
In its most formal description, a theory knows only the basic logical symbols $\land, \lor, \neg,\to,\leftrightarrow,\forall,\exists, =$ and an infinite supply of variables $a,b,c,\ldots, a_1,a_2,,\ldots, a',a'',\ldots$ which can be combined in certain manners forming well-formed formulas. Depending on the topic at hand, some additional symbols may be introduced. In particular, for a set theory (not a sets-and-classes theory), one single additional symbol $\in$ is added to the language. You might say that this additional symbol $\in$ carries no a priori meaning and it is the axioms that provide it with meaning or force us to interpret the (a priori meaningless) variables as our mental idea of sets as some kind of collection and the $\in $ relation as element containment.

But to make our lives easier, we tend to enhance the language by additional symbols, each time specifying a way to get rid of that symbol in order to arrive at the original language of the theory. For example we may introduce a $\ne$ symbol that can appear anywhere the $=$ symbol can appear and define $v_1\ne v_2$ as $\neg(v_1=v_2)$. And we certainly introduce $x\subseteq y$ as a shorthand for $\forall z\,(z\in x\to z\in y)$. Or we might introduce $\exists !$ as a new symbol that can appear anywhere a quantor can appear and  define $\exists!v\, \phi(v)$ as $\exists v\,(\phi(v)\land\forall u\,\phi(u)\to u=v)$ (with some elaboration on $\phi$ and $u$ needed here).
Or we might even have started with less symbols than listed above and introduced some of the this way, e.g., $\phi\leftrightarrow \psi$ as $(\phi\to \psi)\land (\psi\to \phi)$.

Furthermore, we introduce more of the above when motivated by the axioms or  theorems of our theory. Actually, the quoted formulation of the axioms already does that by introducing notations such as $\{x,y\}$. More formally, one could write the Axiom of Pairing as
$$ \forall a\,\forall b\,\exists z\,\forall t\,(t\in z\leftrightarrow (t=a\lor t=b)).$$
Then, using the Axiom of Extensionality, one can show as a theorem that $\exists z$ can be replaced with $\exists!z$, which motivates the new notation $\{a,b\}$ that can appear anywhere a variable can appear - except after a quantor, and the required elimination rule is that each occurance of $\{a,b\}$ (i.e., on either side of a $\in $ or $=$ symbol) is replaced with a new variable $z$, which is introduced suitably nearby, so e.g., $\{a,b\}\in c$ becomes $\exists z\,(\forall t\,(t\in z\leftrightarrow (t=a\lor t=b)\land z\in c)$.
The same can be done with the other notations introduced (i.e., for all those axioms that postulate the existence of a set with specific properties where we can prove that this set is in fact unique), i.e., Separation, Power, Union, and Replacement. And the same applies to some set constants that we can define, such as introducing the symbol $\emptyset$ for the (provably existing and unique) set without element or the symbol $\omega$ for the (provably existing and unique) minimal inductive set.
Note that also the Axioms of Extensionality, Infinity, Regularity, and Choice are (or can be) formulated as existence claims (perhaps surprisingly in case of Extensionality: $ \forall a\,\forall b\,\exists z\,(a=b\leftrightarrow(z\in a\leftrightarrow z\in b))$). But in these cases we do not have uniqueness, and that is why these do not motivate us to introduce new notation.

Of course, we can continue to introduce new notation as long as this happens by definitions that can be properly eliminated just as with the examples so far. For example, we can define singleton set notation $\{a\}$ by defining this as $\{a,a\}$, or define ordered pairs $(a,b)$ as $\{\{a\},\{a,b\}\}$ and then derive the theorem(!) that $(a,b)=(c,d)\leftrightarrow (a=c\land b=d)$.

It is perhaps about time I mention classes.

We can avoid viewing classes as objects of their own in our set theory and introduce them as mere matters of speech. Given any predicate, we may want to introduce new notation that suggestively looks suspiciously like an element-of-a-set relation, but truely isn't. For example, if we are interested in the predicate "is an ordered pair", i.e., $\text{is-a-pair}(x)\iff \exists a\,\exists b\,x=\{\{a\},\{a,b\}\}$, it is desirable to introduce a less clumsy notation such as $x\in\text{Pairs}$. This is formally confusing because on the right of $\in$, only sets (more specifically, variables - but these always stand for sets) are "allowed" so far. So one should not view this as a symbol $\text{Pairs}$ introduced that can occur only (in a "forbidden" way) to the right of $\in$, but rather as a single symbol $\in\text{Pairs}$ that stands for a predicate in postfix notation and happens to use the symbol $\in$ as its first letter. This way, the class $\text{Pairs}$ does not exist as any new kind of object (and it is guaranteed that we cannot even express syntactically that the class is an element of something) and thus there is no way of saying that classes obey any specific axioms that would need to be added. We do have some theorem schemes though that look as if we can perform a few basic operations of set theory on classe, e.g., if $A,B$ are classes, we can speak of $A$ being a subclass of $B$ if $\forall t\,(t\in A\to t\in B)$, or we can define a class $A\cup B$ such that  $$x\in A\cup B\iff x\in A\lor x\in B$$ or a class $A\times B$ such that $$x\in A\times B\iff \exists a\,\exists b\,(x=(a,b)\land a\in A\land b\in B).$$

Just as the Axiom Schema of Separation allowed us to introduce the set builder notation $\{\,u\in X:P(u,p)\,\}$, we can introduce a class builder notation $\{\,u:P(u,p)\,\}$ (i.e., unrestricted comprehension) with the same caveat as above, i.e., we actually only introduce the postfix predicate $\in\{\,u:P(u,p)\,\}$ with the property
$$ x\in \{\,u:P(u,p)\,\}\iff P(x,p).$$

As a final remark: Sometimes we want to express that a class $A$ in fact equals a set $a$. Since only $\in A$ and not $=A$ is allowed yet, this should be introduced by defining
$$a=A\iff \forall t\,(t\in a\leftrightarrow t\in A) $$


