问:
Is it possible to have $\xi$ and $\chi$ independent?
Problem:

Random variables $\xi$ and $\eta$ are independent and equally
Bernoulli-distributed. Let $\chi$ be the indicator that their sum is an
odd number. Is it possible to have $\xi$ and $\chi$ independent?

Solution:

Let $\xi,\eta\sim\text{Bern}(p)$ for some $p\in (0,1)$. The sum $\xi+\eta$ then has $\text{Bin}(2,p)$ distribution and thus $\chi\sim\text{Bern}(2p(1-p))$. If $\xi=0$ and $\xi+\eta$ is odd, then $\eta=1$ so that $\{\xi=0,\chi=0\}$ happens with probability $(1-p)^2$. Similar reasoning gives $\Pr(\xi=1,\chi=0)=p^2,$ and $\Pr(\xi=0,\chi=1)=\Pr(\xi=1,\chi=1)=p(1-p)$.

For independence we must have in particular that $\Pr(\xi=1,\chi=1)=\Pr(\xi=1)\Pr(\chi = 1)$, i.e. $p(1-p)=p\cdot 2p(1-p)$ which is uniquely solved by $p=1/2$. Doing this for the rest of the combinations of outcomes gives the same result.

Question:

I do understand most of the solution except when it's concluded that $\chi\sim\text{Bern}(2p(1-p))$. I don't think I'm sure how the indicator is supposed to tie into this. Does it mean that $\chi=1$ if $\xi+\eta=\text{odd}$ and $\chi=0$ else? If someone could inject some intuition to this I'd be thankful.

答:
$$\begin{array}{c|c|c|c|c}
\xi & \eta & \xi + \eta & \chi & \text{Probability}\\
\hline
0 & 0 & 0 & 0 & (1-p)^2 \\
0 & 1 & 1 & 1 & p(1-p) \\
1 & 0 & 1 & 1 & p(1-p) \\
1 & 1 & 2 & 0 & p^2 \\
\end{array}$$
Therefore, if $\xi$ and $\chi$ are independent, we would have in particular $$\Pr[\chi = 1 \cap \xi = 1] = \Pr[\chi = 1]\Pr[\xi = 1].$$  Is this true?


答:
We know that $S = \zeta + \eta \sim \text{Bin}(2, p)$.

Now
$$\mathbb{P}(S \text{ is odd}) = \mathbb{P}(S = 1) =  \binom{2}{1}p(1-p) = 2p(1-p)\text{.}$$
Thus, $$\chi = \mathbf{I}(S \text{ is odd}) = \begin{cases}
1, & S\text{ is odd} \\
0, & \text{otherewise}
\end{cases}$$  follows a Bernoulli distribution with parameter $\mathbb{P}(\chi = 1) = \mathbb{P}(S \text{ is odd}) = 2p(1-p)$.

ContentLicense: CC BY-SA 4.0

Body:
Problem:

Random variables $\xi$ and $\eta$ are independent and equally
Bernoulli-distributed. Let $\chi$ be the indicator that their sum is an
odd number. Is it possible to have $\xi$ and $\chi$ independent?

Solution:

Let $\xi,\eta\sim\text{Bern}(p)$ for some $p\in (0,1)$. The sum $\xi+\eta$ then has $\text{Bin}(2,p)$ distribution and thus $\chi\sim\text{Bern}(2p(1-p))$. If $\xi=0$ and $\xi+\eta$ is odd, then $\eta=1$ so that $\{\xi=0,\chi=0\}$ happens with probability $(1-p)^2$. Similar reasoning gives $\Pr(\xi=1,\chi=0)=p^2,$ and $\Pr(\xi=0,\chi=1)=\Pr(\xi=1,\chi=1)=p(1-p)$.

For independence we must have in particular that $\Pr(\xi=1,\chi=1)=\Pr(\xi=1)\Pr(\chi = 1)$, i.e. $p(1-p)=p\cdot 2p(1-p)$ which is uniquely solved by $p=1/2$. Doing this for the rest of the combinations of outcomes gives the same result.

Question:

I do understand most of the solution except when it's concluded that $\chi\sim\text{Bern}(2p(1-p))$. I don't think I'm sure how the indicator is supposed to tie into this. Does it mean that $\chi=1$ if $\xi+\eta=\text{odd}$ and $\chi=0$ else? If someone could inject some intuition to this I'd be thankful.


问:
All Accumulation points of a set
Let $M = \{\frac{k}{2^m} \mid k,m \in \mathbb N, k\le m \} $.

I showed that zero is an accumulation point of this set by looking at the sequence $(x_n)=\frac{n}{2^n} \in M$(without zero) which obviously converges towards zero.
But how can I show that this is the only accumulation point?

Can someone help? :)

答:
We have $M\subseteq \left(0,\frac{1}{2}\right]$, so $\overline{M}\subseteq \left[0,\frac{1}{2}\right]$.

Let $0<a\leq \frac{1}{2}$ arbitrary, fixed. We find $m_a\in\mathbb{N}$, so that
\begin{align*}
0<\frac{m_a}{2^{m_{a}}}<a\tag{1}
\end{align*}
We see in (1) that $\frac{m_a}{2^{m_{a}}}\in M$ and only a finite number of elements in $M$ is between $\frac{m_a}{2^{m_{a}}}$ and $a$, namely $$\left|\left\{\left.\frac{k}{2^m}\right|1\leq m<m_a,1\leq k\leq m\right\}\right|=\sum_{m=1}^{m_a-1}m=\frac{1}{2}\left(m_a-1\right)m_a.$$ Thus we can find an $\varepsilon>0$ small enough so that $(a-\varepsilon, a+\varepsilon)\cap M=\emptyset$ and $a$ is not a limit point.

Title: All Accumulation points of a set


问:
Proving: $\lim _{x \rightarrow \infty} \ f(x)=\lim_{t \rightarrow 0^+} \ f(1/t)$
Please advise: is my proof sufficient? or do I need to use $\epsilon$?

Prove that $$
\lim_{x \to \infty} \ f(x)=
\lim_{t \to 0^+} \ f(1/t)$$ if these limits exist

Proof: putting $t=\dfrac{1}{x}$ then as $x\to\infty, \ t\to0^+$ and $x=\dfrac{1}{t}$

Then $\lim_{x \to \infty} \ f(x)=\lim_{t \to 0^+} \ f(1/t)$

答:
This is just exercise 75 in Section 2.7, James Stewart, Calculus Early Transcendentals 7th ed. 2011. Page 143.

Note that this result is true for $x \rightarrow -\infty$ and $t \rightarrow 0^-$ also.

ContentLicense: CC BY-SA 3.0


问:
Secret santa problem
We decided to do secret Santa in our office. And this brought up a whole heap of problems that nobody could think of solutions for - bear with me here.. this is an important problem.

We have 4 people in our office - each with a partner that will be at our Christmas meal.

Steve,
Christine,
Mark,
Mary,
Ken,
Ann,
Paul(me),
Vicki

Desired outcome

Nobody can know who is buying a present for anybody else. But we each
  want to know who we are buying our present for before going to the
  Christmas party. And we don't want to be buying presents for our partners.
  Partners are not in the office.

Obvious solution is to put all the names in the hat - go around the office and draw two cards.

And yes - sure enough I drew myself and Mark drew his partner. (we swapped)

With that information I could work out that Steve had a 1/3 chance of having Vicki(he didn't have himself or Christine - nor the two cards I had acquired Ann or Mary) and I knew that Mark was buying my present. Unacceptable result.

Ken asked the question: "What are the chances that we will pick ourselves or our partner?"

So I had a stab at working that out.

First card drawn -> 2/8
Second card drawn -> 12/56

Adding them together makes 28/56 i.e. 1/2.

i.e. This method won't ever work... half chances of drawing somebody you know means we'll be drawing all year before we get a solution that works.

My first thought was that we attach two cards to our backs... put on blindfolds and stumble around in the dark grabbing the first cards we came across... However this is a little unpractical and I'm pretty certain we'd end up knowing who grabbed what anyway.

Does anybody have a solution for distributing cards that results in our desired outcome?

I'd prefer a solution without a third party..

答:
Edit: I just realized this method does not produce all possible derangements. Each pairing is random, but not as independent as it could be, because this method forces the derangement to be a single cycle.

Dr. Hannah Fry describes in her Book "The Indisputable Existence of Santa Claus" the following method:

You make as many cards as there are people. Each card has the same number twice written on it and each card has a different number. You then turn the cards around, shuffle and arrange them in a line. Then you cut them in half so that you now have two lines of cards with equal numbers each facing each other. Then, still hidden, you shift one of the lines by one. Each person can now pick one pair of cards each from one of the lines. The card from the first line tells them which number they are and the card from the second line tells them whom they are supposed to give a gift to. The only thing that remains is to tell everyone what number they are, best by filling in a list with a mapping from number to name.

For example:

```<code>1.

|1| |2| |3| |4| |5| |6|
|1| |2| |3| |4| |5| |6|

2.

|3| |6| |4| |2| |5| |1|
|3| |6| |4| |2| |5| |1|

3.
|3| |6| |4| |2| |5| |1|
|1| |3| |6| |4| |2| |5|

<span class="math-container">```
```

Title: Secret santa problem

Body:
Edit: I just realized this method does not produce all possible derangements. Each pairing is random, but not as independent as it could be, because this method forces the derangement to be a single cycle.

Dr. Hannah Fry describes in her Book "The Indisputable Existence of Santa Claus" the following method:

You make as many cards as there are people. Each card has the same number twice written on it and each card has a different number. You then turn the cards around, shuffle and arrange them in a line. Then you cut them in half so that you now have two lines of cards with equal numbers each facing each other. Then, still hidden, you shift one of the lines by one. Each person can now pick one pair of cards each from one of the lines. The card from the first line tells them which number they are and the card from the second line tells them whom they are supposed to give a gift to. The only thing that remains is to tell everyone what number they are, best by filling in a list with a mapping from number to name.

For example:

1.

|1| |2| |3| |4| |5| |6|
|1| |2| |3| |4| |5| |6|

2.

