答:
You can see this product as a direct sum. You can take, by example, the natural isomorphism
$$\mathbb{Z} \oplus \mathbb{Z}_n \rightarrow \mathbb{Z}_n \times \mathbb{Z}, x + y \mapsto (x,y)$$

In fact, this is the case whenever you had a finite  product! That's because the direct sum is constructed to be a subgroup of the product
$$\oplus G_i \subset \Pi G_i $$
such that the elements of the direct sum have just finite non-null coordinates. Well, if you have just a finite number of groups, they are equal!

Title: Determining the quotient group $(\mathbb{Z} \times\mathbb{Z})/{\langle(n,n)\rangle}$

Body:
I am trying to classify, $\frac{\mathbb{Z} \times \mathbb{Z}}{\langle(n,n)\rangle}$, $n=1,2,3,\cdots, $ using fundamental theorem of finitely generated abelian groups.

Inspire by @user134824 in Classifying the factor group $(\mathbb{Z} \times \mathbb{Z})/\langle (2, 2) \rangle$, I construct $\varphi: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z}_n$ by $(k,l) \mapsto (k-l,l)$. Then its kernel is $\langle(n,n)\rangle$ and from the 1st isomorphism theorem, $\mathbb{Z} \times \mathbb{Z} \simeq \mathbb{Z}_n \times \mathbb{Z}$.

But how I can understand this from the fundamental theorem of finitely generated abelian groups?


问:
If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.
If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.

I think it is trivial because they are distinct.

So I wonder just saying "Since they are distinct" is enough to prove it?

Of course there could be several more detailed versions but I just want to know that reasoning is true or not.

ContentLicense: CC BY-SA 3.0

Body:
$$
(-1)^3 + (-2)^3 + 3^3 = -1 - 8 + 27 = 18 = 3(-1)(-2)(3)
$$


问:
How to prove $\sum_{1 \leq i,j \leq n} {\frac{x_i x_j}{x_i^2+x_j^2}} \geq 0$ without using integral
Let $x_1,x_2,...x_n$ be nonzero real number. Prove that
$$\sum_{1 \leq i,j \leq n} {\frac{x_i x_j}{x_i^2+x_j^2}} \geq 0.$$

I think this problem by Prove that $E\left(\frac{XY}{X^2+Y^2}\right) \geqslant 0$ for i.i.d. $X$ and $Y$.
But I want to find proof without using integral. How to solve this?

Edit: As @qfwfq pointed out, a proof without using integral is given in Proof that $\sum\limits_{j,k=1}^N\frac{a_ja_k}{j+k}\ge0$.

@Ramil's comment there: I'm also pretty interested in more elementary proof so that even an eighth-or-ninth-grader from high school student could understand it. Is there any way to do without determinants, integrals or any tools of higher (advanced) math? Maybe just using a couple of standard (or not quite standard) inequalities, mathematical induction, algebra

ContentLicense: CC BY-SA 4.0

Body:
First, note that:
$$\int_{-\infty}^0 e^{(x_i^2+x_j^2)t}\, dt=\frac{1}{x_i^2+x_j^2}$$
Therefore
$$\sum_{i,j} \frac{x_i x_j}{x_i^2+x_j^2} =  \sum_{i,j} x_i x_j \int_{-\infty}^0 e^{(x_i^2+x_j^2)t}\, dt $$
$$ = \int_{-\infty}^0 \sum_{i,j} x_i x_j e^{(x_i^2+x_j^2)t}\, dt $$
$$ = \int_{-\infty}^0 \left(\sum_{i} x_i  e^{x_i^2t} \right) \left(\sum_{j} x_j  e^{x_j^2t} \right) \, dt $$
$$ = \int_{-\infty}^0 \left(\sum_{i} x_i  e^{x_i^2t} \right)^2 \, dt \geqslant 0 $$


问:
Iterative policy evaluation algorithm in "Reinforcement Learning" written by Sutton.

I am studying the reinforcement learning using the textbook entitled "Reinforcement Learning An Introduction," written by Richard S. Sutton.

However, I got a weird point in the iterative policy evaluation algorithm, and cannot make sure whether I am wrong or the textbook is wrong. Hence, I am posting this question here to ask about this.

In the picture above, the following statement is stated in the second algorithm "Policy Evaluation."
$$\Delta \gets \max \left( \Delta , \lvert v - V(s) \rvert \right)$$

In my opinion, in order for the algorithm to terminate someday, $\max$ must be replaced with $\min$. However, all the iterative policy evaluation codes I've seen in github sticks to $\max$, not $min$.
I want to make sure what is correct, and if $\max$ is correct, can someone say why it is?

答:
That is not a mistake, the convergence is due to fixed point theorem.

We examine each state and record down what is the maximal update across the states.

We only terminate the procedure when the change in the previous iteration and current iteration is small enough.


问:
Evaluate integral in terms of Gamma function
Need help evaluating the following integral in terms of the gamma function where gamma function is: $$\Gamma(x)=\int_0^\infty e^{-t}t^{x-1}dt.$$  The integral is the following: $$\large{\int_0^\infty t^{x-1}e^{-\lambda t\cos(\theta)}\cos(\lambda t \sin(\theta))dt}$$</p>

答:
This is the same as
$$\frac12\left(\int_0^{\infty}t^{x-1}\exp\left\{-\lambda t e^{i\theta}\right\}dt+\int_0^{\infty}t^{x-1}\exp\left\{-\lambda t e^{-i\theta}\right\}dt\right).$$
Making the obvious change of variables $s=\lambda t e^{\pm i\theta}$ in the 1st and 2nd integrals, we find
$$\frac{\lambda^{-x}}{2}\left(e^{-ix\theta}+e^{ix\theta}\right)\int_0^{\infty}s^{x-1}e^{-s}ds=\lambda^{-x}\cos x\theta\, \Gamma(x).$$</p>


问:
How do I show that a domain is not simply connected?

Let $U$ be a star-shaped domain in $\mathbb{C}$. Prove that the subset obtained from $U$ by removing a finite number of points is not simply connected.

Usually when we want to show something is simply connected we show that any two curves with fixed endpoints are homotopic by finding a suitable homotopy. However, what can you do if you need to show that the domain is not simply connected? Can we somehow show that there doesn't exist a homotopy?

答:
If $\Omega$ is simply connected then $Ind_{\gamma} (z)=0$ for any closed path $\gamma $ in $\Omega$ and any point $z $ not in $\Omega$.

If $z$ is one of the points removed from your region there exists a closed disk $\overline {D} (z,r)$ around $z$ cantained in $U$ and the index of $z$ w.r.t. the circle of radius $r$ around $z$ is not zero. Hence, the region is not simply connected.


问:
How to desgin a fair raffle
I need to organize a raffle and it has to be fair.

Some people have a peice of land that lets say has 22 owners, and in that land they are planning to build 39 apartments (maybe 47 but that's not the point), so they have a contractor that will build it, and in order to decide which owner gets which apartment they want to have a raffle.

They wanted to number all the apartments from 1 to 39 and draw from a hat.

By the way they only get 22 or 24 apartments, the rest go to the contractor as his payment.

Problem is it's not fair, because the first to draw had a 1/39 chance the second has a 1/38 etc.

I taught about using a 2 phase raffle with random number pages.

Phase 1, we select the order in which people raffle using the random numbers, this phase is important in my opinion since not every apartment is the same there are 2 penthouses 2 garden apartments and other factors, so i order to keep everythibg fair, we also raffle the order of the raffle.

One important point about special apartments (penthouses and garden apartments) is that it may be agreed uopn in advance, that the contractor or an owner will get them WITHOUT a raffle, but that is ok since it was agreed upon.

After that is done using a w digit method we raffle for the apartments.

Is that fair enoguh? Is there a way to make a fully 100% fair raffle?

Thabks in advance

答:
Too long for a comment ...

Define $P$=Pr(penthouse on the first draw) = $\frac2{39}$,

$G$=Pr(garden on the first draw) = $\frac2{39}$,

$R$=Pr(regular on first draw) = $\frac{35}{39}$,

What happens in the second draw depends on the first:

Say $P_P$ is Pr(penthouse on the second draw given penthouse on the first) = $\frac{1}{38} \times P = \frac{2}{38\times39}$,

$P_G$ is Pr(penthouse on the second draw given garden on the first) = $\frac{2}{38} \times G = \frac{4}{38\times39}$,

$P_R$  is Pr(penthouse on the second draw given regular on the first) = $\frac{2}{38} \times R = \frac{70}{38\times39}$.

Putting these together,

Pr(penthouse on the second draw regardless of the first draw) = $P_P+P_G+P_R = \frac{76}{38\times39} = \frac2{39}$, the same probability as getting a penthouse on the first draw.

Title: How to desgin a fair raffle


问:
Show that if $H$ is a subgroup of $S_n$, then either every member of $H$ is even or half of them are even
Show that if $H$ is a subgroup of $S_n$ the symmetric group of order $n$, then either every member of $H$
is an even permutation or exactly half of the members are even.

I can see that if $a,b$ are arbitrary even elements in $H$, then their product and any combination of their products will be even, which implies $H$ is full of even permutations, but I'm not sure how to show the half-even half-odd part.

答:
Let $H$ be a subgroup of $S_n$

If every member of $H$ is even permutation, then $H\subseteq A_n$, the subgroup of $S_n$ consisting of all even permutations.

Suppose not, i.e. there is at least one element namely $\sigma_1$ which is an odd permutation. Let $m_e,m_o$ denote number of even, odd permutations in $H$. If $\sigma_1,\sigma_2,\ldots,\sigma_{m_o\ }$ are odd cycles, then $\sigma_1^2,\sigma_1\sigma_2,\ldots,\sigma_1\ \sigma_{m_o}$ are all distinct even permutations, so there are at least that many even permutations as many there are odd ones, which implies $m_e\geq m_o$.

On the other hand, if $\alpha_1,\ldots,\alpha_{m_e\ }$ are even permutations, then $\sigma_1\ \alpha_1,\ldots,\sigma_1\ \alpha_{m_e\ }$ are all odd permutations, so there are at least that many odd permutations as many there are even ones, which gives $m_o\geq m_e$.

Thus, $m_e=m_o=|H|/2$

The result we have proved is: If $H$ is a subgroup of $S_n$, then either $H\subseteq A_n$ or $|H\cap A_n|=|H|/2$


问:
Given $x_{0}$ and $x_{1}$, is there a formula to find $x_n = d\,(2\,x_{n-1} - x_{n-2})$ at a specified $n$?
The question covers the main problem I want to solve: $d$ can be any scalar constant (ideally in between 0 and 1, if that makes the problem easier), as well as $x_{0}$ and $x_{1}$. I tried figuring this out on my own, but if the answer is there it is eluding me (this uses an almost identical sequence, but I have no idea what difference equations are to know if that's relevant). This was inspired after learning about the Hugo Elias algorithm for efficiently generating water ripples, if that helps at all. I would be grateful for any help regarding this problem!

As a side note, I may not know the correct terms for everything, so please feel free to let me know or modify this question with more accurate terminology!

答:
$$x_n = d\,(2\,x_{n-1} - x_{n-2})$$ The chaacteristic equation is
$$r^2= d\,(2\,r - 1)$$ the roots of which being
$$r_1=d-\sqrt{d^2-d} \qquad \text{and} \qquad r_2=d+\sqrt{d^2-d}$$ giving as solution
$$x_n=c_1 \,r_1^n+c_2\,r_2^n$$ Now, applynig the conditions
$$x_0=c_1+c_2\qquad \text{and} \qquad x_1=c_1\, r_1+c_2 \,r_2$$ SOlving the two linear equations
$$c_1=\frac {x_1-x_0\, r_2 } {r_1-r_2 }\qquad \text{and} \qquad c_2=\frac{r_1\, x_0-x_1 }{r_1-r_2 }$$

Just replace $r_1$ and $r_2$ by their expressions (the result, even simple, will not be very nice).


问:
Determining the quotient group $(\mathbb{Z} \times\mathbb{Z})/{\langle(n,n)\rangle}$
I am trying to classify, $\frac{\mathbb{Z} \times \mathbb{Z}}{\langle(n,n)\rangle}$, $n=1,2,3,\cdots, $ using fundamental theorem of finitely generated abelian groups.

Inspire by @user134824 in Classifying the factor group $(\mathbb{Z} \times \mathbb{Z})/\langle (2, 2) \rangle$, I construct $\varphi: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z}_n$ by $(k,l) \mapsto (k-l,l)$. Then its kernel is $\langle(n,n)\rangle$ and from the 1st isomorphism theorem, $\mathbb{Z} \times \mathbb{Z} \simeq \mathbb{Z}_n \times \mathbb{Z}$.

But how I can understand this from the fundamental theorem of finitely generated abelian groups?

答:
You can see that the (finitely generated) Abelian group $(\mathbb Z\times\mathbb Z)/\langle(n,n)\rangle$ is indeed a product of cyclic groups: $\mathbb Z\times\mathbb Z/n\mathbb Z$, just as the fundamental theorem of finitely generated Abelian groups predicts.

However, the fundamental theorem is wholly unsuitable to predict the classification of this group (or any other finitely generated Abelian group*). It can tell you what it might be, up to isomorphism, but to tell exactly what it is, you need to use different tools, e.g. the ones you actually used to derive your result.

*Except for a cyclic group of prime order.


问:
Balls and bins - picking drawing one ball from each bin and calculating probability that all balls are black

Assume we have $3$ bins: bins $L_1, L_2, L_3$, in each bin, there are $3$ balls. Alice is picking one ball from each bin at random. The balls can be either white or black. Alice does not know how many black or white balls are in each bin. She only knows the total number of black balls (denoted as $B$) and white (denoted as $W$) in the bins. Let's assume that the total number of black balls is $6$. We want to compute the probability that all selected balls are black (let's denote it $p$).

Intuitively, the distribution of white and black balls can end up as follows:

Case 1:

$L_1$ has $B = 1 $,

$L_2$ has $B = 2$,

$L_3$ has $B = 3$

or, Case 2:

$L_1$ has $B = 2$

$L_2$ has $B = 2$

$L_3$ has $B = 2$

Case 3:

$L_1$ has $B = 3$

$L_2$ has $B = 3$

$L_3$ has $B = 0$

Thus, the total probability of picking only black balls is
$$ p = (\frac{1}{3} \cdot \frac{2}{3} \cdot 1) + (\frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3}) + 0 = 0.5185$$
Is the above correct?

