问:
problems with $\int _0 ^\infty \frac{dx}{\sqrt[3]{x}(x+1)^2}$
I'm trying to find $\int _0 ^\infty \frac{dx}{\sqrt[3]{x}(x+1)^2}$ using complex analysis.

My first problem is that I have to deal with the derivative of $\sqrt[3]{x}$ while computing the residue at $-1$. Explain, please, how one can compute the derivative at a point of such functions (with several branches).

And the second problem is that I don't know how to show that the  contour integral over an upper semi circle centred at $0$ has a zero limit.

答:
If we start off with:
$$I=\int_0^\infty\frac{1}{\sqrt[3]{x}(x+1)^2}dx$$
and make the substitution $u^3=x$ we get:
$$I=3\int_0^\infty\frac{u}{(u^3+1)^2}du$$now we can factorise: $(u^3+1)^2=(u+1)^2(u^2-u+1)^2$ and so:
$$I=3\int_0^\infty\frac{u}{(u+1)^2(u^2-u+1)^2}du$$
now we can use partial fraction decomposition to get:
$$\frac{1}{(u+1)^2(u^2-u+1)^2}=-\frac{1}{9(u+1)}-\frac{1}{9(u+1)^2}+\frac{u-1}{9(u^2-u+1)}+\frac{1}{3(u^2-u+1)^2}$$
we can then break our integral down into multiple others:
$$I_1=\int_0^n\frac{1}{u+1}du=\ln|n+1|$$
$$I_2=\int_0^n\frac{1}{(u+1)^2}du=1-\frac{1}{n+1}$$
$$I_3=\int_0^n\frac{u-1}{u^2-u+1}du=\frac12\int_0^n\frac{2u-1}{u^2-u+1}-\frac{1}{u^2-u+1}du=\frac12\ln|n^2-n+1|-3^{-1/2}\arctan\left(\frac{2\sqrt{3}n-\sqrt{3}}{3}\right)+\frac{\pi}{2*3^{3/2}}$$
$$I_4=\int_0^n\frac{1}{(u^2-u+1)^2}du=\frac{4\sqrt{3}\left(n^2-n+1\right)\arctan\left(\frac{2\sqrt{3}n-\sqrt{3}}{3}\right)+6n-3}{9(n^2-n+1)}+\frac{2\sqrt{3}\pi+9}{27}$$
now just remember that our original integral, $I$ is the same as:
$$I=\lim_{n\to\infty}\left(-\frac13I_1-\frac13I_2+\frac13I_3+I_4\right)$$

The answer should come out as $I=2*3^{-3/2}\pi$


问:
Meaning behind Multiplication of Matrix and its Transpose
When you multiply a matrix M by its transpose, what exactly does this product represent, what do each value in the cell represent?<br>
I see that a lot of these examples, when a document term matrix  (dtm) is created and then this dtm is multipled by its transpose and they call it the term-term matrix.  what does each cell in this 2-dim matrix represent.
Thanks in advance for your replies. </p>

答: If the rows of $M$ are $m_1, m_2,\ldots, m_n$, then the $i,j^\textrm{th}$ entry of $MM^T$ is $m_i\cdot m_j$, the dot product.</p>

ContentLicense: CC BY-SA 3.0


问:
Compute difficult integral $\int \frac{dx}{2 + x + \sqrt{1 - x^2}}$
To solve the integral
$$I = \int \frac{dx}{2 + x + \sqrt{1 - x^2}}$$
I have tried several things, such as $t = \arcsin x$, because $\cos(\arcsin x) = \sqrt{1 - x^2}$. If I am not wrong, we can conclude with this variable change
$$
I = \int \frac{\cos t\,dt}{2 + \sin t + \cos t}
$$
but if it were correct, how could I go on?

答:
Continue with $x=\sin t$

\begin{align}
&\int \frac{1}{2 + x + \sqrt{1 - x^2}} dx=\int \frac{\cos t}{2 + \sin t + \cos t}dt\\
= &\ \frac12\int \left(1+ \frac{\cos t-\sin t}{2 + \sin t + \cos t}-
\frac{2}{2 + \sin t + \cos t} \right)dt\\
=& \ \frac12t +\frac12 \ln(2 + \sin t + \cos t) -\sqrt2\tan^{-1} \frac{\tan\frac t2+1}{\sqrt2}\\
=&\ \frac12\sin^{-1}x+\frac12 \ln\left(2 + x+ \sqrt{1-x^2}\right) -\sqrt2\tan^{-1} \frac{1+x-\sqrt{1-x^2}}{x \sqrt2}
\end{align}

ContentLicense: CC BY-SA 4.0


问:
Could somebody correct my reasoning for Buffon's needle problem?
I'm getting $1/4$ as the probability in Buffon's needle problem and I'm having some trouble finding my mistake.

The problem statement:

If a needle of length 1 is dropped at random on a surface ruled
with parallel lines at distance 2 apart, what is the probability that the needle will cross one
of the lines?

I reduced the problem to a simpler one:

If a needle of length one is dropped in a region of length two such that its center lies on a horizontal line, creating an angle theta with that line, what is the probability that it will contact a vertical line at the center of said region?

(Visualization)

From simple trig rules, the needle will begin contact when the center is $\frac{\cos(\theta)}{2}$ away from the central line, and likewise ends contact at the same distance.

So we have a total sample space of 2 and a hit-space of $\cos(\theta)$, so we get a probability to hit of $\frac{\cos(\theta)}{2}$. (Assume $0\le\theta\le2\pi$)

Graphing our space over all possible theta we have:

Calculating the probability of placing a needle in the hit region of our sample space is simply the hit area (blue) divided by the total area.

So we get $$P=\frac{4(\frac{1}{2}\cdot1\cdot\frac{\pi}{2})}{4\pi}=\frac{1}{4}$$

We can extend these regions both horizontally and vertically without affecting the probability, so this 'solution' claims the answer to the problem is $1/4$.

What have I done wrong?

答:
That isn't the graph of cosine!!!

The area under the curve of abs(cosx) between zero and 2pi is 4. The total sample space area is 4pi. So the probability is 1/pi as desired.


问:
Pascals Identity Question, $n-1 \choose k-1$
The idea behind pascals identity is $\displaystyle \binom n k = \binom {n-1}k + \binom{n-1}{k-1}$.

I understand that we're splitting up the number of $k$ element sets into the number of $k$ element sets which include $n$ and the number of $k$ element sets which don't include $n$.

$n-1 \choose k$ makes sense to me, however $n-1 \choose k-1$ doesn't make sense, how this would determine the number of $k$ element subsets which include $n$, any explanations would be greatly appreciated.

答:
Let us imagine a kingdom of $n$ people, and one person is the king. Let us now try to choose $k$ people from the kingdom. We may do it in one of two ways: either the king is chosen or he is not. If we do not choose the king, we, therefore, choose $k$ people from $n-1$ people, as the king remains unchosen. However, if the king is chosen, we need yet to choose $k-1$ people from the remaining $n-1$ people. As the two cases were separate, we achieve that we can choose $k$ people from $n$ people as the sum:
$$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$$


问:
Integral containing exponential
I am trying to prove below (Reference: Green's function with application, page 159):
$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i(x-y)\tau}}{\tau^4-a^4}d\tau=\frac{1}{4a^3}(ie^{ia|x-y|}-e^{-a|x-y|})$$

When trying to calculate the integral, I used Wolfram, and I get the following:
$$\int\frac{e^{(x-y)i\tau}}{\tau^4-a^4}d\tau=\frac{1}{4}\sum_{{\omega:a^4-\omega^4=0}}\frac{e^{i(x-y)\omega}Ei(iA(\tau-\omega))}{\omega^3}+const$$
I am not sure how to deal with Ei function for an imaginary number with specified integral boundary.

答:
To calculate this integral use complex analysis.  We make a contour with the real line and a semi-circle through the upper-half of the complex plane.

As the radius of the semi-circle goes to infinity, the integral along this piece of the contour goes to $0.$

Now we just need to examine the residuals.

$\tau^4 - a^4 = (\tau - a)(\tau - ia)(\tau + a)(\tau + ia)$

Two of these poles will be inside the contour, and 2 will be outside the contour.  In order to know which two to pick, lets assume that a has positive real and imaginary parts.

$\oint_\gamma \frac {e^{i|x-y|\tau}}{\tau^4 - a^4} \ d\tau$

$2\pi i \left[\lim_\limits{\tau \to a} \frac {e^{|x-y|\tau}}{\tau^4 - a^4}(\tau - a) + \lim_\limits{\tau \to ia} \frac {e^{|x-y|\tau}}{\tau^4 - a^4}(\tau - ia)\right]$

$2\pi i \left[\frac {e^{i|x-y|a}}{4a^3} + \frac{e^{i|x-y|(ia)}}{4(ia)^3}\right]$

$2\pi i \left[\frac {e^{i|x-y|a}+ie^{-|x-y|a}}{4a^3}\right]$

$2\pi \left[\frac {ie^{i|x-y|a} - e^{-|x-y|a}}{4a^3}\right]$

Title: Integral containing exponential


问:
verifying properties of a certain relation
Problem:Let M be a set of 2 elements or more. Let $P(M)$ be the set of all possible subsets of $M$. We define the relation $R$ on $P(M)$ by saying $(U,V) \in R$ if and only if $U\cap V \neq \oslash$ Which of the following qualities does the relation have: transitive, reflexive, symmetrical.

What I don't understand:
does $(U,V)\in{R} $ is some notation for $(U,V)$ verifying $R$ ? if so what convention of notation is this?

My answer

It's reflexive since $U \cap U = \emptyset \implies U= \emptyset $ and it should be in M since it's the set of all subsets, reciprocally $U = \emptyset \in M \implies U \cap U = \emptyset$ so $U \cap U = \emptyset  \implies (U,U) \in M$

for symmetry it's verified since $U \cap V= \emptyset \iff V \cap U = \emptyset \iff (U,V) \in M $

It's also not necessarily (not sure)transitive since

$U \cap X= \emptyset $and $X\cap V = \emptyset$
is not  equivalent to $V \cap U = \emptyset$ (but I think we should find a counter example here)

Is how I am thinking about the problem correct? Thank you so much.

答:
By definition a relation on a set $A$ is a subset of $A\times A$, so its elements are ordered pairs of elements of $A$. By analogy with such familiar usages as $x\le y$ and $U\subseteq V$ we often write $U\mathrel{R}V$ instead of $\langle U,V\rangle\in R$, but writing $\langle U,V\rangle\in R$ is not a convention: it’s based on the actual definition of a relation.

$R$ is almost but not quite reflexive: for $A\in P(M)$ we have $\langle A,A\rangle\in R$ iff $A\cap A\ne\varnothing$ iff $A\ne\varnothing$. If we restricted $R$ to the non-empty subsets of $M$, it would be reflexive, but on $P(M)$ it is not reflexive.

It is symmetric, since intersection is a commutative operation.

It is not transitive, and you do indeed need a counterexample to demonstrate this. You’ll need to use the fact that $|M|\ge 2$. Let $a$ and $b$ be distinct elements of $M$. Then $\{a\},\{b\},\{a,b\}\in P(M)$, $\{a\}\cap\{a,b\}\ne\varnothing\ne\{a,b\}\cap\{b\}$, and $\{a\}\cap\{b\}=\varnothing$, so $\left\langle\{a\},\{a,b\}\right\rangle\in R$ and $\left\langle\{a,b\},\{b\}\right\rangle\in R$, but $\left\langle\{a\},\{b\}\right\rangle\notin R$.


问:
The composition of $\Delta_0$ fn's $f, g: HF\to HF$ is not necessarily $\Delta_0$?
I am trying to solve the following exercise in Ken Kunen's book The Foundations of Mathematics

The hint vaguely makes sense at the high level but I am unable to follow the proof skeleton. (I am interested in particular in a proof following the given proof skeleton hint, because I am the TA for a class using this text and I want to know if the students will be able to produce a correct proof following the hint; I'm not just interested in a correct proof of the theorem.)

I first established that the basic operators involving pairing and projection of ordered pairs can all be added harmlessly to a $\Delta_0$ formula without affecting it.

Let $f : HF\to HF$ be a function whose graph is $\Delta_1$. Then by
definition there is a $\Delta_0$ formula $\varphi$ such that
\begin{equation}
  f(x)=y\iff \exists z. \varphi(x,y,z)
\end{equation}
Let $x$ be any $HF$ set. Denote by $n_x$ the least natural number $n$ such
that $\exists y,z\in R(n),\varphi(x,y,z)$. Denote by $R_x$ the
disjoint sum of all $R(n)$ for $n\leq n_x$, i.e.
\begin{equation}
R_x = \coprod_{n\leq n_x}R(n) = \left\{ \left\langle n,R(n)
  \right\rangle \mid n\leq n_x\right\}
\end{equation}
Denote by $h: HF\to HF$ the function which sends $x$ to $\left\langle
  x,R_x \right\rangle$. We will prove $h$ is $\Delta_0$.

First, by (6.) of Example II.17.5, $\left\langle x,y \right\rangle=z$
is logically equivalent to a $\Delta_0$ sentence, which Kunen calls
$\operatorname{op}(x,y,z)$. Note also that ``$z$ is an ordered pair'',
$\exists b\in z,\exists x,y\in p.p=\left\langle x,y \right\rangle$ is
then $\Delta_0$ in $z$. This is pointed out in (14.) of Lemma
II.17.9. I call this $\operatorname{op}(z)$ and use the number of
arguments to distinguish them. Note that if $\pi_1, \pi_2$ are the
projection operators onto the first and second argument of an ordered
pair (and return the empty set if the argument is not an ordered pair)
then $\operatorname{op}(z)\land P(x_{1},\dots, x_k,\pi_1(z))$ is logically
equivalent to
$\exists b\in z,\exists x,y\in b, \operatorname{op}(x,y,z)\land P(x_1,\dots,x_k,x)$
which is $\Delta_0$ in the free variable $z$, so I can use the
projection operators freely in a $\Delta_0$ sentence.

By (10.) of Lemma II.17.9, the sentence ``$n$ is a natural number'' is
$\Delta_0$. By abuse of notation, I write $n\in \omega$; but be
careful to recall that $\omega$ does not denote a term in $HF$.
Lastly observe that $\exists ! a\in b. P(x_1,\dots, x_k,a)$ is
logically equivalent to the  $\Delta_0$ sentence $\exists a\in
b.P(x_1,\dots, x_k,a) \land \forall a,a'\in
b. P(x_1,\dots,x_k,a')\land P(x_1,\dots, x_k,a)\implies a=a'$.

So define $h$ by $\psi(x,t) = \operatorname{op}(t)\land \pi_1(t)=x \land
\pi_2(t)=R_x$.  We will show that the relation $s=R_x$ is $\Delta_0$
in $s$ and $x$; from here it is clear that $\psi$ is a $\Delta_0$
definition of $h$.

