问:
Giving an exact description for $E_{R}:=\{\cos(z):\, R<|z|<\infty\}$
I have the following homework problem:

For each $R>0$ prove $$A(0,0,\infty)=\{e^{z}:\, z\in A(0,R,\infty)\}$$
  and give an description of the set $$E_{R}:=\{\cos(z):\, z\in
 A(0,R,\infty)\}$$

Where $$A(z_{0},r_{0},r_{1}):=\{z\in\mathbb{C}:\, r_{0}<|z-z_{0}|<r_{1}\}$$

There is also a hint for the second part: if $\cos(z)=\omega$ for
some given $\omega\in\mathbb{C}$ then $e^{iz}$ must satisfy a certain
quadratic equation.

What I did:

I have proved the first part (which doesn't seem relevant for the
second part).

I have also used the hint: If
$$
\cos(z)=\omega
$$

then
$$
\frac{e^{iz}+e^{-iz}}{2}=\omega
$$

hence
$$
\frac{e^{iz}\cdot e^{iz}+e^{iz}\cdot e^{-iz}}{2}=\omega
$$

Denote $t=e^{iz}$. Then
$$
t^{2}-2\omega t+1=0
$$

Hence
$$
t_{1,2}=\frac{2\omega\pm\sqrt{4\omega^{2}-4}}{2}
$$
$$
t_{1,2}=\omega\pm\sqrt{\omega^{2}-1}
$$

I also noticed that $e^{-iz}$satisfies the same equation hence one
of $e^{iz},e^{-iz}$ is $t_{1}$ and the other is $t_{2}$.

I don't see how this help me since
$$
t_{1}+t_{2}=2\omega
$$

hence
$$
\frac{e^{iz}+e^{-iz}}{2}=\frac{t_{1}+t_{2}}{2}=\omega
$$

so this doesn't give me any new information.

This is where I am stuck, can someone please help me continue this
exercise ?

答:
Solution 1

Let $f(z)=e^{iz},$ so $f(A(0,R,\infty))=A(0,0,\infty)$

Also, $g(z)=\frac12(z+\frac1z)$

Then $\cos z=\frac12(e^{iz}+e^{-iz})=(g\circ f)(z)$

Where foes $g$ map $A(0,0,\infty)?$

Well, generally $g$ maps the points $\{|z|>1\}$ conformally to $\mathbb C\setminus [-1,1]$

(See Bak-Newmann , Riemann Map chapter)
.

Also, the points of the circumference $|z|=1$ are mapped to the interval $[-1,1]$

Therefore, the image of $|z|\geq 1$ is the whole plane, which implies that the image of $A(0,0,\infty)$ is the whole plane.

Solution 2 (using the hint)

You have proved that the image of $A(0,R,\infty) $ through $e^z$ is $A(0,0,\infty)$. The same holds if we replace $e^z$ with $e^{iz}$.

Now we will prove that the image of $A(0,R,\infty)$ through $\cos z$ is the whole plane.

Take a $w\in \mathbb C$. We need to prove that there exists $z\in A(0,R,\infty)$ such that $$\cos z=w\iff$$
$$e^{2iz}-2we^{iz}+1=0\iff $$$$t=e^{iz}, t^2-2wt+1=0$$

Therefore $t=\dfrac{2w\pm\sqrt{4w^2-4}}{2}=w\pm\sqrt{w^2-1}$

This proves already that for a given $w\in\mathbb C$, you can find a $t\in\mathbb C$ which is not zero, therefore it belongs in $A(0,0,\infty)$ . Now, obviously(from the first part of the exercise), for a given $t\in A(0,0,\infty)$ we can find a $z\in A(0,R,\infty)$ such that $e^{iz}=t$ and the proof has finished.


问:
Give example of applications in maths domains or more generally of the use of Borel measure that is not a Radon measure
I'm looking for examples of research domains or themes ,where Borel measure that are not Radon measures are used in some ways.</p>

答:
Geometric measure theory, and all fields where you have to work with Hausdorff measures. Indeed, let $s \in [0,\infty)$ and $\mathcal{H}^s$ denote the $s-$dimensional Hausdorff measure on $\mathbb{R}^n$. Then $\mathcal{H}^s$ is a Borel regular outer measure on $\mathbb{R}^n$, but it is not a Radon outer measure unless $s\ge n$.

Note that, as an example, with Borel regularity of a measure you can apply lots of theory: for example the approximation of outer measures by open and closed sets. The Radon measure allows you to approximate outer measures also with compact sets, instead of closed sets.

ContentLicense: CC BY-SA 4.0


问:
Why should every metric space be open?
I am reading a book called: Introduction to Topology (Mendelson) and there the following statement has been made.

Theorem 6.4:
Let $(X,d)$ be a metric space.
Then $X$ is open.

The question instantly arises: Why would $([0,1], d)$, where $d$ is the euclidean distance not be a metric space?

So I have to take into account, that the book changed the definition of open:

Quote:

A subset $O$ of a metric space is said to be open if $O$ is a
neighbourhood of each of its points.

But surely, any good definition of "Openness" would boil down to the standard definition for the case above?

Am I missing something?

答:
The set $[0,1]$ is an open subset of $[0,1]$ with respect to the Euclidean metric.

It is true that $[0,1]$ is not an open subset of $\Bbb R$, but that's not relevant here.

Title: Why should every metric space be open?


问:
Why do we not define the metric on the sequence space as the larget seperation between the sequence?
Space consists of the set of all (bounded or unbounded) sequences of complex numbers and the metric $d$ defined by

$$d(x,y)=\sum_{j=1}^{\infty}\frac{1}{2^j}\frac{|\xi_j-\eta_j|}{1+|\xi_j-\eta_j|}.$$

where $x=(\xi_j)$ and $y=(\eta_j).$

I just want to ask why $d'(x,y)=\text{sup}_{j\in\Bbb{N}}|x_j-y_j|$ is not suitable for the above space?

What can I think that if we take unbounded sequence  then the sup would be infinity. But I am unable to take down it in an appropriate way.

答:
I just lost in the storm of overthinking...although with the help of@saulspatz...Now it's clear.

If $(x_n)$ is unbounded sequence and $(y_n)$ is identically zero, then

$d'(x,y)=\text{sup}_{n\in\Bbb{N}}|x_n-y_n|=\text{sup}_{n\in\Bbb{N}}|x_n|=\infty$,

which is not a real number therefore $d'$ is not well defined and hence  not suitable for said metric space.


问:
Solving summation $2n+2^2(n-1)+2^3(n-2)+....+2^n$
Can anyone help me with this summation? I tried to use the geometric series on this, but I can't use that.
$$2n+2^2(n-1)+2^3(n-2)+....+2^n$$
I am trying to do this for studying algorithms. Can we get a closed form for this ?</p>

ContentLicense: CC BY-SA 3.0

Body:
Define
$$s_n= \sum_{i=0}^{n-1}2^{i+1}(n-i)$$
and note that you have the relation:
$$S_{n+1}=S_n+2+2^2+\cdots+2^{n+1}=S_n+2^{n+2}-2$$ this comes from the fact that
$$2^{i}(n-i)=2^{i}(n-i+1-1)=2^{i}(n-i-1)+ 2^i$$
Noting that : $S_1=2$ and that the solution of the simple recurrence above is:
$$S_{n+1}=S_1+2^3+\cdots^+2^{n+2}-2n=2^{n+3}-2^3+2-2n$$
Thus
$$S_{n}=2^{n+2}-2^3+2-2(n-1)$$


问:
How can I prove that the first line of a singular Toeplitz matrix is linearly dependant of the others?
Here is a Toeplitz matrix of the form:

\begin{pmatrix}
 a_{0} &   a_1    & a_2 &\cdots&a_n\\
a_{-1} &   a_0    & a_1 &\cdots&a_{n-1}&\\
a_{-2} &   a_{-1} & a_0&\cdots&a_{n-2}\\
\vdots &   \vdots & \vdots &\ddots&\vdots\\
a_{-n} & a_{-n+1} & a_{-n+2}&\cdots&a_0
\end{pmatrix}

We assume that the matrix $A \geq 0$, which means that $(AX, X) \geq 0$ $\forall X \in \mathbb{R}^{n+1}$. We also assume that the determinant of this matrix is $0$.

I know that the $\det = 0$ constraint the values of the elements down the diagonal like in a $2 \times 2$ matrix, the value of the element down the main diagonal should be the product of the element on the main diagonal divided by the other element, i.e $a_{2,1} = (a_{1,1}\times a_{2,2}) \div a_{1,2} $

答:
Since $A$ is not invertible ($\det A=0$) then there must be a vector $x$ such that $x^TA=0$ (this follows since $A^T$ is also singular). Expanding out $$0=x^TA=\sum_{i=1}^nx_i\mathbf{a}_i$$ where $\mathbf{a}_i$ is the $i$th row of $A$; so if $x_1\ne0$ then $\mathbf{a}_1$ can be written as a linear combination of the rest of the rows.

However this is not always possible. For example, $\begin{pmatrix}4&8&5\\2&4&8\\1&2&4\end{pmatrix}$ is of the given form and singular, yet the first row cannot be written as a linear combination of the other rows.


问:
On the reiteration (or repetition) rule in natural deduction: is the rule "obvious" or does it need a proof?

I think the " reiteration rule" could be stated as follows : anything that is true under no assumption or under assumption $A_n$ is also true under any subordinate assumption $A_{n+k}$ with k greater than 0.

Is there a tautology that guarantees this rule?

Is it possible to prove this rule?

Are there logical systems in which the reiteration/repetition rule does not hold?

答:
The reiteration rule follows from a natural deductive proof system which uses an inference rule called "Addition". Specifically the addition rule says from $P$ we can infer $P \lor Q $, for any statement
$Q$.

In some cases reiteration is trivial, for instance if the statement is a premise you can cite the same premise as many times as you want, citing 'premise' as justification. In other cases reiteration is unavoidable, for instance in the proof of $~ \vdash P \to P $ we need to reiterate an assumption.
$$ ~~~~~~~~~~~~~~~~~~~~~~~~~~1.~ P ~~~~~~~~~~~~~~~~~~~~~~~\rm{(Some~Inference ~Rule)}
\\ ~~~~~~~2. ~P \lor \neg P~~~~~~~~~~\rm {~~~Addition ~1. }
\\ ~~~~~~~~~~~~~~~~~~~~~3. ~\neg \neg P ~~~~~~~~~~~~~~~~~~~\rm{Double~ Negation ~1}
\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~4. ~P  ~~~~~~~~~~~~~~~~~~~~~~\rm{Disjunctive ~Syllogism~2, 3} $$
Alternatively you could cite double negation (introduction and then elimination) .


问:
Complementary book for Conway's book "A Course in Functional Analysis"
I am first year graduate student and I have taken Functional Analysis course this semester. The main book of teaching is Conway's book "A Course in Functional Analysis" but the lecturer is notorious for giving exams from questions that are not exercises of the book he's teaching. So I need an accompanying book for functional analysis with lots of cool problems that covers same materials of Conway's book and is on graduate level. Any suggestions would be much appreciated.</p>

答:
As people have recommended in the comments, Rudin is a good book for exercises and as a complementary book, but I would not recommend studying from it. Also, Kreyszig's book is a standard functional analysis textbook: don't underestimate it just because it is considered undergraduate. Graduate students can learn much from this book too.

Two undergraduate level textbooks with cool aspects of functional analysis and nice, pedagogical exercises are the following:

Barbara MacCluer, Elementary Functional Analysis https://www.springer.com/gp/book/9780387855288

Karen Saxe, Beginning Functional Analysis https://www.springer.com/gp/book/9780387952246

Keep in mind that they are aimed at undergraduate students, so they might not be really close to what you are looking for. But, for example, the standard topics of "The big three theorems" or the Hahn-Banach theorem are nicely addressed. Moreover, there are many important exercises included that someone might have missed in their first functional analysis course.

Of course, Conway's book is nice to study from but I cannot not recommend G.K. Pedersen's book "Analysis Now": https://www.springer.com/gp/book/9780387967882

A book with lots of information about standard Banach spaces and their "geometry" and quite a few exercises is N. L Carothers's "A Short Course on Banach Space Theory": https://www.cambridge.org/core/books/short-course-on-banach-space-theory/88B71D535E744B24CA17CA82E670BFF4 but I would only recommend having this "on the side".

And, a book with many, difficult (this is only my opinion of course) exercises on $L^p$ spaces (and measure-theory), Banach spaces, Hilbert spaces and operators between those is Alberto Torchinsky's "Problems in Real and Functional Analysis": https://bookstore.ams.org/gsm-166
The book comes with the solutions to half the exercises. As I said, I think that these are difficult problems so do not fall in the trap where you start solving one problem, get stuck and then fixate on solving it!

And, as mentioned in the page between the first part of the book and the solutions to the exercises:

Don't panic!


问:
Is it analytic ? $f(z)=z^2\bar{z}$
Please find the question and its solution.

My answer gone wrong dont know why, I think the answer marked in the question is wrong.

My try:

I solved for CR equations and replaced x=y=0 and found CR equations satisfied. But not sure why in the solution it says CR equations doesn't satisfy.

Please let me know what should be the correct answer ?

答: You are right: since $u_x(0,0)=u_y(0,0)=v_x(0,0)=v_y(0,0)=0$, $(0,0)$ is a solution of the Cauchy-Riemann equations (and, since the partial derivatives are continuous at $(0,0)$,  $f$ differentiable at $0$).

Title: Is it analytic ? $f(z)=z^2\bar{z}$


答:
This is a trivial example, but consider the counting measure defined over the Borel sigma algebra. As it is a measure defined over the Borel sigma algebra, it is a Borel measure. It is not a Radon measure, as it is not finite on all compact sets.

Evidently, it has many applications. For example in statistics, where it is used to construct empirical measures.

Title: Give example of applications in maths domains or more generally of the use of Borel measure that is not a Radon measure

Body: I'm looking for examples of research domains or themes ,where Borel measure that are not Radon measures are used in some ways.</p>


问:
Is There A Function Of Constant Area?
If I take a point $(x,y)$ and multiply the coordinates $x\times y$ to find the area $(A)$ defined by the rectangle formed with the axes, then is there a function $f(x)$ so that $xy = A$, regardless of what value of $x$ is chosen? </p>

答: Take any desired real value $A$, then from $xy = A$, define $$f: \mathbb R\setminus \{0\} \to \mathbb R,\quad f(x) = y = \dfrac Ax$$ </p>

Title: Is There A Function Of Constant Area?