Q2: In the second case (when each box has an equal number of black balls) Alice has the highest chances to pick only black balls (which is $0.2962$), so this is her "best-case scenario", while the $p = 0.5185$ is the total probability given that Alice cannot predict a-priori how the black balls will be distributed among the bins. Is the total probability considered an "average case" or should the average-case be computed differently?

EDIT:

Following the guidance in the comments, here is an updated version:

Assume we have $n$ balls and $k$ bins. $B$ balls are black, $W = n - B$ balls are white. We assigned the balls uniformly at random into the bins in such a way that each bin has $x = \frac{n}{k}$ elements ($k|n$). Once the balls are in the bins we pick from each bin one ball uniformly at random. What is the probability that all $k$ selected balls will be black?
Let $F$ be an event that all balls are black, and let $B_i$ denote the number of black balls in the $i'th$ bin.
From the total probability we can define it as follows:

$$
Pr(F) = \sum_{b_1 + b_2 + \ldots b_k = B} Pr(F | B_1 = b_1, \ldots B_k = b_k)\cdot Pr(B_1 = b_1, \ldots B_k = b_k)
$$

The first probability component is
$$ Pr(F | B_1 = b_1, \ldots B_k = b_k) = \prod_{i=1}^k \frac{b_i}{x} $$

The second is
$$
Pr(B_1 = b_1, \ldots B_k = b_k) = \frac{B!W!}{b_1!\cdot b_2! \cdot \ldots \cdot b_k! w_1! \cdot w_2! \cdot \ldots \cdot w_k!} \cdot \frac{(x!)^k}{n!} \\
= \frac{B!(n-B)!}{\prod_{i=1}^k b_i!\cdot(x-b_i)!}\cdot \frac{(x!)^k}{n!}
$$

Combining the two together we get
$$
(*) \quad
Pr(F) = \sum_{b_1 + b_2 + \ldots b_k = B} \left( \prod_{i=1}^k \frac{b_i}{x} \cdot \frac{B!(n-B)!}{\prod_{i=1}^k b_i!\cdot(x-b_i)!}\cdot \frac{(x!)^k}{n!}\right)
$$

Is that correct?

Also, there was a lot of discussion in the comments that the problem which I described is equal to asking: given $n$ balls, where $B$ are black and $W$ are white if I pick $k$ balls what is the probability that all of them are black. This probability would be defined as
$$ (**)\quad
Pr(F) = \frac{{B \choose k}}{{n \choose k}}
$$
I totally agree with the given examples, that it is the same value. However, this means the formulas $(*)$ and $(**)$ should be equal. I tried to simplify the formula (*) but I couldn't get what is in $(**)$. Is there something wrong with $(*)$?

答:
Assume the bins are filled randomly.
Then the probability of the cases are the following:

$(3,2,1):\quad 6\times \frac{\binom63\binom30}{\binom93}\frac{\binom32\binom31}{\binom63}=\frac9{14}$

$(2,2,2):\quad 1\times\frac{\binom62\binom31}{\binom93}\frac{\binom42\binom21}{\binom63}=\frac9{28}$

$(3,3,0):\quad 3\times\frac{\binom63\binom30}{\binom93}\frac{\binom33\binom30}{\binom63}=\frac1{28}$

Correspondingly the probability to pick only the black balls is:
$$
\frac9{14}\frac33\frac23\frac13+\frac9{28}\frac23\frac23\frac23=\frac5{21}.
$$

The same probability one would obtain if the balls were drawn directly from a bin with 9 balls:
$$
\frac69\frac58\frac47=\frac5{21},
$$
which was clear also without any calculation.

EDIT:

In general the number of ways to fill the boxes with $b_1,b_2,\dots b_k$ black balls and $w_1,w_2,\dots w_k$ white balls is:
$$
\frac{b!}{b_1!b_2!\cdots b_k!}\frac{w!}{w_1!w_2!\cdots w_k!}
$$


答:
The probability of getting exactly three black balls is given by:

$$
P(\text{3 B}) = \frac{6\choose 3}{9\choose 3} \simeq 0.2380952
$$

and it's the same of course of what Alice had if she extracts the 3 balls directly from a unique bin of 9 balls.

The bin subdivision doesn't affect the probability and it is obvious if you look at the following way. Imagine that Alice extracts one ball at the time, without looking, and "rank" the balls as the order of extraction. The "free" extraction corresponds in selecting balls 1,2,3; the "bins" extraction corresponds in selecting balls 1,4,7 after putting 1,2,3 on the first bin, 4,5,6 on the second and 7,8,9 on the third. It's obvious that the probability of having 3 blacks does not change.

However, if you really insist in starting with bins, things get unnecessarily complicated and tedious, because you have to calculate the probability of each bin composition. So, let's start assuming balls are numbered 1 to 9, and, as above, 1-2-3 belong to the first bin and so on. Next, we think at the "positions" of the white balls (easier to visualize, being less); there are ${9\choose 3} = 84$ different combinations they might assume. Now, we see how many of this combinations belong to the 1/2/3, 2/2/2 or 3/3/0 bin distribution.

Let's start with 2/2/2, which is easier. A white ball must stay in position 1,2 or 3; another in 4,5 or 6 and the last in 7,8 or 9. In total, you have $3\cdot 3\cdot 3 = 27$ combinations and so:

$$
P(2/2/2) = \frac{27}{84} \; .
$$

Now, we consider the 3/3/3 distribution. Now there are just 3 combinations of white balls that allow it, and namely 1-2-3, 4-5-6 and 7-8-9. So we have:

$$
P(3/3/0) = \frac{3}{84} \; .
$$

For the 3/2/1, we see that there are 6 of them, by permuting 1,2 and 3. For the specific 1/2/3 distribution we have that the first white ball must stay either in position 1,2 or 3; the other two must stay between ball 4 and 6, leaving a black ball that can be stay in any three of the remaining spot. So it means that you have $6\cdot 3 \cdot 3 = 54$ combinations for 3/2/1 and so:

$$
P(3/2/1) = \frac{54}{84} \; .
$$

Now we can go ahead and calculate your desired probability:

$$
P(\text{3 B}) = P(2/2/2) \cdot \left(\frac{2}{3}\right)^3 + P(3/2/1)\cdot\frac{1}{3}\cdot\frac{2}{3}
$$

Title: Balls and bins - picking drawing one ball from each bin and calculating probability that all balls are black

Body:
Assume we have $3$ bins: bins $L_1, L_2, L_3$, in each bin, there are $3$ balls. Alice is picking one ball from each bin at random. The balls can be either white or black. Alice does not know how many black or white balls are in each bin. She only knows the total number of black balls (denoted as $B$) and white (denoted as $W$) in the bins. Let's assume that the total number of black balls is $6$. We want to compute the probability that all selected balls are black (let's denote it $p$).

Intuitively, the distribution of white and black balls can end up as follows:

Case 1:

$L_1$ has $B = 1 $,

$L_2$ has $B = 2$,

$L_3$ has $B = 3$

or, Case 2:

$L_1$ has $B = 2$

$L_2$ has $B = 2$

$L_3$ has $B = 2$

Case 3:

$L_1$ has $B = 3$

$L_2$ has $B = 3$

$L_3$ has $B = 0$

Thus, the total probability of picking only black balls is
$$ p = (\frac{1}{3} \cdot \frac{2}{3} \cdot 1) + (\frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3}) + 0 = 0.5185$$
Is the above correct?

Q2: In the second case (when each box has an equal number of black balls) Alice has the highest chances to pick only black balls (which is $0.2962$), so this is her "best-case scenario", while the $p = 0.5185$ is the total probability given that Alice cannot predict a-priori how the black balls will be distributed among the bins. Is the total probability considered an "average case" or should the average-case be computed differently?

EDIT:

Following the guidance in the comments, here is an updated version:

Assume we have $n$ balls and $k$ bins. $B$ balls are black, $W = n - B$ balls are white. We assigned the balls uniformly at random into the bins in such a way that each bin has $x = \frac{n}{k}$ elements ($k|n$). Once the balls are in the bins we pick from each bin one ball uniformly at random. What is the probability that all $k$ selected balls will be black?
Let $F$ be an event that all balls are black, and let $B_i$ denote the number of black balls in the $i'th$ bin.
From the total probability we can define it as follows:

$$
Pr(F) = \sum_{b_1 + b_2 + \ldots b_k = B} Pr(F | B_1 = b_1, \ldots B_k = b_k)\cdot Pr(B_1 = b_1, \ldots B_k = b_k)
$$

The first probability component is
$$ Pr(F | B_1 = b_1, \ldots B_k = b_k) = \prod_{i=1}^k \frac{b_i}{x} $$

The second is
$$
Pr(B_1 = b_1, \ldots B_k = b_k) = \frac{B!W!}{b_1!\cdot b_2! \cdot \ldots \cdot b_k! w_1! \cdot w_2! \cdot \ldots \cdot w_k!} \cdot \frac{(x!)^k}{n!} \\
= \frac{B!(n-B)!}{\prod_{i=1}^k b_i!\cdot(x-b_i)!}\cdot \frac{(x!)^k}{n!}
$$

Combining the two together we get
$$
(*) \quad
Pr(F) = \sum_{b_1 + b_2 + \ldots b_k = B} \left( \prod_{i=1}^k \frac{b_i}{x} \cdot \frac{B!(n-B)!}{\prod_{i=1}^k b_i!\cdot(x-b_i)!}\cdot \frac{(x!)^k}{n!}\right)
$$

Is that correct?

Also, there was a lot of discussion in the comments that the problem which I described is equal to asking: given $n$ balls, where $B$ are black and $W$ are white if I pick $k$ balls what is the probability that all of them are black. This probability would be defined as
$$ (**)\quad
Pr(F) = \frac{{B \choose k}}{{n \choose k}}
$$
I totally agree with the given examples, that it is the same value. However, this means the formulas $(*)$ and $(**)$ should be equal. I tried to simplify the formula (*) but I couldn't get what is in $(**)$. Is there something wrong with $(*)$?


问:
Geodesic Uniqueness in the Hyperbolic Plane
I am studying Hyperbolic Geometry. At this part, I have proved that semicircles and straight lines orthogonals to the real axis are geodesics in the hyperbolic plane. But how I proof that this geodesics are uniques? That it does not exist others geometric places between two different points that minimizes the hyperbolic length? </p>

<p>I have seen a topic related to this, but I do not know nothing about Riemannian Geometry. If someone could indicate to me some references to this proof(or the idea behind) I will be very grateful! </p>

答: You can argue "geometrically" as follows: For two points $p$, $q$ on the imaginary axis it is pretty obvious that the segment $[p,q]$ is the unique shortest curve connecting $p$ and $q$. Now if $p$, $q\in H$ are arbitrary there is a Moebius transformation $T: H\to H$ sending $p$ and $q$ to two points $p'$, $q'$ on the imaginary axis. Since $T$ is a hyperbolic isometry it follows that $p$ and $q$ have a unique shortest connection, and since $T$ (now considered as a map $\bar {\mathbb C}\to \bar {\mathbb C}$) maps circles to circles and preserves angles between curves it follows that this shortest connection lies on a circle intersecting the real axis orthogonally.</p>


问:
Ellipse centre from two points and radius
I might need some help on this problem. I searched online and didn’t find any way to solve it.

Given two points, I need to find the center of the two ellipses of given radius $rx, ry$ that intersects at those two points.

I know the two A and B point coordinates, as well as ellipses radius $rx$ and $ry$ (on $X-$axis and $Y-$ axis respectively).

Is there a way, from those $4$ variables, to calculate the center $C$ and $C'$ of each ellipse?

Here is a little diagram I made to make things clearer.

答: If you divide all coordinates by $r_x$ and $r_y$ (respectively), the ellipses become two unit circles, by two points. The centers are on the mediatrix of the two points, at a distance from the middle given by Pythagoras (if the distance between the points is $d, \sqrt{1-\dfrac{d^2}4}$). You can compute this by vector analysis.


问:
Linear transformation and direct sums
We consider the direct sum $E_1 \oplus E_2$ of vector spaces. Then proof that the transformation $E_1 \oplus E_2 \to E_1$ defined by $(u_1, u_2) \mapsto u_1$ is linear. Also the map $E_1 \to E_1 \oplus E_2$ given by $u \mapsto (u, \mathbf{0})$.

I don't know if my procedure is right. First, name $f$ to be the first linear transformation we have defined. Then $f(\mathbf{0},\mathbf{0})= \mathbf{0}$. Now take $u, v \in E_1 \oplus E_2$, $u = (u_1, u_2)$ and $v = (v_1, v_2)$. $f(u+v) = f(u_1 + v_1, u_2 + v_2) = u_1+v_1 = f(u) + f(v)$. Lastly, let $\lambda \in K$, where $K$ is a field. Then $f(\lambda \cdot u) = f(\lambda \cdot u_1, \lambda \cdot u_2) = \lambda \cdot u_1 = \lambda \cdot f(u)$.

Any comment or correction would be welcomed!

答:
Denote the said map by $T$. Consider any $u,v\in E_1$. Then for any scalar $\alpha$

$T(\alpha u+v)=(\alpha u+v,0)=\alpha (u,0)+(v,0)=\alpha Tu+Tv,$

which shows that $T$ is linear.

Title: Linear transformation and direct sums


问:
Question about the infimum of $|f(x)-x|$ , where $f(x)=x$, $x$ is a fixed point of the nonlinear equation.
I am trying to check if the following property holds for fixed points:
Suppose:

$ f(x)= x $ is given, with solution $x = \theta \gt 0 $

I would like to show :

$ \forall \epsilon \in (0,1), \forall x : 1/\epsilon > |x -\theta| > \epsilon => \inf_x |f(x) - x | > 0  $.

I am not sure if this property holds for the particular case. I would be grateful is someone can give me any idea on the topic.

The best I was able to come up with was that due to the fact that the function is continuous it does not change sign within some vicinity of $\theta$. Because $f(x) = x $ has a solution for $x=\theta$ then we can find $ |f(x) -x|<\epsilon $, when $ x \to \theta $, so $ x-\epsilon<f(x) $ and since $\epsilon$, can be chosen arbitrary small and $x>0$ than $ 0<x-\epsilon<f(x)$ hold, for a small $\epsilon$ vicinity of $\theta$.

