问:
Why does an odd perfect number seemingly "violate" basic inequality rules?
Let $p^k m^2$ be an odd perfect number (OPN) with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

My question is as is in the title:

Why does an OPN seemingly "violate" basic inequality rules?

Denote the sum of divisors of the positive integer $x$ by $\sigma(x)$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the abundancy index of $x$ by $I(x)=\sigma(x)/x$.

Note that it is trivial to prove that
$$\frac{p+1}{p} \leq I(p^k) < \frac{p}{p-1}$$
from which we obtain
$$\frac{2(p-1)}{p} < I(m^2) = \frac{2}{I(p^k)} \leq \frac{2p}{p+1}.$$
This implies that
$$\frac{2}{p+1} \leq \frac{D(m^2)}{m^2} < \frac{2}{p}.$$
Taking reciprocals, multiplying by $2$, and subtracting $1$, we get
$$p-1 < \frac{\sigma(m^2)}{D(m^2)} \leq p,$$
where we note that $(2/p) < p-1$.

Notice then how the OPN seems to "violate" the basic inequality rule

$$(a \leq b) \land (c \leq d) \implies (ac \leq bd).$$

Note that
$$I(m^2) = \frac{D(m^2)}{m^2}\cdot\frac{\sigma(m^2)}{D(m^2)},$$
and that the premise is false while the conclusion is true in the above implication.

答:
On OP's request, I am converting my comment into an answer.

The implication says only that if $a\le b$ and $c\le d$, then $ac\le bd$ (assuming that $a,c$ are non-negative).

It does not say that $ac\le bd$ only if $a\le b$ and $c\le d$.

(It is false that if $ac\le bd$, then $a\le b$ and $c\le d$.

For $(a,b,c,d)=(3,2,2,4)$, for example, we have $ac\le bd$ and $a\gt b$.)


问:
finding the functional square root of an even function
This question was asked earlier Question about proving existence of a function $f$ such that $f \circ f = g$ for an odd function $g$ about the functional square root to an odd function. Now consider an even function $g:\mathbb{R}\rightarrow \mathbb{R}$ where $g(x)>0$ for all $x\in\mathbb{R}$. Does there exist a functional square root $f$ such that $f\circ f=g$? Consider $g(x)=1$ then $f=g$ is a solution. But I suspect in general it does not exist, but I can't seem to reason why.

答:
Consider this even and strictly-positive function $g$:

$$g(x) = \begin{cases}
2&|x|<1.5\\
1 & |x| \ge 1.5
\end{cases}$$

And assume there exists some $f: \mathbb R\to \mathbb R$, $f(f(x)) = g(x)$ for all $x\in\mathbb R$.

Let $a = f(1)$, then

$$f:1\mapsto a \mapsto 2 \mapsto g(a) \mapsto 1.$$

If $|a| < 1.5$ and $g(a) = 2$, then $f(2)$ is simultaneously equal to both $2$ and $1$.

If $|a| \ge 1.5$ and $g(a) = 1$, then $f(1)$ is simultaneously equal to both $a$ and $1$.

Both cases lead to contradiction, so there is no such $f$ for the function $g$.

ContentLicense: CC BY-SA 4.0


问:
Do these definitions of congruences on categories have the same result in this context?
Let $\mathcal{D}$ be a small category and let $A=A\left(\mathcal{D}\right)$
be its set of arrows. Define $P$ on $A$ by: $fPg\Leftrightarrow\left[f\text{ and }g\text{ are parallel}\right]$
and let $R\subseteq P$.

Now have a look at equivalence relations
$C$ on $A$. Let's say that:

$C\in\mathcal{C}_{s}$ iff $R\subseteq C\subseteq P$
  and $fCg\Rightarrow\left(h\circ f\circ k\right)C\left(h\circ g\circ k\right)$
  whenever these compositions are defined;

$C\in\mathcal{C}_{w}$ iff $R\subseteq C\subseteq P$ but now combined with $fCg\wedge f'Cg'\Rightarrow\left(f\circ f'\right)C\left(g\circ g'\right)$
  whenever these compositions are defined.

Then $P\in\mathcal{C}_{s}$ and $P\in\mathcal{C}_{w}$
so both are not empty. For $C_{s}:=\cap\mathcal{C}_{s}$ and $C_{w}:=\cap\mathcal{C}_{w}$
it is easy to verify that $C_{s}\in\mathcal{C}_{s}$ and $C_{w}\in\mathcal{C}_{w}$.

My question is:

Do we have $C_{w}=C_{s}$ here?

It is in fact the question whether two different definitions of 'congruences' both result in the same smallest 'congruence' that contains relation $R\subseteq P$. I ask it here for small categories so that I can conveniently speak of 'relations'
(small sets), but for large categories I have the same question.

Mac Lane works in CWM with $C_{s}$, but is $C_{w}$ also an option?

答:
They are identical. I will suppress the composition symbol for brevity and convenience.

Suppose first that $C \in \mathcal C_w$, and that $f C g$. Since $h C h$ and $k C k$, we have $f C g$ implies $hf C hg$, which in turn implies $hfk C hgk$. Thus $C \in \mathcal C_s$.

Suppose now that $C \in \mathcal C_s$, and that $f C g, f' C g'$. Then we have $ff' C gf'$ (take $h = \operatorname{id}, k = f'$) and $gf'Cgg'$ (take $h = g, k = \operatorname{id}$). By transitivity, $ff'Cgg'$. Thus $C \in \mathcal C_w$.

Therefore, $\mathcal C_s = \mathcal C_w$, and we conclude $C_s = C_w$.

ContentLicense: CC BY-SA 3.0


问:
What would you call these 3D planes?
I want to call these 3D planes 'Cartesian planes':

But maybe that is incorrect as the Cartesian plane is traditionally just the XY plane with a third perpendicular (unnamed) one.

Is there another name for what I'm thinking of?

答: As one of the commenters said, they are called coordinate planes, in analogy to coordinate axes.  Examples of this usage are at https://en.wikipedia.org/wiki/Cartesian_coordinate_system, https://www.usna.edu/Users/oceano/raylee/SM223/Ch12_1_Stewart(2016).pdf, https://math.la.asu.edu/~surgent/mat267/examples/xyz_coordinate_system_2.pdf.</p>

Title: What would you call these 3D planes?


问:
Geometry of an ellipse
I was going through some properties a ellipses and this question comes to my mind:
Suppose you have an ellipse on paper with nothing marked on it and for some reason you need to find the foci. How can you do that with just a pencil and a compass?
Can we do that using this property: "The portion of a tangent to an ellipse between the point of contact and the directrix subtends a right angle at the corresponding focus." ?</p>

答: Assuming you want a ruler and compass construction, there is a nice sequence at http://whistleralley.com/conics/conic_construction/ellipse_parts/.  It shows how to find the center (via conjugate diameters), followed by the axes and then the foci.</p>

Title: Geometry of an ellipse


问:
Proof the sequence $(x_n)=\frac{n}{n+1}$ is contractive
After proving $(x_n)=(\frac{n}{n+1}:n\in \Bbb N^+)$ is convergent, a bounded, monotone, and Cauchy sequence. Next I have to prove that this sequence is contractive. Though I have no idea if it is true or false right now. I read some examples on how to do it, most of it are about defining $x_{n+1}$ first, but I don't know how to do that for this sequence.

So I don't even know how to begin proving this.

Body:
$$|x_{n+1}-x_n| = \left|\frac {n+1}{n+2}-\frac{n}{n+1}\right| = \left|\frac 1{(n+1)(n+2)}\right|$$
hence
$$\frac {|x_{n+2}-x_{n+1}|}{|x_{n+1}-x_{n}|} = \frac {(n+1)(n+2)}{(n+2)(n+3)} = \frac {n+1}{n+3} = 1-\frac 2{n+3}$$

Now, does there exist any $0 < C < 1$ such that the above is bounded above by $C$?


问:
let $x_{n+1}=2x_n^2-1$ and $t=x_0$ then How many $t\in [-1,1]$ exist such that $x_{11}=1$?

let $x_{n+1}=2x_n^2-1$ and $t=x_0$ then How many $t\in [-1,1]$ exist such that $x_{11}=1$ ?

my try ; if  $\cos x=t=x_0$ then we have $x_n=\cos(2^nx)$ and $x_{11}=\cos(2^{11}x)=1$ and also $2^{11}x=2k\pi$ now what ?

答:
You're almost there

$$t = \cos x = \cos\left(\frac{k\pi}{2^{10}}\right)$$

For each $k$ such that $0 \leq k \leq 2^{10}$ - this will give a unique $t$ - After which any value of $t$ obtained can be written as an already found value - hence there will be $2^{10}+1$ distinct values of $t$ corresponding to each $k$


问:
Class Group of $\mathbb Q(\sqrt{-35})$
As an exercise I am trying to compute the class group of $\mathbb Q(\sqrt{-35})$.

We have $-35\equiv 1$ mod $4$, so the Minkowski bound is $\frac{4}{\pi}\frac12 \sqrt{35}<\frac23\cdot 6=4$. So we only need to look at the prime numbers $2$ and $3$.

$-35\equiv 5$ mod $8$, so $2$ is inert. Also, $-35\equiv 1$ mod $3$, so $3$ splits, i.e. $(3)=Q\overline Q$ with $Q=(3,1+\sqrt{-35})$, $\overline Q=(3,1-\sqrt{-35})$. The ideals $Q,\overline Q$ are not principal, because there are no solutions to $x^2+35y^2=12$, i.e. no elements of norm 3.

Now we know that there are at most $3$ elements (or do we?), namely $(1),Q,\overline Q$. Mathematica tells me that the class number is $2$, so $Q$ and $\overline Q$ must be in the same equivalence class and $Q^2$ has to be a principal ideal. But how can I show this?

答:
Note that the ring of integers is $\mathbb Z[(1+\sqrt{-35})/2]$.

If you compute $(3, 1 + \sqrt{-35})^2$, you get $$(9,3 + 3\sqrt{-35}, -34 + 2 \sqrt{-35} ) = (9, 1 + \sqrt{-35}) = ( \dfrac{1-\sqrt{-35}}{2} \dfrac{1 + \sqrt{-35}}{2}, 2 \dfrac{1+\sqrt{-35}}{2}) = ((1 + \sqrt{-35})/2 )$$
(because
$\dfrac{1-\sqrt{-35}}{2}$ and $2$ are coprime, and so generate the ideal $1$).

I find the computation a bit easier by phrasing the factorization of $(3)$
in the following alternative way:
$(3) = (3, (1 + \sqrt{-35})/2) (3,(1- \sqrt{-35})/2)$, and
$$(3,(1+\sqrt{-35})/2)^2 = (9, 3(1+\sqrt{-35})/2,(-17+\sqrt{-35})/2) = ( (1+\sqrt{-35})/2 )$$ is principal.

