问:
name or characterization for the “partition” lattice of integer partitions of some n? (Young lattice row partial order)
Is there a name or characterization for the "partition" lattice of integer partitions of some n?

Young's Lattice depicts the integer partitions of numbers. Often Young diagrams are used in displaying the lattice as in this picture of  Young's Lattice up to n=7. In this Hasse diagram an arrow mean adding 1 box in a connected position.

I'm not asking about this lattice's partial order, but about the partial order that can be imposed  within each (equal rank) row of it.

For example, for $n=6$ the 11 row elements could be encoded as

6, 51, 42, 33, 411, 321, 222, 3111, 2211, 21111, 111111

These elements can be partially ordered. In the Hasse diagram of this order, an arrow means joining two partitions. Eg. since $1+1 = 2$ there is an arrow $411\to 42$. Likewise there is one $411\to 51$.

Here are diagrams I drew for partition lattices of integer partitions for 1 through 7 .

I'm guessing somebody has studied these lattices but I can't guess the right terminology to search for information or references or find that there is some other way of generating them.

These lattices come up in characterizing the monoid of endofunctions (a category) on an $n$ element set, $End(n)$. There is a map from each endofunction $f$ of $End(n)$ to the integer partition lattice $IP(n)$, $ip: End(n) \to IP(n)$. For function composition we have $ip(fg) = ip(gf)$ and  $ip(fg) \ge ip(f)\vee ip(g)$.

(why are there no available tags for words I've bolded in this post?)

答:
As a comment already noticed, your structure is not a lattice in the most commonly used mathematical sense.

There is however a lattice defined on the partition of an integer, introduced in 1973 by Brylawski, and extended into an infinite lattice of all integer partitions in our paper entitled The Lattice of integer partitions and its infinite extension. It has nice connections to the Young lattice.

For $n=7$, we obtain:

See this MO post for more.


问:
Why principal ideal ring has Krull dimension at most 1?
We know PID which is not has a Krull dimension at most one.
But can we say the same thing with PIR（which is not necessarily integral domain）？

We cannot say in PIR that every nonzero prime ideal is maximal, so I wonder we can say the statement.

答:
We can reduce to the case of a PID by modding out as follows. Let $R$ be a PIR and suppose $\newcommand{\q}{\mathfrak{q}} \q_0 \subsetneq \q_1 \subseteq \q_2$ is a chain of prime ideals. Then $R/\q_0$ is a PID, and we have the induced chain of primes $0 \subsetneq \q_1/\q_0 \subseteq \q_2/\q_0$. As you pointed out, nonzero primes are maximal in a PID, so $\q_1/\q_0 = \q_2/\q_0$.

Since $R$ is a PIR we have $\q_1 = (\pi_1)$ and $\q_2 = (\pi_2)$ for some $\pi_1, \pi_2 \in R$. By the above we have $(\overline{\pi_1}) = (\overline{\pi_2})$, where the bar denotes the image in $R/\q_0$. Then $\overline{\pi_2} = \overline{r \pi_1}$ for some $r$ in $R$, so $\pi_2 - r \pi_1 \in \q_0$. Thus $\pi_2 \in (\pi_1) + \q_0 \subseteq (\pi_1)$, and hence $\q_1 = (\pi_1) = (\pi_2) = \q_2$.

ContentLicense: CC BY-SA 4.0


问:
The common difference is equal to the common ratio.
Four numbers are in A.P. The first, the second and the fourth are in G.P. Find the numbers if the common difference is equal to the common ratio.

Let the terms of the A.P. be $a_1,a_1+d,a_1+2d,a_1+3d$ and the terms of the G.P. will be $a_1,a_1+d,a_1+3d.$ We know that $d=q.$ I am not sure what to do from here.

答:
Write $\dfrac {a_1+d}{a_1} = \dfrac{a_1+3d}{a_1+d} = d$.

Now we also have $\dfrac {d}{a_1} = \dfrac {2d}{2a_1} =\dfrac{2d}{a_1+d} = d-1$.

This gives $a_1 = d$, and from $\dfrac {a_1+d}{a_1}=d$ we have $a_1=d=2$.


问:
Why are rotation matrices represented by the unit sphere?
I have heard two statements:

"Although rotation matrices have 9D scalars, they only require 3D elements".

Rotation matrices are represented by the unit sphere.

Although I understand point 1), in the sense that the 3 unitary and 3 orthogonality constraints mean only 3 more scalars are needed to solve for the 9 scalars... what does the space of these 3 scalars look like geometrically?

Would it represent the unit sphere? Why or why not?

If not, how else can rotation matrices be thought of as the "unit sphere", especially considering rotation matrices are 9D matrices and not 3D points.

答: The spin group of Wikipedia rotation paragraph explains well the link between  $\text{SO}(3)$ the group of 3D rotations or special orthogonal group and the sphere.


问:
How to solve this integral in terms of $c$?
Let $c>0$, is there a "closed form" solution of the integral

$$
\int_0^1 \exp\left(-\frac{1}{cx(1-x)}\right) dx
$$

or a special function (of $c$) giving the solution of the integral above? So far, I've only managed to obtain some upper bounds, for example by
$$
\int_0^1 \frac{1}{x^2}\exp\left(-\frac{1}{cx}\right) dx = c (1-e^{-1/c}),
$$
or lower bounds, for example by
$$
\int_0^{1/2}\exp\left( -\frac{1}{cx^2}\right)dx +\int_{1/2}^1 \exp\left( -\frac{1}{c(1-x)^2}\right)dx=2a\left(
\frac{\text{erf}(2/\sqrt{c})-1}{\sqrt{c}}
\right)+e^{-4/c}
$$
where $a=1.77245$ and $\text{erf}(\cdot)$ is the Gauss error function, which might be not so sharp. Any ideas/known results?

EDIT: my primary interest is to determine the rate at which the integral converges to $1$ as $c\to \infty$, thus also approximate solutions/sharp bounds would be appreciated.

答:
With the substitution $x=\frac12(1+\tanh t)$, we obtain $$\int^1_0 e^{-\frac1{c\,x\,(1-x)}}\,dx=\frac12\,\int^\infty_{-\infty}\frac1{\cosh^2 t}\,e^{-\frac4c\,\cosh^2 t}\,dt=\int^\infty_0\frac1{\cosh^2 t}\,e^{-\frac4c\,\cosh^2 t}\,dt,$$ so let's investigate
$$I(x)=\int^\infty_0\frac1{\cosh^2 t}\,e^{-x\,\cosh^2 t}\,dt.$$
Since $$\frac{d}{dt}\frac1{\cosh t}e^{-x\,\cosh^2t}=\left(-\frac{\sinh t}{\cosh^2t}-2x\sinh t\right)e^{-x\,\cosh^2t},$$ we have
$$\int^\infty_0\left(-\frac{\sinh^2 t}{\cosh^2t}-2x\sinh^2 t\right)e^{-x\,\cosh^2t}\,dt=\int^\infty_0\sinh t\,\frac{d}{dt}\frac1{\cosh t}e^{-x\,\cosh^2t}\,dt=-\int^\infty_0e^{-x\,\cosh^2t}\,dt$$ by partial integration. Now we know that
$$1-\frac{\sinh^2 t}{\cosh^2t}=\frac1{\cosh^2t},$$ this means
$$\int^\infty_0\frac1{\cosh^2 t}\,e^{-x\,\cosh^2 t}\,dt=2x\int^\infty_0\sinh^2t\,e^{-x\,\cosh^2 t}\,dt.$$ We make use of the well-known identities
$$\sinh^2t=\frac12(\cosh2t-1)$$ and $$\cosh^2t=\frac12(\cosh2t+1)$$ to get
$$I(x)=x\,e^{-x/2}\int^\infty_0(\cosh2t-1)\,e^{-x/2\,\cosh2t}\,dt.$$ Substituting $t\to t/2$ and taking into account the integral representation $$K_\nu(z)=\int^\infty_0e^{-z\cosh t}\,\cosh(\nu t)\,dt$$ (this is formula 10.32.9 in https://dlmf.nist.gov/10.32), we finally arrive at

$$I(x)=\frac{x}2\,e^{-x/2}(K_1(x/2)-K_0(x/2)).$$
As this almost looked too good to be true, I checked it numerically, with $x=4/c$ and various values of $c$ between $0.1$ and $10$, and found perfect agreement.


问:
If $\vec A=6z\hat i+(2x+y)\hat j-x\hat k$ evaluate $\iint_S \vec A\cdot \hat n\,dS$

Suppose $\vec A=6z\hat i+(2x+y)\hat j-x\hat k$. Evaluate $\iint_S \vec A\cdot\hat n\,dS$ over the entire surface $S$ of the region bounded by the cylinder: $x^2+z^2=9,x=0,y=0,z=0$ and $y=8$. Here $\hat n$ is unit normal vector to surface $S$.

I got this question in Schaum's outline of Vector Analysis. Answer is given $18\pi$. But I don't understand on which plane I would make projection of the surface.

答:
If you can use divergence theorem, that is more straightforward.

Find divergence of the vector field using $\ \nabla \cdot \vec{A}$ and integrate over the cylinder volume in the first octant. Applying divergence theorem will give you flux through the entire closed surface.

In this case, you realize that divergence is simply $1$ so you are left with finding volume of the cylinder in the first octant. It may not even require an integral.

But if you have to do double integral then

Parametrize your surface as $ \ \vec{r}(y, \theta) = (3 \cos \theta, y, 3 \sin \theta)$

$\vec{r'}_{\theta} \times \vec{r'}_y = (3 \cos \theta, 0, 3 \sin \theta)$

So your integral to find flux through cylindrical surface is

$\displaystyle \int_0^{\pi/2} \int_0^8 \vec{A}{(r(y,\theta))} \cdot (3 \cos \theta, 0, 3 \sin \theta) \ dy \ d\theta$

Then you also need to consider planes $x = 0, z = 0, y = 0, y = 8$ with normal vectors as $(-1, 0, 0), (0, 0, -1), (0, -1, 0), (0, 1, 0) \ $ respectively as the question seeks flux through the entire surface.


问:
Simulating a mixed Poisson gives me fewer events than expected.
Referencing section 5.4.3 of the book, Introduction to Probability Models by Sheldon Ross (10th edition), the Mixed Poisson is introduced where the rate, $\lambda$ of a Poisson process is itself drawn from a Gamma distribution. Let this Gamma distribution we're drawing $\lambda$ from be described by $L$. On page 352 we get:

$$E(N(t)|L) = Lt$$

And so we get the intuitive result:

$$E(N(t)) = E(E(N(t)|L)) = E(L)t$$

I'm trying to replicate this with simulation. Let's start with a simple Poisson process. We use Python to simulate inter-arrival exponential distributions and stop when a total time of 1000 units has elapsed. When the rate, $\lambda=5$ per the code below, we get about 5000 events once 1000 units of time have elapsed. No surprises here.

```<code>import numpy as np
period=1000
evnts = 0
t=0
lm = 5
while t<period:
    t_del = np.random.exponential(1/lm)
    t+=t_del
    evnts+=1
print(evnts)
</code>```

Now, I want to extend this simulation to a mixed Poisson as described in Ross. Instead of simulating from a fixed exponential every time, I first simulate the $\lambda$ parameter from a Gamma distribution. I confirmed that the Gamma used does have a mean of 5. I then use that $\lambda$ to simulate an exponential distribution. And the number of events in 1000 units of time drops to about 3000 instead of 5000 as expected from the equation above. What am I missing? Why should the number of events be smaller?

```<code>import numpy as np
period=1000
evnts = 0
t=0
while t<period:
    theta = 0.5
    lm = np.random.gamma(5*theta,1/theta)
    t_del = np.random.exponential(1/lm)
    t+=t_del
    evnts+=1
print(evnts)
</code>```