Define
\begin{align}
  \gamma(x,s):= & \forall p\in s. \operatorname{op}(p)\land \pi_1(p)\in \omega \land
                  \pi_2(p)=R(\pi_1(p))\\
                &\textrm{"s is a set of ordered pairs} \\
                &\textrm{of the form } (n,R(n)) \textrm{ where } n\in \omega"\\
                & \forall p\in s. \forall n'\in \pi_1(p)\exists p'\in
                  s.\pi_1(p')=n'\\
                &\textrm{"if }(n,R(n))\in s,\textrm{ so is
                  }(n',R(n'))\textrm{ for }n'<n" \\
                & \exists ! q\in s.\exists y,z\in \pi_2(q).\varphi(x,y,z)\\
                &\textrm{"in a unique pair }(n,R(n))\textrm{ witnesses }y,z\textrm{ to
    }\varphi \textrm{ exist in }R(n)"\\
\end{align}
However this is not quite right. For this to be correct I would have to establish that the relation
$a=R(n)$ is $\Delta_0$, which is equivalent to saying that the graph
of the function $n\mapsto R(n)$ is $\Delta_0$. But we don't have that, per se. What we have, given in a previous lemma, is that the graph of $n\mapsto \coprod_{m<n}R(m)$ is $\Delta_0$. This  means somehow I need to get hold of the quantity $n_x+1$; but I don't know how to "construct" it in a $\Delta_0$ way from $n_x$ or from $R_x$. I am a bit lost in the face of the unique handicap that I am unable to freely introduce function symbols representing functional relations in virtue of the fact stated in the problem.

答:
The key ideas are

Even if we do not know $R(n)=x$ is $\Delta_0$, $R_x$ contains sufficient information to define itself.

$V_\omega$ is the closure of $\{\varnothing\}$ under the operation $x,y\mapsto x\cup\{y\}$.

Let us consider the conjunction of the following formulas:

$s$ is a function such that $\operatorname{dom} s$ is an ordinal (so $V$ thinks  $\operatorname{dom} s\in\omega$) ,

$s(0)=0$,

For $n\in\operatorname{dom}s$ such that $n+1\in\operatorname{dom}s$, if $x,y\in s(n)$ then $x\cup\{y\}\in s(n+1)$.

For $m=\bigcup\operatorname{dom}s$, there are $y,z\in s(m)$ such that $\phi(x,y,z)$, and

If $k<m$, then no $y,z\in s(k)$ satisfy $\phi(x,y,z)$.

It is easy to see that the listed formulas are all $\Delta_0$

To prove $s=R_x$, it suffices to prove that $s(n)=R(n)$ for each $n\in\operatorname{dom}s$. By induction, assume that $s(n)=R(n)$. For each $a\in R(n+1)$, we have
$a = \bigcup\{\{b\}\mid b\in a\}$, and each $b\in a$ is an element of $R(n)$. Hence we can prove $a\in s(n+1)$ by induction on the size of $a$.


问:
I was wondering what this equation I use looks like written in proper math terms
I'm just going to write out the steps for an equation I use at a restaurant and I hope you in the community wouldn't mind answering by showing how the equation should be written properly.

I use this math process, or equation when dividing the quantity of cashew milk out for bottling at a restaurant I work at.

BEGIN: I have some total amount of cashew milk over 8 quarts, (though the quantity is always unknown and the equation party solves this variable part of the process). I split the total cashew milk in half for drink 1. The rest of the cashew milk goes to making 3 equal quantities of drinks 2, 3, and 4. ALTHOUGH, this half of the cashew milk is not split into 3rds because drink 4 is a coffee drink. This coffee drink is 2 parts cashew milk and 1 part coffee. Though this drink with coffee is not 100% cashew milk the end goal is to make the same quantity of this drink as I do drinks 2 and 3, which are all cashew milk by volume.

The math gets heaviest here

To figure out how much cashew milk the coffee drink (drink 4) needs I add 1 + 1 for the drinks with 100% cashew milk in them each plus 0.666 for the coffee drink which is 2/3rds cashew milk. I divide 0.666 by 2.666 and get 0.249. As a percent this rounds to 25% so I take 25% of the half of the total cashew milk. 0.333 divided by 2.666 tells me how much coffee I need. I divide 1 by 2.666 to get the percentages of drinks 2 and 3 from the total, which is 0.375 or 37.5% of the half of the total.

I hope that is clear and would just like to know if there is a proper equation form this can be written in.

答:
I find this very difficult to follow here's what I understand.  We have an initial volume $M$ of cashew milk.  $M/2$ is set aside for drink $1$, and the remaining $M/2$ is used for drinks $2,3,4$.  Drinks $2$ and $3$ take an equal volume $V$ of the milk, but drink $4$ takes $\frac23V$ so that $$V+V+\frac23V=\frac M2$$ or $$V=\frac{3M}{16}$$  The amount of coffee required $C$ is $$\frac V3=\frac{M}{16}$$

Does this answer your questions?


问:
Write ‘There is exactly 1 person…’ without the uniqueness quantifier
During a lecture today the prof. posed the question of how we could write "There is exactly one person whom everybody loves." without using the uniqueness quantifier.

The first part we wrote as a logical expression was "There is one person whom everybody loves.", ignoring the 'exactly one' part of the question initially. From this he wrote

$L(x,y): x$ loves $y$; domain for $x$ and $y$: $\{\text{people}\}$

$\exists x\forall y: L(y,x)$

Which I understand to mean 'There is a person $x$ such that for all $y$, $x$ is loved by $y$' AKA 'There is a person who is loved by everyone'. I get that part.

The part I don't get is how the expression of 'exactly one'.

$\forall z(\forall y(L(y,z))\to x =z)$

which then creates the joint expression

$\exists x\forall y(L(y,x))\land \forall z(\forall y(L(y,z))\to x=z)$

I just can't seem to understand how $\forall z(\forall y(L(y,z))\to x =z)$ means exactly one here.  I suppose you can take $\forall z$ here to mean 'for any given person', which means the $\forall z$ is considering every person in the world.  This would translate the second expression block to something like, "For any given person $z$, if everybody loves $z$ then $z$ is the same person as $x$".

To me though $\forall z$ generally means for every element in the domain which I see as meaning every person in the world simultaneously, as it seems to for $y$.  Is that just plain wrong?  How can I tell when $\forall$ means 'all' and when it means 'for any (one)'?  In the previous English translation the only reason I was able to translate it (if it's even right) is because I already knew what the statement was suppose to mean.

Is it just that $\forall z$ means that this statement could be true for any element, and if so what's the difference between $\forall z$ and $\exists z$?  Someone told me that $\exists z$ would be redundant because the expression says $x=z$ but how do I know that $x$ and $z$ are automatically the same person if $\exists$ is used for both?

Sorry if this is a bit long with too many questions.  I just wanted to try to make the cause of my confusion as clear as possible so you can help me figure this out.

答: The idea behind the formula is that, to say exactly one person is loved by everybody, you first say that there is such a person $x$ (this is the part you said you understood), and then you say that there isn't a second such person.  That second part is expressed here as saying, if any other $z$ shows up who is also (like $x$) loved by everybody, then this wasn't really another $z$ but just the same $x$.


问:
Prove that the Klein group $V_4$ is isomorphic to $C_2\times C_2$
I'm really struggling trying to prove that the Klein group $V_4$  is isomorphic to $C_2\times C_2$. Can someone please help me?

Body:
Hint:

Table for $V_4$
$$\begin{array}{|c|c|c|c|c|}
\hline
*& e & a & b & c \\ \hline
e & e&a&b&c\\ \hline
a & a& e&c&b\\ \hline
b & b& c&e&a\\ \hline
c & c& b& a& e \\ \hline
\end{array}$$

Table for $C_2\times C_2$

$$\begin{array}{|c|c|c|c|c|}
\hline
+& (0,0) & (0,1) & (1,0) & (1,1) \\ \hline
(0,0) &(0,0)&(0,1)&(1,0)&(1,1)\\ \hline
(0,1) & (0,1)&(0,0)&(1,1)&(1,0)\\ \hline
(1,0) &(1,0)&(1,1)&(0,0)&(0,1)\\ \hline
(1,1) & (1,1)&(1,0)&(0,1)&(0,0) \\ \hline
\end{array}$$


答:
Let $V$ be the volume of milk you have to start.

Then drink $1$ uses $\frac{1}{2} V$ of the milk, and drinks $2,3,4$ together use the other $\frac{1}{2} V$.

Now let's call $C$ the volume of coffee that you will use for drink $4$. Then $2 C$ is the amount of milk used in drink $4$ (since 1/3 of the drink is coffee).

You want drinks 2,3,4 to all have the same volume (which is $3C$ by the previous discussion). So together drinks 2,3,4 have volume $9C$, of which $8C$ is milk. But we also know drinks 2,3,4 together use $\frac{1}{2} V$ of milk. So you get
$$ 8C = \frac{1}{2} V $$
Which tells you $C = \frac{1}{16} V$, which in turn means $2C = \frac{1}{8}V$ is the amount of milk in the coffee drink, and $3C = \frac{3}{16}V$ is the amount of milk used for each drink 2 and 3.

Title: I was wondering what this equation I use looks like written in proper math terms

Body:
I'm just going to write out the steps for an equation I use at a restaurant and I hope you in the community wouldn't mind answering by showing how the equation should be written properly.

I use this math process, or equation when dividing the quantity of cashew milk out for bottling at a restaurant I work at.

BEGIN: I have some total amount of cashew milk over 8 quarts, (though the quantity is always unknown and the equation party solves this variable part of the process). I split the total cashew milk in half for drink 1. The rest of the cashew milk goes to making 3 equal quantities of drinks 2, 3, and 4. ALTHOUGH, this half of the cashew milk is not split into 3rds because drink 4 is a coffee drink. This coffee drink is 2 parts cashew milk and 1 part coffee. Though this drink with coffee is not 100% cashew milk the end goal is to make the same quantity of this drink as I do drinks 2 and 3, which are all cashew milk by volume.

The math gets heaviest here

To figure out how much cashew milk the coffee drink (drink 4) needs I add 1 + 1 for the drinks with 100% cashew milk in them each plus 0.666 for the coffee drink which is 2/3rds cashew milk. I divide 0.666 by 2.666 and get 0.249. As a percent this rounds to 25% so I take 25% of the half of the total cashew milk. 0.333 divided by 2.666 tells me how much coffee I need. I divide 1 by 2.666 to get the percentages of drinks 2 and 3 from the total, which is 0.375 or 37.5% of the half of the total.

I hope that is clear and would just like to know if there is a proper equation form this can be written in.


问:
Let $G$ be a group of order 77. Show that $\operatorname{Aut}(G)$ is abelian but not cyclic.
The proposition is this:
Let $G$ be a group of order 77. Show that $\operatorname{Aut}(G)$ is abelian but not cyclic.

There's also a hint to use Sylow theorems for both $G$ and $\operatorname{Aut}(G)$.

I know that since we have $11\not\equiv 1\mod 7$, and $77=7\times 11$, then every group of order 77 is cyclic, this its automorphism group is abelian.

Furthermore, we know that $\operatorname{Aut}(G)$ is isomorphic to the group of elements of $Z_{77}$ which are invertible under multiplication, thus $\operatorname{Aut}(G)\simeq Z_{60}$.

Now there are only two abelian groups of order 60 (up to isomorphism) and one way to show that $\operatorname{Aut}(G)$ isn't cyclic is to show that it doesn't have an element of order 4, which is equivalent to showing that there's no integer $m$ such that $m^4 \equiv 1 \mod 77$.

I calculated all the squares $\mod 77$ and then managed to show what I wanted, but this is obviously not a fast or smart way of proving what I want to prove. I can't really think of a better way to show this, so any ideas are appreciated!

Thank you for all the help in advance.

答:
Write $C_n$ for the cyclic group of order $n$. The automorphism group of $C_{77}$ is $C_{77}^\times$. (This is a group of order $60$, but it's better not to write $Z_{60}$ for it.)

By the Chinese remainder theorem
$$C_{77}^\times\cong C_{7}^\times\times C_{11}^\times\cong C_6\times C_{10}\cong C_2^2\times C_3\times C_5.$$
Can you convince yourself of the fact that there is no element of order $60$ in there?

Overkill: use the classification of finite abelian groups.


问:
How to evaluate $\lim_{x \to \infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)$?
How can I evaluate  $\lim_{x \to \infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)$?</p>

答: Rationalize. Then divide througout by $\sqrt x$. Then substitute limit.</p>


问:
Finding four square roots of $\small\begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$
This question is about finding the 4(!!) square roots of a 2 by 2 matrix. The method I use is through Diagonalization (or actually doing the "opposite"). I have no problem finding the four roots of a 2 by 2 matrix that has real eigenvalues (and thus real eigen vectors). Heck, I found 8 roots of a 3 by 3 matrix that has real eigen values. The problem I am running into is if the to be square rooted matrix has complex eigenvalues. In my view, the method should work the same, but somehow the algebra is letting me down. Example: Find the four square roots of
$ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$. Note, I found this matrix by squaring $ \begin{bmatrix}2 & -1\\1 & 2\end{bmatrix}$, so obviously this is one root as well as taking opposite of its entries. So here is my work: The characteristic equation of $ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$ is $(3-\lambda)^2+16=0$ from which we find $\lambda=3+4i$ and $\lambda=3-4i$. An eigenvector for $\lambda=3+4i$ is $ \begin{bmatrix}1 \\-i  \end{bmatrix}$ and an eigenvector for $\lambda=3-4i$ is $ \begin{bmatrix}1 \\i  \end{bmatrix}$. So the given matrix can now be diagonalized: $ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$=$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix}$$ \begin{bmatrix}3+4i & 0\\0 & 3-4i\end{bmatrix}$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix}^{-1}$. I found$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix}^{-1}$=$\frac{1}{2i} \begin{bmatrix}i & -1\\i & 1\end{bmatrix}$ which is $ \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$. It's easy to find out that the square roots of $3+4i$ are $2+i,-2-i$, and of $3-4i$ are $2-i,-2+i$. So now I would arrive at my 4 solutions:$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}2+i & 0\\0 & 2-i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}2+i & 0\\0 & -2+i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}-2-i & 0\\0 & 2-i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}-2-i & 0\\0 & -2+i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
When I work out my first candidate, I get a matrix, but when squared it doesn't give me $ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$ and so something isn't right. I don't know what I did wrong and ANY help from the math community is greatly appreciated. I feel that my line of thinking is correct, where did I go wrong? And does anyone happen to know online software that can do matrix multiplications with complex entries, because the TI doesn't work

答:
There should only be two square root matrices.  $\pm\begin{bmatrix} 2&-1\\1&2\end{bmatrix}$

I have been noodling with this a bit.

