问:
$k$-th moment of product of gaussian and sinc
I would like to calculate the following integrals:

$$\int_{-\infty}^{+\infty} \quad x^k\quad \left(\frac{\sin(\pi a x)}{\pi ax}\right)^2\quad \exp(-bx^2)\,dx$$

$$\int_{-\infty}^{+\infty} \quad x^k\quad \left(\frac{\sin(\pi a x\pm\pi)}{\pi ax\pm\pi}\right)^2\quad \exp(-bx^2) \,dx$$

Thanks!

答:
$$\int_{-\infty}^{+\infty} \quad x^k\quad \left(\frac{\sin(\pi a x)}{\pi ax}\right)^2\quad \exp(-bx^2)\,dx$$

can be solved using known integrals involving bessel function of the first kind.

Since
$$\left(\frac{\sin(\pi a x)}{\pi ax}\right)^2=\frac{1}{2ax}J_{\frac{1}{2}}(\pi ax)J_{\frac{1}{2}}(\pi ax)$$

can be written as

$$\int_{-\infty}^{+\infty} \quad\frac{1}{2a} x^{k-1}\,J_{\frac{1}{2}}(\pi ax)J_{\frac{1}{2}}(\pi ax)\, \exp(-bx^2)\,dx$$

That is a known integral from Volume II of "Higher Transcendental Functions". It holds for $k\geq 0, \,a>0,\,b>0$.


问:
Is the following an example of a non-measurable event?
Goodmorning everyone.

I started reading DeGroot and Schervish's 'Probability and Statistics' (4th editon) and wondered,  is there an uncountable union of events, that is not an event?

This question has been answered on this site.
The topic of question is "What is an uncountable union of events?" (sorry ba I can't report the code or the link of the question that has already been answered).

The responses were very illuminating,
however, re-reading the example of an uncountable union of events that is not an event, I'm referring to Arthur's answer, I wonder if the example is adequate. This confuses me.

I report the example referred to in the aforementioned answer and then I ask my question.

Example:
Assuming the Axiom of Choice (which is a very reasonable and common thing to do, but not universal), you can construct unions of events which, if allowed to be events themselves, will have a problematic probability of ocurring. Basically, it can't be 0 and it can't be positive.

To see it in action, let's say your experiment is to pick a point uniformly at random on a circle. Then any point is an event. Using the AoC, we can construct (or more correctly, we can prove the existence of) a set of points A0 on the circle with a special property: Rotating the set along the circle by any rational angle α∈(0∘,360∘) results in a new set of points Aα. None of these Aα have any points in common with any of the others, but together they cover the entire circle.

So, if we were to assign some probability p to picking a point in A0, then by rotational symmetry the same probability should apply to any of the Aα. And since they are all pairwise disjoint, and they together cover the circle, the sum of all those p's should be 1. Thus we have
∑α∈[0,360)∩Qp=1

But if p is 0, the sum is 0, and if p is positive, then the sum is infinite. So it is impossible to assign a probability to this union A0, and therefore we are better off not calling it an event.

Question
Why can't I consider the set of rational points to have zero measure? If I imagine the circle composed of rational and irrational points, why not assign a measure of nothing on the first (rational) and measure equal to 1 on the second (irrational)? so, having admitted the above, am I wrong if I consider the circle as a union of rational and irrational points?

Thanks for any answer or clarification.

Francesco.

答:
Let us divide the answer in two parts:

Consider the space $[0,1]$ with the Lebesgue $\sigma$-algebra and the Lebesgue measure.
We know that there are subsets of $[0,1]$ that are not Lebesgue. One of such subsets is the Vitali set. Let $V$ be the Vitali set. Then $V$ is uncountable and it is not an event. On the other hand, every single point in $V$ is an event (is measurable). So $V$ is an example of an uncountable union of events that is not an event.

(For more details on Vitali set, see, for instance : https://en.wikipedia.org/wiki/Vitali_set)

Arthur answer's in What is an uncountable union of events? includes the "construction" of a non-measurable subset in the circle, very similar to the Vitali subset of $[0,1]$ .

If you want to go deeper and understand the proof of the existence of Vitali set, I suggest you do it, with the usual Vitali set (subset of $[0,1]$). After that, you can read Arthur's answer and see how the same argument is adapetd to the circle.

Remark 1: Regarding your questions at the end.

"Why can't I consider the set of rational points to have zero measure?"

Answer: the set rational points has zero measure

"If I imagine the circle composed of rational and irrational points, why not assign a measure of nothing on the first (rational) and measure equal to 1 on the second (irrational)?"

Answer: Assuming the circle has measure 1, the set of irrational points has measure 1.

"Am I wrong if I consider the circle as a union of rational and irrational points?"

Answer: No, you are not wrong. But Arthur's answer is not about this.

Remark 2: Explaining the "Vitali set" in the circle.

You don't get just one point to perform rational rotations.

Given any angle $\theta \in [0^\circ, 360^\circ)$, let us write $R_\theta$ to indicate the rotation by $\theta$.

You first consider the equivalence relation between points in the circle:
$x \sim y$ iff there a rational angle $\alpha$ such that the $R_\alpha(x)=y$. It is easy to see that such relation is actually an equivalence relation.

Then (using the Axiom of Choice) we choose one and just one element (one point) from each equivalence class and with them we have a set $A$. Note that $A$ has two interesting properties:

For any two rational angles $\alpha, \beta$, if $\alpha \neq \beta$, then $R_\alpha(A) \cap R_\beta(A) =\emptyset$.

If $C$ is the whole circle,
$$C = \bigcup_{\alpha \textrm{ rational angle in } [0^\circ, 360^\circ)} R_\alpha(A)$$

In fact, for 1: Let $\beta'$ be the (rational) angle in $[0^\circ, 360^\circ)$ congruent to $-\beta$ and let $\gamma$ be the (rational) angle in $[0^\circ, 360^\circ)$ congruent to $\alpha-\beta$.
If there is $p\in R_\alpha(A) \cap R_\beta(A)$, then
there is $q=R_{\beta'}(p)$ and  $q \in R_\gamma(A) \cap A$. But it means that $q \in A$ and there is $q_1 \in A$ such that $q=R_\gamma(q_1)$. Since $\alpha \neq\beta$, we have $\gamma \neq 0$ and $q \neq q_1$. But $q \sim q_1$. However, it is not possible, because in $A$ there is only one representative of each equivalence class.

For 2: Just note that every point of the circle is in one equivalence class. It means that, for every $p$ in the circle, there is $q\in A$ and a rational angle $\alpha \in [0^\circ, 360^\circ)$, such that $p=R_\alpha(q)$.

Since the set of rational angle is countable, the measure of the whole circle is finite, and for any rational angle $\alpha \in  [0^\circ, 360^\circ)$, $R_\alpha(A)$ would have the same measure of $A$, we conclude that $A$ is not measurable (it is not an event).

Remark 3: @Francesco, to answer your questions in the comments,
let me give more details regarding the "Vitali set" in the circle.

Let us pick just one point $p$ in the circle. By applying rotations by a rational angle, we produce a countable set of points (why countable? because there are only a countable "quantity" of rational angles). This countable set of points is the equivalence class of $p$.  In this equivalence class you must pick exactly one point to represent the equivalence class.

Is it possible that this equivalence class includes all points in the circle?  The answer is NO, because the circle is an uncountable set of points. So, there are others equivalence classes, all of them countable sets.  In fact, it is known that taking the union of any countable family of countable sets produces a countable set as result. Since the circle is uncountable, we can conclude that there is an uncountable "quantity" of such equivalence classes.

A simple example that one such equivalence does not include the whole circle is:  let us start with the point at $0^\circ$. Any rotation by a rational angle wil produce a point located at a rational angle. So, for instance, the point located at angle ${\sqrt{2}\,}^\circ$ can not be produced from our initial point by a rotation by a rational angle.

Another point that you need to keep in mind is that for any set $S$ and any equivalence relation on $S$, the equivalence classes are disjoint  and the union of all equivalence classes is the $S$ (that is, every point of $S$ is one equivalence class).

ContentLicense: CC BY-SA 4.0


问:
Is it possible to solve this nonlinear first order ODE using elementary methods?
Is there a trick, such as substitution or otherwise to solve this ode using elementary methods?

$$
  y = \left(y'\right)^3 y^2 + 2 x y'
$$

Maple says this is 1st order with linear symmetries. Tracing it, is seems to be using Lie methods which I do not know too well yet or something it calls 1st order, parametric methods which I also did not study.

I wonder if someone can see a simpler method to solve this? This is my attempt.

First solved for $y'$. This gives 3 ode's. Two of them are complex, so I threw these away as I am looking for real solution $y(x)$. The third one turned out to be isobaric. So I solved it using substitution as normally is done. But this gives integral which can't be solved. So my final solution was not verified by Maple and it is too complicated.

Maple gives these solutions (I show only the real ones). The first three solutions below are singular solutions. So the last two ones below is what I am interested in finding for now.

\begin{align*}
y \! \left(x \right) &=
-\frac{2 2^{\frac{1}{4}} 3^{\frac{1}{4}} \left(-x^{3}\right)^{\frac{1}{4}}}{3}\\
 y \! \left(x \right) &=
\frac{2 2^{\frac{1}{4}} 3^{\frac{1}{4}} \left(-x^{3}\right)^{\frac{1}{4}}}{3}\\
y(x) &=0\\
y \! \left(x \right) &= \sqrt{c_1^{3}+2 c_1 x}\\
y \! \left(x \right) &= -\sqrt{c_1^{3}+2 c_1 x}
\end{align*}

Is it possible to solve this ode, without using Lie symmetry methods using something more basic. most likely some smart substitution?

Just looking for hint.

答:
See Kamke, E. Differentialgleichungen, Lösungsmethoden und Lösungen...$\quad$ 1.541.

Let $u(y)=1/y'$, $x=x(y)$.

We diff(%,y) and get
$$(yu'-u)(2y+u^3)=0$$
Then solutions is
$$y^2=2Cx+C^3$$
and
$$32x^3+27y^4=0$$


问:
Determining if a recursive sequence is converging with induction
Consider this sequence

$$ a_0 = 2,$$
$$ a_{n+1} = \frac 1 a_n $$

Now to prove for convergence I simply put in values ($a_1, a_2, a_3$) and got the the sequence has this behaviour: $2, 1/2, 2, 1/2,\dots$. Now since it isn't approaching any value it shouldn't be converging.

But I would like to prove this, so I looked up how to do it and you're supposed to prove monotony and limitations of the sequence. Now to prove monotony we would have to assume that the sequence is falling or raising. But how can I do that here? Since the sequence is changing its value from <1 to >1 in every iteration what should I assume? Is there a way to solve this without the "obvious" answer?

Thank you!

答: If the sequence $\{a_n\}$ is convergent, then $|a_{N+1}-a_N|<\epsilon$. As $a_{2n}=2, a_{2n+1}=\frac{1}{2}$ we have $$|a_{n+1}-a_n|=\frac{1}{2}$$ So there is no $N$ for $\epsilon\leq\frac{1}{2}$.


问:
Is it valid to to multiply an equation with general variable?
Let $y= x^2+x$.

Is it valid/legal to multiply
$(x^2+x)$ by $x/x$?
As $x$ can be $0$ too. So $0$/0 can't be expressed as 1... But on contrary side, $x$ can be $1, 2, -1,$ etc. any possible real number too and can be cancelled to $1$.
So what's the solution?

答:
In such situations, we usually divide the problem in two cases:

Case 1

If $x=0$, then work out using $x=0$.

Case 2

If $x\neq0$, then you can safely divide by $x$.

Hope this helps. Ask anything if not clear :)


问:
Boundary Curves
Planar region D is given by the following conditions
$$25x^2-150x+9y^2\leq 0$$
$$x\leq 3 $$

Determine the boundary curves.

Calculate the mutual intersections of all boundary curves.

I know a method to do this called strokes theorem but there are a few things missing to use that method. Can someone please suggest a way to solve this question or give an answer. Thank you

答:
The region $D$ is drawn in the picture below. The intersection points are self evident.

The function to integrate is missing :)

$$...$$

Title: Boundary Curves


问:
Canonical examples of inner product spaces that are not Hilbert spaces?
That is, what are some good examples of vector spaces which are inner product spaces but in which not every Cauchy sequence converges?</p>

答:
Any non-closed subspace of a Hilbert space will do. For example, the linear span of a Hilbert basis of an infinite-dimensional Hilbert space.

For an explicit example, take $H = \ell^2(\mathbb{R})$, the vector space of square summable sequences of real numbers with pointwise operations and inner product
$$\Bigl\langle \{a_n\},\{b_n\}\Bigr\rangle = \sum_{k=1}^{\infty} a_kb_k.$$
Let $\mathbf{e}_i$ be the sequence that has a $1$ in the $i$th coordinate and $0$s elsewhere. Then $\{\mathbf{e}_i\mid i\in\mathbb{N}\}$ is a Hilbert basis for $H$, and its linear span (the vector space of all almost null sequences) is an inner product space that is not complete.

ContentLicense: CC BY-SA 3.0


问:
Non-monotonic decrease of residuals in Conjugate Gradients:
In some of my numerical programming using conjugate gradient solvers, I noticed an alarming problem: The residuals were not monotonically decreasing to zero, but were sometimes increasing. In this document, the author writes

About fifteen years ago enough experience had been gained with matrix preconditoning that variants on the Conjugate Gradient method had come into wide use. Evolution of this methodology has continued with the introduction of several variations on the basic algorithm. The most popular of these is currently Sonneveld's conjugate gradient squared (CGS) algorithm. This class of methods has a rate of convergence that is generally very good, but is not monotonic. Plots of residual versus iteration count can show oscillations.