答:
To simulate a mixed Poisson you first generate a single Gamma variable $L$. This  value for $L$ gives you the $\lambda$ (rate) parameter, which you use for the duration of your Poisson process. In other words, you have to move the line

```<code>lm = np.random.gamma(5*theta,1/theta)
</code>```

out of your while loop. When you do this, your event count will end up closer to the expected 5000.

Your current code is not simulating a Poisson process, since the interarrival times are not drawn from a single exponential distribution. Rather, each interarrival time is drawn from a different exponential, whose rate parameter $\lambda$ ranges from about $5-\sqrt{10}$ to $5+\sqrt{10}$ (these values are the mean $\pm$ standard deviation). Your code as written will generate many low values for $\lambda$ (which correspond to high interarrival times, which eat up a lot of the period) on virtually every run of your simulated point process; this is what causes the event count to drop. When you simulate properly, the low-$\lambda$ process will occur, but much less often, so low event counts are much less frequent.

Body:
To simulate a mixed Poisson you first generate a single Gamma variable $L$. This  value for $L$ gives you the $\lambda$ (rate) parameter, which you use for the duration of your Poisson process. In other words, you have to move the line

lm = np.random.gamma(5*theta,1/theta)

out of your while loop. When you do this, your event count will end up closer to the expected 5000.

Your current code is not simulating a Poisson process, since the interarrival times are not drawn from a single exponential distribution. Rather, each interarrival time is drawn from a different exponential, whose rate parameter $\lambda$ ranges from about $5-\sqrt{10}$ to $5+\sqrt{10}$ (these values are the mean $\pm$ standard deviation). Your code as written will generate many low values for $\lambda$ (which correspond to high interarrival times, which eat up a lot of the period) on virtually every run of your simulated point process; this is what causes the event count to drop. When you simulate properly, the low-$\lambda$ process will occur, but much less often, so low event counts are much less frequent.


问:
Convolution with shifted arguments

Given two function $f(t)$ and $g(t)$ and denote the convultion of the functions by $h(t)$, that is $$f(t)*g(t)=h(t).$$ What is $f(t-1)*g(t+1)$ in terms of $h(t)$?

Have no idea how to treat this question, everywhere I read, it is said, that this isn't a сonventional way of writing convolution.

The only thing that Im thinking about is, if we denote $T_kf(t)$ as$f(t-k)$ (shifting operator),
we can write:
$$T_1f * T_{-1}g$$

And then what?

答:
Given a function $s\colon\mathbb R\to\mathbb R$ such that $\int_{-\infty}^\infty s(\tau)\,\mathrm d\tau$ exists, note that for any $z\in\mathbb R$ we have
$$
\int_{-\infty}^\infty s(\tau)\,\mathrm d\tau = \int_{-\infty}^\infty s(\tau-z)\,\mathrm d\tau
$$
by substitution.

Hence, for the convolution:
\begin{align*}
f(t) * g(t) &= \int_{-\infty}^\infty f(\tau)\,g(t-\tau)\,\mathrm d\tau \\
&= \int_{-\infty}^\infty f(\tau-z)\,g(t-\tau+z)\,\mathrm d\tau \\
&= f(t-z)*g(t+z).
\end{align*}

For $z=1$ this yields $f(t)*g(t) = f(t-1)*g(t+1)$.

I agree that "$f(t) * g(t)$" can be considered a bad notation, since $f(t)$ is just a real number, not a function. Here we used $f(t)$ as a shorthand for the function $t\mapsto f(t)$ and $f(t-z)$ as a shorthand for the function $t\mapsto f(t-z)$.

In the notation you suggested, this result might be written as
$$
f * g = T_z f * T_{-z} g,
$$
where $(T_z s)(t) := s(t-z)$.

Title: Convolution with shifted arguments


问:
is $(X,\tau)$ where $\tau$ is finite seperable?
is $(X,\tau)$ where $\tau$ is finite seperable? There was a similar question where $X$ is finite and the answer to that is yes. But in the case where $\tau$ is finite, I assume that means the number of subsets of $X$ which are open is finite. If we take the Real numbers and include a singleton as a topology, would that be not seperable?

答:
Every second countable space is separable. So any space where there are only countably many open sets is separble.

In the real line if you are considering a topology in which every singleton is open then the space is discrete and it is  not separable.


问:
Must a complex power series *fail* to be convergent somewhere on its circle of convergence?
My textbook asserts so, but I can't find its proof of the claim. On the other hand, a lecture slide I'm cross-referencing claims that a power series is allowed to be convergent at ALL points of its circle of convergence.

So who is correct?

答:
It seems to me that

$$\sum_n \frac{x^n}{n^2}$$

is convergent everywhere on the circle of convergence. This follows immediately from the triangle inequality and the convergence of $\sum_n \frac{1}{n^2}$.

ContentLicense: CC BY-SA 3.0


问:
Continuous and invertible function
Let $f(x)=\frac{x^7}{15}+9x^3+25x+1$. If I want to prove that it is a surjective function, can I say that, since it is strictly monotone and continuous, then it is surely invertible and so surjective?

答:
Let $y_0 \in \mathbb R$.

Since $f(x) \to \infty$ as $x \to \infty$, there is some number $b$ such that $f(b) >y_0$.

Since $f(x) \to -\infty$ as $x \to -\infty$, there is some number $a<b$ such that $f(a) <y_0$.

The IVT shows that there is $x_0 \in [a,b]$ such that $f(x_0)=y_0$. This gives $f(\mathbb R)= \mathbb R.$

Title: Continuous and invertible function


问:
$K(x_1, x_2,\dots x_n)/K(x_1, x_2,\dots x_r)$ is finite algebraic extension
$K(x_1, x_2,\dots x_n)$ is field extension of $K$ where $x_1, x_2,\dots x_r$ is transcendental basis of $K(x_1, x_2,\dots x_n)$ over $K$.

Then, $K(x_1, x_2,\dots x_n)/K(x_1, x_2,\dots x_r)$ is algebraic extension by definition,

My question is, $K(x_1, x_2,\dots x_n)/K(x_1, x_2,\dots x_r)$ is finite extension?
Atiye Macdonald uses this fact without proof.But it does not seem obvious to me.
Thank you for your help.

答:
If $F$ is a field and $G$ is a finitely generated algebraic field extension, then $G$ is finite. This can be proved by induction with the identity $[F(\alpha,\beta):F]=[F(\alpha,\beta):F(\alpha)][F(\alpha):F]$. Now, $K(x_1,\cdots,x_n)$ is generated by $x_{r+1},\cdots, x_n$ over $K(x_1,\cdots,x_r)$.

Remark: Note that here I'm talking about a field $G$ which is finitely generated as a field extension of $F$, which is different from being finitely generated as an $F$-algebra. Namely, $F(T)$ is not a finitely generated $F$-algebra, because you have too many polynomials to take the reciprocal of. Both are different from being finitely generated as a $F$-vector space.


问:
How to solve the recurrence $()=5(/2)+^3()$ using iteration method
How do we solve the recurrence $()=5(/2)+^3()$ using the Iteration method?

I solved the recurrence using a master method - master method

Now using the iteration method

$()=5(/2)+3() = 5(5T(/4)+(n/2)^3)*log(n/2))+n^3*logn$ $=$ ... $= 5^i(n/2^i)+n^3*log(n) * ∑(n^3*(5^k) *log(n/2^k) ) / 8^k) $

How is it equal to $Θ(n^3log(n))$ ?

答:
Do the substitution:

$$\begin{array} {rcl}
T(n) &=& n^3\log(n) \\
     &+& 5(n/2)^3\log(n/2)\\
     &+& 5^2(n/2^2)^3\log(n/2^2)\\
     &+& 5^3(n/2^3)^3\log(n/2^3)\\
     &+& \dots \\
\\
&=& n^3\log(n) \\
     &+& 5/2^3 n^3 (\log n - \log 2)\\
     &+& 5^2/2^6 n^3 (\log n - 2\log 2)\\
     &+& 5^3/2^9 n^3 (\log n - 3 \log 2)\\
     &+& \dots \\
\\
&=& n^3\log(n) (1 + 5/2^3 + 5^2/2^6 + 5^3/2^9 + \dots) \\
&-& n^3\log(2) (0 + 1\cdot 5/2^3 + 2\cdot 5^2/2^6 + 3 \cdot 5^3/2^9 + \dots )\\
\end{array}$$

That's a geometrics series and an arithmetico-geometric series.  But you don't actually need to calculate them to get the asymptotic behavior, since it is just going to be some positive constants.


问:
How can I find a generalized expresion for $A^{n}$?
I have to find a generalized expresion for $A^{n}$, $\forall n\in \mathbb{N}$
\begin{align}
A=
\begin{pmatrix}
-5 & 3\\
-6 & 7
\end{pmatrix}
\end{align}

Do you have any idea of how can I do that?

Body:
$$\begin{bmatrix} -5 & 3\\ -6 & 7 \end{bmatrix},~~ B=\begin{bmatrix} -2 & 1 \\ -2 & 2 \end{bmatrix}$$ Let $A=I+3B$
Here, $$B^2=2I, B^3=2B, B^4=2^2I,B^5=2^2 B......$$, then

$$A^n=(I+3 B)^n=I+ {n \choose 1}3 B+{n \choose 2} 3^2 B^2+ {n \choose 3} 3^3 B^3+....+{b \choose n}3^nB^n$$
$$A^n=I\sum_{k=0}^{n} (3\sqrt{2})^{2k} {n \choose 2k} + \frac{B}{\sqrt{2}}\sum_{k=0}^{n} (3 \sqrt{2})^{2k+1} {n \choose 2k+1}=P_n I+ Q_n B.$$
The elements of $A^n$ turn out to be  integers.

$$A^n= \begin{bmatrix} P_n & 0 \\ 0 & P_n \end{bmatrix}+ \begin{bmatrix} -2Q_n & Q_n \\ -2Q_n &  2Q_n \end{bmatrix}= \begin{bmatrix} P_n -2Q_n & Q_n \\ -2Q_n &  P_n+2Q_n \end{bmatrix}$$
Where $$P_n=\frac{1}{2}[(1+3\sqrt{2})^n+ (1-3\sqrt{2})^n],$$
$$Q_n=\frac{1}{2\sqrt{2}}[(1+3\sqrt{2})^n - (1-3\sqrt{2})^n].$$


答:
Let the four numbers in A.P. be $a-3d,a-d,a+d,a+3d$. Common difference is $2d$. Again 1st,2nd,4th are in G.P. with $2d$ as common ratio i.e. ${a-d\over a-3d}={a+3d\over a-d}=2d$.

Simplifying $(a-d)^2=(a-3d)(a+3d)$ we get $a=5d$. Putting this into ${a-d\over a-3d}=2d$ we get $d=1\implies a=5$.

Hence the four numbers are $2,4,6$ and $8$.

Title: The common difference is equal to the common ratio.

Body:
Four numbers are in A.P. The first, the second and the fourth are in G.P. Find the numbers if the common difference is equal to the common ratio.

Let the terms of the A.P. be $a_1,a_1+d,a_1+2d,a_1+3d$ and the terms of the G.P. will be $a_1,a_1+d,a_1+3d.$ We know that $d=q.$ I am not sure what to do from here.


问:
Is this statement true or false? Find its negation.

Consider the statement
$$\forall x \in \mathbb{R}\;\forall y \in \mathbb{R}\ x+y>0$$
Is this statement true or false? Also, find its negation.

I think this statement is false because the inequality is not always true; for example, for $y=-x$, the inequality is false.

Can you show me how to write the solution in the proper way?

答:
For the sake of contradiction, assume the statement holds true.

Next, we fix $x\in\mathbb{R}$ and set $y = -x$. Observe that from our contradiction assumption, we have
\begin{align*}
x + y &>0\\
x + (-x)&>0\\
0 &>0
\end{align*}
However, this is impossible and hence it is a contradiction. Therefore, the statement must be false.


问:
How to show that following function is analytic in domain |z|<1 and and domain |z|>1
Consider the following problem:

Show that $f(z)= \sum_{n=0}^{\infty}[ z^{2^{n+1}}/ (1-z^{2^{n+1}})] $ is analytic in domain |z|<1 and domain |z|>1 .

So, I thought of trying to find it's radius of convergence and seeing if i can deduce anything from it.

But, finding the radius of convergence is itself a herculian  task(to me atleast ) for this question and so I think there must be some alternative approaches for this question but could not think of any.

Can you please tell me which results should I use.

No need to fully solve the question , just outlines will be sufficient.

Thanks!

答:
This is not a power series, so radius of convergence is not even defined.

Let $|z| \leq r$ where $r <1$. Then $|\frac {z^{2^{n+1}}} {1-z^{2^{n+1}}}|\leq \frac {r^{2^{n+1}}} {1-r }$. Hence the series converges uniformly on compact subsets of the open unit disk. This makes $f$ analytic in this disk.

I will leave the case $|z|>1$ to you. Just divide numerator and denominator by $z^{2^{n+1}}$.


问:
Canonical examples of inner product spaces that are not Hilbert spaces?
That is, what are some good examples of vector spaces which are inner product spaces but in which not every Cauchy sequence converges?</p>

答:
A slightly different example than already mentioned would be a (sub-) vector space over the rationals. You can take any real or complex Hilbert space, for example, and consider the linear subspace over the rational numbers.

Edit: Sorry for the confusion: It would seem that "inner product space" is usually defined to be over the real or complex numbers, although the definition also works if one uses the rational numbers instead (or any other normed field). So, strictly speaking, the restriction of the ground field to the rational numbers is not an example of an inner product space that is incomplete.


问:
Finding integral in Number Theory paper
$$\frac{2}{\pi}\int_{-\infty}^{\infty}\left( \frac{\sin (\alpha t)}{t} \right)^2e^{i \nu t} \,\mathrm{d}t=\max(0,\alpha-|\nu|)$$

Apparently I found this in Montgomery's paper (1977). I need to know how exactly this happened. It looks strange when I am trying to do it by simple trigonometric expansion and integration by parts.

答:
Note that this is simply the Fourier transform (inverse Fourier transform to be more precise) of
\begin{align}
f(t)^2 = \left(\frac{\sin \alpha t}{t} \right)^2
\end{align}
which is a product. Hence we see that
\begin{align}
\mathcal{F}(f^2) = \mathcal{F}(f)\ast \mathcal{F}(f).
\end{align}
where $\mathcal{F}$ is the Fourier transform and $\ast$ denotes convolution of two functions. Moreover, the Fourier transform of $f(t)$ is a characteristic function. I will let you finish this on your own.

Title: Finding integral in Number Theory paper


答:
Let the four numbers be $a_1, a_2, a_3, a_4$. The common difference $d$ will make the terms equal to $a_1, a_1+d, a_1+2d, a_1+3d$, and the terms of the geometric sequence should be $a_1, a_1+d, a_1+3d$ as you correctly mentioned.

Since $d=r$, $a_1, a_1r, a_1r^2$ are also the terms of the geometric sequence.
So we have $a_1r=a_1+d$, $a_1r^2=a_1+3d$. Solving the first equation for $r$ gives $r=1+\frac{d}{a_1}$. Sustituting $r$ into the second equation gives $a_1(1+\frac{d}{a_1})^2=a_1+3d\Rightarrow a_1(1+\frac{2d}{a_1}+\frac{d^2}{a_1^2})=a_1+3d\Rightarrow a_1+2d+\frac{d^2}{a_1}=a_1+3d$

Simlifying gives $d^2=a_1d\Rightarrow d=0,a_1$.

A common difference of $0$ will give a common ratio of $1$ (since all $4$ terms are the same), so that's an extraneous solution.

The only other option is $d=a_1$.

In that case, substituting it back into $r=1+\frac{d}{a_1}$ gives $r=2$.

Therefore, since $d=r$, $d=r=a_1=2$, and so your only sequence is $2, 4, 6, 8$


答:
Write:

Since for $x=-1$ and $y=-1$, $(-1)+(-1)=-2>0$ is false. So, the given statement is false.

Clearly, the negation is:
$$\exists x,y\in\mathbb{R}\ x+y\leq0$$

DISCUSSION

To show that the statement is false, we just need one counterexample and we are done.

To find the negation, remember that the negative of "for all" is "there exists" and that of $>$ is $\ngtr$ or $\leq$.

Hope this helps. Ask anything if not clear :)

Title: Is this statement true or false? Find its negation.

Body:
Consider the statement
$$\forall x \in \mathbb{R}\;\forall y \in \mathbb{R}\ x+y>0$$
Is this statement true or false? Also, find its negation.

I think this statement is false because the inequality is not always true; for example, for $y=-x$, the inequality is false.

Can you show me how to write the solution in the proper way?


答:
Yes, for every $O \in \tau\setminus \{\emptyset\}$ we pick $x_O \in O$.

By definition, $\{x_O: O \in \tau\setminus\setminus\{\emptyset\}\}$ is countable (as $\tau$ is) and intersects all non-empty open sets so is dense. So $(X,\tau)$ is separable.

Also note that if $\{p\}\in \tau$ and $D$ is dense we must have $p \in D$.

If $X$ is finite or countable, $X$ itself is dense and countable trivially, so then also $(X, \tau)$ is separable, regardless of $\tau$.

Body: is $(X,\tau)$ where $\tau$ is finite seperable? There was a similar question where $X$ is finite and the answer to that is yes. But in the case where $\tau$ is finite, I assume that means the number of subsets of $X$ which are open is finite. If we take the Real numbers and include a singleton as a topology, would that be not seperable?


问:
If $(X,\tau)$ is a $T_1$-space, and every infinite subset of $X$ is dense in $X$, then$ (X,\tau)$ is the closed finite topology.
Let $(X,\tau)$ be a topological space. Prove that $\tau$ is the finite-closed topology on $X$ if and only if (i)$(X,\tau)$ is a $T_1$-space, and (ii) every infinite subset of $X$ is dense in $X$.

I already proved the forward direction but I'm stuck on the backward direction. We know that everyone singleton is closed because of i) and from ii) every open set intersects any infinite set non trivially. Now I need to figure out how to show that every open set are just infinite sets with countably finite points removed, thus the topology is finite-closed.

答: Let $U$ be a non-empty open  set. If $X \setminus U$ has an infinite number of points then  the infinite set $X \setminus U$ cannot be dense. This is because it does not intersect $U$. This contradiction  proves that $X \setminus U$ is finite whenever $U$ is a non-empty open  set.


问:
which of the followings are positive definite:
(CSIR-UGC NET Mathematical Sciences-2011)

Suppose $A,B$ are $n \times n$ positive definite. Then which of the followings are positive definite:

$A+B$

$ABA^{*}$

$A^2+I$

$AB$

答:
case #4 has a simple counterexample (if $A$ is not necessarily symmetric or Hermit):

$A = B = \begin{pmatrix}
1 & 1\\
 -1&1
\end{pmatrix}$,

$x = (0, 1)$.

Since $z^{T}Az= z_1^2+z_2^2 ,A$ is positive defined; however

xTABx = xTAAx = 0


问:
In mathematics, is there a object like this ( ,y) ? one sided ordered pair?
I 'm studying analysis 1 by Prof Terence Tao by myself

The book says that if $x$ and $y$ are any objects then we can define the ordered pair $(x,y)$ to be a new

object and that the ordered pair $(x,y)$ is something like this $\{\{x\},\{x,y\}\}$.

If ordered pair $(x,y)$ is defined to be the set $\{\{x\},\{x,y\}\}$

then there exists something like this $( ,y)=\{\{\},\{ ,y\}\} = \{∅,\{y\}\}$

Is this right??

答:
When you define an ordered pair by $(x,y) = \{\{x\},\{x,y\}\}$ both $x,y$ are supposed to be sets.

Hence $(\emptyset,y) = \{\{\emptyset\},\{\emptyset,y\}\}$ makes sense. But omitting one of the symbol $x,y$ doesn't make sense. At least if you want to define an ordered pair.

You could define the symbol $(,y)$ by $\{\emptyset,\{y\}\}$. However the value seems limited apart to provide an alternate definition of $\{\emptyset,\{y\}\}$.

For the sake of clarity, you can replace a free variable by any set in a first order logic formula, but not omit it.


问:
Intersection of collection of sets.
I can not find the flaw in my reasoning: a collection of sets being a set itself, so, as the empty set is a subset of every set, the empty set is an element of the collection. So the intersection of a collection of set is always empty. Can someone help in correcting me? Thx</p>

答: Yes, $\emptyset$ is a subset of every set. It doesn't follow from this it is an element of every collection of subsets of a set. For instance, it is not an element of $\bigl\{\{1,2\},\{1,3\}\bigr\}$.

Title: Intersection of collection of sets.


答:
You seem to confuse nothing, i.e. what is written here $\to$ "" between double quotes , which is not considered to be neither a mathematical concept nor a useful notation, and the empty set, which is a set.

So

$$\{,x\}$$ and $$\{\emptyset,x\}$$

have little in common and the first expression is not meaningful.

In the theory of formal languages, the notation $\lambda$ is used to denote an empty string.

Title: In mathematics, is there a object like this ( ,y) ? one sided ordered pair?

Body:
I 'm studying analysis 1 by Prof Terence Tao by myself

The book says that if $x$ and $y$ are any objects then we can define the ordered pair $(x,y)$ to be a new

object and that the ordered pair $(x,y)$ is something like this $\{\{x\},\{x,y\}\}$.

If ordered pair $(x,y)$ is defined to be the set $\{\{x\},\{x,y\}\}$

then there exists something like this $( ,y)=\{\{\},\{ ,y\}\} = \{∅,\{y\}\}$

Is this right??


问:
Connection between derivability and invertibility of a function
In my lecture notes it says that if $f'(x)>0$ for all $x\in H \subset\mathbb{R}$, then $f(x)$ is invertible. This specific question raises a more general question: How is the invertibility  of a function linked to the derivative (if it exists)?

答: A continuous function on an interval of real numbers is invertible if and only if it is strictly monotone. (This is a good exercise!) And a differentiable function on an interval is strictly monotone if and only if its derivative is always positive or always negative.</p>


答:
Consider $f : \mathbb R \to \mathbb R$ differentiable.

What is for sure:

If $f^\prime$ is positive or negative then $f$ is a bijection between $\mathbb R$ and $f[\mathbb R]$ as $f$ is monotonous and continuous.

If $f^\prime$ vanishes, $f$ can still be invertible. Example $f: x \mapsto x^3$ with $f^\prime(0) = 0$.

If $f^\prime$ vanishes at a point $x_0$ and has a minimim or maximum at that point, then $f$ isn't invertible.

$f$ can be invertible while $f^\prime(x_n) = 0$ where $\{x_n\}$ is a converging sequence.

Title: Connection between derivability and invertibility of a function

Body: In my lecture notes it says that if $f'(x)>0$ for all $x\in H \subset\mathbb{R}$, then $f(x)$ is invertible. This specific question raises a more general question: How is the invertibility  of a function linked to the derivative (if it exists)?


问:
What is the formula for combinations with identical elements?
Given a set of $n$ objects that has $p_1$ identical objects of one kind, $p_2$ identical objects of another kind, ... and so on until $p_k$ objects of the $k$th kind (so that $p_1 + p_2 + ... + p_k = n$), in how many ways can one select $r$ objects out of the $n$, with order irrelevant, where $r$ is <s>less than the largest</s> greater than or equal to the smallest number of identical objects (so $r \geq \max \{ p_1, ..., p_k \}$)?

E.g. in how many ways can you select two letters from the letters in some word like "chincherinchee"?

I feel I should be able to compute this with factorials in some form, but I can't wrap my head around it.

答:
I think this is about selection with constraints, which is similar to partition with constraints.

We want to select $r$ objects from a population with $t$ various types but we have limits on how many of some/all the types are available, with inclusion/exclusion to cope with multiple cases of limit breaking.

So we partition $r$ into $t$ types freely, then subtract off any cases which might break the constraints, then add back in double counted constraint breaking, etc.

For example choosing $r=5$ letters form "chincherinchee". There are $t=6$ letters $(t_i)=$ (c,h,i,n,e,r) with limits $(\ell_i)=(3,3,2,2,3,1)$.

The basic free selection from those $6$ types is sticks-and-stones (or stars&bars) division into categories, $$\binom{5+6-1}{6-1}=\binom{10}{5}$$

Then to remove cases where we have selected more of a given type than allowed, we repeat this division but with each type in turn allocated a limit-breaking $\ell_i{+}1$, which is removed from the total available. So for example when we consider breaking the limit on letter "n"($c_4$), this means we remove $\ell_4+1=3$ items from general allocation because they are preallocated to break the limit for "n", meaning the options for this case are $\binom{2+6-1}{6-1}=\binom{7}{5}$. So over all possible letters this adjusts the total to
$$\binom{10}{5}-\left(\binom{6}{5}+\binom{6}{5}+\binom{7}{5}+\binom{7}{5}+\binom{6}{5}+\binom{8}{5}\right)$$

However there are now some cases which have been removed twice, where two constraints were broken, which is handled similarly by preallocating enough slots to break two constraints simultaneously, then selecting from the reduced pool. In my example though these are very limited - we could have chosen $3\times$"i"${}+2\times$"r", or $3\times$"n"${}+2\times$"r" - only two options for double constraint breaking (and none for breaking more than two constraints). So we add these $2$ options back in for the result:
$$\binom{10}{5}-\left(3\binom{6}{5}+2\binom{7}{5}+\binom{8}{5}\right)+2$$


问:
How to deal with "If P, then Q" in the description of a set?
Here is the original question:

Given sets $A,B,C$ express the set in terms of A,B,C, using the symbols $\cup$,$\cap$, and $-$. $$F=\{x | x\in A\text{ and } (x\in  B) \rightarrow (x\in C)\}$$

I have found some solution like $A-(B-C),\,A\cap(B \cap C),\,(B \cap C)\cap A$. I do these all by trying, I wonder if there is a regular way to find or to judge whether I have found all the solution. Someone else give a more answer (−)∪(∩) for the same question. Welcome edit if there is something unclear, thanks!

New add: Maybe an explanation for my second and third answers is necessary. Take the second one. If $x\in ∩(∩)$, it is trivially true that $x\in A$ and the both side of → is true because $x\in B, x\in C$. Above is my understanding, or I am wrong? Please point out. And if my second and third answers is right, my question will be different from which is suggested below.

答:
$\phi \to \psi$ is logically equivalent to $\lnot \phi \lor \psi$.

So $$\{x\mid (x \in B) \to (x \in C\}$$ is then just the same as (with $\phi=(x \in B)$, $\psi=(x \in C)$  $$\{x\mid x \notin B \lor x \in C\}$$

which is $$B^\complement \cup C$$

The first clause of the definition of $F$ says $x \in A$ so we must intersect that set with $A$ and we get

$$A \cap (B^\complement \cup C) = (A \cap B^\complement) \cup (A \cap C)$$

(Using distributivity of $\cap$ over $\cup$)
