问:
Boundedness in probability of Stochastic Process
This question is related to some other similar ones I did in the recent past.

Let $X_t$ be a Stochastic Process defined through the equation $$\text{d}X_t=f(X_t,t)\text{d}t+\text{d}W_t,$$
where $f$ is a twice differentiable function such that $f(x)<a<0$ for all $x>k$, and $X_0<k$.

What I am wishing to show is:

For all $\epsilon>0$ there is $M>0$ (which we can WLOG assume to be bigger than $k$) such that for all $t\ge0$ it holds that $$\mathbb{P}[X_t>M]\le\epsilon.$$

What I tried to do is define $\tau_t$ as $\sup\{0<s<t: X_s=k\}$, i.e. the last hitting time of the level $k$ before time $t$, where the sup is set equal to $0$ if the set $\{0<s<t: X_s=k\}$ is empty. This should be a stopping time, even if last hitting times in general are not.

Then, we can manipulate the given probability in this way:
\begin{align*}\mathbb{P}[X_t>M]&=\underbrace{\mathbb{P}[X_t>M,\tau_t=0]}_{=0}+\underbrace{\mathbb{P}[X_t>M,\tau_t\ne0, X_t<k]}_{=0}+\mathbb{P}[X_t>M,\tau_t\ne0, X_t>k]\\
&\le\mathbb{P}[X_t-X_{\tau_t}>M-k,\tau_t\ne0]\\
&\le \mathbb{P}[a(t-\tau_t)+W_t-W_{\tau_t}>M-k, \tau_t\ne0]\\
&\le \mathbb{P}[a(t-\tau_t)+W_t-W_{\tau_t}>M-k].\end{align*}

Since $a<0$ we can bound this probability uniformly over $t$, and the claim should be proven. However, I think I've been too slick with some steps, and I want to make sure they are correct (in particular, the $\tau_t$ seems to be not a stopping time). Can someone help me find any potential errors? Also, is there a more straightforward way to prove this? Maybe I just didn't see an easier solution.

EDIT: the $f$ in my specific problem satisfies the hypothesis $|f(x,t)-k_0|<\theta\cdot |x|+ \mu,$ for some constants $k_0, \theta$ and $\mu$: is there some kind of comparison principle with the absolute value bound? I can't find it online, but if it exists, then I would be done, as the Ornstein-Uhlenbeck process is bounded in the sense I am looking for.

答:
I'll just give some partial answers to the original question, but my work is rather long so I highly doubt I can type it all in just one sit.

Organisation of my post

I'll start by showing some trivial points about $\mathbb{E}(X_t^2)$, to show that "at least we have something".

Then, I'll incorporate two sequences of stopping times to divide the time horizon in a nice way.
From which, I present a tighter bound for $\mathbb{E}(X_t^2)$.

In the end, even though I have a strong feeling that I can well control $\sup \mathbb{E}(X_t^2)$ even without the time-homogeneity of $f$, I'll assume that condition nonetheless to give a sure partial answer.

Remark : To be honest, it's not really the $\mathbb{E}(X_t^2)$ that we will see but rather $\mathbb{E}((X_t)_+^2)$

A weak bound for $\mathbb{E}((X_t)_+^2)$

Theorem 1
If $f(t,x)<0$ for all $x \ge m$, we have:

$$\mathbb{E}((X_t)_+^2) \le 2m^2+2t$$
$\square$

Demonstration 2

By Ito's formula, we have:
$$ (X_t-m)_+^2=2(X_t-m)_+d X_t+ 1_{(X_t -m)\ge 0} dt$$
Thus( by some standard local martingale+Fatou arguments which I omit)
$$ \mathbb{E}((X_t-m)_+^2) \le \underbrace{2\int_{0}^t \mathbb{E}((X_s-m)_+f(s,X_s))ds}_{ \le 0}+ \mathbb{E} ( \int_{0}^t 1_{(X_s -m)\ge 0}ds )$$

Hence $$ \mathbb{E}((X_t-m)_+^2) \le t$$
Thus the conclusion $\square$

Corollary 2
If $f(t,x)<0$ for all $x\ge m$, we have:
$$\mathbb{P}(X_t > M) \le \frac{2m^2+2t}{M^2}$$
for all $M>0$
$\square$

A better bound for $\mathbb{E}((X_t)_+^2)$

1.Some set up

So the above theorem provides a pretty nice bound for $\mathbb{P}(X_t > M)$ when we have some control for the negativity of $f$

Naturally, we should expect a better bound for $\mathbb{P}(X_t > M)$ when we have more control.

I first repeat the initial assumption on $f$,

Condition C1: $f(t,x) < a <0 $ for all $x>k$ $\square$.

Let's define the following two sequences of stopping time with two given real numbers $k<m_1<m_2$ (from now on, we only consider $M>m_2$ )

$\sigma_1:= \inf\{ t \ge 0 : X_t \ge m_1\}$ (because $X_0<k$)

$\tau_n := \inf\{ t \ge \sigma_n : X_t \ge m_2\}$

$\sigma_{n+1} =  \inf\{ t \ge \tau_n : X_t \le m_1\}$

Remark 3: $(\sigma)$ and $(\tau)$ is nothing mysterious, they are just stopping times which delimit the upcrossings and downcrossings of $X$. In particular, $X_{[\tau_n,\sigma_{n+1}]}$ is a down crossing, while $X_{[\sigma_n,\tau_n]}$ is a upcrossing.

Assumption A1 $\mathbb{P}( \tau_n<+\infty) =1 \forall n$

Remark 4: In fact, this assumption is a bit superfluous because the infiniteness of $\tau_n$ will lead to an upper bound for $X_t$. However,  more elaborate work is indeed needed to get rid of that case, I decided to just assume that equality.

Remark 5 If I remember correctly, I have proven that equality under some assumptions in the previous post.

For any t, we define:
$$N(t):= \inf \{ k : \sigma_k >t\}$$

2. Main result

I now present the main theorem for this section

Theorem 6

Under the condition C1 and the assumption A1, we have:
$$ \mathbb{E}( (X_t-m_1)_+^2 ) \le  \mathbb{E}( t- \sigma_{N(t)-1})$$
$\square$

Remark 7: I lost my note on this part, so it is not really the "true" theorem, however, the essential is presented.

3. Demonstration

Again, by Fatou and local martingale arguments, we have:
$$ \mathbb{E}\left( X_{ \sigma_{n+1} \wedge ( t \vee \sigma_n)}- X_{\sigma_n} \right)_+^2 \le  \underbrace{ \mathbb{E}(2\int_{\sigma_n}^{\sigma_{n+1} \wedge ( t \vee \sigma_n)} (X_s-m_1)_+f(s,X_s))ds}_{ \le 0}+ \mathbb{E} ( \int_{\sigma_n}^{\sigma_{n+1} \wedge ( t \vee \sigma_n)} 1_{(X_s -m_1)\ge 0}ds )$$
(Note that: $X_{\sigma_n}= m_1$ )

Thus,
$$ \mathbb{E}( \underbrace{ X_{ \sigma_{n+1} \wedge ( t \vee \sigma_n)}- X_{\sigma_n}}_{ =X_t-X_{\sigma_n} \text{ if } t \in [\sigma_{n},\sigma_{n+1}] \text{ and } 0 \text{otherwise} } )_+^2  \le \mathbb{E}( \underbrace{ \sigma_{n+1} \wedge ( t \vee \sigma_n)-\sigma_n}_{ = t-\sigma_n \text{ if } t \in [ \sigma_n,\sigma_{n+1}] \text{ and } 0 \text{ otherwise}})  $$
By varying $n$ and adding up all the produced inequalities, we impy that:
$$ \mathbb{E}( (X_t-m_1)_+^2 ) \le \mathbb{E}( t- \sigma_{N(t)-1}) $$

4.Discussion

+This bound is indeed close enough.
+The condition C1 is mostly needed to assure in some sense that the assumption A1 is reasonable.

(the continuation in the next post)


问:
Integral $ \int_{-\infty}^{\infty}\frac{x^{2}}{\cosh\left(x\right)}\,{\rm d}x $

I need to compute the improper integral
$$
\int_{-\infty}^{\infty}\frac{x^{2}}{\cosh\left(x\right)}\,{\rm d}x
$$
using contour integration and possibly principal values.  Trying to approach this as I would normally approach evaluating an improper integral using contour integration doesn't work here, and doesn't really give me any clues as to how I should do it.

This normal approach is namely evaluating the contour integral
$$
\oint_{C}{\frac{z^2}{\cosh\left(z\right)}\mathrm{d}z}
$$
using a semicircle in the upper-half plane centered at the origin, but the semicircular part of this contour integral does not vanish since $\cosh\left(z\right)$ has period $2\pi\mathrm{i}$ and there are infinitely-many poles of the integrand along the imaginary axis given by $-\pi\mathrm{i}/2 + 2n\pi\mathrm{i}$ and
$\pi\mathrm{i}/2 + 2n\pi\mathrm{i}$ for
$n \in \mathbb{Z}$.

The residues of the integrand at these simple poles are $-\frac{1}{4}\mathrm{i}\pi^{2}\left(1 - 4n\right)^{2}$ and $\frac{1}{4}\mathrm{i}\left(4\pi n + \pi\right)^{2}$, so that even when we add up all of the poles, we have the sum $4\pi^{2}\mathrm{i}\sum_{n = 0}^{\infty}\,n$, which clearly diverges.

Any hints would be greatly appreciated.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{x^{2} \over \cosh\pars{x}}\,\dd x} =
4\int_{0}^{\infty}{x^{2}\expo{-x} \over 1 + \expo{-2x}}\,\dd x
\\[5mm] = &\
4\sum_{n = 0}^{\infty}\pars{-1}^{n}
\int_{0}^{\infty}x^{2}\expo{-\pars{2n + 1}x}\,\,\,\dd x
\\[5mm] = &\
4\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}^{3}}
\int_{0}^{\infty}x^{2}\expo{-x}\,\,\,\dd x
\\[5mm] = &\
8\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}^{3}} =
-8\ic\sum_{n = 0}^{\infty}{\ic^{2n + 1} \over \pars{2n + 1}^{3}}
\\[5mm] = &\
-8\ic\sum_{n = 1}^{\infty}{\ic^{n} \over n^{3}}\,{1 - \pars{-1}^{n} \over 2}
\\[5mm] = &\
8\,\Im\sum_{n = 1}^{\infty}{\ic^{n} \over n^{3}} =
-4\ic\,\bracks{\on{Li}_{3}\pars{\ic} - \on{Li}_{3}\pars{-\ic}}
\\[5mm] = &\
-4\ic\,\braces{\on{Li}_{3}\pars{\expo{2\pi\ic\bracks{\color{red}{1/4}}}} - \on{Li}_{3}\pars{\expo{-2\pi\ic\bracks{\color{red}{1/4}}}}}
\\[5mm] = &\
-4\ic\bracks{-\,{\pars{2\pi\ic}^{3} \over 3!}
\on{B}_{3}\pars{\color{red}{1 \over 4}}}
\end{align}

The last expression is
Jonqui$\grave{\mrm{e}}$re's Inversion Formula. $\ds{\on{B}_{n}}$ is a
Bernoulli Polynomial. In particular, $\ds{\on{B}_{3}\pars{x} =
x^{3} - {3 \over 2}\,x^{2} + {1 \over 2}\,x}$.

Finally,
$$
\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{x^{2} \over \cosh\pars{x}}\,\dd x} = \bbx{\pi^{3} \over 4} \\
$$

ContentLicense: CC BY-SA 4.0


问:
Irreducible and reducible polynomial
We know that f(x) is irreducible in K[x] if and only if f(x+a) is irreducible in K[x]
... Wherer K is a field and f(x) is a polynomial over K

My question is why does K has to be a field, i know all the non zero elements in K are invertible, but why can't this rule be true in any ring??
Please help

答: You don't actually need $K$ to be a field. This works because the map $K[x] \longrightarrow K[x]$ via $x \mapsto x+a$ is an isomorphism of rings, and $K$ need not be a field for this to be an isomorphism.

Title: Irreducible and reducible polynomial


问:
If $f$ is not continuous on $[1,\infty)$ but $\int_1^{\infty} f(x) \, dx$ converges, does $\int_1^{\infty} \frac{|f(x)|}{x^3} \, dx$ converge?
It is true I think that if $f$ is continuous on $[1,\infty)$ and $\int_1^{\infty} f(x) \, dx$ converges, then $\int_1^{\infty} \frac{|f(x)|}{x^3} \, dx$ converges.

But what if $f$ is not continuous but $\int_1^{\infty} f(x) \, dx$ converges.  Must it necessarily be true that $\int_1^{\infty} \frac{|f(x)|}{x^3} \, dx$ converges?

答:
Both statements are false. Consider $f(x) = e^{x/2} \sin(e^x)$. Then

$$ \int_{1}^{\infty} f(x) \, \mathrm{d}x
\stackrel{(u=e^x)}{=} \int_{e}^{\infty} \frac{\sin u}{\sqrt{u}} \, \mathrm{d}u
= \frac{\cos e}{\sqrt{e}} - \frac{1}{2} \int_{e}^{\infty} \frac{\cos u}{u^{3/2}} \, \mathrm{d}u $$

converges. On the other hand,

$$ \int_{1}^{\infty} \frac{\left| f(x) \right|}{x^3} \, \mathrm{d}x
\stackrel{(u=e^x)}{=} \int_{e}^{\infty} \frac{\left| \sin u \right|}{\sqrt{u} \log^3 u} \, \mathrm{d}u $$

and it is not hard to show that this integral diverges.

Almost the same proof goes if we set $f(x) = e^{x/2} \phi(e^x)$, where $\phi$ is a non-zero $T$-periodic function such that $\int_{0}^{T} \phi(x) \, \mathrm{d}x = 0$. (The above proof used $\phi(x)=\sin x$, but there is a plethora of other examples, such as $\phi(x) = x - \lfloor x\rfloor - \frac{1}{2}$.)


