问:
Does Edwards need to explicitly state $a\in\mathcal{W}$ in his Inverse Mapping Theorem?
In view of the answer to my question: A function is continuously differentiable in an open neighborhood of $x_o$ and differentiable at $x_o$, is it continuously differentiable at $x_o$? it seems to me that in the theorem quoted below, Edwards should explicitly state $a\in\mathcal{W}$.  So I ask: is the situation any different in multivariable calculus?  That is, without explicitly stating that $a\in\mathcal{W},$ is Edwards's theorem correct? Is it not possible to have a nonsingular $f^{\prime}\left(a\right),$ and still have $f$ not $\mathscr{C}^1$ at $a$.

C.H. Edwards's Advanced Calculus of Several Variables

Theorem III 3.3 Suppose that the mapping $f:\mathbb{R}^{n}\to\mathbb{R}^{n}$
is $\mathscr{C}^{1}$ in a neighborhood $\mathcal{W}$ of the point
$a$, with the matrix $f^{\prime}\left(a\right)$ being nonsingular.
Then $f$ is locally invertible--- there exist neighborhoods of $a$
$\mathcal{U}\subset\mathcal{W}$ and $\mathcal{V}$ of $b=f\left(a\right)$,
and a one-to-one mapping $g:\mathcal{V}\to\mathcal{W}$ such that

$$
g\left(f\left(x\right)\right)=x\text{ for }x\in\mathcal{U},
$$

and

$$
f\left(g\left(y\right)\right)=y\text{ for }y\in\mathcal{V}.
$$

In particular, the local inverse of $g$ is the limit of the sequence
$\left\{ g_{k}\right\} _{0}^{\infty}$ of successive approximations
defined inductively by

$$
g_{0}\left(y\right)=a,g_{k+1}\left(y\right)=g_{k}\left(y\right)-f^{\prime}\left(a\right)^{-1}\left[f\left(g_{k}\left(y\right)\right)-y\right]
$$

for $y\in\mathcal{V}.$

答:
In the definitive section I-7, Edwards states:

Let $\mathcal{D}$ be a subset of $\mathbb{R}^{n},$ and $f$ a mapping of $\mathcal{D}$ into $\mathbb{R}^{n}$ (that is, a rule that associates with each point $x\in\mathcal{D}$ a point $f\left(x\right)\in\mathbb{R}^{n}$). We write $f:\mathcal{D}\to\mathbb{R}^{n},$ and call $\mathcal{D}$ the domain (of definition) of $f.$

In order to define $\lim_{\mathbf{x}\to\mathbf{a}}f\left(\mathbf{x}\right),$ the limit of $f$ at $\mathbf{a},$ it will be necessary that $f$ be defined at points arbitrary close to $\mathbf{a},$ that is, that $\mathcal{D}$ contains points arbitrarily close to $\mathbf{a}.$ However we do not want to insist that $\mathbf{a}\in\mathcal{D},$ that is, that $f$ be defined at $\mathbf{a}.$ For example, when we define the derivative $f^{\prime}\left(a\right)$ of a real-valued single-variable function as the limit of its difference quotient at $a.$

$f^{\prime}\left(a\right)=\lim_{x\to a}\frac{f\left(x\right)-f\left(a\right)}{x-a},$

this difference quotient is not defined at $a.$

To my knowledge, Edwards gives no formal definition of neighborhood.  There are a number of similar examples where we assume something to be true for points arbitrarily close to a point $\mathbf{a}$, but not at the point $\mathbf{a}.$ So I left open the possibility that open neighborhood $\mathcal{U}$ of $\mathbf{a}$ allows $\mathbf{a}\notin\mathcal{U}.$

In the Mathematical Appendix to Gravitation and Inertia, by Ignazio Ciufolini and John Archibald Wheeler is found:

A neighborhood of a point $P$ is a set of $\mathcal{S}$ containing an open set containing $P.$

At first, I thought this addresses my question.  But, while composing this answer, it occurred to me that the adjective open does not appear in the term defined as neighborhood.

So my answer is that Edwards should define what he means by open neighborhood $\mathcal{U}$ of $\mathbf{a},$ and if, by that definition, open neighborhood, does not require $\mathbf{a}\in\mathcal{U},$ the Theorem, as stated, is incorrect.  Otherwise, it is acceptable.


问:
Subgroups of $G$ if there is a homomorphism from $G$ to $C_2$
Is there always a subgroup of index 2 in $G$ if there is a homomorphism $f: G \rightarrow C_2$? My inkling is that this is true because in this case $G$ must have a normal subgroup of index 2, and the result follows from Lagrange's theorem. Could someone please help me and provide a solution?

答:
Just to elaborate on the other answer and comments:

Given any two groups $G$ and $H$, there is always a homomorphism from $G$ to $H$, namely the trivial map which sends every element of $G$ to the identity in $H$.

What you want is the following:

Theorem.

$G$ has a subgroup of index $2$ if and only if there is a non-trivial homomorphism $G \to C_2$.

Proof.

Note first that because $C_2$ has only $2$ elements, a homomorphism $G \to C_2$ is surjective if and only if it is non-trivial. Now:

($\Rightarrow$). If $G$ has a subgroup $N$ of index $2$, this subgroup is automatically normal and therefore $G / N$ is a group which has order $2$ and hence is isomorphic to $C_2$. Therefore the projection map $G \to G/N$ gives a surjective homomorphism from $G$ to $C_2$ which as mentioned above, must be non-trivial.

($\Leftarrow$). If there exists a non-trivial homomorphism $\varphi: G \to C_2$, as mentioned above, $\varphi$ must be surjective. Hence by Lagrange, the kernel of $\varphi$ has index $2$.

ContentLicense: CC BY-SA 4.0


问:
How to prove syntactic consequence?
I have a problem like this:

$$ F = \{A\sim B, B \sim C, ¬C∨B\} $$

And I want to check if this is true: $\{A∨C\}\vdash F$.

答:
I will assume for the purposes of this answer that $A \sim B$ means $\text{$A$ if and only if $B$}$.

I'm also going to assume that you want to show that the following is true when interpreted as a sequent.

$$ A \lor C \vdash A \leftrightarrow B, B \leftrightarrow C, \lnot C \lor B $$

Let's rewrite this slightly

$$ A \lor C, C \vdash A \leftrightarrow B, B \leftrightarrow C, B $$

This statement is not a tautology because if $A$ and $C$ are true, but $B$ is false, then all the premises are true, but the conclusion is false (in fact, all the disjuncts are false).

Title: How to prove syntactic consequence?


问:
Find the measure of $\angle G$ in a triangle
In the following figure, $\Delta$ADB, $\Delta$PCB and $\Delta$EFG are right triangles.
$PB=AE$,$AC=CB$.

Question:  What is the value of $\angle G$?

I figured out $\angle A = \angle CPB$ ,  $EF=AC$. What should I do next?

答:
All you need is $AP = PB$ (as $AC = CB$) and so $AP = AE$.

If $\angle GAE = \theta, \angle PAE = 90^0 - 2\theta \implies  \angle APE = 45^0 + \theta$

That leads to $\angle CPG = \angle APE - \angle APC = 45^0$

So $\angle G = 45^0$


问:
Prove that $\cup_{i \in I} A_i$ is an open set.
Let $A \subset \mathbb{R}$. We call A an open set if $\forall a \in A$, $\exists r>0$ such that $(a-r,a+r) \subset A$. Let $A_i$ be a family of open sets indexed by I. Prove that $\cup_{i \in I} A_i$ is an open set.

I think I have lack of understanding of open set, and I don't know how to interpret the definition of union of set with open set to prove this. I do not have solid proof, but here's what I have:

Let $a \in \cup_{i \in I} A_i$. There exists $r > 0$ such that $(a-r, a+r) \subseteq \cup_{i \in I} A_i$. Also, there exists $k \in I$ such that $a \in A_k$.

I don't know how to proceed next. Any help would be really appreciated.

答:
You have to start from what you know, which is that each of the $A_i$ are open.

So, let $a\in\bigcup_{i\in I}A_i$.  Then there exists $i'\in I$ such that $a\in A_{i'}$.  But we know that $A_{i'}$ is open... what does that tell you?


问:
Geometry - length of triangle question
[![enter image description here][1]][1]

Could anyone please help me with this question

I can't seem to solve for the ?.

The answer provided for ? is 1.

Thanks!

答:
Let us guess $\Delta ABC$  is an equilateral triangle with side length of 6.  F and D are the mid points of AC and BC respectively, CD=DB=AF=FC=3. Find a point E between D and B with  DE=1.

If we can prove  EF,AD and CH  intercept at the same point G, then   DE=1 is the solution.

Set some coordinates:

A(0,0),B(6,0),C(3,3$\sqrt3$), D(4.5,1.5$\sqrt3$), E(5,$\sqrt3$),F(1.5,1.5$\sqrt3$)

AE: y=$\frac{\sqrt3}{5}x$ (I)

AD:$\frac{\sqrt3}{3}x$  (II)

FB:  $-\sqrt3/3x$+2$\sqrt3$ (III)

EF:    $-\sqrt3/7x$+12$\sqrt3/7$ (IV)

Solve (I) and (III) to get the intercept point of AE and FB at H($\frac{15}{4},\frac{3\sqrt3}{4}$), then

HC:   $-3\sqrt3x+12\sqrt3$  (V)

Solve (II) and (V) to get the intercept point of AD and HC  at G$_1$(3.6,$\frac{6\sqrt3}{5}$)

Solve (II) and (IV) to get the intercept point of  AD  and EF intercept at G$_2$(3.6,$\frac{6\sqrt3}{5}$)

We  proved AD,EF and HC intercept at the same point  $G_1=G_2=G$.

Now  one can conclude the answer is indeed  DE=1.

Title: Geometry - length of triangle question


问:
Showing $\mathrm{Var}(aX+bY) = a^2\mathrm{Var}(X)+b^2\mathrm{Var}(Y)+2ab\mathrm{Cov}(X,Y)$
I am trying to prove this equation $$\mathrm{Var}(aX+bY) = a^2\mathrm{Var}(X)+b^2\mathrm{Var}(Y)+2ab\mathrm{Cov}(X,Y),$$
where $\mathrm{Cov}(X,Y):=  E(XY)-EX\cdot EX$ and $E[\cdot]$ denotes the mean of a random variable. Also $\mathrm{Var}(X)$ is defined like this: $$\mathrm{Var}(X)=E(X^2)-[E(X)]^2$$ which is the variance of the random variable $X$.

I started this way:

$$\mathrm{Var}(aX+bY) = E[((aX+bY)-E(aX+bY))^2] = E[(aX-bY-aEX-bEY)^2] = $$

Please help me showing that this equals $a^2\mathrm{Var}(X)+b^2\mathrm{Var}(Y)+2ab\mathrm{Cov}(X,Y)$.

答:
The first equality in your proof is wrong - you're not using your definition of $\mathrm{Var}(X)$ correctly. Instead it should go something along the lines of:
$$
\begin{align}
\mathrm{Var}(aX+bY)&={\rm {\rm E}}[(aX+bY)^2]-\left({\rm {\rm E}}[aX+bY]\right)^2\\
&={\rm E}[a^2X^2+b^2Y^2+2abXY]-\left(a{\rm E}[X]+b{\rm E}[Y]\right)^2\\
&=a^2{\rm E}[X^2]+b^2{\rm E}[Y^2]+2ab{\rm E}[XY]-a^2{\rm E}[X]^2-b^2{\rm E}[Y]^2-2ab{\rm E}[X]{\rm E}[Y].
\end{align}
$$
I'm sure you can take it from here.</p>

ContentLicense: CC BY-SA 3.0


问:
Questions about the Perceptron convergence proof
I have some amateur questions about the perceptron convergence proof:

I don't understand where we got $k\gamma$ from, specifically how did we go from $\frac{\theta^{k-1}\theta^*}{\|\theta^*\|}+\gamma$ to $k\gamma$?

How did we go from $\|\theta^{k-1}+y^{(i)}x^{(i)}\|^2$ to $\|\theta^{k-1}\|^2+2y^{(i)}\theta^{(k-1)} x^{(i)}+\|x^{(i)}\|^2$?

Since $\| \theta^{k-1}+y^{(i)}x^{(i)} \|$ is just an expression of this vector's length, I would have thought that the $\| \|$ and the square would have cancelled each other out, thus leaving us with only $(\theta^{k-1})^2+(y^{(i)}x^{(i)})^2$?

(As far as I understand, the length of a vector $(x,y)$ is just $\sqrt{x^2+y^2}$)

答:
You can view it from the perspective of mathematical induction.

We have $$\frac{\theta^{(1)}\cdot \theta^*}{\|\theta^*\|}\ge \frac{\theta^{(0)}\cdot \theta^*}{\|\theta^*\|}  + \gamma = \frac{0\cdot \theta^*}{\|\theta^*\|}  + \gamma=\gamma$$
and  hence by induction hypothesis
$$\frac{\theta^{(k)}\cdot \theta^*}{\|\theta^*\|}\ge \frac{\theta^{(k-1)}\cdot \theta^*}{\|\theta^*\|}  + \gamma \ge (k-1)\gamma   + \gamma=k\gamma$$

For real number $(a+b)^2=a^2+b^2+2ab$. For real vector, $\|a+b\|^2=\langle a+b, a+b\rangle=\|a\|^2 + \|b\|^2 + 2\langle a, b \rangle.$


问:
complex number in polar form with 7th root
write $z=(1-\sqrt{3})^7$ in polar form $z=r(\cos\theta + \iota\sin\theta)$

my method was convert it to exponential form as $r=2^7$ and $\theta=-\frac {7\pi}{3}$ but it says i'm wrong. Can someone point out what did I do wrong in this question?

答:
Since the query has been edited to indicate that the answer has to be in the form $r \left[\cos \theta + i\sin \theta\right]$ which is a complex number, it is reasonable to presume that the intended problem is

$\displaystyle z = \left[1 - i\sqrt{3}\right]^7
= \left\{2 \times \left[\cos (-\pi/3) + i\sin(-\pi/3)\right]\right\}^7$

$\displaystyle =~ \left\{128 \times \left[\cos (-\pi/3) + i\sin(-\pi/3)\right]\right\}.$


问:
What is the maximum number of vertices of degree one that a binary tree with 10 vertices can have?
Full question as the title was a maximum of 150 characters.
A binary tree is a connected graph with no cycles, where each vertex has a degree less than or equal to 3. What is the maximum number of vertices of degree one that a binary tree with 10 vertices can have?