问:
Is the integral $\int_0^1{\frac{dx}{\sqrt[3]{x(e^x-e^{-x})}}}$ convergent?
I need to prove that $\int_0^1{\frac{dx}{\sqrt[3]{x(e^x-e^{-x})}}}$ is convergent. First of all, the only problematic point is zero. I noticed that since $e^x$ is a growing function and $e^{-x}$ is a decreasing function, then $e^x-e^{-x}$ must have only one root- at the point zero. For all $x>0$ $e^x-e^{-x}>0$. But then what? I can't think of a function that is bigger or lower than the integral function. If I try with Taylor series, I get:

$e^x=1+x+\frac{x^2}{2}+...$

$e^{-x}=1-x+\frac{x^2}{2}-...$

Since the problem is at zero, only the lowest degree matters. So $e^{x}\sim 1+x$ and $e^{-x}\sim 1-x$, therefore $\int_0^1{\frac{dx}{\sqrt[3]{x(e^x-e^{-x})}}}\sim \int_0^1{\frac{dx}{\sqrt[3]{2x^2}}}$, which is convergent. Therefore $\int_0^1{\frac{dx}{\sqrt[3]{x(e^x-e^{-x})}}}$ is also convergent. Is my solution correct?

Body:
$$e^x-e^{-x}=2\sinh x\ge2x\qquad(x\ge0)$$
Therefore
$$\int_0^1\frac1{\sqrt[3]{x(e^x-e^{-x})}}\,dx\le\int_0^1\frac1{\sqrt[3]{2x^2}}\,dx<\infty$$


答:
No, $x$ is a fixed odd integer, hence there is unique integer $a\in\mathbb{Z}$ such that: $x=2a+1$.

The claim "Then $x=2a+1$ for all $a\in \mathbb{Z}$" means that if you choose $a=5$ and $a=89$, then $x=11$ and $179$, those are odd numbers but then $x$ is not a fixed odd number.

The right way to say it is: $A:\{x\in \mathbb{Z}\mid \exists a\in\mathbb{Z}: x=2a+1\}$ is the set of odd numbers.

Title: Way of writing this direct proof


答:
In the context of that highlighted sentence, we have been introduced to some fixed odd integer $x$. We do not know its value, but we know that it is odd, so we know that it has to be $2a+1$ for some integer $a$, i.e. we can find an integer $a$ such that $x = 2a+1$.

For example, if $x=3$, then $x = 2a+1$ for $a = 1$.

Title: Way of writing this direct proof


问:
Prove or disprove that the rings $R$ and $S$ are isomorphic.
Prove or disprove that the rings $R$ and $S$ are isomorphic.

$R = \mathbb{Q}[x]/\langle x^4 - 2 \rangle$ and $S = \mathbb{Q}(\sqrt2)[x]/ \langle x^2 - 2x + 1 + \sqrt2 \rangle$.

$\textbf{May use the following fact without a prove:}$

Suppose that $E$ is an extension of a field $F$, $\alpha \in E$ and $\alpha$ is algebraic
over F.

Let $m_{\alpha, F}(x)$ be the minimal polynomial of $\alpha$ over $F$.

Define $\psi_{\alpha, F} : F[x]/ \langle m_{\alpha, F}(x) \rangle \rightarrow F(\alpha)$ which by $\psi_{\alpha, F} (f(x) + \langle m_{\alpha, F}(x) \rangle) = f(\alpha)$, where $f(x) ∈ F[x]$.

Then $\psi_{\alpha, F}$ is well-defined and is an isomorphism of fields.

$\textbf{My Attempt:}$

Want to prove that rings $R$ and $S$ are isomorphic.

Notice that $\mathbb{C}$, $\mathbb{Q}$ and $\mathbb{Q}(\sqrt2)$ are all fields.

Then, $\mathbb{C}$ is an extension of a field $\mathbb{Q}$ and $\mathbb{C}$ is an extension of a field $\mathbb{Q}(\sqrt2)$.

Let $a_1 = i\sqrt[4]{2}$ which it is a root of $x^4 - 2 \in \mathbb{Q}[x]$. So, $a_1$ is algebraic over $\mathbb{Q}$.

Since, $x^4 - 2$ is monic polynomial and $a_1^4 - 2 = 0$. Also, by Eisenstein's criterion, $x^4 - 2$ is irreducible over $\mathbb{Q}$.

So, $m_{a_1, \mathbb{Q}}(x) = x^4 - 2$.

Hence, be the fact above, $\psi_{a_1, \mathbb{Q}} : R \rightarrow \mathbb{Q}(i\sqrt[4]{2})$ is well-defined and is an isomorphism of fields.

Let $a_2 =  1 + i \sqrt[4]{2}$ which it is a root of $x^2 - 2x + 1 + \sqrt2 \in \mathbb{Q}(\sqrt2)[x]$. So, $a_2$ is algebraic over $\mathbb{Q}(\sqrt2)$.

Since, $x^2 - 2x + 1 + \sqrt2$ is monic polynomial and $a_2^2 - 2a_2 + 1 + \sqrt2 = 0$.

Also, $x^2 - 2x + 1 + \sqrt2$ is irreducible over $\mathbb{Q}(\sqrt2)$.

So, $m_{a_2, \mathbb{Q}(\sqrt2)}(x) = x^2 - 2x + 1 + \sqrt{2}$.

Hence, be the fact above, $\psi_{a_2, \mathbb{Q}(\sqrt2)} : S \rightarrow \mathbb{Q}(\sqrt2)(1 + i\sqrt[4]{2})$ is well-defined and is an isomorphism of fields.

Notice that $\mathbb{Q}(\sqrt2)(1 + i\sqrt[4]{2}) = \mathbb{Q}(\sqrt2, 1 + i\sqrt[4]{2}) = \mathbb{Q}(\sqrt2, i\sqrt[4]{2})$.