But the question asks to use $-$ (for which I prefer $\setminus$ as a notation BTW) and then a little thought tells us that $A \cap B^\complement = A-B$ and so we get as answer $$(A-B) \cup (A \cap C)$$

This doesn't correspond directly to any of your suggested answers though.


问:
Why is the flux of a uniform field necessarily zero for any closed surface?
From taking and working in physics, I know that, given some uniform vector field $\vec{V}$, the flux of $\vec{v}$ through any closed surface is necessarily zero. I do not, however, understand, why this is, either intuitively or in a mathematically rigorous way.

For simple surfaces, it can be seen easily. Take, for example, a cube "parallel" to some vector field $\vec{V}$. By quick geometric examination, or argument by symmetry, we can see that  only two sides "contribute" to the total flux, and their area are opposite in direction, thus the total flux is $0$. Similar approaches can be taken for spheres, etc.

For more complex surfaces, however, this is both bizzare and amazing. Why, intuitively and rigorously, is this true?

答: Because the flux is proportional to the projected area, not the area itself (there is a cosine factor), and the surface can in a way be seen as flat. As a closed surface has as many entering as exiting faces, the whole contribution is zero.


问:
Is a Banach Space
Show that the vector space, $P_n$, of all the real polynomial functions of degree less than n, is a Banach Space for any norm define.

I think if I prove that $P_{n}$ is a Banach Space with the norm $|| |p| || = ||(a_0,a_1,...,a_n-1)||$, where $(a_0,a_1,...,a_{n-1})$ is a vector in $R^n$ wich coordinates are the coeficients of  $p$ and $||.||$ is any norm in $R^n$, I can prove that is for any norm define in it.

Thank you

答:
It's a finite dimensional real vector space, so all you need to show is that all norms are equivalent over $\mathbb R^n$ and any norm on $P_n$ will arise as a norm over $\mathbb R^n$ under the isomorphism $\phi : P_n \to \mathbb R^n$ you chose. If you want me to give you the details, ask for them. They can probably be found in your notes if you have any.

P.S. I think $P_n$ being the space of polynomials of degree less than $n$ includes degree $n$ polynomials, at least that's what I always see so I think of it as a standard.

Hope that helps,

Title: Is a Banach Space


问:
Decelerate function to match with the example
There is a system which uses its own decelerate function for animations, and I have to implement it manually to be as close to the sample as it can.

Here is the sample function (sorry, not too sophisticated, but this is all they have provided):

Basic things: Both t and n go from 0 to 1.

I have tried 1-1/t, but it has a very hard slope at beginning:

Anything else I have tried ruined the scale. I think this is a very easy problem to solve, but still can't find the right function yet.

Would be appreciated if you can help me out.

As it turned out, this is not that easy, so I have created a simulator where we can validate the ideas: https://www.desmos.com/calculator/4arts2sk99

I don't know if you can edit and save it, but if you post an answer I will add into that, and then we can create a screenshot as well.

Thanks all who is helping me with this.

答:
welcome to the site !

In my opinion, you can go follow two routes :

Either a "parametric route", coming up with an a priori form for your expression, and then fit the parameters to your data. Maybe visually at first, or using some kind of more precise framework if you are not satisfied (least squares for instance, see https://en.wikipedia.org/wiki/Least_squares)

Or a non parametric one, going from your image to a table of numbers. This free open source software does it quite well http://markummitchell.github.io/engauge-digitizer/


问:
Does $\,n\mid a_1b_1,\,a_2b_2, a_1b_2+a_2b_1\Rightarrow n\mid a_1b_2$?
Imagine that we have two pairs of integers $(a_1,b_1)$ and $(a_2, b_2)$ where

$$ a_1b_1\equiv 0,\,\ a_2b_2\equiv 0,\,\ a_1b_2+a_2b_1\equiv 0\pmod n$$

Does this imply that
$$ a_1 b_2 \equiv 0\pmod n $$

I assume that $a_1 b_1 \equiv 0\pmod n$ is only true if $a_1$ and $b_1$ are divisors of $n$. Using that, I've checked $a_1b_2 = 0$ mod $n$ numerically for a large number of values of $n$ and it seems to hold, but I am not sure how to prove it.

答: Hypotheses $\Rightarrow \dfrac{a_1b_2}n$ & $\dfrac{a_2b_1}n\,$ have sum & product $\in\Bbb Z\,$ thus both are $\in\Bbb Z,\,$ by Rational Root Test


答:
You can try the system of functions

$ f(x) = x^{1/n} $ for any suitable vaule of n where n>1 (Note: n doesn't have to be an integer)

(the higher the value of n the more sharp the change is...)

You can check that $f(0)=0$ AND $f(1)=1$
And the slope of the function is

$f'(x)=1/{nx^{(n-1)/n}}$

Or even better if a inverted parabola with roots 0,√2 is rotated by 45° anticlockwise...

Ie. $ f(x) = ax(x-√2) $ is the rotated by 45°

Ie. $ ay^2 +(2a-2ax-√2)y+ax^2-2ax-√2x =0$ use this if you can put y as a variable as well,if you can't then ...

$y=(-(2a-2ax-√2)+√((2a-2ax-√2)^2-4a(ax^2-2ax-√2x)))/2a$

Hopefully I didn't do any mistakes...

Title: Decelerate function to match with the example

Body:
There is a system which uses its own decelerate function for animations, and I have to implement it manually to be as close to the sample as it can.

Here is the sample function (sorry, not too sophisticated, but this is all they have provided):

Basic things: Both t and n go from 0 to 1.

I have tried 1-1/t, but it has a very hard slope at beginning:

Anything else I have tried ruined the scale. I think this is a very easy problem to solve, but still can't find the right function yet.

Would be appreciated if you can help me out.

As it turned out, this is not that easy, so I have created a simulator where we can validate the ideas: https://www.desmos.com/calculator/4arts2sk99

I don't know if you can edit and save it, but if you post an answer I will add into that, and then we can create a screenshot as well.

Thanks all who is helping me with this.


问:
The Radius of Convergence of $\sum_{n=1}^{\infty} x^n\left( \frac{n}{2n+1} \right)^{2n-1}$
The series is in the form of power series, here $x_{0}=0$. We can apply Root Test(We've concluded to use the Root Test in my previous question).
$\displaystyle\sum_{n=1}^{\infty} x^n\left( \dfrac{n}{2n+1} \right)^{2n-1}$. Let $a_{n}= \left | x^n\left( \dfrac{n}{2n+1} \right)^{2n-1} \right |$, it follows that:
$\displaystyle\lim_{n\to\infty} \displaystyle \sqrt[n]{\left | x^n\left( \dfrac{n}{2n+1} \right)^{2n-1} \right |}$ taking limits and we will get $\left |\dfrac{x}{4} \right|$ it must smaller than $1$ to be converge. So Radius of convergence is $4$. Now we must chech end points $x=4$ and $x=-4$. First, I plug $4$ and I used $n^{th}$ term test and I get $\dfrac{1}{2\mathbb{e}}$. So I concluded that the series diverges when $x=4$
Second, I plug $x=-4$ and because I know it doesn't converge absolutely I used Alternating series test(I know because it is same as $x=4$) So I used again $n_{th}$ term test and not surprisingly I get $\dfrac{1}{2\mathbb{e}}$ so again I concluded that the series also diverges at $x=-4$.
All in all, the series diverges iff $x\in \left (\infty, 4 \right]\cup\left [-4, \infty \right)$ and converges iff $x\in\left (-4,4 \right )$. Are there any gaps in my solution? Please verify. Thanks.

答:
Hint: Observe that the radius of convergence can be calculated using the lim sup given by Hadamard-Cauchy's formula. So if the limit exists, it will squeezed between two limits of ratios given by the ratio test, the lim sup and the lim inf.

Observe that the $1/R = \limsup \limits_{n\to\infty} \sqrt[n]{|c_n|}$,

where $c_n$ is giving by $\sum_{i = 0}^{\infty} c_nx^n$.


问:
Second order nonlinear delay differential equation
I have to solve the following delay differential equation:
$$\ddot{x}(t)+A\sin(\omega x(t-\tau))=0$$
Can someone give me a hint on how to solve this equation?
Thanks</p>

答:
A possible approach is the following. Consider the Green function for the problem
$$
   \ddot G(t) = \delta(t)
$$
that in your case takes the simple form $G(t-t')=(t-t')\theta(t-t')$. Then,
$$
   x(t)=-\int_0^t(t-t')\theta(t-t')A\sin(\omega x(t'-\tau))dt'+x_0+v_0t.
$$
You can solve this equation iteratively, starting with $x(t)=x(0)+\dot x(0)t$, and will recognize that, already at the second order, Bessel functions come out.

If you have the conditions $x_1=x(t_1)$ and $x_2=x(t_2)$ the Green function takes the form
$$
   G(t,t')=-\frac{1}{t_2-t_1}\left[(t'-t_2)(t-t_1)\theta(t'-t)+(t'-t_1)(t-t_2)\theta(t-t')\right]
$$
and you have to cope with a more involved integral but the conclusions do not change.


问:
group action of $\mathbb{F}_q$ on a $\mathbb{F}_p$-vector space.
Question: Let $p, q$ be distinct prime numbers. Let $V$ be a $\mathbb{F}_p$-vector space.  Does there exists a non-trivial $\mathbb{F}_p$-linear action of $\mathbb{F}_q$ on $V$?

My guess in that there doesn't exists a non-trivial action and here is a proof which I have thought of (but at one point I am not so sure):

Suppose the action is non-trivial then for $1 \in \mathbb{F}_q$ there exists a $v \in V$ such that $1\cdot v = w$, where $w \neq v$. Now consider the action $1\cdot (pv) = p(1\cdot v)= p\cdot v$ (since the action is $\mathbb{F}_p$-linear). This implies $p\cdot v = 0$, which is a contradiction because action of $p \in \mathbb{F}_q$ is a vector space isomorphism and by assumption $v \neq 0$.

Is this reasoning correct ?

Thanks in advance !!

答:
A crucial mistake in your argument is the claim that
$p\cdot v= 0$, where $p$ in the left would be an element of the field $\Bbb{F}_q$. The step
$$
1\cdot (pv)=p(1\cdot v)
$$
does follow from the linearity (though $pv=0$ as well as $p(1\cdot v)=0$, so this is not very interesting). But $p(1\cdot v)$ is unrelated to $p\cdot v$.

To see the error consider the case $p=2$. We have for all $v\in V$
$$
2\cdot v=(1+1)\cdot v=1\cdot(1\cdot v),
$$
and there is no way to combine the $1$s on the right hand side.
May be the mistake is that you falsely expected the group operation of $\Bbb{F}_q$, the addition, to magically correspond to the addition in $V$?
In fact, in addition to the rule $0\cdot x=x$ (the neutral element acts trivially) all that the action prescribes is
$$
(a+b)\cdot v= a\cdot (b\cdot v)
$$
for all $a,b\in\Bbb{F}_q$ and all $v\in V$. The requirement that the action should be $\Bbb{F}_p$-linear means that
$$
a\cdot(\alpha v+\beta w)=\alpha(a\cdot v)+\beta (a\cdot w)
$$
for all $a\in\Bbb{F}_q$, $\alpha,\beta\in\Bbb{F}_p$ and $v,w\in V$.

There exist non-trivial linear actions of the additive group $\Bbb{F}_q$ on the vector space $V=\Bbb{F}_p^n$ if the dimension $n$ is large enough. We really only want a cyclic group of order $q$ to act on $V$ by $\Bbb{F}_p$-linear transformations.

Let $d$ be the smallest positive integer such that $q\mid p^d-1$. By Little Fermat
$p^{q-1}\equiv1\pmod q$, so $d$ will be a factor of $q-1$. I claim that $n\ge d$ is all we need.

Consider the extension field $K=\Bbb{F}_{p^d}$. Its multiplicative group is cyclic of order $p^d-1$ so there exists an element $\mu\in K^*$ such that $\mu\neq1$ and $\mu^q=1$.
We can define an $\Bbb{F}_p$-linear action of $\Bbb{F}_q$ on $K$ by declaring that for
all $a\in\{0,1,2,\ldots,q-1\}$ we have
$$\overline{a}\cdot x=\mu^a x.$$
Because $\mu^q=1$, this is well-defined. Because $\mu^{a+b}=\mu^a\mu^b$ it defines an action of $(\Bbb{F}_q,+)$. Because multiplication by any element of $K$ obviously yields an $\Bbb{F}_p$-linear transformation $K\to K$, the action is $\Bbb{F}_p$-linear.

If $n\ge d$, we can choose $V=K\oplus \Bbb{F}_p^{n-d}$, and let $\Bbb{F}_q$ act on $K$ as above, and trivially on $\Bbb{F}_p^{n-d}$.

It is not difficult to show that the condition $n\ge d$ is also necessary for the space $\Bbb{F}_p^n$ to have a linear transformation of order $q$.


问:
Probability of hitting square inscribed in a circle of radius $r$?
In my probability book there was a question of finding the probability of choosing a random point from circle inscribed in circle, which I found out to be $pi$/$4$. After solving this question I thought of its reverse scenerio of the same problem in which square is inside in a circle and I have to find out probability of choosing random point from square inscribed in a circle, this time.

My Calculations so far:\

Let $E$ be the event of choosing point from square.

For square inscribed in a circle with radius $r$ and Area = $\pi r^2$, we have diagonal of square given by: D = $2r$

I can't seem to figure out how do find the area of circle now from this point? because that's the only information I need to find the desired result. i.e: P(E) = Area of square/Area of circle

答:
The diagonal of the square is in fact $2r$. Using Pythagoras, the length of the square, $x$, is given by
$$x^2+x^2=4r^2\implies 2x^2=4r^2\implies x^2=2r^2\implies x=r\sqrt2$$
Hence the area of the square is $(r\sqrt 2)^2=2r^2$

Divide the area of the square by the area of the circle and there's your answer.


问:
Density in Cartesian product.
Let $H = \{0,1\}$ and consider the discrete topology. Now, let $I = [0,1]$ with the usual topology. Let $K$ be a subset of $H\times I$ which is dense. We can express $K$ as the union of $\{0\}\times U_{0}$ and $\{1\}\times U_{1}$, where $U_{0}, U_{1}\subseteq I$. As $K$ is dense, can we assert that $U_{0}$ and $U_{1}$ are also dense subsets of $I$?.

I have a possible proof. Suppose, by contradiction and w.l.o.g. that $U_{0}$ is not dense in $I$. Then, there exists $x_{0}\in I$ and $U_{x_{0}}$ a neighborhood of $x_{0}$ in $I$ such that $U\cap  U_{0} = \emptyset$.

Then, $\{0\}\times U$ is a neigborhood of $(0,x_{0})$ in $H\times I$ and

$$\{0\}\times U\cap K = (\{0\}\times U)\cap (\{0\}\times U_{0}) = \{0\}\cap (U\cap U_{0}) = \emptyset$$

Is it correct?

答: It seems correct. A somehow shorter way is to observe that a subset $K$ of a topological space $X$ is dense if and only if $K\cap U$ is dense in $U$ for all open sets $U\subseteq X$, and $\{0\}\times [0,1]$ and $\{1\}\times [0,1]$ are open.

Title: Density in Cartesian product.


问:
Find a circle $O_{\rho}(0)$ containing all the roots of polynomial $z^{4} - 4z + 1$
Given polynomial
$$z^{4} - 4z + 1$$

How can I find the radius of the circle in which all of its roots lie? I was trying to find the exact forms of the roots, but it's too complicated.

答:
Another approach, that gives a larger bound.

The companion matrix of this polynomial is

$$A=\begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-1 & 4 & 0 & 0
\end{bmatrix}$$

Using Gershgorin's theorem, its eigenvalues (i.e., the roots of your polynomial) lie in the union of the disks $D(O,1)$ and $D(O,5)$, that is in $D(O,5)$. This means for all root $z$, $|z|\le5$.

Note that in this case (because the coefficient of $z^3$ is zero), this is equivalent to the Lagrange bound: given a polynomial $a_nz^n+a_{n-1}z^{n-1}+\dots+a_0$, its roots $z$ all must satisfy $|z|\le\max(1,\sum_{k=0}^{n-1}\left|\frac{a_k}{a_n}\right|)$.


问:
Any arbitrary closed smooth curve bounds a orientable surface?
I've got a question that, given an arbitrary closed smooth curve $C:[0,1]\rightarrow\mathbb{R}^3$, can you always find a orientable surface $\Omega$ which satisfy $\partial\Omega=C[0,1]$ ?

I have no idea on this question, and I suppose that the surface $\Omega$ has no restriction such as “smooth”. Thanks a lot for your help.

答:
Do we know that $C$ does not intersect itself, i.e. that $C$ restricted to the half-open interval $\left[ 0 , 1 \right[$ is injective?

Then the orientable surface exists, can be taken to be compact and connected, and is then called a Seifert surface of the knot.


答:
As pointed out in the comments, you can define $(,x) := \{\emptyset,\{x\}\}$, which you can also write $\{\{\},\{x\}\}$ as some people like the notation $\{\}$ for the empty set.

Again, as pointed out in the comments, it's not clear what the point of this notion would be - ordered pairs are extremely useful to say the least: one uses them to define cartesian products, and from there on, functions, relations (for instance orders, well orderings, ...) and these underly all common mathematical practice.

On the other hand, $(,x)$ encodes nothing more than $x$ itself, that is, you can recover $x$ from $(,x)$ as $\bigcup\bigcup (,x)$. So, while ordered pairs encode two objects in a single object, $(,x)$ encodes exactly the same data as $x$.

But you are allowed to define it if you want, the question is not about its well-definedness, but about its relevance.


问:
Showing that the Klein-four group is associative without checking 64 options

Showing that the Klein-four group is associative without checking 64 options

I don't know how to go about this. I could use the fact it's abelian to eliminate a few duplicates, but many options remain. Is there an easier way of proving that the Klein-four is associative?

答:
Thank you all for your help! I will post here my final answer:

First, we'll write down the Cayley Table:
$\begin{array} {|r|r|}\hline * & e & a & b & c \\ \hline e & e & a & b & c \\ \hline a & a & e & c & b \\ \hline b & b & c & e & a \\ \hline c & c & b & a & e \\ \hline  \end{array}$

From the symmetry along the main diagonal of the table, it's immediate that the operation is Abelian.
We'll check whether $x,y,z\in P$ satisfy $(x*y)*z=x*(y*z)$.

If any of the elements are $e$, the equality is immediately true.

If $x=z$ the equality is immediately true (due to the operation being Abelian).

If $y=x\neq z$, the left side of the equation is $(x*y)*z=e*z=z$. For the right side, we'll define $w=y*z$. So $x*(y*z)=y*w$. By definition, $w$ is the third distinct element so that $w\neq e$. (If $x=y=a$ and $z=c$ then $w=b$). Hence $y*w=z$.

Similarly, it can be proven for $y=z\neq x$.

for $x\neq y\neq z$, $(x*y)*z=z*z=e$ and $x*(y*z)=x*x=e$.

So the equality is true for any $x,y,z$, the operation is associative.


问:
Let $a$ and $n$ be integers with $n>0$.
Show that the additive order of $a$ modulo $n$ is $\frac{n}{\gcd(a,n)}$.
Let $a$ and $n$ be integers with $n>0$.
Show that the additive order of $a$ modulo $n$ is $\frac{n}{\gcd(a,n)}$.

attempt:
Let $a$ and $n$ be integers with $n>0$.
Let $m=\frac{n}{\gcd(a,n)}$.
Then,
\begin{equation*}
ma = \left(\frac{n}{\gcd(a,n)}\right)a =
n\left(\frac{a}{\gcd(a,n)}\right)
\equiv 0 \pmod{n}.
\end{equation*}
Since $d=\gcd(a,n)$ is the greatest positive integer such that $d\mid a$ and $d \mid n$, then $m$ is the smallest positive integer such that $n \mid ma$.

Hence, $m$ is the additive order of $a$ modulo $n$.

Am I true?

答:
A bit more constructively. The order of $\bar a \in \Bbb Z/n\Bbb Z$ is by definition the least positive integer, say $o(\bar a)$, such that $o(\bar a)\bar a=\bar 0$. But, $o(\bar a)\bar a=\overline{o(\bar a)a}$, and hence:

\begin{alignat}{1}
o(\bar a)\bar a=\bar 0 &\iff \overline{o(\bar a)a}=\bar 0 \\
&\iff o(\bar a)a \equiv 0 \pmod n \\
&\iff o(\bar a)a=mn
\end{alignat}

for some $m\in \Bbb Z$. Therefore, the order of $\bar a$ is the least positive integer of the form $n/(a/m)$, and it is then gotten when the denominator is the greatest divisor of $a$ which is also divisor of $n$, namely:

$$o(\bar a)=\frac{n}{\operatorname{gcd}(a,n)}$$


问:
If X is a topological space, U is open in X, A is dense in X, then closure A∩U=closure U
If X is a topological space, U is open in X, A is dense in X, then closure A∩U=closure U

I am stuck on the backward direction. The forward direction is easy since the intersection of A and U is a subset of U so that is true for their closures.

答:
If $x$ is in $\overline U$, and $G$ is an open neighbourhood of $x$, then $U\cap G$ is also open (finite intersection of open sets) and non-empty (as $x \in \overline U$ and $G$ is an open neighbourhood of $x$) and thus intersects the dense set $A$, so $A\cap U$ intersects $G$, as $(U \cap G)\cap A = G \cap (A \cap U)$.

As $G$ was arbitrary, $x \in \overline{A \cap U}$ and the inclusion has been shown.


问:
Create a number by multi by $2$ and divide by $3$ (integer part)
How can I create a given postitive integer $N$ by multi by $2$ and divide by $3$ (integer part) ?

(Write a computer program is allowed)
For example:
$$100 = 2*2*2*2*2*2*2*2*2*2*2/3/3/3/3*2*2$$

(The numbers of $2, 3$ need to be minimum)

($/3$ mean $\text{div}\ 3$, eg: $2/3 = 0, 4/3 = 1...$)

Is solving equation

$$\left \lfloor{\frac {2^a}{3^b} }\right \rfloor *2^c = N$$
help ?

If it help then, how to solve
$$\left \lfloor{\frac {2^a}{3^b} }\right \rfloor *2^c = N$$

such that $a > c, a+b+c$ is minimum.

答:
I've got now an algorithm which reproduces perfectly the lengthes for (only odd so far) $N $ from $1$ to $49$ as given by @HandeBruijn in his answer.

Real efficiency occurs, if the heuristics from small cases $N$ extend to larger values as well, that - when a solution with small set of small parameters is found - , no better solutions (in terms of length) shall occur.

I'll give my Pari/GP code below.

To understand the rationale I'll refer to my first answer, reducing the notation to a smaller vectorial notation.

1) For instance in 3 = 2*2*2*2/3*2/3 (#6) I read from right and count, after each /3 term, the *2 terms and simply store that in a vector, arriving at
$ 3 = T(1; [1,4])$ meaning that beginning at $a_0=1$ we have a term of $2^4$ and then one of $2^1$, separated by $/3$-terms, giving:
$$ T([1,4])=3 \qquad \to \qquad 1 \cdot 2^4 /3 \cdot 2^1 /3 = 3 \tag 1$$

Here, to save notation, I omit the leading "$(1;$"-argument in the $T()$-notation assuming it to be the default in the following.

Note that the length in my expression is the sum of the exponents at $2$ (call it $S$) plus the number of divisions by $3$ (call in $N$) which means $\lambda(3)=(1+4) + 2=7$ while Han de Bruijn gives here one less as (#6), let's call it $\beta$ with $\beta(x)=\lambda(x)-1$.

Another example, 27 = 2*2*2*2/3*2*2*2*2*2*2*2/3/3*2/3*2/3*2*2/3*2/3 (#22) .

My notation looks like
$$ T([1,2,1,1,0,7,4])=27 \qquad \to \qquad 1 \cdot 2^4 /3 \cdot 2^7 /3 \cdot 2^0 /3 \cdots = 27 \tag 2
$$

2) But an even more parsimony notation occurs as possible, when we think in reverse: we deconstruct $a_0=3$ by inverting the $/3$ and the $ \cdot 2^A$ operations.

The reverse of $y = x /3 $ is here underdetermined, we have $x_k = y \cdot 3 + \{0,1,2\}_k$ , that means $3$ solutions. We can then write the example of $a_0=3$ by
$$ U(3;[1,1])=1 \qquad \to \qquad {{3 \cdot 3 +1 \over 2^1} \cdot 3+1\over 2^4}=1
\tag 3$$
The vector $[1,1]$ is here the vector of residues $\pmod 3$ , let's call it $R$ and write $R=[1,1]$ and the $k$'th residue as $r_k = R[k]$

Note that we introduce here two remarkable restrictions:

the ininital value $a_{k-1}$ is assumed to be odd (let's implement the routine for the even numbers later) and

the exponent $A_k$ at $2$ is that value that $ a_{k}={a_{k-1} \cdot 3+ r_k \over 2^{A_k}}$ is odd (that simply means $A_k= \nu_2(a_{k-1} \cdot 3+ r_k)$ where $\nu_2()$ is the 2-adic valuation).

It is noteworthy that if all $r_k=1$ we have simply the Collatz-transformation in the "syracuse"-style. But we do not wat to restrict ourselves to $r_k=1$ only but allow $r_k \in \{0,1,2\}$ as well.

Let's see, how this looks for $a_0=25$ . The coding in the OP's notation is 25 = 2*2*2*2*2*2*2*2/3*2*2/3*2/3/3

Constructing first the $A$-vector notation we get $A=[0,1,2,8]$ and $T(1;A)=25$ . So we have $4$ integer divisions by $3$, which means in reverse, $4$ multiplications by $3$ each with residue $r_k$ to be determined.  First, from $A_1=0$ follows, that $r_1 \in \{0,2\}$ because the numerator in $ {a_{k-1} \cdot 3+ r_k \over 2^0}$ must be odd and since $a_{k-1}$ is odd by definition $r_k \in \{0,2\}$ must be even. In reverse, for all $A_k \gt 0$ we need to have $r_k=1$ and have thus one step of the Collatz/Syracuse-transformation. For $a_0=25$ it comes out that $A=[0,1,2,8]$ and $R=[0,1,1,1]$ so $U(25;R)=1$ is a solution with $\lambda(25)=15$.

Let us as well introduce the notation of $\rho=4$ for the length of the $R$ (resp. $A$)-vector.

Caution: that found solution is likely in most cases not unique,  as we can simply see when we look at the big table in my other answer: the same value $a$ occurs repeatedly in the table thus the coding of any $a$ into a $T()$ or a $U()$-transformations is not unique.

3) This makes a search-routine meaningful.
We take $a_0=25$ as initial value.
We begin with $\rho=1$ and  test $U(a_0;R)\overset?=1 $ with $R=[0]$,$R=[1]$,$R=[2]$. We get no solutions.

$\rho=2$ and test with $R=R_{2,j}=[[0,0],[0,1],[0,2],[1,0],[1,1],...[2,1],[2,2]]_j$  . Again, no solutions. Next:

$\rho=3$ and test with $R=R_{3,j}=[[0,0,0],[0,0,1],...[2,2,1],[2,2,2]]_j$  . Again, no solutions.

And so on. With $\rho=4$ we get solution(s) and can proceed to arbitrary $\rho$.

Time consumption is exponentially growing, so up to $\rho=10$ is easy for modern computer (say, $2$ secs) but up to $\rho=16$ needs already a lot of seconds.

However, heuristics suggest two strong reductions of the searchspace.

it seems, there are solutions containg $r_k=2$ only if there are solutions $r_k \in \{0,1\}$ with same $\rho$ and $\lambda$ already, so actually we need only test $R$ containing $r_k \in \{0,1\}$

it seems, after a short solution with a certain $\rho$ and $\lambda_1$ has been found, no shorter solution with $\lambda_k \lt \lambda_1$ occurs.

Example:

$\rho=4$, $\lambda=15$:

```<code>25 15 4  \\ a0, lambda, rho
   a1vec=[25, 75, 113, 85, 1]
       R=[0, 1, 1, 1]
       A=[0, 1, 2, 8]
25 15 4
   a1vec=[25, 75, 113, 341, 1]
       R=[0, 1, 2, 1]            \\ r_3=2 occur at the same rho & lambda
       A=[0, 1, 0, 10]
25 15 4
   a1vec=[25, 75, 227, 341, 1]
       R=[0, 2, 1, 1]             \\ r_2=2 occur at the same rho & lambda
       A=[0, 0, 1, 10]
</code>```

$\rho=7$, $\lambda=23$:  (non optimal solution : $\lambda \gt 15$)

```<code>25 23 7
   a1vec=[25, 19, 29, 11, 17, 13, 5, 1]
       R=[1, 1, 1, 1, 1, 1, 1]
       A=[2, 1, 3, 1, 2, 3, 4]
25 23 7
   a1vec=[25, 19, 29, 11, 17, 53, 5, 1]
       R=[1, 1, 1, 1, 2, 1, 1]
       A=[2, 1, 3, 1, 0, 5, 4]
... (some more solutions with same rho, lambda)
</code>```

$\rho=9$, $\lambda=28$:  (non optimal solution : $\lambda \gt 15$)

