问:
Difference of the Stirling cycle numbers and the Stirling set numbers
Denote by $\left\langle\!\! \left\langle k\atop  j\right\rangle\!\! \right\rangle$ the second-order Eulerian numbers A340556. Define
$$ \left| n\atop  k\right| = \sum_{j=0}^k \left( \binom{n + j - 1}{2k} - \binom{n + k - j}{2k} \right) \left\langle\!\! \left\langle k\atop  j\right\rangle\!\! \right\rangle .$$
Observe that the difference between the Stirling cycle numbers and the Stirling set numbers can be expressed as:
$$ \left[ n\atop k\right] -  \left\{ n\atop  k\right\} = \left| n\atop n- k\right| \quad (0 \le k \le n)$$

It is surprising that the difference between the two Stirling numbers had no entry in the OEIS -- until today! Now they are A341102. But what we still need is an independent combinatorial interpretation of these numbers. Who knows one?

Who is willing to delight the combinatorists among us with a proof of this identity?

答:
We can start from Concrete Mathematics, page 271, Equations (6.43) and (6.44). They are said to be obtained by induction on $n$, the second order Eulerian numbers are supposed to be defined by their basic recursion. With a slight modification, accounting for the difference in the definition of the second order Eulerian numbers between Concrete Mathematics and the OEIS, we have
$${x \brace x-k}= \sum_j \Big<\Big< \begin {align*} &k\\ j&+1 \end{align*}   \Big>\Big>{x+k-1-j\choose 2k} $$
$${x \brack x-k}= \sum_j \Big<\Big< \begin {align*} & k\\ j&+1 \end{align*}   \Big>\Big>{x+j\choose 2k} $$
Then $$ \left| x\atop  k\right| = {x \brack x-k}-{x \brace x-k}= \sum_j \Big<\Big< \begin {align*} &k\\ j&+1 \end{align*}   \Big>\Big>\left({x+j\choose 2k}-{x+k-j-1\choose 2k} \right) $$
Then $$ \left| x\atop  k\right| = \sum_j \Big<\Big< \begin {align*} &k\\ &j \end{align*}   \Big>\Big>\left({x+j-1\choose 2k}-{x+k-j\choose 2k} \right) $$


问:
Length of a line inside a rectangle
Given the width and height of a rectangle, and equation of a line, how can I efficiently find the total length of the line when it is inside the rectangle?

The rectangle is at the origin of thethe referencial.

Input 800 widht 400 height and a line y=-0.5x + 375 would give me length 838,53.

I have tried intersecting with y=0 and y=H to find the intersection points but it doenst work sometimes. Can someone help me? Thanks

答:
The rectangle has four corners, numbered as $(0,1,2,3)$.
The equation of a straight line divides the plane into a negative ( ≤ ) and a positive ( > ) region.
The line itself is the place where the equation equals zero ( ≤ ). Each of the corners of the rectangle
is in the negative ( ≤ ) or in the positive ( > ) region of the straight line.
A full solution of the problem requires that a number of different cases shall be distinguished.
A picture says more than a thousand words.

Mapping to the binary number system ensures that none of these sixteen cases will be omitted by accidence.
Employing symmetry, it turns out that the 16 cases can be collected into three classes.

```
   binary   3 2 1 0  class
 0  0000    ≤ ≤ ≤ ≤    0
 1  0001    ≤ ≤ ≤ >    1
 2  0010    ≤ ≤ > ≤    1
 3  0011    ≤ ≤ > >    2
 4  0100    ≤ > ≤ ≤    1
 5  0101    ≤ > ≤ >    2
 6  0110    ≤ > > ≤    0
 7  0111    ≤ > > >    1
 8  1000    > ≤ ≤ ≤    1
 9  1001    > ≤ ≤ >    0
10  1010    > ≤ > ≤    2
11  1011    > ≤ > >    1
12  1100    > > ≤ ≤    2
13  1101    > > ≤ >    1
14  1110    > > > ≤    1
15  1111    > > > >    0
```

Now it's only a matter of finding the intersection of two lines for the classes (1) and (2), none for class (0).
Accompanying source code will be available at MSE publications / references 2021.
Here is some output, augmented with numbering of the 3 classes.

ContentLicense: CC BY-SA 3.0

Title: Length of a line inside a rectangle

Body:
The rectangle has four corners, numbered as $(0,1,2,3)$.
The equation of a straight line divides the plane into a negative ( ≤ ) and a positive ( > ) region.
The line itself is the place where the equation equals zero ( ≤ ). Each of the corners of the rectangle
is in the negative ( ≤ ) or in the positive ( > ) region of the straight line.
A full solution of the problem requires that a number of different cases shall be distinguished.
A picture says more than a thousand words.

Mapping to the binary number system ensures that none of these sixteen cases will be omitted by accidence.
Employing symmetry, it turns out that the 16 cases can be collected into three classes.

   binary   3 2 1 0  class
 0  0000    ≤ ≤ ≤ ≤    0
 1  0001    ≤ ≤ ≤ >    1
 2  0010    ≤ ≤ > ≤    1
 3  0011    ≤ ≤ > >    2
 4  0100    ≤ > ≤ ≤    1
 5  0101    ≤ > ≤ >    2
 6  0110    ≤ > > ≤    0
 7  0111    ≤ > > >    1
 8  1000    > ≤ ≤ ≤    1
 9  1001    > ≤ ≤ >    0
10  1010    > ≤ > ≤    2
11  1011    > ≤ > >    1
12  1100    > > ≤ ≤    2
13  1101    > > ≤ >    1
14  1110    > > > ≤    1
15  1111    > > > >    0

Now it's only a matter of finding the intersection of two lines for the classes (1) and (2), none for class (0).
Accompanying source code will be available at MSE publications / references 2021.
Here is some output, augmented with numbering of the 3 classes.

ContentLicense: CC BY-SA 4.0


问:
Finite groups such that $H^1(G,M)=0$ for any simple $G$-module $M$
I'm trying to understand for which finite groups $G$ the augmentation ideal of $\mathbb{F}_2G$ is generated by a single element over $\mathbb{F}_2G$.

I'm reading a paper with a result that implies that such groups have $H^1(G,M)=0$ for any simple $G$-module $M$. This can help me narrow down the search.

What are the finite groups $G$ such that $H^1(G,M)=0$ for any nontrivial simple $\mathbb{F}_2G$-module $M$?

答:
A finite group $G$ has $H^1(G,M)=0$ for all non-trivial simple $kG$-modules for $k$ a field of characteristic $p$ if and only if $G$ has a normal subgroup of order coprime to $p$ and index a power of $p$ [iff $G/O_{p'}(G)$ is a $p$-group, iff $G$ has a normal $p$-complement, iff $G$ is $p$-nilpotent].

For instance the non-abelian group of order 6 is like this mod 2. In other words, the order of the group does not need to be odd, but all of the evenness has to be on top. Every finite group with a cyclic Sylow 2-subgroup is like this.

The relationship between $H^1(G,M)$ and the second-layer composition factors of $P_1 = P(k)$ is Theorem A.15.11 in Doerk–Hawkes. The main claim follows from Theorem B.4.24, since, as you noticed in your followup question, groups like this have only the trivial simple module in the principal block. You will also need a better definition of block, such as Proposition IV.13.3 in Alperin's Local Representation Theory.

Generalizable statements

In case you might want to relax your requirements some day (I don't think you'll want to for this specific problem, but I've already got the references handy): In a $p$-nilpotent group, all the $p$-chief factors live up on top (of the $p'$-part) and so have to act trivially. In general, a $p$-chief factor always lies in the first block, and the vanishing of the cohomology has a bit to do with whether it is complemented. Now a module can be in the first block without being a composition factor of $P_1 = P(k)$, but in a $p$-soluble group, the $p$-chief factors must in fact be composition factors of $P_1 = P(k)$, by Theorem B.6.17.

Doerk, Klaus; Hawkes, Trevor.
Finite soluble groups.
de Gruyter Expositions in Mathematics, 4. Walter de Gruyter & Co., Berlin, 1992. xiv+891 pp. ISBN: 3-11-012892-6
MR1169099

Alperin, J. L.
Local representation theory.
Modular representations as an introduction to the local representation theory of finite groups.
Cambridge Studies in Advanced Mathematics, 11. Cambridge University Press, Cambridge, 1986. x+178 pp. ISBN: 0-521-30660-4; 0-521-44926-X
MR860771
DOI:10.1017/CBO9780511623592


问:
Best books for self-studying differential geometry
Next semester (fall 2021) I am planning on taking a grad-student level differential topology course but I have never studied differential geometry which is a pre-requisite for the course. My plan is to self-study in the summer and this semester so that I do not have to waste a semester taking a differential geometry course which will ruin my schedule.

I am looking for a book that covers all of the following topics (ideally) but at least most of them:

any suggestions would be welcome.

答:
Make sure your vector calculus is fully mastered, and your linear algebra, at least at the level of acing an intro course in each. Solid foundations make all the difference.

My experience with Michael Spivak's Differential Geometry was unpleasant and required a lot of use of his Calculus on Manifolds, which some love but I strongly do not recommend.

A big book meant for physicists is Misner Thorne and Wheeler's Gravitation, which I think is decent for intuition building.
Unfortunately I haven't found a book on Differential Geometry that I actively like; MTW comes closest.


问:
Question about the comparison test for factorial series
I'm trying to determine if $\displaystyle \sum_{n=1}^{\infty} \frac{n!}{n^n}$ converges or diverges using only the comparison test (I am aware I can do this using the ratio test quiet easily).

The explanations I found were mostly:

$$\frac{n!}{n^n}=\frac{1}{n}\cdot\frac{2}{n}\cdot\frac{3}{n}\cdot ... \cdot \frac{n}{n}\leq
\frac{1}{n}\cdot\frac{2}{n}\cdot1\cdot ... \cdot 1= \frac{2}{n^2}$$

My professor does this too:

My question is why do you keep the first two terms? In other words, why do you keep only the $\frac{1}{n}$ and $\frac{2}{n}$? Why not keep $\frac{1}{n}$,$\frac{2}{n}$,$\frac{3}{n}$ instead? Is there a reason that only the first two terms were kept? Thanks!

答:
We know that
$\displaystyle \sum_{n=1}^\infty \frac{k}{n^2}$ converges (indeed it is equal to $\frac{k\pi^2}{6}$). Therefore, if we can show that that $\frac{n!}{n^n} < \frac{k}{n^2}$ for some $k$, we have, by the comparison test, that $\displaystyle \sum_{n=1}^\infty\frac{n!}{n^n}$ converges too.

So why do we take the first two terms?

We cannot simply take the first term, because then our upper bound would be $\frac{1}{n}$. This is no good, since the harmonic series diverges. We have shown that that the proof works with the first two terms. The reason we didn't take the first three terms was that we simply didn't need to! You could have taken the first three terms, and the proof would have been equally as valid, but we might as well take the fewest terms necessary.


问:
Find the restrictions on the angles $\theta$ and $\varphi$ to the spherical surface $x^{2}+y^{2}+z^{2}=4$, z≥1
I have a portion of a sphere defined by $x^{2}+y^{2}+z^{2}=4$,  z ≥ 1, and want to change into spherical coordinates and compute the area of over the surface. For this I need to find the restrictions on the angles $\theta$ and $\varphi$. I know that the radius of the sphere is 2, and the the projection onto the xy plane gives a circle with radius $\sqrt3$.

I am having a hard time imaging in geometrically, how can I find the restrictions on the angles? Any help would be appreciated!

答:
Given radius of the sphere is $2$, its surface can be parametrized in spherical coordinates as,

$x = 2 \cos \theta \sin\phi, y = 2 \sin \theta \sin \phi, z = 2 \cos \phi$

As we need to find surface area of the sphere for $z \geq 1$,

$\cos \phi \geq \frac{1}{2} \implies \phi \leq \frac{\pi}{3}$.

So limits of double integral to find surface area are $0 \leq \theta \leq 2\pi$ and $0 \leq \phi \leq \frac{\pi}{3}$.


问:
Finding initial velocity with only acceleration and distance
I’m having a hard time trying to figure this question out.

With what initial velocity must an object be thrown upward (from ground level) to reach the top of the Washington Monument, which has a height of approximately 555 feet? (Use -32ft/sec^2 as acceleration due to gravity)

答:
Let $ H $ be the height of the monument.

The object reaches the top, when its velocity $ V_1=0 $.

Let  $ V_0 $ be the initial velocity.

You just need and you should, know that

$$V_1^2-V_0^2 = -2g(H-0)= -V_0^2$$

thus

$$\boxed{V_0=\sqrt{2gH}}$$

with $$g=32 \text{ and } H=555$$

Remark :

Other way to get the formula above.
It attains the top, after a time $ T $ such that
$$H=-\frac 12 gT^2+V_0T$$

with

$$V_1=-gT+V_0=0$$
So

$$T=\frac{V_0}{g}$$
and
$$H=-\frac 12 g(\frac{V_0}{g})^2+V_0.\frac{V_0}{g}$$
$$=\frac{V_0^2}{2g}$$


问:
Trying to prove by induction that $\sum_{i=1}^{k} F_{2i}=F_{2k+1}-1$ for all $k∈N$
I am trying to prove by induction the following proposition of the fibonacci sequence:

The fibonacci sequence is defined recursively as: $f_1 = 1$, $f_2 = 1$, and that for all integers $k>=1$, $f_{k+1} = f_{k+1} +f_k $

Propostion I am trying to prove:

$\sum_{i=1}^{k} F_{2i}=F_{2k+1}-1$ for all $k∈N$

So far I have this:

Base case:

For $F_1$:
$\sum_{i=1}^{k} F_{2i}$ = $F_{2(1)}$ = $F_{2}$  This is true.

Induction Hypothesis:

$\sum_{i=1}^{n+1} F_{2i}$ = $F_{2(n+1)}$

I am not sure if what I have so far is correct.

I have brainstormed this also:

That $f_2 = f_3 - 1$. And that $f_2\:+\:f_4\:+\ldots+f_{2n}=f_{2n+1}-1$. By adding $f_{2n+2}$ to both sides, you get
$f_2\:+\:f_4\:+...+f_{2n} + f_{2n+2}= f_{2n+2}+f_{2n+1}-1$. And $f_{2n+2}+f_{2n+1} = f_{2n+3}$.

