问:
Prove that a non-abelian group of order $pq$ ($p<q$) has a nonnormal subgroup of index $q$
So I've come up with a proof for the following question, and I'd like to know if it's correct (as I couldn't find anything online along the lines of what I did).

Question Let $p$ and $q$ be primes with $p<q$.  Prove that a non-abelian group of order $pq$ has a nonnormal subgroup of index $q$, so there there eixists and injective homomorphism into $S_q$

First I prove a Lemma:

If $H\triangleleft G$, then for every conjugacy class $\mathcal{K} \subseteq H$ or $\mathcal{K} \cap H = \emptyset$.

proof:

If $x\in K\cap H$, then $gxg^{-1} \in gHg^{-1}$ for all $g\in G$.  Since $H$ is normal $gHg^{-1}=H$, so that $H$ contains all the conjugates of $x$,i.e., $\mathcal{K} \subseteq H$ (Dummit and Foot, 4ed, pg 127).

Here's my attempt on the question:

proof:

Let $G$ be a group such that $|G|=pq$ and $p<q$.   Then $G \cong \mathbb{Z}_{pq}$, and so $G$ has two subgroups, say $H$ and $K$, of orders $p$ and $q$, respectively.  'Suppose that $H$ is a subgroup in $G$ of order $p$; a subgroup $K$ of order $q$ exists in $G$ by Cauchy (Nicky Hekster).'  Also, by Lagrange, $H=\langle x \rangle$ and $K=\langle y \rangle$ for some $y,x \in G$.  Since $p$ is the smallest prime that divides the order of $G$ and $|G:K|=p$, then $K\triangleleft G$ (by some theorem).

Suppose, by contradiction, that $H\triangleleft G$.  Then for $x^i \in H$, $gx^ig^{-1}=x^k \in H$ for all $g\in G$.  So $H\subseteq \mathcal{O}_{x^i}$ (the conjugacy class of $x_i$).  Since $|x^i|$ divides $p$, then $x_i\not \in K$.  Thus $\mathcal{O}_{x_i} \not \subseteq K$.  Also, since $e=x^p$ and $gx^pg^{-1}=e$, then $e\in \mathcal{O}_{x_i}$.  Thus $\mathcal{O}_{x_i} \cap K \neq \emptyset$.  Therefore $K$ is not normal in $G$, which is absurd.

Now let $\phi: G \rightarrow S_q$ be the permutations representation afforded by the action of $G$ on $A=G/H$ by left multiplication -- say $\pi_g:A\rightarrow A$ defined by $aH\mapsto gaH$ for all $g\in G, \; aH\in A$.  We know that $\ker \pi_g \leq H$ and $\ker \pi_g \triangleleft G$.  Since $|H|=p$, then $|\ker \pi_g|=p$ or $|\ker \pi_g|=1$.  If $|\ker \pi_g|=p$, then $\ker \pi_g$ is not normal in $G$.  Therefore, $|\ker \pi_g|=1$.  Since $\ker \pi_g=\ker \phi$, then $\phi$ must be injective.

Sorry for any typos (and a HUGE 'thank you' to anyone who actually reads this).

答: By Sylow I, there is some $p$-Sylow subgroup $P$ of $G$. If $P$ is normal, then there is a homomorphism $\gamma:G\to \text{Aut}(H)$ defined by conjugation. Since $H$ is cyclic of order $p$, $|\text{Aut}(H)|=p-1$ and therefore $|\gamma(H)|$ must divide $p-1$ and $pq$ so $\gamma$ must be the trivial map. Thus, conjugation is trivial on $H$, and therefore $H\subseteq Z(G)$ (the center of $G$), but then the center is not trivial, so $G$ must be abelian (compute $|G/Z(G)|=p$ or $q$ so $G/Z(G)$ is cyclic and therefore trivial).


问:
Proof of the hockey stick/Zhu Shijie identity $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$
After reading this question, the most popular answer use the identity
$$\sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1},$$
or, what is equivalent,
$$\sum_{t=k}^n \binom{t}{k} = \binom{n+1}{k+1}.$$

What's the name of this identity? Is it the identity of the Pascal's triangle modified.

How can we prove it? I tried by induction, but without success. Can we also prove it algebraically?

Thanks for your help.

EDIT 01 : This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself are highlighted, a hockey-stick shape is revealed.

答:
A Generalization

In this answer, I prove the identity
$$
\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}\tag{1}
$$
Here is a generalization of the identity in question, proven using the Vandermonde Identity
$$
\begin{align}
\sum_{t=0}^n\binom{t}{k}\binom{n-t}{j}
&=\sum_{t=0}^n\binom{t}{t-k}\binom{n-t}{n-t-j}\tag{2}\\
&=\sum_{t=0}^n(-1)^{t-k}\binom{-k-1}{t-k}(-1)^{n-t-j}\binom{-j-1}{n-t-j}\tag{3}\\
&=(-1)^{n-j-k}\sum_{t=0}^n\binom{-k-1}{t-k}\binom{-j-1}{n-t-j}\tag{4}\\
&=(-1)^{n-j-k}\binom{-k-j-2}{n-j-k}\tag{5}\\
&=\binom{n+1}{n-j-k}\tag{6}\\
&=\binom{n+1}{j+k+1}\tag{7}
\end{align}
$$
Explanation:

$(2)$: $\binom{n}{k}=\binom{n}{n-k}$

$(3)$: apply $(1)$ to each binomial coefficient

$(4)$: combine the powers of $-1$ which can then be pulled out front

$(5)$: apply Vandermonde

$(6)$: apply $(1)$

$(7)$: $\binom{n}{k}=\binom{n}{n-k}$

To get the identity in the question, set $j=0$.

A Simpler Proof of the Basic Formula
$$
\begin{align}
\sum_{k=0}^n\color{#C00}{\binom{k}{m}}
&=\sum_{k=0}^n\color{#C00}{\left[x^m\right](1+x)^k}\tag8\\
&=\left[x^m\right]\frac{(1+x)^{n+1}-1}{(1+x)-1}\tag9\\[6pt]
&=\left[x^{m+1}\right](1+x)^{n+1}-1\tag{10}\\[6pt]
&=\binom{n+1}{m+1}\tag{11}
\end{align}
$$
Explanation:

$\phantom{1}\text{(8)}$: use the definition of the binomial coefficient

$\phantom{1}\text{(9)}$: sum the finite geometric series

$(10)$: multiply by $x$ and adjust the exponent of $x$

$(11)$: extract the coefficient using the binomial theorem

ContentLicense: CC BY-SA 4.0


问:
Infection spread on a torus chessboard
In one of his books, Peter Winkler includes the following problem:

A disease is spreading on a $n\times n$ chessboard as follows: if a healthy cell is neighboring at least 2 infected cells, it becomes infected. Using the property that the perimeter of the infected area never increases, it’s easy to prove that it’s impossible to infect the entire chessboard with fewer than $n$ infected cells.

If the chessboard is a torus, the result no longer holds (verified on some instances of $n$). It seems that $n-1$ is the smallest number of infected cells required to infect the chessboard. The perimeter argument can’t be used here. Does anyone know any other ‘invariant’ that can be used?

Note: two cells are neighbors if they share one side.

答:
EDIT: This answer is actually totally wrong! I'm leaving it up so that others do not make the same mistake as I, but here is an example (thanks to Anton Petrunin) which shows that my supposed non-increasing quantity actually can increase.  Initially, three cells are infected, and the restricted perimeter is $10$, but after 3 moves, the restricted perimeter increases to $12$.

Suppose there are only $n-2$ infected cells. Without loss of generality, these infected cells occur in the lower left $(n-1)\times (n-1)$ subgrid of the torus. If not, rotate the torus until one of the columns with no infections is on the right (such a column must exist, since there are at most $n-2$ infections), then rotate until a row with no infections is on top.

The desired invariant quantity is the perimeter of the figure comprising the infected cells only in the lower left $(n-1)\times (n-1)$ subgrid, ignoring wrap-around. Let us call this the "restricted perimeter."

To prove invariance, note that a cell which becomes infected in the top row or right column does not affect the restricted perimeter at all. On the other hand, a cell in the lower left $(n-1)\times (n-1)$ subgrid becoming infected does not increase the restricted perimeter for the same reason as in the non-torus case.

With this definition, the same argument goes through. If there are initially $n-2$ infected cells, then the restricted perimeter is at most $4(n-2)$. But when if the entire torus is infected, the restricted perimeter would be $4(n-1)$.

Here is an example when $n=4$. The left grid is the initial infection, with two infected cells. The restricted perimeter is initially $8$. The right grid is the infection one step later, and the restricted perimeter is still $8$. The edges contributing to the restricted perimeter are highlighted in orange. The topmost orange edge in the second picture might appear strange, since it is between two black squares. But remember, you have to mentally delete the top row and right edge, then consider the perimeter of what remains.

Title: Infection spread on a torus chessboard


问:
Solving the probability difference equation $p_{n} = p(1 - p_{n-1}) + (1 - p)p_{n-1}$ yields a questionable closed-form equation?
I am trying to solve the following difference equation by writing it in closed-form, which represents the probability that a person flips an even number of heads in $n$ tries, with $p$ being the probability that a person flips a head at any given time: $p_{n} = p(1 - p_{n-1}) + (1 - p)p_{n-1}$.

To derive this equation, I did a simple case analysis. There are 2 cases that $n$ flips yields an even number of heads:

odd # of heads $n - 1$ times, then a head: $(1 - p_{n - 1})p$

even # of heads $n - 1$ times, then a tail: $p_{n-1}(1 - p)$

Adding the cases up yields the difference equation I wrote above.

My strategy was to derive the characteristic equation, as this is a non-homogeneous linear equation:

$p_{n} = p - pp_{n-1} + (1 - p)p_{n-1}$

$p_{n} = (1 - 2p)p_{n-1} + p$

$x = (1 - 2p)$

$p_{n} = C(1-2p)^{n}$

We know that $p_{1} = p$:

$C(1 - 2p) = p$

$C = \frac{p}{1 - 2p}$

And plugging back into the original equation, we get:

$p_{n} = \frac{p}{1 - 2p}(1 - 2p)^n = p(1 - 2p)^{n - 1}$.

However, this does not seem to be correct, because if $p$ is $1/2$ (if it is a $1/2$ chance of getting a head on a flip), $p_{n}$ will always be 0.

答:
Just another perspective for this problem.

Treat this as a two-state Markov chain, with state space $\{0,1\}$ corresponding to the number of heads flipped modulo $2$. The transition matrix is
$$P=\begin{pmatrix}1-p & p \\ p & 1-p\end{pmatrix}.$$
The initial distribution is $\lambda=(1,0)$ (with $0$ flips, we always flip an even number of heads!), so that the probability of flipping $0$ heads (mod $2$) after $n$ flips is $(\lambda P^n)_0$.

To compute this, the eigenvalues of $P$ are $1-2p$ and $1$. Now we don't actually need to diagonalise $P$: just note that if we did, then we would find that $$(\lambda P^n)_0=A(1-2p)^n+B,$$ for some suitable constants $A$, $B$ independent of $n$. But finding $A$ and $B$ is very easy: look at the $n=0$ and $n=1$ cases to get
\begin{align*}
1&=A+B \\
1-p&=A(1-2p)+B,
\end{align*}
from which we find $A=B=\frac{1}{2}$.


问:
Getting the kinematic equation $v^2=v_0^2+2a(x−x_0)$
As I've searched through the internet, looking for explanation. I found an answer with transformation of this formula:

$$
x = x_0 + v_0t + \frac{1}{2}at^2 ⟹ x = x_0 + \frac{1}{2}a(t + \frac{v_0}{a})^2 - \frac{v^2_0}{2a}
$$
I don't understand how to get $\frac{1}{2}a(t + \frac{v_0}{a})^2$
Can anyone help please?

答:
Considering that $a$ is constant, we can approach this problem alternatively as follows
\begin{align*}
a(s_{1} - s_{0}) = \int_{s_{0}}^{s_{1}}a\mathrm{d}s = \int_{s_{0}}^{s_{1}}\frac{\mathrm{d}v}{\mathrm{d}t}\mathrm{d}s = \int_{s_{0}}^{s_{1}}\frac{\mathrm{d}s}{\mathrm{d}t}\mathrm{d}v = \int_{s_{0}}^{s_{1}}v\mathrm{d}v = \frac{v^{2}(s_{1}) - v^{2}(s_{0})}{2}
\end{align*}

and the result holds.


问:
deriving Stokes' theorem from substituting this identity
As per requested: deriving Stokes' theorem from substituting this identity

Making substitution with

$ ∇ \times(\phi a) $

where $\phi$ is a scalar field, and $a$ is a vector field. Given that the identity is equal to:

$ ∇ \times(\phi a) = \phi (∇ \times a) + (∇\phi)\times a$

then derive stokes theorem:

$\int_{C}\phi dr = \iint_{S}n \times (∇\phi )dS$

for a surface $S$ with unit normal $n$ that is bounded by the closed curve $C$ , where the direction of $n$ obeys the usual convention.

What I've tried:

following from the hint

$$\nabla \times (\phi \mathbf{a}) \cdot \mathbf{n} = (\nabla \phi \times \mathbf{a})\cdot \mathbf{n} = (\mathbf{n} \times \nabla \phi) \cdot \mathbf{a}$$

substituting this into the theorem above gives,

$\int_{C} \phi \cdot \mathbf{a} dr=\iint_{S} (\mathbf{\hat{n}}\times\nabla\phi)\cdot\mathbf{a}dS$

Then taking out the constant a

$\mathbf{a}\cdot(\int_{C}\phi dr)=\mathbf{a}\cdot(\iint_{S}\mathbf{\hat{n}}\times(\nabla\phi)dS) $

since a is arbitrary we deduce that:
$\int_{C}\phi dr = \iint_{S}n \times (∇\phi )dS$

答:
Hint:

If  $\mathbf{a}$ is an arbitrary constant vector, then

$$\nabla \times (\phi \mathbf{a}) \cdot \mathbf{n} = (\nabla \phi \times \mathbf{a})\cdot \mathbf{n} = (\mathbf{n} \times \nabla \phi) \cdot \mathbf{a}$$


问:
Is the union axiom really needed to prove existence of intersections?
I once asked a while ago how to prove that the intersection $\bigcap S$ of any non-empty set $S$ exists, using the ZFC axioms. The answer involved the union axiom. However, I don't think it is necessary. Since the intersection of $S$ is a subset of an element $s$ of $S$, I think all that is needed is the axiom schema of separation. Is this true? Or is the union axiom essential?