问:
$\hat{f}(\omega)=0$ and $f$ is continuous then $f(t)=0$.
If $f \in L^1(\mathbb{R})$ and its Fourier transform $\hat{f} \in L^1(\mathbb{R})$, then the inverse Fourier transform of $\hat{f}(\omega)$ is defined by: $g(t) = \mathcal{F}^{-1}\{\hat{f}(\omega)\} $.

If the original $f$ is continuous, then $g(t)=f(t)$ for every $t \in \mathbb{R}$.

Question:- If,  $$\hat{f}(\omega)=0$$ then$$ f(t)=\ \mathcal{F}^{-1}\{\hat{f}(\omega)\}$$

$$f(t)= \mathcal{F}^{-1}({0})$$
$$f(t)=0.$$
Is it correct?

答:
Let $f\in L^1(\mathbb{R})\cap C(\mathbb{R})$ and $\hat{f}\in L^1(\mathbb{R})$. The Fejér kernel
$$\begin{equation}
F_{\lambda}(x)=\frac{1}{2\pi}\int \limits_{-\lambda}^{\lambda}\left(1-\frac{|s|}{\lambda}\right)e^{isx}dx
\end{equation}$$

is an approximate identity for $L^1(\mathbb{R})$ thus
$\begin{equation}
\lim \limits_{\lambda \rightarrow \infty} \|F_{\lambda}*f-f\|_1=0
\end{equation}$ where
$$ F_{\lambda}*f(x)=\frac{1}{2\pi}\int \limits_{-\lambda}^{\lambda}\hat{f}(s)\left(1-\frac{|s|}{\lambda}\right)e^{isx}dx.$$

Now since $\hat{f}\equiv 0$ one obtains $f\equiv 0.$ By continouity this means $f(t)=0\ \forall t\in \mathbb{R}.$

Remark:
For $f\in L^1(\mathbb{R})$ the Fourier transform $\hat{f}$ is bounded and uniformly continous, so we can indeed evaluate $\hat{f}$ at any point $\omega_0\in \mathbb{R}.$


问:
What is the probability of the amount of different numbers obtained after random selection?
There are $n$ numbers, and each number is different. Randomly select a number with equal probability each time and record it, and repeat this process $t$ times. Note that the number you have selected before can continue to be selected. That is to say, for each selection, the probability of each number being selected is $\frac{1}{n}$. Finally, the amount of digits obtained is $t$. There may be cases where some digits are equal among these $t$ numbers. Repeated digits are deleted and a set of unique elements is obtained. What is the probability $p_{k}$ that the set size is $k$? The value range of $k$ is of course $1\leq k\leq t$. $n$ is much larger than $t$, so there is no need to consider whether $k$ needs to be smaller than $n$. For example, there are $n=1000$ numbers, select $t=40$ times, and finally get $k=35$ non-repeated numbers. What is the probability formula $p_{k}$ for $k$?

For $k=1$, I think $p_1={(\frac{1}{n})}^{t-1}$. And for $k=t$, I think $p_t=\frac{(t-1)\cdot P_n^{t-1}}{n^t}.$ (The $P_x^y$ is number of permutations).

I'm not sure whether these two cases are correct. For $2\leq k\leq t-1$, I didn't find the right way to calculate.

答:
I will assume that the digits play no role and you just found another example of the coupon collector problem.  The answer to your question is:

$$
P_k (n,t)=\frac {n!}{(n-k)!\,n^t}{t \brace k},
$$
where ${t \brace k}$ is the Stirling number of the second kind. Details can be found here.


问:
What math should a computer scientist take in college?
I'm a computer science major and like many of us we have to take two additional sciences. These two additional science courses are in addition to three semesters of calculus,two semesters of physics, and one semester each of linear algebra and discrete math.

I could take the easy way out and take an algebra based astronomy course, but I want the most bang for my buck as it relates to my major.What other types of math classes should I take? I do enjoy math a lot and have always done well.

答:
I would advise you to take advantage of college to study something sufficiently different if available, that could be even advantageous from a professional point of view:

biology / genomics / bio-modelization / physiology

captors / sensors / signal analysis / astronomy / experimental engineering

linguistics / phonology / ethnography / library science

My reasoning is that you are already accumulating sufficient CS culture to train yourself in nearby areas. You will work with other people with a strong culture in C.S. and math. That's probably only in college that you will have a hands-on experience on something else.

A potential recruiter can appreciate this openness, and most of the topics I quote are developing strong computer science interaction.

ContentLicense: CC BY-SA 2.5


问:
Finding an integral submanifold of an involutive distribution
I've been given the following vector field on the manifold $M = \mathbb{R}^3 \setminus \{\text{$z$-axis}\}$ $$X = (x-2y)\,\frac{\partial}{\partial x} + (2x+y)\, \frac{\partial}{\partial y} + z\, \frac{\partial}{\partial z},\quad Y = -y\,\frac{\partial}{\partial x} + x\, \frac{\partial}{\partial y}$$
It was straightforward to show that they're linearly independent and commute, therefore they define an involutive rank $2$ distribution, say $D$. The question is find and describe its integral submanifolds. I tried the following methods:

I was able to figure out that the form $\omega = - xz\,dx - yz\,dy + (x^2 + y^2)\,dz$ annihilates $X$ and $Y$ and $d\omega = 3x\,dx\,dz + 3y\,dy\,dz$. Since our distribution is involutive, naturally $d\omega\wedge \omega = 0$. But I've been unable to figure out an $\alpha$ such that $d\omega = \omega\wedge\alpha$. This could then help me figure out the integral submanifolds

Alternatively, we can the flows of $X$ and $Y$, which I did after a painstaking computations but I'm not sure how to proceed after that.

Any help would be appreciated.

答:
I'll assume you mean $\require{color}Y=-y\dfrac{\partial}{\partial{\color{red}x}}+x\dfrac{\partial}{\partial y}$ instead of $Y=-y\dfrac{\partial}{\partial{\color{green}z}}+x\dfrac{\partial}{\partial y}$ (which doesn't give $[X,Y]=0$ in fact $[X,Y]\notin\operatorname{span}\{X,Y\}$).  Then using cylindrical polars, we see
$$
X=\rho\frac{\partial}{\partial\rho}+2\frac{\partial}{\partial\theta}+z\frac{\partial}{\partial z},\quad Y=\frac{\partial}{\partial\theta}
$$
So $X,Y$ commutes and the integral submanifolds are given by $z/\rho=\text{constant}$, i.e., (one-sided) cone with the $z$-axis being its axis of symmetry.


问:
Probability of drawing a white balls from the urn having identical white and identical black balls?
In an urn there are $m$ white balls and $n$ black balls.
If a ball is drawn from the urn, then what is the probability that the drawn ball is white, given the condition that all white balls are identical and all black balls are identical?

My approach: Number of cases favorable to the event is $m$, and total number of cases for the sample space is $m+n$. So,
$$
P(\text{white ball}) = \frac{m}{m+n}
$$

But question is given that 'all white balls are identical and all black balls are identical'.

In the case of identical balls, number of cases favorable to the event is $1$ (no of ways of selection of one white ball from the $m$ identical white balls).

And total number of cases for the sample space are $2$ cases (either white ball is selected or black ball is selected). Here,
$$
P(\text{white ball}) = \frac{1}{2}
$$

I know that this $\frac{1}{2}$ is incorrect. Could any one help me out?

答:
It is just a matter of what you mean by "probability". Surely you can assign 0.5 to both cases, but that is not natural(telling us nothing interesting in real life), which means not work with our instinct. So the most natuaral way of defining probability is just assigning equal probability to every "possible cases".

Now I think this is the subtle point here that you are confused of. "Balls are identical" doesn't mean there is only one ball, it means you cannot tell the difference between white balls--the only factor that counts is the color. See? Identical here only justifies our intuition that we SHOULD assign equal probability to all balls, since only color matters.


答:
There are $m + n$ equally likely outcomes of drawing one ball. The number of outcomes that consist of a white ball is $m$. Thus the probability is
$$
P(\text{white ball}) = \frac{m}{m+n},
$$
as you suggested initially. The "identical" modifier is just to suggest that when you count a "white ball", there's no way to distinguish between different white balls.

Title: Probability of drawing a white balls from the urn having identical white and identical black balls?

Body:
In an urn there are $m$ white balls and $n$ black balls.
If a ball is drawn from the urn, then what is the probability that the drawn ball is white, given the condition that all white balls are identical and all black balls are identical?

My approach: Number of cases favorable to the event is $m$, and total number of cases for the sample space is $m+n$. So,
$$
P(\text{white ball}) = \frac{m}{m+n}
$$

But question is given that 'all white balls are identical and all black balls are identical'.

In the case of identical balls, number of cases favorable to the event is $1$ (no of ways of selection of one white ball from the $m$ identical white balls).

And total number of cases for the sample space are $2$ cases (either white ball is selected or black ball is selected). Here,
$$
P(\text{white ball}) = \frac{1}{2}
$$

I know that this $\frac{1}{2}$ is incorrect. Could any one help me out?


问:
The variation of a Ukrainian Olympiad problem: 10982

Given a recursion $a_{n+ 1}= \dfrac{a_{n}}{n}+ \dfrac{n}{a_{n}}$ with $a_{1}= 1.$ Prove that
$$\lim a_{n}^{2}- n= \frac{1}{2}$$
Source: StachMath/@RiverLi _ The limit and asymptotic analysis of $a_n^2 - n$ from $a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n}$

The original problem already has an answer, I'll suggest this way of thinking, which is not mine, but @twelve_sakuya

Let $b_{n}:=a_{n}^{2}- n,$ so
$$a_{n+ 1}= \frac{a_{n}}{n}+ \frac{n}{a_{n}}\Leftrightarrow b_{n+ 1}- \frac{1}{2}= -\frac{n}{b_{n}+ n}\left ( b_{n}- \frac{1}{2} \right )+ \frac{b_{n}}{2\left ( b_{n}+ n \right )}+ \frac{b_{n}}{n^{2}}+ \frac{1}{n}$$
That means $\left | b_{n+ 1}- \dfrac{1}{2} \right |\leq\dfrac{n}{\left | b_{n}+ n \right |}\left | b_{n}- \dfrac{1}{2} \right |+ \dfrac{\left | b_{n} \right |}{2\left | b_{n}+ n \right |}+ \dfrac{\left | b_{n} \right |}{n^{2}}+ \dfrac{1}{n}.$ Therefore, if we can get the evaluations of $\left | b_{n} \right |$ or $\dfrac{\left | b_{n} \right |}{n},$ maybe there exists a number $\beta\in\left ( 0, 1 \right )$ so that
$$\left | b_{n}- \frac{1}{2} \right |\leq\beta^{n}B\left ( n \right )\rightarrow 0\,{\rm as}\,n\rightarrow\infty$$
My friend has no confidence to continue, he also said that is the variation of a Ukrainian Olympiad problem #10982 (I searched and got the result, but I couldn't access it). I need to the help, thanks a real lot !

Body:
Note that, for $n\ge 1$, $$ n^6+10n^5+38n^4+68n^3+57n^2+18n>n^6+10n^5+37n^4+62n^3+46n^2+12n+1$$  and, factorising,
$$n(n+1)^2(n+2)(n+3)^2>(n^3+5n^2+6n+1)^2.$$
Also note that the function $F_n$ such that $$F_n(x)=\frac{n}{x}+\frac{x}{n}$$ is monotonic decreasing for $0<x\le n$.

Now suppose that $$\frac {n+2}{\sqrt {n+1}}\ge a_{n+2}\ge \frac{n+1}{\sqrt n} \tag 1$$ which is true for $n=1$. Then $$F_{n+2}\left(\frac {n+1}{\sqrt {n}}\right)\ge F_{n+2}\left(a_{n+2}\right)\ge  F_{n+2}\left(\frac{n+2}{\sqrt {n+1}}\right)$$
$$\frac {n^3+5n^2+6n+1}{\sqrt {n}(n+1)(n+2)}\ge a_{n+3}\ge \frac{n+2}{\sqrt {n+1}} $$
$$\frac {n+3}{\sqrt {n+2}}\ge a_{n+3}\ge \frac{n+2}{\sqrt {n+1}} $$ Therefore, by induction, Equation (1)  is true for all positive integers $n$.

Let $a_n^2=n+x_n$ for all $n$. Then, for $n\ge 1$, $$\frac{(n+2)^2}{n+1}\ge {a_{n+2}}^2\ge\frac{(n+1)^2}{n} $$  $$\frac{n+2}{n+1}\ge x_{n+2}\ge \frac{1}{n}.$$

We can now use this bound on $x_n$. $${a_{n+1}}^2=\frac{{a_n^2}}{n^2}+\frac{n^2}{{a_n}^2}+2$$
$$n+1+x_{n+1}=\frac{n+x_n}{n^2}+\frac{n^2}{n+x_n}+2$$
$$x_{n+1}=1+\frac{1}{n}+n\left(1+\frac{x_n}{n}\right)^{-1}-n+O\left(\frac{1}{n^2}\right)$$
$$x_{n+1}=1+\frac{1}{n}-{x_n}+\frac{{x_n}^2}{n}+O\left(\frac{1}{n^2}\right)$$
Then $x_n+x_{n+1}=1+O\left(\frac{1}{n}\right)$ and so $x_n+x_{n+1}$ tends to $1$. We also have
$$x_{n+2}=1+\frac{1}{n}-{x_{n+1}}+\frac{{x_{n+1}}^2}{n}+O\left(\frac{1}{n^2}\right)$$ and therefore $$|x_{n+2}-x_{n+1}|=\frac{n-1}{n}|x_{n+1}-x_{n}|+O\left(\frac{1}{n^2}\right)$$
$$|x_{2n}-x_{2n-1}|=\frac{(n-1)n...(2n-3)}{n(n+1)...(2n-2)}|x_{n+1}-x_{n}|+O\left(\frac{1}{n}\right)=\frac{1}{2}|x_{n+1}-x_{n}|+O\left(\frac{1}{n}\right)$$
Then $x_{n+1}-x_n$ tends to $0$ and so $x_n$ tends to $\frac{1}{2}.$


问:
Existence of a real structure on a complex vector space
As defined on Wikipedia here and there, a real structure on a complex vector space is an antilinear involution $\sigma: V\to V$ such that

$\sigma^2=id_V$

$\sigma(\lambda x)=\bar\lambda \sigma(x)$.

My question is : does it exist on every complex vector space ?

If $V$ has a basis $\{e_i\}_{i\in I}$, I can build this one :

$$\sigma(\sum_{i\in I}\lambda_i e_i)=\sum_{i\in I}\bar\lambda_i e_i$$

But the existence of a basis requires the axiom of choice, and I suspect that there should exist a more direct proof for the existence of a real structure.