问:
Number of combination with repetitions containing a subset of all the root element
I am in trouble with this question: Given the set $S$={$a,b,c,d,e,f$} and 10 position to arrange these elements. How many combination with repetitions exist if the only allowed must contain at least one of each elements of $S$?

For example a valid configuration can be {$a,a,b,b,c,d,e,f,f,f$} and
an incorrect one can be {$a,a,a,a,a,b,b,b,b,b$} .

I try to find a solution that can be applicable to a Set of size $m$, with $n$ position to arrange the elements and with $n$>$m$.
My starting assumption is that if I consider the Diophantine equation each unknown of solving equation (fulfilling the request) must have at least $1$ like result. However this does not help me to calculate the number of equations that fulfill the initial question. There is an easy or a tricky way to solve this problem?

答:
You must have heard of a counting technique called stars and bars which gives that the number of ways of placing $n$ identical objects into $k$ distinct bins = $\binom{n+k-1}{k-1}$

Possibly what is bothering you is that here the "objects" are distinct numerals. Then just imagine that the bins are numbered $1$ through $6$ and unnumbered identical objects magically acquire the numeral of the bin they are put in, then it fits the stars and bars format.

Firstly put $1$ object in each bin, $4$ are left to be put, and by stars and bars, # of ways = $\binom{4+6-1}{6-1}$


问:
Checking for isomorphism.
What is given is  the group $R$.

G(R)

Prove that $R$ is isomorphic to $G$ by first giving an formula for a map $ϕ:R→G$, and checking that your map.

答:
Let matrices of $G$ be denoted as $M(\theta)$, in which we just plug in $\theta$ for each entry in the rotation matrix. Just define $\phi : R \to G$ as $\phi(x):=M(x)$. First let's check this is a group homomorphism:

$$\phi(x+y)=M(x+y)=M(x)\cdot M(y)=\phi(x)\cdot\phi(y)$$

which follows from the fact that $M(\theta)$ is a rotation matrix. Note that the identity in $G$ is given by
$$\begin{bmatrix}1 & 0\\ 0& 1\end{bmatrix}=\begin{bmatrix}\cos(0) & -\sin(0)\\ \sin(0)& \cos(0)\end{bmatrix}=\begin{bmatrix}\cos(2\pi n) & -\sin(2 \pi n)\\ \sin(2 \pi n)& \cos(2 \pi n)\end{bmatrix}$$
for any $n \in \mathbb{Z}$. Hence, the kernel of $\phi$ is given as
$$Ker(\phi):=\{2\pi n : n \in \mathbb{Z}\}$$
We want to use the first isomorphism, but first we have to show $\phi$ is a surjection. $\phi$ is surjective if for all $M(\theta) \in G$, there exists some $x \in R$ such that $M(\theta)=\phi(x)$. This is trivial; since $\theta$ is a real number just choose $x=\theta$, then for all $M(\theta)$ we have that there exists an $x$ so that $\phi(x)=\phi(\theta)=M(\theta)$. Since $\phi$ is a surjection we can now use the first isomorphism theorem to assert

$$R/Ker(\phi)=R/H\approxeq G$$

Title: Checking for isomorphism.


问:
$\lim_{R \to \infty} \int_0^R \frac{dx}{(x^2+x+2)^3}$

$$\lim_{R \to \infty} \int_0^R \frac{dx}{(x^2+x+2)^3}$$

Please help me in this integral, I've tried some substitutions, but nothing work.
Thanks in advance!

答:
First complete the square:
$$
x^2 + x + 2 = \left( x^2 + x + \tfrac{1}{4} \right) + \tfrac{7}{4} = \left( x + \tfrac{1}{2} \right)^2 + \left( \tfrac{\sqrt{7}}{2} \right)^2.
$$
Now, make the (inverse) trigonometric substitution:
$$
\tan t = \frac{x + \tfrac{1}{2}}{\tfrac{\sqrt{7}}{2}}.
$$
This choice of ratio is motivated by the sum of square expression above.  As a consequence, we have
$$
x^2 + x + 2 = \left( \tfrac{\sqrt{7}}{2} \tan t \right)^2 + \left( \tfrac{\sqrt{7}}{2} \right)^2 = \tfrac{7}{4} \left( \tan^2 t + 1 \right) = \tfrac{7}{4} \sec^2 t.
$$
Now, the differential is
$$
dx = \tfrac{\sqrt{7}}{2} \sec^2 t \, dt
$$
and the limits of integration become
$$
\begin{align}
x &= 0 &\Longleftrightarrow \quad t(0) &= \arctan \tfrac{1}{\sqrt{7}} \\
x &= R &\Longleftrightarrow \quad t(R) &= \arctan \tfrac{2R + 1}{\sqrt{7}}.
\end{align}
$$
Now, substitute:
$$
\int_0^R \frac{dx}{(x^2 + x + 2)^3} = \int_{t(0)}^{t(R)} \frac{\tfrac{\sqrt{7}}{2} \sec^2 t \, dt}{\left( \tfrac{7}{4} \sec^2 t \right)^3} = \left( \tfrac{2}{\sqrt{7}} \right)^5 \int_{t(0)}^{t(R)} \cos^4 t \, dt.
$$
Can you finish it from here, using power reducing identities?</p>

ContentLicense: CC BY-SA 3.0


问:
How many ways are there to wear $5$ different rings in $4$ different fingers and also the order in which the ring is placed in a finger matters?
so, we know how to solve if the question was only $5$ different rings in $4$ different fingers, which is $4^5$. but what if internal order of rings within the finger matters, is this counted in this answer or we have a different answer?

答:
There are $5!=120$ ways to arrange the rings in order.  Divide them into $4$ groups and place the first group, in order, on the first finger, the second group on the second finger, and so on.

We have five rings to place on $4$ fingers.
By stars and bars, there are $\binom83=56$ ways to do this, so in all we have $$120\cdot 56 = 6720$$ ways.


问:
Is there a matrix that can change the position of an entry?
Suppose that I have a matrix A of size (n x n), and the entry in the i-th row and j-th column is 'a'.  Is there a matrix (also of size (n x n)) that I can multiply A by in order to in a sense "change" the position of a, so that the resulting matrix has in its k-th rown and l-th column the element 'a'?

In other words, I essentially want to be able to move a particular entry in the matrix to any other given position in the matrix.

Edit:  I am trying to use linear transformations, and no I don't care about what happens to the other entries of the matrix.

答:
False, and this can be seen in the 2x2 case. Left multiplication gives
$$\begin{pmatrix}
0 & 0 \\
1 & 0
\end{pmatrix}\begin{pmatrix}
0 & a \\
0 & 0
\end{pmatrix} = \begin{pmatrix}
0 & 0 \\
0 & a
\end{pmatrix}$$
and right multiplication gives
$$\begin{pmatrix}
0 & a \\
0 & 0
\end{pmatrix}\begin{pmatrix}
0 & 0 \\
1 & 0
\end{pmatrix} = \begin{pmatrix}
a & 0 \\
0 & 0
\end{pmatrix}
$$
but to achieve
$$\begin{pmatrix}
0 & a \\
0 & 0
\end{pmatrix}
\mapsto \begin{pmatrix}
0 & 0 \\
a & 0
\end{pmatrix}
$$
would require conjugation (check for yourself).


答:
For me, this is simply a take off my shoes and count problem.

Type of distribution :

Explanation

Number of ways

$5,0,0,0 :$

$4$ ways of choosing which finger, $5!$ ways of permuting the rings.

$T_1 = 4 \times 5!.$

$4,1,0,0 :$

$4$ ways of choosing the heavy finger, $3$ ways of choosing the 2nd finger, $5$ ways of choosing which ring goes on the 2nd finger, $4!$ ways of permuting the rings on the heavy finger.

$T_2 = 4 \times 3 \times 5 \times 4!.$

$3,2,0,0 :$

$4$ ways of choosing the heavy finger, 3 ways of choosing the 2nd finger, $\binom{5}{3}$ ways of selecting which rings go on the heavy finger, $3!$ ways of permuting the rings on the heavy finger, $2!$ ways of permuting the rings on the 2nd finger.

$T_3 = 4 \times 3 \times \binom{5}{3} \times 3! \times 2!.$

$3,1,1,0 :$

$4$ ways of choosing the heavy finger, $3$ ways of choosing the no-ring-finger, $\binom{5}{3}$ ways of selecting the rings on the heavy finger, $2$ ways of selecting the ring on the 1st light finger, $3!$ ways of permuting the rings on the heavy finger.

$T_4 = 4 \times 3 \times \binom{5}{3} \times 2 \times 3!.$

$2,2,1,0 :$

$4$ ways of choosing the no-ring-finger, $3$ ways of choosing the 1-ring finger, $5$ ways of selecting the ring for the 1-ring finger, $\binom{4}{2}$ ways of distributing the rings on the 2-ring fingers, $(2! \times 2!)$ ways of permuting the rings on the 2-ring fingers.

$T_5 = 4 \times 3 \times 5 \times \binom{4}{2} \times 2! \times 2!.$

$2,1,1,1 :$

$4$ ways of choosing the 2-rings-finger, $\binom{5}{2}$ ways of selecting which rings go on the 2-rings finger, $2!$ ways of permuting these 2 rings, $3!$ ways of distributing the remaining rings among the 3 light fingers.

$T_6 = 4 \times \binom{5}{2} \times 2! \times 3!.$

Final answer:
$$T_1 + T_2 + T_3 + T_4 + T_5 + T_6.$$

Edit

I got a total of $6720$ ways.

Title: How many ways are there to wear $5$ different rings in $4$ different fingers and also the order in which the ring is placed in a finger matters?

Body: so, we know how to solve if the question was only $5$ different rings in $4$ different fingers, which is $4^5$. but what if internal order of rings within the finger matters, is this counted in this answer or we have a different answer?


问:
Convergence of independent random variables with different distributions
Let $Y_1,Y_2,...,$ be independent r.v. with distribution $P(Y_k=k)=P(Y_k=−k)=1/2$ for all k. Here, $S_n=Y_1+Y_2+...+Y_n \ \ ∀k>0$.

Does ${S_n/n^{3/2}}$ converge in distribution? If yes, please write the limit distribution.
Note: each Yi has a different distribution

I know that the Law of Large Numbers requires i.i.d. random variables but I don't know how to start given this condition.

答: A simple calculation shows $E\Big(\frac{S_n}{n\sqrt{n}}\Big)=0$ while $$V\bigg(\frac{S_n}{n\sqrt{n}}\bigg)=\frac{n(n+1)(2n+1)}{6n^3}\longrightarrow\frac{1}{3}$$ as $n \longrightarrow \infty$. The latter calculation suggests $\frac{S_n}{n\sqrt{n}}$ won't converge in distribution to the constant random variable $0$ since its limiting variance is non$-$vanishing. Moreover, since $\frac{S_n}{n\sqrt{n}}$ resembles an average of some sort, we may hypothesize that $\frac{S_n}{n\sqrt{n}}$ converges in distribution to a $N(0,1/3)$ random variable. Let's use moment generating function to see this. The moment generating function of $\frac{S_n}{n\sqrt{n}}$ is $$E\bigg(e^{\frac{tS_n}{n\sqrt{n}}}\bigg)=E\bigg(e^{\frac{tY_1}{n\sqrt{n}}}\bigg)\times \ldots \times E\bigg(e^{\frac{tY_n}{n\sqrt{n}}}\bigg)=\prod_{k=1}^n\cosh\bigg(\frac{kt}{n\sqrt{n}}\bigg)$$ You can see here that the right hand side in the above expression  converges to the function $y=e^{x^2/6}$ which is precisely the moment generating function of a $N(0,1/3)$ random variable.


答:
The Statement is False Even in the Continuous Case

If $f(x)=x^3\sin\left(x^5\right)$, then
$$
\begin{align}
\int_1^\infty x^3\sin\left(x^5\right)\,\mathrm{d}x
&=\frac15\int_1^\infty x^{-1/5}\sin(x)\,\mathrm{d}x\tag{1a}\\
&=\frac15\cos(1)-\frac1{25}\int_1^\infty x^{-6/5}\cos(x)\,\mathrm{d}x\tag{1b}
\end{align}
$$
Explanation:

$\text{(1a)}$: substitute $x\mapsto x^{1/5}$

$\text{(1b)}$: integrate by parts

The integral in $(1)$ converges, but
$$
\begin{align}
\int_1^\infty\frac{\left|\,x^3\sin\left(x^5\right)\,\right|}{x^3}\,\mathrm{d}x
&=\frac15\int_1^\infty x^{-4/5}\left|\,\sin(x)\,\right|\mathrm{d}x\tag{2a}\\
&\ge\frac15\sum_{k=2}^\infty\int_{(k-1)\pi}^{k\pi}x^{-4/5}\left|\,\sin(x)\,\right|\mathrm{d}x\tag{2b}\\
&\ge\frac15\sum_{k=2}^\infty2(k\pi)^{-4/5}\tag{2c}\\
&=\frac25\pi^{-4/5}\sum_{k=2}^\infty k^{-4/5}\tag{2d}
\end{align}
$$
Explanation:

$\text{(2a)}$: substitute $x\mapsto x^{1/5}$

$\text{(2b)}$: remove the integral on $[1,\pi)$ and

$\phantom{\text{(2b):}}$ break up the integral on $[\pi,\infty)$ into pieces

$\text{(2c)}$: $\int_{(k-1)\pi}^{k\pi}|\!\sin(x)|\,\mathrm{d}x=2$

$\phantom{\text{(2c):}}$ $x^{-4/5}\ge(k\pi)^{-4/5}$ on $[(k-1)\pi,k\pi]$

$\text{(2d)}$: pull common factors out front

the integral in $(2)$ does not converge.

Motivation for $\boldsymbol{x^3\sin\left(x^5\right)}$

Since $|f(x)|$ is being divided by $x^3$, we want the function to grow at least as fast as $x^3$; thus, the factor of $x^3$.

The average of $|\!\sin(x)|$ over each period is $\frac2\pi$, so the integral of $\left|\sin\left(x^n\right)\right|$ over a large interval should be approximately $\frac2\pi$ times the size of the interval.

