问:
Proof that the orbits of the action of the Galois group on the roots correspond to the irreducible factors
I've spent around 5 hours on this single proof but I still don't fully get it.

Statement (some terms might be unusual/imprecise because I'm translating, sorry about that).

"Let $L$ be a splitting field of a polynomial $f \in K[X]$. The orbits of the left action of $Gal(L|K)$ on $\{1,...,n \}$ correspond exactly to the monic irreducible factors of $f$. In particular, the action is faithful $\iff$ $f$ is irreducible."

I assume the set $\{1,...,n \}$ identifies the roots $\{a_1,...,a_n \} $ of f.

Proof:
Let $A \subset \{1,...n \}$ and set $f_A(X)=\prod_{ i \in A}(X-a_i)$.
$$ \forall \gamma \in Gal(L|K): \gamma (f_A(X))=\prod_{ i \in A}(X-a_{\gamma \cdot i})= \prod_{I \in \gamma A}(X-a_i)=f_{\gamma A}(X)$$
Hence
$$f_A(X) \in K[X] \iff \forall \gamma \in Gal(L|K): \gamma (f_A(X))=f_A(X) \iff  \forall \gamma \in Gal(L|K): \gamma A = A \iff A \,  is \; Gal(L|K) \, invariant$$
End of the proof.

I understand all of the above steps, however I do not understand the proof idea. We pick this polynomial $f_A$. Then we show that this polynomial is in $K[X]$ if and only if $A$ is $Gal(L|K)$ invariant, hence $A$ is an orbit. Is then $f_A$ automatically irreducible over $K$? Why? The biggest problem is that I don't see any correlation with irreducibility in this whole thing.

Thank you so much in advance.

答:
$A$ need not be an orbit in order for $\gamma A = A$; it could be a union of orbits. So the final set of equivalences above shows that any union $A$ of orbits produces a polynomial $f_A\in K[X]$. It follows that if $A$ contains more than one orbit, then $f_A$ is reducible.

Conversely, suppose that $A$ consists of a single orbit, and write $f_A = gh\in K[X]$. Then $g = f_B$ for some $B\subset A$. But this forces $B=A$ since otherwise $\gamma B\ne B$.


问:
Integrating $x^x$ and getting a graph
I've heard many times of functions that cannot be integrated. For example, $x^x$, which is the most common. But what I don't know is how could you, even if the graph has no equation, plot this integral. If anyone could give me a graph, I would be extremely pleased. Or at least, could you tell me how to graph it? Maybe in Wolfram Mathematica?</p>

答:
Here is an interesting thread for you: http://www.physicsforums.com/showthread.php?t=77605

Quote: The integral of x^x between 0 and 1 was calculated by Johann Bernoulli in 1697 using power series (not Euler). The proof appears in "Opera Omnia" vol. 3 (1697) pp. 376 - 381.

ContentLicense: CC BY-SA 3.0

Title: Integrating $x^x$ and getting a graph


问:
Matrix inequality of inverses of sum of two matrices
$A$ and $B$ are two real symmetric non-singular matrices but not necessarily sign definite. Suppose $A+B$ is non-singular. I want to find a $\lambda \in \Bbb R$ such that $$(A+B)^{-1}\leq(A+\lambda I)^{-1}$$ where $A\leq B$ means $A-B$ is negative semi-definite. Does such $\lambda$ exist? How can I find it?

I have tried the identity $(A+B)^{-1}=A^{-1}-A^{-1}(B^{-1}+A^{-1})^{-1}A^{-1}$. But it led me to

$$(A^{-1}+\lambda^{-1} I)^{-1}\leq(A^{-1}+B^{-1})^{-1}$$

Thanks in advance!

答:
No, you cannot always do that. Here's a counterexample. Let $\epsilon > 0$ and
$$
A = \begin{pmatrix} 1 & 0 \\ 0 & - \epsilon \end{pmatrix} \qquad B = \begin{pmatrix} - \epsilon & 0 \\ 0 & 1 \end{pmatrix}.
$$
Then
$$
(A + B)^{-1} = \frac{1}{1-\epsilon}I
$$
and
$$
(A + \lambda I)^{-1} = \begin{pmatrix} \frac{1}{1+\lambda} & 0 \\ 0 & \frac{1}{\lambda - \epsilon} \end{pmatrix}.
$$
Thus
$$
\begin{aligned}
(A + \lambda I)^{-1} - (A+ B)^{-1}  &= \begin{pmatrix} \frac{1}{\epsilon-1} + \frac{1}{1+\lambda} & 0 \\ 0 & \frac{1}{\epsilon-1} + \frac{1}{\lambda-\epsilon} \end{pmatrix} \\
\end{aligned}
$$

But then this latter matrix is positive semidefinite iff its diagonal entries are both non-negative. One can then show that if $\epsilon \in (0, 1)$ no such $\lambda$ exists such that they are both non-negative.

ContentLicense: CC BY-SA 4.0


问:
Why $\frac{y’}{y}$ is equivalent to $(\ln|y|)’$?
I just read a proof process from a textbook about differential equations, it has following content :
$$y’= ay\implies\frac{y’}{y}= a\implies (\ln|y|)’= a$$
I want to know why $\frac{y’}{y}$ is equivalent to $(\ln|y|)’$?

答:
$$y'=ay$$
$$\dfrac {dy}{dx}=ay$$
This is separable:
$$\int \dfrac {dy}{y}=a\int dx$$
$$\ln y=ax+C$$
Differentiate:
$$(\ln y)'=a$$


问:
Using combinations to combinations for probability with dependent events
I am reading a problem about probabilities with dependent events.

There are $3$ people selected from $5$ freshmen, $2$ sophomores and
$4$ juniors. What is the probability that the first $2$ people
selected are freshmen and the third a junior?

I thought that this is solvable by using combinations.

I.e.

There are $\binom{11}{3}$ ways to select $3$ out of $11$ students (I hope that the notation is for combinations and not permutations as I get these 2 notations mixed).

This is $$\frac{11!}{3!\cdot 8!} = 165$$ ways

To choose $2$ freshman from $5$ is: $$\frac{5!}{2!\cdot 3!}$$
To choose $1$ junior from $4$ is: $$\frac{4!}{1!\cdot 3!}$$

So the probability we are looking for is:

$$\frac{\frac{5!}{2!\cdot 3!} \cdot \frac{4!}{1!\cdot 3!}}{165} = \frac{40}{165}=\frac{8}{33}$$

but the answer is wrong.

Is it not possible to use the combinations formula for calculating such a probability with $3$ dependent events?

答:
When a particular order is specified, it is simplest to directly multiply probabilities,

so $\dfrac5{11}\cdot\dfrac4{10}\cdot\dfrac4{9}$

If you want to use combinations, you will have to take an extra step which unnecessarily lengthens the process, viz divide by $\dbinom32$,

thus $\dfrac{\dbinom52\dbinom41}{\dbinom{11}3}\div\dbinom32$


问:
Find the greatest integer $N$ such that...

Find the greatest integer $N$ such that $N<\dfrac{1}{\sqrt{33+\sqrt{128}}+\sqrt{2}-8}$.

The way I did it is this: first, I rewrote the biggest square root as $\sqrt{1+2*16+8\sqrt{2}}$. Then I made the substitution $x=\sqrt{2}$, so this became $\sqrt{16x^2+8x+1}$, which factors to $\sqrt{(4x+1)^2}$ which is equal to $4x+1$. Then I substituted back: $\dfrac{1}{4\sqrt{2}+1+\sqrt {2}-8}$, or $\dfrac 1{5 \sqrt{2}-7}$. This is where I have a problem. I am able to find the answer because I know that $1.4<\sqrt 2 <1.5$. However, is there any way to find this answer without knowing this? Thanks!

答:
Picking up from where you left off, we have:
$$ \begin{align*}
N &= \left\lfloor \dfrac{1}{5\sqrt{2}-7} \right\rfloor\\
&= \left\lfloor \dfrac{1}{5\sqrt{2}-7} \cdot \dfrac{5\sqrt{2}+7}{5\sqrt{2}+7} \right\rfloor\\
&= \left\lfloor \dfrac{5\sqrt{2}+7}{50-49} \right\rfloor\\
&= \lfloor 5\sqrt{2}+7 \rfloor\\
&= \lfloor \sqrt{50}+7 \rfloor\\
&= \lfloor \sqrt{50} \rfloor+7  \\
&= \sqrt{49}+7 \\
&= 14 \\
\end{align*}$$</p>


问:
Prove that $\lim_{(x, y) \to (0, 0)} \frac{xy}{\sin(x^2+y^2)}$ does not exist.
I'm trying to prove that $$\lim_{(x, y) \to (0, 0)} \frac{xy}{\sin(x^2+y^2)}$$ does not exists.

To do so, I'm trying to find two different sequences $\{\bar{a_n}\}$, $\{\bar{b_n}\}$ such that $\{\bar{a_n}\} \to (0, 0), \{\bar{b_n}\} \to (0, 0)$ and $$\lim_{n\to\infty} f(\bar{a_n}) = L_1 \not = L_2 = \lim_{n\to\infty} f(\bar{b_n})$$
Def. Let $f: A\subseteq \mathbb{R^2} \to \mathbb{R}, (x_0, y_0) \in \mathbb{R^2}$ accumulation point of $A.$ We say that $$\lim_{(x, y) \to (x_0, x_0)} f(x, y) = L \Longleftrightarrow$$ if $\forall \{\bar{a_n}\}$ such that $\forall n\in\mathbb{N}, \bar{a_n} \in A, \bar{a_n}\not = (x_0, y_0)$ and $\lim_{n \to \infty} \bar{a_n} = (x_0, y_0),$ then $$\lim_{n \to \infty} f(\bar{a_n}) = L$$

答:
The limits along the lines $y=kx$ exist but are not all equal.

If you want to take sequences, take $a_n^{(k)}=(1/n, k/n)$.


问:
Set theory self taught (check if my logic is okay)
I am a beginner in set theory and I really do not feel confident writing proofs so a help would be much appreciated. I am not a mathematician by any stretch but I always felt like math is an essential tool and so I am setting myself into this adventure of learning it.
Here's where I am struggling:

Problem: let $R$ be a binary relation Proof that

$R^{-1}[R[A]]\supseteq A\cap domR$

with $domR$ being the domain of $R$ and $ranR$ being its range

My proof :) "Don't laugh"

$R^{-1}[R[A]]=\{x:x \; \in  dom R \; and \; \exists y\;(y \in R[A] \; and \; xRy)\}$

which mean

$R^{-1}[R[A]]=\{x:x \; \in  dom R \; and \; \exists y \; \exists x' \; (y \in ran R \; and \; x' \in A  \; and \; \; x'Ry\; and \; \; xRy)\}$
Now let

$x \in A \cap domR $ then

$x \in domR \; and \; \exists y \exists x'(y \in ran R\; and \;x'=x \in A\; and \;xRy) $
Thus

$x \in R^{-1}[R[A]]$

Should I give up and do something more productive with my life?

答:
It's best to start on the right because we want to show the inclusion from right to left. And the proof then follwos from definitions in a simple way:

Let $x \in \text{dom} R \cap A$. So $x \in A$ and $x \in \text{dom} R$ which implies that $xRy$ for some $y$. This $y \in R[A]$ by definition, which then again by definition makes that $xRy$ is a witness to the fact that $x \in R^{-1}[R[A]]$. So we're done showing the inclusion.


问:
Quadratic inequality with Modulus of a Linear Equation : Solution Verification & Clarification Needed
$x^2-|5x-3|-x<2$

For this problem I took two cases where in one case $5x-3 \geq 0$ and in second case $5x-3 \lt 0$.

$1st$ Case  when $5x-3 \geq 0$ :-

For this case I got the solution set as $x \in [0.6,3+2\sqrt{2}]$.

$2nd$ Case  when $5x-3 \lt 0$ :-

For this case I got the solution set as $x \in [-5,0.6]$.

Upon combining both the cases I am getting the solution set as $x \in [-5,3+2\sqrt{2}]$ but in the answer that has been provided for this problem is not including the points and thus the answer given is $x \in (-5,3+2\sqrt{2})$.

What am I doing wrong here that is making my endpoints of the solution inclusive whereas the the solution set provided has got endpoints exclusive. Please help me on this !

Thanks in advance !

答:
Let me forcus on when $5x-3<0$, we have $x < \frac35$ and

$$x^2-|5x-3|-x<2$$

becomes

$$x^2-(3-5x)-x<2$$

$$x^2+4x-5<0$$

$$(x+5)(x-1)<0$$

Hence we have $-5 < x < 1$ and $x < \frac35$, hence taking intersection, we have $-5<x < \frac35$.

To answer the question, why did we exclude $x=-5$, consider $(x+5)(x-1)<0$, if we substute $x=5$ inside, we get $0<0$ which is not true.

If $5x-3 \ge 0$, we have $x \ge \frac35$ and

$$x^2-|5x-3|-x < 2$$

becomes

$$x^2-5x+3-x < 2$$

$$x^2-6x+1 < 0$$

$$\frac{6-\sqrt{36-4}}{2} < x < \frac{6+\sqrt{36-4}}{2} $$

$$3-2\sqrt2< x < 3+2\sqrt2 $$

Taking intersection with $x \ge \frac35$, we have

$$\frac35 \le x < 3+2\sqrt2$$

Taking union, we have $-5 < x < 3+2\sqrt2$.


答: Take $a_n=\left(\frac1n,0\right)$ and your limit will be $0$. But if you take $b_n=\left(\frac1n,\frac1n\right)$, then your limit will be $\frac12$.

Title: Prove that $\lim_{(x, y) \to (0, 0)} \frac{xy}{\sin(x^2+y^2)}$ does not exist.

Body:
I'm trying to prove that $$\lim_{(x, y) \to (0, 0)} \frac{xy}{\sin(x^2+y^2)}$$ does not exists.