As a consistency check, note that $ (9) = (3) (3) = Q \overline{Q} Q \overline{Q}
= Q^2 \overline{Q}^2,$ but also $9  = ( (1+\sqrt{-35})/2) ( ( 1 - \sqrt{-35})/2),$ so we must have $Q^2$ equal to one of $( (1 \pm \sqrt{-35})/2).$

Title: Class Group of $\mathbb Q(\sqrt{-35})$


问:
Combinatorics: interview strategy problem (secretary problem)
This is a probability problem.

Suppose we have n applicants for a job with the following interviewing process:

after interview an applicant is rejected or accepted

if accepted, process is over

if no applicant was selected in the first n-1 interviews, the n-th one is chosen by default.

The problem is to answer some questions about the following hiring strategy:

Reject the first r-1

Accept the first one after r-1 with ranking better than the first r-1

Note that there is an absolute rank of all n applicants so there is a best applicant i.e. one with the best rank, but is unknown. We only know the relative ranks of the interviewed applicants.

We define $P_n(r)$ as the probability of accepting the absolute best ranked using the above strategy.

For example $P_4(2)$ if I am not wrong is 11/24.
If we look at all the permutations of the absolute rank of all the applicants (4!).

1*** (6) best applicant is already rejected here

2*** (6) best applicant will definitely be taken here because nobody is better than the second best than the first best

31** (2) yes

32** (2) no

3412 (1) yes

3421 (1) no

41** (2) yes

42** (2) no

43** (2) no

In total 11/24.

First I have to show that $P_{10}(3) = \frac{2}{10} (\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{9} )$, derive the probability of accepting the applicant with the absolute rank 1 among n applicants at the k-th interview then find a recursive formula of $P_n(r)$ in the form $P_n(r)=A+BP_n(r+1)$.

I have done some research on the problem and it seems it has a classical name "Secretary Problem".

答:
$$P_n(r) = \sum_{k=0}^{n}P(applicant\ k\ is\ selected\ \cap applicant\ k\ is\ the\ best)$$
$$ = \sum_{k=0}^{n}P(applicant\ k\ is\ selected\ |\ applicant\ k\ is\ the\ best)P(applicant\ k\ is\ the\ best)$$
$$ = \sum_{k=r}^{n}P(best\ applicant\ is\ selected\ in\ the\ k^{th}\ interview )\frac{1}{n}$$
$$=\sum_{r}^{n}\frac{r-1}{i-1} \frac{1}{n}$$
This seems to answer my question for $P_{10}(3)$ and the recursive formula can also be derived from the explicit one.


问:
Locus of point at a fixed distance from midpoint of intercepts of a variable line segment with fixed distance

Let A and B be variable points on the x-axis and y-axis respectively such that the line segment AB is in the first quadrant and of a fixed length 2d. Let C be the mid-point of AB and P be a point such that

(a) P and the origin are on the opposite sides of AB and,

(b) PC is a line of length d which is perpendicular to AB.

Find the locus of P.

Source

In this problem, I have drawn the figure containing the line AB and point P and everything. I have also named point as(h, k),thereby applying all suitable equations.Inspite of all that, it is getting clumsy.
Can anyone provide a elegant solution to it?

答:
Let $A = (x, 0), 0 \leq x \leq 2d$ and $\alpha = O\hat{A}B$ (note that $\cos\alpha = \frac{x}{2d}$). Then $B = \left(0, x \tan\alpha\right)$ and $C = (\frac{x}{2}, \frac{x\tan\alpha}{2})$. Because $P$ and the origin are on the opposite sides of $AB$, we define $\vec{v} = (\sin\alpha, \cos\alpha)$ so that $P = C + d\vec{v} = (\frac{x}{2} + d\sin\alpha, \frac{x\tan\alpha}{2} + d\cos\alpha)$.

Also, $\cos\alpha = \frac{x}{2d} \implies \sin\alpha = \frac{\sqrt{4d^2 - x^2}}{2d}$ and $\tan\alpha = \frac{\sqrt{4d^2 - x^2}}{x}$. Substituting in $P$, $P = \left(\frac{\sqrt{4d^2 - x^2} + x}{2}, \frac{\sqrt{4d^2 - x^2} + x}{2}\right)$: $P$ lies on $x=y$.

For the limits (remember that $0 \leq x \leq 2d$): $f(x) = \frac{\sqrt{4d^2 - x^2} + x}{2}$ attains minimum at $x=2d$, with $f(2d) = d$ and maximum at $x=\sqrt2 d$, with $f(\sqrt2 d) = \sqrt2d$.

Therefore, locus of $P$ is the segment from $(d,d)$ to $(\sqrt2 d, \sqrt2 d)$.


问:
union of a Sequence of closed subsets
Let $F_1,F_2,...$ be closed subsets of some topological space $X$. If for each $x \in X$ there is a neighborhood of $x$ (open set containing) such that

$N_x \cap F_j \neq \emptyset$

for only finitely many $j$. Prove the arbitrary union of the $F_j$ is closed in $X$.

Here is my attempt and all I have so far,

So I would like to show the complement is open in $X$, that is,

$X \backslash \bigcup_{j=1}^\infty F_j \subset X$

is open. Let $x \in X \backslash \bigcup_{j=1}^\infty F_j$ be arbitrary, we wish to put $x$ in a neighborhood that is disjoint from $\bigcup_{j=1}^\infty F_j$

So we are guaranteed the existence of a neighborhood $N_x$ of $x$ such that

$N_x \cap F_1,...,N_x \cap F_n$

all have non-trivial intersection. But for $F_{n+1},F_{n+2},...$ we know the intersection with $N_x$ is empty, that is

$N_x \cap F_{n+1},N_x \cap F_{n+2}$

are all empty. Also arbitrary union of open is open and since the $F_j$ are closed their complements in $X$ are open, so can I take the complements of the infinitely many $F_j$ that is $X \backslash \bigcup_{k=n+1}^\infty F_k$ is the open neighborhood of $x$? a bit lost. SO basically is it saying that those finitely many $F_j$ have elements that can get close to $x$ but after the threshold of $n$ the $F_k$ become disjoint from balls around $x$? and they are complements of closed so each is open and arbitrary union of open is open?

答:
You started with  a countable union of $F_i$'s. You are supposed to consider uncountable unions also. However, to make my notations consistent with yours I will stick to your countable union. The same proof works for uncountable unions also. Only a minor change in notation is needed.

$N_x \cap F_1^{c}\cap...\cap F_n^{c}$ is an open set containing $x$ and it is contained in $X \setminus \bigcup_n F_n$. This proves that every point $x$ of $X \setminus \bigcup_n F_n$ is an interior point, so $X \setminus \bigcup_n F_n$ is an open set. Hence, $\bigcup_n F_n$ is closed.

Title: union of a Sequence of closed subsets


问:
$\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}$ without L'Hôpitals Rule.
The limit,

$\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}=6$

is easily found using L'Hôpitals Rule. However, the exercise is to evaluate it without using this method. I've tried multiple substitutions, mainly pythagoric and reciprocal identities, but without success, the indeterminate form $\frac{0}{0}$ keeps appearing. I’ve also tried splitting the limit and doing the substitutions. I'm looking to simplify the expression in order to evaluate the limit directly.

Thanks in advance.

答:
This limit can actually be calculated quite directly. You only need $\tan \frac{\pi}{4}= 1$

Rewrite the expression as follows and set $t=\tan x$ and consider $t\to 1:$

\begin{eqnarray*}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}
& = & \frac{\frac{\cos^2 x}{\sin^2 x}+1-2\tan^2x}{\frac 1{\tan x}- 1} \\
& \stackrel{t=\tan x}{=} & \frac{\frac 1{t^2}+1-2t^2}{\frac 1t-1}\\
& = & \frac{1+t^2 - 2t^4}{t(1-t)}\\
& = & \frac{(1-t)(1+t)(1+2t^2)}{t(1-t)}\\
& = & \frac{(1+t)(1+2t^2)}{t} \stackrel{t\to1}{\longrightarrow}6\\
\end{eqnarray*}


问:
What's going wrong with my cohomology computation?
Use the following $\Delta$ complex for the klein bottle:

we may obtain the chain complex:

$$\mathbb{Z}^2 \rightarrow \mathbb{Z}^3 \rightarrow \mathbb{Z} \rightarrow 0$$

My goal is to find the cohomology of this chain complex with $\mathbb{Z}$ coefficients: so I want to apply $Hom(-, \mathbb{Z})$ to these groups and find the homology groups for the corresponding (co)chain complex. There may be more powerful tools for doing this but I just started cohomology so I need to do it using just the definitions.

Now it is clear that $Hom(\mathbb{Z}^n, \mathbb{Z}) = \mathbb{Z}^n$ (identifying $(a_1, .. a_n)$ with the function that sends $(0,..1,..0)$ to $a_i$ where $1$ is in the $i$th slot)

So our cochain complex is $$0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z}^3 \rightarrow \mathbb{Z}^2$$

I think I'm making mistakes in computing the coboundary maps $\delta^n$.
Given $(a_1, ..., a_n)$ denote $\overline{(a_1, ..a_n)}$ to be the associated element of $Hom(\mathbb{Z}^n, \mathbb{Z})$.

So $\partial_1 : \mathbb{Z}^3 \rightarrow \mathbb{Z}$ is the $0$ map. Notice then that given any $\overline{a} \in \mathbb{Z}$, $\delta^1 (\overline{a}) = \overline{a}(\partial_1) \in Hom(\mathbb{Z}^3, \mathbb{Z})$. Since $\partial_1 = 0$, so is $\delta^1(\overline{a}) = \overline{a}\partial_1$. So this means that $\delta^1$ is the $0$ map as well, right? Is there any mistake in this type of argument?

And similarly, computing $\delta^2 : \mathbb{Z}^3 \rightarrow \mathbb{Z}^2$. $\delta(\overline{(1,0,0)}) =  \overline{(1,0,0)}(\partial_2) \in Hom(\mathbb{Z}^2, \mathbb{Z})$. Notice that $\overline{(1,0,0)}\partial_2(1,0) = \overline{(1,0,0)}((1,1,-1)) = 1$ , and similar $\overline{(1,0,0)}\partial_3(0,1) = 1$ as well, so $\delta^2(\overline{(1,0,0)}) = (1,1)$. Similarly we have that $\delta^2(\overline{(0,1,0)}) = (1,-1)$ and $\delta^2(\overline{(0,0,1)}) = (-1,1)$.

This means that ultimately $\delta^2(x,y,z) = (x + y - z, x-y+z)$ describes $\delta^2$ which feels right, sorta.

So now we now our coboundary map and we may compute the cohomology groups. Let $C$ refer to the original chain complex.

Then $H^1(C; \mathbb{Z}) = ker(\delta^1) / 0 = \mathbb{Z}$ since $\delta^1 = 0$.