So I guess it might not be an error in my code. However, given the structure of the CG algorithm and the principles on which it was based, I cannot understand why the residuals don't decrease monotonically with iteration number. Is it finite arithmetic?

答:
I suppose you want to solve $Ax=b$ with $A$ being a symmetric, positive definite matrix.

Convergence of CG is monotonic -- with respect to the norm that is induced by the $A$-weighted scalar product. That is, if $r_k:=Ax_k-b$ is the current residual, then $\|x_k-x\|_A = \|r_k\|_{A^{-1}}^2 := r_k^*A^{-1}r_k$ is a decreasing sequence.

At the same time $\|r_k\|^2=r_k^*r_k$ need not be monotonically decreasing.

As CG is a stable method, the influence of rounding errors is only recognizable when the norm residuals is close to machine precision.

However, I don't know about the CGS methods you mentioned...

I made an EDIT to my answer to correct a wrong statement. Actually, in the $A$ norm, only the errors decrease monotonically but not the residuals. Thanks to @fieres who pointed this out in his answer.


问:
Can the values of the expressions $\frac{1}{\sqrt{2a+1}},\frac{1}{\sqrt{2a-1}},3(a^2-19)\sqrt{2a-1},18\sqrt{2a+1}$ be in G.P.?
Can the values of the expressions $\dfrac{1}{\sqrt{2a+1}},\dfrac{1}{\sqrt{2a-1}},3(a^2-19)\sqrt{2a-1},18\sqrt{2a+1}$ be in geometric progression (in the given order)?

I am confused by the fact that the expressions aren't defined for every value of $a$. Should I determine the range of $a$ or this won't be necessary in the solution? I mean we will have $$\begin{cases}2a+1>0\\2a-1>0\end{cases}\iff\begin{cases}a>-\dfrac12\\a>\dfrac12\end{cases}\iff a\in\left(\dfrac12;+\infty\right).$$ Can we use this further in the solution in some way? Thank you in advance!

答:
Hint:

WLOG

The terms can be written as $$b,br,br^2,br^3$$

$$\implies br^3\cdot b=br\cdot br^2$$

$$18=3(a^2-19)\implies a^2=?$$

Again to keep $\sqrt{2a\pm1}$ real, $2a\ge$max$(-1,1)$


问:
Proof of $n$ being quadratic residue for primes of the form $4n+1$
I'm trying to prove the following statement:

If $4n+1$ is a prime $p$, then $n$ is a quadratic residue $\bmod p$.

For this, I thought I could evoke the quadratic reciprocity law and deduce:

$$\genfrac(){}{0}{n}{4n+1}\genfrac(){}{0}{4n+1}{n}
=(-1)^{(n-1)\frac{4n+1-1}{4}}=(-1)^{(n-1)n}=1
\\\iff\genfrac(){}{0}{n}{4n+1}=(\genfrac(){}{0}{4n+1}{n})^{-1}
=(\genfrac(){}{0}{1}{n})^{-1}=1$$

with Legendre Symbols, but then it occured to me that $n$ need not be prime.
Looking for a workaround, I found I can deduce that the Jacobi-Symbol must be $1$, but If I
I can deduce that the Jacobi-Symbol must be $1$ but this does not necessarily imply that $n$ is a quadratic residue. How do I work around this?

答:
HINT: $n$ is a quadratic residue if and only if $-1$ is a quadratic residue.

Indeed, if $-1=a^2 \pmod p$, then $$(2^{-1} a)^2 = 4^{-1}a^2 = (4^{-1})(-1) = (4^{-1})(4n) = n \pmod p$$


问:
Rolling dice until each has taken a specific value
I'm facing the following problem.

Let's say I have $N$ dice in hand. I need to calculate how much time I should roll my dice to make all of them equal to some selected (pre-defined) number. Each time I roll the selected number with some dice, I remove these dice from my hand and keep rolling the rest.

Example:
I have $2$ dice and I want to roll sixes. When I get one, I will remove this die and will roll one die instead of two. How many times do I need to roll my dice in order to get sixes in all (to make my hand empty)?

I suppose that the correct answer is (for two dice) ${1\over6} +{1\over6} + {1\over6}\times{1\over6}$, but it seems to be wrong because I tried to implement an algorithm to calculate the probability, in which I ran 1M continuous rolls to calculate the average amount of required rolls.

Any help is appreciated.

答:
I will try to interpret the question in a more formal way. Let $X_1, \ldots, X_n$ be i.i.d. geometric random variables with rate of success $p$ (which is $1/6$ in this case). Find $E[\max\{X_1, \ldots, X_n\}]$.

My approach is to first compute the distribution of $Y = \max\{X_1, \ldots, X_n\}$. Suppose $y \in \mathbb N$ is given.

\begin{align*}
P(Y \le y) & =
\prod_{i=1}^n P(X_i \le y) \\
& = \prod_{i=1}^n \left(1 - P(X_i > y)\right) \\
& = \left(1 - (1 - p)^y\right)^n \\
\therefore P(Y = y) & = (1 - (1 - p)^y)^n - (1 - (1 - p)^{y-1})^n.
\end{align*}
(Note that $P(Y = 1) = p^n$.)
For ease of writing, let $q = 1 - p$.
The expected value of $Y$ is
\begin{align*}
\sum_{y=1}^\infty yP(Y = y) & =
\sum_{y=1}^\infty y\left((1 - q^y)^n - (1 - q^{y-1})^n\right).
\end{align*}
This is the simplest expression I can find. There might be simpler ones, but I haven't found any.


问:
Algebraic extensions $F(\alpha), F(\beta)$ with a $F$-isomorphism yield the same minimal polynomial for $\alpha, \beta$
Suppose we are given two simple algebraic extensions $F(\alpha), F(\beta)$ (over a field $F$) with an $F$-isomorphism $\phi$ of extensions (i.e. fixing $F$) such that $\phi(\alpha)=\beta$.

Then we want to prove that $\alpha, \beta$ have the same minimal polynomial.

I'd like to argue that $F(\alpha) \simeq F[x]/(f(x))$ and $F(\beta) \simeq F[y]/(g(y))$ where $f,g$ are the minimal polys of $\alpha, \beta$ and from this I should get an isomorphism $F[x] \simeq F[y]$ sending $x$ to $y$ that is induced by $\phi$. But I'm not sure if this argument is actually beating around the bush, or if there is a more direct or obvious argument?

答:
We have $
0 = \phi(0) = \phi(f(\alpha)) = f(\phi(\alpha)) = f(\beta)
$
implies that $g$ divides $f$.

By symmetry, $f$ divides $g$ and so $f=g$ (since both are monic).


问:
Combinatorics - Counting Something Similar to Unsigned Stirling Numbers of the First Kind
I want to count something similar to unsigned Stirling numbers of the first kind. Can someone tell me what math describes the quantity I'm interested in?

Suppose I have $n$ people numbered $1, 2, ..., n$ and a sequence of $n$ empty sets. I sequentially choose one person at a time, in the ordering of their numbers, and either place that person in any non-empty set in my sequence or in the first empty set in my sequence. For example, if $n=3$, I start with the sequence of 3 empty sets:

{}{}{}

For person $1$, I must place them in the first empty set:

{1}{}{}

For person $2$, I have two choices. I can either add them to the first set or the second set:

{1, 2}{}{}

{1}{2}{}

For person $3$, if I took the first option for person $2$, I now have two choices:

{1, 2, 3}{}{}

{1, 2}{3}{}

And if I took the second option for person $2$, I now have three choices:

{1, 3}{2}{}

{1}{2, 3}{}

{1}{2}{3}

I'm interested in counting the number of ways that the last person $n$ can end up in each set. In this example, there are 2 ways for person $3$ to end up in the first set, 2 ways to end up in the second set, and 1 way to end up in the third set. What combinatoric function describes the number of ways that the last person $n$ can end up in each set?

答:
In order for item $n$ to end up in set $k$, the first $n-1$ items need to span at least $k-1$ sets. The number of ways $n-1$ objects can span exactly $i$ sets is the Stirling number ${n-1 \brace i }$. Therefore, the number of ways to have item $n$ in spot $k$ is
$$
{n-1 \brace k-1} + {n-1 \brace k} + \dots + {n-1 \brace n-1}
$$


问:
find closest point of a line to another point
Given a line $r$ from $a$ to $b$ and a point $c$, how can I find the point $d$ in $r$ whose distance to $c$ is minumum?

I know how to find the minimum distance between $r$ and $c$, but I'm struggling extending that equation to find $d$.

I think there must be a faster way to find $d$ than generating a sphere centered in $c$ with radius equals to the minimum distance between $r$ and $c$ and then checking the intersection point between this sphere and $r$. I'm also doing this in a programming language.

答: In order to find the point $d$, you can compute the gradient of the line r and find the coordinates on the line where the line carried by the gradient intersects c


问:
Two definitions of the radical of a module over a commutative ring
Let $R$ be a commutative ring and $M$ is an $R$-module. There are two definitions of the radical of $M$ in the literature. First definition is
$$
\operatorname{Rad}(M)=\bigcap \{N \mid N \text{ is a maximal submodule of } M\},
$$
which gives us a submodule of $M$. The second definition is
$$
\operatorname{rad}(M) = \bigcap \{m \subset R \mid  m \text{ is a maximal ideal and }\operatorname{Ann}(M) \subset m\},
$$
which is an ideal. This definition is used, for example, in the book "A term of Commutative Algebra" by Altman and Kleiman.

Is it true that $\operatorname{rad}(M) M = \operatorname{Rad}(M)$?

I see how to show that
$$
\bigcap_{m \text{ is maximal ideal}} mM=\operatorname{Rad}(M),
$$
but I have problems showing that $\operatorname{rad}(M)M = \bigcap_{m \text{ is maximal ideal}} mM$. If this is not correct, what is a counterexample?

答:
As written, the desired equality does not hold, and can in fact fail very acutely! For instance, let $R=\mathbb{Z}$ and $M=\mathbb{Q}$. Then $M$ has no maximal submodules (why?), and so in particular we have $\text{Rad}(M)=M$. However, the annihilator of $M$ is clearly just the ideal $(0)<\mathbb{Z}$, so $\text{rad}(M)$ is the intersection of all maximal ideals of $\mathbb{Z}$, which is $(0)$. Hence $\text{rad}(M)M=\{0\}$, which is certainly not equal to $M$.

For an example where $M$ does have maximal submodules, let $M=\mathbb{Q}\times \mathbb{Z}$, with $R=\mathbb{Z}$ as above. By lemma 2 here, $$\text{Rad}(M)=\text{Rad}(\mathbb{Q})\times\text{Rad}(\mathbb{Z})=\mathbb{Q}\times\{0\},$$ which is not all of $M$. (In particular, $\mathbb{Q}\times(p)$ is a maximal submodule of $M$ for any $p\in\mathbb{Z}$ prime.) However, we again have $\text{Ann}(M)=(0)$ and thus $\text{rad}(M)=(0)$, so this gives another counterexample.


问:
rewriting equation in terms of $y$
From Stewart, Precalculus, 5th ed, P98, Q.45

$$x^2+xy+y^2=4$$

how can I re-write this equation in terms of $y$? I want to put this equation into graphing software but don't know to put $y$ on one side and $x$ on the other.

Any suggestion would be great.

答: It is a quadratic equation in $x$. So, use the standard quadratic solution.</p>

Title: rewriting equation in terms of $y$


问:
How to cut a cube out of a tree stump, such that a pair of opposing vertices are in the center?
I saw this picture of a cube cut out of a tree stump.

I've been trying to craft the same thing out of a tree stump, but I found it hard to figure out how to do it.

One of the opposing vertices pair is on the center of the tree stump:

I've been struggling to find the numbers and angle needed to make the cuts.

Any kind of advice would be greatly appreciated. Thanks for taking your time to read this and I'm sorry for my bad english.

答:
First, form your stump into a cylinder with a height greater than $\frac{3}{\sqrt{2}}\approx 2.121$ times its radius so that the cube will fit inside the cylinder. Here, we assume you can mark given points and lines on the surface of the cylinder as well as cut a plane through $3$ points.

For the first vertex, mark the center of one of the cylinder's circular faces. Next, from this point mark $3$ radii of the circular face spaced $120$ degrees from each other (so that their endpoints create an equilateral triangle).

Now, go down from each of the $3$ vertices of the triangle a distance equal to $\frac{1}{\sqrt{2}}\approx 0.707$ the radius of the cylinder, and mark the points there. If your cylinder has the minimal height-to-radius ratio, these points will be exactly $\frac{1}{3}$ of the way down.

Next, cut $3$ planes through the first vertex and each pair of the next $3$ vertices, forming the first corner of the cube.

Now, mark the $3$ midpoints of the arcs formed by your cuts. These are the next vertices of the cube. Once again, if your cylinder has the minimal height-to-radius ratio, these will be $\frac{2}{3}$ of the way down.

Finally, cut $3$ more planes to finish the shape of your cube. You do not need to mark the last vertex as it is predefined by the intersection of the planes.

When you are all done, the side length of the cube will be $\sqrt{\frac{3}{2}}\approx 1.225$ times the radius of the starting cylinder, so take this into account if you want a specific side length. The angle formed by each face of the cube to the original end of the cylinder is $\arctan(\sqrt{2}) \approx 54.734^{\circ}$.