答:
If $\mathcal{C}$ is a non-empty collection of sets we can pick $C_0 \in \mathcal{C}$ (by that non-emptyness) and define

$$\bigcap \mathcal{C} := \{x \in C_0\mid \forall C \in \mathcal{C}: x \in C\}$$

purely from separation. You might want to show too that the resulting set does not depend on the set $C_0$ we choose first. The union axiom is only used to avoid that last extra proof I think.


答:
Yes, you can do this. However, there is a nice uniformity aspect to the union-based argument which the union-free argument lacks, namely that it does not involve a choice at any step. In the union-free argument we need to pick some element of $S$ to "frame" our intersection. Of course the choice doesn't matter, and first-order logic lets us perform such a choice in the first place, but for some people there's still something annoying about this.

This can be brought out more clearly by an "algebraic" interpretation of the situation. Imagine an expansion of the language of set theory by some new operations corresponding to the various axioms, including:

a unary function for union $f_{union}$ with the intended meaning that $$f_{union}(x)=\bigcup x,$$ and

for each formula $\varphi(x_1,...,x_n,y)$ in the language of set theory with only the displayed free variables, an $(n+1)$-ary separator function $g_{sep, \varphi}$ with the intended meaning that $$g_{sep,\varphi}(a_1,...,a_n,b)=\{c\in b: \varphi(a_1,...,a_n,c)\}.$$

The union-based argument lets us construe the intersection map $x\mapsto \bigcap x$ as a term $\tau$ in this language: it's just $$x\mapsto g_{sep,\psi}(x, f_{union}(x))$$ where $\psi(x,y)\equiv\forall z\in x(y\in z)$.

By contrast, I believe (the argument is pretty tedious sadly) that the union-free argument does not yield such a term, and indeed if we drop the function corresponding to union we can no longer represent intersection as a term. in the resulting more limited language.

So the union-based argument has genuine content that the union-free argument lacks, but this content only shows up when we consider expanded contexts (= set theory + some particular additional operations).

Incidentally, note that $\tau(x)$ is defined even when $x$ is empty: we have $\tau(\emptyset)=\emptyset$. We don't run into the "$\bigcap \emptyset=V$"-issue since we've trapped everything inside the specific set $f_{union}(\emptyset)$. This is kind of cute!

Title: Is the union axiom really needed to prove existence of intersections?

Body: I once asked a while ago how to prove that the intersection $\bigcap S$ of any non-empty set $S$ exists, using the ZFC axioms. The answer involved the union axiom. However, I don't think it is necessary. Since the intersection of $S$ is a subset of an element $s$ of $S$, I think all that is needed is the axiom schema of separation. Is this true? Or is the union axiom essential?


问:
Local connection of closure of open connected subspaces in some metric spaces.
Let $X$ be a metric, connected, locally connected and locally compact space, and let $U$ be an open connected subspace of $X$. Is $\overline{U}$ always locally connected?

I´m writing a thesis about Peano spaces and this would be useful if it were true, but I haven´t been able to prove it or find any counterexamples.

答:
Expanding on the previous answer, it's not true even if $U$ is connected and bounded and $X$ is a continuum.  Just let $X$ be a large, closed disc in the plane and let $U$ be a tubular neighborhood of sin$(\frac{1}{x})$ near $0$.

The spaces you want are the hlc spaces.  Look at examples of Peano curves that aren't hlc for more sophisticated obstructions.


问:
Linear equation of 4 variables
I'm stuck on this Math problem :

How many solutions does the equation

$x_{1} + x_{2} + 3x_{3} + x_{4} = k$

have, where $k$ and the $x_{i}$ are non-negative integers such that $x_{1} \geq  1$, $x_{2} \leq  2$, $x_{3} \leq 1$
and $x_{4}$ is a multiple of 6.

I tried to write the possible cases for $x_{2}$ and $x_{3}$ since they are bounded.

Case $x_{2} = 2$ :

Case $x_{3} = 1$ : $x_{1} + 5 + x_{4} = k$

Case $x_{3} = 0$ : $x_{1} + 2 + x_{4} = k$

Case $x_{2} = 1$ :

Case $x_{3} = 1$ : $x_{1} + 4 + x_{4} = k$

Case $x_{3} = 0$ : $x_{1} + 1 + x_{4} = k$

Case $x_{2} = 0$ :

Case $x_{3} = 1$ : $x_{1} + 3 + x_{4} = k$

Case $x_{3} = 0$ : $x_{1} + x_{4} = k$

But now, I'm stuck. Should I try to resolve all this equations ? Am I in the right direction ? Any help would be grealty appreciated.

EDIT :

I found that the number of solutions will be given by the coefficient $a_{k}$ of $x_{k}$, i.e :

$(x^0+x^1+x^2)(x^0+x^1)(\displaystyle\sum\limits_{k=1}^\infty x^k)(\displaystyle\sum\limits_{k=0}^\infty x^{6k})$

$=(x^0+x^1+x^2)(x^0+x^1)(\frac{1}{(1-x^3)(1+x^3)})(\frac{x}{1-x})$

$=\frac{x+2x^2+2x^3+x^4}{(x^3-1)(x^3+1)(x-1)}$

Now I don't know how to find the $a_{k}$ coefficient.

答: This will give you a start: it's the coefficient of $x^k$ in $$(x+x^2+x^3+\dots)(1+x+x^2)(1+x^3)(1+x^6+x^{12}+\dots)$$</p>

ContentLicense: CC BY-SA 3.0

Title: Linear equation of 4 variables


问:
Shorthand notation for partial?
If I am taking a regular derivative, and I want to show the process in detail, I'll do something of the sort $f'(x) = g'(x) + h'(x) - l'(x) ..... $, etc, using that "prime" notation. </p>

<p>However, what if I wanted to take the partial derivative? Would I still use the prime notation, as long as it's clear what we are treating as a constant? Or is there some other notation for this? </p>

答:
You have a few options. As noted by others $\frac{\partial f}{\partial x}$ and $f_x$ both commonly express the derivative of $f$ with respect to $x$.

Another option is $D_x f$. To explain, $D$ is commonly notation for the differential operator, the function that maps a function to its derivative. For instance, $D(f) = f'$. Then adding the subscript indicates taking a partial derivative, so $D_x(f)=\frac{\partial f}{\partial x}$. Writing it out we can often omit the parentheses to make the notation even lighter, so we get $D_x f$ denoting the partial derivative of $f$ with respect to $x$.

IMO, the notation $\frac{\partial f}{\partial x}$ is universally recognized but cluttered. The notation $f_x$ is very compact but isn't always obvious. So I like the notation $D_x f$, it reinforces the idea that we aren't referring to $f$ itself but its related partial derivative.

Title: Shorthand notation for partial?


问:
Why a matrix with normalized vectors whose elements are iid Gaussian satisfy restricted isometry property.
On page 23, section 4.1, the well-known paper by Candes et al. lays out the way of constructing a matrix $A$ satisfying the restricted isometry property as follows:

Sampling from Gaussian distribution independently for each entry of a column to create that column and then normalizing the column to be the column of $A$.

Question: Why does such a procedure guarantee that the isometry property is preserved for matrix $A$? Also, what would be the restricted isometry constant $\delta_s$?

Definition:  Let $\mathbf{A}$ be an $m\times p$ matrix and let $1 \leq s \leq p$ be an integer. Suppose that there exists a constant $\delta _{s}\in (0,1)$ such that  every $s$-sparse vector $\mathbf{v}$ satisfies

\begin{align}
    (1-\delta _{s})\|\mathbf{v}\|_2^2 \leq \|\mathbf{A}\mathbf{v}\|_2^2 \leq (1+\delta _{s})\|\mathbf{v}\|_2^2
\end{align}

Then, the matrix $\mathbf{A}$ is said to satisfy the $s$-restricted isometry property with restricted isometry constant $\delta_s$.

答:
Note that your matrix $A$ satisfies
$$
A \overset{(d)}{=} \frac{1}{\sqrt{m}} \tilde{A},
$$
where $\tilde{A}$ is a matrix with zero-mean independent columns $A_{:, i}$ satisfying $\| A_{:, i} \|_2 = \sqrt{m}$. By Theorem 5.65 in this paper [1], such matrices are restricted isometries and satisfy with high probability:
$$
(1 - \delta_{s}) \| x\|_2^2 \leq \| Ax \|_2^2 \leq (1 + \delta_{s}) \|x \|_2^2, \;  \quad \text{simultaneously over all $s$-sparse $x$}.
$$
Here, we can drive $\delta_{s}$ arbitrarily close to $0$ by choosing $
m \geq \frac{C}{\delta_s^2} s \log(e p / s),
$
where $C$ is a constant.

[1]: Roman Vershynin, Introduction to the non-asymptotic analysis of random matrices (2011 version).


问:
Is this a Viable/New Proof for Pythagoras Theorem?
A triangle with side lengths $a$, $b$, $c$ and a height ($h$) that intercepts the hypotenuse ($c$) such that $c$ is split into two side lengths, $c = m + n$, we can find Pythagorean theorem using simple trigonometry.

We start by finding the sum of tan ratios that are equivalent,

$$\tan^{-1}\frac{a}{b}+\tan^{-1}\frac{b}{a} = 90^\circ$$
$$\tan^{-1}\frac{h}{m}+\tan^{-1}\frac{h}{n} = 90^\circ$$
$$\frac{a}{b}+\frac{b}{a}=\frac{h}{m}+\frac{h}{n}$$

Using $ab = hc$, Substitute $h = \dfrac{ab}{c}$,

$$\frac{a}{b}+\frac{b}{a}=\frac{ab}{cm}+\frac{ab}{cn}$$

After simplifying, we get:

$$\begin{align}
c&=\frac{ma^2b^2+na^2b^2}{mn\left(a^2+b^2\right)} \\[4pt]
&=\frac{(m + n)(a^2b^2)}{mn\left(a^2+b^2\right)} \\[4pt]
&=\frac{c(a^2b^2)}{mn\left(a^2+b^2\right)} \\[4pt]
mn &=\frac{a^2b^2}{\left(a^2+b^2\right)}
\end{align}$$

Note that $\sqrt{mn}=\dfrac{ab}{\sqrt{a^{2}+b^2}}$ and $h = \dfrac{ab}{c}$, this tells us that $\sqrt{mn} = h$.

$$\begin{align}
\frac{ab}{c}&=\frac{ab}{\sqrt{a^2+b^2}} \\[4pt]
c^2 &= a^2 + b^2
\end{align}$$

答:
This amounts to a well-known similar-triangles argument, splitting the area, that $(a/c)^2+(b/c)^2=1$, which is a bit more direct.

As @DavidK notes, your opening use of trigonometry is invalid, and its conclusion (in fact, something stronger) should instead be deduced as per @WillOrrick's suggestion, giving $\tfrac{a}{b}=\tfrac{h}{m}$ and $\tfrac{b}{a}=\tfrac{h}{n}$, whence $1=\tfrac{h^2}{mn}=\tfrac{(ab)^2}{mnc^2}$, and the Pythagorean theorem is equivalent to $mn=\tfrac{(ab)^2}{a^2+b^2}$. You also use $\tfrac{a^2+b^2}{ab}=\tfrac{hc}{mn}=\tfrac{ab}{mn}$, which completes the proof.

So your strategy amounts to combining two formulae for area with$$u_1=u_2,\,v_1=v_2,\,u_1v_1=1\implies u_1+v_1=\tfrac{u_2+v_2}{u_2v_2}$$(where $u_1=v_1^{-1}=\tfrac{a}{b},\,v_1=\tfrac{h}{m},\,v_2=\tfrac{h}{n}$). But compare $\tfrac{a}{b}=\cot\theta$ with$$\left(\tfrac{a}{c}\right)^2=\tfrac{[\triangle CBH]}{[\triangle ABC]}=\tfrac{m}{m+n}=\tfrac{m/h}{v_1^{-1}+v_2^{-1}}=\tfrac{\cot\theta}{\cot\theta+\tan\theta}.$$Similarly (viz. $\theta\mapsto\tfrac{\pi}{2}-\theta$), $\left(\tfrac{b}{c}\right)^2=\tfrac{\tan\theta}{\cot\theta+\tan\theta}$. The proof technique I pointed to - that of seeing the Pythagorean theorem - notes the sum of these squares is $\tfrac{\color{blue}{blue}+\color{green}{green}}{\text{all}}=1$. Your proof works with fractions $k:=\cot\theta+\tan\theta$ times these, so Einstein's proof is equivalent to$$1=\tfrac{a/b+b/a}{k}=\tfrac{a^2+b^2}{kab}=\tfrac{a^2+b^2}{kch},$$where the last step uses your equation of two formulae for $[\triangle ABC]$, and the proof that $\tfrac{a^2+b^2}{kch}=1$ uses $kh=m+n=c$ by your $AB$-splitting observation that defines $m,\,n$.


问:
Decomposition of a permutation group into a direct product

Let $S_6$ be the symmetric group of a set of six elements. Let $\sigma=(1,3,4,6)$ and $\tau=(1,3,4,6)(2,5)$ be two cycles in $S_6$.

(A) Determine the order of the two permutations and the order of $G:=\langle\sigma,\tau\rangle$.

(B) Determine all the non trivial possible decompositions of $G$ in a direct product of its subgroups.