Context My ultimate purpose is to prove the One-dimensional dominated extension theorem in the complex case.

答:
First, I'd argue that your formula $\sigma(\sum_{i\in I}\lambda_i e_i)=\sum_{i\in I}\bar\lambda_i e_i$ is already an incredibly direct proof of the existence of such a structure.

Second, I don't think there is a 'canonical choice' for $\sigma$ that does not rely on choosing a basis. For example, if $V = \mathbb{C}$, one could consider either of
$$\sigma_1(\mu) = \overline{\mu}$$
or
$$\sigma_2(\mu) = -\overline{\mu}.$$
Both are real structures on $\mathbb{C}$, but they are not the same.

Essentially you have to choose in $V$ a 'real' direction and an 'imaginary' direction, but before choosing a basis there is no canonical way to do so. (The $\sigma_1$ and $\sigma_2$ essentially correspond to choosing either $1$ or $i$ as a complex basis of $\mathbb{C}$.)


问:
Existence of Intermediate field $F$ of $E$ and $G$ such that the $[G: F] < \infty$
if $G$ is an infinite extension of $E$
$$E \subset G $$

Then does there exists an extension $F$ of $E$ such that $G$ is finite extension of $F$

$$E \subset F \subset G$$

I think we need to remove some kind of elements from $G$ to get $F$, but obviously it looks silly as it has not been defined. Hence I need help here.

答:
If we require $G\ne F$ then there does not necessarily exist such a field $F$. Take for example $G=\Bbb R$ and $E=\Bbb Q$. If $F$ is a subfield of $\Bbb R$ with $[\Bbb R:F]<\infty$, then also $[\Bbb C:F]<\infty$. However, as $\Bbb C$ is algebraically closed, the Artin-Schreier theorem implies that $F=\Bbb R$. Hence there does not exist such a field with $F\ne G$.

Edit: For completeness, here is the statement of the theorem:

Let $L$ be an algebraically closed field and $K$ a subfield with $[L:K]<\infty$. Then $L$ is obtained from $K$ by adjoining a squareroot of $-1$, in particular $[L:K]=1$ or $2$.

For a reference, see e.g. here.


问:
Determining poles and their orders without using Laurent series
I'm trying to understand how my book determines the poles of a function and their corresponding order, by just looking at the terms in the denominator and not using Laurent series. Hope somebody can clarify this.

My book teached me that:

Singularities

Singularities are isolated points where the complex derivative of $f(z)$ doesn't exist.

Poles

The most negative power of the term $(z-z_0)$ in the Laurent expansion of $f(z)$ about the singularity $z_0$ is some finite power $(z-z_0)^{-n}$. Then $z_0$ is a pole of order $n$.

But after explaining what a pole is, the book started finding poles and the corresponding order of the pole, just by factoring the denominator of a given function. So I just want to understand why it is that simple.

Consider a function $f(z)$ that has been rewritten as $$f(z)=\frac{z^4+1}{z^2(z-\frac{1}{2})(z-2)}.$$ Then the book determines immediately that the function has poles at $z=0$ of order $2$, $z=1/2$ and $z=2$ of order $1$.

The first issue is determining the singularities. It can easily be seen that the function diverges at these points and thus the derivative can't exist at these points. But is it always this easy? I believe you can construct functions where points of divergence is not the only types of singularities? But anyway, finding singularities is not really what I'm concerned about and it's pretty easy in this case.

After you have found the singularities. How can you tell that the power of the singularity in the denominator corresponds directly to the most negative term in the Laurent Series expanded about that point? In other words, how can you easily see that $1/(z-2)$ will be the highest negative power in the Laurent expansion around $z=2$, and thus $z_0=2$ is a pole of first order?

Is this having something to do with partial fraction decomposition of the denominator?:

$$f(z)=(z^4+1)(\frac{A}{z} +\frac{B}{z^2} +\frac{C}{z-\frac{1}{2}}+\frac{D}{z-2})$$

Because I can't quite see how that would help.

答: Note that, if$$g(z)=\frac{z^4+1}{z^2(z-1/2)},$$then$$f(z)=\frac{g(z)}{z-2}.$$But $g$ is analytic near $2$ and $g(2)\ne0$. So, near $2$, you can write $g(z)$ as$$g(z)=a_0+a_1(z-2)+a_2(z-2)^2+a_3(z-2)^3+\cdots,$$ with $a_0\ne0$, and therefore$$f(z)=\frac{a_0}{z-2}+a_1z+a_2(z-2)+a_3(z-2)^2+\cdots$$So, indeed, $f$ has a pole of order $1$ at $2$. A similar argument works for the poles of $f$ at $0$ and at $\frac12$.


问:
If $\Gamma \vdash \phi$ then $\Gamma[v / c] \vdash \phi[v / c]$
I'm taking a course in Mathematical Logic, and in the lectures there was the proposition:

Suppose $\Gamma \vdash \phi$ and $v$ is a variable that is not occurring in $\Gamma$ or $\phi .$ If $\Gamma \vdash \phi$ then $\Gamma[v / c] \vdash \phi[v / c]$ where $\Gamma[v / c]:=\{\theta[v / c]: \theta \in \Gamma\}$.

Here, I think $\Gamma$ is a set of $\mathcal{L}$-formulas, $\phi$ is an  $\mathcal{L}$-formula, $v$ is an $\mathcal{L}$-variable, and $c$ is an $\mathcal{L}$-constant. The proof of this is left as an exercise, the hint being to use induction on length of proof. I have done the induction steps for the rules $\forall I$, $\to I$, $\to E$, $\land I$, $\land E$, $\bot$, and $RAA$. However, I am struggling to do the $\forall E$ rule.

For example, suppose we have a deduction $D \cdots \forall v_1 \phi(v_1) \vdash \phi(t)$, where $t$ is free for $v_1$ in $\phi(v_1)$. By the induction hypothesis, we then have  $D[v/c] \cdots \vdash \forall v_1 \phi(v_1) [v/c]$, and from this we must deduce $D[v/c] \cdots (\forall v_1 \phi(v_1))[v/c] \vdash \phi(t)[v/c]$. By assumption, $v$ does not occur in $\phi(t)$.

If $v_1 \neq v$, then $v$ is free for $c$ in $\phi(v_1)$, because $v$ does not even occur in $\phi(v_1)$ (as $v$ does not occur in $\phi(t)$). Thus, $\forall v_1 \phi(v_1) [v/c]$ is $\forall v_1 (\phi[v/c])(v_1)$ as variable capture does not happen. As $t$ is still free for $v_1$ in $\phi[v/c]$, we use $\forall E$ rule to deduce $D \cdots \forall v_1 (\phi[v/c])(v_1) \vdash (\phi[v/c])(t)$. Now I should somehow conclude that $(\phi[v/c])(t)$ is equal to $(\phi(t))[v/c]$, but I don't know how to do this.

Furthermore, if $v_1 = v$, then I am totally lost about what to do. I tried to argue that the deduction  $D \cdots \forall v_1 \phi(v_1) \vdash \phi(t)$ somehow implied that $v_1$ appeared in a hypothesis of $D$, thus $v_1$ appears in $\Gamma$, but I don't know if this is even a correct argument, much less make it rigorous.

Could anyone help me with this? Maybe I'm doing something wrong?

答:
Assume your proof ends with $D \cdots \forall v_1 \phi(v_1) \vdash \phi(t)$.

[EDIT: The following paragraph is a correction following a comment from Albert.]

If $v_1=v$, first rename all bound occurences of $v_1$ in the derivation $D \cdots \forall v_1 \phi(v_1)$ to a variable $v_1'\neq v$ which does not yet occur in the proof. All rules remain sound, and $D$ is not affected since it contains no occurences of $v$ (bound or free) by assumption. In this way you obtain a derivation $D \cdots \forall v_1' \phi(v_1')$ where $v_1'\neq v$.

Hence we can now assume $v_1\neq v$. By induction hypothesis we get a proof ending with $\forall v_1 (\varphi[v/c])(v_1)$. As you noted, what we ultimately want to get is a proof of $\varphi(t)[v/c]$. The trick is to apply $\forall E$ instantiating $v_1$ with $t[v/c]$ (instead of $t$). In this way you obtain the formula $\varphi[v/c](t[v/c])$, and this is indeed identical to $\varphi(t)[v/c]$.

The intuition is that $t$ might contain occurences of $c$, and so if you first replace all $c$'s by $v$'s and then all $v_1$'s by $t$'s, then the latter step reintroduces some $c$'s into your formula that you don't want to have. Consider for example the case that $\varphi=P(v_1)$, $t=c$ and your initial proof is $\forall v_1 P(v_1)\vdash P(c)$. The proposition says that you should be able to obtain a proof of $\forall v_1 P(v_1)\vdash P(v)$. Clearly you should not instantiate $v_1$ with $t=c$, but rather with $v=t[v/c]$.


答: The curve you are after is the rectangular hyperbola given by the equation $xy=A$. For instance, in the figure below, you see that the red, purple and blue rectangles are all of unit area ($A=1$).

Title: Is There A Function Of Constant Area?

Body: If I take a point $(x,y)$ and multiply the coordinates $x\times y$ to find the area $(A)$ defined by the rectangle formed with the axes, then is there a function $f(x)$ so that $xy = A$, regardless of what value of $x$ is chosen? </p>


问:
Consistency of the minimizer of the convex loss
This is an exercise from the Stanford Stat 300B course, which I am auditing and don't have the solutions for. It is taken from the following source (Q2.6): https://web.stanford.edu/class/stats300b/Exercises/all-exercises.pdf. I have been stuck on the question for quite a while and think that I am missing some important part.

Let $\mathcal{X}$ be a measurable space and $X_i \sim P$ be i.i.d, where $P$ is a probability distribution on $\mathcal{X}$. Let $\Theta \subset \mathbb{R}^d$ be an open set and let $l :\Theta \times \mathcal{X} \to \mathbb{R}^{+}$ be a loss function. Define the risk functional $R(\theta) := E_P [l(\theta, X)]$, which is the expected loss of a vector $\theta$. Let $ \theta^* := \arg \min_{\theta\in\Theta} R(\theta)$. Define the empirical risk functional as $\hat{R}_n := \frac{1}{n}\sum_1^n l(\theta, X_i)$. Let $ \hat{\theta}_n := \arg \min_{\theta\in\Theta} \hat{R}_n(\theta)$.

We also have the following assumptions:

$\theta \to l(\theta, x)$ is convex

$\nabla^2 R(\theta^*) \succ 0$ i.e. the Hessian of the risk is positive definite at $\theta^*$

$\theta^* \in \text{interior of }\Theta$

There is a function $H : X → R^+$ such that $E[H^2(X)] < \infty$ and the Hessian $\nabla^2l(\theta, x)$ is $H(x)$-Lipschitz in $\theta$, that is, $$|| \nabla^2l(\theta, x) - \nabla^2l(\theta', x)||_{op} \leq H(x)||\theta - \theta'||~~\forall \theta,\theta'\in\Theta$$

Assume that gradients and Hessians can be passed through all expectations and integrals and as many moments of $\nabla l$ as you need.

Also assume the following theorem holds true. If $f$ is convex and satisfies $\nabla^2 f(x)
 - \lambda I\succ 0$ for all $x$ such that $||x - x_0|| \leq c$, then: $$ f(x) \geq f(x_0) + \nabla f(x_0)(x - x_0) + \frac{\lambda}{2}\min(||x-x_0||^2, c||x-x_0||)$$

Need to show that under these assumptions: $\hat{\theta}_n \to \theta^*$ in probability.

I feel hopelessly overwhelmed by all the details. I am not sure how to approach the problem.

答:
Careful: this proof requires one additionnal technical assumption, that is, that for any $\theta$, $\nabla^2 l(\theta,X)$ be $L^1$ (or, equivalently given condition 4, that $\nabla^2 l(\theta^*,X)$ be $L^1$).

Here's how the proof roughly goes (there's a slight technicality, but this is how the proof works if $\Theta$ is bounded): we show (part two) that there exists some integer $N_0$ such that for any $\theta$ and any $n \geq N_0$, with high probability, the inequality $|\hat{R_n}(\theta)-R(\theta)| \leq \epsilon$ holds.

Thus, with high probability, we have $R(\hat{\theta_n})\leq \hat{R_n}(\hat{\theta_n})+\epsilon \leq \hat{R_n}(\theta^*) +\epsilon\leq R(\theta^*)+2\epsilon$.

Now, $R(\theta)$ is convex, and we know that its minimum is reached at $\theta$ in the interior of the domain, and we know that $R(\hat{\theta_n})$ is close to said minimum (with high probability), which will imply (part one) that $\hat{\theta_n}$ is close to $\theta^*$.

$ $

First part: the convex analysis

By assumption, $R(\theta)$ is a convex function, twice differentiable, with $\theta^*$ as a global minimum, and $\nabla^2 R(\theta^*)$ is positive definite. Let $\lambda > 0$ be its smallest eigenvalue.

Moreover, we have $\|\nabla^2 R(\theta)-\nabla^2R(\theta')\|_{op}\leq\mathbb{E}_{x \sim P}[\|\nabla^2 l(\theta,x)-\nabla^2 l(\theta',x)\|_{op}] \leq \mathbb{E}_x[H(x)\|\theta-\theta'\|] \leq C\|\theta-\theta'\|$, that is $\nabla^2 R$ is Lipschitz continuous with constant $C$. Note that $C,\lambda$ are parameters of the problem.

Let $\theta \in \Theta$ (which we assume is convex), then consider $f(t)=R((1-t)\theta^*+t\theta)$. Then $f$ is $C^2$, convex, minimal at zero (thus increasing), $f'(0)=0$ and $f''(0) \geq \lambda \|\theta-\theta^*\|^2$. We want to show that if $f(1)-f(0)$ is small, then so must be $\delta=\|\theta-\theta^*\|^2$.

Indeed, let $\tau = \lambda/C$ (up to choosing a larger $C$, we assume $\tau < 1$). For $t \leq 1$, $f''(t) \geq \delta(\lambda-Ct)$, so for $t \leq \tau$, $f'(t)\geq \delta\lambda t-\delta Ct^2/2$ and thus $f(t)-f(0) \geq \delta \lambda t^2/2-\delta C t^3/6$. In particular $f(1)-f(0) \geq f(\tau)-f(0) \geq \frac{1}{6}\delta \tau^2 (3\lambda-\tau C)=\frac{\delta \tau^2 \lambda}{3}$.