问:
Equivalence of differentiability on regular surface
i need to prove that the two following definitions of differentiability on regular surfaces are equivalent:

Let $S$ be a regular surface. A function $f: S \rightarrow \mathbb{R^n}$ is differentiable if:

i) for any local chart of $S$ like $X: U \rightarrow S$ we have $f \circ X : U \rightarrow \mathbb{R^n}$ differentiable.

or

ii) for every point $p$ in $S$ there exists a local chart $X_p : U_p \rightarrow S$ around $p$ such that $f \circ X_p: U_p \rightarrow \mathbb{R^n}$ is differentiable.

Clearly i implies ii but i'm not seeing how ii implies i. Can anyone help me?

Thx!

答:
The key point is the compatibility requirement for two overlapping charts in $S$. Suppose $X\colon U\to S$ is a chart and $p\in X(U)$. Then there is, by assumption, a chart $X_p\colon U_p\to S$ with $p\in X_p(U_p)$. By definition of a surface,
$$X_p^{-1}\circ X\big|_{X^{-1}(X_p(U_p)\cap X(U))}\colon X^{-1}(X_p(U_p)\cap X(U))\to X_p^{-1}(X_p(U_p)\cap X(U))$$
is a differentiable map from an open subset of $\Bbb R^2$ to another. But then $f\circ X = \big(f\circ X_p\big)\circ \big(X_p^{-1}\circ X)$ is the composition of differentiable functions on a neighborhood of $X^{-1}(p)$ and is differentiable.


问:
Is the set of all nowhere differentiable functions on $C[0,1]$ countable?
I know this is a dense set, but is it countable?</p>

答: No. For example, if $W$ is any nowhere differentiable function, then $W(x)+c$ is as well, for any real number $c$.


答: First project the vector $\vec{ac}$ onto the vector $\vec{ab}$, and then the minimizing point is located at $\vec{a} + \vec{ac}_\parallel$. Here $\vec{ac}_\parallel$ is the component of $\vec{ac}$ in the direction of $\vec{ab}$.

Title: find closest point of a line to another point

Body:
Given a line $r$ from $a$ to $b$ and a point $c$, how can I find the point $d$ in $r$ whose distance to $c$ is minumum?

I know how to find the minimum distance between $r$ and $c$, but I'm struggling extending that equation to find $d$.

I think there must be a faster way to find $d$ than generating a sphere centered in $c$ with radius equals to the minimum distance between $r$ and $c$ and then checking the intersection point between this sphere and $r$. I'm also doing this in a programming language.


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:
There are really two very different kinds of proofs:

Informal proofs are what mathematicians write on a daily basis to convince themselves and other mathematicians that particular statements are correct. These proofs are usually written in prose, although there are also geometrical constructions and "proofs without words".

Formal proofs are mathematical objects that model informal proofs. Formal proofs contain absolutely every logical step, with the result that even simple propositions have amazingly long formal proofs. Because of that, formal proofs are used mostly for theoretical purposes and for computer verification. Only a small percentage of mathematicians would be able to write down any formal proof whatsoever off the top of their head.

With a little humor, I should say there is a third kind of proof:

High-school proofs are arguments that teachers force their students to reproduce in high school mathematics classes. These have to be written according to very specific rules described by the teacher, which are seemingly arbitrary and not shared by actual informal or formal proofs outside high-school mathematics. High-school proofs include the "two-column proofs" where the "steps" are listed on one side of a vertical line and the "reasons" on the other.  The key thing to remember about high-school proofs is that they are only an imitation of "real" mathematical proofs.

Most mathematicians learn about mathematical proofs by reading and writing them in classes. Students develop proof skills over the course of many years in the same way that children learn to speak - without learning the rules first. So, as with natural languages, there is no firm definition of "what is an informal proof", although there are certainly common patterns.

If you want to learn about proofs, the best way is to read some real mathematics written at a level you find comfortable. There are many good sources, so I will point out only two: Mathematics Magazine and Math Horizons both have well-written articles on many areas of mathematics.

Title: What is a proof?


问:
How to show that a Group is not profinite?
I tried to show that $\mathbb{Z}$ is not profinite.
So I know that a topological group $G$ is profinite if $G\cong \underleftarrow{\lim}~ G/H$, with
$H$ all open normal subgroups of $G$. And I know
that there is no surjective map $\mathbb{Z}\longrightarrow\underleftarrow{\lim}~\mathbb{Z}/n\mathbb{Z}$.
So $\mathbb{Z}$ is not profinite with the profinite topology.
But how to show that there is no other topology that $\mathbb{Z}$ is profinite?

答:
Your reference (the book Introduction of Profinite Groups and Galois Cohomology by Ribes) defines a profinite group to be a compact Hausdorff totally disconnected topological group. I'll argue directly from this definition that any profinite group is either finite or has cardinality $\geq 2^{\aleph_0}$.

Suppose $G$ is a profinite group. If $G$ has an isolated point, then every point of $G$ is isolated (since for any  $a\in G$, the map $x\mapsto ax$ is a homeomorphism). But then $\{\{a\}\mid a\in G\}$ is an open cover of $G$, so $G$ is finite by compactness.

Now suppose $G$ has no isolated points (such a space is called perfect). It's a standard fact that a perfect locally compact Hausdorff space has cardinality $\geq 2^{\aleph_0}$. But the proof is a bit easier for a compact Hausdorff totally disconnected space, so I'll outline the argument in this case.

We build a complete binary tree of non-empty clopen sets in $G$ by associating to each finite binary sequence $w$ a clopen set $C_w\subseteq G$, such that for any $w$, writing $w0$ and $w1$ for the sequences obtained by extending $w$ by a $0$ and a $1$, respectively, we have $C_{w0}\subseteq C_w$, $C_{w1}\subseteq C_w$, and $C_{w0}\cap C_{w1} = \emptyset$.

We start by associating the empty sequence to all of $G$. Now given $C_w$, since $G$ has no isolated points, $C_w$ has at least two points, $x$ and $y$. Since $G$ is Hausdorff and totally disconnected, there are two complementary clopen sets such that one contains $x$ and the other contains $y$. We define $C_{w0}$ and $C_{w1}$ by intersecting these two clopen sets with $C_w$.

Now for any infinite binary sequence $s$, we can take the intersection of the clopen sets corresponding to the finite initial segments of $s$: $\bigcap_{n\in \mathbb{N}} C_{s\restriction n}$. By compactness, this intersection is non-empty, so we can pick a point $x_s$ in the intersection. This gives an injective map from the set of infinite binary sequences (equivalently, paths through the complete binary tree we constructed above) to $G$, which shows that $|G|\geq 2^{\aleph_0}$.


问:
Show that $\|XY-YX\|^2\le2\|X\|^2\|Y\|^2$ for diagonal matrix $X$ and any matrix $Y\in M_n$
I am studying on the topic of norm of the commutator and I would like to solve the following problems:

For any matrix $X\in M_n$, $\|X\|=\sqrt{\sum|x_{ij}|^2}=\sqrt{trA^*A}$ , where $A^*$ is the conjugate transpose of $A$(So the field is $\mathbb C$).

Let $X=diag(x_1,x_2,...,x_n)$ be a diagonal matrix in $M_n$.
Show that $\|XY-YX\|^2\le2\|X\|^2\|Y\|^2$ for any matrix $Y\in M_n$.

This is the first part of my problem.
I try the question in this way:

$X=\operatorname{diag}(x_1,x_2,...,x_n)$,
$Y=\begin{bmatrix}y_{11} & y_{12} & ... & y_{1n}\\y_{21} & y_{22} & ... & y_{2n}\\ \vdots & \vdots & \ddots & \vdots \\y_{n1} & y_{n2} & ... & y_{nn}\end{bmatrix}$

then $$XY-YX=\begin{bmatrix}(x_1-x_1)y_{11} & (x_1-x_2)y_{12} & ... & (x_1-x_n)y_{1n}\\(x_2-x_1)y_{21} & (x_2-x_2)y_{22} & ... & (x_2-x_n)y_{2n}\\ \vdots & \vdots & \ddots & \vdots \\(x_n-x_1)y_{n1} & (x_n-x_2)y_{n2} & ... & (x_n-x_n)y_{nn}\end{bmatrix}$$

So $\|XY-YX\|^2=\sum|(x_i-x_j)y_{ij}|^2$ for $1\le i,j\le n$
and I get stuck in this step.

The second part of my problem is:

If $X$ is a normal matrix,  show that the inequality is still true.
I have no idea how to start with the second problem.

答:
The slightly more general version of your question was in fact Conjectured back in 2005 for arbitrary matrices (without the assumption that $X$ is diagonal) by Bottcher and Wenzel. It was proven shortly after by Vong and Jin, and you can see the follow-up paper by the authors of the conjecture here with the proof of the desired fact.

The proof by Bottcher and Wenzel doesn't appear to be simple enough to explain in an MSE post. That being said I will show you how elementary methods can get you a weaker bound that is off by a factor of only $2$.

The matrix norm you mention is called the Frobenius norm, so we will denote it by $\|\cdot\|_F$. To obtain a slightly weaker bound, you can use the fact that $(M_n(\mathbb{C}),\|\cdot\|_F)$ is a Hilbert space (with the Frobenius a.k.a Hilbert-Schmidt inner-product), and therefore the parallelogram law holds. In other words, for any complex $n\times n$ matrices $X,Y$ we have that $$\|XY-YX\|_F^2+\|XY+YX\|_F^2=2\|XY\|_F^2+2\|YX\|_F^2.$$

Another useful fact is that the Frobenius norm is submultiplicative and therefore $$\|XY\|_F\leq \|X\|_F\|Y\|_F,$$ for any $X,Y\in M_n(\mathbb{C})$.

Combining these two facts we can see that for any matrices $X$ and $Y$ $$\|XY-YX\|_F^2\leq 2\|X\|_F^2\|Y\|_F^2+2\|Y\|_F^2\|X\|_F^2=4\|X\|_F^2\|Y\|_F^2.$$

The above bound is weaker than the bound in the question (which is in fact tight), however, it can be proven using elementary results.

Remark 1. I have not used the fact that $X$ is diagonal... perhaps the argument in the proof is easier with this assumption, however, this was not obvious...

Remark 2 As mentioned in the comments the Frobenius norm is unitarily invariant, and therefore if $U\Sigma V$ is the singular value decomposition of $X$ with $\Sigma$ being the diagonal matrix of singular values then $\|X\|_F=\|U\Sigma V\|_F=\|\Sigma\|_F$, in particular, if $X$ was normal then $V=U^*$, and the same thing would essentially hold.


问:
Can any positive real be approximated as $2^m/3^n$ with $(m,n)$ large enough?
Conjecture.

There exist positive integers $(m,n)$ large enough, such that for any positive real number $r$ and a given error $\epsilon$ :
$$
   \left| r - \frac{2^m}{3^n} \right| < \epsilon
$$
There is numerical evidence for this conjecture. I have tried $r = \sqrt{2}$ and $\epsilon = 10^{-3}$.

Below is a little Delphi Pascal program with accompanying output.

But .. can somebody prove the conjecture?

```
program apart;<br>
procedure test(r : double; eps : double);
var
  a : double;
  m,n : integer;
begin
  a := 1;
  m := 0; n := 0;
  while true do
  begin
    if a < r then
    begin
      m := m + 1;
      a := a * 2;
    end else begin
      n := n + 1;
      a := a / 3;
    end;
    if abs(r-a) < eps then Break;
  end;
  Writeln(r,' = 2^',m,'/3^',n,' =',a);
end;<br>
begin
  test(sqrt(2),1.E-3);
end.
```

Output:

```
 1.41421356237310E+0000 = 2^243/3^153 = 1.41493657935359E+0000
```

UPDATE.

The answer by lhf does look like a very concise proof. But for me - as an retired physicist by education - it's a bit beyond comprehension.

Furthermore, it leaves a few issues untouched. One might ask for example whether there are estimates for $m$ and $n$ when $r$ and $\epsilon$ are given.

Note. The question can also be formulated as:
Can any positive real be approximated as $3^m/2^n$ with $(m,n)$ large enough? Which is the same as allowing negative integers with the original formulation. In this form, it shows some resemblance to the (in)famous Collatz problem.

EDIT.

As suggested by the answers, an approach with logarithms could be more effective:

```
program anders;<br>
procedure proef(r : double; eps : double);
var
  a,l2,l3,lr : double;
  m,n : integer;
begin
  l2 := ln(2); l3 := ln(3);
  lr := ln(r); a := 0;
  m := 0; n := 0;
  while true do
  begin
    a := m*l2 - n*l3 - lr;
    if abs(a) < eps then Break;
    if a < 0 then m := m + 1 else n := n + 1;
  end;
  Writeln(r,' = 2^',m,'/3^',n,' =',exp(a)*r);
end;<br>
begin
  proef(sqrt(2),1.E-3);
  proef(sqrt(2),1.E-9);
end.
```

Output:

```
 1.41421356237310E+0000 = 2^243/3^153 = 1.41493657935356E+0000
 1.41421356237310E+0000 = 2^911485507/3^575083326 = 1.41421356125035E+0000
```

The first line in the output is almost identical to the result obtained previously .

The last line in the output shows that the latter approach indeed is more effective.
The error plays the same role in both approaches. Oh well, almost.
Let's take a look at the places where the 'Break's are. First program:
$$
\left| r - \frac{2^m}{3^n} \right| < \epsilon
$$
Second program:
$$
-\epsilon < m\ln(2) - n\ln(3) - \ln(r) < +\epsilon \\
\ln(1-\epsilon) < \ln\left(\frac{2^m/3^n}{r}\right) < \ln(1+\epsilon) \\
-\epsilon < \frac{2^m/3^n}{r} - 1 < +\epsilon \\
\left| r - \frac{2^m}{3^n} \right| < \epsilon.r
$$
So $\epsilon$ in the first program is an absolute error,
while $\epsilon$ in the second program is a relative error.