```<code>25 28 9
   a1vec=[25, 19, 57, 43, 65, 49, 37, 7, 21, 1]
       R=[1, 0, 1, 1, 1, 1, 1, 0, 1]
       A=[2, 0, 2, 1, 2, 2, 4, 0, 6]
25 28 9
   a1vec=[25, 19, 57, 43, 65, 49, 37, 113, 85, 1]
       R=[1, 0, 1, 1, 1, 1, 2, 1, 1]
       A=[2, 0, 2, 1, 2, 2, 0, 2, 8]
... (a lot more solutions with same rho, lambda)
</code>```

$\rho=10$, $\lambda=31$:  (non optimal solution : $\lambda \gt 15$)

remarkable: all $r_k \in \{1,2\}$, no $r_k=0$ !

```<code>25 31 10
   a1vec=[25, 19, 29, 11, 17, 13, 5, 17, 13, 5, 1]
       R=[1, 1, 1, 1, 1, 1, 2, 1, 1, 1]
       A=[2, 1, 3, 1, 2, 3, 0, 2, 3, 4]
... (some more solutions with same rho, lambda)

</code>```

$\rho=12$, $\lambda=36$:  (non optimal solution : $\lambda \gt 15$)

```<code>25 36 12
   a1vec=[25, 19, 57, 43, 65, 49, 37, 7, 11, 17, 13, 5, 1]
       R=[1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
       A=[2, 0, 2, 1, 2, 2, 4, 1, 1, 2, 3, 4]
25 36 12
   a1vec=[25, 19, 57, 43, 65, 49, 37, 7, 11, 17, 53, 5, 1]
       R=[1, 0, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1]
       A=[2, 0, 2, 1, 2, 2, 4, 1, 1, 0, 5, 4]
... (a lot more solutions with same rho, lambda)
</code>```

While the first heuristic is already surprising, the latter heuristic is really astonishing and should either be improved or be dealt with attempts of proving or disproving.

