问:
Why truth table is not used in logic?
One day, I bought Principia Mathematica and saw a lot of proofs of logical equations, such as $\vdash p \implies p$ or $\vdash \lnot (p \wedge \lnot p)$. (Of course there's bunch of proofs about rel&set in later)

After reading these proofs, I suddenly thought that "why they don't use the truth table?". I know this question is quite silly, but I don't know why it's silly either (just my feeling says that).

My (discrete math) teacher says that "It's hard question, and you may not understand until you'll become university student," which I didn't expected (I thought the reason would be something easy).

Why people don't use truth table to prove logical equations? (Except for study logic (ex: question like "prove this logic equation using truth table"), of course.)

PS. My teacher is a kind of people who thinks something makes sense iff something makes sense mathematically.

答:
Your teacher may be referring to the difference between semantics and syntax. A syntactic proof is a finite formal derivation of a sentence from the axioms of a theory using the logical axioms and the rules of inference of a logic. A proof by a truth table is a semantic proof; in allowing truth tables you are tacitly assuming the completeness theorem of propositional logic. Essentially, a priori, we don't know that everything we can prove by studying the models of a theory (i.e. truth tables, in the case of propositional logic) can be proven syntactically, or even for that matter vice versa. It's a non-trivial result in logic, which I will state below:

Definition. Let $T$ be a theory in some logic. We say the logic is sound if every sentence $P$ which can be formally derived from the axioms of $T$ is true in every model of $T$, i.e. $T \vdash P$ implies $T \vDash P$. We say the logic is complete if when a sentence $P$ is true for every model of $T$, $P$ can also be derived formally from the axioms of $T$, i.e. $T \vDash P$ implies $T \vdash P$.

Theorem. Classical propositional logic is sound and complete.

Theorem (Gödel). Classical first-order logic is sound and complete.

And yes, it is that Gödel. There are also incomplete logics; you don't have to go very far to find them, since even second-order logic (under the usual semantics) is incomplete.

Title: Why truth table is not used in logic?


问:
How to show that differential operator can be defined in terms of certain commutator operators
Let $U$ be any open subset of $\mathbb{R}^n$ (or, more general, of some smooth manifold). Define $\mathcal{D}_{-1}(U):=\{0\}$. For any two linear operators $A$ and $B$, the commutator operator $[A,B]$ is given by $A\circ B-B\circ A$. For any $f\in C^\infty(U)$ we define the operator $m_f\in \text{End}(C^\infty(U))$ such that $m_f\ g=f g$. Consider $P\in \text{End}(C^\infty(U))$. By definition we take that $P\in \mathcal{D}_k$ iff $[P,m_f]\in \mathcal{D}_{k-1}$.

My problem is that I want to show that the set, call it $\mathcal{D}_k'$, of all partial differential operators with smooth coefficient functions of order at most $k$ equals $\mathcal{D}_k$ for each $k\in\{0,1,2,\dots\}$.

It's easy to show that $\mathcal{D}'_0=\mathcal{D}_0$. The same goes for the statement that $\mathcal{D}'_k\subset \mathcal{D}_k$ for all $k$. Difficulty lies in the reverse inclusion. I have strong feeling that any $P\in \mathcal{D}_k$ can be represented by

$ P=\sum_{|\alpha|\leq k}\frac{C^\alpha_P(1)}{\alpha!}\partial^\alpha $.

Here $\alpha$ are multi-indices and $C^0_P:=P$ and $C^\alpha_P=[C^\beta_P,f_{\alpha-\beta}]$, where $\beta$ is any multi-index of an order lower than $\alpha$ such that $\alpha-\beta$ is well defined as a multi-index and its order is equal to one. Furthermore for any $f_{\alpha-\beta}$ gives the $(\alpha-\beta)^\text{th}$ coordinate of any $x\in U$: $f_{\alpha-\beta}(x)=x_{\alpha-\beta}$.

However I have not been able to complete the inductive argument. It should be possible to show this with just algebraic properties of differential calculus and the definition given above. But I don't see it.

Kindly appreciated,

Aris

答:
I think I have found a solution. For anybody interested, here are some pointers in what I think to be the right direction.

Consider an operator $P\in \mathcal{D}^k$ and some smooth function $f$ and suppose we are interested in the value of $Pf$ at $x_0$. Then consider the $k^\text{th}$ Taylor expansion $f^k_{x_0}$ of $f$ around $x_0$. The error term in the expansion can be written in the form $E(x)=\sum_{|\beta|=k}g_\beta(x) (x-x_0)^k$, where $g_\beta$ are smooth functions. If you let $P$ work on f written in it's Taylor's expansion and use the inductive properties of class $\mathcal{D}^k$, and evaluate the function each time at $x=x_0$, you will finally get something that looks like what is stated in my problem description and the error term drops out.

ContentLicense: CC BY-SA 3.0


问:
Commuting with a rank deficient matrix
Given a positive definite matrix $X$ of size $n \times n$ and postive semidefinte $Q$ of same size and rank $m \leq n$.

If $X$ and $Q$ commute, i.e, $XQ = QX$ is then $Q X = \alpha Q$ for some scalar $\alpha$?

For instance $m = 1$, the eigenvalue decomposition yields $Q = \lambda u u^T$ for vector $u$ and eigenvalue $\lambda$. For $Q X = \lambda u u^T X$ to be symmetric, it is $u^T X = \alpha u^T$. Can this argument be expanded to $m > 1$?

答:
The answer is no. As a counterexample, consider the diagonal matrices
$$
X = \pmatrix{1\\&2\\&&3}, \quad Q = \pmatrix{1\\&1\\&&0}.
$$

A non-diagonal counterexample:
$$
X = \pmatrix{3&-1&-1\\-1&4&-2\\-1&-2&4},
\quad Q = \pmatrix{2&-1&-1\\-1&2&-1\\-1&-1&2}.
$$

ContentLicense: CC BY-SA 4.0

Title: Commuting with a rank deficient matrix


问:
Lebesgue-Integral as Riemann-Integral
This is a rather basic question on parts of the Lebesgue-Integral formula. Intuitively it should be clear, but I am searching for a mathematical explanation.

If $f: (\Omega, \mathcal{A}) \rightarrow (\mathbb{R}, \mathfrak{B}(\mathbb{R}))$ is a measurable function and $\mu$ a measure on $(\Omega, \mathcal{A})$ and $\mu(\Omega) < \infty$, than we can write the Integral of $f$ wrt $\mu$ as
$$\int f d\mu  = \int f^+ d\mu - \int f^- d\mu
=\int_{0}^{\infty} \mu \{y|f(y) \geq x\} dx + \int_{-\infty}^{0} [\mu \{y|f(y) \geq x\} - \mu(\Omega)] dx.  $$

I don't get how to derive the second part, i.e. where does the expression $\mu \{y|f(y) \geq x\} - \mu(\Omega)$ come from? (I know that this concerns the potentially negative part of the function, but at the moment I am unable to get, how I derive this second expression). Some help would be awesome!

答:
Hint:
$$ \mu( \{y \quad | \quad f(y) < x \}) + \mu (\{y \quad | \quad f(y) \ge x\} = \mu (\Omega)  $$

Title: Lebesgue-Integral as Riemann-Integral


问:
$f(x) < g(x) \Rightarrow \int_{a}^{b} f(x) dx < \int_{a}^{b} g(x)dx \ $. Why not in both directions?
Let $f$ and $g$ be continuous functions on the interval $\left[ a,b\right]$ and $a<b$

$ f(x) < g(x)  \Rightarrow \int_{a}^{b} f(x) dx < \int_{a}^{b} g(x)dx \ $ for $x \in \left[ a,b\right]$

Why doesn't the statement apply in the other direction?

$ \int_{a}^{b} f(x) dx < \int_{a}^{b} g(x)dx \ \not\Rightarrow  f(x) < g(x)$

答: Because this could happen (same reasoning as the other answer):


问:
Proving a given statement in 3D trigonometry, but given NO side lengths

I've found $tan(\alpha)$, $tan(\beta)$ and $tan(\gamma)$ in terms of $h$ and the side lengths
but I'm having trouble proving what's required because these types of questions normally include numerical values for the side lengths
I think the key is using the bottom triangle - if I am correct then it is made up of two 45, 45, 90 triangles
So perhaps I use the ratios of such a triangle? 1, 1, √2

答:
Let S b the area of triangle ABC, we have:

$2S=AC\times AB= AP\times BC$

$\frac{AC^2\cdot AB^2}{AP^2}=BC^2=AB^2+AC^2$

Dividing both sides by $AC^2\cdot AB^2$ we get:

$\frac 1{AP^2}=\frac 1{AB^2}+\frac 1{AC^2}$

Multiplying both sides by $h^2$ we get:

$\frac{h^2}{AP^2}=\frac{h^2}{AB^2}+\frac{h^2}{AC^2}$

Therefore:

$tan^2 \gamma=tan^2 \alpha + tan^2 \beta$


问:
How do I turn this into a sigma function?
I asked a question yesterday where I was being a bit too vague I believe. Here I have detailed it further, to show the proper trickiness with the task at hand.

The n-value starts at 8 and rises by one. I will show the different operations for the differen't n-values below.

$$\lceil\frac{8-(3\times2+1)}{2}\rceil$$

$ans=1$

$$\lceil\frac{9-(3\times2+1)}{2}\rceil$$

$ans=1$

$$\lceil\frac{10-(3\times2+1)}{2}\rceil$$

$ans=2$

$$\lceil\frac{11-(3\times2+1)}{2}\rceil + \lceil\frac{11-(3\times3+1)}{2}\rceil$$
$ans=3$
$$\lceil\frac{12-(3\times2+1)}{2}\rceil + \lceil\frac{12-(3\times3+1)}{2}\rceil + \lceil\frac{12-(3\times3+2)}{2}\rceil$$
$ans=5$

$$\lceil\frac{13-(3\times2+1)}{2}\rceil + \lceil\frac{13-(3\times3+1)}{2}\rceil + \lceil\frac{13-(3\times3+2)}{2}\rceil$$
$ans=7$

$$\lceil\frac{14-(3\times2+1)}{2}\rceil + \lceil\frac{14-(3\times3+1)}{2}\rceil + \lceil\frac{14-(3\times3+2)}{2}\rceil + \lceil\frac{14-(3\times4+1)}{2}\rceil$$
$ans=9$

$$\lceil\frac{15-(3\times2+1)}{2}\rceil + \lceil\frac{15-(3\times3+1)}{2}\rceil + \lceil\frac{15-(3\times3+2)}{2}\rceil + \lceil\frac{15-(3\times4+1)}{2}\rceil + \lceil\frac{15-(3\times4+2)}{2}\rceil$$
$ans=11$

$$\lceil\frac{16-(3\times2+1)}{2}\rceil + \lceil\frac{16-(3\times3+1)}{2}\rceil + \lceil\frac{16-(3\times3+2)}{2}\rceil + \lceil\frac{16-(3\times4+1)}{2}\rceil + \lceil\frac{16-(3\times4+2)}{2}\rceil + \lceil\frac{16-(3\times4+3)}{2}\rceil$$
$ans=13$

So, how do I make this into a sigma function? The answers given on the other questions didn't take into account negative numbers, meaning that the way it was set up would eventually lead to negative numbers decreasing the sum below its actual value (at least, that's how I understood those double sums).

答:
I think the following works:

$$\sum_{k=2}^{\lfloor\frac{n}{3}\rfloor} \sum_{j=1}^{min(k-1,n-3k-1)} \lceil \frac{n-(3  k+j)}{2} \rceil$$

Explanation

The outer sum follows from your $3 \times k$, because you have every $k$ from $2$ to $\lfloor\frac{n}{3}\rfloor$ at least one time in your summation. The inner sum then takes a $k$ and shows which numbers you add behind the $3k$-term. In your summation you have two rules: \

The fraction has to be positive. Therefore the upper summation bound is $\leq n-3k-1$, which gives $\frac{n-(3k+j)}{2} \geq \frac{n-(3k+(n-3k-1))}{2}=\frac{1}{2}$. Every greater $j$ would make it $\leq 0$.

The added term $j$ has to be less than $k$. Therefore the upper summation bound is $\leq k-1$.


问:
Solve $3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$
Solve for $x$ in

$$3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$$

I'm really not good with 4th degree equations but since the 1st term in the RHS looked a simple (a+b)(a-b) application, I tried solving that but I'm really not able to reach to the final answer as it only gets more complicated...
can anyone please help me out?

The equation I got was:

$$x^4-3x^3-22x^2-48x-32=0$$

答:
Expand $$x^4-3 x^3-22 x^2-48 x-32=0$$
Try to factorize $$x^4-3 x^3-22 x^2-48 x-32=\left(a x+b+x^2\right) \left(c x+d+x^2\right)$$
RHS gives $$x^4+x^3 (a+c)+x^2 (a c+b+d)+x (a d+b c)+b d$$
thus we must have

$$
\begin{cases}
a+c=-3\\
ac+b+d=-22\\
ad+bc=-48\\
bd=-32\\
\end{cases}
$$

Final factorization should be
$$\left(x^2-6 x-8\right) \left(x^2+3 x+4\right)=0$$


问:
Prove that if $G$ is abelian, then $H = \{a \in G \mid a^2 = e\}$ is subgroup of $G$

Let $G$ be an abelian group. Prove that $H = \{a \in G \mid a^2 = e\}$ is subgroup of $G$, where
  $e$ is the neutral element of $G$.

I need some help to approach this question.

答:
Hint: To show it is a subgroup you must show the following three things:

$1)$ $e \in H$.

$2)$ If $a$ and $b$ are in $H$ then so is $ab$.

$3)$ If $a$ is in $H$, then so is $a^{-1}$.

To show these three things, think about what it means for a particular element to be in $H$ and try and use this to show that the elements you want to be in $H$ actually are.

Once you have shown this is true, you might like to think about why we needed the group to be abelian, and see if you can find a (non-abelian) group where $H$ is not a subgroup.