Continuing story at:

Can the Stern-Brocot tree be employed for better convergence of $2^m/3^n$?

答:
Let $G= \mathbb Z \log 2 + \mathbb Z \log 3$. Then $G$ is an additive subgroup of $\mathbb R$. Since $\log 2 / \log 3$ is irrational, $G$ cannot be cyclic [1] and so must be dense [2]. Therefore, $\log r$ is arbitrarily approximated by elements of $G$.

[1] If $G = \mathbb Z \theta $, then $\log 2 = a \theta$ and $\log 3 = b \theta$ and so $\log 2 / \log 3 = a/b $ is rational.

[2] See https://math.stackexchange.com/a/889775/589


问:
$\arctan x$ vs $1/\tan x$
Why is $\arctan x = (\tan x)^{-1} = \tan^{-1}x$ but is not equal to $1/\tan x$, even though $\tan^{2}x = (\tan x) \cdot (\tan x)$?

Is there not another way to write $1/\tan x$?

答:
The notation $\tan^2(x)$ "by default" denotes the square of the tangent because it is exceptional to meet the iterated function $\tan(\tan(x))$ that the exponent $2$ might denote.

On the opposite, $\tan^{-1}(x)$ might as well denote the functional inverse ($\arctan(x)$) or the multiplicative inverse ($\cot(x)$), because the two functions are frequently used. I guess that the first interpretation is more frequent, but without guarantee.

The moral of the story: make sure that by context what you mean is unambiguous, or state it explicitly.

Title: $\arctan x$ vs $1/\tan x$


问:
Geometric sum with binomial coefficient
I'm looking for any kind of formula or insight concerning

$$
f(k,N)=\sum_{n=0}^N\binom{n}{k}\lambda^n
$$

There are loads of binomial identities, so maybe I was just missing something, but I couldn't find anything yet.

Addendum:
Further assumptions on $\lambda$, i.e. $\lambda=\exp(2\pi i\alpha)$ yield to the following problem: Gaussian like sum with binomial coefficients

答:
Allow me to change your notation to
$$
f(x,m,n) = \sum\limits_{k = 0}^n {\binom{k}{m}x^{\,k} }
$$

Then a first approach would be through a double o.g.f. as
$$
\eqalign{
  & \sum\limits_{0\, \le \,n} {f(x,m,n)y^n }
  = \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,k\, \le \,n}
 {\left( \matrix{  k \cr   m \cr}  \right)x^{\,k} y^n } }  =   \cr
  &  = \sum\limits_{0\, \le \,k} {\sum\limits_{\,k\, \le \,n}
 {\left( \matrix{  k \cr   m \cr}  \right)\left( {xy} \right)^{\,k} y^{n - k} } }
  = \sum\limits_{0\, \le \,k} {\sum\limits_{\,0\, \le \,j}
 {\left( \matrix{  k \cr   m \cr}  \right)\left( {xy} \right)^{\,k} y^j } }  =   \cr
  &  = {1 \over {1 - y}}\sum\limits_{0\, \le \,k}
 {\left( \matrix{  k \cr   k - m \cr}  \right)\left( {xy} \right)^{\,k} }
  = {{\left( {xy} \right)^{\,m} } \over {1 - y}}\sum\limits_{0\, \le \,k - m}
 {\left( { - 1} \right)^{\,k - m} \left( \matrix{   - m - 1 \cr   k - m \cr}  \right)
\left( {xy} \right)^{\,k - m} }  =   \cr
  &  = {{\left( {xy} \right)^{\,m} } \over {\left( {1 - y} \right)\left( {1 - xy} \right)^{\,m + 1} }} \cr}
$$

An other way is first to express the sum as an infinite sum with the index starting from $0$, as
$$
\eqalign{
  & f(x,m,n) = \sum\limits_{k = 0}^n {\left( \matrix{  k \cr   m \cr}  \right)x^{\,k} }  =   \cr
  &  = \sum\limits_{0\, \le \,k} {\left( \matrix{  n - k \cr   n - k \cr}  \right)
\left( \matrix{  k \cr   k - m \cr}  \right)x^{\,k} }  =   \cr
  &  = \sum\limits_{0\, \le \,j}
 {\left( \matrix{  j \cr   j \cr}  \right)\left( \matrix{  n - j \cr  n - m - j \cr}  \right)x^{\,n - j} }  =   \cr
  &  = x^{\,n} \sum\limits_{0\, \le \,j} {{{\left( {n - j} \right)^{\,\underline {\,n - m - j\,} } }
 \over {\left( {n - m - j} \right)!}}\left( {1/x} \right)^{\,j} }  \cr}
$$
where $x^{\,\underline {\,k\,} }$
represents the Falling Factorial

Then we can convert it into a Hypergeometric function by putting
$$
t_{\,j}  = {{\left( {n - j} \right)^{\,\underline {\,n - m - j\,} } } \over {\left( {n - m - j} \right)!}}
$$
and determine that
$$
t_{\,0}  = {{\left( n \right)^{\,\underline {\,n - m\,} } } \over {\left( {n - m} \right)!}}
 = \left( \matrix{  n \cr   m \cr}  \right)
$$
as well that the ratio is
$$
\eqalign{
  & {{t_{\,j + 1} } \over {t_{\,j} }}
 = {{\left( {n - 1 - j} \right)^{\,\underline {\,n - 1 - m - j\,} } }
 \over {\left( {n - 1 - m - j} \right)!}}{{\left( {n - m - j} \right)!}
\over {\left( {n - j} \right)^{\,\underline {\,n - m - j\,} } }} =   \cr
  &  = {{\left( {n - 1 - j} \right)^{\,\underline {\,n - 1 - m - j\,} } }
 \over {\left( {n - j} \right)^{\,\underline {\,1\,} }
\left( {n - 1 - j} \right)^{\,\underline {\,n - 1 - m - j\,} } }}{{\left( {n - m - j} \right)} \over 1} =   \cr
  &  = {{n - m - j} \over {n - j}} = {{j - n + m} \over {j - n}} \cr}
$$

Therefore
$$
f(x,m,n) = x^{\,n} \left( \matrix{  n \cr   m \cr}  \right){}
_2F_{\,1} \left( {\left. {\matrix{   { - n + m,\;1}  \cr    { - n}  \cr  } \;}
 \right|\;{1 \over x}} \right)
$$

Still another way to express $f(x,m,n)$ is
$$
\eqalign{
  & f(x,m,n) = \sum\limits_{k = 0}^n {\left( \matrix{  k \cr  m \cr}  \right)x^{\,k} }  =   \cr
  &  = \sum\limits_{k = 0}^n {{{k^{\,\underline {\,m\,} } } \over {m!}}x^{\,k} }
  = {{x^{\,m} } \over {m!}}\sum\limits_{k = 0}^n {k^{\,\underline {\,m\,} } x^{\,k - m} }  =   \cr
  &  = {{x^{\,m} } \over {m!}}\sum\limits_{k = 0}^n {{{d^{\,m} } \over {dx^{\,m} }}x^{\,k} }
  = {{x^{\,m} } \over {m!}}{{d^{\,m} } \over {dx^{\,m} }}
\left( {{{1 - x^{\,n + 1} } \over {1 - x}}} \right) \cr}
$$

Title: Geometric sum with binomial coefficient


问:
How to prove this formula $ \sum_{k=2}^n k(k-1)\binom{k}{n} = n(n-1) \cdot 2^{n-2}$?
$$ \sum_{k=2}^n k(k-1)\binom{k}{n} = n(n-1) \cdot 2^{n-2}$$

答:
Observe that for all $ n\geq 2 $, $ k\in\left[\!\left[2,n\right]\!\right] $, we have : $$ k\left(k-1\right)\binom{n}{k}=n\left(n-1\right)\binom{n-2}{k-2} $$

Thus : \begin{aligned} \sum_{k=2}^{n}{k\left(k-1\right)\binom{n}{k}}&=n\left(n-1\right)\sum_{k=2}^{n}{\binom{n-2}{k-2}}\\ &=n\left(n-1\right)\sum_{k=0}^{n-2}{\binom{n-2}{k}}\\ &=n\left(n-1\right)2^{n-2} \end{aligned}


答: Hint: show that $k(k-1) \binom{n}{k}  = n(n-1) \binom{n-2}{k-2}$.

Title: How to prove this formula $ \sum_{k=2}^n k(k-1)\binom{k}{n} = n(n-1) \cdot 2^{n-2}$?

Body: $$ \sum_{k=2}^n k(k-1)\binom{k}{n} = n(n-1) \cdot 2^{n-2}$$


问:
A problem about probability theory and set theory
I'm studying the fact that if $\lambda \in (0,1)$ and $\{Y_{n}\}_{n\geq 1}$ is a sequence of i.i.d. random variables such that $Y_{i}\sim \mathrm{Bernoulli}(\lambda)$ defined in the space $(X, E, P)$, then $X$ is necessarily uncountable.

Approach: Suppose by way of contradiction that  $X$ is a countable set. Then we have $X=\{ x_{n} | n\in \mathbb{N},~x_{n} \in \{0,1\} \}$. Then $\displaystyle X= \bigcap_{n\geq 1} (Y_{n}=x_{n}) \subset T_{k}= \bigcap_{1\leq n \leq k} (Y_{n}=x_{n})$ for all integer $k\geq 1$. Now maybe I need to look at suitable $k$? But how can I continue from here?

Maybe a way to complete the proof, in $X$ either the number of ones or zeros is infinite, assume Wlog the we have infinitely many ones?

答:
Let $x_0 \in X$ be fixed but arbitrary, and define $y_n := Y_n(x_0) \in \{ 0,1 \}$ for $n \in \Bbb{N}$.
For brevity, let $\alpha := \max \{ \lambda, 1 - \lambda \}$.
Note for each $N \in \Bbb{N}$ that $M_N := \bigcap_{n=1}^N \{ x \in X \colon Y_n(x) = y_n \}$
satisfies $\{ x_0 \} \subset M_N$.
By the independence of the $Y_n$, we have
$$
  0
  \leq P(\{ x_0 \})
  \leq P(M_N)
  = \prod_{n=1}^N
      P(Y_n = y_n)
  \leq \prod_{n=1}^N \alpha
  = \alpha^N
  \xrightarrow[N\to\infty]{} 0 .
$$
We have thus shown $P(\{ x_0 \}) = 0$ for all $x_0 \in X$.

Thus, $X$ cannot be countable since otherwise $1 = P(X) = \sum_{x \in X} P(\{ x \}) = 0$.


问:
Why it is opposite for negative integers when it comes to bigger or smaller number?
Example :Why (-5) is smaller than (-2). And can you explain the reason behind it.</p>

答:
What does "big" and "small" mean?

It could mean two different things and in mathematics we have two different methods of dealing with it.

One: Imagine we line all the numbers on a line.   At one point we have $0$ as our central focus point.  To the right we have $1$ and to the right of that we have $2$ and so on and advancing further and further to the right we have all the possitive numbers receding miles and miles in to the distance, the line going forever further and further to the right the numbers being larger and and larger receding to the distance.

To the left of $0$ we have all the numbers, coming from the far distant horizon on the far left approaching from the left and moving to the right to get approach us and our focus point.

Watch!  Here they come!  $-10, -9, -8$ thet are moving to the right to reach us.  $-3, -2, -1$ and finally $0$.  The have arrived.... and there they go.  They are continuing onto the right.  $1$,$2$,$3$, now the are going away from us (us being us standing at the $0$ focus point) ... $8$, $9$, $10$.

In this case the progress of going from left to right is what we mean by being "greater"  $a < b$ means $b$ is further to the right than $a$.

That's all.

$-5 < -2$ because in the march of numbers from left to right $-5$ came first.  And $2 < 5$ because in the march of numbers from left to right $2$ came first.

More formally:  The numbers are divided into to types. Positive and negative.  Mathematicians in the abstract dissection, and distaste of subjective (and therefore fallible) and emotional, often consider these distinctions arbitrary.  But emotionally and subjectively, the  positive numbers that are larger, greater, above, to the right of zero.  These are the numbers that are have direct physical existence. These are the numbers for measuring a positive bank balance, traveling past a point, measuring how long its been since a historical event. And negative numbers that are smaller, less than, below, to the left of zero.    These are the numbers that measure the absence of presence. These measure how far you are in debt, how far you must go to reach a point, a countdown to an event yet to come, how far under the surface you are.

But to the rigorous mathematician:  Some numbers, in particular "$1$" are "greater" than $0$ and others are "less" than $0$ and is arbitrary and abstract.

And to compare two numbers to each other,  if $a -b$ is positive and greater than $0$ we say that $b < a$ (or $a > b$).  For example, which is larger $a=5$ or $b=3$.  Well if we subtract $5-3 = 2$ that is greater than $0$ so $3 < 5$.  And if $a-b$ is negative and less than $0$ we say $a < b$ (or $b > a$).  For example, which is larger  $a=19$ or $b=27$.  Well if we subtract $19 -27 = -8$ that is less ant $0$ so $19 < 27$.

And this works for negative numbers.  Which is larger  $a = -5$ or $b=-3$.  Well if we subtract $-5 - (-3) = -5 +3 = -2$ that is less than $0$ so $-5 < -3$. And if $a=-19$ and $b=-27$ we get $-19 -(-27)=-19+27 = 8$ and that's positive so $-19> -27$.

But, honestly, I think simply thinking $-19$ is further to the right than $-27$ so $-27 > -19$ is the most straightforward way.  I was taught that with a "number line" in the first grade and it hasn't failed my in over 50 years.