Thanks,

答:
If $f$ is continuous and the equation $f(x)=x$ has a unique solution $x=\theta$, then the statement is true:
$$\inf\{ |f(x) - x |: \epsilon \le |x-\theta|\le \epsilon^{-1} \} > 0,\quad \text{for all } \ 0<\epsilon<1$$
Indeed, since the set $\{ x : \epsilon \le |x-\theta|\le \epsilon^{-1} \} $ is compact, the continuous function $|f(x)-x|$ attains its minimum on this set. The minimum is positive since $f(x)\ne x$ for $x\ne \theta$.


问:
If $X$ is irreducible, then so is $f(X)$.
Let $f: X \to Y$ be a continuous map between topological spaces. I'm trying to show that if $X$ is irreducible, then so is $f(X)$.

Let $Y_1, Y_2 \subset f(X)$ two closed sets for the induced topology from $Y$ on $f(X)$ with the property that
$$f(X) = Y_1 \cup Y_2.$$
By definition of the induced topology, there is $Y_1', Y_2' \subset Y$ closed such that
$$Y_i = Y'_i \cap f(X), \quad i = 1, 2.$$
Now by continuity of $f$, we have that
$$f^{-1}(Y'_i) = f^{-1}(Y_i), \quad i = 1, 2, $$
are closed and that
$$X = f^{-1}(Y'_1) \cup f^{-1}(Y'_2).$$
By using the irreducibility of $X$, we deduce that (without lost of generality)
$$f^{-1}(Y'_1) = \varnothing \quad \text{and} \quad f^{-1}(Y'_2) = X.$$
It follows that
$$Y_1 = Y_1' \cap f(X) = \varnothing \quad \text{and} \quad Y_2 = Y_2' \cap f(X) = f(X),$$
which exactly means that $f(X)$ is irreducible.

Does it seems correct ? I feel like I'm missing something but I do not see what..

答: Your proof is correct. Maybe spend some time on the final step: if $f^{-1}[Y_2'] = X$ we conclude that $x \in X \to f(x) \in Y_2' \to f(x) \in Y_2$ so $f[X] \subseteq Y_2 \subseteq f[X]$ so indeed $f[X] = Y_2$ and $f^{-1}[Y'_1]=\emptyset$ implies $\emptyset = Y_1$ as otherwise $y \in Y_1 \subseteq f[X] \to (\exists x \in X: f(x)=y) \to (\exists x \in f^{-1}[Y_1'])$ etc. It's a minor point, but might make the final step a bit clearer. Otherwise, no comments.


问:
Graphically solving system of inequalities
Please check my work:

Jillian is going to college next year and has decided to start working this summer. She makes $20$ dollars an hour working at a farm and $16$ teaching gymnastics. Write $3$ inequalities.

She wants to

Make at most $\$ 800$

Work more than $10$ hours per week

Work less hours as a gymnastics teacher.

If we set $Y$ = gymnastics, $X$ = farm.

Would the following answers fit?

$20x + 16y < 800$ (making at most $800).

$x + y > 10$ (working more than 10 hours a week).

$x > y$ (working more at the farm than the gymastics class).

Graphing these should create an overlap?

答: At most would include equality as well. Hence the first inequality should be $$20x+16y \le 800.$$


问:
Showing that a function space is Banach
Let $D = (0,T) \times (0,\infty)$ and suppose $V := \bigg\{ f \in C(D) \bigg| \|f \|_V := \sup_{t,s}\bigg|\frac{f(t,s)}{1+s^2} \bigg| < \infty\bigg\}$, where $C(D) = \bigl\{f : D \to \mathbb{R} \big|\textrm{ f is continuous component-wise}\bigr\}$. I want to show that $V$ alongwith the norm $\| \cdot\|_V$ forms a Banach space.

I started by considering $\{f_n\}_{n \in \mathbb{N}}$, a Cauchy sequence in $V$, and now I should ideally have a candidate function to which the Cauchy sequence will converge. How do I explicitly construct such a candidate function? I am new to functional analysis, so references would also be greatly helpful.

答: If $(f_n)$ is Cauchy and $\epsilon >0$ then there exists $n_0$ such that $|f_n(t,s)-f_m(t,s)| <\epsilon (1+s^{2})$ for all $n,m \geq n_0$. For any fixed $t$ amd $s$ this implies that $(f_n(t,s))$ is Cauchy. Define $f(t,s)$ as $\lim f_n(t,s)$. I will let you handle the rest.

Title: Showing that a function space is Banach


问:
Finite covering of bounded, open sets
Consider $\Omega\subset \mathbb{R}^n$ open and bounded. Let $c>0$ be given. Is it possible fo find $V_1,...,V_n\subset \Omega$ with $|V_i|\leq c$ for all $i$, $|V_i\cap V_j|=0$ for $i\neq j$ and $ \Omega \subset \bigcup_{i=1}^n V_i$.

I think this can be done by taking $(V_i)_{i=1}^n$ to be a rhombuses. Nevertheless I don't have any reference. I would be thankful for any reference or any other idea.

答:
The definition of totally bounded does not require that $V_i\subset \Omega$. Can I assume this ?

Sorry, I did'nt want to take the same $n$. Assume there are finitely many $V_1,...,V_N$. Furthermore I am also only interested in a special chosen and fixed $c$.

Title: Finite covering of bounded, open sets


问:
How to prove that a tychonoff space $C(X, Y)$ equipped with fine topology can be embedded in a tychonoff cube.
As we know that a topological space is Tychonoff if and only if it can be embedded in a Tychonoff cube. We have that a function space $C(X, Y)$ equipped with fine topology where $X$ is a Tychonoff space and $Y$ also a (Tychonoff) metric space, is a Tychonoff space. How can we then prove that it can be embedded in a hilbert cube?

答: If $Z$ is any Tychonoff space and its weight $w(Z)$ (the minimal (infinite) size of a base for $Z$) is $\kappa$ then $Z$ embeds into the Tychonoff cube $[0,1]^\kappa$ (and the weight of that cube is also $\kappa$). This can be applied to a space of the form $C(X,Y)$, when we know the topology on it is Tychonoff. It can only be embedded into the Hilbert cube if $w(C(X,Y)) = \aleph_0$ which will depend on the specific $X,Y$ of course.


问:
Prove that the functions $h(x)=|f(x)|$ and $k(x)=\max\{f(x),g(x)\}$ are both continuous.
Here is the full question:
Let $X$ be a topological space and $f,g$ two continuous functions $f:X\to \Bbb R$. Prove that the functions $h(x)=|f(x)|$ and $k(x)=\max\{f(x),g(x)\}$ are both continuous.

I am having trouble proving the first part that $h(x)=|f(x)|$ is continuous.

Here's my answer for the second part:

We know the maximum of two functions $f(x)$ and $g(x)$ is defined as $max\{f(x),g(x)\}=\frac{f(x)+g(x)}{2}+\frac{|f(x)-g(x)|}{2}$. Since this is the sum of two continuous functions $f(x)$ and $g(x)$, then by the properties of continuous functions, $k(x)$ is also a continuous function.

Could someone point me in the right direction for the first part?

答:
The function $a(x)=|x|$ is continuous on $\Bbb R$. You can do an appeal to calculus, or apply the pasting lemma to

$$a(x) = \begin{cases} x & x \in [0,+\infty)\\
                       -x & x \in (-\infty, 0]
\end{cases}$$

using that both sets are closed and $x \to x$ and $x \to -x$ are continuous on those domains.
Or use the reverse triangle inequality

$$|a(x)-a(y)|= \left|\,  |x| - |y|\, \right| \le |x-y|$$

to see $a$ is uniformly continuous..

Then $|f| = a \circ f$, a composition of continuous maps.


问:
Use the intermediate value theorem to prove that if $f:[0,1]\to[0,1]$ is continuous, then there exists $c\in[0,1]$ such that $f(c)=\sqrt{c}$
Suppose that $f(0)<f(1)$. Consider now, $f(0)<\sqrt{c}<f(1)$. By the intermediate value theorem, $\exists b\in[0,1]:f(b)=\sqrt{c}$.

Now we need to show that its possible that $b=c$, but this is exactly where I am stuck.

答: Write $g(x)=f(x)-\sqrt x$, $g(0)\leq 0, g(1)\geq 0$, so there exists $c$ such that $g(c)=f(c)-\sqrt c=0$.


问:
Why define simplicial chain groups as functions rather than free Abelian groups?
I am trying to understand why the usual definition of chain groups goes

Define $n$-chains as maps from $n$-simplices to $\mathbb{Z}$ that vanish cofinitely

Prove $C_n$ is free Abelian

Why not start with definition of $C_n$ as free Abelian? I saw this in a bunch of different lecture notes on the internet, but textbooks seem to prefer the definition in terms of maps. I would appreciate if someone could point me to where this simplification fails or what is the "profit" gained in intuition or otherwise? Thanks!

答:
You could do so.

An Abelian group $F$ is called free if it has a basis. This is a subset $B \subset F$ such that each $x \in F$ has a unique representation as finite sum $x = \sum_{i=1}^n n_i b_i$ with $n_i \in \mathbb Z$ and $b_i \in B$. Frequently one wants to start with a set $X$ and find a free Abelian group $F$ having $X$ as a basis. This leads to the following definition:

Given a set $X$, an Abelian group $F(X)$ together with a function $\iota : X \to F(X)$ is called freely generated by $X$ if for every Abelian group $G$ and every function  $f : X \to G$ there exists a unique group homomorphism $\bar f :  F(X)\to G$ such that $f = \bar f \circ \iota$.

This definition via a universal property determines the pair $(F(X),\iota)$ uniquely up to canonical isomorphism. In fact, if $\iota' : X \to F'(X)$ has the same property, then we get unique group homomorphims $\bar \iota' : F(X) \to F'(X)$ such that $\iota' = \bar \iota' \circ \iota$ and  $\bar \iota' : F(X) \to F'(X)$ such that $\iota = \bar \iota \circ \iota'$. Hence $\bar \iota \circ \bar \iota' \circ \iota = \bar \iota \circ \iota' = \iota =  id_{F(X)} \circ \iota$ which implies $\bar \iota \circ \bar \iota' = id_{F(X)}$ via uniqueness. Similarly $\bar \iota' \circ \bar \iota = id_{F'(X)}$.

It remains to prove the existence of such a pair $(F(X),\iota)$. The standard construction is $F(X) = \{ \phi : X \to \mathbb Z \mid \phi(x) \ne 0 \text{ only for finitely many } x \in X \}$ and $\iota(x')(x) = 1$ for $x' = x$, $\iota(x')(x) = 0$ for $x' \ne x$. We do not go into details here, you can consult a textbook. Note that this $F(X)$ is easily seen to be a free Abelian group with basis $\iota(X)$ which justifies the notation freely generated by $X$.

Now let us come to your question. Defining $C_n$ as the free Abelian group generated by the $n$-simplices is not really precise because there are many models for such an object. For example, if you have only one $0$-simplex, you could simply take $C_0 = \mathbb Z$. But you want to obtain a functor $C_n$ and therefore it is not expedient to say "pick any free Abelian group generated by the $n$-simplices" - doing so would involve a variant of the axiom of choice for proper classes. Concerning classes have a look at Can we say," The set of all compact metric spaces"?

This problem can easily be avoided by taking a concrete construction of free Abelian groups, for example that as maps as above. We do not need any choice here.


问:
"There exists a" does it mean At least one or Exactly one...
There exists a minimum weight edge in G which is present in every minimum spanning tree of G.

Does it mean, We only have one minimum weight edge in G and that minimum weight edge should be in every Minimum Spanning Tree of G.

Or does it mean, We have at least one minimum weight edge in every minimum spanning tree of G?

I understand that Grammatically, 'Exists' is singular.
But, some more clarifications would really be helpful.

答: "There exists" ($\exists$) means at least one; for uniqueness, we use $\exists!$.


问:
Linear Algebra: Dimensionality of the Spaces
I was reading Gilbert Strangs book when I got to this quote:

"(Dimension of column space) + (dimension of nullspace) = dimension of $\mathbb{R}^n$)"

I understand that dimension of column space is $r$, and dim null space is $n - r$. However, I do not know why he added $C(A) + N(A)$. I thought the whole point was to add the dims of $C(A) + N(AT)$ to get $m$? Aren't the column and the left null space perpendicular? I do not see why he emphasized this relationship.

答:
If $V$ is finite dimensional, and $L:V\to W$ linear, then the rank-nullity theorem states that the $\rm{rank}(L)+\rm{null}(L)=\rm{dim}V$.

As to your other comment/question, the nullity of a square matrix and it's transpose are the same, by the theorem that the row rank equals the column rank, a.k.a. the rank of the matrix.


问:
How to solve this integral? With Q-function, help with derivation.
I am reading a textbook about pathloss models for wireless communications, namely Wireless Communications by Andrea Goldsmith

and it has some derivation where it goes from (sect 2.10)

$$
C = \frac{2}{R^{2}}\int_{0}^{R}rQ(a+bln\frac{r}{R})dr
$$

to this

$$
C = Q(a) + exp(\frac{2-2ab}{b^{2}})Q(\frac{2-ab}{b})
$$

Stating this is the closed-form solution of the integral.

Where Q is the Q-function(closely related to the complementary error function)

$$
Q(x) = \frac{1}{\sqrt{2\pi}} \int_x^\infty \exp\left(-\frac{u^2}{2}\right) \, du.
$$

I am wondering what the intermediate steps look like? Since evaluating at r = 0 seems invalid? And evaluating at r=R simply provides C = Q(a). Obviously, I am missing some basic stuff about integrals inside integrals, but I'm not sure where to start.

Any help would be appreciated.

答:
$$Q(x) = \frac{1}{\sqrt{2\pi}} \int_x^\infty \exp\left(-\frac{u^2}{2}\right) \, du=\frac{1}{2} \text{erfc}\left(\frac{x}{\sqrt{2}}\right)$$
$$C=\frac 1{R^2}\int_0^R r\,\,\text{erfc}\left(\frac{a+b \log \left(\frac{r}{R}\right)}{\sqrt{2}}\right)\,dr$$

Let
$$r=R e^{\frac{\sqrt{2} }{b}t-\frac{a}{b}}\implies C=\frac{\sqrt{2} }{b}e^{-\frac{2 a}{b}}\int_0^1e^{\frac{2 \sqrt{2} }{b}t} \text{erfc}(t)\,dt$$

For $$I=\int e^{\frac{2 \sqrt{2} }{b}t} \text{erfc}(t)\,dt$$ one integration by parts gives
$$I=\frac{b }{2 \sqrt{2}}e^{\frac{2 \sqrt{2}}{b}t}\, \text{erfc}(t)+\frac b {\sqrt{2\pi}}\int e^{\frac{2 \sqrt{2} }{b}t-t^2}\,dt$$ Complete the square and face the gaussian integral. The result follows and go back to $Q(.)$.