I am really struggling to understand this question and have not seen questions that are similar. I was hoping someone could push me into the right direction. As I understand the problem, we can only use 10 vertices. So would the maximum amount of vertices not be 10? I tried drawing it out and don't see how it can be any larger.

答:
Let $v_1$ be the number of vertices of degree $1$, $v_2$ the number of vertices with degree $2$, and $v_3$ the number of vertices with degree $3$, so that

$$v_1+v_2+v_3=10\,.\tag{1}$$

I’m going to assume that you already know that if a tree has $v$ vertices, it has $v-1$ edges, and I’m going to assume that you know the handshaking lemma. If you put those two facts together, you find that

$$v_1+2v_2+3v_3=2\cdot(10-1)=18\,.\tag{2}$$

Subtract $(1)$ from $(2)$ to find that

$$v_2+2v_3=8\,.\tag{3}$$

You want to make $v_1$ as large as possible, so $(1)$ tells you that you want to make $v_2+v_3$ as small as possible. Experiment with possible values of $v_2$ in $(3)$ to find which one minimizes $v_2+v_3$, and use $(1)$ to find the maximum possible value of $v_1$. (When you’ve done all that, it would be a good idea actually to find a tree with $10$ vertices and the maximum possible number of leaves.)


问:
Show that the sequence $a_n = a_{n-1}(a_{n-1} + \frac{1}{n})$ is unbounded when $a_1 = 1$
Show that the sequence $a_n = a_{n-1}(a_{n-1} + \frac{1}{n})$ is unbounded when $a_1 = 1$

I know intuitively that it is unbounded but how can I show that it is unbounded. In particular I want to show that it is monotone increasing and then show there cannot be an upper bound. However I can't seem to figure out what I should do to show that it is monotone increasing.

答:
If the series has a limit then it must be a stationary point which is

$$a_n=a_{n-1}(a_{n-1}+\frac1{n})|_{n \to \infty}$$

$$a_{\infty}=a_{\infty}^2$$

so it is either $a_{\infty}=0$ or $a_{\infty}=1$.

But for $a_1=1$ it is already $a_2=\frac{3}{2}$ and the series is monotonically increasing, thus it is unbounded. It simply cannot reach either $0$ or $1$.

In order to find the exact value where it becomes unbounded, you need to find the limit of $a_1$ for each $1=a_{n-1}(a_{n-1}+\frac1{n})$ (and similarly $0=a_{n-1}(a_{n-1}+\frac1{n})$). For example find the solution for $a_{1,1}=a_{1},1=a_{1}(a_{1}+\frac1{2})$ then $a_{1,2}=a_{1},1=a_{2}(a_{2}+\frac1{3})$, $a_{1,3}=a_{1},1=a_{3}(a_{3}+\frac1{4})$.

You will get a decreasing series (obviously, right?) $a_{1,m}$ and its limit $a_{1,\infty}$ will be the breaking point bellow which or at which the series will tend to $1$ or $0$, and above which to infinity.

But the solution of $1=a_{1}(a_{1}+\frac1{2})$, $a_1=\frac{\sqrt{17}}{4}-\frac{1}{4} \approx 0.78$ is already less than $1$, proving yet again that your series is unbounded.


问:
Two-penny problem in Casella & Berger
There is the following exercise in Casella:

Two pennies, one with $P(\text{head})=u$ and one with $P(\text{head})=w$, are to be tossed together independently. Define
\begin{align*}
p_0 & =P(0\ \text{heads occur}),\\
p_1 & =P(1\ \text{head occurs}),\\
p_2 & =P(2\ \text{heads occur}).
\end{align*}
Can $u$ and $w$ be chosen such that $p_0=p_1=p_2$? Prove your answer.

I know the solution, but I am not sure how Casella's logic works about sample spaces.

In this exercise, the sample space has to be: $$S = \{HH, HT, TH, TT\}$$ since there are 2 throws of different coins.

However, what does the author exactly mean by $P(H) = u$? If we just use the intuition, it is evident. However, our sample space is NOT one coin. It is the set of 4 outcomes. Thus, IMO when talking about  $P(H) = u$ the author is really talking about $P(\{HH, HT\})$.

So, when the independence comes into play, it is NOT just $P(H \cap T) = P(H)P(T)$, which is intuitive, but incorrect to say. In reality, it is $$P(\{HH, HT\} \cap \{HH, TH\}) = P(\{HH\}) = P(\{HH, HT\})P(\{HH, TH\})$$ Thus, as I understand we are supposed to assign the probabilities of $P(H)$ to $P(\{HH, HT\})$. However, Casella never mentions any kind of similar thought about this anywhere. Moreover, in the text, he casually jumps in between the sample spaces without assigning any probability. For example, in his "4 aces in 4 cards"-example, he jumps in between the sample space of combinations of 4 cards against sample spaces of combinations of three or less cards, but preserves the probability function.

For example, first talking about the probability of 4 aces from 52 cards, unordered, where the sample space is $52\choose 4$. But then, he talks about probability of 3 aces from 52 cards, which is clearly a restricted subspace since only 3 card draws, of size $52 \choose 3$. Then, the author contradicts himself when he says the event '4 aces from 4 cards' a subset of "3 aces from 3 cards", because the sample spaces are different:

Conditional Probability Question, from Casella and Berger

How do I properly think about this?

And if we move from a sample space of, say, 4 outcomes ($S = \{HH, HT, TH, TT\}$), into the reduced space of $\{H, T\}$, how can I be sure that the probability function is actually preserved? E.g. for the first coin toss: $P(H) = P(\{HH, HT\})$?

答:
Consider throwing one coin. Then the sample space for this random experiment, as you know, is ${H,T}$, since there are only two possibilities - heads or tails. Now, when it comes to the problems you mentioned, we can be more explicit, and consider adding phantom coins/cards/die in the following manner:

Suppose the main random experiment you are considering is that of two coins being thrown. Then, consider two separate random experiments -

One real coin and one phantom coin being thrown (both are identical in all respects)

Two real coins being thrown

The sample space for  the second experiment is straightforward - $\{HH, HT,TH, TT \}$. But for the first random experiment, consisting of one real coin and one phantom coin, we write the sample space as - $\{\{HH,HT\},\{TH,TT\}\}$. In this way, the cardinality of the sample space is $2$, which makes sense because there is only one real coin, the other phantom coin is to help us make sense of everything. So when we say $P(H)$, (wrt the first random experiment) we mean $P(\{HH,HT\})$. In a way, it doesn't matter if the second coin is thrown or not, because we are not concerned with its outcome, hence we always cover all possibilities wrt the second, phantom coin.

In a similar vein, when we consider the experiment of drawing three cards from a deck, we can draw one more phantom card. Therefore, when we consider the event of drawing $3$ aces, our $4^{th}$ phantom card can be anything (apart from what we have drawn already, that is). Thus, the event $A$ - drawing 3 aces from 3 cards, has a whole lot of elements, when we consider the $4^{th}$ phantom card. The number of possibilities is exactly $52-3 = 49$. Thus the event that our phantom card is also an ace, is a subset of the event $A$.

Title: Two-penny problem in Casella & Berger


问:
Is $M_n$($\mathbf{C})$ spanned by {$AB-BA$ where A,B in $M_n$($\mathbf{C})$}?
I know that span of  {$AB-BA$ where A,B in $M_n$($\mathbf{R})$} is $n^2$ -1. But what happens if entries are from $\mathbf{C}$? Is $M_n$($\mathbf{C})$ spanned by {$AB-BA$ where A,B in $M_n$($\mathbf{C})$}, with base field $\mathbf{C}$ ?

答:
This has nothing to do with the field.

For convenience, we use the standard notation $[A , B] := AB - BA$.

Let $F$ be any field, and denote by $M_n(F)$ the $F$-algebra of $n \times n$ matrices with entries in $F$. Then the trace $\text{tr}: M_n(F) \to F$ is surjective linear map, and hence its kernel has dimension $n^2 - 1$. We claim that the kernel is equal to $W := \text{span}\{ [A , B] \mid A , B \in M_n(F) \}$. It follows from linearity of the trace and the fact that $\text{tr}(AB) = \text{tr}(BA)$ that $W \subseteq \ker(\text{tr})$. To show the reverse containment, note that

$$[E_{ij} , E_{k \ell}] = \delta_{jk} E_{i \ell} - \delta_{\ell i} E_{k j}$$

where $E_{ij}$ is the matrix with a $1$ in the $i$th row and $j$th column and $0$'s everywhere else.

Therefore for any $i \neq j$, we have that $[E_{i 1} , E_{1 j}] = \delta_{1 1} E_{i j} - \delta_{i j} E_{1 1} = E_{i j}$.

Furthermore, for any $1 \leq i < n$, we have that $[E_{i \  i + 1} , E_{i + 1 \ i}] = \delta_{i + 1 \ i + 1} E_{ii} - \delta_{i i} E_{i + 1 \ i + 1}  = E_{i i} - E_{i + 1 \ i + 1}$.

Hence $\{ E_{i j} \mid i \neq j \} \cup \{ E_{i \ i + 1} \mid 1 \leq i < n \} \subseteq W$. However the former is a linearly independent set containing $n^2 - 1$ elements, so $\dim W \geq n^2  - 1$. Since we already showed that $\ker(\text{tr})$ has dimension $n^2 - 1$ and contains $W$, it follows that $W = \ker(\text{tr})$ as desired.


问:
Proving $[0,1]^{\omega}$ in box topology is not compact
Here is my attempt at proving this, any ideas if it is completely correct?

Let $U^{(0)}=[0,\frac{3}{5}), U^{(1)}=(\frac{2}{5},1]$. Let $x=(x_n)_{n \in \mathbb{N}}$ be an arbitrary sequence of $0's$ and $1's$ in the collection $C$ of all infinite sequences of $0's$ and $1's$. With each sequence $x$ in this collection, associate a sequence of subsets of $[0,1]$ defined by $$U_{nx}= \begin{cases}  U^{(1)}   &\text{ if } x_n=1 \\
          U^{(0)}  & \ \text{if} \ x_n=0\end{cases}$$ Then the collection $S$ of all products of the form $\prod_{n \in \mathbb{N}} U_{nx}$,which are open in $[0,1]^{\omega}$ where $x$ ranges over the set of all binary sequences is an open cover for $[0,1]^{\omega}$ with the box topology.Since for any $k \in \mathbb{N}$ there are at least two elements in $S$ such that the $k$'th factor in the products differ. Now if any element is removed from $S$, one of the infinite sequences $x$ will be removed from $C$, and this sequence will have exactly one coordinate differing from all the other sequences in either a $0$ or $1$ in say the $m$'th slot. So the $m$'th factor in the product of the union of all the elements of this new collection $S'$ will either  be $[0,\frac{3}{5})$ or $(\frac{2}{5},1]$ and hence the collection upon removing the element from $S$ will not cover $[0,1]^{\omega}$. Thus since $S$ is infinite any finite subset of $S$ will certainly not cover the space.

答:
It’s correct but could be stated much more clearly. The key point, which you saw but didn’t really state explicitly, is that for each $x\in C$, the set $\prod_nU_{nx}$ is the only member of $S$ that contains $x$. Your argument for this could be much clearer. Clearly $x\in\prod_nU_{nx}$. If $y\in C\setminus\{x\}$, however, there is an index $k$ such that $y_k\ne x_k$. But then $x_k\notin U_{ky}$, so $x\notin\prod_nU_{ny}$.

$S$ is in fact an irreducible open cover of $[0,1]^\omega$: each member of $S$ contains a point that is not in any other member of $S$, so no proper subset of $S$ covers $[0,1]^\omega$.

Note that essentially the same argument can be used to show that the box product $\{0,1\}^\omega$ is not compact (where $\{0,1\}$ has the discrete topology): here you’re working directly with $C$ as a box product. Since this space is a closed subset of $[0,1]^\omega$, that is another way to see that $[0,1]^\omega$ cannot be compact: if it were, all of its closed subsets would be compact as well.


问:
Inequality in a bounded real sequence
Prove or disprove that for any bounded real sequence $\{x_n\}_{n\in\mathbb{N}}$ there exist two distinct natural numbers $u,v$ such that:
$$|x_u-x_v|\cdot|u-v|\leq 1.$$</p>

答:
Here is a counter-example using Hurwitz's theorem

Let $\phi = \frac{1+\sqrt 5}2$ be the golden ration.
Since the inequality $|\phi - p/q| < 1/3q^2$ has finitely many solution for integers $p,q$,
let $N$ be an integer greater than all the $|p|$ occurring in those solutions.

Let $x_n = 3n\phi \pmod {3N}$

Suppose you find distinct integers $u$ and $v$ such that $|x_u - x_v||u-v| < 1$.
Then there are integers $p_u$ and $p_v$ such that $x_u = 3u\phi - 3Np_u$ and $x_v = 3v\phi - 3Np_v$. So $|3(u-v)\phi-3N(p_u-p_v)||u-v| < 1$, and $|\phi - N(p_u-p_v)/(u-v)| < 1/3(u-v)^2$. Hence $(N(p_u-p_v), u-v)$ is one of the finitely many solutions to the inequality above, hence $N > |N(p_u-p_v)|$. This implies $p_u = p_v$, and we get $|\phi| < 1/3(u-v)^2 \le 1/3 < \phi$, which is a contradiction.

Title: Inequality in a bounded real sequence


问:
More random points on a circle
Please assist with this problem.

Suppose 3 (distinct) points are uniformly and independently distributed on a circle of unit length (smaller than a unit circle!). This is really circle and not disc. Call one of these points $B$. Let $W$ be the minimum distance of the point B to its nearest neighbour, either clockwise or anti-clockwise, whichever is smaller.

Find the pdf of $W$. (Well there's no measure theory for this problem, but I assume this pdf exists. Of course we can see for ourselves by computing the cdf $F_W(w)= P(W \le w)$ 1st and then hope the cdf is absolutely continuous.)