To do so, I'm trying to find two different sequences $\{\bar{a_n}\}$, $\{\bar{b_n}\}$ such that $\{\bar{a_n}\} \to (0, 0), \{\bar{b_n}\} \to (0, 0)$ and $$\lim_{n\to\infty} f(\bar{a_n}) = L_1 \not = L_2 = \lim_{n\to\infty} f(\bar{b_n})$$
Def. Let $f: A\subseteq \mathbb{R^2} \to \mathbb{R}, (x_0, y_0) \in \mathbb{R^2}$ accumulation point of $A.$ We say that $$\lim_{(x, y) \to (x_0, x_0)} f(x, y) = L \Longleftrightarrow$$ if $\forall \{\bar{a_n}\}$ such that $\forall n\in\mathbb{N}, \bar{a_n} \in A, \bar{a_n}\not = (x_0, y_0)$ and $\lim_{n \to \infty} \bar{a_n} = (x_0, y_0),$ then $$\lim_{n \to \infty} f(\bar{a_n}) = L$$


问:
Is every finite field of the form $GF(p^2)$ isomorphic to $(\mathbb{Z}/p\mathbb{Z})[i]$?
I heard that finite fields are either $GF(p)$, or $GF(p^k)$, with $p$ prime and $k > 1$. One obtains a $GF(p^k)$ through an algebraic quotient of polynomials over $GF(p)$; the polynomial ideal used in the quotient should itself be based on an irreducible polynomial. For $k = 2$, the irreducible polynomials are of the form $X^2 - r$, which can be reduced to $X^2 + 1$; and we get spaces isomorphic to $(\mathbb{Z}/p\mathbb{Z})[i]$. But I was wondering if $(\mathbb{Z}/p\mathbb{Z})[j]$, with $j^2 = 1$ like the split-complex numbers, were also a (different) solution, with $X^2 - 1$ as the generator for the ideal. If not, why not?

答:
Concretely: for $f\in \Bbb{F}_p[x]$ irreducible then $\Bbb{F}_p[x]/(f(x))$ is an integral domain with $p^{\deg(f)}$ elements, thus it is a field with $p^{\deg(f)}$ elements. All such fields with $p^{\deg(f)}$ elements are a splitting field of $x^{p^{\deg(f)}}-x\in \Bbb{F}_p[x]$, they are isomorphic, so we can consider the field with $p^{\deg(f)}$ elements as being unique.

For your question: it remains to find an irreducible polynomial of degree $2$.


问:
Manageable project to learn some arithmetic geometry
OK, this may be an unusual question, since I'm asking you for advice, the way I would ask a supervisor were I a grad student.

I am a thirty-something maths teacher with an unresolved attraction to arithmetic geometry, and would like to know more about themes I find fascinating when I read texts about arithmetic geometry, such as:

the local-global principle

the way geometry influences the arithmetic behaviour of Diophantine equations, like in Faltings's theorem

the way scheme theory helps number theory by making a lot of geometric constructions possible, starting from base change

One of the reasons this attraction is unresolved is that I never managed to seriously learn enough algebraic number theory, algebraic geometry, etc. in a "bottom-to-top" way: I have often started to read for instance R. Vakil's The rising sea, D. Eisenbud & J. Harris The geometry of schemes or Q. Liu's Algebraic geometry and arithmetic curves  and find them to be very good texts, but I don't have the stamina to ingest hundreds of pages without a more specific goal.

Because of this there are a number of subjects I'm vaguely familiar with, but don't know in any depth. For instance, I know what a scheme is, I even think I understand in part their raison d'être, but am probably unable to pass an scheme theory exam (and the same goes for elliptic curves, derived categories, class field theory...).

To change this, I would like to try and understand a specific result and working my way from the top down to acquire the needed skills. The result has to be interesting enough to preserve my motivation in the long run, but "small" enough so that I could realistically understand it in a few months. I would love it to really mix number theory and geometry.

So the question is:

what theorem would constitute a reasonable-size project?

Personal facts that might be relevant: my PhD is in low-dimensional topology, so I'm quite comfortable with things like cohomology and Riemann surfaces, and I like it when geometric intuition is useful. Large doses of homological algebra or category theory don't frighten me.

答:
If your interest is arithmetic geometry, a starting point could be to read sections IV.1 and IV.2 in Hartshorne (HH) on the Riemann-Roch theorem and the Hurwitz theorem. These sections are about morphisms of curves $v: C \rightarrow C'$ over any algebraically closed field $k$. In particular the results are valid for $k=\overline{\mathbb{F_q}}$ or $\overline{K}$ where $K$ is a number field. Fulton's book "Algebraic curves" is more elementary than HH and also works over any algebraically closed field. It requires some basic knowledge on commutative algebra. Milne's book "Etale cohomology" is self contained and in this book you will find a proof of the Weil conjectures and Riemann Hypothesis
in general - in particular for a smooth projective curve over a finite field.
Note also that exercise AppC.5.6+7 in HH proves the Weil conjectures for elliptic curves over a finite field. All these books are "elementary" in a sense - you need to know some commutative algebra.


问:
General form of elements in $ \mathbb{Z} [\frac{1+\sqrt{-3}}{2} ] $
What is the general form of elements in $\displaystyle \mathbb{Z} \left[\frac{1+\sqrt{-3}}{2} \right] $?

I'm getting muddled.

Thanks

答:
In general, if $\alpha$ is an algebraic integer (satisfies a monic polynomial with integer coefficients), and its monic irreducible is of degree $n$, then the elements of $\mathbb{Z}[\alpha]$ can be written uniquely as
$$a_0 + a_1\alpha + \cdots + a_{n-1}\alpha^{n-1},\qquad a_i\in\mathbb{Z}.$$
(A simple application of the remainder theorem for polynomials, and the fact that $\mathbb{Z}[\alpha] = {p(\alpha)\mid p(x)\in\mathbb{Z}[x]}$).

So here, the simplest is to note that $\frac{1+\sqrt{-3}}{2}$ satisfies a monic polynomial of degree $2$ over $\mathbb{Z}$.

In this particular case, there is an alternative possible description: every element of $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ can be written uniquely as
$$\frac{a + b\sqrt{-3}}{2},\qquad a,b\in\mathbb{Z},\qquad a\equiv b\pmod{2}.$$


问:
Specify all the bases for which can be explicitly constructed out of the above vectors
$$v_1=
\begin{bmatrix}
  1 \\
 -2 \\
 1  \\
 \end{bmatrix}$$
$$v_2=
\begin{bmatrix}
  2 \\
 -1 \\
 13  \\
 \end{bmatrix}$$
$$v_3=
\begin{bmatrix}
  7 \\
 -9 \\
 -8  \\
 \end{bmatrix}$$
$$v_4=
\begin{bmatrix}
  -4 \\
 8 \\
 -4  \\
 \end{bmatrix}$$
$$v_5=
\begin{bmatrix}
  -6 \\
 7 \\
 9  \\
 \end{bmatrix}$$

(a)  Specify all the bases for which can be explicitly constructed out of the above vectors (just the unique sets, regardless of the order of vectors in each set).

I would really appreciate any help and feedback, thanks in advance!

I am confused that in this question that what is  required to do
is it using all vectors as a basis of a set V? or using such as v1 and v2 as a basis of a set V?

答:
Even if you want to get a basis for the whole vector space (which I will assume is $\Bbb{R}3$) or if you want to get a basis for some subspace $V$, you should think as follows.

Suppose that you have a space $V$. First, look at the dimension of the space (let’s say $n$). That will be the number of elements that you will need to form a linearly independent set that will generate a basis for that space (by the Steinitz exchange lemma).


答:
Let $f$ be the function defined by $f(y)=\log|y|$ for all $y\neq0$. We see that
$$
f(y)=
\begin{cases}
\log(y) &\text{if $y>0$} \\
\log(-y) &\text{if $y<0$} \, .
\end{cases}
$$
If $y>0$, then $$\frac{d}{dy}f(y)=\frac{d}{dy}(\log y)=\frac{1}{y} \, .$$ And if $y<0$, then $$\frac{d}{dy}f(y)=\frac{d}{dy}(\log(-y))=\frac{1}{-y} \cdot -1=\frac{1}{y}$$by the chain rule. Hence, $f'(y)=1/y$ for all $y\neq0$. By contrast, the function $g$ defined by $g(y)=\log y$ has the domain $\mathbb{R}^+$ and so $g'(y)=1/y$ only when $y>0$. Hence, $g$ is not an antiderivative of the function $h$ given by $h(y)=1/y$, since the domain of $h$ is $\mathbb{R} \setminus \{0\}$. So while we have $f'=h$, we don't have $g'=h$.

Title: Why $\frac{y’}{y}$ is equivalent to $(\ln|y|)’$?

Body:
I just read a proof process from a textbook about differential equations, it has following content :
$$y’= ay\implies\frac{y’}{y}= a\implies (\ln|y|)’= a$$
I want to know why $\frac{y’}{y}$ is equivalent to $(\ln|y|)’$?


问:
Why does ${x}^{x^{x^{x^{\,.^{\,.^{\,.}}}}}}$ bifurcate below $\sim0.065$?
When you calculate what ${x}^{x^{x^{x\cdots }}}$ converges to between $0$ and $1$, before approximately $0.065$ the graph bifurcates.

Why does this happen and is there a reason for it happens at that number?

答:
The following reference is recommended reading, if you want to have a more detailed account of what follows:

Exponentials reiterated

The power tower at hand is recursively defined as:
$$
h_0(x) = x \quad ; \quad h_{n+1}(x) = x^{h_n(x)} \quad ; \quad h(x) = \lim_{n\to\infty} h_n(x)
$$
The recursion process is visualized in the picture below:

The $\color{blue}{blue\;lines}$ are the iterands for $n=$ odd; the $\color{red}{red\;lines}$ are the iterands for $n=$ even.
It is seen that splitting up odd and even starts at $x\lt 1$ and that there is a bifurcation point (black dot).
Going to infinity leads to the functional equation:
$$
y = x^y \quad \mbox{with} \quad y = h(x)
$$
The inverse function (where it exists) has an explicit analytical form, which is the black line in the above figure:
$$
x = y^{1/y}
$$
And can be used to establish the rightmost bound of the tower's domain:
$$
\frac{d}{dy}y^{1/y} = e^{\ln(y)/y}\left[-\frac{\ln(y)}{y^2}+\frac{1}{y^2}\right] = 0
\\ \Longrightarrow \quad \ln(y)=1 \quad \Longrightarrow \quad y=e \; ; \; x=e^{1/e}
$$
But in order to answer the question as stated, we have to consider another functional equation, namely:
$$
y = x^{x^y}
$$
Odd and even are separated now, as is required for observing the bifurcation.
There is no neat inverse function of the latter equation. Therefore we shall investigate the two-dimensional
funcion $\,f(x,y)=x^{x^y}-y\,$. An insightful way to do this is make an isoline chart / contour plot of the function.
Lines are darker where the contour levels are higher. The massive black line is the place where $f(x,y)=0$.

Implicit differentiation is employed for finding the derivatives:
$$
df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy \\
df =  x^{x^y}\left[\frac{x^y y \ln(x)}{x} + \frac{x^y}{x}\right] dx + \left[x^{x^y}x^y\ln^2(x)-1\right] dy
$$
If the gradient is zero, then $f(x,y)$ has a critical point at that place:
$$
\begin{cases}
\partial f/\partial x = 0 & y \ln(x)+1 = 0 \\
\partial f/\partial y = 0 & x^{x^y}x^y\ln^2(x)-1=0
\end{cases}
$$
Both equations are essentially the same, because $\,x^{x^y}=x^y=y\,$, giving with $\,x=y^{1/y}\,$:
$$
y \ln(y^{1/y})+1 = \ln(y)+1 = 0 \quad \Longrightarrow \quad y = 1/e \quad ; \quad x = e^{-e}
$$
If we look at the picture with the isolines, then it is evident that the bifurcation must be at this place, with coordinates $\,(x,y)=(e^{-e},1/e)\,$.

EDIT.

It is observed in the last picture that the isolines in the neighborhood of the critical point are like hyperbolas.
To confirm this observation, the second order derivatives at the bifurcation point should be investigated as well.
Let a computer algebra system (MAPLE) do the hard work.
```
> f(x,y) := x^(x^y)-y;
> A(x,y) := diff(diff(f(x,y),x),x);
> B(x,y) := diff(diff(f(x,y),x),y);
> C(x,y) := diff(diff(f(x,y),y),y);
> D(x,y) := diff(diff(f(x,y),y),x);
> a := simplify(subs({x=exp(1)^(-exp(1)),y=1/exp(1)},A(x,y)));
                          exp(-3 + 2 exp(1))
> b := simplify(subs({x=exp(1)^(-exp(1)),y=1/exp(1)},B(x,y)));
                          -exp(-1 + exp(1))
> c := simplify(subs({x=exp(1)^(-exp(1)),y=1/exp(1)},C(x,y)));
                                  0
> d := simplify(subs({x=exp(1)^(-exp(1)),y=1/exp(1)},D(x,y)));
                          -exp(-1 + exp(1))
```
So the Hessian matrix at the crirical point is:
$$
\begin{bmatrix}
a & b \\
b & c
\end{bmatrix} =
\begin{bmatrix}
e^{-3+2e} & -e^{-1+e} \\
-e^{-1+e} & 0 \end{bmatrix}
$$
And the two eigenvalues of the Hessian turn out to be
$$
\lambda_{1,2} = \frac{a+c}{2} \pm \sqrt{b^2 + \left(\frac{a-c}{2}\right)^2} \\ =
\frac{1}{2} e^{-3+2e} \left[1 \pm \sqrt{\left(2\frac{e^{-1+e}}{e^{-3+2e}}\right)^2+1}\right]
$$
Herewith it is clear that one eigenvalue is positive while the other is negative.
Therefore the critical point is a saddle point, as expected.
As has been said, the isolines in the neighborhood of this saddle point are hyperbolas.
The asymptotes of these hyperbolas can be calculated as follows.
$$
\begin{bmatrix} x-e^{-e} & y-1/e \end{bmatrix}
\begin{bmatrix} a & b \\ b & c \end{bmatrix}
\begin{bmatrix} x-e^{-e} \\ y-1/e \end{bmatrix} = 0 \\
e^{-3+2e}\left(x-e^{-e}\right)^2-2\,e^{-1+e}\left(x-e^{-e}\right)\left(y-1/e\right) = 0 \\
\Longrightarrow \quad x = e^{-e} \quad ; \quad y = 1/e + e^{e-2}/2\cdot\left(x-e^{-e}\right)
$$
The asymptotes are tangent to the crossing massive black lines at the critical point.