问:
2019 IMC Day 1 Problem 3
Source: https://www.imc-math.org.uk/?year=2019&section=problems&item=prob3q

Let $f: (-1,1)\rightarrow\mathbb{R}$ be a twice differentiable function such that $2f'(x)+xf''(x)\geq 1$ for $x\in (-1,1)$. Prove that $\int\limits_{-1}^1 xf(x)dx\geq \frac{1}{3}$.

My work: We have by chain-rule

$$\frac{d}{dx}x^2f'(x)=2xf'(x)+x^2f''(x).$$

Then WLOG $x\geq 0$, and we have

$$\begin{aligned}\frac{d}{dx}x^2f'(x)\geq x\Rightarrow x^2f'(x)\geq \frac{x^2}{2}\iff f'(x)\geq \frac{1}{2}\iff f(x)-f(0)\geq \frac{x}{2}\end{aligned}.$$

And so

$$\int\limits_{-1}^1 xf(x)dx=\int\limits_{-1}^1x\left(f(x)-f(0)\right)dx\geq \int\limits_{-1}^1 \frac{x^2}{2}dx=\frac{1}{3}.$$

Question: I was wondering if the work makes sense or if some bits of the inequalities/work need more/less explanation.

答:
Just another way, let $g(x) = x f(x)$.  Then, we have
$$g'(x) = f(x) + xf'(x), \qquad g''(x) = 2f'(x) + xf''(x) \geqslant 1$$

$$\implies g'(x)-g'(0)=\int_0^x g''(t)\,dt \geqslant \int_0^x 1\, dt = x$$

$$\implies g(x) \geqslant \int_0^x (t+g'(0)) \, dt = \frac{x^2}2+g'(0)x$$

$$\implies \int_{-1}^1 g(x)\, dx \geqslant \int_{-1}^1\left(\frac{x^2}2+g'(0)x \right)\, dx = \frac13$$

Title: 2019 IMC Day 1 Problem 3


问:
Derivative at a point
I have a to solve the following question.

Q: Find $(f'(0))^2$ when

$$f\left(x\right)=\cases{\frac{\sin^2\left(3x\right)}{2x}, x\neq0\\0, x=0}$$

Can anyone please guide me as to how to go about such questions. I have tried taking the derivative of the function and then solved it when $x=0$, but I didn't get the correct answer.

Thank you!

答:
Use $$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}$$
What you get is $$f'(0)=\lim_{x\to 0}\frac{\sin^2(3x)}{2x^2}=\lim_{x\to 0}\left(\frac{\sin(3x)}{3x}\right)^2\frac92$$
Can you take it from here?

Title: Derivative at a point


问:
What precisely is a vacuous truth?
Is there a proper and precise definition that goes something like this?

Definition. A statement $S$ is a vacuous truth if ... ...

答:
Contrary to some of the existing answers, I don't have the impression that one typically speaks of a vacuous truth if the statement is a pure implication whose antecedent happens to be false; the usual use of "vacuous truth" occurs in the context of universal claims where the antecedent is always (i.e. for every object) false.

A universally quantified implication is vacuously true iff it is true just because there are no objects to satisfy the antecedent. Formally,

A statement of the form $\forall u (\phi \to \psi)$ (where $u$ is a varible and $\phi$ and $\psi$ are arbitrary formulas) is vacuously true
in a structure $\mathcal{S} = \langle \mathcal{D}, \mathcal{I} \rangle$
iff

for all $a \in \mathcal{D}$ and arbitarary $v$: $\mathcal{S},
 v^{[u \mapsto a]} \not \models \phi$.

Title: What precisely is a vacuous truth?


问:
Is my solution correct in terms of proof? I can't prove directly that $f(x)=4$.
Given  $f:\mathbb{R}\rightarrow \mathbb{R}^+$ and the following conditions:

$f(x+2)\cdot f(x+3)=16$,

$f(x)+f(-x)=8$,

we have to find integral $\int_{-8}^8 f(x-2019)dx$. So, if we plug in $0$, then $f(0)+f(0)=8$. Therefore, $f(0)=4$. Then plug in $-2$ in first equation,  we get $f(0)\cdot f(1)=16$, therefore $f(1)=4$, analogically plug $x=-1$ and I think that $f(x)=4$, so answer will be  $64$ (value after integrating). Is this correct? How can i prove it in a better manner?

Body:
From 1 we have $f(x+1)f(x)=16$, so $\displaystyle f(x+2) = \frac{16}{f(x+1)} = \frac{16}{16/f(x)} = f(x)$. Since $f$ has period $2$ then
$$\int_{-8}^8 f(x-2019)\,dx = \int_{-8}^8 f(x-1)\,dx = 8 \int_0^2f(x-1)\,dx$$
$$ = 8 \int_{-1}^1 f(x)\,dx = 8\int_{-1}^0f(x)\,dx+8\int_{0}^1 f(x)\,dx$$
$$=8 \int_{0}^1f(-x)\,dx + 8\int_{0}^1 f(x)\,dx = 8 \int_0^1f(x)+f(-x)\,dx = 8 \int_0^1 8\,dx = 64$$


问:
Calculus involving a Physics problem
This is the equation that occurs in a particular problem in physics. If earth falls into the sun if it stopped orbiting suddenly.

$\mathrm{d}r = \frac{GM}{2r^2} \mathrm{dt}^2$enter image description here

This is my attempted solution. Is it correct? Also can someone explain how on one side we have only one integral and on the other side we have two? If we integrate one side of the equation, surely we must also integrate the other side as well. No? So how come we integrate one side once and the other side twice.

$\int_{0}^{R}\frac{2r^2}{GM}\mathrm{d}r=\int_{0}^{T}\int\mathrm{d}t^2$

This becomes $\frac{T^2}{2}=\frac{2R^3}{3GM}$

答:
You've misunderstood how to set up second-order differential equations.

A body that falls vertically has $\frac{d^2r}{dt^2}=-\frac{GM}{r^2}$. This has a number of implications you were considering that I review in the paragraph below, but it is this equation alone that we use to solve for $r$ as a function of $t$, e.g. as in @GerryMyerson's link.

Over short enough falls with initially zero speed for $r$ to change little (this is the simplest case to analyze), the acceleration is approximately constant, so $r\approx r_0-\frac{GM}{2r_0^2}t^2$, where $r_0$ is the $t=0$ value of $r$. On the other hand, at small times $dt$ we have$$r\approx r_0+\underbrace{\dot{r}_0}_0dt+\tfrac12\ddot{r}_0dt^2=r-\tfrac{GM}{2r_0^2}dt^2,$$i.e. the change in $r$ is $-\tfrac{GM}{2r_0^2}dt^2$. However, at a later time when $r$ has reduced, $\dot{r}$ has become negative and the elapsed time is no longer negligible, the result is rather different.

Title: Calculus involving a Physics problem


问:
distribution function of time T
an ambulance station is located 30 miles from one end of a 100-mile road. the station services accidents along the entire road. suppose that an accident occurs. suppose that Suppose accidents occur with uniform distribution along the road and suppose that the ambulance travels 60 miles per hour. Let T be the amount of time it takes the ambulance to arrive at the sence with the assumption that the ambulance leaves the station right after the accident occurs.  find the distribution function FT(t) and the density function Ft(t)</p>

答:
Let random variable $X$ be the distance the ambulance has to travel. The time $T$ is then $\frac{X}{60}$. We will find the distribution of $X$. From that it is not hard to find the distribution of $T$. We do $X$ instead of $T$ for two reasons: (i) It is closer to the intuition and (ii) There might as well be something left for you to do.

We will find the cumulative distribution function $F_X(x)$ of the random variable $X$. Recall that $F_X(x)=\Pr(X\le x)$.

First let's knock off the easy stuff. If $x\lt 0$, then $F_X(x)=0$. (The probability that the ambulance has to travel $\le -17$ miles is $0$. Also, if $x\gt 70$, then $F_X(x)=1$ (for sure the ambulance will have to travel fewer than $88$ miles).

So from now on we confine attention to $0\le x\le 70$. Because of the non-central location of the ambulance, we need to break up this interval.

Suppose $0\le x\le 30$. The ambulance has to travel a distance $\le x$ if te accident takes place within $\le x$ miles from the normal position of the ambulance. This is an interval of length $2x$, so the probability the accident takes place in this interval is $\frac{2x}{100}$.

In symbols, $F_X(x)=\frac{2x}{100}$ if $0\le x\le 30$.

Now suppose that $30\lt x\le 70$. Then we have to travel $\le x$ if (i) the accident takes place within $30$ miles of the ambulance (probability $\frac{60}{100}$) or the accident takes place at a distance between $30$ and $x$ from the ambulance (probability $\frac{x-30}{100}$. Add the two probabilities.

Thus if $30\lt x\le 0$, then $F_X(x)=\frac{x+30}{100}$.

Now we have the cdf. Differentiate to get the density. there are points of non-differentiability at $0$, $30$, and $70$, but we won't worry about them We get $f_X(x)=0$ if $x\lt 0$ and also if $x\gt 70$. For $0\le x\le 30$, $f_X(x)=\frac{2}{100}$. For $30\lt x\le 70$, $f_X(x)=\frac{1}{100}$.

Remark: For your homework problem, you can either imitate the analysis, using $T$ or (easier) directly write down the cdf and density of $T$ (they are close relatives of the cdf and density of $X$.)

Title: distribution function of time T


问:
open covering for compact set exist $\delta$ neighborhood covering
Let $K$ be a compact set in metric space $\Bbb{R}^n$.Prove for an open covering $\mathcal{O}_\alpha$ exist finite open covering such that those set not only covers $K$ but also $\delta$ neiborhood of $K$.that is exist some $\delta>0$ such that subcovering $\mathcal{O}_i$ $i = 1,...,n$ convers $K_\delta = \{x:\text{dist}(K,x)<\delta\}$.

I try to solve it type of arguement in  Lebesgue number lemma  as follows:

Let finite covering be $\mathcal{O}_i$ ,first we can refine the covering to be the compact included convering(as my previous post shows here).Then exist a compact set $\bigcup \mathcal{O}_i\subset C$.then we know $C\setminus\mathcal{O}_i = V_i$ is compact set.then we know continuous function $f:K\to \Bbb{R}$ which is

$$x\mapsto 1/n \sum d(x,V_i)$$

is always positive,hence exist minimum value $\delta$ which can be shown any $x\in K_\delta$,exist some $\mathcal{O}_i$ covers it using the reation that average value is always less than maximum value.

Is my proof correct.

答:
I think there might be problems in your last line of reasoning.

First, let me settle some notation. If $O_1,\dotsc, O_n$ is your original finite cover, then let $U_i$ be the compactly included refinement. We have
$K\subset \bigcup_i U_i\subset C\subset \bigcup_i O_i$ where $C$ is the union of the closures of the $U_i$. Also, we put $V_i =C\setminus U_i$, which are each compact.

I think your desired argument goes something like this: if $z\in K_\delta$ we have $d(z,x)<\delta$ for some $x\in K$. On the other hand, $d(x,V_i)\geq\delta$ for some $i$ (since otherwise $f(x)<\delta$, contradicting that $\delta$ was the minimum). Now you want to say $z\in U_i$, since otherwise $z\in V_i$ and we get $d(x,V_i)\leq d(x,z)<\delta$, a contradiction. This only works though if we assume $z\in C$ to begin with. The most we can say is that $z\in (C\setminus U_i)^c = C^c\cup U_i$.

Here's one possible fix: define the $V_i$ to be $U_i^c$ instead of $C\setminus U_i$. After all, we don't need $V_i$ to be compact for $x\mapsto d(x,V_i)$ to be continuous. And we still know $f(x)>0$ for $x\in K$ since otherwise $x\in V_i$ for each $i$, which would in turn mean $x\notin \bigcup U_i\supset K$.

Honestly though, this result can be proven with much less machinery. Let ${\cal O}= \{O_\alpha: \alpha\in A\}$ be an open cover of $K$. For each $x\in K$, there is some element of this collection $O_{\alpha_x} = O_x$ which contains $x$. Also, by openness, there is some $\delta_x>0$ such that $B(x,2\delta_x)\subset O_x$.

Now it's not hard to see that the collection $\{B(x,\delta_x): x\in K\}$ is an open cover of $K$, and hence there are finitely many points $x_1,\dotsc, x_n$ such that
$$K\subset B(x_1,\delta_1)\cup\dotsm\cup B(x_n,\delta_n).$$
Yet now we've given ourselves some wiggle room. Denoting $O_k = O_{x_k}$, we actually have
$$K\subset B(x_1,\delta_1)\cup\dotsm\cup B(x_n,\delta_n)\subset B(x_1,2\delta_1)\cup\dotsm\cup B(x_n,2\delta_n)\subset O_1\cup\dotsm\cup O_n.$$
From here, put $\delta = \min (\delta_1,\dotsc, \delta_n)$. You can easily see by triangle inequality that each point of $K_\delta$ belongs to one of the $B(x_k,2\delta_k)$ balls, and in particular is covered by $O_k$.


Title: Solve $3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$

Body:
Solve for $x$ in

$$3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$$

I'm really not good with 4th degree equations but since the 1st term in the RHS looked a simple (a+b)(a-b) application, I tried solving that but I'm really not able to reach to the final answer as it only gets more complicated...
can anyone please help me out?

The equation I got was:

$$x^4-3x^3-22x^2-48x-32=0$$

Body:
Note that the given equation

$$3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$$

is purposefully structured to allow convenient factorization as follows

$$3x^3=x^4-(\sqrt{18}x+\sqrt{32})^2 -4x^2$$
$$x^4-3x^3-4x^2-(\sqrt{18}x+\sqrt{32})^2 =0$$
$$x^4-(3x+4)x^2-2(3x+4)^2 =0$$
$$(x^2+(3x+4))(x^2-2(3x+4))=0$$


答:
Hint:

You'll find the answer instantly using some elementary asymptotic analysis:

As $\sin u\sim_0 u$, we have
$$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}x=\lim_{x\to 0}\frac{\sin^2 3x}{2x^2}$$
and

$$\frac{\sin^2 3x}{2x^2}\sim_0 \frac{(3x)^2}{2x^2}.$$

Title: Derivative at a point

Body:
I have a to solve the following question.

