问:
Finding the Image of an endomorphism on periodic indefinitely differentiable functions
Let $T$ be a strictly positive real number, and $E$ the set of $T$-periodic functions from $\mathbb{R}$ to $\mathbb{R}$ that are indefinitely differentiable (of class $C^{\infty}$), and $\phi$ an endomorphism of the vector space $(E,+,\cdot)$ such that $\phi(f)=f'$ for all $f\in E$. I have to find the Kernel and Image of $\phi$. I found that $\mathrm{Ker} \, \phi$ is the set of constant functions, however, I'm having trouble finding $\mathrm{Im} \, \phi$. I thought of showing that $\phi$ is surjective and concluding that $\mathrm{Im} \, \phi = E$, but I think this is wrong because if $f'$ is $T$-periodic, that doesn't necessarily imply that $f$ is $T$-periodic. I don't know where to start now, any hints will be welcome.

答: The image of $\phi$ is the set $\left\{f\in E\,\middle|\,\int_0^Tf(x)\,\mathrm dx=0\right\}$. In fact, if $f\in E$ is such taht $\int_0^Tf(x)\,\mathrm dx=0$ and if you define$$F(t)=\int_0^tf(x)\,\mathrm dx,$$then $F'=f$ and $F\in E$. So, $f\in\operatorname{Im}(\phi)$. And if $f\in\operatorname{Im}(\phi)$, then, if $F\in E$ is such that $F'=f$; then$$\int_0^Tf(x)\,\mathrm dx=F(T)-F(0)=0.$$


问:
Least upper bounds and greatest lower bounds: open vs. closed intervals
I tried looking through other questions but couldn't find this one answered. I'm working through Lax and Terrell's 'Calculus with Applications'.

The book asks to state whether the following intervals have GLBs and LUBs and if so, what they are: $(8,10)$, $(8,10]$.

My initial thought was that an open interval can't have a LUB or GLB as they don't include the end points and thus there is always a number arbitrarily closer to the end point that doesn't quite 'reach' it. However, I've also seen some discussion such that, e.g., the LUB of both half-open and open intervals are the same given that $9.99999... = 10$.

Help appreciated here as I want to understand this before proceeding!

答: Every bounded non-empty subset of $\Bbb R$ has a greatest lower bound and a lowest upper bound. In the case of $(8,10)$ and of $(8,10]$, these are $8$ and $10$ respectively. The fact that $10$ belongs to the second interval but not to the first one changes nothing. And the fact that $9,99999999\ldots=10$ also changes nothing.

ContentLicense: CC BY-SA 4.0


问:
Looking for beginner resources.
I'm looking for resources that will get me into the basics of mathematics.

I'm coding a small utility library for manipulating n-dimensional vectors because I need a simple way of looking at data from different angles.

I want to be able to rotate the underlying euclidian space to efficiently change the direction of traversal of multidimensional arrays.

Given the vector [0, 1] I want to acces [y, x] instead of [x, y] or [0,1,2,3,4,5] - [z,x,y,a,b,c]. This is simple for <4 dimensions, but after the third dimension I fail to generalize the concept of rotation matrix generation to higher dimensions.

I mainly need this to access data with no mathematical properties, but it would be intriguing if I could use this to project n-dimensional implicit surface equations onto a 3d space and have it rotate around axes of different dimensions to visualize tesseracts, 4-simplex, 5-cubes, or cartesian products of either.

The problem is I don't know even beginners algebra, no trigonometry nothing really. I started playing around with desmos.com today. I thought a good exercise would be to find a way to compute the square root of x.

I started working backward from $x^{2}$. After an embarrassingly long time working my way through different naive approaches I found $2^{\left(0.5\frac{\log\left(x2\right)}{\log\left(2\right)}\right)}$ works. In retrospect that's pretty obvious, but well.

I also tried to find a formula for calculating the sum of exponential series like $n^{1} + n^{2} + n^{k}$. I had the lucky intuition that it's gotta be < $n^(n-1)$ and $\frac{1}{n-1}\left(n^{\left(n+1\right)}-1\right)-1$ works, but that took about half a day to get there.

The thing is I could have avoided most of the naive steps in the wrong direction If I had at least a basic idea of mathematical concepts. I want to be able to read an expression and get a mental image of the parts it's composed of and their mathematical meaning. I want to be able to write my own n-dimensional rotation matrix generation algorithm, rather than having to read pseudocode implementations and deduce the mathematical meaning of it.

I think Math is an intriguing world and I'd love to come up with my own formulas for existing problems because you learn a whole lot on the way. I also think that having a deeper understanding of mathematics would allow me to write cleaner code and algorithms.

The thing is I don't know where to start. - I know there's the Pérez-Aguila algorithm, but I'd rather come up with my own in order to actually understand why it's working.

Can you recommend a few good books/resources name the topics that I would need to learn about in order to construct such an algorithm? I don't even know what I need to know to start.

答:
there are many ways to start learning this.

https://encyclopediaofmath.org/wiki/Higher-dimensional_geometry.

higher dimensional rotations preserve information. from an information theory perspective they are all equivalent ways to present the same information. projections from n dimensional space to lower dimensional spaces guarantee that there will be information loss.

since your data is finite and discrete you can think of all the operations on your data as actions on a finite group. you can learn group theory. you can learn graph theory. you can learn abstract algebra which includes some group theory and linear algebra. you can learn linear algebra more. you can approach this as a geometry problem where your focus is on projections in n dimensional continuous space. then when you apply that to writing software there will be some hard lessons about continuous vs discrete and pure vs applied mathematics. if you want the applied analytic geometry books then that is one more book to get.

now think about if you are not rotating 90 degrees then you have a lot more math. you need to interpolate. you need a lot of algorithms to solve all the various problems.

a good read about gimble lock and euler angles should get you started before you move on to quarternions. There may be other algorithms for the rotations but I haven't found exactly what I am looking for yet. I may need to attempt to invent something like this. one of the solutions to gimble lock that I found was some javascript coder that was constantly making copies of the 3d object to destroy the old one after the rotation. this way it would never have gimble lock because the new copy has a rotation vector (0,0,0,0) at the start of every transformation (rotation).

Title: Looking for beginner resources.


问:
Why does multicollinearity cause the standard errors of the coefficients to go up?
I understand that multicollinearity is a problem because the stronger the correlation, the more difficult it is to change one predictor without changing another and it becomes difficult for the model to estimate the relationship between each predictor and the response independently because the predictors tend to change in unison.

But why do the standard errors of the coefficients go up in the case of multicollinearity?
Can someone give me some intuition to why this happens?

答: Intuitively, the standard errors represent uncertainty in your coefficient estimates, so it's part of the exact problem you already laid out. Since the model has a harder time assigning the distinct effects of the collinear variables, there's more uncertainty in each of their estimates. That means the estimates are more imprecise, and you have larger standard errors.</p>


问:
Find the probability that the median of the three uniform numbers is less than 0.4.

Three numbers are randomly and independently selected from the interval (0, 1). Determine the probability that the median of the three numbers is less than 0.4.

I know that in order for this to happen, we need at least two values to be < 0.4. I'm just not sure where to go from here and what formula to use. Thanks in advance!

答:
Comment: Let $X$ be the number of observations below $0.4.$
Then $X \sim \mathsf{Binom}(3, 0.4).$
As you say, the median is below $0.4,$ if $X \ge 2.$

Here are three methods of evaluation, the first of which is @Henry's (+1).

(1) You seek
$$P(X \ge 2) = P(X=2)+P(X =3)\\ = 3(.4)^2(.6) + (.4)^3 = 0.352.$$

(2) Also, $P(X \ge 2) = 1 - P(X \le 1) = 0.352;$ computation in R, where pbinom is a binomial CDF.

```<code>1 - pbinom(1, 3, .4)
[1] 0.352
</code>```

(3) Looking at the distribution of the median: Simulation in R of a million iterations: each sampling three independent standard
uniform values and taking their median. One can expect
three place accuracy.

```<code>set.seed(2020)
med = replicate(10^6, median(runif(3)))
mean(med < .4)  # aprx. prop. of medians < .4
[1] 0.351944
</code>```

One can prove analytically that the distribution of the median of three standard uniform random variables $U_i$ is $U_{(2)} \sim \mathsf{Beta}(2,2).$ Here is a histogram of the one million simulated medians along with the density function of this distribution.

```<code>hist(med, prob=T, col="skyblue2")
 curve(dbeta(x,2,2), add=T, col="orange", lwd=2)
 abline(v = .4, col="darkgreen", lwd=2)
</code>```

Note: The simulation result can be compared with the exact answer as follows.

The density of $\mathsf{Beta}(2,2)$ is $f(x) = 6x(1-x),$ for $ 0<x<1.$
Then $$P(U_{(2)} \le 4) = \int_0^{0.4} 6x(1-x)\,dx = 0.352.$$

```<code>diff(pbeta(c(0,.4), 2, 2))
[1] 0.352
</code>```

Body:
Comment: Let $X$ be the number of observations below $0.4.$
Then $X \sim \mathsf{Binom}(3, 0.4).$
As you say, the median is below $0.4,$ if $X \ge 2.$

Here are three methods of evaluation, the first of which is @Henry's (+1).

(1) You seek
$$P(X \ge 2) = P(X=2)+P(X =3)\\ = 3(.4)^2(.6) + (.4)^3 = 0.352.$$

(2) Also, $P(X \ge 2) = 1 - P(X \le 1) = 0.352;$ computation in R, where pbinom is a binomial CDF.

1 - pbinom(1, 3, .4)
[1] 0.352

(3) Looking at the distribution of the median: Simulation in R of a million iterations: each sampling three independent standard
uniform values and taking their median. One can expect
three place accuracy.

set.seed(2020)
med = replicate(10^6, median(runif(3)))
mean(med < .4)  # aprx. prop. of medians < .4
[1] 0.351944

One can prove analytically that the distribution of the median of three standard uniform random variables $U_i$ is $U_{(2)} \sim \mathsf{Beta}(2,2).$ Here is a histogram of the one million simulated medians along with the density function of this distribution.

hist(med, prob=T, col="skyblue2")
 curve(dbeta(x,2,2), add=T, col="orange", lwd=2)
 abline(v = .4, col="darkgreen", lwd=2)

Note: The simulation result can be compared with the exact answer as follows.

The density of $\mathsf{Beta}(2,2)$ is $f(x) = 6x(1-x),$ for $ 0<x<1.$
Then $$P(U_{(2)} \le 4) = \int_0^{0.4} 6x(1-x)\,dx = 0.352.$$

diff(pbeta(c(0,.4), 2, 2))
[1] 0.352


问:
What are logarithms?
I have heard of logarithms, and done very little research at all. From that little bit of research I found out its in algebra 2. Sadly to say, I'm going into 9th grade, but yet I'm learning [calculus!?] and I don't know what a logarithm is! I find it in many places now. I deem it important to know what a logarithm is even though I'm jumping the gun in a sense. My understanding of concepts, is just like that of programming. In the mean time, you know its there, and your ITCHING SO HARD to find out what that is, but nope! For now we use it, tomorrow we learn what it does.

I just know that to identify a logarithm at my level, I just look for a log. :P

答:
It's funny how things come full circle...

Logarithms arise as a natural order within the subset of all numbers as the inverse of exponentiation. If exponentiation is repeated multiplication, then logarithms are repeated division. When this type of number first appeared, it went by the name of logarithmus, literally "ratio-number", and was invented by John Napier and Henry Briggs in the late 16th to early 17th century AD.

From a natural perspective, logarithms describe a pattern of growth in general, or can describe a scale such as pH.

Finally, from the perspective of the representations of quantities themselves, logarithms directly describe the nature of a particular representation of quantities of a given base. Each number can be represented in any base as the sum of unique powers of bases. Base two, the base commonly used to represent quantities in computers, is the sum of $n$ digits which correspond to the sum, $2^0 + 2^1 + 2^2 +...+ 2^n = \sum_{k=0}^{n}{2^n}$, for example; however, this can be generalized to all representations of numbers in any base with base $b$ and digital length $n$ as

$$\sum_{k=0}^{n}{b^k}$$

ContentLicense: CC BY-SA 3.0

Title: What are logarithms?


问:
Proof of angle Bisector Theorem by Vectors
For $\Delta ABC$, let $AD$ is its internal angle bisector, then we know that $$\frac{AB}{AC}=\frac{BD}{CD}$$ and I know that it can be proved by geometry by drawing a line through $C$ parallel to $AD$ and extend $AB$ so that it cuts new line and then use various angles and concept of similar triangle to prove our result. But I was wondering if we can prove it with the help of vectors as well?

If we take position vectors of $A,B$ and $C$ as $\overrightarrow{0},\overrightarrow{b}$ and $\overrightarrow{c}$ respectively , then $\overrightarrow{AD}$ can be taken as $\lambda(\hat{b}+\hat{c})$.  Could someone help me to proceed after that or provide some alternate approach through vectors?

答:
As mentioned, $\vec{AD} \neq \lambda (\hat b+\hat c)$ in general.
I think there's still a decent looking proof using dot product.

As usual, it helps if you sketch a diagram.

Let's suppose $\angle BAD=\angle CAD=\theta$, where $\theta$ is acute, and $\lvert \vec{BD} \rvert:\lvert \vec{CD} \rvert=r:s$. Using the section formula,
$$\vec{AD} = \frac{r}{r+s}\vec{AC}+\frac{s}{r+s}\vec{AB}.$$

Then by dot product,

$$\vec{AD}\cdot\vec{AC}
=\lvert\vec{AD}\rvert\lvert\vec{AC}\rvert \cos\theta
=\frac{r}{r+s}\vert\vec{AC}\vert^2+\frac{s}{r+s}\vec{AB}\cdot\vec{AC}$$

$$\implies\lvert\vec{AD}\rvert \cos\theta
=\frac{r}{r+s}\vert\vec{AC}\vert+\frac{s}{r+s}\vert\vec{AB}\vert\cos 2\theta. \tag{1}\label{1}$$

Similarly, by considering $\vec{AD}\cdot\vec{AB}$, we have
$$\lvert\vec{AD}\rvert \cos\theta
=\frac{r}{r+s}\vert\vec{AC}\vert\cos 2\theta+\frac{s}{r+s}\vert\vec{AB}\vert. \tag{2}\label{2}$$

Combining \ref{1} and \ref{2} gives

\begin{align}
\frac{r}{r+s}\vert\vec{AC}\vert+\frac{s}{r+s}\vert\vec{AB}\vert\cos 2\theta &=\frac{r}{r+s}\vert\vec{AC}\vert\cos 2\theta+\frac{s}{r+s}\vert\vec{AB}\vert\\
\frac{r}{r+s}\vert\vec{AC}\vert(1-\cos 2\theta) &= \frac{s}{r+s}\vert\vec{AB}\vert(1-\cos 2\theta)\\
\frac{\vert\vec{AB}\vert}{\vert\vec{AC}\vert}&=\frac{r}{s}\\
&=\frac{\vert\vec{BD}\vert}{\vert\vec{CD}\vert}.
\end{align}


问:
Isomorphism ${}^*P\otimes_A P \cong {}_A\text{End}(P)$ for $P$ finitly generated and projective.
Let $P \in {}_A \text{Mod}$ be a left $A$-module for $A$ a ring and ${}^*P = {}_A\text{Hom}(P, A)$ the dual which is a right $A$-module for the action defined by
$$fa(p):= f(ap).$$
If $P$ is finitly generated and projective, we can show that ${}^*P\otimes_A P$ is a ring with the unit given by
$$\sum_{i = 1}^nf_i \otimes e_i,$$
where $(e_i, f_i)_{i= 1}^n$ is a finite dual base for $P$. In my course, my teacher said that, for $P$ finitly generated and projective, the previous tensor product is canonically isomorphic to ${}_A\text{End}(P)$ and the isomorphism is given by
$$f:{}^*P\otimes_A P \to {}_A\text{End}(P): f\otimes_A p \mapsto (f(\cdot)p:p' \mapsto f(p')p).$$
But here is my problem: this map does not seem well-defined. For $a\in A$,
$$(fa\otimes_A p)(p') = fa(p')p = f(ap')p = af(p')p$$
and
$$(f\otimes_A ap)(p') = f(p')ap$$
which is not the same when $A$ is not commutative. Where is the mistake ?