Next $H^2 (C; \mathbb{Z}) = ker(\delta^2)/im(\delta^1)$

And here's where the problem is; since $\delta^2(x,y,z) = (x+y-z, x-y+z)$, this means that $ker(\delta^2) = 0$, which means that $H^2(C; \mathbb{Z}) = 0$. But this can't be right, right?

Cos then we would have a sequence $$0 \rightarrow \mathbb{Z} \rightarrow 0$$ which makes no sense.

My suspicion is that I'm confusing direct sums and products or something like that. I'm still slightly confused about that stuff. Or perhaps I'm just going completely in the wrong direction and have no idea what I'm doing.

Please let me know what I'm doing wrong.

答:
You can write the map $\delta^2$ as a matrix:
$$
\begin{pmatrix}
1 & 1 & -1 \\
1 & -1 & 1
\end{pmatrix}.
$$
If you subtract the first row from the second (an elementary row operation), this becomes
$$
\begin{pmatrix}
1 & 1 & -1 \\
0 & -2 & 2
\end{pmatrix}.
$$
This second matrix will have the same kernel and image as the first. Its kernel consists of elements of the form $(0, n, n)$ for $n \in \mathbb{Z}$. That is isomorphic to $\mathbb{Z}$, and that gives you $H^1$. To compute $H^2$, you have to compute $C^3/\textrm{im}\, \delta^2 = \mathbb{Z}^2/\textrm{im}\, \delta^2$. I claim that the image is given by all elements of the form $(a,b)$ where $a$ can be any integer but $b$ must be even. From that, you can get $H^2$.


问:
Is our teacher speaking correct about the probability of a national lottery ticket?
I'm trying to keep the question short.  Today the teacher said something like this.Your chance of winning in the national lottery does not depend on which city you bought the ticket from.  If you want, buy your ticket from city "A" or buy your ticket from city "B".  Your probability of winning does not change.A criticism was made to the teacher: But the people who win are always those who live in city "A". The teacher replied as follows. Because the number of tickets sold in city "A" is more than the number of tickets sold in city "B".  However, no matter which city you buy a ticket from, your chances of winning will always remain the same.After some thought, I made an objection.

I said, this argument is valid only and only if:

If the number of tickets DISTRIBUTED to be sold to city "A" and "B" is the same, then my chances of winning will not change.  Then, the probability of winning for people living in city "A" may be attributed to the over-selling of tickets.  But even this argument is not valid if the winning ticket should only have $1.$
Because the number of distribution and purchase of the ticket to cities is not the same thing.

My question: Is the teacher right?  What am I getting wrong?  Where am I making a mistake?

Thank you!

答:
I think the confusion between conditional and absolute probability - the chance that a particular lottery number is the winning number is the same regardless of which city you purchase it from, and I will demonstrate it

Suppose there are $N$ total numbers, out of which only one is the winning ticket. Let $N_A$ and $N_B$ be the number of tickets that go to cities $A$ and $B$ respectively

Hence Probability that winning ticket goes to city A is $P_A = \frac{N_A}{N}$, and similarly for city B is $P_B = \frac{N_B}{N}$

Now, the probability that someone wins it from city A is

$$\text{Probability winner from City A} = \text{Probability of drawing winning ticket from city A} \times \text{Probability winning ticket is in City A}$$

$$\implies P(A) = \frac{1}{N_A} \times \frac{N_A}{N} = \frac{1}{N}$$

So the effect of "big city" vs "small city" does not really matter


问:
Compute $\int_{|z|=1}\frac{\log z}{z}dz$.
Here is a question about contour integration in complex analysis:

Compute
  $$\int_{|z|=1}\frac{\log z}{z}dz$$

I am not sure if I understand the question since the logarithm must be defined in a simply connected domain $\Omega$ with $0\notin\Omega$ and $1\in\Omega$. Hence, we cannot define $\log$ for every value in the circle $|z|=1$; there must be at least one point where it is not defined. However, if I forget this and consider the principal branch, I can compute
$$\int_{|z|=1}\frac{\log z}{z}dz = \int_{-\pi}^\pi \frac{\log e^{i\theta}}{e^{i\theta}}ie^{i\theta}d\theta = \int_{-\pi}^\pi i\theta id\theta=-\frac{\theta^2}{2}\bigg\vert_{-\pi}^\pi=0$$
Is this right?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\oint_{\verts{z}\ =\ 1}
{\ln\pars{z} \over z}\,\dd z}
\\[5mm] = &\
\lim_{\epsilon\ \to\ 0^{+}}\,\,\left[-\int_{-1}^{-\epsilon}{\ln\pars{-x} + \ic\pi \over x}\,\dd x -
\int_{\pi}^{-\pi}{\ln\pars{\epsilon} + \ic\theta \over \epsilon\expo{\ic\theta}}\,\epsilon\expo{\ic\theta}\ic\,\dd\theta\right.
\\[2mm] &\
\phantom{\lim_{\epsilon\ \to\ 0^{+}}\,\,\left[\,\,\right.}\left.
-\int_{-\epsilon}^{-1}{\ln\pars{-x} - \ic\pi \over x}\,\dd x
\right]
\\[5mm] = &\
\lim_{\epsilon\ \to\ 0^{+}}\,\,\braces{%
-2\pi\ic\ln\pars{\epsilon} + \ic\int_{-\pi}^{\pi}
\bracks{\ln\pars{\epsilon} + \ic\theta}\dd\theta} =
\bbx{0} \\ &
\end{align}


问:
Dominated Convergence Theorem, Probability Theory
I am reading the proof of the dominated convergence theorem

Dominated Convergence Theorem

Let $X$ and $X_{n}$, $n=1,2,3...$ be random variables. Suppose that there's a random variable $Y$ such that

(1)$|X_{n}|\leq Y$ a.s for all $n$

(2)$\lim_{ x\to \infty}$ $X_{n}=X$ either a.s or in probability

(3)$E(Y)< \infty$
Then$$\lim_{ x\to \infty}E(X_{n})=E(X)$$

I have a question about one sentence of the proof

Suppose $X_{n} \to X$ in probability. Since $E(|X_{n}|) \leq E(Y)$, the sequence of expectations is bounded. If it does not converge to $E(X)$, there exists a subsequence $X_{n_{k}}$ such that $E(X_{n_{k}})$ converges to a limit $L \neq E(X)$

My question is that, when we know the expecations are bounded, we only know the sequence has a convergent subsequence, how do we know that the limit of the subsequence is not $E(X)$?

Thanks in advance!

答:
If the sequence $E(X_{n})$ does not converge, then $\liminf_{n} E(X_{n}) < \limsup_{n} E(X_{n})$. So $\liminf_{n} E(X_{n}) \neq E(X)$ or $\limsup_{n} E(X_{n}) \neq E(X)$. For example, suppose $\liminf_{n} E(X_{n}) \neq E(X)$. Then there is a convergent subsequence converging to $\liminf_{n} E(X_{n})$.

Skipping the details, you can just take a convergent subsequence converging to $L \neq E(X)$.


Title: $\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}$ without L'Hôpitals Rule.

Body:
The limit,

$\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}=6$

is easily found using L'Hôpitals Rule. However, the exercise is to evaluate it without using this method. I've tried multiple substitutions, mainly pythagoric and reciprocal identities, but without success, the indeterminate form $\frac{0}{0}$ keeps appearing. I’ve also tried splitting the limit and doing the substitutions. I'm looking to simplify the expression in order to evaluate the limit directly.

Thanks in advance.

Body: $$\frac{\csc^2(x)-2\tan^2(x)}{\cot (x)-1}=2 \tan (x)+\cot (x)+2 \sec ^2(x)-1$$


答:
Your cochain complex is correct and generated by the dual basis elements via $$0 \xrightarrow{\delta_{-1}} \langle v^*\rangle \xrightarrow{\delta_0} \langle a^*,b^*,c^*\rangle \xrightarrow{\delta_1} \langle U^*, L^*\rangle \xrightarrow{\delta_2} 0$$

$\delta_0$ is the $0$-map since $\delta_0v^*$ evaluated on the generators yields $0$, i.e. $$\delta_0v^*(a) = v^*(\partial_1a) = v^*(\partial_1b) = v^*(\partial_1c) = v-v = 0 $$

$\delta_2$ is obviously the $0$-map.

Evaluating $\delta_1$ on the generators $U,L$ gives $$\delta_1a^*(U) = a^*(\partial_2U) = a^*(a+b-c) = 1$$
$$\delta_1a^*(L) = a^*(\partial_2L) = a^*(c+a-b) = 1$$

$$\delta_1b^*(U) = b^*(\partial_2U) = b^*(a+b-c) = 1 $$
$$\delta_1b^*(L) = b^*(\partial_2L) = b^*(c+a-b) = -1 $$

$$\delta_1c^*(U) = c^*(\partial_2U) = c^*(a+b-c) = -1 $$
$$\delta_1c^*(L) = c^*(\partial_2L) = c^*(c+a-b) = 1 $$

Thus, $\delta_1$ is fully determined by $$\delta_1 = \begin{pmatrix} 1 & 1 & -1 \\ 1 & -1 &1 \end{pmatrix}$$

$$H^0(K;\mathbb{Z}) = \ker \delta_0/{\operatorname{im} \delta_{-1}} = \mathbb{Z}/0 =  \mathbb{Z}$$

The kernel of $\delta_1$ is generated by $\langle b^*+c^*\rangle$ and since $\operatorname{im} \delta_0 = 0$, we get $$H^1(K,\mathbb{Z}) = \langle b^*+c^*\rangle \cong \mathbb{Z}$$

Now $\operatorname{im} \delta_1$ is generated by $\langle U^*+L^*, U^*-L^*\rangle$ and since $\ker \delta_{2} = \mathbb{Z}^2$ we get

$$H^2(K,\mathbb{Z}) = {\langle U^*,L^*\rangle}\big/{\langle U^*+L^*, U^*-L^*\rangle} = \langle U^*, U^*-L^*\rangle\big/{\langle 2U^*, U^*-L^*\rangle} = \langle U^*\rangle\big/\langle 2U^*\rangle \cong \mathbb{Z}/2\mathbb{Z}$$

Title: What's going wrong with my cohomology computation?

Body:
Use the following $\Delta$ complex for the klein bottle:

we may obtain the chain complex:

$$\mathbb{Z}^2 \rightarrow \mathbb{Z}^3 \rightarrow \mathbb{Z} \rightarrow 0$$

My goal is to find the cohomology of this chain complex with $\mathbb{Z}$ coefficients: so I want to apply $Hom(-, \mathbb{Z})$ to these groups and find the homology groups for the corresponding (co)chain complex. There may be more powerful tools for doing this but I just started cohomology so I need to do it using just the definitions.