Thus, we have, for any $\theta \in \Theta$, $$\|\theta-\theta^*\|^2 \leq \frac{3}{\tau^2 \lambda}(R(\theta)-R(\theta^*)).$$

$ $

Second part: the probability

We want to find a uniform bound on $\theta \in \Theta$ on the probability that $|\hat{R_n}(\theta)-R(\theta)|\geq \epsilon$. It turns out that the estimate we really can use (in absence of a $L^2$ moment bound for $l$ which is uniform in $\theta$) requires using derivatives, and thus we'll need to set a positive bound $B>0$ and consider only the $\theta$ such that $\|\theta-\theta^*\| \leq B$.

Now, by Taylor, if $\theta_t=(1-t)\theta^*+t\theta$, $$\hat{R_n}(\theta)-R(\theta) = \hat{R_n}(\theta^*)-R(\theta^*)+\nabla (\hat{R_n}-R)(\theta^*)\cdot (\theta-\theta^*) + \frac{1}{2}(\theta-\theta^*)^T\times \nabla^2(\hat{R_n}-R)(\theta^*)\times (\theta-\theta^*) + \int_0^1{(1-t)(\theta-\theta^*)^T \times (\nabla^2(\hat{R_n}-R)(\theta_t)-\nabla^2(\hat{R_n}-R)(\theta^*)) \times (\theta-\theta^*)\,dt}$$.

Let $V(X) = l(\theta^*,X)$, $G(X)=\nabla l(\theta^*,X)$, $L(X)=\frac{1}{2}\nabla^2 l(\theta^*,X)$. Finally, define $W^{\theta}(X) = \int_0^1{(1-t)(\nabla^2 l(\theta_t,X)-\nabla^2 l(\theta^*,X))\,dt}$.

By the weak law of large numbers, we know that there is a $N=N^0_{\epsilon}(\theta^*,B)$ such that for all $n \geq N$:

$P\left(|\hat{V}_n-\mathbb{E}[V]| \geq \frac{\epsilon}{8}\right) \leq \frac{\epsilon}{8}$.

$P\left(\|\hat{G}_n-\mathbb{E}[G]\| \geq \frac{\epsilon}{8(B+1)}\right) \leq \frac{\epsilon}{8}$.

$P\left(\|\hat{L}_n-\mathbb{E}[L]\|_{op} \geq \frac{\epsilon}{8(B+1)^2}\right) \leq \frac{\epsilon}{8}$.

Finally, note that $$\mathbb{E}[\|W^{\theta}\|^2_{op}] \leq \int_0^1{(1-t)^2\mathbb{E}[\|\nabla^2 l(\theta_t,X)-\nabla^2 l(\theta^*,X)\|_{op}^2]\,dt} \leq \int_0^1{(1-t)^2\mathbb{E}[H^2]\|\theta_t-\theta^*\|^2\,dt} = \|\theta-\theta^*\|^2\mathbb{E}[H^2] \int_0^1{(1-t)^2t^2\,dt} \leq C'B^2$$ for some $C' > 0$.

Thus, reasoning coordinate-wise, and recalling that the variance of a $L^2$ random variable $A$ is at most $\mathbb{E}[A^2]$, we have $P\left(\|(\hat{W^{\theta}})_n-\mathbb{E}[W^{\theta}]\|^2 \geq \frac{\epsilon^2}{64(B+1)^4}\right) \leq \frac{64C'B^2(B+1)^4}{n\epsilon^2}$.

So if $n$ is large enough, with probability at least $1-\epsilon/8$ that $|(\theta-\theta^*)^T \times ((\hat{W^{\theta}})_n - \mathbb{E}[W^{\theta}]) \times (\theta-\theta^*)| \leq \frac{\epsilon}{8}$.

It follows from this that with for $n \geq N_{\epsilon}(\theta^*,B)$, for any $\theta \in \Theta$ at distance at most $B$ of $\theta^*$, with probability at least $1-\epsilon/2$, $|\hat{R}_n(\theta)-R(\theta)| \leq \frac{\epsilon}{2}$.

$ $

Part three: the complete proof

Fix $B > \sqrt{\frac{3}{\tau^2 \lambda}}$, some $0 < \epsilon < 1$ and take $n \geq N_{\epsilon}(\theta^*,B)$.

Now choose $\theta$ on the line segment linking $\hat{\theta_n}$ to $\theta^*$ and at distance at most $B$ of $\theta^*$.

Then with probability $1-\epsilon$, the inequalities $\hat{R_n}(\theta^*) \leq R(\theta^*)+\frac{\epsilon}{2}$ and $\hat{R_n}(\theta) \geq R(\theta)-\frac{\epsilon}{2}$ both hold.

Now, $t \in [0,1] \longmapsto \hat{R_n}((1-t)\theta^*+t\hat{\theta_n})$ is convex, minimal at $t=1$, and thus is nonincreasing, and $\hat{R_n}(\theta)$ is some $f(t)$ with $0 \leq t \leq 1$, which implies that $\hat{R_n}(\theta) \leq \hat{R_n}(\theta^*)$.

Therefore, with probability $1-\epsilon$ at least, $R(\theta) \leq \hat{R_n}(\theta) +\frac{\epsilon}{2} \leq \hat{R_n}(\theta^*)+\frac{\epsilon}{2} \leq R(\theta^*)+\epsilon$.

So by part one, for any $\theta \in [\theta^*,\hat{\theta_n}]$ with distance at most $B$ from $\theta^*$, with probability at least $1-\epsilon$, $\|\theta-\theta^*\|^2 \leq \frac{3\epsilon}{\tau^2 \lambda}$. In particular, if we choose the farthest possible $\theta$ from $\theta^*$, (still with probability at least $1-\epsilon$), the inequality $\min(B,\|\hat{\theta_n}-\theta^*\|) \leq \sqrt{\frac{3\epsilon}{\tau^2\lambda}}$ holds. By the definition of $B$, it follows that with probability $1-\epsilon$ at least, $\|\theta^*-\hat{\theta_n}\| \leq \sqrt{\frac{3\epsilon}{\tau^2\lambda}}$, which concludes.


问:
Decomposition of Linear Maps between Tensor Product Spaces into Linear Maps between the Individual Spaces
What are the sufficient and necessary conditions for representing every linear map $\varphi \in L(V_1\otimes\cdots\otimes V_k; W_1\otimes\cdots\otimes W_k)$ by a tensor in $L(V_1;W_1) \otimes\cdots\otimes L(V_k;W_k)$ that acts as follows
$$
A_1 \otimes\cdots\otimes A_k: v_1 \otimes\cdots\otimes v_k \mapsto (A_1 v_1) \otimes\cdots\otimes (A_k v_k)
$$
where $A_j \in L(V_j;W_j),\ j \in\{1,\dots,k\}$.

This question is exactly my question but the answer there doesn't provide details on how the inclusion
$$
L(V_1;W_1) \otimes\cdots\otimes L(V_k;W_k) \hookrightarrow L(V_1\otimes\cdots\otimes V_k; W_1\otimes\cdots\otimes W_k)
$$
fails to be an isomorphism for infinite dimesnional vector spaces. Could someone explain that by an example ?

答:
The standard example is to take
$$ V_1 = V, W_1 = \mathbb{F}, V_2 = \mathbb{F}, W_2 = V $$
for some $\mathbb{F}$ vector space $V$. Then

$$ L(V_1, W_1) = L(V,\mathbb{F}) = V^{*}, \,\,\, L(V_2,W_2) = L(\mathbb{F}, V) \cong V $$ and we have an inclusion
$$ L(V_1, W_1) \otimes L(V_2, W_2) \cong V^{*} \otimes V \hookrightarrow L(V \otimes \mathbb{F}, \mathbb{F} \otimes V) \cong L(V, V) $$
given by $\varphi \otimes v \colon w \mapsto \varphi(w)v. $
Note that an element of $V^{*} \otimes V$ is a finite linear combination of simple tensors $\varphi \otimes v$ which means that in particular, the linear map that it defines on $V$ must be of finite rank. So any linear map on $V$ which is not of finite rank (such as the identity) cannot be in the image of the inclusion.


问:
Image of an intersection is the intersection of the image
I'm studying the proof that path-connectedness implies connectedness. However there is one assertion that I don't understand.

Let $f:[0,1] \rightarrow X$ be a continuous mapping where $X$ is a metric space. Let $U$ and $V$ be two disjoint subsets of X such that $X = U \cup V$.

I don't understand why $f^{-1}(U) \cap f^{-1}(V) = \emptyset $.

Is it because:

$$ \emptyset = f^{-1}(\emptyset) = f^{-1}(U \cap V)=f^{-1}(U) \cap f^{-1}(V)$$

That makes sense but I thought the last equality $f^{-1}(U \cap V)=f^{-1}(U) \cap f^{-1}(V)$ only holds iif $f^{-1}$ is injective. Does that mean $f^{-1}$ is injective or am I missing something?

Or is it because $f$ is continuous?

For a reference to the proof: https://topospaces.subwiki.org/wiki/Path-connected_implies_connected

答: It $t\in f^{-1}(U)\cap f^{-1}(V)$, then $f(t)\in U\cap V$. But we are assuming that $U$ and $V$ are disjoint.


问:
Distributions of random variables dependent on exponential random variables
Let $X$ and $Y$ be two independent random variables such that $X,Y\sim Exp(\lambda)$ Define:

$$W=\min(X,Y)$$
$$Z=\max(X,Y)$$
$$O=Z-W$$ $$M=\mathbf1_{X\le Y}=\begin{Bmatrix}
1 & \text{ if } X\le Y \\
0 & \text{ if } X\gt Y
\end{Bmatrix}$$

Identify the distributions for $W$ and $M$ (i.e. for $w \ge 0$, compute $P(W>w)$ and $P(M=1)$).

So for this question I was just wondering whether I should compute the joint probability density function $P(X>x,Y>y)$ for $W$

If X and Y both follow exponential distributions would it be reasonable to assume $M$ follows a uniform distribution with $P(M=1)=\frac{1}{2}$?

Any help would be appreciated as I am a bit stumped and feel I am missing something obvious

答:
Starting from

$$P(W>w)=P(X>w,Y>w)$$

you will easy realize that

$$W\sim \exp(2\lambda)$$

M is clearly a bernulli with parameter 0.5

I do not know what $Z$ and $O$ are given for...


问:
Developing Kepler's first law from the two-body problem
I'm supposed to develop Kepler's first law from the two-body problem. I've the information that I'd get a differential equation which can't be solved by elementary function but however the orbitals can be found out from equation.

I've been doing a lot research but I still am a bit confused what is the starting point and what is the differential equation I'm supposed to end up with. I would be very pleased if someone could help me.

答:
Starting point

Newtons law of gravity gives for a 2-body system
\begin{align}
\ddot X_1&=-\frac{m_2G(X_1-X_2)}{\|X_1-X_2\|^3}
\\
\ddot X_2&=-\frac{m_1G(X_2-X_1)}{\|X_2-X_1\|^3}
\end{align}

From 2-body to 1-body around center-of-mass

Note that the right side only depends on the difference of the positions, and that the left side does not change if one adds linear functions to the positions. In total, the system is invariant if one adds the same linear function to both. Now consider that for the linear combination $m_1X_1+m_2X_2$ you get
$$
m_1\ddot X_1+m_2\ddot X_2=0
$$
so the center-of-mass moves linearly, this linear motion is computable from the initial values.

To complement this, for the difference of positions one now gets
$$
\ddot X_1-\ddot X_2=-\frac{(m_1+m_2)G(X_1-X_2)}{\|X_2-X_1\|^3}
$$
Set $m=m_1+m_2$. With $m_1X_1+m_2X_2=m(X_0+tV_0)$ and $D=X_1-X_2$ the trjectories of the two bodies are obtained as solutions of the linear system, so
$$
X_1 = (X_0+tV_0) + \frac{m_2}{m}D\\
X_2 = (X_0+tV_0) - \frac{m_1}{m}D\\
$$

Reduction to polar coordinates

Represent $D=X_1-X_2=rS$ in a polar decomposition where $r$ is a positive scalar and $S$ a unit vector. Then per the Leibniz product rule
$$
\ddot rS+2\dot r\dot S+r\ddot S=-\frac{mGS}{r^2}
$$
A first conclusion is that $\ddot S$ lies in the plane of $S$ and $\dot S$, so that this plane is invariant. Let $N$ be a normalized unit vector orthogonal to that plane. The radius vector of a curve on a sphere is always orthogonal to its tangent vector. Together this means that
$$
\dot S=\omega(N\times S)\implies \ddot S =\dot\omega(N\times S)+\omega(N\times \dot S)=\dot\omega(N\times S)-\omega^2 S
$$
with some function $\omega$ which is the angular speed of the orbit around the origin.

The equation of motion thus splits into two components
$$
\left(\ddot r+\frac{mG}{r^2}-r\omega^2\right) S +(2\dot r\omega+r\dot\omega)(N\times S)=0
$$

Keplers second law

The second component has an integrating factor $r$  to get
$$
2\dot r\omega+r\dot\omega=0\implies r^2\omega=c
$$
where $c/2$ is the constant "area velocity".

Keplers first law

The remaining part of the equation of motion can now be formulated in $r$ alone
$$
\ddot r+\frac{mG}{r^2}-\frac{c^2}{r^3}=0
$$
Multiply with $2\dot r$ and integrate
$$
\dot r^2-\frac{2mG}r+\frac{c^2}{r^2}=const.
$$
The last two terms can be completed to a square by adding a constant, so finalize that equation as
$$
\dot r^2+\left(\frac{c}{r}-\frac{mG}{c}\right)^2=\rho^2.
$$
This can be read as a circle equation of constant radius $\rho$, so parametrize it as such
$$
\dot r=\rho\sin\phi,~~~ \frac{c}{r}-\frac{mG}{c}=\rho\cos\phi.
$$
At this point, $\phi$ is just some time-dependent function without a direct relation to the orbit position. But take the derivative of the second equation and insert the first to get
$$
-\rho\dot \phi\sin\phi=-\frac{c\dot r}{r^2}=-\frac{c\rho\sin\phi}{r^2}
\implies \dot\phi=\frac{c}{r^2}=\omega
$$
which means that $\phi$ plus some offset is the angle of the orbit.

The second circle component equation can be transformed into the usual radius parametrization
$$
r=\frac{c}{\frac{mG}{c}+\rho\cos\phi}=\frac{\frac{c^2}{mG}}{1+\frac{c\rho}{mG}\cos\phi}=\frac{R}{1+e\cos\phi}
$$
To see that this is indeed a conic section/quadratic curve in coordinates $x=r\cos\phi$, $y=r\sin\phi$ take the last equation to get
$$
R=\sqrt{x^2+y^2}+ex\\
(R-ex)^2=x^2+y^2\\
R^2=(1-e^2)x^2+2eRx+y^2\\
(e\ne1):~~R^2=\left((1-e^2)x+eR\right)^2+(1-e^2)y^2
$$
Depending on the value of the eccentricity $e$ this gives [$e<1$] an ellipse, [$e=1$] a parabola or [$e>1$] a hyperbola.