|3| |6| |4| |2| |5| |1|
|3| |6| |4| |2| |5| |1|

3.
|3| |6| |4| |2| |5| |1|
|1| |3| |6| |4| |2| |5|

```


问:
Why does rewriting an equation change the limits?
Why does rewriting

$$f(x) = \frac{x^3 - 1}{x -1}$$

into

$$f(x) = x^2 + x + 1$$

change the answer? In the first equation, it's obvious that $x \neq 1$, but when we solve it a little, the $x \neq 1$ completely disappears.

Wouldn't that mean if you rewrite all equations that have multiplication to include variables under the divisor, then wouldn't they all have indeterminants?

答:
It does not change. As pointed out in the other answers, the domain of the first function is $\mathbb{R}\setminus \{1\}$ and  in this domain we do have $\frac{x^3-1}{x-1}=x^2+x+1.$ The difference is that the second one is defined at $x=1$, while the former is not. A simpler case where it also happens is $\frac{x}{x}=1, x\neq 0$.

More generally, we can do this whenever we have a rational function $\frac{f(x)}{g(x)}$, with $g(a)=0$, $g$ is not identically zero, $f(a)=0$ and the multiplicity of $a$ as a zero of $f$ is not less than its multiplicity as a zero of $g$.


问:
Find time to collision in a pursuit

Problem 1.13 from Irodov's Problems in General Physics:

Point A moves uniformly with velocity $v$ so that the vector $v$ is continually "aimed" at point B which in its turn moves rectilinearly and uniformly with velocity $u<v$. Initially, $v⊥u$, and the points are separated by a distance $l$. The time at which the points converge is given as?

If we write $\tau$ as the time to collision, then we can immediately find
$$\int_0^\tau v\cos\theta\,dt=u\tau\Leftrightarrow\int_0^\tau\cos\theta\,dt=\frac{u\tau}{v}\\\int_0^\tau v\sin\theta\,dt=l\Leftrightarrow\int_0^\tau\sin\theta\,dt=\frac{l}{v}$$
Irodov asserts that
$$\int_0^\tau(v-u\cos\theta)dt=l$$
It turns out that if we consider $\vec w=\vec v-\vec u=(v\cos\theta-u,v\sin\theta)$, then $v-u\cos\theta$ is $v\cos\theta-u$ in the basis $(\hat v,\hat v_\perp)$, where $\hat v=(\cos\theta, \sin\theta)$ and $\hat v_\perp=(\sin\theta,-\cos\theta)$ in $(\hat\imath,\hat\jmath)$.

$\vec w=(v-u\cos\theta,-u\sin\theta)$ in $(\hat v,\hat v_\perp)$.

Consider integrating $\vec w$ (I use the integrals of the sine and cosine above in the process)
$$\int_0^\tau\vec w\,dt=\left(\int_0^\tau(v\cos\theta-u)\hat\imath dt\right)+\left(\int_0^\tau(v\sin\theta)\hat\jmath dt\right)=l\hat\jmath$$
Now consider the same but with the coordinates in $(\hat v,\hat v_\perp)$
$$\int_0^\tau\vec w\,dt=\left(\int_0^\tau(v-u\cos\theta)\hat v dt\right)+\left(\int_0^\tau(-u\sin\theta)\hat v_\perp dt\right)=\left(\frac{v^2-u^2}{v}\tau\right)\hat v-\left(\frac{ul}{v}\right)\hat v_\perp$$
If we equate both integrals and take the norm squared, we get
$$\left(\frac{v^2-u^2}{v}\tau\right)^2+\left(\frac{ul}{v}\right)^2=l^2\Leftrightarrow\tau=\frac{l}{\sqrt{v^2-u^2}}$$
But this is wrong, the actual value is
$$\tau=\frac{vl}{v^2-u^2}$$
Can anyone spot the error? Thank you!

答: Let $A(t),B(t)$ denote the positions of the bodies at time $t\geq 0$. Take $B(0)=(0,0)$ and $A(0)=(0,-l)$. The information provided in the problem tells us $\frac{dB}{dt}=\Big$ and $$\frac{dA}{dt}=\frac{v\big(B(t)-A(t)\big)}{||B(t)-A(t)||}$$ If we take $\vec{p}(t)=B(t)-A(t)$ we can say $$\frac{d\vec{p}}{dt}=\Big-v\frac{\vec{p}(t)}{||\vec{p}(t)||}$$ We need to find $\tau$ so that $\vec{p}(\tau)=\vec{0}$ Writing $\vec{p}(t)=\big(x(t),y(t)\big)$ gives us the system $$\frac{dx}{dt}=u-\frac{vx}{\sqrt{x^2+y^2}}$$ $$\frac{dy}{dt}=-\frac{vy}{\sqrt{x^2+y^2}}$$ Converting the polar yields $$\frac{dr}{dt}=u\cos(\theta)-v$$ $$\frac{d\theta}{dt}=-\frac{u\sin(\theta)}{r}$$ Here $(r,\theta)=(l,\pi/2)$ when $t=0$. Dividing these provides us with a separable DE in $r$ and $\theta$: $$\frac{dr}{d\theta}=r\cdot \Bigg(\frac{v-u\cos(\theta)}{u\sin(\theta)}\Bigg)$$ This has solution $$r=\frac{l\big(\sin(\theta)\big)^{\frac{v}{u}-1}}{\big(\cos(\theta)+1\big)^{\frac{v}{u}}}$$ Where $\theta \in [0,\pi/2]$ and $r=||\vec{p}(t)||=0$ when $\theta=0$. Plugging into our formula for $\frac{d\theta}{dt}$ gives us $$\Bigg(\frac{\big(\sin(\theta)\big)^{\frac{v}{u}-2}}{\big(\cos(\theta)+1\big)^{\frac{v}{u}}}\Bigg)d\theta=-\frac{u}{l}dt$$ This means $$\int_{\pi/2}^0\Bigg(\frac{\big(\sin(\theta)\big)^{\frac{v}{u}-2}}{\big(\cos(\theta)+1\big)^{\frac{v}{u}}}\Bigg)d\theta=\int_{0}^{\tau}-\frac{u}{l}dt$$ Solving for $\tau$, $$\tau=\frac{l}{u}\int_{0}^{\pi/2}\Bigg(\frac{\big(\sin(\theta)\big)^{\frac{v}{u}-2}}{\big(\cos(\theta)+1\big)^{\frac{v}{u}}}\Bigg)d\theta$$ I know this isn't the answer you're looking for but it works!

Title: Find time to collision in a pursuit

Body: Let $A(t),B(t)$ denote the positions of the bodies at time $t\geq 0$. Take $B(0)=(0,0)$ and $A(0)=(0,-l)$. The information provided in the problem tells us $\frac{dB}{dt}=\Big<u,0\Big>$ and $$\frac{dA}{dt}=\frac{v\big(B(t)-A(t)\big)}{||B(t)-A(t)||}$$ If we take $\vec{p}(t)=B(t)-A(t)$ we can say $$\frac{d\vec{p}}{dt}=\Big<u,0\Big>-v\frac{\vec{p}(t)}{||\vec{p}(t)||}$$ We need to find $\tau$ so that $\vec{p}(\tau)=\vec{0}$ Writing $\vec{p}(t)=\big(x(t),y(t)\big)$ gives us the system $$\frac{dx}{dt}=u-\frac{vx}{\sqrt{x^2+y^2}}$$ $$\frac{dy}{dt}=-\frac{vy}{\sqrt{x^2+y^2}}$$ Converting the polar yields $$\frac{dr}{dt}=u\cos(\theta)-v$$ $$\frac{d\theta}{dt}=-\frac{u\sin(\theta)}{r}$$ Here $(r,\theta)=(l,\pi/2)$ when $t=0$. Dividing these provides us with a separable DE in $r$ and $\theta$: $$\frac{dr}{d\theta}=r\cdot \Bigg(\frac{v-u\cos(\theta)}{u\sin(\theta)}\Bigg)$$ This has solution $$r=\frac{l\big(\sin(\theta)\big)^{\frac{v}{u}-1}}{\big(\cos(\theta)+1\big)^{\frac{v}{u}}}$$ Where $\theta \in [0,\pi/2]$ and $r=||\vec{p}(t)||=0$ when $\theta=0$. Plugging into our formula for $\frac{d\theta}{dt}$ gives us $$\Bigg(\frac{\big(\sin(\theta)\big)^{\frac{v}{u}-2}}{\big(\cos(\theta)+1\big)^{\frac{v}{u}}}\Bigg)d\theta=-\frac{u}{l}dt$$ This means $$\int_{\pi/2}^0\Bigg(\frac{\big(\sin(\theta)\big)^{\frac{v}{u}-2}}{\big(\cos(\theta)+1\big)^{\frac{v}{u}}}\Bigg)d\theta=\int_{0}^{\tau}-\frac{u}{l}dt$$ Solving for $\tau$, $$\tau=\frac{l}{u}\int_{0}^{\pi/2}\Bigg(\frac{\big(\sin(\theta)\big)^{\frac{v}{u}-2}}{\big(\cos(\theta)+1\big)^{\frac{v}{u}}}\Bigg)d\theta$$ I know this isn't the answer you're looking for but it works!


问:
$\lim_{x \to 0} (e^x - 1)/x^8$ .
I want to know how to solve this limit problem with Maclaurin's series expansion of only $e^x$ in the neighborhood of $x=0$.

答:
You can certainly apply l'Hôpital here, provided you distinguish between the limits from the right and from the left. Since
$$
\lim_{x\to0^+}\frac{e^x}{8x^7}=\infty
$$
you can certainly conclude that
$$
\lim_{x\to0^+}\frac{e^x-1}{x^8}=\infty
$$
Similarly
$$
\lim_{x\to0^-}\frac{e^x}{8x^7}=-\infty
$$
and therefore
$$
\lim_{x\to0^-}\frac{e^x-1}{x^8}=-\infty
$$
Hence the two-sided limit doesn't exist.

What about Taylor (or Maclaurin, if you prefer to call it this way)? The Taylor expansion is $e^x-1=x+o(x)$ so you have, from the right,
$$
\lim_{x\to0^+}\frac{e^x-1}{x^8}=\lim_{x\to0}\frac{x+o(x)}{x^8}=\infty
$$
(you get $\infty$ because you know that the function is positive in a right punctured neighborhood of $0$). Similarly for the limit from the left and you get the same result as before.

Title: $\lim_{x \to 0} (e^x - 1)/x^8$ .


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

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

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

Many thanks!

答:
$8k+3,8k+7$ can be merged into $4c+3$ where $k,c$ are integers

Now, $a^2+5b^2=4c+3\implies a^2+b^2=4c+3-4b^2=4(c-b^2)+3\equiv3\pmod 4,$

But as $(2c)^2\equiv0\pmod 4,(2d+1)^2\equiv1\pmod 4,$

$a^2+b^2\equiv0,1,2\pmod 4\not\equiv3$

Clearly, $a^2+5b^2$ in the question can be generalized $(4m+1)x^2+(4n+1)y^2$ where $m,n$ are  any integers


问:
Challenging Integrals for High School Students
I am now in my last year of high school. We have covered all the techniques useful for indefinite integration that are included in our Maths and Further Maths courses. This includes:

Integration by parts, inspection, substitution, partial fraction decomposition

Integration of regular and inverse trigonometric, regular and inverse hyperbolic, exponential, logarithmic, polynomial functions

I would like to have some challenging integrals to attack that are possible for me to solve at my current level of knowledge. By challenging, I mean integrals similar to the ones in this document. They were generally enjoyable and very satisfying to solve. If you have an integral that you think I could do that is more challenging than those in the aforementioned link, so much the better.

Thank you for your suggestions.

答:
These problems are not challenging but still good to do them

$\int\exp(x)\bigg(\frac{1+x\ln x}{x}\bigg)\mathrm dx$

$\int \sin(101x) \sin^{99}(x)\mathrm dx$

$\int \sqrt{x-\sqrt{x^2-4}} \ \ \mathrm dx$

Play with these they are high school level.


问:
Parametric lines integral
I've been trying to make sense of this for days now, but I can't seem to be able to!

How does this

Let $t$ be a parameter equal to the distance from the beginning of the polygonal
line till the current point along this line. Let $l_{j_{1}}$ and $l_{j_{2}}$ be values of the parameter defining the beginning and the end of the analyzed part of the
polygonal line. The horizontal line approximating a given part of the polygonal
line can be defined as
$$y=\frac{V_{y}}{l_{j_{2}}-{l_{j_{1}}}}, V_{y}=\int_{{l_{j_{1}}}}^{{l_{j_{2}}}} y(t)dt$$
The vertical line can be found similarly
$$x=\frac{V_{x}}{l_{j_{2}}-{l_{j_{1}}}}, V_{x}=\int_{{l_{j_{1}}}}^{{l_{j_{2}}}} x(t)dt$$

Make these straight lines ($L_{i}$)

What should I plug in as $t$ ? Should I plug in the last point's length?

答:
Are you familiar with the concept of the "average value of a function"?

For a function $f(x)$, the "average vale of $f$ on the interval $[a,b]$" is defined as
$$ \mathrm{avg} = \frac{1}{b-a} \, \int_a^b f(x) \, dx $$

If you evaluate this integral, you get a number (a constant), let's call it $c$. Then graphing the horizontal line $y=c$ will show you the average height of the graph between $x=a$ and $x=b$.

This is what is being described in the quote from your post. You have a parameterized curve $r(t) = (x(t), \, y(t))$. Then the quantities:
$$ \frac{1}{j_2-j_1} \; \int_{j_1}^{j_2} x(t) \, dt \hspace{0.5cm} \text{ and } \hspace{0.5cm} \frac{1}{j_2-j_1} \; \int_{j_1}^{j_2} y(t) \, dt $$
are just the average $x$-value and average $y$-value of all the points along the curve between times $t=j_1$ and $t=j_2$. In the picture the horizontal and vertical lines are just plotting the equations $x=\mathrm{const}$ and $y = \mathrm{const}$, where "const" is the value of the appropriate integral.

Title: Parametric lines integral


问:
Why does the proper class construction of Stone–Čech compactification fail?
Wikipedia's article on the Stone–Čech compactification gives several constructions of it, one which is this:

One attempt to construct the Stone–Čech compactification of $X$ is to take the closure of the image of $X$ in $$\prod\nolimits_{f:X\to K} K$$
where the product is over all maps from $X$ to compact Hausdorff spaces $K$. By Tychonoff's theorem this product of compact spaces is compact, and the closure of ''X'' in this space is therefore also compact. This works intuitively but fails for the technical reason that the collection of all such maps is a proper class rather than a set. There are several ways to modify this idea to make it work; for example, one can restrict the compact Hausdorff spaces $C$ to have underlying set $P(P(X))$ (the power set of the power set of $X$), which is sufficiently large that it has cardinality at least equal to that of every compact Hausdorff set to which $X$ can be mapped with dense image.

My question is, why does this construction of the Stone–Čech compactification "fail"?  Is there something illegitimate about an infinite product being a proper class, or about endowing a topology to a proper class?  Or is it about applying the axiom of choice to proper classes?  Or what?

Does the existence of $\prod\nolimits_{f:X\to K} K$ follow from either $NBG$ or $MK$?

答:
I'd like to give you a slightly more general perspective, assuming that you know a bit of category theory.

Let $Top$ be the category of topological space and $\bf CHaus$ the full subcategory of $Top$ given by compact Hausdorff spaces. Let $j\colon\bf CHaus \to Top$ be the inclusion functor.

The problem of finding the stone Čech compactification is equivalent to find a left adjoint $\beta\colon\bf Top \to CHaus$. Indeed, you search for a universal compactification in the following sense:  for any $f\colon X \to i(C) $ where $X \in\mathbf{Top}, C \in\mathbf{CHaus}$, there is a unique extension $ \beta X \to C$.