答:
You're using the wrong action of $A$ on the dual. When you compute $\text{Hom}_A(P, A)$ you've "used up" the left action of $A$ on $P$ already so your left multiplication definition doesn't work; if you check carefully you'll find that $f(ap)$ is not $A$-linear in general. The correct $A$-module structure comes from the $(A, A)$-bimodule structure on $A$, which is to say concretely that the correct action is

$$(fa)(p') = f(p') a$$

and then there's no problem. Of course when $A$ is commutative everything is a bimodule and then your definition is equivalent to this one.


答:
Phew! Thanks to the help of Will Jagy and Gottfried Helms, I have a complete solution!

Fix positive integers $m$ and $n$.
We will study the equation
\begin{equation}
\tag{1}
    mn(y-x)(y+x)=(y-1)(y+m).
\end{equation}
Classification of Small Solutions

We begin by classifying the integer solutions to (1) with small $y$ coordinate.

Case 1: $y>1$

Assume that $y>1$.
Then the right hand side of (1) is positive, so the left hand side of (1) is also positive.
Then $y^2-x^2$ is positive, so $y^2-x^2\geq2y-1$.
This gives
$$(y-1)(y+m)=mn(y^2-x^2)\geq mn(2y-1).$$
Dividing through by $2y-1$ gives
$$\frac{(y-1)(y+m)}{2y-1}\geq mn.$$
Let $f(y)=(y-1)(y+m)/(2y-1)$.
Note that $f$ is increasing on $(\frac{1}{2},\infty)$.
Let $y_0=2mn-m$.
Then
$$f(y_0)=\frac{2mn(y_0-1)}{2y_0-1}<mn\leq f(y)$$
so $y_0<y$.
Thus, $y\geq2mn-m+1$.

Case 2: $-m<y<1$

Assume that $-m<y<1$.
Then the right hand side of (1) is negative, so the left hand side of (1) is also negative.
Then $y^2-x^2$ is negative, so $y^2-x^2\leq2y-1$.
This gives
$$(y-1)(y+m)=mn(y^2-x^2)\leq mn(2y-1).$$
Dividing through by $2y-1$ gives
$$\frac{(y-1)(y+m)}{2y-1}\geq mn.$$
Let $f(y)=(y-1)(y+m)/(2y-1)$.
Note that $f$ is increasing on $(-\infty,\frac{1}{2})$.
Let $y_0=0$.
Then
$$f(y_0)=m\leq mn\leq f(y)$$
so $y_0\leq y$.
Thus, $y=0$.

Case 3: $y<-m$

Assume that $y<-m$.
Then the right hand side of (1) is positive, so the left hand side of (1) is also positive.
Then $y^2-x^2$ is positive, so $y^2-x^2\geq-2y-1$.
This gives
$$(y-1)(y+m)=mn(y^2-x^2)\geq mn(-2y-1).$$
Dividing through by $-2y-1$ gives
$$\frac{(y-1)(y+m)}{-2y-1}\geq mn.$$
Let $f(y)=(y-1)(y+m)/(-2y-1)$.
Note that $f$ is decreasing on $(-\infty,-\frac{1}{2})$.
Let $y_0=-2mn-m+2$.
Then
$$f(y_0)=\frac{(y_0-1)(y_0+m)}{-2y_0-1}\leq\frac{(y_0+1)(y_0+m-2)}{-2y_0-1}=\frac{-2mn(y_0+1)}{-2y_0-1}<mn\leq f(y)$$
so $y_0>y$.
Thus, $y\leq-2mn-m+1$.

Completing the Classification

Theorem 1:
If $(x,y)\in\mathbb{Z}^2$ satisfies (1) then exactly one of the following holds:

$y\geq2mn-m+2$,

$(x,y)=(\pm m,m+1)$ and $n=1$,

$(x,y)=(\pm1,1)$,

$(x,y)=(\pm1,0)$ and $n=1$,

$(x,y)=(\pm m,-m)$,

$(x,y)=(\pm(3m-2),-3m+1)$ and $n=1$,

$y\leq-2mn-m$.

Proof:
We have already shown that either $y\geq2mn-m+1$ or $y=1$ or $y=0$ or $y=-m$ or $y\leq-2mn-m+1$.

Assume that $y=2mn-m+1$.
Then (1) becomes
$$mn(y-x)(y+x)=(2mn-m)(2mn+1).$$
Dividing through by $m$ and reducing modulo $n$ forces $n=1$.
Then $y=m+1$ and solving for $x$ gives $x=\pm m$.

If $y=1$ then solving for $x$ gives $x=\pm1$.

If $y=0$ then solving for $x$ gives $nx^2=1$.
This forces $n=1$ and $x=\pm1$.

If $y=-m$ then solving for $x$ gives $x=\pm m$.

Assume that $y=-2mn-m+1$.
Then (1) becomes
$$mn(y-x)(y+x)=(2mn+m)(2mn-1).$$
Dividing through by $m$ and reducing modulo $n$ forces $n=1$.
Then $y=-3m+1$ and solving for $x$ gives $x=\pm(3m-2)$. $\blacksquare$

Mapping Solutions

We now define functions $f,g\colon\mathbb{R}^2\to\mathbb{R}^2$ that preserve integer solutions to (1).
Our goal is to show that every integer solution to (1) can be obtained via $f$ and $g$ from one of the five solutions listed in Theorem 1.

Let $R=2mn-1$.
Let $f,g\colon\mathbb{R}^2\to\mathbb{R}^2$ be defined by
$$f\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}(2R^2-1)x+2R(R-1)y-2(m-1)R\\(2R^2-1)y+2R(R+1)x-2(m-1)(R+1)\end{bmatrix},$$
$$g\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}(2R^2-1)x-2R(R-1)y+2(m-1)R\\(2R^2-1)y-2R(R+1)x-2(m-1)(R+1)\end{bmatrix}.$$

Lemma 2:
The functions $f$ and $g$ preserve solutions to (1).