4) In my code I use that heuristics, and jump out from the two innermost loops once a solution is found. This code-line, however, can simply be commented-out if the heuristics should be improved.

```<code> \\ flgs: 1 - use residues 0,1 only
 \\       2 - print only running minimal lambda
 \\       4 - find only first solution
 \\       8 - print only short docu
{docsolutions(a0,maxrho=12,flgs=1+4)=my(lambda,rho,minlambda, minlambdarho,Rvectors,a1,a1vec,S,A,Avec,r,rrng);
   if( bittest(flgs,0), rrng=[0,1], rrng=[0,2]);
   minlambda=999;
   for(rho=1,maxrho,
         Rvectors=vector(rho,j,rrng);
         minlambdarho=999;
         forvec(R=Rvectors,
             a1vec=vector(rho+1);a1vec[1]=a0;
             Avec=vector(rho);
             S=0;
             a1=a0;
           for(j=1,rho,
                  r=R[j];
                  a1=3*a1+r;
                  A=valuation(a1,2);
                  a1>>=A;
                  Avec[j]=A;
                  S+=A;
                  a1vec[1+j]=a1;
                       if(a1==1,break())
              );
            if(a1<>1,next()); \\ test did not converge to 1
            if(a1vec[1+rho]==0,next()); \\ test has converged to solution with
                                        \\ smaller rho
            lambda=S+rho;
                    if(bittest(flgs,1),if(lambda>=minlambdarho,next()));
                    minlambdarho=lambda;
            print(a0," ",lambda," ",rho); \\ for compatibility with HdB lambda-1
            if(!bittest(flgs,3),
                 print("   a1vec=",a1vec);
                 print("       R=",R);
                 print("       A=",Avec);
              );
            if(bittest(flgs,2),break(2)); \\ assuming the first solution is the best
                      \\ don't test any further
           );\\ end forvec(R=...
      ); \\ end for(rho=...
      }
</code>```

Perform tests:

```<code>      docsolutions(25,10,1+2) \\ runs short test up to rho=10
</code>```

Some more protocol, with appended time-consumption in milliseconds:

```<code>gettime();docsolutions(25,10,2+8);gettime()

\\ a0 lambda rho   ++ note:lambda is length+1 from Han de Bruijn-doku
25 15 4
25 23 7
25 28 9
25 31 10
%9 = 6536

gettime();docsolutions(25,10,1+2+8);gettime()

\\ a0 lambda rho
25 15 4
25 23 7
25 28 9
%10 = 125

gettime();docsolutions(25,10,1+4+8);gettime()

\\ a0 lambda rho
25 15 4
%11 = 0

a0=1093
gettime();docsolutions(a0,10,2+8);gettime()

\\ a0 lambda rho
1093 31 8
1093 36 10
%13 = 6661

gettime();docsolutions(a0,10,1+2+8);gettime()

\\ a0 lambda rho
1093 31 8
1093 36 10
%14 = 125

gettime();docsolutions(a0,10,1+4+8);gettime()

\\ a0 lambda rho
1093 31 8
%15 = 31

\\ use higher rho / greater length for composition
gettime();docsolutions(a0,12,1+2+8);gettime()

\\ a0 lambda rho
1093 31 8
1093 36 10
%16 = 671

gettime();docsolutions(a0,14,1+2+8);gettime()
\\ a0 lambda rho
1093 31 8
1093 36 10
1093 44 13
%17 = 3276

gettime();docsolutions(a0,16,1+2+8);gettime()
\\ a0 lambda rho
1093 31 8
1093 36 10
1093 44 13
1093 49 15
%18 = 15117
</code>```

The check for $a=\{3,5,7,...,499\}$ as the following

```<code>gettime(); forstep(a0=3,499,2,docsolutions(a0,16,1+4+8));  Str(gettime(),"msecs")
\\a0 l rho   a0 l rho   a0 l rho  a0 l rho  a0 l rho
               3 7 2      5 5 1      7 8 2     9 11 3
   11 14 4    13 9 2     15 17 5    17 12 3    19 20 6
   21 7 1     23 15 4    25 15 4    27 23 7    29 18 5
   31 18 5    33 18 5    35 13 3    37 13 3    39 21 6
   41 21 6
  ....
  481 40 12    483 22 5    485 35 10    487 35 10    489 40 12
  491 40 12    493 22 5    495 40 12    497 22 5     499 40 12
 "27799 msecs"
</code>```

needed 28 secs

Body:
I've got now an algorithm which reproduces perfectly the lengthes for (only odd so far) $N $ from $1$ to $49$ as given by @HandeBruijn in his answer.

Real efficiency occurs, if the heuristics from small cases $N$ extend to larger values as well, that - when a solution with small set of small parameters is found - , no better solutions (in terms of length) shall occur.

I'll give my Pari/GP code below.

To understand the rationale I'll refer to my first answer, reducing the notation to a smaller vectorial notation.

1) For instance in 3 = 2*2*2*2/3*2/3 (#6) I read from right and count, after each /3 term, the *2 terms and simply store that in a vector, arriving at
$ 3 = T(1; [1,4])$ meaning that beginning at $a_0=1$ we have a term of $2^4$ and then one of $2^1$, separated by $/3$-terms, giving:
$$ T([1,4])=3 \qquad \to \qquad 1 \cdot 2^4 /3 \cdot 2^1 /3 = 3 \tag 1$$

Here, to save notation, I omit the leading "$(1;$"-argument in the $T()$-notation assuming it to be the default in the following.

Note that the length in my expression is the sum of the exponents at $2$ (call it $S$) plus the number of divisions by $3$ (call in $N$) which means $\lambda(3)=(1+4) + 2=7$ while Han de Bruijn gives here one less as (#6), let's call it $\beta$ with $\beta(x)=\lambda(x)-1$.

Another example, 27 = 2*2*2*2/3*2*2*2*2*2*2*2/3/3*2/3*2/3*2*2/3*2/3 (#22) .

My notation looks like
$$ T([1,2,1,1,0,7,4])=27 \qquad \to \qquad 1 \cdot 2^4 /3 \cdot 2^7 /3 \cdot 2^0 /3 \cdots = 27 \tag 2
$$

2) But an even more parsimony notation occurs as possible, when we think in reverse: we deconstruct $a_0=3$ by inverting the $/3$ and the $ \cdot 2^A$ operations.

The reverse of $y = x /3 $ is here underdetermined, we have $x_k = y \cdot 3 + \{0,1,2\}_k$ , that means $3$ solutions. We can then write the example of $a_0=3$ by
$$ U(3;[1,1])=1 \qquad \to \qquad {{3 \cdot 3 +1 \over 2^1} \cdot 3+1\over 2^4}=1
\tag 3$$
The vector $[1,1]$ is here the vector of residues $\pmod 3$ , let's call it $R$ and write $R=[1,1]$ and the $k$'th residue as $r_k = R[k]$

Note that we introduce here two remarkable restrictions:

the ininital value $a_{k-1}$ is assumed to be odd (let's implement the routine for the even numbers later) and

the exponent $A_k$ at $2$ is that value that $ a_{k}={a_{k-1} \cdot 3+ r_k \over 2^{A_k}}$ is odd (that simply means $A_k= \nu_2(a_{k-1} \cdot 3+ r_k)$ where $\nu_2()$ is the 2-adic valuation).

It is noteworthy that if all $r_k=1$ we have simply the Collatz-transformation in the "syracuse"-style. But we do not wat to restrict ourselves to $r_k=1$ only but allow $r_k \in \{0,1,2\}$ as well.

Let's see, how this looks for $a_0=25$ . The coding in the OP's notation is 25 = 2*2*2*2*2*2*2*2/3*2*2/3*2/3/3

Constructing first the $A$-vector notation we get $A=[0,1,2,8]$ and $T(1;A)=25$ . So we have $4$ integer divisions by $3$, which means in reverse, $4$ multiplications by $3$ each with residue $r_k$ to be determined.  First, from $A_1=0$ follows, that $r_1 \in \{0,2\}$ because the numerator in $ {a_{k-1} \cdot 3+ r_k \over 2^0}$ must be odd and since $a_{k-1}$ is odd by definition $r_k \in \{0,2\}$ must be even. In reverse, for all $A_k \gt 0$ we need to have $r_k=1$ and have thus one step of the Collatz/Syracuse-transformation. For $a_0=25$ it comes out that $A=[0,1,2,8]$ and $R=[0,1,1,1]$ so $U(25;R)=1$ is a solution with $\lambda(25)=15$.

Let us as well introduce the notation of $\rho=4$ for the length of the $R$ (resp. $A$)-vector.

Caution: that found solution is likely in most cases not unique,  as we can simply see when we look at the big table in my other answer: the same value $a$ occurs repeatedly in the table thus the coding of any $a$ into a $T()$ or a $U()$-transformations is not unique.

3) This makes a search-routine meaningful.
We take $a_0=25$ as initial value.
We begin with $\rho=1$ and  test $U(a_0;R)\overset?=1 $ with $R=[0]$,$R=[1]$,$R=[2]$. We get no solutions.

$\rho=2$ and test with $R=R_{2,j}=[[0,0],[0,1],[0,2],[1,0],[1,1],...[2,1],[2,2]]_j$  . Again, no solutions. Next:

$\rho=3$ and test with $R=R_{3,j}=[[0,0,0],[0,0,1],...[2,2,1],[2,2,2]]_j$  . Again, no solutions.

And so on. With $\rho=4$ we get solution(s) and can proceed to arbitrary $\rho$.

Time consumption is exponentially growing, so up to $\rho=10$ is easy for modern computer (say, $2$ secs) but up to $\rho=16$ needs already a lot of seconds.

However, heuristics suggest two strong reductions of the searchspace.

it seems, there are solutions containg $r_k=2$ only if there are solutions $r_k \in \{0,1\}$ with same $\rho$ and $\lambda$ already, so actually we need only test $R$ containing $r_k \in \{0,1\}$

it seems, after a short solution with a certain $\rho$ and $\lambda_1$ has been found, no shorter solution with $\lambda_k \lt \lambda_1$ occurs.

Example:

$\rho=4$, $\lambda=15$:

25 15 4  \\ a0, lambda, rho
   a1vec=[25, 75, 113, 85, 1]
       R=[0, 1, 1, 1]
       A=[0, 1, 2, 8]
25 15 4
   a1vec=[25, 75, 113, 341, 1]
       R=[0, 1, 2, 1]            \\ r_3=2 occur at the same rho & lambda
       A=[0, 1, 0, 10]
25 15 4
   a1vec=[25, 75, 227, 341, 1]
       R=[0, 2, 1, 1]             \\ r_2=2 occur at the same rho & lambda
       A=[0, 0, 1, 10]

$\rho=7$, $\lambda=23$:  (non optimal solution : $\lambda \gt 15$)

25 23 7
   a1vec=[25, 19, 29, 11, 17, 13, 5, 1]
       R=[1, 1, 1, 1, 1, 1, 1]
       A=[2, 1, 3, 1, 2, 3, 4]
25 23 7
   a1vec=[25, 19, 29, 11, 17, 53, 5, 1]
       R=[1, 1, 1, 1, 2, 1, 1]
       A=[2, 1, 3, 1, 0, 5, 4]
... (some more solutions with same rho, lambda)

$\rho=9$, $\lambda=28$:  (non optimal solution : $\lambda \gt 15$)

25 28 9
   a1vec=[25, 19, 57, 43, 65, 49, 37, 7, 21, 1]
       R=[1, 0, 1, 1, 1, 1, 1, 0, 1]
       A=[2, 0, 2, 1, 2, 2, 4, 0, 6]
25 28 9
   a1vec=[25, 19, 57, 43, 65, 49, 37, 113, 85, 1]
       R=[1, 0, 1, 1, 1, 1, 2, 1, 1]
       A=[2, 0, 2, 1, 2, 2, 0, 2, 8]
... (a lot more solutions with same rho, lambda)

$\rho=10$, $\lambda=31$:  (non optimal solution : $\lambda \gt 15$)

remarkable: all $r_k \in \{1,2\}$, no $r_k=0$ !

25 31 10
   a1vec=[25, 19, 29, 11, 17, 13, 5, 17, 13, 5, 1]
       R=[1, 1, 1, 1, 1, 1, 2, 1, 1, 1]
       A=[2, 1, 3, 1, 2, 3, 0, 2, 3, 4]
... (some more solutions with same rho, lambda)

$\rho=12$, $\lambda=36$:  (non optimal solution : $\lambda \gt 15$)

25 36 12
   a1vec=[25, 19, 57, 43, 65, 49, 37, 7, 11, 17, 13, 5, 1]
       R=[1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
       A=[2, 0, 2, 1, 2, 2, 4, 1, 1, 2, 3, 4]
25 36 12
   a1vec=[25, 19, 57, 43, 65, 49, 37, 7, 11, 17, 53, 5, 1]
       R=[1, 0, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1]
       A=[2, 0, 2, 1, 2, 2, 4, 1, 1, 0, 5, 4]
... (a lot more solutions with same rho, lambda)

While the first heuristic is already surprising, the latter heuristic is really astonishing and should either be improved or be dealt with attempts of proving or disproving.

4) In my code I use that heuristics, and jump out from the two innermost loops once a solution is found. This code-line, however, can simply be commented-out if the heuristics should be improved.

 \\ flgs: 1 - use residues 0,1 only
 \\       2 - print only running minimal lambda
 \\       4 - find only first solution
 \\       8 - print only short docu
{docsolutions(a0,maxrho=12,flgs=1+4)=my(lambda,rho,minlambda, minlambdarho,Rvectors,a1,a1vec,S,A,Avec,r,rrng);
   if( bittest(flgs,0), rrng=[0,1], rrng=[0,2]);
   minlambda=999;
   for(rho=1,maxrho,
         Rvectors=vector(rho,j,rrng);
         minlambdarho=999;
         forvec(R=Rvectors,
             a1vec=vector(rho+1);a1vec[1]=a0;
             Avec=vector(rho);
             S=0;
             a1=a0;
           for(j=1,rho,
                  r=R[j];
                  a1=3*a1+r;
                  A=valuation(a1,2);
                  a1>>=A;
                  Avec[j]=A;
                  S+=A;
                  a1vec[1+j]=a1;
                       if(a1==1,break())
              );
            if(a1<>1,next()); \\ test did not converge to 1
            if(a1vec[1+rho]==0,next()); \\ test has converged to solution with
                                        \\ smaller rho
            lambda=S+rho;
                    if(bittest(flgs,1),if(lambda>=minlambdarho,next()));
                    minlambdarho=lambda;
            print(a0," ",lambda," ",rho); \\ for compatibility with HdB lambda-1
            if(!bittest(flgs,3),
                 print("   a1vec=",a1vec);
                 print("       R=",R);
                 print("       A=",Avec);
              );
            if(bittest(flgs,2),break(2)); \\ assuming the first solution is the best
                      \\ don't test any further
           );\\ end forvec(R=...
      ); \\ end for(rho=...
      }

Perform tests:

      docsolutions(25,10,1+2) \\ runs short test up to rho=10

Some more protocol, with appended time-consumption in milliseconds:

gettime();docsolutions(25,10,2+8);gettime()

\\ a0 lambda rho   ++ note:lambda is length+1 from Han de Bruijn-doku
25 15 4
25 23 7
25 28 9
25 31 10
%9 = 6536

gettime();docsolutions(25,10,1+2+8);gettime()

\\ a0 lambda rho
25 15 4
25 23 7
25 28 9
%10 = 125

gettime();docsolutions(25,10,1+4+8);gettime()

\\ a0 lambda rho
25 15 4
%11 = 0

a0=1093
gettime();docsolutions(a0,10,2+8);gettime()

\\ a0 lambda rho
1093 31 8
1093 36 10
%13 = 6661

gettime();docsolutions(a0,10,1+2+8);gettime()

\\ a0 lambda rho
1093 31 8
1093 36 10
%14 = 125

gettime();docsolutions(a0,10,1+4+8);gettime()

\\ a0 lambda rho
1093 31 8
%15 = 31

\\ use higher rho / greater length for composition
gettime();docsolutions(a0,12,1+2+8);gettime()

\\ a0 lambda rho
1093 31 8
1093 36 10
%16 = 671

gettime();docsolutions(a0,14,1+2+8);gettime()
\\ a0 lambda rho
1093 31 8
1093 36 10
1093 44 13
%17 = 3276

gettime();docsolutions(a0,16,1+2+8);gettime()
\\ a0 lambda rho
1093 31 8
1093 36 10
1093 44 13
1093 49 15
%18 = 15117

The check for $a=\{3,5,7,...,499\}$ as the following

gettime(); forstep(a0=3,499,2,docsolutions(a0,16,1+4+8));  Str(gettime(),"msecs")
\\a0 l rho   a0 l rho   a0 l rho  a0 l rho  a0 l rho
               3 7 2      5 5 1      7 8 2     9 11 3
   11 14 4    13 9 2     15 17 5    17 12 3    19 20 6
   21 7 1     23 15 4    25 15 4    27 23 7    29 18 5
   31 18 5    33 18 5    35 13 3    37 13 3    39 21 6
   41 21 6
  ....
  481 40 12    483 22 5    485 35 10    487 35 10    489 40 12
  491 40 12    493 22 5    495 40 12    497 22 5     499 40 12
 "27799 msecs"

needed 28 secs


问:
Show that no number of the form 8k + 3 or 8k + 7 can be written in the form $a^2 +5b^2$
I'm studying for a number theory exam, and have got stuck on this question.