I am not sure how to prove the proposition by induction properly. I have a general understanding of how induction works and why the proposition is true, but I am having trouble proving it with induction steps. I think I need to somehow get to $f_{2n} = f_{2(n+1)+1}-1$, but I am struggling with that. I haven't practiced much strong induction yet, so I am only looking for a simple proof. To provide context as to why I am doing this exercise: I am planning to use this proof as a base for me to refer back to when doing more complex proofs that also involve the fibonacci sequence.

Thank you for all the help or any proofs provided.

答:
We want to prove $\varphi(n)$ for all $n\ge1$, with $\varphi(n)$ meaning $\sum_{i=1}^nF_{2i}=F_{2n+1}-1$. The principle of mathematical induction states that if $\varphi(1)$ (base case) and $\forall k\ge1(\varphi(k)\to\varphi(k+1))$ (induction step), $\forall n\ge1(\varphi(n))$. Your calculations have proved the base case $F_2=F_3-1$ and the induction step viz.$$\sum_{i=1}^kF_{2i}=F_{2k+1}-1\implies\sum_{i=1}^{k+1}F_{2i}=\sum_{i=1}^kF_{2i}+F_{2k+2}=F_{2k+1}-1+F_{2k+2}=F_{2k+3}-1,$$so you're done.

(The principle is usually stated as starting at $0$; $\varphi(0)$ equates an empty sum, i.e. $0$, to $F_1-1$, so we could have taken an $n=0$ base step. But based on your calculations I started at $n=1$, which is a common definition of $\Bbb N$.)


问:
Cubic Diophantine equations
We know how to solve Diophantine equations of first degree (linear) and of the second degree. What is available for solving Diophantine equations of the third degree? Let us say that we are given a two-variable polynomial equation P(x,y)=0 with integer coefficients, of third degree in both variables. What known methods or theories are used to solve such a general Diophantine equation of third degree?
Added: I am assuming the polynomial is genuine third degree (i.e. is irreducible) and that all powers of x and y may be present (alas second powers may be always eliminated by appropriate substitutions).</p>

答: I'll assume $P(x,y)$ is irreducible over the rationals.  The first thing you'll want to look at is its genus.  In many cases the genus will be $1$, meaning this is an elliptic curve.  Especially if it is Weierstrass form, there are ways to find the integer solutions (e.g. with Sage). If it isn't in Weierstrass form, you can put it in Weierstrass form with a rational transformation, look at the rational points of that transformed curve, and try to see which correspond to integer solutions of the original.

Title: Cubic Diophantine equations


问:
Is $\gcd(a,bc)=\gcd(a,b)\gcd({a\over\gcd(a,b)}, c)$?
Is it true that $\gcd(a,bc)=\gcd(a,b)\gcd({a\over\gcd(a,b)}, c)$?

It is true in quite a few examples that I came up with, e.g.

$a = 18, b = 21, c = 33$

$\gcd(18,21)\gcd({18\over\gcd(18,21)}, 33) = 3 \gcd({18 \over 3}, 33) = 3 \times 3 = 9$

$\gcd(18,693) = 9$

Here is my (probably not very good) attempt at a proof:

For a $b$ where $\gcd(a, b) = 1$

For a $c$ where $\gcd(a, c) = 1$

$\gcd(k_1a, k_1b) = k_1$

$\gcd(k_2a, k_2c) = k_2$

$\gcd(k_1k_2a, k_1k_2bc) = k_1k_2$

Hence, $\gcd(k_1k_2a, k_1b) \gcd({k_1k_2a \over \gcd(k_1k_2a, k_1b)}, k_2b) = k_1 \gcd({k_1k_2a \over k_1}, k_2c) = k_1k_2 = \gcd(k_1k_2a, k_1k_2bc)$

Is this statement true, or did I just stumble across several lucky coincidences?

答:
Yes, by basic gcd laws (distributive, commutative, associative, etc) we have
$$(a,b)\left(\frac{a}{(a,b)},c\right) = (a,(a,b)c) = (a,ac,bc) = (a(1,c),bc) = (a,bc) $$</p>


问:
Salt in tank DEQ. When is salt gone?

One tank contains 300 liters of water mixed with 50 g of salt. 3
liters of clean water are pumped in every minute. The mixed solution
is pumped out at 3 liters per minute. Determine a differential
equation for the amount of salt after $x$ minutes. When is all the
salt gone?

$$y=amount\ of\ salt\ by\ x\ minutes. \frac{salt\ gram}{water\ liter}\\y'=salt\ in - salt\ out\\salt\ in=0\\salt\ out=\frac{3y}{300}\\y'=-\frac{3y}{300}\Rightarrow y'+\frac{1}{100}\cdot y=0\Rightarrow y=e^{\frac{-x-c}{100}}$$

We know that $y(0)=\frac{50}{300}$, which gives $\frac{50}{300}=e^{\frac{-0-c}{100}}\Rightarrow c=100ln(6)$

$$\Rightarrow y=e^{\frac{-x-100ln(6)}{100}}$$

When is $y=0$? $$0=e^{\frac{-x-100ln(6)}{100}}\Rightarrow No\ solutions$$

What am I missing?

答:
I'm posting my edit, the answer, as the comment requested.

$$y=amount\ of\ salt\ by\ x\ minutes. \frac{salt\ gram}{water\ liter}\\y'=salt\ in - salt\ out\\salt\ in=0\\salt\ out=\frac{3y}{300}\\y'=-\frac{3y}{300}\Rightarrow y'+\frac{1}{100}\cdot y=0\Rightarrow y=e^{\frac{-x-c}{100}}$$

We know that $y(0)=\frac{50}{300}$, which gives $\frac{50}{300}=e^{\frac{-0-c}{100}}\Rightarrow c=100ln(6)$

$$\Rightarrow y=e^{\frac{-x-100ln(6)}{100}}$$

When is $y=0$? $$0=e^{\frac{-x-100ln(6)}{100}}\Rightarrow No\ solutions$$

Therefor the salt never truly disappear.

Title: Salt in tank DEQ. When is salt gone?


问:
Solution to Problem 10 in Reed-Simon, Chapter 3, Volume I.

The Question

The problem is from "Methods of Modern Mathematical Physics", Volume I, and it's closely related to the Hahn-Banach theorem. It asks to prove the following:

Corollary 3. Let Z be a subspace of a normed linear space X and suppose that y is an element of X whose distance from Z is d. Then there exists a $\Lambda\in X^*$ so that $||\Lambda||\le1$, $\Lambda(y)=d$, and $\Lambda(z)=0$ for all $z\in Z$.

To me this theorem seems like a sort of a normed space version of the Banach separation theorem for locally convex spaces. A proof of the latter can be found easily, but I couldn't make a connection between that argument and my problem. I gave it a go on my own, but I couldn't find a definitive proof.

My Understanding

Here is what I've come up with:

Let $Y$ be the subspace generated by $y$. We can define a bounded functional $\lambda:Y\longrightarrow\mathbb{C}$ by
\begin{equation}
\lambda(\alpha y) = \alpha d, \quad \forall\alpha\in\mathbb{C}.
\end{equation}
The Hahn-Banach theorem ensures there is a bounded linear functional $\Lambda$ defined on all $X$ which extends $\lambda$ and such that $||\Lambda||_{X^*}=||\lambda||_{Y^*}$ (Corollary 1 of Theorem III.6 in Reed-Simon).

However, there are two difficulties: for starters, I can't convince myself that $||\lambda||_{Y^*}\le1$: in fact
\begin{equation}
\frac{|\lambda(\alpha y)|}{||\alpha y||} = \frac{|\alpha|\cdot|\lambda(y)|}{|\alpha|\cdot||y||} = \frac{d}{||y||} \le \frac{||y-z||}{||y||} \le \frac{||y||+||z||}{||y||} = 1 + \frac{||z||}{||y||},
\end{equation}
where $z\in Z$. More importantly, I can't find a reason why $\Lambda$ should vanish on $Z$.

For completeness, here is the statement of the theorem I used:

Corollary 1. Let X be a normed linear space, Y a subspace of X, and $\lambda$ an element of $Y^*$. Then there exists a $\Lambda\in X^*$ extending $\lambda$ and satisfying $||\Lambda||_{X^*}=||\lambda||_{Y^*}$.

答: Instead, try the following: consider the subspace $E= Z+ Y$, and define $\lambda:E\to \Bbb{C}$ by $\lambda(z+ty)=td$. So, by construction, $\lambda$ vanishes on $Z$. Now, verify $\lambda$ is bounded by $1$ on $E$ and use Hahn-Banach to extend $\lambda$ to $X$. In fact, if $Z$ is closed and $y\notin Z$ then then you can even ensure the extension has norm exactly $1$.


问:
finding the piecewise linear function in a way that the results can be replicated or I can carry it forward to other problems
We have a parking deck that it costs 5cents per minute to park at. After 9:00 p.m it costs $3$ cents per minute to park on the deck.
We assume in this problem that parking starts at $8:00$ p.m.
The objective is to find the piecewise function and graph the equation.
So, the first thing I do is identify my points. with $(0,0) (60,3)$

With that my slope is $m = \frac{3-0}{60-0}$ therefore, $m = 0.05$

I take the point slope equation of y - y$_1$ = m(x - x$_1$)
Which is $y - 0 =0.05(x-0) $

So my first equation is $0.05x$

Next, I need to find out what happens when the time hits 9:00 p.m. More specifically, we go from

$0 \leq x \leq 60$

to

$60 \lt x \leq 120$

So when I look at this equation, right? I notice that 60 is not included, 60 is a time variable for use of the function when it has 5 cents per minute.

What I think I can do :

So, my first thought is to take the points and involve them in the slope, but not in the plotting of points in the equation.  So, I have the points $(60, 3),(120,3.6)$ However, my leftmost point is not included. So, rather than use (60, 3) for the slope equation, I can use (60,1.8) as this is where the function begins, and this is how it scales.

Specifically,  I create a slope $m = \frac{3.6-1.8}{120-60}$

so that  m = 0.03.

Next, I use the plot point line formula so that y - y$_2$= m(x-x$_2$)
But here, my $x_1$ and $y_1$ are different, since the slope itself  is 0.03,and uses the non-inclusive point of (60,1.8) to define its growth. But, it is between the points of (60, 3) and (120, 3.6)

So,

$y - 3 = 0.03(x - 60)$

$y - 3 = 0.03x - 1.8$

Then I add three to both sides

$y = 0.03x + 1.2$

Which is congruent with the answer key in the back of the book. Did I do this correctly?

答:
The answer to the direct question

Did I do this correctly?

is simply "yes", if what matters is the correct answer.

But it's rather longwinded. In each part of the problem you first found a second point on the graph using the given slope and then used that second point to calculate the slope you already knew.

Here is the first version of you question:
I can't find the piecewise linear function given the following set of dependent and independent variables.


问:
Continuous maps $f,g$ such that $\left.f\right|A = \left.g\right|A$. Show that $\left.f\right|\overline{A} = \left.g\right|\overline{A}$

Let $f,g : X \to Y$ be continous maps and $A \subset X$ such that $\left.f\right|A = \left.g\right|A$. Show that $\left.f\right|\overline{A} = \left.g\right|\overline{A}$

According to Wikipedia the restrictions can be represented as graphs the following way $G(\left.f\right|A)= \{(x,f(x)) \in G(f) \mid x \in A\}$ so I'm wondering if this should be proven by showing that $G(\left.f\right|\overline{A})=G(\left.g\right|\overline{A})$ e.g showing that two sets are equal or am I completely wrong here?

答:
Let $X=\Bbb R$ with standard topology and $Y=\Bbb R\cup \{*\}$. Let $\pi\colon Y\to X$, $y\mapsto \begin{cases}0&y=*\\y&y\ne *\end{cases}$. To turn $Y$ into a topological space, declare a set $U\subseteq Y$ open iff $\pi(U)$ is open in $\Bbb R$.
Define $f\colon X\to Y$, $x\mapsto \begin{cases}*&x=0\\x&x\ne 0\end{cases}$ and let $g$ be the inclusion map. For open $U\subseteq Y$, we have that $f^{-1}(U), g^{-1}(U)$ are $\pi (U)$ or $\pi(U)\setminus \{0\}$, so certainly open. Thus $f,g$ are continuous. With $A=\Bbb R\setminus\{0\}$, we have $f|_A=g|_A$, but as $\overline A=X$, we do not have $f|_{\overline A}=g|_{\overline A}$.


问:
An onto map to a Unit Circle $|z|=1$
An onto map from the critical line $$\frac{1}{2}+it , t\in \mathbb{R}$$ to the unit circle $$|z|=1$$

My try-
$$z=  \frac{\frac{1}{2}+it}{\frac{1}{2}-it} $$
$$|z|=1$$
But the map is not onto $z=-1$ is not attained

答: $f: t \mapsto e^{it}$ is an onto map from $\mathbb R$ to the unit circle. $g : \frac 12 + it \mapsto t$ is a bijective map from the critical line to $\mathbb R$. Their composition $f \circ g$ will be the function you're looking for.

Title: An onto map to a Unit Circle $|z|=1$


问:
Determinant of Vector
Is posible obtain the determinant of any vector?.How I will be able to obtain the determinant of any vector $v=[v_1,v_2,\cdots,v_n]\in \mathbb{R}^n$?</p>

答:
If you identify a vector $v \in \mathbb{R}^n$ with the linear mapping $$
  F_v \,:\, \mathbb{R}^n \to \mathbb{R}^n \,:\, x \to (v^T\cdot x, 0, \ldots 0)^T \text{,}
$$
i.e. if you identify the vector $v=(v_1,\ldots,v_n)$ with the matrix $$
  F_v = \begin{pmatrix}v_1 &\ldots &v_n\\ 0&\ldots&0\\\vdots&\ddots&\vdots\\0&\ldots&0\end{pmatrix} \text{,}
$$
you can say that $$
  \det v = \det F_v = \begin{cases}
    v_1 &\text{if $n=1$} \\
    0   &\text{otherwise.}
  \end{cases}
$$
Though whether that's particularly helpful is a different question..</p>

Title: Determinant of Vector


问:
Proof of an identity with e and ln?
So my instructor just wrote up that $x^{\epsilon} = (e^{\ln x})^{\epsilon}$

Hmmm...... That's true, I'm sure, but I couldn't prove it. Can anyone show me the necessary manipulations?