The integral of $\sin(x)$ over each period is $0$, but when multiplied by $x^3$, the integral oscillates between large and small values
$$
\int x^3\sin(x)\,\mathrm{d}x
=-x^3\cos(x)+3x^2\sin(x)+6x\cos(x)-6\sin(x)+C\tag3
$$
If we accelerate the oscillation by using $\sin\left(x^n\right)$, the integral doesn't have time to get big before being cancelled. Letting $u=x^n$, we have
$$
\int x^3\sin\left(x^n\right)\,\mathrm{d}x=\frac1n\int u^{\frac4n-1}\sin(u)\,\mathrm{d}u\tag4
$$
To get convergence on the right side, we need $n\gt4$, so $n=5$ seems a good choice.

An Extension to the Preceding

Consider $f(x)=x^\alpha\sin\left(x^{\alpha+\beta}\right)$. For $\beta\gt1$ and $\alpha\gt-\beta$,
$$
\begin{align}
\int_1^\infty f(x)\,\mathrm{d}x
&=\int_1^\infty x^\alpha\sin\left(x^{\alpha+\beta}\right)\,\mathrm{d}x\tag{5a}\\
&=\frac1{\alpha+\beta}\int_1^\infty x^{-\frac{\beta-1}{\alpha+\beta}}\sin(x)\,\mathrm{d}x\tag{5b}\\
&=\frac1{\alpha+\beta}\cos(1)-\frac{\beta-1}{(\alpha+\beta)^2}\int_1^\infty x^{-\frac{\beta-1}{\alpha+\beta}-1}\cos(x)\,\mathrm{d}x\tag{5c}
\end{align}
$$
Explanation:

$\text{(5a)}$: definition

$\text{(5b)}$: substitute $x\mapsto x^{\frac1{\alpha+\beta}}$

$\text{(5c)}$: integrate by parts

The integral in $\text{(5a)}$ converges absolutely for $\alpha\lt-1$ and the integral in $\text{(5c)}$ converges for $\beta\gt1$ and $\alpha\ge-1$. That is, the integral in $(1)$ converges for $\beta\gt1$ and all $\alpha$.

However,
$$
\begin{align}
\int_1^\infty\frac{|f(x)|}{x^3}\,\mathrm{d}x
&=\int_1^\infty\frac{\left|\,x^\alpha\sin\left(x^{\alpha+\beta}\right)\,\right|}{x^3}\,\mathrm{d}x\tag{6a}\\
&=\frac1{\alpha+\beta}\int_1^\infty x^{-\frac{\beta+2}{\alpha+\beta}}\left|\,\sin(x)\,\right|\mathrm{d}x\tag{6b}\\
&\ge\frac1{\alpha+\beta}\sum_{k=2}^\infty\int_{(k-1)\pi}^{k\pi}x^{-\frac{\beta+2}{\alpha+\beta}}\left|\,\sin(x)\,\right|\mathrm{d}x\tag{6c}\\
&\ge\frac1{\alpha+\beta}\sum_{k=2}^\infty2(k\pi)^{-\frac{\beta+2}{\alpha+\beta}}\tag{6d}\\
&=\frac2{\alpha+\beta}\pi^{-\frac{\beta+2}{\alpha+\beta}}\sum_{k=2}^\infty k^{-\frac{\beta+2}{\alpha+\beta}}\tag{6e}
\end{align}
$$
Explanation:

$\text{(6a)}$: definition

$\text{(6b)}$: substitute $x\mapsto x^{\frac1{\alpha+\beta}}$

$\text{(6c)}$: remove the integral on $[1,\pi)$ and

$\phantom{\text{(6c):}}$ break up the integral on $[\pi,\infty)$ into pieces

$\text{(6d)}$: $\int_{(k-1)\pi}^{k\pi}|\!\sin(x)|\,\mathrm{d}x=2$

$\phantom{\text{(6d):}}$ $x^{-\frac{\beta+2}{\alpha+\beta}}\ge(k\pi)^{-\frac{\beta+2}{\alpha+\beta}}$ on $[(k-1)\pi,k\pi]$

$\text{(6e)}$: pull common factors out front

for $\beta\gt1$, the sum in $\text{(6e)}$ does not converge for $\alpha\ge2$.

A Family of Counterexamples

Thus,
$$
f(x)=x^\alpha\sin\left(x^{\alpha+\beta}\right)\tag7
$$
is a counterexample for $\alpha\ge2$ and $\beta\gt1$.

Title: If $f$ is not continuous on $[1,\infty)$ but $\int_1^{\infty} f(x) \, dx$ converges, does $\int_1^{\infty} \frac{|f(x)|}{x^3} \, dx$ converge?

Body:
It is true I think that if $f$ is continuous on $[1,\infty)$ and $\int_1^{\infty} f(x) \, dx$ converges, then $\int_1^{\infty} \frac{|f(x)|}{x^3} \, dx$ converges.

But what if $f$ is not continuous but $\int_1^{\infty} f(x) \, dx$ converges.  Must it necessarily be true that $\int_1^{\infty} \frac{|f(x)|}{x^3} \, dx$ converges?


问:
Find expression for $dy/dx $ + state where it is valid
hopefully you guys can shed some insight into this question I'm working on.

Given

$xy+y^{2}-e^{x^{2}} = 6$

find an expression for $dy/dx$ and state where it is valid.

So, what I did was differentiate it, which resulted in:

$x+3y-2xe^{x^{2}} = 0 $

Although I am unsure whether this is correct, and I do not understand the state where it is valid. Any help or direction is much appreciated!

Body:
$xy+y^2 - e^{x^{2}}=6---(1)$

Differentiate with respect to $x$ to get

$y+xy'+2yy' - 2xe^{x^{2}}=0$

$y'(x+2y)=2xe^{x^{2}}-y$

$\displaystyle y'=\frac{2xe^{x^{2}}-y}{x+2y}---(2)$

Solving $(1)$ we have

$\displaystyle y=\frac{-x\pm \sqrt{x^2+4e^{x^{2}}+24}}{2}$

Put $y$ in $(2)$

$\displaystyle y'=\frac{2xe^{x^{2}}-\left(\frac{-x\pm \sqrt{x^2+4e^{x^{2}}+24}}{2}\right)}{x+2\left(\frac{-x\pm \sqrt{x^2+4e^{x^{2}}+24}}{2}\right)}$

$\displaystyle y'=\frac{4xe^{x^{2}}+x\mp \sqrt{x^2+4e^{x^{2}}+24}}{{\pm 2\sqrt{x^2+4e^{x^{2}}+24}}}$

Since $\pm 2\sqrt{x^2+4e^{x^{2}}+24}$ is defined for all $\mathbb{R}$, we say that $y'$ is defined on all $\mathbb{R}$.


问:
if linear map onto itself is bijective given that it is injective?
If given $l \in L(V)$ is injective then can we say that it is bijective too?
My reasoning:
by rank nullity theorem $N(l) = \{0\}$, implies $dim(R(l)) = dim(V)$, so $l$ is also surjective

答:
It's a provable theorem that a linear map $T \in \mathcal{L}(V)$ is injective if and only if it is surjective, provided that $V$ is finite-dimensional. The argument is almost exactly as you suggested:

\begin{align*}
\text{$T$ is injective } & \iff \text{Null}(T) = \{0\} \\
& \iff \dim V = \dim \text{Null}(T) + \dim \text{Range}(T) = \dim \text{Range}(T) \\
& \iff \text{Range}(T) = V.
\end{align*}
The third line follows from the second since $\text{Range}(T)$ is a subspace of $V$, so if $\dim V = \dim \text{Range}(T)$, we have $V = \text{Range}(T)$.

This result doesn't generalize to the infinite case. Notice that we used the rank-nullity theorem in the second line, which requires that $\dim V$ be finite.


问:
How to determine that the improper integral $\int_{\pi}^{\infty}\frac{\sin^2(x)}{x}dx$ diverges?
How is the divergence of the improper integral $\int_{\pi}^{\infty}\frac{\sin^2(x)}{x}dx$ shown and determined with the improper integral comparison test or other methods?

答:
We know that $|\sin(x)| \geq \frac12$ for $x \in [2\pi n + \pi/6, 2\pi n + 5\pi/6]$ (this holds for any $n \in \mathbb{N}$; to see why it's true just visualize the unit circle).  Therefore we can estimate
$$
\int_\pi^\infty \frac{\sin^2x}{x} \,dx \geq \sum_{n = 1}^{\infty} \int_{2\pi n + \pi/6}^{2 \pi n + 5\pi/6} \frac{\sin^2x}{x} \,dx \geq \sum_{n=1}^{\infty} \frac{1/4}{2\pi n + 5\pi/6},
$$
and this sum can be seen to diverge by comparison with the harmonic series $\sum_{n=1}^{\infty} \frac1n$.


问:
Does $1-ab\geq (1-a)(1-b)$ hold, for $a,b\in [0,1]$?
Does this inequality hold, for any $a,b\in[0,1]$

$$1-ab\geq (1-a)(1-b)?$$

I'm don't have idea to conclude $1-ab\geq 1-a-b+ab = (1-a)(1-b)$.

Anyone can prove (or disprove) it?

答:
No.  Take $a=0$ and $b=1$ to disprove it.

Addendum:

Now that you changed the inequality, it follows from AM-GM:

$a+b\ge2\sqrt a\sqrt b\ge 2ab\iff 1-ab\ge1+ab-a-b=(1-a)(1-b)$.


问:
The basis for intersection of two subspaces
I know there are other posts on how to do this, but I was wondering if there is a slicker method.

$X = Span\{(1,1,0,-1),(1,2,3,0),(2,3,3,-1)\}$
$Y = Span\{(1,2,2,-2),(2,3,2,-3),(1,3,4,-3)\}$

I want to find a basis for $X \cap Y$

The method I have been trying to use is by letting $v \in X$ and $v \in Y$ and attempting to solve the equation $v - v = 0$. However it seems very long and tedious. I was thinking there may be a better method.

答:
This is a perfect situation where changing the point of view can provide some good insight.  You are trying to find the intersection of two vectors subspaces of $\mathbb{R}^4$  and you are given those spaces as the spans of sets of vectors.  In other words you are given X and Y as the column spaces of matrix $V=[v_i]$ an $W=[w_i]$ (where the columns are the vectors you gave above).  At first glance it's not clear how to find the intersection.  Indeed, what do we do? Write a matrix / take echelon form? But there are two matrices... Hmm... However...

In general - there are straightforward efficient ways to find the null space or column space of a matrix. In this case, thinking about null spaces turns out to help a lot!

Indeed, suppose instead we were give these (same) spaces not as the column space of matrices, but instead as the null space of a matrix,  say $X = \mathrm{Nul }A$ and $Y = \mathrm{Nul }B$, then it would be easy to find the intersection of $X$ and $Y$.  Indeed, we can think of $X$ as solutions to $Ax=0$ and $Y$ as solutions to $Bx=0$, and so to find $X\cap Y$ we just need to combine all of these linear equations into one big system, namely the matrix with $A$ and $B$ stacked on top of each other.  Then if we solve:
$$\begin{bmatrix}A \\ B \end{bmatrix} x = 0$$
then the set of solutions will be $X\cap Y$ so we just need to row-reduce this bigger matrix and then we can read off the null space from that.

Ok, I know what you're thinking - how do we find the matrices $A$ and $B$.   Well this is where some more linear algebra magic can come to the rescue.  So the question is:  Given that $X = Col(V)$ how can you find an $A$ so that $X = Nul(V)$.  Well here's a sneaky trick, the answer is ``Find a basis for the null space of $V^T$ (a null space calculation) and then write these basis vectors as columns of a matrix, call it $P$.  Then by construction we have that $V^TP = 0$.  Now if we take the transpose we get $P^TV = 0$, which means that the columns of $V$ (i.e. our original vectors in your problem) are in the null space of $P^T$.   So $P^T$ is the matrix that you want.''

I don't know how much linear algebra you've seen, and this is surely not the fastest way to do this, but you sound like you'd be interested in learning a bit more and I'd be happy to fill in more details.

In pseudocode if you want to flip this from a "span" problem to a "null space" problem the ideas you want are:

X = col(V);  Y = col(W);

A = transpose ( null  transpose( V ) );

B = transpose ( null  transpose( W ) );

Then $X\cap Y$ = null space of $\begin{bmatrix} A\\ B\end{bmatrix}$


问:
Prove $\sum_{k=0}^n\binom{n}{k}\cos\left(\frac{k\pi}{2} - \frac{n\pi}{4}\right) = \sqrt{2^n}$
While working on a separate problem, I stumbled upon a beautiful fact that I simply couldn't prove.. I suspect it requires some tools/tactics I've not encountered yet. The statement is the following:
$$\sum_{k=0}^n\binom{n}{k}\cos\left(\frac{k\pi}{2} - \frac{n\pi}{4}\right) = \sqrt{2^n}$$

From my humble experience, I've got two suspicions for a successful approach:

Complex number manipulation and working with the real part;

Some kind of Fourier transform... this really just reminds me of it.

I couldn't come up with anything nice using complex numbers and I'm not that familiar with Fourier transforms yet, so I don't have the right intuition there.

Furthermore, this sum is just begging to employ Newton's binomial at some point, but I couldn't transform it to a form where it is applicable.

Any advice is welcome! Thanks in advance!