A. My model:

A.1. The circle is bijective with $[0,1)$, so let's call these 3 points $A,B,C$ s.t. they are iid $\sim \ Unif(0,1)$ (or $[0,1)$ or whatever).

A.2. (Not sure if any measure theory needed here, but they are all distinct $\mathbb P$-almost surely.)

A.3. Let $A$ be the anti-clockwise neighbour and $C$ clockwise.

Question:

Well, I hope to find the pdf of $W$ via its cdf, which I think I'm able to compute if I know what $W$ is. Is it $W = \min\{B-A,A-C\}$? Or is it $W=\min\{Z,U\}$, where $Z$ is the distance from $B$ to $A$ (anti-clockwise of course) and $U$ is the distance from $B$ to $C$ (clockwise of course; which I think is equivalent to saying 'distance from $C$ to $B$ anti-clockwise') ? Or something else? (Note: I asked about $Z$ here, but I hope I made this post self-contained.)

These questions are all related, but I hope I made each self-contained

Random points on a circle

More random points on a circle

Most random points on a circle

Modelling random points on a circle

Remodelling random points on a circle: Arc length between points distributed on circle is uniform?

Title: More random points on a circle

Body:
I'm going to use David K's $X$ and $Y$ but compute $F_W(w)$ differently for $w \in (0,\frac12)$. I don't use $X_m$, $Y_m$.

$$P(W \ge w) = P(X \ge w, 1-X \ge w, Y \ge w, 1-Y \ge w) = P(w \le X \le 1-w, w \le Y \le 1-w)$$

$$= P(w \le X \le 1-w) P(w \le Y \le 1-w)$$

$$= \int_w^{1-w} 1_{(0,1)}(x) dx \int_w^{1-w} 1_{(0,1)}(y) dy$$

$$= (1-2w)^2$$

$$ \iff F_W(w) = 1-(1-2w)^2$$


答: Multiply it by $\frac{i-1}{i-1}$.

Title: How does $\frac{-1}{2(i+1)} = \frac{-(1-i)}{4}$?

Body: How did $\frac{-1}{2(i+1)}$ change to $\frac{-(1-i)}{4}$??

Body: Multiply it by $\frac{i-1}{i-1}$.


问:
Constructing $\mathbb R$ from $\mathbb Q$ using Dedekind Sections/cuts.
I'll try to present my understanding of construction of $\mathbb R$ and request your help in doubts that I have in the same.

So we consider a straight line $\Gamma$ and mark origin (reference point) on it as $O$, we'll measure all distances (length/taxi-cab distances) from $0$. We mark integers on it and call the set of as $\mathbb Z$. This set $\mathbb Z$ does not form a field.
Assuming that we can fill the line using only integers has problems as then nothing lies in between integers $n$ and $n+1$, which makes the line full of "gaps".   We do not have an integral solution to equations like $2x+1=0,2x+3=0$ etc.

Nos. of the form $p/q:q\ne 0$ with $\gcd(p,q)=1$ (Let's call this set the set of rationals and denote it by $\mathbb Q$ and it includes $0$ when $q=0$) can be placed in the gaps and we may hope that now the line will be completely filled  using elements of $\mathbb Q$. It is to be noted that $\mathbb Q$ forms a field.

Now another problem arises. There is no rational number $x$ which solves $x^2=2$. It seems that still there are some gaps in our line $\Gamma$. But why? Suppose that Geometry (in particular right angled triangle) is not known, why does this equation have to be solvable?What happens if real line consists of only $\mathbb Q$.

Now let's try to use Dedekind construction of $\mathbb R$ from $\mathbb Q$. Let $P$ and $Q$ be two mutually exclusive properties exactly one of which must be possessed by every rational number. Let $P$ be the property "less than 2" and $Q$ the property "greater than 2" i.e. if $x\in Q$ then either $x^2\lt 2$ or $x^2\gt 2$. Let $L\subset \mathbb Q$ be such that $L=\{x\in \mathbb Q:x^2\lt 2\}$ and $R=\{x\in \mathbb Q:x^2\gt 2\}$.

It can be shown that $L$ does not have its greatest member and $R$ does not have its least element in it. My question is now without geometric intuition why boundary separating $L$ and $R$ should exist? Why should $x$ exist such that $x^2=2$?

If the above is clear then it can be shown that dedekind cuts/sections corresponding to every $t$ (rational or not), will satisfy field axioms and thus we'll have a new field and name it $\mathbb R$.

Dedekind cuts/sections' definition: This is the way I understand it is that there are two mutually exclusive properties $P$ and $Q$ exactly one of which should be possessed by every rational number. So as above $L$ and $R$ for $x^2=2$ form dedekind sections/cuts. The sets $\{x:x^2\lt 4\}$ and $\{x:x^2\gt 4\}$ do not form dedekind cuts as $2$ belongs to neither of these.

答:
You ask:

Suppose that Geometry (in particular right angled triangle) is not known, why does this equation have to be solvable?What happens if real line consists of only $\mathbb Q$? [...] Without geometric intuition why boundary separating $L$ and $R$ should exist? Why should $x$ exist such that $x^2=2$?

Well first of all let me push back against the question itself. Geometric intuition is exactly what motivates the Dedekind cut construction. The idea is that we're looking for an algebraic structure which captures some very simple geometric properties of the "ideal line" (key amongst them being connectedness); if we stop caring about those geometric properties, then of course there's no particular reason to keep looking for the same sort of structure! There's certainly nothing wrong with not being particularly interested in geometry and considering $\mathbb{Q}$ to be the "right" number system. (If you can say more about why you want to get away from the geometric motivation of $\mathbb{R}$, I can probably be more satisfying with this section.)

That said, though, one of the major themes in mathematics is the power of abstract concepts to clarify concrete ideas. For example, we see this in analytic number theory, where results about integers are proved intuitively and naturally by considering things related to $\mathbb{R}$. So even if geometric intuition isn't something you care about a priori, you should still care about it as a kind of "new-mathematics-generator" in general.

Separately, I think the following theorem will help clarify the "essentially geometric" flavor of the construction of $\mathbb{R}$:

Up to isomorphism, there is exactly one connected ordered abelian group.

Here "abelian group" basically means a collection of things we can add and subtract, where addition is commutative. Think (signed) lengths, as represented by (directed) line segments. "Ordered" means what it sounds like (if a bit technically), and "connected" is the new geometric idea. "Isomorphism" is basically a notion of "sameness" - think about how all the different constructions of the reals (Dedekind cuts, Cauchy sequences, etc.) still wind up producing "basically the same thing."

If we replace the connectedness requirement with something weaker, then suddenly there are tons of possibilities. Besides things like $\mathbb{Q}$ and $\mathbb{Z}$ we also have weirder objects - say, the set of polynomials in a single variable $x$ with rational coefficients, ordered by setting $p\trianglelefteq q\iff \lim_{x\rightarrow\infty} p(x)\le\lim_{x\rightarrow\infty}q(x).$ So connectedness plays a crucial role in pinning down a unique number system for us to work with. Even if you're not wedded to it, or geometric intuition more generally, as a foundational idea, this is a huge sign that it's mathematically significant: it's generally worth understanding simple-sounding hypotheses which suddenly wind up picking out a single object from a gigantic class of objects.


答:
No, $M_n(\mathbb{C})$ is NOT spanned by $\{AB-BA|A,B\in M_n(\mathbb{C})\}$.

$\mathbb{I}_{n\times n}\in M_n(\mathbb{C})$, and $Tr(\mathbb{I}_{n\times n})=n$, $Tr$ denoting the trace. $Tr(AB-BA)=0$ for all $A,B\in M_n(\mathbb{C})$, and $Tr$ being a linear functional, it is zero on the span of $\{AB-BA|A,B\in M_n(\mathbb{C})\}$ as well. So, $\mathbb{I}_{n\times n}\in M_n(\mathbb{C})$ does not lie in the linear span of $\{AB-BA|A,B\in M_n(\mathbb{C})\}$.

Title: Is $M_n$($\mathbf{C})$ spanned by {$AB-BA$ where A,B in $M_n$($\mathbf{C})$}?

Body: I know that span of  {$AB-BA$ where A,B in $M_n$($\mathbf{R})$} is $n^2$ -1. But what happens if entries are from $\mathbf{C}$? Is $M_n$($\mathbf{C})$ spanned by {$AB-BA$ where A,B in $M_n$($\mathbf{C})$}, with base field $\mathbf{C}$ ?


问:
How can we show this set is stable, but not an attractor?
Let $\Omega$ be a topological space and $\tau:\Omega\to\Omega$. $A\subseteq\Omega$ is called stable if for every neighborhood $V$ of $A$, there is a neigheiborhood $U$ of $A$ with $$\tau^n(U)\subseteq V\;\;\;\text{for all }n\in\mathbb N_0\tag1.$$ $A$ is called attractor if $A$ is

forward invariant, i.e. $A=\tau(A)$;

stable; and

there is a neighborhood $U_0$ of $A$ with $U_0\subseteq E(A)$,

where $E(A)$ is the set of all $x\in\Omega$ for which the orbit $$\operatorname{orb}x:\mathbb N_0\to\Omega\;,\;\;\;n\mapsto\tau^n(x)$$ is eventually in every neighborhood of $A$.

Now consider $$g:[0,1]\to[0,1]\;,\;\;\;x\mapsto\begin{cases}\displaystyle x\left(1-\sin^2\frac\pi x\right)&\text{, if }x\ne0\\0&\text{, otherwise}.\end{cases}$$

How can we show that $A:=\{0\}$ is stable, but not an attractor, and $A_n:=[0,1/n]$ is an attractor for every $n\in\mathbb N$?

I'm not sure how we see that $A$ is stable, but not an attractor, but we it's easy to see that $1/n$ is a fixed point of $g$ for every $n\in\mathbb N$, which might be crucial ...

答: Your hunch is correct. Every neighborhood $_0$ of $$ contains some fixpoint $1/n$, and those points are not in $E(A)$, so condition 3 is violated.


问:
Hard definite integral. Does it have a closed form?
Could someone help me with this integral? $$ \int_{-1}^1 \frac{\log \left(\frac{1-y z}{\sqrt{y^2-2 y z+1}}+1\right)}{\sqrt{y^2-2 y z+1}} \, dy$$
with $|z|<1$.

I've tried all methods known... Trigonometric substitutions doesn't work, neither the change of variables $ u = y^2-2 y z+1 $, etc. I also tried using complex integration, but the problem is that the poles are at $ y = z + i \sqrt{1-z^2}$, so they lie on the unit circle.

答:
Define the function $\mathcal{I}:\left(-1,1\right)\rightarrow\mathbb{R}$ via the definite integral

$$\mathcal{I}{\left(p\right)}:=\int_{-1}^{1}\mathrm{d}x\,\frac{\ln{\left(1+\frac{1-px}{\sqrt{x^{2}-2px+1}}\right)}}{\sqrt{x^{2}-2px+1}}.\tag{1}$$

Note: Given fixed but arbitrary $p\in\left(-1,1\right)$, it's relatively easy to show that

$$\forall x\in\left[-1,1\right]:0<1-px\land0<x^{2}-2px+1,$$

so the RHS of $(1)$ above is a proper integral.

Recalling the definition of the real dilogarithm as

$$\operatorname{Li}_{2}{\left(z\right)}:=-\int_{0}^{z}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{x};~~~\small{z\in\left(-\infty,1\right]},\tag{2}$$

we obtain a closed-form expression for $\mathcal{I}$ in terms of the dilogarithm and elementary functions.

Suppose $p\in\left(-1,1\right)$. Using the Euler substitution $-x+\sqrt{x^{2}-2px+1}=t$, we have

$$x=\frac{1-t^{2}}{2\left(p+t\right)}\implies dx=dt\,\frac{(-1)\left(1+2pt+t^{2}\right)}{2\left(p+t\right)^{2}},$$

$$\sqrt{x^{2}-2px+1}=x+t=\frac{\left(1+2pt+t^{2}\right)}{2\left(p+t\right)},$$

and then,

$$\begin{align}
\mathcal{I}{\left(p\right)}
&=\int_{-1}^{1}\mathrm{d}x\,\frac{\ln{\left(1+\frac{1-px}{\sqrt{x^{2}-2px+1}}\right)}}{\sqrt{x^{2}-2px+1}}\\
&=\int_{1+\sqrt{2+2p}}^{-1+\sqrt{2-2p}}\mathrm{d}t\,\frac{(-1)\left(1+2pt+t^{2}\right)}{2\left(p+t\right)^{2}}\cdot\frac{2\left(p+t\right)}{\left(1+2pt+t^{2}\right)}\\
&~~~~~\times\ln{\left(1+\frac{2\left(p+t\right)}{1+2pt+t^{2}}\cdot\frac{p+2t+pt^{2}}{2\left(p+t\right)}\right)};~~~\small{\left[x=\frac{1-t^{2}}{2\left(p+t\right)}\right]}\\
&=\int_{-1+\sqrt{2-2p}}^{1+\sqrt{2+2p}}\mathrm{d}t\,\frac{1}{\left(p+t\right)}\ln{\left(1+\frac{p+2t+pt^{2}}{1+2pt+t^{2}}\right)}\\
&=\int_{-1+\sqrt{2(1-p)}}^{1+\sqrt{2(1+p)}}\mathrm{d}t\,\frac{1}{\left(p+t\right)}\ln{\left(\frac{\left(1+p\right)\left(1+t\right)^{2}}{1+2pt+t^{2}}\right)}\\
&=\int_{-1+\sqrt{2}\sqrt{1-p}}^{1+\sqrt{2}\sqrt{1+p}}\mathrm{d}t\,\frac{\ln{\left(1+p\right)}+2\ln{\left(1+t\right)}-\ln{\left(1+2pt+t^{2}\right)}}{p+t}\\
&=\int_{p-1+\sqrt{2}\sqrt{1-p}}^{p+1+\sqrt{2}\sqrt{1+p}}\mathrm{d}u\,\frac{\ln{\left(1+p\right)}+2\ln{\left(1-p+u\right)}-\ln{\left(1-p^{2}+u^{2}\right)}}{u};~~~\small{\left[t=u-p\right]}.\\
\end{align}$$