$\begin{bmatrix}0 & -i\\i & 0\end{bmatrix} \begin{bmatrix} 2&-1\\1&2\end{bmatrix} = \begin{bmatrix} i&2i\\-2i&i\end{bmatrix}$

Appears to work, and so would

$-\begin{bmatrix} i&2i\\-2i&i\end{bmatrix}$


问:
$\alpha|f(x-y)| = \beta|f(x-y)||g(y)|$ a.e. implies $g(y)$ is constant a.e
Suppose we have two functions $f,g \in L^1$  on $\mathbb{R}$ such that $||f||_1 > 0$ and  for some $\alpha, \beta \in \mathbb{R}$ not both $0$, $\alpha|f(x-y)| = \beta|f(x-y)||g(y)|$ for almost all $x$ and almost all $y$. Does this  imply $g(y)$ is constant almost everywhere?

We could almost just divide by $f(x-y)$ but we have to deal with $f(x-y)$ being $0$. I feel like you should be able to argue that since $||f||_1 > 0$, $f$ is positive on some set $A$ of measure greater than $0$. So for any $y$, we would translate $\alpha|f(x-y)| = \beta|f(x-y)||g(y)|$ by some $x$ such that $x-y$ is in $A$ and thus we can divide by $f(x-y)$ to get that $|g(y)| = \alpha/\beta$ a.e.. But I cant quite formalize this.

答:
Your idea is correct. The formal details are the following:

First of all i assume that $g\geq 0$. Since if you wanna keep the absolute value on $g$ then it is not true, for example take $\alpha=\beta=1$ and $g(y)=\mathbb{1}_{[0,1]}(y)-\mathbb{1}_{[2,3]}(y)$ (here $\mathbb{1}_{A}$ is indicator function on $A$) then the condition
$$\tag{*}\alpha|f(x-y)|=\beta |f(x-y)||g(y)|$$
is true for all integrable $f$, but $g$ is not constant. Also, we need to assume that $\alpha,\beta \neq 0$. Since, if $\beta=0$ this forces immediately $\alpha=0$ because $f$ cannot be $0$ a.e. Now, if $\alpha=0$ and $\beta\neq 0$ with similar argument as before we conlude that $g=0$ a.e. and we are done in this case. So we suppose that $g\geq 0$ and that $\alpha,\beta\neq 0$. Now, let
$$A_y=\{x\in \mathbb{R}:\, (*)\, \text{ holds for the pair }\, (x,y)\}$$
The assumption that $(*)$ holds for almost every $x$ and almost every $y$ means that $$\tag{1}\mu(\mathbb{R}\setminus A_y)=0$$

for almost all $y$, and that
$$\tag{2}\mu(\mathbb{R}\setminus A)=0$$
where $A=\{y\in \mathbb{R}:\, \mu(\mathbb{R}\setminus A_y)=0\}$. Now, let
$B_y=\{x\in \mathbb{R}:\, |f(x-y)|>0\}$
.Then, using the translation invariance of the integral, for all $y$ we have,
\begin{align}
||f||_1&=\int_\mathbb{R} |f(x)|\,d\mu(x)\\
&=\int_\mathbb{R} |f(x-y)|\,d\mu(x)\\
&=\int_{B_y}|f(x-y)|\,d\mu(x)>0
\end{align}
The last inequality implies that $\mu(B_y)>0$ for all $y \in \mathbb{R}$. We claim that for all $y\in A$ there is $x$ such that $|f(x-y)|>0$. Indeed, pick $y\in A$. Then, $\mu(\mathbb{R}\setminus A_y)=0$ and $\mu(B_y)>0$ implies that $A_y\cap B_y\neq 0$. But, now for $x\in A_y\cap B_y$ we have that $(*)$ holds for the pair $(x,y)$ and that $|f(x-y)|>0$ which implies that $|g(y)|=\alpha/ \beta$. Hence, $g$ is constant on $A$ and since $\mathbb{R}\setminus A$ has zero measure it follows that $g$ is constant almost everywhere.


问:
Showing an extension of degree 2 is normal in a different way
Let $F$ be the splitting field of a separable polynomial over $K$, and let $E$ be a subfield between $K$ and $F$. Show that if $[E:K]=2$, then E is the splitting field of some polynomial over $K$.

I've seen solutions that just go through some rote calculation of some irreducible $f(x)$, but I would rather go about it by correlating the index of some subgroup H of $G=Gal(F/K)$ to the degree of $[E:K]$. Since normal subgroups correspond to normal field extensions by FTGT then it shows the extension is normal. I am however having a hard time with expressing this the correct way.

答: Hint. By the fundamental theorem of Galois theory, the degree $[E:K]$ equals the index $[G:H]$. Group theory: what do you know about subgroups of index $2$?


问:
Find the eigenvalues and find a basis for each eigenspace.

Find the eigenvalues and find a basis for each eigenspace.

$$\begin{equation*}
A_{m,n} =
\begin{pmatrix}
0.5 && -0.6 \\
0.75 && 1.1
\end{pmatrix}
\end{equation*}$$

So the characteristic polynomial is det(A - $\lambda$I) =

$$$$\begin{equation*}
A_{m,n} =
\begin{pmatrix}
0.5 - \lambda && -0.6 \\
0.75 && 1.1 - \lambda
\end{pmatrix}
\end{equation*}$$$$

$$= 0.55 - 0.5 \lambda - 1.1 \lambda + \lambda^2 + 0.45 = \lambda^2 - 1.6 \lambda + 1$$

I'm a bit stuck here. Any help?

Eventually I get the roots are $\lambda_1 =  \frac{4}{5} + i\frac{3}{5}$ and $\lambda_2 = \frac{4}{5} - i \frac{3}{5}$

So how do I go from here?

Body:
Solving the quadratic equation, for example $\lambda_2 = 0.8 - 0.6i$ is a root:

$$\begin{align*}
\det[A-(0.8-0.6i)]
&= (0.8-0.6i)^2 -1.6 (0.8-0.6i) + 1\\
&= 0.64 - 0.96i - 0.36 - 1.28 + 0.96i + 1\\
&= 0
\end{align*}$$

Then solving the equation $(A-\lambda_2 I)\mathbf v = 0$,

$$\begin{align*}
\pmatrix{0.5-\lambda_2 & -0.6\\0.75 & 1.1-\lambda_2} \mathbf v &= 0\\
\pmatrix{-0.3+0.6i & -0.6\\0.75 & 0.3+0.6i} \mathbf v &= 0\\
\pmatrix{-0.3+0.6i & -0.6\\0.75(0.3-0.6i) & (0.3+0.6i)(0.3-0.6i)} \mathbf v &= 0 && R_2 \to (0.3-0.6i)R_2\\
\pmatrix{-0.3+0.6i & -0.6\\0.75(0.3-0.6i) & 0.09 + 0.36} \mathbf v &= 0\\
\pmatrix{-0.3+0.6i & -0.6\\0.75(0.3-0.6i) & 0.45} \mathbf v &= 0\\
\pmatrix{-0.3+0.6i & -0.6\\0 & 0} \mathbf v &= 0 && R_2 \to R_2 + 0.75R_1\\
\mathbf v &= k\pmatrix{0.6\\-0.3+0.6i}
\end{align*}$$

Similar for the remaining complex eigenvalues.


问:
Isomorphism Subalgebra
Given, the unit interval $I$ endowed with the Lebesgue measure $\mu$, and let $A$ be the (Boolean) algebra of Jordan measurable subsets of $X$ with respect to $\mu$, (i.e. those sets that satisfying $\mu(\partial(A))=0)$, modulo sets of zero measure.

Is it true that every (nontrivial) (Boolean) subalgebra $B$ of $A$ is isomorphic to $A$.

答:
The answer to your question, as we have already stated in comments, is negative. A nonatomic counterexample is the algebra $B$ generated by intervals of the form $[0,1/n]$ where $n$ is a nonnegative integer. This algebra is countable, while $A$ is uncountable -- every two different intervals $(a,b)$ and $(c,d)$ in $I$ have the symmetric difference of nonzero Jordan measure, hence they are in different equivalent classes after division by the ideal of Jordan null sets. Thus $B$ and $A$ are nonisomorphic.

So far I have encountered only one study of the algebra $A$, or more precisely the algebra $\mathscr{J}$ of all Jordan-measurable sets in $I$ (ie. without dividing by null sets). It was a paper by Schachermayer entitled On some classical measure-theoretical theorems for non-complete Boolean algebras, published in the Dissertationes Mathematicae series in 1982. The paper is rather difficult to read, as it requires some knowledge about vector measures and $C(K)$-spaces. The result in the article states that $\mathscr{J}$ has the Nikodym Boundedness Property and Orlicz-Pettis Property but it fails to have the Grothendieck Property.

If you look for some informations on Jordan measure just read these excelent notes by Terrence Tao: http://terrytao.wordpress.com/2010/09/04/245a-prologue-the-problem-of-measure/.

Title: Isomorphism Subalgebra


问:
Lower bound on clique number of graph through vertex degrees?
Suppose $\Gamma$ is a finite simple graph with $n$ vertices, such as the $i$-th vertex has degree $d_i$. Suppose $c$ is the clique number of $\Gamma$. Is it always true, that $c \geq \sum_{i=1}^n \frac{1}{n - d_i}$?

This inequality definitely holds for small graphs ($n \leq 5$) and for some "nice" classes of graphs (complete graphs and complete bipartite graphs). Also it holds for all graphs without vertices of degree exceeding $\frac{n}{2}$ (because in that case $\sum_{i=1}^n \frac{1}{n - d_i} \leq 2$). However, I do not know, whether the inequality is correct in general.

答: Yes, it is true. Dually, you can show that any graph has an independence number of at least $$\sum_{i=1}^n \frac1{1+\deg v_i},$$ then apply that result to the complement of $\Gamma$. This bound is was first shown by Caro, and Wei, and is proven as follows. Choose a random permutation of the $n$ vertices, and build an independent set $A$ consisting of all vertices $v$ such that none of $v$'s neighbors appear earlier in the permutation (i.e. a greedy algorithm on a random order). The probability a particular vertex $v$ is included in $A$ is the probability that $v$ occurs before all of its neighbors, yielding the expected size of $A$ as the expression above.


问:
Pumping lemma (context-free) of $L = \{a^nb^{\max\{n,m\}}a^m\ |\ n, m ≥ 0\}$
I want to show that $L = \{a^nb^{\max\{n,m\}}a^m\ |\ n, m ≥ 0\}$ is not context-free.

I tried with things like take $w=a^pb^pa^p$ and got that $vwz$ is $a^qb^{p-q}$ but... I don't know, it doesn't seem right to me.

答:
If $p$ is the pumping length, let $w=a^pb^pa^p$. If $L$ is context-free, it is possible to write $w=uvxyz$ in such a way that $|vy|\ge 1$, $|vxy|\le p$, and $uv^kxy^kz\in L$ for every integer $k\ge 0$. Now consider cases.

If $vxy$ lies entirely within the block of $b$s or one of the blocks of $a$s, $uv^2xy^2z\notin L$: if $vxy$ lies in one of the $a$ blocks, $uv^2xy^2z$ has too few $b$s, and if it lies in the $b$ block, $uv^2xy^2z$ has too many $b$s.

If $vxy$ contains both $a$s and $b$s, then $uxz\notin L$, because it contains too few $b$s. (Always remember that it’s possible to pump down, by setting $k=0$, as well as up.)


问:
A consistent set of axioms
Let A be a set of axioms, we say that A is nearly consistent is there is a finite sub set of A (A') for which A\A' is consistent.

I saw the following 2 claims, and want some help on which direction to go (prove or contradiction):

If A and B are nearly consistent then A union B is nearly consistent.

If A and B are nearly consistent then A intersection B is nearly consistent.

For the first I have did the following trying to prove it and got to dead end:

A is nearly consistent then A\A' is consistent (which means there is a for which a doesn't stem from A\A')

B is nearly consistent then B\B' is consistent (which means there is b for which b doesn't stem from B\B')

答:
Claim 1 is false, even if $A$ and $B$ are fully consistent. (In fact, we could even have every sentence in $A$ contradict every sentence in $B$.)

In propositional logic we can whip up a very simple example (see the comments below): fixing a set $\{p_i: i\in\mathbb{N}\}$ of propositional atoms, let $A=\{p_i:i\in\mathbb{N}\}$ and $B=\{\neg p_i: i\in\mathbb{N}\}$.

As an aside, note that as long as we require the sentences involved to be semantically distinct - which is needed to rule out silliness like $A=\{p, p\wedge p, p\wedge (p\wedge p), ...\}$, $B=\{\neg p, \neg p\wedge \neg p, \neg p\wedge(\neg p\wedge \neg p), ...\}$ -  an infinite propositional language is required since there are only $2^{2^n}$ logically distinct propositional sentences in $n$ propositional variables.)

The principle also breaks down in first-order logic, and here a finite-language example exists even when we demand semantic distinctness. One way to do this is the following. Take two sentences $\varphi,\psi$ such that each has an infinite model but $\{\varphi,\psi\}$ is inconsistent. Then we can let $$A=\{\varphi\wedge \eta_i:i\in\mathbb{N}\}\quad\mbox{and}\quad B=\{\psi\wedge\eta_i:i\in\mathbb{N}\},$$ where $\eta_i$ is the sentence "There are at least $i$ many elements in the domain."

Of course you now have to construct such a $\varphi$ and $\psi$, but this is a good exercise. For example, in the language consisting of a single unary relation $U$, we can let

 $\varphi\equiv\exists xU(x)$ and $\psi\equiv\neg\varphi\equiv\forall x\neg U(x)$.

Title: A consistent set of axioms


问:
Find an example of non-locally finite collection
I got stuck on this problem and can't find any hint to solve this. Hope some one can help me. I really appreciate.

Give an example of a collection of sets $A$ that is not locally finite, such that the collection $B = \{\bar{X} | X \in A\}$ is locally finite.

Note: Every element in $B$ must be unique, so maybe there exist 2 distinct sets $A_{1}, A_{2} \in A$, but have $\bar{A_{1}} = \bar{A_{2}}$. So that's why this problem would be right even though $A \subset \bar{A}$.

Thanks everybody.

答:
If someone wants to check my easy proof for metric spaces that would be good.

Let the topological space be a metric space. Here I will use X = $(0,1)$ with the basis of (0,1) being the standard basis of R consisting of open balls, but those balls not contained in (0,1) are removed. Use the absolute value function as the metric for simplicity. (|x-y| = d(x,y) for x,y $\in$ R).