Then, $\psi_{a_2, \mathbb{Q}(\sqrt2)} : S \rightarrow \mathbb{Q}(\sqrt2, i\sqrt[4]{2})$ is well-defined and is an isomorphism of fields.

Then, the inverse of $\psi_{a_2, \mathbb{Q}(\sqrt2)}$ which is $\psi_{a_2, \mathbb{Q}(\sqrt2)}^{-1} : \mathbb{Q}(\sqrt2, i\sqrt[4]{2}) \rightarrow S$ is well-defined and is an isomorphism of fields.

$\textbf{I am now stack on prove that $\sigma : \mathbb{Q}(i\sqrt[4]{2}) \rightarrow \mathbb{Q}(\sqrt2, i\sqrt[4]{2})$ is an isomorphism of fields.}$

Since, isomorphism is transitive. Then, I can prove that $R$ and $S$ are isomorphic.

Body: $$\sqrt2 = -\left( i \sqrt[4]2 \right)^2$$


问:
Closure of a set is contained in the half space
If $X\subset\mathbb{R}^{d}$ lies in the closed half-space $\mathbb{h}^{+}$, then the closure $\text{cl} X$ of $X$ also lies in $\mathbb{h}^{+}$.

I understand the question but I don't have any idea how to prove this. Any hints/ help would be great. Thanks.

答: The closure of a set $X$ is (depending on your textbook: by definition) the smallest closed set containing $X$. In fact, such a smallest closed set is guaranteed to exist because the intersection of arbitrarily many closed sets is closed; and that means that the closure is just the intersection of all closed sets containing $X$. In particular, if $h^+$ is one of the closed sets containing $X$, then $h^+$ is one of the sets being intersected to form the closure, i.e., the closure is a subset of $h^+$.


问:
Probability, Expected Value, Standard Deviation

I did not understand part c. How should I calculate it? Can someone please show it?

答:
I did not understand part c.

after some calculations you get that

$$E(X)=7$$

$$\sigma(X)\approx 2.65$$

They request you to calculate

$$P(7-1.5\cdot2.65\leq X\leq 7-1.5\cdot2.65)=$$

$$=P(3<X<11)=1-2[P(X=2)+P(X=3)]=1-2\left[\frac{1}{36}+\frac{2}{36}\right]=\frac{30}{36}$$


问:
Rank inequality

Let $ A, B \in M_n (\mathbb{C})$ such that $(A-B)^2 = A -B$. Then $\mathrm{rank}(A^2 - B^2) \geq \mathrm{rank}( AB -BA)$.

I tried to apply the basic inequalities without results. How to start? Thank you.

答:
Hint: Let $P:=A-B$, then $A=P+B$, $P^2=P$ and
$$A^2-B^2= (P+B)^2-B^2=P^2+PB+BP=P+PB+BP\\
AB-BA=(P+B)B-B(P+B)=PB-BP$$
Now we can apply a basis transformation using an eigenbasis of $P$ so that the matrix of $P$ becomes diagonal with $k:={\rm rank}(P)$ ones and $n-k$ zeroes, so $PB$ gives the first $k$ columns of $B$ then $n-k$ zero columns, and similarly $BP$ gives the first $k$ rows of $B$ then $n-k$ zero rows.

Title: Rank inequality


问:
Understanding tensor transformation under coordinate rotation formula
Given a Cartesian vector $x \in \mathbb{R}^3$, and suppose we wish to find the coordinates $(x'_1, x'_2, x'_3)$ of $x$ in a new rotated coordinate system  that shares the same origin with original coordinate system.

One can show that $x'_j = x_k C_{ki}$, where  $C_{ki}$ is the $3 \times 3$ matrix of direction cosines, and the proof is relatively easy.

Now, given a second order tensor $T_{ij}$, my textbook defines the transformation under the new rotated coordinate system as $$ T'_{mn} = C_{im}C_{jn}T_{ij}$$

Yet, the above formula is given with no proof, only a reference to Mechanics of Deformable Bodies (1964) book by Sommerfeld, which I don't have access to. I tried to prove it myself with no luck.

Note: I'm a beginner in tensor calculus, so please go easy on me.

答:
Unfortunately the textbooks on tensors get several things mixed up. To do it properly requires introducing tensors in linear algebra (it's usually called multilinear algebra) and there are two definitions to take care of here (and which are related). Then we need manifolds (basically your bodies as in deformable bodies) and vector bundles over manifolds and then it turns out we can bundle up all the tangent vectors into the tangent bundle which is a vector bundle over the manifold; then finally applying the tensor operation, we get the tensor bundle.

Given all this technology it's perhaps not surprising that textbooks still keep to the the indexful notation, pedagogically speaking, it's simply less hassle - even when terms are introduced without enough explanation or motivation.

Now, given a second order tensor $T_{ij}$, my textbook defines the transformation under the new rotated coordinate system as $$ T'_{mn} = C_{im}C_{jn}T_{ij}$$

The mistake here is that you can't be given a second order tensor without its transformation law, the two go together. The transformation law in effect defines the second order tensor. Hence, because it is a definition there is nothing to prove. What's really required is motivation and an explanation why second order tensors are a useful thing and why they are defined the way that they are. Unfortunately most books I've looked at do this pretty poorly, so your confusion is quite common - least wise, I was similarly confused.


问:
Action/character of group representation on dual space
I am somewhat confused how trace is defined on the induced action of a group representation on the dual space. In my case, I think of the action as $u^\star \to u^\star \circ \rho(g^{-1})$ for representation $\rho: G \to GL(V)$. To me, defining the character as ${\rm trace}{\rho(g^{-1})}$ does not clearly show that this definition is consistent with finding out what the action of $\rho(g^{-1})$ does if you view $u^\star$ as a vector of coefficients of the element as a linear combination of dual space basis elements and thinking of the action of the group as acting on such a vector by multiplication on the left. Thank you.