答:
You should know that $\ln(x)$ is the inverse function of $e^x$. That means that $\ln(e^x)=x$ and that $e^{\ln x}=x, x>0.$

Rewriting your expression: We have $(x)^\varepsilon=\left(e^{\ln x}\right)^{\varepsilon}$.

Title: Proof of an identity with e and ln?


问:
Show that $\|AB\|_p \le \|A\|_p \|B\|_p$

Show that $\|AB\|_p \le \|A\|_p \|B\|_p$ for any $p \in [1, \infty)$.

The definition I have is that $\|A\|_p = \underset{x \not= 0}{\sup}\dfrac{\|Ax\|_p}{\|x\|_p} = \underset{\|x\|_p = 1}{\max}\|Ax\|_p$.

According to the definition:
$$\|AB\|_p = \sup_{x \not = 0}\frac{\|ABx\|_p}{\|x\|_p} = \sup_{x \not = 0}\frac{\|ABx\|_p\|Bx\|_p}{\|Bx\|_p\|x\|_p} = \sup_{x \not = 0}\frac{\|Bx\|_p}{\|x\|_p} \frac{\|ABx\|_p}{\|Bx\|_p} \\ \le \sup_{x \not = 0}\frac{\|Bx\|_p}{\|x\|_p} \sup_{Bx \not = 0}\frac{\|ABx\|_p}{\|Bx\|_p} = \|B\|_p\|A\|_p.$$

Now I don't understand the inequality part. Why is $\underset{Bx \not= 0}{\sup}\dfrac{\|ABx\|_p}{\|Bx\|_p} \le \underset{x \not= 0}{\sup}\dfrac{\|Ax\|_p}{\|x\|_p}$?

答:
By your definition, if $x\neq 0$, then
$\frac{\|Ax\|_p}{\|x\|_p}\leq \|A\|_p $.
It follows that
$$\|Ax\|_p\leq\|A\|_p\|x\|_p\qquad \forall x.$$
Using the above,for $x\neq 0$, we have
$$\frac{\|ABx\|_p}{\|x\|_p}\leq \frac{\|A\|_p\|Bx\|_p}{\|x\|_p}\leq \frac{\|A\|_p\|B\|_p\|x\|_p}{\|x\|_p}=\|A\|_p\|B\|_p,$$
which implies $\|AB\|_p\leq \|A\|_p\|B\|_p$.


问:
Computing a singular integral
I would like to derive the exact value of the following integral

$$ I_s= \int_1^\infty r| r^{-2s}-  (r^2-1)^{-s}|d r\qquad \text{with}\qquad 0<s<1.$$

The existence of $I_s$ is warranted since around $r=\infty$ one should observe that

$$  r| r^{-2s}-  (r^2-1)^{-s}|= \frac{1}{r^{2s-1}} \Big|1-(1-\frac{1}{r^2})^{-s}\Big|\sim \frac{s}{r^{2s+1}}$$

and around $r=1$ we have

$$  r| r^{-2s}-  (r^2-1)^{-s}] \sim \frac{2^{-s}}{(r-1)^{s}}.$$

Title: Computing a singular integral

Body: $$I_s=\int_1^\infty\left(\frac{r}{(r^2-1)^s}-\frac{1}{r^{2s-1}}\right)dr=\frac{1}{2-2s}\Big((r^2-1)^{1-s}-r^{2-2s}\Big)\Bigg|_{r\to 1}^{r\to\infty}=\frac{1}{2-2s}.$$


答:
What the comment says is the following:

If $A\subseteq B$ then $\sup\{x:x\in A \}\leq \sup\{x : x\in B \}$.

Consider $x\in\mathbb{R}^n\backslash\{0\}$ and  $N,M\in\mathcal{M}_n(\mathbb{R})$, then we have $\Gamma=\{Mx\in \mathbb{R}^n\}\backslash \{0\}\subseteq \mathbb{R}^n\backslash \{0\}$, therefore $A=\{Ny:y\in \Gamma\}\subseteq \{Nx:x\in \mathbb{R}^n\backslash \{0\}\} =B$ for which $\sup A\leq \sup B$.

Body:
Show that $\|AB\|_p \le \|A\|_p \|B\|_p$ for any $p \in [1, \infty)$.

The definition I have is that $\|A\|_p = \underset{x \not= 0}{\sup}\dfrac{\|Ax\|_p}{\|x\|_p} = \underset{\|x\|_p = 1}{\max}\|Ax\|_p$.

According to the definition:
$$\|AB\|_p = \sup_{x \not = 0}\frac{\|ABx\|_p}{\|x\|_p} = \sup_{x \not = 0}\frac{\|ABx\|_p\|Bx\|_p}{\|Bx\|_p\|x\|_p} = \sup_{x \not = 0}\frac{\|Bx\|_p}{\|x\|_p} \frac{\|ABx\|_p}{\|Bx\|_p} \\ \le \sup_{x \not = 0}\frac{\|Bx\|_p}{\|x\|_p} \sup_{Bx \not = 0}\frac{\|ABx\|_p}{\|Bx\|_p} = \|B\|_p\|A\|_p.$$

Now I don't understand the inequality part. Why is $\underset{Bx \not= 0}{\sup}\dfrac{\|ABx\|_p}{\|Bx\|_p} \le \underset{x \not= 0}{\sup}\dfrac{\|Ax\|_p}{\|x\|_p}$?


问:
Poisson distribution with changing interval
I'm trying to learn how to use Poisson distribution for questions involving changing intervals. I have the following exercise:

Ron gets on average 5 points a day (24 hours) in his computer game. What is the probability for getting zero points during a) one day b) one afternoon (8 hours)?

So for a I have the following formula: $\frac{5^0e^{-5}}{0!}$

However when I try to adjust it to include only one third (8/24 hours), I encounter the situation where I should take a factorial of $\frac{5}{3}$, which is not integer. Do you have any tips how to proceed?

答:
Invoking the reproducibility property of poisson distribution, the number of point get in 8 hours follows a $Po\left(\frac{5}{3}\right)$

Thus

$$\mathbb{P}[X=0]=\frac{e^{-5/3}\left(\frac{5}{3}\right)^0}{0!}=e^{-5/3}$$


问:
Possible combinations of elements from different sets
I have the following problem. Let's start from a simple case. Suppose i have to sets, A and B with 4 and 5 elements respectively. I need to pick a certain number of elements from these two sets that match a certain value. For example, if I need to pick 7 elements, there are
$$
\binom{4}{2} \cdot \binom{5}{5} + \binom{4}{3} \cdot \binom{5}{4} + \binom{4}{4} \cdot \binom{5}{3} = 36
$$
possible combinations (the sum of the $k$'s in $\binom{n}{k}$ is 7). This case is simple. The problem I face is to find a general formula to find all the possible combinations of elements that sum to a given number. So, suppose I have N sets, each with $N_i$ elements. I need to select k total elements. How can i express this using a formula? Any hints or reference? This formula, when provided with 2 sets with 4 and 5 elements should output the previous equation, with the sum of the product of two binomal coefficients. Clearly, a formula with N sets will have a sum of a certain number of factors, each one with N binomial coefficients. What i'm looking for is the $k$'s for these N binomial coefficients. Hope this is clear.

答:
The $k_i$ values (meaning the number of elements you choose from the set $N_i$) come from the weak compositions of $n$. A composition of $n$ is a sequence of positive integers that add up to $n$. For example, the compositions of $3$ are $3$, $2+1$, $1+2$, and $1+1+1$. A weak composition is a composition allowing $0$ as a term. To count weak compositions, we need to bound the number of terms. The weak compositions of $3$ into three terms are $3+0+0,\ 0+3+0,\ 0+0+3,\ 2+1+0,\ 2+0+1,\ 0+2+1,\ 1+2+0,\ 1+0+2,\ 0+1+2$, and $1+1+1$.

The $k_i$ values that appear in your sum will form a weak composition of $k$ into $N$ terms. You can express this sum as $$\sum_{k_1+k_2+\cdots +k_N=k,\ k_i\geq 0}{N_1\choose k_1}{N_2\choose k_2}\ldots{N_N\choose k_N}.$$

The number of weak compositions of $k$ into $N$ terms is ${k+N-1\choose N-1}$, which you can show using the stars-and-bars method. This formula isn't too nice because you still have to compute the individual terms in the sum, but it at least tells you where the $k_i$ values come from.


答:
It is trivial if we argue using ``net''. We need the assumption
that $Y$ is Hausdorff.

Let $x\in\bar{A}$. Choose a net $(x_{\alpha})$ in $A$ such that
$x_{\alpha}\rightarrow x$. Define $y_{\alpha}=f(x_{\alpha})=g(x_{\alpha})$.
Since $f$ is continuous at $x$, we have that $y_{\alpha}\rightarrow f(x)$.
Since $g$ is also continuous at $x$, we have that $y_{\alpha}\rightarrow g(x)$.
Finally, that $Y$ is Hausdorff implies that the limit of the net
$(y_{\alpha})$ is unique. Hence, $f(x)=g(x)$.

If $Y$ is not Hausdorff, in general, we cannot conclude that $f\mid_{\bar{A}}=g\mid_{\bar{A}}$.
Counter-example: Let $X=\mathbb{R}$, equipped with the usual topology.
Let $Y=\mathbb{R}$, equipped with the trivial topology $\{\emptyset,\mathbb{R}\}$.
Define $f(x)=x$ and $g(x)=\begin{cases}
x, & \mbox{if }x<1\\
0, & \mbox{if }x\geq1
\end{cases}.$ Let $A=[0,1)$, then $\bar{A}=[0,1]$. Clearly, $f$ and $g$ are
continuous because the topology of $Y$ is the trivial topology. Moreover,
$f\mid_{A}=g\mid_{A}$. However, $f\mid_{\bar{A}}\neq g\mid_{\bar{A}}$
because $f(1)\neq g(1)$.

Title: Continuous maps $f,g$ such that $\left.f\right|A = \left.g\right|A$. Show that $\left.f\right|\overline{A} = \left.g\right|\overline{A}$

Body:
Let $f,g : X \to Y$ be continous maps and $A \subset X$ such that $\left.f\right|A = \left.g\right|A$. Show that $\left.f\right|\overline{A} = \left.g\right|\overline{A}$

According to Wikipedia the restrictions can be represented as graphs the following way $G(\left.f\right|A)= \{(x,f(x)) \in G(f) \mid x \in A\}$ so I'm wondering if this should be proven by showing that $G(\left.f\right|\overline{A})=G(\left.g\right|\overline{A})$ e.g showing that two sets are equal or am I completely wrong here?


问:
Approximating continuous functions $S^n \to S^n$
I'm trying to check that every continuous function $f:S^n \to S^n$ can be approximated by differentiable ones. Well, by Stone-Weierstrass I can approximate the coordinate functions $f_i:S^n \to \Bbb R$ by differentiable ${\tilde f_i}:S^n \to \Bbb R$. The problem is, of course, I can't guarantee ${\tilde f}:=({\tilde f}_1,\dots,{\tilde f}_{n+1})$ to be in $S^n$. To solve this i thought of projecting ${\tilde f}(p)$ to the point in the sphere that minimizes the distance ${\tilde f}(p)$ to $S^n$, this projection would still approximate $f$ but I see no reason for this projection to be differentiable. Any hint?

Sorry, by approximate I mean $\sup_{x \in S^n} |{\tilde f}(x) - f(x)| < 1$ (I don't know if $1$ is important here)

答: This depends on what you mean by "approximating".  If you mean "up to homotopy", i.e. a continuous deformation, then the result is immediate from the calculation of the homotopy group $\pi_n(S^n)=\mathbb{Z}$.  Thus, every class can be represented by a specific function obtained as a suitable smooth suspension of $z\mapsto z^k$ on the unit circle.</p>


问:
Prove that $\lim_{n\to\infty}\sin(nx)/ \pi x$ is a delta function
I need to prove that $\delta_n(x)=\sin (nx)/\pi x$ is a delta function. That is, to prove that: $$\underset{n\rightarrow \infty}{\lim}\int^\infty_{-\infty}f(x)\frac{\sin( nx)}{\pi x}dx=f(0).$$ For that I made $y=x/n$ and took the limit under the integral since $f(x)$ is supposed to be analytic in $\mathbb{R}$. I have a fundamental limit for the $\sin$ leaving me with $$\int^\infty_{-\infty}f(0)\cdot1 \;dy.$$ But this is equal to $f(0)$ $\iff$ $f(x)$ have contribuition only at $x=0$, but here the only assumption is that $f(x)=0$ when $x\rightarrow \pm \infty$.

What is wrong here?

答: Since $\delta_n(x)=n\delta_1(nx)$,$$\begin{align}\int_{\Bbb R}\delta_n(x)f(x)dx&=\int_{\Bbb R}\delta_1(y)f(y/n)dy\\\implies\lim_{n\to\infty}\int_{\Bbb R}\delta_n(x)f(x)dx&=\lim_{n\to\infty}\int_{\Bbb R}\delta_1(y)f(y/n)dy\\&\stackrel{*}{=}\int_{\Bbb R}\delta_1(y)\lim_{n\to\infty}f(y/n)dy\\&=f(0)\underbrace{\int_{\Bbb R}\delta_1(y)dy}_{1},\end{align}$$provided $f$ has the right behaviour for $\stackrel{*}{=}$ to be legal. For example, while $f$ being a Schwartz function suffices, $f$ being analytic doesn't.


问:
How to sum $\sum_{x=0}^{N-1} \cos{\left( \frac{2\pi x}{N} \right)}^{2}$?
I've tried solving it, but I'm not familiar with these kind of progressions and I couldn't get to a result. Also, I've found that the result is just $N/2$, but I don't get why would that be the answer, as it doesn't work for $N=1$ or $N=2$.

Appreciate your help :)

答:
Not for finite $N$, but a cute and fast way to see that asymptotically the result is true, and thus gives a simple sanity check as to "why [that] would be the answer."