Observe that $0<p-1+\sqrt{2}\sqrt{1-p}<p+1+\sqrt{2}\sqrt{1+p}$. We then obtain the following result:

$$\begin{align}
\mathcal{I}{\left(p\right)}
&=\int_{p-1+\sqrt{2}\sqrt{1-p}}^{p+1+\sqrt{2}\sqrt{1+p}}\mathrm{d}u\,\frac{\ln{\left(1+p\right)}+2\ln{\left(1-p+u\right)}-\ln{\left(1-p^{2}+u^{2}\right)}}{u}\\
&=\int_{p-1+\sqrt{2}\sqrt{1-p}}^{p+1+\sqrt{2}\sqrt{1+p}}\mathrm{d}u\,\frac{2\ln{\left(\frac{1-p+u}{1-p}\right)}-\ln{\left(\frac{1-p^{2}+u^{2}}{1-p^{2}}\right)}+\ln{\left(1-p\right)}}{u}\\
&=\int_{p-1+\sqrt{2}\sqrt{1-p}}^{p+1+\sqrt{2}\sqrt{1+p}}\mathrm{d}u\,\frac{2\ln{\left(\frac{1-p+u}{1-p}\right)}}{u}-\int_{p-1+\sqrt{2}\sqrt{1-p}}^{p+1+\sqrt{2}\sqrt{1+p}}\mathrm{d}u\,\frac{\ln{\left(\frac{1-p^{2}+u^{2}}{1-p^{2}}\right)}}{u}\\
&~~~~~+\int_{p-1+\sqrt{2}\sqrt{1-p}}^{p+1+\sqrt{2}\sqrt{1+p}}\mathrm{d}u\,\frac{\ln{\left(1-p\right)}}{u}\\
&=\int_{\frac{p-1+\sqrt{2}\sqrt{1-p}}{1-p}}^{\frac{p+1+\sqrt{2}\sqrt{1+p}}{1-p}}\mathrm{d}v\,\frac{2\ln{\left(1+v\right)}}{v};~~~\small{\left[u=(1-p)v\right]}\\
&~~~~~-\int_{\left(p-1+\sqrt{2}\sqrt{1-p}\right)^{2}}^{\left(p+1+\sqrt{2}\sqrt{1+p}\right)^{2}}\mathrm{d}v\,\frac{\ln{\left(\frac{1-p^{2}+v}{1-p^{2}}\right)}}{2v};~~~\small{\left[u^{2}=v\right]}\\
&~~~~~+\ln{\left(1-p\right)}\int_{p-1+\sqrt{2}\sqrt{1-p}}^{p+1+\sqrt{2}\sqrt{1+p}}\mathrm{d}u\,\frac{1}{u}\\
&=\int_{-\frac{p-1+\sqrt{2}\sqrt{1-p}}{1-p}}^{-\frac{p+1+\sqrt{2}\sqrt{1+p}}{1-p}}\mathrm{d}x\,\frac{2\ln{\left(1-x\right)}}{x};~~~\small{\left[v=-x\right]}\\
&~~~~~-\int_{\frac{\left(p-1+\sqrt{2}\sqrt{1-p}\right)^{2}}{1-p^{2}}}^{\frac{\left(p+1+\sqrt{2}\sqrt{1+p}\right)^{2}}{1-p^{2}}}\mathrm{d}w\,\frac{\ln{\left(1+w\right)}}{2w};~~~\small{\left[v=(1-p^{2})w\right]}\\
&~~~~~+\ln{\left(1-p\right)}\left[\ln{\left(p+1+\sqrt{2}\sqrt{1+p}\right)}-\ln{\left(p-1+\sqrt{2}\sqrt{1-p}\right)}\right]\\
&=2\int_{-\frac{p+1+\sqrt{2}\sqrt{1+p}}{1-p}}^{-\frac{p-1+\sqrt{2}\sqrt{1-p}}{1-p}}\mathrm{d}x\,\frac{(-1)\ln{\left(1-x\right)}}{x}\\
&~~~~~-\frac12\int_{-\frac{\left(p+1+\sqrt{2}\sqrt{1+p}\right)^{2}}{1-p^{2}}}^{-\frac{\left(p-1+\sqrt{2}\sqrt{1-p}\right)^{2}}{1-p^{2}}}\mathrm{d}x\,\frac{(-1)\ln{\left(1-x\right)}}{x};~~~\small{\left[w=-x\right]}\\
&~~~~~+\ln{\left(1-p\right)}\left[\ln{\left(p+1+\sqrt{2}\sqrt{1+p}\right)}-\ln{\left(p-1+\sqrt{2}\sqrt{1-p}\right)}\right]\\
&=2\left[\operatorname{Li}_{2}{\left(-\frac{p-1+\sqrt{2}\sqrt{1-p}}{1-p}\right)}-\operatorname{Li}_{2}{\left(-\frac{p+1+\sqrt{2}\sqrt{1+p}}{1-p}\right)}\right]\\
&~~~~~-\frac12\left[\operatorname{Li}_{2}{\left(-\frac{\left(p-1+\sqrt{2}\sqrt{1-p}\right)^{2}}{1-p^{2}}\right)}-\operatorname{Li}_{2}{\left(-\frac{\left(p+1+\sqrt{2}\sqrt{1+p}\right)^{2}}{1-p^{2}}\right)}\right]\\
&~~~~~+\ln{\left(1-p\right)}\left[\ln{\left(p+1+\sqrt{2}\sqrt{1+p}\right)}-\ln{\left(p-1+\sqrt{2}\sqrt{1-p}\right)}\right]\\
&=2\left[\operatorname{Li}_{2}{\left(\frac{1-p-\sqrt{2}\sqrt{1-p}}{1-p}\right)}-\operatorname{Li}_{2}{\left(-\frac{1+p+\sqrt{2}\sqrt{1+p}}{1-p}\right)}\right]\\
&~~~~~-\frac12\left[\operatorname{Li}_{2}{\left(-\frac{\left(\sqrt{2}-\sqrt{1-p}\right)^{2}}{1+p}\right)}-\operatorname{Li}_{2}{\left(-\frac{\left(\sqrt{2}+\sqrt{1+p}\right)^{2}}{1-p}\right)}\right]\\
&~~~~~+\ln{\left(1-p\right)}\ln{\left(\frac{p+1+\sqrt{2}\sqrt{1+p}}{p-1+\sqrt{2}\sqrt{1-p}}\right)}.\tag{3}\\
\end{align}$$


问:
minimum lines, maximum points
There are $P$ points in the 2-dimensional plane. Through each point, we draw two orthogonal lines: one horizontal (parallel to x axis), one vertical (parallel to y axis). Obviously, some of these lines conicide. How can we arrange the points such as the total number of lines is minimized?

I started with solving the dual problem (is that the correct name?): suppose we are allowed exactly $L$ lines, what is the maximum number of points we can have?

Suppose the number of different x values is $X$, and the number of different y values is $Y$. Then the number of lines is $L=X+Y$, and the maximum number of points is $P=X*Y$. So we have the following optimization problem:

$maximize(X*Y)$

$such-that: X+Y=L$

AFAIK, the solution to this problem is to take X and Y as close as possible, e.g., if $L$ is even, take $X=Y=L/2$, and then $P=(L/2)^2$.

However, I don't know how to go from this to the original problem. For example, if we have $P=14$ points, then we can arrange them in a 7-by-2 grid and have $L=9$ lines, but we can also arrange them in a 4-by-4 grid with 2 points missing, then we will have only $L=8$ lines. So, how can I find the solution to the original minimization problem:

$minimize(X+Y)$

$such-that: X*Y>=P$

?

答:
You're correct. We will place the points one at a time, in a zigzag greedy fashion, so that no lattice points are wasted. Redefine $n$ to be the number of points. Now for each $i\in \{1,...,n\}$, let $P_i$ be the $i$th point to be placed. Notice that since any integer $i$ lies between two consecutive perfect squares of the form $k^2$ and $k^2 + 2k + 1$ (where $k=\lfloor \sqrt{i} \rfloor$), we can uniquely express $i$ in the form $i=k^2+r$, where $r\in \{0,1,...,2k\}$. We now map each point to the Cartesian plane as follows:

$$ P_i =
\begin{cases}
(k,k)   & \text{if } r = 0 \\
(k+1,r) & \text{if } r \in \{1,...,k\} \\
(r-k,k+1) & \text{if } r \in \{k+1,...,2k\} \\
\end{cases}
$$

For example, since $14=3^2+5$ and $5\in \{4,5,6\}$, we would place the $14$th point $P_{14}$ at $(5-3,3+1)=(2,4)$.

Note that if $L(n)$ is the total number of drawn lines after placing down the $n$ points $P_1,P_2,...,P_n$, and if we express $n$ in the form $n=k^2+r$ where $r\in \{0,1,...,2k\}$ as above, then:
$$ L(n) =
\begin{cases}
2k   & \text{if } r = 0 \\
2k+1 & \text{if } r \in \{1,...,k\} \\
2k+2 & \text{if } r \in \{k+1,...,2k\} \\
\end{cases}
$$

Title: minimum lines, maximum points


问:
Matrix with the same value in all entries — which notation to use?
Pretty trivial but for a matrix

\begin{bmatrix}x&x&\ldots&x&x\\x&x&\ldots&x&x\\\vdots&\vdots&\ddots&x&x\\x&x&\ldots&x&x\end{bmatrix}

with  $N\times N $, is there is simpler notation. I had $A=[x_{i,j}]\in \Bbb R^{N\times N}$ in my mind, but I can't find anywhere were it is used

Edit: I forgot to mention but, I need to work with the elements in the matrix, so like doing scalar multiplication (to do a proof by the principle of mathematical induction question). But it a $3\times 3$ so I figured out I do have to write every single element 9 times in a row. Is there any notation that can make my math simpler?

答: I don't think there is a common notation, as this type of matrix isn't extremely common. However in Statistics I have seen a column or matrix of ones written as a $J$.


问:
Factorization $x^4+px^3+qx^2+r x +s=(x^2+a x +b)(x^2+\bar a x +\bar b)$
Question:

Under what condition, does the quartic polynomial with rational coefficients
$p$, $q$, $r$ and $s$ factorizes as

$$x^4+px^3+qx^2+r x +s= (x^2+a x +b)(x^2+\bar a x +\bar b) $$

with $a$, $b$ complex numbers, along with their conjugates $\bar a $, $\bar b$.

Examples:

$$x^4+2x^3+6x^2+2x+1=( x^2 +(1-i \sqrt3)x +1) (x^2 +(1+i \sqrt3)x +1) $$
$$x^4+2x^3+4x^2+2=( x^2 +(1+i)x +(1-i)) (x^2 +(1-i)x +(1+i))  $$

Note that the symmetry of coefficients leads to such factorization, as seen in the first example; but not exclusively so, as shown by the second example.

Is there any test on the coefficients $p$, $q$, $r$ and $s$ that can be carried out to determine the possibility of such factorization? I reviewed here the discriminate tests on the nature of roots for quartic equations and did not find anything applicable.

答:
Short answer:

Actually, the things I do, do not involve any special manipulation. It just contains simple algebra.

Let $a,b,c,d \in\mathbb R$  then we have,

$$\left(x^2+(a+bi)x+(c+di)\right)\left(x^2+(a-bi)x+(c-di)\right)=x^4+2ax^3+(a^2+b^2+2c)x^2+(2ac+2bd)x+(c^2+d^2).$$

$$x^4+px^3+qx^2+rx+s=x^4+2ax^3+(a^2+b^2+2c)x^2+(2ac+2bd)x+(c^2+d^2)$$

which follows

$$\begin{cases} p=2a \\ q=a^2+b^2+2c \\ r=2ac+2bd \\s=c^2+d^2\end{cases} $$

If $d=0$, then the system of equation becomes extremely simple. I'll leave this case to you.  Here we will work with $d≠0.$

If $d≠0$, then $s-c^2≠0$. We have,

$$\begin{cases} a=\frac p2 \\b^2+2c=q-\frac{p^2}{4} \\pc+2bd=r \\c^2+d^2=s \end{cases}$$

$$\implies \begin{cases}\left( \frac{r-pc}{2d}\right)^2+2c-q+\frac{p^2}{4} =0 \\ c^2+d^2=s \end{cases}$$

Finally we get,

$$\frac{(r-pc)^2}{4(s-c^2)}+2c-q+\frac{p^2}{4} =0$$

$$\color {gold}{\boxed {\color{black}{8 c^3-4qc^2+(2pr-8s)c+(4qs-sp^2-r^2)=0.}}}$$

As it seems, we obtain the cubic equation with respect to $c$.

If all of the coefficients of the cubic equation are real numbers, then it has at least one real root. Therefore, we can always choose $c$ to be a real number.

If you accept $a,b,c,d$ as rational, then we can immediately use the Rational Root theorem.

For this, all we need is to find the factors of this expression $\color{red}{\dfrac{4qs-sp^2-r^2}{8}.}$

We're almost done.  We have the last two restrictions.

After finding real $c$, then we must check the following two cases:

$s-c^2≥0$

$p^2-4q+8c≤0.$

If these conditions hold, then you can easily find the $a,b,d$ from the system of equations.

End of the answer.


问:
True or false? If $f(X)$ is the image of $f$ on the set $X$ then $f\colon X \to f(X)$ must be surjective
Title. It seems obvious to prove because if I suppose $y \in f(X)$ then by definition there is a $x \in X$ so that $f(x)  = y$. But I am slightly uncertain.

答: In general, a map $f$ is surjective if the image of the map is equal to the codomain. By restricting the codomain to the image, we are forcing this property to occur, and thus also making $f$ surjective.


问:
Proving that $\mathfrak{A}$ and $\mathfrak{N}$ are elementarily equivalent
This is an exercise problem in the textbook A Friendly Introduction to Mathematical Logic. The question is as follows:

Let $\mathfrak{N}$ be the usual structure of the language of number theory $\mathcal{L}_{NT}$. Let $\Sigma$ be the theory of $\mathfrak{N}$. Let $\mathcal{L}$ be a language given by $\mathcal{L}=\mathcal{L}_{NT} \cup \{ c \}$. Let $\Theta=\Sigma\cup \{ 0<c, S0<c, SS0<c,\ldots \}$. Let $\mathfrak{A}'$ be a model of $\Theta$. Now we define $\mathfrak{A}=\mathfrak{A}'\upharpoonright_{\mathcal{L}_{NT}}$. Prove that $\mathfrak{A}$ is elementarily equivalent to $\mathfrak{N}$.