Does a left adjoint to the inclusion exist? We will invoke here the

General Adjoint Functor Theorem. If $C$ is complete and locally small and $R: C\to D$ preserve small limits, then $R$ has a left adjoint if and only if it satisfies the solution set condition.

The solution set condition can be stated as follows: for every object $Y \in D$, there exist a small set $I$ of maps $ Y \to R(X_i) $ that is "initial"; this means that for any $Z \in C$ and a map $Y \to R(Z) $, there exist some $i$ and a map $X_i \to Z$ so that

$$ Y \to R(Z) = Y \to R(X_i) \to R(Z) $$

The only if part is simple. If there exist an adjoint $L$, then $\{LY\}$ will be the solution set: for any $Y \to R(Z) $, by adjunction you have $LY \to Z$ , and it is a classical result that
$$ Y \to R(Z) = Y \to RLY \to R(Z) $$

Conclusions. Note that $\bf CHaus$ is stable for products and quotients, which gives that $\bf CHaus$ is complete. It is also locally small, because $\operatorname{Hom}(X, Y) $ is a set for any spaces $X, Y$. The inclusion preserve limits, because both products and quotients in $\bf CHaus$ are computed as in $\bf Top$.

Since hypothesis of GAFT are verified, the existence of the left adjoint is equivalent to find a small solution set. Since for any $f\colon X \to C$ where $C \in\bf CHaus$ this factors through $\overline{\operatorname{Im} f} \in\bf CHaus$, you can take the small solution set of compacts Hausdorff in which $X$ is dense, which is bounded in cardinality.

What I want you to focus on is that size can be a real issue, and there exist examples in which GAFT can't be applied; not because there is some argument involving classes that would do the job, but because the left adjoint does not exist at all. See example 3.1 at the nlab page:

https://ncatlab.org/nlab/show/adjoint+functor+theorem


问:
A question on the divergence theorem

Let $\Omega \subseteq \mathbb{R}^n$ a bounded domain with boundary $\partial \Omega \in C^1$. Then if $w \in C^1(\Omega)$ it holds
$$\int_\Omega \text{div} \ w \ dV = \int_{\partial \Omega} \ w \cdot \nu \ d\sigma$$

This is a classical formulation of the divergence theorem.

My question is: if $w \in C^1_c(\Omega)$ is the hypothesis $\partial \Omega \in C^1$ still crucial?

I think the answer is no since $w = 0$ near the boundary but I'm not very sure how to prove this formally.

EDIT

I think the following is trivial but an equivalent formulation of my question is:

Let $\Omega \subseteq \mathbb{R}^n$ a bounded domain and $w \in C^1_c(\Omega)$. Does it hold $\int_\Omega \text{div} \ w \ dV = 0$ ?

This is equivalent since $w|_{\partial \Omega} = 0$ and the right integral in the divergence theorem becomes exactly $0$

答:
This could be an elegant solution:

Since $\Omega$ is bounded, there exists a ball of radius $R$ big enough such that $\Omega \subseteq B_R(0)$. Then we can extend $w$ to all the ball so that $w = 0$ out of its support.

Now we can apply the usual divergence theorem and we can conclude.

Title: A question on the divergence theorem


问:
Semisimplicity of Lie algebras and non-degeneracy of associated bilinear forms of representations
Let $ \mathfrak{g} $ be a finite-dimensional Lie algebra over a field of characteristic $ 0 $. For a finite-dimensional representation $ \rho $, we define a bilinear form $ B_\rho $ on $ \mathfrak{g} $ by
$$
B_{\rho}(x, y) = \operatorname{Tr}(\rho(x) \circ \rho(y))
\qquad (x, y \in \mathfrak{g}).
$$
$ B_\rho $ is called the bilinear form associated to $ \rho $. Note that the Killing form is the bilinear form associated to the adjoint representation.

If $ \mathfrak{g} $ is semisimple, $ B_\rho $ is non-degenerate for every finite-dimensional faithful representation $ \rho $ of $ \mathfrak{g} $ (Bourbaki, Lie Groups and Lie Algebras, I.6.1, Proposition 1).

Question. Does the converse of the above statement hold? That is, if $ B_\rho $ is non-degenerate for every finite-dimensional faithful representation $ \rho $ of $ \mathfrak{g} $, can we conclude that $ \mathfrak{g} $ is semisimple?

(If in addition we assume that $ \mathfrak{g} $ has a trivial center, the adjoint representation $ \mathfrak{g} $ is faithful, and hence $ \mathfrak{g} $ is semisimple by the assumption and Cartan’s criterion.)

答:
$\DeclareMathOperator\g{\mathfrak{g}}\DeclareMathOperator\h{\mathfrak{h}}$The first preliminary remark is that $\mathfrak{z}(\g)\cap [\g,\g]$ is in the kernel of $B_\rho$ for every rep $\rho$. Indeed, for this we can pass to an algebraic closure, block-trigonalize the representation with irreducible diagonal blocks: on each (irreducible) diagonal block, this central intersection acts by scalars of trace zero, hence by zero.

So, combined with Ado's theorem, the assumption ensures that $\mathfrak{z}(\g)\cap [\g,\g]=0$.
Se can write $\g$ as $\mathfrak{z}(\g)\times\h$. By Ado's theorem choose a faithful rep for $\h$, and take the direct sum with a faithful rep for the central factor so that it acts by strict upper triangular matrices. The direct sum of those reps is a faithful rep of the direct product, with degenerate associated bilinear form unless $\mathfrak{z}(\g)=0$.

Hence, $\g$ has trivial center, and this case is already mentioned by the OP, using the adjoint representation.

(Note that the result trivially implies Ado's theorem, so we can't expect to bypass it.)


问:
Proof of the uniqueness of maximal ideal
Let $R$ be a commutative ring with $1$. Let $M$ be a maximal ideal of $R$ such that $M^2 = 0$. Prove that $M$ is the only maximal ideal of $R$.</p>

答:
A maximal ideal is prime. Let $\mathfrak{m}$ be any maximal ideal and let $x\in M$; since $x^2=0\in\mathfrak{m}$, you get $x\in\mathfrak{m}$. Therefore $M\subseteq\mathfrak{m}$.

Note that the same holds if $M$ is a nil ideal.


问:
Metrics on n-sphere
What are some common metrics on an n-sphere $S = \{x \in \mathbb{R}^n : ||x||_2 = 1 \} $ ?

I think that the angle between vectors is a metric on $S$. What other common metrics are there? Citations also welcome.

Background: I have an estimation algorithm which theoretically works for any metric. However, in practice some metrics may have additional properties which I may be able to make use of algorithmically (e.g. for faster computation). However, the vector-angle-metric is the only one I know of.

答: There are surprisingly few sensible choices to consider. Symmetry considerations motivate any metric $d(x,\,y)$ reduce to one on the centre-$O$ circle passing through $x,\,y$, and further to some function of the angle $\theta\in[0,\,\pi]$ between them, which lies in that circle's plane. The most obvious options are the arc length $\theta$ (which you cited) and the chord length $\sqrt{2(1-\cos\theta)}=2\sin\tfrac{\theta}{2}$.

Title: Metrics on n-sphere


问:
The subspace $C$ of the space $(X,d)$ is compact
Let:

$(X,d)$ - complete metric space.

$K_1,K_2,\dots $ - compact sets in the space $(X,d)$ (sequence $K_1,K_2,\dots$ need not be descending)

$C_n=\{x\in X: \text{dist} (x,K_n)\le \frac 1n \}$ for  $n=1,2,\dots$

$C=\bigcap\limits_{n=1}^{\infty} C_{n}\subset X$

Claim: The subspace $C$ of the space  $(X,d)$ is compact

My attempts:

We want to show that $\bigcap\limits_{n=1}^{\infty} C_{n}$ is compact. Let's establish $V=\{ \bigcup\limits_{\alpha}:\alpha \in A\}$ - cover of $\bigcap\limits_{n=1}^{\infty} C_{n}$. From this $V$ we want to choose finite subcover for $\bigcap\limits_{n=1}^{\infty} C_{n}$. I think that we can use the total limitation of space $C$ with metric $d$ truncated to $C$. However I don't have idea how I can do it.

答:
We assume the following fact without proof:

If the topology of a topological space is induced by a metic, then
a set $K$ is compact iff it is sequentially compact (i.e., every
sequence in $K$ has a subsequence that converges to a point in $K$).

Claim 1: Let $C=\cap_{n}C_{n}$. For any sequence $(x_{n})$ in $C$
and any $\varepsilon>0$, there exists a subsequence $(x_{n_{k}})$
of $(x_{n})$ such that $d(x_{n_{k_{1}}},x_{n_{k_{2}}})<\varepsilon$
for all $k_{1},k_{2}\in\mathbb{N}$.

Proof of Claim 1: Let $(x_{n})$ be a sequence in $C$ and $\varepsilon>0$.
Choose $N_{1}\in\mathbb{N}$ such that $\frac{1}{N_{1}}<\frac{\varepsilon}{4}$.
Since $x_{n}\in C_{N_{1}}$, there exists $y_{n}\in K_{N_{1}}$ such
that $d(x_{n},y_{n})\leq\frac{1}{N_{1}}$. Since $K_{N_{1}}$
is compact, for the sequence ($y_{n})$, there exists a subsequence
$(y_{n_{k}})$ and $y\in K_{N_{1}}$ such that $y_{n_{k}}\rightarrow y$.
Choose $N_{2}$ such that $d(y_{n_{k}},y)<\frac{\varepsilon}{4}$
whenever $k\geq N_{2}$. Let $N=\max(N_{1},N_{2})$. Consider the
subsequence $(x_{n_{k}})_{k\geq N}$. Let $k_{1},k_{2}\geq N.$ We
have that
\begin{eqnarray*}
 &  & d(x_{n_{k_{1}}},x_{n_{k_{2}}})\\
 & \leq & d(x_{n_{k_{1}}},y_{n_{k_{1}}})+d(y_{n_{k_{1}}},y)+d(y,y_{n_{k_{2}}})+d(y_{n_{k_{2}}},x_{n_{k_{2}}})\\
 & \leq & \frac{1}{N_{1}}+\frac{\varepsilon}{4}+\frac{\varepsilon}{4}+\frac{1}{N_{1}}\\
 & < & \varepsilon.
\end{eqnarray*}

Claim 2: Every sequence in $C$ has a convergent subsequence.

Proof of Claim 2: We prove by Cantor's diagonal argument. Let $(x_{n})$
be a sequence in $C$. Choose a subsequence $(x_{1,k})_{k}$ of $(x_{n})$
such that $d(x_{1,k_{1}},x_{1,k_{2}})<\frac{1}{1}$ for any $k_{1},k_{2}\in\mathbb{N}$.
Suppose that a subsequence $(x_{n,k})_{k}$ has been chosen. Invoking
Claim 1 for the sequence $(x_{n,k})_{k}$ and positive number $\varepsilon=\frac{1}{n+1}$,
we choose a subsequence $(x_{n+1,k})_{k}$ of $(x_{n,k})_{k}$ such
that $d(x_{n+1,k_{1}},x_{n+1,k_{2}})<\frac{1}{n+1}$ for any $k_{1},k_{2}\in\mathbb{N}$.
Define $y_{n}=x_{n,n}$. We go to show that $(y_{n})$ is a convergent
subsequence of $(x_{n})$. To be precise, we show all working steps
in detail. By choosing a subsequence $(x_{1,k})_{k}$ of $(x_{n})$,
we mean choosing a strictly increasing function $\theta_{1}:\mathbb{N}\rightarrow\mathbb{N}$
such that $x_{1,k}=x_{\theta_{1}(k)}$. Since $(x_{n+1,k})_{k}$ is
a subsequence of $(x_{n,k})_{k}$, there exists a strictly increasing
function $\theta_{n+1}:\mathbb{N}\rightarrow\mathbb{N}$ such that
$x_{n+1,k}=x_{n,\theta_{n+1}(k)}.$ Now,
\begin{eqnarray*}
 &  & x_{n,n}\\
 & = & x_{n-1,\theta_{n}(n)}\\
 & = & x_{n-2,\theta_{n-1}(\theta_{n}(n))}\\
 & = & \ldots\\
 & = & x_{1,\theta_{2}\circ\theta_{3}\circ\cdots\circ\theta_{n}(n)}\\
 & = & x_{\theta_{1}\circ\theta_{2}\circ\cdots\circ\theta_{n}(n)}.
\end{eqnarray*}
Define $\psi:\mathbb{N}\rightarrow\mathbb{N}$ by $\psi(n)=\theta_{1}\circ\cdots\circ\theta_{n}(n)$.
Observe that $\theta_{n+1}(n+1)\geq n+1>n$, so $(\theta_{1}\circ\cdots\circ\theta_{n})\left(\theta_{n+1}(n+1)\right)>(\theta_{1}\circ\cdots\circ\theta_{n})(n)$,
i.e., $\psi(n+1)>\psi(n)$. Therefore, $\psi$ is a strictly increasing
function. Since $y_n = x_{\psi(n)}$. This shows that $(y_{n})$ is a subsequence of $(x_{n})$.

Let $\varepsilon>0$ be given. Choose $N$ such that $\frac{1}{N}<\varepsilon$.
Let $n\geq N$ and $k\in\mathbb{N}$. Observe that
\begin{eqnarray*}
y_{n+k} & = & x_{n+k,n+k}\\
 & = & x_{n+k-1,\theta_{n+k}(n+k)}\\
 & = & x_{n+k-2,\theta_{n+k-1}\circ\theta_{n+k}(n+k)}\\
 & = & \cdots\\
 & = & x_{n,\theta_{n+1}\circ\theta_{n+2}\circ\cdots\circ\theta_{n+k}(n+k)}.
\end{eqnarray*}
Therefore
\begin{eqnarray*}
d(y_{n+k},y_{n}) & = & d(x_{n,\theta_{n+1}\circ\theta_{n+2}\circ\cdots\circ\theta_{n+k}(n+k)},x_{n,n})\\
 & < & \frac{1}{n}\\
 & < & \varepsilon.
\end{eqnarray*}
This shows that $(y_{n})$ is a Cauchy sequence and hence is convergent
because $(X,d)$ is complete.

Claim 3: If $K$ is compact, $l>0$, then $C_{K,l}=\{x\in X\mid d(x,K)\leq l\}$
is closed.

Proof of Claim 3: Let $(x_{n})$ be a sequence in $C_{K,l}$ and suppose
that $x_{n}\rightarrow x$ for some $x\in X$. We go to show that
$x\in C_{K,l}$. For each $n$, choose $y_{n}\in K$ such that $d(x_{n},y_{n})<d(x_{n},K)+\frac{1}{n}\leq l+\frac{1}{n}.$
Since $K$ is compact, for the sequence ($y_{n})$, there exists a
subsequence $(y_{n_{k}})$ such that $y_{n_{k}}\rightarrow y$ for
some $y\in K$. We have that $d(x_{n_{k}},y_{n_{k}})<l+\frac{1}{n_{k}}$.
Letting $k\rightarrow\infty$, we have $d(x,K)\leq d(x,y)\leq l$.
It follows that $x\in C_{K,l}$.

Claim 4: $C:=\cap_{n}C_{n}$ is compact.

Proof of Claim 4: By Claim 3, $C_{n}$ is closed and hence $C$ is
also closed. Together with Claim 2, it is clear that every sequence
in $C$ has a subsequence that converges to a point in $C$. That
is, $C$ is compact.


问:
Finding non-primes for $n^2+n+17$