If you are familiar with Riemann sums: you can write
$$
\sum_{k=0}^{N-1} \cos\!\left(\frac{2\pi k}{N}\right)^2=
N\cdot \frac{1}{N}\sum_{k=0}^{N-1} \cos\!\left(\frac{2\pi k}{N}\right)^2
=N\cdot \frac{1}{N}\sum_{k=0}^{N-1} f\!\left(\frac{k}{N}\right)
$$
for $f\colon[0,1]\to\mathbb{R}$ defined by $f(x) = \cos(2\pi x)^2$. Now, under mild assumptions easy to check here, you get
$$
\frac{1}{N}\sum_{k=0}^{N-1} f\!\left(\frac{k}{N}\right) \xrightarrow[N\to\infty]{} \int_0^1 f(x)dx
$$
so you just need to compute
$$
\int_0^1 \cos(2\pi x)^2dx = \frac{1}{2}.
$$

This last part follows from (1) $
\int_0^1 \cos(2\pi x)^2dx + \int_0^1 \sin(2\pi x)^2dx = 1
$ (since $\cos^2+\sin^2=1$)
and (2) $\int_0^1 \cos(2\pi x)^2dx = \int_0^1 \sin(2\pi x)^2 dx $ (e.g., via integration by parts).


答:
Make your substitution, then break up your

$$\int_{-\infty}^\infty f\left(\frac{u}{n}\right) \frac{\sin u}{\pi u} d u$$

as $I_1 + I_2,$ where $I_1$ is the integral from $-\sqrt{n}$ to $\sqrt{n}$ and $I_2$ is the rest. The point is that $I_1$ converges to $$\int_{-\infty}^\infty f(0)\frac{\sin u}{\pi u} du = f(0),$$ and $I_2$ converges to $0.$

Title: Prove that $\lim_{n\to\infty}\sin(nx)/ \pi x$ is a delta function

Body:
I need to prove that $\delta_n(x)=\sin (nx)/\pi x$ is a delta function. That is, to prove that: $$\underset{n\rightarrow \infty}{\lim}\int^\infty_{-\infty}f(x)\frac{\sin( nx)}{\pi x}dx=f(0).$$ For that I made $y=x/n$ and took the limit under the integral since $f(x)$ is supposed to be analytic in $\mathbb{R}$. I have a fundamental limit for the $\sin$ leaving me with $$\int^\infty_{-\infty}f(0)\cdot1 \;dy.$$ But this is equal to $f(0)$ $\iff$ $f(x)$ have contribuition only at $x=0$, but here the only assumption is that $f(x)=0$ when $x\rightarrow \pm \infty$.

What is wrong here?


问:
Determining compactness of the set $S=\{\frac{1}{m}+\frac{1}{n}:m\in \mathbb N, n\in \mathbb N\}\cup \{0\}$ in metric space $\mathbb R$
Given a set $S=\{\frac{1}{m}+\frac{1}{n}:m\in \mathbb N, n\in \mathbb N\}\cup \{0\}$.

My attempt:

Claim $1:0$ is a limit point of set $S$.

Proof: Given any $\epsilon \gt 0, \exists N_\epsilon \in \mathbb N: \frac1{N_\epsilon}\lt \epsilon/2$. Choosing $m,n \ge N_\epsilon$,we have $\frac1{n}\le \frac1{N_\epsilon}\lt \epsilon/2$ and $\frac1{m}\le \frac1{N_\epsilon}\lt \epsilon/2 $. Thus $0\lt\frac{1}{m}+\frac{1}{n}\lt \epsilon$. Since $\epsilon \gt 0$ is arbitrary, we conclude that $0$ is a limit point of $S$.

Claim $2$: All elements of set $T=\{1/n: n\in \mathbb N\}$ are also limit points of $S$.

Proof: Given any $\epsilon \gt 0, \exists M_\epsilon \in \mathbb N: \frac1{M_\epsilon}\lt \epsilon\implies \frac1{m}\lt \epsilon \;\;\forall m\ge M_\epsilon\implies -\epsilon\lt \frac1{m}\lt \epsilon$. Hence for any $\frac1{n}\in T$, we have $\frac 1{n}-\epsilon\lt \frac 1{n}+ \frac1{m}\lt \frac1{n}+\epsilon$. Just as previosuly, it follows that all elements of $T$ are limit points of $S$.

Heine-Borel theorem states that a set $S$ in $\mathbb R$ is compact $\iff S$ is closed and bounded. Indeed, in the present case, $S$ is bounded as $2$ is one upper bound.

Since, $1/n \notin S$ for any $n\in \mathbb N$, $S$ is not closed and therefore, $S$ can't be compact.

Is my conclusion correct? Thanks.

Edit: How do I show that $S$ does not have limit points except $0$ and $\{1/n: n\in \mathbb N\}$?

答:
Consider this:

Let $x \not\in S$. We are going to show $x$ can't be a limit point of $S$.  As $S$ is bounded below by $0$ and above by $2$ we don't need so consider $x \le 0$ or $x \ge 2$.

So $x \ne \frac 1m = \frac 1{2m}+\frac 1{2m}$

Now $x > 0$ So there is a minimal $m \in \mathbb N$ so that $\frac 1m < x$.  If $m> 1$ then $\frac 1m < x < \frac 1{m-1}$ and if $m= 1$ then $\frac 1m = 1 < x$

Now $x -\frac 1m > 0$ so there is a minimal $n\in \mathbb N$ so that $\frac 1m + \frac 1n < x$.  And again if $n>1$ then $\frac 1m +\frac 1n < x < \frac 1m + \frac 1{n-1}$ and if $n=1$ then $\frac 1m + \frac 1n = \frac 1m + 1 < x$. (but it's not actually possible for $n = 1$)

It's easy to see that if $m =1$ then $\frac 1m + \frac 1n < x < 2$ so $n > 1=m$, and if $m> 1$ then $\frac 1m +\frac 1n < x < \frac 1{m-1}$ so $\frac 1n < \frac 1{m-1} + \frac 1m = \frac 1{(m-1)m} < \frac 1m$ and $n > m$.  SO $n > m\ge 1$.  So $\frac 1m +\frac 1n < x < \frac 1m +\frac 1{n-1}$

Now pick an $r$ so that $r < \min(x - (\frac 1m +\frac 1n),  (\frac 1m +\frac 1{n-1})- x)$

The $B_r(x)$ can't contain any points of $S$.

So $x$ is not a limit point of $S$ and then only limit points of $S$ (if any) are in $S$ so $S$ is closed.

(I say "if any" because we did nothing to find any limit points and we don't care.  If we actually cared we can see all $\frac 1m$ (including $1$) and $0$ are limit points [as there are infinitely many $\frac 1m + \frac 1n$ is any $B_r(\frac 1m)$] and for any $\frac 1m +\frac 1n\ne \frac 1t; t\in \mathbb N$ we can use a similar argument above is not t limit point.)

====

On the other hand a proof by finite subcover could be easier:

If $0 \in U$, an open set, then all but a finite number of $\frac 1t$ are in $U$.  For each of these finite number $\frac 1m \not \in U$ if $\frac 1m \in V_m$, then all but a finite number of $\frac 1m +\frac 1s \not \in V_m$. If each of these finite number of $\frac 1m + \frac 1{n_m}$ is in an open set $W_{m,n_m}$ then we have covered $S$.

For any open cover, $\mathscr U$, of $S$ we can find a $U\in \mathscr U$  so that $0 \in U$.  Then we can have a finite number of $\frac 1m \not \in U$ and we can find a finite number of $V_m \in \mathscr U$ so that $\frac 1m \in V_m$.  And for each of those we can find a finite number of $n_m$ so that $\frac 1m +\frac 1{n_m}\not \in V_m$ we can find a $W_{m,n_m}$ so that $\frac 1m +\frac 1{n_m} \in W_{m,n_m}$.

And $\{U\} \cup \{V_m\}\cup \{W_{m,n_m}\}$ is a finite subcover of $\mathscr U$.


问:
Spinning a disk and painting one half - probability whole disk is painted after $n$ steps?
This is a follow-up question based on a conjecture of mine of this question (which phrases the problem slightly differently but functionally identical).

Suppose we randomly spin a disk and then paint its right half, repeatedly. What is the probability $p(n)$ that after $n$ iterations the entire disk is painted?

To disambiguate, let $D$ be a unit-size disk centered on $(0, 0)$. At each step we rotate $D$ clockwise around its center by $\alpha \sim \mathcal{U}(0, 360)$ degrees, and then paint the semicircle where $x \geq 0$. We repeat this (with a new random $\alpha$ every time) until the entire disk is painted. Then $p(n)$ is the probability this process stops after $n$ iterations, and the probability we're interested in.

In the previous question it's proven that the expected number of iterations to fully paint the disk is $\sum_{n=1}^\infty p(n)n = 5$.

I conjecture that $p(n) = 2^{-n}(2n - 4)$ for $n \geq 3$ and $0$ otherwise, which perfectly matches my simulated data, and has the right expectation. But I haven't been able to prove it.

So is my conjecture correct and if yes what is a proof?

答:
Let $X_n$ be the nonpainted fraction of the circle after $n$ steps and let $Y_n=1-2X_n$. We have $Y_1=0$ and we easily find that, for $n\ge 1$, $Y_{n+1}=$

$Y_n$ with probability $Y_n/2$

$1$ with probability $Y_n/2$

A uniform random number in $[Y_n,1]$ with probability $1-Y_n$

For $n\ge 2$, let $f_n:(0,1)\to \mathbb{R}_+$ be the probability distribution of $Y_n$ (so that the integral over $(0,1)$ of $f_n$ is the probability that $Y_n\ne 1$). We have $f_2(y)=1$ for all $y$ and our description of $Y_{n+1}$ gives us the recursion:
$$f_{n+1}(y)=\frac{y}{2}f_n(y)+\int_0^y f_n(t) dt$$
This recursion is easily solved and we find (for $n\ge 2$) : $f_n(y)=\frac{n(n-1)}{2^n}y^{n-2}$.

Hence the probability that $Y_n=1$ (i.e. that the circle is fully painted after $n$ steps) is
$$1-\int_0^1 f_n(y)dy = 1- \int_0^1 \frac{n(n-1)}{2^n}y^{n-2} dy=1-\frac{n}{2^{n-1}}$$

So for the probability that exactly $n$ steps are needed (and not less), it is (for $n\ge 3$) :
$$\left(1-\frac{n}{2^{n-1}}\right)-\left(1-\frac{n-1}{2^{n-2}}\right)=\frac{n-2}{2^{n-1}}$$
as you conjectured.


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答: This is not a very specific answer but I was struck with awe when I saw the entanglement between partial differential equations and stochastic processes.</p>

LastEditorDisplayName: anon


问:
Is the adjoint operation WOT-WOT continuous?
This is a well-known property of the Hilbert-space adjoint operator that it is WOT continuous. Is a similar true for Banach spaces? That is, for a given Banach space $X$ is the operation

$\varphi\colon B(X)\to B(X^*)$, $\varphi(T)=T^*$

continuous with respect to WOT-topologies of $B(X)$ and $B(X^*)$? I guess it will be the case for $X$ reflexive.

答:
No.  (Of course it does work for reflexive spaces, as noted by Martin.)

Example.

$X = c_0$, $X^* = l_1$, $X^{**} = l_\infty$.

$T_n : c_0 \to c_0$ is the left-shift by $n$, so that
$$
(a_1,a_2,\dots) \mapsto (a_{n+1},a_{n+2},\dots)
$$
where we deleted the first $n$ coordinates.

Compute $T_n^* : l_1 \to l_1$.  It is the right-shift by $n$, so that
$$
(a_1,a_2,a_3,\dots) \mapsto (0,0,\dots,0,a_1,a_2,\dots)
$$
where we inserted $n$ extra zeros at the beginning.

Then: $T_n \to O$ in the WOT for $B(c_0)$, where $O$ is the zero operator.  If $x=(x_1,x_2,\dots) \in c_0$ and $y = (y_1,y_2,\dots) \in l_1$, then
$$
\left|\langle T_nx , y\rangle\right| = \left|\sum_{k=1}^\infty x_{k+n} y_k\right| \le \|y\|_1 \left(\max_{j \ge n+1} |x_j|\right) \to 0 .
$$

But $T_n^*$ does not converge to $O$ in WOT for $B(l_1)$.  Indeed, take $y \in l_1$ with positive coefficients, and $z = (1,1,\cdots) \in l_\infty$.  Then
$$
\langle z,T_n^* y\rangle = \sum_{k=n+1}^\infty y_{k-n} z_k = \|y\|_1
$$
for all $n$ and does not go to zero.


答: Typically we think of functions being inverses as a path towards simplifying an expression. That is, if I see $e^{\ln(x)}$ in the wild I can just as soon replace it with $x$, because, as we all know $e^{\ln(x)} = x$. However, this equality is just as true in reverse, $x = e^{\ln(x)}$. I can take the simpler $x$ and replace it with the more complicated $e^{\ln(x)}$ to, hopefully, (ab)use some properties of $e$ or of $\ln$ to solve whatever problem is at hand. Your professor has done exactly this.

Title: Proof of an identity with e and ln?

Body:
So my instructor just wrote up that $x^{\epsilon} = (e^{\ln x})^{\epsilon}$

Hmmm...... That's true, I'm sure, but I couldn't prove it. Can anyone show me the necessary manipulations?


问:
Show that $f^{n}(x)\xrightarrow{n} x_0\forall x\in X$
$\text{Let $(X,d) $ be a compact metric space and $f:X\longrightarrow X$ a continuous operator}$

$\text{such that } d\big(f(x),f(y)\big)<d(x,y)\text{ } \forall x,y\in  X, x\neq y.$

$\text{I know that there exists a unique fixed point $x_0\in X: f(x_0)=x_0$},$

$\text{however, how can we  show that:  }$
$$\lim_{n\to\infty}f^n(x)=x_0 \text{ }\forall x\in X$$
$\text{Where }f^n(x):=f\color{black}{\underbrace{ ofofo\dots o f(x)  }_{ \text{$n$ times }  }}.$

$$\bullet \text{ My approach: $X$ is compact, hence $\exists$ subsequence $f^{k_n}(x) :k_n\ge n$ s.t}$$
$$f^{k_n}(x) \longrightarrow a\in X,\text{ $f$ is continuous, thus $f^{k_n +1}(x)\longrightarrow f(a)\implies f(a)=a\implies a=x_0$}$$

$\text{How can i continue from this point on, any ideas?}$

答:
You're almost there: To finish you can use the Uryshon subsequence principle which can be found here: https://www.math.ucla.edu/~hmkhang24/131BHW19/Notes.pdf

Indeed for any $x \in X$ take the sequence $(f^n(x))_k$. Take a subsequence $(f^n_k(x))_k$. Then with this subsequence you use your argument (compactness $\implies$ convergent subsequence which converges to $x_0$).
Then use Uryshon's subsequence principle (also called subsubsequence principle) since you took an arbitrary subsequence of the original sequence.