My attempt:

In order to prove that $\mathfrak{A}$ is elementarily equivalent to $\mathfrak{N}$, we need to prove that theory of $\mathfrak{A}$ is $\Sigma$. Let's take $\phi\in \Sigma$. Then $\phi\in\Theta$. Thus, $\Theta \vdash \phi$ and by Soundness theorem, $\Theta\models\phi$. Since $\mathfrak{A}'$ is a model of $\Theta$, we have that $\mathfrak{A}'\models\phi$. Thus $\mathfrak{A}\models \phi$ as $\phi$ is a $\mathcal{L}_{NT}$ formula. Thus, $\Sigma$ is a subset of theory of $\mathfrak{N}$.

I am having trouble proving the other direction however. If I take a $\phi$ such that $\mathfrak{A}\models\phi$. Then I can say that $\mathfrak{A}'\models \phi$. But I am not really sure how to proceed after this.

Any hints will be appreciated.

答:
HINT: $\Sigma$ is already "as big as possible" - if $\mathfrak{A}'\models\varphi$ for some $\varphi\not\in\Sigma$, there will be a "clash."

In more detail, suppose $\mathfrak{A}'\models\varphi$ but $\mathfrak{N}\not\models\varphi$. Can you think of a sentence related to $\varphi$ which $\mathfrak{N}$ does satisfy instead?

 Its negation, $\neg\varphi$. Remember that by definition of the satisfaction relation we have $$\mathfrak{S}\not\models\eta\quad\iff\quad\mathfrak{S}\models\neg\eta$$ for every structure $\mathfrak{S}$ and every sentence $\eta$.

Now why would that be a problem?

 If $\mathfrak{N}\models\neg\varphi$ then $\neg\varphi\in\Sigma$ so $\mathfrak{A}'\models\neg\varphi$. But this contradicts the assumption that $\mathfrak{A}'\models\varphi$.

Basically, what's going on here is that the theory of a structure is always a maximal consistent theory (with respect to the structure's language, that is), so there's no "room for improvement."


问:
Consider the inequality $9-x^2>|x-a|$ where $a$ is real, then find set of values for $a$ such that at least one negative solution exists
For $x>a$

$$9-x^2 >x-a$$
$$x^2 + x -(9+a)<0$$

So
$$1+4(9+a)<0$$
$$a<\frac{-37}{4}$$

Similarly for $x<a$
$$x^2-x+a-9<0$$
So $a>\frac{37}{4}$

Now in both cases, we need at least one $x<0$

In case one, $a<0$ and $x>a$ so $x$ can be both positive and negative, so I don’t know how to deal with that

Similar problem in case 2.

Another inequality I found was $a>-9$, obtained by solving the quadratic for first case, giving $x\in (\frac{1-\sqrt {37+4a}}{2}, \frac{1+\sqrt {37+4a}}{2})$, and then setting the lower limit to $<0$

I can also find another inequality from the second case, but there are way too many inequalities to choose from and the condition that $x>a$ and $x<a$ is throwing me off.

Also is a downward parabola even possible for these cases? The equation demands it to be downward if $f(x)<0$ for all $x$, but the coefficient of $x^2$ is positive here.

答:
The graphical approach is illuminating - a rough hand drawn graph is sufficient.

As $a$ varies from $A$ to $B$, $y=|x-a|$ intersects the negative branch of $y=9-x^2$. At $A$, $y=|x-a_1|$ is tangent to the parabola at $D$. At $B$, $y=|x-a_2|$ passes through the vertex of parabola, $C=(0,9)$.

Since $\triangle COB$ is right isosceles, $B=(9,0)$.

Slope at $D$ is $-2x=1 \Rightarrow D=(-1/2,35/4)$. Since $\triangle AED$ is right isosceles, $A=(-(35/4+1/2),0)=(-37/4,0)$.

Desired range is $a \in [-37/4,9)$.


问:
I have some $\frac{1}{2i}\left(\frac{1}{z}(1-\frac{i}{2})^{-1}-\frac{1}{z}(1+\frac{i}{2})^{-1}\right)$
For my Radius of convergence, I have that
$|\frac{i}{2}| < 1 \Longrightarrow |z| >1$. Is this a valid radius of convergence?

答: If $|z|<1$, then$$\frac1{1+z^2}=1-z^2+z^4-z^6+\cdots,$$and this series diverges if $|z|\geqslant1$. So, the radius of convergence is $1$.


问:
Is there any onto homomorphism from $\mathbb{C}^*$ to $\mathbb{R}^*$?
$\mathbb{C}^*$ be the multiplicative group of nonzero complex numbers similarly $\mathbb{R}^*$.
If so can you provide an example

答: It seems that no such homomorphism exists. Assume that $\varphi\colon \mathbb{C}^*\to \mathbb{R}^*$ is a homomorhism. Take any $z\in \mathbb{C}^*$ and any $w\in \mathbb{C}^*$ such that $w^2=z$. (Every complex number has a square root.) Then $\varphi(z)=\varphi(w^2)=\varphi(w)^2>0$. Thus $\varphi(z)$ is always positive.


答:
It is common for $J$ or $J_{n}$ to be used as a square ($n \times n$) matrix of all 1s, though not so common that you shouldn't specify that this is what you mean. You can also write $J_{m,n}$ for an $m \times n$ matrix of all ones.  It is also common for ${\bf{j}}$ to be used for a vector of all ones.

These all ones matrices and vectors come up in a variety of situations in algebra and combinatorics; for example

$J_{n}-I_{n}$ is the adjacency matrix for the complete graph $K_{n}$

$J_{m,n}$ is the identity element for the set of $m\times n$ matrices under the Hadamard product;

the proof of Fisher's Inequality for combinatorial designs relies on the fact that the incidence matrix of such a structure must satisfy the equation $B^{T}B=(r-\lambda)I+\lambda J$ (where $r$ and $\lambda$ are parameters of the design).

Title: Matrix with the same value in all entries — which notation to use?

Body:
Pretty trivial but for a matrix

\begin{bmatrix}x&x&\ldots&x&x\\x&x&\ldots&x&x\\\vdots&\vdots&\ddots&x&x\\x&x&\ldots&x&x\end{bmatrix}

with  $N\times N $, is there is simpler notation. I had $A=[x_{i,j}]\in \Bbb R^{N\times N}$ in my mind, but I can't find anywhere were it is used

Edit: I forgot to mention but, I need to work with the elements in the matrix, so like doing scalar multiplication (to do a proof by the principle of mathematical induction question). But it a $3\times 3$ so I figured out I do have to write every single element 9 times in a row. Is there any notation that can make my math simpler?


问:
Operators between $L_p$ spaces.
Let $p \in (1, \infty)$. If $E$ and $F$ are Banach spaces, we define $E \oplus_p F$ to be the usual direct sum equipped with the norm $\|(e,f)\|=(\|e\|^p+\|f\|^p)^{1/p}$.

Now let $(X_j, \mu_j)$ be measure spaces for $j=1,2$.

Suppose that $T_k: L_p(\mu_1) \to L_p(\mu_2)$ are bounded linear maps for $k=1,2$.

Define $T_*: L_p(\mu_1) \to (L_p(\mu_2) \oplus_p L_p(\mu_2))$ by
$$
T_*f=(T_1f, T_2f).
$$
Define $T^*: (L_p(\mu_1) \oplus_p L_p(\mu_1)) \to L_p(\mu_2)$ by
$$
T^*(f,g)=T_1f + T_2g.
$$

Problem: Show that $T_*$ and $T^*$ are bounded linear maps and compute their norms.

Progress: I think I have successfully shown that $T_*$ and $T^*$ are bounded linear with
$$
\| T_*\| \leq (\|T_1\|^p+\|T_2\|^p)^{1/p},
$$
and
$$
\| T^*\| \leq (\|T_1\|^q+\|T_2\|^q)^{1/q},
$$
where $q$ is the Hölder conjugate of $p$, i.e. $\frac{1}{p}+\frac{1}{q}=1$.

Questions: I suspect that we get equality for the norms. Do we? I've tried to prove it but I can't. I feel that I am either missing something obvious or that it might not be true. If we do not get equality, is there any hope to compute the norms?

Hope someone can help me here. Thanks.

答:
You do not get equality in general. Let me explain for $T_*$ only and leave the one for $T^*$ to you - it can be shown in much the same way. The proof for the norm inequality is \begin{align}\|T_*f\|&=\|(T_1f,T_2f)\|\\
&=\sqrt[p]{\|T_1f\|^p+\|T_2f\|^p}\\
&\le\sqrt[p]{\|T_1\|^p\|f\|^p+\|T_2\|^p\|f\|^p}\tag{1}\\
&=\sqrt[p]{\|T_1\|^p+\|T_2\|^p}\|f\|\end{align}
There is only one inequality here, so to address whether the norm of $T_*$ can be achieved one needs to check whether a function $f$ can always be found which makes $(1)$ an equality (at least in the limit); that is $\|T_1f\|=\|T_1\|\|f\|$ and $\|T_2f\|=\|T_2\|\|f\|$. But linear operators need not have their maximum norm on the same function. This can be easily seen if one takes $T_i$ to be matrices.

So to construct a counterexample, take $X_1=X_2=\{1,2\}$, so $L_p(\mu)=\mathbb{R}^2_p$, and $T_1=\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $T_2=\begin{pmatrix}0&0\\0&1\end{pmatrix}$, and $\mathbf{f}=\begin{pmatrix}x\\y\end{pmatrix}$, then $$\|T_*\|=\sup_{\mathbf{f}\ne0}\frac{\|T_*\mathbf{f}\|}{\|\mathbf{f}\|}=\sup\frac{\sqrt[p]{|x|^p+|y|^p}}{\sqrt[p]{|x|^p+|y|^p}}=1$$ $$\|T_1\|=\sup\frac{\|T_1\mathbf{f}\|}{\|\mathbf{f}\|}=\sup\frac{|x|}{\sqrt[p]{|x|^p+|y|^p}}=1=\|T_2\|$$ $$\implies \sqrt[p]{\|T_1\|^p+\|T_2\|^p}=\sqrt[p]{2}$$

Thus one cannot hope to calculate $\|T_*\|$ unless $T_1$, $T_2$ are specified.

Title: Operators between $L_p$ spaces.


问:
How many solutions possible for the equation $x_1+x_2+x_3+x_4+x_5=55$ if
How many solutions possible for the equation $x_1+x_2+x_3+x_4+x_5=55$ if all $x$ are non-negative integer:

No restrictions. The solution is $C(55 + 4, 4) = C(59,4)$ but I fail to see why, can someone explain this to me?

Every $x_k$ is odd.

If $x_1\ge1,x_2\ge2,x_3\ge3,x_4\ge2,x_5\ge1$

答:
You forgot to say that $0\le x_i$ (otherwise there is an infinite number of solutions). This is a standard "choice with repetitions and without order" case (hence the formula, which is one of the basic formulas in combinatorics). In this case, you have 55 "balls" to distribute freely into the five "holes" - the variables.

Here you can write $x_i=2y_i+1$ and solve for $y_i$ instead of $x_i$. You'll get something like $2(y_1+\dots+y_5)+5=55$ - simplify to get $y_1+\dots+y_5=25$ (how?)

Here you use $x_1=y_1+1$, $x_2=y_2+2$ and then again solve for $y_i$.


问:
Absolutely continuous measure which is not equivalent
Does there exist a simple example of a probability measure on $[0,1]$ such that ${\rm spt}\,\mu = [0,1]$ and $\mu \ll \mathcal{L}$ but $\mathcal{L} \not\ll \mu$? Here $\mathcal{L}$ denotes the Lebesgue measure.

答: Let $C$ be  a fat  Cantor set and $\mu (E)=\mathcal L (E\setminus C)$. Then $\mu$ has full support (since $C$ has no interior) and it is absolutely continuous w.r.t. Lebesgue measure but $\mu (C)=0$ whereas Lebesgue measure of $C$ is  $>0$.


问:
GRE problem involving LCD, prime factorization, and sets.
I think this problem was a bit tricky and I'm trying to better think through this. Here is the problem:

Let S be the set of all positive integers n such that $n^2$ is a multiple of both 24 and 108. Which of the following integers are divisors of every integer n in S?

So I'm trying to first figure the LCM of 24 and 108

24's prime factorization is $2^33^1$ and 108's prime factorization is $2^23^3$. So the LCM is $2^3 3^3$ which is 27 * 8 = 216.

We know that $n^2$ is a multiple of 216 or $2^3 3^3$ but this does not mean that every multiple of 216 is a possible value of $n^2$ because some multiples of 216 may not be squares of a number.

Where do I go from here?

答: I understand why 36 is the lowest value in S. The part I am not able to understand is why all the bigger values in S has to be all the integer multiples of 36? Can some one explain that part to me please ?</p>


问:
Note or textbook about algebraic k-theory
I'm reading the Weibel's K-book. It's difficult for me. I have read some $K_0$ with the hlep of Bass' book. But Bass' book is so old. So, are there any more detailed books or notes to recommend.

I appreciate any advice, thanks!

答:
Here are some useful list:

In any field (and any level!), Milnor is a nice choice:

Milnor, John W., Introduction to algebraic K-theory, Annals of Mathematics Studies. No. 72. Princeton, N. J.: Princeton University Press and University of Tokyo Press. XIII, 184 p. $ 6.25 (1971). ZBL0237.18005.

and some other Books:

Rosenberg, Jonathan, Algebraic K-theory and its applications, Graduate Texts in Mathematics. 147. New York, NY: Springer-Verlag. x, 392 p. (1994). ZBL0801.19001.

Inassaridze, Hvedri, Algebraic (K)-theory, Mathematics and its Applications (Dordrecht). 311. Dordrecht: Kluwer Academic Publishers. 438 p. ; (1995). ZBL0836.19001.

Dundas, Bjørn Ian; Goodwillie, Thomas G.; McCarthy, Randy, The local structure of algebraic K-theory, Algebra and Applications 18. London: Springer (ISBN 978-1-4471-4392-5/hbk; 978-1-4471-4393-2/ebook). xv, 435 p. (2013). ZBL1272.55002.