P. S. I don't opt for induction proofs, I'm rather interested in evaluating the sum on the LHS and arriving at the RHS.

答:
You first idea is correct.

$$\begin{align}
\sum_{k=0}^n\binom nk\cos\left(\frac{k\pi}2-\frac{n\pi}4\right)&=
\sum_{k=0}^n\binom nk \Re\left(\exp\left(\frac{k\pi i}2-\frac{n\pi i}4\right)\right)\\
&=\Re\left(\left(e^{-n\pi i/4}\right)\sum_{k=0}^n\binom nk \left(e^{\pi i/2}\right)^k\right)\\
&=\Re\left(\left(e^{-n\pi i/4}\right)\sum_{k=0}^n\binom nk i^k\right)\\
&=\Re\left(\left(e^{-n\pi i/4}\right)(1+i)^n\right)\\
&=\Re\left(\left(e^{-n\pi i/4}\right)\left(\sqrt2e^{\pi i/4}\right)^n\right)\\
&=\sqrt{2^n}
\end{align}$$


问:
Measurable in $\mathbb{C}$
Let $(\Omega, \mathcal{M}, \mu)$ be a measure space. For measurable functions $f,g: \Omega \to \mathbb{C}$ let $f \sim g$ if $f$ is almost everywhere equal to $g$.

Show that $\{x \in \Omega \space| \space f(x) \neq g(x) \}$ is measurable.

I found $\int_E f \space d \mu:=\int_E u \space d \mu \space + i \int_E v \space d \mu$ for $f= u+iv$ with $u,v: \Omega \to \mathbb{R}$ as a definition, and we say that $f$ is measurable if $u$ and $v$ are measurable but I don't really know how to proof this.

答:
First write
$$
\{ x : f(x) \neq g(x) \} = \{ x : (\Re f)(x) \neq (\Re g)(x)\} \cup \{ x : (\Im f)(x) \neq (\Im g)(x)\}
$$
so that it suffices to assume that $f$ and $g$ are real-valued.  Then we can write
$$
\{x : f(x) \neq g(x)\} = \{x : f(x) < g(x) \} \cup \{x : f(x) > g(x) \} = \left( \bigcup_{q \in \mathbb{Q}} \{ x : f(x) < q \} \cap \{ x : g(x) > q \} \right) \cup \left( \bigcup_{q \in \mathbb{Q}} \{ x : f(x) > q \} \cap \{ x : g(x) < q \} \right).
$$
Thus we've expressed the desired set in terms of countable unions and intersections of measurable sets.

Title: Measurable in $\mathbb{C}$


问:
Ratio problem solving- ten machines use 200 kW
Ten machines use 200 kilowatts to operate for 5 days, 10 hours per day. Determine the number of kilowatts used if they work for 3 days, 8 hours per day?

and the answer has to be in kW

updated

would it be

200kw /130hrs = x/80hrs

1600/130 = 130x

123= x

I did fractions with cross multiplying and got x= 71.4 kW but I know I am wrong.

5/3 + 10/8 = 200 kW/ x

40/24 + 30/42 = 200kW/ x

70/24 = 200kW/ x

70x = 5000 kW/70

x= 71.4 kW

答:
The unit of energy is the kilowatt-hour or some such.  Kilowatts are a measure of power.  As such, the duration of operation does not matter.  If ten machines use $200$ kW, they use $200$ kW no matter how long they run.  It is possible the point of the problem is for you to notice the difference between kilowatts and kilowatt-hours.

If the problem means $200$ kWh for $10$ machines, $5$ days, $10$ hours/day, that is $50$ hours of operation, so a $4$ kW power level.  $3$ days, $8$ hours/day is $24$ hours, so they would consume $96$ kWh.


答:
This answer adapted from the AoPS link in comments
$$\sum_{k=0}^n\binom nk\cos(k\pi/2-n\pi/4)$$
$$=\operatorname{Re}\left(\sum_{k=0}^n\binom nk(\cos(k\pi/2-n\pi/4)+i\sin(k\pi/2-n\pi/4))\right)$$
$$=\operatorname{Re}\left(\sum_{k=0}^n\binom nke^{i(k\pi/2-n\pi/4)}\right)$$
$$=\operatorname{Re}\left(e^{-in\pi/4}\sum_{k=0}^n\binom nk i^k\right)$$
$$=\operatorname{Re}\left(\left(\frac{\sqrt2}2(1+i)\right)^{-n}(1+i)^n\right)$$
$$=\left(\frac{\sqrt2}2\right)^{-n}=2^{n/2}$$

Title: Prove $\sum_{k=0}^n\binom{n}{k}\cos\left(\frac{k\pi}{2} - \frac{n\pi}{4}\right) = \sqrt{2^n}$

Body:
While working on a separate problem, I stumbled upon a beautiful fact that I simply couldn't prove.. I suspect it requires some tools/tactics I've not encountered yet. The statement is the following:
$$\sum_{k=0}^n\binom{n}{k}\cos\left(\frac{k\pi}{2} - \frac{n\pi}{4}\right) = \sqrt{2^n}$$

From my humble experience, I've got two suspicions for a successful approach:

Complex number manipulation and working with the real part;

Some kind of Fourier transform... this really just reminds me of it.

I couldn't come up with anything nice using complex numbers and I'm not that familiar with Fourier transforms yet, so I don't have the right intuition there.

Furthermore, this sum is just begging to employ Newton's binomial at some point, but I couldn't transform it to a form where it is applicable.

Any advice is welcome! Thanks in advance!

P. S. I don't opt for induction proofs, I'm rather interested in evaluating the sum on the LHS and arriving at the RHS.


问:
Integral $ \int_{-\infty}^{\infty}\frac{x^{2}}{\cosh\left(x\right)}\,{\rm d}x $

I need to compute the improper integral
$$
\int_{-\infty}^{\infty}\frac{x^{2}}{\cosh\left(x\right)}\,{\rm d}x
$$
using contour integration and possibly principal values.  Trying to approach this as I would normally approach evaluating an improper integral using contour integration doesn't work here, and doesn't really give me any clues as to how I should do it.

This normal approach is namely evaluating the contour integral
$$
\oint_{C}{\frac{z^2}{\cosh\left(z\right)}\mathrm{d}z}
$$
using a semicircle in the upper-half plane centered at the origin, but the semicircular part of this contour integral does not vanish since $\cosh\left(z\right)$ has period $2\pi\mathrm{i}$ and there are infinitely-many poles of the integrand along the imaginary axis given by $-\pi\mathrm{i}/2 + 2n\pi\mathrm{i}$ and
$\pi\mathrm{i}/2 + 2n\pi\mathrm{i}$ for
$n \in \mathbb{Z}$.

The residues of the integrand at these simple poles are $-\frac{1}{4}\mathrm{i}\pi^{2}\left(1 - 4n\right)^{2}$ and $\frac{1}{4}\mathrm{i}\left(4\pi n + \pi\right)^{2}$, so that even when we add up all of the poles, we have the sum $4\pi^{2}\mathrm{i}\sum_{n = 0}^{\infty}\,n$, which clearly diverges.

Any hints would be greatly appreciated.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\!\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{x^{2} \over \cosh\pars{x}}\,\dd x} =
4\int_{0}^{\infty}{\sinh\pars{x} \over \sinh\pars{2x}}\,x^{2}\,\dd x
\\[5mm] \stackrel{x\ =\ t/4}{=}\,\,\,&
{1 \over 16}\int_{0}^{\infty}{\sinh\pars{t/4} \over \sinh\pars{t/2}}\,t^{2}\,\dd t
\\[5mm] = &\
\left.{1 \over 4}\,\partiald[2]{}{\alpha}\int_{0}^{\infty}{\sinh\pars{\alpha t/2} \over \sinh\pars{t/2}}\,\dd t
\,\right\vert_{\,\alpha\ =\ 1/2}
\\[5mm] = &\
\left.{1 \over 4}\,\partiald[2]{}{\alpha}\int_{0}^{\infty}
{\expo{-\pars{1 - \alpha}t/2} -
\expo{-\pars{1 + \alpha}t/2} \over 1 - \expo{-t}}\,\dd t\
\,\right\vert_{\,\alpha\ =\ 1/2}
\\[5mm] = &\
{1 \over 4}\,\partiald[2]{}{\alpha}\left[%
\int_{0}^{\infty}{\expo{-t} -
\expo{-\pars{1 + \alpha}t/2} \over 1 - \expo{-t}}\,\dd t
\right.
\\[2mm] &
\left.\phantom{{1 \over 4}\,\partiald[2]{}{\alpha}}
-\int_{0}^{\infty}{\expo{-t} -
\expo{-\pars{1 - \alpha}t/2} \over 1 - \expo{-t}}\,\dd t
\right]
_{\,\alpha\ =\ 1/2}
\\[5mm] = &\
{1 \over 4}\,\partiald[2]{}{\alpha}\bracks{%
\Psi\pars{1 + \alpha \over 2} - \Psi\pars{1 - \alpha \over 2}}
_{\,\alpha\ =\ 1/2}
\\[5mm] = &\
{1 \over 4}\,\partiald[2]{}{\alpha}\bracks{%
\pi\cot\pars{\pi\,{1 - \alpha \over 2}}}_{\,\alpha\ =\ 1/2}
\\[5mm] = &\
{\pi \over 4}\,\partiald[2]{}{\alpha}\bracks{%
\tan\pars{\pi\alpha \over 2}}_{\,\alpha\ =\ 1/2}
\\[5mm] = &\
{\pi \over 4}\bracks{%
{\pi^{2} \over 4}\sec^{2}\pars{\pi a \over 2}
\tan\pars{\pi a \over 2}}_{\,\alpha\ =\ 1/2}
\\[5mm] = &\
{\pi^{3} \over 8}\
\underbrace{\sec^{2}\pars{\pi \over 4}}_{\ds{2}}\
\underbrace{\tan\pars{\pi \over 4}}_{\ds{1}}
\\[5mm] = &\
\bbx{\pi^{3} \over 4} \\ &
\end{align}


问:
Prove this square matrix has no square root
Prove the matrix with filled with 0s everywhere except 1s just above the main diagonal has no square root.
The matrix looks like this:
enter image description here</p>

答:
(Fill in the gaps as needed. If you're stuck, show your work and explain why you're stuck.)

Hint: Prove that the minimal polynomial is $ A^{n} = 0 $.

Do a proof by contradiction.

Suppose $ A = B^2$.

Then, ...

More generally, the minimal polynomial contains $A^k = 0 $ where $ k > n/2$ as a factor, then $A$ has no square root.


Title: Does $1-ab\geq (1-a)(1-b)$ hold, for $a,b\in [0,1]$?

Body:
Does this inequality hold, for any $a,b\in[0,1]$

$$1-ab\geq (1-a)(1-b)?$$

I'm don't have idea to conclude $1-ab\geq 1-a-b+ab = (1-a)(1-b)$.

Anyone can prove (or disprove) it?

Body:
Hint:

as $1-a\ge 0,1-b\ge 0$
$$1-ab-(1-a)(1-b)=a(1-b)+b(1-a)\ge 0$$


答:
$$\text{RHS} - \text{LHS} = 1+ab -a-b - (1-ab) = 2ab-a-b ≤ 2ab-a^2 - b^2 = -(a-b)^2 ≤ 0.  $$
Thus $\text{LHS} \ge \text{RHS}$.


问:
Integrating Factor of $(x\ln(y) + xy)\mathrm{d}x + (y\ln(x) + xy)\mathrm{d}y$
Last day I try to prove that the equation
$$
(x\ln(y) + xy)\mathrm{d}x + (y\ln(x) + xy)\mathrm{d}y =0
$$
is not exact, really easy task, but I want to go further, I tried to find a integrating factor such that the equation become exact, but is so strange, I try every method that I know, but I get nothing, later I catch some hope with the factor $R = \ln\big(\ln(x+y)\big)$, but then nothing again. There's some method to find this integrating factor? Do you know any? Thank you so much!

答:
Both coefficient terms are separable, so that you can separate terms in $x$ and $y$.
$$
\frac{x\,dx}{\ln x+x}+\frac{y\,dy}{\ln y+y}=0.
$$


问:
Existence of $\delta_p(A)=\int_Ah \space d \lambda$
We are looking at the measure space $(\mathbb{R}^d, \mathcal{P}(\mathbb{R}^d),\delta_p)$, for all $A\subset \mathbb{R}^d$:

$\delta_p(A):=\begin{cases}
1, & \text{if } p \in A \\
0, & \text{otherwise}
\end{cases}$.

Let $\lambda$ be the Lebesgue measure on $\mathbb{R}^d$. Does an integratable function $h: \mathbb{R}^d \rightarrow \mathbb{R}$ exist such that for $A \subset \mathbb{R}^d: \delta_p(A)=\int_Ah \space d  \lambda$?

答:
Let $A = \{ x | x \neq 0, h(x) \ge 0 \}$, then $\int_A h d \lambda = 0$ and so $h = 0$ ae. $[\lambda]$ and similarly
on $B = \{ x | x \neq 0, h(x) \le 0 \}$. In particular, $h = 0$ ae. $[\lambda]$ and so $\int_A h  d \lambda =0 $ for all measurable $A$.


问:
Prove that two maximal $k$-connected sub-graphs of a graph $G$ have at most $k-1$ common vertices.
Assume that the two sub-graphs $A$ and $B$ of graph $G$ such that $A \neq B$.

A maximal $k$-connected sub-graph will not be a "proper" sub-graph for any other $k$-connected sub-graph. Here - $k$ is the size of the minimal vertex cut.

How do I prove this by contradiction?

答: Suppose the two maximal $k$-connected subgraphs share $k$ vertices. Deleting any $k-1$ vertices across the two subgraphs leaves both subgraphs connected and must leave at least one common vertex, so the union of subgraphs remains connected after deleting any $k-1$ vertices – it is $k$-connected and strictly larger than both original subgraphs, a contradiction. Hence the statement is proved.


问:
Inviscid Burgers' equation: drawing the shock

Solve Burgers' equation
$$
\left\{\begin{aligned}
u_{t}+uu_x &=0 \quad \text { for } \quad t>0 \\
u(x, 0) &=u_{0}(x)
\end{aligned}\right.
$$
with $u=u(x,t)$ and the side condition $u(x,0)=-x$.

I am aware that a similar question with initial condition u=x has been asked before, and asked this because I was wondering what the difference would be when the characteristic lines are set to converge.

答:
From implicit function theorem we have the following

$$u_t+uu_x = 0 \implies \frac{dx}{dt}=u$$

So in other words, the slopes of the characteristics depend on the value of $u$. With $u=x$, you can see that the characteristics that start at negative $x$ move left (negative slope) and vice versa for positive $x$. Could you reason out the behavior for $u=-x$ instead?