Then let $A_{N}$ = $\bigcup_{i=0}^{N-1}(\frac{i}{N}, \frac{i+1}{N})$ and A = $\{A_{N}\}_{N=1}^{N=\infty}$

It should be easy to see that $A_{N} = (0, \frac{1}{N})\bigcup(\frac{1}{N}, \frac{2}{N})\bigcup...\bigcup(\frac{N-1}{N}, 1)$

1.) Choose $x_{0} \in (0,1)$. Then a neighborhood U containing $x_{0}$ intersects (0,1), in which case U will intersect infinitely many $A_{N} \in A$.

This is because an open set U containing $x_{0}$ will be the union of open balls B contained in U. Some ball B in U contains $x_{0}$. This ball has a radius $r > 0$.

The existence of a ball B in U containing $x_{0}$ with positive radius r implies that some $N_{0} \in \mathbb{Z^{+}}$ satisfies $\frac{1}{N_{0}} < r$.

Thus B intersects at least 2 balls of radius $\frac{1}{N}$ in $A_{N}$ for all $N > N_{0}$ no matter where x lies in (0,1). Thus B intersects all $A_{N}$ which have $N > N_{0}$. Since B is contained within U, U must intersect the same two balls. As N > $N_{0}$ goes to infinity, any neighborhood U of $x_{0}$ intersects 2 balls of infinitely many $A_{N}$.

Here we note that A is not locally finite since no open set U of X intersects only finitely many $A_{N}$.

2.) Now take the closure of each $A_{N}$.

3.) Then $\overline{A_{N}}$ = (0,1) for each N. (Note 0 and 1 are not elements of X = (0,1)).

Hence the neighborhood U containing $x_{0}$ intersects the only member of $\overline{A} = (0,1)$

Now, to expand this to any metric space X: first expand this proof to a bounded subset of the metric space X. Then use a limiting process to expand to the entire space. Should work great.

Ill do R^n here for example: Just use a non-empty bounded open ball B of radius R' in R^n, instead of (0,1) in R. Approximate the set from the inside with almost disjoint closed cubes of side length r' = R'/N. (r = R/2, R/3, R/4, etc, etc...) Now take the interiors of these cubes to get a collection of disjoint open cubes. Call the collection $C_{N}$. The interior of $\overline{\bigcup{C_{N}}}$ can now be exhausted with almost disjoint closed cubes $C_{N+1}$ such that the $\overline{\bigcup{C_{N}}} = \overline{\bigcup{C_{N+1}}}$ = V. Then as above, any ball B containing x in U will intersect 2 cubes of each $C_{N}$ as long as N is large enough. Such as when R'/N = r' is smaller than 2 times the radius r of B containing x. Then do some sort of limiting process on R' to exhaust the entire set $R^{n}$; $\lim_{R' \to \infty} B \supset X$.


答: On the other hand, 2 is true. If $A\setminus A^\prime,\,B\setminus B^\prime$ are each consistent, so is their intersection, which contains all elements of $A\cap B$ with the possible exception of those in $A^\prime\cup B^\prime$, i.e. finitely many. This makes $A\cap B$ nearly consistent.

Title: A consistent set of axioms

Body:
Let A be a set of axioms, we say that A is nearly consistent is there is a finite sub set of A (A') for which A\A' is consistent.

I saw the following 2 claims, and want some help on which direction to go (prove or contradiction):

If A and B are nearly consistent then A union B is nearly consistent.

If A and B are nearly consistent then A intersection B is nearly consistent.

For the first I have did the following trying to prove it and got to dead end:

A is nearly consistent then A\A' is consistent (which means there is a for which a doesn't stem from A\A')

B is nearly consistent then B\B' is consistent (which means there is b for which b doesn't stem from B\B')


问:
Integral containing exponential
I am trying to prove below (Reference: Green's function with application, page 159):
$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i(x-y)\tau}}{\tau^4-a^4}d\tau=\frac{1}{4a^3}(ie^{ia|x-y|}-e^{-a|x-y|})$$

When trying to calculate the integral, I used Wolfram, and I get the following:
$$\int\frac{e^{(x-y)i\tau}}{\tau^4-a^4}d\tau=\frac{1}{4}\sum_{{\omega:a^4-\omega^4=0}}\frac{e^{i(x-y)\omega}Ei(iA(\tau-\omega))}{\omega^3}+const$$
I am not sure how to deal with Ei function for an imaginary number with specified integral boundary.

答:
@Doug M already posted an answer assuming that $a$ is neither real nor pure imaginary. In this answer, I will tackle the opposite case, i.e., the case with $a > 0$.

If $a > 0$, then the integrand suffers from poles $\pm a$ along the real line, and so, the integral does not converge in ordinary sense. For this reason, we will interpret the integral in Cauchy principal value sense. Then by noting that the imaginary part of the integrand is an odd function, we get

$$ I
:= \frac{1}{2\pi} \operatorname{PV}\! \int_{-\infty}^{\infty} \frac{e^{i(x-y)\tau}}{\tau^4-a^4} \, \mathrm{d}\tau
= \frac{1}{2\pi} \operatorname{PV}\! \int_{-\infty}^{\infty} \frac{\cos((x-y)\tau)}{\tau^4-a^4} \, \mathrm{d}\tau. $$

In particular, the integral depends only on $k := \left| x - y \right|$.

Now by applying the residue theorem applied to the contour integral along the dented semicircular arc

and then letting the outer radius to $\infty$ and the inner radii to $0^+$, we get

\begin{align*}
I
&= \frac{1}{2\pi} \operatorname{PV}\! \int_{-\infty}^{\infty} \frac{e^{ik\tau}}{\tau^4-a^4} \, \mathrm{d}\tau \\
&= \frac{1}{2\pi} \biggl( i\pi \, \mathop{\operatorname{Res}}_{z=a} \, \frac{e^{ikz}}{z^4-a^4} + i\pi \, \mathop{\operatorname{Res}}_{z=-a} \, \frac{e^{ikz}}{z^4-a^4} + 2i\pi \, \mathop{\operatorname{Res}}_{z=ia} \, \frac{e^{ikz}}{z^4-a^4} \biggr) \\
&= \frac{i}{2} \frac{e^{ika}}{4a^3} + \frac{i}{2} \frac{e^{-ika}}{4(-a)^3} + i \frac{e^{-ak}}{4(ia)^3} \\
&= - \frac{1}{4a^3}\bigl( \sin(ak) + e^{-ak} \bigr) \\
&= - \frac{1}{4a^3}\bigl( \sin(a\left|x-y\right|) + e^{-a\left|x-y\right|} \bigr).
\end{align*}

Notice that this coincides with the real part of the proposed answer by OP.

Title: Integral containing exponential


问:
Tangent planes for the function $f(x,y) = 1 - x^2 - y^2$
I have to solve the following statements, finding the tangent planes to $f$ and its points of tangency:

that contains the line in $\mathbb{R}^3$ that passes through the points $(3,0,3)$ and $(0,-3,3)$

and this other statement:

that contains the point $(0,0,2)$

For the first statement, I tried first making the parametric equation and I found that $L(t) = (3-3t, -3t, 3)$, and I also know that the equation of the plane is given by $$p(x,y) = \frac{\partial f}{\partial x}(x_0,y_0)(x-x_0) + \frac{\partial f}{\partial y}(x_0,y_0)(y-y_0) + f(x_0,y_0) = -2x_0 (x-x_0) -2y_0 (y-y_0) + (1 -x_0^2 - y_0^2)$$ but after that, I don´t see what else to do. Could you give me any hint for both statements?

答:
Given

$$
\cases{
f(x,y,z)=x^2+y^2+z-1=0\\
\pi(x,y,z) = a x+b y + z c + d=0\\
p=(x,y,z)\\
p_1=(3,0,3)\\
p_2=(0,-3, 3)
}
$$

Construct $\pi(x,y,z)$ such that it contains the line $L \to p = p_1+\mu(p_2-p_1)$  and is tangent to $f(x,y,z)$.

At the tangency point $p_0$ we have

$$
\cases{
\nabla f(p_0) = \lambda \nabla \pi(x_0)\\
f(p_0) = 0\\
\pi(p_0) = 0\\
\pi(p_1) = 0\\
(p_2-p_1)\cdot\nabla \pi(p_0) = 0\\
\|\nabla \pi(p_0)\|=1
}
$$

Here

$$
\cases{
\nabla f(p_0) = (2x_0,2y_0,1)\\
\nabla\pi(p_0) = (a,b,c)
}
$$

so we have $8$ equations on the $8$ unknowns $x_0,y_0,z_0,a,b,c,d,\lambda$. Solving those equations we have two solutions:

$$
p_0 = \left\{
\begin{array}{ccc}
 (\frac{1}{2} \left(3-\sqrt{13}\right), & \frac{1}{2} \left(\sqrt{13}-3\right), & 3 \sqrt{13}-10) \\
( \frac{1}{2} \left(3+\sqrt{13}\right), & -\frac{1}{2} \left(3+\sqrt{13}\right), & -10-3 \sqrt{13})\\
\end{array}
\right.
$$

and the planes are characterized by

$$
(a,b,c,d) = \left\{
\begin{array}{cccc}
 (-\sqrt{\frac{2}{51} \left(9-\sqrt{13}\right)}, & \sqrt{\frac{2}{51} \left(9-\sqrt{13}\right)}, & \frac{1}{68} \left(\sqrt{102}
   \left(9-\sqrt{13}\right)^{3/2}-8 \sqrt{102 \left(9-\sqrt{13}\right)}\right), & \frac{1}{204} \left(12 \sqrt{102 \left(9-\sqrt{13}\right)}-\sqrt{102}
   \left(9-\sqrt{13}\right)^{3/2}\right)) \\
 (-\sqrt{\frac{2}{51} \left(9+\sqrt{13}\right)}, & \sqrt{\frac{2}{51} \left(9+\sqrt{13}\right)} ,& \frac{1}{68} \left(\sqrt{102}
   \left(9+\sqrt{13}\right)^{3/2}-8 \sqrt{102 \left(9+\sqrt{13}\right)}\right), & \frac{1}{204} \left(12 \sqrt{102 \left(9+\sqrt{13}\right)}-\sqrt{102}
   \left(9+\sqrt{13}\right)^{3/2}\right)) \\
\end{array}
\right.
$$

Attached a plot showing in red the given line and in black the tangentcy points.


问:
Prove that $c^2=a^2+b^2-2ab\cos\left(C-\frac{KS}{3}\right)$ holds on a smooth surface.

Problem: Given a infinitely small geodesic triangle $\triangle ABC$ on a smooth surface, denote the corresponding edges as $a,b,c.$ Prove: the area of $\triangle ABC$ (denote as $S$) and the Gauss curvature on $C$ ( denote as $K$) satisfies

$$c^2=a^2+b^2-2ab\cos\left(C-\frac{KS}{3}\right).$$

Since the triangle is infinitesimal, therefore it is naturally to use the first fundamental form and the Gauss-Bonnet theorem. However, I got stuck at how to handle the $\frac{KS}{3}$ in the equation. Since the coefficient is $\frac{1}{3}$, if using Gauss-Bonnet theorem, we would include new variables $\angle A,\angle B$. Could anyone help me out? Thanks a lot in advance!

答:
Making certain reasonable approximations due to the assumption of infinitesimal triangles, this works on the sphere. For simplification of notation we will work on a unit sphere.

Assume $s=\frac{a+b+c}2$ is very small and $\frac{\min(a,b,c)}{\max(a,b,c)}\ge\lambda\gt0$ is not. That is, $\frac23s\le\max(a,b,c)\le s$ and $\min(a,b,c)\ge\frac23\lambda s$.

Let $\tilde{C}$ be the angle opposite side $c$ in a spherical triangle with sides $a,b,c$, and $C$ be the angle opposite side $c$ in a plane triangle with sides $a,b,c$. Then
$$
\begin{align}
\cos(\tilde C)
&=\frac{\cos(c)-\cos(a)\cos(b)}{\sin(a)\sin(b)}\tag1\\
&=\frac{\left(1-\frac12c^2+\frac1{24}c^4\right)-\left(1-\frac12a^2+\frac1{24}a^4\right)\left(1-\frac12b^2+\frac1{24}b^4\right)+O\!\left(s^6\right)}{\left(a-\frac16a^3\right)\left(b-\frac16b^3\right)+O\!\left(s^6\right)}\tag2\\
&=\frac{a^2+b^2-c^2+\color{#C00}{\frac1{12}c^4-\frac1{12}a^4-\frac1{12}b^4-\frac12a^2b^2}}{2ab\color{#090}{-\frac13a^3b-\frac13ab^3}}+O\!\left(s^4\middle/\lambda^2\right)\tag3\\
&\approx\frac{a^2+b^2-c^2}{2ab}+\frac{\color{#C00}{\frac1{12}c^4-\frac1{12}a^4-\frac1{12}b^4-\frac16a^2b^2-\frac13a^2b^2}}{2ab}\\
&-\frac{a^2+b^2-c^2}{2ab}\frac{\color{#090}{-\frac13a^3b-\frac13ab^3}}{2ab}\tag4\\
&=\cos(C)+\frac{a^2+b^2-c^2}{12}\frac{a^2+b^2-c^2}{2ab}-\frac{ab}6\tag5\\
&=\cos(C)+\frac{(a-b)^2-c^2}{12}\frac{(a+b)^2-c^2}{2ab}\tag6\\
&=\cos(C)-\frac{16\Delta^2}{48\cdot\frac12ab}\tag7\\
&=\cos(C)-\tfrac\Delta3\sin(C)\tag8\\[6pt]
&\approx\cos\left(C+\tfrac\Delta3\right)\tag9
\end{align}
$$
Explanation:

$(1)$: Spherical Law of Cosines

$(2)$: $\cos(x)=1-\frac12x^2+\frac1{24}x^4+O\!\left(x^6\right)$

$\phantom{\text{(2):}}$ $\sin(x)=x-\frac16x^3+O\!\left(x^5\right)$

$(3)$: collect terms and move the error out of the fraction

$(4)$: $\frac{u+\color{#C00}{\mathrm{d}u}}{v+\color{#090}{\mathrm{d}v}}-\left(\frac uv+\frac{\color{#C00}{\mathrm{d}u}}v-\frac{u\,\color{#090}{\mathrm{d}v}}{v^2}\right)=\frac{(u\,\mathrm{d}v-v\,\mathrm{d}u)\,\mathrm{d}v}{v^2(v+\mathrm{d}v)}=O\!\left(s^4\middle/\lambda^2\right)$

$(5)$: apply the Law of Cosines to term $1$ and combine terms $2$ and $3$

$(6)$: factor

$(7)$: Heron's formula

$(8)$: $\Delta=\frac12ab\sin(C)$

$(9)$: $\cos(x+\mathrm{d}x)\approx\cos(x)-\sin(x)\,\mathrm{d}x$

Note that
$$
\begin{align}
\cos\left(C+\tfrac\Delta3\right)
&=\cos(C)\overbrace{\ \cos\left(\tfrac\Delta3\right)\ }^{1-O\left(s^4\right)}-\sin(C)\overbrace{\ \sin\left(\tfrac\Delta3\right)\ }^{\frac\Delta3-O\left(s^6\right)}\tag{10a}\\[6pt]
&=\cos(C)-\tfrac\Delta3\sin(C)+O\!\left(s^4\right)\tag{10b}
\end{align}
$$
so that the total error in the approximation $\tilde{C}\approx C+\frac\Delta3$ is on the order of $s^4$.