Now it is clear that $Hom(\mathbb{Z}^n, \mathbb{Z}) = \mathbb{Z}^n$ (identifying $(a_1, .. a_n)$ with the function that sends $(0,..1,..0)$ to $a_i$ where $1$ is in the $i$th slot)

So our cochain complex is $$0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z}^3 \rightarrow \mathbb{Z}^2$$

I think I'm making mistakes in computing the coboundary maps $\delta^n$.
Given $(a_1, ..., a_n)$ denote $\overline{(a_1, ..a_n)}$ to be the associated element of $Hom(\mathbb{Z}^n, \mathbb{Z})$.

So $\partial_1 : \mathbb{Z}^3 \rightarrow \mathbb{Z}$ is the $0$ map. Notice then that given any $\overline{a} \in \mathbb{Z}$, $\delta^1 (\overline{a}) = \overline{a}(\partial_1) \in Hom(\mathbb{Z}^3, \mathbb{Z})$. Since $\partial_1 = 0$, so is $\delta^1(\overline{a}) = \overline{a}\partial_1$. So this means that $\delta^1$ is the $0$ map as well, right? Is there any mistake in this type of argument?

And similarly, computing $\delta^2 : \mathbb{Z}^3 \rightarrow \mathbb{Z}^2$. $\delta(\overline{(1,0,0)}) =  \overline{(1,0,0)}(\partial_2) \in Hom(\mathbb{Z}^2, \mathbb{Z})$. Notice that $\overline{(1,0,0)}\partial_2(1,0) = \overline{(1,0,0)}((1,1,-1)) = 1$ , and similar $\overline{(1,0,0)}\partial_3(0,1) = 1$ as well, so $\delta^2(\overline{(1,0,0)}) = (1,1)$. Similarly we have that $\delta^2(\overline{(0,1,0)}) = (1,-1)$ and $\delta^2(\overline{(0,0,1)}) = (-1,1)$.

This means that ultimately $\delta^2(x,y,z) = (x + y - z, x-y+z)$ describes $\delta^2$ which feels right, sorta.

So now we now our coboundary map and we may compute the cohomology groups. Let $C$ refer to the original chain complex.

Then $H^1(C; \mathbb{Z}) = ker(\delta^1) / 0 = \mathbb{Z}$ since $\delta^1 = 0$.

Next $H^2 (C; \mathbb{Z}) = ker(\delta^2)/im(\delta^1)$

And here's where the problem is; since $\delta^2(x,y,z) = (x+y-z, x-y+z)$, this means that $ker(\delta^2) = 0$, which means that $H^2(C; \mathbb{Z}) = 0$. But this can't be right, right?

Cos then we would have a sequence $$0 \rightarrow \mathbb{Z} \rightarrow 0$$ which makes no sense.

My suspicion is that I'm confusing direct sums and products or something like that. I'm still slightly confused about that stuff. Or perhaps I'm just going completely in the wrong direction and have no idea what I'm doing.

Please let me know what I'm doing wrong.


问:
Real life usage of Benford's Law
I recently discovered Benford's Law. I find it very fascinating. I'm wondering what are some of the real life uses of Benford's law. Specific examples would be great.</p>

答:
It may or may not be useful for detecting fraud in elections, for example the 2009 Iranian election.

Some links:

http://sbseminar.wordpress.com/2009/06/15/benfords-law-and-the-iranian-election

http://www.fivethirtyeight.com/2009/06/statistical-evidence-does-not-prove.html

ContentLicense: CC BY-SA 2.5

Title: Real life usage of Benford's Law


问:
Symplectic Form Preserved by Orthogonal Transformation
I'm trying to prove that the symplectic form

$$\omega = d(\cos\theta) \wedge d\phi$$

is preserved by the action of $SO(3)$ on $S^2$ where $\phi$ and $\theta$ are spherical polars. Now $SO(3)$ simply acts by

$$\theta \mapsto \theta + \epsilon, \ \phi \mapsto \phi + \delta$$

and writing this diffeomorphism as $F:S^2 \to S^2$ I compute

$$F^*(\omega) = d(\cos(\theta + \epsilon))\wedge d(\phi + \delta) = \cos(\epsilon) d(\cos\theta)\wedge d\phi - \sin(\epsilon) d(\sin \theta)\wedge d \phi$$

Is this correct? It doesn't seem like the form is invariant under $SO(3)$. Perhaps it's only meant to be a local symplectomorphism though.

Am I allowed to claim that it is a local symplectomorphism because it gives the right result in the limit as $\epsilon \to 0$? I think that would be right, because it would mean that the Lie derivative vanishes.

Thanks in advance!

答: How many Euler angles are there?  Can you compute new $\theta$ and $\phi$ in terms of them? Can you represent these transformations as $\theta\mapsto\theta+\epsilon$, $\phi\mapsto\phi+\delta$?


问:
Demonstrate the statement in section 3.I.3 on the decomposition of the Möbius transformations.

I'm wondering how to do this question. I believe the question wants to decompose the following statements together from the text book to find the Mobius function

Thank you in advance (:

答: Compose the 4 transformations you'll get $(az+b)/(cz+d)$ where $a,b,c,d$ were (almost) arbitrary. For the calculation it might help you to notice that for $c\ne 0$ $$\frac{az+b}{cz+d} = \frac{a}{c}+\frac{a(-d/c)+b}{cz+d}$$


问:
Complex numbers iff proof
I let $z_{0}=i$, $z_{1}=1+i$, and for $n \geq2$, defines $z_{n}=z_{n-1}z_{n-2}$. Prove for every non-negative integer n, $Re(z_{n})Im(z_{n})=0$if and only if $3|n$.

What is the best way to do it? I want to use de moivre theorem, but how is it exactly? Is the question for induction, too?

答:
Let $z_n=r_ne^{ia_n\frac{\pi}{4}}$, we see $z_0=e^{i2\cdot \frac{\pi}{4}}$, $z_1=\sqrt 2 e^{i 1\cdot \frac{\pi}{4}}$, so $a_0=2$, $a_1=1$. In general,
$$
z_n=z_{n-1}z_{n-2}=r_{n-1}e^{ia_{n-1}\frac{\pi}{4}}
 r_{n-2}e^{ia_{n-2}\frac{\pi}{4}}
=r_{n-1}r_{n-2}e^{i(a_{n-1}+a_{n-2})\frac{\pi}{4}},
$$
so $a_n=a_{n-1}+a_{n-2}$. Now $a_0=2$, $a_1=1$,
$a_2=3$, $a_3=4$, we prove by induction $a_n$ is even if and only if $n=3k$. Assume this is true
up to $n=3k$, especially $a_{3k}$ even, $a_{3k-1}$ odd. Then $a_{3k+1}=a_{3k}+a_{3k-1}$ is even plus odd so is odd, $a_{3k+2}=a_{3k+1}+a_{3k}$ is odd plus even so is odd, $a_{3k+3}=a_{3k+2}+a_{3k+1}$ is odd plus odd so is even. That concludes the induction.

Now if $3|n$, $a_n$ is even, write $a_n=2m$, so
$$
z_n=r_ne^{i(2m)\frac{\pi}{4}}=r_ne^{im\pi/2},
$$
if $m$ is even this is real (so $Im z_n=0$); and if $m$ is odd then this is pure imaginary (so $Re z_n=0$). In any case $(Im z_n)(Re z_n)=0$.

Title: Complex numbers iff proof


问:
$a ^ x = |x + 2| - |2x + 8|$ has only one solution
So, this question was asked at a town-level olympiad.
$$a ^ x = |x + 2| - |2x + 8|$$
Knowing that there is only one solution, we must find a. So, basically $a ^ x$ is tangent to the some lines. I chose to get rid of the module to be it easier, but I am stuck at the first equation...
If we set $a < -4$, we get  $a ^ x = x + 6$, but I do not know how to continue. Any ideas? I suppose the other cases will be similar, right?

答:
$$a ^ x = |x + 2| - |2x + 8|$$
Maximum value for the right side occurs when $|2x+8|$ equals $0$ $\Rightarrow x=-4$.

Making sure the left side $a^x$ passes through $(-4, 2)$ will do.


问:
Let$A$ be an open, dense set in $\mathbb R^n$. Prove that $A + A = \mathbb R^n$
I have no idea how to go about this one. What I'm trying to prove is that given some $x$ in $\mathbb R^n$ there has to be some $y$ such that both $\frac x 2 + y$ and $\frac x 2 - y$ both are in A.

But I have no idea how to continue. Just hints are appreciated

答:
If $A$ is a dense open set, then $A-\frac x2$ and $\frac x2-A$ are dense open sets, so their intersection is a dense open set, and in particular it's nonempty. Choose a point $y\in(A-\frac x2)\cap(\frac x2-A)$; then $\frac x2+y\in A$ and $\frac x2-y\in A$, so $x=(\frac x2+y)+(\frac x2-y)\in A+A$.

More generally, if $A$ is a nonempty open set in $\mathbb R^n$ and $B$ is a dense subset of $\mathbb R^n$, then $A+B=\mathbb R^n$.

Proof. Consider any point $t\in\mathbb R^n$; we have to show that $t\in A+B$.

Since the mapping $x\mapsto t-x$ is a homeomorphism, $t-A$ is a nonempty open set. Since $B$ is dense, $B\cap(t-A)\ne\emptyset$. Choose a point $b\in B\cap(t-A)$. Then $b\in B$, and $b=t-a$ for some $a\in A$, so $t=a+b\in A+B$.


问:
Sigma-field generated by a sequence of random variables
I encountered the following statement where $X_1, ..., X_n$ are random variables, then
$$
\sigma(X_1, ... , X_n) = \sigma(\cup_{i=1}^n \sigma(X_i)).
$$
I managed to prove the direction where
$$
\sigma(X_1, ... , X_n) \subset \sigma(\cup_{i=1}^n \sigma(X_i)),
$$
but I have no idea where to start for the other inclusion. I encountered a solution for this here but I did not understand what bundled map or coordinate map are. I also did not think the proof was rigorous enough. Can anyone provide something more concrete and easy to understand? Thank you very much!

答: $\sigma(X_i) \subset \sigma(X_1,X_2,...,X_n)$ for each $i$ so $\bigcup_{1\leq i \leq n}\sigma(X_i) \subset \sigma(X_1,X_2,...,X_n)$. This implies that $\sigma (\bigcup_{1\leq i \leq n}\sigma(X_i)) \subset \sigma(X_1,X_2,...,X_n)$ by definition of $\sigma (\bigcup_{1\leq i \leq n}\sigma(X_i))$ as the smallest sigma algebra containing $\bigcup_{1\leq i \leq n}\sigma(X_i)$.


问:
How to show $x\in(0,1]$?
Problem is to show $(0,1]$ is subset of $\cup [\frac{1}{n},1]$.

What I have so far is:

Let $x$ be in $(0,1]$. And I want to show $x\in[\frac{1}{n}, 1]$

答:
For each given $x$, try to find a $n$ such that $x$ is in the interval $[1/n,1]$.