Bonus question, technically the shock in both situations could be chosen to be anything, but how does one pick out the max entropy solution for both $u=x$ and $u=-x$ ?


问:
Is there a way to do this with double integration?
If $0 < h < a$, then the plane $z = a - h$ cuts off a spherical segment of height h and radius b from the sphere $x^2 + y^2 + z^2 = a^2$.

a) Show that $b^2 = 2ah - h^2$

b) Show that the volume of the spherical segment is $V = \frac{1}{6}\pi h(3b^2+h^2)$

Part a is pretty straightforward but I'm looking for how to approach (b). I've seen it done with triple integration. Is it possible to do it for 2 integrals?

What I have so far is $\int_{0}^{2\pi}\int_{0}^{b} (\sqrt{1-r^2} - (a-h))rdrd\theta$.

答:
What you are talking about is spherical cap. Yes you can find with single / double integral. There seems to be some mistake in how you have written it.

You know radius of the circle at any given $z$ is $\sqrt{a^2-z^2}$. You use the knowledge of perimeter of the circle or the area of the circle at a given $z$ to make this a double or single integral.

$\displaystyle 2\pi \int_{(a-h)}^{a} \int_0^{\sqrt{a^2-z^2}} r \, dr \, dz$

or you can write it as

$\displaystyle \pi \int_{(a-h)}^{a} (a^2 - z^2)  \, dz$


问:
Analytical proof required $\binom{n}{r} < (n+1)^r$
$\binom{n}{r} < (n+1)^r$

I need help to prove the above expression using some analytical method.

答:
Hint:

Equality holds at $r = 0$, so inequality starts at $r = 1.$

Simplifying assumption made that $2 \leq n \in \mathbb{Z^+}$ and that $r \in \{1,2,\cdots, n\}.$

With base case of $r=1$ established, do induction on $r$, noting the scaling factor on each side of the inequality.


答:
Define $m=2a-1$ i.e. $m\in [-1 , 1]$ and $n=b-\frac{1}{2}$ i.e. $n\in [-\frac{1}{2} , \frac{1}{2}]$. Is the following true?

$$
\begin{align}
-\frac{1}{2}&\leq mn\leq \frac{1}{2}\\
-ab&\leq (1-a)(1-b)\leq 1-ab
\end{align}
$$


答:
The ODE:  $$(xln(y) + xy)dx + (yln(x) + xy)dy=0~~~~(1) $$
can be re written as $$\int \frac{xdx}{x+\ln x}+\int \frac{ydy}{y+\ln y}=c~~~~(2)$$ by dividing (1) by $(x+\ln x)(y+\ln y)$ (2) is separable yest the integrals are not doable. So $[(x+\ln x)(y+\ln y)]^{-1}$  can be taken as the integrating factor for (1).

Title: Integrating Factor of $(x\ln(y) + xy)\mathrm{d}x + (y\ln(x) + xy)\mathrm{d}y$

Body:
Last day I try to prove that the equation
$$
(x\ln(y) + xy)\mathrm{d}x + (y\ln(x) + xy)\mathrm{d}y =0
$$
is not exact, really easy task, but I want to go further, I tried to find a integrating factor such that the equation become exact, but is so strange, I try every method that I know, but I get nothing, later I catch some hope with the factor $R = \ln\big(\ln(x+y)\big)$, but then nothing again. There's some method to find this integrating factor? Do you know any? Thank you so much!


问:
Square in a finite field
I need to prove: $a$ is a square in $\mathbb{Z}$ iff $a\mod p$ is a square in $\mathbb{F_p}$ for every $p$ prime. First side ($\rightarrow$) is trivial and is derived directly from multiplicativity of modulo. Now I need to prove the other side ($\leftarrow$). I tried taking $p>a$ so $a\equiv a\mod p$, and I know that $a$ is a square so there exists $b<p$ s.t $\space$$b^2\mod p \equiv a$. But, $b$ is not necessarily the root of $a$ in $\mathbb{Z}$, so I'm stuck. I also know that for each $p$, $(\frac{a}{p})=1$ but this doesn't get me anywhere. Any hint would help.

答:
If think the easiest way is to use quadratic reciprocity, that if $a$ is not a square then for $p\ne 2$,
$(\frac{a}p) =\chi(p) $ with $\chi$ a non-trivial Dirichlet character modulo $4a$,

there is $m>0$ such that $\chi(m)=-1$, some prime $p|m$ is such that $-1=\chi(p)=(\frac{a}p)$

Title: Square in a finite field


答:
$$
\binom{n}{r} = \frac{n!}{r!(n-r)!} = \frac{\prod\limits_{k=1}^{n}k}{r!\prod\limits_{k=1}^{n-r}k} = \frac{\prod\limits_{k=n-r+1}^{n}k}{r!} \le \frac{\prod\limits_{k=n-r+1}^{n}n}{r!} = \frac{n^r}{r!} \le n^r \le \sum\limits_{k=0}^{r} \binom{r}{k} n^k = (n+1)^r
$$

Title: Analytical proof required $\binom{n}{r} < (n+1)^r$

Body:
$\binom{n}{r} < (n+1)^r$

I need help to prove the above expression using some analytical method.


Body: $$\binom{n}{r}=\frac{n!}{r!(n-r)!}<\frac{n!}{(n-r)!}=\underbrace{n(n-1)\dots(n-r+1)}_{r-terms}<\underbrace{(n+1)(n+1)\dots(n+1)}_{r-terms}=(n+1)^r$$


问:
If $z_n$ are the zeros of the zeta function, what is the limit of $\Im{(z_n)}$ as $n$ goes to infinity?
Sorry if this question has already been asked, but it's a little difficult to look things up in Google if the statement of the problem is not very simple and involves symbols that Google doesn't recognize.

The question I have regards the zeta function. If $z_n$ is the sequence of non-trivial zeros of the zeta function with positive imaginary part and sorted by ascending imaginary part, what is the limit when $n$ goes to infinity of $\Im{(z_n)}$?

Does this explode out to infinity or is it finite?

Asking for a friend (paper here).
He has derived a new super simple equation whose solution is equivalent to the Riemann hypothesis.

答:
This just reports some (old) empirical results.

Many years ago, in my research group, the same question came and one of my Ph.D. developed a simple empirical correlation $(R^2=0.999991 )$
$$\log \left(\Im\left(\rho _{2^k}\right)\right)\sim a+b \,k^c$$

For $1 \leq k \leq 23$, this gave
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval}
   \\
 a & 2.72774 & 0.02399 & \{2.67752,2.77795\} \\
 b & 0.27581 & 0.00566 & \{0.26396,0.28767\} \\
 c & 1.21848 & 0.00627 & \{1.20535,1.23161\} \\
\end{array}$$

from which the estimate of the imaginary part of the millionth zero is $ 595894$ instead of $600270$.

$$\left(
\begin{array}{ccc}
 n & \text{estimate} & \Im\left(\rho _{10^n}\right) \\
 1 & 50.3377 & 49.7738 \\
 2 & 244.508 & 236.524 \\
 3 & 1436.66 & 1419.42 \\
 4 & 9672.79 & 9877.78 \\
 5 & 72559.8 & 74920.8 \\
 6 & 595894. & 600270. \\
 7 & 5292950 & 4992381
\end{array}
\right)$$

Edit

Using @Qiaochu Yuan's answer, we could inverse

$$\frac{T}{2\pi} \log \frac{T}{2\pi} - \frac{T}{2\pi} + O(\log T)$$ and get
$$\Im\left(\rho _{n}\right)\sim \frac{2 \pi  n}{W\left(\frac{n}{e}\right)}$$ where $W(.)$ is Lambert function.

Using its usual series expansion,
$$\Im\left(\rho _{n}\right)\sim \frac{2 \pi  n}{L_1-L_2+\frac{L_2} {L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\cdots }$$ where $L_1=\log(n)-1$ and $L_2=\log(L_1)$. For $n=10^6$, this would give $600219.$

If you look at the paper by G.Franca and A.LeClair, equation $(163)$ gives sharp bounds
$$\frac{2 \pi  \left(n-\frac{7}{8}\right)}{W\left(\frac{n-\frac{7}{8}}{e}\right)} \leq \Im\left(\rho _{n}\right) \leq \frac{2 \pi  \left(n-\frac{3}{8}\right)}{W\left(\frac{n-\frac{3}{8}}{e}\right)}$$


问:
Space of bounded continuous functions from $\mathbb R$ to $\mathbb C$
Let $V$ denote the space of bounded continuous functions from $\mathbb R$ to $\mathbb C$. Enumerate the rationals by $r_1,r_2,\ldots$. Define the inner product on $V$ by
$$\langle f,g \rangle = \sum \frac{1}{2^k}f(r_k) \overline{g(r_k)}$$

The question is whether $V$ endowed with this inner product becomes a Hilbert space.

It is fairly easy to check the above operation is indeed an inner product, but I got stuck about the completeness part. Any help is appreciated.

答: No, it is not an inner product.  It is not defined when the sum diverges, which it will in all but trivial cases.</p>


问:
Asking for an example of a (number) system failed to exist
Quite a few analysis books I've looked into start with some number system, claim some (reasonably convincing) axioms on it, assume it to be existed, and build analysis from there. Some explain why the number system exists by constructing it from some more basic number system, but I've probably never came across anything about why some number system does not exist.

Could anyone please explain how some number system does not exist by giving an example?

One of my understandings is that it might be due to the contradiction arisen from the axioms that fails the number system to be existed.

According to page 11 of Pugh's Real Mathematical Analysis (2017):

For what if a system satisfying the axioms failed to exist? Then one would be studying the empty set!

My understanding to this sentence above is as follows: let $U$ be a collection that contains everything that exists that we ever need, and $L$ be the list of axioms that are needed to be satisfied by our number system $X$. Define $X$ as a set such that for each $x \in U$, iff $x$ satisfies everything in $L$ simultaneously, then $x \in X$. Since there is a contradiction in $L$, no element in $U$ would be filled into $X$ through these implications and thus we get $X=\emptyset$. And all the subsequent analyses on $X$ are carried through vacuous implications, that is, by assuming there are elements in the empty set $X$.

May I ask if my understandings are on the right track? (it kinda confuses me...)

And here come some other questions:

Is that possible to have a number system that doesn't exist while the axioms on it are consistent?

When we say some number system does not exist, does this mean that the entities in the system, speaking of themselves, literally cannot exist? Or does this mean that those entities assumed to be inside the system indeed can exist, but they are just living in somewhere else and the system is empty? (Or are there other interpretations?)

答:
Here's an answer to your first question. It's also rather topical right now:

We work over $\mathbb{R}$ throughout.

A Normed Division Algebra is a "number system" which generalizes $\mathbb{C}$ in a similar way that $\mathbb{C}$ generalizes $\mathbb{R}$. More precisely:

A division algebra is a a vector space over $\mathbb{R}$ (so it has addition and subtraction, among other things) which also admits multiplication and division (as long as you don't divide by the $0$ vector).

So you might think of a division algebra as a "number system" (since we can add/subtract/multiply/divide) which contains the reals $\mathbb{R}$ in a nice way. Indeed, $\mathbb{R}$ itself and $\mathbb{C}$ are two good examples of division algebras.

A normed division algebra is a division algebra which also admits a norm $x \mapsto |x|$ which tells you how "big" a number is. Again, $\mathbb{R}$ with the absolute value function, and $\mathbb{C}$ with its standard norm $a + bi \mapsto \sqrt{a^2 + b^2}$ are both examples of this object too.

The drama:

There is a theorem due to Hurwitz which says that if you restrict your attention to "finite dimensional" systems, the only such number systems are

$\mathbb{R}$

$\mathbb{C}$

$\mathbb{H}$

$\mathbb{O}$

So if you are interested in "number systems" which contain $\mathbb{R}$, admit a notion of "size", and "aren't too big", you are FORCED into picking one of these systems.

Rather interestingly, you lose something on each stage.

Passing from $\mathbb{R}$ to $\mathbb{C}$, you lose the ability to order your numbers (is $1 < i$?)

Passing from $\mathbb{C}$ to $\mathbb{H}$, you lose commutativity (so $ab \neq ba$ in general)

Passing from $\mathbb{H}$ to $\mathbb{O}$, you lose associativity! That is,
$(ab)c \neq a(bc)$ in general!

What is most interesting, though, is that a few months ago, somebody published an article saying they had found a new algebra, which was like $\mathbb{O}$, but managed to preserve associativity!

That is, they claimed the existence of an $8$-dimensional, associative, normed division algebra.

This is certainly a "number system" of the kind you might be interested in. It is also nonexistent by Hurwitz' Theorem! The paper has since been retracted from the journal, though I believe the author is still fighting the decision, claiming their result is actually true.

So you see, there are cases where mathematicians accidentally study objects which don't exist. There are lots of apocryphal examples as well, but most of them don't fall under the heading of "number systems".

As a rather quick reply to your other questions:

Can a number system fail to exist even when its axioms are consistent?

My instinct says "no". Certainly in the case of a First Order set of axioms, a system exists if and only if the axioms are consistent. This is Godel's Completeness Theorem.

I'm notably less well versed in higher order logics, such as those needed to formalize the completeness of $\mathbb{R}$. In general, completeness/soundness/effectiveness cannot all happen at the same time for higher order logics, and I'm far from an expert in the subtleties of what is and isn't possible. Rather funnily, this is a corollary of Godel's Incompleteness Theorem.

I'm sure if you post another question on this site asking for specifics of completeness of second order logic, you'll find some more specialized logicians than me who are willing to help :)

What does it mean for a system to "not exist"

I think this is better asked to a room of philosophers than a room of mathematicians. For me, it "doesn't exist" if there isn't a model. That is, there is no set equipped with some functions that actually satisfy the axioms.