答: If you pick a basis for $V$ and a dual basis for the dual space $V^{\ast}$, then $\rho(g)$ can be considered matrix-valued, and you will find that the matrix of $g$'s action on the dual space is in fact $\rho^\ast(g):=\rho(g^{-1})^T$. This has the same diagonal entries as $\rho(g^{-1})$, so the traces $\mathrm{tr}\rho^\ast(g)=\mathrm{tr}\rho(g^{-1})$ are the same.


问:
Prove that $C_g(x) = gxg^{-1}$ is an isomorphism
Let $G$ be a group and $g \in G$. Define $C_g : G \to G$ as $C_g(x) = gxg^{-1}$
Prove that $C_g$ is an isomorphism.

Homomorphism:
$C_g(xy) = gxyg^{-1} = gx(gg^{-1})yg^{-1} = (gxg^{-1})(gyg^{-1}) = C_g(x)C_g(y)$

Injective:
Assume $C_g(x) = C_g(y)$

$$gxg^{-1} = gyg^{-1}$$
Left multiply by $g^{-1}$ and right multiply by $g$ to get $x=y$

Surjective:

Let $C_g(x) = y$ for some $y$.
Therefore, $y = gxg^{-1} \therefore x = g^{-1}yg \in G$ since it is the multiplication of elements of a group and, by closure it is part of a group.

I feel that something is wrong on the part about proving surjectivity.

答: You proof about surjectivity is fine, except for this point: "...Let $C_g(x) = y$ for some $y$....": no, you have to prove that for every $y\in G$ there is some $x\in G$ such that $G_g(x)=y$, which is what you have correctly done.


问:
Finding the infinitesimal generator of Ornstein-Uhlenbeck process without using a theorem
Consider the 1-dimensional Ornstein-Uhlenbeck process
$$X(t)=\mu+e^{-\lambda t}(x-\mu)+\frac{\sigma}{\sqrt{2\lambda}}e^{-\lambda t}W^{(e^{2\lambda t}-1)}$$

with mean $\mu+e^{-\lambda t}(x-\mu)$ and variance $\frac{\sigma^2}{2\lambda}(1-e^{-2\lambda t})$. I'd like to show the infinitesimal generator of O-U process is

$$\mathcal{L}f(x)=\lambda(\mu-x)f'(x)dx+\frac{1}{2}\sigma^2f''(x)$$

My idea is first transform the equation by taking derivatives w.r.t. $t$,

$$dX(t)=\lambda\int_{0}^{t}(\mu-X(t))dt+\sigma dW(t),X(0)=x$$

Note that $\int_{0}^{t}e^{\lambda s}dW(s)$ has the same distribution as $(2\lambda)^{-1/2}W^{(e^{2\lambda t}-1)}$, so I can make the adjustment.

Here is a theorem in Oksendal's SDE:

(Theorem 7.5.4) Let $f\in C^2$, then $f\in\mathcal{D}_A$ and
$$\mathcal{A}f=\sum_{i}b_i\frac{\partial{f}}{\partial{x_i}}+\frac{1}{2}\sum_{i,j}(\sigma\sigma)^T_{ij}\frac{\partial^2 f}{\partial x_i\partial x_j}$$.

Using this theorem, we can find the generator immediately. But I'd like to know methods without using this, since we have not covered it in our class yet. We have given the definition of infinitesimal generators:$$\mathcal{L}f=\lim_{h\to 0}\frac{\mathbb{E}[f(W^{(h)})]-f(x)}{h}$$

but using the definition, the calculation will be pretty messy.

A similar question has been asked here before: Infinitesimal Generator for Stochastic Processes, but the answer does not quite solve my problem. Can anyone provide a derivation without using the theorem? Thank you.

答:
Okay, I figured it out by doing Taylor expansion---

Write $\mathbb{E}_x[f(X^t)]$ as the following by Taylor expansion:
$$\mathbb{E}_x[f(X^t)]\simeq\mathbb{E}_x[f(x)]+f'(x)\mathbb{E}_x[x-X(t)]+\frac{1}{2}f''(x)\mathbb{E}_x[(x-X(t))^2]$$

Compute the expectations for the second and third brackets based on expectation and variance of O-U process,
$$\mathbb{E}_x[x-X(t)]=(1-e^{-\lambda t})(x-\mu)$$
$$\mathbb{E}_x[(x-X(t))^2]=x^2+\frac{\sigma^2}{2\lambda}(1-e^{-2\lambda t})+(\mu+e^{-\lambda t}(x-\mu))^2-2x[\mu+e^{-\lambda t}(x-\mu)].$$

We emphasize that $\mathbb{E}_x[f(x)]=f(x)$.

By definition of infinitesimal generator,

$$\mathcal{L}f(x)=\lim_{t\to0}\frac{\mathbb{E}_x[f(X^t)]-f(x)}{t}=\lim_{t\to0}\frac{(1-e^{-\lambda t})(x-\mu)f'(x)+\frac{1}{2}\left (x^2+\frac{\sigma^2}{2\lambda}(1-e^{-2\lambda t})+(\mu+e^{-\lambda t}(x-\mu))^2-2x[\mu+e^{-\lambda t}(x-\mu)]\right )f''(x)}{t}$$

Apply L'Hospital Rule, all exponents goes to $1$, and by noting that $$\frac{d}{dt}((\mu+e^{-\lambda t}(x-\mu))^2)=-2\lambda e^{-2\lambda t}(x-\mu)(\mu(e^{\lambda t}-1)+x),$$
with also some algebra,
$$\mathcal{L}f(x)=\lambda(\mu-x)f'(x)+\frac{\sigma^2}{2}f''(x)$$


问:
Finding the value of $x$ in an exponential form
Solve the following equation using Logarithms and leave your answer to 2 significant figures.

$4^{x}-2^{x+1}-3 = 0$

I tried to change the expression into a quadratic expression but I am stuck on where I found $3 = 2^{x}$.