I wrote that $o(\sigma)=4$ and $o(\tau)=lcm(4,2)=4$. Since $(1,3,4,6)\cdot(1,3,4,6)\cdot(2,5)=(1,3,4,6)^2\cdot(2,5)=$

$(1,3,4,6)\cdot(2,5)\cdot(1,3,4,6)\implies$ $\sigma$ and $\tau$ are permutable, hence $\langle\sigma\rangle\langle\tau\rangle$ has elements of form $\sigma^h\tau^k$.

This means that $\langle\sigma\rangle\langle\tau\rangle=\langle\sigma,\tau\rangle\le G$ (I think that in general $\langle\sigma,\tau\rangle$ has the same form of $\langle\langle\sigma\rangle\langle\tau\rangle\rangle$).
$$|\langle\sigma,\tau\rangle|=\dfrac{|\langle\sigma\rangle|\cdot|\langle\tau\rangle|}{|\langle\tau\rangle\cap\langle\sigma\rangle|}=\dfrac{4\cdot4}{2}=8.$$
I state this because $\langle\sigma\rangle=\{id,(1,3,4,6),(1,4)(3,6),(1,6,4,3)\}$ and $\langle\tau\rangle=\{id,(1,3,4,6)(2,5),(1,4)(3,6),(1,6,4,3)(2,5)\}\implies$ the intersection contains $id$ and $(1,4)(3,6).$

I think that in order to classify the type of decomposition of $G$ I have to look at the normal subgroups of $G$ whose intersection is trivial but I always find an element in the intersection... What did I miss?

答:
The subgroup generated by $\tau$ is cyclic of order $4$; the subgroup generated by $\sigma$ is cyclic of order $4$; you correctly calculate (since $\tau$ and $\sigma$ commute) that $\langle \sigma,\tau\rangle=\langle\sigma\rangle\langle\tau\rangle$, and that the order is $8$.

This is an abelian group. As such, it must be (abstractly) isomorphic to either $C_8$, $C_4\times C_2$, or $C_2\times C_2\times C_2$. The latter has no subgroups of order $4$, so that can't be it. The first has a unique subgroup of order $4$, so that can't be it, either. So that means that $G$ is abstractly isomorphic to $C_4\times C_2$. As such, it definitely has direct product decompositions.

Being abelian, every subgroup is normal. So you are just looking for subgroups with trivial intersection, one of order $4$ and one of order $2$ (those are really the only possibilities, because of the structure theorem for finite groups).

For instance, can we find a $2$-element subgroup $K$ such that $G=\langle \tau\rangle \times K$? Note that $(2,5)=\tau\sigma^{-1}\in G$. What is the intersection of $\langle (2,5)\rangle$ with $\langle \tau\rangle$? With $\langle \sigma\rangle$?  Can we find other subgroups of order $4$ and decompositions?

If you find it difficult to it directly with the permutations, find an explicit isomorphism to $C_4\times C_2$ and work in the latter, then translating back to $G$.


问:
Calculating Probability of getting All Options at least once, with Extra Attempts
I was looking at this question about "Expected time to roll all 1 through 6 on a die" where it was established that the average number of times needed to accomplish this would be 14.7

I understand that that is the average amount of time, but I don't understand how to expand that since the logic doesn't utilize things like combinations or permutations. How would you determine this variant?

"What is the chance that I roll every side on a six-sided dice at least once if I roll it 10 times?"

I'm not certain how to approach this, but I feel like I'm just missing something simple. Should I be using permutations? Or another approach?

答:
For $i=1, \dots, 6$, let $A_i$ be the event "in the first $10$ rolls, we never roll $i$". Then $\Pr[A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5 \cup A_6]$ can be expanded out using the inclusion-exclusion principle as:

the sum of $\Pr[A_1]$ through $\Pr[A_6]$, each of which is $(\frac56)^{10}$;

minus the sum of all $\binom 62 = 15$ probabilities $\Pr[A_i \cap A_j]$, each of which is $(\frac46)^{10}$;

plus the sum of all $\binom 63 = 20$ probabilities $\Pr[A_i \cap A_j \cap A_k]$, each of which is $(\frac36)^{10}$;

minus the sum of all $15$ probabilities $\Pr[A_i \cap A_j \cap A_k \cap A_l]$, each of which is $(\frac26)^{10}$;

plus the sum of all $6$ probabilities $\Pr[A_i \cap A_j \cap A_k \cap A_l \cap A_m]$, each of which is $(\frac16)^{10}$: at this point we're down to cases where we roll only a single number.

minus the probabilitiy of $\Pr[A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5 \cap A_6]$, but that's $0$: at least one value has to be rolled.

This gives us a value of
$$
   6 \left(\frac56\right)^{10} - 15 \left(\frac46\right)^{10} + 20 \left(\frac36\right)^{10} - 15 \left(\frac26\right)^{10} + 6 \left(\frac16\right)^{10} = \frac{101923}{139968}
$$
for $\Pr[A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5 \cup A_6]$, which is the probability that we don't see all six outcomes. So the probability we do see all six outcomes is $1 - \frac{101923}{139968} = \frac{38045}{139968} \approx 0.272$.

It's a complete coincidence that this probability is within $0.006\%$ relative error of $\frac{e}{10}$.


答:
Here is a formulation which gives a role to the average velocity. Let $v_{avg}:=\dfrac{\Delta x}{\Delta t}=\dfrac{x-x_0}{t}$ be the average velocity from time $0$ to $t$. Since this is constant-acceleration motion, we may use the kinematic equation for position to write
$$v_{avg} = \frac{v_0 t+\frac12 at^2}{t}=v_0+\frac{at}{2}.$$
By contrast, the kinematic equation for velocity gives the final velocity as $v=v_0+at$. As such, we may write $v_{avg}=(v+v_0)/2$---that is, the average velocity at constant acceleration is also the average of the initial and final velocities. Invoking constant acceleration once more, we have

$$a=\frac{\Delta v}{\Delta t}=\frac{\Delta v}{\Delta x}\frac{\Delta x}{\Delta t} = \frac{\Delta v}{\Delta x}v_{avg}=\frac{v-v_0}{x-x_0}\frac{v+v_0}{2}=\frac{v^2-v_0^2}{2(x-x_0)}$$
which is the desired equation. (Note how closely this parallels the use of the chain rule, which would apply even if $a$ weren't constant.)

Title: Getting the kinematic equation $v^2=v_0^2+2a(x−x_0)$

Body:
As I've searched through the internet, looking for explanation. I found an answer with transformation of this formula:

$$
x = x_0 + v_0t + \frac{1}{2}at^2 ⟹ x = x_0 + \frac{1}{2}a(t + \frac{v_0}{a})^2 - \frac{v^2_0}{2a}
$$
I don't understand how to get $\frac{1}{2}a(t + \frac{v_0}{a})^2$
Can anyone help please?


问:
countability, intermediate values, and Cantor's sizes of infinity
I understand that there is a difference in the "size" of infinities describing integers and real numbers - formalised by Cantor.

Countability

One way this is explained in tutorials is that the integers are countable, whereas the real numbers are not.

By countability, I understand that there is a 1-to-1 correspondence with the integers. This is why the even numbers have the same size of infinity as the integers themselves.

Intermediate Values

A related concept is the ability to find a value between any two others. For real numbers, I can always find a value between two close values, say 1.001 and 1.002, and this means I can't establish a 1-to-1 correspondence with the integers.

For integers, I can't always find an intermediate value. For example, there is no integer between 2 and 3.

(I don't know the technical term for the existence of in-between-values - continuity? compactness?)

Question

This question is asking for conceptual clarity.

Does countability - eg of integers - imply the first Cantor infinity $\aleph_0$?

Does the ability to find intermediate values - eg for real numbers - imply the second Cantor infinity $\aleph_1$?

I read that the rational $p/q$ are countable. However I can always find an intermediate value between any two different rationals. Does that mean the size of infinity is different to the above two?

I am not a trained mathematician so would appreciate responses that minimise use of terminology.

答: $\Bbb Q$ is a dense order (as the in-between property is named) but still countable (it has a bijection with $\Bbb Z$ or $\Bbb N$) so being a dense order or not is independent of the cardinality $\aleph_0$ or continuum. A bijection need not preserve order and cardinality is just determined by bijections. The fact that $\Bbb R$ has a dense order is not the reason no bijection with $\Bbb Z$ exists. It's more subtle.


问:
Fulton-Harris Lemma 3.35
In the proof of Lemma 3.35 in Fulton--Harris, Representation Theory, it is claimed that the identification $H(\phi^2(x),y)=H(x, \phi^2(y))$ implies that $\lambda$ is a positive real ($\phi^2$ is known to act by a scalar $\lambda$).  I see why $\lambda$ must be real, but I do not see why $\lambda$ must be positive.

I googled this question, and I found that Noam Elkies came across the same subtlety, and resolved it by saying that $H$ can be assumed to be positive-definite, after which it follows that $H(\phi^2(x),x) = H(\phi(x), \phi(x))$, so $$\frac{H(\phi^2(x),x)}{H(x,x)}$$ is positive.  I do not however understand why $H$ can be assumed positive-definite, and I suspect that the reason would involve the same manipulations necessary to prove the desired result in general.

Anyone care to help out?

答:
To elaborate on Alex-omsk's comment, you can construct a $G$-invariant Hermitian form on a representation $V$ by "averaging" over $G$. (This is done in Fulton and Harris in the proof of Proposition 1.5.) That is, for any choice of Hermitian form $H_0$ on the underlying vector space $V$, set

$$
H(v,w) = \sum_{g \in G} H_0(gv,gw).
$$

Then $H$ is a $G$-invariant Hermitian form on $V$. Now you can check that if we choose a form $H_0$ that is positive definite, then $H$ is positive definite as well.

Title: Fulton-Harris Lemma 3.35


问:
Analytic map from punctured disk to punctured plane
If $\mathbb{D}$ is the open unit disk, we know from Liouville's theorem that there cannot exist a 1-1 and onto analytic map (i.e. biholomorphic) from $\mathbb{D}\to\mathbb{C}$? But could there exist one from $\mathbb{D}\setminus\{0\}\to\mathbb{C}\setminus\{0\}$?</p>

答:
I think there is a faster way to prove that. There is an equivalent proposition for removable singularities.

Proposition: Let $f$ be a holomorphic function defined on an open set $G$ that has an isolated singularity $z_0$. Then $f$ is bounded on $ D(z_0,r) \backslash \{z_0\} \subset G$,  for some  $r \iff z_0$ is removable singularity of $f$.

So, let a holomorphic $f: \mathbb{C} \backslash \{0\} \to \mathbb{D} \backslash \{0\}$

Since the range is bounded, as a subset of $\mathbb{D} $, $f$ must have a removable singularity.

Thus $f$ can be analytically extended to $\mathbb{C}$. Being both Entire and Bounded, by Liouville's theorem must be constant.

So, $f$ can only be constant.


答:
Yes, $|\Bbb N|=\aleph_0$. There is a bijection (one-to-one correspondence) between $\Bbb N$ and $\Bbb Q$, so $|\Bbb Q|=\aleph_0$ as well; the usual order on $\Bbb Q$ is simply not relevant to the question of its cardinality. Indeed, if $h:\Bbb N\to\Bbb Q$ is a bijection, we can define a new order on $\Bbb N$ that makes it look just like $\Bbb Q$: for $m,n\in\Bbb N$ let $m\preceq n$ if and only if $h(m)\le h(n)$. Now we have an ordering of $\Bbb N$ with respect to which we always can find a natural number between any two natural numbers. Given $m,n\in\Bbb N$ with $m\prec n$, we know that $h(m)<h(n)$. Let $q=\frac12\big(h(m)+h(n)\big)$; then $q\in\Bbb Q$, and $h(m)<q<h(n)$. The map $h$ is a bijection, so there is a natural number $k$ such that $h(k)=q$, so $m\prec k\prec n$.

It is independent of the usual axioms of set theory whether $|\Bbb R|=\aleph_1$: it is consistent with them that $|\Bbb R|=\aleph_1$, but it is also consistent with them that $|\Bbb R|=\aleph_2$, $|\Bbb R|=\aleph_3$, or indeed that $|\Bbb R|=\aleph_n$ for any positive integer $n$, and there are infinitely many other even larger possibilities.

Title: countability, intermediate values, and Cantor's sizes of infinity

Body:
I understand that there is a difference in the "size" of infinities describing integers and real numbers - formalised by Cantor.

Countability

One way this is explained in tutorials is that the integers are countable, whereas the real numbers are not.

By countability, I understand that there is a 1-to-1 correspondence with the integers. This is why the even numbers have the same size of infinity as the integers themselves.

Intermediate Values

A related concept is the ability to find a value between any two others. For real numbers, I can always find a value between two close values, say 1.001 and 1.002, and this means I can't establish a 1-to-1 correspondence with the integers.

For integers, I can't always find an intermediate value. For example, there is no integer between 2 and 3.

(I don't know the technical term for the existence of in-between-values - continuity? compactness?)

Question

This question is asking for conceptual clarity.

Does countability - eg of integers - imply the first Cantor infinity $\aleph_0$?

Does the ability to find intermediate values - eg for real numbers - imply the second Cantor infinity $\aleph_1$?

I read that the rational $p/q$ are countable. However I can always find an intermediate value between any two different rationals. Does that mean the size of infinity is different to the above two?

I am not a trained mathematician so would appreciate responses that minimise use of terminology.


问:
is $(x+1)^4-x^4$ non-prime for all natural positive integers $x$
Looking at difference between two neighbouring positive integers raised to the power 4, I found that all differences for integer neighbours up to $(999,1000)$ are non-prime.

Does this goes for all positive integers?
And can someone please prove?