For instance, there is no noncyclic group of order $2$. Why does this not exist? Because you can look at a set of size $2$, and try all possible group structures, and see for yourself that none of them work. It's the same idea for other objects that "don't exist", we just need cleverer arguments since we can't check by hand that all (possibly infinitely many) options fail.

I hope this helps ^_^


问:
A square modulo every prime is a square. Proof valid?
As Eric Schneider asked, "Am I mistaken, or does the following (actually) elementary proof work?"

Theorem. Any integer which is a square modulo every prime is a square.

Lemma. For any odd prime $p$, any integer which is a square modulo $p$ is a square modulo every power of $p$.

Proof. Let $a$ be any square modulo $p$. Let $r$ be any integer $\geqslant 1$. Use induction on $r$. (I adopt this approach to avoid a bug, spotted by Ingix, in my earlier proof.) The result is true for $r=1$. Suppose, by the inductive hypothesis, that $a$ is a square modulo $p^{r-1}$. Then $a=x^2\mod p^{r-1}$ for some $x$.

Work modulo $p^r$. If $x=0$ then $a=0=0^2$ modulo $p^r$. Otherwise, for $0\leqslant k<p$, $(kp^{r-1}+x)^2=jp^{r-1}+a\mod p^r$ for some $0\leqslant j<p$.

Suppose two distinct $kp^{r-1}+x$ had the same square. Then
\begin{align*}
(kp^{r-1}+x)^2&=(lp^{r-1}+x)^2\mod p^r\tag{ with $k\ne l$}\\
(k^2-l^2)p^{2r-2}+2(k-l)p^{r-1}x&=0\mod p^r\\
2(k-l)p^{r-1}x&=0\mod p^r\tag{ as $r>1$}\\
2(k-l)x&=0\mod p
\end{align*}
$p$ is an odd prime and $x\ne 0\mod p$, so $p\nmid 2x$, so the last line is false.

Therefore, by the pigeonhole principle, the values of the $p$ distinct expressions $(kp^{r-1}+x)^2$ for $0\leqslant k<p$ are the values $jp^{r-1}+a$ for $0\leqslant j<p$ in some order. In particular, in one case $j=0$, so $a$ is a square modulo $p^r$.

Proof of theorem. Let $a$ be a square modulo every prime. Then, by Lemma 1, $a$ is a square modulo every odd prime-power. Then, by the Chinese remainder theorem, $a$ is a square modulo every odd integer. Then, by Eric Schneider's argument, with $n=2$, but applying it only to $p$ being an odd prime and not to $p=2$, for every odd prime $p$, $p^r\; ||\;a$ for an even number $r$.

Thus either $a=x^2$ or $a=2x^2$ for some integer $x$. Let $p$ be a prime where $p=\pm 3\mod 8$. Then, modulo $p$, $a$ is a quadratic residue modulo $p$ by supposition, but 2 is not a quadratic residue, so $2a$ is not a quadratic residue, so $2a$ is not a square in $\mathbb{Z}$. Thus for every integer $x$, $(2x)^2=4x^2\ne 2a$, so $2x^2\ne a$, so $a$ is a square, as required.

I see no elementary proof of this theorem, with no use of quadratic reciprocity (QR). I wonder if the above qualifies. Chan asked for a proof of this very result but that question was closed as a duplicate of a different question, viz the one where Eric Schneider's answer has been used above. There is Hagen von Eitzen's proof of a similar result, but that relies on QR. ArithmeticGeometer requested a proof without using QR, and the only answer is an advanced proof.

答: That $a$ is a square $\bmod p$ implies that $a$ is a square modulo every $p^k$ whenever $p$ is an odd prime not dividing a. If $p^{2m+1}\| a$ then it fails for $k>2m+1$.


问:
Primitive recursive functions and sets
(i) Show that the set of Fibonnaci numbers is a primitive recursive set.

(ii) Show that there is not a "minimal non-recursive set". That is to show that for all $A$ non-recursive there is a $B \subsetneq A$ such that $B$ is non-recursive.

(iii) Show that tha subset $A \subseteq \mathbb{N}$ is finite if and only if every subset of $A$ is recursive.

My attemp:

(i) I have consider the function definied by recursion $\mathcal{F}: \mathbb{N} \longrightarrow \mathbb{N}$:
$$\mathcal{F}(x) = \left\{ \begin{array}{l}
\mathcal{F}(0) = \mathcal{F}(1)  = 1\\
\mathcal{F}(n+1) = \mathcal{F}(n+2) + \mathcal{F}(n)\end{array}\right.$$
Then I tried to reduce it to a recursion but I fail because I have "two base cases" and recursion only uses one of them. I found on the web a function called paring function. I think is helpfull because turns two naturals into one and is primitive recursive.

(ii) I think I should go togh contradition. want should I suppose? I have two options:

Let A recursive set. Suppose that for all set $B$, $A \subseteq B$ or $B$ is recursive.

Suppose that exists a set $A$ shuch that:

$A$ is recursive or

exists $B$ such that $B \subsetneq A$ and $B$ is non-recursive.

(iii) Let $ A \subseteq \mathbb{ N} $. We test by double implication:

$ (\Rightarrow) $ Suppose $ A $ is finite, we want to see that every subset of $A $ is recursive. For this we consider an arbitrary subset of $ A $, $ B $. Since again this set is finite, we can consider a numbering with indices $ I = \{1,2, \cdots, n \} $:
$$ B = \{a_{i} \}_{i = 1}^{n} \text{ with } i \in I $$
Notice that its characteristic function given by:
$$ \ chi_{B} (x) = \left \{\begin{array}{r c l}
                1 & \text { if } & x \in B \\
                0 & \text { if } & x \not\in B
            \end{array} \right. $$
Let's see that this function is recursive. For this, it should be noted that:
$$ \chi_{B} (x) = \prod_{i = 1}^{n} \left[\chi_{=} (x \,,\, a_{i}) \right] $$
Since $ \chi_{=} $ and the finite product are recursive functions and the function $ \chi_{B} $ is being considered as a composition of these we can say that $ \chi_{B} $ is recursive. Just as you wanted to see.
$( \Leftarrow) $ To test this direction, the counterpossitive asociated with this direction says:
"If $ A $ is an infinite set then there is a subset $ B $ of $ A $ that is non-recursive "
So, suppose $ A $ is infinite, we want to see that there exists a subset of $ A $ that is not recursive. [Here I die!]

Can you help me, I feel I am very near to the solution!

答:
Yes, the pairing function accomplishes what you need.

Just delete any single element $x_0\in A$ from $A$, and the resulting $B=A-\{x_0\}$ will be non-recursive... otherwise you could construct a decision procedure for $A$ by taking the decision procedure for $B$ and then additionally checking that the number in question is not $x_0.$

The forward direction is much simpler: any subset of a finite set is finite, hence recursive. You are correct that the crux of the other direction is to show that any infinite set $A$ has a non-recursive subset. If $A$ is non-recursive, we're done, so assume it is recursive. Then there is a computable bijection $f:\mathbb N\to A.$ (Since $A$ is recursive, it is recursively enumerable. Run the enumeration algorithm till $n$ distinct elements have come off, then return the $n$-th distinct element.) Then let $X$ be any non-recursive subset of $\mathbb N.$ Then $f(X)$ is a nonrecursive subset of $A,$ since from a decision procedure for $f(X)$ you could get a decision procedure for $X$ as follows: given $n$, compute $f(n)$ and then decide if it is in $f(X)$ (which decides whether $n\in X$ faithfully, since $f$ is a bijection).

Title: Primitive recursive functions and sets


问:
Algorithm making of Die Roll for each pair who throws the same number + Mean and Variance
Each member of a group of $n$ players rolls a die.

Find the Mean and Variance of the total score if any pair of the players who throw the same number of scores as that number.

答:
I edited the question to make it more suitable, so hopefully, this question gets unflagged as an improper question!

There are many things wrong with your question. Remember, this is a math site and not a "code" site. If you want the mathematical answer to this question, then only ask this part: "Find the mean and variance of the total score if any pair of players who throw the same number scores that number." Remember to ask on Stack Overflow for "Create an algorithm and a code in R," but even then show what you've done.

However, the approach for the mean and variance part should be along the lines of:

Let's declare variables for better understanding -

Say,

M$i$ = the score of players who actually throw a die 1, 2, . . . ,6

V$i$ = the number of people who throw those die (i) and 1xᵢⱼ (indicator function) and the indicator function can Iff be 1 if the jᵗʰ person throws $i$.

E = Expectation score.

Based on those rules, we can form a simple series function for that:

V$i$ = $\sum_{j=1}^n = 1xᵢⱼ$

So,

Expectation

$E[Mᵢ][Vᵢ] = \frac{(Vᵢ)(Vᵢ-1)}{2}$, then  $E[Mᵢ] = \frac{(n^2-n)}{12}$

Expectation of 1xᵢⱼ = 1/6 (Die Roll: Probability), so 1/6 * (1 - 1/6) = $\frac{5}{36}$

Therefore,

Expectation:
=> (n :: 2 [Pair]) * 1/6 = $\frac{n(n - 1)}{12}$

E[M] = $\frac{(n^2-n)}{12}$, in which  M = $\sum_{i=1}^6 Mᵢ$ (Total Score)

Variance

Variance => V = (n :: 2) * 5/36 =

$\frac{5n(5n - 1)}{72}$


问:
Show that the integral $\int_{0}^\infty\frac{1}{x\sqrt{1+x^2}}dx$ is divergent
I have some trouble with this integral. Actually, I know when the interval ranges from 1 to $\infty$, this integral is convergent. But the difficulty I have encountered is how to show the divergence where the case of interval ranges from 0 to 1.

Body:
Use $x=\tan t$,
$$\int \frac{dx}{x\sqrt{1+x^2}}= \int_{0}^{\pi/2} \csc t dt =\ln(\tan(x/2))_{0}^{\pi/2}=0-\ln 0=\infty$$


问:
Question on combinatorics and the inclusion-exclusion principle
Find the number of n digit-numbers formed using the first 5 natural numbers, that contain the digits '2' and '4', essentially.

I tried attempting this with the inclusion- exclusion principle but got stuck with the first 5 natural numbers condition. I made 2 sets, one with natural numbers where 2 is there and another with numbers with 5 is there and tried to find their intersection, but got stuck while subtracting from the total set, ie, the set of natural n digit numbers that are made from the digits 1 to 5 (wo any restrictions)
hence i have a doubt in how to find the total number of elements in the set of n digit numbers that are made from the first 5 natural numbers.

Can someone please help me with this?

答: The total number of $n$-digit numbers formed from $12345$ is $5^n$. The count of those numbers not containing a $2$ or $4$ is $4^n$ each, and the count of those not containing either is $3^n$. Thus the required answer is $5^n-2\cdot4^n+3^n$.


问:
the set of countable sets of Real numbers
I would like to ask some hints towards the proof that The set of countable sets in $\mathbb{R}$ is equinumerous to the set $\mathbb{R}$</p>

答:
Hint: $\Bbb{R^N\sim(N^N)^N\sim N^{N\times N}\sim N^N}$. Show that there exists a surjection from that set onto the set of countable subsets and use the axiom of choice to conclude there is an injection in the reverse direction.

Note that the axiom of choice has to be used, it is consistent that the axiom of choice fails and there is no bijection between the two sets (but there is still a surjection as above)!


问:
Proving that the theory is $\aleph_{0}-$categoric, giving and example and going to a particular case
Consider the language $\mathcal{L} = \{<,U\}$, where $<$ is a binary relation $U$ is a unary relation. Let $T$ a $\mathcal{L}-$theory given by:
$$\begin{array}{l r}
    \forall_{x}[\neg(x < x)] & \forall_{x,y}\{\neg [(x < y) \land (y < x)]\}\\
    \forall_{x,y,z}\{[(x < y)\land(y < z)] \longrightarrow (x < z)\} & \forall_{x,y}[(x < y)\lor(x = y)\lor(y < x)]\\
    \exists_{w}\{U(w) \land \forall_{x}[(w < x) \lor (w = x)]\} & \exists_{w}\{\neg U(w) \land \forall_{x}[(x < w) \lor (x = w)]\}\\
\end{array}$$
$$\forall_{x,y}\{ (x < y) \rightarrow \exists_{z_{1},z_{2}}[U(z_{1}) \land \neg U(z_{2}) \land (x < z_{1} < y) \land (x < z_{2} < y)]\}$$

1. Describe a model of $T$ which its domain is a subset of $R$ with the usual order.

2. Show that $T$ is $\aleph_{0}-$categoric (Use back and forth argument).

3. Use the last part to show that $\mathcal{M}_{1} \equiv \mathcal{M}_{2}$ but, $\mathcal{M}_{1} \not\cong \mathcal{M}_{2}$, where:

$$\begin{array}{l r}
        \mathcal{M}_{1} = \left([-\sqrt{2},1];<;U^{\mathcal{M}_{1}} = \mathbb{I} \cap [-\sqrt{2},1]\right) & \mathcal{M}_{2} = \left([0,\sqrt{2}];<;U^{\mathcal{M}_{2}} = \mathbb{Q} \cap [0,\sqrt{2}]\right)
    \end{array}$$
I do not understand very well the staments of the theory I would like that we build a group solution in order to work on it.

答:
The theory can be described succinctly as the theory of a dense linear order with endpoints with a dense subset $U$ whose complement is also dense, and such that the least element is in $U$ and the greatest isn't. They give two examples of models in part three, where the domain is a closed interval reals and $U$ represents either the rationals or irrationals.

The back and forth argument is almost identical to the standard argument for dense linear orders. The only difference is that if an element of one model satisfies $U$, the element from the other model that you pair it with must also satisfy $U$ (and similarly for elements not satisfying $U$). There is no issue making this work since both $U$ and its complement are dense.

The categoricity result (and the fact that there are clearly no finite models) implies completeness, so all models are elementarily equivalent. The fact that they are not isomorphic follows from the fact that the rationals are countable but the irrationals aren't.