Alternatively you can argue by contradiction (which amounts to using the subsequence principle). Suppose your sequence does not converge to $x_0$, then there is a subsequence which does not converge to $x_0$, for some $x \in X$. But this subsequence has a convergent subsequence due to compactness and then you use the argument that you gave.


问:
Prove that if a transformation preserves inner product, then it is a linear transformation.
If a transformation T preserves inner product, then T it is a linear transformation.

答: Such a transformation sends $0$ to $0$, preserves the norm, and preserves orthogonality, thus it sends the standard basis  $e_1, \dotsc, e_n$ to an orthonormal frame, $d_1, \dotsc, d_n$ and now write an arbitrary vector $v = \sum a_i e_i.$ It will be mapped to $\sum a_i d_i.$ (since $\langle f(v), f(e_i)\rangle = \langle v, e_i\rangle.$ So, linear.


问:
Circle from $n$-gon circumference${}=1$
To get a better intuition why: $\pi$ is irrational and transcendental I came up with the following analyses. I am a amateur but I give it my best to describe:

Imagine a piece of very flexible thin robe. The robe length is set to 1, while the robe is continuous all point between robe start $0$ and $1$ exists. Tie the robe start and end together, creating a loop. Now we can create polygons with $n$ sides ($n$-gon) with total circumference equal to 1.

To construct the $n$-gon, we require the side length $l(n)$ and the enclosed angle $\alpha(n)$ between $3$ edge points.

$$l(n)=\frac{1}{n}$$
$$\alpha(n)=\frac{2\pi}{n}$$

The $n$-gon's can be constructed with both formulas below. All $n$-gon's have an edge point in the origin:

$$x(n)=\frac{1}{n} \cdot \sum_{k=0}^{n} cos \left( \frac{(2k+1) \ \pi}{n}  \right)$$
$$y(n)=\frac{1}{n} \cdot \sum_{k=0}^{n} sin \left( \frac{(2k+1) \ \pi}{n}  \right)$$

Analysis.

With both equations the $n$-gon's can be drawn [Github]. It is possible to compute the $n$-gon's with $n$ as a continuous number interpolation the edge points. 3 plots have been made:

Top: The polygons with discrete $n$ (red polygons) and continuous $n$.

Middle: A finite number of continuous $n$.

Bottom: An infinite “integrated like” continuous $n$.

The circle diameter can be determined from the $y$-axis, this way we can determine $\pi$ see: [Youtube].

Intuition.

The side lengths $l$ can be constructed with compass and straight edge for rational numbers only, irrationals take infinate time. The angle $\alpha$ can only be determined for constructible polygons [wiki]. Creating a continuous circle edge would require an iterative/statistical process thus, I would say (intuition): we require a number $\pi$ to be irrational, transcendental and random/normal to fill in the gaps.

Question.

Is there formal math describing the method of creating a circle with polygons with constant circumference?

How must I interpret the graph with a continuous $n$-gon? What type of math describes that?

答:
The following method describes how to derive the circle found from $n$-gons with circumference $1$. The $n$-gons can be described as:

$$x(n)=\frac{1}{n} \cdot \sum_{k=0}^{p} \ cos \left( \frac{2k+1}{n}\ \pi  \right)$$

$$y(n)=\frac{1}{n} \cdot \sum_{k=0}^{p} \ sin  \left( \frac{2k+1}{n} \ \pi \right)$$

As demonstrated in the question $n$ can be substituted for non integer continuous values, interpolating the edge points of the $n$-gons. Normally the summation will be till $p=n$ and must remain an integer.

Though, by taking the integral also $k$ can be a continuous number. This will give us the limit values of the $n$-gons:

$$x_c(n)=\frac{1}{n}  \int  cos \left( \frac{2k+1}{n}\ \pi  \right)dk= \frac{1}{2 \pi} \cdot sin\left( \frac{2k+1}{n}\ \pi  \right)+a$$

$$y_c(n)=\frac{1}{n}  \int  sin \left( \frac{2k+1}{n}\ \pi  \right)dk=- \frac{1}{2 \pi} \cdot cos \left( \frac{2k+1}{n}\ \pi  \right)+b$$

This parametrized function gives us a circle with radius: $R=\frac{1}{2 \pi}$:

$$R=\frac{1}{2\pi} =\sqrt{x^{2}+y^{2}}$$

The circumference $C$ of the circle is:

$$C=2  \pi  R=2 \pi \cdot \frac{1}{2\pi}=1$$

We defined the circumference of the $n$-gons $1$ this matches the circumference of the found circle.

This method is intuitive and I am not sure if it's valid. Especially substituting the integers for continuous values.

Title: Circle from $n$-gon circumference${}=1$


问:
How to solve a recursive sequence that is oscillating to a limit point
In class we only learned the "tools" to prove that a sequence that is monotonically increasing or decreasing converges. How to prove that a sequence converges if it is given in recursive form and is increasing at some intervals and decreasing at some intervals that are very hard to "select".
Is it even possible without the standard form?</p>

答:
There isn't one fixed answer to this question - there are many ways we can show a sequence is convergent. However, when we are working with recursively defined sequences, it is often easier to prove that the sequence is a Cauchy sequence.

Once we have a Cauchy Sequence, if we are working with series of real numbers, we can use the General Principle of Convergence which says that every Cauchy Sequence is convergent.

This link gives some really nice explanations and examples.


问:
What is an isolated subgroup?
" As in the case of finite numbers, the infinitesimal numbers form an isolated subgroup  of $R_{inf}$ of $^*R$"page 152

What does this sentence mean? What is an isolated subgroup?

答:
The full quote reads:

As in the case of finite numbers, the infinitesimal numbers form an isolated subgroup $\mathbf{R}_{\text{inf}}$ of $^*\mathbf{R}$; this implies that the factor group $^*\mathbf{R}/\mathbf{R}_{\text{inf}}$ inherits naturally its order relation from $^*\mathbf{R}$.

From this context, I would infer that for an ordered group $(G,\leq)$, a subgroup $H$ is isolated if for all $x,y\in H$, if $z\in G$ with  $x\leq z \leq y$, then $z\in H$. This condition is necessary and sufficient for the ordering  $\leq$ on $G$ to induce an ordering on the quotient group $G/H$. And the infinitesimal numbers certainly  satisfy this condition.

I have never heard the term "isolated" used in this way - I would call such a  subgroup $H$ "convex". The very first Google result for "isolated subgroup" is this page: https://encyclopediaofmath.org/wiki/Isolated_subgroup. Here Encyclopedia of Math gives a different (non-order-theoretic) definition. But it also notes

In the theory of ordered groups, isolated subgroups are sometimes referred to as convex subgroups.

This is a bit odd: it's clear that a convex subgroup of an ordered group is isolated (by the EOM definition), but it seems to me that there are isolated subgroups  of ordered groups  which are not convex  (e.g. $\mathbb{Q}\subseteq \mathbb{R}$  as additive groups). But in any case, it seems relatively clear that "isolated" means "convex" in the context of the paper  you're reading.

Title: What is an isolated subgroup?


问:
The reason for the superiority of Lebesgue measure / Lebesgue integration
I'm learning about Lebesgue measure and Lebesgue integration.

I know that Lebesgue integration is superior to Riemann integration in the sense that we can integrate a much larger class of function with Lebesgue integral.

I would like to ask which is the main reason that creates the superiority of Lebesgue integral?

EDIT

As for me, basically, Lebesgue integration is another way to calculate the integration (or geometrically, the area under the curve), so why just changing the way we compute can make such a huge difference between those 2 kind of integration ?

答:
Riemann integration has trouble with functions which take very different values very rapidly. For example, the indicator function on the rationals is not Riemann integrable, because every dissection of an interval will trap a rational and an irrational between every point of your dissection.

Lebesgue integration, however, is perfectly fine with functions taking very different values rapidly in a wide range of circumstances: all that matters is that we can somehow split up the domain into "not too horrible" (i.e. measurable) sets on which the function is constant, or that we can approximate the function by such "simple" functions. The indicator function on the rationals is fine, for example, because the rationals are a "nice" set (a measurable one), as are their complement.

I should add that I've approached this from a "conceptually why is Lebesgue integration good?" angle; peek-a-boo commented from an "in practice, what properties of Lebesgue integration make it nice to work with?" angle, and I haven't touched that aspect.


问:
Applying repeated doubling to an update Matrix
Given a rule to obtain a new $x,y$ positions from an initial $x_0,y_0$

The rule is
$ \binom{x'}{y'} = \begin{pmatrix}
v_x & -v_y \\
v_y & v_x
\end{pmatrix}
\binom{x}{y}$

$x' = v_x  x - v_y  y$

$y' = v_y  x + v_x y$

$f(x,y) = (v_x x - v_y y, v_y x + v_x y)$

Now let's say I wanted to use this rule to update a position $x_0,y_0$

How can I use repeated doubling to calculate the position after updating it $n$ times without having to manually update the position $n$ times

For instance lets say we started at position $(-10, 0) $ and we wanted to update this position 5 times and our given value for $(u_x,u_y) = (\frac{1}{2},\frac{1}{2})$

after the first update(n=1) the position would be $(-5,-5)$ and then the 5th update(n=5) the position would be $( \frac{5}{4},  \frac{5}{4}) $

How could I calculate the position for $( \frac{5}{4},  \frac{5}{4}) $ without having to calculate n = 2 or n= 3 or n= 4 or simply just using the concept of repeated doubling

答:
Assuming that $v_x$ and $v_y$ are constants, note that the update rule is equivalent to

$$x'+iy' = (v_x + iv_y)(x + iy) $$

where $i$ is the imaginary unit. So, if we write $v_x+iv_y$ in the polar form $re^{i\theta}$, then after $n$ updates, the result will be

\begin{align*}
(v_x+iv_y)^n(x+iy)
&= r^n(\cos(n\theta)+i\sin(n\theta))(x+iy) \\
&= r^n(x\cos(n\theta)-y\sin(n\theta)) + ir^n(x\sin(n\theta)+y\cos(n\theta)).
\end{align*}

Reading out its real and imaginary part will yield the desired coordinates. In doing so, all you need to know is

$$r = \lvert v_x+iv_y\rvert = \sqrt{v_x^2+v_y^2}$$

and

$$\theta = \arg(v_x + iv_y) = \operatorname{arctan2}(v_y, v_x), $$

where $\operatorname{arctan2}$ is the 2-argument arctangent function.

The following is a Python code for this:

```<code>import math

# arguments
(x, y) = (-10, 0);
(v_x, v_y) = (1/2, 1/2);
n = 5;

# computation
r = math.sqrt(v_x**2 + v_y**2);
th = math.atan2(v_y, v_x);
x_n = (r**n)*(x*math.cos(n*th) - y*math.sin(n*th));
y_n = (r**n)*(x*math.sin(n*th) + y*math.cos(n*th));

# print the result
print(x_n, y_n);
</code>```

The outcome will be

```<code>1.2500000000000009 1.2500000000000004
</code>```

which is $(5/4, 5/4)$ ignoring the numerical errors due to rounding.

Body:
Assuming that $v_x$ and $v_y$ are constants, note that the update rule is equivalent to

$$x'+iy' = (v_x + iv_y)(x + iy) $$

where $i$ is the imaginary unit. So, if we write $v_x+iv_y$ in the polar form $re^{i\theta}$, then after $n$ updates, the result will be

\begin{align*}
(v_x+iv_y)^n(x+iy)
&= r^n(\cos(n\theta)+i\sin(n\theta))(x+iy) \\
&= r^n(x\cos(n\theta)-y\sin(n\theta)) + ir^n(x\sin(n\theta)+y\cos(n\theta)).
\end{align*}

Reading out its real and imaginary part will yield the desired coordinates. In doing so, all you need to know is

$$r = \lvert v_x+iv_y\rvert = \sqrt{v_x^2+v_y^2}$$

and

$$\theta = \arg(v_x + iv_y) = \operatorname{arctan2}(v_y, v_x), $$

where $\operatorname{arctan2}$ is the 2-argument arctangent function.

The following is a Python code for this:

import math

# arguments
(x, y) = (-10, 0);
(v_x, v_y) = (1/2, 1/2);
n = 5;

# computation
r = math.sqrt(v_x**2 + v_y**2);
th = math.atan2(v_y, v_x);
x_n = (r**n)*(x*math.cos(n*th) - y*math.sin(n*th));
y_n = (r**n)*(x*math.sin(n*th) + y*math.cos(n*th));

# print the result
print(x_n, y_n);

The outcome will be

1.2500000000000009 1.2500000000000004

which is $(5/4, 5/4)$ ignoring the numerical errors due to rounding.


问:
What is the antiderivative of one over an arbitrary polynomial?
Suppose we have an arbitrary polynomial as follows.
$$p(x):=\sum_{i=0}^n a_nx^n$$
And we want to find the antiderivative of $(p(x))^{-1}$ with respect to $x$.
$$\int\frac{dx}{p(x)}$$
Messing around with an online CAS suggests that this would be equal to
$$C+\sum_{z\in\{r\mid p(r)=0\}}\frac{\ln |x-Z|}{\sum^n_{i=1}ia_iz^{i-1}}$$
Which sort of makes sense to me, and I think I can visualize that, but my question is, how do you derive this result and how true is it? Do you have to employ a form of partial fraction decomposition? The website also says that this only applies to cases where each root is distinct. What happens in the case of repeated roots? Can I extend this to a power series?

答:
The formula follows from a partial fraction decomposition when the polynomial has no repeated roots.