Show that no number of the form $8k + 3$ or $8k + 7$ can be written in the form $a^2 +5b^2$

I know that there is a theorem which tells us that $p$ is expressible as a sum of $2$ squares if $p\equiv$  $1\pmod 4$.
This is really all I have found to work with so far, and I'm not really sure how/if it relates.

Many thanks!

答:
Hint If $x$ is even $x^2 \equiv 0,4 \pmod 8$ while if $x$ is odd we have $x^2 \equiv 1 \pmod 8$.

Thus

$$a^2+5b^2 \in \{0,1,4\} + 5 \cdot \{0,1,4 \} \pmod 8 \,.$$

Actually you can prove that $a^2+5b^2$ is $0,1$ or $2 \pmod 4$.


问:
How to derive transform fourier for general expression of gaussian $e^{-\frac{ax^2}{2}}$ for $a>0$
I am trying to understand how to get general expression for fourier transform for $$\large f(x)=e^{-\frac{ax^2}{2}}$$

if $a=1$ I understand that I can calculate it through the fact that
$$ f'(x)=-xf(x)$$ $$ w \hat f(w) = -\hat f'(w)$$
and because $f$ and $\hat f$ solve the same differential equation they are same function that might different in scalar multiply. so we can get
$$\hat f(w)=\frac{1}{\sqrt{2\pi}}e^{-\frac{w^2}{2}}$$
how do I get general form for $a>0$?

答: $\int e^{-itx} e^{-ax^{2}/2}dx=\frac1  {\sqrt a}\int e^{-ity/\sqrt a} e^{-y^{2}/2}dy$ by the substitution $y=\sqrt a x$. The integral here is the value of the FT in the case $a=1$ with $t$ changed to $\frac  t {\sqrt a}$.


问:
Understanding what $d(f(x))$ means
I know that $d(f(x))$ is small change in $f(x)$. But why is $\int d(f(x))=f(x)$? We know $\int dx$ means adding little bit of $x$. But what does adding little bit of $d(f(x))$ actually mean? Since it is not on the $x$ axis?

Title: Understanding what $d(f(x))$ means

Body:
Let's pick a function for a concrete thought experiment.

Let $f(x) = x^2$ and let's sample it at the integers.

  0   1   4   9  16  25  36  49  64  81 ...

Then list the differentials, like $df(x)$:

   1   3   5   7   9  11  13  15  17    ...

Now, let's take the cumulative sum, like $\int df(x)$:

  0   1   4   9  16  25  36  49  64  81 ... (noting that $f(0)=0$, so we can start at $0$)

$f(x)$ is restored!

Moreover, this is true at any sampling rate, so let the rate go to infinity and every point of $f(x)$ is reconstructed.

$x$
$f(x)$
$df(x)$
$\int df(x)$

0
0

0.01
0.0001
0.0001
0.0001

0.02
0.0004
0.0003
0.0004

0.03
0.0009
0.0005
0.0009

0.04
0.0016
0.0007
0.0016

0.05
0.0025
0.0009
0.0025

0.06
0.0036
0.0011
0.0036

0.07
0.0049
0.0013
0.0049

0.08
0.0064
0.0015
0.0064

0.09
0.0081
0.0017
0.0081


问:
Show this equality containing exp using the Itô Lemma
Let $B_t$ be any standard Brownian motion and define $X_t = \text{exp}(B_t)$. I want to prove the following equation:
$$dX_t = \text{exp}(B_t)dB_t+\frac{1}{2}\text{exp}(B_t)dt$$
This is supposed to follow directly from Itô's formula, but I don't see how it was used in this case.

答:
Denote $dS_t = dB_t$ and $f(x) = e^x$ then from Ito's lemma it follows that $$dX_t = df(t, S_t) = \left(\frac{\partial f}{\partial t} + \mu_t\frac{\partial f}{\partial x}+\frac{1}{2}\sigma^2_t\frac{\partial^2 f}{\partial x^2}\right)dt + \sigma_t\frac{\partial f}{\partial x}dB_t$$
In our case, $\mu_t = 0, \sigma_t = 1$ hence $$dX_t = \frac{1}{2}e^{B_t}dt + e^{B_t}dB_t$$


答:
The additive order of $a$ modulo $n$ is the smallest positive integer $m$ such that $ma \equiv 0 \pmod{n}$.

Now, $ma = kn$ for some integer $k$.
Then, $m = \frac{kn}{a}$.

Let $\gcd(a,n) = d$. Then,
$m = k\frac{\frac{n}{d}}{\frac{a}{d}}$.

Since $\frac{a}{d}$ and $\frac{n}{d}$ are relative prime, then $\frac{a}{d} \mid k$. Hence, the order of $a$ is a multiple of
$\frac{n}{d}$. Since the order must be minimum, then the order of $a$ is
$\frac{n}{d}$.

Is the approach above true?

Body:
Let $a$ and $n$ be integers with $n>0$.
Show that the additive order of $a$ modulo $n$ is $\frac{n}{\gcd(a,n)}$.

attempt:
Let $a$ and $n$ be integers with $n>0$.
Let $m=\frac{n}{\gcd(a,n)}$.
Then,
\begin{equation*}
ma = \left(\frac{n}{\gcd(a,n)}\right)a =
n\left(\frac{a}{\gcd(a,n)}\right)
\equiv 0 \pmod{n}.
\end{equation*}
Since $d=\gcd(a,n)$ is the greatest positive integer such that $d\mid a$ and $d \mid n$, then $m$ is the smallest positive integer such that $n \mid ma$.

Hence, $m$ is the additive order of $a$ modulo $n$.

Am I true?


问:
Is Q(4th root(2)):Q(sqrt(2)) a Galois extension?
I'm having difficulty with this little exercise. I know it is a Galois extension but I don't know how to prove it formally. Could someone help me? Thanks.</p>

答:
$\sqrt[4]{2}$ is a root of $x^2 - \sqrt{2} = p(x)\in \mathbb Q(\sqrt{2})$.

$\mathbb Q(\sqrt[4]{2})/ \mathbb Q(\sqrt{2})$ is a normal extension as the roots of $p$ are $\pm\sqrt[4]{2}$. As $\mathbb Q(\sqrt[4]{2})/ \mathbb Q(\sqrt{2})$ is also separable, it is Galois.


问:
Is $\mathbb{Z}_{(p)}[x] $ a Principal Ideal Domain (PID)?
Let
$$\mathbb{Z}_{(p)}:=\left\{\frac{a}{b}\in \mathbb{Q}:a,b\in\mathbb{Z}, p\nmid b\right\}$$

Is $\mathbb{Z}_{(p)}[x] $ a Principal Ideal Domain (PID)? Use the following theorem:

Let $ f,q\in \mathbb{Z}_{(p)}[x] $ $$\deg(fq)=\deg(f)+\deg(q)$$

My idea to prove $\mathbb{Z}_{(p)}[x]$ is a PID is prove $\mathbb{Z
}_{(p)}$ is a field then it is a PID but I wouldn't be using the previous theorem

Can you help me please?

答:
Consider the ideal $(p,x)$.

This ideal is not whole the ring after localization because does not intersect the set of elements you are inverting.

Moreover, it is not generated by only one element: let suppose that there exists a dipendence relation between $x$ and $p$, you can lift this relation to a dipendence relation in $\mathbb{Z}[x]$, which is absurd because of the $\mathbb{Z}[x]$-analogue of the relation between degrees that you stated in the question.


问:
find the measure of this set
Let $\alpha>0$ and define $A_{\alpha}\subset\mathbb{R}^{2}$ as:
$$A_{\alpha}=\{x>0, y>\alpha x, x^{2}+y^{2}<\alpha^{2}\}$$
The integreation is simple, but I can't find the mistake I keep making. I tried via:

$$m(A_{\alpha})=\int^{\frac{\alpha}{\sqrt{\alpha^{2}+1}}}_{0}\int^{\sqrt{\alpha^{2}-x^{2}}}_{0}dydx=\dots=\frac{1}{2}\alpha^{2}\arcsin\Bigg(\frac{\alpha}{\sqrt{\alpha^{2}+1}}\Bigg)\Bigg(1+\cos\arcsin\Bigg(\frac{\alpha}{\sqrt{\alpha^{2}+1}}\Bigg)\Bigg)$$
and through spherical coordinates, for which:

$$m(A_{\alpha})=\int^{\alpha}_{0}\int^{\frac{\pi}{2}}_{\arctan\alpha}\rho d\theta d\rho=\dots=\frac{\alpha^{2}}{2}\Big(\frac{\pi}{2}-\arctan\alpha\Big)$$

Although they may, I doubt these 2 results are the same but I can't understand which is the correct one, assuming I integrated properly.

答: In the first method the integral w.r.t $y$ is from $\alpha x$ to $\sqrt {\alpha^{2}-x^{2}}$

Title: find the measure of this set


问:
Proving that $\sum_{i=1}^{n} \left(a\check{q}_{i}^{\frac{1}{\alpha}}+\left(1-a\right)\hat{q}_{i}^{\frac{1}{\alpha}}\right)^{\alpha}\geq 1$
I'd like to show that

$$
\sum_{i=1}^{n} \tilde q_{i}=\sum_{i=1}^{n} \left(a\check{q}_{i}^{\frac{1}{\alpha}}+\left(1-a\right)\hat{q}_{i}^{\frac{1}{\alpha}}\right)^{\alpha}\geq 1,
$$

given that $a,\alpha,\check{q}_i,\hat{q}_i\in\left[0,1\right]$ and that $\sum_{i=1}^n\check q=\sum_{i=1}^n\hat q=1$.

If $\alpha=1$, $\sum_{i=1}^n\tilde q=1$, and I believe that this sum is decreasing in $\alpha$. However, I'm not able to establish the sign of the derivative. This derivative is

$$
\frac{d \sum_{i=1}^{n} \tilde q_{i}}{d \alpha}=\sum_{i=1}^{n}\tilde q_{i}\frac{1}{\alpha}\left(-\frac{a \check{q}_{i}^{\frac{1}{\alpha}}\ln \check{q}_{i}+\left(1-a\right) \hat{q}_{i}^{\frac{1}{\alpha}}\ln \hat{q}_{i}}{\tilde{q}_i^{\frac{1}{\alpha}}}+\ln \tilde{q}_{i}\right),
$$

and, as $\check{q}_i,\hat{q}_i\in\left[0,1\right]$ (and, therefore, $\tilde q_i\in\left[0,1\right])$, the first summand in the parenthesis is positive and the second is negative.

答:
The derivative can be written as

$$
\frac{d \sum_{i=1}^{n} \tilde q_{i}}{d \alpha}=\sum_{i=1}^{n}\tilde q_{i}^{\frac{\alpha-1}{\alpha}}\left(-a \check{q}_{i}^{\frac{1}{\alpha}}\ln \check{q}_{i}^{\frac{1}{\alpha}}-\left(1-a\right) \hat{q}_{i}^{\frac{1}{\alpha}}\ln \hat{q}_{i}^{\frac{1}{\alpha}}+\tilde{q}_i^{\frac{1}{\alpha}}\ln \tilde{q}_{i}^{\frac{1}{\alpha}}\right)<0,
$$

where the inequality follows from $f(x)=x\ln x$ being a convex function.


问:
Definition of Lebesgue-Stieltjes measure on $\mathbb R$

Let $F:\mathbb R\to\mathbb R$ be a non-decreasing, left-continuous function. Let $a,b\in\mathbb R$, then define the Lebesgue-Stieltjes measure
    $$ m[a,b]=F(b+)-F(a), \quad m(a,b)=F(b)-F(a+) $$
    $$ m(a,b]=F(b)-F(a+), \quad m[a,b)=F(b)-F(a) $$

I am wondering what the notation F(a+) or F(b+) means.

Since $F$ is left-continuous we could have, e.g. $F(x)=\begin{cases} 1 & ,x\le3\\ 2 & ,x>3\end{cases}$.
Now is $F(3+)=1$ or $F(3+)=2$ ?

Intuitively I would assume that e.g.
  $$m[3,4]=m(3,4)=m(3,4]=m[3,4)=F(4)-F(3)=2-1=1$$
in any case. But I am probably wrong.

答:
The notation $f(a+)$ means "right hand limit," i.e., $\lim_{x\downarrow a} f(x)$.
Similarly you have
$$ f(a-) = \lim_{x\uparrow a} f(x).$$
These limits always exist for a cumulative distribution function since such function have a range contained in $[0,1]$ and are monotone nondecreasing.

In your example, you have $F(3-) = 1$ and $F(3+) = 2.$  Since $F(3) = 1$, $F$ is left-continuous at 3.  Remember, the limits disregard behavior at the target point.


问:
What type of set is the intersection of two spheres?
Let $X$ be a normed space and $S(x, \varepsilon)=\{y\in X: \|x-y\|=\varepsilon\}$ denote the sphere centered at $x\in X$ with radius $\varepsilon>0.$ Consider two different spheres $S_1=S(x_1, \varepsilon_1)$ and $S_2=S(x_2, \varepsilon_2)$ with non-empty intersection. What can we say about the set $S_1\cap S_2$?   Is it true that this set is always homeomorphic to a sphere or a ball of "one dimension less" (or even more dimensions less)?