问:
Hasse Norm Theorem
I am looking to put a section together involving the Hasse Norm Theorem in a piece of work I am writing. As well as the Hasse Norm Theorem itself, Wikipedia also mentions a theorem by Hilbert in 1897 for the special case n = 2 and the case n is prime by Furtwangler 1902. I was wondering if anybody could point me in the direction of any books or online resources concerning these theorems and maybe even their proofs. I have also seen two well known counter examples namely,

$\mathbb{Q}(\sqrt{-3},\sqrt{13})/\mathbb{Q}$ and $\mathbb{Q}(\sqrt{13},\sqrt{17})/\mathbb{Q}$,

by Hasse and Serre, Tate respectively. I would also be grateful if anybody could highlight where I may find literature on these counterexamples.

答:
Hilbert's work is contained in his collected papers, vol I, right after the Zahlbericht. Furtwängler's proofs are available online (see https://www.emis.de/MATH/JFM/ for finding the references). Their approach is highly computational; it took me a very long time to figure out the ideas behind them. But all of these articles are available online, if you want to have a look at them.

The counterexamples you mention were studied in detail by Arnold Scholz; search for
Scholz and "number knots" to find modern literature; the standard reference is Jehne's article "On knots in algebraic number theory" (see https://eudml.org/doc/152174). I also recommend my notes https://www.mathi.uni-heidelberg.de/~flemmermeyer/publ/pcft.pdf and some sections in the book "Der Briefwechsel Hasse - Scholz - Taussky" (https://univerlag.uni-goettingen.de/handle/3/isbn-978-3-86395-253-2).

Title: Hasse Norm Theorem


问:
Notation for parts of an integer partition
If I have an integer partition of $n \in \mathbb{N}$

$$n=j_1 + 2 j_2 +3 j_3 + \dots + ij_i$$

and say the example is

$$10 = 1(2) + 2(0) + 3(1) + 4(0) + 5(1) = 1 +1 +3 + 5$$

what is the notation for the parts of the partition (the "summands"), here $\{1,3,5\}$?
Is there a standard way or should I just define it like this ?

答:
With $m_{i}=m_{i}(\lambda)$ we denote the number of parts of the partition $\lambda$ equal to $i$. For the partition $\lambda$ of $n$ we use the notation: $\lambda=(1^{m_{1}}, 2^{m_{2}},\ldots)$.

For example, we can rewrite the partition $(4,3,3,2,1,1,1)$ of $n=15$ as $(1^{3},2,3^{2},4)$. This is done purely for purposes of brevity.


问:
Prove the following using the properties of matrix
Show that the matrix $$A=\begin{pmatrix} a&b\\c&d\end{pmatrix}$$
satisfies the equation
$$A^2-(a+b)A+ (ad-bc) I =0$$

Please help me prove this. I have hard time doing it.

答:
$$A^2=\begin{bmatrix} a^2+bc & ab+bd\\ ac+cd & ad+bd \end{bmatrix}$$
$$-(a+d)A=\begin{bmatrix} -a^2-ad & -ab-bd\\ -ac-cd & -ad-d^2\end{bmatrix}$$
$$(bc-ad)I=\begin{bmatrix} ad-bc & 0 \\ 0 & bc-ad \end{bmatrix}$$
Add these three results to get $$A^2-(a+d)A+(bc-ad)I=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

Note for OP: the middle co-efficiens has to be $(a+d)$ and not $(a+b)$ as written by you. This may be the reason why you could not get it.


问:
The torus is connected
Let $T^2=S^1 \times S^1$ be a torus, where $S^1 \subset \mathbb{R}^2$ is the unit circle. Is $T^2$ connected? If so, how does one show this?

答: One way:  the torus is the continuous image of the connected space the unit square $S$ under certain identifications $\sim$.  The projection $p:S\to S/\sim \cong T^2$ is a quotient map and hence continuous.  (The identifications are of the opposite sides of the square.)

Title: The torus is connected


问:
Can a set have a complement in intuitionistic ZF?
Does IZF (ZF formulated in intuitionistic logic) prove that for any set $a$, $\{x: x \notin a \}$ does not exist?

答:
Suppose $b$ is the proposed set such that: $$∀x. x \in b \Leftrightarrow x \notin a$$ Then let $$u = a \cup b \\ d = \{x \in u : x \notin x\}$$

Suppose $d \notin u$. Then:

$d \notin a$ because $d \in a ⇒ d \in u$

Therefore $d \in b$ because $d \notin a ⇒ d \in b$

However $d \in b ⇒ d \in u$ which is contradictory

Therefore $¬\ d \notin u$.

Supposing $d \in d$, then $d \notin d$ which is contradictory, so $d \notin d$.

Supposing $d \in u$, then:

$d \in d$ because $d \in u ∧ d \notin d$

But $d \in d$ contradicts the prior $d \notin d$

So $d \notin u$.

But now we have deduced both $d \notin u$ and $¬\ d \notin u$, which is contradictory. So the assumption of $b$ existing is false.

You may have expected that intuitionistic logic would save you from $u$ being a universal set, because you cannot prove $∀ x. x \in u$ without excluded middle. However, when proving negative statements in intuitionistic logic, uses of excluded middle can be reduced to non-contradiction, which is intuitionistically valid reasoning. So if an assumption is classically contradictory, it is probably also intuitionistically so (for simple cases, at least).


问:
$\frac{dy}{dx}=\frac{2y}{2x+1}$ with initial condition of $y(0)=e$ for $x>-\frac{1}{2}$
When I do separation by variables, I get to $\frac{1}{2}\ln|y|=\frac{1}{2}\ln|2x+1|+C$ but then I'm not sure what to do from here.

I want to multiply by $2$, exponentiate, and then solve for $y$. Then I get $y=e^{2c} (2x+1)$

答:
Since $x >-1/2$ we have $|2x+1|=2x+1.$ then we get

$$\ln|y|=\ln(2x+1)+c$$

Since $y(0)=e>0$, we see that

$$\ln y=\ln(2x+1)+c,$$

hence

$$y(x)=e^c(2x+1)$$

From $y(0)=e>0$, we derive $e=e^c$, thus $c=1.$

The solution of the IVP is given by

$$y(x)=e(2x+1).$$


问:
The Torus is not homeomorphic to the unit circle
Let $T^2=S^1 \times S^1$ be a torus in $\mathbb{R}^4$, where $S^1:=\{x \in \mathbb{R}^2 \mid \|x\|_2 = 1\}$ is the unit circle. Does there exist a homeomorphic map $f: T^2 \to S^1$ ?

Intuitively this makes sense (somewhat), since $\mathbb{R}^4 \cong \mathbb{C}^2$ can be seen by considering the isomorphism
$$
\mathbb{R}^4 \to \mathbb{C}^2, \; (a,b,c,d) \mapsto (a+\mathrm{i}b,c+\mathrm{i}d)
$$
but I wonder whether it is possible to explicitely write down a homeomorphism $T^2 \to S^1$. I know that a homeomorphism is a continuous bijection with a continuous inverse map.

If not, I'd be interested in knowing how to start tackling this problem, because right now, I'm clueless as to where to start.

答:
It is well-known that $\pi_1(S^1)\cong\Bbb Z$.  From this it follows by functoriality that $\pi_1(S^1\times S^1)\cong\pi_1(S^1)\times\pi_1(S^1)\cong\Bbb Z\times\Bbb Z\cong\Bbb Z^2$.

Since they have different fundamental groups, they can't be homeomorphic.


问:
Proof of a formula for the period of a $\sin$ function subtracting distances
I would appreciate it if someone could explain this demonstration of the formula for the period of the sine function: $f(x)=\sin ax$. It is as follows:

We want to find the period for this function:

$f(x)=\sin ax$

Since we know the period for the sine function is $2$, if $t=ax$, we
need to find the minimal distance that should exist between $x_1$ and $x_2$
so that when these $x$ values are transformed into $t_1$ and $t_2$, the
distance between them will be $2$:

$|t_1-t_2|=2 ⇔|a(x_1-x_2)|=2 ⇔|a|.|(x_1-x_2)|=2 ⇔ |x_1-x_2|=2/|a|$

I know the easier explanations of the $2/|a|$ formula, but I don't understand the logic behind this one. In particular, I don't understand why $|t_1-t_2|=2$, since we're not dealing with a plain sine function (with a $2$ period), but with $f(x)=\sin ax$ whose period is not $2$.

I would be grateful if someone could explain what the logic behind this demonstration is.

答:
Assume, without loss of generality that $x_1 > x_2.$

Since you accept that the period of $f(x)$ is $(2\pi/a)$,

this means that you accept the goal that $(x_1 - x_2) = (2\pi/a)$

which means that you accept the goal that $(ax_1 - ax_2) = 2\pi$.

The explanation that you excerpted had $t = ax$ which means that $t_1 = ax_1, t_2 = ax_2$.

Therefore, $t_1 - t_2 = 2\pi.$

Addendum

Responding to the OP's comment:

We want to find the period for this function:
$f(x)=\sin ax.$

Without loss of generality, $a > 0.$

In order to find the period of $f(x)$, you have to define what it means for a function to have a period.

Given a generic function $g:\mathbb{R}\to\mathbb{R}$, the specification that $r$ is the period of $g$ denotes the following:

Constraint-1 : For all $y \in \mathbb{R}, g(y) = g(y + r)$.

Constraint-2: $r$ is the smallest possible positive value that satisfies Constraint-1.

For example, suppose $g(y) = \sin(y).$  Then the period of $g$ would be designated as $(2\pi)$, rather than [for example] $(4\pi$).  This means that $(2\pi)$ is the smallest positive value $r$ such that for any real number $y,$

$g(y) = \sin(y) = \sin(y+r) = g(y+r).$

Now, the challenge is to show that the period of $f(x)$ is $(2\pi/a).$  The way to do that is to show that $(2\pi/a)$ satisfies both Constraint-1 and Constraint-2.

Constraint-1:

For any $x \in \mathbb{R},$

$f(x + [2\pi/a]) = \sin(a[x + 2\pi/a]) = \sin(ax + 2\pi) = \sin(ax) = f(x).$

Thus, $(2\pi/a)$ satisfies Constraint-1.

Constraint-2:

Suppose that there exists $s \in \mathbb{R}$, such that

$0 < s < (2\pi/a)$ and

For all $x \in \mathbb{R}, f(x) = f(x + s)$.

Let $u$ denote $(as) \implies u < (2\pi).$

Since $s$ is assumed to be the period of $f$,
for all $x \in \mathbb{R},$

$f(x+s) = f(x) \implies
\sin(a[x+s]) = \sin(ax + u) = \sin(ax).$

This means that for any $x \in \mathbb{R}$,

$\sin(ax + u) = \sin(ax)$.

This means that $u$ satisfies Constraint-1, with respect to the sine function.

This yields a contradiction, since the sine function is known to have a period of $(2\pi)$.  This means that with respect to the sine function, Constraint-2 implies that there can be no positive real number $u < (2\pi)$ that satisfies Constraint-1.

Thus, the assumption that $f(x)$ had some period less than $(2\pi/a)$ yielded a contradiction.  Therefore, $(2\pi/a)$ not only satisfies Constraint-1, but also satisfies Constraint-2.


问:
If $p$ is a prime then all the non trivial subgroups of $G$ with $\lvert G\rvert=p^2$ are cyclic

If $p$ is a prime then all the non trivial subgroups of $G$ with $\lvert G\rvert=p^2$ are cyclic.

I tried looking online where does this result come from, but could not find any direct result. I found that all groups with order $p^2$ are isomorphic to $\mathbb{Z}_{p^2}$ or $\mathbb{Z_p} \times \mathbb{Z_p}$; is this a consequence of this result?

答:
You do not need the classification of groups of order $p^2$. Just use Lagrange theorem to conclude that a nontrivial subgroup (that is, different of the trivial one and $G$) has order $p$.

Now, groups of prime order  $p$ are known to be cyclic (the order of a nontrivial element cannot be $1$, so it is $p$).


问:
MLE for $\mu$ for Normal distribution
I am doing some homework and have a little trouble with the following problem:

Suppose the weight $X$ in pounds of a girl in a certain town is normally distributed with $N(\mu,15^2)$ while the weight $Y$ of a boy in this town is normally distributed $N(1.3 \mu, 20^2).$ The weights of randomly chosen girl and boy are $x = 95$ and $y = 130$ pounds respectively.

Then I have to find the MLE of $\mu$.

However, I know that the MLE of $\mu$ for a normal distribution is simply the sample mean, thus

$$MLE_{\mu} = \frac{1}{n} \sum_{i = 1}^n X_i = \frac{95 + 130}{2} = 112.5$$

Is this simply it?

TIA for any help.

答:
It is not this simple.

You have got two singleton samples from two different normal populations $N(\mu,15^2),N(1.3\mu,20^2)$. The joint distribution of the weight of a boy and girl is

$$l(x,y)=\frac{\exp\left(-\frac{(x-\mu)^2}{2(15^2)}-\frac{(y-1.3\mu)^2}{2(20^2)}\right)}{2\pi(15)(20)}$$

This likelihood function attains maximum at $$\hat\mu_{MLE}=\frac{\frac x{15^2}+\frac{1.3y}{20^2}}{\frac1{15^2}+\frac{1.3^2}{20^2}}$$

Title: MLE for $\mu$ for Normal distribution


问:
Finding $a \pmod c$ if $a \pmod b$ is known
Suppose that:

$Y \pmod B = 0$

$Y \pmod C = X$

I know $B$ and $C$. $Y$ is unknown, it might be an extremely large number, and it does not interest me.

The question is: Is it possible to find $X$, and if so, how?

答:
In the case where $b$ aand $c$ are relatively prime, knowing $y\bmod b$ gives absolutely no information about $y\bmod c$. It could be any of $0,1,2, \dots,c-1$.

In the general case where $b$ and $c$ are not necessarily prime, let $d=\gcd(b,c)$. Knowing $y\bmod b$ tells us what $y\bmod d$ is. But $y\bmod c$ can take on any value compatible with the known value of $y\bmod d$. That is, if $y\bmod d=y_1$, then $y\bmod c$ can take on any of the values $y_1,y_1+d, y_1+2d, \dots,y_1+kd$, where $k=\frac{c}{d}-1$.