Q: Find $(f'(0))^2$ when

$$f\left(x\right)=\cases{\frac{\sin^2\left(3x\right)}{2x}, x\neq0\\0, x=0}$$

Can anyone please guide me as to how to go about such questions. I have tried taking the derivative of the function and then solved it when $x=0$, but I didn't get the correct answer.

Thank you!


问:
Estimating the Shanon entropy of a high-dimensional discrete multivariate random variable by dead simple monte carlo sampling?
I'm interested in estimating Shannon's entropy for a discrete multivariate random variable $X$ that has high dimensionality (i.e. $X=(X1,..,Xn)$, where n is at the hundreds).

I can effectively sample from $X$, as well as evaluate $P(X=x)$. But due to the high dimensionality, I cannot enumerate all possible values to calculate the entropy.

A simple, straightforward Monte Carlo estimator seems to be:

Sample a large number ($m$) of observations $\{o_1,..,o_m\}$ from $X$ and calculate the sample average of their negative log-probability:
$\hat{H}(X)=-\frac{1}{M}\sum_i{\log{p(o_i)}}$.

This approximates $H(X)=-\sum_{x\in X}{p(x)\log{p(x)}}$ which requires summation over all possible values of $X$.

Is this a correct approach? Am I missing something?

Is this estimator biased? and if it is, why?

If this is an unbiased estimator, then what is the issue that is addressed by entropy estimation methods such as those mentioned in this question? Does the bias arise in the setting where we can't evaluate $\log{p(x)}$ and must resort to noisy occurrence counts or histograms?

答:
In general, for any rv $X$ and any function $Y=g(X)$ such that $E[Y^2]$ is finite, we have that the sample average

$$ \frac{1}{M}\sum_M  g(x_i)  $$ is an unbiased consistent estimator of $E[g(X)]$ (hence it converges in probability, cf WLLN)

Then, if you know the probability distribution $p$, then of course the estimator is unbiased (and consistent).

The problem is when we don't know $p$ and we must estimate it from the samples.
As the accepted answered of the linked question points in the flowchart, when one knows $p$ the problem does not arise.


问:
How show that $x_0$ is an accumulation point of the domain of a function
Let the function $\frac{\sin x}{x}$ defined $\forall x\neq 0$. How I can prove that $x_0=0$ is an accumulation point for $A=\{x\in\mathbb{R}: x\neq 0\}$?

答:
By the definition of accumulation point, for every open set in $\mathbb{R}$ (recall you can prove it for elements of the basis, so you can prove this claim for every open interval) that contains $x=0$, you can show that it contains another point of your set $A$.

Another hint may be to construct a sequence of points $x_n \in A$ converging to $0$, for example pick the sequence $x_n =1/n \in A^{\mathbb{N}}$.


问:
Numbers of the form $\sum_{n=0}^{N} 2^{a_n} 3^n$
Just out of curiosity.
Does the numbers of the form
\begin{align}
 \sum_{n=0}^{N} 2^{a_n} 3^n \text{,}
\end{align}
with $a_n \ge 0$ and $N \ge 0$, cover all integers not divisible by 3?

These numbers can never be divisible by three.
It is also easy to discover integers with multiple representations (e.g., $4 = 2^2 3^0 = 2^0 3^0 + 2^0 3^1$).
I am however stuck to prove that they cover all integers not divisible by 3.

Second question: The numbers of the form
\begin{align}
 \sum_{n=0}^{N} 5^{a_n} 3^n \text{,}
\end{align}
with $a_n \ge 0$ and $N \ge 0$, are numbers congruent to $\{1, 4, 5, 8\} \pmod{12}$.
These are the numbers not divisible by 3 and not $\{ 2, 3 \} \pmod{4}$.
Many other such forms may be considered.
What is the name of such numbers (numbers of such a form)?

答:
If $x \le 8$, then all numbers not divisible by $3$ have the desired form as one can check by hand (e.g. $1=1$, $2=2$, $4=4$, $5=2+3$, $7=1+2 \cdot 3$, $8 = 8$).

If $x > 8$, and $x \equiv 1,2,4,5,7,8 \bmod 9$, then let $m = 4,8,1,2,4,2$ respectively, so $m$ is a power of $2$ and $x \equiv m \bmod 3$ but $x \not\equiv m \bmod 9$. Now write

$$x = m + 3y$$

Now, by construction, $y<x$ is not divisible by $3$. By induction, it has the required form, which implies that $x$ also has the required form. Indeed from this proof you see that one can take the power of $2$ to be in the set $\{1,2,4,8\}$.


问:
Does there exist a group with two such subgroups?
Does there exist a group $G$ with the following properties? ($N_G$ is the normalizer)

$\exists H,F\leq G$ such that there is no injection $i:F\to H$ or $i:H\to F$

$N_G(H)/H$ and $N_G(F)/F$ are both abelian

$N_G(F\cap H)/F\cap H$ is not abelian

This problem emerged from a disagreement I had with a professor about wether we can omit the normality condition in the definition of a Universal Abelian Covering but no Algebraic Topology knowledge is required to tackle it of course.

答:
The smallest nonabelian group is already an example: $G=S_3$, $H=S_2$, $F=A_3$.
More generally, for $n\geq 3$, you can take $G=S_n$, $H=S_{n-1}$, $F=C_n$.


问:
Is this the correct procedure for parametric derivative?
I'm practicing parametric and polar derivatives, in some of those i have trouble with the answers because they don't look "nice", i know there is no parameters of whats is supposed to be "nice" but in fact this kind of trouble commontly ends with a very small expression

in particular with this: $$x=e^{2t} \cos^{2}(t)$$  $$y=e^{2t} \sin^{2}(t)$$ find $\frac{dy}{dx}$ for that i use  $\frac{y'}{x'}$
then $$x'=2e^{2t}( \sin^2(t) + \cos(2t))$$ and $$y'=2e^{2t}( \cos^2(t) - \sin(2t))$$
(aplying the product, power and chain rule, etc) after this doing $$\frac {x'=2e^{2t}( \sin^2(t) + \cos(2t))}{y'=2e^{2t}( \cos^2(t) - \sin(2t))}$$
$$\frac {x'=( \sin^2(t) + \cos(2t))}{y'=( \cos^2(t) - \sin(2t))}$$ and beyond this point i miss because in my answer page this must be
$$ \frac {dy}{dx} = \frac {1+\tan(t)}{\cot(t)-1}$$

答:
Use

$$
\begin{align}
x'&=2e^{2t}\cos t\left(\cos t-\sin t\right)\\
y'&=2e^{2t}\sin t\left(\sin t+\cos t\right)\\
\end{align}$$
Then we have
$$
\begin{align}
\frac{2e^{2t}\sin t\left(\sin t+\cos t\right)}{2e^{2t}\cos t\left(\cos t-\sin t\right)}
&=\frac{\sin t\left(\sin t+\cos t\right)}{\cos t\left(\cos t-\sin t\right)}\\
&=\frac{\cos^2 t}{\cos^2 t}\frac{\sin t\left(\sin t+\cos t\right)}{\cos t\left(\cos t-\sin t\right)}\\
&=\frac{\tan t\left(\tan t+1\right)}{1-\tan t}\\
&=\frac{\tan t+1}{\cot t-1}
\end{align}
$$


问:
Finding standard deviation from normal distribution given mean and probability x is between specific interval
In general how can one compute the standard deviation of the normal distribution when the mean and the following is given:
$$ P(x_1 < x < x_2) = \alpha$$

I already tried a lot of things but can't figure it out.

Is there enough information to solve this probem?

答:
If the mean and standard deviation are denoted by $\mu$ and $\sigma$ respectively, then
$$\alpha = P(x_1 < X < x_2) = P\left(\frac{x_1-\mu}{\sigma} < \frac{X-\mu}{\sigma} < \frac{x_2 - \mu}{\sigma}\right).$$
Note that $\frac{X-\mu}{\sigma}$ is a standard normal random variable, so if you knew $\sigma$ the right-hand side could be computed from a normal table or a computer.

If the mean $\mu$ is between $x_1$ and $x_2$, then the right-hand side decreases continuously from $1$ to $0$ as $\sigma \to \infty$, so there is some value of $\sigma$ such that the probability equals $\alpha$. Otherwise the situation is more complicated.


问:
Prove for some $n\in\mathbb{N}$ there are infinitely many primes $p$ , S.T the numbers $p-1,p+1,p+2$ have $n$ different prime factors .
Prove for some $n\in\mathbb{N}$ there are infinitely many primes $p$  , S.T the numbers $p-1,p+1,p+2$ have $n$ different prime factors .

Attempt:

by the fundamental theorem of arithmetic's

we know that
$$p-1=p_1^{a_1}...p_k^{a_n}$$
also $$p+1=q_1^{a_1}...q_k^{a_n}$$
$$p+2=w_1^{a_1}...w_k^{a_n}$$

Noticing that all the factorizations consists of $n$ different prime factors.

I tried proving a stronger case with adding $p$ and its factorization to achieve 4 consecutive numbers. However that led me no where.

I tried to solve it with the Chinese remainder theorem using the abstract idea that there is only one unique $x_0$ solution. Not sure how to continue with this idea.

答:
As pointed out in the comments:
If the question is asking for exactly $n$ factors, then the claim is false for every $n$, for the following simple reason:

If $p \neq 2,$ then $p \equiv 1 \pmod{2},$ thus $p-1$ and $p+1$ are both divisible by $2$, thus can not have different prime factorizations.

However, the statement is true for at least $n$ different prime factors:

Let $p_1,...,p_n, q_1,...,q_n, r_1,...,r_n$ be different odd prime numbers.

Let $P = p_1p_2 \cdots p_n, Q =q_1q_2\cdots q_n$ and $ R= r_1r_2\cdots r_n.$

(1) By the  Chinese Remainder Theorem,
\begin{align*}
x-1 &\equiv 0 &\pmod{P} \\
x+1 &\equiv 0 &\pmod{Q}\\
x+2 &\equiv 0 &\pmod{R} \\
\end{align*}
has a solution.

(2) Let $x\in \mathbb{N}$ be a solution of the congruence equations. Note that  $x_k = x+ k \cdot PQR $ is also a solution for all $k\in \mathbb{N}$.

(3) Now $(x,PQR) =1$, so by Dirichlet's theorem of primes in arithemetic progression you get the result.


答: Hint: Find a homomorphism from $G$ whose kernel is exactly $H$.

Title: Prove that if $G$ is abelian, then $H = \{a \in G \mid a^2 = e\}$ is subgroup of $G$

Body:
Let $G$ be an abelian group. Prove that $H = \{a \in G \mid a^2 = e\}$ is subgroup of $G$, where
  $e$ is the neutral element of $G$.

I need some help to approach this question.


问:
Solve $\int \frac{\sin(x)+\cos(x)}{e^x}~dx$
How should I calculate this integral$$\int\frac{\sin(x)+\cos(x)}{e^x}dx~~~~?$$
I don't know what is the first step, so I tried symbolab calculator but this cannot solve.
Can someone help me to solve this? Thank you in advance.

答:
Note the integral fornula:
$$\int e^{-x}(f(x)-f'(x)) dx=-e^{-x} f(x).$$
So here $$\int e^{-x} (\cos x+\sin x) dx=-e^{-x} \cos x +C$$


问:
Are $\mathbb{N} $ $\times$ $\mathbb{Z}$ and $\mathbb{Z} $ $\times$ $\mathbb{Z}$ similar?
In other words, my question is: Is there any order preserving bijection between $\mathbb{N} $ $\times$  $\mathbb{Z}$ and $\mathbb{Z} $ $\times$  $\mathbb{Z}$, with antilexicographic order:

$$ (a,b)<(c,d)⟺(b<d) \lor (b=d \land a<c) $$

(Two sets are similar if there exists a bijection between them which keeps the order in sets.)

I know that  $\mathbb{N} $ and  $\mathbb{Z} $ are not similar, but still, I don't know what to do with this task...

答: If there where an order-preserving bijection $f:\mathbb N\times\mathbb Z\to\mathbb Z\times\mathbb Z$ then two consecutive elements would go to two consecutive elements, so $f(a,b)=(c,d)$ implies $f(a+1,b)=(c+1,d)$, this would imply that $\mathbb Z$ and $\mathbb N$ are similar.


问:
Does being asymptotic imply the use of Big-O notation and vice-versa?
Suppose that $f(x)\sim g(x)$, then does $f(x)=O(g(x))$? Conversely, suppose that $f(x)=O(g(x))$, then does $f(x)\sim g(x)$?

答:
It is useful to recognize that

"$f(x)=O(g(x))$" is equivalent to "there exists a constant $C$ such that $\left|\frac{f(x)}{g(x)}\right| \le C$ for sufficiently large $x$."

$f(x) \sim g(x)$ is defined as $\frac{f(x)}{g(x)} \to 1$.

From here, we immediately see that $f(x) \sim g(x)$ implies $f(x)=O(g(x))$ (take $C=2$, for instance).

However, the converse does not hold. For example, $f(x)=2x$ and $g(x)=x$ satisfy $\frac{f(x)}{g(x)} \le 2$ but $\frac{f(x)}{g(x)} \not\to 1$.


问:
Prove that if $G$ is abelian, then $H = \{a \in G \mid a^2 = e\}$ is subgroup of $G$

Let $G$ be an abelian group. Prove that $H = \{a \in G \mid a^2 = e\}$ is subgroup of $G$, where
  $e$ is the neutral element of $G$.

I need some help to approach this question.

答:
Hint: To show it is a subgroup, you need to answer yes to the following questions, and justify your answers.

Questions to answer:

$(1)$ What is $e^2\,$? So is $\,e \in H\,$?

(This will prove the identity of $G$ is in $H$.)

$(2)$ If $\,a, b \in H$, then $a^2 = b^2 = e.\;$ So we need to know whether $(ab) \in H$?:

$Yes, ab \in H$ if you can show that $(ab)^2 = e$.