Note. One doesn't need a tower of powers to observe bifurcation.

What is the value of $\,x^0\,$ for $\,x=0\,$? As a physicist by education, I would think that the limit of $\,x^0\,$ for $x\downarrow 0$ is
a bifurcation at $\,(x,y)=(0,1)\,$, as is clearly seen in the picture below.
Actually displayed are graphs of $y=x^{1/N}$ with $N\in\{\color{red}{-2,-4,-6,-8,-20},\color{blue}{+20,+8,+6,+4,+3}\,\}$.

Formulated otherwise it would mean that $\,y=x^0=1\,$ is only defined for $\,x\gt 0\,$; it is not a function at $\,x=0\,$ and we have
Commutativity of iterated limits:
$$
\lim_{x\downarrow 0}\lim_{N\to\infty} \begin{cases} \color{blue}{x^{+1/N} = [0,1]} \\ \color{red}{x^{-1/N} = [1,\infty]} \end{cases} \\
\lim_{N\to\infty}\lim_{x\downarrow 0} \begin{cases} \color{blue}{x^{+1/N} = [0,1]} \\ \color{red}{x^{-1/N} = [1,\infty]} \end{cases}
$$

Body:
The following reference is recommended reading, if you want to have a more detailed account of what follows:

Exponentials reiterated

The power tower at hand is recursively defined as:
$$
h_0(x) = x \quad ; \quad h_{n+1}(x) = x^{h_n(x)} \quad ; \quad h(x) = \lim_{n\to\infty} h_n(x)
$$
The recursion process is visualized in the picture below:

The $\color{blue}{blue\;lines}$ are the iterands for $n=$ odd; the $\color{red}{red\;lines}$ are the iterands for $n=$ even.
It is seen that splitting up odd and even starts at $x\lt 1$ and that there is a bifurcation point (black dot).
Going to infinity leads to the functional equation:
$$
y = x^y \quad \mbox{with} \quad y = h(x)
$$
The inverse function (where it exists) has an explicit analytical form, which is the black line in the above figure:
$$
x = y^{1/y}
$$
And can be used to establish the rightmost bound of the tower's domain:
$$
\frac{d}{dy}y^{1/y} = e^{\ln(y)/y}\left[-\frac{\ln(y)}{y^2}+\frac{1}{y^2}\right] = 0
\\ \Longrightarrow \quad \ln(y)=1 \quad \Longrightarrow \quad y=e \; ; \; x=e^{1/e}
$$
But in order to answer the question as stated, we have to consider another functional equation, namely:
$$
y = x^{x^y}
$$
Odd and even are separated now, as is required for observing the bifurcation.
There is no neat inverse function of the latter equation. Therefore we shall investigate the two-dimensional
funcion $\,f(x,y)=x^{x^y}-y\,$. An insightful way to do this is make an isoline chart / contour plot of the function.
Lines are darker where the contour levels are higher. The massive black line is the place where $f(x,y)=0$.

Implicit differentiation is employed for finding the derivatives:
$$
df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy \\
df =  x^{x^y}\left[\frac{x^y y \ln(x)}{x} + \frac{x^y}{x}\right] dx + \left[x^{x^y}x^y\ln^2(x)-1\right] dy
$$
If the gradient is zero, then $f(x,y)$ has a critical point at that place:
$$
\begin{cases}
\partial f/\partial x = 0 & y \ln(x)+1 = 0 \\
\partial f/\partial y = 0 & x^{x^y}x^y\ln^2(x)-1=0
\end{cases}
$$
Both equations are essentially the same, because $\,x^{x^y}=x^y=y\,$, giving with $\,x=y^{1/y}\,$:
$$
y \ln(y^{1/y})+1 = \ln(y)+1 = 0 \quad \Longrightarrow \quad y = 1/e \quad ; \quad x = e^{-e}
$$
If we look at the picture with the isolines, then it is evident that the bifurcation must be at this place, with coordinates $\,(x,y)=(e^{-e},1/e)\,$.

EDIT.

It is observed in the last picture that the isolines in the neighborhood of the critical point are like hyperbolas.
To confirm this observation, the second order derivatives at the bifurcation point should be investigated as well.
Let a computer algebra system (MAPLE) do the hard work.

> f(x,y) := x^(x^y)-y;
> A(x,y) := diff(diff(f(x,y),x),x);
> B(x,y) := diff(diff(f(x,y),x),y);
> C(x,y) := diff(diff(f(x,y),y),y);
> D(x,y) := diff(diff(f(x,y),y),x);
> a := simplify(subs({x=exp(1)^(-exp(1)),y=1/exp(1)},A(x,y)));
                          exp(-3 + 2 exp(1))
> b := simplify(subs({x=exp(1)^(-exp(1)),y=1/exp(1)},B(x,y)));
                          -exp(-1 + exp(1))
> c := simplify(subs({x=exp(1)^(-exp(1)),y=1/exp(1)},C(x,y)));
                                  0
> d := simplify(subs({x=exp(1)^(-exp(1)),y=1/exp(1)},D(x,y)));
                          -exp(-1 + exp(1))

So the Hessian matrix at the crirical point is:
$$
\begin{bmatrix}
a & b \\
b & c
\end{bmatrix} =
\begin{bmatrix}
e^{-3+2e} & -e^{-1+e} \\
-e^{-1+e} & 0 \end{bmatrix}
$$
And the two eigenvalues of the Hessian turn out to be
$$
\lambda_{1,2} = \frac{a+c}{2} \pm \sqrt{b^2 + \left(\frac{a-c}{2}\right)^2} \\ =
\frac{1}{2} e^{-3+2e} \left[1 \pm \sqrt{\left(2\frac{e^{-1+e}}{e^{-3+2e}}\right)^2+1}\right]
$$
Herewith it is clear that one eigenvalue is positive while the other is negative.
Therefore the critical point is a saddle point, as expected.
As has been said, the isolines in the neighborhood of this saddle point are hyperbolas.
The asymptotes of these hyperbolas can be calculated as follows.
$$
\begin{bmatrix} x-e^{-e} & y-1/e \end{bmatrix}
\begin{bmatrix} a & b \\ b & c \end{bmatrix}
\begin{bmatrix} x-e^{-e} \\ y-1/e \end{bmatrix} = 0 \\
e^{-3+2e}\left(x-e^{-e}\right)^2-2\,e^{-1+e}\left(x-e^{-e}\right)\left(y-1/e\right) = 0 \\
\Longrightarrow \quad x = e^{-e} \quad ; \quad y = 1/e + e^{e-2}/2\cdot\left(x-e^{-e}\right)
$$
The asymptotes are tangent to the crossing massive black lines at the critical point.

Note. One doesn't need a tower of powers to observe bifurcation.

What is the value of $\,x^0\,$ for $\,x=0\,$? As a physicist by education, I would think that the limit of $\,x^0\,$ for $x\downarrow 0$ is
a bifurcation at $\,(x,y)=(0,1)\,$, as is clearly seen in the picture below.
Actually displayed are graphs of $y=x^{1/N}$ with $N\in\{\color{red}{-2,-4,-6,-8,-20},\color{blue}{+20,+8,+6,+4,+3}\,\}$.

Formulated otherwise it would mean that $\,y=x^0=1\,$ is only defined for $\,x\gt 0\,$; it is not a function at $\,x=0\,$ and we have
Commutativity of iterated limits:
$$
\lim_{x\downarrow 0}\lim_{N\to\infty} \begin{cases} \color{blue}{x^{+1/N} = [0,1]} \\ \color{red}{x^{-1/N} = [1,\infty]} \end{cases} \\
\lim_{N\to\infty}\lim_{x\downarrow 0} \begin{cases} \color{blue}{x^{+1/N} = [0,1]} \\ \color{red}{x^{-1/N} = [1,\infty]} \end{cases}
$$


问:
Tally method to build a machine (on paper, Turing Machine
Consider function $q$: For any even integer $x\ge0$ (including $0$): $q(x) = 4x$

I want to design a machine (on paper of course) to compute q under the Tally system. Another restriction is that when the input is odd number, the machine should not halt.

I decided to put this into programming (C++) then conclude the logic there, but the machine can not have any while loops, however; the machine can go to left right and write 0 or 1 on the paper. The paper is basically an infinite tape with boxes, each box is counted as one place that can hold one value $(x, 0, 1)$.

Any one knows how to solve this problem or possibly give me a suggestion to put me on right track?

答:
First, you must choose a method of read input.

For example: in low level Turing Machine(DTM), you can choose unary method of read input, for example if 11111 is written on tape of DTM it means number 5 and 11 for 2.

Second, you must choose a method of write output.

For example: In end of your program, every thing written in your tape of DTM(in unary method) it means your output(in unary method).