答: Hint: it is a difference of squares. Alternatively, by the Factor Theorem, we deduce that the polynomial difference $\ f(x+1) - f(-x)\ $ is divisible by $\ x+1 - (-x) =\, 2x+1.$


问:
Integral of $f(x)f'(x)$
I found a problem that says:

Let $f$ be a real function with continuous derivative on $[a , b]$ such that $f(a)=f(b)=0$. Suppose $\int_a^b f(x)^2 dx =1$. Show that $\int_a^b f(x)f'(x) dx = -\frac{1}{2}$.

The suggested solution was to do integration by parts using $u=f(x)^2$ and $dv=dx$ which leads to
$1=\int_a^b f(x)^2 dx = 2xf(x)^2 |_a^b - \int_a^b 2xf(x)f'(x) dx$. Then since $f(a)=f(b)=0$ that makes the first part go away and we get $1=- \int_a^b 2xf(x)f'(x) dx$ then we can make the conclusion.

But what am I missing? We showed that $\int_a^b xf(x)f'(x) dx = -\frac{1}{2}$ but the problem wanted to show $\int_a^b f(x)f'(x) dx = -\frac{1}{2}$. Was this just a typo?

答:
Yes, it was a typo.

Note that $(f^2(x))'=2f(x)f'(x)$. Then $$\int_a^bf(x) f'(x) dx=\frac12 (f^2(b) -f^2(a))=0.$$

Title: Integral of $f(x)f'(x)$


答: The question is wrong. You don't even have to integrate by parts to see that $\int_a^{b} f(x)f'(x)dx=0$: $\int_a^{b} f(x)f'(x)dx=\int_a^{b} \frac d {dx} (\frac 1  2(f(x)^{2}) dx=\frac 1  2(f(x)^{2})|_a^{b}=0-0=0$.

Title: Integral of $f(x)f'(x)$

Body:
I found a problem that says:

Let $f$ be a real function with continuous derivative on $[a , b]$ such that $f(a)=f(b)=0$. Suppose $\int_a^b f(x)^2 dx =1$. Show that $\int_a^b f(x)f'(x) dx = -\frac{1}{2}$.

The suggested solution was to do integration by parts using $u=f(x)^2$ and $dv=dx$ which leads to
$1=\int_a^b f(x)^2 dx = 2xf(x)^2 |_a^b - \int_a^b 2xf(x)f'(x) dx$. Then since $f(a)=f(b)=0$ that makes the first part go away and we get $1=- \int_a^b 2xf(x)f'(x) dx$ then we can make the conclusion.

But what am I missing? We showed that $\int_a^b xf(x)f'(x) dx = -\frac{1}{2}$ but the problem wanted to show $\int_a^b f(x)f'(x) dx = -\frac{1}{2}$. Was this just a typo?


答: Unlike countability, "intermediate values" is not a concept that applies to a set directly, but only when an order structure is applied on top of a set.  It is possible to order the natural numbers in a way where intermediate values can be found, and it is possible to order the rational numbers in a way where intermediate values cannot be found.</p>


问:
How many zeros does a radical equation (eg, $X^{4/3}-5X^{2/3}+6=0$) have?
I want to know if there is a general rule that will give me the answer.
I'm not talking about crazy expressions under a radical, just simple
variables raised to a fractional exponent like:

$$X^{4/3}-5X^{2/3}+6=0$$

答: For a case like this we can define $y=x^{\frac 23}$ and make the equation $y^2-5y+6=0$ which has two roots in $y$.  Then you can have either square root, so there will be four zeros.  You need to carefully specify the range of equations that is acceptable to get any simple rule.


问:
Well definedness of the correspondence between connections on $B$ and sections of $J^1B$
Let $M$ be a smooth manifold and $\pi\colon B\to M$ a fiber bundle on $M$. A (Ehresmann) connection on $B$ is a subbundle $H$ of $TB$ such that $TB=H\oplus V$, where $V$ is the bundle of vertical vectors.

In different places I've seen claimed that there is a correspondence between connections on $B$ and sections of $J^1B$, in particular the book I'm looking at provides the following recipe to construct a section $\sigma\colon B\to J^1B$ of the jet bundle given a connection $H$:

Consider $b\in B$ and pick a local section $\sigma_b\colon U\subseteq M\to B$ such that $\sigma_b(x)=b$ and $\mathrm{Im}(T_x\sigma_b)=H_b$, then set $\sigma(b)=j^1_x\sigma_b$.

The book claims that this is obviously well defined, meaning that if I pick two different local sections $\sigma_b,\rho_b$ with $\sigma_b(x)=b=\rho_b(x)$ and $\mathrm{Im}(T_x\sigma_b)=H_b=\mathrm{Im}(T_x\rho_b)$, then $j^1_x\sigma_b=j^1_x\rho_b$, but I don't see why that's the case.

Obviously the two sections have tangent maps in $x$ with equal images, but to have the same jet I need all derivatives in local coordinates to agree in $x$, which is a stronger requirement, hence my question: why is this map well defined?

答:
Suppose you have two local sections $\sigma^1, \sigma^2 : U \subseteq M \to B$ satisfying $\sigma^i(x) = b$ and $D_x\sigma^i(T_xM) = H_b$. Since $\sigma^1$ and $\sigma^2$ are both sections of the bundle given by $\pi : B \to M$, they are injective, and so they induce isomorphisms $D_x\sigma^i : T_xM \to H_b$ (counting dimensions), which we can turn into an automorphism $(D_x\sigma^2)^{-1} \circ (D_x\sigma^1)$ of $T_xM$. The question now is, why is this automorphism the identity. In some sense, it's because $H_b$ already encodes how $\sigma^i(x)$ may vary as $x$ varies; since $x$ can only vary in $M$, $\sigma^i(x)$ cannot have a derivative along the direction of the fiber, so its derivative is guided by $H$. This is also why you should have $H$ as a vector bundle on $B$; the sections need a derivative regardless of their value!

You just need the following lemma: assume $m \le n$ are positive integers and $f,g : \mathbb R^m \to \mathbb R^n$ are injective linear maps for which $f(\mathbb R^m) = g(\mathbb R^m)$ and there exists $h : \mathbb R^n \to \mathbb R^m$ linear with $h \circ f = h \circ g$. Then $f=g$. The proof is trivial; we can assume $m=n$ without loss of generality just by replacing $\mathbb R^n$ by $f(\mathbb R^m) = g(\mathbb R^m) \simeq \mathbb R^m$, in which case it is trivial since $f$ and $g$ have the same inverse. Apply this lemma to the sections $\sigma^i : U \to B|_U$ satisfying $\pi|_U \circ \sigma^i = \mathrm{id}_U$, which implies $D_b\pi \circ D_x\sigma^i = \mathrm{id}_{T_xM}$, and you will deduce $D_x\sigma^1 = D_x\sigma^2$.

You can also see it in a less coordinate-patchy way as follows: you define the bundle of vertical vectors as the kernel of the morphism of vector bundles $V = \ker(TB \to \pi^* TM)$ on $B$. Vertical vectors go along the direction of $b \in B$ when $b$ varies along a fiber, since then $\pi(b)$ does not vary. An Ehresmann connection satisfies $TB = H \oplus V$ as vector bundles on $B$, so in particular $H \simeq TB/V \simeq \mathrm{im}(TB \to \pi^* TM) \simeq \pi^*TM$. So an Ehresmann connection is really just a global version of the locally available isomorphism $TB \simeq \pi^*TM \oplus V$ (which exists by trivializing all the bundles involved), and it is therefore not surprising that $\pi^*TM$ (and its global counterpart $H$) and $J^1B$, being two vector bundles on $B$ corresponding to derivatives on $M$, are closely related.

Hope that helps,


问:
Prove that $T$ is an isometric isomorphism of $M^\perp$ onto $H/M$
I am interested in solving Exercise 2.4 from this set of notes I found online. The exercise reads:

Exercise 2.4: Let $M$ be a closed subspace of a Hilbert space, $H$. Define a map $T:M^\perp\to H/M$ by $T(x)=x+M$. Prove that $T$ is an isometric isomorphism of $M^\perp$ onto $H/M$.

I have included my attempt below, and am looking for help regarding the parts I have not been able to  complete. I would also appreciate feedback on tidying up my attempt so far - is there anything which can be better explicated? Or anything that I have missed out? I am quite new to dealing with quotients of Hilbert spaces and things like cosets; is there a standard reference in which quotient space results like this are proven in full?

My attempt: I need to show that (1) $T$ is linear, (2) that $T$ is an isometry and (3) that $T$ is onto. I understand that if $T$ is linear, then additionally being an isometry yields injectivity. Hence, we then only need to show that $T$ is surjective. I show (2) last, since this is the part I am currently stuck on.

(1) Let $f,g\in M^\perp$. Then
$$T(f+g)=(f+g)+M=(f+m)+(g+M)=T(f)+T(g),$$
by the definition of addition of cosets in $H/M$. Thus, $T$ is linear.

(3) By the Hilbert space decomposition of $H=M\oplus M^\perp$, for any $x\in H$ there exists a unique $u_x\in M$ and $v_x\in M^\perp$ such that $x=u_x+v_x$. With this in mind, let $x+M\in H/M$ with $x\in H$ arbitrary. Since the Hilbert space decomposition is unique, we can identify this $x$ by it's corresponding $v_x\in M^\perp$ part in the decomposition. This proves the surjectivity of $T$, since we can map to any $x+M\in H/M$ by $v_x\in M^\perp$.

(2) Let $f\in M^\perp$. Then
$$\|T(f)\|_{H/M}=\|f+M\|_{H/M}=\inf_{m\in M}\|f-m\|_H\le\|f-0\|_H\le\|f\|_H,$$
since $M$ is a linear subspace and contains in particular $m=0$. As for the other direction, I am not so sure. I found the question $Q:\mathbb{M}^{\perp}\rightarrow \mathbb{H}/\mathbb{M}$ is an isometric isomorphism and noticed the initial comment given there. I understand how Pythagoras' Theorem has been used together with the Hilbert space decomposition I used in (3) above. But what I don't understand is how, exactly, it is used with the norm on the quotient space. Being unfamiliar with cosets and quotient spaces, I am also not entirely sure how to translate it to the form of the norm I have here.

Could somebody help me to bridge the gap here, please? How do I translate from the ideas sketched there, to the scenario as I have phrased it here?

答:
If $x \in M$ and $y \in M^{\perp}$ then $\|x+y\|^{2}=\|x\|^{2}+\|y\|^{2}$. Hence, $\|x+M\|^{2}=\inf \{\|x+y\|^{2}: y \in M\} \geq \|x\|^{2}$. This proves that $\|Tx|| \geq \|x\|$.

[I have just used the definition of the norm in $X/M$ and the fact that if each eelment of a set of real numbers is greater than or equal to $c$ then the infimum of the set is greater than or equal to $c$].


问:
I need math help in calculating the variance, to find variance when given range and mean
x is a random variable with range (2,3), mean=9/4 what is the variance? I need to know if what the variance is by only having the mean and range. Please help</p>

答:
You know $1 = p_2 + p_3$ and $9/4 = 2 p_2 + 3 p_3$, so you can figure out that $p_2 = 3/4$. Then it's a matter of computing the variance as
$(2 - 9/4)^2p_2 + (3 - 9/4)^2p_3$.


答:
There is no "general rule" for solving all such equations. But your equation is a quadratic equation in disguise. Such questions mostly appear in pre-calculus exercises.

A rule of thumb is that you can make your equation easier by introducing new variables.

Since $x^{m/n}=(x^{1/n})^m$, you can write
$$
(x^{4/3})=(x^{1/3})^4,\quad (x^{1/3})^2
$$
so that your equation becomes
$$
(x^{1/3})^4-5(x^{1/3})^2+6=0\ \tag{1}
$$

Now if you introduce a new variable $y=x^{1/3}$, your equation becomes
$$
y^4-5y^2+6=0\tag{2}
$$
which looks simpler since there is no "radical" (fraction powers) anymore.

But this is still not easy: it is of power $4$.

But if you observe that $y^4=(y^2)^2$, (2) is the same as
$$
(y^2)^2-5y^2+6=0\tag{3}
$$
where if you introduce another new variable $w=y^2$, you have a quadratic equation
$$
w^2-5w+6=0
$$
which can be solved easily since $w^2-5w+6=(w-2)(w-3)$.

Title: How many zeros does a radical equation (eg, $X^{4/3}-5X^{2/3}+6=0$) have?

Body:
I want to know if there is a general rule that will give me the answer.
I'm not talking about crazy expressions under a radical, just simple
variables raised to a fractional exponent like:

$$X^{4/3}-5X^{2/3}+6=0$$


问:
Existence and uniqueness of solutions to $Ax=b$ in terms of number of equations and number of variables
I roughly know that  $\textbf Ax=b$ where $\textbf A\in\mathbb R ^{m\times n}$ has the following properties:

Existence: $rank(A)=m$ means a solution exists (because the column space would span all of $\mathbb R^m$), $rank(A)<m$ means a solution might not exist

Uniqueness: $rank(A)=n \implies nullity(A)=0 \implies$ any solution is unique, and $rank(A)<n \implies nullity(A)>0 \implies$ any solution, if it exists, is not unique (because since $Ay=0$ for some nonzero y, and if x is a solution to Ax=b, then A(x+y)=Ax+Ay=b+0=b shows that x+y is also a solution).

This works  out and all, but is there a way to think about this in terms of the number of equations and number of variables in the system of equations? Like, for existence, rank(A)=m=the number of equations means a solution exists and for uniqueness, rank(A)=n=the number of variables means any solution is unique. I'm hoping there can be a more intuitive or simpler way to think about it.

答:
There is this general result:

The linear system $Ax=b$  has solutions if & only if the matrix $A$ and the augmented matrix $[A|b]$ have the same rank. If it is the case, this common rank is the codimension of the (affine) subspace of solutions.


问:
Truncation and $\Pi$-types (ii)
I just read Ximei's post Truncation and $\Pi$-types. Can anyone explain in simple terms why ($\star$) implies ($\star\star$) but not vice versa. In other words, I wonder why ($\star$) is stronger than ($\star\star$).

$$||\prod_{x:A}B(x)||\quad(\star)\quad\quad\quad\quad\prod_{x:A}||B(x)||\quad(\star\star)$$

Thanks!