Proof:
The algebraic manipulations are rather involved, so I instead wrote up a proof in Lean 3.

```<code>import tactic.ring

variables {A : Type*} [comm_ring A]

lemma lem0 {a b c d : A} (h : a = b) : c - a = d - b → c = d :=
by { rw h, exact sub_left_inj.mp }

structure sol (m n : A) :=
(x y : A)
(h : m * n * (y - x) * (y + x) = (y - 1) * (y + m))

def R (m n : A) : A := 2 * m * n - 1

def sol.forward {m n : A} (s : sol m n) : sol m n :=
{ x := (2*(R m n)*(R m n)-1)*s.x+2*(R m n)*((R m n)-1)*s.y-2*(m-1)*(R m n),
  y := (2*(R m n)*(R m n)-1)*s.y+2*(R m n)*((R m n)+1)*s.x-2*(m-1)*((R m n)+1),
  h := by { dsimp only [R], apply lem0 s.h, ring } }

def sol.backward {m n : A} (s : sol m n) : sol m n :=
{ x := (2*(R m n)*(R m n)-1)*s.x-2*(R m n)*((R m n)-1)*s.y+2*(m-1)*(R m n),
  y := (2*(R m n)*(R m n)-1)*s.y-2*(R m n)*((R m n)+1)*s.x-2*(m-1)*((R m n)+1),
  h := by { dsimp only [R], apply lem0 s.h, ring } }
</code>```

You just need to check that I correctly copied (1) (sol.h), $f$ (sol.forward), and $g$ (sol.backward). $\blacksquare$

We now give a few lemmas describing the relationship between $f$ and $g$.

Lemma 3:
Let $\overline{(x,y)}=(-x,y)$.
Then $f(\overline{p})=\overline{g(p)}$ for all points $p\in\mathbb{R}^2$.

Proof:
This is clear from the definitions of $f$ and $g$. $\blacksquare$

Lemma 4:
The functions $f$ and $g$ are inverses to each other.

Proof:
We can write $f$ in terms of matrix multiplication as
$$f\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}2R^2-1&2R(R-1)\\2R(R+1)&2R^2-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}-2(m-1)\begin{bmatrix}R\\R+1\end{bmatrix}.$$
Note that the $2\times2$ matrix has determinant $(2R^2-1)^2-4R^2(R^2-1)=1$.
Then the inverse function to $f$ is given by
\begin{align*}
f^{-1}\left(\begin{bmatrix}x\\y\end{bmatrix}\right)&=\begin{bmatrix}2R^2-1&2R(R-1)\\2R(R+1)&2R^2-1\end{bmatrix}^{-1}\left(\begin{bmatrix}x\\y\end{bmatrix}+2(m-1)\begin{bmatrix}R\\R+1\end{bmatrix}\right)\\
&=\begin{bmatrix}2R^2-1&-2R(R-1)\\-2R(R+1)&2R^2-1\end{bmatrix}\left(\begin{bmatrix}x\\y\end{bmatrix}+2(m-1)\begin{bmatrix}R\\R+1\end{bmatrix}\right)\\
&=\begin{bmatrix}2R^2-1&-2R(R-1)\\-2R(R+1)&2R^2-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+2(m-1)\begin{bmatrix}2R^2-1&-2R(R-1)\\-2R(R+1)&2R^2-1\end{bmatrix}\begin{bmatrix}R\\R+1\end{bmatrix}\\
&=\begin{bmatrix}2R^2-1&-2R(R-1)\\-2R(R+1)&2R^2-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+2(m-1)\begin{bmatrix}R\\-(R+1)\end{bmatrix}=g\left(\begin{bmatrix}x\\y\end{bmatrix}\right).
\end{align*}
Thus, $f$ and $g$ are inverses to each other. $\blacksquare$

By Lemma 4, fixed points of $f$ are fixed points of $g$, and vice versa.

Lemma 5:
If $(x,y)$ is a fixed point of $f$ and $g$ then $x=0$.

Proof:
Suppose that $(x,y)$ is a fixed point of $f$ and $g$.
Then
$$\begin{bmatrix}2R^2-1&2R(R-1)\\2R(R+1)&2R^2-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}-2(m-1)\begin{bmatrix}R\\R+1\end{bmatrix}=f\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}.$$
Rearranging terms and dividing through by 2 gives
\begin{equation}\tag{2}\begin{bmatrix}R^2-1&R(R-1)\\R(R+1)&R^2-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=(m-1)\begin{bmatrix}R\\R+1\end{bmatrix}.\end{equation}
Note that the $2\times2$ matrix has determinant $(R^2-1)^2-R^2(R^2-1)=1-R^2$.
If $R=1$ then $(m,n)=(1,1)$, so (2) becomes
$$\begin{bmatrix}0&0\\2&0\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}.$$
This forces $x=0$.
Now suppose that $R>1$.
Then the $2\times2$ matrix in (2) is invertible, so (2) has a unique solution.
In other words, $f$ and $g$ have a unique fixed point $p$.
However, Lemma 3 shows that $\overline{p}$ is also a fixed point of $f$ and $g$.
Then $p=\overline{p}$, so $x=0$.
Alternatively, you can solve (2) explicitly and get $(x,y)=(0,(m-1)/(R-1))$. $\blacksquare$

Lemma 6:
The functions $f,g\colon\mathbb{R}^2\to\mathbb{R}^2$ have no fixed points satisfying equation (1).

Proof:
Let $(x,y)$ be a fixed point of $f$ and $g$.
By Lemma 5, $x=0$.
However, setting $x=0$ in (1) gives
$$mny^2=(y-1)(y+m).$$
The quadratic $(mn-1)y^2-(m-1)y+m$ has discriminant $(m-1)^2-4m(mn-1)<0$. $\blacksquare$

We can now prove the main theorem.

Theorem 7:
Any integer solution to (1) can be mapped via $f$ and $g$ to one of the five solutions listed in Theorem 1.

Proof:
Since the quadratic $(mn-1)y^2-(m-1)y+m$ has negative discriminant, the graph of (1) consists of the two curves $x=\pm\sqrt{y^2-((y-1)(y+m)/(mn))}$.
Since $f(-1,1)=(1-2mR,1-2m(R+1))$ has negative $x$-coordinate, $f$ and $g$ do not swap the two curves.
By Lemma 6, $f$ and $g$ do not flip either of the two curves, and each shift each curve either up or down.
Since $f(-1,1)=(1-2mR,1-2m(R+1))$ has negative $y$-coordinate, $f$ shifts the left curve downward.
Then by Lemma 4, $g$ shifts the left curve upward.
Then by Lemma 3, $f$ shifts the right curve upward and $g$ shifts the right curve downward.

It remains to show that $f$ and $g$ can't jump over the gap $-2mn-m+1\leq y\leq2mn-m+1$.
By Lemmas 3 and 4, it suffices to show that if $y_0=2mn-m+1$ and $x_0=\sqrt{y_0^2-((y_0-1)(y_0+m)/(mn))}$ then the $y$-coordinate of $g(x_0,y_0)$ is at least $-2mn-m+1$.
In other words, we need to prove the inequality
$$(2R^2-1)y_0-2R(R+1)x_0-2(m-1)(R+1)\geq-2mn-m+1.$$
Rearranging terms gives the equivalent inequality
$$x_0\leq\frac{(2R^2-1)y_0-2(m-1)(R+1)+2mn+m-1}{2R(R+1)}.$$
Simplifying the right hand side gives the equivalent inequality
$$x_0\leq2mn-m.$$
To prove this, we directly compute
$$y_0^2-x_0^2=\frac{(y_0-1)(y_0+m)}{mn}=\frac{(2mn-m)(2mn+1)}{mn}=4mn-2m+2-\frac{1}{n}\geq4mn-2m+1=y_0^2-(y_0-1)^2.$$
Then $x_0\leq y_0-1$ as desired. $\blacksquare$

Consequences

We can combine Theorems 1 and 7 to make conclusions about all integer solutions to (1).

Theorem 8:
Let $(x,y)\in\mathbb{Z}^2$ satisfy (1).

$y\equiv0,1\pmod{m}$.

If $n\neq1$ and $y\equiv0\pmod{m}$ then $x\equiv y\equiv m\pmod{2}$.

If $n\neq1$ and $y\equiv1\pmod{m}$ then $x\equiv y\equiv1\pmod{2}$.

Either way, if $n\neq1$ then $x\equiv y\pmod{2}$.

Proof:
Theorem 7 states that any integer solution to (1) can be mapped via $f$ and $g$ to one of the five solutions listed in Theorem 1.
Note that $f$ and $g$ preserve the value of $y\pmod{m}$, $x\pmod{2}$, and $y\pmod{2}$.
Since each of the five solutions listed in Theorem 1 satisfy the theorem statement, all integer solutions to (1) satisfy the theorem statement. $\blacksquare$

Theorem 9:
Let $m,n,x,y,z$ be positive integers such that $y=m(z-1)+1$ and $yz-1=n(y^2-x^2)$.

If $n\geq2$ then $x,y,z$ are all odd.

If $(m,n)\neq(1,1)$ then $z$ is indivisible by 4.

Proof:
Note that
$$mn(y^2-x^2)=m(yz-1)=y(mz)-m=y(y+m-1)-m=(y-1)(y+m)$$
so our previous work applies.
Furthermore, we are in the $y\equiv1\pmod{m}$ case.
By Theorem 8, if $n\geq2$ then $x\equiv y\equiv1\pmod{2}$.
Then the equation $yz-1=n(y^2-x^2)$ forces $z\equiv1\pmod{2}$.

It remains to show that if $m\geq2$ and $n=1$ then $z$ is indivisible by 4.
We know that $(x,y)$ can be mapped via $f$ and $g$ to one of the five solutions listed in Theorem 1.
Since we are in the $y\equiv1\pmod{m}$ case, and since $m\geq2$, we know that $(x,y)$ can be mapped via $f$ and $g$ to $(\pm m,m+1)$ or $(\pm1,1)$ or $(\pm(3m-2),-3m+1)$.
Since $(x,y)$ is in the first quadrant, we know that $(x,y)$ can be obtained from $(1,1)$ or $(m,m+1)$ via repeated application of $f$ (we can discard $(3m-2,-3m+1)$ since $g(m,m+1)=(3m-2,-3m+1)$).
The solution $(1,1)$ has $z=1$.
The solution $(m,m+1)$ has $z=2$.

We now check how applying $f$ changes $z$.
Applying $f$ has the effect
$$y\mapsto(2R^2-1)y+2R(R+1)x-2(m-1)(R+1).$$
In terms of $z$, this is
$$z\mapsto\frac{(2R^2-1)(m(z-1)+1)+2R(R+1)x-2(m-1)(R+1)-1}{m}+1$$
which simplifies to
$$z\mapsto(2R^2-1)z+4nRx-4(m-1)nR.$$
In particular, the value of $z\pmod{4}$ does not change. $\blacksquare$

Title: An infinite family of generalized Pell equations

Body:
Let $m,n,x,y,z$ be positive integers such that $y=m(z-1)+1$ and $yz-1=n(y^2-x^2)$.

If you fix $m$ and $n$ then this is a generalized Pell equation in $x$ and $z$ (i.e., a quadratic Diophantine equation of the sort that can be solved by this website).

I have 2 specific conjectures, which I have checked computationally for $m,n\leq100$, $z\leq10^7$.

Conjecture 1: If $n\geq2$ then $x,y,z$ are all odd.

Conjecture 2: If $(m,n)\neq(1,1)$ then $z$ is indivisible by $4$.

A proof or counterexample of either of these conjectures would be much appreciated.

The table below gives the first four $(x,y,z)$ solutions for fixed $(m,n)$:

$$\begin{array}{c|c|c|c}&n=1&n=2&n=3\\\hline m=1&
\begin{array}{c}(1,1,1)\\(1,2,2)\\(1,3,3)\\(1,4,4)\end{array}&
\begin{array}{c}(1,1,1)\\(5,7,7)\\(29,41,41)\\(169,239,239)\end{array}&
\begin{array}{c}(1,1,1)\\(9,11,11)\\(89,109,109)\\(881,1079,1079)\end{array}\\\hline m=2&
\begin{array}{c}(1,1,1)\\(2,3,2)\\(23,33,17)\\(64,91,46)\end{array}&
\begin{array}{c}(1,1,1)\\(167,193,97)\\(32397,37409,18705)\\(6284851,7257121,3628561)\end{array}&
\begin{array}{c}(1,1,1)\\(439,481,241)\\(211597,231793,115897)\\(101989315,111723697,55861849)\end{array}\\\hline m=3&
\begin{array}{c}(1,1,1)\\(3,4,2)\\(69,85,29)\\(287,352,118)\end{array}&
\begin{array}{c}(1,1,1)\\(417,457,153)\\(200993,220177,73393)\\(96878209,106124761,35374921)\end{array}&
\begin{array}{c}(1,1,1)\\(1053,1117,373)\\(1215161,1288873,429625)\\(1402294741,1487358181,495786061)\end{array}
\end{array}$$

In the comments, Will Jagy observed that if $n\geq2$ then the $x$ values satisfy a linear recurrence $x_{n+2}=Cx_{n+1}-x_n$ (and you can do something similar for $y$ and $z$). The table below gives the values of $C$:

$$\begin{array}{c|c|c|c|c|c|c|c|c|c}&n=2&n=3&n=4&n=5&n=6&n=7&n=8&n=9&n=10\\\hline m=1&6&10&14&18&22&26&30&34&38\\\hline m=2&194&482&898&1442&2114&2914&3842&4898&6082\\\hline m=3&482&1154&2114&3362&4898&6722&8834&11234&13922\\\hline m=4&898&2114&3842&6082&8834&12098&15874&20162&24962\\\hline m=5&1442&3362&6082&9602&13922&19042&24962&31682&39202\\\hline m=6&2114&4898&8834&13922&20162&27554&36098&45794&\\\hline m=7&2914&6722&12098&19042&27554&37634&49282&&\\\hline m=8&3842&8834&15874&24962&36098&49282&&&\\\hline m=9&4898&11234&20162&31682&45794&&&&\\\hline m=10&6082&13922&24962&39202&&&&&\end{array}$$

Notice the symmetry!

Gottfried Helms observed the formula $C=4(2mn-1)^2-2$ for $m,n\geq2$.

When $m=1$, it looks like $C=4n-2$ for $n\geq2$.

This answer proves the $m=1$ case of conjecture 2, using methods from Diophantine approximation.

Body:
Phew! Thanks to the help of Will Jagy and Gottfried Helms, I have a complete solution!

Fix positive integers $m$ and $n$.
We will study the equation
\begin{equation}
\tag{1}
    mn(y-x)(y+x)=(y-1)(y+m).
\end{equation}
Classification of Small Solutions

We begin by classifying the integer solutions to (1) with small $y$ coordinate.

Case 1: $y>1$

Assume that $y>1$.
Then the right hand side of (1) is positive, so the left hand side of (1) is also positive.
Then $y^2-x^2$ is positive, so $y^2-x^2\geq2y-1$.
This gives
$$(y-1)(y+m)=mn(y^2-x^2)\geq mn(2y-1).$$
Dividing through by $2y-1$ gives
$$\frac{(y-1)(y+m)}{2y-1}\geq mn.$$
Let $f(y)=(y-1)(y+m)/(2y-1)$.
Note that $f$ is increasing on $(\frac{1}{2},\infty)$.
Let $y_0=2mn-m$.
Then
$$f(y_0)=\frac{2mn(y_0-1)}{2y_0-1}<mn\leq f(y)$$
so $y_0<y$.
Thus, $y\geq2mn-m+1$.

Case 2: $-m<y<1$

Assume that $-m<y<1$.
Then the right hand side of (1) is negative, so the left hand side of (1) is also negative.
Then $y^2-x^2$ is negative, so $y^2-x^2\leq2y-1$.
This gives
$$(y-1)(y+m)=mn(y^2-x^2)\leq mn(2y-1).$$
Dividing through by $2y-1$ gives
$$\frac{(y-1)(y+m)}{2y-1}\geq mn.$$
Let $f(y)=(y-1)(y+m)/(2y-1)$.
Note that $f$ is increasing on $(-\infty,\frac{1}{2})$.
Let $y_0=0$.
Then
$$f(y_0)=m\leq mn\leq f(y)$$
so $y_0\leq y$.
Thus, $y=0$.

Case 3: $y<-m$

Assume that $y<-m$.
Then the right hand side of (1) is positive, so the left hand side of (1) is also positive.
Then $y^2-x^2$ is positive, so $y^2-x^2\geq-2y-1$.
This gives
$$(y-1)(y+m)=mn(y^2-x^2)\geq mn(-2y-1).$$
Dividing through by $-2y-1$ gives
$$\frac{(y-1)(y+m)}{-2y-1}\geq mn.$$
Let $f(y)=(y-1)(y+m)/(-2y-1)$.
Note that $f$ is decreasing on $(-\infty,-\frac{1}{2})$.
Let $y_0=-2mn-m+2$.
Then
$$f(y_0)=\frac{(y_0-1)(y_0+m)}{-2y_0-1}\leq\frac{(y_0+1)(y_0+m-2)}{-2y_0-1}=\frac{-2mn(y_0+1)}{-2y_0-1}<mn\leq f(y)$$
so $y_0>y$.
Thus, $y\leq-2mn-m+1$.

Completing the Classification

Theorem 1:
If $(x,y)\in\mathbb{Z}^2$ satisfies (1) then exactly one of the following holds:

$y\geq2mn-m+2$,

$(x,y)=(\pm m,m+1)$ and $n=1$,

$(x,y)=(\pm1,1)$,

$(x,y)=(\pm1,0)$ and $n=1$,

$(x,y)=(\pm m,-m)$,

$(x,y)=(\pm(3m-2),-3m+1)$ and $n=1$,

$y\leq-2mn-m$.

Proof:
We have already shown that either $y\geq2mn-m+1$ or $y=1$ or $y=0$ or $y=-m$ or $y\leq-2mn-m+1$.

Assume that $y=2mn-m+1$.
Then (1) becomes
$$mn(y-x)(y+x)=(2mn-m)(2mn+1).$$
Dividing through by $m$ and reducing modulo $n$ forces $n=1$.
Then $y=m+1$ and solving for $x$ gives $x=\pm m$.

If $y=1$ then solving for $x$ gives $x=\pm1$.

If $y=0$ then solving for $x$ gives $nx^2=1$.
This forces $n=1$ and $x=\pm1$.

If $y=-m$ then solving for $x$ gives $x=\pm m$.

Assume that $y=-2mn-m+1$.
Then (1) becomes
$$mn(y-x)(y+x)=(2mn+m)(2mn-1).$$
