答: If $g$ exists, then $f'=g'e^g$ and therefore $\frac{f'}f=g'$. So, $\frac{f'}f$ has a primitive, and therefore the integral of $\frac{f'}f$ along any closed path is $0$. But if you take $r>0$ so small that $f$ has no zero on $D(z_0,r)\setminus\{z_0\}$, then, by the Argument Principle, $\oint_{|z-z_0|=r}\frac{f'}f$ is $-2\pi i$ times the number of poles of $f$ on $D(z_0,r)$, which is not $0$.

Title: the nessesary condition of branch of logarithm

Body:
Let $U$ be an open subset of $\mathbb{C}$.Let $z_0$ be a point in U,and suppose that f is a meromorphic function on U with a pole at $z_0$.Prove that there is no holomorphic function g:U$\setminus${$z_0$}$\rightarrow$$\mathbb{C}$,such that $e^{g(z)}$=f(z)for all z$\in$U$\setminus${$z_0$}

my idea:as we know that if a simply connected domain which does not contain the origin,it can be chosen as a branch of logarithm,however,this is just sufficient condition,and we don't know what the image of f is like,thus how to deal with?this really confuses me.thanks!


问:
How can I solve the following differential equation by reducing it to first order?
I'm trying to solve the following differential equation

$ \displaystyle y''+ \left( 1+ \frac{1}{y} \right)y'^{2} =0 \tag*{} $

The book in which I found this question says that It can be solved by reducing it to a first order differential equation but I can't seem to find a way to do so.

答:
Switching variables, the equation write
$$-\frac {x''}{[x']^3}+ \left( 1+ \frac{1}{y} \right)\frac 1{[x']^2}=0$$ and assuming $x'\neq 0$
$$x''-\left(1+\frac{1}{y}\right) x'=0$$ Now, let $p=x'$ and you face a very simple equation.
$$p=C\, y\, e^y$$

ContentLicense: CC BY-SA 4.0


问:
Solve $x^3+2x+1=2^n$ over positive integers?
I just have proved that $x^3+2x+1$ can not be divided by $2^n$ ($n>3$), for $x=8k+t$ and $t=0,1,2,3,4,6,7$, but for $x=8k+5$, I have no way to deal with this case, may someone give me some hints for this question? It seems that $(x,n)=(1,2)$ is the unique solution. Also what I know is that $x$ must divide $2^n-1$.

Note. Here both  $x$ and $n$ are positive integers.

答:
Use the Quadratic Reciprocity - https://en.wikipedia.org/wiki/Quadratic_reciprocity (The supplementary laws using Legendre symbols):
\begin{align}
\left(\frac{-2}{p}\right) = (-1)^{\frac{p^2+4p-5}{8}}=\begin{cases}
1 & p\equiv 1,3~\text{(mod 8)}\\
-1 & p\equiv 5,7~\text{(mod 8)}
\end{cases}
\end{align}

\begin{align}
 {}
\end{align}

As you mentioned, for $n\ge 3$, we must have $x\equiv 5 ~~\text{(mod 8)}$. Now note that
\begin{align}
 x^3+2x+1 \equiv 1 ~~ \text{(mod 3)}.
\end{align}
You can directly check it by using Fermat's little theorem. Thus, we have \begin{align}
x^3+2x+1 = 2^n\equiv (-1)^n\equiv 1 ~~\text{(mod 3)},
\end{align}
which implies that $n$ is even. Let $n=2m$. By adding $2$ to both sides, we obtain
\begin{align}
(x+1)(x^2-x+3) = 2^{2m}+2.
\end{align}
Now consider $x^2-x+3$. Clearly, it is an odd number, and its prime factor $p$ must divide $(2^m)^2+2$. It then follows that $p=8k+1$ or $p=8k+3$ due to the quadratic reciprocity. This implies that
\begin{align}
  x^2-x+3 \equiv \text{1 or 3}~~\text{(mod 8)}.
\end{align}

However, because $x\equiv 5 ~~\text{(mod 8)}$, we have
\begin{align}
x^2-x+3 \equiv 25-5+3\equiv 7~~\text{(mod 8)},
\end{align}
which is a contradiction.


问:
Order of convergence of a sum
Let $(X_t)_{t\geq 0},\;X_0=0$, be a positive stochastic process such that
\begin{align*}
\mathbb{E}\left[\sum_{n=1}^{\infty}X_t^n\right]=\sum_{n=1}^{\infty}\mathbb{E}[X_t^n]<\infty.
\end{align*}
Assume that $\mathbb{E}[X_t]=O(t),\;t\to 0$. Clearly for each fixed $N\in\mathbb{N}$,
\begin{align*}
\sum_{n=1}^{N}\mathbb{E}[X_t^n]=O(t).
\end{align*}
Does we also have the following:
\begin{align*}
\sum_{n=1}^{\infty}\mathbb{E}[X_t^n]=O(t)?
\end{align*}</p>

答:
(This is a revised version which addresses the mention (silently added to the question) that the limit to consider is when $t\to0$, not when $t\to\infty$. As regards the other modification, that in contradiction to the original question there would be only one process $(X_t)$, I refuse to address it, for reasons explained in the comments.)

No. Try $\mathbb E[X_t^n]=t\mathbf 1_{nt\leqslant1}$.

ContentLicense: CC BY-SA 3.0

Title: Order of convergence of a sum


问:
How to do part (i) of the following high-school level problem of vectors?
Question.

$ABC$ is a triangle whose vertices have position vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ respectively, relative to an origin in the plane $ABC$. Show that an arbitrary point $P$ on the segment $AB$ has position vector
$$\rho\mathbf{a}+\sigma\mathbf{b},$$
where $\rho\geqslant0$, $\sigma\geqslant0$ and $\rho+\sigma=1$. Give a similar expression for an arbitrary point on the segment $PC$, and deduce that any point inside $ABC$ has position vector
$$\lambda\mathbf{a}+\mu\mathbf{b}+\nu\mathbf{c},$$

where $\lambda\geqslant0$, $\mu\geqslant0$, $\nu\geqslant0$ and $\lambda+\mu+\nu=1$.

Sketch the region of the plane in which the point $\lambda\mathbf{a}+\mu\mathbf{b}+\nu\mathbf{c}$ lies in each of the following cases:

(i) $\lambda+\mu+\nu=-1$, $\lambda\leqslant0$, $\mu\leqslant0$, $\nu\leqslant0$;

(ii) $\lambda+\mu+\nu=1$, $\mu\leqslant0$, $\nu\leqslant0$.

I am stuck on part (i). Though I can do the rest, I do not think that I fully understand the core concept of this question. I just failed to figure out how to make $\lambda+\mu+\nu=-1$.

My Attempt. I tried to implement and extend the ideas introduced in the parts before the sketch questions. So for the point on segment AB part, I had $\vec p=(1-\lambda)\vec a+\lambda \vec b$ for $0\le\lambda\le 1$. Then for the point within triangle ABC part (inclusive), I had $(1-\mu)\vec p+\mu \vec c$ for $0\le\mu\le 1$; upon substituting $\vec p$, I had $(1-\mu)(1-\lambda)\vec a+(1-\mu)\lambda\vec b+\mu\vec c$ for which if I add all the vector scalars I would have $1$. I was thinking that maybe I could change the last term $\mu\vec c$ into $-\mu\vec c$ so that the scalars could sum to $-1$, but that would make the region not on the plane ABC. This is as far as I went.

Comment. I'd rather hear some hint than simply pointing out the way to do it. Thanks a lot in advance!

答: A hint for looking at it is to transfer the negative sign from the $\lambda, \mu, \nu$ to $\vec{a}, \vec{b}, \vec{c}$. Can you see which region would be represented?


问:
Solving limit without using l'hopitals rule
Find the limit as $h \to 0$ for $\frac{f(x)-f(x-h)}{h}$ at $x=0$, where $f(x) = x + |x|$ without using L'hopital rule.

Once I plug in the numbers and try to work it down I get brought to $\frac{-(-h + |-h|)}{h} = \frac{h-h}{h}$ which is $\frac{0}{h}$ and then if go to plug in $0$ for $h$ then I would get $0/0$ which is indeterminate. How should I approach this problem to get the answer?

答:
Two mistakes you are making:

one:  $|-h|\ne h$.  $|-h| =\begin{cases} h& h \ge 0\\-h& h < 0\end{cases}$.

two: And as $\frac 0h = 0$ for all $h \ne 0$, you have $\lim_{h\to 0} \frac 0h = \lim_{h\to 0} 0$.   The $0$ has already "canceled" the $h$ out so there is nothing to "plug in".

$\lim_{h\to 0^+} \frac {f(0) -f(0-h)}h = \lim_{h\to 0^+} \frac {-(-h + |h|)}h=\lim_{h\to 0^+} \frac {-(-h + h)}h =\lim_{h\to 0^+} \frac 0h =\lim_{h\to 0^+} 0 = 0$.

And $\lim_{h\to 0^-} \frac {f(0)-f(0-h)}h = \lim_{h\to 0^-} \frac {-(-h + |h|)}h=\lim_{h\to 0^-} \frac {-(-h - h)}h=\lim_{h\to 0^-} \frac {2h}h = \lim_{h\to 0^-} 2 = 2$.

(Note, you don't plug in $h = 0$ to get $\frac {2h}h=\frac 00$ because we aren't evaluating $\frac {2h}h$ when $h=0$.  We are taking the limit as $h$ is near but not equal to $0$.  So you don't plug in $h = 0$ for $\lim \frac 0h$ either.)

so the left hand limit is not equal to the right hand limit and the $\lim_{h\to 0} \frac {f(0) - f(0-h)}h$ does not actually exist.


问:
Does this IVP have a unique solution for all $x \in \mathbb R$
Is $\displaystyle {dy\over dx}=\sin(y)$ with initial conditions $y(X)=Y$ guaranteed to have a unique solution for all $x\in\mathbb R$? </p>

答:
Artem's answer is complete, though I thought I would give a few more details in case you want to learn a bit more about existence/uniqueness questions. The key part of what I have to say is the last paragraph, in which I introduce my favourite example of non-unique solutions.

First of all, the existence of solutions to an initial value problem is (perhaps surprisingly) complicated machinery. Both existence theorems I know, the Picard-Lindelöf aka Cauchy-Lipschitz existence/uniqueness theorem and the Cauchy-Peano existence theorem require some real analysis. In particular, the proofs I know set up an iteration scheme to find better and better approximations to the solution to the IVP, and show that this iteration scheme has to converge to a function. The proof of uniqueness is relatively simple, using something called Gronwall's inequality (I can give more details on this if you are interested).

In particular, a special case of Picard-Lindelöf says that if you consider an IVP of the form $y' = f(x, y)$, $y(x_0) = y_0$, and the function $f$ is continuous in $x$ and $\frac{\partial}{\partial y}$ exists and is continuous, then there exists a local solution to the initial value problem, and this local solution is unique.
(A local solution is one defined for a small interval of $x$ values, $x \in (x_0 -\epsilon, x_0 + \epsilon)$ for some small value $\epsilon >0$. The theorem doesn't tell us what $\epsilon$ is, just that there is one. You can, in principle, work through the proof to get an estimate on how big $\epsilon$ is, but this is quite difficult.)

In particular, your IVP satisfies this requirement: here, $f(x,y) = \sin(y)$. This does not depend on $x$ so is obviously continuous in $x$. $\partial_y f$ exists and is continuous (even $C^\infty$ smooth as Artem points out). Thus, the theorem applies and we have a unique local solution.

Now, you are left with two questions: are these solutions defined for all $x\in \mathbb{R}$? are they globally unique?

Again, Artem gave the answer to the first question. There is a solution for all $x \in \mathbb{R}$ because, if you had a solution defined for a finite interval $x \in (a,b)$, say with $b$ finite, then any solution $y(x)$ would have to leave the domain in which $f(x,y)$ satisfies the conditions of Picard-Lindelöf as $x \to b^-$. Now, your function $f(x,y) = \sin(y)$ satisfies the conditions for all $x$ and for all $y$... thus, if $b$ were finite, $y(x)$ would have to go to $\pm \infty$ as $x \to b^-$. Artem gave two arguments to show that $y(x)$ can't actually blow up in finite time.

This is the key fact about existence of solutions: the only way a solution to an IVP can cease to exist is if you have "finite time blow-up". An example is the following IVP: $y' = y^2$, $y(0) = 1$. Note that this satisfies the requirements of Picard-Lindelöf. We can find an explicit solution by $y(x) = \frac{1}{1-x}$. This is only defined for $x < 1$. The reason the solution can't be extended to $x=1$ and beyond is that $y(x)$ does not have a limit as $x \to 1^-$.

Key example of non-uniqueness below

OK, we are now left with global uniqueness. This follows because $\sin(y)$ is actually continuously differentiable with respect to $y$ everywhere. Here is my favourite example of what can go wrong. I really think that understanding this example is key to understanding what is going on in ODE, so I highly recommend you think about it.
The IVP is
$$
y' = 3 y^{\frac{2}{3}}, y(x_0) = y_0.
$$
Note that this is of the form $y' = f(x,y)$ for $f(x,y) = 3 y^{\frac{2}{3}}$. This function $f(x,y)$ is continuous in both variables, so Cauchy-Peano's existence theorem tells us that we always have existence of a solution to the IVP. Unfortunately, Picard-Lindelöf is only verified when $y \ne 0$, so we are guaranteed local uniqueness of solutions ONLY when $y \ne 0$. When $y=0$, all bets are off.

I will now illustrate how we don't have global uniqueness of solutions.

Then, we can do like Ron did, and use separation of variables. We obtain the following:
[
y(x) = (x + C)^3.
]
Note that for the initial value $y = 0$, you get other solutions... notably by taking the constant solution $y(x) = 0$.
You can calculate however, that you get many other solutions by taking
$$
y(x) = \begin{cases} (x + C)^3 &\text{ for } x < -C \\
0 &\text{ for } -C \le x \le C' \\
(x+C')^3 &\text{ for } x > -C'.
\end{cases}
$$
In particular, for the initial condition $y(0) = 1$, you can find infinitely many global solutions that pass through this point BUT all these solutions that are different from each other globally are the same when you consider them in a small neighbourhood of the initial condition $x=0$, $y=1$.