If I'm given $n^2+n+17$ and I'm asked to find some $n\in \mathbb N$ such that the polynomial is not prime, is there any reasoning I can use to quickly find a counter-example?

It took a good minute for me to guess one counter-example, $n=17$. But is there a way I could have found a counter example without some lucky trial-and-error?

答: $n=17$ is a good choice, because if $17$ divides $n$, then $17$ divides $n^2$ and $n^2+n+17$.

Title: Finding non-primes for $n^2+n+17$


问:
Chain Rule for Multivariable change of variables
Given the partial differential equation
$$
\frac{\partial^2 F(x,t)}{\partial t^2} = \frac{\partial^2 F(x,t)}{\partial x^2} + g(F)
$$

for a given solution $F = F(x, t)$  of this equation, show that the function
$$
F_1 = F(x \cosh \beta + t\sinh \beta, t \cosh \beta + x \sinh \beta),
$$
where $\beta$ is an arbitrary constant, is also a solution. Note that $g(F)$ is just some function, not particularly important.

I can't even really decipher what this question means. I've tried using MV chain rule but don't understand how to apply it here. I've seen the solution, which goes:
$$
\frac{\partial F_1}{\partial x} = \cosh\beta\frac{\partial F}{\partial x} + \sinh\beta\frac{\partial F}{\partial  t},
$$
and similarly for $\frac{\partial F_1}{\partial t}$, which implies
$$
\begin{align}
\frac{\partial^2 F_1}{\partial x^2} &= \cosh^2\beta\frac{\partial^2 F}{\partial x^2} + \sinh^2\beta\frac{\partial^2 F}{\partial  t^2} + 2\cosh\beta \sinh\beta\frac{\partial^2 F}{\partial x\partial t}\label{1} \tag{1}\\
\frac{\partial^2 F_1}{\partial t^2} &= \sinh^2\beta\frac{\partial^2 F}{\partial x^2} + \cosh^2\beta\frac{\partial^2 F}{\partial  t^2} + 2\cosh\beta \sinh\beta\frac{\partial^2 F}{\partial x\partial t}\label{2}\tag{2}
\end{align}
$$
Then equations are subtracted and we get the answer, which I understand. I do not understand how we get \eqref{1} and \eqref{2} though?

Can anyone chuck us some assistance? It's not a homework problem, I have the solution to this past paper question but don't understand how to do it :-(

答:
I think the confusion may be that when you write:
$$
\frac{\partial F_1}{\partial x} = \cosh\beta\frac{\partial F}{\partial x} + \sinh\beta\frac{\partial F}{\partial  t},
$$
you are leaving out the arguments of $\frac{\partial F}{\partial x}$ and $\frac{\partial F}{\partial t}$. Because of the chain rule, these have to be evaluated not at $(x,t)$, but rather at $(x \cosh \beta + t \sinh \beta, x \sinh \beta + t \cosh \beta )$. So although it may look uglier, the more correct way to write this is:
$$ \begin{align*}
\frac{\partial F_1}{\partial x}(x,t) &= \cosh\beta\frac{\partial F}{\partial x}(x \cosh \beta + t \sinh \beta, x \sinh \beta + t \cosh \beta ) \\ &\phantom{=} + \sinh\beta\frac{\partial F}{\partial  t}(x \cosh \beta + t \sinh \beta, x \sinh \beta + t \cosh \beta )
\end {align*}
$$
Now when you take the derivatives of this (to get the second partial derivatives), you will have to use the chain rule again, and you should get the formulas you mentioned.


问:
differential of an integral (composition of functions)
I'm sorry, despite of all the hints here I don't manage to put it together.
Given $\frac{\mathrm{d}}{\mathrm{d}x}\int_{0}^{x}e^{(x-t)\,A}\,b(t)\,\mathrm{d}t$

As solution $b(x) + \int_{0}^{x}A\,e^{(x-t)\,A}\,b(t)\,\mathrm{d}t$ is sugged.

Apparently the chain rule applies somehow. But I don't know which functions are chained. Is it $\frac{\mathrm{d}}{\mathrm{d}x}\left[F(x)-F(0)\right]$ ? Doesn't seem right to me...

答:
Ah, I was terrified by the formula, but basically I just got to plug in my values:

Solution:

$\begin{align}f(x,x)\,\frac{\mathrm{d}}{\mathrm{d}x}(x) - f(x,0)\,\frac{\mathrm{d}}{\mathrm{d}x}(0) + \int_{0}^{x}\frac{\mathrm{d}}{\mathrm{d}x}f(x,t)\\[12pt]
 = e^{0}\,b(x)\,1-e^{x\,A}\,b(0)\,0 + \int_{0}^{x}A\,e^{(x-t)\,A}\,b(t)\,\mathrm{d}t \end{align}$


问:
Multivariable Riemann Integral
Working currently on multivariable integrals, I got stucked. Let $R=[0,1]^3 \subset \Bbb R^3$. Suppose $f:R \to \Bbb R$ is defined by
$$\begin{align*}
    \displaystyle f(x,y,z)=\left\{\begin{array}{cl}1&\text{if }x+y+z \leq 1\\0&\text{else}\end{array}\right.\hspace{.25in}
\end{align*}$$
I wanna now show that $f$ is integrable (in the sense of Riemann) but I don't know how to do it. I guess I should show that the upper and the lower integral are the same, but I don't know how to start..

答: Break up the domain into tiny cubes.  Except for points along the plane $x+y+z=1$, the upper and lower contribution will be the same, while the total contribution of cubes containing points on the plane will go to $0$ as cubes shrink..

Title: Multivariable Riemann Integral


问:
How to prove that there is an isomorphism between a free module and a direct sum of copies of $R$?
The definition for a free $R$-module is the following:

An $R-$module $F$ is free if there exists a linearly independent generating set $B=\{b_i\}_{i\in I}$, called a basis. Note that in this case $F$ is isomorphic to a direct sum of copies of $R$. The isomorphism
$F \to \bigoplus_{i \in I} R_i$ sends a linear combination $\sum_{i\in I}r_ib_i$ to the tuple $(r_{i})_{i\in I}$.

Now I am trying to understand this definition. Let's say we call the isomorphism $\sigma$ and so for some element in $F$, $m=r_{1}b_{1}+r_{2}b_{2}$, this map gives us the following: $\sigma(r_{1}b_{1}+r_{2}b_{2})=(r_{1},r_{2})$. The first step is to show that $\sigma$ is injective. Thus, we need to show that if $\sigma(r_{1}b_{1}+r_{2}b_{2})=(0,0)$, it follows that $r_{1}b_{1}+r_{2}b_{2}=0$.

But I am stuck because I do not know where to start since we have one coordinate on the left-hand side of the equation and two coordinates on the right. Where do I go from here? Thank you.

答: The key here is linear independence. This is required for the map $r_1 b_1 + r_2 b_2 \mapsto (r_1, r_2)$ to be well-defined. How do we know we can't write $r_1 b_1 + r_2 b_2 = s_1 b_1 + s_2 b_2$ with different coefficients? Well by definition, $b_1, b_2$ are linearly independent if and only if $r_1$ and $r_2$ are unique. So we have a well-defined map. This is the key step. Injectivity is automatic since $r_1 b_1 + r_2 b_2 \mapsto (0, 0)$ if and only if $(r_1, r_2) = (0, 0)$ if and only if $r_1 b_1 + r_2 b_2 = 0b_1 + 0b_2 = 0$.


问:
Laplace formula for integral of exponential of function
I have to calculate the following limit:
$$\lim_{N\rightarrow \infty }\frac{1}{N}\log \sqrt{\beta J N} \int_{-\infty}^{\infty} dy e^{-NF(y)} $$
where $\beta, J$ are constants such that $\beta>1/J$ and $F(x)$ has a minimum in $\bar{x}$. Then he wants to use the Laplace Formula so one has to rescale the function in the minimum and rewrite the limite above as
$$ \lim_{N\rightarrow \infty }\frac{1}{N}\log \sqrt{\beta J N} e^{-NF(\bar{x})} \int_{-\infty}^{\infty} dy e^{-N(F(y)-F(\bar{x}))}=-F(\bar{x})$$
So my question is: what is the Laplace formula? There's something one can get when there is integrals as $\int_{-\infty}^{\infty} dy e^{-NF(y)}$ and $F(x)\sim x^2$ ? How can have the last equality?

For completeness: the function $F$ is $$F(x)= \frac{\beta J x^2}{2}-\log \cosh [\beta J x]$$

答:
The idea behind the Laplace approach to evaluate the asymptotics of such integral is that most of the integral will be coming from a region near the minimum, here near $\bar{x}$, and so you can Taylor-expand your integral your function $F(y)-F(\bar{x})=\frac{1}{2}F''(\bar{x})(x-\bar{x})^2+\mathcal{O}(x-\bar{x})^3$ around the minimum. If you stop at the quadratic term, you can exactly evaluate the integral in closed-form (widen the integration now back to the entire real line to obtain the asymptotics, since in the limit $N\to\infty$ the region away from $\bar{x}$ will contribute negligibly). You final result for the asymptotics would be

$$\frac{1}{2}\sqrt{\frac{2\pi}{F''(\bar{x})}}\frac{1}{N^{3/2}}\log\left(\beta J N\right)\,e^{-NF(\bar{x})}.$$


答:
Adding on slightly to J. W. Tanner's excellent answer, if we have the quadratic
$$ax^2+bx+c$$
then a value for $x$ that makes this composite (non-prime) will be the lowest common multiple of $a,b,c$ (assuming all the coefficents aren't $1$).

If any of the coefficients are negative, then I suppose you'd need to check if this input is indeed positive.

Title: Finding non-primes for $n^2+n+17$

Body:
If I'm given $n^2+n+17$ and I'm asked to find some $n\in \mathbb N$ such that the polynomial is not prime, is there any reasoning I can use to quickly find a counter-example?

It took a good minute for me to guess one counter-example, $n=17$. But is there a way I could have found a counter example without some lucky trial-and-error?


问:
All the product topologies
I wonder what are all the topologies on the product of spaces that satisfy certain properties. For example, let $(X_i,\mathcal{T}_i)_{i\in I}$ be a family of topological spaces and let $X:=\prod_{i\in I}X_i$ be the Cartesian product of these sets. Also, let $\pi_i:X\rightarrow X_i,i\in I$ be the canonical projections. Now, the Tychonoff's topology on the set $X$ satisfies these two properties:

For all $i\in I$, the mapping $\pi_i$ is continuous.

For all $i\in I$, the mapping $\pi_i$ is open.

It is easy to prove that Tychonoff's topology is the coarsest topology on the set $X$ that satisfies these two properties. The box topology is another topology on the set $X$ that satisfies these two properties. I want to know if the box topology is, as the opposite, the finest topology on the set $X$ that satisfies these two properties. I also want to know (if my previous assumption is false) what is some characterization of all the topologies on the set $X$, beside these two, that satisfy these two properties.

答:
If we take the box product on $\Bbb R^\omega$ for example, in which the "diagonal"
$D = e[\Bbb R]$ is not open (where $e: \Bbb R \to \Bbb R^\omega$ maps $x$ to the constant sequence with value $x$.) we can add $D$ to the box topology and generate a strictly larger topology that still obeys 1 and 2. (1 because it's finer than the product topology and 2 because $\pi_n[D]=\Bbb R$ is not going to "spoil" the opennness.). So I think the answer to your question is "no" in general (it's trivially true for a product of discrete spaces, as the box topology is then also discrete).

Title: All the product topologies


问:
Defining ordered pairs using only first-order logic
The usual definition of an ordered pair $\langle x,y \rangle$ is $\{\{x\},\{x,y\}\}$, but how may one define that using purely first-order logic contructs (variables, logical operations, quantifiers) and the $\in$ relation?

After reading the wiki, which defines the properties:

$x$ is the first coordinate of $p$: $\forall A[A \in p \rightarrow x \in A]$

$x$ is the second coordinate of $p$: $\exists A[A \in p \land x \in A] \land \forall A \forall B[A \in p \land B \in p \rightarrow (A \neq B \rightarrow x \notin A \lor x \notin B)]$

I arrived at the following definition:

$$
\langle x,y \rangle = p \leftrightarrow \forall A[A \in p \rightarrow x \in A] \land \exists A[A \in p \land y \in A] \land \forall A \forall B[A \in p \land B \in p \rightarrow (A \neq B \rightarrow y \notin A \lor y \notin B)]
$$

However, this definition would also consider $\{\{x\},\{x,y\},\{x,z\}\}$ or $\{\{x\},\{x,y,z\}\}$ as representing that ordered pair, and as such, it is not unique.

So, how may one define ordered pairs using only first-order logic such that only the set $\{\{x\},\{x,y\}\}$ represents the ordered pair $\langle x,y \rangle$?

答:
Here is a more or less mechanical way to do the expansion.  First, $p = \langle x, y \rangle = \{ \{ x \}, \{ x, y \} \}$ is equivalent to:
$$\forall a, a \in p \leftrightarrow (a = \{ x \} \vee a = \{ x, y \}).$$
Now, we can (by manual recursion) replace $a = \{ x \}$ with:
$$\forall b, b \in a \leftrightarrow b = x.$$
Similarly, replace $a = \{ x, y \}$ with:
$$\forall b, b \in a \leftrightarrow (b = x \vee b = y).$$
Therefore, substituting in, we get that $p = \langle x, y \rangle$ is equivalent to:
$$\forall a, a \in p \leftrightarrow [(\forall b, b \in a \leftrightarrow b = x) \vee (\forall b, b \in a \leftrightarrow (b = x \vee b = y))].$$

Note that a generalization of this procedure (formalized) comes up in model theory, under the name of "conversative extension by definition".  (However, in general, in expanding a formula not of the form variable = definition, we might need to introduce some more variables.  For example, if you wanted to expand $\langle x, y \rangle \in z$ instead, then you would first need to expand this to $\exists p, p = \langle x, y \rangle \wedge p \in z$, and then proceed to expand $p = \langle x, y \rangle$ as above.)


问:
Does $a_n:=\left(1+\frac1{n+c}\right)^n$ converge to $e$, for $c\in\Bbb{N}$?
In my lecture, we defined $e$ as
$$\lim_{n\to\infty}\left(1+\frac 1n\right)^n = \sum_{k=0}^\infty \frac 1{k!} = e$$

While playing around with those definitions, I stumbled onto
$$a_n := \left(1 + \frac{1}{n +c}\right)^n$$ where $c \in \mathbb{N}$. From plotting it, it seems like it converges to $e$ as well, but slower if $c$ gets larger, but I can't find a proof for it.

Is my assumption true, and if so how would I go about proving it?

Body:
For the simple scheme you are trying, the fastest possible convergence is $c = - \frac{1}{2},$  so that the sequence becomes
$$ b_n = \left( \frac{2n+1}{2n-1} \right)^n  $$

This decreases to $e.$

$$ e \approx 2.71828182845904523536028747135266249775724709369995$$

With $n \geq 268$ we $b_n < 2.718285$

With $n \geq 1780$ we $b_n < 2.7182819$

With $n \geq 12023$ we $b_n < 2.71828183$

267    2.718285006005104
268    2.71828498233662

1779    2.718281900038067
1780    2.718281899955931

12022    2.718281830026634
12023    2.718281829999742

19639    2.718281829072112
19640    2.718281828999756

40778    2.718281828526699
40779    2.718281828499953

45450    2.718281828512961
45451    2.718281828459387

Name
$$ c_n =  \left( 1 + \frac{1}{n} \right)^n $$

Alright, from the Taylor series for log, we find
$$ b_n \approx e \; \cdot \; \;  e^{\left( \frac{1}{12n^2} \right)}   $$

while
$$ c_n    \approx \frac{ e}{e^{\left( \frac{1}{2n} \right)} }  $$
is a worse approximation


问:
Number of arrangements of letters with repetition permitted under stated constraints
I have a lot of difficulty with probability so I'd like to know if I am thinking correctly:

Considering an alphabet with $21$ consonants and $5$ vowels, how many words with $4$ letters exist:

(i) Exactly $1$ vowel and allowing repetitions

I did $5 \cdot 21 \cdot 21 \cdot 21 \cdot 4$ (the $4$ is to vary the position of the vowel)

(ii) Starting and ending with x, having at least one vowel and allowing repetitions

I did $1 \cdot 5 \cdot 26 \cdot 1 \cdot 2 - 5$ (the $2$, to vary the positions of the vowel and the other letter and the $-5$ to exclude cases where the vowel is repeated so the order doesn't matter)

(iii) Having exactly one x letter and not allowing repetitions

$1 \cdot 25 \cdot 24 \cdot 23 \cdot 4$ (the $4$, to vary the position of the x)

Is my logic right?

答:
(i) Your answer is correct.

(ii) I assume you mean the words starting and ending with 'x'. Consider this array: x _ _ x . We can fill in the blanks with any letter from the alphabet, as long as the word has at least 1 vowel. We count the number of ways to do this by first counting all permutations of 2 letters with repetitions, then subtracting the permutations that include no vowels (hence, made entirely of consonants). The correct answer is:
$$26 \times 26 - 21 \times 21 $$
(iii) Your answer is correct.


问:
Questions on Spivak's proof that making x close to a is the same as making $x - a$ close to 0.

On my first attempt, I started with the left side of the question, $\lim\limits_{x \rightarrow 0} f(x)$. I defined it as $L$. I was befuddled after 20 minutes.

On my second attempt, I defined the right side of the question, $\lim\limits_{x \rightarrow a} f(x-a)$. Again, I got befuddled for another 20 mins.

I didn't commence with  $\lim\limits_{x \rightarrow a} f(x) = L$ because it doesn't appear in the question. Isn't it more natural to start with what appears in the question, $\lim\limits_{x \rightarrow 0} f(x)$ or $\lim\limits_{x \rightarrow a} f(x-a)$? How do I do this?

I know $|y|$ means absolute value of y, but what does $y$ represent here? $y$ doesn't appear in the question.

Spivak "let $g(x) = f(x - a)$". Then $g(y) = f(y - a)$. Not $f(y + a)$?

Spivak, Calculus 2008 4 edn.  His website's errata lists no errata for these pages. I haven't MathJax'ed this to avoid typos.

答:
Let's go through the proof again, a bit more carefully.

Suppose ${\lim_{x\to 0}f(x)=L}$. We will show ${\lim_{x\to a}f(x-a)}$ also ${=L}$.

By definition of the first limit, ${\forall\ \epsilon>0\ \exists\ \delta>0\ |\ |x|<\delta\Rightarrow |f(x)-L|<\epsilon}$.

Let ${y=x+a}$. Notice that ${|x|=|(x+a)-a|=|y-a|}$. So we overall now have, with our new variable $y$ defined, that equivalently ${\forall\ \epsilon>0\ \exists\ \delta>0\ |\ |y-a|<\delta\Rightarrow|f(y-a)-L|<\epsilon}$ (all we have done is write the first limit definition with $y$ instead of $x$ being used). This is equivalent to saying
$$
\lim_{y\to a}f(y-a)=L
$$
which is what we wanted to show.

More generally, if you have some function ${c(x)}$ continuous at ${x=a}$, you have that
$$
\lim_{x\to c(a)}f(x)=\lim_{x\to a}f(c(x))
$$


问:
Find a delta given an interval
So the question is the following:

We consider $f(x) = x^2$ on the interval $[0,100]$. Let $\epsilon > 0$ be given. Find a number $\delta > 0$ so that if the length of an interval $I \in [0, 100]$ is less than delta, then $f(u) - f(l) < \epsilon$, where $f(u), f(l)$ are max and min of $f$ on $I$, respectively.

It's probably really simple but I just can't seem to wrap my head around it. Thanks!

答:
Let me clean up your question a bit as follows.

Consider the function $f(x)=x^2$ on the closed interval $J=[0,100]$. Let $\epsilon>0$. Show that there exists $\delta >0$ such that for any closed interval $I\subset J$ with length less than $\delta$, $f(a)-f(b)<\epsilon$ where
$$
f(a)=\max\{f(r):r\in I\},\quad f(b)=\min\{f(r):r\in I\}.
$$

(This question is about the uniform continuity of $f$ on a compact interval. )

Back to your question, notice that
\begin{align}
|f(x)-f(y)|
&=|x+y||x-y|\\
&\le |x-y+2y|\cdot |x-y|\\
&\le (|x-y|+2|y|)|x-y|\\
&\le (100+2\cdot 100)|x-y|<301|x-y|,\quad x,y\in I
\end{align}

If you let $\delta=\frac{\epsilon}{301}$, then
$$
|f(x)-f(y)|< \epsilon
$$
for all $x,y\in I$ with $I$ of length $\delta$.

Title: Find a delta given an interval


问:
Short question about submanifolds
Which of these subsets $M \subset \mathbb{R}^2$ is a submanifold?

a. $M = \{(x,y)\,|\, x^2+y^2=1\}$

b. $M = \{(x,y)\,|\, x=0 \mbox { oder } y= 0\}$

c. $M = \mathbb {Q}^2$

I'm kind of sure that a is a submanifold and that c is not. I'm not sure about b but I think that it is a submanifold. Is this correct?

答:
No. (b) is not: consider the origin $(0,0)$.

If you take a neighborhood of the origin, it is not locally an open interval of $\mathbb{R}$.

Title: Short question about submanifolds


问:
How do constants differ from numbers?
Although I scanned the other version that's not Early Transcendentals, in Calculus Early Transcendentals 7th ed 2011, James Stewart never defines "constant". It first appears on p. 3 but not in this context, and contextually it first appears on p. 27 when he defines "coefficients of a polynomial".

Why not just call "constants" numbers? Isn't "constant" superfluous?  Aren't numbers constant?

Kindly see green underline. Why not call $a$ a positive number, instead of "positive constant"?

答:
I'll try to explain it the best I can.

All numbers "are constant", since a $2$ is a $2$ always. A $2$ is not some variable value that can change.

But you need to understand that, in mathematics, there are way more things than numbers: sets, functions, etc. and quite anything can be defined to be constant (tought often "fixed" is the word used to describe them too), so no, constant isn't equal to number.

In your text, it's obvious that $a\in\mathbb{R}$ so it's a number and hence it's constant, so it's irrelevant if they say positive number of positive constant (constant is more correct in my opinion, but in this case it's not important at all).

To sum up, numbers are constants, but a constant does not have to be a number necessarily.

Now, seeing the difference between a variable and a constant, variables are entities that don't have a fixed value. When a function, for example $f(x)=ax$ is defined, notice how the $f$ has an $x$ inside the pharenthesis. That means $x$ is a variable that, depending of its value, will give different outcomes through $f$. But the $a$ isn't inside the $f$'s pharenthesis, that's because it's a fixed value, to be said, $f$ does not depend on variations of $a$ (yes, different values of $a$ will also give different results through $f$, but $f$ is defined depending only on $x$, so to it, $a$ is fixed).

I hope it's a bit clearer for you, comment if necessary.

Title: How do constants differ from numbers?


问:
Does "All additive functions are linear" imply every set of reals has the Baire property?
This is a follow-up to my previous question. In that question, I asked whether "All additive functions are linear" imply every set of reals is measurable. The answer is no. So, a natural follow-up is, does "All additive functions are linear" imply every set of reals has the Baire property? Just to be clear, the background theory is ZF set theory without the axiom of choice.</p>

答:
No. It follows from "Every set of reals is Lebesgue measurable", which is known by the work of Shelah to be independent from the Baire proprety. This was shown by Shelah to be consistent in his paper

Shelah, Saharon, On measure and category, Isr. J. Math. 52, 110-114 (1985). ZBL0602.03013.

It's almost the same proofs as those that show that the Baire property implies continuity also work for Lebesgue measurability, and I am guessing they will also work for a number of other properties which may very well be independent as well.


问:
Multiplication of 1-forms
I decided to learn differential geometry by myself, and I am confused with the definition of the multiplication of 1-forms. According to the book that I am currently reading [1], given two 1-forms $\omega$ and $\nu$ on $T\mathbb{R}^2$, and vectors $V_1,V_2\in\mathbb{R}^2$, the operation
$$ \omega\wedge\nu(V_1,V_2)=\begin{vmatrix}\omega(V_1)&\nu(V_1)\\
\omega(V_2)&\nu(V_2)\end{vmatrix}$$
defines the multiplication of the 1-forms $\omega$ and $\nu$.

In Exercise 3.14 of [1] the author states

Show that, for any 1-forms $\omega$ and $\nu$ on $T\mathbb{R}^3$, there exists constants $c_1$, $c_2$ and $c_3$ such that
$$ \omega\wedge\nu=c_1 dx\wedge dy+c_2 dx\wedge dz + c_3 dy\wedge dz .$$

Why is this true?

[1] D. Bachman "A geometric approach to differential forms", Birkhauser, 2nd ed. 2012.

答:
Let $\omega = a dx + b dy + c dz$ and $\nu = p dx + q dy + r dz$. Then just compute using the determinant formula you gave:

$$ \begin {align*}
\omega \wedge \nu (V,W) &= \left| \begin{array}{cc} \omega(V) & \nu(V) \\ \omega(W) & \nu(W) \end{array} \right| \\
&= \left| \begin{array}{cc} av_1+bv_2+cv_3 & pv_1+qv_2+rv_3 \\ aw_1+bw_2+cw_3 & pw_1+qw_2+rw_3 \end{array} \right| \\
&= (av_1+b_2+cv_3)(pw_1+qw_2+rw_3) - (pv_1+qv_2+rv_3)(aw_1+bw_2+cw_3)
\end {align*} $$

After you multiply it out, some terms cancel, and what's left factors as:

$$ (aq-bp)(v_1w_2-v_2w_1) + (ar-cp)(v_1w_3-v_3w_1) + (br-cq)(v_2w_3-v_3w_2) $$

Notice that $v_1w_2-v_2w_1 = dx \wedge dy(V,W)$, and similarly $v_1w_3-v_3w_1 = dx \wedge dz(V,W)$ and $v_2w_3-v_3w_2 = dy \wedge dz(V,W)$.

Title: Multiplication of 1-forms


问:
Finding the solution of cosx and sinx via Operator Method
$$ (D^2 + 1)y = x + 2\cos x + \sin x$$
I could'nt apply the operator $D$ due to nominator is $0$..

答:
$$(D^2 + 1)y_p= 2\cos x$$
Since $e^{ix}= \cos x +i \sin x \implies \cos x = \Re e^{ix}$:
$$(D^2 + 1)y = 2\Re e^{ix}$$
$$y_p=2\Re \dfrac 1 {D^2+1}e^{ix}$$
$$y_p=2\Re e^{ix}\dfrac 1 {(D+i)^2+1}1$$
$$y_p=2\Re e^{ix}\dfrac 1 {D(D+2i)}1$$
$$y_p=2\Re \dfrac {e^{ix}}{2i}\left (\dfrac 1 D-  \dfrac 1{D+2i} \right)1$$
$$y_p=\Re \dfrac {e^{ix}}{i}\left (x-  \dfrac 1{2i} \right)$$
Since we have $e^{ix}= \cos x +i \sin x$, we take the real part and get this:
$$y_p=x \sin x +  \dfrac 1{2} \cos x$$
This gives you ( the cos function is absorbed by the homogeneous solution of the DE):
$$y_p=x \sin x$$
Do the same for :
$$(D^2 + 1)y_p = \sin x$$
$$y_p=\Im \dfrac 1 {D^2+1}e^{ix}$$
$$y_p=\Im e^{ix} \dfrac 1 {(D+i)^2+1}1$$
Where $\Im$ is the imaginary part.


问:
$(3+x)(2+x) = 0 \pmod 3$ Is Modular Arithmetic like any normal equation?
Let's say we have, for the integer x, this modular arithmetic equation

$ (3+x)(2+x) = 0 \pmod 3 $

Is it like any normal equation where you can say either $(3+x) = 0 \pmod 3$ or $(2+x) = 0 \pmod 3$ and then you solve it normally by saying either $x = 3k$ or $x = 3k + 1$ or nope?

I personally think I can do that because it's only logical that $xy = 3k$ then either 3 divides $x$ or 3 divides $y$ but still I want confirmation.

答: Yes, but this is because $3$ is prime. Notice that your statement "3 divides xy implies 3 divides x or 3 divides y" isn't true if you replace 3 with, say, 4 or 6.