Can you find a $n$ for every $x$? Then you have proven your theorem.

Title: How to show $x\in(0,1]$?


答: Take $x\in (0,1)$. There exists $n>0$ such that $0<1/n<x$ ( because the sequence $(1/n$ goes to 0, as $n\to\infty$). Then $x$ is in $[1/n,1]$.

Title: How to show $x\in(0,1]$?

Body:
Problem is to show $(0,1]$ is subset of $\cup [\frac{1}{n},1]$.

What I have so far is:

Let $x$ be in $(0,1]$. And I want to show $x\in[\frac{1}{n}, 1]$


问:
Is there a name for statements about set inclusion?
A statement like the following is called an equality:
$$ 2 + 2 = 4 $$

In contrast, a statement like the following is called an inequality:

$$ 2 + 2 < 5 $$

One can refer to groups of these statements with phrases like "We list several important equalities below" or "Recall the inequality on the previous page".

Is there a name for statements like this, indicating inclusion in a set?
$$ x \in \mathbb{X} $$

Specifically, it would be nice to know if there is a name that can be used to refer to groups of these statements.

I considered "definition" but if one has defined the member of the set (say, $x$ from the previous example) is a different way already, and one wishes to express that $x$ is a member of some, also previously defined, set, then "definition" doesn't work. One could the generic term "lemma" or use "statement" as I have been in this question, but I'm curious if there is another, more specific term.

The question of whether there is a similar name for statements like this, indicating exclusion from a set arises naturally, as well:

$$ y \notin \mathbb{X} $$

答: Maybe "membership" for $x\in X$.

Body: Maybe "membership" for $x\in X$.


问:
Identity matrix and its relation to eigenvalues and eigenvectors
Kindly help me understand this statement made by my prof.

The identity matrix I has the property that any non zero vector $V$ is an eigenvector of eigenvalue $1$.

My assumption of this statement is that the column vector (1,1) multiplied by the identity matrix is equal to the identity matrix. But the confusing part is when he says "...any non zero..". This is implying we can use other values that don't equal one. I believe the eigenvalue would change in light of the different non-$1$ values.

答:
Every vector is an eigenvector of the identity matrix

First remember that even in the simple case where we have distinct eigenvalues and eigenvectors, then eigenvectors are not unique because if $v$ is an eigenvector, then so is every $\lambda v$ on that line.

But even if we require normal eigenvectors, this is not enough for uniqueness, because a single eigenvalue can have two distinct eigenvectors. And when that happens, any linear combination of those two eigenvectors is also an eigenvector because:

$$A(v_1 + v_2) = Av_1 + Av_2 = \lambda v_1 + \lambda v_2 = \lambda(v_1 + v_2)$$

So if for example there are two such distinct eigenvectors, then any value in that plane, and not a line, is also an eigenvector, so we could just rotate around those vectors and maintain normality with infinitely many choices.

Have a look at the concepts of:

algebraic multiplicity: the order of a root of the characteristic polynomial

geometric multiplicity: the dimension of the linear subspace associated with a given eigenvalue

and the key relationship:

$$1 \le \text{algebraic} \le \text{geometric} \le \text{matrix dimension}$$

So from this, we see that in the case of the identity matrix, there is a single eigenvalue of 1 with:

algebraic multiplicity: n

geometric multiplicity: n

and therefore the entire space is an eigenvector.


问:
Choose between Combinations & Permutations
What is the most natural way to choose either permutations or combinations when encountering a particular combinatorial problem?

For example, in the following problem:

Given $n$ antennas of which $m$ are defective and  $n-m$ are functional, how many linear 'orderings' are there in which no two defectives are consecutive?

What seems to be the natural way of doing this (P or C)?

答:
lets change the question to daily situation for understanding.
consider n students in which m are boys so we have n-m girls.how many linear orders are possible if no two boys sit consecutive?
[In our example consider 7 students in which there are 2 boys then n-m=5 so we have 5 girls]
GGGGG arrangement of n-m girls
since no boy should sit consecutively they can sit beside each girl
BGBGBGBGBGB [so there are 6 places for boys 5+1=n-m+1]
so we need m positions out of n-m+1 [we need 2 position out of 6 positions available]
possible combiantions are( n-m+1Cm)</p>


问:
Class Group of $\mathbb Q(\sqrt{-35})$
As an exercise I am trying to compute the class group of $\mathbb Q(\sqrt{-35})$.

We have $-35\equiv 1$ mod $4$, so the Minkowski bound is $\frac{4}{\pi}\frac12 \sqrt{35}<\frac23\cdot 6=4$. So we only need to look at the prime numbers $2$ and $3$.

$-35\equiv 5$ mod $8$, so $2$ is inert. Also, $-35\equiv 1$ mod $3$, so $3$ splits, i.e. $(3)=Q\overline Q$ with $Q=(3,1+\sqrt{-35})$, $\overline Q=(3,1-\sqrt{-35})$. The ideals $Q,\overline Q$ are not principal, because there are no solutions to $x^2+35y^2=12$, i.e. no elements of norm 3.

Now we know that there are at most $3$ elements (or do we?), namely $(1),Q,\overline Q$. Mathematica tells me that the class number is $2$, so $Q$ and $\overline Q$ must be in the same equivalence class and $Q^2$ has to be a principal ideal. But how can I show this?

答:
Let $\alpha=\frac{1+\sqrt{-35}}{2}$, then in fact, $(3)=Q\overline{Q}=(3,\alpha)(3,\alpha+2)$. In the class group then, $[Q]+[\overline{Q}]=[\mathcal{O}_K]$, so $[Q]$ and $[\overline{Q}]$ are inverses. So the class group is generated by $Q$, and we just need to show that $Q^2$ is principal.

The minimal polynomial of $\alpha$ is $f(x)=x^2-x+9$, and recall that $f(\beta)=\text{Nm}(\beta-\alpha)$. In particular, $f(0)=9=3^2$. The element $(\alpha)$ is sent to $0$ under the evaluation $\alpha\mapsto 0$, corresponding to the ideal $Q$, so $(\alpha)=Q^2$, and the order of $[Q]$ divides $2$. You've already shown that $Q$ can't be principal, so we're done.

Title: Class Group of $\mathbb Q(\sqrt{-35})$


问:
Extension of Monotone Convergence Theorem and Dominated Convergence Theorem
A very simple question: Do the Monotone Convergence Theorem and Dominated Convergence Theorems hold for signed measures? In particular, if $\mu$ is a signed measure and $f_{n} \to f$ pointwise a.e., is it true that
$$
\lim_{n \to \infty} \int f_{n} d \mu= \int f d \mu
$$
if the conditions of the either of the theorems are met? If so, how would one prove and extension of these theorems?

答:
Let $X$ be the underlying space. Let $P, N$ be a Hahn decomposition of $X$. Let $\mu^{+}, \mu^{-}$ be the corresponding Jordan decomposition of $\mu$. Then
$$
\int_{X} f_{n} d\mu = \int_{P} f_{n} d\mu^{+} - \int_{N} f_{n} d\mu^{-}.
$$
You can then apply whatever theorem you like to each integral on the right-hand side separately.


问:
Multilateration with complex signal behavior
I'm trying to solve the TDoA Multilateration problem, in three dimensions. That is, given the coordinates of 4 receivers in 3D space, the velocity of some signal, and the time at which each receiver "saw" the signal, I want to determine the coordinates of the signal source. To sum things up:

Known:

Coordinates of 4 observers (call them A, B, C, D)

Time of arrival of signal at each observer (call them $t_A$, $t_B$, $t_C$, $t_D$).

Velocity of the signal (call it $v$).

Unknown:

Time of signal emission (call it $t_i$)

Coordinates of signal source (call it $(x_i, y_i, z_i)$).

My Problem:

This is, in general, not a trivial problem so solve. It can, of course, be described by a nonlinear system. Many "standard" methods, however, quickly fail (Newton's method, etc.). My case is further complicated by the following:

In my specific situation, the signal in question is a light signal moving through a medium with a "complex" index of refraction. That is, the velocity of light depends upon depth, following the function:

$$n = 1.3 + 0.5(1 - e^{-0.02d})$$

Where $n$ is the index of refraction, and $d$ is depth.

For a simple refractive index, $n = \frac{c}{v}$, where $c$ is the speed of light (in a vacuum), and $v$ is the speed of light in the medium.

The goal here is to implement an algorithmic solution (i.e. in code), so as to process a large amount of signals quickly, determining the coordinates of each signal source as efficiently as possible.

How can this be done?

答:
I change notations since yours at not very clear to me.

I worked this problem many years ago with the problem of sound velocity in deep water (my code was working on-line and had to be very fast).

Normally, for the observer $i$, the equation  write
$$\sqrt{(X-x_i)^2+(Y-y_i)^2+(Z-z_i)^2 }=v_i(Z)\, (t_i-T)$$ where

$(X,Y,Z)$ are the coordinates of the source

$T$ is the time at which the signal emitted

$(x_i,y_i,z_i)$ are the coordinates of the  observer $i$

$t_i$ is the time at which the  observer $i$ received the signal

$v_i$ is the average (in the integral sense) of the velocity of the signal between the source and the observer

This makes four variables $(X,Y,Z,T)$ and I am not sure that four observers would be sufficient (except if there is absolutely no noise).

After trying many things without success, I decided to use the following iterative procedure : fix the velocity at a given value and solve the  problem. From the results, update the $v_i(Z)$ and repeat again and again until the system stabilizes.

Edit

In my answer to this question, everything as based on the fact that, for observer $i$, the equation was
$$(X-x_i)^2+(Y-y_i)^2+(Z-z_i)^2 =\big[v_i(Z)\, (t_i-T)\big]^2$$ while, in fact, we want is to minimize
$$\text{SSQ}=\frac12\sum_{i=1}^n \Big[\sqrt{(X-x_i)^2+(Y-y_i)^2+(Z-z_i)^2 }-v_i(Z)\, (t_i-T)\Big]^2$$ which is a "bit" different.

