问:
Two complex numbers have a product of $30$ and positive integers as real parts. Write all possible combinations of such numbers.

Two complex numbers have a product of $30$ and positive integers as real parts. Write all possible combinations of such numbers.

$(3,10), (5,6),(2,15)$ and $(1+ i\sqrt(29),1-i\sqrt(29)$ are few solutions. Is there any neat way to find all? I tried by assuming numbers to be $m+ia$ and $n+ib$ but this was does not seems neat.

答:
Assume that $n,m$ are positive integers. We have that
$$(m+ia)(n+ib)=mn-ab+i(an+bm)=30$$
implies $an=-bm$, that is $a=-bm/n$. It follows that
$$mn-ab=mn+\underbrace{b^2\frac{m}{n}}_{\geq 0}=30$$
Therefore $mn\le 30$ and $b=\pm\sqrt{\frac{n}{m}(30-mn)}$.
So by choosing all couples of positive integers $(m,n)$ such that $mm\leq 30$ you will find all possible solutions for your problem:
$$m+ia=m\mp i\sqrt{\frac{m}{n}(30-mn)}\quad,\quad
n+ib=n\pm i\sqrt{\frac{n}{m}(30-mn)}.$$


问:
confused about the definition of subgroup
(2.1) Proposition. Let G be a group and let H be a nonempty subset of G. Then H is a subgroup if and only if the following two conditions are satisfied.
(1) If $a, b \in H$ then $a b \in H$.
(2) If $a \in H$ then $a^{-1} \in H$.

(2.2) Remarks. (1) Conditions (1) and (2) of Proposition 2.1 can be replaced by the following single condition.
$(1)^{\prime}$ If $a, b \in H$ then $a b^{-1} \in H$

Indeed, if $(1)^{\prime}$ is satisfied then whenever $a \in H$ it follows that $e=$ $a a^{-1} \in H$ and then $a^{-1}=e a^{-1} \in H .$ Thus $a \in H$ implies that $a^{-1} \in H$ Also, if $a, b \in H$ then $b^{-1} \in H$ so that $a b=a\left(b^{-1}\right)^{-1} \in H .$ Therefore, $(1)^{\prime}$ implies (1) and (2). The other implication is clear.

I do not understand the remark. Here is what I think. If (1)' is satisfied, whenever $a \in H$,we treat a as b, then $a a^{-1}$=$e\in H$, then using multiplication left cancelation law to get $a^{-1}=e a^{-1} \in H .$ Since $ b \in H$, the same reason to get $b^{-1} \in H$. b=$\left(b^{-1}\right)^{-1}$ for any $b \in H$(H is a group since it is a subgroup). Therefore, (1)′ implies (1) and (2). What is another implication

答: The other implication is that $(1)$ and $(2)$ together imply $(1')$. Indeed, if $a$ and $b$ belong to $H$, then by $(2)$ also $b^{-1}$ is in $H$, and by $(1)$ also $ab^{-1}$ is in $H$, which is condition $(1')$.

ContentLicense: CC BY-SA 4.0


问:
$x^2\ast y = y = y \ast x^2$ implies commutativity?
I would appreciate some help with this task:

Let be $S$ a non-empty set equipped with an operation $\ast :S\times S\to S$ such that $\ast$ is associative and has the property $x^2\ast y = y = y \ast x^2$ for every $x,y\in S$ (here, $x^2 = x\ast x)$. I have to show that $(S,\ast)$ is an abelian group. In other words, I have to show that there is an identity element $e$ such that $x\ast e = e\ast x = x$ for all $x$ and the commutativity $x\ast y = y \ast x$, for all $x,y \in S$

Note that, assuming there exists such element $e$, we have by virtue of the property  $x^2\ast y = y = y \ast x^2$ :
$$ x^2 = x^2\ast e = e  \quad \forall x\in S$$
This means that, in case an identity element exists, every element of $S$ is involutive. But how can I assert the existence of $e$ and, even more, the commutativity?.

Any hint is welcome. Thanks in advance.

答:
Identity element: We have to show $\exists\,e\in S\ \forall\,x\in S:\ ex=x=xe$.

Let $y\in S$ fixed, arbitrary. We have for all $x\in S$:
\begin{align*}
y^2x=x=xy^2\tag{1}
\end{align*}
We see from (1) that $y^2$ is an identity element $y^2=e$. We can also show it is unique: Let $z \in S$ fixed, arbitrary. We have
\begin{align*}
z^2e=\color{blue}{e=}&\color{blue}{ez^2}\\
&\color{blue}{ez^2=e^2z^2=z^2}=z^2e^2=z^2e
\end{align*}
and uniqueness $e=z^2$ of the identity element follows.

Inverse element: We have to show $\forall\, x\in S\ \exists\,y\in S:\ xy=e=yx$.

We have for $x\in S$:
\begin{align*}
x^2e=e=ex^2\qquad\Rightarrow\qquad x^2=e
\end{align*}
showing that $x$ is its own inverse for all $x\in S$.

Commutativity:

Let $x,y\in S$ fixed, arbitrary. We have since $z^2=e$ for all $z\in S$:
\begin{align*}
\color{blue}{xy}&=(xy)e=(xy)(yx)^2=(xy)(yx)(yx)\\
&=x(yy)x(yx)=xex(yx)=x^2(yx)=e(yx)\color{blue}{=yx}
\end{align*}

and the claim follows.


答:
If you represent your complex numbers as
$$w=m+iq \,,~\,~\,~\,~ z=n+ir  \,,$$
for natural numbers $m,n$ (and real numbers $q,r$), then your hypotheses equivocates to
$$mn-qr =30\,,\,~\,~\,~ mr + nq =0\,.$$
It follows from the first that $qr$ is an integer, and from the second equation that it is negative or $q=r=0$; multiplying the second through by $q$ and $r$ also shows that $q^2$ and $r^2$ are both rational numbers. Explicitly (solving them algebraically), that means the possible combinations take the form
$$w=m\pm i\sqrt{m^2-30\frac{m}{n}}\,,~\,~\,~\,~ z=n\mp i\sqrt{n^2-30\frac{n}{m}}\,,$$ with the caveat that $mn\le 30$ (and, a fortiori, the imaginary parts of $w$ and $z$ are of opposite signs).

Body:
Two complex numbers have a product of $30$ and positive integers as real parts. Write all possible combinations of such numbers.

$(3,10), (5,6),(2,15)$ and $(1+ i\sqrt(29),1-i\sqrt(29)$ are few solutions. Is there any neat way to find all? I tried by assuming numbers to be $m+ia$ and $n+ib$ but this was does not seems neat.


问:
Ranking of value based on given quartiles of a range
May be a noob question here, but my query is:

I have a range of values of which I have quantile values:
example:
q = [-10, 0.25, 0.5, 0.75, 10]

I have a score value say .85 which i wish to rank but the ranking should be influenced by the quantiles as well.

I tried min max scaling, but obviously it was not influenced by the mid range values.

My second approach was piecewise approach:

initialize empty range

for each quantile

calculate min and max, calculate range, divide by 25 to get step

build an array with the new step, min and max

add current array to main range.

for the derived range, figure out at which percentile does the input value exist.

i know this may be a bit coding related as well so please forgive in advance.

Basically i need is a rising value of ranking as below in image.

x axis: values in the original range

y axis: ranking percentile.

答:
This is how I interpret your problem: You have a list of increasing values:
$$
\{ x_i \} = \{x_0, x_1, x_2, \ldots, x_n \}\qquad \text{with} \qquad x_n \leq x_{n+1}
$$
and another list which just counts "how far we are in the list before the $i$th entry" in percentages,
$$
\{ y_n \} = \left\{ \frac{0}{n}\cdot 100~\% ,\quad \frac{1}{n}\cdot 100~\%
, \quad \frac{2}{n}\cdot 100~\%, \quad \ldots ,\quad \frac{n-1}{n}\cdot 100~\%, \quad 100\%  \right\}
$$
and the only task is to do linear interpolation between the points $(x_i, y_i)$ ($i=0,1,2,\ldots, n$). This is really easy because we know the values for these points for every index $i$. Wikipedia has a pretty clear article about this: https://en.wikipedia.org/wiki/Linear_interpolation#Linear_interpolation_between_two_known_points

In our case, the difference between the different values of $y$ is a constant $\frac{100~\%}{n-1}$ so the expressions simplify a little bit.

So basically what you want to do is to define a function like this:
$$
f(x) =
\begin{cases}
y_0 + \frac{100~\%}{n-1}\frac{1}{x_1-x_0} (x-x_0), \qquad x\leq x_1 \\
y_1 + \frac{100~\%}{n-1}\frac{1}{x_2-x_1} (x-x_1), \qquad x_1 < x\leq x_2 \\
y_2 + \frac{100~\%}{n-1}\frac{1}{x_3-x_2} (x-x_2), \qquad x_2 < x\leq x_3 \\
\vdots \\
y_{n-1} + \frac{100~\%}{n-1}\frac{1}{x_n-x_{n-1}} (x-x_{n-1}), \qquad x_{n-1} < x\leq x_n \\
\end{cases}
$$
Does this help you?


问:
What's the asymptotic distribution of $p^n$ (powers of primes)?
We know by the prime number theorem that
$\lim_{n\to\infty}\frac{\pi(n)}{n\,/\ln n} = 1$
An even better approximation is
$\lim_{n\to\infty}\frac{\pi(n)}{\int_2^n\frac{1}{\ln t}\mathrm{d}t} = 1$.

Is there a similar formula that approximates the distribution of natural numbers of the form $p^n$ where $p$ is a prime? That is, an approximation of
$$\pi'(x)=\left|\,\Pi'\cap\{1,\ldots,x\}\,\right|
\qquad\mbox{ where }\quad
\Pi'=\{p^n\;|\;p\mbox{ is prime}, n\in\mathbb{N}\}$$

(There is already a question that asks if $\lim_{n\to\infty}\frac{\pi'(n)}{n}=0$, but I'm interested in a more precise approximation.)

答:
The same asymptotics hold, and the intuitive explanation for that is that the number of prime powers up to $x$ are negligible in comparison to the number of primes up to $x$. In fact, the usual proofs of the prime number theorem actually prove the asymptotics for $\pi'$ first, and then one shows that these coincide with the asymptotics of $\pi$.

Since $\pi'(x)$ for a real $x$ counts the number of primes $p$ up to $x$, the number of prime squares $p^2$ up to $x$, the number of prime cubes $p^3$ up to $x$, and so on, we can write
$$\pi'(x)=\pi(x)+\pi(x^{1/2})+\pi(x^{1/3})+\cdots+\pi(x^{1/m})$$
where $m=\log(x)+1$. Now, using the prime number theorem, we get
$$\pi'(x)-\pi(x)\sim x^{1/2}/\log(x^{1/2})+\cdots+x^{1/m}/\log(x^{1/m})$$
which can be bounded above by $x^{1/2}/\log(x^{1/2})\cdot (\log x+1)\leq x^{1/2}/\log(x^{1/2})\cdot 2\log x=4 x^{1/2},$ and thus $\pi'(x)=\pi(x)+O(x^{1/2})$, so $\pi'(x)$ is asymptotic to $x/\log x$.

ContentLicense: CC BY-SA 3.0


答:
Taking norms, we get
$$\sqrt{m^2+a^2}\sqrt{n^2+b^2} = 30$$
where all quantities involved are real. The left hand side is clearly larger than or equal to $\sqrt{m^2}\sqrt{n^2}=mn$ (since the question requires that $m,n>0$). Therefore $30\geq mn$, now there are only a few cases to check.

Title: Two complex numbers have a product of $30$ and positive integers as real parts. Write all possible combinations of such numbers.


问:
Show that $\ker(\phi) \cong \mathbb{Z}^2$
Let $\phi(x,y,u,v):=(x+2y+u, y-3v)$ from $\mathbb{Z}^4 \rightarrow \mathbb{Z}^2$

How do I show that?
$\ker(\phi) \cong \mathbb{Z}^2$

I have tried to set $y=3v$ and then $x+2 \cdot 3v+u = 0 \Leftrightarrow u=-x-6v$ but cannot get any further.

答: You can choose any $x\in \mathbb{Z}$ and $v\in \mathbb{Z}$, and then set $u=x+6v$ and $y=3v.$ Then $\phi(x,y,u,v)=0.$ Thus the kernel is $\mathbb{Z}^2.$


问:
Continuous time Stochastic Process stopping time measurability
Let $\{X_t,\mathcal{F}_t;0\leq t < \infty\}$ be continuous time stochastic processes and $T$ be $\{\mathcal{F}_t\}_{0\leq t < \infty}$ stopping time. How to prove $X_T$ is $\mathcal{F}_T$ measurable?</p>

答: We need additional hypothesis that $X_t$ is progressively measurable and $T$ is finite a.s-P. With this hypothesis we can prove that $X_{\min \{T,t\}}$ is progressively measurable and hence that $X_T$ is $\mathcal{F}_T$ measurable. It is a theorem in chapter 1 of the book "Stochastic Calculus and Brownian Motion" by Karatzas and Shreve.</p>


答: So you have proved that $\ker(\phi)=\{(x,3v,-x-6v,v)\mid x,v\in \mathbb{Z}\}$. Now consider $f:\ker (\phi)\to \mathbb{Z}^2$ defined by $f(x,3v,-x-6v,v)=(x,v)$, clearly $f$ is a bijection, and it is also easy to prove that is an homomorphism (w.r.t addition, I suppose).

Title: Show that $\ker(\phi) \cong \mathbb{Z}^2$

Body:
Let $\phi(x,y,u,v):=(x+2y+u, y-3v)$ from $\mathbb{Z}^4 \rightarrow \mathbb{Z}^2$

How do I show that?
$\ker(\phi) \cong \mathbb{Z}^2$

I have tried to set $y=3v$ and then $x+2 \cdot 3v+u = 0 \Leftrightarrow u=-x-6v$ but cannot get any further.