问:
A family of sets such that the each subfamily has a different union
If it helps anything, please assume that everything below is finite.

Let $\mathcal A$ be a family of subsets of a set $X$. I want to consider the following independence condition (C) on $\mathcal A$.

(C).  The function $\bigcup: 2^\mathcal A\to2^X$ is an injection.

In other words, (C) says that each subfamily of $\mathcal A$ has a different union.

Every pairwise disjoint family $\mathcal A$ satisfies this condition. If I put together pairwise disjoint pieces, I can recognize them in the union.

But there are other examples. For $|\mathcal A|=1,$ (C) simply says that $\mathcal A\neq \{\varnothing\}$ because

if $\mathcal A=\{\varnothing\},$ then $\varnothing,\{\varnothing\}\subseteq\mathcal A,$ and $\bigcup\varnothing=\bigcup\{\varnothing\}=\varnothing,$ but $\varnothing\neq\{\varnothing\},$ so $\bigcup$ is not injective;

if $\mathcal A = \{Y\},$ $Y\neq\varnothing,$ then the only subfamilies of $\mathcal A$ are $\varnothing$ and $\{Y\}$, and they have different unions.

For $|\mathcal A|=2$, $\mathcal A=\{Y,Z\},$ (C) says that $Y\setminus Z\neq\varnothing\neq Z\setminus Y.$ (In other words, neither of the sets is a subset of the other.) Indeed,

if $Y\setminus Z=\varnothing,$ then $\bigcup\{Y,Z\}=\bigcup\{Z\}$ (and the same for $Z\setminus Y=\varnothing$);

otherwise, the unions of $\varnothing,\{Y\},\{Z\},\{Y,Z\}$ are all different.

For $|\mathcal A|=3,$ I already don't see a simple condition equivalent to (C). Is there one in the general case? (C) is of course a simple condition in terms of formulation, but it is quite a complex one in terms of the "little conditions" it comprises. I'm not certain that a condition I'm asking for exists, but perhaps it does, and someone here can help me see it?

答:
The condition that
$$\tag1 A\not\subseteq\bigcup(\mathcal A\setminus\{A\})\qquad\text{for all }A\in \mathcal A$$
is necessary to distinguish $\mathcal A$ from $\mathcal A\setminus\{A\}$ and sufficient as it implies that $A\in \mathcal B\subseteq \mathcal A$ iff $A\subseteq \bigcup\mathcal B$.

For each $A\in \mathcal A$, $(1)$ allows us to select an element $a(A)\in A\setminus \bigcup(\mathcal A\setminus\{A\})$ that works as a "sensor" for $A$. That is, we can restrict everything from $X$ to $\{a(A)\mid A\in\mathcal A\}$ and by that end up in the simple case of disjoint singleton sets.
In other words, an equivalent condition is:

There exists a subset $X'\subseteq X$ such that $A\cap X'$ is a singleton for every $A\in\mathcal A$ and $A_1\ne A_2$ implies $A_1\cap X'\ne A_2\cap X'$.


答:
By constant, what it means is the number is fixed. Just as @Alejandro Bergasa Alonso said, $2$ is always $2$, $3$ is always $3$. If someone says "${f(x) = \sin(ax)}$ where $a$ is some constant" that just means that $a$ is some fixed, unchanging number.

However, the "$x$" in the expression ${f(x) = \sin(ax)}$ is not a constant, since it varies. We can plug whatever number we like into it. It's a variable.

Usually, whether you can treat a particular symbol as a constant or variable will depend completely upon context. But the key bit is it's called a variable since it varies, and it's called a constant because it's fixed and doesn't change.

Title: How do constants differ from numbers?

Body:
Although I scanned the other version that's not Early Transcendentals, in Calculus Early Transcendentals 7th ed 2011, James Stewart never defines "constant". It first appears on p. 3 but not in this context, and contextually it first appears on p. 27 when he defines "coefficients of a polynomial".

Why not just call "constants" numbers? Isn't "constant" superfluous?  Aren't numbers constant?

Kindly see green underline. Why not call $a$ a positive number, instead of "positive constant"?


问:
Complement of the product of Bernstein set and meager set
Assume $M\subset\mathbb R$ be a meager set with cardinality $\mathfrak c.$ I want to construct a Bernstein set $B$ such that $\mathbb R\setminus(B\cdot M)$ has cardinality $\mathfrak c$ , $B\cdot M=\{b\cdot m\mid b\in B,  m\in M\}$. It is an easy transfinite induction to construct a Bernstein set but I want an extra condition.

I do not know if I need a set-theoretical assumption for that or not. More precisely, It is possible or not. I hope I can find an answer or hint since I spent  plenty of time but I did not get any thing at least  possible or not.

答:
My idea wold be to do this, following up on my comments: assume MA. We are given a meagre set $M$ and want to construct a Bernstein set $B$ such that $\Bbb R\setminus M\cdot B$ is size continuum.

Enumerate all relevant perfect subsets of $\Bbb R$ as $\{P_\xi: \xi < \mathfrak{c}\}$.

We will construct points $\{(x_\zeta, y_\zeta, z_\zeta)\mid \zeta < \mathfrak{c}\}$  by transfinite recursion; at stage $\xi < \mathfrak c$ we pick distinct

$$x_\xi, y_\xi \in P_\xi \setminus (\{x_\zeta \mid \zeta < \xi\} \cup \{y_\zeta \mid \zeta < \xi\} \cup \{0\})\tag{1}$$

while ensuring

$$z_\xi \notin \bigcup\{ M \cdot x_\zeta\mid \zeta < \xi\}\tag{2}$$

$(1)$ follows the standard construction of a Bernstein set and works for the same reasons, while $(2)$ can be done as the sets $M \cdot x_\zeta$ are also meagre and so fewer than $\mathfrak{c}$ cannot cover $\Bbb R$ by MA.

In the end $B:=\{x_\xi: \xi < \mathfrak{c}\}$ will be Bernstein, and $(2)$ ensures the condition about $\Bbb R\setminus M \cdot M$ as the $z_\xi$ witness the size of it (I think); I see that I haven't built in assurances for them to be all distinct yet... So this is almost OK, but not quite.

If a condition like MA is at all necessary I do not know yet. It's late and I'll call it a night.


答:
As J. W. Tanner says, it works when the modulus is prime.  The integers modulo a prime number, $\mathbb{Z}_p$ are a field so most familiar arithmetic rules hold.  One thing to watch is that something which does not look like $0$ might be $0$  e.g. $3$ in your example.  Another is that $>$ cannot be defined in a way that retains the familiar behaviour.

It is quite different when the modulo is not prime.  E.g. in modulo $6$, $2 \times 3 = 0$ so your cancellation reduction may fail.  These are called "zero divisors".

Title: $(3+x)(2+x) = 0 \pmod 3$ Is Modular Arithmetic like any normal equation?

Body:
Let's say we have, for the integer x, this modular arithmetic equation

$ (3+x)(2+x) = 0 \pmod 3 $

Is it like any normal equation where you can say either $(3+x) = 0 \pmod 3$ or $(2+x) = 0 \pmod 3$ and then you solve it normally by saying either $x = 3k$ or $x = 3k + 1$ or nope?

I personally think I can do that because it's only logical that $xy = 3k$ then either 3 divides $x$ or 3 divides $y$ but still I want confirmation.


问:
Why this space is not a complete space with this norm
Show that the space $C_0(\mathbb{R})$ of all the real continuous functions $f:\mathbb{R} \to \mathbb{R}$ with compact support is not a complete space with the norm $||f||= \sup_{t∈ \mathbb{R}}|f(t)|$.

Thank you

答: Let $f(x) = \frac 1{1+x^2}$, define $f_n\colon \mathbb R \to \mathbb R$ by $f_n(x) = f(x)$ for $x \in [-n,n]$, $f_n(x) = 0$ for $x \not\in [-n-1, n+1]$, affine-linear in between. Show that $(f_n)$ is Cauchy, but not convergent to some compact support function (as its limit in $\ell^\infty(\mathbb R)$ is $f$).</p>


问:
To prove the logarithm laws with differentiation, why did James Stewart commence with a as a positive constant, rather than y as a positive number?
Kindly see the green underline.  Why didn't Stewart just commence with $y$ as a positive number? Why define $a$ as "a positive constant", then in the last line replace "a by any [positive] number y"?

I know that the domain of logarithms is {positive real numbers}.

Stewart, Calculus  7th ed 2011. Not Early Transcendentals. p. 422 scanned.

答:
Lots of narratives in Stewart's book on calculus should be understood with context. That book is not a rigorous real analysis textbook. You would have many troubles if you read it seriously line by line. If you want to learn real analysis from scratch so that every single word has a precise meaning, you shall try another one.

Going back to your question, the symbol $y$ is mentioned in the red box as a positive real number. So the author may consider it clear to not mention it explicitly again.


问:
What does multiplying two permutations do to their order?
when multiplying different permutations together, I noticed that the product will always be of the same order.

Let's suppose these two permutations:

σ₁ = (12)(3) and σ₂ = (132)

That means that σ2σ1 = (32)(1), and σ1σ2 = (13)(2).

Obviously, these products are two different permutations. Yet, they are of the same order and this seems to be true for all products of permutations.

Is there a general rule for the order of a product of two permutations or does anyone have an explanation about what's going on here?

答: If your permutation $\sigma$ is a product of disjoint cycles $\sigma_1, \dots,\sigma_n$, then the order of the product $\operatorname{ord}(\sigma)=\operatorname{lcm}(\operatorname{ord(\sigma_1),\dots,\operatorname{ord}(\sigma_n)})$, where $\operatorname{lcm}$ is the lowest common multiple.


问:
showing an extension is unramifed by decomposition groups
Let $F=\mathbb{Q}(\sqrt{-5})$. I can prove that $F(\sqrt{5})/F$ is an unramified extension in three ways, using any of these two questions Reference request and proof verification about using a generalized version of a theorem , Counter example relating the ramification and also by Different ideal. I know that the two linked questions are almost the same, and they should be counted as one.

I want also to prove it by decomposition and inertia group. But I can not finish the idea. Assume $K=\mathbb{Q}(\sqrt{-5},{i})$ and $F=\mathbb{Q}(\sqrt{-5})$. Let $\mathfrak{P}\mid\mathfrak{p}\mid p$ be primes respectively in $K$, $F$ and $\mathbb{Q}$. Suppose on contrary that $e(\mathfrak{P}\mid\mathfrak{p})=2$. We have two cases $e(\mathfrak{p}\mid p)=2$ or $e(\mathfrak{p}\mid p)=1$.

First suppose that we have $e(\mathfrak{p}\mid p)=2$, then we have $e(\mathfrak{P}\mid p)=4$. By considering the multiplicativity of ramification index and the degree of extensions, this implies that $p$ ramifies in both of the $\mathbb{Q}(\sqrt{5})$ and $\mathbb{Q}(\sqrt{-1})$. So $p$ divides both of $5$ and $-4$, which is obviously impossiblee.

Now suppose that we have $e(\mathfrak{p}\mid p)=1$, then $p$ is unramified in $\mathbb{Q}(\sqrt{-5})$. So $p$ does not divides $-20$, so $p$ does not divides none of $5$ and $-4$. So $p$ is unramified in both of the $\mathbb{Q}(\sqrt{5})$ and $\mathbb{Q}(\sqrt{-1})$. I can not go further.

答:
Given a prime $$\mathfrak{P} \subset O_{EF}$$ If $$e(\mathfrak{P}\cap O_F,F/\Bbb{Q})=1$$ then
$$e(\mathfrak{P},EF/E)=1$$
so that $$e(\mathfrak{P},EF/\Bbb{Q})=e(\mathfrak{P},EF/E)e(\mathfrak{P}\cap O_E,E/\Bbb{Q})=e(\mathfrak{P}\cap O_E,E/\Bbb{Q})$$

To show it from the Galois groups, for $E/\Bbb{Q},F/\Bbb{Q}$ both Galois and $E\cap F=\Bbb{Q}$ then $Gal(EF/\Bbb{Q})=Gal(E/\Bbb{Q})\times  Gal(F/\Bbb{Q})$ so you want to prove that $I(\mathfrak{P},EF/\Bbb{Q})$ is a subgroup of $I(\mathfrak{P}\cap O_E,E/\Bbb{Q})\times I(\mathfrak{P}\cap O_F,F/\Bbb{Q})$ which is more or less immediate as $g\times h$ sends $\mathfrak{P}$ to itself and acts trivially on the residue field implies that $h$ sends $\mathfrak{P}\cap O_F$ to itself and acts trivially on the residue field.


问:
A closed set inside a compact is compact
So I know that if $(X, \tau)$ is compact and $A$ is a closed set of $\tau$, then $A$ is compact. However if $(X, \tau)$ is not compact and $A$ is a compact subset of $X$, and $B \subset A$ where $B$ is a closedset, then $B$ is a compact of $A$. Is it a compact of $(X, \tau)$ though?

Note: When I talk about $A$ as a topological space, it has the topology induced by $\tau$

答:
Being compact doesn't depend upon the larger space. That is: is $Z\subset W$ and if $\tau$ is a topology on $W$ such that $(W,\tau)$ is compact, then $Z$ is compact with respect to the subspace topology if and only if it is a compact subset of $W$.

So, since $A$ is compact and $B$ is a closed subset of $A$, $B$ is compact too.


问:
How to prove these pseudo-random number generators don't repeat until running through the entire set?
I just asked this question to find an algorithm for a pseudo-random number generator that generates 100% unique numbers for a bounded set of inputs. There are two answers I've seen. First is this one, which comes from here, and relies on quadratic residues:

```<code>b = (x, o) => x >= o ? x : (x <= o / 2) ? (x * x) % o : o - ((x * x) % o)

b32 = x => b(x, 4294967291)
b16 = x => b(x, 65519)
b8 = x => b(x, 251)

f32 = (x, o) => b32((b32(x) + o) ^ 1542469173)
f16 = (x, o) => b16((b16(x) + o) ^ 42703)
f8 = (x, o) => b8((b8(x) + o) ^ 101)
</code>```

Second was an answer to the question, and relies on Linear congruential generators.

```<code>b = (m, a, c) => x => (a * x + c) % m
m = 2 ** 16
f32 = b(m, 97, (m >> 1) - 1)
</code>```

The question I have is, how can you prove either the main idea in the first link, given by this image:

Or how can you prove that the linear congruential generator one, that either of them generate a complete set of unique numbers before repeating. That is, for $2^{32}$ values, it will loop through seemingly randomly to the lay observer, all of the $2^{32}$ values, without ever encountering the same number twice. Same goes for any bounded set of numbers.

Is there a proof for this? If so, what does it look like? If it's far too involved, then what is the gist of it? If there's no proof, well then why not? :) I tested it on some numbers in a bounded set and they were all unique, is that the only proof? Wondering because it would be cool to visualize how this is even possible, I can't imagine it currently.

答:
I'll not prove them instead, I'll show you how to generate with a block cipher and will show a little proof that the output will be unique and visit all numbers from $0$ to $2^{32}-1$

A block cipher is a family of permutations. Formally;

$$E_K(P) := E(K,P): \{0,1\}^{k} \times \{0,1\}^n \rightarrow \{0,1\}^n$$ where the $k$ is the key size and $n$ is the block size. A key is expected to selects a random permutation from all possible premutations from $\{0,1\}^n \rightarrow \{0,1\}^n$, that is there are $(2^n)!$ permutations that we select only $2^k$ of them, enough theory.

There is a block cipher called Skip32 that has a 32-bit block size and 80-bit key, though not secure. Now how to generate all number;

Select a uniform random, key $k$

Set a 32 bit counter $counter=0$

output $r = E(k,counter)$

if $counter = 2^{32}-1$  then exit

$counter++$

return step 3.

Since a block cipher is a permutation, it will output unique values for each input, remember, we need the reverse for the decryption. Counter visits all possible values between $0$ and $2^{32}-1$ then the output will be a random order of these values.

On the bonus side, you have $2^{80}$ different ( expected) random number generator for each of the keys.

If you want the numbers between $1$ to $2^{32}$ than just increment the output.

If you don't want the 0 but in the range $1$ to $2^{32}-1$ then a maximal LFSR with length 32 can be a solution, too.

The bonus, the cipher has javaScript implementation, too. And you can turn into a random number generator, very esaily.


答:
Ultimately, the choice of symbol doesn't matter. It doesn't matter if you put $a$, or ${y}$, or ${☺}$. The author just changed symbol to try and get you to think a bit about the context.

To start with, $a$ was picked first and fixed. $a$ was picked first. Then we showed ${\ln(ax) = \ln(a) + \ln(x)}$ ($x$ is some varying number, we needed it to be a variable so we could use Calculus and take a derivative). The point is, $a$ was picked and fixed at the start.

But the choice of $a$ at the start was arbitrary. This means for all positive numbers ${a,x}$: ${\ln(ax) = \ln(a) + \ln(x)}$. The author chooses to replace $a$ with $y$ to try and symbolise the fact that this choice of $a$ was arbitrary, and could be anything. $y$ is more commonly used to mean some potentially varying quantity. There's no significance behind the change in symbol apart from this contextual one.

Hope that helped.

Title: To prove the logarithm laws with differentiation, why did James Stewart commence with a as a positive constant, rather than y as a positive number?

Body:
Kindly see the green underline.  Why didn't Stewart just commence with $y$ as a positive number? Why define $a$ as "a positive constant", then in the last line replace "a by any [positive] number y"?

I know that the domain of logarithms is {positive real numbers}.

Stewart, Calculus  7th ed 2011. Not Early Transcendentals. p. 422 scanned.


问:
How many 10 letter 'words' can be made with no repeated letters and at least 8 consonants?
Need a little bit of help with this one.

How many 10 letter 'words' can be constructed which have no repeated letters and at least 8 consonants?

where 'word' is defined as any combination of the 26 letters of the english alphabet, and by consonant, (obviously) a letter of the english alphabet that are not either a,i,u,e, or o.

This is where I am at:

We need to choose the positions for
the 8 consonants which is $c(10,8)$

Then we need to choose the 8
consonants which I think is
$p(21,8)$? (not sure about this one)

Choose the remaining 2 letters: We have already used 8 letters so we only have 18 letters to choose from so $p(18, 2)$

What else needs to be calculated? Do we need to choose the position of the remaining 2 letters?

答:
I think the best way would be to separate it into 3 non-overlapped groups:

Words with exactly 8 consonants

Words with exactly 9 consonants

Words with exactly 10 consonants

Logic would be similar with Arturo. Arvin comments wouldn't work, because you count same words several time. Once when the consonant is one of the 8, and again, when it happens to be one of the two remaining letters.

For the 8 consonants:

Select 8 positions - C(10,8)

Select the 8 consonants out of the 21, in order - P(21,8)

Select the 2 vowels in order - P(5,2)

So answer is:
c(10,8) * p(21,8) * p(5,2) = (10!/(2!*8!)) * (21!/13!) * (5!/3!) = 45 * (21!/13!) * 20

Similarly, do the same for 9 and 10:

9 consonants:  C(10,9) * P(21,9) * P(5,1) = 10 * (21!/12!) * 5 = 50 * (21!/12!)

10 consonants: C(10,10) * P(21,10) * P(5,0) = (21!/11!)

Sum the 3 answers.


问:
How can I visualize $\lim\limits_{x \rightarrow \pm \infty} f(x) = \lim\limits_{t \rightarrow 0^{\pm}} f(1/t)$?
I'm not asking about the proof, that I already understand. I'm itching to understand this graphically. I use James  Stewart's graphs. As you can see, I wrote $1/t$. Then what?

Also I'm trying to visualize how Formula 8 shifts to Formula 9 below, but the underlying difficulty feels the same?

答:
Borrowing the same notation from equn's 8-9, I think the piece you want to look at is $$nx=1 \:.$$

Since you know the proof, the limit cases are familiar: If $x$ gets really small, then $n$ must get very large. Flipping this on its head, we can also say $1/x$ gets very large as $n$ gets large, etc. This accounts for why the limits toggle between 0 and infinity, depending on which variable is being limited.

Point is, $nx=1$, if you dont mind a variable switch $n \rightarrow y$, is simply the hyperbola $$y=\frac{1}{x}$$ in the Cartesian plane. There's not much more to it, graphically. When you walk the limit $x \rightarrow \infty$, you're also walking $y\rightarrow 0$, and vice versa.

EDITS:

(I was not pedantic about signs, only magnitudes.)

(And of course the numerator $1$ need not just be $1$ for the limits to work.)


问:
Counting minimum elements needed such that their sum covers the whole finite space.
In $\mathbb{F}_p$, how to find a subset with smallest cardinality such that the sum between its pairs cover $\mathbb{F}_p\setminus\{0\}$.
So for a subset with cardinality $n$ there are ${n\choose 2}$ pairs and correspondingly ${n\choose 2}$ sums. I want $n$ to be smallest such that those ${n\choose 2}$ sums cover $\mathbb{F}_p\setminus\{0\}$.

For example in $\mathbb{F}_7$, let $A=\{0,1,2,4\}$,
The sums between its pairs covers $\mathbb{F}_7\setminus\{0\}$ since
\begin{align*}
0+1=1,0+2=2,1+2=3,0+4=4,1+4=5,2+4=6\;.
\end{align*}
I want to know if over large $p$ similar set can be easily constructed and what would be the size of that set (I expect it to be close to $\sqrt{2p}$). Any similar literature is also appreciated.

答:
In this thread I answered a similar question for sumsets. Some of results mentioned in the thread can be applicable to this question. Namely, lower size sumsets bounds for $\Bbb Z$ provide lower bounds for $f_+(n)$  (see the beginning of Qiaochu Yuan’s answer for the definition of $f_+(n)$).:

it seems that in this direction was mainly investigated the Z counterpart of the problem, for which is devoted a lot of papers, in which the lower bound was more and more improved by complicated techniques, see, for instance [Mos], [GN], [Hor], [Yu].

I am checking the mentioned papers, but I already can say that the results of [Mos] imply that $f_+(n)>\sqrt{\tfrac {2n-2}{1-.0197}}$ for sufficiently large $n$.

I also found that another my colleague, Oleg Pikhurko dealt with a related stuff in his paper [Pih]. In particular, he noted that Mrose [Mro] constructed a set $S\subset [0, 10t^2+8t]$ of size $7t + 3$ such that $S + S\supset [0, 14t^2 + 10t − 1]$. In fact, $S$ is the union of ﬁve disjoint arithmetic progressions, see the description at p. 6 of Pikhurko’s paper.

Since sums of  distinct elements of $S$ cover the discrete segment $[0, 14t^2 + 10t − 1]$ but a number $8t^2+4t-2$, see the last paragraph of [Pih, p.6], we can add one element to the set $S$ and obtain a set $S’$ such that sums of distinct elements of $S’$ cover the discrete segment $[0, 14t^2 + 10t − 1]$, which follows $f_+(14t^2 + 10t)\le 7t +4$.

This fact provides an upper bound for $f_+(n)$ as follows. Pick the smallest natural $t$ such that $14t^2+10t\ge n$. Then $n\ge 14(t-1)^2+10(t-1)+1=14t^2-18t+5$. Solving the respective quadratic equation, we see that for $n\ge 4$ we have $t\le\tfrac {9+\sqrt{56n-199}}{28}$ and so $$f_+(n)\le 7t+4\le \frac{25+\sqrt{56n-199}}{4}.$$

References

[GN] C. Sinan Güntürk, Melvyn B. Nathanson A new upper bound for finite additive bases, Acta Arithmetica 124 (2006), 235-255.

[Hor] G. Horváth An improvement of an estimate for finite additive bases, Acta Arithmetica 130 (2007), 369-380.

[Mos] L. Moser On the representation of $1,2,\dots, n$ by sums, Acta Arithmetica 6 (1960), 11-13.

[Pih] Oleg Pikhurko Dense Edge-Magic Graphs and Thin Additive Bases.

[Yu] Gang Yu Upper bounds for finite additive $2$-bases, Proc. Amer. Math. Soc. 137 (2009), 11-18.


问:
$\left(\sum_{i=1}^{n} x_{i}^{2}\right)^{\frac{1}{2}} \leq \sum_{i=1}^{n} x_{i}$
I was wondering if for $x_i\in\mathbb{R}_{\geq0}$ the inequality
$$
\left(\sum_{i=1}^{n} x_{i}^{2}\right)^{\frac{1}{2}} \leq \sum_{i=1}^{n} x_{i}
$$
Holds. If so, is there a name for it?

My attempt

$$
\sum_i^n i = \frac{n(n+1)}{2}
$$
$$
\sum_i^n i^2 = \frac{n(n+1)(2n+1)}{6}
$$
Since the right hand side is $O(n^{3/2})$ and the left hand side is $O(n^2)$, and n is a positive integer, the inequality should hold.

答:
Yes.  It is the triangle inequality (for right triangles).

To explain.  Consider a point $x=(x_1,x_2, \cdots, x_n) \in \mathbb{R}_{\ge 0}^n$.

$$|x|= \left(\sum_{i=1}^n x_i^2\right)^{1/2}$$ is the diagonal of a hyperparallelapiped. The sum of the "legs" is $\sum_{i=1}^n x_i$.

We can prove it by induction on $n$.

$$\sqrt{x_1^2 + x_2^2} \le x_1+x_2. \quad (*)$$  Also known as $(x_1+x_2)^2=x_1^2+x_2^2+2x_1x_2 \ge x_1^2+x_2^2$.

Assume $$\sqrt{x_1^2+ \cdots x_{n-1}^2} \le \sum_{i=1}^{n-1} x_i$$

Let $y=(x_1,x_2,\cdots,x_{n-1})\in \mathbb{R}_{\ge 0}^{n-1}$.

$$|y|= \left(\sum_{i=1}^{n-1} x_i^2\right)^{1/2}$$

Now apply $(*)$ to $\{|y|,x_n\}$.


答: I don't know whether there's a name for it, but note that\begin{align}\left(\sum_{i=1}^nx_i\right)^2&=\sum_{i=1}^nx_i^{\,2}+\overbrace{\sum_{i\ne j}x_ix_j}^{\geqslant0}\\&\geqslant\sum_{i=1}^nx_i^{\,2}\end{align}and that therefore$$\sum_{i=1}^nx_i\geqslant\sqrt{\sum_{i=1}^nx_i^{\,2}}.$$

Title: $\left(\sum_{i=1}^{n} x_{i}^{2}\right)^{\frac{1}{2}} \leq \sum_{i=1}^{n} x_{i}$

Body:
I was wondering if for $x_i\in\mathbb{R}_{\geq0}$ the inequality
$$
\left(\sum_{i=1}^{n} x_{i}^{2}\right)^{\frac{1}{2}} \leq \sum_{i=1}^{n} x_{i}
$$
Holds. If so, is there a name for it?

My attempt

$$
\sum_i^n i = \frac{n(n+1)}{2}
$$
$$
\sum_i^n i^2 = \frac{n(n+1)(2n+1)}{6}
$$
Since the right hand side is $O(n^{3/2})$ and the left hand side is $O(n^2)$, and n is a positive integer, the inequality should hold.


问:
An intuitive explanation for the ecological fallacy
I believe that it is called the ecological fallacy. People say that one cannot apply population-wide statistics to individuals of that population. So, just because some trait exists in a higher proportion in population A than it does in B, that doesn't mean that an individual from pop. A is more likely to have that trait than an individual from pop. B.