For convenience, I shall note $D_i=\sqrt{(X-x_i)^2+(Y-y_i)^2+(Z-z_i)^2 }$

So, now, in order to polish the solution (and this step is important), we need to solve for $(X,Y,Z,T)$ the four equations
$$\frac{\partial \text{ SSQ}}{\partial X}=\sum_{i=1}^n\frac{(X-x_i) (D_i-v_i(Z) (t(i)-T))}{D_i}=0$$
$$\frac{\partial \text{ SSQ}}{\partial Y}=\sum_{i=1}^n\frac{(Y-y_i) (D_i-v_i(Z) (t(i)-T))}{D_i}=0$$
$$\frac{\partial \text{ SSQ}}{\partial Z}=\sum_{i=1}^n (D_
i-v_i(Z) (t_i-T)) \left(\frac{Z-z_i}{D_i}-(t_i-T) v_i'(Z)\right)=0$$
$$\frac{\partial \text{ SSQ}}{\partial T}=\sum_{i=1}^n v_i(Z) (D(i)-v_i(Z) (t(i)-T))=0$$


问:
If the number of positive integer $n$ for which $n \leq 1991$ and $6$ is a factor of $n^2 + 3n + 2$ is given by $K$, Find the value of $\sqrt{K - 32}$
I tried by factorising $n^2 + 3n + 2$ into $(n + 1)(n + 2)$ and tried by
assuming $ n = 2k + 1 $ and $ n = 2k $ for odd and even respectively and puting them in place of n

For odd : $(2k + 2)(2k+ 3)$ $\Rightarrow$ $2k + 3$ should be divisible by 3

For even :
$(2k + 1)(2k + 2)$ $\Rightarrow$ $2k + 1$ should be divisible by 3

But I can't proceed any further... Can anyone explain how to go through this question?

答:
Observe that:

$$(n+6)^2 + 3(n+6) + 2 = n^2+12n+36+3n+18+2 = (n^2+3n+2) + 12n+36+18$$
$$ = (n^2+3n+2) + 6(2n+6+3).$$

Therefore, if $n^2+3n+2$ is divisible by $6$, then $(n+6)^2 + 3(n+6) + 2$ is as well. If $n^2+3n+6$ is not divisible by $6$, $(n+6)^2 + 3(n+6) + 2$ is also not divisible by $6$.

Find which $n$ where $1 ≤ n ≤ 6$ the statement is true for. Then, since all natural numbers have remainders from $1$ to $6$ ($0$) when divided by $6$, you can extend this to all natural numbers.


问:
$\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}$ without L'Hôpitals Rule.
The limit,

$\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}=6$

is easily found using L'Hôpitals Rule. However, the exercise is to evaluate it without using this method. I've tried multiple substitutions, mainly pythagoric and reciprocal identities, but without success, the indeterminate form $\frac{0}{0}$ keeps appearing. I’ve also tried splitting the limit and doing the substitutions. I'm looking to simplify the expression in order to evaluate the limit directly.

Thanks in advance.

答:
The denominator $\cot x-1$ resembles $t-1$ too much to not be taken into consideration.

You surely know that
$$
\csc^2x=1+\cot^2x,\qquad \tan x=\frac{1}{\cot x}
$$
so the numerator becomes, with $\cot x=t$,
$$
1+t^2-\dfrac{2}{t^2}=\dfrac{t^4+t^2-2}{t^2}
$$
Never mind if you don't see the factorization $t^4+t^2-2=(t^2-1)(t^2+2)$; you do know that a factor $t-1$ can be found! With synthetic division you find
$$
\begin{array}{r|rrrr|r}
& 1 & 0 & 1 & 0 & -2 \\
1 & & 1 & 1 & 2 & 2
\\\hline
& 1 & 1 & 2 & 2 & 0\end{array}
$$
and thus $t^4+t^2-2=(t-1)(t^3+t^2+2t+2)$ and you can rewrite your limit in the form
$$
\lim_{x\to\pi/4}\frac{\cot^3x+\cot^2x+2\cot x+2}{\cot^2x}=6
$$
If you had seen $t^4+t^2-2=(t^2-1)(t^2+2)=(t-1)(t+1)(t^2+2)$, you'd get the equivalent form
$$
\lim_{x\to\pi/4}\frac{(\cot x+1)(\cot^2x+2)}{\cot^2x}
$$


答:
Hint: for positive integer $n$, $(n+1)(n+2)$ is divisible by $6$ if and only if $n$ is not divisible by $3$.

This is because $(n+1)(n+2)$ is even, and if $n$ is not divisible by $3$, one of $ n+1, n+2$ would be. If $n$ is divisible by $3$, the situation reverses.

Title: If the number of positive integer $n$ for which $n \leq 1991$ and $6$ is a factor of $n^2 + 3n + 2$ is given by $K$, Find the value of $\sqrt{K - 32}$

Body:
I tried by factorising $n^2 + 3n + 2$ into $(n + 1)(n + 2)$ and tried by
assuming $ n = 2k + 1 $ and $ n = 2k $ for odd and even respectively and puting them in place of n

For odd : $(2k + 2)(2k+ 3)$ $\Rightarrow$ $2k + 3$ should be divisible by 3

For even :
$(2k + 1)(2k + 2)$ $\Rightarrow$ $2k + 1$ should be divisible by 3

But I can't proceed any further... Can anyone explain how to go through this question?


问:
Check whether the following series is uniformly convergent OR not?
Check whether the following series is uniformly convergent  OR not?

$\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}, x\in [0,\infty).$

My attempt

$\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}=x^4\sum_{n=1}^\infty \frac{1}{(1+x^2+x^4)^n}=( \frac{1}{(1+x^2+x^4)}).(\frac{x^4}{1-(\frac{1}{1+x^2+x^4})})=\frac{x^4}{(x^2+x^4)}$

$|\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}-\frac{x^4}{(x^2+x^4)}.1|=|\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}-\frac{x^4}{(x^2+x^4)}.\sum_{n=1}^\infty \frac{1}{2^n}|=|\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}-\sum_{n=1}^\infty \frac{x^4}{(x^2+x^4)}. \frac{1}{2^n}|=|\sum_{n=1}^\infty (\frac{x^4}{(1+x^2+x^4)^n}-\frac{x^4}{(x^2+x^4)}. \frac{1}{2^n})|$
How do I proceed further?
Is there any theorem helps to solve it faster?
I couldn't apply weirstrauss M-Test.

Body:
Hint

$$\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}=\frac {x^2}{1+x^2}$$


问:
Visual Explanation of the Dominated Convergence Theorem
From a pedagogical perspective, is there a way to explain to the Dominated Convergence Theorem using visual examples? This may sound tasteless for mathematicians but many times, engineering students do not have exposure to the ways of thinking that mathematicians are comfortable with.

I was hoping that with examples of plots of sequence of functions $f_n(x)$, there must be an easy way of explaining this beautiful result.

Your help and creative ways to explaining this theorem are appreciated.

答:
Here's a simple example that you can easily visualize.
$$
f_{n}(x) =
\begin{cases}
1 & \text{if $x \in [n,n+1)$,}
\\
0 & \text{otherwise.}
\end{cases}
$$
The domain of $f_{n}$ is $[0,\infty)$.

There's no dominating function that is integrable, but the sequence converges to $0$ pointwise. As $n$ increases, the strictly positive part of $f_{n}$ escapes to $\infty$, and the dominating convergence theorem fails.

Now take any integrable function $g > 0$ on $[0,\infty)$. If we change the above example to
$$
f_{n}(x) =
\begin{cases}
g(x) & \text{if $x \in [n,n+1)$,}
\\
0 & \text{otherwise,}
\end{cases}
$$
then the strictly positive part of $f_{n}$ asymptotically disappears (fast enough) before it escapes to $\infty$. So the dominating convergence theorem holds.

You can understand the role of a dominating function as a constraint that prevents the above escaping behavior.


问:
What do we know about the series $\sum_{k=1}^\infty \frac{1}{x+k}$?
If possible, I would like to find a closed-form expression for
$$
f(x)=\sum_{k=1}^\infty \frac{1}{x+k}
$$
The series is so simple (possibly deceptively so) that I'm sure it has been studied before somewhere, but I don't know what to call it and where to find it.

What is this series called so I can learn more? I know I can expand it into a double series by expanding $\frac{1}{1-x/k}$ for $|x|<k$ and $\frac{1}{1-k/x}$ for $|x|>k$ and working on the partial series, which I did before in the special case of $|x|<1$, and the result yields a sum over Bernoulli numbers that I'm not familiar with.

Is a nice closed-form solution for this series known?

答:
Consider $$f_n(x)=\sum_{k=1}^n \frac{1}{x+k}=\psi(n+x+1)-\psi (x+1)=H_{n+x}-H_x$$ where appear the digamma function and the genarlized harmonic numbers.

Use the asymptotics
$$H_p=\gamma +\log (p)+\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^4}\right)$$ and make $p=n+x$. Then continuing with Taylor series
$$f_n(x)=\log(n)-\psi
   (x+1)+\frac{2x+1}{2n}-\frac{6 x^2+6 x+1}{12 n^2}+O\left(\frac{1}{n^3}\right)$$


问:
Proof for interchangeable sum and integrals involving Leibniz test
I would be grateful for some help or hints with this proof.
Let $ \left ( X, \Omega , \mu  \right )  $ be a measure space and $ f_n $ a decreasing sequence of $\mu$-integrable functions, converging pointwise to $0$,

Prove that $ \sum_{n \geq 0} (-1)^{n}\int f_n = \int \sum_{n \geq 0} (-1)^{n}f_n $
I have a hint to use Leibniz proof for alternating series

So here is what I've done so far:

For the right-hand side:

Let $ \int \sum_{n \geq 0} (-1)^{n}f_n = \int S_n $, with $ S_n = \sum_{n \geq 0} (-1)^{n}f_n $

Note that $ f_n\xrightarrow[n\rightarrow \infty]{} 0 $, and
$ f_0 \geq f_1 \geq f_2 \geq f_3 \geq ... \geq $

So we have $f_0 \geq f_0-f_1+f_2\geq f_0-f_1+f_2-f_3+f_4 \geq ...\geq$ and $f_0-f_1 \leq f_0-f_1+f_2-f_3 \leq ... \leq$

Then $S_{2n}$ is monotone increasing and $S_{2n+1}$ is monotone decreasing and note that $S_{2n+1}=S_{2n}+f_{2n+1}\geq S_{2n}$

Note also that $lim_{n \to \infty}(S_{2n+1}-S_{2n})=lim_{n \to \infty}f_{2n+1}=0$ and therefore the sum converges and $S_{2n}<S<S_{2n+1}$
This implies that the $ \int \sum_{n \geq 0} (-1)^{n}f_n = \int S $ is finite.

We also note that the left-hand side is $\sum_{n \geq 0}(-1)^{n}\int f_n = \int f_0 + \int f_2 + ... + \int(-f_1) + \int(-f_3) +...$ = $\sum_{n \geq 0}\int f_{2n} + \sum_{n \geq 0} \int (-f_{2n+1}) $

I'd like to apply Fubini's theorem that for general $f_n$, if $\int \sum_{n \geq 0}|f_n| < \infty $ or $ \sum_{n \geq 0}\int |f_n| < \infty $ then $ \sum_{n \geq 0} (-1)^{n}\int f_n = \int \sum_{n \geq 0} (-1)^{n}f_n $ but Leibniz test isn't enough to prove that $\sum_{n \geq 0}\int |f_n| < \infty$. I'm stuck here and I don't know how to finish. I feel like I'm close either totally wrong. Thanks in advance

答:
You are close, not totally wrong. If $S_n$ is defined as
$$S_n:=\sum_{k=0}^n (-1)^kf_k,
$$
then you may have some of your inequalities pointing the wrong direction. I find that the $(S_{2n})$ are monotone decreasing, the $(S_{2n+1})$ are monotone increasing, and that $S_{2n+1}=S_{2n} - f_{2n+1}\le S_{2n}$. Nonetheless, your argument shows that the sequence $(S_n)$ converges pointwise to a limit $S$.