问:
Total claim amount from the sum of different Poisson distributions
I got the following question:

Let $S$ be the total claim amount.  Find $P(S=s) ,s= 0,...,6$, if $S = Y_1 + 2Y_2 + 3Y_3, Y_ j∼Poisson(j)$.

I don't get why they first add $Y_1$ and $Y_2$ and then $Y_3$. Multiplying column 2 and 3 doesn't result in column 4, it gives different numbers. And also, why is the distribution of $2Y_2$ given by $\frac{2^{(x/2)}}{(x/2)!}$ (and similar for $3Y_3$)? I included the link of the answer below:

Answer given by the notes

答:
Think in t his way:

$Y\sim Po(2)$

find the distribution of

$Z=2Y$

By definition you get

$$P(Z=z)=P(2Y=z)=P(Y=z/2)=\frac{e^{-2}2^{z/2}}{(z/2)!}$$

This is the distribution of $2Y_2$

similarly for $3Y_3$

Now, as an example, to calculate

$$P(S=0)=P(Y_1=0,Y_2=0;Y_3=0)=e^{-6}$$

$$P(S=2)=P(Y_1=2;Y_2=0;Y_3=0)+P(Y_1=0;Y_2=1;Y_3=0)=e^{-6}[2+\frac{1}{2}]$$

same results as in your notes....is it clear how to proceed?

Let's focus for example on $S=3$, as you stated in your comment:

the result is $P(S=3)=e^{-6}[5+\frac{1}{6}]$ and it is given by summing the numbers in the blue circles: $2+\frac{1}{6}+3$ and multyplied by $e^{-6}$ above in the first row.

this can be achieved without the distribution of $2Y_2$ and $3Y_3$ they calculated but simply summing all the interested events.

For example, to get $S=3$ you can have

$\{y_1,y_2,y_3\}=$

$\{3,0,0\} \xrightarrow{\text{with probability}}e^{-6}[\frac{1}{3!}\times1 \times 1]$

$\{1,1,0\} \xrightarrow{\text{with probability}}e^{-6}[1\times2 \times 1]$

$\{0,0,1\} \xrightarrow{\text{with probability}}e^{-6}[1\times1 \times 3]$

Concluding

$$P(S=3)=e^{-6}[\frac{1}{6}+2+3]=e^{-6}\times 5\frac{1}{6}$$

using the "strange" notation of your notes.


问:
Finding a recurrence for $\int(c+\operatorname{sn}u)^{-n}\,du$
Consider the set of integrals
$$I_n=\int_0^{F(\varphi,m)}(c+\operatorname{sn}u)^{-n}\,du=\int_0^{\sin\varphi}\frac1{(c+t)^n\sqrt{(1-t^2)(1-mt^2)}}\,dt$$
This is the series denoted $\sigma_n$ in Byrd and Friedman's Handbook of Elliptic Integrals for Engineers and Physicists. They give formulas that suffice to compute $I_n$ for $n\le2$ (330.00–04, .50, .51).

However, in my work I have come across the need to evaluate $I_3,I_4$ and higher indices of $I_n$. I cannot see a way to convert these integrals into other integrals covered in B&F's "reduction of integrals of elliptic functions" section, and I have a feeling that I will need another recurrence to handle them. How can I derive a recurrence for $I_n$ that allows me to solve $I_3,I_4$ and so on, or can I solve it by any other means? More generally, how did Byrd and Friedman come up with all those recurrences in the first place?

答:
Using the standard reduction formula (B&F 250.02) applied to the algebraic form of $I_n$ gives for all $n\ge2$
$$I_n=\frac1{(n-1)(1-c^2)(1-mc^2)}\left(\frac1{c^{n-1}}-\frac{\operatorname{cn}u\operatorname{dn}u}{(c+\operatorname{sn}u)^{n-1}}+\sum_{j=1}^4C_jI_{n-j}\right)$$
where
$$\begin{align}
C_1&=c(2n-3)(2c^2m-m-1)\\
C_2&=-(n-2)(6c^2m-m-1)\\
C_3&=2cm(2n-5)\\
C_4&=-m(n-3)
\end{align}$$
provided that $c^2\ne1$ and $c^2\ne\frac1m$. Otherwise, B&F 250.03 applies:
$$I_n=\frac1{c(2n-1)(2c^2m-m-1)}\left(\frac{\operatorname{cn}u\operatorname{dn}u}{(c+\operatorname{sn}u)^n}-\frac1{c^n}+\sum_{j=1}^3D_jI_{n-j}\right)$$
where
$$\begin{align}
D_1&=(n-1)(6c^2m-m-1)\\
D_2&=-2cm(2n-3)\\
D_3&=m(n-2)
\end{align}$$
It is perhaps because of the need to distinguish between $P(c)\ne0$ and $P(c)=0$ (where $P$ is the polynomial under root), as well as the complicated factors involved, that B&F does not list recurrence relations for the $I_n$ here.


问:
Determining the Ring of Integers
Let $a,b>1$ be two relatively prime integers with $3$ dividing the product $ab$. Set $f(X) = X^3 - ab^2$. Pick a root $\alpha \in \mathbb{C}$ and consider the extension $K = \mathbb{Q}(\alpha)$. How does one show that,
$$ \mathbb{Z}_K = \mathbb{Z}[\alpha,\tfrac{\alpha^2}{b} ]$$
At first I tried to compute the discriminant of the basis $(1,\alpha,\tfrac{\alpha^2}{n})$ and checking if it is square-free. But it is not.

The second approach I tried was to somehow show that $$\mathbb{Z}[\alpha,\tfrac{\alpha^2}{b} ]$$ is a Dedekind domain by localizing at maximal ideals and showing they are all PIDs. The problem is it is not clear why $\mathbb{Z}[\alpha,\tfrac{\alpha^2}{b} ]$ is a ring of dimension $1$, and if we somehow accept that, then it is not clear what that maximal ideals of this ring are.

答:
As stated, this is not quite true. For example, if $a=p^3$ for a prime $p$ then your ring is $\mathbb Z[p\sqrt[3]{b^2},p^2\sqrt[3]{b}]$ which is certainly not integrally closed. Because of this counterexample, the only interesting case of this problem is when $a$ and $b$ can be chosen to be squarefree, so we will assume this going forward.

Let $\alpha = \sqrt[3]{ab^2}$, $\beta = \sqrt[3]{a^2b} = \alpha^2/b$, so that $1,\alpha,\beta$ is a $\mathbb Q$-basis for $K$. First we obviously have the inclusion $\mathbb Z[\alpha,\beta] \subseteq \mathcal O_K$. Now for $x,y\in K$, let $T(x,y) = \text{tr}_{K/\mathbb Q}(xy)$. Let $A$ be the largest subset of $K$ for which $T(\mathcal O_K,A)\subseteq \mathbb Z$. Then $\mathcal O_K\subseteq A$. If $\gamma = r + s\alpha+t\beta \in \mathcal O_K$, then the conditions $T(1,\gamma), T(\alpha,\gamma), T(\beta,\gamma)\in \mathbb Z$ give $r\in \frac{1}{3}\mathbb Z, s\in \frac{1}{3ab} \mathbb Z, t\in \frac{1}{3ab}\mathbb Z$, so $$A = \frac{1}{3}\mathbb Z + \frac{1}{3ab}\mathbb Z\alpha + \frac{1}{3ab}\mathbb Z\beta.$$ In this way, one can prove the more general fact that if $\alpha$ is a primitive element, then $A= \frac{1}{f'(\alpha)}\mathbb Z[\alpha]$.

So we need to determine "which denominators are allowed." We can see that the possible denominators that can appear come from factors of $a$, and factors of $b$ (since $3$ is known to be a factor of $ab$). Let $p$ be a prime dividing $a$. Then in $\mathbb Z_p$ we have $\alpha^3=ab^2=up$ for some $u\in \mathbb Z_p^\times$, so $L:=\mathbb Q_p(\alpha)$ is a totally ramified extension of $\mathbb Q_p$ and $\alpha$ is a uniformizer. Via general theory, then, $\mathcal O_L = \mathbb Z_p[\alpha]$. In this way, one sees that denominators involving $p$ (in the expression for $A$) cannot appear in an element of $\mathcal O_K$. The same argument works for a prime $q$ dividing $b$. So $\mathcal O_K = \mathbb Z[\alpha,\beta]$.

It's actually relatively doable to compute the ring of integers without the assumption that $3$ divides $ab$. For this, we will WLOG assume $3$ doesn't divide $ab$ and $a,b\equiv 1\bmod 3$ since we can achieve this by negating $a,b$ if necessary.

Let's look at our local argument with the prime $3$ specifically. In this case, $L$ is still totally ramified, but it is less clear what the uniformizer is since $3$ no longer divides $ab^2$. If $ab^2\not \equiv 1\bmod 9$, then $$f(x-1) = x^3-3x^2+3x-1+ab^2$$ is Eisenstein at $3$, so $\mathcal O_L = \mathbb Z_3[\sqrt[3]{ab^2}+1]$ and we see as before that there can be no denominators containing a $3$. If $a^2b\not \equiv 1\bmod 9$, the same argument works.

So in fact we have shown that $\mathcal O_K = \mathbb Z[\alpha,\beta]$ except possibly in the case where $ab^2\equiv a^2b\equiv 1\bmod 9$. This is equivalent to $a\equiv b\bmod 9$. Even in this case, we know $\mathcal O_K\subseteq \frac{1}{3}\mathbb Z[\alpha,\beta]$. Another thing we can use to our advantage in this case is that $\mathcal O_K$ is in fact a ring, so we must have $\mathcal O_K\cdot \mathcal O_K\subseteq \mathcal O_K \subseteq \frac{1}{3}\mathbb Z[\alpha,\beta]$, and in particular if $J=3\mathcal O_K$ is an ideal of $\mathbb Z[\alpha,\beta]$, then $J^2 \subseteq 3\mathbb Z[\alpha,\beta]$. So if $\gamma  = r + s\alpha + t\beta \in J$, we must have $$3\mathbb Z[\alpha,\beta] \ni \gamma^2 \equiv r^2-st + (s^2-rt)\beta + (t^2-rs)\alpha \bmod 3.$$

It is easy to show that all coefficients are $0\bmod 3$ iff $r\equiv s\equiv t\bmod 3$. Thus $\mathcal O_K \subseteq \mathbb Z[\alpha,\beta,\frac{1+\alpha+\beta}{3}]$. In fact, $\mathcal O_K = \mathbb Z[\alpha,\beta,\frac{1+\alpha+\beta}{3}]$. To see this, let $n=\frac{ab^2-1}{9}$. Let $$\delta=\frac{1+\alpha+\alpha^2}{3} = \frac{1+\alpha+\beta}{3} + \frac{b-1}{3}\beta.$$ Then $\delta(\alpha-1) = 3n$, so $\delta$ is a root of $f(\frac{3n}{x}+1) = \frac{27n^3}{x^3}+\frac{27n^2}{x^2}+\frac{9n}{x}-9n$, so $\delta$ is a root of $g(x) = x^3-x^2-3nx-3n^2$, and in particular $\mathcal O_K = \mathbb Z[\alpha,\beta,\delta]$.

So when $a,b\equiv 1\bmod 3$, we have shown that $\mathcal O_K = \mathbb Z[\alpha,\beta]$ except in the exceptional circumstance where $a \equiv b\bmod 9$, and have handled that case appropriately to show that $\mathcal O_K=\mathbb Z[\alpha,\beta,\delta]$.

Title: Determining the Ring of Integers


问:
Find subgroup of $S_{10}$ isomorphic to $G=\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/6\mathbb{Z}$
The only subgroup of $S_{10}$ with the same number of elements as $G$ which I can recall is $S_4$. Is this the subgroup?

答: The group $G=\mathbb Z/4\mathbb Z \times \mathbb Z/6\mathbb Z$ is generated by the two commuting elements $(1,0)$ and $(0,1)$ of orders $4$ and $6$, respectively. In $S_{10}$ we can find a $4$-cycle and a $6$-cycle on disjoint sets to make them commute. Hence, consider the subgroup of $S_{10}$ generated by the cycles $(1\ 2\ 3\ 4)$ and $(5\ 6\ 7\ 8\ 9\ 10)$.


问:
On the pH scale, each unit change in pH represents a tenfold increase in acidity or alkalinity.
Trying to solve similar type equation to this.

On the pH scale, each unit change in pH represents a tenfold increase in acidity or alkalinity. According to the diagram, vinegar is how many times as acidic as pure water?

答:
Hint Each unit decrease in pH represents a tenfold increase in acidity. In other words the increase in acidity was:

$$10^{n-(n-1)}=10^{1}$$

Use some algebra to generalize this.


问:
$\frac{1}{u_n}W_n-v_n\Rightarrow W$ implies $u_n \to +\infty$
Suppose $(X_n)_n$ are i.i.d. random variables and let $W_n = \sum_{k=1}^n X_k$. Assume that there exist $u_n>0 , v_n \in \mathbb{R}$ such that

$$\frac{1}{u_n}W_n-v_n\Rightarrow W$$

where $W$ is not degenerate. Show that

$$u_n\to \infty , \frac{u_n}{u_{n+1}}\to 1$$

What happens to $u_n$ if $W$ is degenerate?

Hint: You may need to consider ${u_{2n}}/{u_n}$.

Is the following attempt, for the first part, true?

In order to remove $v_n,$ we can consider $\frac{1}{u_n}\sum_{k=1}^n(X_{2k+1}-X_{2k})$ which converges in distribution to a non-degenerate random variable $Y.$ So we can suppose that $\frac{1}{u_n}W_n$ converges in distribution to $W$.