Tip: use the fact that $G$ is abelian, and so $(ab)^2 = a^2b^2$

(If $ab\in H$, this would prove that $H$ is closed under the group operation of $G$).

$(3)$ If $a \in H$, is $a^{-1} \in H$?

(If so, this would prove that $H$ is closed under taking inverses.)


问:
Variance of $2$-d random points projected onto a line
Assume that we take normally distributed random points in $\mathbb{R}^2$, which follow $\mathcal{N}(\mu, \sigma^2)$, where $\mu = (1,1)^T$ and $\sigma^2 = 0.4$.

Now suppose that we project these points to the line $y = x$ and thus each point $(a,b)$ gets mapped to $\left(\frac{a+b}{2}, \frac{a+b}{2}\right)$.

Seeing the initial $2$-d points as $1$-d points onto $y=x$, the new mean would be:

$$
\tilde{\mu} = 1
$$

My question is: what would be the new variance $\tilde{\sigma}$ for the projections?

答:
The projection of any d-dimensional normal distribution with d*s^2 variance onto a line has s^2 variance!

The probability density function of a d-dimensional normal distribution centered at 0 with d*s^2 variance is just the product of d normal probability distribution functions of s^2 variance.

If we project that onto a single dimension, and integrate with respect to each other dimension over -inf to inf, we see the terms go to 1, and we're left with our singular dimensional normal distribution function multiplied by our delta squared term.

I am rushed, I can add math later, hopefully this can help you now.


问:
Does a mixed NE in complete information induce a mixed BNE in incomplete information
So far, I always checked if there is a mixed Bayesian Nash Equilibrium in static games with incomplete information if there was at least one mixed Nash Equilibrium in static games with complete information and I am wondering if there is a better way to quickly check if there is a mixed BNE.

So, is there a proposition which states that a mixed Nash Equilibrium in static games with complete information induces a mixed Bayesian Nash Equilibrium in static games with incomplete information or another proposition from which one can quickly see if there exists a BNE?

答:
As Bayes-Nash equilibria are Nash equilibria (Transform a Bayesian game into a extremely large normal-form game), Nash's theorem guarantees their existence. If we have a finite incomplete information Bayesian game then a mixed strategy Bayes-Nash equilibrium exists.

We also have that in a Bayesian game with continuous strategy spaces and types. If the strategy and type sets are compact, and payoff functions are continuous and concave in own strategies, then a pure strategy Bayes-Nash equilibrium exists.


问:
Proof Explanation in Hatcher, Theorem 3.43
Here is the link to the book: http://pi.math.cornell.edu/~hatcher/AT/AT.pdf

In Theorem 3.43, checking the commutativity of the diagram is left to the reader. The first square commutes trivially, but I have no idea of the second and the third  squares. I also can't see where the $[B]$ in the diagram came from. Thanks in advance.

Edit. This theorem is a generalization of a theorem in Bredon's Topology and Geometry, Theorem VI.9.2, and the proof is similar too. I understood the proof in Bredon's but I think it is nontrivial to apply the same proof directly.

答:
OK I just tried to work through this so here's my attempt at commutativity. Actually as Bredon points out, the square only commutes up to sign, but that's actually good enough.

We asume that $A$ and $B$ keep the orientation of $\partial M$.
Pick $[\phi] \in H^k(M, A)$, where $\phi \in C^k(X)$ is zero on $C_k(A)$. I'll assume that the fundamental classes $[M] \in H^k(M, \partial M), [A] \in H^k(A, \partial A)$ and $[A] \in H^k(A, \partial A)$ are represented by cycles $\mu \in C^k(M), \beta \in C^k(A)$ and $\alpha \in C^k(A)$ respectively.

We observe that $\alpha + \beta$ represents $[\partial M]$ : First note that $[\partial \left( \alpha + \beta\right)] = [\partial A] - [\partial A]$ (the orientations of $\partial A$ and $\partial B$ differ by a sign) so $[A] + [B]$ indeed represent a class in $H_{n-1}(\partial M)$. Second, this class restricts to a local orientation of $\partial M$ on any interior point of $A$ or $B$, so by the isomorphism $H_{n-1}(\partial M) \cong H_{n-1}(\partial M , \partial M - x)$ for $x \in \partial M$ this class represents $[\partial M]$.

We denote the lower horizontal map in the relevant square $\delta$. Now lets take a look at the result of going down first and then right
$$
\delta ([M] \frown [\phi]) = [ \partial (\mu \frown \phi ) ] = [ (-1)^k (\partial \mu \frown \phi - \mu \frown \delta \phi) ] = (-1)^k  [ \partial M ] \frown [\phi]
$$
(since $\delta \phi = 0$). Now we consider going right first and then down (in two steps). The first two maps only consist of restricting $\phi$, first to $C_k(\partial M)$ and the to $C_k(B)$, so the result of this map is
$$
[B] \frown \left[\phi|_{C_k(B)}\right] = [\beta \frown \phi] = [(\alpha + \beta) \frown \phi] = [\partial M] \frown [\phi]
$$
And we are done.


问:
Prove if a = b, then f(a) = f(b) for any function f (with natural deduction)
I want to be able to prove that for all functions f, that $a = b \to f(a) = f(b)$.  That this is true is obvious, but I'm not sure how to formally prove it using only the rules of inference in first-order logic (using natural deduction).  I'm guessing I need to reference a formal definition for a function, where $f \subset A \times B$ and satisfies:

$$ (\forall x \in A)(\exists y \in B)((x,y) \in f \wedge (\forall z \in B)((x,z) \in f\to y=z))$$

However, I don't have any clue where to begin with this.  Can anyone help?  Please only give hints/answers using natural deduction as the deductive system.

答:
I know it's been a while since you asked your question, but I suppose the answer you're looking for is this one. Basically, you want to prove that $F : X \rightarrow Y$ implies $\forall x' \in X \forall x'' \in X [x'=x'' \rightarrow F(x') = F(x'')]$. As you correctly stated, $F : X \rightarrow Y$ can be translated in predicate logic as $\forall x[x \in X \rightarrow \exists y[y \in Y \land (x,y) \in F \land \forall z[z \in Y \land (x, z) \in F \rightarrow y=z]]]$. As a sidenote, the expression $\forall x' \in X \forall x'' \in X [x'=x'' \rightarrow F(x') = F(x'')]$ is equivalent to $\forall x'  \forall x'' [(x' \in X \land x'' \in X \land x'=x'') \rightarrow F(x') = F(x'')]$.

The first step in our proof is to declare $x'$ and $x''$ as arbitrary variables. As a consequence, the goal of our proof turns into $(x' \in X \land x'' \in X \land x'=x'') \rightarrow F(x') = F(x'')$. The second step is to hypothetically assume $x' \in X$, $x'' \in X$ and $x'=x''$. As a consequence, the goal turns into $F(x') = F(x'')$. Hence:

PREMISE 1: $\forall x[x \in X \rightarrow \exists y[y \in Y \land (x,y) \in F \land \forall z[z \in Y \land (x, z) \in F \rightarrow y=z]]]$

PREMISE 2: $x' \in X$

PREMISE 3: $x'' \in X$

PREMISE 4: $x'=x''$

GOAL: $F(x') = F(x'')$

We apply universal instantiation upon PREMISE 1 to derive $x' \in X \rightarrow \exists y[y \in Y \land (x',y) \in F \land \forall z[z \in Y \land (x', z) \in F \rightarrow y=z]]$ and $x'' \in X \rightarrow \exists y[y \in Y \land (x'',y) \in F \land \forall z[z \in Y \land (x'', z) \in F \rightarrow y=z]]$. Since we have $x' \in X$ (PREMISE 2) and $x'' \in X$ (PREMISE 3), we use modus ponens and conclude $\exists y[y \in Y \land (x',y) \in F \land \forall z[z \in Y \land (x', z) \in F \rightarrow y=z]]$ and $\exists y[y \in Y \land (x'',y) \in F \land \forall z[z \in Y \land (x'', z) \in F \rightarrow y=z]]$.

Now, we should use existential instantiation to get rid of the existential quantifiers in both statements. Therefore, we introduce the variables $y'$ and $y''$ and assume the premises $y' \in Y \land (x',y') \in F \land \forall z[z \in Y \land (x', z) \in F \rightarrow y'=z]$ and $y'' \in Y \land (x'',y'') \in F \land \forall z[z \in Y \land (x'', z) \in F \rightarrow y''=z]$. Notice that the first expression means the same as $y' = F(x')$ and the second expression means $y'' = F(x'').$

At this point, we use universal instantiation at the uniqueness part of both expressions, substituting $z$ in the first expression by $y''$ and $z$ in the second expression by $y'$. Hence, $y' \in Y \land (x',y') \in F \land [y'' \in Y \land (x', y'') \in F \rightarrow y'=y'']$ and $y'' \in Y \land (x'',y'') \in F \land [y' \in Y \land (x'', y') \in F \rightarrow y''=y']$.

Since we have $y' \in Y \land (x',y') \in F$ and since $x'=x''$ (PREMISE 4), then, by modus ponens, we conclude $y' \in Y \land (x'',y') \in F$. Since we have from the uniqueness part of the second statement that $[y' \in Y \land (x'', y') \in F \rightarrow y''=y']$, then we conclude by modus pones $y''=y'$. Since $y'=F(x')$, since $y''=F(x'')$ and since $y'=y''$, we conclude $F(x')=F(x'')$. The same reasoning could have been applied upon statement 2 and the uniqueness part in statement 1 to reach the same result.