Comparing Heron's formula
$$
\Delta=\sqrt{s(s-a)(s-b)(s-c)}\tag{11}
$$
and L'Huillier's formula
$$
\tan\left(\frac S4\right)=\sqrt{\tan\left(\frac{s}2\right)\tan\left(\frac{s-a}2\right)\tan\left(\frac{s-b}2\right)\tan\left(\frac{s-c}2\right)}\tag{12}
$$
using $\tan(x)=x\!\left(1+O\!\left(x^3\right)\right)$, and noting that $K=1$ on the unit sphere, we get
$$
KS=\Delta+O\!\left(s^4\right)\tag{13}
$$


问:
The Halting Problem -- why do we do the opposite?
I don't really get why testhalt doesnt exist based on the proof that some program, ie turing, called on testhalt does the opposite of testhalt.
Below is a definition for testhalt

And one for turing.

How come we absolutely must, in turing, do the opposite of testhalt? (meaning loop if testhalt halts, for example)

答:
The point isn't that $\mathsf{TURING}$ does the opposite of $\mathsf{TESTHALT}$, but rather that - when fed a judiciously chosen input - it does the opposite of itself. Since this is blatantly impossible, $\mathsf{TURING}$ can't exist and therefore $\mathsf{TESTHALT}$ can't exist either.

Specifically, consider what happens when we feed $\mathsf{TURING}$ its own index $\tau$. We get that $\mathsf{TURING}(\tau)$ halts iff $\mathsf{TESTHALT}(\tau,\tau)$ says "no" (by definition of $\mathsf{TURING}$) iff $\mathsf{TURING}$ does not halt on input $\tau$ (by assumption on $\mathsf{TESTHALT}$). This self-contradictory behavior ("I halt on input $\tau$ ifff I don't halt on input $\tau$") is what concludes the argument.


问:
Base Number Addition (base 6)
So, I'm studying base number addition (AoPS book) and I'm kind of stumped. Any help would be thankful. So the question was:

Perform the following base number addition: $ 1_6 + 5_6 = ? $

So initially, I thought it was 6 but turns out that I'm wrong. Here's my work. So I did $1_6 = 1*6^0$ and $5_6 = 5*6^0$ So I did $(5+1) * 6^0$ and my answer was 6 but it's actually 10. Can someone explain why?

答: Notice $1+5=6$. But $6_6$ is not the proper way to represent the number! $12346_6$ is not a proper way, just like you can't put a $10$ in one digit in base $10$. If you recall how bases work, if you wanted to represent $6$ in base $6$, this is $\boxed{10_6}$, since $6=1*6^1+0*6^0$.

Title: Base Number Addition (base 6)


问:
A subgroup of order $12$ for $SL(2, 3)$
Let $G$ be $SL(2, 3)$, the group of $2×2$ matrices with determinant $1$ and with entries in $\mathbb{Z/3Z}$.

Assume that $H$ is a subgroup of order $12$ in $SL(2, 3)$. Argue that $H$
must contain an element of order $2$, all eight elements of order $3$, and
eight elements of order $6$, and hence $H$ cannot exist.

Answer:

I am trying to show that the existence of element $A=\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$ is neccseray in $H$, but I have no clues yet. Also, The product of this element with any element of order $3$ is an element of order $6$.
If we can prove that $A$ and all eight elements of order $3$ are necessary, we will finish the proof.

Thank you for your answers.

答:
By Cauchy's theorem, $H$ contains an element of order $2$ and an element of order $3$.

The cardinality of $SL_2(\mathbf{F}_3)$ is $24$, so the index of $H$ is $2$. In other words, for each element of $SL_2(\mathbf{F}_3)$, either it is in $H$ or its square is.

Can you now show that this forces all order $3$ elements of $SL_2(\mathbf{F}_3)$ to be in $H$?

Title: A subgroup of order $12$ for $SL(2, 3)$


答:
It's only a question of how one represents numbers. The underlying value, say $8$, can be represented in many ways, whether it's $8_{10}$, or $1000_2$, or $4_{10} + 4_{10}$: the underlying quantity is the same concept of "the number eight".

The "base", or "radix", is the number of distinct symbols you can use, when representing numbers.

For example, decimal (base 10) has 10 symbols: $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. Binary (base 2) has 2 symbols $\{0, 1\}$.

You'll notice that no matter the base, the "carry" always happens for powers of the radix, which are always written as $1, 10, 100, 1000$, etc. Each power of the radix looks like what a power of $10$ would look like in base $10$. In particular, the radix is always written as $10_n$ in base $n$.

One way to think about this is with each digit existing on a clock of radix hours. When you hit the "0" hour again, you have a carry, and need to start counting on a new clock (representing a new digit).

Also, you can think of "base 1" (a single symbol), as the way you would count numbers on a dice, or on cards. Every new element needs a new digit.

So what happens in your case ? Well with $5_6 + 1_6$, you run out of symbols on your $6$-hour clock. So you add (carry) a new digit, and get $10_6$, representing a power of the radix, here, $6_{10}^1 = 6_{10}$.

Title: Base Number Addition (base 6)

Body:
So, I'm studying base number addition (AoPS book) and I'm kind of stumped. Any help would be thankful. So the question was:

Perform the following base number addition: $ 1_6 + 5_6 = ? $

So initially, I thought it was 6 but turns out that I'm wrong. Here's my work. So I did $1_6 = 1*6^0$ and $5_6 = 5*6^0$ So I did $(5+1) * 6^0$ and my answer was 6 but it's actually 10. Can someone explain why?


问:
What are good ways of understanding a permutation group from a set of generators?
I'm trying to understand the structure of a Rubik's Cube-style puzzle I'm playing with; I have an expression of the solutions as the permutation group generated by four elements of $S_{16}$, each a simple cycle. My first inclination is to check all words of a given length (I think I can check all 8-element words easily, and maybe all 9- or 10-element words with a little effort) to see which ones move as few pieces as possible and then try and use that to understand more of the structure (for instance, I already know that there are elements of odd sign, so it's not a subgroup of $A_{16}$); above and beyond that, though, what tricks and techniques should I be aware of?  In particular, are there any good ways of finding useful/meaningful expressions of my group as some sort of wreath product, or of understanding its quotient in $S_{16}$ (assuming it's even normal!)?  I'm presuming that understanding $S_{16}/G$ will provide a sense of the constraints that any valid position has to satisfy...

EDIT: For anyone who's curious enough and has better tools to play with it than I do, the four generators are the cycles: $\alpha = (1\quad 2\quad 3\quad 7\quad 11\quad 10\quad 9\quad 5)$ , $\beta = (2\quad 3\quad 4\quad 8\quad 12\quad 11\quad 10\quad 6)$, $\gamma = (5\quad 6\quad 7\quad 11\quad 15\quad 14\quad 13\quad 9)$, and $\delta = (6\quad 7\quad 8\quad 12\quad 16\quad 13\quad 14\quad 10)$.  I wouldn't be surprised to learn it's all of $S_{16}$, but that still leaves a lot of obviously-open questions from a puzzle perspective (minimal-length words to swap two elements and the like), and any good suggestions on ways of finding particular elements of the group (where to look for interesting commutators and conjugates, etc.) would be welcome; I have a little background from doing some basic Rubik's-Cube stuff (for instance, my first look was at the element $(\alpha^2\beta^2)^2$ ), but not much.

答:
GAP is reasonably well suited to such problems.  Here is a sample session for your problem:

```<code>gap> g:=Group( (1,2,3,7,11,10,9,5), (2,3,4,8,12,11,10,6),
> (5,6,7,11,15,14,13,9), (6,7,8,12,16,13,14,10) );;
gap> StructureDescription(g);
"S16"
</code>```

To find short move sequences that leave lots of points stable, you could use your idea of just checking short sequences:

```<code>gap> f := FreeGroup("a","b","c","d");;
gap> for w in f do
> p := MappedWord(w,GeneratorsOfGroup(f),GeneratorsOfGroup(g));
> if p <> () and NrMovedPoints(p) < 5 then Print(p," ",w,"\n"); fi;
> od;
( 5,10)( 7,12) d^-2*a^-2*d^2*a^2
</code>```

Which means that you un-apply δ twice, then un-apply α twice, then δ twice, then α twice.  In other words, the commutator of α and δ is a "nice" move.