Patiently waiting for your help mates

答:
Just take the log base 2 on both sides.
You would get $x=log_2 3$.


问:
Evaluating $\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \cdots}}}}$

$$x =\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \sqrt{5+ \sqrt{6 - \cdots}}}}}}$$

Find $x$.

I am not sure how to proceed.

Is this a sort of Arithmetico-Geometric Progression? Will this converge at a point?

Any help would be sincerely appreciated.

答:
I doubt a closed form exists for this number, but I'll show convergence and give some bounds on the values.

Let $f_m^n = \sqrt{m\pm\sqrt{(m+1)\mp \sqrt{\cdots\pm \sqrt n}}} $
be the the version where we go from $m$ to $n$ (don't take the $\pm$'s too seriously). We have the recursive expression
$$
f_m^n = \begin{cases}
\sqrt{m + (-1)^{m+1}f_{m+1}^n} & m<n \\
\sqrt n & m=n
\end{cases}
$$

Realness and general bounds:

Let $3 \le m < n$ with $n$ large enough. I claim that
$$
f_m^n - \sqrt{m-1} \in \begin{cases}
[0.5, 0.7] & m \textrm{ odd} \\
[-0.4, -0.6] & m \textrm{ even}
\end{cases}
$$
These constants can surely be refined, and maybe the ranges can be made to depend on $m$ as well to sharpen it further. Furthermore, it shouldn't be hard to show that $f_m^\infty-\sqrt m \to \pm \frac12$ for $m\to \infty$.

In particular everything is real in these cases (which is easy to check). We can show it by backwards induction on $m$. You can check that the base case $m=n-1$ holds for $n\ge8$.

For $m<n-1$ we must check for odd $m$ that
$$
f_m^n \in \left[ \sqrt{m + \sqrt{m} - 0.4}, \sqrt{m + \sqrt{m} - 0.6} \right]
\subset \left[ \sqrt{m-1} + 0.5, \sqrt{m-1} + 0.7 \right]
$$
and for even $m$ that
$$
f_m^n\in \left[ \sqrt{m - \sqrt{m} - 0.7}, \sqrt{m - \sqrt{m} - 0.5} \right]
\subset \left[ \sqrt{m-1} - 0.6, \sqrt{m-1} - 0.4 \right]
$$
These are quadratic inequalities, so we can run the numbers and see that they all hold for $m\ge3$.

Bounds on $f_1$:

Assume for now that $f_m^n$ converges as $n\to\infty$, which I'll prove below. Then the same bounds as above hold for $f_m := \lim_{n\to\infty}f_m^n$. Let's get some better bounds on $f_3$. We have
$$
f_3 = \sqrt{3+\sqrt{4-f_5}}
\in \left[\sqrt{3+\sqrt{4-\sqrt{4}-0.7}}, \sqrt{3+\sqrt{4-\sqrt{4}-0.5}}\right]
\approx [2.035, 2.055]
$$
We could get increasingly better bounds by using a higher term than $f_5$. Assuming the principal squareroot for complex numbers, we get approximately
$$
f_2 = \sqrt{2-f_3}
\in [0.186i, 0.235i]
$$
This finally gives us bounds on $f_1$. If $f_1=x+iy$, then $f_1$ lies on the curve $x^2-y^2=1$ and between the values approximately $1.00430 + 0.0928i$ and $1.00681 + 0.117i$.

Convergence:

We want to use that
$$
\lvert \sqrt{a+x} - \sqrt a \rvert \le \frac{|x|}{2}
$$
or
$$
\sqrt{a+x} = \sqrt a + \varepsilon\left(\frac{|x|}{2}\right)
$$
where $\varepsilon(x)$ means some number that is smaller than $x$ in absolute value.

This holds for all $x$ as long as $a\ge4$. We only need to show convergence for one value of $m$, so choose $m_0$ large enough that $(f_m^n)^2 \ge 4$ for all $n \ge m \ge m_0$ (we can do this using the bounds above). We unravel the radical from the inside out and get inductively (with the base case $k=0$ evident):
$$ \begin{split}
f_{n-k}^n
&= \sqrt{(n-k) \pm f_{n-k+1}^{n}} \\
&= \sqrt{(n-k) \pm f_{n-k+1}^{n-1} + \varepsilon\left(\frac{\sqrt n}{2^{k-1}}\right)} \\
&= \sqrt{\left(f_{n-k}^{n-1}\right)^2 + \varepsilon\left(\frac{\sqrt n}{2^{k-1}}\right)} \\
&= f_{n-k}^{n-1} + \varepsilon\left(\frac{\sqrt n}{2^{k}}\right) \\
\end{split}
$$
In particular
$$
f_{m_0}^n = f_{m_0}^{n-1} + \varepsilon\left(\frac{\sqrt n}{2^{n-m_0}}\right)
$$
which means that
$$
f_{m_0} = \sum_{n=m_0}^\infty \varepsilon\left(\frac{\sqrt n}{2^{n-m_0}}\right)
$$
which is absolutely convergent.


问:
Why the recurrence relation of Markov chain can be written as the following?
I am reading E-Li-Vanden-Eijnden's Applied Stochastic Analysis, and the author claims:

The recurrence relation of $\boldsymbol{\mu}_n=\boldsymbol{\mu}_{n-1}\boldsymbol{P}$, where $\boldsymbol{\mu}_n$ is the distribution of $X_n$, and $\boldsymbol{P}$ is the transition matrix, can be written as $$\mu_{n,i}=\mu_{n-1,i}\left(1-\sum_{j\in S,j\neq i}p_{ij}\right)+\sum_{j\in S, j\neq i}\mu_{n-1,j}p_{ji}$$
where $S$ is the state space, $p_{ij}=P(X_n=j|X_{n-1}=i)$, and $p_{ji}=P(X_n=i|X_{n-1}=j)$.