答:
This fact actually corresponds in a very natural way with a classical fact of (Zermelo-Fraenkel) set theory: namely, that the axiom of choice is equivalent to the statement that a Cartesian product of any family of non-empty sets is itself non-empty. Now, in homotopy type theory, for any type $C$, the type $||C||$ should be thought of as containing precisely the information of whether $C$ is inhabited, and nothing more. In other words, there is a canonical map $C\to||C||$ that collapses all of the internal structure of $C$ except whether $C$ inhabited. If we read "inhabited" as "non-empty", then the statement $(\star)$ can be thought of as expressing that $\prod_{x:A}B(x)$ is non-empty, while the statement $(\star\star)$ can be thought of as expressing that $B(x)$ is non-empty for every $x:A$.

This gives us a very natural "classical" analogue of $(\star)$ in Zermelo-Fraenkel set theory, which we will denote $(\bullet)$; this is the assertion that a Cartesian product of family of sets $(B_x)_{x\in A}$, indexed by a set $A$, is non-empty. Similarly, the "classical" analogue of $(\star\star)$, which we will denoted $(\bullet\bullet)$, is the assertion that $B_x$ is non-empty for every $x\in A$. In ZF set theory, $(\bullet)$ always implies $(\bullet\bullet)$, since the Cartesian product of any family of sets that includes the empty set is itself empty. However, as stated above, the assertion that $(\bullet\bullet)$ implies $(\bullet)$ in general is equivalent to the axiom of choice. As you can see, this gives a very precise analogue of the situation with $(\star)$ and $(\star\star)$ in homotopy type theory. With this in mind, let me fill in the details of the proof given in the question you linked to; hopefully this will clarify what's going on.

Recall that, for any $x:A$, the type $||B(x)||$ is a mere proposition, meaning that we have $a=b$ for any $a,b:||B(x)||$. We claim also that the dependent product of a family of mere propositions is itself a mere proposition, whence $\prod_{x:A}||B(x)||$ is a mere proposition as well. To see this, note that, for any $f,g:\prod_{x:A}||B(x)||$, showing $f=g$ amounts by function extensionality to showing $f(x)=g(x)$ for any $x:A$. But each $f(x),g(x):||B(x)||$, and hence, since $||B(x)||$ is a mere proposition, we have $f(x)=g(x)$ for each $x:A$, as desired.

Now, the type $||\prod_{x:A}B(x)||$ has the following constructor: for any element $f:\prod_{x:A}B(x)$, there is an element $|f|:||\prod_{x:A}B(x)||$. This constructor allows us to formulate the recursion principle of $||\prod_{x:A}B(x)||$:

If $C$ is any mere proposition and $\alpha:\left(\prod_{x:A}B(x)\right)\to C$, then there is an induced element $\overline{\alpha}:||\prod_{x:A}B(x)||\to C$ satisfying $\overline{\alpha}(|f|)\equiv \alpha(f)$ for any $f:\prod_{x:A}B(x)$.

In particular, since $\prod_{x:A}||B(x)||$ is a mere proposition, to define a element of type $$||\prod_{x:A}B(x)||\to\prod_{x:A}||B(x)||,$$ it suffices to define an element of type $\prod_{x:A}B(x)\to\prod_{x:A}||B(x)||$. We claim that such an element is given by the expression $\lambda f.\lambda x.|f(x)|$, where $|\cdot|$ is the constructor for $||B(x)||$. Indeed, if $f$ is of type $\prod_{x:A}B(x)$, then $f(x)$ is of type $B(x)$ for any $x:A$, and so we have $|f(x)|:||B(x)||$ for each $x:A$, as desired.

On the other hand, there is no such way of constructing an element of type $$\prod_{x:A}||B(x)||\to||\prod_{x:A}B(x)||,$$ without some additional axioms. The reason for this is that, a priori, there is no way of "lifting" an element of $||B(x)||$ to an element of $B(x)$. In particular, if we wanted to construct an element of $||\prod_{x:A}B(x)||$, we would first need to construct an element of $\prod_{x:A}B(x)$; if we only have an element $f:\prod_{x:A}||B(x)||$ on hand, then we're able to get elements of $||B(x)||$ from elements $x:A$, but there's no way to to get elements of $B(x)$ from those elements of $||B(x)||$, and so we're stuck.

Title: Truncation and $\Pi$-types (ii)


问:
How do I find $\lim_{x\to 3} (x-3)\cot(\pi x)$ without using l'hopitals rule?
While doing some problems, I got stuck on this limit:
$$\lim_{x\to 3} (x-3)\cot(\pi x)$$
L'hospitals rule wasn't formally introduced in the lecture yet, so I'm not supposed to use it. I honestly don't know how to approach this, so any help would be appreciated!

答: Hint: $\sin (3\pi -y)=\sin y$.  Hence, $\sin (\pi x)=-\sin (\pi (x-3))$. Now use the fact that $\frac {\sin (\pi (x-3))} {x-3} \to \pi$.


问:
Determine conjugate function
Let $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=e^x$. Determine $f^*(y)$.

I try to use some inequalities to get supremum but it is impossible. Seemingly, I must consider some cases of $x$.

答:
$$f^*(y)=\textstyle\sup_x \langle y, x \rangle - e^x.$$
Since the supremum involves a smooth function we can examine the derivative:
$$\frac{d}{dx}(yx-e^x) = y-e^x = 0\quad\Longrightarrow\quad x=\log y~~\text{if}~~y>0.$$
So if $y>0$, at least,
$$f^*(y) = y\log y - y.$$
If $y=0$, then the supremum is $0$; if
$y<0$, then the supremum is $+\infty$. (In both cases, let $x\rightarrow-\infty$.) So the final answer is
$$f^*(y)=\begin{cases} y\log y - y & y > 0 \\ 0 & y = 0 \\ +\infty & y < 0 \end{cases}$$
If you don't like extended-valued functions just drop the $y<0$ case and state that
$$\mathop{\textrm{dom}}(f^*) = \mathbb{R}_+.$$</p>

Title: Determine conjugate function


答:
Substitute $x=t+3$, where $t \to 0$.
$$\lim_{t \to 0} t \cot ( \pi (t+3))$$
But now
$$\cot ( \pi t + 3 \pi) = \cot ( \pi t) = \frac{\cos( \pi t)}{ \sin ( \pi t)} = \frac{ \pi t}{ \sin ( \pi t)} \cdot \frac{\cos( \pi t)}{  \pi t}$$
Hence
$$\lim_{t \to 0} t \cot ( \pi (t+3)) = \lim_{t \to 0} t  \cdot \frac{ \pi t}{ \sin ( \pi t)} \cdot \frac{\cos( \pi t)}{  \pi t} = \frac{1}{\pi}$$

Title: How do I find $\lim_{x\to 3} (x-3)\cot(\pi x)$ without using l'hopitals rule?

Body:
While doing some problems, I got stuck on this limit:
$$\lim_{x\to 3} (x-3)\cot(\pi x)$$
L'hospitals rule wasn't formally introduced in the lecture yet, so I'm not supposed to use it. I honestly don't know how to approach this, so any help would be appreciated!


答:
Say that in your equation, the exponents of your unknown $x$ are fractions
$$\frac{m_1}{n_1},\ldots,\frac{m_k}{n_k}.$$

Let $N$ be the least common multiple of $n_1,\ldots,n_k$, and let $N = n_ic_i$. Then
$$\frac{m_i}{n_i} = \frac{m_ic_i}{n_ic_i} = \frac{m_ic_i}{N}.$$
That means that you can rewrite your equation as an equation in which every radical is an $N$th root. Letting $y=x^{1/N}$, you can then turn your equation into an equation in $y$ which has no radicals, just integral powers. That equation will be an equation of degree $M=\max\{c_1m_1,\ldots, c_km_k\}$, and so has at most $M$ solutions for $y$. When $N$ is odd, each solution for $y$ will give you a unique solution for $x$, yielding at most $M$ solutions for $x$.

If $N$ is even then at least one of the $n_i$ is even, which means that your $x$ must be nonnegative. So nonnegative values of $y$ will give one corresponding solution, but negative values of $y$ will not correspond to solutions of the original equation. In any case, there is at most one $x$ value per $y$-value, so again you get at most $M$ solutions for $x$.


问:
Euler’s gamma functions and exponential integrals
I am trying to solve these integrals,

where $ (C_i,S_i) ≡ (\cos λp_i, \sin λp_i)$
I have done $D_{SS}$. But in $D_{SC}$, I encounter some divergences.
The author says this in the paper:

The integration of λ in Eqs. (C15)-(C18) seems straightforward as one can express the trigonometric
functions to exponential functions and then convert the integrals to Euler’s gamma functions. It
is worth mentioning, however, that one should handle the branch cut singularities in the gamma
functions carefully

Could any of you point me to some resources regarding solving these types of integrals? Where can I read up on exponential integrals and gamma functions?

Apologies if this is not the right place to ask this question.

答:
If you intend to follow the process described in your quotation, you will want

the exponential forms of sine and cosine (alternatively, (3) here and (4) here, or alt. ref.)

partial fraction decomposition (probably, although one might sneak around this requirement), and

the gamma function, et seq. (alt. ref., alt. ref.)

For instance, using the exponential forms of sine and cosine in the integrand of $D_{SC}$, we obtain
$$ D_{SC} = \frac{4}{\pi} \int_0^{\infty} \frac{-1}{16 \lambda^2} \left( \mathrm{e}^{-\mathrm{i}p_1 \lambda} - \mathrm{e}^{\mathrm{i}p_1 \lambda} \right)
\left( \mathrm{e}^{-\mathrm{i}p_2 \lambda} - \mathrm{e}^{\mathrm{i}p_2 \lambda} \right)
\left( \left( \mathrm{e}^{-\mathrm{i}p_3 \lambda} + \mathrm{e}^{\mathrm{i}p_3 \lambda} \right) - \frac{\mathrm{i}\left( \mathrm{e}^{-\mathrm{i}p_3 \lambda} - \mathrm{e}^{\mathrm{i}p_3 \lambda} \right)}{p_3 \lambda} \right)
\left( \left( \mathrm{e}^{-\mathrm{i}p_4 \lambda} + \mathrm{e}^{\mathrm{i}p_4 \lambda} \right) - \frac{\mathrm{i}\left( \mathrm{e}^{-\mathrm{i}p_4 \lambda} - \mathrm{e}^{\mathrm{i}p_4 \lambda} \right)}{p_4 \lambda} \right) \,\mathrm{d}\lambda
$$

The remainder of this process is a herculean effort.  Are you sure you wouldn't be happier evaluating these integrals using the Cauchy integral formula?


问:
Show $\limsup_{x\to \infty} \sin\left((x+t)^2\right)-\sin\left(x^2\right)=2$ for $ t \ne 0$
My messing around on desmos strongly suggests to me that this is true, yet I have been been unable to prove it. I have tried to explicitly find a sequence $(x_n)$ such that $\lim x_n =\infty$ and $ \lim \sin\left((x_n+t)^2\right)-\sin\left(x_n^2\right) = 2$ without success. Any help would be much appreciated.

答:
First we write

$$\sin((x+t)^2) - \sin(x^2) = 2 \cos(x^2 + tx + (t^2/2)) \sin(tx + (t^2/2)).$$

So we need $x$ such that $x^2 + tx + (t^2/2)$ is close to an integer multiple of $2\pi,$ and $tx + (t^2/2)$ is close to an integer multiple of $2\pi,$ plus $\pi/2.$

Let $x_k = \sqrt{2\pi k - \pi/2}$. By Fejér's theorem, the sequence $y_k = tx_k + t^2/2$ is equidistributed $\mod 2\pi$ (see for instance this MathOverflow question), so there is a subsequence that gets arbitrarily close to $\pi/2 \mod 2\pi$. For that subsequence $\{x_{k_i}\}_{i \ge 0}$, we have that $x_{k_i}^2 + tx_{k_i} + t^2/2 = 2\pi k_i - \pi/2 + tx_{k_i} + t^2/2$ gets arbitrarily close to $0$ $\mod 2 \pi$. This concludes the claim.


问:
Confused about fourier coefficients of a lipshitz continuous $2\pi$ periodic function
This is only part of a larger question.

I've already shown that if $f$ is $2\pi$ periodic, then $$c_n = \frac{1}{4\pi}\int_{-\pi}^\pi\left(f(x)-f(x+\pi/n)\right)e^{-inx}dx.$$

I'm asked to then show that if $f$ is also lipschitz continuous then there is some constant $M$ such that $|c_n|\leq \frac{M}{|n|}$.

Using the lipschitz property, we then have $$\frac{1}{4\pi}\int_{-\pi}^\pi |f(x)-f(x+\pi/n)|e^{-inx}dx\leq \frac{1}{4\pi}\int_{-\pi}^\pi M\left\vert \frac{\pi}{n}\right\vert e^{-inx}dx =  \frac{M}{4|n|}\int_{-\pi}^\pi e^{-inx}dx.$$

However, $\int_{-\pi}^\pi e^{-inx}dx = 0$, so unless the fourier coefficients are all $0$, I'm pretty sure that I'm missing something.

Any thoughts would be really appreciated.

Thanks in advance!

答: Your inequalities don't even make sense since there is no ordering for complex numbers. To correct your argumnet all you have to do is to insert absolute value sign for $e^{-inx}$.


问:
Complex combinatorics with colors

In the picture above, I've created an example. We've got eight squares being chosen from, and five squares choosing, $8 \choose 5$. It may look like only four squares, but that rightmost square is actually two squares, sharing a spot.

So, in this method of combination, two squares can occupy the same spot. However, no more than two squares can occupy the same spot.

However, there is more: The different colors don't have the same reach. So, for example, the color blue can exist anywhere along the 8 squares. The color green can exist anywhere along the inner 6 squares. The color red can exist anywhere along the inner 4 squares. The subtractor is always $2n$, because whatever
$n$ is, it is subtracted from both ends.