Now do some clean-up of the ad-hoc constants. Comparing the parametrizations one has $c^2=mGR$, $\rho=e\sqrt{\frac{mG}{R}}$.


问:
Find a matrix $B$ such that $B^* B=A$ for a given Hermitian $A$
Let

$$A=\left[
                           \begin{array}{ccc}
                             4 & 0 & 0 \\
                             0 & 1 & i \\
                             0 & -i & 1
                           \end{array}
                         \right].$$

Find a matrix $ B $ such that $B^*B$ =$A$
(star means conjugate transpose of $B$).

I think that $A$ is hermitian, and so $A^* =A$
Also, We can edit this as $BB^*=A$.
But I couldn't solve it completely.

答:
Let $A$ be an $n\times n$ Hermitian matrix. There must therefore be an orthonormal basis of eigenvectors corresponding to real eigenvalues. Write this as
$$A = \sum_k \lambda_k P_{v_k},$$
where $\lambda_k\in\mathbb R$ and $P_{v_k}\equiv v_k v_k^*$ is the orthoprojection onto the vector $v_k$.

Let $B$ be some $m\times n$ matrix whose SVD reads
$$B = \sum_k s_k (u_k w_k^*),$$
for some $s_k\ge0$ and $u_k\in \mathbb C^m, w_k\in\mathbb C^m$ and $\{u_k\}_k,\{w_k\}_k$ sets of orthonormal vectors.

Then, $B^* B$ reads
$$B^* B = \sum_k s_k^2 (w_k w_k^*) = \sum_k s_k^2 P_{w_k}.$$
If we want $B^* B=A$, we thus need $\sum_k\lambda_k P_{v_k}=\sum_k s_k^2 P_{w_k}$.
Being $\{v_k\}_k,\{w_k\}_k$ both sets of orthonormal vectors, this is only possible if, up to some relabeling, $\lambda_k=s_k^2$.

If $A$ (and thus $B^* B$) is non-degenerate, then we can also immediately conclude that $P_{v_k}=P_{w_k}$. More generally, there might be degenerate eigenvalues. In this case, the uniqueness of the spectral decomposition of a matrix tells us that the eigenspaces of $A$ and $B^* B$ corresponding to the same eigenvalue must be the same.

In conclusion, if $B^*B=A$ for $A$ Hermitian, then $B$ must have the form
$$B = \sum_k \sqrt{\lambda_k} (w_k v_k^*),$$
for some choice of orthonormal vectors $\{w_k\}_k\subset\mathbb C^m$.


问:
Is there a precise way to state that a structure does not satisfy a theory because its signature is different?
A structure $\mathcal{S}$ is a triple $<\sigma, D, I>$ containing a signature $\sigma$, which determines what functions and relations the structure is able to interpret, a domain $D$, which specifies what objects are in consideration, and an interpretation $I$, which assigns truth values to predicates, functions, and relations.

Yesterday I learnt that the $\vDash$ and $\not\vDash$ symbols are only defined for structures when the signature of the structure (on the LHS), at least contains the symbols of the theory or sentence (on the RHS).

$$ \mathcal{S}\vDash T$$

This got me wondering: It is a true statement to say that a structure $\mathcal{S}$ does not satisfy a theory $T$, if the structure does not contain in its signature at least the syntax used by the theory $T$ (for instance, it may not contain a predicate the theory $T$ uses).

However, with the way $\vDash$ is defined, we cannot write $\mathcal{S}\not\vDash T$ from this observation.

A question then arises: Is there any way to express that a structure does not satisfy a sentence or theory, because its signature doesn't match, in more precise terminology than natural language?

答:
You have hit on an interesting kind of question that has been asked since the beginning of the field of mathematical logic.

The general question is: When we define some relation $≈$ between objects of type $S$ and objects of type $T$, we are usually interested in asking whether $x≈y$ only when $x∈S$ and $y∈T$. So what should we do in the other case?

There are two main 'camps' of preference for the answer. Everyone agrees that if we represent $≈$ as a set of pairs from $S×T$, and define "$x≈y$" to be simply "$⟨x,y⟩∈S×T$, then of course nothing is related by $≈$ outside of $S×T$.

The first camp sees no reason not to consider $≈$ as nothing more than that representing set. That is precisely why Noah says "For what it's worth I disagree with your initial claim, although of course this will come down to a convention: I would say that "$⊭$" is exactly the complement of "$⊨$"".

The second camp prefers to consider the expression "$x≈y$" as well-typed iff one has shown that $x∈S$ and $y∈T$. If one has not done so, then one is not permitted to write that expression. Note that this is a meta-level statement; we cannot within the foundational system itself write down some sentence saying that $x≈y$ is well-typed! This is also called strict typing, and such restrictions on what we can write make it correspond more to what we intuitively feel is sensible to write. This is precisely why Noah also said "I also think that expositions ought not to crucially exploit pathologies when there's a natural way to avoid this.". However, be aware that strict typing comes with a rather heavy price; we no longer have a nice clean recursive definition of well-formed expressions! So from the perspective of foundations of mathematics it is best to first see how things are done in a standard FOL system first (as Noah showed) before looking at other systems with stricter typing rules, even if the latter may feel more natural.

For a very simple example, we define $x/y$ for $x,y∈ℝ$ such that $y≠0$. But the cleanest FOL formalization of reals would also define $x/0$ for all $x∈ℝ$. Surprised? Well, if not you cannot have $∀x∈ℝ\ ( x≠0 ⇒ 1/(1/x)=x )$!

By the way, this divide also arises in programming (unsurprisingly), where one kind of programming languages has (mostly) static typing and you simply cannot use objects in ways not allowed by their types, whereas the other kind has (mostly) dynamic typing and you can willy-nilly attempt to use an object in (almost) any way you like (and it may give a run-time error instead of a compile-time error).


问:
Number of functions from $A$ to $A$
We have the set $A=\{1,2,3,4,5\}$. what is the number of functions from $A$ to $A$?

I think the domain of the function should contain all the elements of $A$ but range of the function may have any elements of $A$ or not having them. so for each of the numbers from $1$ to $5$ we have two possibilities $(1)$ existing in the range of function. $(2)$ not existing in the range. so the number of functions from $A$ to $A$ is $2^5=32$. is my justification right?

答:
No, you must take account of which element is mapping to which.

Each element of the domain can map to one of $5$ elements so the number of possible functions is $5^5$.

Title: Number of functions from $A$ to $A$


问:
Limit of exponential of sines and cosines
The task is to calculate
$$\lim_{x\rightarrow \infty} e^{x \ln (\sin(1/x)+\cos(1/\sqrt x)}.$$
Since $\lim_{u\rightarrow 1}\ln u\approx u-1$, the original limit can be rewritten as
$$
\lim_{x\rightarrow \infty} e^{x  (\sin (1/x)+\cos ((1/\sqrt{x})-1)} = \lim_{x\rightarrow \infty} e^{x (1/x -1 -1 )} = e.
$$

However, the solution manual says $\sqrt{e}$ is the correct answer. Can you help me figure out why? Thank you.

答:
Use Maclaurin's expansions:

$\sin\frac1x=\frac 1 x+o\bigl(\frac 1x\bigr)$,

$\cos \frac1{\sqrt x}=1-\frac1{2x}+o\bigl(\frac 1x\bigr)$ (from MacLaurin's expansion at order $2$ and substitution).

We deduce that the expansion of the exponent is
\begin{align}x \ln \biggl(\sin\Bigl(\frac{1}{x}\Bigr)+\cos\biggl(\frac{1}{\sqrt x}\biggr)\biggr)
&=x\biggl(\ln\Bigl(\frac1x+1-\frac1{2x}+o\Bigl(\frac 1x\Big)\biggr)=x\biggl(\ln\Bigl(1+\frac1{2x}+o\Bigl(\frac 1x\Bigr)\Bigr)\biggr)\\
&=x\biggl(\frac1{2x}+o\Bigl(\frac 1x\Big)\biggr)=\frac12+o(1).
\end{align}
Is it clear now?


问:
Using the surface of the unit sphere to calculate $\int_{\mathbb R^3}\chi_{\{x_3>0\}}\chi_{B^c_{1/n}(0)}\exp(-\lvert x\rvert^2)/\lvert x\rvert dx$
I'm struggling with finding a proof idea for this integral from an integration theory practice exam.

Suppose
$$f_n:=\frac{\exp(-\lVert x\rVert_2^2)}{\lVert x\rVert_2}\chi_{\{x_3>0\}}(x)\chi_{B^c_{1/n}(0)}(x),$$
show that
$$\int_{\mathbb R^3}f_n d\lambda_3 = \pi\exp(-n^{-2}).$$

Hint The surface area of $\mathbb S^2$ is $\omega_3=4\pi$.

Geometrically, we are integrating over the top half-space, excluding a $1/n$ sphere, so naturally, I switched to spherical coordinates. This left me with

$$\int_{\mathbb R^3}f_nd\lambda_3 = \int_{1/n}^\infty\int_0^{\pi/2}\int_0^{2\pi}r\sin(\theta)\frac1r\exp(-r^2)d\varphi d\theta dr= 2\pi\int_{1/n}^\infty\exp(-r^2)dr,$$
which is not sufficient, even if it it correct.

Because of the hint, I'm surely missing a different perspective on the problem. The only relation I can think of between the surface and the volume is the divergence theorem (Gaussian integral theorem), but I don't see how that could be applied here.

How can this be shown using the hint, and what proof idea is behind it?

答:
Your argument is correct, but the determinant you obtain after transformation is $r^2 \sin(\theta)$ leaving you with the expression:

$$2 \pi\int \limits_{1/n}^{\infty} r\cdot exp(-r^2)dr=-2 \pi\frac{\exp(-r^2)}{2}|_{r=1/n}^{r=\infty}.$$
which can be seen using the $u-$substitution $u=r^2.$


问:
can a continuous or differentiable function have consecutive same values over a partition [a,b]? except constant function
Can a continuous or differentiable function have consecutive same values over a partition [a,b], which means it has a graph containing a horizontal segment?</p>

答:
Without loss of generality, take $[a,b]=[-1,1]$.

For a continuous example, take the function
$$
f(x)=x1\{x>0\}
$$
This is the "rectified linear unit".

For a differentiable example, take:
$$
g(x)=x^31\{x>0\}
$$
The derivative of this function at $x=0$ is zero.

The answer to your question is no, if you require that the function is real analytic (a real analytic function for which all derivatives are zero at a point is constant everywhere).


问:
Is the following relation $(x_1,x_2)\mathcal{R}(y_1,y_2) \iff x_1\geq y_1 \text{ and } x_2\geq y_2$ complete?
Let $\mathcal{R}$ denote a relation between vectors in $x, y \in \mathbb{R_+^2}$.

The relation is called complete if $\forall x, y \in \mathbb{R_+^2}$ we have $x\mathcal{R}y$ or $y\mathcal{R}x$.

The following relation is given $$(x_1,x_2)\mathcal{R}(y_1,y_2) \iff x_1\geq y_1 \text{ and } x_2\geq y_2.$$

If we take the vectors $x=(2,2)$ and $y = (3,1)$ then we have $\Big(\text{not } [x\mathcal{R}y]\Big)$ since $2 = x_1 < y_1 = 3 $ and also $\Big(\text{not } [y\mathcal{R}x]\Big)$ since $1 = y_2 < x_2=2 $.

Hence $\mathcal{R}$ is not a complete relation. Is this reasoning correct?

答:
Its neither $x<y$ nor $x>y$.
Thus the ordering relation is not total.


问:
Convergence of double summation series under diagonalization

I am stuck at 2.8.7b. The author suggested the strategy in the proof of theorem 2.8.1. But I can't see how the strategy can be deployed. As attached, the proof of theorem 2.8.1 involves 2 variables whereas 2.8.7b involves only 1 variable.

I just don't know how. I can't see how $s_{nn}$ is related in this case. If I tried to minus $d_k$ from $s_{kk}$, it's just another triangle sum.

答:
Since $t_{nn}$ converges, it is a Cauchy sequence. Hence,
\begin{align}
\forall\epsilon\in\mathbb{R}^+\exists n_1\in\mathbb{N}\forall n,m\geq n_1\rightarrow\lvert t_{nn}-t_{mm}\rvert<\epsilon
\end{align}
Hence,
\begin{align*}
\forall\epsilon\in\mathbb{R}^+\exists n_1\in\mathbb{N}\forall n\geq 2n_1,\\
\left\lvert\sum_{k=2}^{n} d_k-\sum_{i,j=1}^{n}a_{ij}\right\rvert&\leq\left\lvert\sum_{i,j=2}^{n}\lvert a_{ij}\rvert-\sum_{i,j=2}^{\left\lfloor\frac{n}{2}\right\rfloor}\lvert a_{ij}\rvert\right\rvert\text{ (see footnote)}\\
&=\left\lvert t_{nn}-t_{\left\lfloor\frac{n}{2}\right\rfloor\left\lfloor\frac{n}{2}\right\rfloor}\right\rvert\\
&<\epsilon
\end{align*}
Therefore,
\begin{align}
d_{kk}-s_{nn}\to 0\\
s_{nn}\to\mathbb{S}
\end{align}
By algebraic limit theorem, $d_{kk}\to$ S.

Footnote: This inequality can be visualized as such. $s_{nn}$ is the sum of a square sized with $n^2$ elements. $d_{kk}$ is the sum of elements in the upper left triangle. Hence, the difference of two will be the sum of elements in the lower right triangle. By triangle inequality, they are less than the sum of absolute terms. However, I can't come up with a Cauchy sequence that possesses this property. So I turn to $t_{nn}-t_{\left\lfloor\frac{n}{2}\right\rfloor\left\lfloor\frac{n}{2}\right\rfloor}$, which turns out cover slightly more than the lower right triangle.


问:
Prove if $n^2$ is even, then $n$ is even.
I am just learning maths, and would like someone to verify my proof.

Suppose $n$ is an integer, and that $n^2$ is even.  If we add $n$ to $n^2$, we have $n^2 + n = n(n+1)$, and it follows that $n(n+1)$ is even.  Since $n^2$ is even, $n$ is even.

Is this valid?

答:
Yes, it's valid.  That said I wouldn't say "it follows that n(n+1) is even", because n(n+1) is just even.

So, you have  n$^2$+n=n(n+1) and n(n+1) as even.  So, we have that n$^2$+n equals the sum of an even integer n$^2$, and some integer n.  So, n is either odd or even.