Lluis-Puebla, Emilio; Loday, Jean-Louis; Gillet, Henri; Soulé, Christophe; Snaith, Victor, Higher algebraic (K)-theory: an overview, Lecture Notes in Mathematics. 1491. Berlin etc.: Springer-Verlag. ix, 164 p. (1992). ZBL0746.19001.


问:
$f:[-1,1]\to\Bbb{R}$ be continuous and $\int_{-1}^{1}f(t)dt=1$ then find $\displaystyle \lim_{n\to \infty} \int f(t) \cos^2nt \mathrm{d}t$.[NBHM2008]
let $f:[-1,1]\to \mathbb{R}$ be a continuous function such that $\int_{-1}^{1}f(t)dt=1$ then find $\displaystyle \lim_{n\to \infty} \int f(t) \cos^2nt   \mathrm{d}t$.

My method: I have solved the problem by choosing $f(t) =\frac12$. And the answer I get is also $\frac12$. But I want to know, without choosing this type of example, is there any other general proof?
Thanks in advance

答: $\int f(t)\cos^{2}(nt)dt=\frac  1 2  \int f(t)[1+\cos (2nt)]dt$. By Riemann Lebesgue Lemma $\int f(t) \cos (2nt)dt \to 0$, so the answer is $\frac 1  2$.


答:
ok so as you have said:

Case 1: $x>a$

$$9-x^2 >x-a$$
$$x^2 + x -(9+a)<0$$
Now, $x^2 + x -(9+a)$ is an upward facing parabola and we need to find the region of where it is below the x axis; because after all, that's where it is less than zero. For an upward facing parabola to have a region below the x-axis it must have two real roots, and hence the discriminant must be greater than zero. $\implies b^2-4ac> 0 \implies 1+4(9+a)>0 \implies a>\dfrac{-37}{4}$. So when $a>\dfrac{-37}{4}$ that upward facing parabola has two real roots and hence there exists a region below the x-axis. But this is not the only thing we need. We need this region below the x-axis to have at least one negative value of x. Notice the x coordinate of the vertex of this parabola = $ \dfrac{-b}{2a}= \dfrac{-1}{2}$, which is less than zero. That means at least one of the two roots is negative, and at least one negative value exists in the solution set of: $x^2 + x -(9+a)<0$ iff $a>\dfrac{-37}{4}$.

But we also said $x$ is strictly greater than $a$ initially before removing the absolute value. This means for at least one negative solution, $a$ must be strictly less than zero, because if $a$ exceeds zero, then $x$ must also and then there will be no negative solutions for x... So for $x>a, a\in (\dfrac{-37}{4},0).$

Case 2: $x<a$

$$9-x^2 >-(x-a)$$
$$x^2 - x +(a-9)<0$$

Again discriminant $D>0 \implies a<\dfrac{37}{4}$, but here vertex of parabola is positive = 0.5. This means at least one root is positive. To make sure that the other root is negative we need the parabola to be below the $x$ axis when $x=0.$ So $0^2 - 0 +(a-9)<0 \implies a<9.$ We also need to make sure this negative root is less than $a$ because our initial constraint was $x<a.$ So when does the line x =a, meet the parabola at the negative root? $a^2 - a +(a-9)=0 \implies a = -3 \implies a\in [-3,9)$

So $a \in (\dfrac{-37}{4},0) \cup [-3,9] \implies a \in (\dfrac{-37}{4},9)$

Title: Consider the inequality $9-x^2>|x-a|$ where $a$ is real, then find set of values for $a$ such that at least one negative solution exists

Body:
For $x>a$

$$9-x^2 >x-a$$
$$x^2 + x -(9+a)<0$$

So
$$1+4(9+a)<0$$
$$a<\frac{-37}{4}$$

Similarly for $x<a$
$$x^2-x+a-9<0$$
So $a>\frac{37}{4}$

Now in both cases, we need at least one $x<0$

In case one, $a<0$ and $x>a$ so $x$ can be both positive and negative, so I don’t know how to deal with that

Similar problem in case 2.

Another inequality I found was $a>-9$, obtained by solving the quadratic for first case, giving $x\in (\frac{1-\sqrt {37+4a}}{2}, \frac{1+\sqrt {37+4a}}{2})$, and then setting the lower limit to $<0$

I can also find another inequality from the second case, but there are way too many inequalities to choose from and the condition that $x>a$ and $x<a$ is throwing me off.

Also is a downward parabola even possible for these cases? The equation demands it to be downward if $f(x)<0$ for all $x$, but the coefficient of $x^2$ is positive here.


问:
Approximation of an integral
)
I can't get over a step in my teacher's exercise.

$$I(x) = 2\int_{0}^{1} \frac{y^3 + 4y^2}{y^2+4y+5} dy = 2\int_{0}^{1} y dy  -5\int_{0}^{1} \frac{2y}{y^2+4y+5} dy$$

Probably it's very easy..

Thank you :)

Title: Approximation of an integral

Body: $\textbf{Hint: } y^3+4y^2 = y(y^2+4y+5)-5y$


问:
Estimating coefficient in linear regression
Given:
$y = b_0 + b_1x$

I am wondering what is the explanation behind this formula for estimating the $ b_1$ coefficient:

$$ b_1 = \frac{\sum_{i=1}^n( x_i-\bar{x})(y_i-\bar{y})}{
  \sum_{i=1}^n( x_i-\bar{x})^2 }
$$

What are the steps to derive this formula?

Part 1 Update -March 18 2021:

When  tried to substitute $\bar{y} - b_1\bar{x}$ for $b_0$ in

$$
b_0 \bar{x} + b_1 \overline{x^2} = \overline{xy}
$$
I got stuck with $b_1$ in both sides of the equations.

$$
 b_1 \overline{x^2} = \overline{xy}-(\overline{x}
\bar{y} - b_1\overline{x^2})
$$

Can you please guide me in further derivation steps. Thanks

Part 2 Update

With another help from @MartinVesely, I realized that this should be:

$$b_0 \bar{x} + b_1 \overline{x^2} = \overline{xy}$$

$$ ((\bar{y} - b_1\bar{x})\bar{x}) + b_1 \overline{x^2} = \overline{xy}$$

$$(\bar{x}\bar{y} - b_1(\bar{x})^2) + b_1 \overline{x^2} = \overline{xy}$$

$$( - b_1(\bar{x})^2) + b_1 \overline{x^2} = \overline{xy} - \bar{x}\bar{y} $$

$$b1( -(\bar{x})^2 +  \overline{x^2}) = \overline{xy} - \bar{x}\bar{y} $$

$$b1= \frac{\overline{xy} - \bar{x}\bar{y} }{   \overline{x^2} -(\bar{x})^2} $$

答:
A derivation of the formula is done with the least square method.

Firstly write down a function $L = \sum_{i=1}^n (y_i - b_0 - b_1 x_i)^2$. This is a sum of squared differences between actual output data $y_i$ and output given by a regression line.

Our goal is to minimize a difference between actual data and theregression line. This means that we need to calculate first derivatives with respects to $b_0$ and $b_1$:

$$
\frac{\partial L}{\partial b_0} = -\sum_{i=1}^n 2(y_i - b_0 - b_1 x_i)
$$

$$
\frac{\partial L}{\partial b_1} = -\sum_{i=1}^n 2x_i(y_i - b_0 - b_1 x_i)
$$

Now, by setting $\frac{\partial L}{\partial b_0}$ and $\frac{\partial L}{\partial b_1}$ equal to zero and dividing by -2 we have

$$
\sum_{i=1}^n (y_i - b_0 - b_1 x_i) = 0
$$

$$
\sum_{i=1}^n x_i(y_i - b_0 - b_1 x_i) = 0
$$

Rewriting leads to
$$
\sum_{i=1}^n (y_i - b_0 - b_1 x_i) = \sum_{i=1}^n y_i - b_1\sum_{i=1}^n x_i - nb_o = 0
$$

$$
\sum_{i=1}^n x_i(y_i - b_0 - b_1 x_i) = \sum_{i=1}^n x_iy_i - b_1\sum_{i=1}^n x_i^2 -b_0\sum_{i=1}^n x_i  = 0
$$

Now, if we divide both eqautions by $n$ and rearranging them, we have
$$
b_0 + b_1 \bar{x} = \bar{y}
$$

$$
b_0 \bar{x} + b_1 \overline{x^2} = \overline{xy},
$$

where $\bar{x}$ is average of $x_i$ values (similarly for $y_i$) and $\overline{xy}$ is average of products $x_iy_i$.

Clearly $b_0 = \bar{y} - b_1\bar{x}$. After substituing this to the other equation we get
$$
b_1 = \frac{\overline{xy} -\bar{x}\bar{y}}{\overline{x^2}-(\bar{x})^2}.
$$

Since $\overline{xy} -\bar{x}\bar{y}$ is covariance of $x$ and $y$ and $\overline{x^2}-(\bar{x})^2$ is variance of $x$ we have your formula, because
$$
\frac{1}{n}\sum_{i=1}^n (x_i - \bar{x})^2
$$
is variance of $x$ and
$$
\frac{1}{n}\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})
$$
is covariance of $x$ and $y$.


问:
Isomorphism between vector space of all linear transformations
If $V$ and $W$ are vector spaces, is there an isomorphism between $L(W, V^*)$ and $L(V,W^*)$? If so, how do I find this? ($L(V,W)$ is the vector space of linear transformation from $V$ to $W$). Any element in $L(W, V^*)$ maps elements in $w$ to a linear functional on $V$; I assume this linear functional can be further evaluated at elements in $V$, but am still very confused about how to construct an isomorphism if it exists.

答:
Your intuition is correct. Take $f:W\rightarrow V^*$ linear. Then consider $\pi(f)(v)(w):=f(w)(v)$, where $\pi(f): V\rightarrow W^*$, and $\pi(f)(v)\in W^*$.

Intuitively we just exchanged the two inputs, and from this it should be clear that this "input permutation map" $\pi$ is the desired isomorphism.


问:
Matrix representation of the adjoint of an operator, the same as the complex conjugate of the transpose of that operator?
Since I'm not taking summer classes I decided to do some self learning on more advanced mathematics, and I've found myself stuck on this problem:

I have to show that for any operator $\hat{A}$ the matrix representation of the adjoint $\hat{A}\dagger$ is given as the complex conjugate of the transpose of the matrix representation of the operator $\hat{A}$.

So basically show that
$\underline{A}\dagger = (\underline{A^T})^*$

I've done many examples on the textbook using actual matrices, but I don't really know how to prove this in terms of operators, any hint or where to start, or some guidance?
Thank you

答:
I assume that the definition of adjoint that you have is the following:

in a complex vector space $V$ endowed with an inner product $\langle\,|\,\rangle$, the adjoint $\hat{A}^\star$ of the (linear) operator $\hat{A}$ is the unique operator with the property that
$$\langle \hat{A}\psi | \varphi\rangle = \langle \psi | \hat{A}^\star\varphi\rangle,\qquad \forall \psi, \varphi\in V.$$

If that is the case, then a way to prove your claim easily is to fix an orthonormal base $(e_i\ :\ i=1\ldots N)$ and observe than any operator is uniquely determined as a complex linear combination of the operators $\hat{A}_{ij}$, defined by this property:
$$\hat{A}_{ij}e_i=e_j\quad \text{and}\quad \hat{A}_{ij}e_k=0\quad \forall k\ne i.$$
This operator corresponds to a matrix of all zeros except for a $1$ at $(i, j)$ place. Now since the operation $\hat{A}\to \hat{A}^\star$ is (conjugate-)linear, you only have to check that the adjoint $\hat{A}^\star_{ij}$ is represented by the transpose of this matrix, and this is a matter of applying the definition.

The above was more or less a hint, see if it convinces you otherwise leave a comment.


问:
Weak Law of Large Numbers Confusion
If I have a sequence of IID random variables $\{X_i\}_{i=1}^{n}$ with $\mathbb{E}[X_i]=\mu$  then the WLLN implies that $\bar{X}_n$ converges to $\mu$. This is where my confusion lies. It is my understanding that RV are deterministic functions $X:\Omega\rightarrow \mathbb{R}$, where the random component comes from $\omega\in\Omega$. If we are defining $\bar{X}_n(\omega)=\frac{1}{n}\sum_{i=1}^{n}X_i({\omega})$, if $X_i$ are iid then it is not assumed that $X_i(\omega)=X_j(\omega)$ so have an average of different values. When we apply the weak law of large numbers in real life we take the average of a sequence of iid random variables, but doesn't the WLLN require that the RV's be evaluated in the same $\omega$, why are we able to assume that this holds. Am I understanding the WLLN wrong?

答:
To clarify a bit more about the role of the $\omega$ here: in this setup generally the space $\Omega$ is defined as a space of sequences and each $X_i$ is the $i$-th projection of each $\omega$.

A concrete way to define it: set a probability measure in $\mathbb{R}$, then this measure will represent the law of each $X_i$. Then the infinite product space $\Omega:=\mathbb{R}^{\infty }:=\prod_{n\in \mathbb N }\mathbb{R}$ can be given with a probability measure induced by the individual probability measures of each $\mathbb{R}$ probability space.

Then, as advised at the beginning, each $\omega\in \Omega$ is just an arbitrary real valued sequence. Then when you evaluates something like $\Pr[\bar X_n\in A ]$ you are measuring a set (of sequences) in $\Omega$. Then informally the weak law of large numbers say us that as $n$ increases then the amount of sequences in $\Omega$ such that the mean of it $n$-th first values are far away of $\mu$ becomes arbitrarily small in measure (i.e., in probability). Formally it assert that

$$
\forall \epsilon >0,\, \forall \delta >0,\,\exists N\in \mathbb N,\, \forall n\geqslant N: \Pr [|\bar X_n-\mu|\geqslant \delta ]<\epsilon
$$

where $\mu:=\operatorname{E}[X_1]$ and $\Pr $ is the probability measure in $\Omega $.

In summary: each $\omega $ represent intuitively a "random sample" (a sequence of values), and remember that $\Pr [X\in A]:=\Pr (\{\omega \in \Omega : X(\omega )\in A\})=\Pr(X^{-1}(A))$, in this case $X=|\bar X_n-\mu|$ and $A=[\delta ,\infty )$, so we are measuring the probability that a random sample belongs to some subset of $\Omega $, in this case we are measuring the probability that, given a sequence, the mean of it $n$-th first values deviates from $\mu$ more than a given amount.