答:
Given $x_1, x_2, \alpha, \mu$ (i.e. the mean), solve $$\alpha = \frac{1}{\sqrt{2\pi\sigma^2}}\int_{x_1}^{x_2}\exp\left(-\frac 1 2 \frac{(x - \mu)^2}{\sigma^2}\right)\,\mathrm dx$$
for $\sigma$. However, there exists no closed-form solution. That is, the solution has to be found numerically. The R code below examplifies the procedure:

```<code>x1 = 1
x2 = 10
mu = 5
alpha = 0.5

uniroot(
  function(sigma) alpha - pnorm(x2, mu, sigma) + pnorm(x1, mu, sigma),
  lower = 0, upper = 1e6
)
</code>```

Result: 6.652907. Indeed, $$P(1\leq X \leq 10) = \frac{1}{\sqrt{2\pi\cdot6.652907^2}}\int_{1}^{10}\exp\left(-\frac 1 2 \frac{(x - 5)^2}{6.652907^2}\right)\,\mathrm dx  = 0.4999997 \approx 0.5 = \alpha.$$

Title: Finding standard deviation from normal distribution given mean and probability x is between specific interval

Body:
In general how can one compute the standard deviation of the normal distribution when the mean and the following is given:
$$ P(x_1 < x < x_2) = \alpha$$

I already tried a lot of things but can't figure it out.

Is there enough information to solve this probem?

Body:
Given $x_1, x_2, \alpha, \mu$ (i.e. the mean), solve $$\alpha = \frac{1}{\sqrt{2\pi\sigma^2}}\int_{x_1}^{x_2}\exp\left(-\frac 1 2 \frac{(x - \mu)^2}{\sigma^2}\right)\,\mathrm dx$$
for $\sigma$. However, there exists no closed-form solution. That is, the solution has to be found numerically. The R code below examplifies the procedure:

x1 = 1
x2 = 10
mu = 5
alpha = 0.5

uniroot(
  function(sigma) alpha - pnorm(x2, mu, sigma) + pnorm(x1, mu, sigma),
  lower = 0, upper = 1e6
)

Result: 6.652907. Indeed, $$P(1\leq X \leq 10) = \frac{1}{\sqrt{2\pi\cdot6.652907^2}}\int_{1}^{10}\exp\left(-\frac 1 2 \frac{(x - 5)^2}{6.652907^2}\right)\,\mathrm dx  = 0.4999997 \approx 0.5 = \alpha.$$


问:
Explicit Fuchsian group of surface of genus $g$
I am learning hyperbolic geometry on my own. I want to learn Fuchsian groups.

Let $\Sigma_g$ be a geuns $g$ Riemann surface with $g \geq 2$, $\pi_1(\Sigma_g)$ denotes the fundamental group of $\Sigma_g$. Now $\mathbb{H}/\pi_1(\Sigma_g)$ can be thought as $\Sigma_g$ where an element of $\pi_1(\Sigma_g)$ can be thought of as a biholomorphism of upper half plane $\mathbb{H}$. That is equivalent to finding the explicit description of genus $g$ surface as the upper half plane modulo group of Deck Transformation.

My question: Can I study (or get) Fuschian group from the construction? Is Fuchsian group related to it?

I want to study Fuchian group. Please help me.

Thanking in advanced.

答:
Fuchian groups are (by definition) discrete subgroups of PSL($2$, $\mathbb{R}$), and this is isomorphic to Aut($\mathbb{H}$), the group of biholomorphic maps from $\mathbb{H}$ to $\mathbb{H}$. As you mentioned, there is a connection between subgroups of automorphisms that act properly discontinuously induce a quotient that is a covering map $\mathbb{H} \rightarrow \Sigma_g$.

One of the requirements for a properly discontinuous action is that the subgroup of Aut($\mathbb{H}$) acting must be a discrete subgroup, and hence you get the Fuschian groups. Now, the Riemann surfaces that are covered by the half-plane are called hyperbolic surfaces, because you can induce the metric using the covering map.

This gives you a relation between Hyperbolic Riemann Surfaces and Fuschian Groups.

The book "Riemann Surfaces", by Hershell M. Farkas and Irwan Kra study some relations about these groups and the corresponding Riemann Surface in the chapter IV, specially IV.5. This may be a good starting point.


问:
Lebesgue-Integral as Riemann-Integral
This is a rather basic question on parts of the Lebesgue-Integral formula. Intuitively it should be clear, but I am searching for a mathematical explanation.

If $f: (\Omega, \mathcal{A}) \rightarrow (\mathbb{R}, \mathfrak{B}(\mathbb{R}))$ is a measurable function and $\mu$ a measure on $(\Omega, \mathcal{A})$ and $\mu(\Omega) < \infty$, than we can write the Integral of $f$ wrt $\mu$ as
$$\int f d\mu  = \int f^+ d\mu - \int f^- d\mu
=\int_{0}^{\infty} \mu \{y|f(y) \geq x\} dx + \int_{-\infty}^{0} [\mu \{y|f(y) \geq x\} - \mu(\Omega)] dx.  $$

I don't get how to derive the second part, i.e. where does the expression $\mu \{y|f(y) \geq x\} - \mu(\Omega)$ come from? (I know that this concerns the potentially negative part of the function, but at the moment I am unable to get, how I derive this second expression). Some help would be awesome!

答:
Hint In addition to @AsemAbdelraouf 's hint, apply Fubini's theorem to the integral
\begin{equation}
\int_{\mathbb R}\int_{\Omega} \left(\mathbf{1}_{0\le x\le f(y)}-\mathbf{1}_{f(y)<x<0}\right) d x d\mu(y)
\end{equation}

Title: Lebesgue-Integral as Riemann-Integral


问:
$a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$

Q. If $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$, prove that $a = b = c = 0$ ($a$, $b$, $c$ and $p$ are rational and $p$ is not a perfect cube.)

My approach:

Solving the quadratic, I get:

$p^\frac{1}{3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}$

Case 1: If the $b^2 - 4ac$ is a perfect square, I get the LHS as irrational and the RHS as rational, which is a contradiction.

Case 2: If $b^2 - 4ac$ is not a perfect square, $b =  \pm \sqrt{b^2 - 4ac} - 2cp^\frac{1}{3}$

Here, the LHS is rational and the RHS is irrational, contradiction again. (Edit: The answer of @GNUSupporter has the proper proof.)

So the equation is not quadratic and $c = 0$.

$a + bp^\frac{1}{3} = 0$

$-\dfrac{a}{b} = p^\frac{1}{3}$

This is a contradiction and hence $b = 0$ and $a = 0$

Is there any other way to solve this?

答:
I have somewhat weird way to see this. Consider the system
\begin{align*}
a + b p^{1/3} + c p^{2/3} & = 0\\
cp + a p^{1/3} + b p^{2/3} & = 0\\
bp + cp p^{1/3} + a p^{2/3} & = 0
\end{align*}
or
$$\begin{pmatrix} a & b & c \\
cp & a & b \\
bp & cp & a \end{pmatrix} \begin{pmatrix} 1 \\ p^{1/3} \\ p^{2/3} \end{pmatrix} =0.
$$

So the coefficient matrix has zero determinant, i.e.
$$a^3 + b(b^2-3ac)p + c^3 p^2=0.$$
Now we can proceed with infinite descent.

(The essence is until here, and below is just some calculations.)

Note that we can assume that $p$ is an integer; If $a + b (n/d)^{1/3} + c (n/d)^{2/3}=0$ then
$$ad^2 + bd\cdot  d^{2/3} n^{1/3} + c d^{4/3}n^{2/3}=0,$$
i.e.
$$ad^2 + bd\cdot   (d^2n)^{1/3} + c (d^2n)^{2/3}=0$$ so we are reduced to the integer $p$ case.

Let $q$ be a prime factor or $p$. One can assume that $q^3 \not\mid p$; in this case $q$ factor is absorbed into coefficients $b$ and $c$. Assume $(a, b, c)$ is a nontrivial integer solution. We have two cases;

Case 1: Let $p = qN$ with $q\not\mid N$. Then
$$a^3 + b(b^2-3ac)qN + c^3 q^2N^2=0,$$
i.e. $a = qA$. Then
$$q^2A^3 + b(b^2-3qAc)N + c^3 qN^2=0,$$
i.e. $b = qB$. Then again
$$qA^3 + B(q B^2-3Ac)qN + c^3 N^2=0,$$
i.e. $q|c$, i.e. $c = qC$, and
$$A^3 + B(B^2-3AC)qN + C^3 q^2 N^2 = A^3 + B(B^2-3AC)p + C^3 p^2 =0.$$
Thus, if $(a, b, c)$ is an integer solution then $(a/q, b/q, c/q)$ also is an integer solution; this descent cannot be done infinitely since $a, b, c$ are finite, i.e. a contradiction.

Case 2 : Let $p = q^2 N $ with $q\not\mid N$. Then
$$a^3 + b(b^2-3ac)q^2 N + c^3 q^4 N^2=0.$$
One can assert $a = q A$, then
$$q A^3 + b(b^2-3qAc) N + q^2c^3 N^2=0,\quad \mathbf{(**)}$$
i.e. $b = qB$. Thus
$$ A^3 + B(q B^2-3Ac)q N + q c^3  N^2=0,$$
i.e. $A$ can be divided by $q$ once again. Let $A = qA'$ to have
$$ q^2 A'^3 + B( B^2-3A'c)q N + c^3 N^2=0,$$
i.e. $c = qC$,
$$ q A'^3 + B( B^2-3qA'C) N + q^2 C^3 N^2=0 \quad \mathbf{(**)}$$
Compare two equations marked by (**);
If $(A, b, c)$ satisfies
$$q A^3 + b(b^2-3qAc) N + q^2c^3 N^2=0 $$ then we have another integer solution $(A/q, b/q, c/q)$. So, again by infinite descent, there is no such $(a, b, c)$.

This method also works for $p^{1/4}$ case.

I think I have never seen the matrix of the form
$$\begin{pmatrix} a & b & c \\
cp & a & b \\
bp & cp & a \end{pmatrix} $$
or its variants. Are there any reference?


答:
Hints:

Prove that if $\,G\,$ is an abelian group then $\,G^n:=\{ g^n\;;\;g\in G\}\,$ is a subgroup of G .

Further hint: $\,(xy)^n=x^ny^n\,$ ...what about inverses?


问:
How show that $x_0$ is an accumulation point of the domain of a function
Let the function $\frac{\sin x}{x}$ defined $\forall x\neq 0$. How I can prove that $x_0=0$ is an accumulation point for $A=\{x\in\mathbb{R}: x\neq 0\}$?

答:
First, know your definitions:

We define the open ball centred around $x$; denoted $B(x;r)$ in $\Bbb R$ as the set $\bigg\{y \in\Bbb R\ \bigg|\  |x-y|< r\bigg\}$, where $r>0$. In higher dimensions with different metrics, the definition is similar.

This is represented in $\Bbb R$ by the set $(x-r, x+r)$.

Now, $x_0$ is an accumulation point of a set $A$ if every open ball centred around $x_0$ contains a point in $A$ that is distinct from $x_0$. In other words, you need to show that $$A\cap B(0; r) \neq \emptyset$$ for each $r>0$. Clearly, this is the case, as $A=\Bbb R/ \{0\}$, since $0$ is the only missing point from $A$, we can find a point in any non-empty ball that is non-zero, and hence in $A$.


问:
Expressing solution of matrix equations in matrix form
I want to find $x_{11}$, $x_{12}$, $x_{21}$, $x_{22}$, $a$, and $b$ that solve the following system of equations
\begin{align*}
\begin{bmatrix}x_{11} & x_{12}\\x_{21} & x_{22}\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}& =
\begin{bmatrix}\gamma_{11} & \gamma_{12}\\\gamma_{21} & \gamma_{22}\end{bmatrix}
\begin{bmatrix}x_{11} & x_{12}\\x_{21} & x_{22}\end{bmatrix}\begin{bmatrix}a\\0\end{bmatrix},
\\[2ex]
\begin{bmatrix}x_{11} & x_{12}\\x_{21} & x_{22}\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}& =
\begin{bmatrix}\gamma_{11} & \gamma_{12}\\\gamma_{21} & \gamma_{22}\end{bmatrix}
\begin{bmatrix}x_{11} & x_{12}\\x_{21} & x_{22}\end{bmatrix}\begin{bmatrix}0\\b\end{bmatrix},
\\[2ex]
\begin{bmatrix}x_{11} & x_{12}\\x_{21} & x_{22}\end{bmatrix}\begin{bmatrix}\theta_{2}\\\theta_{1}\end{bmatrix}& =
\begin{bmatrix}\alpha_{1}\\\alpha_{2}\end{bmatrix}.
\end{align*}
All parameters ($\gamma_{11}$, $\gamma_{12}$, $\gamma_{21}$, $\gamma_{22}$, $\theta_1$, $\theta_2$, $\alpha_1$, and $\alpha_2$) are non-zero and $\gamma_{11}\gamma_{22}-\gamma_{12}\gamma_{21}\neq0$.

Treating the system as 6 separete equations, I can express the solution as
\begin{align*}
x_{11} &  =\frac{a\gamma_{12}}{1-a\gamma_{11}}x_{21},\\
x_{12} &  =\frac{b\gamma_{12}}{1-b\gamma_{11}}x_{22},\\
x_{21} &  =\frac{1}{\left(  a-b\right)  \theta_{2}}\left(  \frac{\gamma
_{11}\alpha_{2}-\gamma_{21}\alpha_{1}}{\gamma_{11}\gamma_{22}-\gamma
_{12}\gamma_{21}}-b\alpha_{2}\right),  \\
x_{22} &  =\frac{1}{\left(  b-a\right)  \theta_{1}}\left(  \frac{\gamma
_{11}\alpha_{2}-\gamma_{21}\alpha_{1}}{\gamma_{11}\gamma_{22}-\gamma
_{12}\gamma_{21}}-a\alpha_{2}\right),
\end{align*}
with $a$ and $b$ distinct roots of
$$1-\left(\gamma_{11}+\gamma_{22}\right)  z+\left(\gamma_{11}\gamma_{22}-\gamma_{12}\gamma_{21}\right)  z^{2}=0.$$

I would like to represent the solution for $\begin{bmatrix}x_{11} & x_{12}\\x_{21} & x_{22}\end{bmatrix}$ in matrix form. Is this even possible? Any suggestion would be highly appreciated.

答:
Let $G=[\gamma_{ij}]$ and $X=[x_{ij}].$

As the OP already figured out themselves, $a$ and $b$ are the roots of $1-\mathrm{tr}(G)\,z+\det(G)\,z^2=0.$ If we substitute $z=\lambda^{-1}$ and multiply with $\lambda^2,$ we get $\lambda^2-\mathrm{tr}(G)\,\lambda+\det(G)=0.$ This means that $a^{-1}$ and $b^{-1}$ are the eigenvalues of $G,$ and it might make sense to diagonalize $G.$ So let $$G = P \begin{pmatrix} a^{-1} & 0 \\ 0 & b^{-1}\end{pmatrix}P^{-1} = PDP^{-1} $$ with an invertible matrix $P,$ and let us see what we get.
$$
X\begin{pmatrix} 1 \\ 0\end{pmatrix}
=PDP^{-1}X\begin{pmatrix} a \\ 0\end{pmatrix}
$$
or
$$
P^{-1}X\begin{pmatrix} 1 \\ 0\end{pmatrix}
=DP^{-1}Xa\begin{pmatrix} 1 \\ 0\end{pmatrix}
=aDP^{-1}X\begin{pmatrix} 1 \\ 0\end{pmatrix}
$$
which means
$$
\left(I-aD\right)P^{-1}X\begin{pmatrix} 1 \\ 0\end{pmatrix} = 0
$$
This can easily be fulfilled iff the lower left element of $P^{-1}X$ is $0.$

From  $$
X\begin{pmatrix} 0 \\ 1\end{pmatrix}
=PDP^{-1}X\begin{pmatrix} 0 \\ b\end{pmatrix}
$$
we get
$$
\left(I-bD\right)P^{-1}X\begin{pmatrix} 0 \\ 1\end{pmatrix} = 0
$$
which means that the upper right element of $P^{-1}X$ must be $0.$ If we put all this together, we have $P^{-1}X=C$ with a diagonal matrix $C,$ which in turn means that $X=PC.$

Now we use the last of the three given equations to figure out the diagonal elements of $C.$ In other words, we figure out by which factor the initial choice of the eigenvectors of $G$ must be scaled to get the columns of $X.$

We want $X\theta = \alpha$ or $PC\theta = \alpha$ or $C\theta = P^{-1}\alpha$ with $\theta = (\theta_2,\theta_1)^T$ (sic!) and $\alpha=(\alpha_1,\alpha_2)^T.$

From this, we can conclude
$$
C= \mathrm{diag}\left(P^{-1}\alpha\right)
\,
\mathrm{diag}\left(\theta_2^{-1},\,\theta_1^{-1}\right)
$$
Therefore, we get
$$
X=
P
\,
\mathrm{diag}\left(P^{-1}\alpha\right)
\,
\mathrm{diag}\left(\theta_2^{-1},\,\theta_1^{-1}\right)
$$


问:
Is every group isomorphic to the automorphism group of some magma?
I believe that magma isomorphism is defined as $\phi(x*y)=\phi(x)*'\phi(y)$. The automorphism group is the set of bijective isomorphisms from the elements of the magma to itself, under the operation of function composition. I take a magma to be just a set $X$ together with any operation from $X^2$ to $X$.

The reason I suspect that this might be the case is that:

I know that groups are closely related to permutations, which is exactly what automorphisms are. If a permutation leaves something invariant, then it is not hard to generally show that its inverse does so too. The axioms of closure and inverse are met, and the requirement of identity is obviously met. Function composition is always associative, and so these functions do indeed form a group.

These groups seem to potentially have the quite complex structure, and so few visible constraints that it just seems reasonable that any group could be like the automorphism group of a magma, but I don't believe that I am experienced enough in the ways of abstract algebra yet to prove this.

I'm just looking for a better way for myself to think of groups in general, that's all.

Thanks!

答:
Yes, and in fact every group is the automorphism group of some commutative monoid; see Hagen von Eitzen's answer here. His construction proceeds by showing first that every group is the automorphism group of some graph, then constructing a commutative monoid out of the graph with the same automorphism group; it's quite a nice construction.

The first sentence of the abstract of this paper claims that every group is the automorphism group of a lattice, which is more or less an idempotent commutative monoid (depending on the precise definitions).


问:
Pre-calc algebraic method for predicting symmetry
I am looking for an algorithm that can be used on any equation that contains polynomials containing x and y to determine if reflective or rotational symmetries exist. If it is possible, I would like a method that works for all algebraic equations to make it slightly more general. I can graph any function to quickly look at the graph for a visual confirmation of symmetry but that is not precise and it does not tell me exactly if the symmetry I see is on the y axis or very close epsilon to the y axis where visual methods fail. Another concern would be that I can not use the visual obviously method on an algebraic proof that symmetry does exist. What book and what chapter should I start looking for the area of mathematics that is concerned with such mathematical methods.

I did not find this on stack exchange with a basic search.

答:
I figured it out on youtube. negate x and check for equality with the original polynomial. same technique for negating y. negating both is the rotation around the origin.

example: $y=x^2=(-x)^2 \Rightarrow y=0\mathrm{\;is\;a\;line\;of\;symmetry\;for\;the\;plot\;of\;}y=x^2$


答:
Closure

If $a,b \in H $ then $abab=abba=ab^2a=aea=a^2=e \in H$

Associativity

If $a,b,c \in H$ then $(ab)c=a(bc)$ because $G$ is a group

Identity

$e\in H$ because $ee=e^2=e$

Inverse

Let $a \in H$ Then  $a^2=e=aa^{-1}=a^{-1}a$ Multiply by $a$ then $a^{-1}aa=aa^{-1}a \rightarrow a^{-1}e=ea \rightarrow a^{-1}=a \in H$. The object $a^{-1}\in G$ exists because $G$ is a group


问:
Partition 100 people, 4 from each country into 4 groups with conditions
This is a problem from the $2005$ All-Russian Olympiad. Problem is as follows:

$100$ people from $25$ countries, four from each country, sit in a circle.
Prove that one may partition them onto $4$ groups in such way that no
two countrymen, nor two neighboring people in the circle, are in the
same group.

My approach was to try and find an algorithm that can monotonically reduce the number of bad states. However, I haven't been able to figure out a way to do this for every possibility. Does anyone have any hints on how to approach this problem?

I've tried to start with an arbitrary partition where each partition only has 1 person from each country, and preserve that with each move.

I also claim that there is always a country for which not all of its representatives' neighbors are in a single partition. Proof idea: Suppose not. WLOG, say that country $1$ has all of its representatives' neighbors in the first partition. So if $1_i$ is the representative from country $1$ in partition $i$, we have $1_1$ and its neighbor, WLOG from country $2$, is also in group $1$. So we have $1_1, 2_1$. Then $2_1$'s other neighbor must also be in partition $1$ or we are done. Continuing in this way, we find that all the representatives are in partition $1$, contradiction.

Using this, I try to decrease the number of "bad" pairs (i.e. neighbors or same country) in each partition monotonically. However, I haven't been able to figure out how to do so.

In particular I'm stuck on cases like the following: country $1$ has its representatives in partitions $1-4$ s/t the neighbors of $1_2,1_3,1_4$ are all in the first partition. $1_1$'s neighbors are in partition $1$ and another partition, say $2$. Trying to swap the placements of country $1$ representatives seems to only increase number of "bad" pairs, and trying to swap the neighbors may also increase the number of bad pairs.

Is there an observation that I'm missing here, or is this approach not the right way to go about it?

答:
This questions seems to be very similar to graph theory problems and as @MisraLavrov has showen, its solvable with methods from graph theory. One example of those methods is the concept of vertex-coloring, where adjacent (by an edge connected) vertices have to be colored different. In the following, I will explain how to translate your problem to graph theory and how to solve it there.

Graph

Let $G=(V,E)$ be a Graph with 100 vertices (= people).

Group problem

In graph theory you can reduce sorting people in groups to finding vertex-colorings. That means, in a proper coloring each vertex represents a person and the colour its group. Here only 4 different colours are allowed.

Neighbour problem

Algorithm

Now we build a graph-circle out of $G$ by connecting each vertex to exactly two others. Therefore you start at a vertex, connect it to an unconnected and repeat this. If you get to the last unconnected vertex, just connect it to the first.

Explanation

The edges solve the neighbour problem, because in vertex-colorings adjacent vertices have to be colored different.

Countryman problem

Algorithm

To complete the graph, you make always an edge between vertices (people) from the same country.
Explanation

The edges between people from the same country prevents them for getting colored identically.

Solving the group problem

When the graph is finished, you start coloring. Therefore are many different ways and algorithms known and easily found in the internet (for example [1]). Although vertex-coloring is NP-complete, this kind of graph can be colored faster.

Internet research

[1] https://www.researchgate.net/publication/292100744_Graph_coloring_algorithms

I hope this helps and gives an insight to translating general mathematical problems into graph theory problems.


问:
Weyl constructions for finite groups
Let $G$ be a finite group.

Is there a complex finite dimensional irreducible representation $V$ such that all irreducible ones are submodules of $V^{\otimes n}$ for some $n \in \mathbb{N}$?

If not, what's the smallest group $G$ for this to fail?

答:
It's a nice exercise in character theory to show that for $G$ a finite group and $V$ a representation, every irreducible representation occurs in $V^{\otimes n}$ for some $n$ iff $V$ is faithful (faithfulness is obviously necessary and sufficiency is the interesting direction).

So your question is equivalent to asking when $G$ has a faithful irreducible representation. A necessary condition, by Schur's lemma, is that the center $Z(G)$ must be cyclic (since it embeds into $\mathbb{C}^{\times}$), so $C_2 \times C_2$ (the smallest non-cyclic group) is the smallest counterexample.

Edit: Regarding the nice exercise above, here is the argument I had in mind.

Proposition: Let $G$ be a finite group and $V$ a faithful finite-dimensional representation of it. Then every irreducible representation occurs in $V^{\otimes n}$ for some $n$.

Proof. Let $W$ be some irreducible and consider the characters $\chi_W, \chi_V$. We want to show that

$$\langle \chi_W, \chi_V^n \rangle = \frac{1}{|G|} \sum_{g \in G} \overline{\chi_W}(g) \chi_V(g)^n$$

is nonzero for some $n$. The idea of this proof is to consider what happens to this sum when $n$ is large: it's dominated by the terms where $|\chi_V(g)|$ is largest and all the other terms exponentially decay in comparison. Using the fact that $\chi_V(g)$ is the sum of the eigenvalues of $\rho_V(g)$, which are roots of unity, $|\chi_V(g)|$ has maximum value $\dim V$ and takes this maximum value iff all the eigenvalues of $\rho_V(g)$ are the same iff $\rho_V(g)$ is a scalar multiple of the identity (and the scalar must be some root of unity).

In the simplest case the only $g$ with this property is the identity; this happens, for example, in David Speyer's answer explaining a version of this proof in response to this MO question because he added a copy of the trivial representation. In that case we get that the sum above is asymptotically $\frac{(\dim V)^n}{|G|} + O(c^n)$ where $|c| < \dim V$ and hence must be eventually positive, and with a little more effort we could give an upper bound on the smallest $n$ at which this occurs.

Instead of doing that we can just restrict our attention to the values of $n$ such that if $\rho_V(g)$ is a scalar multiple of the identity then $\rho_V(g)^n$ is the identity (at worst this means restricting our attention to $|G| \mid n$). If there are $k$ different elements of $g$ acting by a scalar then we get, for these restricted values of $n$, that the sum above is asymptotically $k \frac{(\dim V)^n}{|G|} + O(c^n)$ where $|c| < \dim V$, and again the conclusion follows. $\Box$

Less analytic and more algebraic arguments are possible that would generalize, for example, to fields of characteristic $0$. This argument has the benefit of generalizing, with some modification, to the case that $G$ is a compact group (although we need to consider irreducibles in $V^{\otimes n} \otimes (V^{\ast})^{\otimes m}$ in this case), as I learned from David Speyer on MO here.

Title: Weyl constructions for finite groups


问:
How to calculate points on a line relative to their rectangular containers?

Given 2 rectangles with known widths and heights of w1, h1 and w2, h2, and a connecting straight line from the centers of each rectangle, with known starting point x1,y1, and known ending point x2,y2, how can I calculate the points on the line where it passes the edge of each rectangle (as defined by x1a,y1a, and x2a,y2a)?

EDIT:
Adding some simple Javascript code to render the 2 rectangles, the line, and plot the point for (x1a, y1a).