For smaller puzzles, GAP supplies a function Factorization which you could normally use to answer these questions quickly.  For whatever reason it was not working well in this particular example.  Normally one would just ask:

```<code> gap> Factorization( g, (1,2) );
</code>```

and it would give you (a short) move sequence that switched the first and second tiles.

Body:
GAP is reasonably well suited to such problems.  Here is a sample session for your problem:

gap> g:=Group( (1,2,3,7,11,10,9,5), (2,3,4,8,12,11,10,6),
> (5,6,7,11,15,14,13,9), (6,7,8,12,16,13,14,10) );;
gap> StructureDescription(g);
"S16"

To find short move sequences that leave lots of points stable, you could use your idea of just checking short sequences:

gap> f := FreeGroup("a","b","c","d");;
gap> for w in f do
> p := MappedWord(w,GeneratorsOfGroup(f),GeneratorsOfGroup(g));
> if p <> () and NrMovedPoints(p) < 5 then Print(p," ",w,"\n"); fi;
> od;
( 5,10)( 7,12) d^-2*a^-2*d^2*a^2

Which means that you un-apply δ twice, then un-apply α twice, then δ twice, then α twice.  In other words, the commutator of α and δ is a "nice" move.

For smaller puzzles, GAP supplies a function Factorization which you could normally use to answer these questions quickly.  For whatever reason it was not working well in this particular example.  Normally one would just ask:

 gap> Factorization( g, (1,2) );

and it would give you (a short) move sequence that switched the first and second tiles.


问:
On the smallest size of a set of multiples in modular arithmetic
Let $m$ be a positive integer. Let $X_m = \{1,\dots,m-1\}$ and $Y \subseteq X_m$. Let us say that $Y$ covers $X_m$ if all elements of $X_m$ have at least one multiple in $Y$ mod $m$. For example, if $m=6$, then $2$ is a multiple (mod $6$) of $1, 2, 4,$ and $5$. Likewise, $3$ is a multiple of $1, 3,$ and $5$. And since $\{1,2,4,5\}\cup\{1,2,5\} = X_m$, we have that $\{2,3\}$ covers $X_6$, which is the smallest possible choice (together with $\{3,4\}$). I am interested in bounding the size $\gamma(m)=|Y|$ of the smallest subset that covers $X_m$, as a function of $m$.

Clearly, $\gamma$ is not a monotone function, since $\gamma(p)=1$ for every prime $p$ (any number in $X_p$ is multiple of all others mod p). Nonetheless, the worst-case value for $\gamma$ seems to grow with some regularity. Experiments show that the first time that $\gamma(m)=2$ is at $m=6$, that $\gamma(m)=3$ is at $m=30$, and that $\gamma(m)=4$ is at $m=210$. From there, one may speculate that $\gamma(m)$ hits $k$ only when $m = p_k$, where $p_k$ is the $k^{th}$ primorial number (i.e., the product of the first $k$ prime numbers). In such case, the next milestone would be $\gamma(2310)=5$.

My question: Is there a known theorem in modular arithmetics that implies these observations? Now, even if $\gamma$ is unrelated to primorial numbers, I am actually interested in bounding $\gamma(m)$ by some function growing slowly in $m$ (typically, sublogarithmic). If these questions are more commonly formulated in different ways, please comment on this aspect as well.

答:
We have $\gamma(m) = \omega(m)$, the number of distinct prime factors of $m$, therefore we have
$$\limsup_{m \to \infty} \frac{\gamma(m)\log \log m}{\log m} = 1\,,$$
which is a wee bit better than a logarithmic bound.

To see that $\gamma(m) = \omega(m)$, let
$$m = \prod_{\kappa = 1}^{k} p_{\kappa}^{e_{\kappa}} \tag{1}$$
with the $p_{\kappa}$ being distinct primes, and $e_{\kappa} \geqslant 1$ for all $\kappa$.

For each $1 \leqslant r \leqslant k$, there is by the Chinese remainder theorem an $n_r$ such that
$$n_r \equiv 1 \pmod{p_r^{e_r}} \qquad\text{and}\qquad n_r \equiv 0 \pmod{p_{\kappa}^{e_{\kappa}}}$$
for all $\kappa \neq r$.

Since all multiples of $n_r$ are divisible by $p_{\kappa}^{e_{\kappa}}$ if $\kappa \neq r$, this means a $Y$ covering $X_m$ must contain a $y_r \in X_m$ with
$$\prod_{\kappa \neq r} p_{\kappa}^{e_{\kappa}} \mid y_r$$
for each $r \in \{1, \dotsc, k\}$. Thus $\gamma(m) \geqslant k = \omega(m)$.

On the other hand, the $k$-element set
$$Y = \{ p_r^{e_r-1}\cdot n_r : 1 \leqslant r \leqslant k\}$$
covers $X_m$.

Let $1 \leqslant a \leqslant m-1$, and let $r$ be the smallest $\kappa$ such that $p_{\kappa}^{e_{\kappa}}$ does not divide $a$. Then $p_r^{e_r-1}\cdot n_r$ is a multiple of $a$. First, by multiplying with suitable powers of the other primes dividing $m$, we find a multiple $a'$ of $a$ such that $p_r^{e_r}$ is the only prime power on the right hand side of $(1)$ that doesn't divide $a'$. Multiplying $a'$ with a suitable power of $p_r$ we obtain a multiple $a'' = b\cdot p_r^{e_r-1}\cdot n_r$, $p \nmid b$, of $a$. Multiplying with a modular inverse of $b$ modulo $p$ we see that $p_r^{e_r-1}\cdot n_r$ is indeed a multiple of $a$, so $Y$ covers $X_m$.


问:
Show that $\operatorname{Lip}_\alpha $ is not an homogeneous Banach space
This is an exercise of An Introduction to Harmonic Analysis of Yitzhak Katznelson:

Let $\operatorname{Lip}_\alpha(\mathbb{R}/2\pi\mathbb{Z}),\, \alpha \in (0,1)$ the Banach space of $2\pi$-periodic continuous functions such that
$$
r(f):=\sup_{t\in \mathbb{R}/2\pi \mathbb{Z},\,h\neq 0}\frac{|f(t+h)-f(t)|}{|h|^{\alpha }}<\infty
$$ with norm $\|f\|_{\operatorname{Lip}_\alpha }=\|f\|_\infty +r(f)$. Show that $\operatorname{Lip}_\alpha $ is not an homogeneous Banach space.

The statement to be proved is equivalent to show that there exists some $f\in \operatorname{Lip}_\alpha(\mathbb{R}/2\pi \mathbb{Z})$ such that
$$
\limsup_{\tau \to 0 }r(f_\tau -f)> 0
$$
for $f_\tau(x) :=f(x -\tau )$. I'm not sure how to continue with this exercise and I don't know some function that works as an example, some help will be appreciated.

答:
I will use $a$ instead of $\alpha$. Let $T$ be the interval $[0,2\pi]$.
Let $f$ be a function of the shape

$f(t)=t^a$ for $t\in[0,1]$,

$f$ is continuous and smooth on $(0,2)$ with $f'(2)=0$,

$f$ is zero on $[2,2\pi]$.

We identify $f:T\to \Bbb R$ (with $f(0)=f(2\pi)$) with a $2\pi$-periodic function on $\Bbb R$.

The quantity inside the supremum defining $r(f)$ is bounded, because (we may and do assume $h>0$, and then) the function
$$
\begin{aligned}
T^2&\to\Bbb R\ ,\\
(t,h)&\to\frac 1{h^a}(f(t+h)-f(t))&&\text{ for }t\in T\ ,\ h\in(0,2\pi]\ ,\\
(t,0)&\to\lim_{h\searrow0}
\frac 1{h^a}(f(t+h)-f(t))=\delta_0(t)&&\text{ for }t\in T\ ,
\end{aligned}
$$
is bounded. (It is continuous on a compact set obtained from $T^2$ obtained by removing a "small open (quarter)disc" with center in $(0,0)$. Then the control around $(0,0)$ comes from $(t+h)^a-t^a\le h^a$. This holds since for $t=0$ or $h=0$ this is clear, and else by $a$-homogenity it is enough to show $(1+h)^a\le 1+h^a$, i.e. $1+h\le 1+\frac 1ah^a\le (1+h^a)^{1/a}$.)

So $f$ is an element in the given Banach space. Let now $\tau>0$ be the number used for the shift of $f$. The shifted function $f_\tau$ is zero in a neighborhood of $0$, so $f-f_\tau=f$ in this small neighborhood. So for the special $t=0$ we have
$$
\begin{aligned}
r(f_\tau-f)
&\ge
\sup_{0< h}\frac{|(f_\tau-f)(0+h)-(f_\tau-f)(0)|}{h^a}
\\
&=
\limsup_{0< h}\frac{|f(0+h)-f(0)|}{h^a}
\\
&=
\sup_{0< h}\frac{h^a}{h^a}
\\
&=
1\ .
\end{aligned}
$$


问:
Can a martingale converge in $L^1$ but not almost surely?
In general, almost sure convergence and convergence in $L^1$ are independent modes of convergence in the sense that neither implies the other. For martingales, at least one implication, namely that almost sure convergence implies convergence in $L^1$ remains false. How about the other one? Does convergence of a martingale in $L^1$ imply almost sure convergence?

答: Another, somewhat shorter proof: As $L^1$ convergence is equivalent to uniform integrability and convergence in probability, almost sure convergence follows from the convergence theorem for uniformly integrable martingales.


问:
Dominated convergence theorem by Jensen inequality
Define functional $\psi(x_n):=\limsup_n x_n$ on a sequence of positive numbers $x_1,x_2...$. We can check that $\psi$ is convex.

Then if $f_n \to f$ pointwise, by Jensen's inequality,
$$\limsup_n \int |f_n-f|\le \int \limsup_n|f_n-f|=0$$

This proof is obviously wrong because I used no dominance conditions for $f_n$. But where did make a mistake? Is it because $\limsup_n|f_n-f|\neq 0$?

答:
You're trying to say that if $\phi$ is a convex function $\mathbb R^{\mathbb N} \to \mathbb R$ then
$$ \phi\left(\left\{\int g_n(x) \mu(\text{d}x)\right\}_n\right) \leq \int \phi(\{g_n(x)\}_n) \mu(\text{d}x) .$$
for a measure $\mu$ on $X$ and any sequence of functions $g_n:X \to \mathbb R$. This isn't the statement of Jensen's inequality and more importantly isn't true.


问:
Finding the maximal $t$ satisfying a family of inequalities
Given $c \in (0,1)$, find the maximal positive $t$ satisfying the following:
$$\forall n \in \{1,2,\ldots \}: 1+\frac{c}{n+(1-c)} \le \left(1+\frac{1}{n+t}\right)^{c}$$

My progress thus far:

A special case is $c = \frac{1}{2}$: the inequality simplifies to $3t-1 \le n(1-4t)$, which implies $t \le \frac{1}{4}$. $t=\frac{1}{4}$ actually works.

The inequality $\left(1+\frac{1}{n+t}\right)^{c} \le \frac{c}{n+t}$ shows $t \le 1-c$.

答:
The correct $t$ is $\frac{1-c}{2}$, as I was able to guess in the comments using vadim123's calculations.

Proof: define $f(x)=\ln(1+\frac{1}{\frac{1}{x}+t})^c - \ln(1 + \frac{c}{\frac{1}{x}+1-c})$. We are given that $f(\frac{1}{n}) \ge 0$ for any $n \ge 1$.

Simplifying $f$, we find $f(x)=c \ln(1+\frac{x}{1+tx})-\ln(1+\frac{cx}{1+x(1-c)})$.

The first derivative is $f'(x)=c (\frac{1}{(1+tx)(1+(t+1)x)}-\frac{1}{(1+x)(1+(1-c)x)})$

If $t=\frac{1-c}{2}$, the first derivative is positive for positive $x$, since it amounts to $(1+x(1-c))(1+x) \ge (1+x\frac{1-c}{2})(1+x\frac{3-c}{2})$, which upon expanding it becomes $x^2(1-c^2) \ge 0$. This shows that $f$ is increasing, hence $f(\frac{1}{n}) \ge f(0) = 0$.

In general, $f'(x) \ge 0$ iff $x(1-c-t(t+1))\ge 2t+c-1$ (assuming $x \ge 0$). Assume $1-c\ge t > \frac{1-c}{2}$ also works. Since $ 2t+c-1 > 0$, $f$ is necessarily decreasing in a small right neighborhood of $0$, so $f(\frac{1}{n})$ is negative for big enough $n$ since $f(0)=0$.


问:
Equivalent norms in $\mathcal{C}([0, 1], \mathbb{R})$
Let $E = \mathcal{C}([0, 1], \mathbb{R})$ seen as a vector space.

Let $F$ be a subspace of $E$ such that there exists a real number $A > 0$ verifying: $\forall g \in F, \forall x \in [0, 1], |g(x)| \leq A (\int_{0}^{1} g^{2})^{1/2}$.

How can we show that $F$ is a finite-dimensional vector space with $\dim(F) \leq A^{2}$ ?

I though about using an orthonormal set of vectors (for the usual scalar product) but I could not make it work.

Thank you for your help.

答:
Suppose $\{g_1,g_2,...,g_n\}$ is orthonotmal on $F$ w.r.t $L^{2}$ inner product. Then $ |\sum\limits_{k=1}^{n} a_ig_i(x)|\leq A \|\sum\limits_{k=1}^{n} a_ig_i\|_2=A\sqrt {\sum\limits_{k=1}^{n} |a_i|^{2}}$ for each $x$. This implies that $\sqrt {\sum\limits_{k=1}^{n} |g_i(x)|^{2}} \leq A$. Squaring and integrating we get $n \leq A^{2}$.

I have used the following basic fact: $\sup \{\sum\limits_{k=1}^{n} a_ib_i: \sum\limits_{k=1}^{n} |a_i|^{2} \leq 1\}=\sqrt {\sum\limits_{k=1}^{n} |b_i|^{2}}$.


问:
Let $H$ be the set of all permutations $\alpha\in S_5$ satisfying $\alpha(2) = 2$. Is $H$ a group?
Let $H$ be the set of all permutations $\alpha\in S_5$ satisfying $\alpha(2) = 2$. Which of the
properties (closure, associativity, identity, inverses) does $H$ enjoy
under composition of functions?

Solution:

Let $\alpha, \beta, \gamma \in H$.

Closure: For any $\alpha, \beta \in H$, we have $\alpha\beta(2)=\alpha(\beta(2))=\alpha(2)=2 $. Thus composition is closed.

Associativity: it's obvious that $(\alpha \beta) \gamma(2)=\alpha (\beta \gamma)(2)$

Identity: I guess it is the same as for $ S_5 $. Since it leaves the rest of the elements fixed and especially, at $ 2 $

Inverses: Since every permutation in $ H $ leaves $2$ fixed, its inverse also. So for every $ \alpha \in H $ there is an inverse.

I am right?

答:
Your proof is fine.

Here's another . . .

It suffices to show that $H$ is a subgroup of $S_5$, since every subgroup of a group is itself a group.

I will use the one-step subgroup test.

Since the identity in $S_5$ fixes $2$ trivially,  we have $H\neq\varnothing$.

Let $\sigma\in H$. Then $\sigma\in S_5$ such that $\sigma(2)=2$; so, in particular, $\sigma\in S_5$. Hence $H\subseteq S_5$.

Let $\sigma,\tau\in H$. Then $\sigma(2)=2$ and $\tau(2)=2$. Since $\tau$ is a bijection on $\{1,\dots, 5\}$ that fixes $2$, so is its inverse $\tau^{-1}\in H$ (because $$\begin{align}
\tau^{-1}(2)&=\tau^{-1}(\tau(2))\\
&=(\tau^{-1}\tau)(2)\\
&={\rm id}_{S_5}(2)\\
&=2).
\end{align}$$ We have

$$\begin{align}
(\sigma\tau^{-1})(2)&=\sigma(\tau^{-1}(2))\\
&=\sigma(2)\\
&=2,
\end{align}$$

but the composition of two bijections is a bijection, so $\sigma\tau^{-1}\in H$.

Hence $H\le S_5$.

Hence $H$ is a group.


问:
Understanding the proof that $3X^3 + 4Y^3 + 5Z^3 = 0$ has no non-trivial solutions in $\mathbb{Q}$