I understand the intuition behind this equation: the first term on RHS gives the total probability of not making a transition from state $i$, and the second term gives the probability of transitioning from one state $j\neq i$ to state $i$. However, I do not know how to prove this rigorously.

My current attempt is to follow how I understand the equation, by separating them into two parts using conditional probability, but I am showing little progress. Can anyone give a proof of this equation? I will update the post if I yield further progress. This might be a naive question, but I am new to Markov chains. Thank you.

Edit: Thanks to someone in the chat,

$\Bbb P(X_n = i) = \sum_j \Bbb P(X_n = j \mid X_{n-1} = j) \cdot \Bbb P(X_{n-1} = j) = \Bbb P(X_n = i \mid X_{n-1} = i) \cdot \Bbb P(X_{n-1} = i) + \sum_{j\neq i} \Bbb P(X_n = j \mid X_{n-1} = j) \cdot \Bbb P(X_{n-1} = j)$

$\sum_j \Bbb P(X_n = j \mid X_{n-1} = i) = 1$ so $\Bbb P(X_n = i \mid X_{n-1} = i) = 1 - \sum_{j \neq i} \Bbb P(X_n = j \mid X_{n-1} = i)$


问:
How to solve this type of integration?
I had the following integration: $$\int_0^z w^{m-1}K_0(2\frac{m}{\Omega}\sqrt{w})dw.\tag{1}$$

Can i use the following formula from Table of integrals to solve the integral in eq. $(1)$

$$\int_0^{\infty}x^{\mu}K_v(ax)dx=2^{\mu-1}a^{-\mu-1}\Gamma\left(\frac{1+\mu+v}{2}\right)\Gamma\left(\frac{1+\mu-v}{2}\right).$$
Any help in this regard is highly appreciated.

答:
It is much more complicated for finite values of $z$ since it will be (assuming $m>0$).

Let $w=\frac{ \Omega ^2}{4 m^2}t^2$ to face
$$I=2 \left(\frac{\Omega ^2}{4m^2}\right)^m\int  t^{2 m-1} K_0(t)\,dt$$
$$\int  t^{2 m-1} K_0(t)\,dt=$$ $$\frac{t^{2 m} \left(2 m K_0(t) \, _1F_2\left(1;m,m+1;\frac{t^2}{4}\right)+t K_1(t)
   \, _1F_2\left(1;m+1,m+1;\frac{t^2}{4}\right)\right)}{4 m^2}$$

For $t=0$ the value is zero.

Title: How to solve this type of integration?


问:
Is $(Q,.)$ an abelian and cyclic group?
My teacher said that an example of an abelian group that is not cyclic is $(Q/\{0\},.)$.

So I was curious why did he use $Q/\{0\}$ instead of just $Q$. Does using $Q$ make it cyclic?

答:
As pointed out, $\Bbb Q \color{red}{\setminus} \{0\}$ was used since $(\Bbb Q, \cdot)$ is not a group because $0$ has no multiplicative inverse.

$(\Bbb Q \color{red}{\setminus} \{0\}, \cdot)$ is indeed abelian since the usual multiplication of rationals is commutative. However, it is not cyclic.

To see this, suppose that $r \in \Bbb Q \color{red}{\setminus} \{0\}$ was a generator. Then, $r^n = 2$ and $r^m = 3$ for some integers $n$ and $m$.

By considering $2^m$ and $3^n$, arrive at a contradiction.

Note the slash that I have used. $\Bbb Q \color{red}{\setminus} \{0\}$ denotes the set of non-zero rationals. The slash which you have used has a different meaning.

Title: Is $(Q,.)$ an abelian and cyclic group?


问:
Are Unions of simply connected sets still simply connected?
Is this true? I would like to use this fact to make a different proof more intuitive and readable, but don't see how to justify it.</p>

答: A circle is not simply connected, but you can write it as an union of two half-circles, each of which is simply connected.</p>


问:
A basis for a subspace of $\mathbb R^3$
I have the following question:

Find the basis of the following subspace in $\mathbb R^3$:
  $$2x+4y-3z=0$$

This is what I was given. So what I have tried is to place it in to a matrix $[2,4,-3,0]$ but this was more confusing after getting the matrix $[1,2,-3/2,0]$. This was done to get a leading 1. Now I solved for $x,y,z$.

$$1x + 2y - \frac3 2z = 0$$ from the matrix. Then
$x = \frac3 2z - 2y$
so
$$\left(\frac3 2z-2y, 2y, \frac3 2z\right)$$

Now I have $$(x,y,z) = z(3/2, 0, 3/2) + y(-2, 2, 0) \implies
{(3/2,0,3/2),(-2,2,0)}$$

which has two dimensions.

But this solution that I am giving is wrong. The actual answer according to my text book is ${(3,0,2),(-2,1,0)}$. Can any one please tell me where I messed-up? Thanks.

答:
I think there are a little confusions, I'll show it step by step (even the most basic ones), but of course this is the kind of thought that after dealing with math for some time becomes "automatic", so is just a way to understand what is going on. Consider the set:

$$S=\{(x,y,z)\in \Bbb R^3 : 2x+4y-3z=0\}$$

So what this means is that if $(x,y,z) \in S$, then it must be true that $2x+4y-3z =0$, so in truth this is saying that "being an element of $S$ implies satisfying this equation" (of course the converse is also true, but we all know that).

Now, satisfying $2x+4y-3z=0$ implies satisfying:

$$z=\frac{2x+4y}{3}$$

So, since being a point of $S$ implies satyisfying that equation and since satisfying that equation implies that the last coordinate is related to the first two in this way, this chain of implications leads us to the conclusion that being a point  of $S$ implies that it has the following form:

$$p \in S\Longleftrightarrow \ \exists \ x,y\in \Bbb R \ : \ p=\left(x,y,\frac{2x+4y}{3}\right)$$

And because of the operations defined in $\Bbb R^3$ we have that $p$ must be of the form:

$$p = x\left(1,0,\frac{2}{3}\right)+y\left(0,1,\frac{4}{3}\right)$$

This means that being an element of $S$ is the same as being linear combination of those two vectors, so $S$ is the span of those vectors, and since they are linearly independent the definition of basis implies that a basis for $S$ is just the set of those two vectors.