```<code>let x1 = 75;
let y1 = 40;

let x2 = 375;
let y2 = 340;

let w1 = 150;
let h1 = 80;

let w2 = 150;
let h2 = 80;

// Create objects
let objects = document.getElementById("objects");

let rect1 = document.createElement("div");
rect1.classList.add("rect");
rect1.style.top = "0px";
rect1.style.left = "0px";
rect1.style.width = `${w1}px`;
rect1.style.height = `${h1}px`;

objects.appendChild(rect1);

let rect2 = document.createElement("div");
rect2.classList.add("rect");
rect2.style.width = `${w2}px`;
rect2.style.height = `${h2}px`;
rect2.style.top = "300px";
rect2.style.left = "300px";

objects.appendChild(rect2);

// Insert the path
let line = document.createElementNS(
  "http://www.w3.org/2000/svg",
  "path"
);

line.setAttribute("stroke", "#000");
line.setAttribute("fill", "none");
line.setAttribute("stroke-width", "2");
line.setAttribute("d", `M${x1},${y1} ${x2},${y2}`);

document.getElementById("svg").appendChild(line);

// Update the point

let x1a = x1 - ((h1 / 2) * (x2 - x1)) / (y2 - y1);

let point1 = document.createElement("div");
point1.classList.add("point");

// y coordinate of the point is just the height of the rectangle given it crosses
// the bottom edge of the rectangle
point1.style.top = `${h1}px`;

point1.style.left = `${x1a}px`;
objects.appendChild(point1);
</code>```

The point does not show up in the right position, as it has the wrong x1a co-ordinate.

Screenshot of rendering (blue dot is the x1a, y1a point):

答:
The equation of the line connecting the two centers is

$$(x_2-x_1) y-(y_2-y_1)x=y_1x_2-x_1y_2$$

Then, for the case in the diagram, substitute $y=y_{1a}= y_1-\frac{h_1}2$ and $y=y_{2a}=y_2+\frac{h_2}2$ to obtain the respective

$$x_{1a}=x_1 -\frac{h_1}2 \frac{x_2-x_1}{y_2-y_1}, \>\>\>\>\>
 x_{2a}=x_2 +\frac{h_2}2 \frac{x_2-x_1}{y_2-y_1} $$

(Note that there are four cases to consider depending on the relative sizes of the widths and heights.)


问:
How to prove that: $\phi^{2}(0) \leq \|\phi\|^2_{L^{2}}+\|\phi'\|^{2}_{L^{2}}$
Let $\phi$ be a function and $\phi \in C^{\infty}(\mathbb{R}_{+},\mathbb{R})$ with compact support and $\mbox{supp }{\phi} \subset [0, \infty)$.

I want to prove that: $$\phi^{2}(0) \leq \|\phi\|^2_{L^{2}}+\|\phi'\|^{2}_{L^{2}}.$$

Someone give an indication that I should begin with:

$$\phi(x)-\phi(0)=\int_{0}^{x}{\phi'(t)}\mbox{dt}$$ and then to prove that:

$$\phi^{2}(0) \leq \int_{0}^{\infty}{|2\phi(x)\phi'(x)|dx} \mbox{ . }$$

Thanks :)

答:
First notice $2\phi \phi' = (\phi^2)'$:
$$\phi^2(\infty) - \phi^2(0) = \int^\infty_0 2\phi(x)\phi'(x)\,dx.$$
Compactly supportedness drives $\phi\to 0$ when $x\to \infty$, so:
$$
\phi^2(0) = \left|\int^\infty_0 2\phi(x)\phi'(x)\,dx\right| \leq \int_{0}^{\infty}|2\phi(x)\phi'(x)|\,dx
$$
Lastly simply using $2ab\leq a^2 + b^2$:
$$
\int_{0}^{\infty}|2\phi(x)\phi'(x)|\,dx \leq \int_{0}^{\infty}(|\phi(x)|^2 + |\phi'(x)|^2)\,dx = \|\phi\|^2_{L^{2}}+\|\phi'\|^{2}_{L^{2}}.
$$</p>


问:
If G is finite cyclic group then why is it true that for every divisor of the cardinality of $G$, there only exists at most one subgroup?
Question: If $G$ is finite cyclic group then why is it true that for every divisor of the cardinality of $G$, there only exists at most one subgroup with such cardinality?

I am struggling to see a way to explain this properly. I realise that this group $G\cong \mathbb{Z}/n\mathbb{Z}$ and every subgroup of a cyclic group is again cyclic.

My attempt thus far:

Suppose for a contradiction that for some $d<n$ such that $\exists p,q\in G$ such that $ord(p)=ord(q)=d$ with the property that $\langle p \rangle \neq\langle q\rangle$ but now how do I proceed further? (I realise this eventually boils down to finding two elements with order $d$ such that one is not in the other's orbit.)

Moreover I saw that the converse is also true and thus could someone please give me some feedback for my attempt, too?

Claim I also tried to prove: Suppose for every $d<n=|G|$ there only exists one subgroup with cardinality $d$ then show $G$ is cyclic.

My attempt to this claim: Suppose $G$ is not cyclic then by structure theorem that $G\cong \mathbb{Z}/p_1\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}/p_n\mathbb{Z}$ where $p_1|\cdots|p_n$ and $n\geq 2.$ Now clearly, since $n\geq 2$ there are at least two subgroups with order $p_1$ and thus we reached a contradiction (for instance, since $\mathbb{Z}/p_1\mathbb{Z}$ is embedded in $\mathbb{Z}/p_2\mathbb{Z}$)

Many thanks in advance for providing any feedback to either question! :)

答:
Hints: if $G=\langle g \rangle$, $|G|=o(g)=n$, and $d \mid n$, then $o(g^{\frac{n}{d}})=d$. Conversely, if $|\langle h \rangle|=d$, and $h=g^i$, then by Lagrange $d \mid n$. And $n|id$, whence $\frac{n}{d}|i$, say $i=m \cdot \frac{n}{d}$. So $h=(g^{\frac{n}{d}})^m$. Over to you to fill in the details. Note that this shows that for every divisor $d$ of $|G|$ there is precisely one subgroup of order $d$.