....

Two:  Sometimes we talk about the magnitude of something.   If we stand at $0$ and we look to see how far away a number is.  $19$ is $19$ away but $27$ is bigger and  further away.  But look in the other direction.  $-3$ is $3$ away but $-5$ is further away.  Look at $32$ and $-64$.  $32$ is $32$ away in the positive direction but $-64$ is $64$ away in the negative direction.  Well $64$ is more than $32$ so shouldn't is some way we should say $-64$ is bigger than $32$?

Yes, and this is ignore whether numbers are positive or negative and just being concerned with their magnitude.

We can this distance away from $0$ the absolute value and $|-64| = 64$ and $|32| = 32$ and $|-64| > |32|$ so $-64$ is larger in magnitude than $32$ if we disregard positivity/negativity.

....

But mathematicians need to be consistant.

If One:  $b$ is further to the right of $a$ and $a$ is further to the right of $b$ we say $b$ is greater than $a$ and $a$ is less than $b$.  And this is what mathematicians mean when they say a number is greater than another.  In order to be perfectly clear, it's best to say "greater" and not "bigger" or "larger" (although "more" is okay).

If Two:  we want to talk about  how far something is from an origin of $0$ and strip away all reference to direction.  We talk of "absolute value" or "magnitude" or "normalizing a value".  We might informally talk of $-64$ being "bigger" than $32$ but we shouldn't and we should always stress we are talking about it being bigger "in magnitude".


答:
$x^2+xy+y^2=4$

$\displaystyle x^2+2(x)(\frac y 2)+(\frac y 2)^2+\frac {3y^2} 4=4$

$\displaystyle (x+\frac y 2)^2 = 4-\frac {3y^2} 4$

$\displaystyle x+\frac y 2 = \pm \frac{\sqrt{16-3y^2}}2$

$\displaystyle x= \frac{-y\pm\sqrt{16-3y^2}}2$

This is a method known as completing square.

Title: rewriting equation in terms of $y$

Body:
From Stewart, Precalculus, 5th ed, P98, Q.45

$$x^2+xy+y^2=4$$

how can I re-write this equation in terms of $y$? I want to put this equation into graphing software but don't know to put $y$ on one side and $x$ on the other.

Any suggestion would be great.


问:
Asymptotic inner product of correlated random vectors
Suppose $\mathbf{x}$ and $\mathbf{y}$ are N-dimensional non-white complex random vectors independent of each other i.e., covariance matrices $\mathbf{C_{xx}}\neq\mathbf{I}$, $\mathbf{C_{yy}}\neq\mathbf{I}$, $\mathbf{C_{xy}}=\mathbf{0}$. Is there any results on asymptotic orthogonality $\mathbf{x}^H\mathbf{y}=\mathbf{u}^H\mathbf{C_{xx}}^{\frac{1}{2}}\mathbf{C_{yy}}^{\frac{1}{2}}\mathbf{v}=0$ where $\mathbf{u}$ and $\mathbf{v}$ are independent white vectors. Thanks.</p>

答:
I Think using the Theorem 3.4 in [1] (page 45), we can say:

$\boldsymbol x^H\boldsymbol y\xrightarrow{a.s.}tr(\boldsymbol C_{xx}^{\frac{1}{2}}\boldsymbol C_{yy}^{\frac{1}{2}})$

[1] Couillet R, Debbah M. Random matrix methods for wireless communications. Cambridge University Press; 2011 Sep 29.


问:
Solving the equation $(x^2-7x+11)^{(x^2-13x+42)} = 1$
I have the following equation. $$(x^2-7x+11)^{(x^2-13x+42)} = 1$$ The question is to find the number of positive integer values possible for $x$. The answer is $6$. But I am only able to find $4 ( 2,5,6,7)$. What are the other two possible values?

答:
Anything of the form $a^b=1$ can be solved by taking 3 cases.

a=1

b=0;a≠0

a=-1;b=2n (n belongs to Integers)

Thus for,

a=1;

x²-7x+11=1

x²-7x+10=0

i.e. x={2,5}

For b=0;

x²-13x+42=0

(x-6)(x-7)=0

x={6,7}

For a=-1;

x²-7x+11=-1

x²-7x+12=0

x={3,4}

But for this, we have to check if b is even or not.

Therefore for x=3,

3²-13*3+42

=9-39+42

=12=even.

Also for x=4,

4²-13*4+42

=16-52+42

=6=even.

Thus we see that b is even in both cases.

So the correct values for x are, x={2,3,4,5,6,7}.


问:
Counting how many items can be weight on a scale if we pick the weights optimally
I am reading about a problem that states the following:

Suppose you have a balance and are allowed to choose the weights for
its functionality. The objective is to pick the weights in a way that
maximizes the number of items we can check by the scale.   Given $n$
weights all of which are different how many different ways are there
to arrange them on the scales of the balance? Assuming that the two
sides of the balance are indistinguishable what conclusion can we make
about the max range of items that we can use the scale to weigh?

The solution logic goes as follows:

Each weight can be placed on the left scale, the right scale or on the
table. These are 3 options so for $n$ weights we have $3^n$
configurations of the balance.   One of these is symmetric between
left and right scale (when no weights are on the scales). The rest are
asymmetric between the left and right scale. Which means the same
configuration is counted twice.