问:
Doubt in finding an inverse of an ideal

Background

I'm currently working on problem 8.7 from Alaca & Williams' Intro. Alg. Number Theory. The problem states the following:

Show that $\langle3,1+2\sqrt{-5}\rangle\mid\langle1+2\sqrt{-5}\rangle$ in $\mathbb Z[\sqrt{-5}]$. Determine the integral ideal $A$ such that
$$\langle3,1+2\sqrt{-5}\rangle A=\langle1+2\sqrt{-5}\rangle.$$

My progress

Proving that one ideals divides another is immediate from the fact that to divide is to contain. However finding the ideal $A$ is my problem.

My first thought was using the following exercise which asks for the inverse of the ideal $\langle3,1+2\sqrt{-5}\rangle$. So let's assume that I've found it, right? Suppose $J$ is it's inverse, then
$$J\langle3,1+2\sqrt{-5}\rangle A=A=J\langle1+2\sqrt{-5}\rangle.$$
That sounds like a plan until we try to find $J$.

The problem

I had a really nice hunch with the ideal $\langle3,1-2\sqrt{-5}\rangle$, multiplying we get

\begin{align*}
\langle3,1+2\sqrt{-5}\rangle\langle3,1-2\sqrt{-5}\rangle&=\langle9,3-6\sqrt{-5},3+6\sqrt{-5},10\rangle=\langle 1\rangle,
\end{align*}
since $1=10-9$ is in the product ideal. Therefore since $1$ is in the ideal, it is the whole ring and we conclude that both ideals are inverses.

However, I found a worksheet which claims that the inverse ideal is $\frac13\langle3,1+\sqrt{-5}\rangle$ and by multiplying these ideals we get
$$\langle3\rangle\langle3,1+\sqrt{-5},1+2\sqrt{-5},-3+\sqrt{-5}\rangle.$$
We see that the second ideal is trivial since
$$1+\sqrt{-5}-3-(-3+\sqrt{-5})=1$$
and therefore the product equals a principal ideal, and by a known result (Example 2.8 in K. Conrad's notes) we see that the fractional ideal is an inverse as well.

My questions

Is my first approach correct? Solving the equation by finding the inverse?

The two ideals $\langle3,1-2\sqrt{-5}\rangle$ and $\frac13\langle3,1+\sqrt{-5}\rangle$ seem to be inverses to the ideal in the question. However, one is fractional and the other isn't. What's happening here? Unique factorization of ideals should guarantee that inverses are unique, shouldn't it?

答:
$A$ has norm $21/3 = 7$ so it contains $7$ (it is one of the two prime ideals above $7$).

The inverse of $(3,1+2\sqrt{-5})$ is $\frac13(3,1-2\sqrt{-5})$, and $\langle3,1+2\sqrt{-5}\rangle\langle3,1-2\sqrt{-5}\rangle=(3)$,

as it is $\sqrt{-5}$ not $\sqrt5$.

Title: Doubt in finding an inverse of an ideal


问:
Definition of Tensor Product of V-spaces
I'm reading a book on tensors and it says the following:

I'm still a bit confused on Definition 1.1
I never really learned what a free group was. I was just basically told to think of it as a group on letters. Anyways I'd like to understand what a quotient of a free abelian group is and also maybe a specific example or two. Someone told me that the definition is really never used and is just for technical purposes and to just forget about it. What does it mean that it is generated by elements of that form? I noticed that if I set the right hand side to 0 then we basically have bilinearity. I think an example or two or three could illuminate this.

Thanks

答:
The free abelian group generated by some set $A$ consists of formal expressions of the form

$$ \sum_{i=1}^n m_i a_i $$
where the $m_i$ are integers, and $a_i \in A$ (and $n$ can be any positive integer). Think of it like a vector space, where $A$ is the basis, but instead of having real number coefficients, you are only allowed integer coefficients.

For example, if $A = \{a,b\}$ only has two elements, then every element of the free abelian group generated by $A$ is of the form $ma + nb$ for integers $m$ and $n$.

For another example (but where $A$ is infinite), consider $A = \{1,x,x^2,x^3,\dots\}$, thinking of "$x^k$" as formal symbols. Then the free abelian group generated by $A$ is the set of polynomials with integer coefficients, since any polynomial is a (finite) sum of terms which look like $c \, x^k$ for some integer $c$.

The example for the tensor product considers the free abelian group generated by $M \times N$. This means a generic element looks like
$$ \sum_{i=1}^n k_i \, (m_i,n_i) $$
where $n$ is some positive integer, $k_i$ are integers, and $m_i \in M$, $n_i \in N$. Again, using the vector space analogy, it's like a vector space (with integer coefficients), where the whole infinite set $M \times N$ is the basis.

Now, a "quotient" of a free abelian group informally means you take certain elements and declare them to be equal to zero.

For example, when $A = \{a,b\}$, if you quotient by the subgroup generated by $a-b$, that means you are setting $a-b=0$, which would imply $a=b$. So in the quotient, since $a=b$, all elements can just be expressed as $m a$ for some integer $m$.

In the tensor product example, the elements of the form $(i)$, $(ii)$, and $(iii)$ are just ensuring that in the quotient, you have the relations:
$$
\begin {align*}
(m_1+m_2,n) &= (m_1,n) + (m_2,n) \\
(m,n_1+n_2) &= (m,n_1) + (m,n_2) \\
(rm,n) &= (m,rn)
\end {align*}
$$


问:
Identity involving Differential Operators
I am trying to prove the following

Let $F$, $\Phi$ and $\phi$ be arbitrarily differentiable functions of $x$. I have come across the following identity in some literature I am reading which I need help in proving:

$$\frac{1}{\phi^{r-1}}D^{r-1}\left[\frac{(F-\Phi)^r}{\phi}\right]=D_{(r-1)}\left[\left(\frac{F-\Phi}{\phi}\right)^{r}\right],\qquad r=1,2,\ldots$$

where $\displaystyle D^r=\frac{d^r}{dx^r}$ and
$$D_{(r)}=\left(D-\frac{\phi'}{\phi}\right)\left(D-2\frac{\phi'}{\phi}\right)\cdots\left(D-r\frac{\phi'}{\phi}\right)$$
and $D_0$ is just the identity.

I can see this for the case $r=2$:
$$\begin{align*}
\frac{1}{\phi}D\left[\frac{(F-\Phi)^{2}}{\phi}\right]&=\frac{1}{\phi}\left[\frac{\phi D(F-\Phi)^{2}-(F-\Phi)^{2}\phi'}{\phi^{2}}\right]\\
&=\left(D-\frac{\phi'}{\phi}\right)\left[\frac{(F-\Phi)^{2}}{\phi^{2}}\right]\\
&=D_{(1)}\left[\left(\frac{F-\Phi}{\phi}\right)^{2}\right]\end{align*} $$

Its just an application of the quotient rule and some algebra. But how would I prove this for an arbitrary positive integer $r$?

答:
HINT $\rm\displaystyle\ \ \ \ D + n\ \frac{f'}f\ =\:\ f^{\:-n}\ D\ f^{\:n} $

Below are some papers on general techniques.

Steven Roman, Operational Formulas, Lin. Multilin. Algebra, 1982, 1-20.

L. Carlitz, A theorem on differential operators, Amer. Math. Monthly, 83(5) 1976, 351-354.


问:
Finding Conditional Density of $V|U$
$X$ and $Y$ are independent standard normal random variables. Let $U=X$ and $V=cX+Y\sqrt{1-c^2}$, where $c$ is some constant. Find the conditional density of $V|U$.

To find the conditional density, I first find the joint density. I have $f_{U, V}(u,v)=\frac{1}{2\pi}e^{-\frac{u^{2}}{2}}\cdot e^{-\frac{1}{2}\left(\frac{v-cu}{\sqrt{1-c^{2}}}\right)^{2}}\cdot\frac{1}{\sqrt{1-c^{2}}}$ for the joint density with $\frac{1}{\sqrt{1-c^{2}}}$ as the Jacobian. To find the marginal density of $U$, I have: $\int_{-\infty}^{\infty}f_{U,V}\left(u,v\right)dv$. But I do not know how to simplify this expression. Am I on the right track? If so, how do I find the marginal density?

答:
how do I find the marginal density?

Really? $U=X$ thus $U\sim N(0;1)$

Thus it is immediate to derive that

$$f_{V|U}(v|u)\sim N(cu;1-c^2)$$

Title: Finding Conditional Density of $V|U$


问:
Product of increasing functions is integrable in two dimensions

We say $f:[0,1]\rightarrow\mathbb{R}$ is increasing if $f(x_1)\le f(x_2)$ whenever $x_1<x_2$. If $f,g:[0,1]\rightarrow \mathbb{R}$ are increasing and non-negative, show that the function $h(x,y)=f(x)g(y)$ is integrable over $[0,1]^2$.

I'm using the Darboux definition of integration, so I want to prove that for any $\epsilon>0$, there exists a partition $P$ of $Q$ such that $U(f,P)-L(f,P)<\epsilon$. Equivalently, there exists a partition $P$ of $Q$ such that $$\left|\sum_Rv(R)(M_R(f)-m_R(f))\right|<\epsilon,$$ where $M_R(f)$ is the supremum of the values of $f$ inside the rectangle $R$, and $m_R(f)$ is the corresponding value for infimum. (Here, $R$ ranges over all subrectangles formed by the partition $P$.)

Toward that end, I tried to take the partition $P$ to be $[0,\dfrac1n,\dfrac2n,\ldots,1]\times[0,\dfrac1n,\dfrac2n,\ldots,1]$. Since the function $h(x,y)$ is increasing in $x$ and $y$, the maximum of any square is attained at the top-right corner, and the minimum at the bottom-left corner. Therefore, $\left|\sum_Rv(R)(M_R(f)-m_R(f))\right|$ is equal to $$\dfrac{1}{n^2}\left(\sum_{i=1}^{n}f(\dfrac in)g(1)+\sum_{i=1}^{n-1}f(1)g(\dfrac in)-\sum_{i=0}^{n-1}f(\dfrac in)g(0)-\sum_{i=1}^{n-1}f(0)g(\dfrac in)\right)$$

I'm not sure if this is the right partition to take. Does this sum approach $0$ as $n\rightarrow\infty$?

Body:
$$\sum_{R} M_R(f)v(R)=\sum_{1\leq i\leq n\\1\leq j\leq n}f(\frac{i}{n})g(\frac{j}{n})\frac{1}{n^2}\leq\sum_{0\leq i\leq n\\0\leq j\leq n}f(\frac{i}{n})g(\frac{j}{n})\frac{1}{n^2}.$$

$$\sum_{R} m_R(f)v(R)=\sum_{0\leq i\leq n-1\\0\leq j\leq n-1}f(\frac{i}{n})g(\frac{j}{n})\frac{1}{n^2}.$$

$$\sum_{R} M_R(f)v(R)-\sum_{R} m_R(f)v(R)\leq\sum_{0\leq i\leq n\\0\leq j\leq n}f(\frac{i}{n})g(\frac{j}{n})\frac{1}{n^2}-\sum_{0\leq i\leq n-1\\0\leq j\leq n-1}f(\frac{i}{n})g(\frac{j}{n})\frac{1}{n^2}\\=\frac{1}{n^2}(f(\frac{0}{n})g(\frac{n}{n})+f(\frac{1}{n})g(\frac{n}{n})+\dots+f(\frac{n}{n})g(\frac{n}{n})+f(\frac{n}{n})g(\frac{n-1}{n})+\dots+f(\frac{n}{n})g(\frac{0}{n}))\leq\frac{1}{n^2}(2n+1)f(1)g(1)\to 0(n\to\infty).$$


问:
What's the difference between arccos(x) and sec(x)
My question might sound dumb, but I don't really see why the graphics of $\arccos(x)$ and $\sec(x)$ are different.

As far as I know $\arccos(x)$ is the inverse cosine function $(\cos(x)^{-1})$ and $\sec(x)$ = $\frac{1}{\cos(x)}$ (source Wolfram|Alpha).

So why aren't they equal?

Thanks in advance.

答:
The notation $\cos^{-1}(x)$ means the same thing as $\arccos(x)$. If you wanted to talk about $\sec(x)$, which is $1/\cos(x)$, you would write $(\cos(x))^{-1}$.

Yes, this is very confusing - it's inconsistent with the common use of
$$\cos^n(x)$$
to mean $(\cos(x))^n$ when $n\geq 2$; it is just a historical accident that notational conventions turned out this way. All the more reason to just never ever write "$\cos^{-1}(x)$" and use one of the options that is unambiguous: $\arccos(x)$ if you want to say something like $$\cos(\arccos(x))=x,$$
and either $\sec(x)$ or $\dfrac{1}{\cos(x)}$ if you want to say something like
$$\sec(x)\cdot \cos(x)=1\quad\text{when }\cos(x)\neq 0.$$


问:
Schur's lemma and $\mathbb{Z}_2$
Representation theory is painfully new to me, but I am trying to understand simple examples of Schur's (second) lemma. My understanding of Schur's lemma is that, for a given complex and irreducible matrix representation $D$, if a matrix $Q$ commutes with $D(g)$ for every group element $g$, then the matrix $Q$ is proportional to the identity.

In what follows, I think my confusion may be because I am over-simplifying Schur's lemma. Consider the group $\mathbb{Z}_2$, with the irreducible representation $D: \mathbb{Z}_2 \to GL(2, \mathbb{C})$:

$$D(e) = \begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}, \,\, D(p) = \begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}$$

It feels like I can construct a counterexample to Schur's lemma: the matrix $D(p)$ commutes with $D(e)$ and $D(p)$, but it is not proportional to the identity.

Why isn't the above a counterexample? What am I misunderstanding? Heightening my confusion, this website (in example 1) even claims that that the only matrix commuting with $D(e)$ and $D(p)$ is proportional to the identity, but that seems false to me given that $D(p)$ commutes with both $D(e)$ and $D(p)$.

答: Your representation $D$ is not irreducible!  Indeed, you can see this by just carrying out the proof of Schur's lemma for it: any eigenspace of $D(p)$ will be a subrepresentation of $D$.  Explicitly, $D(p)$ has eigenvalues $1$ and $-1$ with eigenvectors $(1,1)$ and $(1,-1)$, respectively, so the spans of each of these eigenvectors are subrepresentations.