Now, this doesn't make sense to me. An individual coming from a population where a certain trait is more common, is more likely to have that trait than an individual coming from a population where said trait is less common, right?

I guess this depends on how one mathematically defines "common". The mean and the average are a bit scary statistical quantifications. I see how "the average" is problematic, but a weighted average is surely quite illuminatory of the real likelihoods?

答:
I should imagine two different notions maybe mixed up here:

First, "one cannot apply population-wide statistics to individuals of that population" is correct if that statistics is not ergodic. Suppose you measure the Fast Blood Sugar of a population (one population, not two or more) and you realize that the average FBS of the healthy people is 80 with a standard deviation of 20, with 95% confidence. It means that if you take blood sample from a healthy individual and measure FBS, it most probably is within the range of 60-100. The fact is this is not quite right. For example, my FBS has been ~110 for the past two decades, and I am healthy sugar-wise! For me, an individual, the normal FBS is 110. If my blood sugar drops below 80, I would probably have a problem that I should look into it, if my doctor actually acknowledges the fact that "one cannot apply population-wide statistics to individuals of that population."

Second, statistics drawn from two populations with distinct traits (that affect the statistics) must be treated with respect to the population that the statistics is drawn from. In Bayesian interpretation of the probability, I suppose it constitutes the notion of apriori (or prevalence) knowledge about the population. However, it by itself is not enough to draw any conclusion.


问:
find the smallest positive multiple $n$ of 1994 such that the produc of the divisors of $n$ is $n^{1994}$

Find the smallest positive multiple $n$ of 1994 such that the produc of the divisors of $n$ is $n^{1994}$

Attempt

Let $n\in \mathbb{N}$ and denote by $d(n)$ the number of divisors of $n$ and $p(n)$ the product of all the divisors of $n$ including $n$.

Consider the factorization of $n$ given by $n=p_1^{\alpha_1} p_2^{\alpha_2}\cdots p_n^{\alpha_n}$ for $\alpha_i\in \mathbb{N}$. Is known the fact that $p(n)=n^{d(n)/2}=n^{1994}$ it is
$d(n)=997 \cdot 2^2$ in addition $n=997 \cdot 2 \cdot k\, $  for some  $k\in \mathbb{N}$.

since $n$ is the smallest multiple of $1994$ with $997 \cdot 2^2$ divisors then
$n$ should be $2^{998} \cdot 997$

Any comment or corretion was helpful

答:
Your idea is really good, but at the end you want the smallest $k$ such that $d(997\cdot 2\cdot k)=997\cdot 2^{2}$. You said that $k=2^{997}$ does the job, but this  doen't, because $d(997\cdot 2^{998})=(1+1)(998+1)\neq 997\cdot 2^{2}$.

To fix this, note that if you write $k=2^{m}\cdot p$, where $p$ doen't have factors of $2$, then
$$d(\color{blue}{997}\cdot \color{red}{2\cdot 2^{m}}\cdot p)=\color{blue}{(1+1)}\cdot\color{red}{(m+2)}\cdot d(p)$$
so you want $(m+2)\cdot d(p)=2\cdot 997$. Here are two possibilities: $m+2=2$ of $m+2=997$.

If $m+2=2$ then $m=0$, therefore $d(p)=997$, and the lowest $p$ possible in this case is $p=3^{996}$.

If $m+2=997$ then $m=995$, therefore $d(p)=2$, and the lowest $p$ possible in this case is $p=3$.

So the candidates are $997\cdot 2\cdot 3^{996}$ and $997\cdot 2^{996}\cdot 3$, but clearly the last is smaller, so $n=997\cdot 2^{996}\cdot 3$.


答:
As you correctly identify, since the product of the divisors of $n$ is $n^{1994}$, $n$ must have $1994$ factor pairs, i.e. $3988$ divisors. We also know that $n$ must be divisible by prime $997$, as must the number of divisors. So to get this we need a factor of $p^{\large{\color{red}{996}}}$ which you narrowly miss, and obviously this should use $p=2$ to get as small as possible.

However you have also missed the second factor $2$ in the number of divisors, which we can arrange with a multiple of $3$ to get the smallest $n$. So $n = 2^{996}\cdot 3 \cdot 997$, which has $(996+1)(1+1)(1+1) = 3988$ divisors as required, adding one to each prime exponent.

Title: find the smallest positive multiple $n$ of 1994 such that the produc of the divisors of $n$ is $n^{1994}$

Body:
Find the smallest positive multiple $n$ of 1994 such that the produc of the divisors of $n$ is $n^{1994}$

Attempt

Let $n\in \mathbb{N}$ and denote by $d(n)$ the number of divisors of $n$ and $p(n)$ the product of all the divisors of $n$ including $n$.

Consider the factorization of $n$ given by $n=p_1^{\alpha_1} p_2^{\alpha_2}\cdots p_n^{\alpha_n}$ for $\alpha_i\in \mathbb{N}$. Is known the fact that $p(n)=n^{d(n)/2}=n^{1994}$ it is
$d(n)=997 \cdot 2^2$ in addition $n=997 \cdot 2 \cdot k\, $  for some  $k\in \mathbb{N}$.

since $n$ is the smallest multiple of $1994$ with $997 \cdot 2^2$ divisors then
$n$ should be $2^{998} \cdot 997$

Any comment or corretion was helpful


答:
Indeed he could have phrased the argument like this:

"Let $y$ be a positive real number. Let $f$ be the function defined by $f(x) = \ln(xy)$. (Here $x$ can be any positive real number.) Notice that $f'(x) = \frac{1}{x}$. It follows that $f(x) = \ln(x) + C$ for some constant $C$. Plugging in $1$ for $x$ reveals that $C = \ln(y)$. Thus, $\ln(xy) = \ln(x) + \ln(y)$."

That does seem slightly simpler to me. Perhaps, pedagogically, Stewart thought the use of the letter $a$ would make it easier for students to compute the derivative of $\ln(ax)$, as it is more similar to notation used previously in the textbook.


问:
approximate $\int_{a}^{b}f(x)f''(x)dx$

Given $f\in C^7[a,b]$ $x_k=a+kh$ where $h=\frac{b-a}{n},k=-1,0,\ldots,n+1$

Find a way to approximate $f''(x_k)$ with $f(x_{k-1}),f(x_k),f(x_{k+1})$ and then find a way to approximate $\int_{a}^{b}f(x)f''(x)dx$

I've found that $f''(x_k)=1/h^2(f(x_{k-1})-2f(x_k)+f(x_{k+1}))+h^2/24(f^{(4)}(C_1)+f^{(4)}(C_2))$

But I didn't know how to solve the other part(I know simpson rule and trapeze rule)

答:
If $g \in C^2([a,b],\mathbb{R})$, then by the trapezoidal rule we have
$$ \int_a^b g(x)dx = \frac{h}{2} \sum_{j=0}^{n-1}  \left[ g(x_j) + g(x_{j+1}) \right] + O(h^2), \quad h \rightarrow 0, \quad nh = b-a > 0.$$
If $f \in C^4(\mathbb{R},\mathbb{R})$, then by Taylor's formula we have
$$ f''(x) = \frac{f(x+h) - 2f(x)+f(x-h)}{h^2} + O(h^2), \quad h \rightarrow 0, \quad h \not = 0.$$
We will now combine these results. If $f \in C^4(\mathbb{R},\mathbb{R})$, then
$$ g(x) = f(x)f''(x) = f(x)\frac{f(x+h) - 2f(x)+f(x-h)}{h^2} + O(h^2), \quad h \rightarrow 0, \quad h \not = 0.$$
If $[a,b] \subset \mathbb{R}$, then we certainly have
$$ \int_a^b f(x) f''(x) = \frac{h}{2} \sum_{j=0}^{n-1} \left[f(x_j) \frac{f(x_{j+1}) - 2f(x_j)+f(x_{j-1})}{h^2} + f(x_{j+1}) \frac{f(x_{j+2}) - 2f(x_{j+1})+f(x_{j})}{h^2} \right]+ O(h^2), \quad h \rightarrow 0, \quad nh = b-a.$$

There are some concerns though. The original problem statements does not stipulate that $f$ is defined outside $[a,b]$, but makes explicit references to the value of $f$ at $a-h$ and at $b+h$. I have circumvented this issue by assuming that $f \in C^4(\mathbb{R}, \mathbb{R})$. If $f$ is not defined outside $[a,b]$, then do the partial integration
$$ \int_a^b f(x) f''(x) = [f(x)f'(x)]_a^b - \int_a^b f'(x)^2 dx$$
and use the midpoint method to evaluated the last integral exploiting that
$$ f'(t) = \frac{f(t+h/2) - f(t-h/2)}{h} + O(h^2), \quad h \rightarrow 0, \quad t \pm h/2 \in [a,b].$$
Assuming that $f \in C^7$ is unnecessary for the portion of the problem that you have presented. It may be necessary for the rest of the problem.

Title: approximate $\int_{a}^{b}f(x)f''(x)dx$


答:
Compactness is a property that a topological space has on its own, it is not relative to some other space in which it is embedded, unlike closedness or openness. Therefore, in this case, $B$ is a closed subset of $A$, which is compact, so $B$ is also compact, and this fact is not influenced from $B$ being a subspace of $A$ or $X$. To clarify, recall the definition of compactness:

A topological space $(X,\tau)$ is compact if, for every collection $\mathcal A$ of open subsets of $X$ whose union is $X$ ($\mathcal A$ is thus called an open cover of $X$), there exist finitely many elements of $\mathcal A$ whose union is still $X$ (we say that $\mathcal A$ has a finite subcover).

Now, if you have a subset $Y$ of $X$, you can define a similar notion of compactness for $Y$:

We say that $Y$ is a compact subset of $(X,\tau)$ if, for every collection $\mathcal A$ of open subsets of $X$ whose union contains $Y$, there exist finitely many elements of $\mathcal A$ whose union still contains $Y$.

Actually, it turns out that, if we equip $Y$ with the subspace topology induced by $(X,\tau)$, the two notions are equivalent, that is:

$Y$ is a compact subset of $(X,\tau)\iff Y$, with the topology induced by $(X,\tau)$, is compact.

You can try to prove this last result by yourself as an exercise, you need to use the definition of induced topology. I hope this clarifies why compactness is a property of a space, that does not change if you embed it in some other space.

Title: A closed set inside a compact is compact

Body:
So I know that if $(X, \tau)$ is compact and $A$ is a closed set of $\tau$, then $A$ is compact. However if $(X, \tau)$ is not compact and $A$ is a compact subset of $X$, and $B \subset A$ where $B$ is a closedset, then $B$ is a compact of $A$. Is it a compact of $(X, \tau)$ though?

Note: When I talk about $A$ as a topological space, it has the topology induced by $\tau$


问:
Why $F(\alpha_1)∩F(\alpha_2)=F$ is false
$\def\Q{\mathbb Q}$Let $\alpha_1$ and $\alpha_2$ be conjugates over a field $F$ such that $\alpha_2 \not\in F(\alpha_1)$. Is true that  $F(\alpha_2)\cap F(\alpha_2)=F$.

Here is my attempt:

Is false. Let $F= \Q(\sqrt 2)$. We know that $\sqrt[3]2$ , $\omega\sqrt[3]2$, $\omega^2\sqrt[3]2$ are roots of the polynomial $x^3-2$. But my doubt is how to prove that $\Q(\sqrt2,\sqrt[3]2)∩\Q(\sqrt2,\omega \sqrt[3]2)≠\Q(\sqrt2)$?

答:
I'm not sure your example works, but here's one that does:

Consider any field $F$ and any polynomial $f(x)\in F[x]$ which has Galois group with at least one pair of elements $g,h$ which do not generate the group or each other. Then the fixed fields of $g$ and $h$ intersect in a proper extension of $F$. By the Primitive Element Theorem, these fixed fields are simple, i.e. we have $\alpha_1,\alpha_2$ such that they are $F(\alpha_1)$ and $F(\alpha_2)$ respectively.

For a concrete example, let $F=\mathbb Q$ and $f(x)=(x^2-2)(x^2-3)(x^2-5)$. This has Galois group $C_2\times C_2\times C_2$, so any pair of distinct nonzero elements satisfies the desired condition. Choosing the automorphisms
$$a+b\sqrt2+c\sqrt3+d\sqrt5\mapsto a-b\sqrt2+c\sqrt3+d\sqrt5$$
$$a+b\sqrt2+c\sqrt3+d\sqrt5\mapsto a+b\sqrt2-c\sqrt3+d\sqrt5$$
gives fixed fields $\mathbb Q(\sqrt3,\sqrt5)$ and $\mathbb Q(\sqrt2,\sqrt5)$ which intersect in $\mathbb Q(\sqrt5)$. If you'd like, you can show that $\mathbb Q(\sqrt3,\sqrt5)=\mathbb Q(\sqrt3+\sqrt5)$ and $\mathbb Q(\sqrt2,\sqrt5)=\mathbb Q(\sqrt2+\sqrt5)$.


问:
$(A-B) \cup (B-A) = (A\cup B) - (A \cap B)$
So it is as it seen in the title. Here is why I am stuck:

Let $x \in [(A-B) \cup (B-A)] \Longrightarrow x \in (A-B)$ or $x \in (B-A) $ $\Longrightarrow$ ($x \in A$ and $x \not \in B$) or ($x \not \in A$ and $x \in B$).

Here is where I am stuck. Can I product these parenthesis? Like:

(x in A or x not in A) and ( x in A or x in B) and (x not in B or x not in A) and (x not in B or x in B).

And please, how I can learn write with symbols? Cant find the button. So sorry.

答:
Instead of "multiplying the parentheses" you can see each case separately:

If $x\in A\text{ and }x\not\in B$, then $x\in A\cup B$ (because $A$ is a subset of $A\cup B$ and $x\in A$) and $x\not\in A\cap B$ (because $A\cap B$ is a subset of $B$ and $x\not\in B$), therefore $x\in (A\cup B)-(A\cap B)$.

If $x\not\in A\text{ and }x\in B$, then $x\in A\cup B$ (because $B$ is a subset of $A\cup B$ and $x\in B$) and $x\not\in A\cap B$ (because $A\cap B$ is a subset of $A$ and $x\not\in A$), therefore $x\in (A\cup B)-(A\cap B)$.

In both cases you conclude that $x\in (A\cup B)-(A\cap B)$, which proves that every element of $(A-B)\cup (B-A)$ is and element of $(A\cup B)-(A\cap B)$.

The only thing left to prove that these two sets are equal is the converse, i.e that every element of $(A\cup B)-(A\cap B)$ is an element of $(A-B)\cup (B-A)$.