I am reading Casell's book 'Lectures on Elliptic Curves' and I am trying to understand a Lemma on page 86 that is used to prove that $3X^3 + 4Y^3 + 5Z^3 = 0$ (Selmer's Curve) has no non-trivial solutions in $\mathbb{Q}$, and thus is a counter-example to the Hasse Principle.

I understand everything in the proof apart from the last line. I don't understand why, if you have two points on the curve $X^3 + Y^3 + dZ^3 = 0$ that are conjugate, then it follows the line joining them intersects the curve at a point in $\mathbb{Q}$ that is not $(1,-1,0)$.

Is there some theorem about conjugate points on curves that is used here? Or is there some way to easily show this in this case?

答:
Let us make the computations explicitly when starting with $(u,v,w)$,
thus writing down explicitly a point $(x_3,y_3,z_3)$ algebraically in therms of $(u,v,w)$.

Let $F=\Bbb Q(\rho)$ be the field obtained by adjoining the primitive third root of one.
Let $(u,v,w)$ with $uvw\ne 0$ be a point on the curve $aU^3+bV^3+cW^3=0$.
It is a good idea to introduce
$$
\begin{aligned}
A &= au^3\ ,\\
B &= bv^3\ ,\\
C &= cw^3\ ,\\
0 &= A+B+C\ .
\end{aligned}
$$
and then as in the lemma the point $(x,y,z)$ over $F$ on the curve $X^3+Y^3+dZ^3=0$ with components
$$
\begin{aligned}
x &= A+\rho  \; B +\rho^2\; C\ ,\\
y &= A+\rho^2\; B +\rho\;   C\ ,\\
z &= -3uvw\ne 0\ ,\\[3mm]
&\qquad\text{ it is on the curve because of }\\[3mm]
x+y &= 2A +(\rho+\rho^2)(B+C)=2A+(-1)(-A)=3A\ ,\\
\rho x+\rho^2 y &= 2C +(\rho+\rho^2)(A+B)=2C+(-1)(-C)=3C\ ,\\
\rho^2 x+\rho y &= 2B +(\rho+\rho^2)(A+C)=2B+(-1)(-B)=3B\ ,\\
x^3+y^3 &=(x+y)(\rho x+\rho^2 y)(\rho^2 x+\rho y)=3^3\; ABC=3^3\; \underbrace{abc}_{=d}\;(uvw)^3 =-dz^3\ .
\end{aligned}
$$
The curve $X^3+Y^3+dZ^3=0$ has the $\Bbb Q$-rational point $(1,-1,0)$, so
it is an elliptic curve. Note the symmetric role of $X,Y$ in the equation.

So if $[x':y':z']\sim(x'/z', y'/z')$ is a rational point for it, then $[y':x':z']\sim(y'/z',x'/z')$ is also a rational point.
(We use two coordinates in the affine $(X,Y)$ plane for points with $z\ne 0$.)

We can also "twist" such an $F$-rational point, by replacing $x'$ by $\rho^j x'$ or/and (in the same time)
$y'$ by $\rho^k y'$.
So let us take two points $P,Q$ from this list, and compute $-(P+Q)$.

Digression. The line through $P=[x:y:z]$ and $\bar P=[\bar x:\bar y:\bar z]=[y:x:z]$ in the affine space $(X,Y)$ has the equation
$$
\begin{aligned}
&\left(Y-\frac yz\right) =m\left(X-\frac xz\right)\ ,\qquad\text{ where }
\\
m
&=\frac{y/z-\bar y/z}{x/z-\bar x/z}
=\frac {y-\bar y}{x-\bar x}
=\frac {y-x}{x-y}
=-1\ .
\end{aligned}
$$
So it is a bad idea to use it to produce a "new " point, since the two equations in the affine $(X,Y)$-plane:
$$
\left\{
\begin{aligned}
X+Y &= \frac 1z(x+y)\ ,\\
X^3 + Y^3 &= -d
\end{aligned}
\right.
$$
give only the two intersection points $P,\bar P$, the third one
lives in the projective space and is the "infinity point" (with $Z=0$).
End of Digression.

So let us use the Cassel's points

$P_1=[x:\rho y:z]=[x/z:\rho y/z:1]\sim(x/z,\rho y/z)=(x_1,y_1)$, and

$P_2=[y:\rho^2 x:z]=[y/z:\rho^2 x/z:1]\sim(y/z,\rho^2 x/z)=(x_2,y_2)$.

The line connecting them has the equation
$$
\begin{aligned}
&\left(Y-y_1\right) =m\left(X-x_1\right)\ ,\qquad\text{ where }
\\
m
&
:=\frac{y_2-y_1}{x_2-x_1}
=\frac{\rho^2 x/z-\rho y/z}{y/z-x/z}
=\frac{\rho^2 x-\rho y}{y-x}
=\frac
{A(\rho^2-\rho)+C(\rho-\rho^2)}
{B(\rho^2-\rho)+C(\rho-\rho^2)}
\\
&=\frac{A-C}{B-C}\in\Bbb Q\ ,
\\
n &:= y_1-mx_1
=\frac 1z(\rho y-mx)
=3C\cdot \frac{A-B}{B-C}\in \Bbb Q\ .
\end{aligned}
$$
The corresponding system is
$$
\left\{
\begin{aligned}
Y &= mX + n\ ,\\
X^3 + Y^3 &= -d
\end{aligned}
\right.
$$
We insert $Y$ from the first equation into the second equation, obtain
$$
X^3 +  m^3X^3 +3m^2nX^2 +\dots =0\ .
$$
Two solutions are known, the $X$-component of the used points, which are $x_1=x/z$ and $x_2y/z$,
so the third solution $x_3$ is obtained using Vieta:
$$
x_3=-\frac xz-\frac yz -\frac{3m^2n}{1+m^3}\ .
$$
The corresponding $y_3$ is obtained by inserting $x_3$ in the equation of the used line,
$y_3=mx_3+n$.
Above, note that $-\frac xz-\frac yz=-\frac 1z(x+y)=\frac 1{3uvw}\cdot 3au^ 3 =\frac {au^2}{vw}\in\Bbb Q$.
So $x_3,y_3\in\Bbb Q$. This answers the question.

$\square$

But it may be important to really have a point $(x_3,y_3)$ explicitly.
It turns out, that after the denominators were made clean, the following point is a point on
$X^3+Y^3+dZ^3=0$:
$$
\begin{aligned}
x_3 &= (A-B)^3 -9AB^2\ ,\\
y_3 &= (B-A)^3 -9A^2B\ ,\\
z_3 &= z(A^2+AB+B^2)\ .\\[3mm]
&\qquad\text{Then:}\\
-dz_3^3 &= -dz^3\cdot (A^2+AB+B^2)^3
\\ &
= -(abc)(-3uvw)^3\cdot (A^2+AB+B^2)^3
\\
&= 27\;ABC\cdot (A^2+AB+B^2)^3\ ,\\[3mm]
x_3^3 + y_3^3 &= (x_3+y_3)\Big(\ x_3^2 - x_3y_3 +y_3^2 \ \Big)\ ,\\
(x_3+y_3) &=-9AB^2-9A^2B=-9AB(A+B)=9ABC\ ,\\
&\qquad\text{And indeed, after computations:}\\
x_3^2 - x_3y_3 +y_3^2 &= 3(A^2+AB+B^2)^3\ .
\end{aligned}
$$


问:
Lie algebra and rotation
I am trying to find the rotation about the x axis in three dimensions by the lie algebra:
$$R(\theta) = I + A$$
where $$A = \theta_{x}J_{x}+\theta_{y}J_{y}+\theta_{z}J_{z}$$
Since we want rotation about x axis
$$A = \theta_{x}J_{x}$$
Not just:
$$R(\theta) = R(\frac{\theta}{N})^{N} = e^{J\theta} = cos(\theta_{x}) + J_{x}sin(\theta_{x})$$
So that
$$\begin{pmatrix}
cos\theta_{x} & 0 &0 \\
0 &cos\theta_{x}  & sin\theta_{x}\\
0 & -sin\theta_{x} &cos\theta_{x}
\end{pmatrix}$$

But this is wrong... Why?

答: $e^{\theta J}=\cos\theta I+\sin\theta J$ is not true. You seem to be assuming $J$ is a square root of $-I$, but actually it squares to the diagonal matrix ${\rm diag}(0,-1,-1)$, which is why your upper left corner is wrong.

Title: Lie algebra and rotation


问:
Finding four square roots of $\small\begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$
This question is about finding the 4(!!) square roots of a 2 by 2 matrix. The method I use is through Diagonalization (or actually doing the "opposite"). I have no problem finding the four roots of a 2 by 2 matrix that has real eigenvalues (and thus real eigen vectors). Heck, I found 8 roots of a 3 by 3 matrix that has real eigen values. The problem I am running into is if the to be square rooted matrix has complex eigenvalues. In my view, the method should work the same, but somehow the algebra is letting me down. Example: Find the four square roots of
$ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$. Note, I found this matrix by squaring $ \begin{bmatrix}2 & -1\\1 & 2\end{bmatrix}$, so obviously this is one root as well as taking opposite of its entries. So here is my work: The characteristic equation of $ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$ is $(3-\lambda)^2+16=0$ from which we find $\lambda=3+4i$ and $\lambda=3-4i$. An eigenvector for $\lambda=3+4i$ is $ \begin{bmatrix}1 \\-i  \end{bmatrix}$ and an eigenvector for $\lambda=3-4i$ is $ \begin{bmatrix}1 \\i  \end{bmatrix}$. So the given matrix can now be diagonalized: $ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$=$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix}$$ \begin{bmatrix}3+4i & 0\\0 & 3-4i\end{bmatrix}$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix}^{-1}$. I found$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix}^{-1}$=$\frac{1}{2i} \begin{bmatrix}i & -1\\i & 1\end{bmatrix}$ which is $ \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$. It's easy to find out that the square roots of $3+4i$ are $2+i,-2-i$, and of $3-4i$ are $2-i,-2+i$. So now I would arrive at my 4 solutions:$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}2+i & 0\\0 & 2-i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}2+i & 0\\0 & -2+i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}-2-i & 0\\0 & 2-i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}-2-i & 0\\0 & -2+i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
When I work out my first candidate, I get a matrix, but when squared it doesn't give me $ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$ and so something isn't right. I don't know what I did wrong and ANY help from the math community is greatly appreciated. I feel that my line of thinking is correct, where did I go wrong? And does anyone happen to know online software that can do matrix multiplications with complex entries, because the TI doesn't work

答:
Your work seems correct.  Doing the matrix multiplications for your four candidates

in Wolfram Alpha gives these square roots:  $\pm\begin{bmatrix}2&-1\\1&2\end{bmatrix}$ and $\pm\begin{bmatrix}i&2i\\-2i&i\end{bmatrix}$.


问:
Do I need uniform convergence for this exercise? (for composition of operators)
I have the following exercise:  Suppose that $X$ is a Banach space and that $T_{n}$ & $S_{n}$ are bounded linear operators on $X$. Prove that if $T_{n}$ & $S_{n}$ are strongly convergent to $T$ & $S$ respectively then $T_{n} \circ S_{n}$ is strongly convergent to $T \circ S .$

But when I try to apply the definitions I am not sure how to follow. I know that if $S_n$ converges uniformly $S$, $||S_n(x)-S(y)||\leq \varepsilon $ if $||x-y||\leq \delta$ for any $\delta$, so I can put instead of $x$ and $y$, $T_n(x)$ and $T(x)$ respectivelly and that finishes the proof, but I dont have the uniform convergence, so I think that there should be another way to approach this, or do I need the uniform convergence?

答: Uniform Boundeness Principle tells you that $\|T_n\|$ and $\|S_n\|$ are bounded. Now use the inequality $$\|T_nS_nx-TSx\|\leq \|T_n(S_nx-Sx)\|+\|(T_n-T)\|Sx\|$$ $$\leq \|T_n\|(\|S_nx-Sx\|+\|(T_n-T)Sx\|$$.


问:
How can I prove this inequality? $\frac{p_k}{p_{k+1} + 1} < \frac{\ln(\ln(Primorial(k)))}{\ln(\ln(Primorial(k+1)))}$
How can I prove the following inequality? Or what are some tips that would help me prove it myself?

$$\frac{p_k}{p_{k+1} + 1} < \frac{\ln(\ln(Primorial(k)))}{\ln(\ln(Primorial(k+1)))}$$     for $k >= 3$, where $p_k$ is the $k$th prime, and $Primorial(k)$ is the product of the first $k$ primes

If we can't prove it for all $k >= 3$, can we at least prove it is assymptotically true?

答:
Not an answer, more like two long comments ..

Function $f(x)=\frac{x}{\ln{x}}$ is ascending for large $x$'s (easy to show, check the $1$st derivative). This means that for large enough $n$'s
$$\frac{p_n}{\ln{p_n}}\leq \frac{p_{n+1}+1}{\ln{(p_{n+1}+1)}}$$
Now:
$$\frac{p_{n}}{p_{n+1}+1}=
\frac{p_{n}}{\ln{p_{n}}}
\cdot\frac{\ln({p_{n+1}+1)}}{p_{n+1}+1}
\cdot\frac{\ln{p_{n}}}{\ln({p_{n+1}+1)}}\leq \frac{\ln{p_{n}}}{\ln{(p_{n+1}}+1)} \tag{1}$$

Chebyshev function
$$\vartheta (p_{n})=\sum\limits_{k=1}^n\ln{p_k}=\ln{(\operatorname{Primorial}(n))}$$
has the property (section 1.1) that
$$\lim\limits_{n\to\infty}\frac{\vartheta (p_{n})}{p_n}=1$$
which means that for large enough $n$'s
$$p_n(1-\varepsilon)<\vartheta (p_{n}) < p_n(1+\varepsilon) \Rightarrow\\
\ln{p_n}+\ln{(1-\varepsilon)}<\ln{(\ln{(\operatorname{Primorial}(n))})} < \ln{p_n}+\ln{(1+\varepsilon)}$$
and
$$\frac{\ln{p_n}+\ln{(1-\varepsilon)}}{\ln{p_{n+1}}+\ln{(1+\varepsilon)}}<
\frac{\ln{(\ln{(\operatorname{Primorial}(n))})}}{\ln{(\ln{(\operatorname{Primorial}(n+1))})}}<
\frac{\ln{p_n}+\ln{(1+\varepsilon)}}{\ln{p_{n+1}}+\ln{(1-\varepsilon)}}$$
or
$$\frac{\ln{(\ln{(\operatorname{Primorial}(n))})}}{\ln{(\ln{(\operatorname{Primorial}(n+1))})}} \sim \frac{\ln{p_n}}{\ln{p_{n+1}}} \tag{2}$$

Now compare $(1)$ and $(2)$ ...


问:
Is there a "nice" rep-tile of order $6$?
A planar set is said to be a rep-tile if it can be tiled by congruent shapes, each similar to the original. If there are $k$ such shapes, each scaled down by a factor of $\sqrt{k}$, it is said to be rep-$k$. Rep-tiles exist for $k=2,3,4,5$:

More examples (not all triangular) are at Wikipedia.

From Michael Beeson's paper here, we know that triangular rep-$k$ tiles exist if and only if $k$ is a square, a sum of two squares, or three times a square. (This yields $k=2,3,4,5,8,9,10,12,13,16,17,18,20,25,26,27,\ldots$)

The Gosper curve is a rep-tile of order $7$.

I have found what looks like an instance of a rep-$6$ tile here:

However, this is a pretty atrocious shape. Are there any polygonal rep-$6$ tiles? I'd settle for simply connected, in a pinch.

答:
If I understood the problem correctly, a possible (though boring) solution would be to take a rectangle of width $\sqrt{3}$ and height $\sqrt{2}$ and divide it like this:

This way we get six smaller rectangles of height $\sqrt{2}/2 = \sqrt{3}/\sqrt{6}$ and width $\sqrt{3}/3 = \sqrt{2}/\sqrt{6}$, which are obviously congruent to the original rectangle, thus forming a polygonal rep-$6$ tile.

More generally, for any pair of integers $m,n$ the rectangle with sides of length $\sqrt{m}, \sqrt{n}$ is a rep-$(m\cdot n)$ tile. In particular, taking $m=k, n=1$ we obtain rep-tiles for any positive $k$.


问:
Roots of polynomial/exponential function
Let $n\in \mathbb{N}$ and $\sigma \in (0,1]$ be fixed.
I am interested in finding the roots of the function
\begin{equation}
f(x) = \big(x^2-n^2-\sigma n^2 x\big)e^{x} - \big(x^2-n^2+\sigma n^2 x\big)e^{-3x};
\end{equation}
on $[0,n^2]$, namely solving $f(x) = 0$ for $x \in [0,n^2]$.

I would be happy with an approximation formula, which clearly indicates the dependence on $n \in \mathbb{N}$.