Note that the denominator in your formula, $\sum_{i=1}^nia_iz^{i-1}$, is just the derivative of $p(x)$ evaluated at the root, $p'(z)$.  Since $p(x)$ and $p'(x)$ have a common root if and only if $p(x)$ has multiple roots, then the formula would require you to divide by $0$ if you had any multiple roots. So the formula cannot work for polynomials with multiple roots, at least as is.

Suppose first that $p(x)$ is monic and has pairwise distinct roots, $z_1,\ldots,z_n$, and the partial fraction decomposition is
$$\frac{1}{p(x)} = \frac{A_1}{x-z_1}+\cdots + \frac{A_n}{x-z_n}.$$
If we do the operation on the right hand side, we get
$$\frac{1}{p(x)} = \frac{\sum_{j=1}^n A_j\prod_{i\neq j}(x-z_i)}{p(x)}.$$
so that means that
$$\sum_{j=1}^n A_j\prod_{i\neq j}(x-z_i) = 1.$$

Using the Heaviside cover-up method, we see by plugging in $z_t$ that
$$A_t = \frac{1}{\prod_{i\neq t}(z_t-z_i)}.$$
But notice as well that
$$p'(x) = \sum_{j=1}^n\left(\prod_{i\neq j}(x-z_i)\right),$$
so that
$$p'(z_t) = \prod_{i\neq t}(z_t-z_i).$$
That is, $A_t = \frac{1}{p'(z_t)}$.
Therefore, we can write the partial fraction decomposition as:
$$\frac{1}{p(x)} = \frac{1}{p'(z_1)(x-z_1)} + \cdots + \frac{1}{p'(z_n)(x-z_n)}.$$
Integrating, we get
$$\begin{align*}
\int\frac{1}{p(x)}\,dx &= \int\left(\sum_{i=1}^n \frac{1}{p'(z_i)(x-z_i)}\right)\,dx\\
&= \sum_{i=1}^n\left(\int\frac{1}{p'(z_i)(x-z_i)}\,dx\right)\\
&= \sum_{i=1}^n\left(\frac{\ln|x-z_i|}{p'(z_i)}\right) + C,
\end{align*}$$
which is exactly your formula.

If $p(x)$ is not monic, but has leading coefficient $a_n\neq 0$, then letting $P(x) = \frac{1}{a_n}p(x)$, then $P(x)$ is monic with the same roots as $p(x)$. So if $z_1,\ldots,z_n$ are the roots, then
$$\begin{align*}
\int \frac{1}{p(x)}\,dx &= \int \frac{1}{a_nP(x)}\,dx = \frac{1}{a_n}\int \frac{1}{P(x)}\,dx\\
&=\frac{1}{a_n}\left( \sum_{i=1}^n \frac{\ln|x-z_i|}{P'(z_i)}\right)+C\\
&= \frac{1}{a_n}\left(\sum_{i=1}^n\frac{\ln|x-z_i|}{\frac{1}{a_n}p'(z_i)}\right)+C\\
&= \frac{1}{a_n}\left(\sum_{i=1}^n\frac{a_n\ln|x-z_i|}{p'(z_i)}\right)+C\\
&= \left(\sum_{i=1}^n\frac{\ln|x-z_i|}{p'(z_i)}\right) + C.
\end{align*}$$
So you get the same formula if $p(x)$ is not monic.


问:
Why use the same basis for domain and codomain of linear operators?
(All of this over finite dimensional vector spaces) In Artin, Algebra, Chapter 4, section 3., the author says "When we are dealing with linear operators. It is clear that we want to pick only one basis for $V$". Why is it clear? In the page before, it was proved that for any linear transformation you can choose basis in domain and codomain so it's matrix takes the block form $\begin{bmatrix}
I & 0\\
0 & 0
\end{bmatrix}$ where $I$ is the identity matrix. This at first sounds very desirable to work with. If we use the same basis, the best we can do is Jordan Canonical form, which is nowhere as simple. So why restrict ourselves to use the same basis? It doesn't seem "clear" at all.

答:
When dealing with linear operators you want to take sums, scalar multiples, compositions. Having new bases every time would be a nightmare.

Even with a single operator $T$, saying that its composition with itself, $T^2$, is again $\begin{bmatrix}I&0\\0&0\end{bmatrix}$ at the cost of furnishing yet a new basis, gives you no benefit.

Also, the Jordan form is a lot better than the Smith form: having the eigenvalues and their multiplicities visible, as well as some eigenvectors, is way better than hiding them in a contrived basis.


问:
Prove that certain elements are not in some ideal
I have the following question:

Is there a simple way to prove that

$x+1 \notin \langle2, x^2+1\rangle_{\mathbb{Z}[x]}$ and

$x-1 \notin \langle2, x^2+1\rangle_{\mathbb{Z}[x]}$

without using the fact that $\mathbb{Z}[x]/\langle2, x^2+1\rangle$ is an integral domain?

Thanks for the help.

答: $x-1 \in I \implies x-1=2p+q(x^2+1) \implies q=0$ (reasoning on degree) so that you have $x-1=2p$ which is clearly impossible.</p>


问:
Find all subgroups of GL(3,7) with a Sylow 3-subgroup of order 27
I want to use GAP to study all subgroups of GL(3,7) with a Sylow 3-subgroup of order 27. How can I do it? Thanks.</p>

答:
Just so that the question is answered, this group is small enough that one can easily compute all subgroups of the group $G$, and select those of a suitable order. This is easiest done in an isomorphic permutation group.

```<code>gap> G:=GL(3,7);;
gap> iso:=IsomorphismPermGroup(G);;p:=Image(iso);
<permutation group of size 33784128 with 2 generators>
gap> u:=List(ConjugacyClassesSubgroups(p),Representative);;
gap> u:=Filtered(u,x->Size(x) mod 27=0 and Size(x) mod 81<>0);;
gap> Length(u);
163
gap> u:=List(u,x->PreImage(iso,x));;
</code>```

finds 163 classes of subgroups.  If the group gets bigger and calculating all subgroups becomes infeasible, one could try to compute the potential Sylow subgroups first (from the subgroups of a Sylow subgroup of $G$), and use IntermediateSubgroups to find all subgroups above, and finally fuse under the conjugaction action:

```<code>gap> s:=SylowSubgroup(p,3);
<permutation group of size 81 with 4 generators>
gap> m:=MaximalSubgroupClassReps(s);
[ <permutation group with 3 generators>, <permutation group with 3 generators>
    , <permutation group with 3 generators>,
  <permutation group with 3 generators> ]
</code>```

Fuse under full group:

```<code>gap> m:=List(SubgroupsOrbitsAndNormalizers(p,m,false),x->x.representative);
[ <permutation group of size 27 with 3 generators>,
  <permutation group of size 27 with 3 generators>,
  <permutation group of size 27 with 3 generators> ]
gap> int:=List(m,x->IntermediateSubgroups(p,x).subgroups);;
gap> int:=Unique(Concatenation(int));;
gap> int:=Filtered(int,x->Size(x) mod 81<>0);;
</code>```

Again eliminate conjugates.

```<code>gap> int:=List(SubgroupsOrbitsAndNormalizers(p,int,false),x->x.representative);;
gap> Length(int)+Length(m);
163
</code>```

Body:
Just so that the question is answered, this group is small enough that one can easily compute all subgroups of the group $G$, and select those of a suitable order. This is easiest done in an isomorphic permutation group.

gap> G:=GL(3,7);;
gap> iso:=IsomorphismPermGroup(G);;p:=Image(iso);
<permutation group of size 33784128 with 2 generators>
gap> u:=List(ConjugacyClassesSubgroups(p),Representative);;
gap> u:=Filtered(u,x->Size(x) mod 27=0 and Size(x) mod 81<>0);;
gap> Length(u);
163
gap> u:=List(u,x->PreImage(iso,x));;

finds 163 classes of subgroups.  If the group gets bigger and calculating all subgroups becomes infeasible, one could try to compute the potential Sylow subgroups first (from the subgroups of a Sylow subgroup of $G$), and use IntermediateSubgroups to find all subgroups above, and finally fuse under the conjugaction action:

gap> s:=SylowSubgroup(p,3);
<permutation group of size 81 with 4 generators>
gap> m:=MaximalSubgroupClassReps(s);
[ <permutation group with 3 generators>, <permutation group with 3 generators>
    , <permutation group with 3 generators>,
  <permutation group with 3 generators> ]

Fuse under full group:

gap> m:=List(SubgroupsOrbitsAndNormalizers(p,m,false),x->x.representative);
[ <permutation group of size 27 with 3 generators>,
  <permutation group of size 27 with 3 generators>,
  <permutation group of size 27 with 3 generators> ]
gap> int:=List(m,x->IntermediateSubgroups(p,x).subgroups);;
gap> int:=Unique(Concatenation(int));;
gap> int:=Filtered(int,x->Size(x) mod 81<>0);;

Again eliminate conjugates.

gap> int:=List(SubgroupsOrbitsAndNormalizers(p,int,false),x->x.representative);;
gap> Length(int)+Length(m);
163


问:
$\vdash (\lnot \chi \to \lnot \theta) \to (\theta \to \chi) $
For $\theta, \chi$ $\mathcal{L}$-formulas in predicate logic, I'm trying to prove: $\vdash (\lnot \chi \to \lnot \theta) \to (\theta \to \chi) $.

I've managed to prove $\vdash \lnot\lnot \theta \to \theta$, which was the hint for this problem. I have already proved $\vdash (\theta \to \chi) \to (\lnot \chi \to \lnot \theta) $. My attempt was to substitute $\lnot \chi$, $\lnot \theta$ into $\theta, \chi$ in $\vdash (\theta \to \chi) \to (\lnot \chi \to \lnot \theta) $. This gives $\vdash (\lnot \chi \to \lnot \theta) \to (\lnot \lnot \theta \to \lnot \lnot \chi) $. Finally, I was trying to show $\vdash (\lnot \lnot \theta \to \lnot \lnot \chi) \to (\theta \to \chi)$, but was not having much luck. Is there a nice way of doing this?

Furthermore, how does one go about doing things like this? I have seen natural deduction proofs that are very complicated, or are written in a way such that I suspect that there is some motivation that I am surely missing. But I can't figure out how to get the intuition/motivation for natural deduction proofs in general. My attempts purely seem to be brute-forcing, which is not so good.

答:
I have seen natural deduction proofs that are very complicated, or are
written in a way such that I suspect that there is some motivation
that I am surely missing. But I can't figure out how to get the
intuition/motivation for natural deduction proofs in general.

The following proof is based on a form of natural deduction. Simpler than most, it may help. It uses direct proof (lines 8 and 9) and proof by contradiction (line 6).

Note: '=>' is left associative here, so I am proving the equivalent of [~A => ~B] => [B => A].

Screen shot from my proof checker:


问:
Uniqueness of Duals in a Monoidal Category
Given a monoidal category ${\cal C}$, and $X \in {\cal C}$, we define a left dual of $X$ to be an object $X^*$ such that there exist morphisms $\epsilon:X^* \otimes X \to I$, and $\eta:I \to X \otimes X^*$, for $I$ the identity of the category, satisfying certain axioms, see here for details.

When is the left of a dual unique up to isomorphism, and when is this isomorphism unique.

答: This is much less clear than claimed by the previous post. See this stackexchange post for the full proof.</p>


问:
What is the basis of the trivial vector space
Why do we say that the set {$\vec 0$} has zero dimensions? I ask this because dimension, as defined in linear algebra, is the cardinality of a basis.

A basis with any actual elements, even {$\vec 0$} does not make sense as a basis as it violates the uniqueness property of creating elements in the space, and it has non trival relations.

So what exactly is the basis of {$\vec 0$}? If it were a set with no elements or the empty set, then how could we possibly construct $\vec 0$? Is there a way to create $\vec 0$ using a set containing nothing?

答: Because $\sum_{x \in \varnothing}x = 0$, and $\varnothing$ is our basis.


问:
Prove that $\lim_{n\to\infty}\sin(nx)/ \pi x$ is a delta function
I need to prove that $\delta_n(x)=\sin (nx)/\pi x$ is a delta function. That is, to prove that: $$\underset{n\rightarrow \infty}{\lim}\int^\infty_{-\infty}f(x)\frac{\sin( nx)}{\pi x}dx=f(0).$$ For that I made $y=x/n$ and took the limit under the integral since $f(x)$ is supposed to be analytic in $\mathbb{R}$. I have a fundamental limit for the $\sin$ leaving me with $$\int^\infty_{-\infty}f(0)\cdot1 \;dy.$$ But this is equal to $f(0)$ $\iff$ $f(x)$ have contribuition only at $x=0$, but here the only assumption is that $f(x)=0$ when $x\rightarrow \pm \infty$.

What is wrong here?

答:
I thought that it might be instructive to present a way forward that uses a regularization of the Dirac Delta.  To that end we proceed.

PRELIMINARIES:

Let $\displaystyle \delta_n(x)=\frac1{2\pi}\int_{-n}^ne^{ikx}\,dk$.  Then, we can write

$$\delta_n(x)=\begin{cases}\frac{\sin(nx)}{\pi x}&,x\ne0\\\\\frac n\pi&,x=0\tag1\end{cases}$$

The function  $\delta_n(x)$ has the following properties:

For each $n$, $\delta_n(x)$ is an analytic function of $x$.

$\lim_{n\to \infty} \delta_n(0)= \infty$

$\left|\int_{-\infty}^x \delta_n(x')\,dx'\right|$ is uniformly bounded.

$\lim_{n\to \infty}\int_{-\infty}^{x}\delta_n(x')\,dx'=u(x)$, where $u$
is the unit step funciton.

For each $n>0$, $\int_{-\infty}^\infty \delta_n(x)\,dx=1$

While heuristically $\delta_n(x)$ "approximates" a Dirac Delta when $n$ is "large," the limit of $\delta_n(x)$ fails to exist.  However, if we interpret this limit in the distributional sense, then $\lim_{n\to\infty}\delta_n(x)\sim\delta(x)$.  We will now show that this is indeed that case.

ANALYSIS:

Let $\phi(x)\in S$ where $S$ is the Schwarz Space of functions.