Now, write this program:

```<code>if number is odd
   loop forever
else
   read one '1' of tape and change it to 'A' and go to end of tape and write '111' after '#'
   go back
   if do not find any '1' befor of 'A'
       remove '#'
       change all of the 'A' to '1'
</code>```

Body:
First, you must choose a method of read input.

For example: in low level Turing Machine(DTM), you can choose unary method of read input, for example if 11111 is written on tape of DTM it means number 5 and 11 for 2.

Second, you must choose a method of write output.

For example: In end of your program, every thing written in your tape of DTM(in unary method) it means your output(in unary method).

Now, write this program:

if number is odd
   loop forever
else
   read one '1' of tape and change it to 'A' and go to end of tape and write '111' after '#'
   go back
   if do not find any '1' befor of 'A'
       remove '#'
       change all of the 'A' to '1'


问:
Condensation points and Lebesgue-zero measure set

Definition. $x \in \mathbb{R}$ is a condensation point of a subset $A \subseteq \mathbb{R} \iff$ the intersection of every neighbourhood of $x$ with $A$ is uncountable.

Can one construct a subset $N \subseteq \mathbb{R}$ that has lebesgue measure zero, such that every point in $\mathbb{R}$ is a condensation point of $N$ ?

答:
Let $I_n = [a_n, b_n]$ be an enumeration of the set of intervals with rational endpoints. For each $n$ let $C_n \subseteq I_n$ be a cantor set contracted to $I_n$. To do so, if $C$ is the cantor set and $l_n$ is the length of $I_n$, we let $C_n = (l_nC) + a_n$ where $l_nC = \{l_n x\;|\;x \in C\}$. Then, recall that the cantor set has Lebesgue measure 0, denoted $|C| = 0$.

Then, by countable subadditivity we have that $$\left| \bigcup_{n \in \mathbb{N}} C_n \right| \leq \sum_{n \in N} |C_n| = \sum_{n \in N} |C| = 0.$$ Moreover, if $p \in \mathbb{R}$ and $(p - \epsilon, p + \epsilon)$ is contained in some neighborhood of $p$, we know that there are distinct rational numbers $q, q' \in (p - \epsilon, p + \epsilon)$ with $q < q'$. Then, $[q, q'] = I_k$ for some $k$ and $[q, q'] = I_k \subseteq (p - \epsilon, p + \epsilon)$. By construction, $C_k \subseteq I_k \subseteq (p - \epsilon, p + \epsilon)$. Since $C_k$ is in bijection with $C$, which is uncountable, we find that $p$ is a condensation point of $\bigcup_{n \in \mathbb{N}} C_n $.

Hence the set $\bigcup_{n \in \mathbb{N}} C_n $ is what we're looking for.


问:
Confusion with finding the range on complex number functions
The question is about finding the range and domain of $f(z)=e^z$, and I have read the solution and the reason why many times but I just don't understand it. My professor simply taught domain and range with real numbers and gave us complex number question which I don't exactly know how to do, so here's the answer:

The complex exponential function has natural domain $\mathbb{C}$; hence $f$ has natural domain $\mathbb{C}$. If $z=x+iy$ with $x, y∈\mathbb{R}$ then $e^z=e^x e^{iy} = e^x(\cos y+i\sin y)$. Since $e^x$ can take any positive value, and $y$ can take any value in $\mathbb{R}$ including all those in $(−π, π]$ then $f(z)$ can take any non zero modulus and all arguments. Hence $f(z)$ has the range $\mathbb{C}\setminus\{0\}$.

I think I have a clear understanding on most of it other than the reason why the range would be $\mathbb{C}\setminus\{0\}$. Could you also please tell me a way to find the range for all complex functions too? That would make my life so much easier!

答:
$e^x > 0$, view it as the non-zero radius.

$e^{iy}$ is a point on the unit circle, it decides the direction.

Given any non-zero point, find the radius $r$ and the angle $\theta$, we can solve $e^x=r$ and let $y$ be $\theta$.

There is no solution for $e^xe^{iy}=0$, since $e^x$ and $e^{iy}$ are non-zero, we can divide them by their multiplicative inverses and get a contradiction of $1=0$.


问:
Show an objective function is convex with respect to multiple matrix variables
Before go through the details, you can just first consider a toy example $(m,n,N,d_1,d_2)=(2,3,4,1,1)$ which is the simplest case.

I'm solving an optimization problem numerically(probably with Newton's method or BFGS) and proving convergence,
\begin{equation}
 \begin{aligned}\label{}
    & \underset{F\in \mathbb{R}_{m,d_1} G\in \mathbb{R}_{n,d_2}}{\text{minimize}}
    & & \phi(F,G) =\|(L\otimes F)^TVG\|_F^2\\
    & \text{subject to}
    & & \sum_{i=1}^N \|A_i^Tf_j\|_2^2=1 , \ j = 1, \ldots, d_1.\\
    &&& \sum_{i=1}^N \|A_ig_j\|_2^2=1 , \ j = 1, \ldots, d_2.
  \end{aligned}
\end{equation}
where $A_i$'s are known $m$ by $n$ full rank matrices (rank = min{m,n}), and $f_i$, $g_i$ are columns of $F$, $G$ respectively.

$L$ is a known $N$ by $N$ matrix. $V = [A_1^T\  A_2^T\ \ldots \ A_N^T ]^T$

$\otimes$ is the kronecker product (or tensor product)

Define $H=[A_1\  A_2\ \ldots \ A_N ]^T$.

I want to show $\phi$ is convex first, I know it is convex w.r.t F and G separatively, but how to show $\phi$ is convex w.r.t F and G jointly (or nonconvex)?

My thought is to define $x = [\text{vec}(F)^T \text{vec}(G)^T]^T$ and show the hessian matrix(second order derivative w.r.t x) $\succ 0$ (or at least $\succeq$)

After some computations, the hessian would be
\begin{bmatrix}P_{11}&  P_{12}\\ P_{12}^T & P_{22} \end{bmatrix}
where $P_{11}=I\otimes(H(M\otimes GG^T)H^T)$ and $P_{22}=I\otimes(V^T(M\otimes FF^T)V)$

$P_{12}=\sum_{i}\sum_jM_{ij}\{(F^TA_jG)\otimes A_i+ [(F^TA_i)\otimes(A_jG)]K_{n,{d_2}}\} $
where $M=L^TL$ and $K_{m,n}$ is commutation matrix of dimension $mn$ by $mn$

Next step is to show $P_{11}-P_{12}P_{22}^{-1}P_{12}^T\succ 0$ (according to thm 1 in link)

Is there any way to simplify $P_{12}$(I mean replace the double sum with some function of $H$, $V$, and $M$)?

Or is there other way to prove convex or nonconvex?

答:
Consider the simplest case, $m = n = d_1 = d_2 = 1$. Then $\phi = L^2V^2F^2G^2$, which is neither jointly convex nor concave in $F$ and $G$ (unless $L$ and/or $V$ are zero, which can't happen due to full rank assumption). Things do not become convex as any of these dimensions increase.

Furthermore, the constraints are nonlinear equality constraints, hence non-convex.

Verdict: This is not a convex optimization problem.


问:
Concentrated schemes are closed under finite gluings
Let $X$ be a scheme. Assume that $X = \cup_i X_i$ is finite open covering, such that the $X_i$ as well as their intersections $X_i \cap X_j$ are concentrated, i.e. quasi-compact and quasi-separated (see Daniel Murfet's notes on concentrated schemes as well as EGA IV.1). Does it follow that $X$ is concentrated? I'm almost embarrassed to ask this question.

Clearly $X$ is quasi-compact, as a finite union of quasi-compact subspaces. In order to show that $X$ is quasi-separated, it suffices to show that for open affines $U \subseteq X_i$ and $V \subseteq X_j$ their intersection $U \cap V$ is quasi-compact. Well this is clear for $i=j$ since $X_i$ is quasi-separated. But what happens for $i \neq j$? Somehow we have to use that $X_i \cap X_j$ is quasi-compact.

Here is a more abstract idea: We have to show that $\Delta : X \to X \times X$ is a quasi-compact morphism. This can be checked locally on the base, hence it suffices to prove that the morphisms $X_i \cap X_j \to X_i \times X_j$ are quasi-compact. Well, $X_i \cap X_j$ is quasi-compact, hence by (EGA IV, 1.1.2 (v)) it would suffice that $X_i \times X_j$ is separated. But we don't assume that the $X_i$ are separated, they are just quasi-separated.

Note that actually EGA IV.1.2.7.c proves the claim, but I don't understand the proof. Here is my translation: "We know that in order to prove that $\Delta : X \to X \times X$ is quasi-compact, it suffices to prove that the inverse images $X_i \cap X_j$ of the $X_i \times X_j$ are quasi-compact (1.1.1)". Here, (1.1.1) is just the definition of a quasi-compact morphisms, after which it is remarked that it suffices to prove this with respect to an affine(!) open covering. But here, we just have an open covering consisting of quasi-compact schemes. This seems to be a gap?

Or does the following statement hold (which we could then apply to $\Delta$)? If $f : X \to Y$ is a morphism and $Y$ is covered by quasi-compact open subsets $Y_i$ such that each $f^{-1}(Y_i)$ is quasi-compact, then $f$ is quasi-compact. I doubt that this is true without any assumption on $Y$.

Edit. Sorry I think now I can answer my question. The treatment of quasi-compact morphisms is better in EGA I (1970), §6. Specifically, in order to show that $X_i \cap X_j \to X_i \times X_j$ is quasi-compact, it suffices that $X_i \cap X_j$ is quasi-compact and $X_i \times X_j$ is quasi-separated by Prop. 6.1.5. (v), and we have both. Note that EGA VI.1.2.7 is reproduced as EGA I.6.1.12, but still the end just refers to the definition of quasi-compact morphisms, which doesn't seem to suffice?

So what about a more direct proof, showing directly that (with the notation above) $U \cap V$ is quasi-compact? One can spell out the proof above as follows (still quite tricky):

$X_i \cap X_j \to (X_i \cap X_j) \times (X_i \times X_j)$ is a base change of the diagonal of $X_i \times X_j$, which is quasi-compact, hence it is also quasi-compact. Since $U \cap V$ is the preimage of the quasi-compact open subscheme $(X_i \cap X_j) \times (U \times V)$, it is quasi-compact.

(Usually I delete a question when I solve it quickly before an answer arrives, but this time I won't because it has already some upvotes and therefore seems to be interesting for others, too. Besides, I would like to know if you also agree that the EGA proof has a gap.)

答:
The problem is roughly the result in the picture below from EGA I.6.1.12 p.296 (Springer 1971).

I claim that the condition d) is not sufficient to imply a) (so EGA I.6.1.12 is not correct as stated). As a counterexample, let take $X$ to be the infinite dimensional affine space over a field with double origin, as Exercise 10.11 in Algebraic geometry I by Görtz-Wedhorn:and we take all $U_{\alpha}$ to be the quasi-compact scheme $X$ itself, then d) holds but a) does not. See also a comment in [Görtz-Wedhorn, Remark 10.2 (4)] for a related issue on quasi-compactness.


问:
Prove that $\tan(x)\tan(x+r) = \frac{\tan(x+r) - \tan(x)}{\tan r} -1$
$\tan(x)\cdot \tan(x+r) = \dfrac{\tan(x+r) - \tan(x)}{\tan r} -1$

I tried L.H.S = $\dfrac{\sin(x)\cdot \sin(x+r)}{\cos(x)\cdot \cos(x+r)}$
= $\dfrac{\sin(x+r- x)}{\sin(r)}\dfrac{\sin(x)\cdot\sin(x+r)}{\cos(x)\cdot \cos(x+r)}$

= $\dfrac{\sin(x+r- x)}{\sin(r)}\dfrac{\sin(x)\cdot\sin(x+r)}{\cos(x)\cdot\cos(x+r)}$

= $\dfrac{\sin(x)\cdot\sin(x+r)}{\sin(r)}\dfrac{\sin(x+r)\cos(x) - \cos(x+r)\cos(x)}{\cos(x)\cdot \cos(x+r)}$

= $\dfrac{\sin(x)\cdot\sin(x+r)}{\sin(r)}\left(\tan(x+r) -\tan(x)\right)$

I get stuck here; what else is there to do?

Edit: using Tavish's hint, we can get it directly.

$$\tan(x+r -x) = \dfrac{\tan(x+r) - \tan(x)}{1 + \tan(x+r)\tan(x)}$$
$$1 + \tan(x+r)\tan(x)= \dfrac{\tan(x+r) - \tan(x)}{\tan(r)}$$
$$\tan(x+r)\tan(x) = \dfrac{\tan(x+r) - \tan(x)}{\tan(r)} -1$$

答:
Hint:

Use $$\tan(a-b)=\frac{\tan a -\tan b}{1+\tan a \tan b} $$ with $a=x+r, b=x$.


问:
What mathematical term for this kind of "solution functions"?
Let $F$ a function of two variables and $x,y,z$ complex variables.

Let an equation $F(x,y)=z$ be given. Then a function or multifunction (correspondence) $H$ with $x=H(y,z)$ does exist.

1.) What is the mathematical term for calling $H$?

2.) In which classes of functions can $H$ be calculated explicitly (symbolically / in closed form)?

I already know, for algebraic functions $F$ e.g., $H$ is also algebraic. An algebraic $H$ can be calculated in closed form if it can be presented by radicals or by other explicit representations (Bring radicals etc.).

$H$ is a solution of the equation. Is it a kind of inverse relations, a kind of projection, an evaluation map or what?

I need to name $H$ to talk about it in general.

答:
$H$ is the explicit function for $x$.

I have never seen a general theory of analytically invertible functions.


问:
What is a proof?
I am just a high school student, and I haven't seen much in mathematics (calculus and abstract algebra).

Mathematics is a system of axioms which you choose yourself for a set of undefined entities, such that those entities satisfy certain basic rules you laid down in the first place on your own.

Now using these laid-down rules and a set of other rules for a subject called logic which was established similarly, you define certain quantities and name them using the undefined entities and then go on to prove certain statements called theorems.

Now what is a proof exactly? Suppose in an exam, I am asked to prove Pythagoras' theorem. Then I prove it using only one certain system of axioms and logic. It isn't proved in all the axiom-systems in which it could possibly hold true, and what stops me from making another set of axioms that have Pythagoras' theorem as an axiom, and then just state in my system/exam "this is an axiom, hence can't be proven".

EDIT : How is the term "wrong" defined in mathematics then? You can say that proving Fermat's Last Theorem using the number theory axioms was a difficult task but then it can be taken as an axiom in another set of axioms.