A good exercise to think about is how this matches your physical intuition. (Not this specific example, but a similar one): consider a cylindrical barrel that has a small leak in the bottom. Toricelli's law says that the speed of the water coming out is proportional to $\sqrt{h}$. Thus, the height of water in the barrel satisfies an equation like $h' = - C \sqrt{h}$, where $C$ is some positive constant. Of course, this equation only makes sense for $h \ge 0$... but it exhibits similar properties to the example I just gave. What does this mean? local existence of solutions tells you that you can tell a possible future and a possible past from the initial value. Local uniqueness would say that there is only one past and only one future possible from this state. Now, if the barrel is empty at some moment in time, you can't tell the past anymore. It could have emptied just before you looked at it, or it could have emptied two months ago, or maybe there never was anything in the barrel to begin with.

OK, I probably got carried away. I hope some of this was useful.


问:
By differentiation or otherwise, prove that $(c(z))^2 +(s(z))^2 = 1$
Assume that the two power series $s(z) = \sum a_nz^n$ and $c(z) =\sum b_nz^n$ are convergent for all $z\in \mathbb{C},$ and that they satisfy the relations $s'(z) = c(z), c'(z) =-s(z).$ Deduce the identities
$$ a_n = -a_{n-2}/(n(n-1)) , b_n=-a_{n-2}/(n(n-1))
$$(To me, it looks like there is a typo for $a_n$)
If further $s(0)=0, c(0) =1,$ determine $s(z)$ and $c(z)$ completely. By differentiation or otherwise, prove that $(c(z))^2 +(s(z))^2 = 1$

To me it looks like $s(z) = sin(z)$ and $c(z) = cos(z)$ but I don't know what to do with this information.

Also while taking the derivate of the power series in an attempt to deduce the identities, I have
$(n+1)a_{n+1}=b_n$ and it does not seem like I have the correct identity

答:
If $s(0) = 0$ and $c(0)=1$ then you can prove your conjecture $s = \sin$ and $c = \cos $ by considering the function
$$
 F(z) = (s(z) - \sin(z))^2 + (c(z) - \cos(z))^2 \, .
$$
We have $F(0) = 0$ and
$$
 F'(z) = 2(s(z) - \sin(z))(c(z)-\cos(z)) + 2(c(z) - \cos(z))(-s(z) + \sin(z)) = 0
$$
so that $F$ is identially zero. This implies $s(x) = \sin(x)$ and $c(x) = \cos(x)$ for all $x \in \Bbb R$, and then  $s = \sin$ and $c = \cos $ because of the identity theorem for holomorphic functions.


问:
Determine whether the given set as follows is linearly independent or not.
Let $V=\mathcal{F}[0,1]$ be a vector space of functions defined on $[0,1]$ over $\Bbb R$.
Let $S = \left\{\frac{1}{x^2-4},\frac{1}{x^2+3x+2},\frac{1}{x^2-x-2}\right\} \subset V$.
Determine whether $S$ is linearly independent or not.

Attempt:
For the simplify, write $f_1 = \frac{1}{x^2-4},f_2 = \frac{1}{x^2+3x+2},$ and $f_3=\frac{1}{x^2-x-2}$.
Then, $S=\{f_1,f_2,f_3\}$.

The linear independence or dependence of these vectors is determined by whether the vector equation
\begin{equation*}
k_1f_1+k_2f_2+k_3f_3=0 \ldots (1)
\end{equation*}
can be satisfied with coefficients that are not all zero. To see whether this is so, let us rewrite $(1)$ in its function form
\begin{equation*}
\frac{k_1}{x^2-4} + \frac{k_2}{x^2+3x+2} + \frac{k_3}{x^2-x-2} = 0 \ldots (2)
\end{equation*}
or, equivalently, as
\begin{align*}
\frac{k_1(x+1) + k_2(x-2) + k_3(x+2)}{(x+2)(x+1)(x-2)} &= 0 \\
\Leftrightarrow \frac{x(k_1+k_2+k_3) + (k_1-2k_2+2k_3)}{(x+2)(x+1)(x-2)} &= 0.
\end{align*}
Since this equation must be satisfied by all $x \in [0,1]$, then each coefficient must be zero.
Thus, the linear dependence or independence of the given functions in $S$ hinges on whether the following linear system has a nontrivial solution:
\begin{align*}
k_1+k_2+k_3 &= 0 \\
k_1 - 2k_2 + 2k_3 &= 0 \ \ \ \ldots (t) \\
0k_1 + 0k_2 + 0k_3 &= 0 \\
\end{align*}
By direct calculation, we get $(k_1,k_2,k_3) = (-4t,t,3t)$ with $t \in \Bbb R$. That is, the solution is depend on $t$ and the system $(t)$ has nontrivial solutions.

Hence, $S$ is linearly dependent. Alternatively, we can show that $(t)$ has nontrivial solutions
by showing that the coefficients matrix has determinant zero, and thus, $S$ is linearly dependent.

Am I true?

答:
Yes, your approach is correct. Note that solving the system $(t)$ is a waste of time, since it is a homogeneous system of $2$ linear equations with $3$ variables. Any homogeneous system of linear equations with more variables than equations has nontrivial solutions.
Another approach would be to note that$$\frac1{x^2-4}=\frac{1/4}{x-2}+\frac{-1/4}{x+2},$$that$$\frac1{x^2+3x+2}=\frac1{x+1}+\frac{-1}{x+2},$$and that$$\frac1{x^2-x-2}=\frac{1/3}{x-2}+\frac{-1/3}{x+1}.$$So, each of your rational fractions belongs to$$\left\{\frac a{x-2}+\frac b{x+1}+\frac c{x+2}\,\middle|\,a+b+c=0\,\right\}$$whose dimension is at most $2$ (actually, it is equal to $2$, but that's not important) and therefore they cannot be linearly independent.


问:
Question about the gradient of a function
We consider a function of class $C^1(\mathbb{R}^2)$, by searching a max or min between the points of $S=\{(x,y) \in \mathbb{R}^2 \;\; | \;\; x^2+y^2=1\}$, if $(x_0,y_0)$ is one of them, is it true that there exists $\lambda \in \mathbb{R}$ such that $\nabla f(x_0,y_0)=2\lambda (x_0,y_0)?$ Why?

答:
Assuming that you want the extremum of some $f(x,y)$, we can parameterize the locus as $x=\cos t,y=\sin t$.

Then by the chain rule, the stationary points of $f$ are given by

$$\frac{df}{dt}=\nabla f\cdot(-\sin t,\cos t)=0$$ and these are such that $$\nabla f=\lambda(\cos t,\sin t).$$

In other words, the derivative of the function when you remain in the locus is zero when the gradient is normal to the locus, i.e. parallel to a normal to the locus.

This is a quite general property.


问:
Explicitation of an equation
We consider the function $f(x,y)=\arctan(x^2-y^2-1)$. Why into the neighborhood of $(1, 0)$, there exists a $x=g(y)$ but not a $y=h(x)$ that make explicit the equation $f(x,y)=0$? I think that the existance of $x$ follows from Dini's theorem, is it right?

Thank you

答:
The map $t \mapsto \tan t$ is a bijection from $(-\pi/2,\pi/2)$ onto $(-\infty, \infty)$ and $\tan(0)=0$. Therefore $f(x,y) = 0$ is equivalent to $r(x,y) = x^2-y^2-1 = 0$ in a neighborhood of $(1,0)$.

$r(x,y)=0$ is equivalent to

$$x_{\pm} = \pm \sqrt{1 + y^2}.$$ Around $x=1$, there is only one solution which is $x=\sqrt{1 + y^2}$. Which proves as requested the existence of $x=g(y)$.

On the other end the equation $y^2 = x^2-1$ has no solution for $x^2 \lt 1$. Therefore, the equation $f(x,y)$ can not be solved as $y = h(x)$ around $(1,0)$.

Title: Explicitation of an equation


问:
Implicit Differentiation???

Assume that $y$ is a function of $x$. Find $y' = \dfrac{dy}{dx}$ for $(x-y)^2 = x + y - 1$.

I've worked out the problem multiple times, but I continue to get a different answer than the correct answer. First I multiply out the $(x-y)^2$ to $(x-y)(x-y)$ and work it out from there; after that I take the derivative of both sides, etc.

This is supposed to be the correct answer:

$$y' = \frac{2y-2x+1}{2y-2x-1}$$

But I keep getting:

$$y' = \frac{2x-2y-1}{2x-2y+1}$$

Help please? Thank you!

答:
This question was answered in the comments; this CW answer intends to remove the question from the Unanswered queue.

As coffeemath remarked:

Your answer is the same as what you're "supposed to get". Multiply top and bottom each by $(-1)$.

Cheers!

Title: Implicit Differentiation???


答:
Hint : Let
$$e : z \mapsto c(z) + i s(z)$$

You get $e'(z)=ie(z)$. Can you proceed further ?

Body:
Assume that the two power series $s(z) = \sum a_nz^n$ and $c(z) =\sum b_nz^n$ are convergent for all $z\in \mathbb{C},$ and that they satisfy the relations $s'(z) = c(z), c'(z) =-s(z).$ Deduce the identities
$$ a_n = -a_{n-2}/(n(n-1)) , b_n=-a_{n-2}/(n(n-1))
$$(To me, it looks like there is a typo for $a_n$)
If further $s(0)=0, c(0) =1,$ determine $s(z)$ and $c(z)$ completely. By differentiation or otherwise, prove that $(c(z))^2 +(s(z))^2 = 1$

To me it looks like $s(z) = sin(z)$ and $c(z) = cos(z)$ but I don't know what to do with this information.

Also while taking the derivate of the power series in an attempt to deduce the identities, I have
$(n+1)a_{n+1}=b_n$ and it does not seem like I have the correct identity


问:
Will or will not sum always equals sum of averages after being rounded up?
Let's say we have $n$ values $x_1, x_2,..., x_n$ and their average $\bar{x}$.

My question is: given $x_i$ and $\bar{x}$ have 4 (or $m$) decimal places, will $\sum_{i=1}^{n}x_i$ always equal $\bar{x} * n$ if both of them are rounded to 3 (or $m - 1$) decimal places or less? Is it provable if yes?

答:
No. Let's take more than $10$ values so that the absolute error can grow by at least one order of magnitude. For instance, take $n=100$ values, out of which $51$ will be $0$ and $49$ will be $0.0001$. Now $\overline{x}=0.0049/100=0.000049\approx 0$ to $4$ decimals. Note now $\sum x_i=0.0049\approx 0.005$ to $3$ decimals, but $n\overline{x}=0$.

The bottom line is: by multiplying $n\overline{x}$ you magnified the rounding error of $\overline{x}$ by the factor $n$, so even if the error previously was smaller than $0.5\cdot 10^{-m}$ it can grow bigger than $0.5\cdot 10^{-(m-1)}$


问:
What is the number of seating plans?
I am reading a book and there was a topic regarding number of sitting arrangements in a wedding.

The information provided is:

107 people.

11 tables.

Each table can accommodate up to 10 people.

The order in each table does matter but the order of tables does not.

The author finally said that "there are about $11^{107}$ possible seating plans."

Isn't this incorrect since it allows for repetition? My answer would be that the number of seating plans is $110 \choose x_1,...,x_{11}$ such that $x_i = 10$ with $i \in$ {1,...,11}.

Can anyone suggest a correct solution to this problem?

P.S. The book is called "Algorithms to Live By" by Brian Christian and Tom Griffiths.

答:
$11^{107}$ would be the answer if you could assign each person any of the $11$ tables. That means that the tables have no limit on the number of people seated at them, that the seating order around the table does not matter, but the order of the tables does.

When everyone is seated there are $3$ empty seats. It is easiest to introduce $3$ dummy people to fill those seats, and remove them afterwards. So we simply have $110$ people, of which $3$ are identical dummies.

There are $110!$ ways to order all the people. The first $10$ go in order around the first table, the next $10$ around the next table, and so on. Some of these seating arrangements are considered the same, so you now just have to work out how often each arrangement has been overcounted.

The order of the $11$ tables do not matter, so this gives a factor of $11!$ by which we overcounted. Also, the three dummy people that stand in for the empty seats must also be considered the same for a factor of $3!$. This leaves a total of $\frac{110!}{11!3!}$ arrangements.

Note that if we only care about the relative positions of the people seated around a table, then rotation of a table seating does not matter. If that is the case, then you would divide by a further factor of $10^{11}$. Even in this case, the $11^{107}$ estimate is far too low.

Title: What is the number of seating plans?


问:
Properties of normalizer of (Sylow $p$-subgroup of the stablizer of a point)
(The original is a proof exercise but apparently contain print errors so I had to edit it to make sense, thus I'm not sure about its validity, which greatly hindered my attempt)

Exercise 2.4.13: Let $G$ be a permutation group on $\Sigma$, fix arbitary $a\in\Sigma$, let $H=\mathrm{Stab}_G(a)=\{g\in G\mid ga=a\}$, $P$ is a Sylow $p$-subgroup of $H$, $\Delta=\{b|b\in Ga, Pb=b\}$ is points fixed by $P$ in orbit $Ga$, does $N_G(P)$ necessarily acts transitively on $\Delta$?

Attempt to solve:

Result from Exercise 2.4.12 says when $G$ is a permutation group on $\Sigma$, $P$ is a Sylow $p$-subgroup of $G$, $a\in\Sigma$, if $p^m\mid |Ga|$ then $p^m\mid |Pa|$. In this result replace $G$ with $H$, $a$ with $b$, then $Pb=b\Leftrightarrow|Pb|=1\Rightarrow p\nmid|Hb|$, and because $|Pb|\mid|P|=p^k, P\leq H$ so $p\nmid|Hb|\Rightarrow |Pb|=1$. Thus $b\in\Delta \Leftrightarrow p\nmid|Hb|$.