The key to completing the proof is to notice (and prove) that each $S_n$ is non-negative. So now you can use dominated convergence to conclude
$$\lim_{n\to\infty} \int S_n=\int S$$
which is the same as saying
$$
\lim_{n\to\infty}\sum_{k=0}^n(-1)^k \int f_k = \int\lim_{n\to\infty}\sum_{k=0}^n(-1)^kf_k$$


问:
How to solve this integral for a hyperbolic bowl?
$$\iint_{s} z  dS $$ where S is the surface given by $$z^2=1+x^2+y^2$$ and $1 \leq(z)\leq\sqrt5$ (hyperbolic bowl)</p>

答:
A related problem. Note that,

$$ z=\sqrt{ 1+x^2+y^2 } \implies z_x=\frac{x}{\sqrt{ 1+x^2+y^2 }},\quad z_y=\frac{y}{\sqrt{ 1+x^2+y^2 }} $$

$$ \iint_{s} z  dS = \iint_{D} z  \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2} dA $$

$$ = \iint_{D} \sqrt{1+(x^2+y^2)}\sqrt{{\frac {1+2\,{x}^{2}+2\,{y}^{2}}{1+{x}^{2}+{y}^{2}}}}\,dxdy  $$

$$ =\iint_{D} \sqrt{{{1+2\,{x}^{2}+2\,{y}^{2}}{}}}\,dx dy  $$

Now, $D\equiv \left\{ x^2+y^2 \leq 4  \right\}$. To see this notice that

$$  1 \leq z\leq\sqrt5 \implies 1 \leq \sqrt{1+x^2+y^2} \leq\sqrt5 \implies x^2+y^2\leq 4.  $$

So, we can use polar coordinates as

$$ = \int_{0}^{2\pi}\int_{0}^{2} \sqrt{1+2 r^2}\,r\, dr d\theta = \frac{26\pi}{3} . $$

Added: if you want to parametrize the surface, you go this way,

$$ x=r\cos(\phi),\quad y = r\sin(\phi), \quad z^2 = 1+x^2+y^2= 1+r^2 \implies z=\sqrt{1+r^2}. $$

You can write it in a vector form as

$$ \textbf{T}(r,\phi)= r\cos(\phi)\textbf{i}+ r\sin(\phi)\text{j}+ \sqrt{1+r^2}\, \text{k} $$

$$ \implies  T_r = \cos(\phi)\textbf{i}+ \sin(\phi)\text{j} + \frac{r}{\sqrt{1+r^2}} \text{k}, $$

$$ T_\phi = -r\sin(\phi)\textbf{i} + r\cos(\phi)\text{j}+ 0 \,\text{k}.$$


问:
Question on vectors addition in case of regular hexagon .
There is a regular hexagon given to you ABCDEF.It is asked to find the sum of AB + AC+AD+AE+AF vectors.

I am not getting is how will we make the directions of vectors or you can say head and tail of the hexagon vectors.

Done till now :

1 I have AF vectors as the resultant using polygon law.

2Then taken vectors AC,AD,AE and AF(This one is already made as well).

$ What I saw done online and doubts I got.

Q They took OC vector direction to make AC = AO+OC.OC direction is not given then how can they say that.Maybe it is CO as well.

If my doubt for OC vector is cleared.I think I can solve it.

Please do share if you have some other method to solve it.

Please help.

答:
If $O$ is the centre of the regular hexagon

$$\begin{aligned}
AB + \dots +AF&= 5AO + OB +\dots +OF\\
&=5AO-OA+OA+ OB+\dots +OF\\
&=6AO
\end{aligned}$$

As the sum
$$OA+\dots +OF$$ vanishes.


问:
Some true-or-false questions about matrices
I have some true/false questions about matrices that I got wrong and I'm trying to figure out why:

If AB = BA and if A is invertible, then A is always square.

Answer: I said true because the only matricies that can have AB = BA are square ones right? What does the invertible part have to do with this?

Eigenvectors of an nxn matrix A forms a basis for straight real numbers to the power of n.

I said True but I am not sure why this is false.

答:
Assume that $A$ is $m \times n$. Then $B$ has to be $n \times k$ and $AB$ is $m \times k$. If we look at the converse product $BA$, we see that $k=m$ and the product is $n \times n$. Since $AB=BA$, we get $m=k=n$, so $A$ is square. You are right that the assumption that $A$ is invertible is not necessary here. However, it is actually sufficient in itself. We only talk about inverses for square matrices, so if it is invertible, it is square.

This one is not generally true. Not at all matrices have a basis of eigenvectors. The ones that do are called diagonalizible.


问:
Uniform convergence of series $ \frac{x^{n}}{n}$

Does $\displaystyle \sum_{n=1}^{\infty}\frac{x^{n}}{n}$converge uniformly on interval $[0,1)$?

I've tried to use tail convergence test, but I've got stuck. How should I proceed? Or Is there another way to solve it?

答: Prove by contradiction. If the series is uniformly convergent then there exists $N$ such that $\sum\limits_{k=N_1}^{N_2}\frac {x^{n}} n <\epsilon$ whenever $x \in [0,1)$ and $N_2 >N_1>N$. Let $x$ increase to $1$ in this to see that $\sum\limits_{k=N_1}^{N_2}\frac 1 n\leq \epsilon$ whenever $N_2 >N_1>N$. But this is false since $\sum \frac  1n$ is divergent.


问:
Application of Open Mapping Theorem
This was stated without proof in the complex analysis text I am reading (Complex Made Simple by Ulrich, page 107). I'm sure it's easy, but I'm tired and need a little help.

Let $f$ be nonconstant and holomorphic in some region $V$ and assume $f'$ is nonconstant. Define

$$\Omega = \left\{ \frac{f(z)-f(w)}{z-w}: z,w\in V, z\neq w \right\}.$$

Why is $\Omega$ open? I don't see how to use the hypothesis that $f'$ is nonconstant.

To be clearer: I tried something like the answers below. But Ulrich specifically mentions that the hypothesis that $f'$ is nonconstant is needed in addition to the hypothesis that $f$ is nonconstant. So I feel I'm missing some subtle point about the derivative.

Edit: After reading the answers below, the key point is $f'$ nonconstant implies the difference quotient is nonconstant.

答:
If you let $$
  g_w(z) = \frac{f(z) - f(w)}{z-w}
$$
then $g_w$ is holomorphic on $V \setminus \{w\}$. Since $V \setminus \{x\} = V \cap \{x\}^C$ is the intersection of open sets, it's open. By the open mapping theorem $g_w(V \setminus \{w\})$ is thus open. Now observe that $$
  \Omega = \bigcup_{w \in V} g_w\left(V \setminus \{w\}\right)
$$
shows that $\Omega$ is a union of open sets and thus open.</p>

Title: Application of Open Mapping Theorem


问:
Best Critical Region for Poisson Random Variable
Let $X_1,\ldots, X_{10}$ be a random sample from a Poisson distribution with mean $\theta$. Show that the best critical region against the hypothesis $H_0: \theta = 0.1$ and $H_1: \theta = 0.5$ is $C = \{(x_1,\ldots,x_n)| \sum_{i=1}^{10} x_i \geq3\}$.

My problem is getting the sum to be greater than equal to 3. I get that $\sum_{i=1}^{10} x_i \geq \frac{4-ln(k)}{ln(5)}$ by the Neyman-Pearson theorem, but I don't understand why this implies this is greater than or equal to 3.

答:
I will give you an outline and some relevant computations from R.

Let $S = \sum_{i=1}^{10} X_i.$ Under $H_0,$ we have $S \sim \mathsf{Pois}(\lambda = 1)$ and under $H_1$ we have $S \sim \mathsf{Pois}(\lambda = 1).$

The Neyman-Pearson theorem show that we should reject $H_0$ for large $S.$
Because one would expect larger values of $S$ if $H_1$ is true, it makes
sense that the critical region should be of the form $S \ge c,$ but we need
to do computations involving Poisson distributions in order to know what
significance level $\alpha$ is implied by a specific critical value $c.$
The statement of your problem suggests using $c = 3.$

The significance level is
$$\alpha = P(\mathrm{Rej}\,H_0) | H_0\,\mathrm{True}) = P(S \ge 3|\lambda=1)\\
= 1 - P(S \le 2|\lambda=1) \approx 0.08,$$
which agrees with @lonzaleggiera's second comment. Computation in R, where ppois is a Poisson CDF:

```<code>1 - ppois(2, 1)
[1] 0.0803014
</code>```

Then the power $\pi$ of the test with this specified critical region is as follows:

$$\pi = P(\mathrm{Rej}\,H_0) | H_1\,\mathrm{True}) = P(S \ge 3|\lambda=5)\\
= 1 - P(S \le 2|\lambda=5) \approx 0.875.$$

```<code>1 - ppois(2, 5)
[1] 0.875348
</code>```

In the figure below the blue bars correspond to the null distribution
$\mathsf{Pois}(1)$ and the brown bars to the alternative distribution
$\mathsf{Pois}(5).$ The critical region is to the right of the dotted vertical
line. The significance level is given by the sum of the heights of the blue bars
in the critical region and the power is given by the sum of the heights of the
brown bars.