If $W$ is non-degenerate then there exist $x \in \mathbb{R};|\phi_W(x)|<1,$ since $\frac{1}{u_n}W_n-v_n\Rightarrow W$ then there exist $k \in \mathbb{N};|\phi_{X_1}(\frac{x}{u_k})|<1$ which means that $X_1$ is not degenerate, if $(u_n)_n$ is bounded from above then there exist a subsequence $(u_{k_n})$ such that $W_{k_n}$ converges in distribution.
Let $(u_{q_n})_n$ be an arbitrary subsequence, since $X_1$ is not-degenerate then $\sum_{l=1}^{q_n}X_l=W_{q_n}$ doesn't converges in distribution, so $u_{q_n}$ is not bounded from above and we can extract a subsequence from $u_{q_n}$ diverging to $+\infty.$

In case $W$ is degenerate, is it possible to know the behavior of $u_n$?

答:
In order to finish the solution we should show that $\frac{u_{n+1}}{u_n} \to 1$.

As $\frac{X_{n+1}}{u_n} =\frac{X_{1}}{u_n} $ and $u_n \to \infty$ we have $\frac{X_{n+1}}{u_n} \to 0$ in distribution. We have

$\frac{\sum_{k=1}^{n+1} X_k - X_{n+1}}{u_n} =  \frac{\sum_{k=1}^n X_k}{u_n} \to W$,  $\frac{X_{n+1}}{u_n} \to 0$, thus $\frac{\sum_{k=1}^{n+1} X_k}{u_n} \to W$. But we know that  $\xi_n = \frac{\sum_{k=1}^{n+1} X_k}{u_{n+1}} \to W$. Put $c_n = \frac{u_{n+1}}{u_n}$. Thus $\xi_n \to W$ and $c_n \xi_n = \frac{u_{n+1}}{u_n} \frac{\sum_{k=1}^{n+1} X_k}{u_{n+1}} = \frac{\sum_{k=1}^{n+1} X_k}{u_{n}} \to W$.

So we know that $\xi_n \to W$, $c_n \xi_n \to W$ in distribution, $c_n >0$, $W$ is nondegenerate and we want to show that $c_n \to 1$.

Let us prove if by contradiction. Suppose that there is $c_{n_k} \to c \in [0,1) \cup (1, \infty]$. Instead of $c_{n_k}$ we will write $c_n$.

According to Skorokhod's representation theorem we may assume W.L.O.G. that $\xi_n \to W$ a.s. (and still $c_n \xi_n \to W$ in distribution).

Case 1. $c = \infty$. As $\xi_n \to W$ a.s., $|W| < \infty$, $c_n \to \infty$, we have $|c_n \xi_n(\omega)| \to (+\infty)$ for $\omega: W(\omega) \ne 0$. But $|c_n \xi_n| \to |W| < \infty$ in distribution.  Thus $W = 0$ a.s. It's contradiction.

Case 2. $c=0$. As $\xi_n \to W$ and $c_n \to 0$, we have $c_n W_n \to 0$, but $c_n W_n \to W$.  Thus $W = 0$ a.s. It's contradiction.

Case 3. $c \in (0,1) \cup (1,\infty)$. As $\xi_n \to W, c_n \xi_n \to W, c_n \to c$, we have $W = cW$ in distribution and hence $\frac{1}{c}W = W$ in distribution.

We see that it's sufficient to show that $W = dW$ in distribution with $d \in (0,1)$ (and $d=c$ or $d = \frac1{c}$) is impossible.

As $W =dW$, we have $$W = dW = d(dW) = d^2 W = \ldots = d^n W $$
in distribution.

Put $\eta_n = d^n W$. Thus $\eta_n \to 0$, because $d^n \to 0$. But $\eta_n = W$ in distribution. Thus $W=0$. It's contradiction.

Hence we got a contradiction in every case, q.e.d.


问:
Hamiltonian diffeomorphisms generated by periodic Hamiltonian function
Let $\phi_{t}$ be a family of Hamiltonian diffeomorphisms on a symplectic manifold $(M,\omega)$, generated by a family of Hamiltonians $H(x,t)$. I'm trying to show that the family of Hamiltonians is periodic with period $1$ with respect to $t$ (in other words $H(x,t) = H(x,t+1)$ for all $x$) if and only if the family of diffeomorphisms satisfies $\phi_{t+1} = \phi_{t} \circ \phi_{1}$.

I know from the definition that $\frac{d}{dt}\phi_{t}(x) = X_{H}(\phi_{t}(x))$, where $X_{H}$ is the Hamiltonian vector field and that $i_{X_{H}}\omega = dH$. The symplectic form is compatible with this vector field in the sense that $\omega(X_{H}(x), \cdot) = dH$.

I tried to differentiate the equation $\phi_{t+1}=\phi_{t} \circ \phi_{1}$ and I obtain the following:
$$X_{H}(\phi_{t+1}(x)) = X_{H}(\phi_{t}(\phi_{1}(x))) \cdot X_{H}(\phi_{1}(x))$$
But I don't see how to proceed or finish the proof for either of the implications. How can I prove this?

答:
For one implication, assume that $H(x,t)=H(x,t+1)$. To prove that $\phi_{t+1}=\phi_{t}\circ\phi_{1}$, you can show that both sides solve the initial value problem
$$
\begin{cases}
\frac{d}{dt}\psi_{t}=X_{H_t}\circ\psi_{t}\\
\psi_{0}=\phi_1
\end{cases}.
$$
Indeed, on one hand we have
$$
\frac{d}{dt}\phi_{t+1}=X_{H_{t+1}}\circ\phi_{t+1}=X_{H_t}\circ\phi_{t+1},
$$
using that $H_{t+1}=H_t$. On the other hand we have
$$
\frac{d}{dt}\phi_{t}\circ\phi_{1}=X_{H_t}\circ\phi_t\circ\phi_1.
$$

The other implication doesn't seem true to me. Information about the flow tells you something about the Hamiltonian vector field, which does not determine the Hamiltonian function uniquely. For instance, take $$(\mathbb{R}^{2},dx\wedge dy)\ \ \text{and}\ \ H(x,y,t)=y+t.$$ Then
$
X_{H}=\partial_{x}
$
is time-independent, so its flow satisfies $\phi_{t+1}=\phi_t\circ\phi_1$. But $H(x,y,t+1)\neq H(x,y,t)$.


问:
Let A ⊂ B ⊂ C. If C/A and C/B are normal extensions then B/A is also normal.
Let A ⊂ B ⊂ C. If C/A and C/B are normal extensions then B/A is also normal. Is that true? If it is not I couldn’t find a contra-example for that? Can someone help me ?</p>

答: Try for example the splitting field of $X^3-2$ over $\mathbb{Q}$. That should provide a counter example.


问:
I wish to change my old algorithm by another old algorithm RATCONVERT

I have an idea for my old problem was mentioned hereFRACTIONS. I saw one relationship between it and another one so I tried to optimize that given problem by similar method. For example,

$$\frac{1}{6}\rightarrow \frac{2}{7}\rightarrow \frac{3}{8}\rightarrow \frac{4}{9}\rightarrow \frac{5}{10}:=\frac{1}{2}\rightarrow \frac{2}{3}$$

I observe
$$\frac{2}{3}= \frac{1}{1+ \frac{1}{2}}$$
$$\frac{1}{6}= \frac{1}{2\cdot 3}\rightarrow \frac{1}{2}$$
Because $\dfrac{1}{a}$ is always in the desired form, and what we need to do is finding $\dfrac{1}{a}\rightarrow \dfrac{1}{b}$ by the greatest divisor of $6$ is $3.$ This way is useful to create lowest terms with continued-fractions.