So far it is clear.
So we would have for e.g. ${1gram, 2gram}$:

```<code>Left, Right, Table
  -    -      (1,2) (--)
  1    -      (2)
  -    1      (2)  (*)
  2    -      (1)
  -    2      (1)  (*)
  1    2      (-)
  2    1      (-)  (*)
  1,2  -      (-)
  -    1,2    (-)  (*)
</code>```

So we would have all the marked with $*$ as duplicates and the $--$ i.e. all on the table as the symmetric one we exclude since no weights are on the scales.

But then it states:

The total number of different configurations is $\frac{3^n - 1}2 + 1$

Where is the $+1$ coming from?

Going over an example for e.g. $n = 2$ for example $1gram, 3gram$

We can measure items of ${1gram, 2gram, 3gram, 4gram}$ which is a total of $\frac{3^2 - 1}2 =4$ and not $5$ as the solution claims.

Am I misunderstanding something here?

答: The +1 term comes from weighing zero grams, where there are no weights on either side.</p>


问:
Understanding a proof that a Hermitian matrix has an orthonormal basis whose diagonal contains its eigenvalues
Context. Accoring to pg. 36 of Principles of Quantum Mechanics:

My confusion surrounds the nature of the portions of the proof highlighted in red.

Notation. It seems to me the proof potentially conflates two distinct matrices and kets.

First, we have the $n$-dimensional matrix $\Omega$ vs. the $n-1$-dimensional matrix $\Omega_{n-1}$, which is the lower right subset of $\Omega$ (as outlined in the proof).

Next we have the $n-1$-dimensional $|\omega_2 \rangle_{n-1}$, which can be expressed as $(\omega_2, 0, \ldots, 0)$ against the basis $\{| \omega_2 \rangle_{n-1}, \ldots \}$. This should be contrasted with $|\omega_2 \rangle_n$, which the author implicitly asserts (but unless I'm missing something: doesn't establish) can be expressed as $(0, \omega_2, 0, \ldots, 0)$ against the basis $\{|\omega_1 \rangle, |\omega_2\rangle_n, \ldots \}$.

Problem. It seems to me that the author has established only that

$$
\Omega_{n-1} | \omega_2 \rangle_{n-1} = \omega_2 |\omega_2 \rangle_{n-1}
$$

But why does that imply that

$$
\Omega_{n} | \omega_2 \rangle_{n} = \omega_2 |\omega_2 \rangle_{n}
$$

..which is what it seems the author actually needs to show in order for his proof to work?

答:
I agree with your misgivings about the proof presented. I believe that the proof can be clarified with a more careful consideration of the relationship between an operator and its matrix relative to some basis.

It is not specified in the statement or proof, but I'll assume that $\Omega$ is an operator on the space $\Bbb C^n$. In the first part of the proof, the author constructs the orthonormal basis
$$
\mathcal B_1^n = \{|\omega_1\rangle, V^1_{\perp 1}, \dots, V^{n-1}_{\perp1}\}.
$$
The superscript $n$ merely indicates that this is a basis for the $n$-dimensional space $\Bbb C^n$. Relative to this basis, $[\Omega]_{\mathcal B_1^n}$ has the form indicated in Equation (1.8.12). This implies that the set
$$
U_1 = |\omega_1\rangle^\perp = \operatorname{span}\{V^1_{\perp 1}, \dots, V^{n-1}_{\perp1}\} \subset \Bbb C^n
$$
is an invariant subspace of $\Omega$. In other words, for any vector $|v\rangle \in U_1$, $\Omega |v\rangle$ will also be an element of $U_1$. With that in mind, we may consider the restriction of $\Omega$ to this subspace, which I will denote as $\Omega|_{U_1}: U_1 \to U_1$. This simply refers to the map defined by
$$
\Omega|_{U_1} |v \rangle = \Omega|v\rangle,
$$
with the caveat that $|v\rangle$ is required to be an element of $U_1$. From $\mathcal B_1^n$, we "inherit" a basis for $U_1$:
$$
\mathcal B_1^{n-1} = \{V^1_{\perp 1}, \dots, V^{n-1}_{\perp1}\}.
$$
I claim that the matrix of $[\Omega|_{U_1}]_{\mathcal B_1^{n-1}}$ is equal to the "boxed submatrix" indicated in the proof. I would suggest that you try to prove that this is the case for yourself. Because $\Omega|_{U_1}$ is also Hermitian, we consider an eigenvalue $\omega_2$ of $\Omega|_{U_1}$ to construct a new orthonormal basis of $U_1$,
$$
\mathcal B_{2}^{n-1} = \{|\omega_2 \rangle, V_{\perp1,2}^1,\dots,V^{n-2}_{\perp1,2}\}.
$$
Note that each of these vectors from the basis are elements of $U_1$ and hence elements of $\Bbb C^n$. In particular, this eigenvector $|\omega_2\rangle$ satisfies
$$
\Omega|_{U_1} |\omega_2 \rangle = \Omega|\omega_2 \rangle = \omega_2 |\omega_2 \rangle.
$$
That is, $|\omega_2\rangle$ is not only an eigenvector of the restricted operator $\Omega|_{U_1}$, but also of the full operator $\Omega$ over $\Bbb C^n$.
The matrix of $\Omega|_{U_1}$ relative to this basis has the block-diagonal form
$$
[\Omega|_{U_1}]_{\mathcal B^{n-1}_{2}} =
\pmatrix{
\omega_2 & 0 & \cdots & 0\\
0 & \star & \cdots & \star\\
\vdots & \vdots & \ddots & \vdots\\
0 & \star & \cdots & \star }.
$$
Now, this basis can be extended to the orthonormal basis
$$
\mathcal B_2^{n} = \{|\omega_1\rangle,|\omega_2 \rangle, V_{\perp1,2}^1,\dots,V^{n-2}_{\perp1,2}\}.
$$
I claim that the matrix $[\Omega]_{\mathcal B_2^n}$ has the above-mentioned matrix $[\Omega|_{U_1}]_{\mathcal B_2^{n-1}}$ as a submatrix. Thus, the matrix of $\Omega$ relative to $\mathcal B_2^n$ has the form
$$
[\Omega]_{\mathcal B_2^n} =
\pmatrix{
\omega_1 & 0 & 0 & \cdots & 0\\
0 & \omega_2 & 0 & \cdots & 0\\
0 & 0 & \star & \cdots & \star\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & \star & \cdots & \star }.
$$


问:
Natural log of a negative number
My teacher told me that the natural logarithm of a negative number does not exist,
but

$$\ln(-1)=\ln (e^{i\pi})=i\pi$$

So, is it logical to have the natural logarithm of a negative number?

答: Just a different point of view by Felix Klein, Elementary Mathematics from an advanced standpoint:

Title: Natural log of a negative number


问:
Conditional Operator Question
T(x,y) = Student x has taken class y

I need to translate the following statement into a quantified expression:

Everyone besides Sam has taken at least two different classes.

My initial thought was:

$$\forall x \exists y \exists z [(x\neq Sam) \leftrightarrow (T(x,y)\land T(x,z)\land (y \neq z)) $$

But I am being told that

$$\forall x \exists y \exists z [(x\neq Sam) \rightarrow (T(x,y)\land T(x,z)\land (y \neq z)) $$

Is the correct answer.

I am a little confused why the biconditional doesn't work here. The way I see it is, if there are two sets of students, those who have taken at least two classes and those who have not, then if you know a student has not then it must be Sam and only Sam and, if you know a student has taken at least two classes you know that it is not Sam.

答:
The problem is that there are two ways to read this sentence in English.

Take the set of students excluding Sam and call it $S$. Everyone in $S$ has taken at least two classes. We aren't commenting on Sam though.

Take the set of all students and call it $S$. Everyone in $S$ has taken at least two classes, except for Sam, for whom it is not the case that they have taken at least two classes.

Your translation captures the meaning of (2). The translation using implication captures the meaning of (1).

In ordinary conversation, rules like Grice's maxims apply, which can rule out weak interpretations like (1) that are less informative than competing interpretations like (2).

When introducing folks to logic though, people will frequently present you with exercises where assumptions that are unremarkable ordinarily are not supposed to be translated into the resulting symbolic claim.

Title: Conditional Operator Question


答: "besides" $\neq$ "except". I would agree with the suggested solution that "besides" perhaps pragmatically implicates, but does not logically entail, "only those who are not": A sentence like "Everyone besides Sam has taken at least two different classes; Sam himself even has taken five" is not contradictory, so the reverse direction of the implication $(x \neq Sam) \leftarrow \ldots$ indeed should not hold.

Title: Conditional Operator Question

Body:
T(x,y) = Student x has taken class y

I need to translate the following statement into a quantified expression:

Everyone besides Sam has taken at least two different classes.

My initial thought was:

$$\forall x \exists y \exists z [(x\neq Sam) \leftrightarrow (T(x,y)\land T(x,z)\land (y \neq z)) $$

But I am being told that

$$\forall x \exists y \exists z [(x\neq Sam) \rightarrow (T(x,y)\land T(x,z)\land (y \neq z)) $$

Is the correct answer.

I am a little confused why the biconditional doesn't work here. The way I see it is, if there are two sets of students, those who have taken at least two classes and those who have not, then if you know a student has not then it must be Sam and only Sam and, if you know a student has taken at least two classes you know that it is not Sam.


问:
Max of algebraic equations
I am going through Dynamic Programming and Optimal Control text by Dimitri Bertsekas, on p. 18 - he derives these equations ...

$$\begin{equation}\begin{aligned}
\text{open-loop probability of win} &=
\max(p_w^2(3-2p_w), p_wp_d+p_w^2(1-p_d)) \\
&= p_w^2+p_w(1-p_w)\max(2p_w,p_d)
\end{aligned}\end{equation}\tag{1}\label{eq1}$$

and then...

$$\begin{equation}\begin{aligned}
\text{value of information} &=
p_w^2(2-p_w)+p_w(1-p_w)p_d \\
&- p_w^2-p_w(1-p_w)\max(2p_w,p_d) \\
&= p_w(1-p_w)\min(p_w,p_d-p_w)
\end{aligned}\end{equation}\tag{2}\label{eq2}$$

Please can someone help explain:

In equation 1 - how one would go about deriving line# 2 from line# 1, and

In equation 2 - how and why max turns into min?

Where:
$p_w$ and $p_d$ are probabilities of winning and drawing the game respectively.

Title: Max of algebraic equations

Body:
$\max(a+b,a+c) = a+\max(b,c)$, and $\max_i \{k f_i\} = k \max_i \{f_i\}$ for $k\ge 0$

$\max_i\{f_i\} = -\min_i \{-f_i\}$


问:
Is it possible that a clock's three hands divide the clock face into 3 equal parts?
And is there a proof/disproof out of intuition? Here we ignore the structure of the clock and suppose the hands move with constant speed.</p>

答:
In a 12 hour period, which I will take to be the unit of time, the hands start out and end aligned. The minutes hand overtakes the hours hand $11$ times during this period, and the seconds hand does so $12\times60-1=719$ times; these give their relative speeds with respect to the hours hand in revolutions per unit time. Note that $11$ and $719$ are relatively prime, which means there is no intermediate moment in time $0<t<1$ at which the hands are all three aligned: this would meen $a=11t$ and $b=719t$ are integers, so $719a=11b$ is a common multiple of $11$ and $719$, therefore a multiple of $\operatorname{lcm}(11,719)=11\times719$, which would force $t$ to be integer (but it isn't).

Now if $t$ is a time at which the relative positions of all hands are all multiples of $\frac13$ of a revolution, then by uniformity of their motion they will all be aligned again at time $3t$. By what we've just seen this means that $t$ is a multiple of $\frac13$ unit of time. But with the unit of time being 12 hours, $\frac13$ unit of time is precisely $4$ hours, at which points in time the minutes and seconds hands are aligned. So it never happens that the hands are precisely in relative positions $0,\frac13,\frac23$ of a revolution.


问:
True or false: $ U^{\bot }\bot V$ then $ U\bot V^{\bot }$
We know that:

$C(A^{T})^{\bot}=N(A) $

$N(A^{T})^{\bot}=C(A) $

C-column space, N - nullspace

How these statements can help me to prove:

$ U^{\bot }\bot V$ then $ U\bot V^{\bot }$ ?

I think it is true beacuse:

$ (U^{\bot })^{\bot }=U $

Thanks for any tip

答:
The statement is false. For example, consider the sets
$$
U = \{(x,y,0):x,y \in \Bbb R\}\subset \Bbb R^3, \quad V = \{(x,0,0):x \in \Bbb R\} \subset \Bbb R^3.
$$
However, the statement will hold if we have the additional fact that $\dim(U) = \dim(V)$.


答:
No one invented all of mathematical notation at once and consistently, and to answer your “Why?” isn’t easy. Over time, mathematicians proposed or invented notations that they felt worked and used them in publications. Sometimes conflicting notations caught on and became conventional in modern mathematics. Most of the time, context makes things clear, and we eventually don’t even notice the inconsistencies.

The uses of superscripted numbers like $^{-1}$ and $^2$ to mean different things in different contexts are just one example. Another is simple grouping and juxtaposition: $f(x+y)$ rarely means $f\times (x+y)$, but $2(x+y)$ usually does mean $2\times(x+y)$, but $z(x+y)$, well, it likely depends (on the context).

Another example: What does $\left({1\  2\  3\atop 2\ 3\ 1}\right)$ mean? Is it an element of the permutation group $S_3$, or is it a $2$ by $3$ matrix? It depends on the context.

If you want to go down a rabbit hole with this, check out “A history of mathematical notations,” by Florian Cajori, available here and elsewhere.

Title: $\arctan x$ vs $1/\tan x$

Body:
Why is $\arctan x = (\tan x)^{-1} = \tan^{-1}x$ but is not equal to $1/\tan x$, even though $\tan^{2}x = (\tan x) \cdot (\tan x)$?

Is there not another way to write $1/\tan x$?


问:
The equation $ax+bu+cs=ayz+bvw+crt$ also has no integers solutions for any integers $a,b,c$
Assuming that the system of equations:

$$x=yz$$
$$u=vw$$
$$s=rt$$

has no integers solutions. How I can prove that the equation $$ax+bu+cs=ayz+bvw+crt$$ also has no integers solutions for any integers $a,b,c$

答:
$$ax+bu+cs=ayz+bvw+crt\\
a(x-yz)+b(u-vw)+c(s-rt)=0$$

Plugging in any specific integers for $x,y,z,u,v,w,r,s,t$ yields specific integer values $x-yz,u-vw,s-rt$.  It really doesn't matter what these values are, they just become coefficients for $ae+bf+cg=0$ which is always solvable in integers (e.g., $a=b=c=0$).  So the implication in the question is false as a starting point (i.e., solutions need not exist for the original "system" in order for the equation to have solutions).

After that, note that the original "system" is somewhat meaningless as there are no constraints given on the listed variables aside from "this system doesn't have solutions".


问:
Intersection of polytopes
Given two polytopes, $\mathcal{S}$ and $\mathcal{U}$, they can be described as the intersection of finitely many halfspaces, i.e., $\mathcal{S} := \{x \in \mathbb{R}^n \mid A x \leq b \}$ and $\mathcal{U} := \{ x \in \mathbb{R}^n \mid C x \leq d \}$, where $A \in \mathbb{R}^{p \times n}$, $C \in \mathbb{R}^{q \times n}$ have full column rank, $b \in \mathbb{R}^p$ and $d \in \mathbb{R}^q$.

Is anyone aware of conditions on $A$, $C$, $b$ and $d$ guaranteeing the nonemptiness of the intersection set, i.e., $\mathcal{S} \cap \mathcal{U} \neq \emptyset$?

答:
Let $m$ and $n$ be natural numbers, $k=1,\dots, m$,  $a_k=(\alpha_{k1},\alpha_{k2},\dots, \alpha_{kn})\in\Bbb R^n$, and $\beta_k\in\Bbb R$. Given a system  $$(\alpha_k, x)\le \beta_k\,  \label{1}\tag{1}$$ of linear inequalities, there is a general problem to find conditions of its compatibility, that is when has at least one solution. In [Cha], beginning from §37 are listed most known compatibility conditions.