Is mathematics as rigorous and as thought-through as it is believed and expected to be? It seems to me that there many loopholes in problems as well as the subject in-itself, but there is a false backbone of rigour that seems true until you start questioning the very fundamentals.

答:
Proving something is no more than just saying why something is true mathematically. There are many ways to prove something. Here are the thre most used:

Logic: Using established rules and thinking to get to a coherent answer.

Contradiction: Trying to prove that what we are proving is false. If this fails, what we are trying to prove is true.

Alternate thinking: Putting the problem in a different way to clarify previously unseen things.

Advanced Mathematical proofs generally come in abstract and obscure ways of rationalization. But a proof can come as simple as some simple logic.

Title: What is a proof?


问:
Proving the parallelity of two lines in a convex Pentagon
Can someone show me how to solve this Geometry Problem? I feel like there is a trick behind it that I don’t see

The convex pentagon $ABCDE$ has a circumference and $\overline{AB} = \overline{BD}$ applies. Let the point $P$ be the intersection of the diagonals $AC$ and $BE$. The straight lines $BC$ and $DE$ intersect at point $Q$.
Prove that the line $PQ$ is parallel to the diagonal $AD$.

答:
$AB = BD$ implies that their corresponding circular arcs are equal too, so $EB$ is the angle bisector of $\angle \, AED$ and thus $\angle \, AEB = \angle \, DEB$. Let $R$ be the intersection point of $BE$ and $AD$. Then, after a short angle chasing, triangles $AEP$ and $BEQ$ are similar, so
$$\frac{EP}{EQ} = \frac{EA}{EB}$$
and triangles $AER$ and $BED$ are similar, so
$$\frac{ER}{ED} = \frac{EA}{EB}$$

Consequently,
$$\frac{EP}{EQ} = \frac{EA}{EB} = \frac{ER}{ED}$$
which by the Thales' intercept theorem yields
$$RD \, || \, PQ$$ and since $R$ lies on $AD$ you get
$$AD \, || \, PQ$$


问:
Find center and commutant of the group.
We have $G = \{\begin{pmatrix}
\alpha & \beta\\
0 & 1\\
\end{pmatrix} |\ \alpha \neq 0\}$ in $GL_2(K)$ ($K$ - field)

I want to find its center and commutant.
$$
Z(G) = \{g \in G \ | \ \forall h \in G \ : \ gh = hg\}-center \\ [G, G] = <[g,h] \ | g, h \in G> - commutant
$$

I tried to do this kind of computations:
Let's see what we have when we multiply elements from group:

$$
\begin{pmatrix}
\alpha & \beta\\
0 & 1\\
\end{pmatrix} * \begin{pmatrix}
\alpha_1 & \beta_1\\
0 & 1\\
\end{pmatrix} = \begin{pmatrix}
\alpha\alpha_1 & \alpha\beta_1+\beta\\
0 & 1\\
\end{pmatrix}
$$
And:
$$
\begin{pmatrix}
\alpha_1 & \beta_1\\
0 & 1\\
\end{pmatrix}*
\begin{pmatrix}
\alpha & \beta\\
0 & 1\\
\end{pmatrix} = \begin{pmatrix}
\alpha\alpha_1 & \alpha_1\beta +\beta_1\\
0 & 1\\
\end{pmatrix}
$$
That gives us some equality for center. (i.e $(\alpha-1)\beta_1 - \beta(\alpha_1-1) = 0$). And I couldn't get any statements for $\alpha, \beta$ we needed that we need for beeng in center.
So what about commitant I tried to find normal subgroups in our group (cuz we know that commutant is normal subgroup of our group), but could't get any good things there.

Can someone help me with a solution of this problem?

答:
If $|K|>2$ the center is trivial and $[G,G]$, which is more commonly called the "derived group", is the subgroup with the upper left entry equal to $1$.  If $|K|=2$, the center is the whole group and the derived group is trivial.

There are two cases: $|K|>2$ and $|K|=2$.  We treat the more general case first.

From the formulas you derived when checking the center, you can see that if $\begin{pmatrix}
\alpha & \beta\\
0 & 1\\
\end{pmatrix}$
is central, then for all $\alpha_1$ and $\beta_1$ you need $\alpha\beta_1 +\beta = \alpha_1\beta + \beta_1$.  For this to hold when $beta_1=0$ for all values of $\alpha_1$ clearly requires $\beta=0$ as $|K|\ne 2$.  That leaves $\alpha\beta_1=\beta_1$, so $\alpha=1$.  This shows the center is trivial.

A general commutator is of the form:

$\begin{pmatrix}
\alpha^{-1} & -\beta\alpha^{-1}\\
0 & 1\\
\end{pmatrix}\begin{pmatrix}
\gamma^{-1} & -\delta\gamma^{-1}\\
0 & 1\\
\end{pmatrix}
\begin{pmatrix}
\alpha & \beta\\
0 & 1\\
\end{pmatrix}
\begin{pmatrix}
\gamma & \delta\\
0 & 1\\
\end{pmatrix}$

Multiplying this out, we get $\begin{pmatrix}
1 & \delta\gamma^{-1}+\beta\alpha^{-1}\gamma^{-1}-\delta\alpha^{-1}\gamma^{-1}-\beta\alpha^{-1}\\
0 & 1\\
\end{pmatrix}$

If $char K \ne 2$, setting $\alpha=1$, $\gamma=-1$, $\beta=0$, and $\delta=\epsilon/2$, we get any matrix of the form $\begin{pmatrix}
1 & \epsilon\\
0 & 1\\
\end{pmatrix}$, and these clearly form a subgroup, so that is the derived group.

If $char K = 2$, take $\alpha=\delta=1$, $\gamma\ne 1$, $\gamma\ne 0$ and $\beta=\frac{\epsilon\gamma}{1-\gamma}$ to get the same.  Note that we use $|K|>2$ to select a $\gamma\notin \{0,1\}$.

If $|K|=2$ then the group has order $2$.  All such groups are abelian, so in that case, the center is the whole group and the derived group is trivial.

Title: Find center and commutant of the group.


问:
The union of any directed set of the closed sets of algebraic closure operator is closed?
In "Notes on Lattice Theory" J. B., there is a theorem such that

Let $\Gamma$ be a closure operator on a set $X$. The following are equivalent.

(1) $\Gamma$ is an algebraic closure operator.

(2) The union of any up-directed set of $\Gamma$-closed sets is $\Gamma$-closed.

(3) The union of any chain of $\Gamma$-closed sets is $\Gamma$-closed.

and A subset $D$ of an ordered set $P$ is said to be up-directed if for every $x,y \in D$ there exists $z \in D$ with $x \leq z$ and $y \leq z$.

The definition of algebraic closure operator

A closure operator $\Gamma$ on a set $X$ is said to be algebraic if for every $B \subseteq X$,
$$\Gamma(B) = \bigcup \{\Gamma(F): F\ \text{is a finite subset of}\ B\}.$$

How to proof this theorem? like (1) $\Rightarrow$ (2), i cant figure out whats the up-directed set of closed set here.

答:
For 1 to 2, let $\mathscr{C}$ be a an up-directed family of $\Gamma$-closed sets.
(This means that $\Gamma(C)=C$ for all $C \in \mathscr{C}$ and $C_1,C_2 \in \mathscr{C}$ implies that there is a $C_3 \in \mathscr{C}$ that contains them both (the up-directed part). Let $\hat{C}= \bigcup \mathscr{C}$ and we have to show $\hat{C}$ is closed.

Now take any finite $F \subseteq \hat{C}$. So for any $f \in F$ we have some $C_f \in \mathscr{C}$ so that $f \in C_f$. Now we apply the up-directedness (which we can extend to finitely many sets by induction) and find some $C_0 \in \mathscr{C}$ such that $C_f \subseteq C_0$ for all $f \in F$ and so $F \subseteq C_0$. It follows that $\Gamma(F) \subseteq \Gamma(C_0) = C_0 \subseteq \hat{C}$ (by $\Gamma$-closedness of $C_0$ and monotonicity of $\Gamma$ under inclusion).

So this holds for all finite subsets of $\hat{C}$ and now using the fact that $\Gamma$ is an algebraic closure:

As $$\Gamma(\hat{C}) = \bigcup \{ \Gamma(F)\mid F \subseteq \hat{C} \text{ finite }\}\} \subseteq \hat{C} \subseteq \Gamma(\hat{C})$$

so $\hat{C}$ is $\Gamma$-closed.

(2) implies (3) is trivial as a chain of subsets is an up-directed family.


答: If the question is to find all bases of $\mathbb{R}^3$, you need to find all subsets of 3 vectors that are linearly independent. The naive way to do this is to check all $\binom{5}{3} = 10$ different subsets and test linear dependence for all of them. However, $v_4 = -4v_1$, which means that any basis you get using $v_4$ is identical to a basis you get using $v_1$. This means you only have to check 4 different sets -- that's $\{v_1, v_2, v_3\}$ and $\{v_1, v_2, v_5\}$ and $\{v_1, v_3, v_5\}$ and $\{v_2, v_3, v_5\}$.

Body:
$$v_1=
\begin{bmatrix}
  1 \\
 -2 \\
 1  \\
 \end{bmatrix}$$
$$v_2=
\begin{bmatrix}
  2 \\
 -1 \\
 13  \\
 \end{bmatrix}$$
$$v_3=
\begin{bmatrix}
  7 \\
 -9 \\
 -8  \\
 \end{bmatrix}$$
$$v_4=
\begin{bmatrix}
  -4 \\
 8 \\
 -4  \\
 \end{bmatrix}$$
$$v_5=
\begin{bmatrix}
  -6 \\
 7 \\
 9  \\
 \end{bmatrix}$$

(a)  Specify all the bases for which can be explicitly constructed out of the above vectors (just the unique sets, regardless of the order of vectors in each set).

I would really appreciate any help and feedback, thanks in advance!

I am confused that in this question that what is  required to do
is it using all vectors as a basis of a set V? or using such as v1 and v2 as a basis of a set V?


问:
How understand the metric of $S^2$ by definition of a spherically symmetric spacetime
I have a doubt linked with the definition of spherically symmetric manifold: when we talk about the isometry group on $\mathbb{R}^3$, considering that its orbit are 2-dimensional spheres (from the definition of  spherically symmetric spacetime), what means this in the view of defining the metric in $S^2$ as the metric induced by $\mathbb{R^3}$.

I mean: the orbits are the equivalence classes induced by an equivalence relation for which $p\sim q\iff Rp=q$, with $R\in SO(3)$. If we now consider the quotient space of the manifold $M$ (the spacetime) with respect to this relation then we have that if $N=M/\sim\implies M\cong N \times S^2$.

Now what I have not understand is: if I can consider $S^2$ embedded in $\mathbb{R}^3$, what is the link of the fact that the orbits of the isometries of $\mathbb{R^3}$ are isomorphic to $S^2$, with the fact that we can consider in $S^2$ a metric proportional to the euclidean (so a metric that must be a multiple of the metric of a 2-sphere, as suggested in https://en.wikipedia.org/wiki/Spherically_symmetric_spacetime)?

Also in my book in fact seems that these two facts are in a relation of implication! Thanks in advance!

答:
I'm not entirely sure I understand your precise question, but I'll attempt to clear up a couple of things.

As people pointed out in the comments, the isometry group of $\mathbb{R}^3$ acts transitively on $\mathbb{R}^3$ (meaning there is exactly one orbit, namely all of $\mathbb{R}^3$). Indeed, translations are isometries, and any point can be mapped to any other point using a translation.

On the Wikipedia page you link to, it says

A spherically symmetric spacetime is a spacetime whose isometry group contains a subgroup which is isomorphic to the rotation group SO(3) and the orbits of this group are 2-spheres (ordinary 2-dimensional spheres in 3-dimensional Euclidean space). The isometries are then interpreted as rotations and a spherically symmetric spacetime is often described as one whose metric is "invariant under rotations". The spacetime metric induces a metric on each orbit 2-sphere (and this induced metric must be a multiple of the metric of a 2-sphere).

This does not mention the full group of isometries of $\mathbb{R}^3$, but only the subgroup $\textrm{SO}(3)$, which consists of all rotations fixing the origin.
We have a spacetime on which $\textrm{SO}(3)$ acts, and the orbits are spheres. The claim is that the induced metric on each orbit is a multiple of the usual metric on the sphere. This is not entirely obvious, and I believe this claim is what you're asking about.
Here it is, rephrased:

Claim. If $\textrm{SO}(3)$ acts transitively and isometrically on $S^2$, the metric on $S^2$ is a multiple of the usual one.

To understand why this claim is true, you should learn about homogeneous spaces (https://en.wikipedia.org/wiki/Homogeneous_space). An example of a document in which this claim is proved is "Homogeneous Metrics on Spheres" by Kerin and Wraith (which I admittedly found by simply googling). Quoting from page 64:

By the above it follows that [...], and hence there is a unique homogeneous metric on $\frac{\textrm{SO}(n+1)}{\textrm{SO}(n)} \cong S^n$ up to scaling by a positive constant.


问:
Solving a combinatorial game with sums of natural numbers
I found this Olympiad style combinatorics problem. Does anybody know how to solve it?

Let $n\geq2$ be a natural number.
An $n\cdot n$ grid is drawn on a board and each field with one of the numbers $-1$
or $+1$ labeled. Then the $n$ row sums and also the $n$ column sums
is calculated and the sum $S_n$ of all these $2n$ sums is determined.

(a) Show that for no odd number $n$ there is a label with $S_n = 0$.

(b) Show that if $n$ is an even number, then there are at least six different labels
with $S_n = 0$.

答:
The (a) part is easy. First, note that $S_n = 2 A_n$, where $A_n$ is the sum of all $n²$ numbers in the grid. Then, note that if $n$ is odd, then also $A_n$ must be odd. To see this, you can consider the invariant (an Olympiad concept) of the parity of $A_n$. Whatever the label is, when you consider $A_n$ you are adding up $n²$ odd numbers (+1 or -1), so the sum must be odd. Finally, if $A_n$ is odd it cannot be zero, so also $S_n$ cannot be 0.

The part (b) is constructive. You should find at least 6 labels for which $A_n$ is zero. Try starting from a 2x2 board.

$$\begin{bmatrix} +1 & -1 \\ +1 & -1 \end{bmatrix}, \
\begin{bmatrix} +1 & -1 \\ -1 & +1 \end{bmatrix}, \
\begin{bmatrix} +1 & +1 \\ -1 & -1 \end{bmatrix}$$

and so on...


问:
Most 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 $M$ be the minimum distance between any 2 of the points.

Find the pdf of $M$. (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 $F_M(m)= P(M \le m)$ 1st and then hope the cdf is absolutely continuous.)

My model: 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).

Question: Well, I hope to find the pdf of $M$ via its cdf, which I think I'm able to compute if I know what $M$ is. What is $M$?

I think $M=\min\{X,Y,1-Y,1-X$ $,|X-Y|,1-|X-Y|\}$, where $X$ and $Y$ are the anti-clockwise distances from $B$ to, resp, $A$ and $C$. Somehow $X$ and $Y$ are iid Unif(0,1).