Clearly one root is $x=0$.
A way to proceed for finding the other root(s?) could of course be to factor the polynomials appearing in $f$; namely,
\begin{equation}
f(x) = \big(x-\big(n^2\sigma-n\sqrt{n^2\sigma^2+4}\big)\big)\big(x-\big(n^2\sigma+n\sqrt{n^2\sigma^2+4}\big)\big)e^x - \big(x-\big(-n^2\sigma-n\sqrt{n^2\sigma^2+4}\big)\big)\big(x-\big(-n^2\sigma+n\sqrt{n^2\sigma^2+4}\big)\big)e^{-3x}.
\end{equation}
Though, I don't see clearly how one may proceed from here.

答:
Find approximate solution of the equation
$$\left(x^2-n^2-\sigma n^2 x\right)e^{x} - \left(x^2-n^2+\sigma n^2 x\right)e^{-3x}=0\tag{1}$$
divide by $e^x$.
$$\left(x^2-n^2-\sigma n^2 x\right) - \left(x^2-n^2+\sigma n^2 x\right)e^{-4x}=0$$
$e^{-4x}$ is almost zero (for $x=4$ we have $e^{-4x}\approx 10^{-7}$)

The equation becomes
$$x^2-\sigma n^2x-n^2=0$$
the positive root is
$$x= \frac{n^2 \sigma +n \sqrt{n^2 \sigma ^2+4}}{2}\tag{2}$$

Examples

$n= 2,\sigma = \frac{1}{2}$

$(1)$ gives $x\approx 3.236074894631493$

$(2)$ gives $x\approx 3.23606797749979$

$n= 6,\sigma = \frac{5}{12}$

$(1)$ gives $x\approx 17.10468635614927$

$(2)$ gives $x\approx 17.10468635614927$


问:
compute $P(X\le 2 \textrm{ or } 3<X<8)$ For exponentially distributed x
Suppose X is exponentially distributed with $\lambda=2$.

I want to compute $P(X\le 2 \textrm{ or } 3<X<8)$.

The probability density function of $x$ is $f(x)=2e^{-2x}$.

Then I will find its distribution function as $F(x)=1-e^{-2x}$.

I don’t know exactly, but I think that $P(X\le 2 \textrm{ or } 3<X<8)$ can be calculated as follows:

$$P(X\le 2 \textrm{ or } 3<X<8) = P(X\le 2)+ P(3<X<8)-P(X\le 2, 3<X<8)$$

$$P(X\le 2)=F(2)= 1-e^{-4}$$

$$P(3<X<8)=F(8)-F(3)= e^{-16}-e^{-6}$$

But what is $P(X\le 2, 3<X<8)$?

Or what is the correct way to solve this question?

答: As $X$ is random variable i.e. in particular usual function, then $\{\omega\colon X(\omega)\leqslant 2\} \cap \{\omega\colon 3<X(\omega)< 8\}=\emptyset$, so $P(X\le 2, 3<X<8) = 0$.


问:
Part (a) of Exercise 5.14 in Chapter II of Hartshorne: Show that $\bigoplus_{n\geq 0}\Gamma(X,\mathcal{O}_X(n))$ is the integral closure of $S$.
I'm doing part (a) of Exercise 5.14 in Chapter II of Hartshorne: Let $k$ be an algebraically closed field, and $X\subseteq\mathbb{P}_k^r$ a connected, normal and closed subscheme. Let $S$ be the homogeneous coordinate ring, and $S'=\bigoplus_{n\geq 0}\Gamma(X,\mathcal{O}_X(n))$. Then we have to show that $S$ is a domain and $S'$ is its integral closure.

Following the hint of Hartshorne, I proved that $X$ is integral, and that $S$ is a domain. He then proposes to consider the sheaf of rings $\mathcal{S}=\bigoplus_{n\geq 0}\mathcal{O}_X(n)$, and show that it is a sheaf of integrally closed domains. And this is the point where I'm failing to see how to proceed.

In this post, Brian Ng is using the following argument: for $\mathfrak{p}\in\operatorname{Proj S}$, we have $$\mathcal{S}_\mathfrak{p}\cong\bigoplus_{n\geq0}S(n)_{(\mathfrak{p})}\cong\{\frac{s}{b}\ |\ b\notin\mathfrak{p}\text{ homogeneous, every non-zero homogeneous part of $s$ has degree $\geq\deg b$}\}\subseteq S_{\mathfrak{p},\text{ hom.}}$$
where $S_{\mathfrak{p},\text{ hom.}}$ is the localization of $S$ at the homogeneous elements not in $\mathfrak{p}$. With this I agree. However, he then sais that this is integrally closed as $S_{\mathfrak{p},\text{ hom.}}$ is (because $X$ is normal), and that therefore that $S'=\Gamma(X,\mathcal{S})$ is integrally closed to, and I don't see how exactly we may conclude this. First of all, $X$ being normal only implies that $X_\mathfrak{p}\cong S_{(\mathfrak{p})}=(S_{\mathfrak{p},\text{ hom.}})_0$ is integrally closed. Furthermore, how does one conclude from the stalks being integrally closed that every ring of sections of the sheaf is integrally closed?

答:
Suppose $x\in S_1$ is a linear form on $\Bbb P^n$ not in $\mathfrak{p}$. I claim $\mathcal{S}_\mathfrak{p}\cong S_{(\mathfrak{p})}[x]$: as $\mathfrak{p}\in D(x)$, we can write $(\mathcal{O}_X(n))_\mathfrak{p} = (\mathcal{O}_X(n)|_{D(x)})_\mathfrak{p}$ since restriction commutes with stalks. By the identification of $D(x)$ with $\Bbb A^n$, we have  $\mathcal{O}_X(n)|_{D(x)}$ is the sheaf associated to the $S_{(x)}$-module $x^nS_{(x)}$, and the prime ideal $\mathfrak{p}$ is associated to the prime ideal $\mathfrak{p}'=\mathfrak{p}_{(x)}$. So $$(\mathcal{O}_X(n)|_{D(x)})_\mathfrak{p} = (\widetilde{x^nS_{(x)}})_{\mathfrak{p}'}=x^nS_{(\mathfrak{p})}$$ and the direct sum $\mathcal{S}_\mathfrak{p}$ is $\bigoplus_{n\geq0} x^nS_{\mathfrak{p}}=S_\mathfrak{p}[x]$. As $S_{\mathfrak{p}}$ is integrally closed, $\mathcal{S}_\mathfrak{p}$ is too.

As for why integrally closed in all stalks means integrally closed for every open, define a sheaf $\mathcal{T}$ by the rule $\mathcal{T}(U)$ is the integral closure of $\mathcal{S}(U)$, which works because integral closure commutes with localization. Then $\mathcal{S}$ is naturally a subsheaf of $\mathcal{T}$ with the same stalks, which implies they must be equal.


答:
You're almost there.  Note that the events $\{X \le 2\}$ and $\{3 < X < 8 \}$ are mutually exclusive.

$$P(X \le 2 \text{ or } 3 < X < 8) \\
= P(\{X \le 2\} \cup \{3 < X < 8\})
\\ = P(X \le 2) + P(3 < X < 8) \\= (1 - e^{-4}) + (e^{-6} - e^{-16})
\\ = 1 - e^{-4} + e^{-6} - e^{-16}$$

Note that you've inverted the sign when calculating $P(3 < X < 8)$.

Title: compute $P(X\le 2 \textrm{ or } 3<X<8)$ For exponentially distributed x

Body:
Suppose X is exponentially distributed with $\lambda=2$.

I want to compute $P(X\le 2 \textrm{ or } 3<X<8)$.

The probability density function of $x$ is $f(x)=2e^{-2x}$.

Then I will find its distribution function as $F(x)=1-e^{-2x}$.

I don’t know exactly, but I think that $P(X\le 2 \textrm{ or } 3<X<8)$ can be calculated as follows:

$$P(X\le 2 \textrm{ or } 3<X<8) = P(X\le 2)+ P(3<X<8)-P(X\le 2, 3<X<8)$$

$$P(X\le 2)=F(2)= 1-e^{-4}$$

$$P(3<X<8)=F(8)-F(3)= e^{-16}-e^{-6}$$

But what is $P(X\le 2, 3<X<8)$?

Or what is the correct way to solve this question?


问:
Can you find the resultant force between these two vectors?
Determine the magnitude of the resultant force on an object if force $A$ is pulling the object with $150$ lbs of force and force $B$ is pulling with $300$ lbs, and the angle between the two forces is $110^\circ$.</p>

答:
use this formula of triangle  $a^2=b^2+c^2-2bc\cdot \cos A $ where a,b,c are sides of triangle.

so resultant force $$\vec R^2={150}^2+{300}^2-2\cdot150\cdot 300 \cos 70^\circ$$
$$\vec R=\sqrt {22500+90000-30781.81}$$
$$\vec R=285.86\,lbs$$

There was typo in the formula which I have now corrected.
a^2=b^2+c^2 - 2bc * cos A


答:
$X<2$ is an event.  So too is $3 <X <8$.  Since the intervals do not overlap, they are mutually exclusive.

So $P(X <2 \text{ or }3 <X<8) = P(X<2)+P(3<X<8)$ which is $P(X<2) +P(X<8)-P(X\le 3)$.  Since the exponential distribution is continuous, this is also equal to   $P(X\le 2) +P(X\le 8)-P(X\le 3)$ and you can calculate each of these terms from the cumulative distribution function and gives $1-e^{-
4}+e^{-6}-e^{-16}$

Meanwhile $P(X <2,3 <X<8)$ is read as  $P(X <2 \text{ and  }3 <X<8)$ and since the events are mutually exclusive, this is $0$


问:
Application of Arzela-Ascoli Thrm where $(f'_n)$ is uniformly bounded
We have:

Let $(f_n)$ be a sequence of differentiable functions on the interval $[0, 2]$

Assume that $(f_n)$ converges pointwise to a function $f$ on $[0, 2]$

Assume there exists positive real $M$ st $\forall n,$ $ \sup_{x\in[0,2]}\{|f'_n(x)|\}=M$

Prove that there is a strictly increasing sequence $(n_k)$ such that $(f_{n_k})$ converges uniformly on $[0,2]$.

I know that the Arzela-Ascoli theorem has the conclusion I want but to use it I need

$(f_n)$ to be uniformly bounded

$(f_n)$ to be uniformly equicontinuous

How would you fulfill those two conditions?

So following suggestions I have:

For 1:

Here we WTS $\forall n \in \mathbb{N}, x \in [0,2] |f_n(x)| \leq Y$ for some Y.

Pick arbitrary $n \in \mathbb{N}$ and $x_0 \in [0,2]$ then since $f_n \to f$ pointwise, $(f_n(x_0))_{n=1}^{\infty}$ is bounded thus $\forall n' \in \mathbb{N} |f_{n'}(x_0)| \leq Y.$ Then since $n\in \mathbb{N}$ we see that $f_n(x_0) \leq Y$ as required.

Then for 2:

let $\epsilon >0,\delta = \epsilon/M$ then $\forall x,y \in [0,2]$ and $n \in \mathbb{N}$ where $|x-y| < \delta,$ apply MVT to see $|f_n(y) -f_n(x)| = |f'_n(c)(y-x)| \leq M|(y-x)| \leq \epsilon$ as required.

Also as a followup do the condition applied here mean limit $f$ of $f_n$ is continuous?

答:
You can use the bound on the derivative to get a uniform Lipschitz constant, which is more than enough for uniform equicontinuity; this follows from the mean value theorem or (if you prefer overkill) the fundamental theorem of calculus.

For the uniform bound, recall that a sequence of numbers that converges is bounded, so use that to get a bound on, say, $f_n(0)$. Then apply the uniform Lipschitz constant to propagate that bound across $[0,2]$.


问:
calculating the expected value of a random variable on two different domains
i have some information about the CDF of a random variable X.

when $0 \le x < 3$:

$F(x) = \frac{1}{9}(2 x^2 - \frac{x^3}{3})$

when $x \ge 3$:

$F(x) = 1$

i need to calculate the expected value of this variable.

it seems like i need to calculate two integrals: one for the first domain using $x \cdot f(x)$ and one for the second domain (3 to infinity) using the same definition of $f(x)$, but not multiplying by x.

but when i calculate the value of this second integral, it's divergent.

what am i doing wrong? any suggestions? thank you in advance.

答: You are confusing the CDF for the denisty. The density is $F'$ which is $0$ for $x \geq 3$. So the expected value is just $\int_0^{3} \frac d {dx} \frac 1 9 (2x^{2}-\frac {x^{3}} 3)dx=1$.


问:
problems with $\int _0 ^\infty \frac{dx}{\sqrt[3]{x}(x+1)^2}$
I'm trying to find $\int _0 ^\infty \frac{dx}{\sqrt[3]{x}(x+1)^2}$ using complex analysis.

My first problem is that I have to deal with the derivative of $\sqrt[3]{x}$ while computing the residue at $-1$. Explain, please, how one can compute the derivative at a point of such functions (with several branches).

And the second problem is that I don't know how to show that the  contour integral over an upper semi circle centred at $0$ has a zero limit.

答:
Let $\operatorname{Log}(\cdot)$ denote the branch of the complex logarithm such that $\operatorname{Im}(\operatorname{Log}(z)) \in [0,2\pi)$. In particular, the branch cut of $\operatorname{Log}(\cdot)$ is $[0, \infty)$. Then we consider the function

$$ f(z) = \frac{e^{-\frac{1}{3}\operatorname{Log}(z)}}{(z+1)^2}. $$

Note that, if $x > 0$, then

$$ f(x + 0^+i) := \lim_{\epsilon \to 0^+} f(x + i\epsilon) = \frac{1}{\sqrt[3]{x}(x+1)^2} $$

and

$$ f(x - 0^+i) := \lim_{\epsilon \to 0^+} f(x - i\epsilon) = \frac{e^{-2\pi i/3}}{\sqrt[3]{x}(x+1)^2}. $$

Now, we consider the keyhole contour, where the inner radius is $\epsilon < 1$ and the outer radius is $R > 1$.

Integrating $f$ along this contour, applying the residue theorem, and letting the width of the "gap" to $0^+$, we have

$$ \int_{\epsilon}^{R} (f(x + 0^+i) - f(x - 0^+i)) \, \mathrm{d}x - \int_{\lvert z \rvert = \epsilon} f(z) \, \mathrm{d}z + \int_{\lvert z \rvert = R} f(z) \, \mathrm{d}z
= 2\pi i \, \mathop{\mathrm{Res}}_{z=-1} \, f(z). $$

Now we make several observations:

If $\lvert z \rvert = r \neq 1$, then by the reverse triangle inequality,

$$ \left| f(z) \right| = \frac{e^{-\frac{1}{3}\operatorname{Re}(\operatorname{Log}(z))}}{\left|z + 1\right|^2} \leq \frac{e^{-\frac{1}{3}\log r}}{\left| r - 1 \right|^2} = \frac{1}{\sqrt[3]{r}\left| r - 1 \right|^2}. $$

So it follows that

$$ \left| \int_{\lvert z \rvert = r} f(z) \, \mathrm{d}z \right| \leq \frac{2\pi r}{\sqrt[3]{r}\left| r - 1 \right|^2} = \frac{2\pi r^{2/3}}{\left| r - 1 \right|^2}. $$

Note that this converges to $0$ either as $r \to 0^+$ or as $r \to \infty$.

On $\mathbb{C}\setminus[0,\infty)$ and for any $\alpha \in \mathbb{C}$, we have

$$ \frac{\mathrm{d}}{\mathrm{d}z} e^{\alpha \operatorname{Log}(z)}
= \frac{\alpha}{z} e^{\alpha\operatorname{Log}(z)}
= \alpha e^{(\alpha-1)\operatorname{Log}(z)}, $$

which essentially replicates the well-known formula $(z^{\alpha})' = \alpha z^{\alpha-1}$. In particular, the Taylor series for $g(z) = \exp\bigl(-\frac{1}{3}\operatorname{Log}(z)\bigr)$ about $z=-1$ begins with

$$ g(z) = 1 - \tfrac{1}{3} e^{-\frac{4}{3}\pi i}(z + 1) + \cdots, $$

and so, we obtain

$$ \mathop{\mathrm{Res}}_{z=-1} \, f(z) = - \frac{1}{3} e^{-\frac{4}{3}\pi i} = \frac{1}{3} e^{-\frac{1}{3}\pi i}. $$

Combining altogether, as $R\to\infty$ and $\epsilon \to 0^+$, we get

$$ \bigl( 1 - e^{-\frac{2}{3}\pi i} \bigr) \int_{0}^{\infty} \frac{\mathrm{d}x}{\sqrt[3]{x}(x+1)^2}
= \frac{2\pi i}{3} e^{-\frac{1}{3}\pi i}. $$

Rearranging, we get

$$ \int_{0}^{\infty} \frac{\mathrm{d}x}{\sqrt[3]{x}(x+1)^2} = \frac{2\pi i}{3\left( e^{\frac{1}{3}\pi i} - e^{-\frac{1}{3}\pi i} \right)} = \frac{\pi}{3\sin(\pi/3)} = \frac{2\pi}{3\sqrt{3}}. $$