问:
Visualizing conformal map from unit disk to upper half plane
I understand that
$$w=\frac{z-i}{z+i}$$
maps the open upper half plane $\mathbb{H}$ to the open unit disk $\mathbb{D}$, so
$$z=i\frac{1+w}{1-w}$$
maps from $\mathbb{D}$ to $\mathbb{H}$. My question is how? Are there any good GIFs or visuals that could show which points map to where?

I am asking because ultimately I want to determine how the right half unit disk would map using this same map.

答: I have been looking for any GIF which illustrates this transformation, but haven't found any... So I decided to create my own. I made this GIF with a little python code. I have uploaded this code I just made here. I hope you like it!</p>


问:
Proving that $SU(n)$ is connected
We know that, for $U(n)$, every unitary matrix has an orthonormal basis of eigenvectors, with eigenvalues of the form $e^{i \theta}$. We can decompose it in its eigenbasis and define a path such that $f:I \longrightarrow U(n)$ that satisfies the following: $f(0) = U$ and $ f(1) = I$. Thus any $2$ elements $U$ and $V$ of $U(n)$ can be connected to each other by a continuos path.

Can we do a similar procedure to prove that $SU(n)$ is connected? If so how?

答: Let $A,B\in SU(n)$ and consider a path $\gamma:[0,1]\to U(n)$ joining $A$ to $B$. As you've pointed out, the determinant of a unitary matrix $V$ is a complex number of norm $1$ and therefore $R_{V}= \begin{pmatrix}\frac1{\det V}&0\\ 0& I_{n-1}\end{pmatrix}$ is unitary. Now the path $\rho(t)=R_{\gamma(t)}\gamma(t)$ is in $SU(n)$ and it joins $A$ to $B$.

Title: Proving that $SU(n)$ is connected


问:
Why $\lim\limits_{z\to\infty}\frac{P(z)}{Q(z)}=\lim\limits_{y\to\infty}\frac{P(iy)}{Q(iy)}$?
While reading about the concept of $L$-stability I came across a result that said that for a rational (complex) function $R(z)=P(z)/Q(z)$ where $P,Q$ are polyomials we have $$\lim\limits_{z\to\infty}\frac{P(z)}{Q(z)}=\lim\limits_{y\to\infty}\frac{P(iy)}{Q(iy)},$$ where $y$ is real and $i^2=-1$. In words, the limit when moving on the real axis to infinity is the same as the limit when moving to infinity on the imaginary axis. I don't really know much about complex analysis but I was wondering why this is the case.

答: If $\deg P(z)>\deg Q(z)$, both limits are infinity. If $\deg P(z)<\deg Q(z)$, both limits are $0$. Otherwise, both limits are equal to$$\frac{\text{leading coefficient of }P(z)}{\text{leading coefficient of }Q(z)}.$$More generally, in each case there is the limit $\lim_{z\to\infty}\frac{P(z)}{Q(z)}$, and so the direction that you take as you go to $\infty$ doesn't matter.


问:
Product of connected spaces
You have two connected topological spaces $(A,B)$. Prove that $A\times B$ is also connected.

I understand that I have to prove that there is a point in $B$ (call it $b$), that makes $A\times\{b\}$ homeomorphic to $A$ making it connected to $A\times B$. Then prove that $\{a\}\times B$ is connected in $A\times B$. But I don't really know where to being with this. If you could help that would be appreciated.

答:
Suppose $A,B$ are connected. To show that $A\times B$ is also connected, it suffices to show that any continuous map $f: A\times B\to \{0,1\}$ is constant. Let's do this right away.

Consider some continuous map $f: A\times B\to \{0,1\}$, and fix $(a_1,b_1) \in A\times B$. Now define $g: A\to\{0,1\}$ so that $g:a\mapsto f(a,b_1)$, and $h:B\to\{0,1\}$ so that $h: b\mapsto f(a_1,b)$. Since $A,B$ are connected, $g$ and $h$ must be necessarily constant. So, $f(a_1,b) = f(a,b_1) = f(a_1,b_1)$ for all $(a,b)\in A\times B$.

Now, fix some other $(a_2,b_2)\in A\times B$, i.e. $(a_1,b_1) \ne (a_2,b_2)$. Repeat the same process as above to obtain $f(a_2,b) = f(a,b_2) = f(a_2,b_2)$ for all $(a,b)\in A\times B$.

In $f(a_1,b) = f(a_1,b_1)$, put $b = b_2$ to get $f(a_1,b_1) = f(a_1,b_2)$. In $f(a,b_2) = f(a_2,b_2)$ put $a = a_1$ to get $f(a_2,b_2) = f(a_1,b_2)$.

This gives $$f(a_1,b_1) = f(a_2,b_2)$$
The choice of $(a_1,b_1)$ and $(a_2,b_2)$ was arbitrary, so indeed $f(a_1,b_1) = f(a_2,b_2)$ for all $(a_1,b_1), (a_2,b_2) \in A\times B$. $f$ is a constant function. Done! $\ \ \blacksquare$

This can be generalized! If $A_1,A_2,\ldots,A_n$ are connected, then so is $A_1\times A_2\times \ldots\times A_n$. You can see this in at least two ways. One, try using induction on $n$, and the result we proved above for $n=2$. $n=2$ is also the base case of induction. Alternatively, you may prove this from scratch by a similar procedure, i.e. selecting $(a_1^{(k)},a_2^{(k)},\ldots,a_n^{(k)}) \in A_1\times A_2\times \ldots\times A_n$ for $k=1,2,\ldots,n$. This is more cumbersome and might be difficult to follow, but the essence remains the same!

Food for Thought:

At this point, I wonder if this can be extended to arbitrary products - i.e. if $\{A_i\}_{i\in I}$ constitute a collection of connected sets, is $\prod_{i\in I} A_i$ also connected? (The answer is Yes).

Title: Product of connected spaces


问:
Why is the following a probability measure?

Suppose $(\Omega,\mathcal F,P)$ is a probability space. Set $P^*(A)=\inf\{P(B):A\subseteq B,B\in\mathcal F\}$ for any $A\subseteq\Omega$. Take $A_0\notin \mathcal F$ such that $P^*(A_0)>0$ and define $P_0(A\cap A_0)=\dfrac{P^*(A\cap A_0)}{P^*(A_0)}$ for any $A\in\mathcal F$. Show that $P_0$ is a probability on $\mathcal F\cap A_0:=\{A\cap A_0:A\in\mathcal F\}$.

Countable additivity is what is turning out to be tricky to prove. Clearly I want to show that if $A_n,n\geq1$ are such that $A_n\in\mathcal F$ for all $n$ and for $k\neq l$, $A_k\cap A_l\cap A_0=\phi$ then $P^*(\cup_n(A_n\cap A_0))=\sum_n P^*(A_n\cap A_0)$. Note that $P^*$ is an outer measure and so combining with countable subadditivity of $P^*$, we only need to show that $P^*(\cup_n(A_n\cap A_0))\geq\sum_n P^*(A_n\cap A_0)$.

So what I tried was, I took any $B\in\mathcal F$ such that $B\supset \cup_n (A_n\cap A_0)$ and I wanted to show $P(B)\geq \sum_n P^*(A_n\cap A_0)$. I thought that if I define $B_n=B\cap A_n\in\mathcal F$ then $B_n$ would be disjoint and I could write $P(B)\geq\sum_n P(B_n)$, and $B_n\supset A_n\cap A_0$, but the issue is that $B_n$ are not necessarily disjoint.

Any help is appreciated.

答:
I have a suggestetion which might help you. Define an outer measure

$$\nu(A):=P^*(A\cap A_0) \quad \text{for all} \quad A\subseteq \Omega.$$

Since an outer measure is a measure when restricted to its $\nu$-measurable sets, it will be enough to show that any $A\in \mathcal{F}$ satisfies

$$ \nu(D)= \nu(D\cap A)+\nu(D\cap A^c) $$

for all $D\subseteq \Omega$. By definition of $P^*$, for all $\epsilon>0$ there exists $D\cap A_0 \subseteq D_\epsilon \in \mathcal{F}$ such that $P(D_\epsilon)< P^*(D\cap A_0)+\epsilon$. Hence,

$$ P(D_\epsilon) \overset{A\in \mathcal{F}}{=} P(D_\epsilon \cap A)+P(D_\epsilon\cap A^c)\geq $$

$$  \geq P^*(D\cap A_0 \cap A)+P^*(D\cap A_0 \cap A^c). $$
Hence, $ P^*(D\cap A_0)\geq P^*(D\cap A_0 \cap A)+P^*(D\cap A_0 \cap A^c) $, which shows that $A$ is $\nu$-measurable and implies countable additivity.


问:
Questions about completeness axiom
I'm now going through the completeness axiom. What is it used for? What can you do with it? Why is it called completeness?
And beside that, how can you proof this theorem?
Suppose that S is a nonempty subset of R and k is an upper bound of S. Then k is the least upper bound of S if and only if for each $\epsilon > 0$ there exists $s \in S$ such that $k - \epsilon < s$.
I tried picking a random $\epsilon$, but then I come to the point $k-s<-\epsilon$..</p>

答:
completeness axiom is ''every non empty subset of R which is bounded above has the least upper bound''
or,
''every non empty subset of R which is bounded below has the greatest lower bound''
The above two are axioms so there is no need to prove it, if any one wants to prove 1st of axioms then he has to use 2nd or for trying to prove 2nd then must aply 1st.</p>

Title: Questions about completeness axiom


问:
Existence and unicity ODE
I've been reading "A text book on ordinary differential equations" of Sahir Ahmad & Ambrossi. On page 37 we have the following excercise:

"Explain why $x'+\dfrac{\sin(t)}{e^t+1}x=0$ cannot has solution $x(t)$ such that $x(1)=1$ and $x(2)=-1$."

As far I know, $f(t,x)$ is continuos everywhere since $e^t\ne -1$ for every $t$, and $\frac{\partial f}{\partial x}$ doesn't depends on $x$, so its constant and therefore continous, then the existence and unicity theorem must apply on neighborhood of those points. Where am I wrong? Can anyone explain  me? Im very stuck af this assignament.

答:
Hint:

$$\left|\frac{\sin(t)}{1+e^t}x-\frac{\sin(t)}{1+e^t}y\right|\leq |x-y|,$$
for all $x,y\in\mathbb R$.

Title: Existence and unicity ODE


问:
Probability of an Event A succeeding, if it constantly repeats another Event B when failing, which calls Event A on its' failiure
Essentially, I have events A and B. $P(A) = \frac{1}{2}$, and $P(B) =\frac{5}{9}$. Event A is called first, and if it fails, event B is called. If event B fails, event A is called again. How would I go about calculating the exact probability of event A? I know I can approximate it using something along the lines of: $$\sum_{i=0}^n\left(\frac{1}{2}\cdot\frac{4}{9}\right)^i\cdot\frac{1}{2}$$ But I'm certain there is a much simpler way of arriving at the exact answer (which I happen to know is $\frac{9}{14}$, I just don't know the proper method of calculating it).

答:
$x = \frac{1}{2} + \left(\frac{1}{2} \times \frac{4}{9} \times x\right) \implies $

$\frac{14}{18}x = \frac{1}{2} \implies x = \frac{18 \times 1}{14 \times 2}.$

The idea behind the 1st line above, is that if events $A$ and $B$ both fail, you are right back where you started.


问:
A travelled inequality found by discriminant

Given three real numbers $x, y, z$ so that $1\leq x, y, z\leq 8.$ Prove that
$$\sum\limits_{cyc}\frac{x}{y}\geq\sum\limits_{cyc}\frac{2x}{y+ z}$$

I found this inequality by discriminant, we realised the homogenous, the generality of this problem. We assume $y= 1$ that leads $x^{2}+ x^{3}- zx- zx^{4}- 3z^{2}x^{2}+ z^{2}x^{4}+ z^{3}+ z^{3}x^{3}+ z^{4}- z^{4}x\geq 0.$ By jp.Wolfram|Alpha_ https://ja.wolframalpha.com/input/?i=discriminant%5Bx%5E2%2Bx%5E3-zx-zx%5E4-3z%5E2x%5E2%2Bz%5E2x%5E4%2Bz%5E3%2Bz%5E3x%5E3%2Bz%5E4-z%5E4x%2Cx%5D%3C%3D0 _of course $constant= 8$ is not the best here I think the one way that we can deal it is using substitution in $x, y, z$ but inhomogenous. I need to the help, thanks a real lot

Body:
Eliminating denominators, we arrive to the inequality $\;f(x,y,z)\ge 0,\;$ where
$$\begin{align}
&f(x,y,z) = x^3 y^3 + y^3 z^3 + x^3 z^3 - 3 x^2 y^2 z^2
\\[4pt]&
 + x^2 y^4 + y^2 z^4 + z^2 x^4 - x^4 y z - x y^4 z - x y z^4.
\end{align}\tag1$$
Since $(1)$ is homogenius and has the rotational symmetry, then WLOG there are two possible cases:

$\mathbf{1=x\le y \le z}.\;$ Assume $\;y=1+u,\; z=y+v,\;$ then
$$\begin{align}
&f(1,1+u,1+u+v)=u^4(v^2+v)+u^3(5v^3+11v^2+7v+2)
\\[4pt]&
+u^2(9v^4+30v^3+33v^2+15v+4)
+u(7v^5+30v^4+44v^3+25v^2+4v)
\\[4pt]&
+2v^6+10v^5+18v^4+14v^3+4v^2 \ge0,
\end{align}$$
and the case is proved.

$\mathbf{1=z\le y\le x\le 1+p}.\;$ Assume $\;y=1+pv,\; x=1+p,\;$ then

$$\begin{align}
&g(p,v)=\dfrac1{p^2}f(1+p,1+pv,1) = (p^4+p^3)v^4+(p^4+7p^3+7p^2+2p)v^3
\\[4pt]&
+(3p^3+12p^2+9p+4)v^2+(-p^3-p^2+p-4)v+2p+4.
\end{align}\tag2$$
If $\;v=\dfrac1p,\;p\to\infty,\;$ then the given inequality does not satisfy.

If $\;\mathbf{p=7},\;$ then
$$g(7,v) = 18 -389v+1684v^2+5159v^3+2744v^4
$$$$
=(18+43v)(1-12v)^2 +13(3v-14v^2)^2 + 7v^2+59v^3+196v^4 \ge 0,$$
and the case is proved.

The greatest value of $\;p\;$ can be defined from the system $\;g'_v =g=0,\;$
with the numerical solution $\;p\approx7.07031\,16307,\; v\approx 0.08248\,18190.\;$

The best constant is 8.07031 16307.