We will now evaluate the limit

$$\begin{align}
\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x)\phi(x)\,dx=\lim_{n\to \infty}\int_{-\infty}^\infty \frac{\sin(nx)}{\pi x}\phi(x)\,dx\tag1
\end{align}$$

Integrating by parts the integral on the right-hand side of $(1)$ with $u=\phi(x)$ and $v=\int_{-\infty}^{nx}\frac{\sin(x')}{\pi x'}\,dx'$ reveals

$$\begin{align}
\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x)\phi(x)\,dx&=-\lim_{n\to \infty}\int_{-\infty}^\infty \phi'(x)\int_{-\infty}^{nx}\frac{\sin(x')}{\pi x'}\,dx'\,dx\tag2
\end{align}$$

Using Property 3 in the Preliminaries section, there exists a number $C$ such that $\left|\phi'(x)\int_{-\infty}^{nx}\frac{\sin(x')}{x'}\,dx'\right|\le C\,|\phi'(x)|$.  Inasmuch as $C|\phi'(x)|$ is  integrable, the Dominated Convergence Theorem guarantees that

$$\begin{align}
\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x)\phi(x)\,dx&=-\lim_{n\to \infty}\int_{-\infty}^\infty \phi'(x)\int_{-\infty}^{nx}\frac{\sin(x')}{\pi x'}\,dx'\,dk\tag3\\\\
&=-\int_{-\infty}^\infty \phi'(x)\lim_{n\to \infty}\left(\int_{-\infty}^{nx}\frac{\sin(x')}{\pi x'}\,dx'\right)\,dx\\\\
&=- \int_{-\infty}^\infty \phi'(x)\underbrace{u(x)}_{\text{Unit Step}}\,dx\\\\
&=-\int_0^\infty \phi'(x)\,dx\\\\
&=\phi(0)
\end{align}$$

Therefore, in the sense of distributions as given by $(3)$, we assert that $\lim_{n\to\infty}\delta_n(x)\sim \delta(x)$ as was to be shown!


问:
Role of wavelength in the representations of the translation group
In the representation theory of the translation group we have $U^p(x)=e^{ipx}$ where $p=2\pi/\lambda$. I know in quantum mechanics this ends up being momentum somehow. And we can also see that $p$ is like an angular speed on the complex plane. Is this basically the intuition behind these infinite representations? Kind of like, okay all the representations reach $x$ but at different speeds? Is there some other way to think of this intuitively?

答:
I think you malformed your operator. In the position representation, the translation operator is
$$
U^p(\Delta x) = e^{i \Delta x ~ \hat p/\hbar }= e^{\Delta x ~ \partial_x},
$$
which then amounts to Lagrange's shift operator,
$$
U^p(\Delta x)  f(x) = f(x+\Delta x).
$$
Here, $\Delta x$ is a constant number not acted upon by $\partial_x$.

(I wouldn't go there, but if you really wished to consider $U^p(  x) $ instead,
you see this is $ e^{x\partial_x}= e^{\partial/\partial(\log x)}$, so that
$$
U^p(x) f(x) = e^{\partial/\partial(\log x)} f(e^{\log x})= f(ex),
$$
a mere scale transformation.)


问:
$f:\mathbb D\rightarrow \mathbb D$ be holomorphic with $f(1/2)=0$ and $f(0)=1/2$
Let $f:\mathbb D\rightarrow \mathbb D$ be holomorphic with $f(1/2)=0$ and $f(0)=1/2,$ where $\mathbb D=\{z:|z|<1\}.$ Then which of the following statements are correct?

(a) $|f'(0)|\leq 3/4,$

(b) $|f'(1/2)|\leq 4/3,$

(c) $|f'(1/2)|\leq 4/3$ and $|f'(0)|\leq 3/4,$

(d) $f(z)=z$ for $z\in \mathbb D.$

by Schwarz-Pick Lemma $|f'(z)|\le {1-|f(z)|^2\over 1-|z|^2}$ we get $|f'(0)|\le {3\over 4}$ and $|f'(1/2)|\le {4\over 3}$ so  so $a,b,c$ are true, but I don't know how to prove or disprove $d$ could anyone tell me? here $f(0)\ne 0$ so I am not able to use Schwarz-Pick Lemma directly.

Body: How about $f(z)=\frac{2z-1}{z-2}$?


问:
Proof of sum of binomial distribution
I have two stochastic variable $X\sim B(n,p)$ and $Y\sim B(m,p)$, I need to prove that $Z=X+Y \sim B(n+m,p)$.

Why is $$\sum_{k=0}^{u} \binom{n}{k} \binom{m}{u-k} = \binom{n+m}{u}$$

答: If you are selecting $(n+m)$ items, $u$ at a time, then, for each selection, you will end up with $k$ of the items, coming from the first $n$, where $k$ must be in $\{0,1,\cdots, u\}$.  Of course, I am assuming that $u \leq n$ and $u \leq m$.

Title: Proof of sum of binomial distribution


答:
I don't know the actual context of that quote, but my guess
is that it refers to maps from a vector space to itself. In
that case, the composition of these maps makes sense. Using
one basis for the vector space associates a square matrix to
the maps where composition of maps corresponds to product
of matrices. If there is a different basis used for the
codomain, then this association is not possible. Given this
context, it makes little sense to use another basis for the
same vector space.</p>

Title: Why use the same basis for domain and codomain of linear operators?

Body:
(All of this over finite dimensional vector spaces) In Artin, Algebra, Chapter 4, section 3., the author says "When we are dealing with linear operators. It is clear that we want to pick only one basis for $V$". Why is it clear? In the page before, it was proved that for any linear transformation you can choose basis in domain and codomain so it's matrix takes the block form $\begin{bmatrix}
I & 0\\
0 & 0
\end{bmatrix}$ where $I$ is the identity matrix. This at first sounds very desirable to work with. If we use the same basis, the best we can do is Jordan Canonical form, which is nowhere as simple. So why restrict ourselves to use the same basis? It doesn't seem "clear" at all.


问:
Understanding a specific process of finding the derivative of $x^TAx$
I am referring to @copper.hat's response to : Derivative of Quadratic Form. I do not have the reputation to reply directly. My goal is to find a way to better differentiate and understand these functions for the purpose of learning the gradient. I will try to write everything here to make this question independent.

Let $Q(x) = x^TAx$. $x∈R^n$ and $A∈R^{nxn}$

Fitting to $Q(x+h)−Q(x)$, $Q(x)$ expands to $(x+h)^TA(x+h)$ which expands further to $x^TAx+x^TAh+h^TAx+h^TAh-x^TAx$ simplifying to $x^TAh+h^TAx+h^TAh$.

How is it that this changes to $x^TAh+h^TAx$ in the reply on the linked post, dropping the $h^TAh$?

I see the references to $|h^TAh|≤∥A∥∥h∥^2$ however Googling "Cauchy Scwarz" gives very broad results and I'm having trouble understanding.

I've looked up many ways to find the gradient of matrix functions (ex: $b^TX^TXb$ , $1/2x^TAx+b^Tx$) and this process seems to be the most intuitive for me.

答:
The derivative by  a vector $v$ is the Gateaux derivative in the direction $v,$ meaning we take an extra variable $t,$ near a point $x_0$ write out the function evaluated at $x_0 + t h$ and see what happens.   Your
$$  x^T A x = (x_0 + tv)^T A (x_0 + t v) = x_0 A x_0 + t \; (x_0^T Av + v^T A x_0) + t^2 \; v^T A v $$
As a 1 by 1 matrix is symmetric, we have $v^T A x_0 =x_0^T Av $ and
$$  x^T A x  = x_0^T A x_0 + 2t \; x_0^T Av + t^2 \; v^T A v $$
and its derivative at $t=0$ is the scalar $$2 x_0^T Av \;.$$
This is the dot product of $v$ with $2Ax_0,$  so the gradient is $2Ax_0$ when written as a column vector


问:
If $A$ is a square matrix and $k$ is positive, then the set $\{v \in \mathbb{R}^n \mid v^t A v \le k\}$ is closed
At first I thought I could define a norm $||v||=v^tAv$ and then conclude that the set is exactly the closed ball centred at origin with radius $k$. But that function is not a norm since $||v||=0$ does not imply $v=0$.

Later I tried to define a function $f(v)=v^tAv$ and prove that it is continuous (thus the set is the inverse image of the closed set $[0,k]$). However, I don't know how to expand expressions like $v^tAv$ (in order to understand how a sequence $v_n^tAv_n$ converging to $v^tAv$ behaves). I know these expressions are pretty common in linear algebra, but I rarely had to deal with sequences of this kind.

I started with a simple example in two dimensions: if $v=(x,y)$ and $A=(a_{ij})$ then
$$v^tAv=x^2a_{11}+xy(a_{12}+a_{21})+y^2a_{22}$$

Is there a clever way to approach the problem? Thanks in advance

答:
mmmm maybe we can try with Lagrange's multipliers? $f(x)=v^tAv$ and $F(x)=v^tAv-K$ and some how show that the maximum reaches when F(x)=0 and is in the set.
for every none symmetric matrix we know that $B=A+A^t$ is symmetric and for symmetric B we know that for $g(v)=<v,Bv>$ we have that $D_g(v) = 2v^tB$...


问:
How to convert a matrix of decision variables in a vector for solving a linear programming problem?
I want to solve this problem :

find the minimum of the function F = sum(sum(A - Q*E)),

were A is an i x j matrix, Q is an i x k matrix and E is an k x j matrix. This is basically a least absolute deviation optimisation, where A are the data, Q are know coefficients and I'm searching for the matrix E.

Apparently it is possible to solve with linear programming, I read that I need to vectorize the matrix E, but I don't know what to do at this point... Can anyone give me some help please ? It will be great !

答:
I assume that your double sum is the sum over rows then columns, or vice versa.

If we let $M=A-Q E$ then each element of $M$ is a linear function of the elements E, and so their sum is also a linear function.

Now you have specified no constraints on the values of $E$, so the problem is a simple algebra problem.  I'll try to guess your actual question as one of the two following:

1)  We want the elements of $E$ to be nonnegative.  This gives us a simple linear program.

2) The sum is over the absolute values of the elements of $M$.  (This is probably what you are asking about.)  Here we introduce one variable for the absolute value of each element of $M$ and constrain it to be greater than the corresponding variable of $M$ and its negative. Minimizing the sum will force the value to be equal to the absolute value.

For example, if we name the matrix of absolute values $N$ then

[N_{1,1} \ge A_{1,1} - \sum_k A_{1,k} E_{k,1}]
and
[N_{1,1} \ge -A_{1,1} + \sum_k A_{1,k} E_{k,1}]
and  so forth.  Your objective will be the sum of the elements of $N$.

If you mean something else, please be clearer.


问:
Square of the squeezing operator
What is the square of the squeezing operator $S(z)=\exp[\frac12\left((z(a^{\dagger})^{ 2}−z^\ast a^2\right)]$?

I mean, with $z \in \mathbb{R}$, what is
$S(Z)S(Z)$?
Is there any formula ?

答:
Indeed, it is a double squeeze, with $z\to 2z$, since the two operators in the exponents commute with each other (!), so you can simply CBH-add them,
$$
S(z)S(z)= S(2z).
$$

You may also see this (in overkill, but good to know for more elaborate combinations with the displacement operator) by squaring the middle line matrix of (3.5) in Fisher, R. A., Nieto, M. M., & Sandberg, V. D. (1984). "Impossibility of naively generalizing squeezed coherent states",  Physical Review D29 (6), 1107.

The idea is to map the oscillator operators $a^\dagger a^\dagger$, $-aa$ and $a^\dagger a +1/2$ to the generators of SU(1,1), and then work out the combination of the exponentials as 2x2 matrix multiplications, in the easier-to-handle faithful 2x2 matrix representation of SU(1,1). Any group relation holding for this matrix rep will then hold for the oscillator realization!

(See a simple application in this.)

Title: Square of the squeezing operator


问:
Use DeMoivre's Theorem to prove $ \cos 5x = 16 \cos^5x - 20\cos^3x+5\cos x$
I need to prove the following equalities using DMT:

$ \cos 5x  = 16 \cos^5x$$ - 20\cos^3x$$+5\cos x$

and

$ \sin 5x  = 16 \sin^5x$$ - 20\sin^3x$$+5\sin x$

Can someone help me with this question?

(Attempt: $\cos5x+i\sin5x=(\cos x+i\sin x)^5$)

Body:
By DeMoivre's Theorem:

$$\begin{equation}\begin{aligned}
cos\,5x &= cos\,5x + i\,sin\,5x \\
&= (cos\,x + i\,sin\,x)^5 \\
&= cos^5\,x + 10\,cos^3\,sin^2\,x\,i + 5\,cos\,x\,sin^4\,x\,i^4\\
&= cos^5\,x - 10\,cos^3 x\,sin^2\,x + 5\,cos\,x\,sin^4\,x \\
&= cos^5\,x - 10\,cos^3\,x (1 - cos^2\,x) + 5\,cos\,x (1 - cos^2\,x)^2 \\
&= 16\,cos^5\,x - 20\,cos^3\,x + 5\,cos\,x \\
\end{aligned}\end{equation}$$


问:
Counting permutation of duplicate items
I am not really sure I fully understand the formula for finding the number of permutation of duplicate items.

The formula is:

$$\frac{n!}{p!q!r!\cdots}$$

Where do the factorials in the denominator exactly come from?

If we asssume the string "ANNA" and we want the count of the permutation of duplicate items.

We have $4$ characters so since we have $4$ options for the first character, $3$ for the second, $2$ for the third and $1$ for the last we have $4!$ different permutations.

But some of the characters are duplicates.

We have "A" and "N" repeated, meaning it adds more permutations to the outcome.

If $p=2$ is the number of occurences of "A" and if $q = 2$ is the number of occurences for "N"  then would the $p!$ and $q!$ essentially be selecting $1$ item at a time from $2$ (or $p$) duplicates? So essentially are the denominators the binomial $\binom{2}{1}$ (or generally $\binom{p}{1}$ where $p$ are the number of repeated characters) where order does not matter?

Is my understanding correct?

答:
The first part giving $4!=24$ different permutations of $4$ distinct objects is fine.

Now we consider the word $ANNA$ where we have two letters of one kind $A$ and two letters of another kind $N$.

Let $x$ denote the number of different permutations we can obtain from the letters $A,A,N,N$.