Beyond that my efforts were fruitless.

答:
Here is a sketch proof that $N_G(P)$ acts transitively on $\Delta$. Let $b \in \Delta$. Then there exists $g \in G$ with $ga=b$.

Note that $P$ and $gPg^{-1}$ are both Sylow $p$-subgroups of ${\rm Stab}_G(b)$, so there exists $h \in {\rm Stab}_G(b)$ with $hgP(hg)^{-1} = P$. Then $hg \in N_G(P)$ with $hga=b$.


问:
Help Understanding Fields
I came across this problem in a Linear Algebra text today:

Let $u$ and $v$ be distinct vectors in a vector space $V$ over a field $F$. Prove that $\{u,v\}$ is linearly independent if and only if $\{u+v,u-v\}$ is linearly independent.

Working on ($\Rightarrow$), I must show that

$$c_1(u+v) + c_2(u-v) = 0 \implies c_1 = c_2 = 0 \text{, where } c_1,c_2 \in F.$$

Using linear independence of $\{u,v\}$ I arrive at the equation $c_1 + c_1 = 0$. However, if $F = \mathbb{Z}_2$ then $1 + 1 = 0$. Also, if $c_1$ and $c_2$ were "integers" then in a field of characteristic $2$ I have

$$c_1 + c_1 = \underbrace{(1 + \cdots + 1)}_{c_1 \text{ times}} + \underbrace{(1 + \cdots + 1)}_{c_1 \text{ times}} = (1 + 1) + \cdots + (1 + 1) = 0 + \cdots + 0 = 0$$

However, I'm assuming that they're integers and I'm not sure (as for example $\mathbb{R}$ is a field which isn't only integers) that $x + x = 0$ in any field with characteristic $2$. Can someone clarify?

Also, I see often the restriction that $F$ be of characteristic not equal to $2$. Why is that?

My knowledge of fields is very limited so any references would also be appreciated.

Thank-you.

答: The statement you are trying to prove is, indeed, not true if the field has characteristic $2$. If $u=(1,0)$ and $v=(0,1)$ then $u$ and $v$ are linearly independent but $u+v=u-v=(1,1)$. </p>

Title: Help Understanding Fields


问:
How to show this sequence is unbounded?
$S_n=1+1/2+1/3+....+1/n$,for all n=1,2,3,...,
The question is to show it is divergent. I am thinking if I can prove it is unbounded, i.e., not bounded above, then I prove it. But I don't know how to say it is not bounded above rigorgosly.

答:
By Integral convergence test for series, $\sum_1^\infty \frac{1}{n}$  have same behavior with
$\int_1^\infty \frac{1}{x}dx$ and we know that $\int_1^\infty \frac{1}{x}dx=\infty$ so
$\sum_1^\infty \frac{1}{n}=\infty$ that is unbounded.

Title: How to show this sequence is unbounded?


问:
Possible counter-proof of the constant rule of integration
$$∫(2x+10)^{-1}dx$$

$$=\frac{1}{2}∫(x+5)^{-1}dx$$

$$=\frac{1}{2}\ln|x+5|+C$$

but here is the same integral, yet different answer

$$∫(2x+10)^{-1}dx$$

Let $u = 2x+10\implies \frac{du}{dx} = 2 \implies dx = du/2$

$$\implies \frac{1}{2}∫u^{-1}du$$

$$=\frac{1}{2}\ln|u| + C$$

$$=\frac{1}{2}\ln|2x+10| + C$$

Yet,

$\frac{1}{2}\ln|2x+10|$ does not equal $\frac{1}{2}\ln|x+5|$

Please let me know if I made any mistakes. I have been seeing this over and over in my DE class and it keeps bugging me

答:
They are still equal up to a constant (as usual with integration), since
$$\log(2x+10)=\log(2(x+5))=\log(x+5)+\log2$$
by the laws of logarithms.

Another nice example of this phenomenon is $\int\sin x\cos x\,dx$, which can work out to be $-\frac14\cos2x$ (if you apply the trigonometric identity for $\sin2x$) but also $-\frac12\cos^2x$ working another way (by parts). Indeed, you should be able to find a constant $C$ such that $-\frac14\cos2x = -\frac12\cos^2x+C$ for all $x$ (by applying an appropriate trigonometric identity).

Why can you get different answers?

Remember that when you "work out" $\int f(x)\,dx$, what you're actually doing is searching for a primitive, i.e.,  function $F$ which satisfies $F'(x) = f(x)$. These are not uniquely determined, i.e., there are many possible functions $F$ with the property $F'=f$, but you can prove (using the mean-value theorem) that any two primitives must be equal up to a constant. (This should make sense intuitively because a constant only shifts the graph of $y=f(x)$ up and down, which does not affect the gradient of the tangent at any point).


问:
Find a matrix with given row and column sums
Please forgive my intrusion. I've been working for days on this problem and it's vexing me. It doesn't seem to have a solution and I could really use some help. I have a "math square" (not a magic square) that looks like this when filled in

$$
\begin{array}{ccccc|c}
9 & 4 & 8 & 4 & 7 & 32 \\
7 & 9 & 15 & 7 & 5 & 43 \\
3 & 2 & 9 & 10 & 9 & 33 \\
5 & 3 & 5 & 6 & 4 & 23 \\
\hline
24 & 18 & 37 & 27 & 25 & 131 \\
\end{array}
$$

However, the puzzle when empty looks like this.

$$
\begin{array}{ccccc|c}
x & x & x & x & x & 32 \\
x & x & x & x & x & 43 \\
x & x & x & x & x & 33 \\
x & x & x & x & x & 23 \\
\hline
24 & 18 & 37 & 27 & 25 & 131 \\
\end{array}
$$

I need an equation of some sort to fill in the unknowns ($X$'s) and recreate the missing values.

There appears to be some symmetry with the puzzle as the columns and row all add up to be the same which in this case is $131$. If you separate them by every other row to perhaps break it down to make it easier to solve you get. You could also do this with the columns but for simplicity I haven't written it here.

$$
\begin{array}{ccccc|c}
9 & 4 & 8 & 4 & 7 & 32 \\
3 & 2 & 9 & 10 & 9 & 33 \\
\hline
12 & 6 & 17 & 14 & 16 & 65 \\
\end{array}
\dots
\begin{array}{ccccc|c}
7 & 9 & 15 & 7 & 5 & 43 \\
5 & 3 & 5 & 6 & 4 & 23\\
\hline
12 & 12 & 20 & 13 & 9 & 86   \\
\end{array}
$$

using these givens in the puzzle is allowed, but not the $X$'s

If it is indeed unsolvable I would like to know but if it can what missing component can be add to achieve that goal?

Your help is very much appreciated and thank you!

Regards,

Tony

p.s. sorry about the formatting.

答:
As this answer notes:

If you don't need integers, just put $r_ic_j/T$ in position $(i,j)$, where $r_i$ is the sum for row $i$, $c_j$ the sum for row $j$ and $T$ the total.

But even if you need integers, you can execute the following step for every cell: write $\left \lfloor{r_ic_j/T}\right \rfloor$ into $(i,j)$. This takes $n \times m$ steps. Then iterate over the matrix and substract the value of $(i, j)$ from ${r_i}$ and ${c_j}$ both. This also takes $n \times m$ steps. Since rounding down decreases the value by less than one the sum of all cells is at most $n \times m$ less than the desired total. Also, the new $r_i < m$ and $c_j < n$ for the same reason.

After this process, pick some $i | r_i > 0$  and a $j | c_j > 0$ and increase $(i,j)$ by one and decrease both $r_i$ and $c_j$ by one. Repeat this until finished -- this takes at most $n \times m$ steps because the total grows by one at every step. Thus, this algorithm is also $O(n \times m)$. Also note no cell will be increased more than $\min(n, m)$. Thus if the sum values were significantly bigger than $\min(n, m)$, the ratio of adjacent cells will be very close to the relevant sum ratio which is sometimes more desirable than the greedy filling of the matrix.


问:
Is $A=B$ if $A=\{a,b\}$ and $B=\{a,b,∅\}$?
I know that $A∩∅=∅$. But what about $A∩\{∅\}$?

Shouldn't  $A∩\{∅\}=\{∅\}$.

If $A=\{a,b\}$ and $B=\{a,b,∅\}$ then  we can conclude $A=B$?

答:
I will try to give a more informal and intuitive clarification. There is an important distinction between the set that contains nothing (which is the empty set specified as $\varnothing$) and the concept we call nothing (which doesn't have a symbol in set theory). To clarify that, we can rewrite $\varnothing$ as $\{\}$. You can then clearly see that $\{\}$ is not the same as nothing; it is something containing nothing. An analogy that is often made in mathematical textbooks is the following: The empty set is analogous to an empty shoebox. Therefore, a set containing the empty set, i.e. $\{\{\}\}$ or $\{\varnothing \}$ is a shoebox containing an empty shoebox. Thus, the empty set is not nothing, it is the set that holds nothing.

To come back to your question: $A \cap \varnothing = \varnothing $. That is true. But why is that? What does a set containing something (i.e. $A$) have in common with a set containing nothing? Well, nothing, regardless of how we specify $A$. And because the intersection of sets is always a set, we get a set containing nothing. And what is a set containing nothing? It is the empty set! Now why is that different to $A \cap \{\varnothing\} = \{\varnothing \}$? Because now you are comparing two sets that hold something. One set holds something unspecified (i.e. $A$), while the other set holds a set containing nothing (i.e. the empty set). And what do both sets have in common? Well, that depends on the specification of $A$. If $A$ contains the empty set, i.e. if $\varnothing \in A$, then we get $A \cap \{\varnothing\} = \{\varnothing \}$. If not, then we get $A \cap \{\varnothing\} = \varnothing$. And, as we said before, $\varnothing \neq \{ \varnothing \}$.

I hope that gives you a more intuitive explanation to a distinction that deludes a lot of people starting with elementary set theory.


问:
Show the existence of a weighted shift operator that converges to a unilateral weighted shift operator
Let $H$ be an infinite dimensional, separable complex Hilbert space with orthonormal basis $(e_n)_n$.

Let $T$ be the unilateral forward weighted shift. i.e. $Te_n=t_n e_{n+1}$. And $t_n = (gcd(n, 2^n))^{-1}$

Show that there exist weight shifts $(W_n)_n$ such that $\lim_{n \rightarrow \infty } W_n= T $ with respect to the same orthonormal basis and if $W_n$ is of order $m$, then  $W_n^{m-1 } \ne 0 =W_n^m$.

Conclude that the spectral radius $\rho:B(H) \rightarrow \mathbb{R}$ is not continuous.

I'm don't have any clues how to construct such weighted shift $W_n$.

I know that $T^me_k=T^{m-1}(t_ke_{k+1})=T^{m-2}t_k\cdot t_{k+1}e_{k+2}=\dots=(t_k\cdot t_{k+1}\cdots t_{k+m-1})e_{k+m}$.

Not sure if this helps.

Thanks in advance!

答:
I am only answering with a hint.

Set $W_n:H\to H$ as $W_ne_k=t_ke_{k+1}$ when $k$ is NOT divided by $2^n$ and $W_ne_k=0$ when $k$ is divided by $2^n$. Show that $W_n^{2^n}=0$ and that $\|W_n-T\|=\frac{1}{2^n}\to0$. Justify why $\rho(W_n)=0$ for all $n$.


问:
Compute the last 1000 digits of $1 + 50 + 50^2 + 50^3 + ... + 50^{999}$
What is a good way to find the last 1000 digits of:

$ 1 + 50 + 50^2 + 50^3 + ... + 50^{999} $

Or more generally, the last n digits of

$ \sum_{k=0}^{n-1} 50^k $

答:
Introduction

As Daniel above guesses, the last $n$ digits of the number $$
1 + 50 + ... + 50^{n-1}
$$

has something to do with the expansion of $\frac 1{49}$. Here's how you get it.

On the expansion of $\frac 1{49}$

Well, first we note that $\frac{10^{42}-1}{49}$ is an integer, and for no $0<k<42$ is $\frac{10^k-1}{49}$ an integer. This is the same as saying that the repetend at the end of $\frac 1{49}$ is of length $42$.

This is a standard thing that can be checked, but to make it easier : we know that $\phi(49) = 42$ where $\phi$ is the Euler totient function. Therefore, by Euler's theorem , $49$ always divides $10^{42}-1$. However, it is easy to prove (probably from group theory) that if $10^k-1$ divides $49$ for $k<42$ then $k$ must divide $42$. Therefore, it is sufficient to check for $k$ divisors of $42$. In fact, it's enough to check that each of $10^{6}-1, 10^{14}-1$ and $10^{21}-1$ are not multiples of $49$, which one can show by using the iterative procedure coming from the identity $10^{k+1} \equiv 10 (10^k) \pmod{49}$. (so you don't need to do long division)

Formula for the last $n$ digits

Now, we note that if $n \leq 42$ then :
$$
10^{42} \equiv 50^n \pmod{10^n}
$$

therefore elementary transformations lead to :
$$
\frac{10^{42}-1}{49} \equiv \frac{50^n-1}{49} \pmod{10^n}
$$

which works because $49$ is coprime to $10^n$ and because both sides of the $\equiv$ are integers. Therefore, the last $n$ digits of $\frac{50^n-1}{49}$ equal the last $n$ digits of $\frac{10^{42}-1}{49}$, or the last $n$ digits in the repetend of $\frac 1{49}$ when expressed as a decimal.

Now, if $n> 42$, write $n = 42q+r$ where $q>0$. We note that $\frac{10^{42(q+1)}-1}{49}$ is an integer. Therefore, beginning from $$
10^{42(q+1)} \equiv 50^n \pmod{10^n}
$$

we get :
$$
\frac{10^{42(q+1)} - 1}{49} \equiv \frac{50^n-1}{49} \pmod{10^n}
$$

However, note that $\frac{10^{42(q+1)-1}}{49}$ can be retrieved as follows : you take the repetend of $\frac{1}{49}$ , a total of $q$ times. Then to the front of this you attach the last $r$ digits of the same repetend. You are done.