Dividing through by $m$ and reducing modulo $n$ forces $n=1$.
Then $y=-3m+1$ and solving for $x$ gives $x=\pm(3m-2)$. $\blacksquare$

Mapping Solutions

We now define functions $f,g\colon\mathbb{R}^2\to\mathbb{R}^2$ that preserve integer solutions to (1).
Our goal is to show that every integer solution to (1) can be obtained via $f$ and $g$ from one of the five solutions listed in Theorem 1.

Let $R=2mn-1$.
Let $f,g\colon\mathbb{R}^2\to\mathbb{R}^2$ be defined by
$$f\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}(2R^2-1)x+2R(R-1)y-2(m-1)R\\(2R^2-1)y+2R(R+1)x-2(m-1)(R+1)\end{bmatrix},$$
$$g\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}(2R^2-1)x-2R(R-1)y+2(m-1)R\\(2R^2-1)y-2R(R+1)x-2(m-1)(R+1)\end{bmatrix}.$$

Lemma 2:
The functions $f$ and $g$ preserve solutions to (1).

Proof:
The algebraic manipulations are rather involved, so I instead wrote up a proof in Lean 3.

import tactic.ring

variables {A : Type*} [comm_ring A]

lemma lem0 {a b c d : A} (h : a = b) : c - a = d - b → c = d :=
by { rw h, exact sub_left_inj.mp }

structure sol (m n : A) :=
(x y : A)
(h : m * n * (y - x) * (y + x) = (y - 1) * (y + m))

def R (m n : A) : A := 2 * m * n - 1

def sol.forward {m n : A} (s : sol m n) : sol m n :=
{ x := (2*(R m n)*(R m n)-1)*s.x+2*(R m n)*((R m n)-1)*s.y-2*(m-1)*(R m n),
  y := (2*(R m n)*(R m n)-1)*s.y+2*(R m n)*((R m n)+1)*s.x-2*(m-1)*((R m n)+1),
  h := by { dsimp only [R], apply lem0 s.h, ring } }

def sol.backward {m n : A} (s : sol m n) : sol m n :=
{ x := (2*(R m n)*(R m n)-1)*s.x-2*(R m n)*((R m n)-1)*s.y+2*(m-1)*(R m n),
  y := (2*(R m n)*(R m n)-1)*s.y-2*(R m n)*((R m n)+1)*s.x-2*(m-1)*((R m n)+1),
  h := by { dsimp only [R], apply lem0 s.h, ring } }

You just need to check that I correctly copied (1) (sol.h), $f$ (sol.forward), and $g$ (sol.backward). $\blacksquare$

We now give a few lemmas describing the relationship between $f$ and $g$.

Lemma 3:
Let $\overline{(x,y)}=(-x,y)$.
Then $f(\overline{p})=\overline{g(p)}$ for all points $p\in\mathbb{R}^2$.

Proof:
This is clear from the definitions of $f$ and $g$. $\blacksquare$

Lemma 4:
The functions $f$ and $g$ are inverses to each other.

Proof:
We can write $f$ in terms of matrix multiplication as
$$f\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}2R^2-1&2R(R-1)\\2R(R+1)&2R^2-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}-2(m-1)\begin{bmatrix}R\\R+1\end{bmatrix}.$$
Note that the $2\times2$ matrix has determinant $(2R^2-1)^2-4R^2(R^2-1)=1$.
Then the inverse function to $f$ is given by
\begin{align*}
f^{-1}\left(\begin{bmatrix}x\\y\end{bmatrix}\right)&=\begin{bmatrix}2R^2-1&2R(R-1)\\2R(R+1)&2R^2-1\end{bmatrix}^{-1}\left(\begin{bmatrix}x\\y\end{bmatrix}+2(m-1)\begin{bmatrix}R\\R+1\end{bmatrix}\right)\\
&=\begin{bmatrix}2R^2-1&-2R(R-1)\\-2R(R+1)&2R^2-1\end{bmatrix}\left(\begin{bmatrix}x\\y\end{bmatrix}+2(m-1)\begin{bmatrix}R\\R+1\end{bmatrix}\right)\\
&=\begin{bmatrix}2R^2-1&-2R(R-1)\\-2R(R+1)&2R^2-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+2(m-1)\begin{bmatrix}2R^2-1&-2R(R-1)\\-2R(R+1)&2R^2-1\end{bmatrix}\begin{bmatrix}R\\R+1\end{bmatrix}\\
&=\begin{bmatrix}2R^2-1&-2R(R-1)\\-2R(R+1)&2R^2-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+2(m-1)\begin{bmatrix}R\\-(R+1)\end{bmatrix}=g\left(\begin{bmatrix}x\\y\end{bmatrix}\right).
\end{align*}
Thus, $f$ and $g$ are inverses to each other. $\blacksquare$

By Lemma 4, fixed points of $f$ are fixed points of $g$, and vice versa.

Lemma 5:
If $(x,y)$ is a fixed point of $f$ and $g$ then $x=0$.

Proof:
Suppose that $(x,y)$ is a fixed point of $f$ and $g$.
Then
$$\begin{bmatrix}2R^2-1&2R(R-1)\\2R(R+1)&2R^2-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}-2(m-1)\begin{bmatrix}R\\R+1\end{bmatrix}=f\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}.$$
Rearranging terms and dividing through by 2 gives
\begin{equation}\tag{2}\begin{bmatrix}R^2-1&R(R-1)\\R(R+1)&R^2-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=(m-1)\begin{bmatrix}R\\R+1\end{bmatrix}.\end{equation}
Note that the $2\times2$ matrix has determinant $(R^2-1)^2-R^2(R^2-1)=1-R^2$.
If $R=1$ then $(m,n)=(1,1)$, so (2) becomes
$$\begin{bmatrix}0&0\\2&0\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}.$$
This forces $x=0$.
Now suppose that $R>1$.
Then the $2\times2$ matrix in (2) is invertible, so (2) has a unique solution.
In other words, $f$ and $g$ have a unique fixed point $p$.
However, Lemma 3 shows that $\overline{p}$ is also a fixed point of $f$ and $g$.
Then $p=\overline{p}$, so $x=0$.
Alternatively, you can solve (2) explicitly and get $(x,y)=(0,(m-1)/(R-1))$. $\blacksquare$

Lemma 6:
The functions $f,g\colon\mathbb{R}^2\to\mathbb{R}^2$ have no fixed points satisfying equation (1).

Proof:
Let $(x,y)$ be a fixed point of $f$ and $g$.
By Lemma 5, $x=0$.
However, setting $x=0$ in (1) gives
$$mny^2=(y-1)(y+m).$$
The quadratic $(mn-1)y^2-(m-1)y+m$ has discriminant $(m-1)^2-4m(mn-1)<0$. $\blacksquare$

We can now prove the main theorem.

Theorem 7:
Any integer solution to (1) can be mapped via $f$ and $g$ to one of the five solutions listed in Theorem 1.

Proof:
Since the quadratic $(mn-1)y^2-(m-1)y+m$ has negative discriminant, the graph of (1) consists of the two curves $x=\pm\sqrt{y^2-((y-1)(y+m)/(mn))}$.
Since $f(-1,1)=(1-2mR,1-2m(R+1))$ has negative $x$-coordinate, $f$ and $g$ do not swap the two curves.
By Lemma 6, $f$ and $g$ do not flip either of the two curves, and each shift each curve either up or down.
Since $f(-1,1)=(1-2mR,1-2m(R+1))$ has negative $y$-coordinate, $f$ shifts the left curve downward.
Then by Lemma 4, $g$ shifts the left curve upward.
Then by Lemma 3, $f$ shifts the right curve upward and $g$ shifts the right curve downward.

It remains to show that $f$ and $g$ can't jump over the gap $-2mn-m+1\leq y\leq2mn-m+1$.
By Lemmas 3 and 4, it suffices to show that if $y_0=2mn-m+1$ and $x_0=\sqrt{y_0^2-((y_0-1)(y_0+m)/(mn))}$ then the $y$-coordinate of $g(x_0,y_0)$ is at least $-2mn-m+1$.
In other words, we need to prove the inequality
$$(2R^2-1)y_0-2R(R+1)x_0-2(m-1)(R+1)\geq-2mn-m+1.$$
Rearranging terms gives the equivalent inequality
$$x_0\leq\frac{(2R^2-1)y_0-2(m-1)(R+1)+2mn+m-1}{2R(R+1)}.$$
Simplifying the right hand side gives the equivalent inequality
$$x_0\leq2mn-m.$$
To prove this, we directly compute
$$y_0^2-x_0^2=\frac{(y_0-1)(y_0+m)}{mn}=\frac{(2mn-m)(2mn+1)}{mn}=4mn-2m+2-\frac{1}{n}\geq4mn-2m+1=y_0^2-(y_0-1)^2.$$
Then $x_0\leq y_0-1$ as desired. $\blacksquare$

Consequences

We can combine Theorems 1 and 7 to make conclusions about all integer solutions to (1).

Theorem 8:
Let $(x,y)\in\mathbb{Z}^2$ satisfy (1).

$y\equiv0,1\pmod{m}$.

If $n\neq1$ and $y\equiv0\pmod{m}$ then $x\equiv y\equiv m\pmod{2}$.

If $n\neq1$ and $y\equiv1\pmod{m}$ then $x\equiv y\equiv1\pmod{2}$.

Either way, if $n\neq1$ then $x\equiv y\pmod{2}$.

Proof:
Theorem 7 states that any integer solution to (1) can be mapped via $f$ and $g$ to one of the five solutions listed in Theorem 1.
Note that $f$ and $g$ preserve the value of $y\pmod{m}$, $x\pmod{2}$, and $y\pmod{2}$.
Since each of the five solutions listed in Theorem 1 satisfy the theorem statement, all integer solutions to (1) satisfy the theorem statement. $\blacksquare$

Theorem 9:
Let $m,n,x,y,z$ be positive integers such that $y=m(z-1)+1$ and $yz-1=n(y^2-x^2)$.

If $n\geq2$ then $x,y,z$ are all odd.

If $(m,n)\neq(1,1)$ then $z$ is indivisible by 4.

Proof:
Note that
$$mn(y^2-x^2)=m(yz-1)=y(mz)-m=y(y+m-1)-m=(y-1)(y+m)$$
so our previous work applies.
Furthermore, we are in the $y\equiv1\pmod{m}$ case.
By Theorem 8, if $n\geq2$ then $x\equiv y\equiv1\pmod{2}$.
Then the equation $yz-1=n(y^2-x^2)$ forces $z\equiv1\pmod{2}$.

It remains to show that if $m\geq2$ and $n=1$ then $z$ is indivisible by 4.
We know that $(x,y)$ can be mapped via $f$ and $g$ to one of the five solutions listed in Theorem 1.
Since we are in the $y\equiv1\pmod{m}$ case, and since $m\geq2$, we know that $(x,y)$ can be mapped via $f$ and $g$ to $(\pm m,m+1)$ or $(\pm1,1)$ or $(\pm(3m-2),-3m+1)$.
Since $(x,y)$ is in the first quadrant, we know that $(x,y)$ can be obtained from $(1,1)$ or $(m,m+1)$ via repeated application of $f$ (we can discard $(3m-2,-3m+1)$ since $g(m,m+1)=(3m-2,-3m+1)$).
The solution $(1,1)$ has $z=1$.
The solution $(m,m+1)$ has $z=2$.

We now check how applying $f$ changes $z$.
Applying $f$ has the effect
$$y\mapsto(2R^2-1)y+2R(R+1)x-2(m-1)(R+1).$$
In terms of $z$, this is
$$z\mapsto\frac{(2R^2-1)(m(z-1)+1)+2R(R+1)x-2(m-1)(R+1)-1}{m}+1$$
which simplifies to
$$z\mapsto(2R^2-1)z+4nRx-4(m-1)nR.$$
In particular, the value of $z\pmod{4}$ does not change. $\blacksquare$


问:
Spherical triple integral for volume enclosed by $(x^2+y^2+z^2)^2=a^2(x^2+y^2-z^2)$
I need to find the volume of the figure that is bounded with surface

$$(x^2+y^2+z^2)^2=a^2(x^2+y^2-z^2)$$

using spherical coordinates $x=r\cos\phi \sin\theta, y=r\sin\phi \sin\theta, z=r\cos\theta$. My main problem is to find bounds for $\theta$ and $\phi$. For $r$ I easily get bounds $0\le r\le a\sqrt{1-2\cos^2 \theta}$. If I'd had two surfaces i would probably get bounds for theta  from another of them, but here not sure. Any help is appreciated.

答:
In spherical coordinates, the surface is $r= a\sqrt{1-2\cos^2\theta}$ with the range $\theta\in [\frac\pi4, \frac{3\pi}4]$, as seen below

The volume is then

\begin{align}
V& =2\pi \int_{\frac\pi4}^{\frac{3\pi}4}\int_0^{a\sqrt{1-2\cos^2\theta}}r^2\sin\theta drd\theta \\
& = \frac{2\pi}3 a^3\int_{\frac\pi4}^{\frac{3\pi}4}
(1-2\cos^2\theta)^{3/2}\sin\theta d\theta \\
 & = \frac{2\pi}3 a^3 \cdot \frac{3\pi}{8\sqrt2}=
 \frac{\sqrt2\pi^2}{8}a^3\\
\end{align}


答:
As volume is symmetric with respect to coordinate planes, then in first octant there is $1/8$ of volume. For angles we have restriction $1-2\cos^2 \theta = -\cos 2\theta \geqslant 0$
$$V=8\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin \theta d\theta \int\limits_{0}^{\frac{\pi}{2}}d\phi\int\limits_{0}^{a\sqrt{-\cos 2\theta}}r^2dr=\frac{4\pi a^3}{3}\int\limits_{0}^{\frac{\pi}{4}}(1-2 \sin^2 t)^{3/2}d(\sin t)=\frac{\pi^2 a^3}{4\sqrt{2}}$$

Title: Spherical triple integral for volume enclosed by $(x^2+y^2+z^2)^2=a^2(x^2+y^2-z^2)$

Body:
I need to find the volume of the figure that is bounded with surface

$$(x^2+y^2+z^2)^2=a^2(x^2+y^2-z^2)$$

using spherical coordinates $x=r\cos\phi \sin\theta, y=r\sin\phi \sin\theta, z=r\cos\theta$. My main problem is to find bounds for $\theta$ and $\phi$. For $r$ I easily get bounds $0\le r\le a\sqrt{1-2\cos^2 \theta}$. If I'd had two surfaces i would probably get bounds for theta  from another of them, but here not sure. Any help is appreciated.


问:
Find multivariable limit: as $(x,y) \to (0,0)$ of $ \frac {x\sin y-y\sin x} {x^2+y^2}$
How do I evaluate the following limit?

$$ \lim_{(x,y) \to (0,0)} \frac {x\sin y-y\sin x} {x^2+y^2}$$

So far I tested the limit along the paths in which $y=mx$ and I got zero. However I have no clue how I can prove the limit exists or find the limit's value.

答:
$\textbf{Hint:}$

$$\frac{|x\sin y - y\sin x|}{x^2+y^2} = \frac{|xy|}{x^2+y^2}\left|\frac{\sin y}{y}-\frac{\sin x}{x}\right| \leq \frac{1}{2}\left|\frac{\sin y}{y}-\frac{\sin x}{x}\right|$$

By AM-GM inequality.


问:
Looking for a paper by Vizing (1976) introducing the list chromatic number
I am studying about the list chromatic number in graph theory, and Vizing (1976) and Erdős–Rubin–Taylor (1979) are credited for independently introducing this concept.

I was able to locate the 1979 paper, but I'm unable to find the 1976 paper anywhere online. I have seen it cited variously as:

V. G. Vizing, Färbung der Ecken eines Graphen mit vorgeschriebenen Farben,
Metody Diskretn. Anal. 29 (1976), 3–10 (Russian). Zbl 0362.05060

V. G. Vizing, Färbung der Ecken eines Graphen mit vorgeschriebenen Farben, Diskret. Analiz, Novosibirsk 29 (1976), 3–10 (Russian).

V. G. Vizing, Coloring the vertices of a graph in prescribed colors, Diskret. Analiz 29 Metody Diskret. Anal. v Teorii Kodov i Shem (1976), 3–10, 101 (Russian). MR0498216

and with a few other variations too.

Does anyone know if this paper is available somewhere that can be (easily) accessed?

答:
Not yet an easy access, but I've found several restricted ones. I'm adding them in case you have specific access.

I started from another article of Vizing, "On connected list colorings of graphs", where he cites himself as:

V.G. Vizing, Vertex coloring of a graph with assigned colors, Metody Diskret. Analiz. (Novosibirsk) 29 (1976) 3–10 (in Russian)

I tried to locate the journal rather than the article. I found a Google Books reference, but without access (or at least I couldn't find how). However the article seems to have been uploaded by UCLA. So I went to their website: they do have it on paper and online, but the online access requires a UCLA authorization. If you go straight to the HathiTrust Digital library, they also say that the access is restricted to UCLA.

I also found a hard copy in some US libraries (Alberta University, Kansas City, University of Michigan).

Depending on your location, your local university might be the safest bet.


问:
Two-torus embeddable knots