How would I compute $M$'s distribution?

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?

答:
If the minimum distance is at least $x$, then the three points, encountered in clockwise order and with the first taken to be at $0$, are at positions $(0, y, y+z)$, where $y \ge x$, $z \ge x$, and $y+z\le 1-x$.  The volume of this in configuration space is
$$
V(x)=\int_{y=x}^{1-2x}dy\int_{z=x}^{1-x-y}dz=\int_{y=x}^{1-2x}(1-2x-y)dy=(1-2x)y-\frac{1}{2}y^2\bigg\vert_{x}^{1-2x}=(1-2x)^2-\frac{1}{2}(1-2x)^2-(1-2x)x+\frac{1}{2}x^2=\frac{1}{2}(1-2x)^2-(1-2x)x+\frac{1}{2}x^2=\frac{1}{2}(1-3x)^2.
$$
There are two equally probable cases, corresponding to the orderings $ABC$ and $ACB$, so the full probability that the minimum distance is at least $x$ is $(1-3x)^2$; the probability that it is $\le x$ (the cdf of $M$) is $1-(1-3x)^2$; and the pdf of $M$ is the derivative of this, which is $$p(x)=6(1-3x)^2.$$

Title: Most random points on a circle


问:
Chance of winning when you are ahead a flip coin game
Two players, $A$ and $B$, alternately and independently flip a coin and the player wins only when he gets a head, and the other player gets the tail. Each round they will flip coin at the same time. If the coin is fair, and let’s assume there is infinitely round until one win. At the end, either player $A$ or $B$ will win, so the probability of player $A$ win plus probability of player $B$ win should equal to $100\%$.

The question I tried to solve is this:  In the above question, each win counts as one point, each loss counts as $-1$ point, whoever gets $3$ points first wins. (Game over when score is 3: -3).Now assume player $A$ wins one point at sometime. So the score are player A : player B is 1: -1; Then, what’s the probability player $A$ wins the game after that. Thanks in advance for any help.

答:
I assume from the title of the question, that you want to know the chance A will win given that he went first and has already flipped heads.  Please rephrase the problem if that's not what you wanted.

The chance is either $2/3$ or $3/4$ depending on the form of the game.  If there are "rounds", and you only win by flipping a head and your opponent flipping a tail on the same round, then it's $3/4$.  If any run of a heads by one player and a tails by the other player wins, then it's $2/3$.

If there are rounds, and you flipped first, then if the opponent flips tails you win that 50% and the other 50% you are back to even, so your total chance of winning is 50% plus half of the other 50%, or $3/4$.

If there are not rounds, you will win if your opponent flips tails, lose if your opponent flips head and you next flip tails, and be back to where you started if you both flip heads.  So 50% to win, 25% to lose, and 25% to get back to where you started (which again will apportion its final results 2:1 in your favor), will give you a $2/3$ chance of winning.

Interesting is to analyze the chance the first player wins.  If you are using "rounds", this is obviously $1/2$ before you start.  But if any heads-then-tails sequence wins, the first player has a $5/9$ chance of winning.  For consider the first two flips:

H T  --- first player wins
H H  --- we know the first player wins $1/3$ of the time with this start (since someone who just flipped heads has a $2/3$ chance)
T H --- we know the first player wins $1/3$ of the time with this start
T T ---no one has a heads yet, so we are back where we started.

As before we can ignore the $1/4$ of the time you get back to where you started, and the first player wins $1/3+1/9+1/9=5/9$ of the rest of the time.


问:
Plotting the graph of $x(t)$ when the graph of $x'(t)=f(x)$ is given
I am not sure how to present this problem clearly. Please feel free to suggest edits to this problem so that this will be useful to someone who will look at this in the future.

Here's my question:

Suppose $x'(t)=f(x)$ and suppose the graph of $f(x)$ is given. How can we draw the graph of $x(t)$ using the graph of $f(x).$

What I have tried so far: I have found equilibrium points and their stabilities using the graph of $f(x).$ But I don't know how this can be used when plotting $x(t).$

Thanks in advance!

答: COMMENT. You can use the second derivative (approximate and in green color below) to have an approximate shape of your function (in black color below). Unfortunately, I believe it is not possible to find out the corresponding market points, two extremunms (maximum and minimum) and three inflection ones because you don't have the analytic expression of your derivative $x'(t)$ but only the graph.


问:
Find the values of $k$ and $p$
When $kx^{3}+ px^{2}- x+ 3$ is divided by $x- 1,$ the remainder is $4.$

When $kx^{3}+ px^{2}- x+ 3$ is divided by $x- 2,$ the remainder is $21.$

Find the values of $k$ and $p.$

I've been stuck trying to solve this, so any help would be appreciated.

答: By the Remainder theorem, when we divide a polynomial $f(x)$ by $x-a$ where $a$ is a constant, the remainder is given by $f(a)$. Now, If $f(x)=kx^3+px^2-x+3$ as stated above, then $f(1)=4$ and $f(2)=21$. Hence we get $k+p+2=4$ and $8k+4p+1=21$. It then follows that $k=3$ and $p=-1$.

Title: Find the values of $k$ and $p$


问:
Convergence of series comparison

To determine if this series converges i was thinking to compare it to the series of $\frac{1}{n}$, is that a good way to handle this type of problem?

答: I would use the integral test, as $$\int_{2}^{\infty} \frac{8}{n\sqrt{\ln n}} dn$$ doesn't look too bad to integrate with the right substitution.  </p>

Title: Convergence of series comparison


问:
A generalized problem with $3$ complex numbers forming an equilateral triangle
There are many problems involving relations between complex numbers and the geometrical configuration of their corresponding affixes. Here is a generalized problem of the special case $n=2:$

Find the natural numbers $n>1$ such that the following implication holds: given distinct $a,b,c \in \mathbb{C}$, with $(a-b)^n+(b-c)^n+(c-a)^n=0,$ the triangle determined by the affixes of $a,b,c$ is equilateral.

The problem is converted into: $z_1+z_2+z_3=z_1^n+z_2^n+z_3^n=0$, where $z_1:=a-b$ and so on. For $n=2$, this is a standard exercise, which implies: $|z_1|=|z_2|=|z_3|.$

I proved that, for $n\in \{2,4,5,7\}$, the implication does hold, and for $n\in \{3,6,8,10\}$, the implication fails. Of course, one can keep checking the next cases of $n$, but I am wondering whether there is a straightforward, rigurous method of finding all those $n$-s.

I was able to reduce the problem to this one: (to be explicit, I keep the $z_i$ notation and I denote $z:=z_2/z_3$, after the elimination $z_1=-z_2-z_3$)

Find the natural numbers $n>1$ such that any solution of $(-z-1)^n+z^n+1=0$, where $z\in \mathbb{C}, z\notin \{0,-1\}$, have absolute value $1$.

Why $z \neq -1?$ Because otherwise, $z_1=0$, then $a=b$. Thus, it is clear that $z_i\neq0,$ and from here we need $z \neq 0$. Also, one can notice that $z \neq 1$.

For example, if $n=3$, we get: $z^2+z=0$, with no solutions in the above domain.

For $n=7$, we get: $z(z+1)(z^2+z+1)^2=0,$ and this is a valid solution of $n$, since if we divide this polynomial to $z(z+1)$, all roots of the quotient polynomial lie on the unit circle, as wanted.

P.S. After trying some small $n$-s, I realize that it has something to to with the primitive root of order $3$ of unity, $\epsilon^3=1$, maybe this could help, as it appears as a root in the valid cases $n\in \{2,4,5,7\}$.

答:
From $|z_2|=|z_3|$ and $z_2=zz_3,$ we conclude $|z|=1.$ From $|z_1|=|z_3|$ and $z_1=-z_2-z_3,$ we conclude $|z+1|=1.$ If we put this together, we get $z\in\{\epsilon,\bar\epsilon\}$ where $\epsilon$ and $\bar\epsilon$ are the solutions of $z^2+z+1=0.$

We have to check if $p(z)=(-z-1)^n+z^n+1=0$ has any other solutions but $0,-1,\epsilon,\bar\epsilon.$

$0$ and $-1$ are roots of $p$ if and only if $n$ is odd. Furthermore, it can easily be verified (using the derivative of $p$) that they are simple roots in this case.

We also know that $\epsilon$ and $\bar\epsilon$ have the same multiplicity as roots of $p,$ because the coefficients of $p$ are real. If we put all this together and scale the polynomials such that the coefficients of $z^{n-1}$ match, we get the following equations that must hold if the choice of $n$ enforces the triangle to be equilateral:
\begin{eqnarray}
nz(z+1)(z^2+z+1)^{(n-3)/2} &=& (z+1)^n -z^n-1  \text{ for odd } n \\
2(z^2+z+1)^{n/2} &=& (z+1)^n +z^n + 1  \text{ for even } n
\end{eqnarray}
For odd $n\geq 5,$ the coefficient of $z^3$ in $nz(z+1)(z^2+z+1)^{(n-3)/2}$ after expansion is $\frac{(n+3)n(n-3)}{8}$, while the coefficient of $z^3$ in $(z+1)^n -z^n-1$ is $\binom n3.$ Those two match only if $n\in\{5,7\}.$

For even $n\geq 6,$ the coefficient of $z^3$ in $2(z^2+z+1)^{n/2}$ after expansion is $\frac{(n+8)n(n-2)}{24}$, while the coefficient of $z^3$ in $(z+1)^n +z^n+1$ is $\binom n3$ again. Those two do not match for any value of $n\geq 6.$

The values you have found are the only ones that enforce the triangle to be equilateral.


问:
Why is the angle that maximizes $a \cos(\theta) + b \sin(\theta)$ in the direction of $(a, b)$?
I'm wondering if there is an intuitive explanation to the result above. Mathematically, it is easy to prove: If we let $x = a \cos(\theta) + b \sin(\theta)$, then setting $\frac{\mathbb{d}x}{\mathbb{d}\theta} = 0$ tells us that $\theta = \arctan(\frac{b}{a})$. Thus $\theta$ is the angle between the positive x axis and the vector $(a, b)$. Another way to show this is to notice that $x = (a, b) \cdot (\cos(\theta), \sin(\theta)) = ||(a,b)||\cos(\alpha)$. If $\cos(\alpha) = 1$, then $x$ will be maximized and will be in the direction of $(a, b)$, since $\alpha = 0$.

At the same time, the answer just seems so simple that I can't help but wonder whether there's a more obvious way to look at it. Trigonometry is not really inherent to the problem, $\cos(\theta)$ and $\sin(\theta)$ can also be replaced by the components of some vector $\vec{v}$.

答:
Per the C-S inequality

$$a \cos\theta + b \sin\theta
\le \sqrt{(a^2+b^2)(\cos^2\theta +\sin^2\theta)}
= \sqrt{a^2+b^2}
$$

where the equality occurs if $(\cos\theta, \sin\theta) $ and $(a,b)$ are in the same direction.


答:
I'm not exactly sure what is $X, Y$. But i assume it's $BA$, $BC$ and $1-X$ is $AB$. $1-Y$ is $CB$, similarly, $|X-Y|$ is $AC$, $1-|X-Y|$ is $CA$.

Given those, the goal is to find a circle with radius $r$ and three points $A, B, C$ on the circle, the minimum interval length is the random variable $M$.

using the same arc length idea. we can simplify this minimum interval problem into the minimum angle.

First, it should be easy to prove. the minimum can't be the sum of two intervals.
Namely $M \neq BC+CA, CA+AB, AB+BC$, since some of the $1-?$ is the sum of two other intervals, E.g $1-BA = AC+CB$

We can simplify $M = r*min(\alpha, \beta, \gamma)$, Since $r$ is a constant, we can remove it for easy of analysis.

Also, as $\alpha, \beta, \gamma \sim U(0, 2\pi)$.

$P(min(\alpha, \beta, \gamma) \geq z) = P(\alpha \geq z)P(\beta \geq z)P(\gamma \geq z) = (2\pi-z)^3$

The CDF would be $P(min(\alpha, \beta, \gamma) \leq z) = 1 - P(min(\alpha, \beta, \gamma) \geq z)$

Title: Most random points on a circle

Body:
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 $M$ be the minimum distance between any 2 of the points.

Find the pdf of $M$. (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 $F_M(m)= P(M \le m)$ 1st and then hope the cdf is absolutely continuous.)

My model: 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).

Question: Well, I hope to find the pdf of $M$ via its cdf, which I think I'm able to compute if I know what $M$ is. What is $M$?

I think $M=\min\{X,Y,1-Y,1-X$ $,|X-Y|,1-|X-Y|\}$, where $X$ and $Y$ are the anti-clockwise distances from $B$ to, resp, $A$ and $C$. Somehow $X$ and $Y$ are iid Unif(0,1).

How would I compute $M$'s distribution?

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?


问:
How to prove that $\phi=f:S\rightarrow \mathbb{S}^2$ is a bijection without using $h$?
If $f:\mathbb{R}^3\rightarrow \mathbb{R}$ given by $f(a,b,c)=e^{a^2}+e^{b^2}+e^{c^2}$ defining $h:\mathbb{R}\rightarrow \mathbb{R}$ by $h(t)= f(ta, tb, tc)$, we have that $h'(t)>0$ give us all conditions to define a bijection (diffeomorphism) $\phi=f:S\rightarrow \mathbb{S}^2$. Would someone explain this bijection please? And also help me to find another way to prove this? ($S=\{(a, b, c)\in \mathbb{R}^3: f(a, b, c)=a\}\; \text{with}\; a>3$)

答:
It's a slightly confusingly-worded proof, but the essential idea is to show that for every $(x,y,z)\ne 0$, the half line $\{ t(x,y,z) \mid t\ge 0\}$ intersects the surface $S$ at precisely one point. This is what the argument with $h$ shows: since $h(t)$ ranges from 3 to $+\infty$ as $t$ ranges from $0$ to $+\infty$, it must equal $a$ for at least one value of $t$. Since $h'(t)>0$, $h$ is monotonically increasing, and so it equals $a$ for at most one value of $t$. Therefore $h(t)$ equals $a$ for $precisely$ one value of $t$. For this value $t^*$ of $t$, $t^*(x,y,z)$ is an element of $S$.