问:
Fraction of probability terms
I repost the following question, since it has been deleted.

Let $A$ and $B$ be independent events in a probability space which are
each less than certain. Determine the value of the following quantity.
\begin{align*}
     \frac{1 - P(A \cup B)}{(1 - P(A))(1 - P(B))} \end{align*}

I hope you find this question interesting as well.

答:
We should know that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$,

where $P(A\cap B)=P(A)\cdot P(B)$, if $A$ and $B$ are independent events.

So we replace $P(A\cup B)$ and multiply out the brackets at the denominator. Finally we compare the terms at the numerator and at the denominator.

$$\begin{align*}
   \ldots= \frac{1 - P(A)-P(B)+P(A)\cdot P(B)}{(1 - P(A))(1 - P(B))}=\frac{1 - P(A)-P(B)+P(A)\cdot P(B)}{1-P(B)-P(A)+P(A)\cdot P(B)}=1
\end{align*}
$$

Moreover the exercise tells us that the events $A$ and $B$ are in probability space which are each less than certain. That means that $P(A),P(B)<1\Rightarrow 1-P(A),1-P(B)>0$. So the denominator is not zero, which is a necessary condition for a valid fraction.

Title: Fraction of probability terms


问:
Correspondence between divergent integrals and analythic functionals
The Dirac delta function can be defined the following way:

$$\delta(x)
= \frac{1}{2\pi}\int_{-\infty}^\infty e^{itx}\, dt={\displaystyle {\begin{cases}\frac1{2\pi}\int_{-\infty}^\infty dt,&{\text{if }}x=0\\0,&{\text{if }}x\ne 0.\end{cases}}}$$

The later formula formally follows from the Fourier transform definition.

We also know that $\int_{-\infty}^\infty f(x)\delta(x)dx=f(0)$

It seems that if the expression under integral has a singularity at a point, the integral may gain wonderful changes there.
Some other examples:

$$\delta''(x)= -\frac{1}{2\pi}\int_{-\infty}^\infty t^2 e^{itx}\, dt={\begin{cases}-\frac1{2\pi}\int_{-\infty}^\infty t^2 dt,&{\text{if }}x=0\\0,&{\text{if }}x\ne 0.\end{cases}}$$

And $\int_{-\infty}^\infty f(x)\delta''(x)dx=f''(0)$

Another example $(x\in\mathbb{R})$:
$$\delta(ai+x)= \frac{1}{2\pi}\int_{-\infty}^\infty e^{-at} e^{itx}\, dt={\begin{cases}\frac1{2\pi}\int_{-\infty}^\infty e^{-at} dt,&{\text{if }}x=0\\0,&{\text{if }}x\ne 0.\end{cases}}$$

And we know that $\int_{-\infty}^{\infty}f(x)\delta(ai+x)dx=f(-ai)$

So, given a divergent integral $S$, an expression $$\phi(x)={\begin{cases}S,&{\text{if }}x=0\\0,&{\text{if }}x\ne 0.\end{cases}}$$

What can we say about the functional $\int_{-\infty}^\infty f(x)\phi(x)dx$? Can we, knowing the functional, find the corresponding divergent integral $S$?

It seems to me, the expressions under the Fourier transform are similar to the algebra of linear operators.

答:
Distributions don't work this way. $\delta$ and its derivatives are not functions, that's not what $\delta(x)
= \frac{1}{2\pi}\int_{-\infty}^\infty e^{itx}\, dt$ means.

What it means is that $$\delta(x) =\lim_{T\to \infty} \frac{1}{2\pi}\int_{-T}^T e^{itx}\, dt$$ where the limit is in the sense of distributions, which means that for all $\phi \in C^\infty_c(\Bbb{R})$ $$\phi(0)=\langle \delta,\phi \rangle=\lim_{T\to \infty} \int_{-\infty}^\infty \phi(x)(\frac{1}{2\pi}\int_{-T}^T e^{itx}\, dt) dx$$

Then

$$\phi^{(k)}(0)=\langle (-1)^k\delta^{(k)},\phi \rangle=\lim_{T\to \infty} \int_{-\infty}^\infty \phi(x)(\frac{1}{2\pi}\int_{-T}^T (-it)^k e^{itx}\, dt) dx$$

For $a\in \Bbb{C}$ the convergence of $\delta(x-a) =\lim_{T\to \infty} \frac{1}{2\pi}\int_{-T}^T e^{-i at} e^{itx}\, dt$ in the sense of analytic functionals follows the same idea.


问:
$\cos(t)$ as a function of (t-2)?
As the title suggests,
I'm doing a Laplace transform problem using the $t$-shift theorem, and I've almost got it, I just can't work out how to transform $\cos(t)$ into a function of $(t-2)$, I saw a trig identity

$$\cos(x)\cos(y) = \frac{1}{2} [\cos(x - y) + \cos(x + y)]$$

But I can't quite work out how to re-arrange it,
any help much appreciated.

Thanks.

答:
$\cos t = \cos((t-2)+2)$ is a function of $t-2$. If you prefer to express it as a linear combination of $\sin (t-2)$ and $\cos (t-2)$, you can indeed use a trig identity:

$\cos (x+y) = \cos x \cos y - \sin x \sin y$ will work (with $x=t-2$ and $y=2$).

Title: $\cos(t)$ as a function of (t-2)?


问:
How to calculate the matrix for an object mirroring over a point?
I have found the following task:
Give the homogeneous 4x4 matrix describing a object mirroring over the point  $P( 1 , 3 , 0 )$ and there solution $\left[\begin{array}{cccc} -1 & 0 & 0 & 2 \\ 0 & -1 & 0 & 6 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$
But how to they come to that solution?

Body:
The reflection of $(x,y,z)$ about $P(1,3,0)$ is $(x_1,y_1,z_1)$ iff the mid-point of $(x,y,z)$ and $(x_1,y_1,z_1)$ is $P$.
Thus$$\frac{x+x_1}2=1\\\frac{y+y_1}2=3\\\frac{z+z_1}2=0$$from where we get the reflection map $(x,y,z)\mapsto(x_1,y_1,z_1)=(2-x,6-y,-z)=(2,6,0)+(-x,-y,-z)$.


答:
Let $F$ be an antiderivative of the function $ \frac{ \sin t}{e^t+1}.$ Then the general solution of the differential equation is given by

$$x(t)=C e^{-F(t)},$$

where $C \in \mathbb R.$

From $1=x(1)= Ce^{-F(1)}$ we get $C>0.$

But from $-1=x(2)=Ce^{-F(2)}$ we get $C<0.$

Contradiction !

Title: Existence and unicity ODE

Body:
I've been reading "A text book on ordinary differential equations" of Sahir Ahmad & Ambrossi. On page 37 we have the following excercise:

"Explain why $x'+\dfrac{\sin(t)}{e^t+1}x=0$ cannot has solution $x(t)$ such that $x(1)=1$ and $x(2)=-1$."

As far I know, $f(t,x)$ is continuos everywhere since $e^t\ne -1$ for every $t$, and $\frac{\partial f}{\partial x}$ doesn't depends on $x$, so its constant and therefore continous, then the existence and unicity theorem must apply on neighborhood of those points. Where am I wrong? Can anyone explain  me? Im very stuck af this assignament.


问:
Probability where there is dependent event
question:

A lot contains $10$ items out of which $3$ items are defective. If three items are chosen at random without replacement find probability that only first one is defective?

My approach: Since selection of items is concerned with order I cannot use combination formula. I can get the answer by doing $(3/10)(7/9)(6/8) = 7/40$. But the events are dependent which means i shouldn't be allowed to use multiplication theorem. Please help.

答: Note that $7/9$ or $6/8$ is not the probablity that a ball is not defective in the original ten items, which indicates that you have not used the multiplication theorem. You have used the conditional probablity relation$$P(\text{B1 defective and B2 ok and B3 ok})=P(\text{B3 ok | B1 defective and B2 ok})P(\text{B2 ok | B1 defective})P(\text{B1 defective})$$


问:
Closed set in product topology which is not in any product $\sigma$-algebra
Definition On the Cartesian product $Y\times Z$ let $\mathcal{B}\otimes\mathcal{C}$ be the $\sigma$-algebra generated by the set of all "rectangles" $B\times C$ with $B\in\mathcal{B}$ and $C\in\mathcal{C}$. Then $\mathcal{B}\otimes\mathcal{C}$ is called the product $\sigma$-algebra on $Y\times Z$.

Definition The collection $2^\mathbb{N}$ of all subsets of a countably infinite set is said to have cardinality
$c$, or the cardinality of the continuum.

Show that for some topological spaces $(X,\mathcal{S})$ and $(Y,\mathcal{T})$, there is a closed set $D$ in $X\times Y$ with product topology which is not in any product $\sigma$-algebra $\mathcal{A}\otimes\mathcal{B}$, for example if $\mathcal{A}$ and $\mathcal{B}$ are the Borel $\sigma$-algebras for the given topologies.

Hint: Let $X=Y$ be a set with cardinality greater than $c$, for example, the set $2^I$ of all subsets of $I:=[0,1]$. Let $D$ be the diagonal {$(x,x):x\in X$}. Show that for each $C\in\mathcal{A}\otimes\mathcal{B}$, there are sequences {$A_n$} $\subset\mathcal{A}$ and {$B_n$} $\subset\mathcal{B}$ such that $C$ is in the $\sigma$-algebra generated by {$A_n\times B_n$}$_{n\geq1}$. For each $n$, let $x=_nu$ mean $x\in A_n$ if and only if $u\in A_n$. Define a relation $x\equiv u$ iff for all $n,x=_n u$. Show that this is an equivalence relation which has at most $c$ different equivalence classes, and for any $x,y,$ and $u$, if $x\equiv u$, then $(x,y)\in C$ if and only if $(u,y)\in C$. For $C=D$ and $y=x$, find a contradiction.

答:
Let $X=Y$ be the set of all functions from $I$ into {0, 1}. Let $X$ have the product topology. Then the basis element of $X$ is in the form of $\prod_{i\in I}U_i$ where $U_i=$ {0} or {1} for finitely many $i$ and $U_i=$ {0, 1} for all the remaining $i$'s. It happens that an open set of $X$ is also in this form. $\mathcal{A}$ is generated by $\prod_{i\in I}U_i$. By taking countable intersection, $\prod_{i\in I}U_i$ with countably many $U_i=$ {0} or {1} is also in $\mathcal{A}$. Taking complement and union do not produce more elements. Thus each member of $\mathcal{A}$ is in the form of $\prod_{i\in I}U_i$ with countably many $U_i=$ {0} or {1}.

Note that $\mathcal{A}=\mathcal{B}$. $\mathcal{A}\otimes\mathcal{B}$ is generated by $\prod_{i\in I}U_i\times\prod_{j\in I}U_j$ with countably many $U_i=$ {0} or {1}. Making complement, intersection and union do not produce new members. Thus each $C\in\mathcal{A}\otimes\mathcal{B}$ is in the form of $\prod_{i\in I}U_i\times\prod_{j\in I}U_j$ with countably many $U_i=$ {0} or {1}. Take out the $U_i$'s with $U_i=$ {0} or {1}. Order the index $i$ with positive integers. Let $f(i)$ be the corresponding integer for $i$. Then $f(i)$ is injective. Similarly, take out the $U_j$'s with $U_j=$ {0} or {1}. Order the index $j$ with positive integers. Let $g(j)$ be the corresponding integer for $j$. Let $A_n=\prod_{i\in I}U_i$ with $U_i=$ {0, 1} for all $i$ except $i=f^{-1}(n)$. Let $U_{f^{-1}(n)}$ of $A_n$ equal to the $U_{f^{-1}(n)}$ of the first coordinate of $C$. Similarly, let $B_n=\prod_{j\in I}U_j$ with $U_j=$ {0, 1} for all $i$ except $j=g^{-1}(n)$. Let $U_{g^{-1}(n)}$ of $B_n$ equal to the $U_{g^{-1}(n)}$ of the second coordinate of $C$. Then $C$ is in the $\sigma$-algebra generated by {$A_n\times B_n$}$_{n\geq1}$.

For each $n$, let $x=_nu$ mean $x\in A_n$ if and only if $u\in A_n$. $x=_nu$ iff $x$ and $u$ have the same coordinate at $i=f^{-1}(n)$. $x\equiv u$ iff $x$ and $u$ have the same coordinates at $i=f^{-1}(n)$ for all $n$. This is an equivalence relation. There are at most $c$ different equivalence classes since it is either 0 or 1 at these countably many coordinates.

Note that $(x,y)\in C$ iff the $i=f^{-1}(n)$ coordinates of $x$ and the $j=g^{-1}(n)$ coordinates of $y$ are in $\pi_i(A_n)$ and $\pi_j(B_n)$, respectively, where $\pi_i$ is the coordinate projection function. For any $x,y,$ and $u$, if $x\equiv u$, then $(x,y)\in C$ if and only if $(u,y)\in C$. This holds since the $y$-part has been taken care and $x\equiv u$ guarantees the coordinate requirement of $x$-part.

For $C=D$ and $y=x$, obviously $(x,y)\in C$. Let $x\equiv u$. Then $(u,y)\in C$. This is a contradiction. The definition of $D$ requires strict diagonal, i.e., all the uncountably many coordinates in $I$ should be the same, but $x\equiv u$ only requires countably many of them are same.

$D$ is diagonal, so it is closed in $X\times Y$. But $D$ is not in any product $\sigma$-algebra $\mathcal{A}\otimes\mathcal{B}$. If it is in, a contradiction occurs as is shown.


问:
Solution of a trigonometric linear equation with the method of the added angle
This morning I have done for my students of an high school an exercise with the method of the added angle. I not write all the steps but the principals.
The equation is:

$$\bbox[5px,border:3px solid #FF7F50]{-2\sin x+\cos x = 1} \tag 1$$

Surely $x=2\Bbb Z \pi$ it is solution of the $(1)$. In fact it is an identity. Being $$\tan \varphi=-\frac 12 \implies \varphi =-\arctan \frac 12, \quad A=\sqrt{5}$$

Now the $(1)$ becomes

$$\bbox[5px,border:3px solid #AA7F50]{\sqrt 5 \sin \left(x-\arctan \frac 12\right)=1} \tag 2$$

Hence I will have the other two solutions:

$$x=\arctan \frac 12+\arcsin\left( \frac{\sqrt{5}}5\right)+2\Bbb Z\pi$$ and $$ \quad x=\pi +\arctan \frac 12-\arcsin\left( \frac{\sqrt{5}}5\right)+2\Bbb Z\pi \tag 3$$

The solution of my textbook are $x=k\pi$ and $x=\alpha+2k\pi$ where $\cos \alpha =-3/5$ and $\sin \alpha =-4/5$.