Examples

Let's take some examples to show how easy this is. The expansion of $\frac 1{49}$ is :
$$
\frac 1{49} = 0.\overline{020408163265306122448979591836734693877551}
$$

(The expansion can be done using long division, but I'll mention it : There's a Vedic sutra that can be used to calculate reciprocals faster. It is explained briefly on this page. Also note how the start has powers of $2$ : this pattern is quite easy to write down as well, but isn't efficient)

Let $S_n = 1+50 + 50^2 + ...+50^{n-1}$.

The last $13$ digits of $S_{13}$ are the last $13$ digits of the repetend i.e. $6734693877551$.

The last $34$ digits of $S_{34}$ are the last $34$ digits of the repetend i.e. $3265306122448979591836734693877551$.

The last $47$ digits of $S_{47}$ : we have $47 = 42 \times 1 + 5$, so take the repetend once and add the last five digits of the repetend in front :
$$
77551020408163265306122448979591836734693877551
$$

The last $100$ digits of $S_{100}$ : we have $100 = 42 \times 2 + 16$, so take the repetend twice and add the last $16$ digits of the repetend in front :
$$
1836734693877551020408163265306122448979591836734693877551020408163265306122448979591836734693877551
$$

and so on.


问:
Can someone explain to me what are these 2 statements talking about?
I have to prove that these 2 statements are equivalent, but I can't even understand them.

There exist $\epsilon_0>0$ such that for all $k\in\mathbb N$, there exist $n_k\in\mathbb N$ such that $n_k\geq k$ and $|x_{n_k}-x|\geq\epsilon_0$?

There exist $\epsilon_0>0$ and a subsequence $(x_{n_k})$ of sequence $(x_n)$ such that $|x_{n_k}-x|\geq\epsilon_0$ for all $k\in\mathbb N$.

Thank you.

答:
This looks like an exercise to get you thinking about the logic of these kinds of statements, and to get used to reading them - you are likely to encounter many more of similar form. It is therefore worth putting in some effort to decode them. Sometimes it is easiest to work from the end, because this is the target statement, and the rest is needed to frame it accurately in context.

So in the first one, this is aiming at $|x_{n_k}-x|\geq\epsilon_0$, which we can interpret as $x_{n_k}$ is not close to $x$ - so we are trying to find terms of the sequence which are further away from $x$ than $\epsilon_0$.

We are indexing these terms with positive integers $n_k$, and we want an infinite number of them. We could do that in different ways - here it is done by setting $n_k \ge k$, so we can't choose $n_k=1$ every time, because the sequence eventually has to grow.

This is an interesting way of doing it, because it doesn't specify that the $n_k$ are growing all the time, rather it says that they eventually get bigger than any definite integer. So the sequence $n_k$ could begin $10, 9, 8, 7, 6, 6 \dots$ but by the time we reach $n_{11}$ the value has to be at least 11.

Note that this example does not give you a subsequence, but you should be able to show that a subsequence does satisfy the condition $n_k \ge k$, hence that a subsequence defined as in $2$  - which we can see is also looking for elements of the original sequence which are not close to $x$ - satisfies all the conditions of definition 1.

The real work is showing that you can pick an increasing sequence of integers from the $n_k$ in definition 1, which will index your subsequence. You will need to use the observation that the $n_k$ grow eventually. One approach is to assume you have picked the first $r$ numbers and  to show that you can then pick the $(r+1)th$.


问:
Counterexample to Tonelli's theorem
Does there exist a $C^{\infty}$ function $f: \mathbb{R}^2 \to \mathbb{R}$ such that $\int dx \int dy \,f(x,y)$ diverges to infinity but $\int dy \int dx\, f(x,y)$ converges?

Note that if $f$ is non-negative this is impossible by Tonelli's theorem.

答: Yes. Define $f_0(x,y)=\exp\left(\frac{1}{(x-1/2)^2+(y-1/2)^2-1/16}\right)$ whenever $\small (x-1/2)^2+(y-1/2)^2<1/16$ and $f_0=0$ otherwise. Then $f(x,y)=\sum_{k\ge 0} (k+1) (f_0(x-k,y-k)-f_0(x-k-1,y-k))$ should do, since $\int f dx=0$ for all $y$ thus $\int dy \int f dx=0$, but $f_y(x)=\int fdy$ is a nonnegative period $1$ function with positive integral in $x\in(k+1/4,k+3/4)$ for all $k\in \mathbb N$ thus $\int dx \int fdy=\infty$. Since $f_0\in C_c^\infty$ and $\text{supp}f\in [0,1]^2$, one trivially have $f\in C^\infty$.

Title: Counterexample to Tonelli's theorem


问:
First half of the Double Arrow is homeomorphic to $(0,1)$ in the Lower Limit Topology
What can we say about a locally compact Hausdorff space whose every open subset is sigma compact?

My question is induced from these problem and from Henno Brandsma's answer.

I want to show in double arrow space the subspace $(0,1)\times\{1\}$ is homeomorphic to $(0,1)$ in lower limit topology or the other name is Sorgenfrey line generated by $[a,b)$ type intervals.

Take double arrow space $[0,1]\times \{0,1\}$

My Attempt to show homeomorphism as follows:

Define $$\phi:(0,1)\times\{1\}\to (0,1)\\ (x,1)\mapsto x$$

from double arrow space to lower limit topology.

First of all it is clearly a bijection.

And continuous because take any open set in lower limit topology on $(0,1)$
as $[a,b)$

then $\phi^{-1}([a,b))=(a,\{0\}, b,\{1\})\cap (0,1)\times \{1\}$
it is open in subspace topology.

Is this way correct? And I cannot show $\phi $ is an open map because I cannot estimate general type of basic open sets in subspace of double arrow space.

答:
I'll call $(0,1) \times \{1\}$ in the subspace topology $\Bbb S_1$ for short.

A basic open subset of a point $(x,1)$ in $\Bbb S_1$ is of the form $\langle A,B\rangle \cap \Bbb S_1$, where $\langle A,B\rangle$ is a basic open interval of Double Arrow, where $A < (x,1) < B$ (the definition of the order topology, and the definition of subspace topology are applied here).

If $A = (a,0)$, with $a \le x$, and $B=(b,0)$ or $B=(b,1)$ with $x < b$ we get that $\langle A,B\rangle \cap \Bbb S_1 = [a,b) \times \{1\}$

If $A =(a,1)$ with $a < x$ and $B=(b,0)$ or $(b,1)$ with $x < b$ we get that
$\langle A,B\rangle \cap \Bbb S_1 = (a,b) \times \{1\}$

In all cases the map $\phi: ((0,1), \mathcal{T}_s) \to \Bbb S_1, x \to (x,1)$
sends basic elements of $\mathcal{T}_s$ (the Sorgenfrey topology on $(0,1)$, which is homeomorphic to the full Sorgenfrey line by any order homeomorphism from $(0,1) $ to $\Bbb R$) to open subsets of $\Bbb S_1$, so $\phi$ is bijective (obvious) and open. As $\phi^{-1}[B]$ is also open for every basic open set of $\Bbb S_1$, as we saw, $\phi$ is also continous. We can check openness and continuity by base elements alone; this is standard.

It follows that $\Bbb S_1$ is homeomorphic to the Sorgenfrey line.


问:
What is a starfinite set?
In the book Field Arithmetic by Fried and Jarden, the following definition is given on p. 273:

Consider an enlargement of a higher order structure that contains both $P$ and $K$. Call the elements of $\mathcal{P}^*_{\mathrm{fin}}(P)$ the starfinite subsets of $P^*$.

What is a starfinite set? I can't understand the definition.

答:
A set $A \subseteq \mathbb{N}$ is finite iff it is contained in $\{0, \ldots, n\}$ for some $n \in \mathbb{N}$.

Applying transfer, an internal set $A \subseteq {}^*\mathbb{N}$ is $*$-finite iff it is contained in $\{0, \ldots, \xi\}$ for some $\xi \in {}^* \mathbb{N}$.

Edit: equivalently, a $*$-finite set is an element of ${}^* \mathcal{P}_\mathrm{fin}(\mathbb{N})$, where $\mathcal{P}_\mathrm{fin}(\mathbb{N})$ is the collection of finite subsets of $\mathbb{N}$.

Title: What is a starfinite set?


问:
Optimisation: Show Formula is True
If postal regulations state that for a cylindrical parcel the sum of the length and circumference must not exceed 3m, a) show that for a cylindrical parcel whose sum of the length and circumference is 3m the volume V of the cylinder is given by $V=\frac{3C^2-C^3}{4\pi}m^3$, where C is the circumference; b) hence, find the dimensions of the cylindrical parcel which has maximum volume.

For reference, optimisation problems, in general, are not my problem. It's this specific question, and this specific formula (part a) that I can't seem to parse :(

Title: Optimisation: Show Formula is True

Body:
$$V=\pi r^2 h$$
$$C=2\pi r \quad r=\frac{C}{2\pi} \to V=\pi \left(\frac{C}{2\pi}\right)^2 h=\frac{C^2}{4\pi} h$$
$$C+h=3 \quad h=3-C \to V=\frac{C^2}{4\pi} (3-C)=\frac{3C^2-C^3}{4\pi}$$


问:
If the following piecewise function is continuous
If the following piecewise function is continuous, what is the value of $\alpha$

$
f(x,y)=
\begin{cases}
 arctan(\dfrac{|x|+|y|}{x^2+y^2})&\text{if}\, (x,y)\ne(0,0)\\
 \alpha&\text{if}\, (x,y)=(0,0)
\end{cases}
$

First I think I need to evaluate the limit of this function but I couldn't find the following limit

$$\lim\limits_{(x,y)\to(0,0)}arctan(\dfrac{|x|+|y|}{x^2+y^2})  $$

I checked this limit from wolfarm calculator but that says this limit does not exist.
So what can I do here ? please help

答: Since $|x|+|y| \geq \sqrt {x^{2}+y^{2}}$ we see that $f(x,y) \to \arctan(\infty)= \frac{\pi} 2 $ as $(x,y) \to (0,0)$. Hence $f$ is continuous iff $\alpha =\frac {\pi} 2$


答: Hint: transform into polar coordinates, $x=r \cos \theta$ and $y = r \sin \theta$.

Body:
If the following piecewise function is continuous, what is the value of $\alpha$

$
f(x,y)=
\begin{cases}
 arctan(\dfrac{|x|+|y|}{x^2+y^2})&\text{if}\, (x,y)\ne(0,0)\\
 \alpha&\text{if}\, (x,y)=(0,0)
\end{cases}
$

First I think I need to evaluate the limit of this function but I couldn't find the following limit

$$\lim\limits_{(x,y)\to(0,0)}arctan(\dfrac{|x|+|y|}{x^2+y^2})  $$

I checked this limit from wolfarm calculator but that says this limit does not exist.
So what can I do here ? please help


答:
Note that
$$\lim_{(x,y) \to (0,0)} \dfrac{|x|+|y|}{x^2+y^2} = +\infty.$$

Then, as $\arctan$ is a continuous function,
$$\lim_{(x,y) \to (0,0)} \arctan\left(\dfrac{|x|+|y|}{x^2+y^2}\right) = \arctan(+\infty) = \dfrac{\pi}{2}.$$

Then, $\alpha$ must be $\pi/2$.

Title: If the following piecewise function is continuous


问:
Locus of the intersection point of angle bisectors of angles made by 2 fixed lines
"$AB$ and $CD$ are two fixed straight lines and a variable straight line cuts them at $X$ and $Y$ respectively. The angular bisectors of $\angle AXY$ and $\angle CXY$ meet at $P$. Find the locus of $P$."

First things first, if the question states $\angle CXY$, it is clear that their angle bisectors meet on line $AB$ and thus, $P$ lies on $AB$. I believe the question wanted to state "$\angle AXY$ and $\angle CYX$" instead. So I considered the question to be stated as,

"$AB$ and $CD$ are two fixed straight lines and a variable straight line cuts them at $X$ and $Y$ respectively. The angular bisectors of $\angle AXY$ and $\angle CYX$ meet at $P$. Find the locus of $P$."

My attempt:

It is trivial to note that had $AB$ and $CD$ been parallel, the locus would have been $\angle XPY=90^{\circ}$. Then I tried considering non-parallel lines.

Assume that $\angle AXY≥\angle BXY$, and $\angle CYX≥\angle BYX$ such that $\angle AXY= 2x$ and $\angle BXY= 2y$. Then, if $P$ lies on the side whose sum of interior angles$>180^{\circ}$, clearly $\angle XPY=180-(x+y)$. The same may be done for all combinations of relations of supplementary pairs $(\angle AXY,\angle BXY)$ and $(\angle CXY,\angle ∠DXY)$.

However, this is an EXTREMELY piece-wise definition (and subject to errors) and as a result, I'm not satisfied with it; there must be a way to refine my answer to produce a much more accurate and concise one. In short, I have 2 primary questions:

If my method of solving is correct, how do I refine my answer?

If I need to re-start from scratch, how do I start?

Thanks for the help!

Source: Challenge and Thrill of Pre-College Mathematics

*Note: This may be marked a duplicate as I've seen the same question before. However as I couldn't find a satisfactory explanation there, I decided to post my own question.

答:
As you have already considered the scenario where $AB\parallel CD$, i am considering these two straight lines two be non parallel.

Let $AB$ and $CD$ meet at point $Q$. Divide this into two cases.

Case $1:$  We get point $Q$ by extending $AB$ and $CD$ through points $A$ and $C$ respectively.

Then, in $\triangle QXY$, the internal bisectors of $\angle QXY$ and $\angle QYX$ meet at $P$.Hence, $P$ is the incentre of $\triangle QXY$ and thereafter the locus of $P$ is the bisector of $\angle Q$.

Case $2:$  We get point $Q$ by extending $AB$ and $CD$ through points $B$ and $D$ respectively.