So, to clarify my illustration: Above here, the number being chosen from, or the element set, is 8. The number of choosing is 5. Those 5 squares have 3 different colors. Those colors come with restrictions. It is also possible for two different squares, whether of the same color or a different one, to occupy the same spot.

So, is there an equation or formula that can do this: there are a lot of variables here, so I understand if it is too complex for any general formula. The variables are: the number of squares in the main set (which is 8 here), the number of squares in the sub-set (which is 5 here), the number of colors (which is 3 here) and the individual subtractors for the different colors.

However, the number of subtractors is always the same the number of colors. The colors are different precisely because they have different restrictions. Blue has the subtractor $0$, green has the subtractor $2$ and red has the subtractor $4$.

EDIT: So what I'm looking for is a general formula that gives me the number of combinations given these restrictions. The input values are the length of the row, the number of colors and the number of squares. The colors carry with them a value, which is the value subtracted from both sides of the row. So, is there a formula for this?

答:
There are 18,948 possible colorings for your example.

This is based on the assumption that

There are three colors.

There are eight spaces.

We must choose and color five spaces from these eight.

Each space can be chosen up to two times.

When a space is chosen twice, order does not matter. In this case, [red, blue] is equivalent to [blue, red]. (I wasn't sure whether you wanted this rule, so I guessed. Feel free to clarify.)

Colors follow the specific palette rule you specify (the first color can be used everywhere; the next color can be used in all but the outer two squares; the next color can be used in all but the outer four; and so on.)

I calculated this by obtaining a recursive formula for counting two quantities:

$f(k,m)$ counts how many ways there are to color exactly $2k$ blocks according to the above rules, when you can make $m$ selections. By coloring the middle two squares and working your way out, you can write this as a recursive sum in terms of $f(k-1, m)$, $f(k-1, m-1)$, $f(k-1, m-2)$, $f(k-1, m-3)$.

$q(k, m, N)$ counts how many ways there are to color $N$ blocks when you can make $m$ selections according to the above rules— but we drop the palette requirement and instead allow all $k$ colors throughout. By coloring the middle square (if $N$ is odd) or middle two squares (if $N$ is even) and working your way outward in pairs, you can similarly get a recursive formula for $q$.

Then if $S$ is the total number of blocks, $k$ is the total number of colors, and $m$ is the total number of selections to be made, your total counting formula is:
$$\sum_{j=0}^{2N} q(k, j, S-2(k-1)) \cdot f(k-1, m-j) $$

which in your specific case $(k=3; m=5; N=8)$ is:

$$18948 = 78 + 864 + 3588 + 6804 + 5670 + 1944 + 0 + 0$$

Details follow.

You can build up a full solution by relaxing some of the constraints.

We can start from a simpler problem: suppose you have $k$ colors and $2k$ blocks. The middle 2 blocks can be colored by all $k$ colors. The two beyond that can be colored with $k-1$ colors. The two beyond that can be colored with $k-2$ colors, and so on until the outer two can only be colored with one remaining color.

We can find the function $g(k,m)$ which counts how many ways there are to color such a system if you have $m$ spaces left to color, and $k$ colors left in your palette. (Unlike in the original problem, each square can be picked at most once.)

We color from the inside outward, which gives us a recursive coloring procedure. First we consider the innermost 2 spaces. We decide whether we will pick 0, 1, or 2 of them to color. Then we decide what color they will be. We deduct the number of spaces we've chosen from our number $m$ of remaining spaces, then proceed to the next outermost spaces.  This gives us the recursive step of our recursive relationship where at each step we're considering a pair of squares: to pick $m$ out of $2k$ spaces to color, first decide whether to pick zero, one, or two of the middle spaces; color each one with one of the $k$ colors; then color the remaining outer spaces.

If, during our coloring process, we run out of spaces to color ($m>2k$), we have failed. If we color too many $(m<0)$, we have failed. If we run out of spaces just as we run out of colors, we successfully finish $(m=k=0)$. These give us the base case of our recursive relationship.

Altogether, we have:

$$g(k,m) = \begin{cases}0 & \text{if }m < 0\\ 1 & \text{if }m=0\text{ and }k=0\\0 & \text{if }m>0\text{ and }k=0\\ \sum_{i=0}^2 k^i {2 \choose i} g(k-1,m-i)&\text{otherwise}\end{cases}$$

Here, $k$ is the number of colors left in the palette; $m$ is the number of spaces we can choose; $i$ is the number of spaces you've chosen to color from the pair of spaces you're considering this round. The recursive step formula is more simply written

$$g(k-1, m) + 2k\,g(k-1, m-1) + k^2\, g(k-1,m-2).$$

Starting from smaller values and working your way up, you can build a table of values of $g$ for each choice of $k$ and $m$; these will count the possible ways of choosing and coloring $m$ of the $2k$ spaces according to the palette constraints of the game.

Now how to modify this to achieve the formula for your more complicated game? If the innermost color can appear in more than two spaces—e.g. the red color in your example can appear in four spaces—then we must adjust the first term in our recurrence relation. Let $h(k,m,N)$ denote the number of ways to pick $m$ of the available spaces and color them with a palette of $k$ colors according to the rules, when the rarest color has $N$ available spaces.

Then $h(k,m,N)$ is just $$\sum_{j=0}^N {N \choose j} k^j \cdot g(k-1,m-j)$$
which recurses into the formula for $g$ that we've already described.

What if we want to allow each space to be selectable up to two times? Well, naively we could add $g(k,m-i)$ and $g(k,i)$ together ("choose $i$ spaces the first time, then choose $m-i$ potentially overlapping spaces) but this would result in double-counting; more finesse is required.

Assuming that when two colors are in the same space, the order doesn't matter (so, [red blue] in the same space is the same as [blue, red]), we have the following counting results:

When you have two spaces under consideration and each space can be colored up two twice, and your palette has $k$ colors:

You can pick just one space and color it. There are $2k$ possibilities.

You can pick two spaces to color. In particular, you could pick two different squares, with $k^2$ possible colorings. Or you could pick two of the same square— either both in the left square or both in the right— with $k(k+1)/2$ colorings each. In total, there are $k^2+2k(k+1)/2 = k^2+k^2+k = k(2k+1)$ colorings when you select two spaces.

You can pick three spaces to color. Two will necessarily be in a single square [$k(k+1)/2$ options there] and one will be alone [$k$ options]. Exchanging whether the double-occupancy is on the left or right doubles the number of possibilities. In total, there are $k^2(k+1)$ coloring possibilities.

You can pick four spaces to color. There are $k(k+1)/2$ for each double-occupied space, which can be picked independently of the other double-occupied space. Hence there are $k^2(k+1)^2/4$ possibilities.

You can pick zero spaces to color. It's allowed; there's just one way to do that.

Hence the number of ways to pick and color $m$ positions out of $2k$ squares, where each square has two indistinguishable spaces, and you have a palette of $k$ colors, is given by the recurrence:

$$\begin{align*}f(k-1, m)&=f(k-1, m) \\ &\;+\; 2k\,f(k-1, m-1)\\ &\;+\; k(2k+1)\,f(k-1, m-2)\\ & \;+\; k^2(k+1)\,f(k-1, m-3) \\&\;+\; k^2(k+1)^2/4\, f(k-1,m-4).\end{align*}$$

Title: Complex combinatorics with colors


问:
Ratio of segments formed by intersection of three cevians
ABC is a triangle and P some point inside it. Cevians AP, BP, CP cross the opposite sides at A',B' and C' respectively.

What is the minimum value the following expression can accomplish:

(1+2*PA'/PA)(1+2*PB'/PB)(1+2*PC'/PC) ?

The solution is 8, when each ratio is 1/2 (meaning that is a centroid) but I'm having trouble reaching that.

Now from the ratio of areas APB, BPC, CPA and the whole triangle ABC we get respectively:

Pa/P = A'P/A'A, Pb/P = B'P/B'B, Pa/P = C'P/C'C, and summing this i get

1 = PA'/(PA' + PA) + PB'/(PB' + PB) + PC'/(PC' + PC)

Does from here directly follows some relation to the wanted expression fully algebraically, and if so how? Or should I still consider them as being in the triangle?

Replacing PA'/PA=x PB'/PB=y PC'/PC=z I get

1 = 1/(1 + 1/x) + 1/(1 + 1/y) + 1/(1 + 1/z), and the wanted expression as (1+2x)(1+2y)(1+2z)

I also tried:

4 = 1 + x/(1+x) + 1 + y/(1+y) + 1 + z/(1+z)

4 = (1+2x)/(1+x) + (1+2y)/(1+y) + (1+2z)/(1+z)

答:
Starting from the following from the question I managed to push to the solution.

Replacing PA'/PA=x PB'/PB=y PC'/PC=z I get

1 = 1/(1 + 1/x) + 1/(1 + 1/y) + 1/(1 + 1/z) [1], and the wanted expression as

(1+2x)(1+2y)(1+2z)

[1] can be written as 1 = xy + yz + xz + 2xyz. [3]

Also from [1] and H-A 1 <= (x+1)/4 + (y+1)/4 + (z+1)/4
that is 1 <= x + y + z [4]

Now using [3] and [4] we have

(1+2x)(1+2y)(1+2z) = 1 + 2x + 2y + 2z + 4xy + 4xz + 4yz + 8xyz = 1 + 2x + 2y + 2z + 4 >= 7

I did it on the paper, but then found mistake and it dropped from 8 to 7 :D hahaha


问:
Logical snail along a straight line
A snail crawled along a straight line for 6 minutes while separate people watched its progress. Each of these people watched snail for  1 minute and during this minute the snail rolled exactly by one foot. It is known that the snail was always watched by at least one person. What is the maximum possible distance the snail could crawl during the 6 minutes?
I tried thinking about sum of overlapped time which turns out to be total number of people - 6. But it was of no help. I think the answer is 6 feet and tried to prove by contradiction but couldn't.</p>

答:
Assuming that there is someone watching the snail for the first minute of the period, and someone watching the snail for the last minute, the maximum snail movement is $10$ feet.

The pattern for this is that the snail moves one foot to start with, then stops for a while (with two people watching the pause), then moves two feet - in the middle of which the watcher changes, and so on through the time period. The final minute can only see one foot of movement by the assumption that there is a watcher for that whole period. (If you allow an initial and final watcher to count their minute of watching outside the six minute period, you can squeeze in two more feet of movement, but that's not how I read the setup).

To minimize the snail speed during the dashes, you can allow $12$ seconds for each foot travelled (each green block in the image is a foot of travel), with the pauses lasting 48 seconds.

Title: Logical snail along a straight line


问:
Terence Tao, Analysis I, Ex. 5.5.2: Entry point needed
Terence Tao, Analysis I, 3e, Exercise 5.5.2:

Let $E$ be a non-empty subset of $\mathbb{R}$, let $n \ge 1$ be an
  integer, and let $L < K$ be integers. Suppose that $K/n$ is an upper
  bound for $E$, but that $L/n$ is not an upper bound for $E$. Without
  using Theorem 5.5.9, show that there exists an integer $L < m \le K$
  such that $m/n$ is an upper bound for $E$, but that $(m-1)/n$ is not
  an upper bound for $E$. (Hint: prove by contradiction, and use
  induction. It may also help to draw a picture of the situation.)

This answer does not do the trick for me, since I'm looking for the proof by contradiction, equipped only with the set of propositions and theorems covered so far.

What I'm actually looking for is an entry point to the proof, since I find it quite hard to negate the assumption.

My attempt is to induct over $n$. Assume I have been able to prove the statement for $n = 1$, is the following a valid negation of the assumption?

For all integers in the range $(L, K]$, at least one of the following statements holds:

$(m-1)/(n+1)$ is not an upper bound and $m/(n+1)$ is not an upper bound.

$(m-1)/(n+1)$ is an upper bound and $m/(n+1)$ is not an upper bound.

$(m-1)/(n+1)$ is an upper bound and $m/(n+1)$ is an upper bound.

If I were able to prove that there is an integer in the given range such that none of the above holds, am I done?

Theorem 5.5.9 (Existence of least upper bound). Let $E$ be a non-empty subset of $\mathbb{R}$. If $E$ has an upper bound, (i.e.,
  $E$ has some upper bound $M$), then it must have exactly one least
  upper bound.

I went through another loop considering this comment of Steve Kass, but I ended up wondering why the proof needs induction at all:

Assume for all integers $m$ where $L < m \le K$ we have that $m/n$ and $(m-1)/n$ are upper bounds of $E$. Then $(m-1)/n = L/n$ is also an upper bound. But $L/n$ is no upper bound, by assumption. Assume $m/n$ and $(m-1)/n$ are no upper bounds. Then $m/n = K/n$ is also no upper bound. But $K/n$ is an upper bound, by assumption.

Hence, there is an integer in the range $(L, K]$ s.t. $m/n$ is an upper bound for $E$, and $(m-1)/n$ is not.