I was wondering if there were any known resources containing information on two-torus embeddable knots? I have checked the knot atlas with no luck, and it seems that many other cites only reference the figure eight knot.

Is there anyway to compute or algorithmically generate two-torus embeddable  knots? I am specifically interested in obtaining several of these and examining their respective Alexander polynomial.

Thanks

答:
Take any knot diagram, choose two points encircled by the knot:

These represent the positions of the holes in the two-torus.

Imagine placing two cylinders vertically on the plane at the positions of your dots.

Return your knot to three dimensions, and onto the two-torus.

QED

Title: Two-torus embeddable knots


问:
On the slope of a strictly increasing function
If $f$ is a strictly increasing function, shouldn't $f'$ be always positive and never zero? Apparently there's this situation where the derivative can be $0$ if it's only at discrete points and not an interval.

How is that possible, when strictly increasing functions are, by definition, never of a zero slope?

答:
A function $f : \mathbb R \to \mathbb R$ is strictly increasing if, for every $b > a$, it follows that $f(b) > f(a)$.  This of course is satisfied by $f(x) = x^3$, since if $b > a$ implies $$f(b) - f(a) = b^3 - a^3 = (b - a)(a^2 + ab + b^2) = \frac{1}{4}(b-a)\left((b-a)^2 + 3(b+a)^2\right).$$  Since $b-a > 0$ and no square is negative, all factors are positive, thus $f(b) - f(a) > 0$.

However, $f'(x) = 3x^2$ so $f'(0) = 0$.


答:
IF the lub of a set is in the set then it is the maximum of the set.
IF the glb of a set is in the set then it is the minimum of the set.

(0, 1), (0, 1], [0, 1), and [0, 1] all have glb 0 and lub 1.

(0, 1) has neither a minimum nor a maximum.

(0, 1] has no minimum but has maximum 1.

[0, 1) has no maximum but has minimum 0.

[0, 1] has minimum 0 and maximum 1.

Title: Least upper bounds and greatest lower bounds: open vs. closed intervals

Body:
I tried looking through other questions but couldn't find this one answered. I'm working through Lax and Terrell's 'Calculus with Applications'.

The book asks to state whether the following intervals have GLBs and LUBs and if so, what they are: $(8,10)$, $(8,10]$.

My initial thought was that an open interval can't have a LUB or GLB as they don't include the end points and thus there is always a number arbitrarily closer to the end point that doesn't quite 'reach' it. However, I've also seen some discussion such that, e.g., the LUB of both half-open and open intervals are the same given that $9.99999... = 10$.

Help appreciated here as I want to understand this before proceeding!


问:
If $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$ then $y=2^k$ and $x=1$
Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. Is it necessarily the case that $x=1$ and $y=2^k$?

Equivalently (I prove equivalence at the end): Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two positive divisors of $(2^k-1)(2^km+1)$ which average to $m(2^k-1)+1$. Is it necessarily the case that $m=1$ and that these  two divisors are $2^k-1$ and $2^k+1$?

I've tested this up to $y\leq10^{10}$ but I haven't been able to make much progress with standard number theoretic techniques.

If $k=1$ then there are infinitely many solutions of the form $x=y-1$.

Let $(1)$ be the initial version of the problem and let $(2)$ be the supposedly equivalent version of the problem.

$(2)\implies(1)$: Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. We can write $y=m(2^k-1)+1$ for some $m\geq1$. Then
$$2^ky-1=2^k(m(2^k-1)+1)-1=(2^k-1)(2^km+1)$$
so $y-x$ and $y+x$ are two positive divisors of $(2^k-1)(2^km+1)$ which average to $y=m(2^k-1)+1$. By $(2)$, $y-x=2^k-1$ and $y+x=2^k+1$. Then $x=1$ and $y=2^k$.

$(1)\implies(2)$: Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two positive divisors of $(2^k-1)(2^km+1)$ which average to $m(2^k-1)+1$. Let $y=m(2^k-1)+1$. We can write the two divisors as $y-x$ and $y+x$ for some $0<x<y$. Thus,
\begin{align*}
y-x&\bigm|2^ky-1,\\
y+x&\bigm|2^ky-1,
\end{align*}
since $2^ky-1=(2^k-1)(2^km+1)$. Manipulating these divisibility relations shows that
\begin{align*}
y-x&\bigm|2^kx-1,\\
y+x&\bigm|2^kx+1,
\end{align*}
where $\gcd(2^kx-1,2^kx+1)=1$. Then $\gcd(y-x,y+x)=1$ so $y^2-x^2\bigm|2^ky-1$. We clearly have $2^k-1\bigm|y-1$. By $(1)$, $x=1$ and $y=2^k$. Then $m=1$ and the two positive divisors were $2^k-1$ and $2^k+1$.

答:
I've finally solved it!

My solution to this question proves a generalized version.

To deduce this special case, set $z=2^k$, note that $z$ is divisible by $4$, and use Theorem 9 to conclude that $y=z$ and $yz-1=y^2-x^2$.
Then $y=2^k$ and $x=1$ as desired.


问:
Is a finite tiling of 45-45-90 triangles uniquely determined from the resulting union (up to trivial flips)?
Suppose I take a union of nonoverlapping $1-1-\sqrt{2}$ triangles in the plane:

The same shape can be tiled in another way, by flipping two triangles joined together into a square:

In general, are these the only such decompositions possible? That is, given two planar configurations of finitely many nonoverlapping $1-1-\sqrt{2}$ triangles whose unions are the same, are they necessarily related by some number of "square flipping" operations?

In the infinite case, this is not true, as seen by the following tilings of an infinite quadrant:

答: While it feels a little like cheating, the following configurations seem to provide a counter-example to the stated conjecture:


答:
Whether or not $(8,10)$ has a l.u.b. depends on what space/set you're working with to begin with, i.e. what the "parent space" is.

If you're working in $X = (8,10)$ then the set $(8,10)$ has no l.u.b. in $X$.

Another example is the set $A =$ {$x^2 < 2: x \in \mathbb{Q}$} has no l.u.b. in the parent set $\mathbb{Q}$. However, if the parent set was $\mathbb{R}$, then of course the l.u.b of $A$ would be $\sqrt{2}$.

So we could conclude that a set $S \subseteq T$ has a l.u.b. in the set $T\ $ if:

$\ T$ is complete and $\exists s \in T$ which is an upper bound of the set $S$.


问:
What is meant by "tangent space at identity of a Lie group is canonically isomorphic to its Lie algebra"?
I need to understand what is meant by "tangent space at identity of a Lie group is canonically isomorphic to its Lie algebra" to understand the definition of adjoint representation.

Could any one tell me in detail what is going on in that definition?
Thank you.

答:
I can do much more than that, here. Let's start at the beginning.

A differential manifold $G$ is a continuum that has been given enough infrastructure to allow for meaningful notions of continuity, contiguity (by its topological structure), and differentiation (by its differential structure); so that one can talk about connected, continuous, differentiable paths. If $t ∈ ℝ ↦ g(t) ∈ G$ is such a path, then by both definition and construction, $g'(t)$ is tangent to the curve at $g(t)$, and $g'(t) ∈ T_{g(t)}G$. That's effectively how tangent spaces are defined and constructed. For each $g ∈ G$, the tangent space $T_gG$ is the set of all tangents to all differentiable curves passing through $g$. So, it's natural to denote the generic element of $T_gG$ as $\dot g$, which I'll do here. Together, the family $TG ≡ \{(g,\dot g): g ∈ G, \dot g ∈ T_gG)\}$ is also called the tangent bundle of $G$.

Similarly, for a function $φ: G → G'$, between differential manifolds $G$ and $G'$, one can define the differential $φ_*: TG → TG'$ by $d/dt (φ(g(t)) = φ_*(g(t))[g'(t)]$, which acts linearly on tangent vectors $g'(t)$ at $g(t)$ and the operation has the signature
$$φ_*(g(t)): T_{g(t)}G → T_{φ(g(t))}G'.$$

I prejudiced the issue by using $G$ to denote a generic differential manifold (one normally uses the letter $M$ to denote manifolds). All of the above applies generically to all manifolds, not just group manifolds. But, if $G$ is also a Lie group, then by definition that means it is a differential manifold which also happens to have group operations on it that are, themselves, differentiable.

My favorite definition - and one of the simplest - is that a Lie group $G$ is a differential manifold containing a distinguished element $e ∈ G$ and a differentiable operation $g,h ∈ G ↦ g/h ∈ G$ such that $g/e = g$, $g/g = e$ and $(g/k)/(h/k) = g/h$, for all $g,h,k ∈ G$. The group operations can be defined from this as $g^{-1} ≡ (g/g)/g [= e/g]$ and $gh ≡ g/h^{-1}$, and you can verify that all the usual properties of groups are satisfied and that the group axioms, themselves, are equivalent to the three quotient axioms above under the correspondence $g/h ↔ g h^{-1}$ (and this equation, in turn, is also a consequence of the three quotient axioms since $g h^{-1} = g/(h^{-1})^{-1}$ and $(h^{-1})^{-1} = e/(e/h) = (h/h)/(e/h) = h/e = h$).

One can define the "translation" maps $L(g) h = g h = R(h) g$, for any $g, h ∈ G$. Their derivatives effectively extend the group operation to the tangent bundle $TG$ by way of the following:
$${d \over dt} (g h(t)) = g h'(t) ≡ L(g)_*(h(t))[h'(t)],$$
$${d \over dt} (g(t) h) = g'(t) h ≡ R(h)_*(g(t))[g'(t)].$$
As algebraic operations, they have the following signatures:
$$\dot g ∈ T_gG ↦ \dot g h ∈ T_{gh}G,$$
$$\dot h ∈ T_hG ↦ g \dot h ∈ T_{gh}G.$$
where $g,h ∈ G$. It satisfies all the expected properties; e.g. $e \dot g = \dot g$, $(g \dot h)k = g(\dot h k)$, $(\dot g h)k = \dot g (hk)$, and so on.

Curiously, mathematicians go out of their way to avoid this economy of notation, so that things get needlessly obscured as a result. The only explanation I can find for that is that they're doing so intentionally for job security - the "Egyptian Scribe" Creed: ("use hieroglyphs, instead of the Phoenician writing, to keep everyone else out of our field"): a form of Academese. "What is Academese?" Thought Co. https://www.thoughtco.com/academese-prose-style-1688963. Or else, the only other explanation I can think of is that there is simply a blind spot in the entire community with respect to the creed "keep it simple".

In fact, the same convention can also be applied more broadly. If a group $G$ acts differentiably on a manifold $M$, then that means one has a differentiable operation
$$g ∈ G, m ∈ M ↦ gm ∈ M$$
such that $em = m, g(hm) = (gh)m$, for all $g,h ∈ G$. Then one can equally well extend the group operation to the tangent bundle $TM$ with the signatures
$$\dot m ∈ T_mM ↦ g \dot m ∈ T_{gm} M$$
for $g ∈ G, m ∈ M$ and even define a map from $TG$ to $TM$ by
$$\dot g  ∈ T_gG ↦ \dot g m ∈ T_{gm} M$$
for $g ∈ G, m ∈ M$. A similar observation holds, if $G$ acts on $M$ to the right.

As we've already seen, this applies to the case where the manifold $M$ is the group $G$, itself: $G$ acts on itself on both the left and the right. So, the group operation applies to both the left and the right on the tangent bundle $TG$ ... which also means that $G$ acts to both the left and right on $TG$.

Sometimes, the right action is taken as a quotient $R(h) g = g/h$, so that the order of operations is preserved; i.e. $R(hk)g = g/(hk) = (g/k)/h = R(h)R(k)g$. I've seen both conventions used. That should be kept in mind. Otherwise, if it is defined as a product on the right, then one would instead have $R(hk) = R(k)R(h)$.

All the expected rules apply, provided one stays within the restrictions laid out by the signatures of the operations. Thus, for instance, we have the same rules for products and quotients as you would for matrix algebra:
$${d \over dt} (g(t) h k(t)) = g'(t) h k(t) + g(t) h k'(t) ∈ T_{g(t)hk(t)}G,$$
$${d \over dt} g(t)^{-1} = -g(t)^{-1} g'(t) g(t)^{-1} ∈ T_{g(t)^{-1}}G.$$

Every group acts on itself in a third way, by operating on the left as a product and at the same time on the right as a quotient. That's the internal automorphism representation
$$Int(g) h = g h g^{-1}.$$
You can verify that is satisfies the required properties $Int(e) g = g$, $Int(gh)k = Int(g)Int(h)k$ to qualify as an action of $G$ on itself.

The differential of this $Ad(g) = Int(g)_*$ is given its own name
$$Ad(g) \dot h = g \dot h g^{-1} ∈ T_{ghg^{-1}}G,$$
and is called the adjoint action of $G$ on $TG$. Curiously, mathematicians don't define it this broadly - not on the entire tangent bundle, but only on $T_eG$. But it can be, and we will do so.

If $g$ and $h$ commute (that is, $gh = hg$), then $T_hG$ is closed under $Ad(g)$: $Ad(g): T_hG → T_hG$. In particular, since $e$ commutes with everyone, then $T_e G$ is closed under $Ad(g)$ and it produces a bona fide operation on $T_e G$:
$$Ad(g) \dot e = g \dot e g^{-1} ∈ T_{geg^{-1}}G = T_eG.$$

The Lie algebra $Lie(G)$ of $G$ is one and the same as the tangent space $Lie(G) ≡
T_eG$ at the identity $e$. Because $Lie(G)$ is closed under $Ad(g)$, with $Ad(g)$ being differentiable, then $Ad(g(s)) \dot e$ forms a differentiable path in $T_eG$ (noting that $T_eG$ is, itself a manifold, as are all tangent spaces). So, one can differentiate $Ad(g(s))$, too
$$Ad_*(g(s))[g'(s)] \dot e = {d \over ds} (g(s) \dot e g(s)^{-1}) ∈ T_eG.$$

This operation also applies more broadly, to tangent spaces $T_hG$ ... provided that $h$ commutes with all of the $g(s)$'es:
$$Ad_*(g(s))[g'(s)] \dot h = {d \over ds} (g(s) \dot h g(s)^{-1}) ∈ T_hG,$$
if $g(s) h = h g(s)$ for all $s$.

If this were a matrix algebra, we would be able to write
$$\begin{align}
‟{d \over ds} (g(s) \dot h g(s)^{-1}) & = g'(s) \dot h g(s)^{-1} - g(s) \dot h g(s)^{-1} g'(s) g(s)^{-1} \\
& = (g'(s) g(s)^{-1}) (g(s) \dot h g(s)^{-1}) - (g(s) \dot h g(s)^{-1}) (g'(s) g(s)^{-1}) \\
& = [g'(s) g(s)^{-1}, g(s) \dot h g(s)^{-1}]”;
\end{align}$$
note my use of Air Quotes. In particular, if $g(0) = e$, then writing $g'(0) = \dot e$, we would have
$$‟\left({d \over ds} (g(s) \dot h g(s)^{-1})\right)_{s=0} = [\dot e, \dot h]”.$$

But since there is no matrix algebra undergirding $T_eG$ (it was defined as the tangent space of a manifold, not as a matrix algebra), then this correspondence can only be taken as a motivation for defining the Lie bracket in the abstract by way of the equation
$${d \over ds} \left(g(s) \dot e g(s)^{-1}\right) = [g'(s) g(s)^{-1}, g(s) \dot e g(s)^{-1}],$$
for all $\dot e ∈ T_eG$ which would actually apply generally, irrespective of whether $s = 0$ or not, and whether $g(s) = e$ or not. Since both $g'(s) g(s)^{-1} ∈ T_eG$ and $g(s) \dot e g(s)^{-1} ∈ T_eG$, then this is actually an operation on $T_eG$. That the operation is actually well-defined is actually already assured, for the most part, by the way tangent spaces were defined in the first place.

The usual definition assumes $s = 0$ and that $g(0) = e$, so that, with these assumptions, we can just write:
$$\left({d \over ds} (g(s) \dot e g(s)^{-1})\right)_{s=0} = [g'(0), \dot e].$$

Additional Remarks Added On Edit:
Some additional notes on the alternate definitions for Lie algebras:

A vector field $X$ is defined on a manifold $M$ as a differentiable map with the signature $m ∈ M ↦ X(m) ∈ T_mM$.

For each Lie group $G$, and each tangent vector $\dot e ∈ T_eG$, both of these are differentiable vector fields:
$${\dot e}_L: g ∈ G ↦ g \dot e ∈ T_gG,$$
$${\dot e}_R: h ∈ G ↦ \dot e h ∈ T_hG,$$
satisfying the properties
$${\dot e}_L(gh) = (gh) \dot e = g (h \dot e) = g {\dot e}_L(h),$$
$${\dot e}_R(gh) = \dot e (gh) = (\dot e g) h = {\dot e}_R(g) h.$$
Any vector field that is covariant on the left $X(gh) = gX(h)$ can be normalized to the form $X(g) = g X(e)$, with the result being that $X = {X(e)}_L$. The same applies in the other direction: if $X$ is covariant on the right $X(gh) = X(g)h$, then it normalizes to the form $X = {X(e)}_R$.

The actual conversion is done by taking the quotient on the left or right: $g\backslash X(g) = X(e)$ if $X$ is covariant on the left, or $X(h)/h = X(e)$, if it is covariant on the right. The operations themselves $g\backslash (⋯)$ and $(⋯)/h$ have names: Cartan-Maurer forms and can be defined much more broadly than mathematicians usually define them, with the signatures:
$$g\backslash (⋯): T_hG → T_{g\backslash h}G,$$
$$(⋯)/h: T_gG → T_{g/h}G.$$
As functions of $g$, when restricted to the signature $T_gG → T_eG$, they map the tangent bundle $TG$, itself, to the Lie algebra $T_eG$.

So, by construction, there is one vector field of each type, for each tangent vector at $T_eG$, so either set of vector fields can be taken as a proxy for $T_eG$, itself. The main significance of this remark is that there is already a way to define Lie brackets for vector fields on manifolds; and the vector fields, with this Lie bracket, can be taken as an alternate way to define the Lie algebra for the Lie group $G$.

The manifold-theoretic Lie bracket arises as a result of the fact that the process of going from curve to tangent vector (by differentiating to get $m'(t) ∈ T_{m(t)}M$ on a differential manifold $M$) can be reversed, with vector fields, by solving the differential equation $d/dt(x(t)) = X(x(t))$. This produces a family of curves, called the flow of $X$. Once you convert vector fields to flows, then you have curves.

A vector field can effectively be defined by the way it acts on real-valued functions over the manifold in such a way that if $f: M → ℝ$, then $d/dt (f(x(t))) = x'(t) [f] = X(x(t)) [f]$, where $x(t)$ is the flow of $X$.

In general, for any differentiable path $x(t)$ on a differential manifold $M$, one has the operation defined by $x'(t)[f] = d/dt (f(x(t)))$. That extra detail is actually how tangent vectors on a differential manifold are actually defined and constructed in the first place.

The Lie bracket $[X,Y]$ of two vector fields $X$ and $Y$ is defined by $[X,Y][f] = X[Y[f]] - Y[X[f]]$, which serves to define $[X,Y]$ as a vector field.

The Lie algebraic definition of the Lie bracket matches the manifold-theoretic definition of the Lie bracket for the corresponding vector field: $[{\dot e}_0, {\dot e}_1]_L = [{\dot e}_{0L}, {\dot e}_{1L}]$. The sign is reversed for the other direction, $[{\dot e}_0, {\dot e}_1]_R = -[{\dot e}_{0R}, {\dot e}_{1R}]$, so there is a preference for vector fields covariant on the left over fields covariant on the right, when defining Lie algebras in terms of such vector fields.


答:
"strictly increasing functions are, by definition, never of a zero slope": that is not the definition of a strictly increasing function. The correct one says

$$x_0<x_1\implies f(x_0)<f(x_1).$$

Geometrically, this can be expressed as "the slope of any chord is positive", but not as "the slope of any tangent is positive".

Title: On the slope of a strictly increasing function

Body:
If $f$ is a strictly increasing function, shouldn't $f'$ be always positive and never zero? Apparently there's this situation where the derivative can be $0$ if it's only at discrete points and not an interval.

How is that possible, when strictly increasing functions are, by definition, never of a zero slope?


问:
Smith normal form and elementary divisors
Say I have the matrix $\begin{pmatrix}16&16&8\\8&6&2\\3&4&2\end{pmatrix}$. To give a structure for a module homomorphism whose representation is given by this matrix, I must reduce the matrix to a Smith normal form. Now, in order to calculate the Smith normal form, I have to calculate the elementary divisors. Are they $2,8$ in this case?

I think no. This is because, though the first elementary divisor is given by $\frac{d_1}{d_0}=\frac{2}{1}=2$, but the second  elementary divisor is just $\frac{d_2}{d_1}=\frac{2}{2}=1$ in my opinion. This is because, the gcd of all the $2\times 2$ minors of the matrix is $2$. Should we only consider the principal minors while calculating the second elementary divisor? Kindly elaborate. Thanks beforehand.

答:
Reduce the matrix to its normal form by following the usual algorithm. In this case, we would go through the following steps:
$$
\pmatrix{16&16&8\\8&6&2\\3&4&2} \to
\pmatrix{3&4&2\\8&6&2\\16&16&8} \to \\
\pmatrix{1&4&2\\6&6&2\\8&16&8} \to
\pmatrix{1&0&0\\6&-18&-10\\8&-16&-8}\to\\
\pmatrix{1&0&0\\0&-18&-10\\0&-16&-8} \to
\pmatrix{1&0&0\\0&18&10\\0&16&8} \to\\
\pmatrix{1&0&0\\0&10&18\\0&8&16} \to
\pmatrix{1&0&0\\0&2&2\\0&8&16} \to\\
\pmatrix{1&0&0\\0&2&2\\0&0&8} \to \pmatrix{1&0&0\\0&2&0\\0&0&8}.
$$
Consequently, conclude that the invariant factors are $2,8$.

If you insist on using the formula with the minors, note that we have $d_1 = 1 \neq 2$. Indeed, the $3,1$ entry and $3,3$ entries are relatively prime, which means that the entries (the "first order minors") are relatively prime.

For the second factor, we compute the cofactor matrix to be
$$
\pmatrix{4&-10&14\\0&8&-16\\-16&32&-32}.
$$
The greatest common factor of these entries is $2$, which gives us the invariant factor $2/1 = 2$.

Finally, we compute the matrix of the whole matrix to be $-16$, which gives us the invariant factor $-16/2 = -8$. We can equivalently say that the invariant factor is simply $8$.


问:
If $A=A^{op}$ is $A^*=A$ in $A-mod$?
Consider a finite-dimensional algebra $A$ over a field $k$. We have the dualization functor $*: Mod-A \to A-Mod$ given by

$$M^*=Hom_k(M,k).$$

This gives us a left $A$-module $A^*$. In general, it does not have to be isomorphic to $A$ as a left module.

But we may consider $A^*$ as a right module over the opposite algebra $A^{op}$. What if $A^{op}$ is isomorphic to $A$? Given a fixed isomorphism $A\cong A^{op}$, is there a way to give a canonical isomorphism $A^*\cong A$ as left $A$-modules (=right $A^{op}$-modules)? This seems like such a triviality, but I have been lost in all the isomorphisms and reversing multiplication orders for a few hours now.

答:
I'm not sure if this statement is true. The semi-canonical choices of $k$-linear morphisms $A \to A^*$ that I can come up with fail to be left $A$-module homomorphisms.

Nevertheless I will describe here my attempt. Let us choose a basis $e_1, \dots, e_n$ for $A$ and let us simply spell out the condition for a $k$-linear morphism $\psi: A \to A^*$ to be a morphism of left $A$-modules. As a basis for $A^*$ we choose $\{\hat{e}_i\}_{i=1}^n$, where
\begin{aligned}
\hat{e_i}(e_j) := \begin{cases} 1 & i = j; \\ 0 & \text{otherwise.}\end{cases}
\end{aligned}
Let $\phi: A \to A^{\rm op}$ be the given involution of $A$, i.e. $\phi(1) = 1$ and $\phi(ab) = \phi(b)\phi(a)$. As you indicated, we then make $A^*$ into a left $A$-module by
\begin{aligned}
(a \cdot f)(b) := f(b\phi(a)).
\end{aligned}
Let's spell out what this is in coordinate. The multiplication $\mu: A \otimes A \to A$ can be written out in coordinates as
\begin{aligned}
e_i e_j = \mu(e_i \otimes e_j) = \sum_{k} \mu_{ijk} e_k.
\end{aligned}
Similarly the involution $\phi: A \to A^{\rm op}$ can be written in coordinates as
\begin{aligned}
\phi(e_i) = \sum_l \phi_{il} e_l.
\end{aligned}
The left $A$-action on $A^*$ thus becomes
\begin{aligned}
(e_i \cdot \hat{e}_j)(e_k) = \hat{e}_j(e_k\phi(e_i)) = \sum_m \phi_{im} \hat{e}_j(e_k \cdot e_m) = \sum_m \phi_{im} \mu_{kmj},
\end{aligned}
or in other words
\begin{aligned}
e_i \cdot \hat{e}_j = \sum_m \sum_k \phi_{im} \mu_{kmj} \hat{e}_k.
\end{aligned}
Now let us assume there is a $k$-linear isomorphism $\psi: A \to A^*$ given by $\psi(e_i) = \sum_j \psi_{ij} \hat{e}_j$ such that $\psi$ is an isomorphism of left $A$-modules. Then $\psi$ should satisfy the equation
\begin{aligned}
\psi(ab) = a \cdot \psi(b),
\end{aligned}
or in other words
\begin{aligned}
\psi(ab)(c) = \psi(b)(c\phi(a)).
\end{aligned}
Writing both sides out in coordinates, one obtains
\begin{aligned}
\psi(e_ie_j)(e_k) = \psi(\sum_{l}\mu_{ijl}e_l)(e_k) = \sum_l \mu_{ijl} \psi_{lk}
\end{aligned}
and
\begin{aligned}
\psi(e_j)(e_k\phi(e_i)) = \sum_m \phi_{im} \psi(e_j)(e_k e_m) = \sum_m \sum_n \phi_{im} \mu_{kmn} \psi(e_j)(e_n) = \sum_m \sum_n \phi_{im} \mu_{kmn} \psi_{jn}.
\end{aligned}
So all in all we need to find $\psi$ satisfying
\begin{aligned}
\sum_l \mu_{ijl} \psi_{lk} = \sum_m \sum_n \phi_{im} \mu_{kmn} \psi_{jn}.
\end{aligned}
We see that taking $\psi_{ij} = \delta_{ij}$ could work whenever we have the identity
\begin{aligned}
\mu_{ijk} = \sum_m \phi_{im} \mu_{kmj},
\end{aligned}
which seems like a very strange condition to me. Another attempt could be taking $\psi_{ij} = \phi_{ij}$, which probably also won't work in general since the condition $\phi(ab) = \phi(b)\phi(a)$ translates into the condition that
\begin{aligned}
\sum_l \mu_{ijl} \phi_{lk} = \sum_m \sum_n \phi_{jn} \phi_{im} \mu_{nmk};
\end{aligned}
the above condition would thus for example be satisfied if $\mu_{nmk} = \mu_{kmn}$ were to hold, which again seems like a strange condition.

I wouldn't be surprised at this point when someone comes up with an explicit counterexample.

Title: If $A=A^{op}$ is $A^*=A$ in $A-mod$?


问:
Compatible PDEs
If we have an overdetermined system of pdes what does one have to check to be sure that they are compatible? Suppose each pde is derived from a different Hamiltonian and we have that the Poisson brackets of any pair of these Hamiltonians vanish (i.e. the Hamiltonians are in involution) then how does that information help?

An example I have in mind is:

Suppose we have some function $u=u(t,x)$ -- assume it to be very well-behaved. And some Hamiltonians as functionals of $u$ say $H_1, H_2,...$ such that each pair are in involution. And they lead to PDEs $$u_{t_1}=f_1(x, u, u_x,...)\\ u_{t_2}=f_2(x, u, u_x,...)\\ \vdots$$

答:
Let us consider for simplicity two times, $t_1$ and $t_2$. We can evolve from the point $(0,0)$ in the 2D "time"-space to another point $(T_1,T_2)$ by different means.

One way, for example, is to use the 1st evolution equation to arrive to the point $(T_1,0)$ and then use the 2nd evolution equation to go from $(T_1,0)$ to $(T_1,T_2)$ (with fixed $T_1$!).

But we could also first use the 2nd equation to go to the point $(0,T_2)$ and then the 1st to arrive to $(T_1,T_2)$. Compatibility means that we have to obtain the same result for $u(T_1,T_2,x)$ in the end.

Now, infinitesimally (when $T_1,T_2\rightarrow0$) this amounts to checking the equality of mixed derivatives $u_{t_1t_2}$ calculated in two different ways: as $\partial_{t_2}f_1$ with $u_{t_2}$ computed using the 2nd equation and as $\partial_{t_1}f_2$ with $u_{t_1}$ computed from the 1st equation. When both equations are hamiltonian, it is then sufficient for the Hamiltonians to be in involution (this is an easy consequence of the Jacobi identity for the Poisson bracket).