Another example:
$$\frac{2}{3}\rightarrow \frac{3}{4}\rightarrow \cdots \rightarrow \frac{11}{14}\neq \frac{11}{15}$$
Because
$$\frac{11}{15}= \frac{1}{1+ \frac{1}{2+ \frac{1}{1+ \frac{1}{3}}}}$$
and
$$\frac{1}{6}< \frac{1}{3}< \frac{1}{2}$$
And for complex counter-example:
$$\frac{2}{3}\rightarrow \frac{25}{26}\neq \frac{25}{34}= \frac{1}{1+ \frac{1}{2+ \frac{1}{1+ \frac{1}{3+ \frac{1}{2}}}}}$$
On the other hand, seems like it involving toRATCONVERT, I can't do it, at least, I completed the algorithm, I will show you in the environment C++:

```<code>#include <iostream>
#include <math.h>
using namespace std;
int main() {
    long long k, m, q;
    cin>>k>>m;
    long long r1=m, r2=k;
    int v1=0, v2=1;
    while (true) {
        if (v2>=sqrt(m/2)) {
            cout<<0<<", "<<0;
            break;
        }
        else {
            if (r2<sqrt(m/2)) {
                cout<<v2/sqrt(v2*v2)*r2<<", "<<sqrt(v2*v2);
                break;
            }
            else {
                q=round(r1/r2);
                r1=r1-q*r2;
                v1=v1-q*v2;
                swap(r1, r2);
                swap(v1, v2);
            }
        }
    }
}
</code>```

So $u= \dfrac{Kv- r}{M}$ satisfied the conditions, how to change it to solve this problem, I need to the help, very much thank you.

答:
Assuming that we are given four positive integers $a,b,c,d$ satisfying $$\gcd(a,b)=\gcd(c,d)=1,\qquad \frac ab\lt \frac cd\lt 1,\qquad \frac{bc-ad}{d-c}\in\mathbb Z$$I'll show a way to reduce the number of the steps from $\dfrac ab$ to $\dfrac cd$ when $D(a/b)\gt D(c/d)$ where $D(x/y):=y-x$.

In order to reduce the number of steps, we want to find the smallest positive integer $c$ such that there is an integer $k\ge 2$ satisfying $b+c\equiv a+c\equiv 0\pmod k$.

Since $b-a\equiv 0\pmod k$, we see that $k$ has to be a divisor of $b-a$.

For each such $k$, let $c_k:=k\left\lfloor\dfrac ak+1\right\rfloor-a$ where $c_k$ is the smallest positive integer such that $c\equiv -a\pmod k$, and $\lfloor x\rfloor $ denotes the greatest integer less than or equal to $x$.

Now, what we want is $\min(c_k)$ for which we have $$\dfrac ab\rightarrow \dfrac{a+\min(c_k)}{b+\min(c_k)}=\dfrac{\dfrac{a+\min(c_k)}{k'}}{\dfrac{b+\min(c_k)}{k'}}$$where $c_{k'}=\min(c_k)$.

Here, we can say that $\min(c_k)$ is smaller than the second smallest positive divisor $\min(k)$ of $b-a$ since $$c_k=k\left\lfloor\dfrac ak+1\right\rfloor-a\lt k\bigg(\dfrac ak+1\bigg)-a=k.$$

Also, since $c_k\ge k-a$, we can ignore $k$ such that $k-a\ge \min(k)$.

Example :

If $(a,b,c,d)=(50,127,2,3)$, then since $D(a/b)\gt D(c/d)$ holds, we can reduce the number of the steps using the above method. The divisors larger than $1$ of $b-a=77$ are $k=7,11$ and $77$. Here, $k-a\ge\min(k)=7$ holds for $k=77$, so we can ignore $k=77$, and have $\min(c_7,c_{11},c_{77})=\min(c_7,c_{11})=\min(6,5)=5$ to get $\dfrac{50}{127}\to \dfrac{50+5}{127+5}=\dfrac{5}{12}$. Now, $D(5/12)\gt D(c/d)$ holds, so we can again reduce the number of the steps. Since $7$ is the only divisor larger than $1$ of $12-5=7$, we have $\dfrac{5}{12}\to \dfrac{5+c_7}{12+c_7}=\dfrac{5+2}{12+2}=\dfrac 12$ where we finally have $D(1/2)=D(2/3)$. In conclusion, we have$$\dfrac{50}{127}\to\bigg(\dfrac{50+5}{127+5}=\bigg)\dfrac {5}{12}\to\bigg(\dfrac{5+2}{12+2}=\bigg)\dfrac 12\to\frac 23$$where the number of the steps is $3$ which is smaller than $x=104$ satisfying $\dfrac{a+x}{b+x}=\dfrac cd$.


问:
Prove this following integral inequality
Prove that $$\int_{1}^{\infty}\frac{(\sqrt{u+b})du}{u^2 + b} > \frac{\pi}{2}, b > 0$$

I know it is necessary to show the work done, but i already tried by parts and by substitution, and it does not help in nothing. My guess is that it will involves trigonometric functions, but i don't know how to get rid of the b in the integrals. Any tip?

答:
Here's a partial answer / an idea for an approach.

For $b \geq 0$, let
$$\ell(b) = \int_1^\infty \frac{\sqrt{u+b}}{u^2+b} \; \mathrm{d}u$$

We see that
$$\ell'(b) = \int_1^\infty \frac{\partial}{\partial b} \frac{\sqrt{u+b}}{u^2+b} \; \mathrm{d}u = \int_1^\infty \frac{u^2 - 2u - b}{2\sqrt{b+u}(b+u^2)^2} \; \mathrm{d}u \geq \int_1^\infty \frac{u^2 -2u - b}{2(b+u^2)^{5/2}}\; \mathrm{d}u,$$
and we can actually do this last integral! It turns out to be
$$\int \frac{u^2 - 2u - b}{2(b+u^2)^{5/2}} \; \mathrm{d}u = \frac{b (2-3 u)-u^3}{6 b \left(b+u^2\right)^{3/2}} + c,$$
so
$$\ell'(b) \geq \int_1^\infty \frac{u^2 - 2u - b}{2(b+u^2)^{5/2}} \; \mathrm{d}u = \frac{\frac{1}{\sqrt{b+1}}-1}{6 b}.$$

Now, for any $b > 0$, we have

$$\ell(b) = \ell(0) + \int_0^b \ell'(t) \; \mathrm{d}t \geq \ell(0) + \int_0^b \frac{\frac{1}{\sqrt{t+1}}-1}{6 t} \; \mathrm{d}t = 2 + \frac{1}{3} \log \left(\frac{2}{1 + \sqrt{1 + b}}\right).$$

For $b \leq 38$ this is enough to show $\ell(b) > \pi/2$. With a better bound on $\ell'$ we might be able to improve this.


问:
What's the correct way to write a multiplication between scalar and vector?
The rule of matrix multiplication states that, the number of columns of the left operand equals the number of rows of the right operand.

$M*N$ if M has $n$ columns and N should have $n$ rows.

Following this convention, the natrual way to write a multiplication between a vector and a scalar is to put the vector on the left side --- taking scalar as a 1 by 1 matrix.

however, I found that, quite often people do not follow above rule: using eigen decomposition as an example:

wiki of eigen decomposition

$A\upsilon=\lambda\upsilon$

Is there any rule-of-thumb to guide us when we should put scalar on the left side ?

答:
Scalar multiplication and matrix multiplication are 2 separate operations. Even though they have the same word "multiplication" in them - they are completely different.

Matrix multiplication isn't commutative - so you have to put the right matrix on the right side, it's not about conventions. Scalars are commutative and you can put them on either side.

I don't think there's a written convention per se - people simply got used  to putting coefficients before other terms. If you put a scalar on the right, depending on the field you're working in some people reading your expressions may stop and think "hugh, wait, are we working with non-commutative algebra?" for a moment. Also some people may think "hugh, is this a scalar or am I missing something?". It may take some extra brain-cycles for a reader, so I'd keep scalars on the left, but it probably won't be a tragedy if you put them on the other side.

While it's possible to imitate scalar multiplication using $1\times n$ or $n \times 1$ matrices - that's not what it is in its essence. Again - these are different operations and only one of them is commutative.


问:
Adjoining different prime roots results in non-isomorphic number fields?
Let $n \ge 2$ and $p,q$ be two different primes. By Eisenstein the polynomials $X^n-p$ and $X^n-q$ are irreducible. My intuition tells me that $\mathbb{Q}(\sqrt[n]{p})$ and $\mathbb{Q}(\sqrt[n]{q})$ are not isomorphic. But for $n>3$ I didn't come up with a proof but the idea of using the discriminant. Is the statement correct and if so how does one prove it?

Is a general formular for the ring of integers known? Even in case $n=2$ this is not always $\mathbb{Z}[\sqrt[n]{p}]$ (in case $p \equiv 1 \pmod 4$ it is bigger, namely $\mathbb{Z}[(1+\sqrt{p})/2]$). Using the knowledge of the ring of the integers one could compute the diskriminants of the number fields and compare them.

While formulating my question I got this older question suggested. It's quite similar but I'd still like to keep my question since it's a bit more generall (in my question $n$ is not required to by prime nor to be different from $p$ and $q$).

答:
Why does your argument for $n \leq 3$ not work for $n > 3$?
If such fields were isomorphic for distinct primes $p$ and $q$ then the field $K := \mathbf Q(\sqrt[n]{p})$ is also $\mathbf Q(\sqrt[n]{q})$ since an isomorphism with an abstract $n$th root would work with real $n$th roots by looking at a real embedding of $K$, which does exist. The point is that those $n$th roots both have degree $n$ over $\mathbf Q$.

Here is the key point: if $r \in \mathbf Q$ and $x^n - r$ is irreducible where $n > 1$, then for the field $K = \mathbf Q(\sqrt[n]{r})$ we have ${\rm Tr}_{K/\mathbf Q}(\sqrt[n]{r}) = 0$.  If also $K = \mathbf Q(\sqrt[n]{s})$ for another $s \in \mathbf Q$, then ${\rm Tr}_{K/\mathbf Q}(\sqrt[n]{s}) = 0$ too.

If you set up a hypothetical $\mathbf Q$-linear relation
$$
\sqrt[n]{q} = a_0 + a_1\sqrt[n]{p} + a_2\sqrt[n]{p}^2 + \cdots + a_{n-1}\sqrt[n]{p}^{n-1}
$$
apply ${\rm Tr}_{K/\mathbf Q}$ to both sides.  Because $\mathbf Q(\sqrt[n]{p}) = \mathbf Q(\sqrt[n]{q})$, all terms have trace $0$ except the constant term, which has trace $na_0$. This has to be $0$, so $a_0 = 0$. Then divide by $\sqrt[n]{p}$ on both sides and take the trace again (the trace of $\sqrt[n]{q/p}$ from $K$ down to $\mathbf Q$ is again $0$) to see $a_1 = 0$ and so on all the way up, until finally you have $\sqrt[n]{q} = a_{n-1}\sqrt[n]{p}^{n-1}$. Now $a_{n-1}$ is necessarily nonzero. Since $n \geq 2$, take $n$th powers to get $q = a_{n-1}^np^{n-1}$, which is impossible by comparing the power of $q$ on both sides.


问:
If $fg$ is continuous at $a$ then $g$ is continuous at $a$.

Suppose that $f$ and $g$ are defined and finite valued on an open interval $I$ which contains $a$,
that $f$ is continuous at $a$, and that $f(a) \neq 0$.
If $fg$ is continuous at $a$ then $g$ is continuous at $a$.

$\underline{Attempt}$

Since $f$ is conituous at $a$ and $fg$ continuous at $a$,

$$\lim_{x\to a}f(x)=f(a)  \text{ and } \lim_{x\to a}f(x)g(x)=f(a)g(a)$$

so

$$\lim_{x\to a} {f(x)g(x)} = \lim_{x\to a}f(x) \lim_{x\to a}g(x)=f(a)\lim_{x\to a}g(x)=f(a)g(a)$$

since $f(a) \neq0$

$$\lim_{x\to a}g(x)=g(a)$$

$\therefore g$ is continuous at $a$

答:
Your proof is not correct. You are assuming existence of $\lim_{ x \to a} g(x)$ but you have to prove the existence of this limit.  Write $g(x)$ as $\frac 1 {f{(x)}} {g(x)f(x)}$ observing that $f(x) \neq 0$ if $|x-a| $ is small enough. Now you can see that the limit exists and equal $\frac {f(a)g(a)} {f(a)}=g(a)$.

[There exists $\delta >0$ such that $|x-a| <\delta$ implies  $|f(x)-f(a)| <\frac  {|f(a)|} 2$. So  $|x-a| <\delta$ implies  $|f(x)| >|f(a)| -\frac  {|f(a)|} 2=\frac  {|f(a)|} 2>0$ and so $f(x) \neq 0$].


答:
This is just a matter of notational conventions. Usually the axioms of a vector space are formulated by writing scalar multiplication in the form
$$\lambda \cdot v$$
where $v \in V$ and $\lambda$ belongs to the ground field $K$. The reason is that we usually understand that in the product $\mu \cdot \lambda$ of elements of $K$ we have a first factor $\mu$ and a second factor $\lambda$. In a field (whose multiplication is commutative) the order of factors seems to be irrelevant (because $\mu \cdot \lambda =  \lambda \cdot \mu$), but in a ring $R$ (whose multiplication is in general non-commutative) the order is essential. This applies for example to the ring of $n\times n$-matrices over a field. One of the axioms of a vector space is
$$(\mu \cdot \lambda) \cdot v = \mu \cdot (\lambda \cdot v)$$
which is mnemonically easier than the the same formula written via scalar multiplication from the right
$$v \cdot (\mu \cdot \lambda) = (v \cdot \lambda) \cdot \mu .$$
Okay, for a field this does not make much difference since it says the same as
$$v \cdot (\lambda \cdot \mu) = (v \cdot \lambda) \cdot \mu .$$
But note that the concept of a vector space can be generalized to that of a module over a ring $R$ and here the order makes a difference. In fact, one distinguishes between left and rght $R$-modules. For left $R$-muodules one usually writes scalar mutliplication as $\lambda \cdot v$, for right $R$-modules as $v \cdot \lambda$. See here.

Now let us come to the the core of your question. The matrix product $A \bullet B$ is usually defined for an $m\times n$ matrix $A$ and an $n\times p$ matrix $B$, i.e. we require that the number of columns of $A$ is equal to the number of rows of $B$. As you say, a scalar $\lambda$ can be regarded the $1 \times 1$ matrix $(\lambda)$ . Thus the following two expressions are defined:
$$(\lambda) \bullet A \text{ for } 1 \times n \text{ matrices } A \tag{1} $$
$$A \bullet (\lambda) \text{ for } n \times 1 \text{ matrices } A \tag{2} $$
In $(1)$ $A$ is called a row vector, in $(2)$ a column vector.

It therefore depends on your favorite notation: If you regard elements of $K^n$ as row vectors, you have to use $(1)$, if you regard them as column vectors, you have to write $(2)$.

Anyway, this is only relevant if you insist by all means to understand the scalar product of $\lambda$ and $A$ as a matrix product. Usually for $A = (a_{ij})$ one simply defines
$$ \lambda \cdot (a_{ij}) = (\lambda \cdot a_{ij}) .$$
Doing so it does not matter if you regard elements of $K^n$ as row vectors or as column vectors.

Title: What's the correct way to write a multiplication between scalar and vector?

Body:
The rule of matrix multiplication states that, the number of columns of the left operand equals the number of rows of the right operand.

$M*N$ if M has $n$ columns and N should have $n$ rows.

Following this convention, the natrual way to write a multiplication between a vector and a scalar is to put the vector on the left side --- taking scalar as a 1 by 1 matrix.

however, I found that, quite often people do not follow above rule: using eigen decomposition as an example:

wiki of eigen decomposition

$A\upsilon=\lambda\upsilon$

Is there any rule-of-thumb to guide us when we should put scalar on the left side ?


问:
Solving a third order polynomial matrix equation
I want to solve the following matrix equation with respect to the matrix variable $\mathbf{X}$ which is a real symmetric positive definite matrix. The given matrix $\mathbf{A}$ is real and symmetric, "a" is a scalar, and $\mathbf{I}$ is the identity matrix of the appropriate size.

$$\mathbf{X}^{-1}\mathbf{X}^{-1} + \big(\mathbf{X}-\mathbf{I}\big)^{-1} +a\mathbf{X}= \mathbf{A} $$

If we ignore the first term in this equation, there is a closed form solution in the following paper, eq. (14)  and Lemma 2.1 (I need a wise closed form solution like this one!)
http://www.optimization-online.org/DB_FILE/2009/09/2409.pdf

答:
Hint: Try to show that if $A$ has the eigendecomposition $V\Lambda V^T$ then $X$ must have the eigendecomposition $VDV^T$ where $\Lambda = D^{-2}+(D-I)^{-1}-aD$. In other words, $X$ must have the same eigenvectors as $A$ and for each eigenpair $(\lambda,v)$ of $A$, we have that $(d,v)$ is an eigenpair of $D$ where $\lambda = \dfrac{1}{d^2}+\dfrac{1}{d-1}-ad$.

From there, you simply need to solve the equation $\lambda = \dfrac{1}{d^2}+\dfrac{1}{d-1}-ad$ for $d$ in terms of each eigenvalue $\lambda$ of $A$.


问:
Monotone numerical scheme and entropy solution to conservation laws
A monotone numerical scheme is when I have two sequences $v=(v_j)_j$, $w=(w_j)_j$, and $v^{(n)} \ge w^{(n)}$ , then $v^{(n+1)} \ge w^{(n+1)}$ , where $ v_j^{(n)}$ is an appoximation of $u(x_j,t_n)$. Or in other words if I have a $2k+1$ Explicit difference scheme, then the derivatives at time $t_n$ are positive.

Why is a monotone numerical scheme a good approximation for an entropy solution? Is it because it ensures the existence of TVD schemes, which prevent oscillation, or does it have a direct relationship to the Entropy solution?

答:
There is a theorem that links conservative monotonous schemes for conservation laws to entropy solutions:

Theorem. If a numerical method for the conservation law $u_t + f(u)_x = 0$ is conservative, consistent and monotonous, then

The scheme is $L^\infty$ stable, and $\|u_h\|_\infty \leq \|u_0\|_\infty$;

The scheme is entropic;

The scheme is convergent (towards the entropy-satisfying weak solution).

Such a scheme is of order 1, at most.


问:
Vectors of complex or hypercomplex numbers?
Is there a field which studies vectors of complex numbers or vectors of hypercomplex numbers in general?
Is it useful to think of a mono audio signal as a vector of complex numbers where imaginary part encodes the timing information? Similarly, is it useful to think of a grayscale image as a quaternion vector where imaginary components encode the pixel position? Can this kind of logic be fruitful in any way?

Source: https://arxiv.org/abs/2006.08321

答:
Is there a field which studies vectors of complex numbers or vectors of hypercomplex
numbers in general?

Not a dedicated field as such; the machinery for $\Bbb C^n$ or $\Bbb H^n$ is basically the same as for $\Bbb R^{jn}$, where $j\in\{1,\,2,\,4\}$ depending on what you're doing, but with some slight adjustments such as inner products being sesquilinear rather than bilinear.

Is it useful to think of a mono audio signal as a vector of complex numbers where imaginary part encodes the timing information?

Or even as a complex-valued function, depending on whether you'll use a Fourier series or Fourier transform to switch from time space to frequency space (or vice versa).

is it useful to think of a grayscale image as a quaternion vector where imaginary components encode the pixel position? Can this kind of logic be fruitful in any way?

It's neither useful nor fruitful. At least with audio signals complex numbers help encode phase information and facilitate the solving of relevant differential equations. But quaternions are only worth using if you'll multiply them, because the only difference between $\Bbb H$ and $\Bbb R^4$ is that we define a multiplication operation on the former. Are you going to multiply two colours to get a third colour? I don't think so. In fact, the quaternion you get won't even be part of your $3$-dimensional subset in general.