Suppose the $2$ letters $A$ are replaced by $2$ new letters, distinct from each other and from all other letters being permuted. These $2$ new letters may be permuted in $2!$ ways; hence the number of permutations of the new set of objects is $x\cdot 2!$.

Next suppse the $2$ letters $N$ are replaced by $2$ new letters, distinct from each other and from all other letters being permuted. These $2$ new letters may be permuted in $2!$ ways; hence the number of permutations of the new set of objects is $x\cdot 2! \cdot 2!$.

Since the number of permutations of $4$ distinct letters is $4!$ we obtain
\begin{align*}
x\cdot 2!\cdot 2!&=4!\\
\color{blue}{x}&\color{blue}{=\frac{4!}{2!2!}}
\end{align*}

Doing the same approach with $p$ equal objects of one kind, $q$ equal objects of another kind, and so on, we obtain in general
\begin{align*}
\frac{n!}{p!q!\cdots}
\end{align*}
with $p+q+\cdots =n$.

Title: Counting permutation of duplicate items


问:
$ x_n = 1 + 2a + 3a ^ 2 + \cdots + na ^ {n-1}$ has any three consecutive terms coprime
Let $ a \in \mathbb N $. For $ n \in \mathbb N $, let $$ x_n = 1 + 2a + 3a ^ 2 + \cdots + na ^ {n-1}. $$

$ x_1 = 1, x_2 = 1 + 2a, \ldots $.

For $ n \in \mathbb N $, show that $ \gcd (x_n, x_ {n + 1}, x_ {n + 2}) = 1 $.

Any help, suggestion on how to start solving that exercise.

答: Start by looking at  $x_{n+1} - x_n$ and $x_{n+2} - x_{n+1}$.


问:
Approximating continuous functions $S^n \to S^n$
I'm trying to check that every continuous function $f:S^n \to S^n$ can be approximated by differentiable ones. Well, by Stone-Weierstrass I can approximate the coordinate functions $f_i:S^n \to \Bbb R$ by differentiable ${\tilde f_i}:S^n \to \Bbb R$. The problem is, of course, I can't guarantee ${\tilde f}:=({\tilde f}_1,\dots,{\tilde f}_{n+1})$ to be in $S^n$. To solve this i thought of projecting ${\tilde f}(p)$ to the point in the sphere that minimizes the distance ${\tilde f}(p)$ to $S^n$, this projection would still approximate $f$ but I see no reason for this projection to be differentiable. Any hint?

Sorry, by approximate I mean $\sup_{x \in S^n} |{\tilde f}(x) - f(x)| < 1$ (I don't know if $1$ is important here)

答: You're fine, as long as your approximation satisfies $|f-g|<1$. The function $\pi\colon \mathbb R^{n+1}-\{0\}\to S^n$, $\pi(x)=x/|x|$, is smooth.</p>


问:
Doubt about the Grothendieck universes
If $U$ is a grothendieck universe, by definition:

$I \in U$ and $\forall i\in I$ $x_i \in U$ $\Rightarrow \bigcup_{i \in I}x_i \in U$.

Is $\forall i,j \in I$: $i \neq j \Rightarrow x_i \neq x_j$ and underlying assumption of these property?

N/A: I'm reading Basic Category Theory Vol 1 of Francis Borceux.

Edit: If the answer is yes, I interpret this property as if $I \in U$ then every union of $|I|$-elements of $U$ belongs to it. By the other side, if the answer is no, I think about it as if $I \in U$ then every union of $\mathbb{at}$ $\mathbb{most}$ $|I|$-elements of $U$ belongs to it. That is the origin of my question.

答:
No, there's no particular reason to assume that, the notion of union makes sense even if there are some redundancies.

Of course, you can reduce to that situation because you can remove the redundancies, but there's no need to; it would make for a more complicated statement, both to assume, and to work with.

Title: Doubt about the Grothendieck universes


问:
How to sum $\sum_{x=0}^{N-1} \cos{\left( \frac{2\pi x}{N} \right)}^{2}$?
I've tried solving it, but I'm not familiar with these kind of progressions and I couldn't get to a result. Also, I've found that the result is just $N/2$, but I don't get why would that be the answer, as it doesn't work for $N=1$ or $N=2$.

Appreciate your help :)

答:
$$\sum_{x=0}^{N-1} \cos^2{\left( \frac{2\pi x}{N} \right)}=\sum_{x=0}^{N-1}
\frac{1+\cos\left( \frac{4\pi x}{N}\right)}{2}
=\boxed{\frac N2}+\frac12\sum_{x=0}^{N-1}\cos\left( \frac{4\pi x}{N}\right)\tag1$$

Now let compute the latter sum:
$$
\sum_{x=0}^{N-1}\cos\left( \frac{4\pi x}{N}\right)=
\operatorname{Re}\sum_{x=0}^{N-1}\exp\left( \frac{4\pi x}{N}i\right)
=\operatorname{Re}\frac{1-\exp\left(4\pi i\right)}{1-\exp\left(\frac{4\pi}{N}i\right)}=0,\tag2
$$
where we applied the formula for the sum of geometric progression.

Observe that the last equality is valid only for $N>2$, since for $N=1,2$ the denominator degenerates to $0$.
More generally
$$\sum_{x=0}^{N-1}\exp\left( \frac{2K\pi x}{N}i\right)=0$$
for all $N $ except for the divisors of $K $.


问:
Characteristics of semilinear PDE $xu_x+yu_y+xyu_z=0$
Remark:
Initially, I have incorrectly copied over the problem. The edit below shows the correct one and how I checked that the function really does satisfy the "new" pde.

I have a pde of the following form

$$xu_{x}+yu_{y}+xyu_{z}=0$$

with the boundary condition $$u(x,y,0)=x^2+y^2.$$

I solved the pde with the method of characteristic and got the solution
$$u(x,y,z)=x^2+y^2-2(\frac{x}{y}+\frac{y}{x})z.$$

However, the other part of this problem asks me to explain the following.

We know that $u$ is constant along characteristics and we know that $u(x,y,0)\geq0$ then why is $u(1,1,1)=-2$.

What I was thinking is that even though we know that $u$ is going to be constant on each characteristic the fact that $u(1,1,1)=-2$ means that no characteristic curve begging on $(x,y,0)$ will pass through the point $(1,1,1)$. Am I correct in thinking this is the answer or is there something else that I forgot about.

EDIT:
Just to check that the solution really does satisfy the pde:

$$x(2x-2\frac{z}{y}+2\frac{yz}{x^2})+y(2y-2\frac{z}{x}+2\frac{xz}{y^2})-2xy(\frac{x}{y}+\frac{y}{x})=$$
$$=x^2-\frac{xz}{y}+\frac{yz}{x}+y^2-\frac{yz}{x}+\frac{xz}{y}-x^2-y^2=0$$ Which really does satisfy the equation.

答:
$$xu_x+yuy+xyu_z=0$$
Charpit-Lagrange system of characteristic ODEs :
$$\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{xy}=\frac{du}{0}$$
A first characteristic equation comes from $\frac{du}{0}\:$finite $\implies\:du=0$.
$$u=c_1$$
A second characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{y}$ :
$$\frac{y}{x}=c_2$$
A third characteristic equation comes from
$\frac{dx}{x}=\frac{dy}{y}=\frac{ydx+xdy}{yx+xy}=\frac{dz}{xy}\quad\implies\quad \frac12 d(xy)=dz$
$$xy-2z=c_3$$
The general solution of the PDE expressed on the form $c_1=F(c_2,c_3)$ is :
$$\boxed{u(x,y,z)=F\left(\frac{y}{x}\:,\:xy-2z\right)}$$
$F(X,Y)$ is an arbitrary function of two variables.

CONDITION : $u(x,y,0)=x^2+y^2$
$$x^2+y^2=F\left(\frac{y}{x}\: , \: xy\right)$$
With $X=\frac{y}{x}$ and $Y=xy$ $\quad\implies\quad$ $x^2=\frac{Y}{X}$ and $y^2=XY$.
$$F(X,Y)=x^2+y^2=\frac{Y}{X}+XY$$
So the function $F(X,Y)$ is known now. We put it into the above geheral solution where $X=\frac{y}{x}$ and $Y=xy-2z$.
$$u=\frac{xy-2z}{\frac{y}{x}}+\frac{y}{x}(xy-2z)$$
$$u=x^2+y^2-2z\left(\frac{x}{y}+\frac{y}{x} \right)$$
This is exactly the result that you obtained.


问:
Radical equation $x- 5\sqrt{x} + 4 = 0$
Solve: Radical equation $x- 5\sqrt{x} + 4 = 0$

I think I understand what I had did wrong. Instead of squaring a product of a number I was squaring the equation.

Title: Radical equation $x- 5\sqrt{x} + 4 = 0$

Body: Hint: $\ 0 = \sqrt{x}^2 - 5 \sqrt{x} + 4 = (\sqrt{x}-4)(\sqrt{x}-1)\,\Rightarrow\,\sqrt{x} = \,\ldots$


问:
Different Solutions to Heat Equation Confusion
I'm currently studying Partial Differential Equations and I'm really now starting to understand why they are quite difficult to understand and solve.

(Please forgive any LaTeX formatting issues. I'm still very much a newcomer to the langauge)

Recently, I've learned about the Heat/Diffusion Equation:
$$
u_t = ku_{xx}
$$

And I've come across the fact that two different types of solutions appear but, to me, they have very different structure. The first solution that I've come across is this:
$$
u(x,t) = \sum_{n=t}^\infty A_n\,\sin\left(\frac{n\pi x}{L}\right)e^{\frac{-n^2\pi^2 kt}{L^2}},\; \text{where}\;\;\;  A_n = \frac{2}{L}\int_{0}^L f(x)\sin\left(\frac{n\pi x}{L}\right)\,dx
$$ And the second solution is this:
$$
u(x,t) = \frac{1}{\sqrt{4k\pi t}} \int_{-\infty}^{\infty} e^{\frac{-(x-y)^2}{4kt}} f(x)\,dx
$$

So from here, I've got a few questions:

One of the things that I took away from my ODE (Ordinary Differential Equations) class was that there was basically an algorithm to solve the ODEs I was doing. I would always get a solution that was a the sum of a particular and homogenous solution. Then I would use the initial and boundary value conditions to solve for the constants that arise either due to integration or from looking at the problem through the eyes of Linear Algebra.

My question here is does this way of thinking not apply when in the realm of PDEs? Do the initial and boundary conditions change form or structure of solutions?

On my PDE journey, I have come across multiple different solution approaches. For first order PDEs, in essence, I really only know the Method of Characteristics/Geometric Method and then some algebraic manipulation and substitutions for other first order linear PDEs so that the Method of Characteristics/Geometric Method can be used. For second order PDEs, I am aware of Separation of Variables, something with Fourier Series, Fourier Transform and Laplace Transform (maybe?), and then the "general" solutions to the heat and wave equations. However, I am absolutely clueless when it comes to knowing when to use a particular solution method for a given problem.

What are the conditions that have to be met? Why choose one method over another?

I can't seem to find answers to these questions. It again, seems to come down to the initial and boundary conditions but I'm not sure

These are my primary sources that I have used to learn PDEs. I've tried textbooks but they are just really dense, especially Strauss' book which is my class book. The only exception is Peter J Olver's PDE book but it still is somewhat confusing at times:

https://www.youtube.com/playlist?list=PLlXfTHzgMRUK56vbQgzCVM9vxjKxc8DCr

https://www.youtube.com/playlist?list=PLMrJAkhIeNNR2W2sPWsYxfrxcASrUt_9j

https://www.youtube.com/channel/UC5luRTUBc0w4TAs-bKk1ijQ/videos

https://www.youtube.com/playlist?list=PLGCj8f6sgswntUil8yzohR_qazOfYZCg_

Third-order Linear Parabolic PDE

(The third link actually follows my class quite closely and has been the most helpful but still does not answer my 2nd question.)

Any help at all, especially with question 2 would be greatly appreciated! Thank you very much Math Stack Exchange community!

答: It's still true that the general solution of a linear PDE is a particular solution + a solution of the homogeneous equation.  The difference is that while for ODE's there are only finitely many linearly independent solutions of the homogeneous equation, for PDE's there can be infinitely many. So rather than a finite linear combination of those solutions, you may want to consider a series (e.g. a Fourier series).</p>


问:
Prove $\gcd(4,10)≠1$
I am trying to write a proof that shows that $\gcd(4,10)≠1$ using only the definition of gcd below.

$$\gcd(a,b) = \min(\{k∈\mathbb N : k = ax + by \text{ for some } x,y ∈ \mathbb Z\}) $$

Here is what I have so far:

By the definition,

Let $S$ be the set where $\gcd(4,10) = \min(\{k∈\mathbb N : k = 4x + 10y \text{ for some }x,y∈\mathbb Z\}).$

I am trying to prove that $\gcd(4,10)=2$ by proving that $\gcd(4,10)\ne1$ and that $2∈S$. I am doing this because $1∈\mathbb N$ and $1$ is the smallest value of $\mathbb N$. If $1$ is not an element of the set $S$, and $2$ is an element of the set, then $2$ is the minimum value of the set because it is the smallest element of $\mathbb N$ possible. And so, $\gcd(4,10) = 2$. I can prove that $2∈S$ by simply plugging in $4x+10y=2$ with $x=-2$ and $y=1$.

So, I need to prove first that $1≠\gcd(4,10)$. To do so, I am trying to show that $1$ is not an element of the set $S$, since $1≠4x+10y$ where both $x$ and $y$ are integers.
(This is where I am, stuck I am not sure how to prove $1≠4x+10y$ properly).

To prove  $1≠4x+10y$, I have some ideas that I came up with so far.

All integer multiples of $4$ and $10$ will result in an even number and $1$ is odd. I don't this is enough though, because I would then have to somehow prove that multiples of $4$ and $10$ are all even. I also have the idea that the smallest integers are $0,1,-1$, and inputting them into $4x+10y$ will not give $1$.

I am looking for a stronger way to prove that $\gcd(4,10)≠1$ by only using the set $S$. Any suggestions or solutions would really help me out a lot.

答: Hint: All elements of $S$ are even: $4x+10y=2(2x+5y)=2z$.

Title: Prove $\gcd(4,10)≠1$