If n were odd, then we would have n$^2$+n would equal the sum of the even number n$^2$ and the odd number n.  The sum of an odd number and an even number is always an odd number.  So, then n(n+1) would equal an odd number.  But, n(n+1) is not an odd number.  Since we have a contradiction, by the rule of negation introduction, we have that n is not odd.  Since every integer is either even or odd, it follows by the (often derived) rule of disjunctive syllogism, that n is even.


答:
Up to $10^8$, there are $4$ counterexamples: $41124, 230867, 358267, 37539572$.

This was possible thanks to Thomas Andrews, where they write a formula for $f(a)$.

I obtained counterexamples by running the formula in PARI/GP:

```<code>forcomposite(a=2,10^8,s=0;fordiv(a,d,s+=d*eulerphi(a/d));if(s%(a-1)==1,print([a])))
</code>```

Title: Show that $\sum_{k=1}^{a-1}(k,a)\equiv0\pmod{a-1}\iff a $ is prime

Body:
Denote $\gcd(a,b)$ as $(a,b)$

Let $a\ge 2$, can it be shown that

$$(1,a)+(2,a)+\cdots+(a-1,a)\equiv0\pmod{a-1}$$

Satisfy if and only if $a$ is prime?

Clearly if $a$ is prime then $(1,a)+(2,a)+\cdots+(a-1,a)\equiv1+1+\cdots +1\equiv a-1\equiv0\pmod{a-1}$

Claim's hold true upto $10^4$ source code Pari GP

```<code>forcomposite(a=2,10000,if(sum(k=1,a-1,gcd(k,a))%(a-1)==0,print([a])))
</code>```