答:
Let $A \subseteq \mathbb Z$ defined by
$$ A = \{ m \in \mathbb Z \cap [L, K] \mid m/n \text{ is an upper bound of $E$}\} $$
Then $A$ is a non-empty ($K \in A$ by assumption) bounded below ($L$ is a lower bound) subset of $\mathbb Z$. As such, it has a least element ($A -L\subset \mathbb N$ is non-empty and $\mathbb N$ is well-ordered). Let $m$ be this least element. Then $m/n$ is an upper bound of $E$, but $(m-1)/n$ isn't, as $m$ is the least element of $A$. As $L/n$ is not an upper bound, we have $L < m \le K$ as wished.</p>


问:
How to generate random numbers from density function if it's known that if $X_i$ are iid and $X_i \sim F$, then $F(x_i) \sim U[0,1]$
The given density function:
$$f(x) = \frac{1}{2}\phi(x, 0, 1) + \frac{1}{10}\sum^4_{j=0}\phi(x; (j/2)-1, 1/10) $$
where $\phi$ is normal distribution.

```<code>f <- function(x){
  dnorm(x)/2 +
    dnorm(x,- 1, 1/10)/10 +
    dnorm(x, 1/2 - 1, 1/10)/10 +
    dnorm(x, 2/2 - 1, 1/10)/10 +
    dnorm(x, 3/2 - 1, 1/10)/10 +
    dnorm(x, 4/2 - 1, 1/10)/10
}
</code>```

It is known that if random variables $X_i$ are iid, and $X_i \sim F$, then $F(X_i)\sim U[0,1]$
I tried to define the cumulative function:

```<code>F <- function(x){integrate(f, lower = -Inf, upper = x)}$value
</code>```

And then to find zeroes of function $F(x)-\text{runif}(x)$

```<code>zeros <- function(x){uniroot(F(x)-runif(x), c(-10,10))}
</code>```

But the program R gives me an error:

```<code>Error in uniroot(F(x) - runif(x), c(-10, 10)) :
  f() values at end points not of opposite sign
</code>```

Anyway, I am at loss as to the way, this problem could be possibly solved. Does anybody got any insights?

答:
Looks to me as if you have a mixture with six different densities, mixing coefficients $\pi_k$ where $\pi_1 = 1/2$ and the other five are $1/10$. What you might want to do to generate $X$ is to select one of these densities at random, using these probabilities, then generate $X$ from that one:

Select $i$ from $\{1, 2, 3, 4, 5, 6\}$ with probabilities
$\{1/2, 1/10, 1/10, 1/10, 1/10, 1/10 \}$.

Return $X$ from $f_i$.


问:
Equivalence of definition of Markov chain
Here are two definitions of MC. On Durrett's text, it is proved that $(2)$ implies $(1)$. However, I am not sure whether $(1)$ implies $(2)$.

A sequence of random variables is said to be a Markov chain if
\begin{align*}
P(B|X_1,\cdots, X_n) = P(B|X_n)\quad a.s. \quad B\in \sigma(X_{n+1}), \forall n\geq 2
\end{align*}

Let $(S,\mathcal{S})$ be the state space. A function $p: S\times \mathcal{S}\rightarrow \mathbb{R}$ is said to be a transition probability if for each $A\in \mathcal{S}$, $p(\cdot, A)$ is measurable and for each $x\in S$, $p(x,\cdot)$ is a probability measure. And the distribution of $X_1,X_2,\cdots $ is
\begin{align*}
P(X_j\in B_j,0\leq j\leq n)=\int_{B_0}\mu(dx_0)\int_{B_1}p(x_0,dx_1)\cdots\int_{B_n}p(x_{n-1},dx_n)
\end{align*}

My question is that, the second definition restricts the conditional distribution $P(X_{n+1}|X_n)$ to be the same for all $n\in \mathbb{N}$ (It is proved that $P(X_{n+1}\in B|X_n)=p(X_n,B)$ so we treat $p(\cdot, B)$ as the conditional distribution), while the first definition seems do not have this restriction. Does anyone have any idea?

Thanks in advance!

答: You are correct: the second definition is for a time homogeneous Markov chain, the first allows non-homogeneity in time.


问:
Complex series and sequences
Question: Let $\ell_2(\mathbb{C})$ be the set of all complex sequences $(z_n)\in\mathbb{C}$ such that $\sum_{n=0}^{\infty} |z_n|^2$ is convergent, i.e.,
$$\ell_2(\mathbb{C}) = \left\{ (z_n)_n\;:\; \sum_{n=0}^{\infty} |z_n|^2 \;\;\text{is convergent}\right\}.$$
If $(z_n),(w_n)\in\ell_2(\mathbb{C})$, show that $\displaystyle{\sum_{n=0}^{\infty} z_n\overline{w}_n}$ is convergent.

My attempt: Let $(z_n),(w_n)\in\ell_2(\mathbb{C})$. It follows from the Schwarz inequality that
$$\left|\sum_{n=0}^{\infty} z_n\overline{w}_n\right|^2 \leq \left(\sum_{n=0}^{\infty} |z_n|^2\right)\left(\sum_{n=0}^{\infty} |w_n|^2\right).$$
Since $(z_n),(w_n)\in\ell_2(\mathbb{C})$, then the both series $\displaystyle{\sum_{n=0}^{\infty} |z_n|^2}$ and  $\displaystyle{\sum_{n=0}^{\infty} |w_n|^2}$ converge and, therefore, the product $\displaystyle{\left(\sum_{n=0}^{\infty} |z_n|^2\right)\left(\sum_{n=0}^{\infty} |w_n|^2\right)}$ converges. Since such a product converges and the above inequality holds, then we can conclude from the comparison test that $\displaystyle{\sum_{n=0}^{\infty} z_n\overline{w}_n}$ is convergent.

答:
Your attempt is okay, assuming that you can use the intended version of the Cauchy-Schwarz inequality.

Alternatively, you can work with the partial sums and then take the limit.

The infinite sum in the conclusion defines an inner product of the two elements in $\ell_2$.

Title: Complex series and sequences


问:
Need help deriving the Schwartz inequality.
I am trying to prove the Schwartz Inequality (through an exercise) in the form of
$$
\sum_{i=1}^{n}x_iy_i \le \sqrt{\sum_{i=1}^{n}x_i^2}\sqrt{\sum_{i=1}^{n}y_{i}^2}
\tag{1}
$$
through a method of knowing that there is no $\lambda$ for $x_1 =\lambda y_1$ & $x_2 =\lambda y_2$  through which we can easily make out that
$$
 0 < \sum_{i=1}^{n}(\lambda y_i -x)^2
\tag{2}
$$
or
$$
0< \lambda^2 (\sum_{i=1}^{n}y_i^2) - 2\lambda (\sum_{i=1}^{n}x_iy_i) + (\sum_{i=1}^{n}x_i^2)
\tag{3}
$$
No problem there. For the next couple of steps, I am going to denote $(\sum_{i=1}^{n}y_i^2)$ by $a$, $(\sum_{i=1}^{n}x_iy_i)$ by $b$ and $(\sum_{i=1}^{n}x_i^2)$ by $c$ (to reduce time writing this),
for example, $(3)$ is written as
$$
0 < \lambda ^2(a)-2\lambda(b) +c
\tag{4}
$$
which we can proceed to make into
$$
0<\lambda ^2-2\lambda(ba^{-1})+ca^{-1}
\tag{5}
$$
Now, since from $(3)$, we can see that $0=\lambda ^2(a)-2\lambda(b)+c$ has no solution for $\lambda$ as I already stated that there is no $\lambda$ for $x_1 =\lambda y_1$ & $x_2 =\lambda y_2$ , so we can say that for $(5)$,

$$
 (ba^{-1})^2-4(ca^{-1}) < 0
\tag{6}
$$
$(6)$ is derived from $b^2-4c<0$ (here is where I made my mistake, I don't know why I didn't use $b^2-4ac$ sorry for this stupid mistake, from here I go on to try to make $a = 1$) for cases where we don't have a solution. Bringing back the summations, we get
$$
(-2(\sum_{i=1}^{n}x_iy_i)(\sum_{i=1}^{n}y_i^2)^{-1})^2-4(\sum_{i=1}^{n}x_i^2)(\sum_{i=1}^{n}y_i^2)^{-1} < 0
\tag{7}
$$
which is as far as I got. I tried doing
$$
4(\sum_{i=1}^{n}x_iy_i)^2(\sum_{i=1}^{n}y_i^2)-4(\sum_{i=1}^{n}x_i^2)(\sum_{i=1}^{n}y_i^2)^{-1} < 0
\tag{8}
$$
but I don't know how to get to $(1)$. So, did I do something wrong? Can someone elaborate on what I am meant to do? I am so sorry for the stupid question, I am really tired and (even though that's not an excuse,) I feel like I made a mistake somewhere and I really want to know the answer by the end of today. Thank you so much.

I am sorry if I am making any stupid mistakes, I am tired but I can't sleep without knowing how to do this.

答:
Go back to
$0 < \lambda ^2(a)-2\lambda(b) +c$.

Using the original quadratic formula,
the discriminant is
$d = (2b)^2-4ac
=4(b^2-ac)
$.

For the quadratic
to have no real roots,
$d < 0$
so
$b^2-ac < 0$
so
$b^2 < ac$.


问:
Surjectivity of $f:\mathbb{S}^{1}\longrightarrow\mathbb{S}^{1}$ via universal covering of $\mathbb{S}^{1}$
Let $f:\mathbb{S}^{1}\longrightarrow\mathbb{S}^{1}$ be given by $f(x,y) = (2xy,x^{2}-y^{2})$. Prove that $f$ is a closed map and $f$ is surjective. Use the universal covering $\phi:\mathbb{R}\longrightarrow\mathbb{S}^{1}, \phi(x)=(\cos(2\pi x),\sin(2\pi x))$ to prove surjectivity.

For the first part, note that $\mathbb{S}^{1}$ is a compact space, so every closed subset is compact. But $f$ is continuous, so it maps compact sets to compact sets, so $f$ is closed. For the second part, I don't know how to proceed. Thanks for any help.

答:
The universal covering is helpful, but you can avoid to use it.

Identifying $\mathbb R^2$ with $\mathbb C$ via $(x,y) = x + iy$, we see that $S^1 = \{ z \in \mathbb C \mid \lvert z \rvert = 1\}$. Let $\tau : S^1 \to S^1, \tau(x+iy) = y + ix$ (which is the same as $\tau(z) =  i \overline z$) be the map flipping real and imaginary part. It is a homeomorphism, thus $f$ is surjective if and only $\bar f = \tau \circ f$ is surjective.

Clearly $\bar f(z) = z^2$. Since each $w \in \mathbb C$ has a square root $z_w$ (if $w \ne 0$, then it has in fact exactly two square root differing by the factor $-1$) and $\lvert z_w \rvert = \sqrt{\lvert w \rvert}$, we see each $w \in S^1$ has the form $\bar f(z_w) = z_w^2$. Thus $\bar f $ is surjective.


问:
Prove that a number is the product of two primes under certain conditions.
Suppose $p$ be the smallest prime dividing $n \in \mathbb{Z}^+$. Prove that if $n$ is not a perfect square and that $p<n<p^3$, then $n$ must be the product of two primes.

Clearly, the smallest $p$ can be is $2$ and $n$ must be an integer that is greater than $1$ in order to be divisible by a prime. I know that the Fundamental Theorem of Arithmetic (FTA) guarantees that every positive integer greater than $1$ is the product of two or more primes. So is it enough to argue that by the FTA, $n$ is the product of two primes?

答:
If $p|n$ and $p < n < p^3$ then $1 < \frac np < p^2$ and $\frac np$ is an integer.

Consider what prime factors can divide $\frac np$.  No prime less than $p$ as $p$ was the smallest prime dividing $n$.

If $p|\frac np$ then we $\frac n{p^2} < p$ but $n$ has no non trivial factors less than $p$ so $\frac n{p^2} =1$ and $n = p^2$.  But $n$ is not a perfect square.

So there is a prime $q > p$ so that $q|\frac np$.  But then $\frac n{pq} < \frac {p^2}q=p\frac pq < p*1 =p$.  But $n$ has no non trivial factors less than $p$.  So $\frac n{pq} = 1$ and $n =pq$ and $pq$.


答:
If $p^3 > n$ then
$p > n^{1/3}$
so
$\dfrac{n}{p}
\lt \dfrac{n}{n^{1/3}}
=n^{2/3}
$.

If another prime
$q > p$ divides $n$,
then
$q | \dfrac{n}{p}
$
so
$q \lt \dfrac{n}{p}
\lt n^{2/3}
$
and
$q > p > n^{1/3}$.

Therefore
$\dfrac{n}{pq}
\lt n^{1/3}$.

No other prime can divide
$\dfrac{n}{pq}$
since that is less than
$n^{1/3}$
and no prime smaller than $p$
divides $n$.

Therefore
$n = pq$.

Title: Prove that a number is the product of two primes under certain conditions.

Body:
Suppose $p$ be the smallest prime dividing $n \in \mathbb{Z}^+$. Prove that if $n$ is not a perfect square and that $p<n<p^3$, then $n$ must be the product of two primes.

Clearly, the smallest $p$ can be is $2$ and $n$ must be an integer that is greater than $1$ in order to be divisible by a prime. I know that the Fundamental Theorem of Arithmetic (FTA) guarantees that every positive integer greater than $1$ is the product of two or more primes. So is it enough to argue that by the FTA, $n$ is the product of two primes?


问:
Proving a general formula for "tangent vectors" (actually $k$-th partial derivatives) when changing coordinates
I want to prove that the order of contact of two functions $f, g: M \to N$ between manifolds is well defined.

Definition. Let $f, g : M^{n} \to N^{m}$ be smooth functions. We say $f$ has order of contact $k$ at $p \in M$ with $g$, and write $f \sim_{k} g$ at $p \in M$ if there exist charts $\varphi: U \to \mathbb{R}^n$ and $\psi: V \to \mathbb{R}^m$ around $p \in M$ and $f(p) \in N$ such that $f(p) = g(p)$ and all the derivatives of $f$ and $g$ in the charts up to order $k$ coincide at $p$, that is:

$$\frac{\partial^{|\alpha|}}{\partial r^{\alpha}}(r^j \circ \psi \circ f \circ \varphi^{-1})(\varphi(p)) = \frac{\partial^{|\alpha|}}{\partial r^{\alpha}}(r^j \circ \psi \circ g \circ \varphi^{-1})(\varphi(p))$$

for every multi-index $\alpha = (\alpha_1, \cdots, \alpha_n)$ (where $|\alpha| = \alpha_1 + \cdots + \alpha_n$) and every $1 \leq j \leq m$, where we define:

$$\frac{\partial^{|\alpha|}}{\partial r^{\alpha}} \doteq \frac{\partial^{|\alpha|}}{\partial r^{\alpha_1} \cdots \partial r^{\alpha_n}}$$

and $r^j: \mathbb{R}^m \to \mathbb{R}$ is the projection onto the $j$-th coordinate and $\frac{\partial}{\partial r^{k}}$ is just the ordinary (Euclidean) $k$-th partial derivative.