Title: A basis for a subspace of $\mathbb R^3$


问:
Irreducibility of unitary representation associated with a positive definite function
I have a positive definite function $u$ (with $u(e)=1$) on $S_4$. Theory says it must be of the form:
$$u(\sigma)=\langle\xi,\rho(\sigma)\xi\rangle$$
for some unitary representation $\rho:G\rightarrow B(\mathsf{H})$ and unit vector $\xi\in\mathsf{H}$.

Suppose that:
$$\sum_{\sigma\in S_4}u(\sigma)=\left\langle\xi,\sum_{\sigma\in S_4}\rho(\sigma)\xi\right\rangle=0.$$

Can I conclude that $\rho$ is irreducible?

答:
We can answer this question or complex representations of finite groups.

Note that $p=\frac{1}{n!}\sum \rho(g)$ projects a rep $V$ onto its invariants $V^G$. Suppose $d=\dim V^G$ and $U$ is a complementary invariant subspace (containing all nontrivial irreps) to $V^G$ with $r=\dim U$. Then $n!p$ has block-diagonal form $n!I_d\oplus 0_r$, and then $w=\sum u(g)=n!\|p(\xi)\|^2$. Thus, your sum $w$ is $0$ exactly when the trivial-rep component of $\xi$ vanishes, or $p(\xi)=0$.


答:
When you found $2^x=3$, by the very definition, it follows that
$$\boxed{x=\log_23}$$

NOTE:

Logarithms are defined as
$$a^b=c\iff \log_ac=b$$
Here, $a$ is the base. For $a=10$, we call it common logarithm, simply denoted by $\log x$ ($=\log_{10}x$). For $a=e$, we call it natural logarithm denoted by $\ln x$ ($=\log_ex$).

Since you are a beginner to the concept, I recommend you to check out some easy introduction to them, such as by mathsisfun.com or brilliant.org.

Hope this helps. Ask anything if not clear :)

Title: Finding the value of $x$ in an exponential form

Body:
Solve the following equation using Logarithms and leave your answer to 2 significant figures.

$4^{x}-2^{x+1}-3 = 0$

I tried to change the expression into a quadratic expression but I am stuck on where I found $3 = 2^{x}$.

Patiently waiting for your help mates


问:
Proving a certain surface is complete
Let's consider the following abstract surface on the unit disk $\mathbb{D}=\left\{(u,v) \in \mathbb{R}^{2} \ | \ u^2+v^2 < 1\right\}$:
$$S = \left\{\mathbb{D}, \left\{E=G=\frac{4}{(1-u^2-v^2)^2}, \ F = 0\right\}\right\}$$
($E,G,F$ denote the coefficients of the first fundamental form.)

Is $S$ necessarily complete? I have tried using arguments related to Gaussian curvatures to show that $S$ must be complete (Using the Brioschi Formula, we can get that the Gaussian curvature $K$ is always equal to $-1$)....but I got stuck on it. Any help will be greatly appreciated!

答:
You are missing the square in the denominator of $E=G=\dfrac4{(1-u^2-v^2)^2}$.

By Hopf-Rinow theorem, we just need to show there is a point $p\in\mathbb{D}$ for which $\mathbb{D}$ is geodesically complete at $p$, i.e., you can extend geodesics infinitely starting at $p$ in every direction.  The point $p=0$ (i.e., $(u,v)=(0,0)$) is a convenient choice here.  Then noting Euclidean reflections across diameters of $\mathbb{D}$ are isometries, we know the geodesics through $0$ are the diameters.  So it suffices to show the length of a Euclidean radius is infinite in this metric:
$$
\int_0^1\frac{2\,\mathrm{d}r}{1-r^2}=\infty
$$

Title: Proving a certain surface is complete


问:
Proving that $a \cos A+b\cos B+c \cos C$ is the semi-perimeter of $\triangle ABC$

With each symbol having its usual meaning, prove that in $\triangle ABC$,
$$a \cos A+b\cos B+c \cos C=s$$

where $s$ is the semi-perimeter.

I started by applying the rule of sines:

$a=k\sin A$

$b=k\sin B$

$c=k \sin C$

where $k$ is a constant

so, we have:

$k(\sin A\cos A+\sin B\cos B+\sin C\cos C)$

manipulating this,  we get:

$\frac{k}{2}(\sin 2A+\sin 2B+\sin 2C)$

$\frac{k}{2}(2\sin(A+B)\cos(A-B) + \sin 2C)$

$\frac{k}{2}(2\sin C\cos(A-B)+\sin 2C)$

$\frac{k}{2}2\sin C(\cos(A-B)+\cos C)$

But this doesn't seem to be yielding desired results. Could someone help me out with this? Or maybe tell me a better approach to this problem?

答:
What is true is the relation $a\cos A+b \cos B+c\cos C=\dfrac{2rs}{R}$ where $r$ is the inradius, $R$ is the circumradius and $s$ is the semi perimeter of the triangle.