You could do something like this with CMYK, provided you use a coordinate transformation that lets each component be unbounded, but it'd still be pointless. If for example you learn yellow times pink is green (which wouldn't even stay true if you switched to another transformation), is that useful? No, not at all.


问:
If $ n= a_0 + a_1p+ \cdots +a_rp^r $, then show that $\frac{\sum a_i}{ p-1} \leq \frac{\log(n)}{\log(p)} + 1$
Write , for prime number $p$ and natural number $n \in \mathbb{N} $
$$ n= a_0 + a_1p+ \cdots +a_rp^r , \hspace{5mm} 0 \leq a_i \leq p-1$$

Then show that $$ \frac{\sum a_i}{ p-1} \leq \frac{\log(n)}{\log(p)} + 1$$

[My attempt]

I find $$ \frac{1}{p-1} = \sum_{k=1}^{\infty}p^{-k}$$

Then, we show that $$ \frac{\sum a_i}{ p-1} = \sum \sum_{k=1}^{\infty} a_ip^{-k} \leq \frac{\log(n)}{\log(p)} + 1$$
but I don't know how to prove this.

Or $$ \frac{\sum a_i}{ p-1} \leq \frac{\log(n)}{\log(p)} + 1 \Longleftrightarrow \log(p) \frac{\sum a_i}{ p-1} \leq \log(n) + \log(p) \Longleftrightarrow p^{\frac{\sum a_i}{ p-1}} \leq np$$

Therefore, we show that $$ p^{\frac{\sum a_i}{ p-1} - 1} \leq n = a_0 + a_1p+ \cdots +a_rp^r $$

but I'm stuck here.

Thank you for your attention.

答: I think that there is an easier way to prove your claim: observe that $\frac{\log(n)}{\log(p)}=\log_p(n)$. If we assume that $r$ is maximal such that $a_r\neq 0$, then $\log_p(n)\geq r$. Hence, the claim would follow if we proved that $\frac{\sum a_i}{p-1}\leq r-1$, or equivalently that $\sum_{i\leq r} a_i\leq (r+1)(p-1)$. But this follows from the fact that $a_i\leq p-1$ for every $i$.


问:
Mathematical symbol is "relative compliment"?
A bit of a dumb question, but it's blocking my work at the moment so I'm hoping it will be tolerated !

I have a research paper describing an approach to Convex Quadratic problems (D. GoldFarb and A. Idnani 20 Oct. 1981).

Unfortunately it's a scanned copy (I bought online and downloaded) and I cannot quite read one of the symbols for sure (I am not an expert on this sort of maths).

I wander if anyone can confirm if this symbol is "Relative compliment" ? Which is what I suspect. I have circled the symbol in red:

答:
Yes, the complement of a set is written with the symbol $\setminus$ :

$$B\setminus A=\{x \in B| x \not\in A\}.$$

See here for details.


答: Yes, the idea is that in b case A becomes smaller and smaller (hence K-A becomes bigger and bigger) until all constraints are satisfied (a case)</p>

Title: Mathematical symbol is "relative compliment"?

Body:
A bit of a dumb question, but it's blocking my work at the moment so I'm hoping it will be tolerated !

I have a research paper describing an approach to Convex Quadratic problems (D. GoldFarb and A. Idnani 20 Oct. 1981).

Unfortunately it's a scanned copy (I bought online and downloaded) and I cannot quite read one of the symbols for sure (I am not an expert on this sort of maths).

I wander if anyone can confirm if this symbol is "Relative compliment" ? Which is what I suspect. I have circled the symbol in red:


问:
What is the integral of three orthonormal basis functions?
The hyperspherical harmonics, given by:

$Z_{l,m}^n(\omega,\theta,\phi)=(-i)^l\frac{2^{l+1/2}l!}{2\pi}\sqrt{(2l+1)\frac{(l-m)!}{(l+m)!}\frac{(n+1)(n-l)!}{(n+l+1)!}}\sin^l(\omega/2)C_{n-l}^{l+1}(\cos(\omega/2))P_l^m(\cos\theta)\exp(im\phi)$

(where $C_{n-l}^{l+1}$ is a Gegenbauer polynomial, and $P_l^m$ is an associated Legendre function), form an orthonormal basis for an expansion of functions on the hypersphere.

As such, the following identity is true:

$\int_\Omega Z_{l,m}^n(\omega,\theta,\phi)Z_{l',m'}^{n'}(\omega,\theta,\phi)d\Omega=\delta_{n,n'}\delta_{l,l'}\delta_{m,m'}$

My question is whether or not the following identity is also true:

$\int_\Omega Z_{l,m}^n(\omega,\theta,\phi)Z_{l',m'}^{n'}(\omega,\theta,\phi)Z_{l'',m''}^{n''}(\omega,\theta,\phi)d\Omega=\delta_{n,n',n''}\delta_{l,l',l''}\delta_{m,m',m''}$

Note: $\Omega$ is the entire domain, i.e., $\omega\in[0,\pi], \theta\in[0,\pi], \phi\in[0,2\pi]$, and $d\Omega=\frac{1}{2}\sin^2(\omega/2)\sin(\theta)d\omega d\theta d\phi$

答:
I don't know whether or not this is true in your particular case, but I'd like to point out that this isn't abstractly true: that is, there isn't a general argument about orthonormal functions that will get you from the first identity to the second. For example, some appropriate normalization of the functions $\cos nx$ are orthonormal on $[-\pi, \pi]$, but it's false that

$$\int_{-\pi}^{\pi} \cos n_1 x \cos n_2 x \cos n_3 x \, dx$$

is zero whenever the $n_i$ are distinct: I think it should be nonzero if $n_1 + n_2 = n_3$, for example.


问:
proving congruence question

Let $k\in \mathbb{N}.$ How do I prove that:$$\left(2^{{(3}^k)}\right)^2-2^{{(3}^k)}+1\equiv0\ \left(\bmod\ 3\right)$$

I have tried to use Fermat's little theorem, to prove that: $(2^k)^3\equiv 2^k\pmod{3}$, but from there I don't know how to proceed.

答:
Hint:

We know that $2 \equiv -1 \pmod{3}$.

$(-1)^r \equiv -1 \pmod{3}$ if $r$ is odd. Try to identify the suitable $r$ in your context. From here, hopefully you can complete the task.

Title: proving congruence question


问:
setting the value of a variable in such equation to have a specific output
Assume I have, for example, $x = 0.7 + 0.7i$ and the equation as below:
$$\tag{1}
(0.5+0.5i)y + (0.5 - 0.5i)x = 0
$$

in that case $y$ has the same magnitude of $x$. At the same time, if we swap $x$ and $y$ in  equation $(1)$, we will also have the same magnitude of $x$ and $y$. By this I mean to make the equation as:
$$
(0.5+0.5i)x + (0.5-0.5i)y = 0
$$
Now, if we have changed the above equation into:
$$\tag{2}
(0.5+0.5i)y + (0.5-0.5i)x = 0.5
$$
It means the $0$ becomes a constant such as $0.5$, What should be $x$ and $y$ in function of the constant $0.5$ in order to keep the property of first equation, which is having the same magnitude for $x$ and $y$ in both cases of  keeping them as it is or swapping them?.

答:
You previously had 0.5y + 0.5x = 0 and 0.5iy - 0.5ix = 0

Now you have (1+i)y + (1-i)x = 1
Hence x+y = 1 and iy - ix = 0

In both cases it guarantees x and y to have the same magnitude.

This is also symmetric set of equations, however if you want to keep the same magnitude, then you enforce x and y to have the same real argument equals 0.5

This is a bit intuitive since you change real argument on RHS.


问:
Determine domain of double integrals
So I have a double integral $ \int_D e^{x/y} \; dA $ at the Domaine $\{(x,y) : 1 \leq y \leq 2 \text{ and } y \leq x \leq y^3 \}$

My question is: how do I translate this to the integration bounds for the two integrals.

I have written then domain in geogebra and saw where the two areas intersected, and this gives me: $\{1 \leq x \leq 8 \text{ and } 1 \leq y  \leq 2\}$

Is this the correct method? I am asking because I'm getting some wierd results when calculuating the integrals.
Hope you can help me!

答:
That is not the correct limits. Please note that your region is $y \leq x \leq y^3$ and $1 \leq y \leq 2$.

So your integral is

$\displaystyle \int_{1}^{2} \int_y^{y^3} e^{x/y} \, dx \, dy$

Here is the sketch (you need to integrate over the region shaded in black) -

Title: Determine domain of double integrals


问:
Find homomorphism where $\ker(\phi) \cong \mathbb{Z}^3$
I need to find a homomorphism where $\ker(\phi) \cong \mathbb{Z}^3$ with $$\phi:\mathbb{Z}^4\rightarrow\mathbb{Z}^2$$

I have set $\phi(x,y,u,v)=(x,x)$

Is this a correct answer?

答:
Yes, your answer is correct, since $${\rm ker}(\phi) = \{(0,y,u,v);u,v,y \in \mathbb{Z}\}$$

Simply put $\psi(0,y,u,v) = (y,u,v)$, then $\psi$ is isomorphism between ker$(\phi)$ and $\mathbb{Z}^3$


问:
Is mutual independence not commutative? Necessity of directed edges in dependency digraph.
Is there a point in saying something is mutually independent of something instead of saying these two things are mutually independent?

Because that is the language used in the Lovasz Local Lemma: $A_i$ is mutually independent of all the events $\{A_j\ :\ (i,j)\notin E\}$

I'm reading about the Lovasz Local Lemma and I don't understand why the graph needs to have directed edges. If there is an edge $(i,j)$ it means $i$ is dependent on $j$. Wouldn't that imply there should also be an edge $(j,i)$?

Maybe a simple example that shows the necessity of directed edges could help.

In my understanding mutual independence doesn't care about the order, but it is a binary function on sets of events.

$f:A\times A\mapsto\{0,1\}$ where $A$ is a subset of the power set of all elementary events. But the dependency graph's vertices are individual events so if we wanted to have a simple graph completely describing all the dependencies and independencies, we would need one vertex per set of events, which would be a lot of vertices (if $A_1,\dots,A_n$ are our elementary events from the lemma, then we would need $2^n$ vertices instead of just $n$ as it is the case in the Lemma), so using directed edges each elementary event has its own set of dependent events.

I guess it comes down to how much information is in a digraph versus a simple graph. There are $2^{n\choose 2}$ possible simple graphs in $n$ nodes, and that quantity squared = $2^{n(n-1)}$ directed graphs.

答:
Your understanding is right: mutual independence is, as you say, "commutative". The point here, rather, is that a "dependence digraph" (as defined in the note you linked in the comments) need not capture the exact dependence structure between the given events. Sure, it must contain every edge (undirected, or directed both ways) corresponding to dependent events; but we can also "forget" about the independence of some pairs of events, i.e. add further (directed) edges. As you might expect, throwing away information like this never yields a better bound when applying the Lovász Local Lemma: if some $x_1,\dots,x_n$ satisfies the condition of the lemma for such a directed dependency graph, then it also satisfies the condition for the undirected version, where we delete all the "extra" directed edges.

What's the point of this? In the note it is stated that in some applications, the digraph structure (i.e. throwing away information) arises naturally. But they don't give an example of this, and I don't know any, either – it would be nice if someone shared one.


问:
How many functions $f:\{1,2,3,4\}→\{1,2,3,4\}$ satisfy $f(1)=f(4)$?
I just need a hint or a way to think a about this problem: $f(1)$ can be $1, 2, 3, 4$ and $f(4)$ can be $1,2,3,4.$</p>

答: $f(1)=f(4)$ can take values from $\{1,2,3,4\}$ which ain't distinct and also $f(2)$ and $f(3)$  takes values from $\{1,2,3,4\}$ which can be distinct.


问:
If the derivative is 0, prove $f$ is constant without connectedness
Given a continuous function $f$: U $\to\mathbb{R}$ where U is an open subset of $\mathbb{R}^n $and suppose that $\nabla $$f$ =0, it is well known that if U is connected then $f$ is constant, and connectedness of U is necessary, for example, if U=(0,1)$\cup $(1,2), define $f$ to be 1 on (0,1) and 0 on (1,2), then$ \nabla $$f$=0 on U but $f$ is not constant. My question is that without the connectedness of U and instead if we add another condition that $f$ is 0 on$ \partial$U and that $f$ is continuous on $\bar{U}$, does $f$ constant on U now? I think this is correct and managed to give a proof of this and I want to know if my result and the proof is right. Any comment is appreciated and thank you all in advance!

Here is my proof:

we first suppose U is connected then $f$ is constant on U,say, $f$$\equiv a$. Since $f$ is continuous near $\partial$U, $a$ must be 0. Next observing that U is locally connected, the boundary of a connected component is in fact contained in the boundary of U.Thus $f$ is 0 on the boundary of a connected component of U, so by the first step, $f$ is 0 on a connected component. Finally, if U is not connected, we can write U as a union of its connected components, and since $f$ is 0 on every connected component of U,  $f$ is 0 on U.

答:
One objection could be that if $U$ does not have a boundary. Think of $(0,1) \cup (2,3)$ as a subset of the topological space $(0,1) \cup (2,3)$. Then you counterexample works, but your proof does not. Although
I admit it is a bit pathological.


问:
How many fours can be formed from eight sports teams?
Suppose we have eight teams, let them be A, B, C, D, E, F, G and H. Two teams can form a pair, e.g. AB (BA is exactly the same pair). Then, pairs can make up into fours. For instance, AB, CD, EF and GH is an example of a four. Again, the order of pairs in the four does not matter.

The question is: how many different fours can be formed?

At school, we were solving problems like "In how many ways can you arrange 8 different objects (no identical ones)?" It's 8! = 40320. Here, however, you may arrange/combine them in any way you wish. In the described above situation with sports teams, you cannot use any letter more than once. E.g. BC and CD cannot coexist in the same four.

Basically, the answer is not that important for me; I just wonder which formula should I use or where I can read some information or maybe watch any Youtube videos about combinatorics problems where certain objects don't 'like' the other ones (combinations have restrictions).

答: There are $\binom 84$ ways to choose the $4$ teams that make up your four.  Once you've selected $4$ teams, there are $3$ distinct ways to pair them up within the four.  Thus, the answer is $210=3 \times \binom 84$.


问:
Thesis in Machine Learning BUT...

Context

I am a final year Bachelor of Mathematics student and next semester I will write my Bachelor thesis.

My interests are in Machine Learning (ML) and I will do a master in ML next year. More specific sub-fields I like are

Deep Learning

Computer Vision

Natural Language Processing

Reinforcement Learning

And my interests outside of ML and mathematics include

Self-driving cars (e.g. Tesla)

Rocket and space exploration

more vaguely, I find tech interesting as a whole.

Question

I am looking for a thesis which would bring me as close as possible to the field of ML. Do you have topic recommendations ?

BUT my Bachelor is in Mathematics therefore I shall not write a thesis in Computer Science as it would not be accepted by my study director.

Some thoughts

I know some people who were in my case. One of them for instance discovered and proved some convergence results in the context of Gradient Descent. Maybe this will inspire you with your answers.

Thanks in advance!

答:
There is lots of to do in machine learning when it comes to optimization. You can take a closer look to gradient descent and other competitive algorithms that may play the same (or even better) role in ML.

I don't know about you analysis and probability skills, but obviously it is very popular to work with MCMC, time series etc.

Another direction would be more towards statistics: Bayesian networks etc.

I do not know well RL, however if you look for inspirations in NLP you should look at Stanford or Edinburgh thesis. There is lots of good work happening there, and I am sure some papers are more theoretical. When it comes to computer vision, you can look at optical flow. Recently it is very popular and it is related to self-driving cars.

I assume that even if you are UG student passionate about ML, you are still missing a great deal of experience in such topic. I believe it is ok not to start immediately ML projects without solid foundations or relevant experience. I studied Mathematics and Computer Science. Even if I took couple of advanced,pure maths courses, they are extremely helpful later on in other fields, such as ML. Don't be worried about doing not precisely ML-focused dissertation. It may be later on even beneficial to look at problems from different angles.

Title: Thesis in Machine Learning BUT...


答:
To give a function between these sets you need to give the data: for each number ${1,2,3,4}$ give a number in that same set to map it to.

This gives you a way to count all the functions. Now once you have decided where to send $1$, you have 4 choices for each of $\{2,3,4\}$ to map it to. However, the image of $4$ is decided by the image of $1$. Do you see how to proceed?

Title: How many functions $f:\{1,2,3,4\}→\{1,2,3,4\}$ satisfy $f(1)=f(4)$?

Body: I just need a hint or a way to think a about this problem: $f(1)$ can be $1, 2, 3, 4$ and $f(4)$ can be $1,2,3,4.$</p>


问:
Proving that a groupoid is a group, knowing the following properties
(G, ·) is a groupoid. Prove that if it has the following properties it is also a group:
$$1) (a · b) · c = a · (b · c), (\forall)a, b, c \in G;$$
$$2) (∃)u ∈ G : u · a = a · u = a, (∀)a ∈ G;$$
$$3) (∀)a ∈ G(∃)a ∈ G : a · a' = u \lor a' · a = u$$

I'm confused. Aren't the above mentioned properties all we need to prove that it's a group? What else am I  supposed to do?

答:
Take a look to the property 3: it states take every element has a left or a right inverse, while in your definition every element must have a two-sided inverse.

In general, in a (non-commutative, of course) monoid, if an element has a right inverse, this is not also its left inverse (see In a non-commutative monoid, is the left inverse of an element also the right inverse?), but in you case it is different, as every element has a left or a right inverse (some a right inverse, some a left inverse).

Just for fun: the following listing can be used in SPASS (a theorem prover) to prove that actually properties $1-3$ imply that every element has a two-sided inverse.