For your second question/claim: let $\varphi$ denote the Euler function, then if there is exactly one subgroup of order $d$ for every $d \mid n$, implies $G$ has at most $\varphi(d)$ elements of order $d$ for each $d\mid n$. But $\sum_{d\mid n}\varphi(d)=n$, so $G$ must contain exactly $\varphi(d)$ elements of order $d$ for each $d\mid n$, in particular, $G$ contains an element of order $n$.

Note Here is a different proof of the fact that if the finite group $G$ has the property
$(*)$ for each $d \mid |G|$ there is exactly one subgroup of order $d$
then $G$ is cyclic. Because of $(*)$ every Sylow subgroup is normal and hence $G$ is nilpotent, that is the direct product of its Sylow subgroups. If $|G|$ has more than one prime factor, by induction (and basically the Chinese Reaminder Theorem) we are done. So we can assume that $G$ is a $p$-group for some prime $p$. Pick a $g \in G$ of maximal order and assume $M=\langle g \rangle \lt G$. Let $x \in G-M$. Then, since $G$ is a $p$-group, we can find an element of $o(x)$ in $\langle g \rangle$. But owing to $(*)$, this forces $\langle x \rangle \subseteq M$, contradicting $x \notin M$. So $G=M$ is cyclic.


问:
can any property of the natural numbers be proven by induction?
1-if the set of natural numbers is constructed as in ZF system, is any proved property of N can also be proved using induction?

in other words, does the induction principle have 'limitation' in the sense that
there are some properties of N which can not be proven with it?

2- Does having 'limitation' on the induction principle depends on the system in which the set N is constructed?

答:
Any proof $P$ that proves that something is true for all numbers can trivially be converted into an inductive proof $P'$:

Base: Repeat proof P to conclude $\forall n \ P(n)$. Now plug in $0$ for $n$ to get $P(0)$

Step: Let $k$ be some arbitrary number and assume $P(k)$. Again, repeat proof P to conclude $\forall n \ P(n)$. Plug in $k+1$ for $n$ to get $P(k+1)$

This, of course, seems like a very cheeky answer ... and in some ways it is. But, I think there is a real question here as to what really counts as an inductive proof, i.e. to what extent the induction 'does the work'.


问:
Jacobian linearization of trigonometric functions
If I have to linearize a nonlinear trigonometric system around the origin $(0,0)$:
$$\dot{x_1} = x_2$$
$$\dot{x_2} = \cos(x_1)$$

I can apply the small angle approximation to find the matrix A:
$$ A = \pmatrix{0&1\\1&0} $$

However, if I apply Jacobian linearization and take the partial derivatives of $\dot{x}$, I get:
$$ A =  \pmatrix{0&1\\-\sin(x_1)&0}$$

Evaluated at the origin, I get:
$$ A = \pmatrix{0&1\\0&0}  $$

Seeing as how the two different linearization techniques yield different results, is one more valid than the other?

答:
Your small angle approximation is wrong and your Jacobian linearization is incomplete.

If $x_1 \approx 0$ then $\cos(x_1) \approx 1$ and so the system

$$
\begin{align}
\dot{x_1} &= x_2 \\
\dot{x_2} &= \cos(x_1) \\
\end{align}
$$

behaves close to $(0,0)$ approximately like

$$
\begin{align}
\dot{x_1} &= x_2 \\
\dot{x_2} &= 1 \\
\end{align}
$$

If you put this in matrix form:

$$
\begin{pmatrix}
\dot{x}_1 \\
\dot{x}_2
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} \begin{pmatrix}
x_1 \\
x_2
\end{pmatrix} + \begin{pmatrix}
0 \\
1
\end{pmatrix}
$$

This is the "small angle approximation" and as you can see this is essentially the same as the Jacobian linearization. Remember that the Jacobian linearization of a nonlinear system $\dot{x} = f(x)$ at $x_0$ is

$$
\dot{x} = f(x_0) + A(x - x_0)
$$

where $A$ is the Jacobian matrix of $f$ at $x_0$. If you insert your system and your point you get:

$$
\begin{pmatrix}
\dot{x}_1 \\
\dot{x}_2
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
-\sin(0) & 0
\end{pmatrix} \begin{pmatrix}
x_1 - 0 \\
x_2 - 0
\end{pmatrix} + \begin{pmatrix}
0 \\
\cos(0)
\end{pmatrix}
$$

which is just:

$$
\begin{pmatrix}
\dot{x}_1 \\
\dot{x}_2
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} \begin{pmatrix}
x_1 \\
x_2
\end{pmatrix} + \begin{pmatrix}
0 \\
1
\end{pmatrix}
$$

so the same as the small angle approximation.

Your mistake was that you neglected the fact that you linearized at a non-equilibrium point. Usually we linearize at an equilibrium and then $f(x_0) = 0$, so the constant term can be neglected.

But you linearize at $(0, 0)$ which is not an equilibrium point of your system because $\dot{x_2} \neq 0$ at $x_1 = 0$, so the system is not at rest at the origin.


答:
No, see e.g. https://en.wikipedia.org/wiki/Goodstein%27s_theorem

It depends. For any “natural” definition, one can show an isomorphism in ZF(C). However, there are also nonstandard models of natural numbers (satisfying the Peano axioms) which are non-isomorphic in ZF(C).

Title: can any property of the natural numbers be proven by induction?

Body:
1-if the set of natural numbers is constructed as in ZF system, is any proved property of N can also be proved using induction?

in other words, does the induction principle have 'limitation' in the sense that
there are some properties of N which can not be proven with it?

2- Does having 'limitation' on the induction principle depends on the system in which the set N is constructed?


问:
Rules of inferences problem
I have an exercise that I have to prove the validity of the following arguments using rules of inference. But there is an exercise I have no idea how to prove it at all.

Problem shows as below.

$P\to Q$

$\neg P\to R$

$S\to \neg R$

$S\to \neg Q$

Conclusion $S\to T$

The exercises I have practiced before have at least one premise. However, in this exercise, it seems that no premise can be used directly. So I have no idea how to prove it by rules of inference. Can anybody give me some hints? Thanks.

PS. The following is detail of the exercise before replacing sentences with notations.

"If Superman were able to prevent evil, he would do so."

"If Superman were unable to prevent evil, he would be impotent."

"If Superman exists, he is not impotent."

"If Superman exists, he does not prevent evil."

Therefore

"If Superman exists, it is impossible to learn logic."

答:
First let's simplify
$$(p\to q)\land(\neg p\to r)\land(s\to\neg r)\land(s\to\neg q)$$
$$\equiv(p\to q)\land(\neg p\to r)\land(r\to\neg s)\land(q\to\neg s)$$
$$\equiv(p\to \neg s)\land(\neg p\to\neg s)$$
$$\equiv(\neg p\lor \neg s)\land(p\lor\neg s)$$
$$\equiv(\neg s\lor\neg s)$$
$$\equiv\neg s$$
So now we have
$$\neg s\to(s\to l)$$
$$\equiv s\lor\neg s\lor l$$
$$\equiv\mbox{tautology}$$

Title: Rules of inferences problem


问:
Is convergence itself necessary for Riemann´s rearrangement theorem?
I have been studying series recently and I´ve gone through the proof of Riemann´s rearrangement theorem among other things. I was given the following related but different statement to prove as an exercise:

A divergent series $\sum{a_k}$ has a convergent rearrangement if and only if the sum of the positive and negative terms diverge to $+\infty$ and $-\infty$ respectively and $\lim_{k\to \infty}{a_k}=0$

While trying to prove that the condition was sufficient I wondered: "Can I not just appeal to Riemann´s rearrangement theorem?" However it states that any conditionally convergent series can be rearranaged to converge to any real number or diverge. The statement I´m trying to prove assumes $\sum{a_k}$ diverges. From what I can tell however, in the proof of Riemann´s rearrangement theorem, the convergence itself of the series doesnt seem to be used to prove the theorem, but rather the fact that the sum of the positive and negative terms diverge to $+\infty/-\infty$ and that $\lim_{k\to \infty}{a_k}=0$.

So my question is: Is the convergence itself necessary to prove Riemann´s theorem?

答: No, it is not. Actually, a generalization of Riemann's rearrangement theorem is this: If $\sum_{k=1}^\infty a_k$ is a series of real numbers such that the sum of the positive $a_k$'s is $\infty$, the sum of the negative $a_k$'s is $-\infty$ and that $\lim_{k\to\infty}a_k=0$, then, given $m,M\in\Bbb R\cup\{\pm\infty\}$ with $m\leqslant M$, there is a rearrangement $\sum_{k=1}^\infty a_{b(k)}$ of the series such that$$\limsup_{n\in\Bbb N}\sum_{k=1}^na_k=M\quad\text{and}\quad\liminf_{n\in\Bbb N}\sum_{k=1}^na_k=m.$$In particular, when $m<M$, the new series is divergent.


问:
For a polynomial $p(z)$, prove there exist $R>0$, such that if $|z|=R$, then $|p(z)|\geq |a_n|R^n/2$
For a polynomial $p(z)=a_0+\ldots+a_nz^n$, prove there exist $R>0$, such that if $|z|=R$, then $|p(z)|\geq |a_n|R^n/2$.

I get $|p(z)|\geq |a_n|R^n$, which makes that factor of $1/2$ useless. So I must have done something wrong.

$$|\frac{\partial^n{f}}{\partial z^n}(P)|\leq \frac{\sup_{z\in \overline{D}(P,r)} |f(z)|k!}{r^k}$$.
This is the Cauchy estimate.

$$|\frac{\partial^n{p}}{\partial z^n}(z)| = |a_n n!|$$
$$|\frac{\partial^n{p}}{\partial z^n}(0)|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|n!/R^n$$
$$|a_n n!|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|n!/R^n$$
$$R^n |a_n|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|$$

There exist a $z_0$, such that $|z_0|>0$ and $$\sup_{\zeta\in \overline{D}(0,|z_0|)} |p(\zeta)| = |p(z_0)|$$.

This is saying there exist a disk centered at the origin, such that $|p|$ is maximum on some point on the boundary.

Let $R=|z_0|$, we have
$R^n |a_n|\leq |p(z_0)|$

Edit:
I know what I got wrong. $z_0$ is just one point on the circle $|z| = R$. One can chose another point $z_1$ on the circle, such that $p(z_1)<p(z_0)$.

Body:
Thanks for Qiaochu Yuan's hint. The solution:

Let $p(z) = q(z) + a_nz^n$.

Lemma: Let $c\neq 0$. $|cz^n| > |q(z)|$ for all $z\in \{z : |z|\geq r_c\}$, where $r_c$ is sufficiently large.

Proof:
If $|z|=r$, then $|cz^n| = |c|r^n$.
$$|q(z)| = |\sum_{i=0}^{n-1} a_iz^i| \leq \sum_{i=0}^{n-1}|a_iz^i| = \sum_{i=0}^{n-1}|a_i|r^i \leq \sum_{i=0}^{n-1} \max(|a_i|) r^i =\max_{i=0}^{n-1}(|a_i|) \sum_{i=0}^{n-1} r^i$$
when $r > 1$,
$$\max_{i=0}^{n-1}(|a_i|) \sum_{i=0}^{n-1} r^i \leq \max_{i=0}^{n-1}(|a_i|) (n-1) r^{n-1}$$

If $r > \max(\frac{\max_{i=0}^{n-1}(|a_i|)(n-1)}{|c|},1)$
$$\max_{i=0}^{n-1}(|a_i|) (n-1) r^{n-1} \leq |c|r^n$$

$|p(z)| = |a_n z^n+q(z)|\geq |a_n z^n| - |q(z)| \geq |a_n z^n| - |c z^n| = (|a_n|-|c|)|z^n|$

Let $c = |\frac{a_n}{2}|$. For all $z$ such that $|z|=R>r_{a_n/2}$, this shows $|p(z)| \geq |a_n|R^n/2$.


问:
Number of pendant vertices
Let $G=(V,E)$ be a simple connected graph with only one cycle

If $G$ has four vertices of degree $2$, five vertices of degree $3$, three vertices of degree $4$, one vertex of degree $5$, how many pendant vertices does $G$ have if $\Delta = \text{max}\{\text{deg}(v)|v\in V\} = 5$?

My approach was to take the unique cycle (not necessarily on $6$ vertices like below) and open it:

to then get a tree.

We are essentially dividing a vertex in two, thus there are $|V|=4+5+3+1+\color{red}{1}+p=14+p$ vertices, where $p$ is how many pendant vertices there are in the original graph.

As per the edges $\sum\limits_{v \in V} \text{deg}(v) = 4\cdot 2 + 5\cdot 3 + 3\cdot 4 + 1\cdot 5 \color{red}{-1 +1} +p= 40+p=2|E|$

Then since we are working with a tree $2|V|=2|E|+2$, and $28+2p=40+p+2$, $p=14$.

I could be wrong though, I do not feel confident about my approach. How does one go about working this problem?

答:
I think your approach is perfectly fine (I would say it is clever actually). With the very same argument, one can show that if a simple connected graph $G$ has exactly one cycle, then $|V(G)| = |E(G)|$ (because when we open a cycle and get a tree, we increment the number of vertices by $1$ while keeping the number of edges same). Having said that, we can write

$$\sum\limits_{v \in V(G)}d(v) = 4\cdot 2 + 5\cdot 3 + 3\cdot 4 + 1\cdot 5 +p= 40+p=2|E(G)|=2|V(G)|$$

Then, since $|V(G)| = 4+5+3+1+p = 13+p$, we get $40+p = 2(13+p) \implies p = 14$ as you suggested. But as I said, the result $|V(G)| = |E(G)|$ when $G$ has only one cycle uses the very same argument that you used to find your result directly, so I am not adding much to your solution.

Title: Number of pendant vertices


答:
"Approximation" in our case means Taylor's theorem, i.e.
$$
f(x) = f(0) + Df(0)x
$$
where $f(x) := (x_2, \cos(x_1))$. This is where $Df(0)$, i.e. your second matrix appears. So this is a linearization in the normal sense. This type of linearization is actually useful when we want to determine stability of an equilibrium point.