Theorem 1. The system \eqref{1} is compatible iff an equality $\sum_{k=1}^m \lambda_k\alpha_k=0$ with non-negative scalars $\lambda_k$ implies an inequality $\sum_{k=1}^m \lambda_k\beta_k \geq 0$.

Theorem 3. (S. N. Chernikov). The system \eqref{1} of rank $r$ is compatible iff the matrix of the system has a minor of order $r$
$$\delta=\left|\begin{matrix}
\alpha_{i_1j_1} & \alpha_{i_1j_2} & \dots & \alpha_{i_1j_r} \\
\alpha_{i_2j_1} & \alpha_{i_2j_2} & \dots & \alpha_{i_2j_r} \\
\dots & \dots & \dots & \dots \\
\alpha_{i_rj_1} & \alpha_{i_rj_2} & \dots & \alpha_{i_rj_r} \\
\end{matrix}
\right|\ne 0,$$
such that
$$\frac 1{\delta}\left|\begin{matrix}
\alpha_{i_1j_1} & \alpha_{i_1j_2} & \dots & \alpha_{i_1j_r} & \beta_{i_1}\\
\alpha_{i_2j_1} & \alpha_{i_2j_2} & \dots & \alpha_{i_2j_r} & \beta_{i_2}\\
\dots & \dots & \dots & \dots & \dots \\
\alpha_{i_rj_1} & \alpha_{i_rj_2} & \dots & \alpha_{i_rj_r} & \beta_{i_r}\\
\alpha_{ij_1} & \alpha_{ij_2} & \dots & \alpha_{ij_r} & \beta_{i_r}\\
\end{matrix}
\right|\ge 0,$$
for all $i\ne i_1, i_2,\dots, i_r$.

Theorem 2. (§38). The system \eqref{1} is not compatible iff its matrix contains $p$ rows, containing a $p\times (p-1)$-matrix
$$\underline{A}=
\begin{pmatrix}
\alpha_{i_1j_1} & \alpha_{i_1j_2} & \dots & \alpha_{i_1j_{p-1}} \\
\alpha_{i_2j_1} & \alpha_{i_2j_2} & \dots & \alpha_{i_2j_{p-1}} \\
\dots & \dots & \dots & \dots \\
\alpha_{i_pj_1} & \alpha_{i_pj_2} & \dots & \alpha_{i_pj_{p-1}} \\
\end{pmatrix}$$
such that $\delta_i\delta_{i+1}<0$ for $i=1,\dots, p-1$ and $\sum_{i=1}^p \beta_{k_i}|\delta_i|<0$ [I guess here should be $\beta_{i}$ instead of $\beta_{k_i}$. AR], where $\delta_i$ is a minor of order $p-1$, obtained from $\underline{A}$ by removing its $i$-th row.

References

[Cha] V. S. Charin, Linear transformations and convex sets, Kyiv: Vyshcha shkola, 1978, in Russian.

Title: Intersection of polytopes


问:
Find all walks that visit all natural numbers exactly $l$ times, and are formed by repeating a finite sequence of steps
I am interested in integer "walks" $x_n\in\mathbb{Z}$, $n\geq0$, on the grid of integers with the following properties:

Given integer parameters $l\geq1,\ \ s_0,s_1,...,s_{d-1}$, :

The "steps" are  $x_n=x_{n-1}+s_k$, where $k\in\mathbb{Z}_{d}\ $, $k=n\ \text{mod} \ d$.

Denote by $v[k,n]$ the number of "visits" at integer $k$ after $n$ steps of the walk. That is, $v[k,n]=\vert \{m | m\leq n\ , x_m=k \} \vert$. Then $x_n$ must obey $\underset{n\rightarrow\infty}\lim v[k,n]=l$ for all $k\in\mathbb{N}=\{0,1,...\}$

In words, I am interested in walks obtained by repeatedly taking steps whose size and direction are from a finite sequence, looped-over again and again; and the walk needs to visit all non-negative integers exactly $l$ times. Notice that I don't restrict the initial location $x_0$ (it could be any number in $\mathbb{Z}$).

I also define $u[k]$ to be the number of steps taken, from the first visit to $k$ to the last visit to $k$: $u[k]=n_2-n_1$,  where $v[k,n_1]=1,\ v[k,n_2]=l,\ x_{n_1}=x_{n_2}=k$.

My question: given integers $l\geq 1,d\geq 1, S\geq 1, U\geq 0$, is there a simple known way to find (=generate), or at least count, all starting positions $x_0$ and sequences $s_0,s_1,...,s_{d-1}$ of length $d$ that give rise to a walk with the above properties, with the restrictions $\forall k \in \mathbb{Z}_d\ |s_k|\leq S$ and $\forall k\in\mathbb{N}\ \ u[k]\leq U$?

To answer the question, I tried to look into lattice paths, but I think they use a different formalism.

EDIT1: I think that this is a necessary condition: $l=\frac{d}{\sum_{i=0}^{d-1} s_i} $

答:
Let me add the restriction that $d$ be a small as possible to obtain the same path. I.e., there is no divisor $d'$ of $d$ such that $s_k = s_{k - d'}$ for all $k \ge d'$. Otherwise, you can generate "new" patterns that work simply by repeating shorter working patterns several times.

Let $T = \sum_{k\in \Bbb Z_d} s_k$. If $T \le 0$, then you will only ever visit a finite number of positive integers, so we can assume $T > 0$.

If all of the $s_k \ge 0$, then you once you leave a number, you will never return. So the only possible solutions are
$$ d = l, x_0 \le 0, s_k =\begin{cases}0,&\quad k < d-1\\1,&\quad k = d-1\end{cases}$$
(If $x_0$ is sufficiently negative, the $1$ entry can also occur earlier in the $s_k$ list, but there can only be a single $1$ and the rest $0$.)

So assume there are negative $s_k$ and let $m = \left|\sum_{s_k < 0} s_k\right|$. After first visiting any particular integer $x$, in $\left\lceil\frac mT\right\rceil$ additional rounds (i.e., repetitions of the entire $s_k$ sequence), the action will have moved away from $x$ and will no longer be able to return to it or any point below it. So in this case, it is a matter of examining $\left\lceil\frac mT\right\rceil$ repetitions of the pattern to see if it visits the first $m$ points $l$ times. For a given $S, U$, there are only a finite number of patterns possible, so they can each be tested.


问:
Divergence of $\sum_{n=2}^\infty\frac{1}{(\ln n)^x}$ for $x>1$
How can we show that the series
$$\sum_{n=2}^\infty\frac{1}{(\ln n)^x}$$
diverges for $x>1$ ?

The book gives the following hint:
consider
$$\sum_{k=1}^\infty\sum_{n=M_{k-1}+1}^{M_k}\frac{1}{(\ln M_k)^x}$$
where $\ln M_k=k$ and note that $M_k-M_{k-1}=e^{-1}(e-1)M_k$; hence show that the series diverges.

But I really can't figure out what this hint means.

答:
Note that by l'Hôpital
$$\lim_{n\to\infty} \frac{\ln n}{n^{\frac{1}{x}}} = \lim_{n\to\infty} \frac{\frac{1}{n}}{\frac{1}{x}n^{\frac{1}{x}-1}} = \lim_{n\to\infty} \frac{x}{n^{\frac{1}{x}}} = 0.$$
Thus, for every $x>1$ there is some $N\in\mathbb{N}$, such that for all $n>N$,
$$\frac{\ln n}{n^{\frac{1}{x}}}<1$$
or equivalently $\ln n < n^{\frac{1}{x}}$.

Thus
$$\sum_{n=2}^\infty \frac{1}{(\ln n)^x} = \sum_{n=2}^N\frac{1}{(\ln n)^x} + \sum_{n=N+1}^\infty\frac{1}{(\ln n)^x} > \sum_{n=2}^N\frac{1}{(\ln n)^x} + \sum_{n=N+1}^\infty\frac{1}{n}=\infty.$$


问:
Relationship between the center and commutator subgroup of a group of nilpotency class 3
I am new to the theory of nilpotent groups. I am dealing with the nilpotent groups of class 3. I want to know is there any relationship between the commutator subgroup $[G, G]$ of the group $G$ and the center $Z(G)$ of the group $G$.  I have a strong belief that $[G, G]^{3}\subseteq Z(G)$ when 3 divides the order of the group $G$. But I could not find any literature for that. Can anyone please help me with this. Thank you, in advance.

答:
This does not hold even in a $3$-group.

Note that in any group $G$, using the commutator convention $[a,b]=a^{-1}b^{-1}ab$, we have
$$[x^r,y^s] \equiv [x,y]^{rs}[x,y,x]^{s\binom{r}{2}}[x,y,y]^{r\binom{s}{2}}\pmod{G_4}$$
where $G_n$ is the $n$th term of the lower central series of $G$, and $\binom{a}{2}=\frac{a(a-1)}{2}$ for any integer $a$. So you would be asserting that if $G$ is of class three, then for all $y\in G$ and $c\in [G,G]$,
$$[c^3,y] = [c,y]^3[c,y,c]^{\binom{3}{2}}=[c,y]^3$$
is trivial. In particular, $[x,y,x]$ and $[x,y,y]$ would both have to be of exponent $3$ for any $x,y\in G$.

This is definitely not true.

By a result of Remeslennikov and of Jónsson (independently), a variety of nilpotent groups of class $3$ is determined by a 4-tuple of nonnegative integers $(m,n,p,q)$  satisfying

$n|\frac{m}{\gcd(m,2)}$;

$p|m$;

$q|\frac{m}{\gcd(m,6)}$;

$p|3q$;

corresponding to the identities $$x^m=[x,y]^n=[x,y,z]^p=[x,y,y]^q=[x,y,z,w]=1.$$
(See here for citations). That means that no other identities will hold universally for groups of nilpotency class $3$. In particular, if we require $q\gt 3$ then it is not the case that $[c,y]^3=1$ for all $c\in [G,G]$. For example, we can take
$m=n=p=27$ and $q=9$. Then the corresponding relatively free group or rank 2 would not satisfy the identity you want.

This would be the group generated by $x$ and $y$, consisting of all elements of the form
$$x^a y^b [y,x]^c[y,x,x]^d[y,x,y]^e,\qquad a,b,c\in\mathbb{Z}/27\mathbb{Z},\qquad d,e\in\mathbb{Z}/9\mathbb{Z}$$
with multiplication given by
$$\Bigl(x^a y^b [y,x]^c[y,x,x]^d[y,x,y]^e\Bigr) \Bigl(x^{\alpha} y^{\beta} [y,x]^{\gamma}[y,x,x]^{\delta}[y,x,y]^{\epsilon}\Bigr)\\ =
x^{a+\alpha} y^{b+\beta} [y,x]^{c+\gamma+b\alpha} [y,x,x]^{d+\delta+c\alpha + b\binom{\alpha}{2}}[y,x,y]^{e+\epsilon+c\beta+b\alpha\beta+\alpha\binom{b}{2}}$$
with exponents taken modulo the appropriate power of $3$. In this group, neither $[y,x,y]$ nor $[y,x,x]$ are of exponent $3$, hence $[y,x]^3$ is not central.

Your "strong suspicion" only holds if $G_3$ has exponent $3$, and for $3$-groups of class $3$, the exponent of $G_3$ can be arbitrarily large.


问:
A surjective, continuous and closed map is a quotient map
I want to prove that a surjective, continuous and closed map is a quotient map. I know that a surjective map $p: X \longrightarrow Y$ is a quotient map if every set $V \subset Y$ is open $\Longleftrightarrow$ $p^{-1}(V)\subset X$ is open. Please note that $p^{-1}$ is the pre-image

Since we have continuity, one only has to prove the implication $\Longleftarrow$

Let $V\subset Y$ so that $p^{-1}(V)\subset X$ is open. $X\backslash p^{-1}(V)$ is closed and since $p$ is a closed map, $p(X\backslash p^{-1}(V))$ is closed. Here is where I get confused. If I knew that $p$ was injective, then I could keep going like this:
$$p(X\backslash p^{-1}(V))=p(X)\backslash p(p^{-1}(V))$$
and since $p$ is surjective:
$$p(X)\backslash p(p^{-1}(V))=Y\backslash V$$
and then we'd be done, because $Y\backslash V$ is closed and therefore $V$ is open. However, as I don't know if $p$ is injective I can't follow this path. Could someone help me?

答:
Since $p$ is surjective, you know that $pp^{-1}(A) = A$ for any $A \subset Y$. Thus

$$
V^c = p(p^{-1}(V^c)) = p(p^{-1}(V)^c)
$$

is closed and we are done.


问:
System of Diophatine equations $x_1,\ x_2,\ x_3,\ \ldots,\ x_n,\ y \in \mathbb{Z^+}$
Let $a_1,\ a_2,\ a_3,\ \ldots,\ a_n$ be distinct positive integers.
Find $x_1,\ x_2,\ x_3,\ \ldots,\ x_n,\ y \in \mathbb{Z^+}$
such that: $$\left\{\begin{array}{rl}(x_1,x_2,\ldots,x_n)&=1\\ a_2x_1+a_3x_2+\cdots+a_nx_{n-1}+a_1 x_n & =y x_2\\ \cdots\cdots\cdots\cdots & = \cdots\cdots \\ a_n x_1+a_1 x_2+ \cdots + a_{n-1} x_n&=y x_n \end{array}\right.$$

How to solve this system equation ?

I am not good at diophantine equation and the problem above is so hard.

答:
By inspection $x_1=x_2=\ldots=x_n=1,y=\sum a_i$ is a solution.

The system is underspecified, though, so there are other solutions, e.g. for $n=2$
$$
x_2=a_2, y=a_1+x_1
$$
is a solution for any choice of $x_1$ with $(x_1,a_2)=1$.


问:
Find a rational number that is greater than $t$ where $t^2<2$
I have trouble in proving this, that is if $t\in Q\land t^2<2$, then there exists another rational number $q$ such that $q>t\land q^2<2$. May I ask how to find this rational number?

答:
If $t$ is negative or zero, we can use $1$ as bigger.

If $t$ is already positive, we can write it as $ t = \frac{p}{q}$ with positive integers $p,q.$  The requirement that $  \left( \frac{p}{q} \right)^2 < 2$ tells us that
$$  p^2 - 2 q^2 < 0 $$
We choose
$$  w = \frac{3p+4q}{2p+3q}= \frac{3t+4}{2t+3}  .   $$
As
$$  \left( 3p+4q  \right)^2 - 2 \left( 2p+3q \right)^2 = p^2 - 2 q^2 < 0$$
we see that $w^2 < 2$
At the same time
$$  w - t  = \frac{3p+4q}{2p+3q}  - \frac{p}{q} = \frac{-2(p^2 - 2 q^2)}{q(2p+3q)}  > 0 $$

The reasons for doing this: every quadratic form, in this case $x^2 - 2 y^2,$  has an automorphism group; we use a matrix that generates the group, which is  a recipe once we find the first nontrivial Pell solution, $3^2 - 2 \cdot 2^2 = 1$

$$
\left(
\begin{array}{cc}
3&2 \\
4&3 \\
\end{array}
\right)
\left(
\begin{array}{cc}
1&0 \\
0&-2 \\
\end{array}
\right)
\left(
\begin{array}{cc}
3&4 \\
2&3 \\
\end{array}
\right) =
\left(
\begin{array}{cc}
1&0 \\
0& -2 \\
\end{array}
\right)
$$

note: many binary quadratic forms also have an automorphism of negative determinant, in this case we could negate one variable while leaving the other fixed. The entire group is defined once we add a single instance of that onto the infinite cyclic group  of automorphisms with positive determinant


问:
Determining if a recursive sequence is converging with induction
Consider this sequence

$$ a_0 = 2,$$
$$ a_{n+1} = \frac 1 a_n $$

Now to prove for convergence I simply put in values ($a_1, a_2, a_3$) and got the the sequence has this behaviour: $2, 1/2, 2, 1/2,\dots$. Now since it isn't approaching any value it shouldn't be converging.

But I would like to prove this, so I looked up how to do it and you're supposed to prove monotony and limitations of the sequence. Now to prove monotony we would have to assume that the sequence is falling or raising. But how can I do that here? Since the sequence is changing its value from <1 to >1 in every iteration what should I assume? Is there a way to solve this without the "obvious" answer?

Thank you!

Body:
Let us solve $A_n A_{n+1}=1$, take $A_n=\frac{B_{n+1}}{B_{n}}$, then
We get $B_{n+2}=B_{n}$. Next, let $B_n=x^n$, then we get $x^2=1 \implies x=\pm 1$
We can write $$B_n=P +Q(-1)^n \implies A_n=\frac{P+Q(-1)^{n+1}}{P+Q(-1)^{n}}=\frac{R+(-1)^{n+1}}{R+(-1)^{n}}.$$ $$A_0=2 \implies \frac{R-1}{R+1}=2 \implies R=-3$$
Hence $$A_n=\frac{-3-(-1)^n}{-3+(-1)^n}$$

Therefore $A_n$ does not converge but it oscillates:$2,1/2,2,1/2,...$


问:
About the perfect residue fields of henselian valuations
Let $(K,v)$ be a henselian valued field. i.e., $v$ is a henselian valuation on a field $K$. If the residue field of $(K,V)$ is perfect (every its algebriac extension is separable), is it true that $K$ is perfect?
Thanks for any help

答: $\Bbb{F}_p[[t]]$ is Henselian.

Body: $\Bbb{F}_p[[t]]$ is Henselian.


问:
Converting cartesian equation to polar issue
I have tried converting the equation $ y = x^3 $ to a polar equation using the following steps but the graph of the resulting polar equation also appears to include $ y=-x^3 $.

$ r^2 = x^2 + y^2 $

$ r^2 = x^2 + (x^3)^2$

$ r^2 = r^2\times cos(\theta)^2 + r^6cos(\theta)^6 $

$ 1 = cos(\theta)^2+r^4\times cos(\theta)^6$

$ \frac{sin(\theta)^2}{cos(\theta)^6}=r^4$

$ r = (\frac{sin(\theta)^2}{cos(\theta)^6})^{1/4}$

However, the graph of this is not $ y = x^3 $.

I am wondering why converting a cartesian equation to a polar equation does not appear to work this way. I understand that $ y=x^3$ can be easily converted to polar by substituting $ y=rsin(\theta)$ and $x=rcos(\theta)$ into $y=x^3$ but I don't understand why I do not get the same answer when substituting $y = x^3$ for $x$ in $r^2 = x^2+y^2$.

Thanks.

答:
$$r=\sqrt{x^2+x^6}$$ is fine, as the RHS is positive. Then

$$\tan\theta=-\frac{x^3}x=-x^2$$ and

$$r=\sqrt{-\tan^3\theta-\tan\theta}=\sqrt{-\dfrac{\sin\theta}{\cos^3\theta}}.$$


问:
Check if two lines intersect
I have a non-linear line with these coordinates:

```<code>(1,50)(2,40)(3,45)(4,40)(5,60)(6,30)
</code>```

When I draw the line passing through each of these points, noted line A, I can visually see that another straight line drawn from last to first point only - from (1,50) to (6,30) - noted line B, would intersect/cross line A.

My goal is to be able to mathematically check that line B does NOT intersects with line A, is there some equation or approach I could go with to check that?

EDIT:
Sorry for the misleading "line" word I use, I should have used curve. Here is the representation of what I meant (credits to Joffan):

答:
Assuming that Line A is a piecewise linear curve and your issue is illustrated by:

(line A blue, line B orange), you can essentially use the basic line intersection equations to check whether each line segment in line A intersects line B between the limits of interest or not.

In this particular case, since the points of line A are in monotonic (here, ascending) $x$ order and line A starts on one side of line B and finishes on the other, you have a shortcut; there will definitely be an intersection somewhere. Also with the same restriction on the order of points on line A, you can check dot product for each line from $a_1$ to successive points $a_i$ against line B (which is $a_1$ to $a_n$) and check if the sign changes, indicated that the point is now the other side of line B, localizing the intersection (and finding multiple intersections if occurring). If you have intermediate points that are outside the $x$-limits of the points that define line B, this may not work:

Title: Check if two lines intersect


问:
A matrix polynomial equation when power goes to infinity.
$$A=\pmatrix{\frac2{10}&\frac3{10}\\\frac3{10}&\frac5{10}},\\B=\pmatrix{2&3\\4&5},\\XA+XA^2+XA^3+...=B$$ what is $X$?
I think it is kind of like the transition matrix, but the sum of each row is not 1. So how should I get started?

答:
You can try to calculate the matrix power of A and evaluate the sum which has to converge against a Matrix $ S $. So consider $ \sum\limits_{k=1}^\infty A^k=: S $.
Than you only have to solve $ X\cdot S=B $.


问:
which of the followings are true for bijective functions
which of the followings are true:-

1. There is a continuous bijection from $\mathbb{R}^2\to \mathbb{R}$.

2. There is a bijection between $\Bbb{Q}$ and $\mathbb{Q}\times \mathbb{Q}$.

Can somebody help me please.I am totally stuck on it.

答: An approach from an algebraic topologist wannabe might be the following: suppose we have a continuous bijection $\psi: \mathbb{R}^{2}\to \mathbb{R}$, then restrict it to $[0,1]^{2}$ and pick an open disc inside it (this choice is arbitrary, it's enough an open connected subset). You have, by continuity, that such a disk is sent homeomorphically (compact$\to$Hausdorff) to an open interval in $\mathbb{R}$ but this violates the Invariance of Domain.</p>


问:
Is the identity map the only map from the positive integers to itself that simultaneously preserves multiplication, order, and primes?
Is the identity map the only map $f$ from the positive integers $\mathbb{Z}_{+}$ to itself that satisfies all three of the following properties?

$f(mn)=f(m)f(n)$ for all positive integers $m$ and $n$ (which implies that $f(1)=1$)

$f(m)<f(n)$ whenever $m<n$ (which implies that $f$ is injective)

$f(p)$ is prime for any prime number $p$

The above three properties (actually, for property 2, the weaker assumption that $f$ is injective suffices) imply that applying $f$ to any positive integer will not change the prime signature, because all that happens with $f(n)$ is that each prime $p$ dividing $n$ is replaced with $f(p)$ while maintaining the same exponent.

Here are some examples that suggest that the answer might be "Yes" (because they only satisfy properties 1 and 3 and injectivity, but are not order-preserving on all positive integers, although they are order-preserving on prime numbers):

Let $f(p)$ be the next prime following $p$, with $f$ extended to composite numbers using (complete) multiplicativity. Then, $8<9$, while $f(8)=f(2^3)=f(2)^3=3^3=27>25=5^2=f(3)^2=f(3^2)=f(9)$.

Let $f$(the $n$th prime) be the $(2n)$th prime. Then, $27<32$, while $f(27)=f(3^3)=f(3)^3=7^3=343>243=3^5=f(2)^5=f(2^5)=f(32)$.

Let $f(p)$ be the $p$th prime. Then, $f(8)=27>25=f(9)$, as in the first example.

Let $f$(the $n$th prime) be the $(n^2)$th prime. Then, $3<4$, while $f(3)=7>4=2^2=f(2)^2=f(2^2)=f(4)$.

But of course, specific examples (using inductive reasoning) are not enough to conclude that the answer is in fact "Yes".

答:
Start with $f(2) = p$ for some prime $p$. Then, we need a choice for $f(3)$. We certainly need that whenever $3^q > 2^l$ then $f(3^q) > f(2^l)$ i.e. $f(3)^q > f(2)^l$ i.e. $f(3) > f(2)^{\frac lq}$.

However, we also need that whenever $3^x < 2^y$ then $f(3) < f(2)^{\frac yx}$.

Therefore, we are constrained to have $\frac lq < \log_{f(2)} f(3) < \frac yx$ for all pairs of (positive, doesn't really matter) rationals $\log_2 3 > \frac lq$ and $\log_2 3 < \frac yx$. By density of rationals in the reals, we are forced to have $\log_{f(2)} f(3) = \log_2 3$, and therefore $f(3) = f(2)^{\log_2 3}$.

Now, we can write this as $f(3) = 3^{\log_2 f(2)}$.

I don't think there's anything special about $2$ and $3$ in the argument above, so doing it for any pair of primes gives you that $f(q) = f(p)^{\log_p q}$ for any pair of primes $p$ and $q$.

This is severely restrictive of all values of $f$,since given just $f(2)$ we get all the other values. So we have a partial answer : everything is decided by $f(2)$, since $f(q) = f(2)^{\log_2 q}$.

Let us write this in beneficial fashion : for the existence of such a function, we require that for all primes $p,q$ we have , using the fact that $\frac{\ln p}{\ln q} = \log_q p$, the relation $\ln f(r) \ln s = \ln f(r) ln s$ for any primes $r,s$.

Now, the point is that we are expecting that $f(q)$ is an integer only when $f(2) = 2$ so that $f(q) = q$. But this is not a "given", since we don't know if $f(2)$ could be a prime so that $f(q)$ were in fact a valid prime integer for every $q$!

To answer this question, we need to get to "quadratic relations between logarithms of algebraic numbers".

Now, usually when we prove that things involving logarithms aren't integers and so on, we usually break them into "relations" involving logarithms. Then we use results known for logarithms and exploit them for our purposes.

Now, there are nice theorems for "linear" relations involving logarithms. These lead to the transcendence of numbers such as $\log_2 3$, and $2^{\sqrt 2}$ (try to see that if they were rational, it breaks into a linear relation between logarithms).

However, the relation we have , of the form $\ln q \ln r = \ln s \ln t$, is called a quadratic relation between logarithms. It turns out that for quadratics , we have a series of conjectures, along with a nice theorem which I state first!

First, let us use the six exponentials theorem. We call a subset $\{z_1,...,z_n\}$ of complex numbers as rationally independent if for any rationals $r_1,...,r_n$ we have $r_1z_1+...+r_nz_n \neq 0$ if at least one of the $r_i$ is non-zero.

Six Exponentials Theorem : Suppose that $x_1,x_2,x_3,y_1,y_2,y_3$ are complex numbers such that :

$x_1,x_2,x_3$ are rationally independent over the integers

$x_1,y_1$ are rationally independent over the integers.

Furthermore, each of $e^{x_i}$ and $e^{y_j}$ are algebraic numbers (read this and transcendental numbers up).
Then the matrix :
$$
\begin{pmatrix}x_1&x_2&x_3 \\ y_1 & y_2 & y_3\end{pmatrix}
$$
is of rank $2$. In particular, at least one of :
$$
x_1y_2 \neq x_2y_1, x_1y_3 \neq x_3y_1,x_2y_3 \neq x_3y_2
$$
is true.

Let us see how to use this. Suppose that $f(2) = p \neq 2$. Note that automatically, $f(s) \neq s$ for all primes $s$. Choose a prime $r$ such that $f(r) \neq p, r \neq p$. Choose a prime $s$ so that $s,f(s) \notin \{2,p,r,f(r)\}$. Note that $\ln 2 , \ln r , \ln s$ are linearly independent over the rationals (using an argument via prime factorization) and $\ln 2 , \ln f(r)$ are also rationally independent. Consequently, let the $x_i$ be $\ln 2 , \ln r,\ln s$ and $y_i$ be $\ln p ,\ln f(r),\ln f(s)$. Then, all conditions are satisfied (the third one trivially, since integers are algebraic).

As a conclusion, one of $\ln 2 \ln f(r)  = \ln r \ln p$, $\ln 2 \ln f(s) = \ln s \ln p$ or $\ln f(r) \ln s= \ln f(s) \ln r$ must be false. But none of these can be false! Consequently, we get that $f(2) = 2$, for that's the only way in which I can violate such a situation which provided our $x_i$ and $y_i$.

Next we use the conjecture :

Four exponentials conjecture : (Read up all unknown terminology) let $a,b,c,d$  each be logarithms of some complex numbers, with $\frac ab$ and $\frac cd$ non-zero and irrational. Then $\frac ab \neq \frac cd$.

Sounds logical : it's saying that a relation $ad = bc$ can't exist between "incompatible" logarithmic quantities $a,b,c,d$.

To use this for our problem : we know that if $f(3) = q$ a prime, then $\ln 2 \ln q = \ln 3 \ln p$. But note that if $p \neq 2$ then $q \neq 3$ and $\frac{\ln 2}{\ln p}$ and $\frac{\ln 3}{\ln q}$ are irrational (use prime factorization), so they are not equal by the conjecture, contradicting the premise. Thus $f(2)=2$, $f(3) = 3$ and we can do this for any pair of primes.

Next, another conjecture of Schaunel, which is also nice. We start with $n$ complex numbers which are linearly independent over the rational numbers i.e. no rational combination of them is zero. To perform this argument, we first choose an $r$ prime with $r \neq p, f(r) \neq p$ so that  $\ln 2, \ln r , \ln p$ and $\ln f(r)$ (we are assuming $p \neq 2$) so that they are all linearly independent over the rationals (once again, you can use prime factorization to get this result).

The conjecture is, stated in simple words :

If there are $n$ complex numbers linearly independent over the rationals, say $z_1,...,z_n$ then consider the subset $\{z_1,...,z_n,e^{z_1},...,e^{z_n}\}$. There is a subset $S$ consisting of $n$ of these elements, say $\{s_1,...,s_n\}$, such that if I take any polynomial in $n$ variables $f(x_1,...,x_n)$ with rational coefficients, then $f(s_1,...,s_n)\neq 0$. This is equivalent to saying that the transcendence degree of the extension of the rationals by all of $\{z_1,...,z_n,e^{z_1},...,e^{z_n}\}$ is at least $n$.

Now, do this with our subset ,and because each of the exponentials of elements of  $\{\ln 2 , \ln r, \ln f(r) , \ln p\}$ are $2,r,f(r),p$ which are already rational so they don't add to the transcendence degree of the extension.  Thus, using the conjecture, these four elements are therefore algebraically independent! But then, the polynomial $f(w,x,y,z) = wx-yz$ cannot be zero at $w=\ln 2, x = \ln f(r), y = r, z = \ln p$, and therefore $\ln 2 \ln f(r) \neq \ln r \ln f(p)$, a contradiction. So $f(2) = 2$.

All these results/conjectures are very helpful in solving these existence problems, as I have shown.

Links :

1 . The Six Exponentials theorem

2 . The Four exponentials theorem

3 . Schanuel's conjecture


问:
An estimate involving Polar Coordinates
I encountered the following computation in a paper:

The highlighted $t$ is mysterious to me, how did it end up there? For some guidance: the first line is fundamental theorem of calculus and change of variables. The last equality is change of variables and fubini. Thanks!

答:
Jensen’s inequality should do the trick. Replacing $dr$ with the probability measure $dr/t$ on the interval $[0,t]$ and using convexity of $x \mapsto x^2$ furnishes the inequality
$$ \left( \int_{[0, t]} f(r) \frac{dr}{t} \right)^2 \leq \int_{[0, t]} f(r)^2 \frac{dr}{t}. $$
Pulling out the factors of $t$ and rearranging gives the desired inequality,
$$ \left( \int_{[0, t]} f(r) dr \right)^2 \leq t \int_{[0, t]} f(r)^2 dr$$
where $f(r) = \nabla u (r \xi) \cdot \xi$.

Title: An estimate involving Polar Coordinates


问:
Subgroups of $\mathbb{Z}^k$ of finite index $n$
I want to describe all subgroups  in $\mathbb{Z}^k$ of finite index $n$.
I have solved it for the case $k=2$. In $\mathbb{Z}^2$, each subgroup of index $n$ corresponds to a matrix $\left( \begin{array}{a}
a & b \\
0 & d \end{array} \right)$ with $a, b, d \in \mathbb{Z}$ and $ad=n; a,d>0; 0 \le b<d$. Can someone give a hint how to go about it in the general case? For the case $k=3$, my guess is that each subgroup corresponds to a matrix $\left( \begin{array}{a}
a & b & c \\
0 & d & e \\
0 & 0 & f \end{array} \right)$ where $a, b, c, d, e, f \in \mathbb{Z}$ and $adf=n$; a, d, f >0; $0 \le b <d; 0 \le c <f; 0 \le e<f$. Is this right?</p>

答:
If $\Gamma<\Bbb Z^k$ has index $n$, then $n(\Bbb Z^k/\Gamma)=(0)$. Therefore $\Gamma$ contains the subgroup $(n\Bbb Z)^k$.

This shows that there's a bijection between the wanted subgroups and the subgroups of
$$\Bbb Z^k/(n\Bbb Z)^k\simeq(\Bbb Z/n\Bbb Z)^k$$
of index $n$.

Then one can simply apply the structure theorem for finite abelian groups.