Then, in $\triangle QXY$, the external bisectors of $\angle QXY$ and $\angle QYX$ meet at $P$. Hence, $P$ is the excentre of $\triangle QXY$ and thereafter the locus of $P$ is the bisector of $\angle Q$.

Thus, the locus of $P$ is the line which internally bisects the angle formed by the lines $AB$ and $CD$.


问:
Using the estimated 96% interval, [0.48, 0.54], what conclusion would we reach if we test H0: P = 0.56 versus H1: P $\neq$ 0.56 at α = 0.04?
ADDITIONAL INFO: According to a poll, (51%) of surveyed employees agree with the statement that the people
who run most companies are "honest and ethical."

Already found the sample size to equal 1167

Found z= -3.44

Concluded to reject the null and accept the alternative, since -3.44 less than or equal to -2.05

Is my answer correct?

答:
It is essentially correct, though you should show how you found $1167$ and $0.0145306$ and $-3.44$

When you have done that, you may be able to see that your argument is essentially that the confidence interval $[0.48, 0.54]$ does not cover the hypothesis value of $0.56$


问:
What is my mistake in this determinant of order $n$?
Consider real numbers $a_i,b_i\in\mathbb{R}$ and the following determinant

$$ \alpha_n= \begin{vmatrix} a_1 + b_1 & b_1 & b_1 & b_1&\cdots & b_1 \\
b_2 & a_2 + b_2 & b_2 & b_2 & \cdots & b_2 \\
 b_3 &   b_3 & a_3 + b_3 & b_3 & \cdots & b_3 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
b_n & b_n & b_n &  b_n & \cdots & a_n + b_n\end{vmatrix} $$
My attempt was to substitute column 1 with column 1 - column 2 $(C_1 \to C_1 - C_2)$. Then do the same with others columns: $C_2 \to C_2 - C_3$,..., $C_n\to C_n - C_1$. After all these transformations, arise this determinant
$$ \alpha_n =\begin{vmatrix}
a_1   &0& 0 &0&\cdots & -a_1 \\
-a_2 &a_2 & 0 & 0 & \cdots & 0 \\
0 &  -a_3 & a_3  & 0 & \cdots &0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 &  0 & \cdots & a_n \end{vmatrix}  $$
So it seems that $\alpha_n$ only depends on $a_i$. However, computing the case $n=2$, gives
$$ \alpha_2 = \begin{vmatrix}
a_1 +b_1  &b_1 \\ b_2 & a_2 + b_2 \end{vmatrix} = a_1a_2 + a_1b_2 + a_2b_1. $$
What am I doing wrong? How can I compute a closed form for $\alpha_n$? Thank you in advance.

答:
Note that while performing row or column operations on determinants, it is necessary to preserve atleast one column/row, we cannot change all of them at once.

In your case, you cannot do $C_n \to C_n-C_1$.


答:
I will work in radius instead of circumference because it is the typical convention, so for part (a).

We need two equations.
The symbols I will use is V = volume, r = radius, h = height/length.

For the volume of the cylinder:
$V = \pi r^2 h$

For the relationship between the circumference and height:
$h = 3-2\pi r$

Then we can substitute equation 2 into equation 1, and obtain equation 3.

$V = \pi r^2 (3-2 \pi r)$

This is your answer for part (a), just substitute the equation $ 2 \pi r = C$.

Meanwhile for part (b), using equation 3

$V = 3 \pi r^2 - 2\pi^2 r^3$

Differentiate it,

$\frac{dV}{dr} = 6\pi r - 6\pi ^2r^2$

And to find the point of maximum volume, the gradient is $0$, $\frac{dV}{dr} = 0$

$6\pi r - 6\pi ^2r^2 = 0$

$r-\pi r^2 = 0$

$r(1-\pi r) = 0$

So,

$r = 0$ or $ 1-\pi r = 0$

$r = 0 $ corresponds to the minimum volume. (can find the second order derivative to prove this)

$r = \frac{1}{\pi}$ corresponds to maximum volume.

So this is your solution, and to find C or h, simply substitute r into,

$ C = 2 \pi r$ and $h = 3-2\pi r$

hope this helps :)

Title: Optimisation: Show Formula is True

Body:
If postal regulations state that for a cylindrical parcel the sum of the length and circumference must not exceed 3m, a) show that for a cylindrical parcel whose sum of the length and circumference is 3m the volume V of the cylinder is given by $V=\frac{3C^2-C^3}{4\pi}m^3$, where C is the circumference; b) hence, find the dimensions of the cylindrical parcel which has maximum volume.

For reference, optimisation problems, in general, are not my problem. It's this specific question, and this specific formula (part a) that I can't seem to parse :(


Body:
Hint:

$$\frac{|x|+|y|}{x^2+y^2}\ge\frac{|x|+|y|}{(|x|+|y|)^2}=\frac1{|x|+|y|}\to\infty.$$


问:
Describing the topology of the set $\{e^{i\theta} : -\pi<\theta \leq \pi\}$
The question asks to describe some topological features of the set $\{e^{i\theta} : -\pi<\theta \leq \pi\}$.

More precisely, to explain whether the set is open, closed, neither or both and also whether or not it is bounded or if its connected?

After some research, I came to the conclusion that the graph would just be of a unit circle. Although, based on that knowledge, how can I know whether or not this set is open, closed, or neither?

答:
The set is of course exactly the unit circle $S^1=\{z \in \Bbb C\mid |z|=1\}$: all $e^{i\theta}$ have length 1 and all argument values from $(-\pi,\pi]$ occur (so no gaps, or skipped points).

The set $S^1$ is the continuous image of $[-\pi,pi]$ under the continuous function $t \to e^{it}$, so as closed intervals are compact and connected in $\Bbb R$, so is $S^1$ in $\Bbb C$. Being compact, it's also closed there (and by consequence not open, as $\Bbb C$ is connected, although it's clear geometrically that no open ball in $\Bbb C$ sits inside the circle, so its interior is empty).


问:
Is there a general expression for the adjoint representation of $U(N)$ or $u(N)$?
At least for low values of $N$ like $2$ or $3$ and such I would like to know if there are explicit matrices known giving the representation of $u(N)$ or $U(N)$ in the adjoint?  </p>

<p>(..a related query: Is it for the Lie group or the Lie algebra of U(N) that it is true that the weight vectors in the fundamental/vector representation can be taken to be N N-vectors such that all have weight/eigenvalue 1 under its Cartan and the ith of them has 1 in the ith place and 0 elsewhere and for the conjugate of the above representation its the same but now with (-1)?..I guess its for the u(N) since they are skew-Hermitian but would still like to know of a precise answer/proof..)  </p>

答:
I've only recently started learning about group theory so it would be worth double checking the following.

As I understand it, when we say something (for example a field, $X$) transforms under the adjoint rep of a group (say $U(N)$), the transformation can involve any matrix, $g$, belonging to the group $U(N)$. The adjoint rep doesn't put any constraints on the matrices we choose from $U(N)$, rather it tells us how to use those matrices in a transformation. When we say $X$ transforms in the adjoint rep of $U(N)$ it simply means it transforms as $X \rightarrow X' = gXg^{-1}$. The matrix $g$ can be any matrix belonging to $U(N)$. Note: $g^{-1}$ is the inverse of the matrix $g$. So when we say 'an object is in the adjoint representation of $U(N)$' this is a statement about how to act the matrices of $U(N)$ on an that object. It doesn't say anything about which matrices we choose from $U(N)$.


问:
How to show that given a catagory $\scr C$ and two terminal objects $X$ and $Y$ in $\scr C$, they are isomorphic.
I know that there are unique morphisms (functions in the categories of sets) $\delta_X: X \to Y$ and $\delta_Y: Y \to X$, but I am unable to show, either diagrammatically, or otherwise, that $\delta_Y \circ \delta_X=1_X$ and $\delta_X \circ \delta_Y=1_Y$.

答: Since $X$ is a terminal object, there is a unique morphism $A\to X$ for any object $A$ in the category. In particular, there must be a unique morphism $X\to X$. Now $\delta_Y\circ\delta_X$ is one such morphism. On the other hand, every object has the identity morphism, so $1_X:X\to X$ is another such morphism. By uniqueness, they must equal. Similarly, we have $\delta_X\circ \delta_Y=1_Y$.


问:
How to find all subsets $Y \subset \mathbb R$ such that $Y \times Y \subset \{(x , y) : x^2 + 4y^2 \leq 1\}$
How to find all subsets $Y \in \mathbb R$  such that $Y \times Y  \subseteq \{(x , y) : x^2 + 4y^2 \leq 1\}$

Can anyone please give some idea to proceed?

I think we have to consider all the squares whose  diagonals fall on $y = x$ in  the ellipse $x^2 + 4y^2 = 1$.

But I can not understand how to see them  properly.

Can anyone please help me ?

答:
Define $E:=\{(x,y) \mid x^2 + 4y^2 \le 1\}$.

If $A \subseteq R$ and $A \times A$ is a subset of $E$, then for any $a \in A$, $(a,a) \in E$ so $a^2 + 4a^2 \le 1$ and hence $5a^2 \le 1$ and so $|a| \le \frac{1}{\sqrt{5}}$ and it follows that $A \subseteq [-\frac{1}{\sqrt 5}, \frac{1}{\sqrt 5}]$

Conversely, suppose that $A \subseteq [-\frac{1}{\sqrt 5}, \frac{1}{\sqrt 5}]$.

Let $a \in A$ so that $|a| \le \frac{1}{\sqrt 5}$ and so $a^2 \le \frac15$ and $4a^2 \le \frac45$ and hence $(a,a)$ obeys the condition $x^2 + 4y^2 \le 1$ and so $A \times A \subseteq E$.

So the answer is as Kavi hinted at in the comments

All $Y$ that are subsets of $[-\frac{1}{\sqrt 5}, \frac{1}{\sqrt 5}]$.


问:
Condition that a family of unitary and antiunitary operators actually consists of unitary operators only
I am currently reading a book on linear operators in quantum mechanics in which the following statement is made in a chapter on propagators.

Let $\mathcal{F} = \{ U(t,s) \: | \: s,t \in \mathbb{R} \}$ be a family of operators in a Hilbertspace $\mathcal{H}$, each of which is either unitary or antiunitary. Furthermore, the following conditions apply to all $r,s,t \in \mathbb{R}$:

$U(t,s) U(s,r) = U(t,r)$.

The map $(s,t) \mapsto U(s,t)$ is strongly continuous in $\mathbb{R}^2$.

Then all $U(t,s) \in \mathcal{F}$ are unitary.

The following definitions are used: A bounded operator $U : \mathcal{H} \to \mathcal{H}$ is called unitary if it is surjective and $(U x , U y) = (x,y)$ holds for all $x,y \in \mathcal{H}$. $U$ is called antiunitary if it is surjective and $(U x , U y) = (y,x)$ is valid for all $x,y \in \mathcal{H}$.

The surjectivity of every $U(t,s) \in \mathcal{F}$ is clear, since any unitary or antiunitary operator is surjective. Furthermore we have that $U(t,s)$ is injective, since $|| U x || = || x||$ applies. Therefore $U(t,s)$ is bijective and using item 1 we find its inverse operator to be $U(t,s)^{-1} = U(s,t)$.

To show the condition $(U(s,t) x , U(s,t) y) = (x,y)$ for all $U(s,t) \in \mathcal{F}$ and all $x,y \in \mathcal{H}$, I honestly do not have a bright idea. Approximating an arbitrary element $U \in \mathcal{F}$ by a sequence unitary operators $U(t_n, s_n) \in \mathcal{F}$ in the strong operator topology (if such a sequence exists) will not work, since the set of unitary operators is not closed in the set of linear and bounded operators (with respect to SOT).

I am grateful for every hint.
Thanks in advance,
GordonFreeman

答:
Take a $v\in H$ and for aesthetic purposes let $\|v\|=1$. Consider the map
$$\Bbb R^2\to \Bbb R,\qquad (t,s)\mapsto \|iU(t,s)v - U(t,s)iv\|$$
The value at $(t,s)$ is $0$ if $U$ is unitary and $2$ if $U$ is anti-unitary, no other values are taken. Since the map is continuous you have that the pre-images of $0$ (resp $2$) are open and closed.

The only open and closed sub-sets of $\Bbb R^2$ are $\emptyset$ and $\Bbb R^2$. Since $U(1,1)=\Bbb 1$ is unitarythe pre-image of $0$ cannot be $\emptyset$, so $U(t,s)$ is unitary for all $(t,s)$.


问:
What is the half-derivative of zeta at $s=0$ (and how to compute it)?
[Update 3:] I gave a new partial answer following the ansatz in question Q3. I leave the other parts of the question untouched, they are also partially answered in specialized other questions in MSE.

I'm trying to understand the concept of fractional derivatives and am fiddling with the examples at wikipedia. The a'th derivative of a monomial in x, where a can be fractional is accordingly $$ {d^a \over dx^a} x^m = { \Gamma(1+m) \over \Gamma (1+m-a) } x^{m-a} $$.

Q1: But what happens for some function $f(x)$ if I want to evaluate the half-derivative at zero?

Originally I'm interested in the fractional derivatives of the zeta at zero. Because I thought that the monomial-halfiterative is the most easy one to understand I tried first the power-series expression of the zeta-function
$$ \zeta(x) = - {1 \over 1-x} + \sum_{k=0}^\infty w_k x^k $$ where $w_k$ are some coefficients beginning with $w_0=0.5, w_1=0.081... , w_2=-0.0031... , \cdots $

But if I want to find the (1/2)'th derivative at $x=0$ I need definitions how I should handle the fractional powers of zero.

Q2: How can I evaluate the fractional derivative of the leading fraction ${1 \over 1-x}$? Can I do better than to express the fraction by its power series and do the derivations termwise at the monomials?

Q3: Or can I do something like with the integer derivatives of the zeta at $s=0$ where I express it as the Dirichlet-series having the logarithms in the numerators?