```<code>begin_problem(monoid).

list_of_descriptions.
name({**}).
author({**}).
status(unsatisfiable).
description({**}).
end_of_list.

list_of_symbols.
    functions[(b,0), (P,2)].
end_of_list.

list_of_formulae(axioms).

formula(forall([x],equal(P(x,b),x))).
formula(forall([x],equal(P(b,x),x))).
formula(forall([x],forall ([y],forall([z],equal(P(P(x,y),z),P(x,P(y,z))))))).
formula(forall([x],exists([y],or(equal(P(x,y),b),equal(P(y,x),b))))).

end_of_list.

list_of_formulae(conjectures).

formula(forall([x],exists([y],and(equal(P(x,y),b),equal(P(y,x),b))))).
end_of_list.

end_problem.
</code>```

Now, let's try to solve your problem.
Let $a\in G$ and suppose $a$ is right-invertible, i.e., exists $a'\in G$ such that $aa'=u$. Clearly, $aa'$ is left- and right-invertible, now, let us consider $a'a$. From $3$, $a'a$ is either left- or right-invertible. So suppose $a'a$ right-invertible, and let $b_l$ and $b_r$ a left and a right inverse of $aa'$ and $a'a$, respectively. Then

$$
u=b_l(aa')=(b_l a)a'
$$
and
$$
u=(a'a)b_r=a'(ab_r)
$$

so $a'$ has both left (i.e., $b_l a$) and right (i.e., $ab_r$) inverse, and then it is invertible (this is a standard fact in a monoid) and $b_l a=ab_r$. Clearly, if $a'a$ is actually left-invertible, the argument is similar. Now, back to $a$.

From above, we have

$$
u=(b_l a)a'=(ab_r)a'=a(b_r a')
$$

and

$$
u=a'(ab_r)=a'(b_l a)=(a' b_l) a
$$

and then also $a$ has both left (i.e., $a'b_l$) and right (i.e., $b_r a'$) inverse, and then it is (two-sided) invertible. As $a$ is generic, we have proved that $G$ is a group (of course, you have to work by yourself the case in which $a'a$ is left-invertible).

Body:
Take a look to the property 3: it states take every element has a left or a right inverse, while in your definition every element must have a two-sided inverse.

In general, in a (non-commutative, of course) monoid, if an element has a right inverse, this is not also its left inverse (see In a non-commutative monoid, is the left inverse of an element also the right inverse?), but in you case it is different, as every element has a left or a right inverse (some a right inverse, some a left inverse).

Just for fun: the following listing can be used in SPASS (a theorem prover) to prove that actually properties $1-3$ imply that every element has a two-sided inverse.

begin_problem(monoid).

list_of_descriptions.
name({**}).
author({**}).
status(unsatisfiable).
description({**}).
end_of_list.

list_of_symbols.
    functions[(b,0), (P,2)].
end_of_list.

list_of_formulae(axioms).

formula(forall([x],equal(P(x,b),x))).
formula(forall([x],equal(P(b,x),x))).
formula(forall([x],forall ([y],forall([z],equal(P(P(x,y),z),P(x,P(y,z))))))).
formula(forall([x],exists([y],or(equal(P(x,y),b),equal(P(y,x),b))))).

end_of_list.

list_of_formulae(conjectures).

formula(forall([x],exists([y],and(equal(P(x,y),b),equal(P(y,x),b))))).
end_of_list.

end_problem.

Now, let's try to solve your problem.
Let $a\in G$ and suppose $a$ is right-invertible, i.e., exists $a'\in G$ such that $aa'=u$. Clearly, $aa'$ is left- and right-invertible, now, let us consider $a'a$. From $3$, $a'a$ is either left- or right-invertible. So suppose $a'a$ right-invertible, and let $b_l$ and $b_r$ a left and a right inverse of $aa'$ and $a'a$, respectively. Then

$$
u=b_l(aa')=(b_l a)a'
$$
and
$$
u=(a'a)b_r=a'(ab_r)
$$

so $a'$ has both left (i.e., $b_l a$) and right (i.e., $ab_r$) inverse, and then it is invertible (this is a standard fact in a monoid) and $b_l a=ab_r$. Clearly, if $a'a$ is actually left-invertible, the argument is similar. Now, back to $a$.

From above, we have

$$
u=(b_l a)a'=(ab_r)a'=a(b_r a')
$$

and

$$
u=a'(ab_r)=a'(b_l a)=(a' b_l) a
$$

and then also $a$ has both left (i.e., $a'b_l$) and right (i.e., $b_r a'$) inverse, and then it is (two-sided) invertible. As $a$ is generic, we have proved that $G$ is a group (of course, you have to work by yourself the case in which $a'a$ is left-invertible).


问:
Impact of copula choice on quantiles (sum of random variables)
I am trying to get my head around the impact of different dependence structures (copulas) on the risk (quantiles) of a sum of dependent random variables (with arbitrary marginals).

In a multivariate normally distributed setting (Gauss copula + Gauss marginals) everything works (for me) intuitively. Increasing the correlation between the variables under joint normality for example also increases the 99% quantile of the sum of the random variables.

However, things become quite unintuitive (for me at least) when deviating from normality (either with the dependence structure or the marginals). For exmaple, if I use a Gauss-Copula on Pareto marginals increasing the dependence results in lower 99% quantiles than for the case with a zero correlation parameter. I am aware that the Gauss Copula is unable to capture tail-dependence but I still find it hard why the uncorrelated case produces higher quantiles than the correlated one. For less heavy-tailed marginals (such as a student-t) the effect is more similar to the Gaussian marginal case.

Any hint on why an increase in dependence of the Gauss copula results in an decrease of (upper-tail) quantiles would be appreciated (and/or good resources with regard to the impact of dependence structures on diversification / risk quantiles).

Thanks / Best

答:
Say you have two random variables $X$ and $Y$ and their sum $S=X+Y$. Then the variance of $S$ is $$Var[S] = Var[X] + Var[Y] + 2Cov(X,Y) = Var[X] + Var[Y] + 2\rho \sqrt{Var[X]}\sqrt{Var[Y]}$$ where $\rho$ is the correlation between $X$ and $Y$. From this you can conclude that increased correlation will lead to increased variance. Put simply, you have discovered that increased variance will not always lead to increased (tail) quantiles.

Actually, this is only incidentally related to $S$ being a sum or the structure of dependence between $X$ and $Y$. The pertinent fact is that only for a few distributions of $S$ there is a tight connection between Variance and tail behaviour. A well known class with this property are the elliptical distributions, normal and student-t among them. For those Variance determines quantiles. See proposition 6.1.3 in Embrechts, Frey, Mc Neil, "Quantitative Risk Management" which is the go-to source for these kind of questions.

In general the connection between Variance and tail quantiles is much looser and you have freedom to wiggle both around. I did explain in some detail the connection between Expected Shortfall and Variance in this answer.

For exactly this reason Variance is quite useless or even misleading as a risk measure for non-elliptical distributions. Furthermore, correlation is of very limited use to express dependence in theses cases.


答: The number of functions that satisfy this condition is the same as the number of functions $f:\{1,2,3\}\to \{1,2,3,4\}$ because since $f(4)$ and $f(1)$ are one and the same, they can be counted together.</p>


问:
Graph with loops but without multiple lines
In the book Graph Theory by Frank Harary we have the following statement

If both loops and multiple lines are permitted,  we have a pseudograph.

My question is as follows.What if we dont have multiple lines but we have loops only?. Can we call them, even then, pseudograph or some special name or nothing

答:
Well, this is a bit tricky because of different naming systems.

According to https://en.wikipedia.org/wiki/Graph_(discrete_mathematics)#Graph "Sometimes, graphs are allowed to contain loops, which are edges that join a vertex to itself. For allowing loops, the above definition must be changed by defining edges as multisets of two vertices instead of two-sets. Such generalized graphs are called graphs with loops or simply graphs when it is clear from the context that loops are allowed."

According to https://en.wikipedia.org/wiki/Multigraph "For some authors, the terms pseudograph and multigraph are synonymous. For others, a pseudograph is a multigraph that is permitted to have loops. "

So, for instance at my graph theory courses I was taught that it is ok for ${graph}$ to have loops, but once they have multiple edges then I would call it pseudograph.


问:
Scale parameter definition
I have seen two definitions of scale parameter of a distribution: Let $X$ be random variable and $\{{P_{\theta, X }: \theta \in \Theta \subset \mathbb{R}\}}$ be a parametric family of distribution of $X$ indexed by $\theta$

$$1)~~ \theta \text{   is a scale parameter iff} \rightarrow \text{ the distribution of } cX \text{ is in } \{{P_{\theta, X }: \theta \in \Theta \subset \mathbb{R}\}} \text{ for every positive constant c}$$

For example Let $X$ ~ $LogNormal(\theta, 1)~, \theta \in (-\infty, \infty)$ then $cX$ ~ $LogNormal(\theta^*,1)$ where $\theta^* = \theta + ln(c) \in (-\infty, \infty)$. With this definition $\theta$ is a scale parameter.

For the second definition it is required that $\theta$ is positive:

$$2) ~ \theta > 0 \text{ is a scale parameter } iff \rightarrow \text{ the distribution of } X/\theta \text{ is independent of } \theta$$

If $X$ ~ $LogNormal(\theta, 1)~, \theta \in (0, \infty)$ (We restrict the parameter space) then $X/\theta $~ $LogNormal(\theta + ln(1/\theta), 1)$. With this definition $\theta$ is not a scale parameter.

I was wondering which of these two is the correct definition of a scale parameter? Or it depends on the book and the author?

答:
Your (1) and (2) are saying similar but different things.  (1) is saying that changing $\theta$ affects the scale of the distribution in some unstated way but keeps the distribution in the same parametric family, while (2) is saying the scale of the distribution is directly proportional to $\theta$.  Personally I would regard  the version in (1) as potentially leading to confusion.  As an illustration, with Gaussian distributions with mean $0$, the standard deviation is a scale parameter in both definitions but the variance does not meet the condition in (2) (physically it is likely to be in the wrong units) and so I would not describe it as a scale parameter.