By the sine rule, we can write the length of the sides as $a=2R\sin A,b=2R\sin B,c=2R\sin C$ therefore we have $$a\cos A+b \cos B+c\cos C=R(\sin 2A+\sin 2B+\sin 2C)$$

It is well known that in a triangle, we have $\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B \sin C$

So we have $a\cos A+b \cos B+c\cos C=4R\sin A\sin B \sin C$

Again, by the sine rule, re writing $\sin A=\dfrac{a}{2R}$ and similarly for other sides, we obtain $a\cos A+b \cos B+c\cos C=\dfrac{abc}{2R^2}$

It is also well known that $abc$ and $\Delta$ are related by the relation $\Delta=\dfrac{abc}{4R}$, so substituting the value into the equation we obtain the value as $\dfrac{2\Delta}{R}$, but since $\Delta=rs$, we have $$a\cos A+b \cos B+c\cos C=\dfrac{2rs}{R}$$

Helpful: Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle


答:
The semi-perimeter formula you're trying to use here is not true
$$a\cos{A}+b\cos{B}+c\cos{C} \ne s $$
If each symbol has its usual meaning in $\Delta ABC$

$$\cos{A} = \frac{b^2+c^2-a^2}{2bc}$$

$$a \frac{b^2+c^2-a^2}{2bc} + b  \frac{a^2+c^2-b^2}{2ac} + c  \frac{a^2+b^2-c^2}{2ab} $$

$$\frac{ a^2(b^2+c^2-a^2)+b^2(a^2+c^2-b^2)+c^2(a^2+b^2-c^2) }{2abc}$$

$$\frac{ 2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4}{2abc}$$

Simplifying this and checking it out doesn't yield the semi-perimeter $s$

Title: Proving that $a \cos A+b\cos B+c \cos C$ is the semi-perimeter of $\triangle ABC$

Body:
With each symbol having its usual meaning, prove that in $\triangle ABC$,
$$a \cos A+b\cos B+c \cos C=s$$

where $s$ is the semi-perimeter.

I started by applying the rule of sines:

$a=k\sin A$

$b=k\sin B$

$c=k \sin C$

where $k$ is a constant

so, we have:

$k(\sin A\cos A+\sin B\cos B+\sin C\cos C)$

manipulating this,  we get:

$\frac{k}{2}(\sin 2A+\sin 2B+\sin 2C)$

$\frac{k}{2}(2\sin(A+B)\cos(A-B) + \sin 2C)$

$\frac{k}{2}(2\sin C\cos(A-B)+\sin 2C)$

$\frac{k}{2}2\sin C(\cos(A-B)+\cos C)$

But this doesn't seem to be yielding desired results. Could someone help me out with this? Or maybe tell me a better approach to this problem?


问:
Questions about Proof of Theorem 2.7, Rudin's RCA
I know several questions related to this already exist on this site, but I didn't find those particularly helpful. In some places, I wish to clarify my line of thought, and in others, I need some help.

$U$ is open, so $C = U^c$ is closed. $\overline G, \overline{W_p}$ are closed by definition. So, $C\cap \overline G\cap \overline{W_p}$ is also closed. We know that $\overline G$ is compact, and that $C \cap \overline G\cap \overline{W_p} \subset \overline G$. Can we use this to conclude that $C \cap \overline G\cap \overline{W_p}$ is compact, since a closed subset of a compact set is also compact?

I understand that $$V = G\cap W_{p_1}\cap\ldots\cap W_{p_n}$$ is open, since it is a finite intersection of open sets. I must show that $\overline V$ is compact. How does the author conclude $$\overline V \subset \overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$$
I think this is how: $\overline V$ is the smallest closed set containing $V$, in that every other closed set containing $V$ must be a superset of $\overline V$. $V \subset \overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$ is verified easily, since $A\subset \overline A$ for every set $A$, and inclusion is preserved under intersections. Is that right?

How does $\overline V \subset \overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$ help us conclude that $\overline V$ is compact? Here's what I'm thinking: $\overline G$ is compact, and $\overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}} \subset \overline G$. Moreover, $\overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$ is closed, since it is an intersection of closed sets. A closed subset of a compact set is also compact, so $\overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$ is compact. We can apply a similar argument to $\overline V$ and $\overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$ to conclude that $\overline V$ is compact - correct?

Not a technical doubt as such, but could you provide some intuition as to why the author chose to prove the theorem in the way he did? In particular, what may have motivated bringing $W_p$'s etc. in the picture?

Proof attached for reference:

Thanks a lot!

答:
Your thoughts are correct. The key is that closed subsets of compact sets are compact (by the way, this is true also in non-Hausdorff spaces). Here are some comments.

What you do in 4. is too complicated. You know that $\overline G$ is compact and $\overline V \subset \overline G$, hence $\overline V$ is compact.

The intuition in Rudin's proof is this: There is no reason why set $\overline G$ should be contained in $U$. Therefore he must shrink the set to obtain something contained in $U$. This is done by introducing the $W_p$ for $p \in C$ which are open sets containing $K$, but not containing $p$ in their closure. The $U_p = G \cap W_p$ are open sets containing $K$ and the $K_p = \overline G \cap \overline{W_p}$ are closed subsets of $\overline G$, hence compact. Therefore $K \subset U_p \subset K_p$ and $K'_p = C \cap K_p$ is the part of $K_p$ not contained in $U$. It is again compact. By construction their intersection does not contain any point of $C$. Since they are subsets of $C$, their intersection is empty. This implies that we find finitely many $K'_{p_i}$ with empty intersection. Then clearly the intersection of the $K_{p_i}$ is contained in $U$.


问:
If I have three points, is there an easy way to tell if they are collinear?
Points $(a,b)$, $(m,n)$, and $(x,y)$ are selected at random. What is the quickest/easiest way to tell if they are collinear? At first I thought it was a matter of comparing slopes but that doesn't appear to be enough.</p>

答:
This is an explanation of $\textbf{Ma Ming}$'s excellent idea, in case it was unclear.

Consider a triangle with vertices at points $(a,b)$, $(m,n)$, and $(x,y)$, and observe that the area $A$ of the triangle is zero if and only if the points are collinear. Now, it can be shown that

$$A=\frac{1}{2}\left|\det\pmatrix{1&1&1\\a&m&x\\b&n&y}\right|=\frac{1}{2}\left|a(n−y)+m(y−b)+x(b−n)\right|.$$

Hence we just have to check whether the expression $a(n−y)+m(y−b)+x(b−n)$ is non-zero or not.