Title: Schur's lemma and $\mathbb{Z}_2$


问:
Does $1-ab\geq (1-a)(1-b)$ hold, for $a,b\in [0,1]$?
Does this inequality hold, for any $a,b\in[0,1]$

$$1-ab\geq (1-a)(1-b)?$$

I'm don't have idea to conclude $1-ab\geq 1-a-b+ab = (1-a)(1-b)$.

Anyone can prove (or disprove) it?

答:
I approached this question geometrically.

Desmos for fun

Start with a square of side length 1. Divide the square along one side in the ratio $a:(1-a)$, and the other side $b:(1-b)$. By considering the relevant areas given by $(1-ab)$ and $(1-a)(1-b)$, it should be obvious the inequality is true.


问:
Independence of connectives in intuitionistic logic
Consider the intuitionistic propositional logic with $\neg, \vee, \land, \to$ as primitive connectives. My question is, can any three of these connectives be used to define the remaining one? So, I am asking for four proofs that the remaining connective can or can't be defined in terms of the other three.

答:
No three intuitionistic connectives can be used to define the fourth one. McKinsey [1] showed this using nigh-trivial semantic proofs: in modern terminology, we would say that he constructs Heyting algebras as counterexamples. Despite its age, the article remains very readable, and answers your question thoroughly.

Here I'll present some proof-theoretic arguments which make use of cut-elimination, and show that in particular disjunction and negation are independent of the other connectives. Unlike semantic proofs, these generalize straightforwardly to the first-order case (in fact, note that the independence of the quantifiers $\forall$ and $\exists$ follows from my answer to your previous, related  question). For the other two connectives (conjunction, implication) I present a variant of McKinsey's proofs: while the independence of conjunction and implication can be shown proof-theoretically as well, the methods I know require quite a bit of symbol-pushing.

We will use the following property of the intuitionistic propositional calculus, which you can prove by induction on (cut-free) derivations (see also [2], Theorem 4.2.4).

Disjunction property: if $\Gamma$ of does not contain any disjunction connectives, and $\Gamma \vdash M \vee N$, then we can find a proof of either $\Gamma \vdash M$ or $\Gamma \vdash N$.

I. Disjunction $\vee$ is independent of the other connectives $\neg, \rightarrow, \wedge$:

Suppose that we can find a formula $Q$ in the the $\{\neg, \rightarrow, \wedge\}$ fragment such that $Q \vdash M \vee N$ and $M \vee N \vdash Q$ holds for the atomic propositions $M,N$.

Since $Q \vdash M \vee N$ and $Q$ does not contain disjunctions, we have either $Q\vdash M$ or $Q\vdash N$ by the disjunction property.

But of course the same fails for $M \vee N$: neither $M \vee N \vdash M$ nor $M \vee N \vdash N$ are provable for atomic $M,N$. (Why? Cut $M \vee N \vdash M$ against the usual proof of $N \vdash M \vee N$ to obtain a proof of $N \vdash M$. By cut-elimination this would also have a cut-free proof, but that's impossible: what would be the last step?!)

Hence the supposed definition $Q$ cannot exist.

II. Negation $\neg$ is independent of the other connectives $\rightarrow, \vee, \wedge$:

The connectives $\{\rightarrow, \vee, \wedge\}$ build provable formulas from provable propositional atoms: if $\vdash A$ and $\vdash B$, then for any formula $Q$ made up of $A,B$ and the connectives $\{\rightarrow, \vee, \wedge\}$ we have $\vdash Q$. However, $\neg$ can be used to build unprovable formulae from provable atoms: if $\vdash A$ and $\vdash \neg A$ both hold, then we can extract a proof of $A \vdash$ (by looking at the last step of a cut-free proof of $\vdash \neg A$), so by cut we'd have a proof of the empty sequent $\vdash$. But that is impossible (what would be the last step of its cut-free proof?), so if $A$ is provable then $\neg A$ is not provable. Hence we cannot express $\neg$ in the $\{\rightarrow, \vee, \wedge\}$ fragment.

The set of numbers $\{1,2,3,4,6,9,12,18,36\}$ forms a Heyting algebra when equipped with the operations $\gcd$ (greatest common divisor, interpreting conjunction), $\mathrm{lcm}$ (least common multiple, interpreting disjunction), and the following implication operation $\Rightarrow$:

```<code>=>|  1  2  3  4  6  9 12 18 36
--+---------------------------
1 | 36 36 36 36 36 36 36 36 36
2 |  9 36  9 36 36  9 36 36 36
3 |  4  4 36  4 36 36 36 36 36
4 |  9 18  9 36 18  9 36 18 36
6 |  1  4  9  4 36  9 36 36 36
9 |  4  4 12  4 12 36 12 36 36
12|  1  2  9  4 18  9 36 18 36
18|  1  4  3  4 12  9 12 36 36
36|  1  2  3  4  6  9 12 18 36
</code>```

All you need to check is that if $z$ divides $x \Rightarrow y$ then $\gcd(z,x)$ divides $y$, and vice versa (admittedly, this is the tedious/annoying part of the semantic method, but you can cut this particular verification short using prime factors).

III. Conjunction $\wedge$ is independent of the other connectives $\neg, \vee, \rightarrow$.

The set $H = \{1,12,18,36\}$ is clearly closed under least common multiples. Referring to the table above, we can check that it is closed under $\Rightarrow$ as well. Since $1 \in H$, it follows that $H$ is closed under $\neg$ too. So, if conjunction was definable in the fragment $\{ \neg, \vee, \rightarrow \}$, then our set would be closed under greatest common divisors as well. But $\gcd(12,18) = 6 \not\in H$.

IV. Implication is independent of the other connectives $\neg, \vee, \wedge$.

Take the Heyting algebra above. Clearly the set $S = \{1,6,12,36\}$ is closed under both $\gcd$ and $\mathrm{lcm}$. Moreover, it is closed under negation, since $1 \Rightarrow 1 = 36$ and $x \Rightarrow 1 = 1$ for any $x > 1$ in $S$. So if implication was definable in the $\{ \neg, \vee, \wedge \}$ fragment, it would preserve membership in $S$. But $12 \Rightarrow 6 = 18 \not\in S$.

[1] J.C.C. McKinsey. Proof of the independence of the primitive symbols of Heyting’s calculus of propositions. The Journal of Symbolic Logic, 4(04), 155–158., 1939. doi:10.2307/2268715

[2] A. S. Troelstra, H. Schwichtenberg: Basic Proof Theory, 2nd edition, 2000. ISBN 9781139168717

edit Simplified the wording of some of the arguments, added links and references

Body:
No three intuitionistic connectives can be used to define the fourth one. McKinsey [1] showed this using nigh-trivial semantic proofs: in modern terminology, we would say that he constructs Heyting algebras as counterexamples. Despite its age, the article remains very readable, and answers your question thoroughly.

Here I'll present some proof-theoretic arguments which make use of cut-elimination, and show that in particular disjunction and negation are independent of the other connectives. Unlike semantic proofs, these generalize straightforwardly to the first-order case (in fact, note that the independence of the quantifiers $\forall$ and $\exists$ follows from my answer to your previous, related  question). For the other two connectives (conjunction, implication) I present a variant of McKinsey's proofs: while the independence of conjunction and implication can be shown proof-theoretically as well, the methods I know require quite a bit of symbol-pushing.

We will use the following property of the intuitionistic propositional calculus, which you can prove by induction on (cut-free) derivations (see also [2], Theorem 4.2.4).

Disjunction property: if $\Gamma$ of does not contain any disjunction connectives, and $\Gamma \vdash M \vee N$, then we can find a proof of either $\Gamma \vdash M$ or $\Gamma \vdash N$.

I. Disjunction $\vee$ is independent of the other connectives $\neg, \rightarrow, \wedge$:

Suppose that we can find a formula $Q$ in the the $\{\neg, \rightarrow, \wedge\}$ fragment such that $Q \vdash M \vee N$ and $M \vee N \vdash Q$ holds for the atomic propositions $M,N$.

Since $Q \vdash M \vee N$ and $Q$ does not contain disjunctions, we have either $Q\vdash M$ or $Q\vdash N$ by the disjunction property.

But of course the same fails for $M \vee N$: neither $M \vee N \vdash M$ nor $M \vee N \vdash N$ are provable for atomic $M,N$. (Why? Cut $M \vee N \vdash M$ against the usual proof of $N \vdash M \vee N$ to obtain a proof of $N \vdash M$. By cut-elimination this would also have a cut-free proof, but that's impossible: what would be the last step?!)

Hence the supposed definition $Q$ cannot exist.

II. Negation $\neg$ is independent of the other connectives $\rightarrow, \vee, \wedge$:

The connectives $\{\rightarrow, \vee, \wedge\}$ build provable formulas from provable propositional atoms: if $\vdash A$ and $\vdash B$, then for any formula $Q$ made up of $A,B$ and the connectives $\{\rightarrow, \vee, \wedge\}$ we have $\vdash Q$. However, $\neg$ can be used to build unprovable formulae from provable atoms: if $\vdash A$ and $\vdash \neg A$ both hold, then we can extract a proof of $A \vdash$ (by looking at the last step of a cut-free proof of $\vdash \neg A$), so by cut we'd have a proof of the empty sequent $\vdash$. But that is impossible (what would be the last step of its cut-free proof?), so if $A$ is provable then $\neg A$ is not provable. Hence we cannot express $\neg$ in the $\{\rightarrow, \vee, \wedge\}$ fragment.

The set of numbers $\{1,2,3,4,6,9,12,18,36\}$ forms a Heyting algebra when equipped with the operations $\gcd$ (greatest common divisor, interpreting conjunction), $\mathrm{lcm}$ (least common multiple, interpreting disjunction), and the following implication operation $\Rightarrow$:

=>|  1  2  3  4  6  9 12 18 36
--+---------------------------
1 | 36 36 36 36 36 36 36 36 36
2 |  9 36  9 36 36  9 36 36 36
3 |  4  4 36  4 36 36 36 36 36
4 |  9 18  9 36 18  9 36 18 36
6 |  1  4  9  4 36  9 36 36 36
9 |  4  4 12  4 12 36 12 36 36
12|  1  2  9  4 18  9 36 18 36
18|  1  4  3  4 12  9 12 36 36
36|  1  2  3  4  6  9 12 18 36

All you need to check is that if $z$ divides $x \Rightarrow y$ then $\gcd(z,x)$ divides $y$, and vice versa (admittedly, this is the tedious/annoying part of the semantic method, but you can cut this particular verification short using prime factors).

III. Conjunction $\wedge$ is independent of the other connectives $\neg, \vee, \rightarrow$.

The set $H = \{1,12,18,36\}$ is clearly closed under least common multiples. Referring to the table above, we can check that it is closed under $\Rightarrow$ as well. Since $1 \in H$, it follows that $H$ is closed under $\neg$ too. So, if conjunction was definable in the fragment $\{ \neg, \vee, \rightarrow \}$, then our set would be closed under greatest common divisors as well. But $\gcd(12,18) = 6 \not\in H$.

IV. Implication is independent of the other connectives $\neg, \vee, \wedge$.

Take the Heyting algebra above. Clearly the set $S = \{1,6,12,36\}$ is closed under both $\gcd$ and $\mathrm{lcm}$. Moreover, it is closed under negation, since $1 \Rightarrow 1 = 36$ and $x \Rightarrow 1 = 1$ for any $x > 1$ in $S$. So if implication was definable in the $\{ \neg, \vee, \wedge \}$ fragment, it would preserve membership in $S$. But $12 \Rightarrow 6 = 18 \not\in S$.

[1] J.C.C. McKinsey. Proof of the independence of the primitive symbols of Heyting’s calculus of propositions. The Journal of Symbolic Logic, 4(04), 155–158., 1939. doi:10.2307/2268715

[2] A. S. Troelstra, H. Schwichtenberg: Basic Proof Theory, 2nd edition, 2000. ISBN 9781139168717

edit Simplified the wording of some of the arguments, added links and references


答: The notation $\cos^{-1}(x)$ is very confusing. it does not mean $(\cos(x))^{-1}=1/\cos(x)$, but rather $\arccos(x)$, which is quite a different function!

Title: What's the difference between arccos(x) and sec(x)

Body:
My question might sound dumb, but I don't really see why the graphics of $\arccos(x)$ and $\sec(x)$ are different.

As far as I know $\arccos(x)$ is the inverse cosine function $(\cos(x)^{-1})$ and $\sec(x)$ = $\frac{1}{\cos(x)}$ (source Wolfram|Alpha).

So why aren't they equal?

Thanks in advance.


问:
Why Jacobson radical is 'radical'?
Nilradical of ring R, $\text{nil}(R)$ is intersection of prime ideals, is radical of $(0)$.
So I feel the name  'radical' is natural.

But Jacobson radical, that is an intersection of maximal ideals, seem not to be radical of a certain ideal.
Why Jacobson radical is nevertheless called 'radical'? Thank you.

答: The term "radical" here doesn't refer to the radical of some ideal, but to the general concept of a "radical of a ring" which is in some way "a set of bad elements". For nilradical, "bad" means nilpotent; for Jacobson radical, "bad" means annihilating all simple left modules.</p>

Title: Why Jacobson radical is 'radical'?


问:
How do we know that $\{x: x\text{ is a set and }x\text{ has at least one element}\}$ has any elements at all?
I've been going through introductory set theory notes and came across the following.

First, it is possible for a set to be an element of itself. An example of a set which is an element of itself is
$$A =\{x: x\text{ is a set and }x\text{ has at least one element}\}$$
This set contains itself, because it is a set with at least one element.

How do we know the set $A$ has at least one element? Could someone help understand this

答:
$A$ contains the set $\{\varnothing\}$, so it has at least one element and thereby contains itself.

However, note that $A$ is defined by an unrestricted comprehension, which is not allowed in ZFC. It becomes a proper class.