[update]: concerning Q3, I've now used the alternating-zeta version and assumed, that
$$ {d^{1/2} \over dx^{1/2}} {1 \over k^x}={d^{1/2} \over dx^{1/2}} \exp(x(-\log(k))) =(-\log(k))^{1/2} {1 \over k^x} = i \cdot (\log(k))^{1/2} {1 \over k^x} $$
and then I set $x=0$ and evaluate the alternating series $$ \eta(0)^{(1/2)} \underset{\mathfrak E}{=} \sum_{k=0}^\infty \left((-1)^k log(1+k)^{1/2}\cdot i \right)\sim - 0.347006596200 \cdot i $$ where $\mathfrak E $ means Euler-summation of the divergent series. However, even if that result is meaningful this does not yet help much because I've now no further idea how I could use the Euler's zeta/eta-conversion-formula here. (I have just developed the conversion scheme for the integer derivatives, but that transforms to an infinite series if this is at all generalizable to fractional indexes)

[update2]: I tried the Riemann-Liouville-formula for the half-derivative as given by @J.M. in MSE, but the result is inconclusive. First, I need to handle zeros in the denominators, and second, if I replace them by limiting expressions with $\epsilon \to 0$ the numerical integration seems to diverge either to $-\infty$ or $- i \infty$ depending on whether I approach zero from positive or from negative values. So I need some help even for this...

答:
Here is possibly an answer. I followed the path at Q3 in my above question  with a final procedure of divergent summation. The ansatz is to express the problem as a sum of a function evaluated at consecutive arguments, just in the sense of the Euler-MacLaurin-summation method, which requires a power series for the function, which gives the "general term" for the series.

While we don't have a Taylor series for the general term $ \sqrt{-\ln(1+x)} $ around $x=0$ we can rewrite this a bit and get a related series:
$$ \sqrt{\ln(1+x)} = \sqrt{x \ln(1+x)/x} = \sqrt x \sqrt{ {\ln(1+x) \over x}} $$
and for the second root-expression we have a nice Taylor-series

$ \displaystyle \qquad \small f(x)=\sqrt{ {\ln(1+x) \over x}} = \sum_{k=0}^{\infty} a_k x^k
\\   \\ \small \qquad \qquad = 1 - 1/4 x + 13/96 x^2 - 35/384 x^3 + 6271/92160 x^4 - 2211/40960 x^5 + O(x^6)  $

We get then

$ \displaystyle \qquad \small \sqrt{\ln(1+x)} = \sum_{k=0}^{\infty} a_k x^{0.5+k } $

Now I assume, that it is possible to express formally the sum of that function for k over all consecutive integers by the double sum

$ \displaystyle \qquad \small \sum_{k=0}^\infty  \sqrt{ - \ln(1+k)} = \sqrt{-1} \cdot \sum_{k=0}^{\infty} \left( a_k \sum_{x=0}^\infty x^{0.5+k } \right) = i \sum_{k=0}^{\infty} a_k \zeta(-0.5-k) $

and the goal to evaluate the fractional half-derivative of $\zeta$ at zero were then, using a procedure $\small \mathfrak N$ for strongly divergent series,

$ \displaystyle \qquad \small  \zeta(0)^{(1/2)} \underset{\mathfrak N}{=} i \sum_{k=0}^{\infty} \left( a_k \zeta(-0.5-k) \right) \qquad \sim - 0.200824660892 \  i $

Now conjectured:
$$   \zeta(0)^{(1/2)} \underset{\mathfrak N}{=} i \sum_{k=0}^{\infty} \left( a_k \zeta(-0.5-k) \right) \\   \zeta(0)^{(1/2)} \sim - 0.200824660892 \  i $$

The applicability and the correctness of the above sequence is not sure/not proven. Heuristics show, that the method applied to the fractional derivatives of the $\eta$ gives the expected results but for instance for the non-alternating zeta series we might need some integral as summand like in the Ramanujan-summation-method.

So far I can only hope that someone else does crosscheck the method and the recieved value.


问:
Name of matrix operation of [ A[0] dot B[0], A[1] dot B[1] ] from 2x2 matrices A, B
Please advise how this matrix operation is called, and what is the numpy operation for it. np.inner, np.dot does not create dot products from 2x2 matrices.

答:
This is the diagonal of the matrix product $AB$, which can be obtained in NumPy using numpy.diagonal and numpy.matmul.

For example

```<code>import numpy as np
A = np.matrix('0 1 2; 3 4 5')
B = np.matrix('0 -3; -1 -4; -2 -5');
np.diagonal(np.matmul(A,B)) # or (A*B).diagonal()
</code>```

yields