In the conventional parametrisation of $X \sim \mathrm{LogNormal}(\mu,\sigma^2)$ with density $f(x) =\frac 1 {x\sigma\sqrt{2\pi}}\ \exp\left(-\frac{\left(\log_e( x)-\mu\right)^2}{2\sigma^2}\right)$, I do not think you can helpfully say $\mu$ is a scale parameter: doubling $\mu$ does not double the scale of the distribution of $X$. Similaraly with your $\theta$.

You could for example reparametrise, say to use $\alpha$ and $\beta$  calling this $X \sim \mathrm{LogNormal}^*(\alpha,\beta)$, and have a density $g(x)=\frac 1 {x\sqrt{2\pi\log_e(\beta)}}\ \exp\left(-\frac{\left(\log_e( x/\alpha)\right)^2}{2\log_e(\beta)}\right)$ where $\alpha$ is a scale parameter meeting both your conditions. As an illustration, $X$ has mean $\alpha \sqrt{\beta}$ and variance $\alpha^2\beta(\beta-1)$, so doubling $\alpha$ quadruples the variance, as you might hope for a scale parameter.   In terms of the conventional parametrisation, $\alpha= e^\mu$ and $\beta = e^{\sigma^2}$, and so there is a bijection $\mathbb R \to \mathbb R^+$ between $\mu$ and $\alpha$, with one determining the other.

My conclusion from this is that I think you can say that the log-normal distribution family is a scalable family of distributions, since if you simply change the scale then you still have a log-normal distribution, and this is what your (1) is trying to say. But I would also conclude that $\mu$ and your $\theta$ are not scale parameters for the log-normal distribution, and your (2) demonstrates that.

Title: Scale parameter definition


问:
Space of Katetov-Functions with sup-metric is complete metric space
Let (X,d) be a metric space. $f: X\longrightarrow\mathbb{R}$ is called "Katetov map" iff : $$∀,,∈:|()−()|≤(,)≤()+()$$
The set of all Katetov-maps on X is denoted by $E(X)$.

Prove that $E(X)$ turns into a complete metric space if endowed with the  sup-metric.

Since Katetov-maps don't necessarily have to be bounded, I have no idea how to prove the statement. Can anyone help me please?

NB: I tried to use that if $f_n$ is a Cauchy-sequence in $E(X)$, then $f_n(x)$ is a Cauchy-sequence in $\mathbb{R}$ for every fixed x and thus converges. But how do I show that $f_n$ uniformly converges to $f$?

Many thanks!

答:
First, we show that d is indeed a metric on E(X). We use a different approach to the (supposed) metric d.

Note that we can embed X in E(X) via $x \mapsto \delta_x$ with $\delta_x: X \to \mathbb{R}: y \mapsto d(x,y)$. For a Katetov-map f we extend the metric on X to a metric on $X \cup \{f\}$ by defining $d(x,f)=f(x)$ (which is a metric, since (i) $d(f,x)=f(x)=0 \Rightarrow f(y) = d(x,y) \forall y \in Y \Rightarrow f = \delta_x \Rightarrow f=x$ and (ii)Katetov-maps are exactly those that fulfil the triangle inequality).

We now show that being a Katetov-map is equivalent to fulfilling $|f(x)-d(x,y)| \leq f(y) \ \forall y \in X$. We define -- as proposed in the question -- the metric d on E(X) as $d(f,g) = \sup |f(x)-g(x)|$ which is always finite, since $|f(x)-d(x,y)| \leq f(y)$ and $|g(x)-d(x,y)|\leq g(y)$ hence $|f(x)-g(x)| \leq f(y)+g(y)$. Therefore $f-g$ is a bounded function and d is a well defined metric on E(X).

We define f as the pointwise limit of the $f_n$.
Since $f_n$ is Cauchy in $E(X)$, we know that $\forall \epsilon > 0 \exists N_0 \in \mathbb{N} \ st \ \forall n,m \geq N_0 : d(f_n,f_m) \leq \epsilon$. Now suppose we find $x$ and $n \geq N_0$ with $d(f_n(x), f(x)) = \epsilon + \delta, \delta > 0$. But then for $ m \geq \max \{N_0, N_x \}$ ($N_x$ being the number st $\forall n \geq N_x: d(f_n(x), f(x)) < \epsilon$) it holds that $d(f_n(x), f_m(x)) \geq d(f_n(x), f(x)) - d(f_m(x), f(x)) \geq \delta >0$ hence $(f_n)$ is not Cauchy. Therefore $d(f_n, f) \rightarrow 0$ and f still fulfils the inequalities necessary to be a Katetov-map.

Edited: Proof that d is a metric on $E(X)$.


问:
A parallelogram is inscribed in a rectangle
A parallelogram is inscribed in a rectangle such that its vertices are the midpoints of the sides of the rectangle. If the area of the parallelogram is 60, and one of the dimensions of the rectangle is 12, what is the other dimension of the rectangle?

答:
Hint:

Knowing the area of the parallelogram is $60$, what is the area of the rectangle?


问:
Chordal Graphs and Minimal Separators. Confusion over Proof.
I am studying some Graph Theory and am confused by the proof of the following statement:

A connected graph $G$, which is not complete, is chordal (has no chordless cycles of length $\geq{4}$) if and only if each of its minimal (with respect to set inclusion) separators is a clique.

Now by my understanding a separator of $G$ is a set $X$ of vertices of G that separates $G$, that is $G-X$ is disconnected. A $\textit{minimal}$ separator of $G$ is a such a set that does not contain another separator as a proper subset.

Now the bit of the proof that adresses the reverse implication (if each minimal separator is a clique then $G$ is chordal), reads:

Suppose now that every minimal separator in $G$ is a clique. Assume $G$ has a chordless cycle $(v_{1},...,v_{k})$ with k $\geq{4}$. Let X be a minimal set separating $v_{1}$ from $v_{3}$. Obviously, $X$ must contain $v_{2}$ and at least one vertex from $v_{4},...,v_{k}$ since otherwise $G-X$ contains a path connecting $v_{1}$ to $v_{3}$. By our assumption, $X$ is a clique. On the other hand $v_{2}$ has no neighbours among $v_{4},...v_{k}$. This contradiction shows that $G$ has no chordless cycles of length $4$ or more i.e. $G$ is chordal.

Unless I am missing something, we have not assumed that $X$ is a clique. We have assumed that all minimal separators of $G$ are cliques. $X$ is a set that separates $v_{1}$ from $v_{3}$, without any proper subset of $X$ doing the same. That does not mean that a proper subset of $X$ does not separate another pair of vertices of $G$ right? In other words, $X$ being a minimal $(v_{1},v_{3})-separator$ does not imply that $X$ is a minimal separator of $G$.

Am I missing something here?

Many thanks.

答:
In order for this result to be true you need the following definition of minimal separator (taken from wikipedia):

Let $S$ be an $(a,b)$-separator, that is, a vertex subset that separates two nonadjacent vertices $a$ and $b$. Then S is a minimal $(a,b)$-separator if no proper subset of $S$ separates $a$ and $b$. More generally, $S$ is called a minimal separator if it is a minimal separator for some pair $(a,b)$ of nonadjacent vertices.

With this definition, the proof is fine - a minimal $(v_1,v_3)$-separator is a minimal separator.

With the definition you might intuitively expect - a set which separates the graph but has no proper subset which does - the result would not be true, as shown by the following graph. The only minimal sets which separate are the singleton sets of red vertices, all of which are trivial cliques.


答:
Simply denote each pair of team a match

I think the answer is $105$. First, there are $ \binom {8}{2}$ ways to select the first match, then from the remaining $6$ teams, there are $ \binom {6}{2}$ ways to select the second match and similarly $ \binom {4}{2}$ ways to select the third match. Accordingly, the last match is determined.

But each way we draw the four matches is repeated $ 4! =24$ times, thus the answer is:

$ \binom{8}{2} \times \binom{6}{2} \times \binom{4}{2} :24 = 105$

Title: How many fours can be formed from eight sports teams?

Body:
Suppose we have eight teams, let them be A, B, C, D, E, F, G and H. Two teams can form a pair, e.g. AB (BA is exactly the same pair). Then, pairs can make up into fours. For instance, AB, CD, EF and GH is an example of a four. Again, the order of pairs in the four does not matter.

The question is: how many different fours can be formed?

At school, we were solving problems like "In how many ways can you arrange 8 different objects (no identical ones)?" It's 8! = 40320. Here, however, you may arrange/combine them in any way you wish. In the described above situation with sports teams, you cannot use any letter more than once. E.g. BC and CD cannot coexist in the same four.

Basically, the answer is not that important for me; I just wonder which formula should I use or where I can read some information or maybe watch any Youtube videos about combinatorics problems where certain objects don't 'like' the other ones (combinations have restrictions).


问:
Do we count only distinct roots in Descartes' rule of signs?
Descartes' rule of signs states that numbet of positive roots of a polynomial $P(x) = a_nx^n + ... + a_1x + a_0$ is not greater than the number of sign changes of coefficients.

By Gauss we know that number of complex roots of $P(x)$ is $n$ and therefore, number of real (in this case positive) roots is certainly $\leq n$ and we can count same roots more than once if $P(x)$ indeed has multiple roots.

Okay, suppose that number of signs changes of $P(x)$ is $k$ and also suppose that we have set $S$ of all positive roots of $P(x), S = \{x_1, x_2, ..., x_m\}, m \leq n$ and some $x_i, x_j$ can be equal. Now, take that same set and eliminate those equal elements (eliminate all $x_i = x_j$ when $i \neq j$), that new set call $S'$ and let's say $|S'| = r.$

Does Descartes' rule of signs tells us that $m$ cannot be greater than $k$ or that $r$ cannot be greater than $k$?

Does this rule counts multiple roots more than once or it counts just disctint roots?
I am asking this because the geometric proof given here https://math.hmc.edu/funfacts/descartes-rule-of-signs/ has sense to me if we count only distinct roots (because that's number of intersections of polynomial with the $x$-axis) but no otherwise. Thanks

I saw in the comments that the rule counts multiple roots (say $x_0$) more than once. Since the proof given in the link is geometric, does polynomial bounces near the crossing point $(x_0, 0)$?

How exatcly the number of $x_0$'s we counted reflect on the behaviour of $P(x)$ near that point? ($)

My hypothesis is that if that number is even polynomial has parabolic shape near that point and if it is odd polynomial has shape simillar to the curve $x^3$ (and since the parity of sign changes and positive roots (counted with multiplicity) are equal, then this does not change number of ups and downs of our curve - now proof has sense but please answer #)

答:
It looks like two things are interchanged in OP.

$\color{blue}{\text{Descartes rule of signs}}$ is not about the sign of $P(x).$

It is about the $\color{blue}{\text{number of sign changes of coefficients}}$, which are ordered accordingly to the powers of $x.$

In the present exercise, the coefficients are $$\underbrace{1}_+,\underbrace{7}_+,\underbrace{-4}_-,\underbrace{-1}_-,\underbrace{-7}_-.$$ Sign change appears only once, it is between $7$ and $-4.$ Therefore, $P(x)$ has exactly one positive root.

As for eventual negative roots, we count the sign changes between coefficients of the polynomial $P(-x).$ The coefficients are $1,-7,-4,1,-7.$ This time, three changes appear. Therefore, our polynomial $P(x)$ has three roots with negative real part. They can be all real, or one real and a pair of complex conjugate roots.

As for your additional question:

Example

Consider $P(x)=x^2-2x+1.$ Here are two sign changes. Accordingly to Descartes rule of signs, the polynomial has two roots with positive real part. These roots can be both positive (equal or not), or it is a pair of complexe conjugate roots. Descartes rule says nothing more.

Since it is a quadratic polynomial, $P(x)$ has no other roots.

The same answer (by Descartes rule) would be given for the polynomials $x^2-2x+4$ or $x^2-2x+\frac{1}{2},$ in general, for $x^2-2x+c$ with $c>0.$

Therefore, the geometric interpretation that you try to associate to the polynomial, is not related to the Descartes rule of signs.


答: Use this picture to calculate the area of the rectangle. Once you have that, the other dimension is just $\frac{\text{Area}}{12}$.

Body: A parallelogram is inscribed in a rectangle such that its vertices are the midpoints of the sides of the rectangle. If the area of the parallelogram is 60, and one of the dimensions of the rectangle is 12, what is the other dimension of the rectangle?


问:
does a matrix like this exist?
Question:

Does a matrix $A \in M_{3 \times 3}(F)$ exist s.t. $A^4=
\begin{bmatrix} 0&0&1\\0&0&0\\0&0&0\end{bmatrix}$

What I thought:

I think it doesn't. How do you start a proof of such a thing (prefer hints at first).
Thx

答:
It's easy to compute $(A^4)^2=O$ (the null matrix).

I believe you already know the following result, but I'll prove it anyway, as it's useful in many situations.

If $\lambda$ is an eigenvalue of $A$ (in the complex numbers), then $\lambda^n$ is an eigenvalue of $A^n$: just do it by induction. The base step $n=1$ is obvious; then, if $v$ is an eigenvector for $\lambda$, you have, for $n>1$,

$$
A^nv=A(A^{n-1}v)=A(\lambda^{n-1}v)=\lambda^{n-1}Av=\lambda^nv.
$$

In our case this means that the only eigenvalue of $A$ is $0$. So the characteristic polynomial of $A$ has a very simple form. Apply Hamilton-Cayley, to conclude that $A^3=\dots$ and get a contradiction.

Title: does a matrix like this exist?