Title: Weak Law of Large Numbers Confusion


问:
Big O for error terms
The following link from wikipedia explains the Big O notation really good. I have only one problem, which is to formalize the usage of Big O notation for error terms in polynomials. In the  example give here  we have
$$
e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...=1+x+\frac{x^2}{2!}+\mathcal{O}(x^3)=1+x+\mathcal{O}(x^2)
$$
as $x\rightarrow 0$. Now we find a similar notation also for the error terms in taylor polynomials. I would like to understand why this is right just formaly.

1.) Why is $\frac{x^3}{3!}+\frac{x^4}{4!}+... = \mathcal{O}(x^3)$

Is this because
$$
\frac{x^3}{3!}\frac{1}{x^3}+\frac{x^4}{4!}\frac{1}{x^3}+\frac{x^5}{5!}\frac{1}{x^3}+...=\frac{1}{3!}+\frac{x}{4!}+\frac{x^2}{5!}+...\leq M, \quad x\rightarrow 0
$$
for some $M$ that has to be bigger than $\frac{1}{3!}$ or can one show this in a different way formaly?

2.) Why is $\frac{x^2}{2!}+\mathcal{O}(x^3) = \mathcal{O}(x^2)$?

I appreciate your help! :)

答:
The intuitive answer to why

$$\frac{x^3}{3!}+ \frac{x^4}{4!} + \cdots = \mathcal{O}(x^3)$$

as $x \to 0$ is that $x^3$ is the term that goes to $0$ at the lowest rate. All higher-order terms go to $0$ faster, so the expression will "at worst" approach $0$ like $x^3$. More precisely, the reason in that

$$|x^3| \geq C|x^4 + x^5 + \cdots |$$
for $x$ small enough. To see this, observe that for $|x|<\frac{1}{2}$, we have

$$\left| x^4 + x^5 + \cdots \right| = \left| \frac{x^4}{1-x} \right| = \left| \frac{x}{1-x} \right| |x^3| \leq C |x^3|$$

since $\frac{x}{1-x}$ is bounded for $|x| < \frac{1}{2}$.

Title: Big O for error terms


问:
Let $U=\{p \in \mathcal{P}_4(\mathbb{R}):p''(6)=0\}$. Find a basis of $U$
Let $U=\{p \in \mathcal{P}_4(\mathbb{R}):p''(6)=0\}$. Find a basis of $U$. I just have a question about how to show that the list that I found is a basis.

Here is what I found:

Let $p \in U$, then $p(x)=a+bx+cx^2+dx^3+ex^4$ with $a,b,c,d,e \in \mathbb{R}$. We determine now $a,b,c,d,e$ using second derivative of $p$ at $6$.

We have that $p''(6)=2c+36d+432e=0 \iff c=-18d-216e$. Thus, each polynomial $p$ of $U$ is of the form: $p(x)=a+bx+d(x^3-18x^2)+e(x^4-216x^2)$. So, the list $(1,x,x^3-18x^2,x^4-216x^2)$ might be a basis of $U$. $(1,x,x^3-18x^2,x^4-216x^2)$ is clearly linearly independent, since they all have different degrees.

But, I don't really see how to show that each $p \in U$ is linear combination of $(1,x,x^3-18x^2,x^4-216x^2)$. Does it follow directly after I found $a,b,c,d,e$ using second derivative?

答:
Your argument is fine: Form $p(x)=a+bx+d(x^3-18x^2)+e(x^4-216x^2)$ you can deduce that ever polynomial in $U$ is a linear combination of your basis; and the linar independence follows form the degree argument.

Another approach is to identificate every polynomial of degree less or equal than 4, with a vector in $\Bbb R^5$ though its coefficientes:
$$a+bx+cx^2+dx^3+ex^4+fx^5\in{\cal P}_4(\Bbb R) \longleftrightarrow (a,b,c,d,e,f)\in\Bbb R^5.$$
So you can take your basis of $U$, convert into $\Bbb R^5$ vectors and try to show that the range of the matrix formed by these vectors has range 4.


答:
When we write $f(x)=O(x^n)$ as $x\to 0$, what this means is that there exist $M>0$ and $\delta>0$ such that for all $|x|<\delta$, we have $\left|f(x)\right|\leq M|x|^n$. Equivalently, there must exist a $\delta>0$ such that
\begin{align}
\sup_{0<|x|<\delta}\left|\frac{f(x)}{x^n}\right|<\infty.
\end{align}
In plain English: the function divided by $x^n$ must be bounded in some punctured neighborhood of the origin. This is just definition.

We have $f(x):= \sum_{k=3}^{\infty}\frac{x^k}{k!}=x^3\left(\sum_{k=3}^{\infty}\frac{x^{k-3}}{k!}\right)$. In this case we are very fortunate because ANY $\delta>0$ works, and there is no need to show the existence of a specific one. This is because if $|x|\leq \delta$ then
\begin{align}
|f(x)|\leq |x^3|\cdot \left(\sum_{k=3}^{\infty}\frac{\delta^{k-3}}{k!}\right)
\end{align}
If you want to pattern-match with the definition, then the bracketed term here is $M$, and this is clearly finite because the entire series defining the exponential converges (absolutely) at every point (heck if one wants a very crude estimate then we can bound the series by $\frac{e^{\delta}}{\delta^3}$, which is clearly finite).

As for 2, I think you should atleast write down your attempt at a proof based on your understanding of the definitions.

Title: Big O for error terms

Body:
The following link from wikipedia explains the Big O notation really good. I have only one problem, which is to formalize the usage of Big O notation for error terms in polynomials. In the  example give here  we have
$$
e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...=1+x+\frac{x^2}{2!}+\mathcal{O}(x^3)=1+x+\mathcal{O}(x^2)
$$
as $x\rightarrow 0$. Now we find a similar notation also for the error terms in taylor polynomials. I would like to understand why this is right just formaly.

1.) Why is $\frac{x^3}{3!}+\frac{x^4}{4!}+... = \mathcal{O}(x^3)$

Is this because
$$
\frac{x^3}{3!}\frac{1}{x^3}+\frac{x^4}{4!}\frac{1}{x^3}+\frac{x^5}{5!}\frac{1}{x^3}+...=\frac{1}{3!}+\frac{x}{4!}+\frac{x^2}{5!}+...\leq M, \quad x\rightarrow 0
$$
for some $M$ that has to be bigger than $\frac{1}{3!}$ or can one show this in a different way formaly?

2.) Why is $\frac{x^2}{2!}+\mathcal{O}(x^3) = \mathcal{O}(x^2)$?

I appreciate your help! :)


问:
Prove $\frac{a^{2}}{b^{2}} +\frac{b^{2}}{c^{2}} +\frac{c^{2}}{a^{2}} +\frac{15abc}{4}\geq \frac{27}{4}$ for $a^{2}+b^{2}+c^{2}+abc=4$

Prove:
$$\frac{a^{2}}{b^{2}} +\frac{b^{2}}{c^{2}} +\frac{c^{2}}{a^{2}} +\frac{15abc}{4}\geq \frac{27}{4}$$ for $a,b,c>0$ such that $$a^{2}+b^{2}+c^{2}+abc=4.$$

I tried to note $a=\cos A$,... because of a known identity, I used $s=\frac{a^{2}}{b^{2}} +\frac{b^{2}}{c^{2}} +\frac{c^{2}}{a^{2}} \geq \frac{a}{b} +\frac{b}{c} +\frac{c}{a}$ and $s\geq \frac{a}{c} +\frac{b}{a} +\frac{c}{b}$, and then I added them. Another suggestion? Please!
I also noted $s=a+b+c,p=ab+bc+ac, r=abc$. The condion is equivalent with $s^2-2p+r=4$. Using the 2 inequalities which I mentioned above $\implies s\ge(a+b)(b+c)(a+c)/(2abc)-1$, and then is enought to prove that $2sp+15r^2\ge33r$, but I don't know if the last one is correct inequality. You can try to prove that. I'm desperate.

答: With the following link $a^2+b^2+c^2+abc=4 \Rightarrow abc \leq 1$ thus the maximum value is $1$ and we assume that $a=1,b=1,c=1$. $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{15abc}{4}$, we substitut $1$ we get exactly $\frac{27}{4}$, this is the maximum value that we could get. If $a,b,c$ is less that $1$ we could not reach $\frac{27}{4}$.So $\mathbf{a,b,c=1}$.


问:
What is a proof of Euler's formula $e^{ix}=\cos(x)+i \sin(x)$ that doesn't use differentiation or De Moivre's theorem, if it exists at all?
The Taylor series expansions of $e^x$, $\sin(x)$ and $\cos(x)$ are the standard way to prove it, but does there exist an elementary way that doesn't use limits, calculus or De Moivre's theorem?

答: The number $e$ is transcendental, so it seems impossible to obtain any non-trivial result about it that does not involve infinite processes (limits, derivatives, integrals...). If you want to define the exponential function for irrational arguments using the basic operations, it indeed seems difficult to avoid polynomials or rational functions, and these must be of infinite degree. Finally, to establish the theorem, you need some definition of the trigonometric functions that is not based on the complex exponential. For this you need to resort to well-axiomatized geometry and to establish the equivalence with statements in calculus. In the end, it boils down to computing the arc length along a circle, hence establish $\pi$, another transcendental number.


问:
Matrix with the same value in all entries — which notation to use?
Pretty trivial but for a matrix

\begin{bmatrix}x&x&\ldots&x&x\\x&x&\ldots&x&x\\\vdots&\vdots&\ddots&x&x\\x&x&\ldots&x&x\end{bmatrix}

with  $N\times N $, is there is simpler notation. I had $A=[x_{i,j}]\in \Bbb R^{N\times N}$ in my mind, but I can't find anywhere were it is used

Edit: I forgot to mention but, I need to work with the elements in the matrix, so like doing scalar multiplication (to do a proof by the principle of mathematical induction question). But it a $3\times 3$ so I figured out I do have to write every single element 9 times in a row. Is there any notation that can make my math simpler?

答:
Assuming that $x \neq 0$, it's a rank-$1$ matrix. Thus, to emphasize its rank-$1$-ness, I would use

$$\color{blue}{x \,{\Bbb 1}_n {\Bbb 1}_n^\top}$$


问:
what will happen to the slope angle if coordinate points changes?
Here is the figure:

Now the slope angle is calculated as:

slope angle \begin{align}
\theta = \tan^{-1}(\frac{Y2 - Y1}{X2 - X1})
\end{align}

Now if I change the notations of point A and B as:

then which angle will the $\theta$? Will it be same or 180 minus the previous $\theta$?

Please let me know what angle will I get as per second figure.

答:
There are two bits of fundamentals that may be causing you some confusion.

One, the angle of inclination is the angle measured by rotating counter-clockwise from the horizontal.  Suppose you have a horizontal line $\ell$.  The angle that you must rotate counter-clockwise to make $\ell$ parallel to your given line is the angle of inclination.

Two, a little bit of work dealing with algebra:  $$\frac{a-b}{c-d}=\frac{-(b-a)}{-(d-c)}=\frac{b-a}{d-c}.$$
As it appears, it does not matter which set of coordinates you denote as $A$ or $(x_1,y_1)$ or what have you.


问:
Why is the distribution of values of this random matrix product (seemingly) independent of dimension?
I'm investigating the behavior of the value $|x A y|$ where $x, y \in \mathbb{C}^N$ have unit 2-norm and are uniformly sampled from the unit ball, and $A \in \mathbb{C}^{N \times N}$ has elements sampled from the complex standard normal distribution.