Now the half line $\{ t(x,y,z) \mid t\ge 0\}$ also intersects the sphere $\mathbb{S}^2$ at precisely one point (obviously). The mapping $f:S\to \mathbb{S}^2$ given by $f(p) = \frac{p}{|p|}$ maps the point in the half line intersecting $S$ to the point intersecting $\mathbb{S}^2$, and so is injective. It's also surjective - take any $(x,y,z)\in \mathbb{S}^2$ and apply the above argument to find a $t^*(x,y,z)\in S$. Then $f(t^*(x,y,z)) = (x,y,z)$. Therefore $f$ is bijective.


问:
Has anyone studied this operator?
I've been studying a particular unary operator on the commutative ring $\mathbb{Z}/n\mathbb{Z}$. The operator is:

$\downarrow(x) = y\pmod{n}$ iff $n \equiv y \pmod{x}$, where $0< x,y \le n$.

The operator isn't distributive on addition or multiplication, so its use is probably fairly limited. I've proven a couple fairly trivial results, but I'm interested in seeing if there are more resources available.

One of the fairly trivial results:
If $n$ is composite and $n=pq$ and WLOG $p \le q$, then if $q\ge kp+k$ I can show that

$(\downarrow(q-k))\mid p$.

答: Yes. If the modulus is prime $\, n = p,\,$ then iterating the map $\ x\mapsto p\ {\rm mod}\ x\ $ yields Gauss's algorithm for computing inverses mod $\,p.\,$ By this variant of the Euclidean algorithm, Gauss deduced the prime divisor property, i.e.  if a prime divides a product then it divides some factor, which immediately yields uniqueness of prime factorizations (fundamental theorem of arithmetic).</p>

Title: Has anyone studied this operator?


问:
Determine convergence of sequence
Does the following sequnce converge or diverge? $$\sum_{n=1}^\infty\frac{n}{3^n}$$

How are these kinds of problems written down and proven? From my understanding, this will converge to $0$ because the denominator grows at a faster rate than the numerator. Is that all I can do about this problem or is there some more detailed way to solve it?

答:
Hint
If your series got strictly positive terms if

$$
\frac{u_{n+1}}{u_n} \underset{n \rightarrow +\infty}{\rightarrow} \ell<1
$$
then $\sum u_n $ converges

Title: Determine convergence of sequence


问:
Does $\text{Tr}A \leq \text{Tr}B \implies \text{Tr}PA \leq \text{Tr}QB$?
Let $A\in M_n$ and $B\in M_m$ be complex positive semi-definite matrices such that $\text{Tr}A\leq \text{Tr}B$. Let $P\in M_n$ and $Q\in M_m$ be projection matrices with the same nullity, i.e., $\ker(P)$ and $\ker(Q)$ have the same dimension. Then, can one conclude that

$$ \text{Tr}PA \leq \text{Tr} QB \,\, ?$$

答: No, we cannot conclude that (not even for $m=n$ and $P=Q$).  Pick $A=diag(1,0)$, $B=diag(0,1)$ and $P$ the projection onto the first coordinate, then $PA=A$ and $PB=0$, so $tr(PA)=1>0=tr(PB)$.


问:
Fourier transform on functions on the unit sphere bounded by the Fourier transform of measure
Let $f$ be a non-negative simple function on the unit sphere. So that the absolute value of the Fourier transform of $f(y)$ on the unit sphere is $$|\widehat{fd\sigma}| = |\int_{\mathbf{S}^n}\sum_{i=1}^N\alpha_i1_{A_i}e^{-i2\pi\langle x, y\rangle} d\sigma(x)| = |\sum_{i=1}^N\alpha_i\widehat{1_{A_i}d\sigma}|.$$ Now, isn't that true if we have $0 < \alpha_i \leq 1,$ for all $i \in \{1,...,N\}$ then we have $$|\sum_{i=1}^N\alpha_i\widehat{1_{A_i}d\sigma}| \leq |\sum_{i=1}^N\widehat{1_{A_i}d\sigma}| = |\widehat{d\sigma}|?$$
EDIT: Then we would have for general bounded $f = f_1^+ - f_1^- +if_2^+ - if_2^-$ that
$$ ||\widehat{fd\sigma}||_q  \leq 4\max_{f_1^+,f_1^-,f_2^+,f_2^-}||\widehat{f_i^j\sigma}||_q \leq 4||f||_{\infty}||\widehat{d\sigma}||_q$$ via Minkowski and a standard limit process.

答:
As the Fourier transform on the sphere is real valued, it follows that $$|\sum_{i=1}^N\alpha_i\widehat{1_{A_i}d\sigma}(y)| =|\alpha\widehat{1_{A_1}d\sigma(y)} + \beta\widehat{1_{A_2}d\sigma(y)}|,$$ for some $A,B$ with $A \cup B = \mathbf{S}^n$ and for some $0 \leq \alpha,\beta \leq 1.$
Above $$A_1 := \{ x \in \mathbf{S}^n | \cos{\langle x, y \rangle}  \geq 0 \}$$ and
$$A_2 := \{ x \in \mathbf{S}^N | \cos{\langle x,y \rangle}  < 0 \}.$$
Now, there at least exist a point wise maximizer because
$$|\alpha\widehat{1_{A_1}d\sigma(y)} + \beta\widehat{1_{A_2}d\sigma(y)}| \leq \max_{A_1,A_2}|\widehat{1_{A_i}d\sigma(y)}| := g(y)$$ This means that for all bounded positive functions $f$ it holds that $|\widehat{fd\sigma}(y)|$ is bounded point wise by $||f||_{\infty}g(y).$ However, the above function $g$ is not dominated point wise by $|\widehat{d\sigma}|.$


问:
Inequality with two variables and a conditional equation
Can someone help me solve this inequality?

Let $x$ and $y$ be real numbers with $(x + 1) (y + 2) = 8$.
Prove:
$$(xy - 10)^2 \geq 64$$
Furthermore, determine all pairs $(x,y)$ of real numbers for which equality holds.

答:
Here's a quick algebraic proof: Note that clearly $y \neq -2$ (otherwise the product would be 0). Thus we get $x = \frac{8}{y+2} - 1 = \frac{6-y}{y+2}$.
Now we have $xy = \frac{y(6-y)}{y+2}$ and some quick sketching and checking extrema points should give the inequality. However we can plug in and prove it directly: the condition to prove becomes

$$(xy-10)^2 \geq 8^2 \iff (xy-18)(xy-2) \geq 0 \iff \left( \frac{y(6-y)}{y+2} - 18\right)\left(\frac{y(6-y)}{y+2} - 2\right) \geq 0$$

The expression can be rewritten as
$$\frac{1}{(y+2)^2}(-y^2 + 6y - 18y - 36)(-y^2 +6y - 2y - 4) \geq 0$$
and now this immediately becomes

$$\frac{1}{(y+2)^2}(y+6)^2(y-2)^2 \geq 0,$$

which is clearly true. $\square$

We have equality when $y=-6$ or $2$, so $(x,y) = (-3,-6)$ or $(1,2)$.


问:
Polynomials that induce the zero function mod $n$

Which polynomials induce the zero function mod $n$?

In particular:

What is the polynomial of least degree that induces the zero function mod $n$?

What is the monic polynomial of least degree that induces the zero function mod $n$?

These are not vacuous questions because of the following general result:

If $r$ is the maximum exponent in the prime factorization of $n$, then $x \mapsto x^{r+\lambda (n)}-x^r$ is the zero function mod $n$. [Wikipedia]

Here, $\lambda$ is the Carmichael function.

When is $x^{r+\lambda (n)}-x^r$ the monic polynomial of least degree that induces the zero function mod $n$?

Fermat's theorem implies that $x^n-x$ is the answer for $n$ prime: all polynomials that induce the zero function mod $n$ are a multiple of $x^n-x$.
How can this be generalized to composite $n$?

Here are some other examples:
$$
\begin{array}{rll}
n & L_n: \text{least degree} & M_n: \text{least degree monic}
\\ 2 & x^2+x
\\ 3 & x^3-x
\\ 4 & 2(x^2+x) & x^4-x^2
\\ 5 & x^5-x
\\ 6 & 3(x^2+x) & x^3-x
\\ 7 & x^7-x
\\ 8 & 4(x^2+x) & x^4+2x^3+3x^2+2x = x(x+1)(x^2+x+2)
\\ 9 & 3(x^3-x) & x^8-x \quad (???)
\\ 10 & 5(x^2+x) & x^5-x
\\ 11 & x^{11}-x
\\ 12 & 6(x^2+x) & x^4+5x^2+6x = x(x+1)(x^2-x+6)
\\ 13 & x^{13}-x
\\ 14 & 7(x^2+x) & ???
\\ 15 & 5(x^3-x) & x^5-x
\\ 21 & 7(x^3-x) & ???
\\ 24 & 12(x^2+x) & x^4+2x^3+11x^2+10x = x(x+1)(x^2+x+10)
\end{array}
$$

It seems clear that $L_{2m}=m(x^2+x)$, because $x^2+x$ is always even.
More generally,

Is $L_{pq} = qL_p$ and $M_{pq}=L_q$ for $p<q$ primes?

If $n=pm$ and $p$ is smallest prime divisor of $n$, then is $L_{pm}=mL_p$?

Corrections and additions welcome.
Please collect partial results as answers.

答:
Which polynomials induce the zero function mod $n$?

As proved in another answer, the polynomials that induce the zero function mod $n$ are precisely those that, when written in the form
$$f(x) = \sum_{k} c_k x^{\underline k},$$
satisfy the property that $c_k$ is divisible by $n\,/\gcd(n, k!)$, for each $k$. (Here, $x^{\underline k}$ is the "falling factorial": $x^{\underline k} = x(x-1)\cdots(x-k+1)$.)

That answers the first question, and to answer the other questions:

What is the polynomial of least degree that induces the zero function mod $n$?

I'm assuming you don't accept polynomials like $f(x) = 0$ or $f(x) = n$ or $f(x) = nx + 2n$, etc, i.e. you want the coefficients to be nonzero mod $n$. When $k$ is less than the smallest factor of $n$, we get $\gcd(n, k!) = 1$, which means that $c_k$ is a multiple of $n$. We can take $c_k$ nonzero mod $n$, when $\gcd(n, k!) > 1$, which happens when $k$ is the smallest prime factor of $n$, say $p$. Then we need $c_k$ to be a multiple of $n / \gcd(n, k!) = n / p$, so we can take $c_k$ to be $n/p$ and all other coefficients $0$, so that $f(x) = (n/p) x^{\underline p} = (n/p) x(x-1)\cdots(x-p+1)$. Other polynomials of degree $p$ may also be possible.

This gives the following table, as in the question:

$$
\begin{array}{rll}
n & L_n\text{ from question} & L_n\text{ from this answer}
\\ 2 & x^2+x & (2/2)x^{\underline 2} = x^2 - x
\\ 3 & x^3-x & (3/3)x^{\underline 3} = x(x-1)(x-2) = x^3 - 3x^2 + 2x
\\ 4 & 2(x^2+x) & (4/2)x^{\underline 2} = 2(x^2 - x)
\\ 5 & x^5-x & (5/5)x^{\underline 5} = x^5 - 10x^4 + 35x^3 - 50x^2 + 24x
\\ 6 & 3(x^2+x) & (6/2)x^{\underline 2} = 3(x^2 - x)
\\ 7 & x^7-x & (7/7)x^{\underline 7} = …
\\ 8 & 4(x^2+x) & (8/2)x^{\underline 2} = 4(x^2 - x)
\\ 9 & 3(x^3-x) & (9/3)x^{\underline 3} = 3(x^3 - 3x^2 + 2x)
\\ 10 & 5(x^2+x) & (10/2)x^{\underline 2} = 5(x^2-x)
\\ 11 & x^{11}-x & (11/11)x^{\underline{11}} = …
\\ 12 & 6(x^2+x) & (12/2)x^{\underline 2} = 6(x^2 - x)
\\ 13 & x^{13}-x & (13/13)x^{\underline{13}} = …
\\ 14 & 7(x^2+x) & (14/2)x^{\underline 2} = 7(x^2 - x)
\\ 15 & 5(x^3-x) & (15/3)x^{\underline 3} = 5(x^3 - 3x^2 + 2x)
\\ 21 & 7(x^3-x) & (21/3)x^{\underline 3} = 7(x^3 - 3x^2 + 2x)
\\ 24 & 12(x^2+x) & (24/2)x^{\underline 2} = 12(x^2 - x)
\end{array}
$$

so you see it matches in each case (is congruent mod $n$). It's not too hard to prove that for prime $p$, $x^{\underline p}$ is the same as $x^p - x$ modulo $p$.

What is the monic polynomial of least degree that induces the zero function mod $n$?

For this we need $c_d = 1$ where $d$ is the degree of the polynomial, and this must be a multiple of $n / \gcd(n, d!)$, so $n / \gcd(n, d!)$ must be $1$, or in other words $\gcd(n, d!) = n$ which means that $n$ divides $d!$. This is called the Kempner function. So we can take $c_d = 1$ and all other $c_k$ to be $0$.

This gives the following table, for the examples in the question:

$$
\begin{array}{rll}
n & M_n\text{ from question} & M_n\text{ from this answer}
\\ 2 & x^2+x & x^{\underline 2}
\\ 3 & x^3-x & x^{\underline 3}
\\ 4 & x^4-x^2 & x^{\underline 4} = x^4 - 6 x^3 + 11 x^2 - 6 x
\\ 5 & x^5-x & x^{\underline 5}
\\ 6 & x^3-x & x^{\underline 3} = x^3 - 3 x^2 + 2 x
\\ 7 & x^7-x & x^{\underline 7}
\\ 8 & x^4+2x^3+3x^2+2x & x^{\underline 4}
\\ 9 & x^8-x \quad (???) & x^{\underline 6}
\\ 10 & x^5-x & x^{\underline 5}
\\ 11 & x^{11}-x & x^{\underline{11}}
\\ 12 & x^4+5x^2+6x & x^{\underline 4}
\\ 13 & x^{13}-x & x^{\underline{13}}
\\ 14 & ??? & x^{\underline 7}
\\ 15 & x^5-x & x^{\underline 5}
\\ 21 & ??? & x^{\underline 7}
\\ 24 & x^4+2x^3+11x^2+10x & x^{\underline 4}
\end{array}
$$

which answers some of the $???$s in the question, corrects one of them ($M_9$ has degree $6$, not $8$), gives the same polynomials mod $n$ for some $n$, and some truly different (though they can be checked to work: so $M_n$ is not unique as a polynomial, though the polynomial function (mod $n$) itself is unique).

When is $x^{r+\lambda (n)}-x^r$ the monic polynomial of least degree that induces the zero function mod $n$?

As "the monic polynomial of least degree" is not unique (see the examples just above), the relevant question is to ask when the degree matches. For this to happen, $r + \lambda(n)$ must be the Kempner function of $n$ (let's call it $K(n)$), i.e. the smallest $k$ such that $n$ divides $k!$. This happens when either:

$n = 4$, or

$n = 12$, or

$n = p_1 p_2 \ldots p_m$ a product of distinct primes and $p_m - 1$ is divisible by $p - 1$ whenever $p$ is any of the primes $p_1 < p_2 < \ldots < p_m$. (This case includes the cases of $n$ being prime, $n$ being $2p$, or $3p$, or $2 \cdot 3 \cdot p$, or $5p$ where $p \equiv 1 \pmod 4$, or $11p$ where $p > 11$ and $p \equiv 1 \pmod{10}$, etc.)

The proof is as follows. First, for prime powers $n = p^r$, note that $\lambda(n) = p^{r-1}(p-1)$ while $K(n) \le rp$ (it's equal when $r \le p$, and smaller for larger $r$). So for them to be equal, we need $r + p^{r-1}(p-1) \le rp$ which happens only when $r=1$ or for $(p, r) = (2, 2)$. Next, when $n$ has prime factorization $n = p_1^{e_1} \cdots p_m^{e_m}$, the Carmichael function $\lambda(n) = \operatorname{lcm}(\lambda(p_1^{e_1}), \ldots, \lambda(p_m^{e_m}))$. Meanwhile, the RHS $K(n)$ is $\max(K(p_1^{e_1}), \ldots, K(p_m^{e_m}))$. Now if the maximum on the right is attained at a certain $J$, then the same inequality gives that either $e_J = 1$ and $r = 1$ so $n$ is a product of distinct primes and the property above needs to hold so that the LCM of the values $(p_j - 1)$ doesn't exceed $p_m - 1$, or else the special case of $J = 1$, $p_1 = 2$, $e_1 = 2$, and other primes if any are less than $2^2$ so $n$ is either $2^2 = 4$ or $2^2 \cdot 3 = 12$.

To answer the final two questions (and the comment before them):

It seems clear that $L_{2m}=m(x^2+x)$

Yes we can take $m(x^2 + x)$; note that our above method gives the polynomial $(2m/m)x^{\underline 2} = m(x^2 - x)$ which is the same mod $2m$.

Is $L_{pq} = qL_p$ and $M_{pq}=L_q$ for $p<q$ primes?

For $L_p$ we can take $(p/p)x^{\underline p}$ and for $L_{pq}$ we can take $(pq/p) x^{\underline p}$, so yes for the first part. Similarly, for the second part, we can take $L_q = x^{\underline q}$ and $M_{pq}$ to be $x^{\underline q}$ as well, as $pq$ divides $q!$ when $p < q$.

If $n=pm$ and $p$ is smallest prime divisor of $n$, then is $L_{pm}=mL_p$?

Yes, as above, for $L_p$ we take $x^{\underline p}$ and for $L_{pm}$ we take $(pm/p) x^{\underline p}$.