```<code>array([ -5, -50])
</code>```

However, this does compute the off-diagonal entries of the matrix product as well and hence isn't very efficient. Describing the entries of your resulting vector as
$$
v_i = \sum_j a_{ij} b_{ji}
$$
you can obtain the same result more efficiently using numpy.einsum as

```<code>np.einsum('ij,ji->i', A, B)
</code>```

Body:
This is the diagonal of the matrix product $AB$, which can be obtained in NumPy using numpy.diagonal and numpy.matmul.

For example

import numpy as np
A = np.matrix('0 1 2; 3 4 5')
B = np.matrix('0 -3; -1 -4; -2 -5');
np.diagonal(np.matmul(A,B)) # or (A*B).diagonal()

yields

array([ -5, -50])

However, this does compute the off-diagonal entries of the matrix product as well and hence isn't very efficient. Describing the entries of your resulting vector as
$$
v_i = \sum_j a_{ij} b_{ji}
$$
you can obtain the same result more efficiently using numpy.einsum as

np.einsum('ij,ji->i', A, B)


问:
Let $A$ be a $n \times n$ matrix such that $A^3 = O$, but $A^2 \ne O$. Show that $\{I,A,A^2\} \subset M_{n \times n}(\Bbb R)$ is linearly independent.
Let $A$ be a $n \times n$ matrix such that $A^3 = O$, but $A^2 \ne O$. Show that $\{I,A,A^2\} \subset M_{n\times n}(\Bbb R)$ is linearly independent.

Attempt:
Is the $O$ above denotes the zero matrix in $M_{n \times n}(\Bbb R)$?

Let $k_1,k_2,k_3 \in \Bbb R$ such that
\begin{equation*}
k_1I + k_2A + k_3A^2 = O \ldots (1)
\end{equation*}
By adding both sides of $(1)$ by $I$, we have
\begin{equation*}
(k_1+1)I + k_2A + k_3A^2 = I \ldots (2)
\end{equation*}
Now, multiplying both sides on the right by $A$, we have
\begin{equation*}
(k_1+1)A + k_2A^2 + k_3A^3 = A
\end{equation*}
or, equivalent, as
\begin{equation*}
(k_1+1)A + k_2A^2 = A.
\end{equation*}

I got stuck there. Any ideas? Thanks in advanced.

答:
Hint :

From $aA^2+bA+cI_n=0$, multiply by $A^2$. What do you get ?


答:
Suppose that $xI+yA+zA^2=0$, we deduce that $A^2(xI+yA+zA^2)=xA^2=0$, we deduce that $x=0$ since $A^2\neq 0$.

We also have $A(yA+zA^2)=yA^2=0$, we deduce that $y=0$ and $zA^2=0$ implies $z=0$.

Title: Let $A$ be a $n \times n$ matrix such that $A^3 = O$, but $A^2 \ne O$. Show that $\{I,A,A^2\} \subset M_{n \times n}(\Bbb R)$ is linearly independent.

Body:
Let $A$ be a $n \times n$ matrix such that $A^3 = O$, but $A^2 \ne O$. Show that $\{I,A,A^2\} \subset M_{n\times n}(\Bbb R)$ is linearly independent.

Attempt:
Is the $O$ above denotes the zero matrix in $M_{n \times n}(\Bbb R)$?

Let $k_1,k_2,k_3 \in \Bbb R$ such that
\begin{equation*}
k_1I + k_2A + k_3A^2 = O \ldots (1)
\end{equation*}
By adding both sides of $(1)$ by $I$, we have
\begin{equation*}
(k_1+1)I + k_2A + k_3A^2 = I \ldots (2)
\end{equation*}
Now, multiplying both sides on the right by $A$, we have
\begin{equation*}
(k_1+1)A + k_2A^2 + k_3A^3 = A
\end{equation*}
or, equivalent, as
\begin{equation*}
(k_1+1)A + k_2A^2 = A.
\end{equation*}

I got stuck there. Any ideas? Thanks in advanced.


问:
Changing order of integration for $\int_{1}^{2}\int_{2-x}^{\sqrt{2x-x^2}}f(x,y)dydx$
I need to change integration order of

$\int_{1}^{2}\int_{2-x}^{\sqrt{2x-x^2}}f(x,y)dydx$

The region is bounded between $1\leq x\leq2$ and $0\leq y\leq1$

The upper limit, $y=\sqrt{2x-x^2}$ in terms of x is:

$x=1+\sqrt{1-y^2}$ or $x=1-\sqrt{1-y^2}$

How to determine which of them is the correct upper limit, by the graph, or by algebraic way?

答: It depends of your choice; indeed the outer integral limit has to be always fixed, and for varying the inner limits. If you want, you can integrate with $dy $ being the the outer, this is, $dx dy$ but you need to keep the outer integral from 0 to 1, and the inner function as a function of the outer one, since when you integrate the inner in respect to $x$ in this case, $x$ will vanish and you will be with just a function in terms of $y$.


问:
Topological linear space
Let $E$ be a topological linear space. A subset $M$ of $E$ is said to be bounded if $\forall U\in I(0)$ there exist a scalar $\lambda$ such that $M\subseteq \lambda U$, where $I(0)$ denotes the family of neighbourhoods of $0$. How to prove that each point of $E$ is bounded?

答:
Suppose $M = \{x\}$ for some $x \in E$.

Let $m: \Bbb F \times E \to E$ be the continuous scalar multiplication map, and it's well known that $m(0,x)=0$. So if $U \in I(0)$, we can find $V$ open in $\Bbb F$ such that $cx \subseteq U$ for all $c \in V$, by continuity.

Pick any $c \neq 0$ in this $V$ (I need that $\Bbb F$ is not discrete so that this is possible) and define $\lambda=\frac{1}{c}$. Then $M \subseteq \lambda U$ as $cx \in U$ implies $x = \lambda(cx) \in \lambda U$.

Title: Topological linear space


答: Let $x\in E$, set $M=\{x\}$. Let $(\lambda_n)\subset\mathbb{F}$ be any sequence of scalars such that $\lambda_n\to0$. Since $E$ is a TVS, we have that $\lambda_nx\to0$ in $E$, so for any $U\in I(0)$ there exists $n_{0,U}\in\mathbb{N}$ so that $\lambda_nx\in U$ for all $n\geq n_{0,U}$. In particular, $x\in \frac{1}{\lambda}U$, where $\lambda=\lambda_{n_{0,U}}$, i.e. $M\subset \frac{1}{\lambda}U$.

Title: Topological linear space

Body: Let $E$ be a topological linear space. A subset $M$ of $E$ is said to be bounded if $\forall U\in I(0)$ there exist a scalar $\lambda$ such that $M\subseteq \lambda U$, where $I(0)$ denotes the family of neighbourhoods of $0$. How to prove that each point of $E$ is bounded?


问:
When dealing with an improper antiderivative, why do you take the limit from the left?
The example I have is $$\left[\tan x+\frac{1}{\cos x}\right]_0^{\frac{\pi }{2}}$$

In my mind, you treat it as a limit as $x \rightarrow \frac{\pi}{2}$

However, apparently it its $$\lim _{x\to \left(\frac{\pi }{2}\right)^-}\left(\tan x+\frac{1}{\cos x}\right)$$ because "you approach $\frac{π}{2}$ from the left".

Why is that so? The source is here: https://www.youtube.com/watch?v=TJzAp_NSsbM&t=0s (timestamp 4:45)

答: Suppose that you want to compute$$\int_0^{\pi/2}\frac{1+\sin(t)}{\cos^2(t)}\,\mathrm dt.\tag1$$Note that this is an improper integral, since $\frac{1+\sin(t)}{\cos^2(t)}$ is undefined at $\frac\pi2$. So, $(1)$ actually means$$\lim_{x\to\pi/2^-}\int_0^x\frac{1+\sin(t)}{\cos^2(t)}\,\mathrm dt\tag2$$and therefore, since $\tan+\frac1\cos$ is a primitive of $\frac{1+\sin}{\cos^2}$, $(2)$ is equal to$$\lim_{x\to\pi/2^-}\left[\tan(t)+\frac1{\cos(t)}\right]_{t=0}^{t=x}.$$


答:
Because the integral
$$\int_{0}^{a} \frac{1}{1-\sin{x}}dx$$
is not defined on the interval $[0,a]$ with $a > \pi/2$, so you are considering it on the interval $[0,a]$, $a < \pi/2$, and letting $a \to \pi/2^{-}$.

Title: When dealing with an improper antiderivative, why do you take the limit from the left?

Body:
The example I have is $$\left[\tan x+\frac{1}{\cos x}\right]_0^{\frac{\pi }{2}}$$

In my mind, you treat it as a limit as $x \rightarrow \frac{\pi}{2}$

However, apparently it its $$\lim _{x\to \left(\frac{\pi }{2}\right)^-}\left(\tan x+\frac{1}{\cos x}\right)$$ because "you approach $\frac{π}{2}$ from the left".

Why is that so? The source is here: https://www.youtube.com/watch?v=TJzAp_NSsbM&t=0s (timestamp 4:45)


问:
diagonals of n-gon
Can anyone please walk me through how to solve problems like this?

1) how many diagonals does an n-gon have?

2) can a kindgom in which 77 roads lead out of each city, have exactly 10000 roads?

答: There are $n$ ways to select $n$ the first vertex. To make a diagonal, we have to choose another vertex out of of the possible $n-3$ vertices (All vertices excluding the first vertex and its $2$ neighbor). However $n(n-3)$ double counts the number of diagonals so the answer should be $\frac{n(n-3)}{2}$.

Title: diagonals of n-gon


问:
Find $y$-Lipschitz constant
$$f(x,y)=x^3e^{-xy^2}, 0\leq x\leq a, y\in \mathbb R, a>0$$

I need to find $K>0$ such that $$|f(x,y_1)-f(x, y_2)|\leq K|y_1-y_2|$$ for all $0\leq x\leq a$ and $y_1,y_2\in \mathbb R$

I did this $$|x^3e^{-xy_1^2}-x^3e^{-xy_2^2}|=x^3|e^{-xy_1^2}-e^{-xy_2^2}|\leq a^3|e^{-xy_1^2}-e^{-xy_2^2}|$$

now I don't know what to do.

答:
The Mean Value Theorem says that the $y$-Lipschitz constant would be
$$
\sup_{\substack{x\in[0,a]\\y\in\mathbb{R}}}\left|\frac{\partial}{\partial y}f(x,y)\right|
$$
So we compute
$$
\begin{align}
\frac{\partial}{\partial y}f(x,y)
&=-2x^4ye^{-xy^2}\\
&=-2x^{7/2}x^{1/2}ye^{-xy^2}\\
&=-2x^{7/2}ue^{-u^2}\\
\left|\frac{\partial}{\partial y}f(x,y)\right|&\le2a^{7/2}\frac1{\sqrt{2e}}
\end{align}
$$
where we find the maximum of $ue^{-u^2}$ by taking its derivative and solving
$$
(1-2u^2)e^{-u^2}=0
$$
to get $u=1/\sqrt2$. Thus, the maximum of $ue^{-u^2}$ is $1/\sqrt{2e}$.

Therefore,
$$
|f(x,y_1)-f(x, y_2)|\le a^{7/2}\sqrt{2/e}\,|y_1-y_2|
$$

Title: Find $y$-Lipschitz constant


问:
Is $\langle X,Y\rangle / m_{a} \langle X,Y\rangle$ a one-dimensional $K$ -vector space?
This is motivated by one of my previous posts, $ \mathfrak{m}_{a}=\left\langle X-a^{2}, Y-a^{3}\right\rangle $ is maximal ideal. In short,

Let $K$ be a field, and let $R$ be the ring $K[X, Y] /\left\langle X^{3}-Y^{2}\right\rangle$. For any element $a$ of $K \setminus \{0\},$ show that the ideal $\mathfrak{m}_{a}=\left\langle X-a^{2}, Y-a^{3}\right\rangle$ makes $I / m_{a} I$ into a one-dimensional $K$ -vector space, where $I$ is the ideal of $R$ generated by $X$ and $Y$.

To do this (I know this holds, but not the solution), I want to show that there is a unit of $R$ (i.e. element of $K$) inside $\left\langle X-a^{2}, Y-a^{3}\right\rangle$. I tried this using $X^3 - Y^2 = 0$. That is, $(X-a^2 + a^2)^3 - (Y-a^3 + a^3)^2 = 0$, and then I expanded this out. The result is a 'polynomial' in the variables $X-a^2, Y-a^3$. So I just ended up showing $0 \in \left\langle X-a^{2}, Y-a^{3}\right\rangle$, which is obvious anyway (but I can't help but think this step will be useful somewhere). I tried messing with this expression to induce a unit inside $\left\langle X-a^{2}, Y-a^{3}\right\rangle$, but I've had no luck. What are the possible next steps? Is what I did even necessary?

Edit: It seems as though trying to show that there is a unit inside $\left\langle X-a^{2}, Y-a^{3}\right\rangle$ is not a good approach (the claim is not true). Now I'm even more lost. Does anyone know how to proceed?

答:
As I pointed out in my comment, if we want to be precise, we should say that $I=(x,y)$ and $\mathfrak{m}_a=(x-a^2,y-a^3)$, where $x,y$ are the classes of $X,Y$ modulo $X^3-Y^2$, since we work with ideals of $R$.

However, this does not really matter here. Let $I'=(X,Y),$ $\mathfrak{m}'_a=(X-a^2,Y-a^3)$.  I claim that both ideals $I'$ and $\mathfrak{m}'_a $ contain $X^2-Y^3$. This is clear for $I'$. Moreover, note that $\mathfrak{m}'_a$ is the set of polynomials such that $P(a^2,a^3)=0$, so $X^3-Y^2\in\mathfrak{m}'_a$.

That said, we get that $I=I'/(X^3-Y^2),\mathfrak{m}_aI=\mathfrak{m}'_aI' /(X^3-Y^2)$ and we have a canonical isomorphism $I/\mathfrak{m}_aI\simeq I'/\mathfrak{m}'_aI'$.

Hence, we can forget about the classes modulo $X^2-Y$ of $R$ and work with ideals of $K[X,Y]$ instead.

So to simplify notation, I will get rid of the $'$  in the arguments above.

Since $\mathfrak{m}_a=(X-a^2,Y-a^3)$ is the set of polynomials such that $P(a^2,a^3)=0$, any polynomial $P\in K[X,Y]$ may be written as $\alpha+ M$, where $\alpha\in K$ and $M\in\mathfrak{m}_\alpha$ (more precisely, $\alpha=P(a^2,a^3),$ and $M=P-P(a^2,a^3)$).

Taking into account that an element of $I=(X,Y)I$ has the form $XP+YQ,P,Q\in K[X,Y]$, we see that an element of $I/\mathfrak{m_a}I$ may be represented by a polynomial of the form $\alpha X+\beta Y$, where $\alpha,\beta \in K$.

Now, $\mathfrak{m}_a I$ contains $(Y-a^3)X-(X-a^2)Y=a^2(Y-aX)$. Since $a\neq 0,\mathfrak{m}_a I$ contains $Y-aX$. Thus, any element of   $I/\mathfrak{m_a}I$ may be represented by an element of the form $\gamma X$, where $\gamma\in K$. This proves that this $K$-vector space is spanned by the class of $X$, and has dimension $\leq 1$. To prove that this vector space is not zero, we should prove that $X$ does not lie in $\mathfrak{m}_a I$. But $\mathfrak{m}_a I\subset \mathfrak{m}_a$. Since the value of $X$ at $(a^2,a^3)$ is $a^2\neq 0$, we are done.


问:
Find the x-coordinate of the stationary points of the curve and determine the nature of these stationary points.
The equation of a curve is $y=x^2e^{-x}$.

Find the x-coordinate of the stationary points of the curve and determine the nature of these stationary points.

Show that the equation of the normal to the curve at the point where $x=1$ is $e^2x+ey = 1+e^2$.

This is the full question I am having difficulty solving, I simply don't know where to begin. I moved back to my home country because of covid and now I am doing self-studying I don't know how to solve this any help wold be great and much appreciated.

答:
$y=x^2e^{-x} \implies y'(x)=2xe^{-x}-x^2e^{-x} \implies y'(1)=e^{-1}.$
So the slope of normal at $x=1$ is $m=-1/y'(1)=-e$. So the equation of line having slope $-e$ and passing through the point $(1,e^{-1})$ is
$$y-e^{-1}=-e(x-1) \implies ey+e^2 x=1+e^2$$


问:
Why this calculation true? (solution added)
Given the following density function $$
f_{(X, Y)}(x, y)=\frac{1}{10} e^{-0.5(y+3-x)}, \quad 5 \leq x \leq 10, \quad y \geq x-3
$$
I am asked to find $P(X<Y)$ but I dont understand the limits of $y$ in the solution:
$$
P(X<Y)=\int_{x=5}^{10} \int_{y=x}^{\infty} 0.1 e^{-0.5(y+3-x)} d y d x=\left.0.1 \int_{x=5}^{10}(-2) e^{-0.5(y+3-x)}\right|_{x} ^{\infty}
$$

why this is not $$
P(X<Y)=\int_{x=5}^{10} \int_{y=x-3}^{\infty} 0.1 e^{-0.5(y+3-x)} d y d x=...$$

答:
why this is not...

Because the request is exactly $P(Y>X)$, that is the integral over the domain upper the line $X=Y$

see the picture below


问:
$\operatorname{tr}(BA)$ where $A$ is low rank and has a subset of eigenvalues from $B^{-1}$
Most of the question is in the title.
Suppose that I have $B$ a square $N\times N$ positive definite matrix, and $A$ a low-rank square matrix with its non-zero eigenvalues being a subset of the spectrum of $B^{-1}$. Is there a simple way to compute the trace : $\mathrm{tr}(BA)$?

My intuition is that if $M$ is the rank of $A$, then $B$ and $A$ can share all $N$ eigenvectors and that one can diagonalize $A$ and $B$ and get automatically $\mathrm{tr}(BA) = \sum_{i:\lambda_i>0} \frac{\lambda_i}{\lambda_i} = M$ where $\lambda_i$ are the eigenvalues of $A$. But I am not sure how to obtain the proper ordering etc.

Thanks for your help!

答:
Ok here is a more concrete attempt.
So if the eigenvalues of $B^{-1}$ are $\Lambda_B = \{\lambda_i\}_{i=1}^N$, I know that the eigenvalues $\Lambda_A$ of $A$ are a subset of $\Lambda_B$ and the rest are $0$.
So I can rewrite $B=W D_B^{-1}W^\top$ where $D_B=\text{diag}(\Lambda_A, \Lambda_{B}/\Lambda_{A})$ and $W$ are the corresponding eigenvectors.
Similarly I can rewrite $A$ as $A=WD_AW^\top$ where $D_A=\text{diag}(\Lambda_A, 0)$

Which means I can rewrite
\begin{align}
\text{tr}(BA) =& \text{tr}(WD_B^{-1}W^\top WD_AW^\top) = \text{tr}(W^\top W D_B^{-1}D_A)\\
=& \text{tr}(D_B^{-1}D_A) = \sum_{i=i}^M \frac{\lambda_i}{\lambda_i} = M.
\end{align}

Is there anything wrong with my reasonning?


问:
Characterization of faithfully flat homomorphisms
Let $A \to B$ be a homomorphism of commutative rings. Why are the following conditions equivalent?

$A \to B$ is faithfully flat.

$A \to B$ is injective, flat and $B/A$ is a flat $A$-module.

This should be elementary, but at the moment I don't see how to prove it. I know the usual characterizations of faithfully flat homomorphisms (which can be found in Atiyah-Macdonald for example).