You can find a very pedagogical exposition of these matters in this book (Section 1.2). I guess it may also be extremely helpful to you in future.

Title: Compatible PDEs


问:
Proving the Archimedean Property of reals and NIP (Nested Interval Property) using MCT
I'm doing exercise 2.4.4 in the book Understanding Analysis by Stephen Abbott. I'd like to ask, if my proof (especially part (a) of the question) is rigorous and technically correct.

(a) In a previous section, we used the Axiom of Completeness(AoC) to prove the Archimedean property of real numbers $\mathbf{R}$. Show that the Monotone Convergence Theorem can also be used to prove the Archimedean property without making any use of AoC.

(b) Use the Monotone Convergence Theorem to supply a proof for the Nested Interval Property that doesn't make use of the AoC.

These two results would suggest that we could have used the Monotone Convergence Theorem in place of the AoC as our starting axiom for building a proper theory of the real numbers.

Proof.

(a) The Archimedean property states that, $\mathbf{N}$ is an unbounded set that sits in $\mathbf{R}$. Given any real number $x$, there exists $n \in \mathbf{N}$ such that $x < n$.

By contradiction, let us assume that $\mathbf{N}$ is bounded subset of $\mathbf{R}$. Then, there exists an upper bound $x \in \mathbf{R}$, such that $n < x$ for all natural numbers $n \in \mathbf{N}$. The natural numbers are recursively defined using the sequence $x_{n+1} = 1 + x_n$. This is a monotonically increasing sequence. Hence, $\mathbf{N}$ is monotonic increasing and bounded, by monotone convergence theorem, the sequence $(x_n)$ is convergent. Let $\lim x_{n+1} = \lim x_{n} = x$. Taking limits on both sides, we have:
\begin{align*}
    x &= 1 + x \\
    0 &= 1
\end{align*}

This is a false statement and violates Peano's axioms. Moreover, we cannot find an interval $V_\epsilon(x)$ such that the values of the sequence $(x_n)$ will eventually lie in this interval. So, $(x_n)$ is not a convergent sequence $\implies$ $(x_n)$ is unbounded $\implies$ $\exists n \in \mathbf{N}$, such that $x < n$ for all reals $x$.

(b) The nested interval property states that the real number line $\mathbf{R}$ contains no gaps. I reproduce the statement of NIP for completeness.

For each $n \in \mathbf{N}$, assume that we are given a closed interval $I_n = [a_n, b_n] = \{x \in \mathbf{R}: a_n \le x \le b_n\}$. Assume also that each $I_n$ contains $I_{n+1}$. Then, the resulting nested sequence of closed intervals
\begin{align*}
    I_1 \supseteq I_2 \supseteq I_3 \supseteq I_4 \ldots
\end{align*}

have a non-empty intersection; that is
\begin{align*}
    \bigcap_{n=1}^{\infty} I_n \ne \phi
\end{align*}

Since, $I_n \supseteq I_{n+1}$, that is $[a_n,b_n] \supseteq [a_{n+1},b_{n+1}]$, the sequence consisting of the left hand end-points $a_1,a_2,a_3,\ldots, a_n$; $(a_n)$ is monotonically increasing. Moreover, the sequence $(a_n)$ is bounded above by $b_n$. By the Monotone Convergence Theorem, the sequence is convergent and $\lim_n (a_n) = s = \sup \{a_n : n \in \mathbf{N}\}$.

Since, $s$ is an upper bound for the sequence $(a_n)$, $a_n \le s$ for all $n \in \mathbf{N}$. Moreover, as $s$ is the least upper bound $s \le b_n$ for all $n \in \mathbf{N}$. Consequently, $a_n \le s \le b_n$ for all $n \in \mathbf{N}$. Thus, $s \in I_n$ for all $n \in \mathbf{N}$, and the countable intersection of these intervals is non-empty.

答:
What you’ve done is mostly correct, but there are a few fairly minor problems. First, your derivation of the contradiction in (a) could be made clearer. Once you have $x_{n+1}=1+x_n$ and $\lim_{n\to\infty}x_n=x$, you can write

$$x=\lim_{n\to\infty}x_n=\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}(1+x_n)=1+\lim_{n\to\infty}x_n=1+x\,;$$

this makes it very clear how you deduce that $x=1+x$ and hence that $0=1$. You should stop there: the remainder of that last paragraph is unnecessary. Moreover, part of it is false: it is simply not true that there is an $n\in\Bbb Z^+$ such that $x<n$ for all reals $x$. It isn’t even true that there is an $n\in\Bbb Z^+$ such that $x\le n$ for all reals $x$: just take $x=n+1$ to get a counterexample.

The idea in (b) is fine, but some of what you’ve written is rather sloppy. For instance, you write that

… the sequence $(a_n)$ is bounded above by $b_n$.

Here you are using the letter $n$ to mean two different things in the same statement: in $(a_n)$ it’s an index that runs over all positive integers, and in $b_n$ it refers to some specific index. What you mean is that for each $n\in\Bbb Z^+$, the sequence $\langle a_k:k\in\Bbb Z^+\rangle$ is bounded above by $b_n$. In particular, it is bounded above, so it converges to some $s=\lim_{n\to\infty}a_n=\sup\{a_n:n\in\Bbb Z^+\}\in\Bbb R$. The remainder of that last paragraph is fine.


问:
Catalan numbers in polygons
I'm stuck on such problem: triangulation of the $n$-gon is division of said $n$-gon into $(n-2)$ triangles whose sides are either sides of the $n$-gon or certain non-intersecting diagonals. How many triangulations of the $n$-gon are there? How many of these are there such that every triangle has at least one side that is also a side of the $n$-gon?

The first part is easy - I've managed to deliver a Catalan number recursive expression and then found a formula for $n^{\text{th}}$ Catalan number which is $$ C_{n} =  \frac{1}{n+1}  {2n \choose n}.$$

What about the second part? Shall it be done by subtracting some set from $n$-th Catalan number ? What should it look like?

Thank you in advance :)

答:
I don’t have a solution, but I do have an answer that illustrates a useful tool for attacking such problems.

It’s not hard to find pictures online of the $14$ triangulations of a hexagon, and here are the $42$ triangulations of a heptagon. If we let $t_n$ be the number of triangulations such that every triangle shares at least one side with the $n$-gon, we find that $t_3=1$, $t_4=2$, $t_5=5$, $t_6=12$, and $t_7=28$. Looking up the sequence $1,2,5,12,28$ in OEIS, we get A045623 as the first hit. If $a_n=t_{n+3}$ for $n\ge 0$, that sequence is listed as the number of $1$s in all compositions of $n+1$, but in the COMMENTS it is noted that $a_n$ is indeed our $t_n$. The FORMULA section of the entry offers the recurrence

$$a_{n+1}=2a_n+2^{n-1}$$

for $n>0$, with $a_0=1$ and $a_2=2$; this translates to

$$t_{n+1}=2t_n+2^{n-4}$$

for $n>3$, with $t_3=1$ and $t_4=2$. It also offers the closed form $a_n=(n+3)2^{n-2}$ for $n\ge 1$ and $a_0=1$, which translates to $t_n=n2^{n-5}$ for $n\ge 4$ and $t_3=1$; this can of course be derived from the recurrence.

I’ve not yet seen an argument for either the recurrence or the closed form, and I may not find much time to think about. If I do come up with one, however, I’ll add it (if no one has beaten me to it).

Title: Catalan numbers in polygons


问:
Polynomial $f(X) g(Y)- g(X) f(Y)$ irreducible in $F[X,Y]$?
Let $F$ be a field and $f(X), g(X) \in F[X]$ two coprime
non constant polynomials.

Question: Is the polynomial $P(X,Y):= f(X)  g(Y)- g(X)  f(Y)$
irreducible in the ring $F[X,Y]$ of polynomials in two variables?

The claim is clearly wrong if $f$ and $g$ are not coprime, so it's neccessary assumption. A possible approach I used was to divide $P(X,Y)$ by $f(X)$ and think
about the polynomial $g(Y)-S \cdot f(Y) \in F[S][Y]$ where
$S:= g(X)/f(X)$ and because the extension of $F$ by $F(S)$ is
transcendental we can regard $S$ as a variable. I know that
$g(Y)-S \cdot f(Y) $ is irreducible in $F[S][Y]$ but not know
how it helps to answer the question if $P(X,Y)$ is irreducible in $F[X,Y]$.

答:
In the ring $F(X)[Y]/(X-Y)$, you have
$$P(X,Y)\bmod (X-Y)= f(X)  g(X)- g(X)  f(X)=0,$$
hence $P(X,Y)$ is divisible by $X-Y$.


问:
How to prove □(∀x□φ → □φ) → (◇∀x□φ → ◇□φ) in s5?
By S5, I mean the system consisting of the following axioms:

K: □(φ→ψ) → (□φ→□ψ)

T: □φ → φ

5: ◇φ → □◇φ

I believe the result I want to prove follows from K alone, but I cannot figure out how. If anyone can offer an explanation I would greatly appreciate it.

答:
In the link you posted in the comments, it is claimed that $$\square (\forall x \square \varphi \rightarrow \square \varphi)\rightarrow (\lozenge \forall x \square \varphi \rightarrow \lozenge \square \varphi)$$ is a theorem of S5 modal propositional logic. Well, that's not exactly right, because this is not a propositional sentence (as indicated by the presence of the quantifier $\forall x$). What is meant is that this sentence is a substitution instance of the propositional theorem $$(\star)\quad\square (P \rightarrow Q) \rightarrow (\lozenge P \rightarrow \lozenge Q)$$
with $\forall x\square \varphi$ substituted for $P$ and $\square \varphi$ substituted for $Q$ in $(\star)$.

Now it should be intuitively clear that $(\star)$ is already a theorem of the modal logic K, since it is valid on every Kripke frame. Suppose we're at a world $w$ at which $\square (P\rightarrow Q)$ is true. If $\lozenge P$ is true, then $P$ holds at a world $w'$ accessible from $w$. But then $(P\rightarrow Q)$ holds at $w'$, so $Q$ holds at $w'$, so $\lozenge Q$ holds at $w$.

Ok, that was a semantic proof of $(\star)$ on Kripke frames. How would you give a formal proof? I'll give a sketch. Rewriting $\lozenge$ as $\lnot \square \lnot$, we want to prove
$$(\star')\quad\square (P \rightarrow Q) \rightarrow (\lnot \square \lnot P \rightarrow \lnot \square \lnot Q).$$
By propositional logic, $(P\rightarrow Q)\rightarrow (\lnot Q\rightarrow \lnot P)$, so by necessitation and K,   $\square(P\rightarrow Q)\rightarrow \square(\lnot Q\rightarrow \lnot P)$. Now if we assume $\square(P\rightarrow Q)$, by Modus Ponens we get $\square(\lnot Q\rightarrow \lnot P)$, so by K again, $\square \lnot Q \rightarrow \square \lnot P$. Taking the contrapositive again gives $\lnot \square \lnot P \rightarrow \lnot \square \lnot Q$, as desired.


问:
How to derive continuous-time Linear Quadratic Gaussian state estimator?
In a Linear Quadratic Gaussian problem, one can prove a separation principle, thereby separating the objective of control design from state estimation, when the linear system has process and measurement noise. The continuous-time solution involves solving for two dual Riccati differential equations for controller and state estimator, respectively.

Most of the papers and lectures I've read spend a lot of time proving the separation principle, as well as the optimality of the linear feedback. However, maybe it's a limitation in my understanding, I don't see any proof/derivation of how the continuous-time state estimator relates to the process/measurement noise covariances and the Riccati equation. The discrete-time variant, specifically Kalman Filter, proof is aplenty, but I don't like to rely on any discussion in discrete-time.

Even more to the basic, I would like to understand precisely even the problem formulation of the continuous-time LQG state estimator. What is it trying to minimize, and what are the "manipulated variables"? (i.e. what is the cost function?)

答:
Thanks to @Pait's comment, I've found my answer in the Kalman-Bucy filter. To answer my latter questions, an optimal linear state estimator has two objectives:

Ensure that the state estimates are unbiased ($E[e(t)-\hat{e}(t)]=0$)

Minimize the variance of the state estimation error

By restricting the state estimation to linear in nature, objective #1 automatically confers the Luenberger observer structure (all time-varying):

$$\dot{x}(t)=Ax+Bu+L(y-C\hat{x})$$

Then, given a symmetric positive-definite matrix, $M(t)$, we can construct a cost function as:

$$J=E[(x-\hat{x})M(t)(x-\hat{x})^T]=tr[M(t)P(t)]$$

where $P(t)=E[(x-\hat{x})(x-\hat{x})^T]$ is the state estimate error covariance matrix.

Objective #2 is then re-stated to minimize the cost function, $J$, by selecting the Kalman gain, $L(t)$. It is then possible to show that the optimal solution involves solving for the Riccati equation for $P(t)$, then use it to set the optimal Kalman gain.

The full derivation can be found in Athens, Tse, A direct derivation of the optimal linear filter using the maximum principle, 1967., which is well-written and easy to follow.


问:
Properties of the cone of positive semidefinite matrices
The set of positive semidefinite symmetric real matrices forms a cone. We can define an order over the set of matrices by saying $X \geq Y$ if and only if $X - Y$ is positive semidefinite. I suspect that this order does not have the lattice property, but I would still like to know which matrices are candidates for the meet and join of two matrices.

In other words, let $P$ be the cone of positive semidefinite matrices. Is there a nice characterization of the set $(X+P)\cap (Y+P)$, for two given matrices? What are the minimal points in this intersection?

答:
Geometric Interpretation. Consider a positive definite matrix $A$. It defines the ellipsoid
$${\cal E}_A = \{u: u^T A u \leq 1\}.$$
Note that the correspondence between $A$ and ${\cal E}_A$ is one-to-one. Moreover, $A\succeq B$ if and only if ${\cal E}_B$ contains ${\cal E}_A$. Using this fact, we can give the following characterization.