问:
Finding the lowest sum of the sequence possible given some conditions
Say we have an infinite sequence of natural numbers $A$ such that no $k$ subsequences can be found adjacent such that the average of the elements in any subsequence is equal for all $k$ subsequences. Sorry about my poor description, an example would be that $\{2, 3, 4, 1\}$ wouldn't work for $k=2$ because $\{2, 3\}$ and $\{4, 1\}$ are adjacent and both their averages are $\frac{5}{2}$. $\{2, 3, 10, 4, 1\}$ would work however because $\{2, 3\}$ and $\{4, 1\}$ are no longer adjacent. Anyway, my question is: which sequence that follows this has the lowest sum for $k$? If that's too general, then which for $k=2$? Honestly, any information on the behavior of this sequence would be great. To me, it looks like for $k=2$ is $\{1, 2, 1, 3, 1, 2, 1, 4 ...\}$ but I have no idea how to prove that it is. Thanks.</p>

答:
Let's try to sharpen the question:

Your example shows that you are considering non-overlapping subsequences of $A$, i.e., you partition $A$ into fragments of fixed length. So let's try some definitions:

For any sequence $s$, let $W_k(s)$ be the sequence whose $n$th term (counting from 0) is given by $s_{kn} + s_{kn+1} + s_{kn+2} + ... + s_{kn+k-1}$. Note that since the average of $k$ numbers is just their sum divided by $k$, we can consider sums instead of averages. ($W$ is for window: A fixed-length fragment).

For any sequence $s$, define $\Sigma s$ to be the sequence whose $n$th term is the sum of the first $n$ terms of $s$.

For two sequences $s$ and $t$, we say that $s < t$ if there is some $m \in N$ such that $s_n < t_n$ for all $n > m$ (I.e., $s<t$ "in the limit").

So, for any $k$ you want to find the sequence $A$ that satisfies the following:

$W_k(A)$ has no adjacent terms that are equal.

If $B$ is another sequence for which $W_k(B)$ has no repeated terms,
$\Sigma A < \Sigma B$ (in the sense of definition 2).

If all this is correct, my candidate for minimum at $k=2$ would be the following periodic sequence (I have bracketed the period):

$$A = [ 1, 1, 1, 2, ] 1, 1, 1, 2, ...$$

This matches the condition, since $W_2(A) = 2, 3, 2, 3, ...$.

I believe it's pretty easy to see how to construct similar sequences for $k > 2$. A more interesting problem would be if you misunderstood the conditions, and the adjacent subsequences are overlapping rather than disjoint. (E.g., terms $s_0s_1, s_1s_2, s_2s_3, ...$).


问:
Find examples that f(x) is discontinuous at $x_0$ $\in$ X but 1.$|f(x)|$ is continuous at $x_0$.
$Find$ examples that $f(x)$ is discontinuous at $x_0$$\in X$  $but$

$1$.$|f(x)|$ is continuous at $x_0$.

$2$.$f^2(x)$ is continuous at $x_0$.

For first one I am thinking about $f(x)$ = \begin{cases}1 &x\geq0\\-1 & x<0\end{cases}

but can't bring any example for $second$ one.

答:
Draw the graph and see what happens, how $|f|$ can be continuous even though $f$ isn't: They meet the $y$-axis at exact opposite values.

Make it so the two horizontal lines are further apart. Make it so that they are sloped instead of horizontal. Make it so they are curved instead of straight. There are plenty of ways to make more examples.

Here are a few concrete formulas:
$$
x\mapsto\cases{-2& if $x<0$\\2& otherwise}\\
x\mapsto\cases{x-1& if $x<0$\\1-x& otherwise}\\
x\mapsto\cases{x^2-1& if $x<0$\\x^2+3x+1& otherwise}
$$


问:
Is there a known formula for the number of $k^{\text{th}}$ power residues modulo $2^n$?
We define a $k^{\text{th}}$ power residue modulo $n$ to be an integer $a$ coprime to $n$ such that there exists an integer $x$ satisfying $$x^k\equiv a\pmod{n}.$$ A fundamental question that we can ask is:

Given fixed integers $k\ge 2$ and $n\ge 2,$ how many $k^{\text{th}}$ power residues modulo $n$ exist?

A variant of the Chinese remainder theorem for $k^{\text{th}}$ powers shows that this is the product of the answers when the moduli are the the maximal prime powers in the prime factorization of $n.$ For powers of odd primes $p,$ this is solved by a variant of Euler's criterion (see section 6.5 of Vinogradov's Elements of Number Theory).

But what about when the modulus is a power of $2$? Unfortunately, Vinogradov's  proof in the odd prime case uses primitive roots, which does not exist modulo powers of $2$ that are greater than $4=2^2$. In the quadratic case, a paper of Stangl provides a formula on p.287-288. His method uses the difference of squares factorization and other ideas that do not readily extend to higher powers, at least not as far as I can tell.

Please let me know if there is a formula for the number of $k^{\text{th}}$ power residues modulo $2^n$ for fixed integers $k\ge 2$ and $n\ge 3$.

The answer might have to do with the structure of $(\mathbb{Z}/2^n \mathbb{Z})^{\times}$ but I'm not sure.

答:
Here is the full proof based on ArturoMagidin's comments on the original post (this proof appears in a book that I am writing and a paper that I wrote that is currently under review). Let $k\ge 2$ and $n\ge 3$ be fixed integers. As I wrote in the proof here, the invertible elements modulo $2^n$ are given by the $\varphi(2^n) = 2^{n-1}$ distinct elements of $$S=\left\{\pm 5^k : k\in [2^{n-2}]\right\},$$ where $[t]$ denotes $\{1,2,\ldots, t\}$. We will show that
$$|S_k (2^n)| =
\begin{cases}
    \hfill \dfrac{2^{n-2}}{(k, 2^{n-2})} &\text{ if } 2\mid k\\
    \hfill 2^{n-1} &\text{ if } 2\nmid k
\end{cases}.
$$
where $S_k (n)$ denotes the set of reduced $k^{\text{th}}$ power residue classes modulo $n$, meaning those residue classes that are coprime to $n$.

We will use the fact that, for $k\ge 2$, the $k^{\text{th}}$ power map modulo an integer $n\ge 2$ is bijective on the set of reduced residue classes modulo $n$ if and only if $(k,\varphi(n))=1.$ Let $k=2^j m,$ where $m$ is odd. If $j=0,$ then the exponent $k$ is odd and so coprime to $2^n.$ By the aforementioned fact, all integers coprime to $2^n$ are in $S_k (2^n),$ so $|S_k (n)| = 2^{n-1}.$ Now we can assume that $j\ge 1.$ We can think of the $k^{\text{th}}$ power map as taking a power of $m$ of each element of $S$ followed by taking a power of $2^j$ of each element of $S.$ Again by the aforementioned fact, the first map is irrelevant because it is a bijection on $S.$ What really matters is the application of the second map to $$S=\left\{\pm 5^k : k\in [2^{n-2}]\right\}.$$ Since $j\ge 1,$ we can remove any negative signs and we get the set $$T=\left\{ (5^k)^{2^j} : k\in [2^{n-2}]\right\}.$$ Not every $(5^k)^{2^j}$ is necessarily distinct modulo $2^n$ so we have to determine to amount of duplication, or rather the number of distinct elements defined as that will be the cardinality of $S_k (2^n)$. We rewrite each element as $(5^{2^j})^k$ for $1\le k\le 2^{n-2}.$ By the standard proof of the fact that there is no primitive root modulo powers of $2$ (which proves that odd integers have order less than or equal to $2^{n-2}$ modulo $2^n$), we find that $$(5^{2^j})^{2^{n-2}} \equiv (5^{2^{n-2}})^{2^j}\equiv 1^{2^j} \equiv 1\pmod{2^n},$$ and higher powers of $5^{2^j}$ cycle through lower powers. So all elements generated by the powers of $5^{2^j}$ are in $T,$ in addition to every element of $T$ being a power of $5^{2^j}.$ Thus, the number of distinct elements of $T$ modulo $2^n,$ and therefore the cardinality of $S_k (2^n),$ is $$|S_k (2^n)| =\text{ord}_{2^n}(5^{2^j})=\frac{\text{ord}_{2^n}(5)}{(2^j, \text{ord}_{2^n}(5))}=\frac{2^{n-2}}{(2^j, 2^{n-2})}=\frac{2^{n-2}}{(k, 2^{n-2})}.$$


问:
property of extremum of two variable function
What can we say about $u_{xx}(x_0,y_0)$, $u_{yy}(x_0,y_0)$ and $u_{xy}(x_0,y_0)$ if we know $u$ reaches its maximum at $(x_0,y_0)$? Especially, what is the sign of each term? </p>

答:
When a local maximum is attained:

$H(x_0,y_0) = u_{xx}(x_0,y_0)  \,u_{yy}(x_0,y_0)  - (u_{xy}(x_0,y_0) )^2 \geq 0 $.

$u_{xx}(x_0,y_0) \leq 0$, $u_{yy}(x_0,y_0) \leq 0 $.

We only know $|u_{xy}(x_0,y_0)| \leq \sqrt{u_{xx}(x_0,y_0)  \,u_{yy}(x_0,y_0) }$, but sign we can't tell much, could be either case.


问:
Tensor product anticommutative?
The tensor product is anticommutative? I mean, is the following true?
$$A B dt\otimes dr=-A B dr\otimes dt$$ with $A$ and $B$  smooth functions.

My doubt comes out for the fact that if I have $(Adt+Bdr)\otimes (Adt+Bdr)=A^2dt\otimes dt+B^2dr\otimes dr+ABdt\otimes dr+ABdr\otimes dt$

Can I cancel the cross products each other?

答:
First, don't call the tensor product a "cross product", as that will inevitably be very confusing. Secondly, no, the tensor product is not anticommutative in general. In fact, the tensor product is defined as being "only bilinear and associative". It is the most general algebra of the kind (you generate a tensor algebra by giving a vector space to a so-called "free functor" that turns it into a tensor algebra in a standard way).

There are interesting algebraic quotients of the tensor algebra (doing an algebraic quotient is like injecting a property into another space using equivalence classes); one of which IS anticommutative for the tensor product (then called the exterior, or outer, product), called the exterior algebra.

However, your example looks like an exterior algebra example (often used for integration), so you might be in a context where you have an anticommutative outer product rather than a tensor product.

Title: Tensor product anticommutative?


问:
Verification of an integral reduction for $I_n=\int_{0}^{2}(2x-x^2)^n\,dx$

Find the reduction formula for the integral $$I_n=\int_{0}^{2}(2x-x^2)^n\,dx.$$

I wanted to verify my formula for this integral. Please check for mistakes.

My Solution:

Since $2x-x^2$ is symmetric about $x=1$ in interval $[0,2]$, we can write the integral as $$I_n=2\int_{0}^{1}x^n(2-x)^n\, dx.$$

Using $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)\,dx$, we get:

$$I_n=2\int_{0}^{1}(1-x)^n(1+x)^ndx=2\int_{0}^{1}(1-x^2)^{n}\,dx.$$

Let
\begin{align*}
x&=\sin\theta\\
dx&=\cos\theta\, d\theta.
\end{align*}

Thus $$I_n=2\int_{0}^{\pi/2}\cos^{2n+1}\theta\, d\theta.$$

Now we know that $$2k\int_{0}^{\pi/2}\cos^k\theta\,d\theta=2(k-1)\int_{0}^{\pi/2}\cos^{k-2}\theta\, d\theta.$$

Putting $k=2n+1$, we get:

$$(2n+1)I_n=2nI_{n-1}\implies I_n=\dfrac{2nI_{n-1}}{2n+1}.$$

Please check this solution for any mistakes. Also please help in finding $\lim_{n\to \infty}I_n.$

Thank you

答:
As for the $\lim_{n\to\infty}I_{n}$, we can use the Dominated Convergence Theorem.

In this case, it suffices to check that

The sequence of functions $\{f_n\}$ given by $f_{n}(x) = (2x-x^2)^n$ converges pointwise to some function $f$ on the interval $[0,2]$.

There exists some integrable function $g$ such that for all $x \in [0,2]$ and for all $n$ we have $|f_n(x)| \le g(x)$.

If both conditions are met we can conclude that $$\lim_{n\to\infty}I_{n} = \lim_{n\to\infty}\int_0^2 f_n(x)\,dx = \int_0^2\lim_{n\to\infty}f_n(x)\,dx = \int_0^2 f(x)\,dx.$$

Considering the pointwise limit, we have that for $x\in [0,2]$,
\begin{align*}
\lim_{n\to\infty}(2x- x^2)^n &= \begin{cases} 1, &x = 1\\
0, &\text{else}
\end{cases}\\
&= 0\quad \text{a.e.}
\end{align*}
which means the first condition is satisfied. Furthermore, for $x\in [0,2]$, $|(2x-x^2)^{n}| \leq 1,$ meaning the second condition is also satisfied.  Therefore, we have that $$\lim_{n\to\infty}I_{n} = \int_{0}^{2} 0\,dx = 0.$$


问:
Is $(B' \cap C' \cap A) \cup (C \cap A) \cup (B \cap A) = A$ always true, or only sometimes?

Is this true, always, or do exceptions exist?
$$(B' \cap C' \cap A) \cup (C \cap A) \cup (B \cap A) = A$$

I can't seem to find any... but Mathematica tells me it's only true sometimes...

What do you think?

Thanks

答:
For any sets $X$ and $Y$, it is always true that $$X = (X\cap Y)\cup (X\cap Y').$$
Your formula is a special form of this one, with $X = A$ and $Y = B\cup C$.

I tried your formula in Wolfram too, and it says it's true for any $A, B, C$. Perhaps a typo?


问:
Determine if it is associative
I have problems with exercise

Determine which of the following binary operations is associative:

The operation $\cdot$ on $\mathbb{Q}$ defined by $a\cdot b = \displaystyle \frac{a+b}{5}$

I have to prove that $(a*b)*c=a*(b*c)$

$(a*b)*c= (\displaystyle \frac{a+b}{5}) *c $
$= \displaystyle\frac{\displaystyle\frac{a+b}{5}+c}{5}$

How can I continue?

Thanks

答:
No, the binary operation is not associative.

$(a*b)*c= (\displaystyle \frac{a+b}{5}) *c $
$= \displaystyle\frac{\displaystyle\frac{a+b}{5}+c}{5}$

can be simplified as $(a+b+5c)/25$

while $a*(b*c)= (5a+b+c)/25$

Title: Determine if it is associative


问:
Counting interpretation of $\sum^{n}_{i=0}\binom{n}{i}i^n$
$$\sum^{n}_{i=0}\binom{n}{i}i^n$$

I can't find a counting interpretation of this formula. I thought of using the binomial theorem but I don't think it applies here. So, I am not sure how to begin.

答:
I will provide a proof of the interpretation that the sum you've given counts the number of functions from $[n]$ to a subset of $[n]$ summed over all possible subsets(mentioned by @metamorphy) and also $[n] = \{1,2,...,n\}$.Consider a function $$ f : [n] \to [i] $$ Each element of $[n]$ has $i$ choices to map to. So there are $i^n$ functions in this case. Also there are $\dbinom{n}{i}$ subsets of $[n]$ of cardinality $i$ from the definition of binomial coefficients. Summing over all possible subsets you get    that the number of functions  from $[n]$ into a subset of $[n]$ is
$$\sum_{i=0}^{n}  \binom{n}{i}i^n$$