Update #1

Sequence1 for composite $a$,  $[a,\sum_{k=1}^{a-1}(k,a)\equiv\pmod{a-1}]$

```<code>[4, 1][6, 4][8, 5][9, 4][10, 8][12, 6][14, 12][15, 2][16, 2][18, 11][20, 14][21, 4][22, 20][24, 7][25, 16][26, 24][27, 2][28, 22][30, 18][32, 18][33, 8][34, 32][35, 14][36, 27][38, 36][39, 10][40, 23][42, 30][44, 38][45, 12][46, 44][48, 4][49, 36][50, 47][51, 14][52, 46][54, 30][55, 26][56, 39][57, 16][58, 56][60, 5][62, 60][63, 24][64, 3][65, 32][66, 54][68, 62][69, 20][70, 5][72, 64][74, 72][75, 28][76, 70][77, 44][78, 66][80, 36][81, 56][82, 80][84, 21][85, 44][86, 84][87, 26][88, 71][90, 32][91, 54][92, 86][93, 28][94, 92][95, 50][96, 84][98, 10][99, 48][100, 24]
</code>```

Edit: link for MO post which asks for such infinitely many composite number.

Body:
Up to $10^8$, there are $4$ counterexamples: $41124, 230867, 358267, 37539572$.

This was possible thanks to Thomas Andrews, where they write a formula for $f(a)$.

I obtained counterexamples by running the formula in PARI/GP:

forcomposite(a=2,10^8,s=0;fordiv(a,d,s+=d*eulerphi(a/d));if(s%(a-1)==1,print([a])))


问:
Prove that $s_{n+1}=s_n^2-2$ diverges.
Let $s_1=5$ and for $n\geq 1$, $s_{n+1}=s_n^2-2$. Do the following steps to show that $(s_n)$ diverges.

1.If $\lim s_n = a$ then $a^2 -2 = a$.

$a=\lim s_{n+1}=\lim {s_n^2-2}=\lim {s_n^2}-\lim (2)=(\lim {s_n})^2-\lim (2)=a^2-2$

2. If $\lim s_n = a$ then a has to be either $−1$ or $2$.

From (1), we have

$a=a^2-2$

$a^2-a-2=0$

$a=−1, 2$

3. Show that $\lim s_n$ can neither be $−1$ nor $2$. (Hint: write down the first few terms of $(s_n)$ to get some insights.)

$s_n\leq s_{n+1}$ for all $n$

Proof, by induction:

$P_1$ is true; at $n=1$, $5=s_1\leq s_2=5^2-2=23$

Suppose that $s_n\leq s_{n+1}$ for some $n$

By the assumption we have $s_{n+1}=s_n^2-2\leq s_{n+1}^2-2=s_{n+2}$

As a result, $s_n$ is increasing, so $s_n\geq 5$ for all $n$. Now since $s_n\geq 5$, this implies that $\lim s_n \geq 5$ which implies that it cannot be $-1$ or $2$.

4. Combine these steps to show that $\lim s_n$ does not exist.

Is there any further step to do here or just combine the 2 and 3?

Are my answers correct, please? Thank you in advance.

答:
In step $4$,
you can just do:

Assume $\text{lim}s_n = a $ exists. Then by step $1,2$, $a$ should be either $-1$ or $2$.
However, step $3$ showed that $s_n \to \infty$. So it's contradiction.

Therefore $\text{lim} s_n $ doesn't exist.

This is what I got. How is it? I have a different solution to step$3$ though.

Cheers math 447 bro.

Title: Prove that $s_{n+1}=s_n^2-2$ diverges.


问:
How to determine the logical form of a statement?
I'm having trouble understanding how to determine the logical form of a statement.

Here's my current understanding (I'm not sure if all of them are correct though):

p is a propositional variable means that p can be any proposition, so the Truth Value of p has 2 possibilities: T or F(but we don't know which one yet until we assign p to be a certain statement). I think it's kind of analogous to numerical variables, like let x be a natural number, then the "evenness" of x has 2 possibilities: even or odd (but we don't know which one yet until we assign a value to x, like 4 or 123...)

A logical form is an expression consist of the component statement variables and the logical connectives, and the truth table of logical form shows all the possible truth values of the form for all combinations of Truth Values of the statement variables. I like to think of the truth table for the logical form to be what all statements of that logical form mean logically (replacing the propositional variables with the respective component statements, of course).

If my understanding of propositional variable and logical form is correct, then here's what I don't understand: when determining the logical form of a statement, the general procedure is to first identify the component statements and give them names as variables, then translate the whole statement into an expression consist of the propositional variables and appropriate logical connectives. For example, say "the car is black and the bike is white". When identifying the component statement, the solution always says that denote p:= the car is black and q:= the bike is white. How can that work?

I thought what I should do is to completely remove the component statements and replace them with 2 propositional variables p and q, which results in "p and q", and then set up the truth table for that logical form and get to the logical form which is p∧q. If we denote p and q to be the component statements, then p and q are not propositional variables anymore, right? therefore, there will be no truth table for the initial statement (because "p and q" now is not a logical form, it's a statement so it can only have a single truth value). Then how can we determine what is the logical connective for the initial statement's logical form?

I guess I also want to ask is that can there be a truth table for a statement? I thought truth tables only exist for logical forms.*

答:
The notions of logical form and truth table really only make sense when speaking about formulas where the component statements are undetermined and independent, which means each component can be true and false  without influencing the truth value of other components. In particular, the truths value of the components are not fixed a priori.

This reading of (atomic) components as undetermined and independent is made explicit by the use of variables: If I use different variables $p$ and $q$ in a formula, I signal to the reader that all $4$ combinations of truth values for $p$ and $q$ can occur.

Sometimes people use descriptive statements such as "the car is black" or "the bike is white" as components. This is indeed a source of confusion, as it might suggests that the truth values are fixed -- we only have to look at the world to find out whether the car is black, or the bike is white and so on. In this fixed-value interpretation, it may make sense to say that the statement "the car is black and the bike is white" has the linguistic form of a conjunction, but from a logical perspective it is actually trivial: Either the statemenent is true, in which case it is equivalent to $\top$ (or any other tautology) or it is false and therefore equivalent to $\bot$ (or any other contradiction).

So long story short, if you are asked to determine the logical form of "The car is black and the bike is white" it is implicitly assumed that the components "The car is black" and "the bike is white" do not have a fixed truth value (and are independent), i.e. they behave like variables. Keeping this in mind, it is still fine to use "the car is black" as a variable name instead of something less descriptive like "p".


问:
Number Theory Squares
Determine all natural numbers $n ≥ 2$ for which
$$n=a^2+b^2$$
holds, where a is the smallest divisor of n other than 1 and b is any divisor of n.

答:
We obviosly have $a$ is prime. Since $a\mid n = a^2+b^2$, we have $a\mid b$, so $b = a k$.

But $b \mid n = a^2+b^2$, so $b \mid a^2$. Then $a k \mid a^2$, so $k \mid a$. Since $a$ is prime, $k=1$ or $k=a$.

If $k=1$ we have $b=a$, then $n=2a^2$, but since $a$ is the smaller non-trivial divisor, $a=2$, so $n=8$.

If $k=a$ we have $b=a^2$, then $n=a^2+a^4=a^2(1+a^2)$ that is even, so again $a=2$ and $n = 2^2+ 4^2 = 20$.

So the only solutions are $n = 8, 20$.

Title: Number Theory Squares


问:
What is a basis for the space of $n\times n$ Hermitian matrices?
So I was working on a specific problem related to Hermitian matrices. If we let $H_n$ denote the set of n x n Hermitian matrices. We're told that $H_n$ is a real vector space under matrix addition and scalar multiplication by a real number. I don't understand why though, because what would prevent you from adding two imaginary numbers?

Then we're told to write a basis for $H_2$. Now for a Hermitian matrix, it's equal to its conjugate transpose, so a basis would be:

$$\left(\begin{array}{ccc}
0 & 1 \\
1 & 0\\
\end{array}\right) , \left(\begin{array}{ccc}
0 & 0 \\
0 & 1\\
\end{array}\right), \left(\begin{array}{ccc}
1 & 0 \\
0 & 0\\
\end{array}\right), \left(\begin{array}{ccc}
0 & i \\
-i & 0\\
\end{array}\right)$$

Which should be the same as the basis for a symmetric matrix, correct?

From there though, we have $sH_n$ represent skew-Hermitian matrices. I believe the basis for $sH_n$ is:

Edit: updated my basis here.

$$\left(\begin{array}{ccc}
0 & -1 \\
1 & 0\\
\end{array}\right), \left(\begin{array}{ccc}
0 & i \\
i & 0\\
\end{array}\right)$$

because the diagonals would have to be zero right, to make the conjugate transpose of A be equal to -A?

From there though, they ask if $sH_n$ is a real vector space, and I'm not sure what to answer. It's real if we only use reals?

I'm asked the same thing for $U_n$, where $U_n$ represents n x n unitary matrices. I'm not even sure how to come up with a basis for this set??

Edit: Technically, the question says "Is $U_n$ a real vector space? Write a basis if possible."

I suppose I couldn't write a basis because $U_n$ is not a real vector space then. How would I show that though?

Thanks in advance for your help.

答:
In the general case, a standard basis is given (up to normalisation) by
$$E_{ii},\qquad E_{ij}+E_{ji}, \qquad i(E_{ij}-E_{ji}),$$
for all $i,j=1,...,n$ with $i<j$, and where I used the notation $E_{ij}\equiv e_i e_j^*$, where $e_i$ is the unit vector that is nonzero only in its $i$-th component: $(e_i)_j =\delta_{ij}$ and $(E_{ij})_{k\ell}=\delta_{ik}\delta_{j\ell}$.

Clearly all these matrices are Hermitian. To verify that form an orthogonal basis, we just need to check that
$$\mathrm{Tr}(E_{ii} (E_{jk}+E_{kj}))
=\mathrm{Tr}(E_{ii}(E_{jk}-E_{kj})))
=\mathrm{Tr}(E_{ii}E_{jk})) = \delta_{ij}\delta_{ik} = 0, \qquad (\text{as } j<k), \\
\mathrm{Tr}[(E_{ij}+E_{ji})(E_{k\ell}-E_{\ell k})]
= 0, \\
\mathrm{Tr}[(E_{ij}+E_{ji})(E_{k\ell}+E_{\ell k})]
=2(\delta_{jk}\delta_{i\ell} + \delta_{j\ell}\delta_{ik}), \\
\mathrm{Tr}[(E_{ij}-E_{ji})(E_{k\ell}-E_{\ell k})]
=2(\delta_{jk}\delta_{i\ell} - \delta_{j\ell}\delta_{ik}).$$

See also this related question.


答:
$a,b$ and $n$ cannot all be odd and so, in all cases, $n$ is even and $a=2$. Then $$n=4+b^2.$$
Since $b$ is a factor of $n$ it is a factor of $4$ and so $b=1,2\text { or }4$.

The solutions are $a=2$ and $n=8\text{ or }20$.

Title: Number Theory Squares

Body:
Determine all natural numbers $n ≥ 2$ for which
$$n=a^2+b^2$$
holds, where a is the smallest divisor of n other than 1 and b is any divisor of n.


问:
Differentiating $ y= xe^{1\over x} $
Can someone please help me? I'm trying, but I really can't find the second derivative of $y= xe^{1/x}$. Thanks!</p>

答:
Thi is a hint to solve this kind of problem: you take the logarithm of the function and you have:
$$\log y= \log(xe^{\frac{1}{x}})=\frac{1}{x}+\log x$$
so derive it:
$$\frac{y'}{y}=-\frac{1}{x^2}+\frac{1}{x}$$
Thus
$$y'=y(-\frac{1}{x^2}+\frac{1}{x})=-\frac{e^{\frac{1}{x}}}{x}+e^{\frac{1}{x}}$$

Deriving again with chain rule :
$$y''=-\frac{e^{\frac{1}{x}}} {x^2}+ \frac{e^{\frac{1}{x}}}{x^2}+\frac{e^{\frac{1}{x}}}{x^3}= \frac{e^{\frac{1}{x}}}{x^3}$$

Title: Differentiating $ y= xe^{1\over x} $


问:
Minimising surface area of a rectangular parallelepiped
We want to construct a rectangular parallelepiped with Volume $2L$, one of its sides measures $10$cm, let $x$ and $y$ be the two dimensions in dm, of this box. (Note: $1$dm = $10$cm)

1)Prove that the surface of this box is $S(x) = 2x+4+4/x$.

2)What are the dimensions such that the quantity of matter used to fabricate it is minimal?

I think I misunderstood something; the given isn't really clear. Can I get some help?

答:
One of the sides is $10$ cm i.e $1$ dm (decimeter). Other two sides are $x$ and $y$ dm.

So volume of box $V = 2 = xy \implies y = \frac{2}{x}$

Surface area $S = 2(xy + y + x)$ (as one of the sides is $1$).

Writing $y$ in terms of $x$,

$S = 4 + 2x + \frac{4}{x}$.

Now to find $x$ that minimizes the surface area, differentiate $S$ with respect to $x$ and equate to zero.

Can you take it from here?


问:
Prove that $b-1 \mid a \iff b-1\mid s$
Let $a=(a_na_{n-1}…a_o)_b$ and $s=\sum_{i=0}^na_i$ Show that if $b-1 \mid a \iff b-1\mid s $.

I’ve found this problem in the number theory book that i read.

I think if the Base was $10$, it would be more simple, however i don’t know how to prove this.

答:
You have

$$a=a_nb^n + a_{n-1}b^{n-1} + ... + a_1b + a_0$$

and because $b \equiv 1$ modulo $b-1$, you get
$$a \equiv a_n + a_{n-1} + ... + a_1 + a_0\quad [b-1]$$
i.e.
$$a \equiv s \quad [b-1]$$

The result follows.

Title: Prove that $b-1 \mid a \iff b-1\mid s$


问:
How to evaluate double limit of multifactorial $\lim\limits_{k\to\infty}\lim\limits_{n\to 0} \sqrt[n]{n\underbrace{!!!!\cdots!}_{k\,\text{times}}}$
Define the multifactorial function $$n!^{(k)}=n(n-k)(n-2k)\cdots$$ where the product extends to the least positive integer of $n$ modulo $k$. In this answer, I derived one of several analytic continuations of this function to the real numbers, which is as follows $$x!^{(k)}=k^{x/k}\Gamma\left(1+\frac xk\right)\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{\sin(\pi(x-i))\cot(\pi(x-i)/k)/k}.$$ The limit of original interest $$F(k)=\lim_{x\to0}\,(x!^{(k)})^{1/x}=\left[\frac k{e^\gamma}\prod_{i=1}^{k-1}\left(\frac{\Gamma(1+i/k)}{ik^{-i/k}}\right)^{\pi(-1)^i\cot\frac{\pi i}k}\right]^{1/k}$$ is a straightforward consequence of the above result. Out of curiosity I plotted $F(k)$ and it appears that the double limit $$m=\lim_{k\to\infty}\lim_{x\to0}\,(x!^{(k)})^{1/x}$$ exists, converging rapidly to around $0.852$ when $k\in(s-1/2,s+1/2)$ for all positive odd integers $s$, but converging much slower to the same value when $k\in(s+1/2,s+3/2)$.

We can rewrite the limit as \begin{align}m&=\lim_{k\to\infty}\exp\left(\frac{-\gamma+\log k+\pi\sum\limits_{i=1}^{k-1}(-1)^i\cot\frac{\pi i}k\log\left(\frac{\Gamma(1+i/k)}{ik^{-i/k}}\right)}k\right)\\&=\lim_{k\to\infty}\exp\left(\frac\pi k\sum\limits_{i=1}^{k-1}(-1)^i\cot\frac{\pi i}k\log\frac{\Gamma(1+i/k)}{ik^{-i/k}}\right)\\&=\exp\left(\lim_{k\to\infty}\frac\pi k\sum\limits_{i=1}^{k-1}(-1)^i\cot\frac{\pi i}k\log\frac{k^{i/k}\Gamma(i/k)}k\right)\end{align} using L'Hopital on $(-\gamma+\log k)/k$. The term inside the exponential looks very much like a Riemann sum but I'm not sure where to go after that. It seems that none of the terms in the logarithm can be split additively to evaluate the limit as each component by itself is divergent.

Is there a closed form for this double limit?

答:
Proposition:

$$\boxed{\;\ln m=-\gamma\ln2+\frac{\ln^22}{2}\;}$$

Explanation (sketch of the proof): for $i,k\to\infty$ such that $i/k=x$ with finite $x>0$ the expression under the sum $\sum_{i=1}^{k-1}$ tends to $0$ and therefore this range of $i$ does not contribute to the limit. We may therefore restrict the summation to the values  with $i/k\ll 1$. Expand everything in $i/k$:
$$\cot\frac{\pi i }{k}\to \frac{k}{\pi i },\qquad \ln\frac{\Gamma(i/k)}{k}\to -\ln i,\qquad \ln k^{i/k}=i\frac{\ln k}{k}.$$
The third piece should not contribute to the limit $k\to\infty$ because of the cutting factor $\frac{\ln k}{k}$. The first two produce the following expression in the exponential:
$$\sum_{i=1}^\infty (-1)^{i+1}\frac{\ln i}{i}=-\gamma\ln2+\frac{\ln^22}{2}.$$
There admittedly remain some details to be filled for the above to become a rigorous proof, however I leave it to whoever is interested.


问:
System of differential equations: why does it solve it the example?
consider the system with conditions: $\quad y_1(0) = 1,\quad y_2(0) = -1,\quad y_3(0) = 0,\quad y_4(0) = 2$

$\begin{align}
&{y_1}'= y_1 + 6\,y_2\\
&{y_2}'= y_2 + 6\,y_3\\
&{y_3}'= y_3 + 6\,y_4\\
&{y_4}'= y_4
\end{align}$

what can be written as matrix: $A = \left(\begin{array}{cccc} 1 & 6 & 0 & 0\\ 0 & 1 & 6 & 0\\ 0 & 0 & 1 & 6\\ 0 & 0 & 0 & 1 \end{array}\right)$

Thus the solution: $\exp(A\,x)\,y_0 = e^{x}\,\underbrace{\left(\begin{array}{cccc} 1 & 6\,x & 18\,x^2 & 36\,x^3\\ 0 & 1 & 6\,x & 18\,x^2\\ 0 & 0 & 1 & 6\,x\\ 0 & 0 & 0 & 1 \end{array}\right)}_{B} \,\left(\begin{array}{c}y_1(0)\\y_2(0)\\y_3(0)\\y_4(0)\end{array}\right) = B\,\left(\begin{array}{c}1\\-1\\0\\2\end{array}\right) $

Now to my question: how does it solve the system above? May sound stupid, because I understand solving something without knowing what. For example: what's $y_1, y_2,y_3,y_4$ in the solution matrix? The 1st, 2nd, 3rd, 4th row or entry? But even then I can't put it together.

Also: How is the starting condition vector built?  What would be the difference between $y_1(0) = 1$ and $y_1(1) = 1$

May be essential questions but these Systems are really kicking me off.

答:
We are given the system

$$\begin{align}
&{y_1}'= y_1 + 6\,y_2 = 1\, y_1 + 6\, y_2 + 0 \,y_3 + 0\, y_4\\
&{y_2}'= y_2 + 6\,y_3 = 0 \,y_1 +1\, y_2 + 6\, y_3 + 0\, y_4\\
&{y_3}'= y_3 + 6\,y_4 = 0\, y_1 + 0\, y_2 + 1\, y_3 + 6\, y_4\\
&{y_4}'= y_4 = 0 \,y_1 + 0 \,y_2 + 0\, y_3 + 1\, y_4
\end{align}$$

This can be compactly written as

$$y' = A y =  \left(\begin{array}{cccc} 1 & 6 & 0 & 0\\ 0 & 1 & 6 & 0\\ 0 & 0 & 1 & 6\\ 0 & 0 & 0 & 1 \end{array}\right) \begin{pmatrix} y_1\\y_2\\y_3\\y_4 \end{pmatrix}$$