问:
3×3 determinant into cross product
I'm trying to implement the InCircle test with this determinant from page 36:
http://graphics.stanford.edu/courses/cs268-16-fall/Notes/torino.pdf

Several authors of their implementations use cofactors to get the 3x3 matrices and then convert calculating the area of the triangle with a 3×3 determinant to calculating the cross product:

\begin{vmatrix}
x_A & y_A & 1\\
x_B & y_B  & 1\\     x_C & y_C & 1\\
\end{vmatrix}

converts to:

\begin{vmatrix}
x_B - x_A & x_C - x_A\\
y_B - y_A & y_C - y_A
\end{vmatrix}

or (b.x - a.x)(c.y - a.y) - (b.y - a.y)(c.x - a.x) in code.

Someone suggested that it can be done with this: https://en.wikipedia.org/wiki/Dodgson_condensation, but I can't make it work. Which rule allows that transformation to happen? Please help.

You can see an example of an implementation here on page 10 (functions TriArea and InCircle):
https://www.researchgate.net/profile/Dani_Lischinski/publication/262235495_Incremental_Delaunay_Triangulation/links/5af699330f7e9b026bceff5b/Incremental-Delaunay-Triangulation.pdf

答: Swap two columns and then two rows, then use Dodgson condensation, take a transpose, and change the sign of off-diagonal elements:$$\begin{align}\left|\begin{array}{ccc}x_A&y_A&1\\x_B&y_B&1\\x_C&y_C&1\end{array}\right|&=-\left|\begin{array}{ccc}x_A&1&y_A\\x_B&1&y_B\\x_C&1&y_C\end{array}\right|\\&=\left|\begin{array}{ccc}x_B&1&y_B\\x_A&1&y_A\\x_C&1&y_C\end{array}\right|\\&=\left|\begin{array}{cc}x_B-x_A&y_A-y_B\\x_A-x_C&y_C-y_A\end{array}\right|\\&=\left|\begin{array}{cc}x_B-x_A&x_A-x_C\\y_A-y_B&y_C-y_A\end{array}\right|\\&=\left|\begin{array}{cc}x_B-x_A&x_C-x_A\\y_B-y_A&y_C-y_A\end{array}\right|.\end{align}$$

Title: 3×3 determinant into cross product


问:
Galois cohomology of projective linear group
I am currently trying to compute Galois cohomology $H^{1}(\overline{k}/k, PGL_2(\overline{k}))$. As far as I know these cocycles correspond to isomorphism classes of smooth genus-$0$ curves over $k$. Any curve of genus $0$ is either isomorphic to $\mathbb{P}^{1}(k)$ or to a quadric $ax^2+by^2+cz^2=0$ which has no $k$-points. How to understand this from general tools such as exact sequences etc.?

For instance, sequence $0\longrightarrow k^{*} \longrightarrow GL_2{(k)}\longrightarrow PGL_2(k)\longrightarrow 0$ gives rise to a long exact sequence of cohomology. However I don't know how to obtain the result I am interested in.

I will be very grateful for any help or reference.

答:
As Alex Wertheim has mentioned in the comments, the key idea here is twisted forms. The (informal) general idea is the following: suppose we are given a category $\mathcal C_K$ of structures defined over a field $K$ (for example, $\mathcal C_K$ may be affine $K$-varieties, projective $K$-varieties, $K$-algebras, etc.). Let $L/K$ be a Galois extension with Galois group $G$, and let $\mathcal C_L$ be the corresponding category of structures defined over $L$. Then $G$ acts on each object of $L$ by automorphisms.

The main theorem is the following. Given an object $X$ of $\mathcal C_K$, we can obtain its base change $X_L$, an object of $\mathcal C_L$. Then there will be a bijection $$\{K\text{-isomorphism classes of }\mathcal C_K\text{ objects isomorphic to }X_L\text{ after base change}\}\longleftrightarrow H^1(G,\text{Aut}_{\mathcal C_L}(X_L))$$ where $\text{Aut}_{\mathcal C_L}(X_L)$ has the $G$-conjugation action induced by the action of $G$ on $X_L$. The objects on the left hand side of this correspondence are the so-called "twisted forms." This correspondence probably easiest to see in the case of affine varieties, or equivalently, (reduced f.g.) commutative $K$-algebras. Given an affine $K$-variety $X$ with coordinate ring $K[X]$ and its base change $X_L$ with coordinate ring $L[X]$, a $1$-cocycle $H^1(G,\text{Aut}_L(L[X]))$ canonically defines a semilinear action on $L[X]$. The theory of faithfully flat descent says that the category of $L$-algebras with semilinear $G$-actions is equivalent to the category of $K$-algebras, and so this canonically gives a $K$-algebra which becomes isomorphic to $L[X]$ after base change. A good reference for this fact about twisted forms is Springer's book on algebraic groups.

To explain the correspondence between conics and cocycles, we apply this to the category of projective varieties over $K$ and $X = \mathbb P^1_K$. The left-hand side of our correspondence is then simply the $K$-isomorphism classes of smooth curves of genus $0$, since we are base changing to an algebraically closed field. The right hand side of the correspondence is $H^1(G,\text{Aut}_{\bar K}(\mathbb P^1_{\bar K})) = H^1(G, \text{PGL}_2(\bar K))$. Tracing through the bijection explicitly will tell you which conic corresponds to which cocycle.

As an example, let's consider the simple case of $\mathbb C/\mathbb R$. Then $\mathbb P^1_{\mathbb R}$ embeds in $\mathbb P^2_{\mathbb R}$ as the conic $C : y^2-xz=0$ via $[s:t]\mapsto [s^2:st:t^2]$. Now one checks that there are only two cocycles in $H^1(G,\text{PGL}_2(\mathbb C))$, namely the trivial cocycle and the one which sends complex conjugation to $\left(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\right)$. This is the automorphism of $\mathbb P^1$ given by $[s:t]\mapsto [-t:s]$ which corresponds to the automorphism of $C$ given by $[x:y:z]\mapsto [z:-y:x]$. To find the conic in $\mathbb P^2_{\mathbb R}$ corresponding to this cocycle, we need to find a conic for which complex conjugation interchanges $x$ and $z$ and negates $y$. So we make the change of variables $x = x'+iz'$, $y=iy'$, $z=x'-iz'$. Then complex conjugation interchanges $x$ and $z$ while negating $y$. The equation of the resulting conic is $$0 = xz-y^2 = x'^2+y'^2+z'^2.$$ So the conic $C' : x'^2 + y'^2 + z'^2 = 0$ is the desired conic, which is an instance of the sort of conic you mention in your question.

As Alex also mentioned, this is all tied up with quaternion algebras, but I won't get too far into that here. The reason is basically that $\text{PGL}_2(\bar K)$ is also the automorphism group of the matrix algebra $M_2(\bar K)$. So 4-dimensional central simple algebras defined over $K$ are also classified by $H^1(G,\text{PGL}_2(\bar K))$. A conic $ax^2+by^2+cz^2$ will correspond to a quaternion algebra easily expressible in terms of the $a,b,c$.


问:
How to find the absolute minimum/maximum of this function over the unit ball?
I am confused by this exercise.
Fin the absolute minimum and absolute maximum (if they exist) of the function
$$
f(x, y) = \log(4- x^2- y^2)
$$
on the unit ball by using a parametrization.

The domain of the function is $x^2 + y^2 <4$. The Weierstrass theorem guarantees that a minimum (or a maximum) exists.
By using the standard parametrization of the unit ball
$$
\begin{cases}
x =\cos\theta\\
y=\sin\theta
\end{cases}
$$
we have
$$F(\theta) = f(\cos\theta,\sin\theta) = \log (4 -\cos^2\theta -\sin^2\theta) = \log (3).$$

It seems to me a little bit strange that we obtain a constant.

Could anyone please help me to understand what I am doing wrong? How to proceed?

Thank you in advance!

答: The unit ball is $\{(r\cos \theta, r \sin \theta): 0\leq \theta \leq 2\pi, 0\leq r \leq 1\}$. In this domain $f(x,y)=\log (4-r^{2})$. Logarithm is an increasing function so $f$ varies from $\log \, 3$ to $\log \, 4$   So, the  minimum is  $\log \, 3$ and the maximum is $\log \, 4$.


问:
Differentiate an equation
I'm learning about implicit differentiation and i'm stuck at understanding the concept of "differentiating an equation".

If I define $ f(x) = x^2$ and $g_1(y) = 9$ and $g_2(y) = 9 + y^2$

Then I define 2 equations:

$(1)$ $f(x) = g_1(y)$ (i.e. $x^2 = 9$)

Then I can't differentiate both side of this equation with respect to x (ortherwise I will get: $2x = 0$)

$(2)$ $f(x) = g_2(y)$ (i.e. $x^2 = 9 + y^2$)

Then I can differentiate both side of this equation with respect to $x$ to get: $ 2.x = 2.y.\frac{dy}{dx}$

Could you please explain me the difference between the 2 cases ? And in general, in which condition could I differentiate both side of an equation concerning more than 1 variable ?

Thank you very much for your help!

答:
$x^2=9$ is an equation that has a finite number of real solutions ($2$) for $x$. Differentiating this would not make much sense.

An equation like $x^2-1 = (x-1)(x+1)$ holds true for all real $x$. You can differentiate this and both sides will turn out to be equal.

In general, you can differentiate both sides of the equation $f_1(x,y)=f_2(x,y)$ iff this equation holds true for all real $x,y$ in a certain interval of the real numbers, containing infinitely many points.

Title: Differentiate an equation


问:
How to estimate the temperature of Keelung when it is modeled by a radical?
The problem is as follows:

The expression from below represents a model made by satellite based on the weather patters in Keelung

$$T=\sqrt[81^{3^{n}}]{\left[\sqrt[3]{8^{3^{3^{n+1}}}}\right]^{3^{3^{n}}}}$$

Assuming in the Central Weather Bureau the value of $4T$ in Celsius scale represented the temperature of Keelung at $\textrm{6:00 a.m}$. Find the temperature which will be given at $\textrm{2:00 p.m}$ using the information given by the CWB which reports that at that time the temperature will be increased $7°C$ with respect that of $\textrm{6:00 a.m}$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&17°C\\
2.&13°C\\
3.&11°C\\
4.&15°C\\
\end{array}$

From what I'm understanding in this problem the whole thing is get the value of $T$, so as with that you can get $4T$, but for this to happen it is required to simplify the expression given in the heading.

$$T=\sqrt[81^{3^{n}}]{\left[\sqrt[3]{8^{3^{3^{n+1}}}}\right]^{3^{3^{n}}}}$$

For this I'm getting:

$\left(\left[\sqrt[3]{8^{3^{3^{n+1}}}}\right]^{3^{3^{n}}}\right)^{3^{4^{3^{n}}}}$

To attack this part I concentrated in what's inside the bracket hence:

$8^{3^{3^{n+1}}}=8^{3^{3^n\cdot 3}}$

$\left(8^{3^{3^n\cdot 3}}\right)^{\frac{1}{3}}=8^{3^{3^n}}$

Since it is mentioned

$\left[2^{3^{3^{3^n}}}\right]^{3^{3^{n}}}=2^{3^{\left({3^{3^n}}\cdot 3^{3^{n}}\right)}}$

Here I'm assuming the number of nested radical would match both $3^{3^n}$ inside and outside the parenthesis, there are three numbers out so it will cover the same length from top to bottom, but is this rationale correct?.

This is reduced to (noting this part is only the exponent):

$$2^{3^{\left({3^{2\cdot 3^n}}\right)}}$$

But that's how far I went with the reduction. From then on I came stuck. Can someone help me here? which sort of manipulation should be done to solve this problem?. All that I know so far is they ask $4T$.

Body:
$$T=\sqrt[81^{3^{n}}]{\left[\sqrt[3]{8^{3^{3^{n+1}}}}\right]^{3^{3^{n}}}} = \left(\left(\left(8^{3^{3^{n+1}}}\right)^{3^{-1}}\right)^{3^{3^{n}}}\right)^{81^{-3^{n}}}$$

Looking at the exponent of $8$, we have:

$$(\log_8 T = ) \quad 3^{3^{n+1}}\cdot 3^{-1}\cdot 3^{3^n} \cdot 81^{-3^n} = 3^{3^{n+1} -1+3^n-4\cdot 3^n} = 3^{-1}$$

Hence $T = 8^{3^{-1}} = 2$.


问:
How do I solve this question on Hooke's law?
A particle of mass $2$ kg is being held in equilibrium on a smooth slope by a horizontal force $p$ and a light elastic spring. The spring has modulus of elasticity $10$N and is attached to the particle and also to the slope $1.5$m up the slope from the particle and also to the slope is inclined at $25°$ and the force $p$ is of magnitude $5$N find the two possible natural lengths of the spring.

答:
It may be helpful to draw a free body diagram, annoted with all the relevant angles. There are $4$ forces involved:

The normal reaction force from the slope

Tension or thrust in the spring (depending on whether's it's stretched or compressed)

The particle's weight

The horizontal force $p$.

Does that help? Please tell me if you require any more help.


问:
Cleverly finding Primitive Roots
From Number Theory we have the theorem that:

There are primitive roots mod $n$ if and only if $n = 1,2,4,p^k, 2p^k$, where $p$ is an odd prime.

The question is, for such general $n$ given as above (with $n \neq p$), how does one find primitive roots without having to brute-force? Perhaps the Chinese Remainder Theorem is relevant? I'm excluding the case $n=p$ because it already has a well-written answer Finding a primitive root of a prime number .

答:
Thanks to everyone who commented! From the comments, the answer can be found:

For $n = p^k$, firstly find a primitive root modulo $p$ as in Finding a primitive root of a prime number. Then, use that if $g$ is a primitive root mod $p$, then $g$ or $g+p$ is a primitive root mod $p^k$ for any $k \in \mathbb{Z}_{\ge 1}$. Then for $n = 2p^k$, denoting by $g'$ the primitive root for $p^k$, either $g'$ or $g'+p^k$ is a primitive root mod $2p^k$.

But if anyone wants to post their own methods or proofs of the propositions used (I actually don't know the proofs), please do so!

Title: Cleverly finding Primitive Roots