$C \succeq A$ and $C \succeq B$ if and only if ${\cal E}_C \subset {\cal E}_A \cap {\cal E}_B$.

Similarly,

$C \preceq A$ and $C \preceq B$ if and only if ${\cal E}_C \supset {\cal E}_A \cup {\cal E}_B$.

Now a matrix $C$ is a minimal matrix s.t. $C \succeq A$ and $C \succeq B$ if and only if ${\cal E}_C \subset {\cal E}_A \cap {\cal E}_B$ and there is no ellipsoid “sandwiched” between ${\cal E}_C$ and ${\cal E}_A \cap {\cal E}_B$. It is easy to see that that happens iff $\partial{\cal E}_C$ intersects $\partial{\cal E}_A$ by a $k$-dimensional ellipsoid and   $\partial{\cal E}_C$ intersects $\partial{\cal E}_B$ by an $n-k$ dimensional ellipsoid. This can be stated in terms of matrices $A$, $B$, and $C$.

Consider a minimal $C$ s.t. $C \succeq A$ and $C \succeq B$. Then there are two subspaces $U$ and $V$ with ${\mathbb R}^n = U \oplus V$ such that $u^T Cu = u^T Au$ for $u\in U$ and $u^T Cu = u^T Bu$ for $u\in V$; in particular, $\operatorname{rank}(C-A) + \operatorname{rank}(C-B) \leq n$.

Visualization. The following animation shows ellipses ${\cal E}_C$ inscribed in ${\cal E}_A \cap {\cal E}_B$ in two dimensions. Every ellipse ${\cal E}_C$ corresponds to a minimal matrix $C$ ($C \succeq A$ and $C \succeq B$)

And this animation shows minimum ellipses ${\cal E}_C$ containing ${\cal E}_A \cup {\cal E}_B$.

I'm afraid that there is no more explicit characterization of sets $\{C: C \succeq A \text{ and } C \succeq B\}$ and $\{C: C \preceq A \text{ and } C \preceq B\}$.

Correspondence between “meet” and “join” matrices. Note that if $C_1$ is a “join” then $C_2 = A+B-C_1$ is a “meet” and vice versa. That follows from the fact that $C_2 \preceq A$ iff $A+B-C_1 \preceq A$ iff $B \preceq C_1$; similarly, $C_2 \preceq B$ iff $A \preceq C_1$.

Characterization for $2\times 2$ matrices. If $A$ and $B$ are $2\times 2$ matrices (s.t. $A\not\preceq B$ and $A\not\succeq B$) then meet and join matrices $C$ must satisfy the following equations $\det(C-A) = 0$ and $\det(C-A) =0$. The set of symmetric matrices that satisfy this system of equations forms a one dimensional curve in the space of matrices. Let us write
$$C_{xyz} = \begin{pmatrix}x&y\\y&z\end{pmatrix}.$$ Then the set

$$\{(x,y,z): \det(C_{xyz} - A) = \det(C_{xyz} - B) = 0\}$$
is a hyperbola (that lies in a plane in ${\mathbb R}^3$). Points on one branch of the hyperbola correspond to join matrices; points on the other branch correspond to meet matrices.

Notes. Note that by changing the basis we can always assume that $A=I$ and $B$ is a diagonal matrix, but I don't think that this observation leads to a very explicit characterization of  $\{C: C \succeq A \text{ and } C \succeq B\}$. In particular, $C$ does not have to be a diagonal matrix. For example, let
$$A=\begin{pmatrix} 1& 0\\0& 1\end{pmatrix}\quad B =\begin{pmatrix} 2& 0\\0& 1/2\end{pmatrix}.$$ Then the following matrices $C$ are minimal matrices greater ($\succeq$) than $A$ and $B$:
$$C=\begin{pmatrix} 2& 0\\0& 1\end{pmatrix}\quad C=\begin{pmatrix} 3& 1\\1& 3/2\end{pmatrix} \quad C=\begin{pmatrix} 3& -1\\-1& 3/2\end{pmatrix}.$$
(The set of all such matrices $C$ forms a one dimensional curve in the space of all $2\times 2$ matrices.)


问:
Sum of modulo for first n natural numbers where divisor is irrational
Given integer n and float z find sum s of modulo z of first n natural numbers i.e 1%z + 2%z + ... + n%z. Note that z is irrational number!

$$ s=\sum_{i=1}^n {(i\mod z)} \\
z \ is \ irrational
$$

For my concrete case z is $$ {\sqrt 2+1}$$

I found similar problem: https://www.geeksforgeeks.org/find-sum-modulo-k-first-n-natural-number

For my case n could be bigger than 10^100, so naive linear time solution would not work.

答:
I ended up reducing the problem to the
$$
s=\sum_{i=1}^n {(i\mod z)} = \sum_{i=1}^n(i * z) - \sum_{i=1}^n\lfloor i*z \rfloor \\
z \ is \ irrational
$$

Solving first term is trivial and the second one could be solved using the Beatty sequence:
How to find $\sum_{i=1}^n\left\lfloor i\sqrt{2}\right\rfloor$ A001951 A Beatty sequence: a(n) = floor(n*sqrt(2)).


问:
If $\sum_{n=1}^\infty | x'(x_n) | < \infty$ for all $x' \in X'$ then $\sup_{\|x'\| \leq 1} \sum_{n=1}^\infty | x'(x_n) | < \infty$.
I have this question to solve, but I have no idea how to solve:

a) Let $X$ a normed space and $\{ x_n \}_{n \in \mathbb{N}} \subset X$ a sequence with $\sum_{n=1}^\infty | x'(x_n) | < \infty$ for all $x' \in X'$. Show that $\sup_{\|x'\| \leq 1} \sum_{n=1}^\infty | x'(x_n) | < \infty$.

b) Let $K \subset C([a,b])$ be $K = \{ f \in C([a,b]) : \int_{a}^{(a+b)/2} f(x) \mathrm{d}x = \int_{(a+b)/2}^{a} f(x) \mathrm{d}x \}$. Show that $K$ is a closed vector subspace and find $x' \in C([a,b])'$ so that $\ker (x')=K$.

Thanks for any help :)

答:
Define a linear map
$T:X'\to\ell^1$, by
$$
T(x')=(x'(x_n))_n.
$$
The Closed Graph Theorem can then be used to show that $T$ is bounded, from where one deduces that
$$
\infty > \|T\| =
\sup_{\|x'\|\leq1} \|T(x')\| = $$$$=
\sup_{\|x'\|\leq1} \sum_n|x'(x_n)|.
$$


问:
Simple functional equation in two variables
In one dimension the equation
$$
\frac{1}{1 + f^2(x)}=\frac{\left(\frac{\cos \phi}{\sqrt{1}} x +\sin \phi \right)^2}{1 + x^2}, \ x\in \mathbb{R}
$$
has the solution
$$f(x) =\frac{x-\frac{\cot \phi}{\sqrt{1}}}{1+ x \frac{\cot \phi}{\sqrt{1}}}, \ x\in \mathbb{R}$$

But, what is a solution of the equation in two dimensions
$$
\frac{1}{1 + f^2(x,y)+f^2(y,x)}=\frac{\left(\frac{\cos \phi}{\sqrt{2}} (x+y) +\sin \phi \right)^2}{1 + x^2+y^2}, \ x,y\in \mathbb{R}.\tag{1}
$$
Note: The function
$$f(x,y) =\frac{x-\frac{\cot \phi}{\sqrt{2}}}{1+ (x+y) \frac{\cot \phi}{\sqrt{2}}}$$
is close but not quite.

Any guess or method to derive a solution?

答:
Using the ansatz $f(x,y)=\frac{P(x,y)}{Q(x,y)}$, where $P,Q$ are polynomials of degree $1$ leads to four possible solutions such that your equation is satisfied. One of them is
$$f(x,y)=\frac{\frac{\cos\phi}{\sqrt{2}}+ x \frac{-1-\sin \phi}{2}+ y \frac{1-\sin \phi}{2}}{(x+y) \frac{\cos \phi}{\sqrt{2}}+\sin \phi}.$$


问:
Quaternion's multiplicative group as a subgroup of $\mathrm{GL}(4,\mathbb{R})$
Thinking about $\mathbb{C}$:

One can view the multiplicative group of complex numbers to be the matrices in $\mathrm{GL}(2,\mathbb{R})$ such that commutes with the matrix $I=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ a quick calculation shows that those are the matrix of the form $\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$.

One can also view the quaternion's multiplicative group as the matrices in $\mathrm{GL}(2,\mathbb{C})$ that satisfies $IX=\bar{X}I$ the same calculation as above shows that those are the matrices of the form $\begin{pmatrix} a & b \\ -\bar{b} & \bar{a} \end{pmatrix}$.

How can i find a matrix that indentifies by a law of the form above the multiplicative group of $\mathbb{H}$ in $\mathrm{GL}(4;\mathbb{R})$?

My attempts:

I've tryed to use the fact that in the matrix representation of complex numbers the conjugation is the transpose operation together with the properties of block matrix under transposition and substituing in the complex matrices the 2x2 blocks that corresponds to the complexes entries, but i didn't succeded.

If someone wants i can give more datails of my attempt. The candidate matrix was a the real-blocks matrix \begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix} but i am not sure it is enought.

Any hint or suggestion please?

Body:
Note that your representation of $\Bbb C$ in $\mathcal M_2(\Bbb R)$ yields a corresponding representation of $GL(\Bbb C,2)$ in $GL(\Bbb R,4)$. In particular, we can take the matrix
$$
\pmatrix{a_{11} + b_{11}i & a_{12} + b_{12}i\\
a_{21} + b_{21}i & a_{22} + b_{22}i}
$$
and replace each complex number with the corresponding element in $M_{2}(\Bbb R)$ to get the block-matrix
$$
\left(
\begin{array}{cc|cc}
a_{11} & -b_{11} & a_{12} & -b_{12}\\
b_{11} & a_{11} & b_{12} & a_{12}\\
\hline
a_{21} & -b_{21} & a_{22} & -b_{22}\\
b_{21} & a_{21} & b_{22} & a_{22}
\end{array}
\right).
$$
Now, we can apply this change to your representation of $\Bbb H$ to go from a subset of $GL(\Bbb C,2)$ to a subset of $GL(\Bbb R,4)$. In other words, we get the map
$$
a_{1} + a_{2}i + b_{1}j + b_{2}k \to
\pmatrix{a_{1} + b_{1}i & a_{2} + b_{2}i\\
-a_{2} + b_{2}i & a_{1} - b_{1}i} \to\\
\left(
\begin{array}{cc|cc}
a_{1} & -b_{1} & a_{2} & -b_{2}\\
b_{1} & a_{1} & b_{2} & a_{2}\\
\hline
-a_{2} & -b_{2} & a_{1} & b_{1}\\
b_{2} & -a_{2} & -b_{1} & a_{1}
\end{array}
\right).
$$

If we define the matrix $I \in GL(\Bbb R,4)$ to be
$$
I = \pmatrix{0&0&1&0\\0&0&0&1\\-1&0&0&0\\0&-1&0&0}.
$$
The "quaternions" in $GL(\Bbb R,4)$ are each matrices that satisfy $I X = X^{\Gamma_1} I$, where $X^{\Gamma_1}$ is a partial transpose of $X$. In particular,
$$
\pmatrix{A&B\\C&D}^{\Gamma_1} = \pmatrix{A^T & B^T\\C^T & D^T}
$$

Writing out the equation in terms of the entries of $X$ yields
$$
IX =
\pmatrix{x_{31} & x_{32} & x_{33} & x_{34}\\
x_{41} & x_{42} & x_{43} & x_{44}\\
-x_{11} & -x_{12} & -x_{13} & -x_{14}\\
-x_{21} & -x_{22} & -x_{23} & -x_{24}} =\\
\pmatrix{x_{11} & x_{21} & x_{13} & x_{23}\\
x_{12} & x_{22} & x_{14} & x_{24}\\
x_{31} & x_{41} & x_{33} & x_{43}\\
x_{32} & x_{42} & x_{34} & x_{44}} I =
\pmatrix{-x_{13} & -x_{23} & x_{11} & x_{21}\\
-x_{14} & - x_{24} & x_{12} & x_{22}\\
-x_{33} & -x_{43} & x_{31} & x_{41}\\
-x_{34} & -x_{44} & x_{32} & x_{42}}
= X^\Gamma  I.
$$
The $a_1$ elements give us the relations
$$
x_{13} = -x_{31}, \quad x_{42} = -x_{42},
$$
the $a_2$ elements give
$$
x_{11} = x_{33}, \quad x_{22} = x_{44},
$$
the $b_1$ elements give
$$
x_{14} = -x_{41}, \quad x_{23} = -x_{32},
$$
and the $b_2$ elements give
$$
x_{21} = x_{34}, \quad x_{43} = x_{12}.
$$
From there, we could use the additional fact that the quaternions satisfy $JX = XJ^{\Gamma_2}$, where
$$
\pmatrix{A & B\\C & D}^{\Gamma_2} = \pmatrix{A & C\\B & D}
$$
and
$$
J = \pmatrix{0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0}.
$$


问:
Is a closed ball in dual space $E'$ of a normed space $E$ compact with respect to the topology of compact convergence on $E'$?
Let $E$ be a normed vector space and $\sigma_c(E',E)$ denote the topology of compact convergence on $E'$. It's easy to show that $\sigma_c(E',E)$ is Hausdorff.

Now let $\delta>0$ and $$V_\delta:=\{\varphi\in E':\left\|\varphi\right\|_{E'}\le\delta\}.$$ Clearly, the subspace topology $\left.\sigma_c(E',E)\right|_{V_\delta}$ is Hausdorff as well.

But how can we show that $V_\delta$ is $\sigma_c(E',E)$-compact? It is clearly $\left\|\;\cdot\;\right\|_{E'}$-bounded. Does this already imply the claim?

答:
The answer is yes. The proof is contained essentially in the following fact:

If $E$ is a LCS and $A \subset E'$ is equicontinuous, then the topology of totally bounded convergence and the topology of pointwise convergence coincide on A.

Indeed, assume that this is true. Then $V_\delta$ is equicontinuous and compact in the topology of pointwise convergence due to Banach-Alaouglu theorem. Thus, it is compact in the topology of convergence on totally bounded sets. A fortiori it is compact in the topology of compact convergence.

To prove the foregoing fact fix $f \in A$ and a totally bounded set $K \subset E$. We are going to construct a finite set $F = \{x_1, \dots, x_n\} \subset E$ s.t. $(f + 1/3\;F^\circ) \cap A \subset (f + K^\circ) \cap A$. Let $U \subset E$ be a neighbourhood of $0$ s.t. $A \subset U^\circ$. Since $K$ is totally bounded there exists a finite set $F \subset E$ s.t. $F + 1/3 \; U \supset K$. Assume that $g \in (f + 1/3\; F^\circ) \cap A$. Then $g-f \in 1/3\; F^\circ$ and, therefore, $|(g-f)(x)| \le 1/3$ for all $x \in F$. Now for all $x \in K$ there exists $x' \in F$ s.t. $x-x' \in 1/3\; U$. Therefore
$$|(g-f)(x)| \le |g(x) - g(x')| + |g(x') - f(x')| + |f(x') - f(x)| \le 1,$$
because $f,g \in U^\circ$ and $x-x' \in 1/3\;U$. Thus, $g \in (f + K^\circ) \cap A$.

Therefore, each neighbourhood of $f$ in $A$ in the topology of totally bounded convergence contains a smaller neighbourhood of $f$ in the topology of pointwise convergence. It easily follows that the topologies coincide.


问:
Sequence of Continuous Functions converging pointwise
Basically, I'm wondering if the following function (defined over all the reals) is continuous:

$f(x)=\sum_{n=1}^{\infty}\frac{1}{n^2+x^2}$

The graph seems to be continuous, but I can't think of any way to prove that the sum itself is continuous.

Edit: Is there a way to do so with the M-test because I haven't covered Hyperbolic Identities

答:
HINT.
Consider the Weierstrass product for the $\sinh$ function:
$$
\frac{\sinh z}{z}=\prod_{n=1}^{+\infty}\left(1+\frac{z^2}{\pi^2 n^2}\right)
$$
then take logarithmic derivatives of both sides. We get
$$
\coth z-\frac{1}{z}=\sum_{n=1}^{+\infty}\frac{2z}{z^2+\pi^2 n^2}
$$
By some easy substitutions, we have
$$
\sum _{n=0}^{\infty } \frac{1}{n^2+x^2}=\frac{1}{2} \left(\frac{1}{x^2}+\frac{\pi  \coth (\pi  x)}{x}\right)
$$

Can you continue from here?

To use M-test, we have
$$
\frac{1}{n^2+x^2} \leq \frac{1}{n^2}
$$
Can you continue from here?