We now solve for the matrix exponential or matrix exponential (note that there are many approaches to solve a system, like eigenvalues/eigenvectors, diagonalization (when possible), Putzer's Algorithm...)

$$e^{Ax} = e^x\left(\begin{array}{cccc} 1 & 6\,x & 18\,x^2 & 36\,x^3\\ 0 & 1 & 6\,x & 18\,x^2\\ 0 & 0 & 1 & 6\,x\\ 0 & 0 & 0 & 1 \end{array}\right)$$

Once we have the matrix exponential and are given initial conditions (they could have chosen anything), we can write the solution as

$$y(x) = e^{Ax} y_0 = e^x\left(\begin{array}{cccc} 1 & 6\,x & 18\,x^2 & 36\,x^3\\ 0 & 1 & 6\,x & 18\,x^2\\ 0 & 0 & 1 & 6\,x\\ 0 & 0 & 0 & 1 \end{array}\right)\left(\begin{array}{c}1\\-1\\0\\2\end{array}\right) = \left(
\begin{array}{c}
 72 e^x x^3-6 e^x x+e^x \\
 36 e^x x^2-e^x \\
 12 e^x x \\
 2 e^x \\
\end{array}
\right)$$

This means

$$\begin{align} y_1(x) &= 72 e^x x^3-6 e^x x+e^x\\ y_2(x) &= 36 e^x x^2-e^x \\ y_3(x) &= 12 e^x x \\y_4(x) &=  2 e^x \end{align}$$

We can verify this solution, for example

$$y_1' = y_1 + 6 y_2$$

Calculating the LHS

$$y_1' = 72 e^x x^3+216 e^x x^2-6 e^x x-5 e^x$$

Calculating the RHS

$$y_1 + 6 y_2 = 72 e^x x^3-6 e^x x+e^x + 6(36 e^x x^2-e^x) = 72 e^x x^3+216 e^x x^2-6 e^x x-5 e^x$$

The LHS $= $ RHS, that checks out and this looks like a valid solition - as will the three other results.


问:
Convergence of integral and summation for Time taken for complete revolution around vertical circle
I want to find Time taken to complete Vertical circular motion by Particle of mass $m$

So I proceed as follows

Applying work energy theorem

$$-mgR(1-\cos(\theta)=\frac12 mv^2-\frac12 mu^2$$

$$\implies v^2=u^2-2gR(1-\cos\theta)$$

$$\implies \omega=\frac{\sqrt{u^2-2gR(1-\cos\theta)}}{R}$$

$$\implies \frac {d\theta}{dt}=\frac{\sqrt{u^2-2gR(1-\cos\theta)}}{R}$$

$$\implies \frac {d\theta}{\sqrt{u^2-2gR(1-\cos\theta)}}=\frac{dt}{R}$$

Let T be the time taken for complete revolution

Now, Integrating both sides

$$\int_{0}^{2\pi}\frac {d\theta}{\sqrt{u^2-2gR(1-\cos\theta)}}=\int_{0}^{T}\frac{dt}{R}$$

$$\implies T=2R\int_{0}^{\pi}\frac{d\theta}{\sqrt{u^2-4gR\sin^2\theta}}$$

$$\implies T=\frac{2R}{u}\int_{0}^{\pi}\frac{d\theta}{\sqrt{1-\frac{4gR}{u^2}\sin^2\theta}}$$

On evaluating this elliptic integral we get,

$$T=\sum_{n=0}^{\infty} \left(\frac{2}{u}\right)^{2n+1}\left(\frac{(2n-1)!!}{2^n n!}\right)^2 (gR)^n$$

Here $T$ denotes Time taken for complete revolution. Block will complete full circle only if $u\geq \sqrt{5gR}$. So my question is $T$ will be Real only when $u\geq \sqrt{5gR}$ so This condition should exist in integral as well as summation for $T$ but I'm not able to see it. Also Wolfram alpha evaluates integral only if $u\geq \sqrt{5gR}$

答:
Both the integral and the series converge if and only if $u > 2 \sqrt{g R}$, but your formula for $T$ is correct only if $u \geq \sqrt{5 g R}$. This discrepancy can be understood as follows:

The first condition only ensures that the particle has enough energy to reach the height $2R$ (it can be found by equating $2 R m g$ and $\frac{1}{2} m u^2$).

But your formula for $T$ can only be correct if the particle slides along the loop at all times. This only happens if the sum of the centrifugal force and the radial component of the gravitational force is always non-negative, i.e. $m R \omega^2 + m g \cos (\theta) \geq 0$. Using the conservation of energy to replace $\omega$ we can rewrite this inequality as $\frac{u^2}{g R} \geq 2 - 3 \cos(\theta)$. For $\theta = \pi$ the right-hand side is maximal, so we obtain the second (stronger) condition $\frac{u^2}{g R} \geq 5 ~ \Leftrightarrow~ u \geq \sqrt{5gR}$.

For $2 \sqrt{gR} < u < \sqrt{5gR}$ the particle has enough energy to reach the top of the loop, but it will detach itself from the track before that happens. Since your entire calculation is based on the energy conservation law (which contains less information than the original equations of motion), the stronger condition for the applicability of your result never appears and must be found separately.


问:
Is the function $f(x)= x(\pi-x)$ $2\pi$-periodic?
Is the function defined as follows:
$$
f(x) = \left\{\begin{align}0 \ \ -\pi \le x \le 0 \\ x(\pi - x) \ \ 0 \le x \le \pi \end{align}\right.
$$
$2\pi$-periodic or not?

My attempt:

The function is defined on $[-\pi, \pi]$. First we must check that if $x$ is in $[-\pi,\pi]$, $x + 2\pi$ remains in $[-\pi, \pi]$, which is wrong since

$-\pi\leq x \leq \pi$

$\implies\pi\leq x + \pi\leq 3\pi$

$\implies  x + 2 \pi\in [\pi,3\pi] \text{ and not in } [-\pi, \pi].$

So the $f$ is not $2\pi$-periodic.

答: It is not $2\pi-$periodic since $f(x)=0$ with $x\in[-3\pi, -\pi]$ an $f(x)\not=0$ $\forall x\in[-\pi, \pi]$.


问:
Approximating the initial value problem $x'(t)=\sqrt{|x(t)|}, x(0)=0$ by solutions of ODEs with smooth right hand side
Consider the initial value problem $$x'(t) = \sqrt{|x(t)|}, \ x(0) = 0.$$ Note that the right hand side is not Lipschitz at $x=0$. There are infinitely many solutions to this, e.g. $x_0\equiv 0$ and for any $r > 0$, $$x_r(t) = \begin{cases} 0 &t\leq r,\\  \frac{1}{4} (t - r)^2 & t > r.\end{cases}$$ My question is: If we pick one of these solutions, call it $x$, can we find a $C^0_{loc}$-approximation of $f(x):=\sqrt{|x|}$ by $C^{\infty}$-functions $f_{\varepsilon}(x)$ such that the solutions of an appropriate initial value problem (which are then necessarily unique) for the ODEs $x_{\varepsilon}'(t) = f_{\varepsilon}(x_{\varepsilon}(t))$ converge to $x$ as $\varepsilon \to 0$?

答:
A continuous, but not smooth idea. Restrict the derivative and thus get a finite Lipschitz constant via $f_ε(x)=\sqrt{\max(ε,|x|)}$. This will initially grow as $\sqrtε·t$ until at $t=\sqrtε$ the level $ε$ is reached and the solution continues with the quadratic. This will not converge to the zero segment on $[0,r]$.

So in a second try, set $f_ε(x)=\sqrt{\max(ε,|x|)}-c\sqrtε$, $c\in (0,1)$. The intercept of $(1-c)\sqrtε·t=ε$ is at $\frac{\sqrtε}{1-c}$. So if this is set to $r$, then $1-c=\frac{\sqrtε}r$, so that in total
$$
f_ε(x)=\sqrt{\max(ε,|x|)}-\sqrtε+\frac{ε}r.
$$

I'm quite optimistic that smooth variants, for instance based on $|x|\approx\sqrt{ε^2+x^2}$, could be constructed in a similar fashion.


问:
Matrices AB, BA eigenspaces
Take two matrices $A$ $n\times m$ and $B$ $m\times n$. They both have a nonzero eigenvalue $\lambda$. How do you prove that the dimension of the eigenspaces of $AB$ and $BA$ corresponding to $\lambda$ are equal?

i.e., $\dim E_\lambda(AB) = \dim E_\lambda(BA)$.

Attempt: If you take $L$ independent eigenvectors $v_1, \ldots, v_n$ of $AB$, they get $Bv_1, \ldots, Bv_n$ eigenvectors in $BA$. This is apparently important in the proof, but I can't understand where to use this fact.
Please help.

答:
The key point is that your $Bv_1,\ldots,Bv_n$ are necessarily linearly independent. Because if
$$
c_1Bv_1+\cdots+c_nBv_n=0,
$$
then applying $A$ we get
$$
0=c_1ABv_1+\cdots+c_nABv_n=c_1\lambda v_1+\cdots+c_n\lambda v_n=\lambda(c_1v_1+\cdots+c_nv_n).
$$
As $\lambda\ne0$, you get that $Bv_1,\ldots,Bv_n$ are linearly independent. </p>

<p>So far we have shown that $\dim E_\lambda(AB)\leq\dim E_\lambda(BA)$. But of course now we can get the reverse inequality by starting with $BA$ instead of $AB$. </p>

Title: Matrices AB, BA eigenspaces

Body:
The key point is that your $Bv_1,\ldots,Bv_n$ are necessarily linearly independent. Because if
$$
c_1Bv_1+\cdots+c_nBv_n=0,
$$
then applying $A$ we get
$$
0=c_1ABv_1+\cdots+c_nABv_n=c_1\lambda v_1+\cdots+c_n\lambda v_n=\lambda(c_1v_1+\cdots+c_nv_n).
$$
As $\lambda\ne0$, you get that $Bv_1,\ldots,Bv_n$ are linearly independent.

So far we have shown that $\dim E_\lambda(AB)\leq\dim E_\lambda(BA)$. But of course now we can get the reverse inequality by starting with $BA$ instead of $AB$.


问:
Find the general solution to the differential equation $(x+1)\frac{d^2y}{dx^2} + x\frac{dy}{dx} - y = (1+x)^2$
I'm having a problem with the following differential equation, from a Sixth Term Examination Paper question (STEP 3 2011 Q1). The problem arises with me trying to apply a technique in an earlier part of the question to a later part.

The third part of the question (iii) asks to find a general solution to the differential equation:
$$(x+1)\frac{d^2y}{dx^2} + x\frac{dy}{dx} - y = (1+x)^2$$

An earlier part of the question (ii) asks to find a solution to a similar differential equation:
$$(x+1)\frac{d^2y}{dx^2} + x\frac{dy}{dx} - y = 0$$
Using the substitution $y = ze^{-x}$, where $z$ is some function of $x$. I managed to get this to work, and using this substitution the equation turns into

$$\frac{d^2z}{dx^2} - \frac{x+2}{x+1}\frac{dz}{dx} = 0$$

Which is fairly straightforward to solve using methods such as the integrating factor, as this is just a first order differential equation in $\frac{dz}{dx}$.

However, applying this to the third part of the question (iii) I assumed that the equation would just turn into

$$\frac{d^2z}{dx^2} - \frac{x+2}{x+1}\frac{dz}{dx} = (1+x)^2$$

Since the $(1+x)^2$ on the right hand side does not depend on $y$.

However, this is wrong and the $(1+x)^2$ turns into $(1+x)e^x$. I don't know if I am just missing some step and rushing to the substitution but I don't see how to get this $(1+x)e^x$ function, and this solution from The Student Room skips over the step.

Please can you show me how the substitution results in $(1+x)e^x$?

答:
$$(x+1)\frac{d^2y}{dx^2} + x\frac{dy}{dx} - y = (1+x)^2$$
If you change the variable then:
$$y=ze^{-x}$$
$$y'=z'e^{-x}-ze^{-x}$$
$$y''=z''e^{-x}-2z'e^{-x}+ze^{-x}$$
So that the differential equation:
$$(x+1)\frac{d^2y}{dx^2} + x\frac{dy}{dx} - y = (1+x)^2$$
Becomes:
$$(x+1)(z''e^{-x}-2z'e^{-x}+ze^{-x}) + x(z'e^{-x}-ze^{-x})- ze^{-x} = (1+x)^2$$
You have to multiply the whole equation by  the exponential $e^x$ since it's not zero on the RHS this explains the presence of the exponential on RHS.
$$(x+1)(z''-2z'+z) + x(z'-z)- z =\color {red}{e^x} (1+x)^2$$
$$(x+1)z''- z'(x+2) =\color {red}{e^x} (1+x)^2$$
When you change the variable in the homogeneous equation you have on RHS:
$$(x+1)(z''e^{-x}-2z'e^{-x}+ze^{-x}) + x(z'e^{-x}-ze^{-x})- ze^{-x} = 0$$
$$e^{-x}((x+1)(z''-2z'+z) + x(z'-z)- z) = 0$$
$$(x+1)(z''-2z'+z) + x(z'-z)- z = e^x \times 0$$
$$(x+1)z''-z'(x+2) = 0$$


问:
Euclid's Division Lemma Extended to Negative Integers Conflict
My textbook states that Euclid's Division Lemma can be extended to all integers with the following information:

Let a and b be any two integers with b ≠ 0. Then, there exist unique
integers q and r such that

a = bq + r, where 0 ≤ r < |b|

Suppose I take a = -16 and b = 3. In this case, q = -5 and r = -1. Won't this contradict the fact that r must be greater than or equal to 0?

Thanks in advance.

Body:
Suppose I take a = -16 and b = 3. In this case...

$q=-6$ and $r=2$.

And if $16=3q_1+r_1=3q_2+r_2$ for some $r_1,r_2\in\{0,1,2\}$, then $r_1-r_2=3(q_2-q_1)$ and $-3<r_1-r_2<3$. Since $r_1-r_2$ must be a multiple of $3$, $r_1=r_2$ and, therefore, $q_1=q_2$.


答:
Call problem $A$ the one you have drawn: the tracks may only provide a normal force radially inwards. Call problem $B$ the similar problem except that cart and track are replaced by bead on a wire. In $A$, the cart may fall off the track. In $B$, the 'cart' may not fall off the 'track'.

For problem $A$:

Minimum initial speed to reach height of $2R$: at least $\sqrt{4gR}$ (exact number not relevant to the arguments here)

Minimum initial speed to complete a loop: $\sqrt{5gR}$.

For problem $B$:

Minimum initial speed to reach height of $2R$: $\sqrt{4gR}$.

Minimum initial speed to complete a loop: $\sqrt{4gR}$.

Your equations of motion are those of problem $B$, since you have fixed $r=R$. The problems are identical when $u>\sqrt{5gR}$, when the normal force of the track is inwards over the whole loop. The problems are not identical if $u<\sqrt{5gR}$.

You get sensible answers (real and finite integral) for $\sqrt{4gR}<u<\sqrt{5gR}$, because in $B$, the cart does complete the loop. We know that in $A$, the cart does not for $u<\sqrt{5gR}$, so we have found sensible answers to a different problem.

Your elliptic integral has a singularity at $\sqrt{4gR}$. Wolfram agrees. This indicates $T \to \infty$ in problem $B$, as $u \to \sqrt{4gR}$: the bead takes infinitely long to reach the top of the loop.

Title: Convergence of integral and summation for Time taken for complete revolution around vertical circle

Body:
I want to find Time taken to complete Vertical circular motion by Particle of mass $m$

So I proceed as follows

Applying work energy theorem

$$-mgR(1-\cos(\theta)=\frac12 mv^2-\frac12 mu^2$$

$$\implies v^2=u^2-2gR(1-\cos\theta)$$

$$\implies \omega=\frac{\sqrt{u^2-2gR(1-\cos\theta)}}{R}$$

$$\implies \frac {d\theta}{dt}=\frac{\sqrt{u^2-2gR(1-\cos\theta)}}{R}$$

$$\implies \frac {d\theta}{\sqrt{u^2-2gR(1-\cos\theta)}}=\frac{dt}{R}$$

Let T be the time taken for complete revolution

Now, Integrating both sides

$$\int_{0}^{2\pi}\frac {d\theta}{\sqrt{u^2-2gR(1-\cos\theta)}}=\int_{0}^{T}\frac{dt}{R}$$

$$\implies T=2R\int_{0}^{\pi}\frac{d\theta}{\sqrt{u^2-4gR\sin^2\theta}}$$

$$\implies T=\frac{2R}{u}\int_{0}^{\pi}\frac{d\theta}{\sqrt{1-\frac{4gR}{u^2}\sin^2\theta}}$$

On evaluating this elliptic integral we get,

$$T=\sum_{n=0}^{\infty} \left(\frac{2}{u}\right)^{2n+1}\left(\frac{(2n-1)!!}{2^n n!}\right)^2 (gR)^n$$

Here $T$ denotes Time taken for complete revolution. Block will complete full circle only if $u\geq \sqrt{5gR}$. So my question is $T$ will be Real only when $u\geq \sqrt{5gR}$ so This condition should exist in integral as well as summation for $T$ but I'm not able to see it. Also Wolfram alpha evaluates integral only if $u\geq \sqrt{5gR}$


问:
Find a spanning set of the smallest possible size for the subspace of $P_3$ consisting of polynomials with roots at $x=0, x= \pm1$

Find a spanning set of the smallest possible size for the subspace of $P_3$ (the polynomials of degree $3)$ consisting of polynomials with roots at $x=0, x= \pm1$. How many elements does it have?

Let $p \in P_3$. Now $p(x)=a_3x^3+a_2x^2+a_1x+a_0$ if the roots of $p$ are $0, \pm1$ then $$p(x)=(x-0)(x-1)(x+1)=(x-0)(x^2-1)=x^3-x$$ but I'm not sure what is the set spanning this? In general for $P_3$ the one spanning set would be the set $\{1,x,x^2,x^3\}$, but I'm not sure I see what the spanning set would be for $x^3-x$ is this just the singleton $\{1\}$?

答:
Bernard has already given the answer in his comment, but let me elaborate a little more on it: let $p\in P_3$ with roots $0$ and $\pm 1$. Then $p$ is of the form
$$p(x) = q(x)(x^3-x)$$
for some polynomial $q$. But $x^3 -x$ is already of degree 3, so $q$ must be constant. Therefore all the polynomials you are looking for are a scalar multiple of $x^3-x$, so $\{x^3-x\}$ is a basis, in particular a spanning set.


问:
If $f$ is a Morse function, then so is $f \circ \phi^{-1}$, where $\phi: U \rightarrow \mathbb{R}^k$ is the coordinate chart.
I am trying to show: if when $f^\prime = 0$, then $f^{\prime\prime} \neq 0 \Leftrightarrow (f \circ \phi^{-1})^\prime = 0$, $(f \circ \phi^{-1})^{\prime\prime} \neq 0$.

But the problem is, because $(f \circ \phi^{-1})^\prime = f^\prime(\phi(x))\phi^\prime(x)$, when $(f \circ \phi^{-1})^\prime = 0$, it could be $f^\prime = 0$ or $\phi^\prime = 0$. Can I simply regard $\phi$ as a change of coordinate function hence its Jacobian is nonsingular, so $\phi^\prime \neq 0$?

$(f \circ \phi^{-1})^{\prime\prime}
= (f^\prime(\phi(x))\phi^\prime(x))^\prime
= f^{\prime\prime}(\phi(x))\phi^\prime(x)\phi^\prime(x) + f^\prime(\phi(x))\phi^{\prime\prime}(x)$

Assuming $f^\prime(\phi(x))=0, \phi^\prime \neq 0$:
$(f \circ \phi^{-1})^{\prime\prime} = f^{\prime\prime}(\phi(x))\phi^\prime(x)\phi^\prime(x) \neq 0$

Assuming $\phi^\prime = 0:
(f \circ \phi^{-1})^{\prime\prime}  = 0$ Failed.

Thank you very much!

答:
This is just a calculus question. The given condition "coordinate chart" is strong. I found this in my calculus book: Advanced Calculus of Several Variables, Edwards

Definition: Let $\phi: U \rightarrow \mathbb{R}^n$ be a $\mathscr{C}^1$ mapping defined on an open subset $U$ of $\mathbb{R}^k (k \leq n)$. Then we call $\phi$ regular if the devrivatire matrix $\phi^\prime(u)$ has maximal rank $k$, for each $u \in U$.

Definition: Given $p \in M$, there exists an open set $U \subset \mathbb{R}^k (k<n)$ and a regular $\mathscr{C}^1$ mapping $\phi: U \rightarrow \mathbb{R}^n$ such that $p \in \phi(U)$, with $\phi(U^\prime)$ being an open subset of $M$ for each open set $U^\prime \subset U$. Then the mapping $\phi$ is called a coordinate patch for $M$ provided it is injective.

Certainly, a full rank matrix can't be an empty matrix.


问:
How to simplify $\frac{\frac{\cos x}{2\sqrt{x}} + \sqrt{x}\sin x}{\cos^2x}$, step by step?
I want to simplify this expression

$$\frac{\frac{\cos(x)}{2\sqrt{x}} + \sqrt{x}\sin(x)}{\cos^2(x)}$$

and I know this is the answer

$$\frac{\cos(x)+2x \sin(x)}{2\sqrt{x}\cos^2(x)}$$

How can I get there step by step? Thanks in advance!

答:
$$\frac{\frac{\cos(x)}{2\sqrt{x}} + \sqrt{x}\sin(x)}{\cos^2(x)}$$
The LCM of the numerator terms is $\space 2\sqrt{x}\space$ so the expression becomes $$\dfrac{\frac{\cos(x)}{2\sqrt{x}}
 + \frac{2x\sin(x)}{2\sqrt{X}}}{\cos^2(x)}$$
A denominator in a numerator-fraction is part of the denominator of the whole fraction

$$\frac{\cos(x)+2x \sin(x)}{2\sqrt{x}\cos^2(x)}$$