答: To go the other way, suppose $f\colon A\to B$ is faithfully flat. The map $f\otimes B\colon A\otimes_AB \to B\otimes_AB$ is injective since it has a section, namely the multiplication map from $B\otimes_AB\to B$ given by $b\otimes b'\mapsto bb'$. Which means it is injective. The original map must therefore be injective, since it is so after a faithfully flat base change.


问:
regular expression
I would like to write the regular expression for the set of
              all binary strings where
              there are no three consecutive 0's.

The following strings are part of the language:

a) strings of the form $1,11,111,...$

b) strings $\{ 0, 01,011,0111,...\}$

d) strings $\{00, 001,0011,00111,..\}$

So,the regular expression is:
$$ (1+01+001)^*$$

Is this a correct regular expression for the given language?

答:
A more precise expression would be

0+00+(1+01+001)*

so you will be able to accept 0 or 00...

Title: regular expression


问:
Why doesn't minimizing the Lagrangian derivative with regards to the multiplier work?
Given a typical optimization problem for a scalar function of a vector $x$ with a set of equality constraints
$$
\min_x f(x)\\
\text{s.t.}\quad g(x)=\mathbf{0}
$$
To solve the above, one would define a Lagrangian function with a vector of multipliers for each of the constraints as follows
$$
\mathcal{L}(x,\lambda) = f(x)+g(x)^T\lambda
$$
Using subscripts to denote differentiation, the solution $(x^*,\lambda^*)$ is found when
$$
\mathcal{L}_x(x^*,\lambda^*)=0
$$
When solving using numerical method, one would use the Hessian to solve the following systems of equations to calculate a step
$$
\left(\begin{matrix}\mathcal{L}_{xx} & g_x^T \\ g_x & 0\end{matrix}\right)
\left(\begin{matrix}s_x \\ s_\lambda \end{matrix}\right) =
\left(\begin{matrix}-\mathcal{L}_x \\ -g \end{matrix}\right)
$$
such that the system will proceed to a new point $(x+s_x,\lambda+s_\lambda)$. In my experience with solving this linear system, setting the initial $\lambda$ to be large is usually enough to make it the solver run smoothly.

Now, since the solution requires $\mathcal{L}_x=0$, in order to accelerate the process I am tempted to try solve the following problem
$$
\min_\lambda \frac{1}{2}|\mathcal{L}_x|^2
$$
for which the minimizing $\lambda$ is found at
$$
\lambda = -(g_xg_x^T)^{-1}g_xf_x
$$
So this $\lambda$ would be calculated first, and after that, $\mathcal{L}_{x}$ and $\mathcal{L}_{xx}$. But when I tried this trick, the solver failed miserably. Why is that so?

答:
If there are fewer constraints than variables, then $g: \mathbb{R}^K \rightarrow \mathbb{R}^J$ with $J<K$. If the constraints are not redundant then $rk(g_x(x))=J$.

The optimal values satisfy the first order conditions
$$ f_x(x)+g_x(x)^T\lambda=0,$$
a system of $K$ equations with $K+J$ unknown $(x,\lambda)$.

After inserting your expression for $\lambda$ into the first order conditions, you obtain the reduced system of $K$ equations and $K$ unknown:
\begin{equation} \tag{1}
(I_K-g_x^T(g_xg_x^T)^{-1}g_x)f_x(x)=0.
\end{equation}

This system has an infinite number of solutions. Indeed, the matrix $P:=g_x^T(g_xg_x^T)^{-1}g_x$ is an orthogonal projector onto $col(g^T_x)$, and $rk(P)=J<K$. The matrix $I_K-g_x^T(g_xg_x^T)^{-1}g_x$ is an orthogonal projector onto $col^\perp (g^T_x)$ and the rank of this matrix is $K-J$.

If you consider only the system (1), your solution is undetermined, but if you exploit the information comprised in the $J$ constraints $g(x)=0$ and include them to the system (1) of $K-J$ independent equations, this would be helpful to reduce the indeterminatness of the solution, and to find possibly a finite number of solutions.


问:
Find the probability $\mathbb{P}(U\leq z|U+M=y)$ given $U,M\sim\mathcal{N}(0,1)$
I came here to verify whether my solution is correct or not. I'm not sure if what I'm doing is valid:

Let $U,M$ such that $U,M\sim\mathcal{N}(0,1)$. Assume $U,M$ are statistically independent. Define:

$$\Phi(x)\triangleq\int_{-\infty}^{x}f(t)\ \text{d}t\qquad f(t)\triangleq \frac{1}{\sqrt{2\pi}}e^{-t^2/2}$$

Find the probability $p\triangleq\mathbb{P}(U\leq z|U+M=y)$.

My Attempt:

$$p\underset{(\star)}=\int_{-\infty}^{\infty}\mathbb{P}(U\leq z|U+M=y,M=m)f(m)\ \text{d}m\underset{(\star\star)}=\int_{-\infty}^{\infty}\mathbb{P}(U\leq z|U=y-m)f(m)\ \text{d}m = \\= \int_{y-z}^{\infty}f(m)\ \text{d}m=1-\Phi(y-z)$$

What I'm most unsure about is $(\star)$. $(\star\star)$ is justified (I hope) since $U,M$ are statistically independent.

Thanks!

答:
You can write the solution more simply as :

\begin{align}
P(U\leq z | U+M =y )=&P(y-M\leq z)= P(M\geq y-z)=1-P(M\leq y-z),
\end{align}

and using the CDF for $M$ you have the result.


问:
Left inverse matrix for two different matrices?
Consider two matrices $\mathbf{D}_1$ and $\mathbf{D}_2$ in $\mathbb{R}^{m\times n}$ $(m>n)$ of full column rank, is there a matrix $\mathbf{D}^{-}$ in $\mathbb{R}^{n\times m}$ satisfying $\mathbf{D}^{-}\mathbf{D}_1=\mathbf{I}_n$ and $\mathbf{D}^{-}(\mathbf{D}_1+\mathbf{D}_2)=\mathbf{I}_n$?  Note that $(\mathbf{D}_1+\mathbf{D}_2)$ is a matrix of full column rank. Many thanks! and if so, any way to prove it?

答:
Note that if $\mathbf{D}^-$ were to exist, we should have

$$
\mathbf I_n = \mathbf{D}^-(\mathbf{D}_1+\mathbf{D}_2) = \mathbf{D}^-\mathbf{D}_1+\mathbf{D}^-\mathbf{D}_2 = \mathbf I_n + \mathbf{D}^-\mathbf{D}_2,
$$

hence

$$
\mathbf{D}^-\mathbf{D}_2 = 0.
$$

But since $\mathbf{D}_2$ has full rank, this implies $\mathbf{D}^- = 0$, a contradiction.


问:
Discrete Math - Calculating $\sum_{k}^{n}$
I have a question regarding on how to calculate expressions like (I need to write it without $\sum$ and without $k$):

$$\sum_{k=2}^{25}\binom{25}{k}\binom{k}{2}$$

And also like:

$$\sum_{k=1}^{10}k\binom{10}{k}\binom{20}{10-k}$$

Unfortunately I didn't learn how to do it, but this is a part of the material for my exam today.

Thank you so much for your help !!

Appreciate that!

Body:
$$S=\sum_{k=0}^{n}\binom{n}{k}\binom{k}{2}$$
$$S=\frac{1}{2}\sum_{k=0}^{n} (k^2-k)\frac{n!}{k! (n-k)!}$$
$$S=\frac{1}{2}\sum_{k=0}^{n} \frac{n(n-1)(n-2)!}{(k-2)!(n-2-(k-2))!} $$
$$S=\frac{n(n-1}{2}\sum_{k=0}^{n} {n-2 \choose k-2}$$
Let $k-2-j$, then
$$S=\frac{n(n-1}{2} \sum_{j=-2}^{n-2} {n-2 \choose j}=\frac{n(n-1}{2} \sum_{j=0}^{n-2} {n-2 \choose j}=n(n-1)2^{n-3}$$
So requried $$\sum_{k=2}^{25}\binom{25}{k}\binom{k}{2}=\sum_{k=0}^{25}\binom{25}{k}\binom{k}{2} =25.24.2^{22}$$


问:
How to differentiate $ \intop_{0}^{t}\intop_{0}^{t}e^{\frac{-tx}{y^{2}}}dxdy $
Let

$ F\left(t\right)=\intop_{0}^{t}\intop_{0}^{t}e^{\frac{-tx}{y^{2}}}dxdy $

Prove that $ F'\left(t\right)t=2F\left(t\right) $.

Im pretty sure that I should not try to calculate this integral directly. Im not sure how to differentiate it by $ t $ correctly. I'd appreaciate some help.

Thanks in advance.

答:
Let $x=tX$, $y=tY$ and you get
$$
F(t)=\int_0^1\int_0^1 t^2 e^{-X/Y^2}\,\mathrm{d}X\,\mathrm{d}Y=t^2\cdot\text{constant}
$$
so the result follows (as long as that constant is finite, which it is).


问:
When are Categorical Properties of Maps Preserved Under An Adjoint Isomorphism?
Suppose that we a functor $F : \mathcal{A} \to \mathcal{B}$. Then we say that a functor $G : \mathcal{B} \to \mathcal{A}$ is right-adjoint to $F$ if $\operatorname{Hom}_{\mathcal{B}}(F(A),B) \cong \operatorname{Hom}_{\mathcal{A}}(A,G(B))$ for all objects $A \in \mathcal{A}$, $B \in \mathcal{B}$.

I am interested in whether the "adjoint isomorphism" $\operatorname{Hom}_{\mathcal{B}}(F(A),B) \cong \operatorname{Hom}_{\mathcal{A}}(A,G(B))$ necessarily preserves any categorical properties of the maps in either Hom set.

That is to say, if we explicitly label the adjoint isomorphism

$$
\vartheta : \operatorname{Hom}_{\mathcal{B}}(F(A),B) \overset{\cong}{\to}\operatorname{Hom}_{\mathcal{A}}(A,G(B)),
$$

then can we say if, or when it is the case, that a map $(F(A) \overset{T}{\to} B) \in \operatorname{Hom}_{\mathcal{B}}(F(A),B)$ has categorical property X if and only if $(A \overset{\vartheta(T)}{\to} G(B)) \in \operatorname{Hom}_{\mathcal{A}}(A,G(B))$ has the same categorical property X?

By this question, we know that the adjoint isomorphism does not neccesarily preserve epimorphisms, so we can exclude this from being a candidate property X. However are there any other properties that are conserved?

Failing this, can we impose extra conditions (exactness?) on $F$ and $G$ that would allow certain properties to be conserved?

For context, I am typically concerned with functors between module categories, and in particular induction / restriction (or similar adjoint pairs).

答:
(Note that in the definition, there should be a "naturally in $A,B$")

The answer is that usually they don't, and the context of induction/restriction adjunctions is the perfect place to put all of these to the test and see that they mostly fail.

For instance say you have a ring map $A\to B$ and are looking at extension/restriction of scalars; I will let $U$ denote the forgetful functor.

Then

$f$ can be an isomorphism without its adjoint being one.

Example : take $A\to UB$ : this is clearly not an isomorphism in general (it is one if and only if $A\to B$ is one); however its adjoint is $B\cong B\otimes_A A\to B$ which is an isomorphism.

Dually, if you take $UB\to UB$, this is an isomorphism of $A$-modules, but its adjoint $B\otimes_A B\to B$ is not necessarily one (it will be if $A\to B$ is a localization or a quotient for instance, but not for general ring maps)

As you can see, there's no reasonable general condition you can impose on $A\to B$ for this to hold, and "$f$ is an isomorphism" is one of the most basic categorical property.

As pointed out in the comments, in general (if $F$ is the left adjoint) the adjoint of $f: FM\to N$ is given by the composite $M\to UFM\to UN$, and so it factors through $\eta_M$. So the properties you are interested in will have to be behaved well with respect to $\eta_M$, which does not depend on $f$ and its categorical properties, which explains why many of these things fail : $\eta$ typically does not have these properties (epimorphism, monomorphism, isomorphism, etc.)

(and dually for $\epsilon$)

So you'll want to look for properties that $\eta$ has,that are preserved by $G$ and stable under composition (or dually, that $\epsilon$ has, that are preserved by $F$ and stable under composition), and these are rare without stringent restrictions on your adjunction (e.g. one of the functors is full/faithful/fully faithful which is somewhat rare)

This isn't a full answer because I can't say "there are none", but ultimately it depends on what restrictions you're ready to impose on your functors. If you want to stay reasonably general, there are basically none that will hold.

There is an important counterexample to what I just said, though ! Suppose $\mathcal{A,B}$ are pointed, that is, they have a zero object. Then it is preserved by the left and the right adjoint, and so we get that a morphism $f$ is $0$ (i.e. factors through the zero object) if and only if its adjoint does.

In the case of additive categories, this can also be seen by the fact that $\hom(F(A),B)\cong \hom(A,G(B))$ is automatically an isomorphism of abelian groups; and so preserves the $0$ morphism, but it is true in the more general pointed case.


问:
Finding an explicit isomorphism between $\mathbb{Z}^{\times}_n$ and $\mathbb{Z}^{\times}_{2n}$
For an odd integer $n$, find an explicit isomorphism between $\mathbb{Z}^{\times}_n$ and $\mathbb{Z}^{\times}_{2n}$.

How do I do this? I don't really know where to start. I can easily find bijections but am yet to find a structure preserving one.

答:
The following solution is concrete in the extreme.  Look at the particular example $n=15$.  Then the elements of $\mathbb{Z}^\times_n$ can be thought of as the numbers $1$, $2$, $4$, $7$, $8$, $11$, $13$, and $14$, under multiplication modulo $15$.  Similarly, the elements of

 $\mathbb{Z}^\times_{2n}$ can be thought of as the numbers $1$, $7$, $11$, $13$, $17$, $19$, $23$, $29$ under multiplication modulo $30$.  Note that these numbers are all odd.

What should our attempted isomorphism $\phi$ do to the elements of $\mathbb{Z}^\times_n$?  The idea is to send $k$ to something that is as "like" $k$ as possible. Obviously $\phi$ should send $1$ to $1$.  What should $\phi(2)$ be?  The only element of
 $\mathbb{Z}^\times_{2n}$ that is congruent to $2$ modulo $n$ is $2+15$, so it is reasonable to send $2$ to $17$.

In general, suppose that $1 \le k \le 14$, and $k$ is relatively prime to $15$. Then if $k$ is odd, we let $\phi(k)=k$.  And if $k$ is even, we let $\phi(k)=k+15$.

It is easy to see that $\phi$ is a bijection. It remains  to show that $\phi(ij)=\phi(i)\phi(j)$.  This divides naturally into $3$ cases: (i) $i$, $j$ both odd; (ii) $i$, $j$ both even; and (iii) one is odd and the other is even.

Out of impatience, we do them all together.  Note that by the definition of $\phi$, we have $\phi(ij) \equiv \phi(i)\phi(j) \pmod{n}$.  But also, again from the definition, $\phi(ij) \equiv \phi(i)\phi(j) \pmod{2}$, since on each side we have an odd number. It follows that $\phi(ij) \equiv \phi(i)\phi(j)\pmod{2n}$, which is exactly what we want.

The idea generalizes immediately to any odd $n$.  I do not know what notation you use for the elements of $\mathbb{Z}^\times_n$.  I will assume that you use the notation $i/(n)$ for the equivalence class of $i$.

Look at the equivalence class $i/(n)$, where $1 \le i \le n-1$.  If $i$ is odd, let $\phi(i/(n))=i/(2n)$.  If $i$ is even, let $i'=i+n$, and let $\phi(i/(n))=i'/(2n)$.  It is straightforward to show that $\phi$ is an isomorphism from $\mathbb{Z}^\times_n$ to $\mathbb{Z}^\times_{2n}$.

Now that you have a concrete picture of what is happening, you can begin to understand more abstract solutions, and appreciate their elegance.


问:
Invariant sets in dynamical systems: a characterization.
I'm currently being introduced to Dynamical Systems in class. Given a set $X$, and maps $S_t \colon X \to X$ for $t \geq 0$, we call $(X,\{S_t\}_{t \geq 0})$ a dynamical (or semi-flow) system if
$$\begin{align*}
(i) & \quad  S_0 = Id_X, \\
(ii) & \quad S_t \circ S_s = S_{t+s}, \quad  \text{for all } t,s \geq 0,
\end{align*}$$
where $\circ$ denotes composition of functions.

I've been asked to prove the equivalence between the following facts, where $D \subset X$ is an arbitraty set:
$$\begin{align*}
(a) & \quad D \text{ is invariant, that is, } S(t)(D) = D \text{ for all } t \geq 0, \\
(b) & \quad D = \bigcup\{u(t) \, : \, t \in \mathbb{R}, \, u \colon \mathbb{R} \to D \text{ solution}\},
\end{align*}$$
where $u \colon \mathbb{R} \to D$ is said to be a solution iff
$$S(t)(u(\tau)) = u(t+\tau), \quad \text{for all } t \geq 0, \, \tau \in \mathbb{R}.$$

I've succesfully proven $(b) \implies (a)$. However, I'm struggling with $(a) \implies (b)$. Note that it is always true that
$$D \supset \bigcup\{u(t) \, : \, t \in \mathbb{R}, \, u \colon \mathbb{R} \to D \text{ solution}\},$$
so Im trying to prove the converse relation by supossing $(a)$.

I know that, in the case of ODEs, the previous equivalence ($S$ can be an evaluation map) is proven by picking a solution of the ODE whose inital condition is an arbitraty point in $D$. Can we do anything similar now? Note that the existence of solution with certain initial conditions isn't obvious for abstract dynamical systems.

答: Finally found the answer, as it is proven in Lemma 21.2 of Sell & You. They prove that, in invariant sets, there exist solutions with specified initial conditions on the set.</p>