Please, do can be that the $(3)$ are equivalent to $x=2k\pi$ and $x=\alpha+2k\pi$ where $\cos \alpha =-3/5$ and $\sin \alpha =-4/5$?

答:
Putting together the two comments we have that the initial equation is equivalent to
$$\sqrt 5 \sin\left(x - \arctan \frac12\right) = -1$$
Leading to the solutions
$$ x = \arctan \frac12 - \arcsin \frac{\sqrt 5}5 + 2k \pi = 2k \pi$$
and
$$x = \pi + \arcsin \frac{\sqrt 5}5 + \arctan \frac12 + 2k \pi=\underbrace{\pi + 2 \arcsin \frac{\sqrt 5}5}_{\alpha}+ 2 k \pi.\tag{1} \label{1}$$
Now let $\alpha = \pi + 2\arcsin \frac{\sqrt 5}5$. We get
$$\sin \alpha = -\sin\left( 2\arcsin \frac{\sqrt 5}5\right) = -2 \sin\left( \arcsin \frac{\sqrt 5}5\right)\cos\left(\arcsin\frac{\sqrt 5}5\right)=-\frac45,$$
and similarly for the cosine of $\alpha$.

Note that $k \pi$ is not a solution for odd $k$.


问:
How do I continue this Convolution Computation
$$
x(t):=\begin{cases}
1&\text{if $0<t<T$}\\ \\
0&\text{if otherwise}
\end{cases}
\qquad\text{and}\qquad
h(t):=\begin{cases}
t&\text{if $0<t<2T$}\\ \\
0&\text{if otherwise}
\end{cases}
$$
Compute $(x*h)(t)$.

My approach:
We first note that :$$\displaystyle x(\tau)=\begin{cases}
1&\text{if $0<\tau<T$}\\ \\
0&\text{if otherwise}
\end{cases}\qquad\text{and}\qquad\displaystyle h(t-\tau)=\begin{cases}
t-\tau&\text{if $0<t-\tau<2T$}\\ \\
0&\text{if otherwise}
\end{cases}$$ We observe that :
$$
0<\tau<T
$$
$$
t-2T<\tau<t
$$
Thus, $(x*h)(t)\in\mathcal{R}[\max(0,t-2T),\min(t,T)]$ (i.e. Riemann integrable). Therefore :
\begin{align*}
(x*h)(t)&:=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)\;\text{d}\tau\\
&=\int_{\max(0,t-2T)}^{\min(t,T)}t-\tau\;\text{d}\tau\\
&=\left. t\tau-\frac{\tau^{2}}{2}\right|_{\max(0,t-2T)}^{\min(t,T)}\\
&=\left[t\min(t,T)-\frac{\min^{2}(t,T)}{2}\right]-\left[t\max(0,t-2T)-\frac{\max^{2}(0,t-2T)}{2}\right]\\
&=t[\min(t,T)-\max(0,t-2T)]-\frac{1}{2}[\text{min}^{2}(t,T)-\text{max}^{2}(0,t-2T)]
\end{align*}
How do I go from here?

答:
You will probably have to discusses different cases according to the value of $t$.

It is slightly easier with the other paramatrization (or change of variable) and  the same arguments (if there is no mistake)
$$ (x\ast h)(t) = \int_{\mathbb{R}} x(t-\tau) \ h(\tau)\ d\tau = \int_{\max(t-T,0)}^{\min(t,2T)} x(t-\tau) \ h(\tau)\ d\tau = \int_{\max(t-T,0)}^{\min(t,2T)} \tau\ d\tau
= \left[ \frac{\tau^2}{2}\right]_{\max(t-T,0)}^{\min(t,2T)} $$
One may use the Heaviside "function" $\theta$ (but this is in fact just a rewriting...)
$$(x\ast h)(t) =  \frac{\big(2\ T - (2T -t)\ \theta(2T -t) \big)^2}{2} - \frac{\big( (t-T)\ \theta (t-T) \big)^2}{2}$$
The result is a function defined "part" by "part" by simple polynomials so there has to be a discussion as for $x$ and $h$, if $t$ sthg, then sthg.

What you probably want is the usual interpretation of convolution. If we considered $\int_{\mathbb{R}^2} f(x)\ g(y)\ dx\ dy$, the domain of integration is the plan. Now if we have a parametrized curve $\gamma: \mathbb{R} \to \mathbb{R}^2$ we considered $\int_{\mathbb{R}} f\big(\gamma_1(t)\big)\ g\big(\gamma_2(t)\big)\ \lambda_{\gamma(t)}\big(\gamma'(t)\big) \ dt $. ($\gamma'\in \mathbb{R}^2$ is a vector so I introduced a linear form $\lambda$ that depends on the point where are at). The "domain" of integration now is the trajectory of the path and the convolution product can be understood as a particular case. In comparison to the first integral, $x$ is restricted to the first component $\gamma_1$ of $\gamma$ while $y$ is restricted to the second component. For a convolution product $(f\ast g)(t)$ and a given $t$, the trajectory is the subset
$$\Gamma_t := \left\lbrace (x,y)\in \mathbb{R}^2,\ x\in \mathrm{supp}(f),\ y\in \mathrm{supp}(g),\ x+y =t\right\rbrace $$
but the condition $y=t-x$ (at $t$ fixed) in the plan $\mathbb{R}^2$ is precisely a "downward" line.

Let us look at the behavior of the integrand $f(x)\ g(y)$ with $(x,y)\in \Gamma_t$. If we moved only "horizontally" (only $x$ varies, doesn't happen if we want to stay in $\Gamma_t$), only $f$ varies and similarly if we move "vertically". A set of the form $x$ fixed, $y$ arbitrary corresponds to a vertical line and $\left\lbrace x\in \mathrm{supp}(f),\ y\in \mathbb{R} \right\rbrace $ to a vertical band. Support of $f(x)\ g(y)$ thus lies on a rectangle $\mathrm{supp}(f)\times \mathrm{supp}(g)$. For the convolution product, we thus have to look at the intersection $\Gamma_t$ with this rectangle.

Back to the example, $f$ is the function $x$ in the original post, just a cut-off function and $g$ is $h$ the unit monomial of order 1 (on a certain interval). With my parametrization, we put the support of $f:=x$ on the horizontal axis and that of $g:=h$ on the vertical axis. With my parametrization of the integral, the variable $x$ is "restricted" to $(t-\tau)$ while $y$ is "restricted" to $\tau$

So we should interpret $\tau$ as the vertical position, and just look at the portion given by the intersection of $\Gamma_t$ (downward line, 45°) with the rectangle (cartesian product of the supports), and integrate $\tau$ only for $\tau$ in the interval corresponding to vertical position corresponding to this portion. Sorry for this tedious description: make a drawing!!!


问:
If $XY-ZW=I_{n}$ and $XW+ZY=0$ show that $YX-ZW=I_{n}$ , $WX+YZ=0$ and $\det(XY)\geq 0$
$$X,Y,Z,W \in M_{n}(\mathbb{R})$$ with $n>2$, $$n\in\mathbb{N}$$
If $XY-ZW=I_{n}$ and $XW+ZY=0$ show that $YX-ZW=I_{n}$ , $WX+YZ=0$ and $\det(XY)\geq 0$

I know that $XW=-ZY$ but I have no idea how to continue and prove the above. Could you help me please ?

答:
Incorporating the correction suggested by Stinking Bishop, we want to show that if $X,Y,Z,W \in M_{n}(\mathbb{R})$ are such that
$$
\left\lbrace
\begin{align*}
&XY-ZW=I
\qquad\qquad\qquad
\\[4pt]
&XW+ZY=0
\\[4pt]
\end{align*}
\right.
$$
then
$$
\left\lbrace
\begin{align*}
&YX-WZ=I
\qquad\qquad\qquad
\\[4pt]
&WX+YZ=0
\\[4pt]
&\det(XY)\ge 0
\\[4pt]
\end{align*}
\right.
$$
Assume the hypothesis.

Following the approach outlined in Stinking Bishop's comment, if $A,B$ are given by
\begin{align*}
A&=X+iZ
\qquad\qquad\;\;\;\;\;\,
\\[4pt]
B&=Y+iW
\\[4pt]
\end{align*}
then
\begin{align*}
&
AB=(X+iZ)(Y+iW)
\\[4pt]
\implies\;&
AB=(XY-ZW)+i(XW+ZY)
\\[4pt]
\implies\;&
AB=I
\\[4pt]
\implies\;&
BA=I
\\[4pt]
\implies\;&
(Y+iW)(X+iZ)=I
\\[4pt]
\implies\;&
(YX-WZ)+i(WX+YZ)=I
\\[4pt]
\implies\;&
\begin{cases}
YX-WZ=I\\[4pt]
WX+YZ=0\\
\end{cases}
\\[4pt]
\end{align*}
which establishes the first two parts of the desired conclusion.

It remains to show $\det(XY)\ge 0$.

If $X$ or $Y$ is singular, then $\det(XY)=\det(X)\det(Y)=0$, and we're done.

So assume $X,Y$ are nonsingular.

From the equation $XW+ZY=0$, we get $W=-X^{-1}ZY$, hence
\begin{align*}
&
XY-ZW=I
\qquad\qquad\qquad\qquad\qquad\,
\\[4pt]
\implies\;&
XY-Z\bigl(-X^{-1}ZY\bigr)=I
\\[4pt]
\implies\;&
XY+ZX^{-1}ZY\bigr)=I
\\[4pt]
\implies\;&
XY+\bigl(ZX^{-1}ZX^{-1}\bigr)\bigl(XY\bigr)=I
\\[4pt]
\implies\;&
(XY)(I+T^2)=I
\\[4pt]
\end{align*}
where $T=ZX^{-1}$.

From $(XY)(I+T^2)=I$, we get that $I+T^2$ is nonsingular.

Since $I+T^2$ is nonsingular, we get that $\det(I+iT),\det(I-iT)$ are nonzero.

Since $T$ is a real matrix, we get that $\det(I+iT),\det(I-iT)$ are complex conjugates.

It follows that $\det(I+T^2) > 0$, hence from $(XY)(I+T^2)=I$, we get $\det(XY) > 0$.

This completes the proof.


答:
The relation:
$$
\arctan\frac12=\arcsin\frac1{\sqrt5}=\frac12\arcsin\frac45.\tag1
$$
will help you to compare the results.

Substituting the above values you will find that your result is "almost correct". The spoiler below shows which error gave rise to the deviation from the correct result.

To justify the second equality in (1) one observes the following facts
which hold for $\phi=\arcsin\frac1{\sqrt5}$:
$$
0<\frac1{\sqrt5}<\frac1{\sqrt2}\implies 0<\phi<\frac\pi4\implies 0<2\phi<\frac\pi2$$
and
$$
\sin2\phi=2\cos\phi\sin\phi=\frac45.
$$

Obviously you are using the method of the added angle in the form:$$A\sin x+B\cos x=C(\cos\alpha\sin x+\sin\alpha\cos x)=C\sin(x+\alpha)\tag2$$with$$C=\sqrt{A^2+B^2};\quad\cos\alpha=\frac AC;\quad\sin\alpha=\frac BC.\tag3$$It follows from (3) that $\tan\alpha=\frac BA$ but one should not conclude from this that $\alpha=\arctan\frac BA$ by the following reason. The range of $\arctan x$ is $\left[-\frac\pi2,\frac\pi2\right]$ whereas the equation (3) determines the angle $\alpha$ uniquely in the range $(-\pi,\pi]$. Particularly in your example with$$\cos\alpha=-\frac2{\sqrt5};\quad \sin\alpha=\frac1{\sqrt5},$$the angle $\alpha$ resides in the II quadrant whereas $\arctan\left(-\frac12\right)$ resides in the IV quadrant. Therefore the correct expression for the angle is:$$\alpha=\arctan\left(-\frac12\right)+\pi=\pi-\arctan\frac12,$$and the equation to solve becomes:$$\sqrt5\sin\left(x+\pi-\arctan\frac12\right)=\sqrt5\sin\left(\arctan\frac12-x\right)=1,\tag4$$which solution is:$$x=2\pi k+\begin{cases}\arctan\frac12-\arcsin\frac1{\sqrt5};\\\arctan\frac12+\arcsin\frac1{\sqrt5}-\pi,\end{cases}\tag5$$i.e. differs from your expression by the position of the term $\pi$. Finally with the help of (1) the equation (5) can be written as:$$x=2\pi k+\begin{cases}0;\\\arcsin\frac45-\pi.\end{cases}\tag6$$ It remains to observe that $\alpha=\arcsin\frac45-\pi$ indeed satisfies the relations $\sin\alpha=-\frac45,\,\cos\alpha=-\frac35$. Another issue is the error in the textbook solution (if you cited it correctly): the second solution is $2\pi k$, not $\pi k$.

Title: Solution of a trigonometric linear equation with the method of the added angle

Body:
This morning I have done for my students of an high school an exercise with the method of the added angle. I not write all the steps but the principals.
The equation is:

$$\bbox[5px,border:3px solid #FF7F50]{-2\sin x+\cos x = 1} \tag 1$$

Surely $x=2\Bbb Z \pi$ it is solution of the $(1)$. In fact it is an identity. Being $$\tan \varphi=-\frac 12 \implies \varphi =-\arctan \frac 12, \quad A=\sqrt{5}$$

Now the $(1)$ becomes

$$\bbox[5px,border:3px solid #AA7F50]{\sqrt 5 \sin \left(x-\arctan \frac 12\right)=1} \tag 2$$

Hence I will have the other two solutions:

$$x=\arctan \frac 12+\arcsin\left( \frac{\sqrt{5}}5\right)+2\Bbb Z\pi$$ and $$ \quad x=\pi +\arctan \frac 12-\arcsin\left( \frac{\sqrt{5}}5\right)+2\Bbb Z\pi \tag 3$$

The solution of my textbook are $x=k\pi$ and $x=\alpha+2k\pi$ where $\cos \alpha =-3/5$ and $\sin \alpha =-4/5$.

Please, do can be that the $(3)$ are equivalent to $x=2k\pi$ and $x=\alpha+2k\pi$ where $\cos \alpha =-3/5$ and $\sin \alpha =-4/5$?