I'm trying to prove this definition does not depend on charts, that is, if the above is satisfied in charts $\varphi = (x^1, \cdots, x^n), \psi = (y^1, \cdots, y^m)$, then it's also satisfied in any other charts $\tilde{\varphi} = (\tilde{x}^1, \cdots, \tilde{x}^n), \tilde{\psi} = (\tilde{y}^1, \cdots, \tilde{y}^m)$. I've been able to prove it is well defined if $k = 1$ or $k = 2$ by using the chain rule in the form of the following formulas (where I'm using Einstein notation):

$$\frac{\partial}{\partial y^i} = \frac{\partial x^{\ell}}{\partial y^i} \frac{\partial}{\partial x^{\ell}}$$

$$\frac{\partial^2}{\partial y^{i} \partial y^j} = \frac{\partial^2 x^{\ell}}{\partial y^i \partial y^j} \frac{\partial}{\partial x^{\ell}} + \frac{\partial x^{\ell}}{\partial y^j} \frac{\partial x^k}{\partial y^i} \frac{\partial}{\partial x^k}$$

and I think the general case has to be done by induction, but I'm having trouble actually using the inductive hypothesis, since I'd have to obtain a relation between the $k$-th partial derivatives in relation to different charts (which in turn I'd also have to use induction to obtatin, but I'm failing to see the pattern in order to think of the formula I'd want to prove for this general case). I'd really appreciate some help on this! Thanks in advance.

答:
It's not really necessary to write down an explicit formula. Instead one can note that every term in the expression for $\frac{\partial^p f^j}{\partial y^{i_1}\cdots\partial y^{i_p}}$ is of the form
$$
a\frac{\partial^q f^j}{\partial x^{k_1}\cdots\partial x^{k_q}}
$$
where $a\in C^\infty M$ does not depend on $f$ and $q\le p$. Taking another partial derivative (w.r.t. $x$ or $y$)of such a term will produce more terms of the same form, and the order of the partial derivatives of $f$ will increase by at most $1$.


问:
Show $\mathbb{Z}[\sqrt{6}]$ is a Euclidean domain
I'm attempting to modify the proof the $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain to prove a similar result for $\mathbb{Z}[\sqrt{6}]$. The idea is to prove that $\mathbb{Q}[\sqrt{6}]$ is Euclidean which will then give the result for  $\mathbb{Z}[\sqrt{6}]$. According to Dummit and Foote's Abstract Algebra this should have norm $d(a+b\sqrt{6})=|a^2-6b^2|$. The result for $\mathbb{Z}[\sqrt{2}]$ relies on the fact that for any element $x$ in $\mathbb{Q}[\sqrt{2}]$ there is an element $x'$ in $\mathbb{Z}[\sqrt{2}]$ with $d(x-x')<1$, which is then used to show that for $x=qy+r$ we have
$$
d(r)=d(x-qy)=d\left(y\left(\frac{x}{y}-q\right)\right)=d(y)d\left(\frac{x}{y}-q\right)<d(y)
$$
But in this case, I'm having a hard time showing that we can find an element with $d(x-x')<1$ for $x\in\mathbb{Q}[\sqrt{6}], x'\in\mathbb{Z}[\sqrt{6}]$. Consider $x=a+b\sqrt{6}\in\mathbb{Q}[\sqrt{6}]$. The furthest this element can be away from any $x'=a'+b'\sqrt{6}\in\mathbb{Z}[\sqrt{6}]$ is when $|a-a'|=\frac{1}{2}=|b-b'| \Rightarrow$
$$
d(x-x')=\left|\left(\frac{1}{2}\right)^2-6\left(\frac{1}{2}\right)^2\right|=\frac{5}{4}>1
$$
What can I do to get around this issue?

答:
Decades after leaving college, I had a chance to audit an introductory course in abstract algebra. I find the subject very interesting. But I am not a proper maths student by any stretch of imagination. Please go easy on someone who had ten weeks' learning on the subject. You mathematicians will certainly find my journey way too verbose, but this is a possible thought process of a newbie and a non-mathematician. It might help you to understand and to have some compassion on students who are struggling badly.

Engineers are visual thinkers. Sketches help me a lot. Allow me to recap. An integral domain is said to be a Euclidean domain if there are a well defined norm $N(a) \leq N(ab)$ and a division algorithm $a=qb+r$ such that $N(r)<N(b)$. Just to cover a few definitions. Such division algorithm works in a Euclidean domain $R$ for all $a,b \in R, b \ne 0$, there are a quotient and a reminder $q,r \in R$. For elements in a quadratic integer ring, $a \in R = \{x+y \sqrt{d}:x,y \in \mathbb{Z} \}$ with $d$ square free, a common choice of norm is $N:R \setminus \{0\} \rightarrow \mathbb{Z}_{\geq 0}$ by setting $a \mapsto |x^2-dy^2|$.

As Chris had suggested, to show $\mathbb{Z}[\sqrt{d}]$ a Euclidean domain, direct application of Pythagorean theorem works only for $\mathbb{Z}[\sqrt{-2}], \mathbb{Z}[\sqrt{-1}], \mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{3}]$. In the previous statement, discriminant $d \in \mathbb{Z} \setminus \{ 0,1 \}$ is square-free such that it has no repeated prime factors. Equivalently, if $p \in \mathbb{Z}$ and $p^2|d$, then $p^2 = 1$.

Let $\mathbb{Z}[\sqrt{3}]=\{s+t\sqrt{3}: s,t \in \mathbb{Z}\}$ and $\mathbb{Q}[\sqrt{3}]=\{s+t\sqrt{3}: s,t \in \mathbb{Q}\}$. Suppose the nearest quadratic integer in $\mathbb{Z}[\sqrt{3}] \subset \mathbb{Q}[\sqrt{3}]$ from point $a/b \in \mathbb{Q}[\sqrt{3}]$ is point $q=m+n \sqrt{3}$ with $m,n \in \mathbb{Z}$. The horizontal and vertical components of the displacement $r/b=a/b-q$ are $x,y \in \mathbb{Q}$ where $|x| \leq \frac{1}{2}$ and $|y| \leq \frac{1}{2}$. Therefore, the maximum possible value of $N(r/b)=N(x+y\sqrt{3})=|x^2-3y^2| \leq \frac{3}{4} < 1$ and we are done. Note that division algorithm does not insist on uniqueness. If a point $a/b$ sits right at the boundary, there could be be more than one, at most four, points which are the "nearest". An example in $\mathbb{Z}$ is $\frac{7}{2}=3+\frac{1}{2}=4-\frac{1}{2}$.

In $\mathbb{Z}[\sqrt{-1}]$, the norm is simply the square of the usual Euclidean distance between points $q$ and $a/b$, which measures what consider to be the nearest Gaussian integer in this case. But the analogue of physical distances fails as $d$ increases.

Hence, the same approach fails to show $\mathbb{Z}[\sqrt{6}]$ is a Euclidean domain, as s/he had pointed out that $N(r/b)=N(x+y\sqrt{6})=|x^2-6y^2| \leq \frac{3}{2} \nless 1$.

Given that $0 \leq x^2 \leq \frac{1}{4}$, I follow Will Jagy's suggestion and split $0 \leq 6y^2 \leq \frac{3}{2}$ into four cases. It essentially considers two more lattice points as possible quotients $q', q''$, in additional to $q$, to the division algorithm for different values of $y$. I had additional inspiration from math_explorer https://artofproblemsolving.com/community/c1285h1036927. Note that the magic numbers $1$ and $\frac{5}{4}$ were arrived backwards. They will be shown in the cases. In order to make my life easier, I will consider only positive values of $x$ with $0 \leq x \leq \frac{1}{2}$, while allowing both positive and negative values for $y$. If $x$ were negative, similar arguments as the follows apply by switching cases 2 and 3.

Case 1: $0 \leq 6y^2 < 1$. Point $q$ has the coordinates $(m,n)$ and $0 \leq x^2 \leq \frac{1}{4}$ as shown earlier. The maximum possible difference occurs when $x^2=0$ and $6y^2=1^-$. Therefore $|x^2-6y^2|<1$ for all $x,y$ and we are done.

Case 2: $1 \leq 6y^2 < \frac{5}{4}$. Now we consider another lattice point to the right, says point $q'=(m+1,n)$. Using this as the new datum for point $a/b$, the horizontal component is now $-1+x$. Therefore, $\frac{1}{4} \leq (1-x)^2 \leq 1$. The maximum possible difference occurs when $(1-x)^2=\frac{1}{4}$ and $6y^2 = \frac{5}{4}^-$. Therefore $|(1-x)^2-6y^2|<1$ for all $x,y$ and we are done.

Case 3: $\frac{5}{4} < 6y^2 \leq \frac{3}{2}$. Now we consider yet another lattice point to the left, says point $q''=(m-1,n)$. Using this as the new datum for point $a/b$, the horizontal component is now $1+x$. Therefore, $1 \leq (1+x)^2 \leq \frac{9}{4}$. The maximum possible different occurs when $(1+x)^2=\frac{9}{4}$ and $6y^2 = \frac{5}{4}^+$. Therefore $|(1+x)^2-6y^2|<1$ for all $x,y$ and we are done.

Case 4: $6y^2 = \frac{5}{4} \implies y=\sqrt{\frac{5}{24}} \notin \mathbb{Q}$, therefore we can ignore this case.

Additional note for Case 3. At $6y^2 = \frac{3}{2}$, it has a comfortable margin from $(1+x)^2=1$ where $|1-\frac{3}{2}|=\frac{1}{2}<1$. In fact, this approach works for $\mathbb{Z}[\sqrt{7}]$ too, since $|1-\frac{7}{4}|=\frac{3}{4}<1$. However, this approach fails for $\mathbb{Z}[\sqrt{11}]$. Gerry Myerson $\mathbb{Z}[\sqrt{11}]$ is norm-euclidean pointed out that Oppenheim (1934) had a solution by considering more nearby lattice points as candidate point $q$. Now, what I will need to do is to make an attempt on $\mathbb{Z}[\sqrt{19}]$.

Links to bigger images:

Hasse diagram from rng to ED https://i.stack.imgur.com/jUcJX.png

$\mathbb{Z}[\sqrt{3}]$ https://i.stack.imgur.com/q4cli.png

$\mathbb{Z}[\sqrt{6}]$ https://i.stack.imgur.com/9QZuV.png

$q=(m,n)$ https://i.stack.imgur.com/X7WyX.png

$q'=(m+1,n)$ https://i.stack.imgur.com/zy8wL.png

$q''=(m-1,n)$ https://i.stack.imgur.com/Vipwd.png


问:
New Elementary Function?
In the February 2000 issue of FOCUS magazine, a short article suggests that the Lambert W function could be introduced into curriculum as a new elementary function saying: "... a case can be made for according it equal respect with the traditional transcendentals of calculus." As the inverse of $xe^x$, Lambert W is easy to understand, its properties are rather straight-forward, and it has found use in a wide range of applications.

Are there other functions you think are good candidates to introduce more widely into mathematics curriculum that are interesting, easy to understand, and broadly applicable?

答:
There is a class of differentiated gamma functions (DGFs) that basically can be expressed via derivatives of Gamma function.
They include the Gamma function, Polygamma function, and Hurwitz Zeta function further generalizes the class to fractional orders.

The thing about this class of functions is that they behave like "nearly-elementary".
These are the arguments for this point:

DGFs generalize the Bernoulli polynomials to negative orders, particularly, $\psi^{(1)}(x)=B_{-1}(x)$ if we define Bernoulli polynomials via Hurwitz Zeta. With power functions it is the reciprocal functions that generalize them to negative orders, and considered elementary.

The antiderivative of the reciprocal function $\frac1x$ is logarithm, an elementary function. The antidifference of the same function is polygamma $\psi(x)$, a DGF.

$\psi(x)-\psi(-x)$ is elementary, while $\psi(x)+\psi(-x)$ is a DGF. The Taylor series for these functions differ only in parity. Also, $\frac1{\pi}\psi(1-\frac x\pi)-\frac1{\pi}\psi(\frac x\pi)=\cot x$.

Due to Gauss's digamma theorem, the digamma function of a rational argument may be expressed in terms of Euler's constant and a finite number of elementary functions:$$\psi\left(\frac{r}{m}\right) = -\gamma -\ln(2m) -\frac{\pi}{2}\cot\left(\frac{r\pi}{m}\right) +2\sum_{n=1}^{\left\lfloor \frac{m-1}{2} \right\rfloor} \cos\left(\frac{2\pi nr}{m} \right) \ln\sin\left(\frac{\pi n}{m}\right)$$

Given the above considerations, the DGFs look like "all-but-elementary" or "nearly elementary" to me.

Title: New Elementary Function?


问:
Matrix equation $X+AX-XA^2=0$ has only zero solution
Let $A$ be a $n\times n$ complex matrix, with even numbers as eigenvalues. Show that the matrix equation $X+AX-XA^2=0$ has only zero solution. My attempt is just consider an eigenvector $v$ of $A$, then $Xv+AXv-4m^2Xv=0.$ Then what to do? Or just $v'(X+AX-XA^2)v=0$. But $A$ is not symmetric, how to proceed?

答:
Hint: If $\mu, \lambda$ are eigenvalues of $A$, then we can find a vector $v$ for which $Av = \lambda v$ and a vector $w$ for which $A'w = \bar \lambda w \implies w'A = \lambda w'$. Let $\Phi$ denote the linear map defined by
$$
\Phi(X) = X + AX + XA^2.
$$
The matrix $X = vw'$ is an eigenvector of $\Phi$ assocaited with the eigenvalue $1 + \lambda + \mu^2$.