What we know for sure: A paper by Jussi Vaisala contains a short proof of the following:

Lemma 2.2: The intersection of two spheres in a finite dimensional normed space $X$ of dimension $n\geq 3$ is a connected set. Furthermore, if $X$ is   strictly convex, then this intersection   is either a singleton or homeomorphic to an $(n-2)$-sphere.

My guess is that the second claim of the Lemma still holds for $n=2$, but was omitted from it because of the intersection  not being connected.
So, this should pretty much settle  my question for finite dimensional strictly convex spaces.

Without the assumption of $X$ being strictly convex we may have some more possibilities. Consider $X=\mathbb{R}^3$  equipped with the $\|⋅\|_\infty$ norm. The spheres are in fact cubes in $\mathbb{R}^3$ and the intersection of two cubes may additionally be (homeomorphic to) a square, which is the closed unit ball  of $(\mathbb{R}^2, \|⋅\|_\infty)$, or   (homeomorphic to) a line segment, which is the closed unit ball  of $(\mathbb{R}, \|⋅\|_\infty)$.

So, what can we say if we remove the strict convexity assumption? Is there a counterexample where the intersection is neither a sphere or a ball? I couldn't find any. Additionally what can be said for infinite dimensional normed spaces?

I couldn't find any literature regarding this topic and probably for a good reason since Vaisala mentions that while $S_1\cap S_2$ being connected was known by Novikoff in 1955,   $S_1\cap S_2$ being homeomorphic to the $(n-2)$-sphere seemed to be a new result [sic] at the time he published it (2010).

答:
The given link is no longer active but I have found the article:

MR2971762
Väisälä, Jussi
Regular simplices in three-dimensional normed spaces.
Beitr. Algebra Geom. 53 (2012), no. 2, 569–570.

https://doi.org/10.1007/s13366-012-0098-2


问:
why $\lim \inf \int f_n = 1?$.
Im posting this problem due to inactive user

taken from here

Here is the outlines

let $f_n = \chi_{[n,n+1]}$. Then $\lim \inf f_n = 0$, and $\lim \inf \int f_n = 1$.

My question is that  why  $\lim \inf \int f_n = 1?$.

My attempt :Here $$f_n =\begin{cases} 1 \ \text{if }\ x\in  [n,n+1] \\ 0 \ \text{if} \ x \notin [n, n+1] \end{cases}$$

I can easily see that  inf $f_n=0$  and sup $f_n=1$

Im not getting why  $\lim \inf \int f_n = 1?$.

答:
Hint:

Try calculating $\int f_n$ for a couple of values of $n$...

Title: why $\lim \inf \int f_n = 1?$.


答: This is a very basic way I am suggesting you. Assume that, for example $8k+3=a^2+5b^2$ for some integers $k,a,b$. And assume that $a$ and $b$ are both even. What will you get? $$a^2+5b^2=4q^2+5(4q'^2)\\\\\ a^2+5b^2=8k+3$$ So $4|3$ which is a contradiction. Another assumption. Let $a=(2q+1),b=(2q'+1)$ for some integers $q$ and $q'$. So $$a^2+5b^2=(2q+1)^2+5(2q'+1)^2\\\\\ a^2+5b^2=8k+3$$ So $4\mid3$ which is a contradiction again. Continue to have two contradiction for other two cases.</p>

Title: Show that no number of the form 8k + 3 or 8k + 7 can be written in the form $a^2 +5b^2$

Body:
I'm studying for a number theory exam, and have got stuck on this question.

Show that no number of the form $8k + 3$ or $8k + 7$ can be written in the form $a^2 +5b^2$

I know that there is a theorem which tells us that $p$ is expressible as a sum of $2$ squares if $p\equiv$  $1\pmod 4$.
This is really all I have found to work with so far, and I'm not really sure how/if it relates.

Many thanks!


问:
Density in Cartesian product.
Let $H = \{0,1\}$ and consider the discrete topology. Now, let $I = [0,1]$ with the usual topology. Let $K$ be a subset of $H\times I$ which is dense. We can express $K$ as the union of $\{0\}\times U_{0}$ and $\{1\}\times U_{1}$, where $U_{0}, U_{1}\subseteq I$. As $K$ is dense, can we assert that $U_{0}$ and $U_{1}$ are also dense subsets of $I$?.

I have a possible proof. Suppose, by contradiction and w.l.o.g. that $U_{0}$ is not dense in $I$. Then, there exists $x_{0}\in I$ and $U_{x_{0}}$ a neighborhood of $x_{0}$ in $I$ such that $U\cap  U_{0} = \emptyset$.

Then, $\{0\}\times U$ is a neigborhood of $(0,x_{0})$ in $H\times I$ and

$$\{0\}\times U\cap K = (\{0\}\times U)\cap (\{0\}\times U_{0}) = \{0\}\cap (U\cap U_{0}) = \emptyset$$

Is it correct?

答: No need for contradiction. If $U$ is open, non-empty in $[0,1]$, so is $\{0\} \times U$ in $H \times I$, so this set intersects $K$, by definition in some $(0,x)$ with $x \in U_0$. So $U_0$ intersects $U$ and so is dense. The same goes for $U_1$, using the open $\{1\} \times U$ instead.

Title: Density in Cartesian product.


问:
An integral involving determinant.
I learned the following identity (without proof) from the book "Lectures on Elliptic Partial Differential Equations" by L. Ambrosio et.al.
$$\int_\Omega \det(A+\nabla\varphi)=\det(A)|\Omega|,$$
where $A$ is $n\times n$ matrix, $\varphi\in C^2_0(\Omega)$ and $\Omega\subset\mathbb{R}^n$ is bounded smooth domain.
After some consideration, I have an elegant proof of this result (using divergent theorem and Hadamard identity). My question is:

What is the name of this identity? I believe that it should be discovered by some mathematician(s), thus should be named by the mathematician(s).

How other people proved this result?

答:
We write
$$
A=\left(
\begin{array}
[c]{c}
A^{1}\\
\vdots\\
A^{n}
\end{array}
\right)  \text{,}
$$
and $M=\left(  M^{1},\ldots,M^{n}\right)  $, where $M^{i}$ is the $\left(
n,i\right)  $-cofactor of $\det(A+\nabla\varphi)$. Applying the Hadamard
identity to $f:x\mapsto Ax+\varphi(x)$, we get $\operatorname{div}M=0$. By the
divergence theorem,
\begin{align*}
&  \int_{\Omega}\det\left(
\begin{array}
[c]{c}
A^{1}+\nabla\varphi^{1}\\
\vdots\\
A^{n-1}+\nabla\varphi^{n-1}\\
\nabla\varphi^{n}
\end{array}
\right)  =\int_{\Omega}\sum_{i=1}^{n}\partial_{i}\varphi^{n}M^{i}=\int
_{\Omega}\nabla\varphi^{n}\cdot M\\
&  =\int_{\Omega}\left(  \operatorname{div}\left(  \varphi^{n}M\right)
-\varphi^{n}\operatorname{div}M\right)  \\
&  =\int_{\Omega}\operatorname{div}\left(  \varphi^{n}M\right)  \,\mathrm{d}
x=\int_{\partial\Omega}\varphi^{n}M\cdot\nu\,\mathrm{d}\sigma=0\text{.}
\end{align*}
Therefore
\begin{align*}
&  \int_{\Omega}\det\left(  A+\nabla\varphi\right)  =\int_{\Omega}\det\left(
\begin{array}
[c]{c}
A^{1}+\nabla\varphi^{1}\\
\vdots\\
A^{n-1}+\nabla\varphi^{n-1}\\
A^{n}+\nabla\varphi^{n}
\end{array}
\right)  \\
&  =\int_{\Omega}\det\left(
\begin{array}
[c]{c}
A^{1}+\nabla\varphi^{1}\\
\vdots\\
A^{n-1}+\nabla\varphi^{n-1}\\
A^{n}
\end{array}
\right)  +\int_{\Omega}\det\left(
\begin{array}
[c]{c}
A^{1}+\nabla\varphi^{1}\\
\vdots\\
A^{n-1}+\nabla\varphi^{n-1}\\
\nabla\varphi^{n}
\end{array}
\right)  \\
&  =\int_{\Omega}\det\left(
\begin{array}
[c]{c}
A^{1}+\nabla\varphi^{1}\\
\vdots\\
A^{n-1}+\nabla\varphi^{n-1}\\
A^{n}
\end{array}
\right)  \text{.}
\end{align*}
Similarly, we can apply Hadamard identity and the divergence theorem
to the map
$$
x\mapsto Ax+\left(
\begin{array}
[c]{c}
\varphi^{1}(x)\\
\vdots\\
\varphi^{n-1}(x)\\
0
\end{array}
\right)
$$
to eliminate $\nabla\varphi^{n-1}$ from the right hand side. Repeating the
above argument, we eliminate $\nabla\varphi^{n-2}$, $\ldots$, $\nabla\varphi
^{1}$, and get
$$
\int_{\Omega}\det\left(  A+\nabla\varphi\right)  =\det(A)\left\vert
\Omega\right\vert \text{.}
$$

Title: An integral involving determinant.


问:
If $f(z)=z^me^{g(z)}P(z)$ is of order $\rho$ then $P(z)$ has order less than $\rho$?
Suppose that $f$ is an entire function of order $\rho$. Using Weierstrass factorization theorem we can write $f(z)=z^me^{g(z)}P(z)$ where $g$ is an entire function and $P$ is the canoncial product of the zeros of $f$. I read in a book that the oder of $P$ is smaller or equal then $\rho$. Can someone explain me why this is the case?

答: Hint: $|z^{n}P(z)| <e^{ |z|^\epsilon}$ for $|z|$ sufficiently large. This gives $|f(z)z^{n}P(z)| <e^{|z|^{\rho}+|z|^{\epsilon}}<e^{|z|^{\rho+\epsilon}}$ for $|z|$ sufficiently large. This is true for each $\epsilon >0$ so we are done.


答:
By definition
\begin{align}
F(a^+)\triangleq \lim_{\substack{x\to a \\ a<x }}F(x)
\end{align}
that is, the number $F(a^+)$ is the number such that
$$
\Big(\forall \;\epsilon>0 \Big)\Big(\exists\; \delta>0 \Big) \bigg[\Big(x\in D_F \Big)\wedge \Big(0<x-a<\delta\Big) \implies \big|F(x)-F(a^+) \big|<\epsilon \bigg]
$$</p>

Title: Definition of Lebesgue-Stieltjes measure on $\mathbb R$

Body:
Let $F:\mathbb R\to\mathbb R$ be a non-decreasing, left-continuous function. Let $a,b\in\mathbb R$, then define the Lebesgue-Stieltjes measure
    $$ m[a,b]=F(b+)-F(a), \quad m(a,b)=F(b)-F(a+) $$
    $$ m(a,b]=F(b)-F(a+), \quad m[a,b)=F(b)-F(a) $$

I am wondering what the notation F(a+) or F(b+) means.

Since $F$ is left-continuous we could have, e.g. $F(x)=\begin{cases} 1 & ,x\le3\\ 2 & ,x>3\end{cases}$.
Now is $F(3+)=1$ or $F(3+)=2$ ?

Intuitively I would assume that e.g.
  $$m[3,4]=m(3,4)=m(3,4]=m[3,4)=F(4)-F(3)=2-1=1$$
in any case. But I am probably wrong.


问:
If $(X,\tau)$ is a $T_1$-space, and every infinite subset of $X$ is dense in $X$, then$ (X,\tau)$ is the closed finite topology.
Let $(X,\tau)$ be a topological space. Prove that $\tau$ is the finite-closed topology on $X$ if and only if (i)$(X,\tau)$ is a $T_1$-space, and (ii) every infinite subset of $X$ is dense in $X$.

I already proved the forward direction but I'm stuck on the backward direction. We know that everyone singleton is closed because of i) and from ii) every open set intersects any infinite set non trivially. Now I need to figure out how to show that every open set are just infinite sets with countably finite points removed, thus the topology is finite-closed.

答:
Let $O$ be an open set in $(X,\tau)$, so $O \in \tau$.

If $O=\emptyset$, $O \in \tau_{fc}$, as required.

If $O=X$, likewise, $O \in \tau_{cf}$.

In the final case $O$ is non-empty and is disjoint from $O^\complement$ by definition, and by $(ii)$, $O$ must intersect every infinite set. So $O^\complement$ is not infinite, hence $O^\complement$ is finite, i.e. $O \in \tau_{cf}$.

QED


问:
What are left and right singular vectors in SVD?
Let $USV^T$ be a singular value decomposition of matrix $A$. In the textbook "Linear Algebra and Its Applications" by D. C. Lay et. al., where SVD is introduced, it says that "the columns of $U$ in such a decomposition are called left singular vectors of $A$, and the columns of $V$ are called right singular vectors of $A$." But it does not make any connections with the eigenvectors of $A^T\!A$. It also says that "the matrices $U$ and $V$ are not uniquely determined by $A$.

But in this web page it says that "the eigenvectors of $A^T\!A$ make up the columns of $V$, the eigenvectors of $AA^T$  make up the columns of $U$."

I'm confused about the relationship between the left and right singular vectors (that is columns of $U$ and $V$) and the eigenvectors of $A^T\!A$ and $AA^T$. Any clarification is appreciated.

答:
Let $A=UDV^*$. Then $$A^*A=VDU^*UDV^*=VD^2V^*\implies A^*AV=VD^2$$
$$AA^*=UDV^*VDU^*=UD^2U^*\implies AA^*U=UD^2$$
When a matrix is right-multiplied by a diagonal matrix, each column is multiplied by the diagonal term: $$[\mathbf{b}_1,\ldots,\mathbf{b}_n]\begin{pmatrix}\sigma^2_1&\\&\ddots\\&&\sigma_n^2\end{pmatrix}=[\sigma^2_1\mathbf{b}_1,\ldots,\sigma^2_n\mathbf{b}_n]$$

Hence the right-hand expressions are exactly the eigenvalue equations, $$A^*A\mathbf{v}_i=\sigma_i^2\mathbf{v}_i,\qquad {AA}^*\mathbf{u}_i=\sigma_i^2\mathbf{u}_i$$