问:
MLE for $\mu$ for Normal distribution
I am doing some homework and have a little trouble with the following problem:

Suppose the weight $X$ in pounds of a girl in a certain town is normally distributed with $N(\mu,15^2)$ while the weight $Y$ of a boy in this town is normally distributed $N(1.3 \mu, 20^2).$ The weights of randomly chosen girl and boy are $x = 95$ and $y = 130$ pounds respectively.

Then I have to find the MLE of $\mu$.

However, I know that the MLE of $\mu$ for a normal distribution is simply the sample mean, thus

$$MLE_{\mu} = \frac{1}{n} \sum_{i = 1}^n X_i = \frac{95 + 130}{2} = 112.5$$

Is this simply it?

TIA for any help.

答:
The trick to find $\hat{\mu}$ is to expand the exponent, waste any quantity not depending on $\mu$ and factorize the exponent in order to have again a perfect square (adding and subtracting a customized quantity).

$$L\propto \exp\left\{-\frac{1}{2}\left[ \frac{(x-\mu)^2}{15^2} + \frac{(y-1.3\mu)^2}{20^2}\right]\right\}=$$

$$=\exp\left\{ -\frac{1}{2}\left[ \mu^2\left(\frac{1}{15^2}+\frac{1.3^2}{20^2}   \right) -2\mu \left( \frac{x}{15^2}+\frac{1.3y}{20^2} \right)\right]  \right\}=$$

$$=\exp\left\{ -\frac{1}{2\frac{1}{\frac{1}{15^2}+\frac{1.3^2}{20^2}}}\left[ \mu^2 -2\mu \frac{ \frac{x}{15^2}+\frac{1.3y}{20^2}} {\frac{1}{15^2}+\frac{1.3^2}{20^2}}\pm ?\right]  \right\}$$

now if you look at the quantity in the square brackets,  you will recognize an incomplete square expression. After completing it, in the brackets you will have an expression like

$$[\mu-\theta]^2$$

and thus the MLE is the quantity near $2\mu$, corresponding to the quantity written by @Shubham Johri (+1)

Title: MLE for $\mu$ for Normal distribution


问:
How could I prove that $n + n^2O (\ln n) = O ( n^2\ln n)$?

How could I prove that $n + n^2O (\ln n) = O ( n^2\ln n)$?

I don't know how to handle the $n^2$ factor on the left side.

答:
If $g(n)\in O(\ln n)$, then for some $m\in\Bbb N,k>0$, we have $|g(n)|\le k\ln n$ for all $n\ge m$. Thus $|n^2g(n)|\le kn^2\ln n$ for all $n\ge m$, which gives $n^2g(n)\in O(n^2\ln n)$. Or concisely, $n^2 O(\ln  n)\subseteq O(n^2\ln n)$.

Similarly we can show that $O(n^2\ln n)\subseteq n^2O(\ln n)$ by observing that $|f(n)|\le k n^2\ln n$ implies $|f(n)/n^2|\le k\ln n$ for sufficiently large $n$, or $f(n)/n^2\in O(\ln n)$.

Can you complete?


答:
Alternative method:

$$\cos x=1+2\sin x\implies1-\sin^2x=1+4\sin x+4\sin^2x\implies\sin x(4+5\sin x)=0.$$

So the solutions are

$$\sin x=0,\cos x=1+2\cdot0=1$$
$$x=2k\pi$$ and

$$\sin x=-\frac45,\cos x=1+2\left(-\frac45\right)=-\frac35$$
$$x=\arctan\frac43+(2k+1)\pi.$$


问:
Finding $a \pmod c$ if $a \pmod b$ is known
Suppose that:

$Y \pmod B = 0$

$Y \pmod C = X$

I know $B$ and $C$. $Y$ is unknown, it might be an extremely large number, and it does not interest me.

The question is: Is it possible to find $X$, and if so, how?

答:
No; more information is needed. To see this, suppose that $B=2$ and $C=5$ and suppose that we know that $Y \bmod 2 = 0$ and we want to figure out $X = Y \bmod 5$. The possibilities for $X$ are not unique and depend on $Y$:

Since $2$ is a factor of $10$, we could have $Y=10$, which yields $X=0$.

Since $2$ is a factor of $12$, we could have $Y=12$, which yields $X=2$.

Since $2$ is a factor of $14$, we could have $Y=14$, which yields $X=4$.

Since $2$ is a factor of $16$, we could have $Y=16$, which yields $X=1$.

Since $2$ is a factor of $18$, we could have $Y=18$, which yields $X=3$.


问:
A conjugacy class of more than average size must have character value zero for some irreducible character.
I found the following claim here and wondered how to go about proving it?

Let $C$ be a conjugacy class of $G$ of more than average size, so $\frac{\#{G}}{\#{C}} <$ Number of Conjugacy Classes.

Show that there is an irreducible character $\chi$ of $G$ over $\mathbb{C}$ such that $\chi = 0$ on $C$.

答:
Proposition Let $Cl_G(g)$ be a conjugacy class of the finite group $G$ with $|C_G(g)| \lt k(G)$, the number of conjugacy classes of $G$. Then $\chi(g)=0$ for some $\chi \in Irr(G)$.

Proof By the Second Orthogonality Relation we have $|C_G(g)|=\sum_{\chi \in Irr(G)}|\chi(g)|^2$, hence $1 \gt \frac{1}{k(G)}\sum_{\chi \in Irr(G)}|\chi(g)|^2$. Let $G$ have exponent $n$, and put $K=\mathbb{Q}(e^{\frac{2\pi i}{n}})$, then by a theorem of R. Brauer (see Isaacs, CTFG, (10.3) Theorem) every $\chi \in Irr(G)$ is afforded by a $K$-representation. $\mathcal{G}=Gal(K/\mathbb{Q})$ acts on $Irr(G)$ by $\chi^{\sigma}(g)=\chi(g)^{\sigma}$ for every $\sigma \in \mathcal{G}$. Let $\mathcal{I}_1, \cdots ,\mathcal{I}_t$ be orbits of $Irr(G)$ under the action of $\mathcal{G}$. The geometric-arithmetic mean formula yields
$$1 \gt \frac{1}{k(G)}\sum_{\chi \in Irr(G)}|\chi(g)|^2 \geq (\prod_{\chi \in Irr(G)}|\chi(g)|^2)^{\frac{1}{k(G)}}=(\prod_{j=1}^t\prod_{\chi \in \mathcal{I}_j}|\chi(g)|^2)^{\frac{1}{k(G)}}$$ and hence
$$1 \gt \prod_{j=1}^t\prod_{\chi \in \mathcal{I}_j}|\chi(g)|^2 \geq 0.$$
Since complex conjugation commutes with any element of $\mathcal{G}$, each of the factors $\alpha_i=\prod_{\chi \in \mathcal{I}_j}|\chi(g)|^2$ is fixed by $\mathcal{G}$ and is an algebraic integer, each $\alpha_i$ is a rational integer. This can only happen when at least one of the $\alpha_i$'s equals zero. Hence $\chi(g)=0$ for some $\chi \in Irr(G)$.


问:
Why we can write $\frac{\partial \psi}{\partial x}+ \frac{\partial \psi}{\partial y}\frac{dy}{dx}$ as $\frac{d\psi(x,y)}{dx}$?
I'm reading Boyce/Diprima/Meade's book, in this example:

It's not clear to me why we can write $\frac{\partial \psi}{\partial x}+ \frac{\partial \psi}{\partial y}\frac{dy}{dx}$ as $\frac{d\psi(x,y)}{dx}$, I think we can write $\frac{d \psi(x,y)}{d y}\frac{dy}{dx}$ as $\frac{d \psi(x,y)}{d y}$ due to the chain rule, so we get:

$$\frac{d \psi(x,y)}{dx}+ \frac{d \psi(x,y)}{d y}$$

I computed this and obtained the original differential equation, so I might be on the right path. But I still don't get the part where we write it as

$$\frac{d\psi(x,y)}{dx}$$

Can you help me?

答:
This is actually a notation problem in calculus.  Unfortunately, partial differentials suffer from a terrible notation problem.  The actual truth about partial differentials is very straightforward: since $\psi$ is a function of independent variables $y$ and $x$, the total change of $\psi$ is the combination of the change in $\psi$ due to $x$ and the change in $\psi$ due to $y$.  This can be written straightforwardly as:

$$ d\psi = \partial_x \psi + \partial_y \psi$$

Here, the subscripts show which variable is allowed to change.

We can get from this to the equation as follows:

(1) Divide both sides by $dx$:

$$ \frac{d\psi}{dx} = \frac{\partial_x \psi}{dx} + \frac{\partial_y \psi}{dx} $$

(2) Multiply the right-hand term by $\frac{dy}{dy}$ (allowable since it is equivalent to 1):

$$ \frac{d\psi}{dx} = \frac{\partial_x \psi}{dx} + \frac{\partial_y \psi}{dx} \frac{dy}{dy} $$

(3) Use the commutative law to rearrange the rightmost term:

$$ \frac{d\psi}{dx} = \frac{\partial_x \psi}{dx} + \frac{\partial_y \psi}{dy} \frac{dy}{dx} $$

(4) Rewrite using modern calculus notation:

$$ \frac{d\psi}{dx} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y}\frac{dy}{dx} $$

Another note on the solution.

A more straightforward way to solve it, in my opinion, is to simply recognize the product rule.

The original equation is:

$$ 2x + y^2 + 2xy\frac{dy}{dx} = 0 $$

Multiplying through by $dx$ yields

$$ 2x\,dx + y^2\,dx + 2xy\,dy = 0 $$

Integrating both sides gives:

$$ \int \left(2x\,dx + y^2\,dx + 2xy\,dy\right) = \int 0 $$

We can break the integral at +s.  In this case, let's just break it at the first plus.

$$ \int 2x\,dx + \int \left(y^2\,dx + 2xy\,dy\right) = \int 0 $$

You can tell that the middle integral is the result of the product rule.  $u = y^2$ and $v = x$.  So that means the middle integral is just $xy^2$.

So the result is:

$$ x^2 + xy^2 = C $$

The whole transition through $\psi$ seems excessive, but it depends on what they are trying to teach.


问:
Finding $f$ such that $\forall x,y \in \mathbb{R},f(x+y)f(x-y)=(f(x))^2$
Find all the possible continuous functions $f : \mathbb{R} \to \mathbb{C}$ such that $|f|=1$ and $$\forall x,y \in \mathbb{R},f(x+y)f(x-y)=(f(x))^2.$$
By taking suitable values for $x,y$ we have $f(0)=1,f(2x)=(f(x))^2,$ also $x \to e^{icx},c \in \mathbb{R}$ verifies the above functional equation. Tried to find $f(nx)$ in term of $f(x),$ but unfortunately didn't find anything.

Any suggestions?

答:
It is easy to observe that $f(x) = \exp({\iota c(x)})$ where $c$ is a real valued funciton with domain same as $f$.

Now plugin the expression to get
$$ 2c(x) = c(x-y) + c(x+y) + 2k\pi $$
Now if we put $y=0$ we can easily see $k=0$ and as you observed the solutions of the equations are $ax+b$


问:
A logarithm base that converges to the sum of the harmonic series at infinity?
This is my first question on this platform, so i'm sorry if my writing & explanations are a bit off.

It is known that:
$$
\lim_{N \to \infty} \left (  \left (\sum_{n=1}^{N}\frac{1}{n} \right ) - \ln(N) \right )  = 0.5772156649... = \gamma
$$
The definition of the Euler–Mascheroni constant.

I will define the following function:
$$
f(x) = \lim_{N \to \infty} \left (  \left (\sum_{n=1}^{N}\frac{1}{n} \right ) - \log_{x}(N) \right )
$$

My question is, is there a well-known constant for $x$ that satisfies $f(x)=0$ ?

From simple numerical evaluation, I was able to see that $x \approx 2.563$, but I am not capable of even thinking if there is an analytic way to solve this.

答:
For $ N>1$ and $ x\ne 1$,

$$f(x)=\sum_{n=1}^N\frac 1n-\log_x(N)=$$

$$\sum_{n=1}^N\frac 1n-\ln(N)+\ln(N)(1-\frac{1}{\ln(x)})$$

the limit when $ N\to+\infty $ is infinite.


问:
Why does an exponential function eventually get bigger than a quadratic
I have seen the answer to this question and this one.

My $7$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don't like the way the question is written.

答:
..it might be just me, but I think we all are reading this question wrong. This is 7th grade, so I don't think the teacher is asking for a specific proof of that assertion. The answer might be way simpler.

Notice that the question asks how you know, not why this happens.

"How do you know an exponential expression will be larger than any quadratic expression?"

The answer for that would be:

"An exponential expression will be larger than any quadratic expression when the base is larger than 1".

My rationale is that this "feels" a lot like "how do you know a number is divisible by 3?", to which the answer would be "when the sum of its digits is also divisible by 3".


问:
Lipschitz constant problem
Let $u^{}$ a function in $\operatorname{Lip}_{k}(\Omega)$, that is the set of lipschitz continuous functions from $\Omega \subset \mathbb{R}^n$ to $\mathbb{R}$ with lip constant $\le k$. Denote by $\left[u^{}\right]$ the lipschitz constant of $u$.

Let $v \in \operatorname{Lip}(\Omega)$ and for $t \in[0,1]$ define
$$
v_{t}=u^{}+t\left(v-u^{}\right)
$$
Then I would have to get that $\left[v_t\right] < k$ but I cannot reach this.

What I tried is the following, by subadditivity of Lipschitz constants and homogeneity and by knowing $\left[u^{}\right] < k$:

$$\left[v_t\right] \le \left[u^{}\right] + t\left[v-u^{}\right] < k + t\left[v-u^{}\right]$$

Then I should be able to conclude from the last passage that it is strictly less than $k$. If $t$ is fixed in $(0,1)$ and the lipschitz constant $\left[v-u^{}\right]$ is always non-negative, how can I conclude that $k + t\left[v-u^{}\right] < k$, getting $\left[v_t\right]<k$? Thanks!

答:
Restate $v_t = u + t(v - u)$ as $v_t = t v + (1 - t)u$

Then for $w \in \Omega$ we have $v_t(w) = tv(w) + (1-t)u(w)$

So, $|v_t(w)| = |tv(w) + (1-t)u(w)| \le t|v(w)| + (1-t)|u(w)| \le tk|w| + (1 - t)k|w| = k|w|$

I.e. $\left[v_t\right] \le k$.

Title: Lipschitz constant problem