The first approach I would call incoherent, especially since you leave $x_2$ untouched. This is not how linearization in analysis is understood. Linear maps have the property that they vanish in $0$. So what we do, is to always approximate
$$
x \mapsto f(x) - f(0)
$$
by a linear map since the former vanishes in $0$, too. This is the idea behind Taylor's theorem. We just accept, that this is actually an affine approximation of $f(x)$ since it simply has no major ramifications in practice.

But what you did, is to find a linear map that approximates $f(x)$ only. But since $f(0) \neq 0$ and $A0 = 0$, this might yield huge errors. To my knowledge (which is of course limited), the first one has no application whatsoever, but I would love to be corrected if I were wrong.

Title: Jacobian linearization of trigonometric functions

Body:
If I have to linearize a nonlinear trigonometric system around the origin $(0,0)$:
$$\dot{x_1} = x_2$$
$$\dot{x_2} = \cos(x_1)$$

I can apply the small angle approximation to find the matrix A:
$$ A = \pmatrix{0&1\\1&0} $$

However, if I apply Jacobian linearization and take the partial derivatives of $\dot{x}$, I get:
$$ A =  \pmatrix{0&1\\-\sin(x_1)&0}$$

Evaluated at the origin, I get:
$$ A = \pmatrix{0&1\\0&0}  $$

Seeing as how the two different linearization techniques yield different results, is one more valid than the other?


问:
Find singularities and their type of $\frac{1}{z(e^z-1)}$
I'm a little bit confused about finding singularities, poles etc. So let's take an example of $\frac{1}{z(e^z-1)}$ at $z=0$.
To do this problem, I've expanded $e^z-1$:
$$e^z-1=z+\frac{z^2}{2!}+\frac{z^3}{3!}+...$$
And so
$$\frac{1}{z(e^z-1)}=\frac{1}{z^2(1+\frac{z}{2!}+\frac{z^2}{3!}+...)}$$
I know that if
$$\lim_{z\rightarrow z_0} (z-z_0)^M f(z)=A_{-M} \ne 0$$
$$\lim_{z\rightarrow z_0} (z-z_0)^{M+k} f(z)=0$$
then we have a pole of order $M$ at $z_0$.

So I take $m=1$:
$$\lim_{z\rightarrow 0} \ z^1 \frac{1}{z^2(1+\frac{z}{2!}+\frac{z^2}{3!}+...)}=\infty $$

Take $m=2$:
$$\lim_{z\rightarrow 0} \ z^2 \frac{1}{z^2(1+\frac{z}{2!}+\frac{z^2}{3!}+...)}=1=A_{-2}=A_{-M} $$

$m=3$:
$$\lim_{z\rightarrow 0} \ z^3 \frac{1}{z^2(1+\frac{z}{2!}+\frac{z^2}{3!}+...)}=0 $$

So it seems that it is an isolated singularity and the function is meromorphic.

I don't understand what this $\infty$ means for $m=1$. Shouldn't this be equal to $A_{-1}$? Or it's only true for a pole, so $m=M$?

答:
So it seems that it is an isolated singularity and the function is meromorphic.

Yes, the function has an isolated singularity at $z=0$, which is a pole of order 2, because  $\lim_{z\to 0}=z^2f(z)=1$, as $z$ approaches $0$, where $f(z)=\dfrac{1}{z(e^z-1)}$, i.e.

\begin{eqnarray*}
\lim_{z\to 0}z^{2}f(z) &=&\lim_{z\to 0}\frac{z}{e^{z}-1}=\lim_{z\to 0}\frac{1}{e^{z}}=1=A_{-2} \end{eqnarray*}

and for $k=1,2,\ldots$ we have

\begin{eqnarray*}
\lim_{z\to 0}z^{2+k}f(z) &=&\lim_{z\to 0}z^{k}\times\lim_{z\to 0}z^{2}f(z)=0\times 1=0=A_{-2-k}.
\end{eqnarray*}

I don't understand what this $\infty$ means for $m=1$.

It just means that $zf(z)$ behaves like $1/z$ as $z$ tends to $0$.

Shouldn't this be equal to $A_{-1}$?

No, $A_{-1}=-1/2$ rather than $\infty$, because it is the coefficient of $z^{-1}$ in the expansion as a Laurent series of
$f(z)$, which is

\begin{eqnarray*}
f(z) &=&z^{-2}-\frac{1}{2}z^{-1}+\frac{1}{12}-\frac{
1}{720}z^{2}+\ldots,\tag{1}\\ \\
&:=&A_{-2}z^{-2}+A_{-1}z^{-1}+A_{0}+A_{2}z^{2}+\ldots \qquad z\neq 0 \tag{2}.
\end{eqnarray*}

Moreover, by definition, the coefficient of $z^{n}$ in $(1)$ is  $A_{n}$. So, as computed by you, $A_{-2}=1$. The same result is obtained if we use $(2)$ to find the limits $\lim_{z\to 0}z^2f(z)$ and $\lim_{z\to 0}z^3f(z)$. We get $\lim_{z\to 0}z^2f(z)=1$ and $\lim_{z\to 0}z^3f(z)=0$. Please note that $A_{-1}$ is the coefficient of $z^{-1}$ in $(1)$ and not in $zf(z)$:

\begin{eqnarray*}
zf(z)&=&z^{-1}-\frac{1}{2}+\frac{1}{12}z-\frac{
1}{720}z^{3}+\ldots \tag{3} .
\end{eqnarray*}


问:
Show factorization $ x^{2n}-1=(x^2-1) \prod_{k=1}^{n-1}(x^2-2x \cos \frac{\pi k}{n} + 1) $
I'm interested in how to show that
$$
x^{2n}-1=(x^2-1) \prod_{k=1}^{n-1}(x^2-2x \cos \frac{\pi k}{n} + 1)
$$
I've seen this equality too often, but have no idea how to derive it. I've tried the following:

$$
x^{2n}-1=(x^n-1)(x^n+1)
$$

We all know that $x^n-1=(x-\xi^0)(x-\xi^1) \dots (x-\xi^{n-1})$,  where $\xi=\exp(i \cdot \frac{2 \pi}{n})$.

Here $x-\xi^0$ gives us the desired $x-1$(for $x^2-1$ in RHS).

But the problem is, we can not do the same with $x^n+1$, because as I understand, there is no such general factorization of $x^n+1$. Firstly, it depends whether $n$ is even or not. Secondly, if for the sake for simplicity $n$ is odd, then we can decompose the polynomial such a way:
$$x^n+1=(x+1)(1-x+x^2-x^3 \dots -x^{n-2}+x^{n-1})$$

I don't see here anything to continue with. Any suggestions?

答:
Hint

You can use the same idea in the factorization of $x^n-1$.

Think of
$$x^n=-1=e^{i(2k+1)\pi}$$


问:
Does a polynomial with real coeffients have a root greater than $1$ iff the sum of its coefficients are greater than $0$?
Take a polynomial $p$ with real coefficients.

$$p(x) = \sum_{k=0}^n a_kx^k$$

It's "well-known" that $1$ is a root if and only if the sum of the coefficients is $0$. $$p(1) = 0 \iff\sum_{k=0}^n a_k = 0$$

Instead, take the case where the sum of the coefficients is positive. Does this this imply that there is real root greater than $1$?

$$ \sum_{k=0}^n a_k > 0 \Rightarrow \exists x_0 > 1 \text{ such that } p(x_0) = 0 $$

I believe this is true, and I expect it's also a fairly standard/well-known result. But I haven't been able to find a reference.

The result where the coefficients sum to zero is apparently a straight-forward application of the polynomial remainder theorem. I'm not sure this is the case for the inequality version.

答: No, this is false. Take the polynomial $x^2+1$.


问:
Is there an isomorphism of $F[x]/\langle x^2+a\rangle$ to $F[x]/\langle x^2+x+b\rangle$ just as fields?
Question

If $F$ is a field of characteristic $2$, can there be irreducible polynomials $x^2+a$ and $x^2+x+b$ over $F$ such that $F[x]/\langle x^2+a\rangle$ is isomorphic to $F[x]/\langle x^2+x+b\rangle$ just as fields?

My Attempt

I know they cannot be isomorphic over $F$ because, if they were so, let $u$ be the representative of $x$ in $F[x]/\langle x^2+a\rangle$, so that $u^2=a$, and let $v$ be the representative of $x$ in $F[x]/\langle x^2+x+b\rangle$, so that $v^2=v+b$, then we can suppose, without loss of generality, that $F[u]=F[v]$ is the same field with $F$ as common subfield, and there would exist $\alpha,\beta\in F$ such  that $v=\alpha+\beta u$, so:
$$v+b=v^2=(\alpha+\beta u)^2=\alpha^2+2\alpha\beta u+\beta^2a=\alpha^2+\beta^2a,$$
so that $v\in F$, a contradiction.

答:
Describing what I think is actually the same example as in Atticus Stonestrom's fine answer, but in different words. I think of this answer as an application of Lüroth's theorem.

Consider the field of rational functions $L=\Bbb{F}_2(u)$. It has the subfields
$$
K_1=\Bbb{F}(u^2)
$$
and
$$
K_2=\Bbb{F}(u^2+u).
$$
Lüroth says that both $K_1$ and $K_2$ are isomorphic to the function field $F=\Bbb{F}_2(t)$. Indeed, this follows from the fact that both $u^2$ and $u^2+u$ are transcendental over $\Bbb{F}_2$ given that $u$ is transcental.

It also follows that $L$ is a quadratic extension $K_1(u)=K_2(u)$ of both $K_1$ and $K_2$, and that the minimal polynomials of $u$ over $K_1$ and $K_2$ have the required forms.

So we get the required example from $L$, when we identify both $K_1$ and $K_2$ with $F$.
First by mapping $t\mapsto u^2$ then by mapping $t\mapsto u^2+u$. The minimal polynomial of $u$ over $K_1$ is then $x^2+t$, but the minimal polynomial of $u$ over $K_2$ is $x^2+x+t$. All because $t$ means different elements of $L$ in the two cases.


问:
Area of a circle is $A = \pi r^2$. Is it possible that both $A$ and $r$ are perfect integers.
Can you produce an example where both the area of a circle and it's radius are integers?</p>

答: $$r\neq 0\;,\;\pi r^2\;,\;r\in\Bbb N\implies \pi\in\Bbb Q\;,\;\text{which is false}$$</p>


问:
Geometric series and Fourier transform
In my book solution example, I have this signal

discrete square wave

I don't understand how the right hand side is attained from the expression on the left
$$\sum_{m=p-M}^{p-1}e^{-imk\omega_0}=e^{iMk\omega_0}\sum_{m=0}^{M-1}e^{-imk\omega_0}$$

Where the last step is accomplished by changing variables and simplifying using the fact that
$$\omega_0 = \frac{2\pi}{p}$$

Is there some book can I read to understand things like that?

答:
Using index shifting property that states:
$$\sum_{s=p}^{N}f(x)=\sum_{s=p+M}^{N+M}f(x-M)$$

Beginning from the R.H.S and using shift of $(M-p)$:

$$R.H.S=\sum_{m=p-M}^{p-1}e^{-imk\omega_0}=\sum_{m=p-M+\bf{M-p}}^{p-1+\bf{M-p}}e^{-ik\omega_0(m-(\bf{M-p}))}=\sum_{m=0}^{M-1}e^{-imk\omega_0}e^{iMk\omega_0}e^{-impk\omega_0}$$

$\because e^{iMk\omega_0}$ is a constant and $e^{-impk\omega_0}=e^{-i2\pi mk}=1$
$$\therefore L.H.S=\sum_{m=p-M}^{p-1}e^{-imk\omega_0}=e^{iMk\omega_0}\sum_{m=0}^{M-1}e^{-imk\omega_0}=R.H.S $$

Title: Geometric series and Fourier transform


问:
Let $a,b\in\mathbb{R}$ and $0 < a < b$. Discuss the convergence of the sequence $x_n:=(a^n+b^n)^{1/n}$ i need steps how to do that
Let $a,b\in\mathbb{R}$ and $0 < a < b$. Discuss the convergence of the sequence
$x_n:=(a^n+b^n)^{1/n}$

my approach to this question is:

while $n \rightarrow\infty$, $1/n \rightarrow 0$,
then $x_n\rightarrow1$ for all $a,\,b \in\mathbb{R}$.
now, by definition:
$$\begin{align}
|x_n-a|&<\varepsilon\\
\end{align}
$$

Body:
1st $(a^n+b^n)^{1/n}=b[[1+(\frac{a}{b})^n]^{(\frac{b}{a})^n}]^{(\frac{a}{b})^n\frac{1}{n}}\to b e^0=b$

2nd $\qquad b\le (a^n+b^n)^{1/n}\le b\cdot 2^{1/n}$ so by the Squeezing Theorem ...


问:
Numbers of the form $\sum_{n=0}^{N} 2^{a_n} 3^n$
Just out of curiosity.
Does the numbers of the form
\begin{align}
 \sum_{n=0}^{N} 2^{a_n} 3^n \text{,}
\end{align}
with $a_n \ge 0$ and $N \ge 0$, cover all integers not divisible by 3?

These numbers can never be divisible by three.
It is also easy to discover integers with multiple representations (e.g., $4 = 2^2 3^0 = 2^0 3^0 + 2^0 3^1$).
I am however stuck to prove that they cover all integers not divisible by 3.

Second question: The numbers of the form
\begin{align}
 \sum_{n=0}^{N} 5^{a_n} 3^n \text{,}
\end{align}
with $a_n \ge 0$ and $N \ge 0$, are numbers congruent to $\{1, 4, 5, 8\} \pmod{12}$.
These are the numbers not divisible by 3 and not $\{ 2, 3 \} \pmod{4}$.
Many other such forms may be considered.
What is the name of such numbers (numbers of such a form)?

答: Erdos asked a similar question some time ago (about “3-smooth representations” of the natural numbers). See here for a discussion on both of your questions.</p>