When I plot an empirical histogram of the magnitude of this product for varying dimension $N$, it seems that the distribution of values is independent of $N$. For example, in Julia:

```<code>julia> using Random, Plots, LinearAlgebra
julia> CF = Complex{Float32}
julia> histogram(norm.([normalize(randn(CF, n))' * randn(CF, (n, n)) * normalize(randn(CF, n)) for _ in 1:2000, n in [1, 8, 64, 512]]), layout = (4, 1))
</code>```

gives this result:

Is there a proof or at least an intuition on why this would be the case? I tried reading a bit on random matrix theory but it was a bit too advanced for me and I'm guessing what's going on here isn't that complicated.

答:
I think I've got it.

The value $x A y$ is equal to $\sum_{i,j \in [N]} x_i A_{i,j} y_j$. The elements of $A$ are random values taken from the complex standard normal distribution, so it's already obvious that $x A y$ is a zero-mean symmetric complex normal distribution.

We can consider the individual elements of $x$ and $y$ as random variables as well. The constraint that $\|x\| = \|y\| = 1$ means that the individual elements of $x$ and $y$ considered as random variables have expectations of their magnitudes $\mathbb{E}[|x_i|] = \mathbb{E}[|y_i|] = \sqrt{\frac{1}{N}}$, and $x_i, y_i$ being independent means $\mathbb{E}[|x_i y_i|] = \frac{1}{N}$.

So,

$x A y \sim \sum_{i,j \in [N]} x_i y_j \mathcal{CN}(0, 1) \sim \sum_{i \in [N^2]}\frac{1}{N} \mathcal{CN}(0, 1) = \sum_{i \in [N^2]} \mathcal{CN}(0, \frac{1}{N^2}) = \mathcal{CN}(0, 1)$

Which means $|xAy|$ is the magnitude of a complex standard normal distribution, which is distributed according to the Rayleigh distribution, and which is independent of $N$.


问:
$ X>0, E(X)=1, E(X^2)=b, \forall a\in (0,1): P(X>a)\geq \frac{(1-a)^2}{b} $
Suppose that
$ X>0, E(X)=1, E(X^2)=b $

And We should prove for every $ a $ such that $ 0 < a < 1 $ the following statement:

$ P(X>a)\geq \frac{(1-a)^2}{b} $

This is a preliminary course of probability and we learned only basic formulas and inequalities of Markov and Chebyshev's. I would be happy if you keep the answer simple as much as possible.
Thanks.

答:
Write
$$
1=\mathsf{E}X=\mathsf{E}X1\{X>a\}+\mathsf{E}X1\{X\le a\}.
$$
Since $\mathsf{E}X1\{X\le a\}\le a$ and $\mathsf{E}X1\{X>a\}\le \sqrt{\mathsf{E}X^2\mathsf{P}(X>a)}$ by Cauchy-Schwartz,
$$
1-a\le \sqrt{\mathsf{E}X^2\mathsf{P}(X>a)},
$$
which implies the result.


问:
Prove that the equation $3x-\sin(2x)=0$ has only one solution
I`m trying to prove that this equation has only one solution$$3x-\sin(2x)=0$$
I need some advice how to do that, Thanks!

EDIT

答: By intermediate value theorem it clearly has at least one solution (or just check by plugging in $x=0$). Now the derivative of $3x-\sin(2x)$ is $3-2\cos(2x)>0$, hence the function is strictly increasing, hence it can cross the x-axis only once, ergo only one solution.</p>


答:
Let $\mathbb{1}_A$ be the indicator function of set $A$. Then for any $a\in(0,1),$ it is easy to see that
$$X \mathbb{1}_{\{X>a\}}\ge X-a.$$ Taking the expectation of both sides yields
$$E[X\mathbb{1}_{\{X>a\}}]\ge E[X]-a=1-a. \qquad(1)$$

Moreover, by the Cauchy-Schwarz inequality,
$$E[X \mathbb{1}_{\{X>a\}}]\le \sqrt{E[X^2]}\sqrt{P\{X>a\}}. \qquad(2)$$

Finally, from $(1)$ and $(2)$, we deduce

$$\sqrt{E[X^2]}\sqrt{P\{X>a\}}\ge 1-a \implies P\{X>a\}\ge \frac{(1-a)^2}{E[X^2]}=\frac{(1-a)^2}{b}.$$

Title: $ X>0, E(X)=1, E(X^2)=b, \forall a\in (0,1): P(X>a)\geq \frac{(1-a)^2}{b} $

Body:
Suppose that
$ X>0, E(X)=1, E(X^2)=b $

And We should prove for every $ a $ such that $ 0 < a < 1 $ the following statement:

$ P(X>a)\geq \frac{(1-a)^2}{b} $

This is a preliminary course of probability and we learned only basic formulas and inequalities of Markov and Chebyshev's. I would be happy if you keep the answer simple as much as possible.
Thanks.


问:
The coincidence of permutations
Let $K$ be a field and let $V$ be a finite-dimensional $K$-vector space.

(a)Assume that $K$ is infinite. Show that $V$ is not the union of finitely many proper linear $K$-subspaces.

(b)Assume that $K$ is finite and $V$ is non-zero. Let $S$ be the set of affine hyperplanes of $V$. Let $g:V \longrightarrow \mathbb{R} $ be a function. The Radon transform $Rg:S \longrightarrow \mathbb{R} $ is defined by $(Rg)(\zeta)=\sum_{x \in \zeta}g(x)$ for $\zeta \in S$. Show that $Rg=0$ implies $g=0$.

(c)Let $v_1 , \cdots , v_n , w_1 , \cdots , w_n \in V$. Assume that for every $K$-linear map $f:V \longrightarrow K,(f(v_1) , \cdots , f(v_n))$ and $(f(w_1) , \cdots , f(w_n))$ coincide up to permutation of the indices. Deduce that $(v_1 , \cdots , v_n)$ and $(w_1 , \cdots , w_n)$ coincide up to permutation of the indices. Here we make no assumptions on $K$.

Assume (a) and (b) is proved.
For (c), in the infinite case, by subtacting $v_1$ to all vectors, this is equivalent to prove:
'If $w_1 , \cdots , w_n \in V$ are such that for each $f\in V^*$ one of the $f(w_j)=0$, then one of the $w_j=0$.'
Then we'll have $V^*=\cup_{i=1}^nAnn(w_i)$, and by (a) there exists a $j$ s.t. $Ann(w_j)=V^*$.

With this hint I tried to prove the finite case but this method is now no longer useful. I suppose this needs to use (b) but I don't know how to use it. How to solve the finite case of (c) ?

答:
For the case when $K$ is finite, we can prove by induction on $n$. If $n=1$ this is trivial. If $v_i= w_j$ for some $i, j$, then we can apply induction hypothesis to conclude the proof. So we may assume that $v_i\neq w_j$ for any $i, j$ and we are going to find a contradiction. Define $g:V\to \mathbb{R}$ by $g(v_i)=1$, $g(w_j)=-1$, and $g(x)=0$ for other $x\in V$.
Then for any affine hyperplane $\xi$,
$$Rg(\xi)=\#\{v_i\mid v_i\in \xi\}-\#\{w_j\mid w_j\in \xi\}.$$
Recall that $\xi$ is defined by $\{x\in V\mid f(x)=a\}$ for some $f\in V^\vee$ and $a\in K$.
Then the assumption implies that $Rg(\xi)=0$ for each $\xi.$
This implies that $g=0$, a contradiction.

Title: The coincidence of permutations


问:
Altitude lines dont appear to be perpendicular to triangle sides in botched orthocenter calculation attempt
I am trying to calculate the orthocenter of this triangle with vertices  v1 = (42.9269, -307.8938), v2 = (-3816.060, 5723.502), v3 = (1766.106, 1973.672). My math seems to be correct, but the altitudes of the triangle do not appear to be perpendicular to the sides of the triangle.

The algorithm:

Find the slope $(dv_1v_2)$ for line between vertices $v_1$ and $v_2$

Find the slope $(dv_1v_3)$ for line between vertices $v_1$ and $v_3$

Calculates the negative reciprocal of both  $dv_1v_2$ and $dv_1v_3$ as these should be the slopes of the triangle altitudes $(m_{alt})$

Calculate $y$-Intercepts of both triangle altitudes from vertice 2 and 3: $I_{alt2}, I_{alt3}$

Find the intersection of these altitudes

Given the equations for the altitude lines:

$$
y = m_{alt2}\cdot x + I_{alt2} \\
y = m_{alt3}\cdot x + I_{alt3}
$$
Where $I_{alt}$ is the intercept of the altitude line given by:
$$
I_{alt3} = v_{3y} - m_{alt3} \cdot v_{3x}\\
I_{alt2} = v_{2y} - m_{alt2} \cdot v_{2x}
$$

And $m_{alt}$ is the slope of the altitude line
. Then the equation for the intersection of these two altitude lines is:
$$
\begin{bmatrix}
    - m_{alt3} & 1  \\
    - m_{alt2} & 1
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix} =
\begin{bmatrix}
I_{alt3} \\ I_{alt2}
\end{bmatrix}
$$

I wrote a code to compute it in R:

```<code>v1 <- c(42.9269, -307.8938)
v2 <- c(-3816.060, 5723.502)
v3 <- (1766.106, 1973.672)
dv1v2 <- (v2[2] - v1[2]) / (v2[1] - v1[1])
dv1v3 <- (v3[2] - v1[2]) / (v3[1] - v1[1])
dv3B <- -1 / dv1v2
dv2A <- -1 / dv1v3

interceptV3B <- v3[2] - dv3B * v3[1]
interceptV2A <- v2[2] - dv2A * v2[1]
A <- matrix(c(-dv3B, 1, -dv2A, 1), byrow = TRUE, nrow = 2)
B <- matrix(c(interceptV3B, interceptV2A), byrow = TRUE, nrow = 2)
orthocenter <- solve(A) %*% B
</code>```

For the given points, my results for the orthocenter are (1431.956 1759.878). An online calculator verified the results.

Here is an image I plotted, followed by an image from an online calculator that matches.

My results (yellow is orthocenter):

Online calculator results match mine (red is orthocenter):

None of these altitudes look perpendicular to me!

This is what I expect an altitude to look like, every altitude is 90 degrees in this image:

I verified that the altitude and side between vertices is perpendicular with the formula: $angle = atan(abs((slope1-slope2)/(1+slope1*slope2)))$. This shows that the angle is $\frac{pi}{2}$, yet none of the altitudes look 90 degrees, so I don't believe it.

Why don't these altitude lines look 90 degrees to their opposite sides? Looking at the image, I would think the orthocenter would be closer to $x = 0$

Orthocenter calculator for reference: https://www.mathportal.org/calculators/analytic-geometry/triangle-calculator.php

答:
Correct drawing:

Correct drawing, angle at $v_3$ is about $86^o$


问:
Calculating the roots of a complex polynomial
the problem is the following:

Given a complex polynomial P = z4 - $\alpha$, with $\alpha \in \mathbb {C} $.  We know that $ e^{i\frac{2\pi}{3}} $ is a complex root of 2P.

Select all the correct options:

(A)  1 is a complex root of P

(B)  $ e^{i\frac{\pi}{6}} $ is a complex root of P

(C)  - $\sqrt{3}$ - $i$ is a complex root of 2P

(D)  $\sqrt{2}$ + $ i\sqrt{2}$ is a complex root of 2P

(E)  None of the above

I think that subtracting $ \alpha $ to z4 and then multiply by 2 will not change the number of complex roots (which is 4). So I divided 2$\pi$ by 4 to get the angle and calculated the other roots using the $ e^{i\frac{2\pi}{3}} $. After this I had to exclude (C) and (D) because the modulus didn't match: | - $\sqrt{3}$ - $i$ | = |$\sqrt{2}$ + $ i\sqrt{2}$| = 2 $\neq$ 1

Then I calculated the complex number 2P wich is equal to $ (e^{i\frac{2\pi}{3}})^4 $ and from that I got P and its complex roots.
I finally excluded (A) and (B) because 1 is a positive real number and therefore the only option is $ e^{i0 + 2k\pi} $ (which isn't a root of P) and $ e^{i\frac{\pi}{6}} $ is not a root of P either.

After this, there's only one option left. Option (E)

Can you tell me if my logic is correct? And if not, what am I missing here?

Thank you

答:
$\exp\left(i\frac{2\pi}3\right)$ is a root of the polynomial $2P$

$\implies$ $\exp\left(i\frac{2\pi}3\right)$ is a root of the polynomial $P$

$\implies$ $\left(\exp\left(i\frac{2\pi}3\right)\right)^4-\alpha=0$

$\implies$ $\alpha=\exp\left(i\frac{8\pi}3\right)=\exp\left(i\frac{2\pi}3\right)$

$\implies$ $P=z^4-\exp\left(i\frac{2\pi}3\right)$

$\implies$ Option (B) is true.


答:
Let $V,W$ be complex, finite dimenstional Hilbert spaces, $A \colon V \to W$ a linear operator. Let $(e_1, \ldots, e_n)$ and $(f_1, \ldots, f_m)$ be orthonormal bases of $V$ and $W$ respectively. If then $(\alpha_{ij})$ is the matrix representation of $A$ with respect to these bases, we have
$$ Ae_i = \sum_j \alpha_{ij}f_j, \quad 1 \le i \le n $$
Or, as $(f_j)$ is orthonormal
$$ (Ae_i, f_j) = \alpha_{ij}, \quad 1 \le i \le n, 1 \le j \le m $$
Let $(\beta_{ij})$ be the matrix representation of $A^\dagger$, we have then, as above
$$ A^\dagger f_i = \sum_i \beta_{ij}e_j $$
or $$ (A^\dagger f_i, e_j) = \beta_{ij}, \text{ each $i,j$} $$
We get, by definition of the adjoint, that
\begin{align*}
  \beta_{ij} &= (A^\dagger f_i, e_j)\\
             &= (f_i, Ae_j)\\
             &= \overline{(Ae_j, f_i)}\\
             &= \overline{\alpha_{ji}}
\end{align*}
So $(\beta_{ij})_{i,j} = (\bar \alpha_{ji})_{i,j}$, as wanted.</p>

Title: Matrix representation of the adjoint of an operator, the same as the complex conjugate of the transpose of that operator?

Body:
Since I'm not taking summer classes I decided to do some self learning on more advanced mathematics, and I've found myself stuck on this problem:

I have to show that for any operator $\hat{A}$ the matrix representation of the adjoint $\hat{A}\dagger$ is given as the complex conjugate of the transpose of the matrix representation of the operator $\hat{A}$.

So basically show that
$\underline{A}\dagger = (\underline{A^T})^*$

I've done many examples on the textbook using actual matrices, but I don't really know how to prove this in terms of operators, any hint or where to start, or some guidance?
Thank you


问:
why for $f(x) =1/(1+1/x)$ the curve pass through the origin
HI thank's for the future replies.
My question is simple : why for $f(x) =1/(1+1/x)$ the curve pass through the origin so it means $f(0)=0.$
And why $0^{-1}$ is infinite? it's what's telle geogebra application.

答: For $f(x)$ as you defined it, the value $f(0)$ is undefined.


答:
I think we need to go in the opposite direction to get an understanding of why to choose (yes we actually get to choose) to expand the real exponential function to have the property that $$e^{i\phi} = \cos(\phi) + i\sin(\phi)$$

In the analysis of one complex variable there is introduced the concept of analytic continuation. As far as I remember this was the first good explanation of why to choose to expand the exponential function in this way.

If we do not put any particular restrictions on how well behaved a function shall be when we expand it's domain, then a lot of weird things can happen. The degrees of freedom of how to expand functions is basically limitless. There exist some really beautifully and disturbingly complicated functions in the wild.

But if we want to preserve well behavedness in the sense of complex analyticity then we will need to choose the one which obeys De Moivre.

If we disregard differentiation altogether, then we suddenly have no particular reason to define the real valued exponential function to be smooth at all, and there's no reason as to why the number $e$ would be special. Remember what makes the number $e$ special is that $$\mathfrak D(e^{x}) = e^x$$

We could use base two instead and expand from integers to reals with a floor function or something. Without differentiability, why should our function be smooth?

Title: What is a proof of Euler's formula $e^{ix}=\cos(x)+i \sin(x)$ that doesn't use differentiation or De Moivre's theorem, if it exists at all?

Body: The Taylor series expansions of $e^x$, $\sin(x)$ and $\cos(x)$ are the standard way to prove it, but does there exist an elementary way that doesn't use limits, calculus or De Moivre's theorem?


答: First question:The function can be written as $f(x) =\frac{x} {x+1}$. So $f(0)=\frac{0}{0+1}=0$

Body:
HI thank's for the future replies.
My question is simple : why for $f(x) =1/(1+1/x)$ the curve pass through the origin so it means $f(0)=0.$
And why $0^{-1}$ is infinite? it's what's telle geogebra application.