R code for figure:

```<code>s = 0:12;  pdf.0 = dpois(s, 1);  pdf.1 = dpois(s, 5)
hdr = "Null and Alternative Distn's"
plot(s-.05, pdf.0, type="h", col="blue", lwd=3, ylab="PDF", xlab="s", main=hdr)
 lines(s+.05, pdf.1, type="h", col="brown", lwd=3)
 abline(h=0, col="green2")
 abline(v = 2.5, lwd=2, lty="dotted")
</code>```

Body:
I will give you an outline and some relevant computations from R.

Let $S = \sum_{i=1}^{10} X_i.$ Under $H_0,$ we have $S \sim \mathsf{Pois}(\lambda = 1)$ and under $H_1$ we have $S \sim \mathsf{Pois}(\lambda = 1).$

The Neyman-Pearson theorem show that we should reject $H_0$ for large $S.$
Because one would expect larger values of $S$ if $H_1$ is true, it makes
sense that the critical region should be of the form $S \ge c,$ but we need
to do computations involving Poisson distributions in order to know what
significance level $\alpha$ is implied by a specific critical value $c.$
The statement of your problem suggests using $c = 3.$

The significance level is
$$\alpha = P(\mathrm{Rej}\,H_0) | H_0\,\mathrm{True}) = P(S \ge 3|\lambda=1)\\
= 1 - P(S \le 2|\lambda=1) \approx 0.08,$$
which agrees with @lonzaleggiera's second comment. Computation in R, where ppois is a Poisson CDF:

1 - ppois(2, 1)
[1] 0.0803014

Then the power $\pi$ of the test with this specified critical region is as follows:

$$\pi = P(\mathrm{Rej}\,H_0) | H_1\,\mathrm{True}) = P(S \ge 3|\lambda=5)\\
= 1 - P(S \le 2|\lambda=5) \approx 0.875.$$

1 - ppois(2, 5)
[1] 0.875348

In the figure below the blue bars correspond to the null distribution
$\mathsf{Pois}(1)$ and the brown bars to the alternative distribution
$\mathsf{Pois}(5).$ The critical region is to the right of the dotted vertical
line. The significance level is given by the sum of the heights of the blue bars
in the critical region and the power is given by the sum of the heights of the
brown bars.

R code for figure:

s = 0:12;  pdf.0 = dpois(s, 1);  pdf.1 = dpois(s, 5)
hdr = "Null and Alternative Distn's"
plot(s-.05, pdf.0, type="h", col="blue", lwd=3, ylab="PDF", xlab="s", main=hdr)
 lines(s+.05, pdf.1, type="h", col="brown", lwd=3)
 abline(h=0, col="green2")
 abline(v = 2.5, lwd=2, lty="dotted")


问:
What is the probability that an event will happen when the probability decreases exponentially?
So I am just a middle student who started studying probability in my stats class and I see the questions on this site are little advanced so I don't hope this one is too basic.

Anyways lets say that the probability of an event happening was some number(like 1%), and each time a trial is run the of the event happening decreases by half(so 1% probability for the first trial, .5% probability for the second, .25% probability, and so on).
What is the probability that the event will happen after 1 trial, 10 trials, 100 trials, and an infinite amount of trials?

答:
Note that if $p,q\gt0$, then
$$
\begin{align}
(1-p)(1-q)
&=1-(p+q)+pq\\
&\gt1-(p+q)
\end{align}
$$
Inductively, we can show that if $p_k\gt0$ and $n\gt1$, then
$$
\prod_{k=1}^n\left(1-p_k\right)\gt1-\sum_{k=1}^np_k
$$
and therefore,
$$
1-\prod_{k=1}^n\left(1-p_k\right)\lt\sum_{k=1}^np_k
$$

Assuming the events are independent, the probability that the event occurs at least once in the first $n$ trials is
$$
1-\underbrace{\prod_{k=1}^n\overbrace{\left(1-\frac{0.01}{2^{k-1}}\right)}^{\substack{\text{probability that}\\\text{the event does}\\\text{not occur on}\\\text{trial $k$}}}}_{\substack{\text{probability that the event}\\\text{does not occur in the}\\\text{first $n$ trials}}}\lt0.02
$$
So the probability that the event occurs at all is less than $2\%$.
$$
\begin{array}{c|l|l}
n&1-\prod(1-p_k)&\sum p_k\\\hline
1&0.01&0.01\\
2&0.01495&0.015\\
3&0.017412625&0.0175\\
4&0.018640859219&0.01875\\
5&0.019254208682&0.019375\\
10&0.019847903640&0.01998046875\\
100&0.019867047111&0.02\\
\infty&0.019867047111&0.02
\end{array}
$$


答:
Note $AD=2AO$.

$AOCB$ is a parallelogram. So $AC=AO+AB$

$AOEF$ is a parallelogram. $AE=AO+AF$

$ABOF$ is also a parallelogram. So $AB+AF=AO$

Summing, $$AB+AC+AD+AE+AF=4AO+2(AB+AF)=6AO$$

Title: Question on vectors addition in case of regular hexagon .

Body:
There is a regular hexagon given to you ABCDEF.It is asked to find the sum of AB + AC+AD+AE+AF vectors.

I am not getting is how will we make the directions of vectors or you can say head and tail of the hexagon vectors.

Done till now :

1 I have AF vectors as the resultant using polygon law.

2Then taken vectors AC,AD,AE and AF(This one is already made as well).

$ What I saw done online and doubts I got.

Q They took OC vector direction to make AC = AO+OC.OC direction is not given then how can they say that.Maybe it is CO as well.

If my doubt for OC vector is cleared.I think I can solve it.

Please do share if you have some other method to solve it.

Please help.


答:
Note that since we have a regular hexagon,

$$\begin{align}
AB + AC + AD + AE + AF &= (AB + AF) + (AC + AE) + AD \\
\\
&= (AB + BO) + [(AB + BC) + (AF + FE)] + AD \\
\\
&= AO + [(AB + BC + AF) + FE] + AD \\
\\
&= AO + [(AB + BC + CD) + AO] + AD \\
\\
&= AO + [AD + AO] + AD \\
\\
&= AO + 2AO + AO + 2AO \\
\\
&= 6AO
\end{align}$$


问:
When is an element of a free module over a principal ideal domain contained in a basis?
I'm trying to show the following:

Let $R$ be a principal ideal domain and let $M$ be free $R$-module of rank $n$. Let $Y=\{y_1,\ldots,y_n\}$ be a basis of $M$ and $x\in M$ with $x=y_1a_1+\cdots+y_na_n$. Then $x$ is contained in a base of $M$ if and only if $\gcd(a_1,\ldots,a_n)=1$.

I have shown that if $x$ belongs to a basis of $M$, then $\gcd(a_1,\ldots,a_n)=1$.

This is my idea for the other direction:

If $n=1$, then $a_1\in R^*$, and we are done. Now suppose $n\geq 2$. Since $\gcd(a_1,\ldots,a_n)=1$, there are $s_1,\ldots,s_n\in R$ such that $a_1s_1+\cdots+a_ns_n=1$, then using this equality I tried to get some non-zero $y\in M$ such that $\langle x\rangle\cap \langle y\rangle=0$. Supposing the existence of such $y$,  using Zorn's lemma you can get a maximal nonzero submodule $N$ such that $\langle x\rangle\cap N=0$, then I showed that $\langle x\rangle\oplus N=M$, and since $M$ is free and $R$ a principal ideal domain $N$ is free and has a basis $\{x_1,\ldots,x_m\}$, and thus $\{x,x_1,\ldots,x_m\}$ would be a basis of $M$.

My question is: is there such an $y$? Also, I would like to know whether this idea is too complicated, and if so, I want to know alternative approaches to it.

Thanks

答:
I like to prove things like this as algorithmically as possible,
using matrices and row and column operations that are as elementary
as possible.

I construct an invertible matrix $A$ such that $A \matrix(a_1,\ldots,a_n)^T$ = $(1,0,\ldots,0)^T$. Let $B$ be the linear map corresponding to $A$ relative to the basis $\{y_1,\ldots,y_n\}$.
Then the desired basis is $\{x = B^{-1}y_1,\ldots,B^{-1}y_n\}$.

If $n = 1$, take $A = \matrix(1/a_1)$.

If $n = 2$, let $c = \gcd(a_1,a_2) = 1$ and choose $s_1$ and $s_2$ such that
$s_1 a_1 + s_2 a_2 = c$. Take
$$A = \begin{bmatrix} s_1 \ \ \ \ s_2 \\ -a_2/c \ \ a_1/c \end{bmatrix}.$$
Then $A \matrix(a_1,a_2)^T = \matrix(c,0)^T$. In this step,
$c = 1$, but we write it as $c$ for use in the induction step.
$A$ is invertible because its determinant is $1$.

If $n > 2$, use induction on the number of nonzero entries in $\matrix(a_1,\ldots,a_n)$.

If there is only $1$ nonzero entry, then it is a $\gcd$ of the entries and
in fact $1$ by the choice made in the previous step. Apply the most
elementary row operation of interchanging rows to move the nonzero entry
to the first entry. Row interchange operations expressed in matrix
form correspond to multiplication on the left of the matrix $A$ constructed
in previous stages of the induction. The product remains invertible.

If there are $2$ or more nonzero entries, first interchange rows so that
$a_1$ and $a_2$ are nonzero (this will actually make the row interchange
in the above step never occur). Apply the $n = 2$ step to the first
$2$ coordinates (more precisely, to the first $2$ rows) (augment the
$2 \times 2$ matrix with a $(n-2) \times (n-2)$ identity matrix).
Now $c = \gcd(a_1,a_2)$ is not necessarily a unit, but $A$ is
invertible because we divided its second row by $c$ to make its determinant
$1$. Most importantly, reducing
$\matrix(a_1,a_2,a_3,\ldots)^T$ to $\matrix(c,0,a_3,\ldots)^T$
leaves the $gcd$ of the entries unchanged, so the induction works.

All the matrices constructed correspond to not-always-elementary row
operations when they are used to multiply column vectors on the left.
Some are non-elementary since they use a linear combination of rows
while for pure elementary row operations only the special linear
combination $unit * row_1 + r_2 * row_2$ is permitted.
If $R$ is Euclidean, then the Euclidean algorithm essentially gives
a construction of the $2 \times 2$ $A$ as a product of matrices for
elementary row operations, so the final $A$ is also such a product.
I don't know of any practical algorithms for constructing the
$2 \times 2$ $A$ for PIDs that are not Euclidean.

This is the dual of a special case of the algorithm for reduction to
Hermite normal form.  In the general case of it, we start with an
arbitrary matrix instead of a column vector, and do reversible (left)
row operations on it to get a triangular matrix (with some complications
for the matrix not being square).  Hermite normal form is what you
can naturally get by doing reversible (right) column operations
instead. There is also Smith normal form.  It is what you can naturally
get by doing both reversible row operations and reversible column
operations -- a diagonal matrix.  All of the elementary theory of
modules over PIDs can be read off from these matrices.


问:
Questions about Euler Theorem
I am not a student of mathematics, and I have never studied number theory. I was looking at RSA recently. Euler Theorem is used here. I don't quiet understand $1(mod\quad n)$ in Euler Theorem$a^{\phi (n)}≡1\pmod n$.  Are $1\pmod n$ and $1\; mod\; n$ the same ?   Isn't $1\; mod\; n$ always one?

Why is the $ed=1\pmod n$ equal to $d=e^{-1}\;mod\;\phi(n)$?

答:
No. The issue is what that notation means. You're thinking of mod in terms of the operation. Rather, what this denotes is a congruence.

Specifically, if

$$a \equiv b \pmod n$$

what this notation means is that $n \mid a-b$. That is, $n$ evenly divides $a-b$. Or, even more clearly, there is some integer $k$ such that $a-b = kn$. Some examples and why they hold:
$
\newcommand{\con}[4]{#1 \equiv #2 \!\!\!\! \pmod{#3} \;\;&\text{ because } \;\; #1 - #2 = #3 \cdot #4 \\}
$
\begin{align*}
\con 5 2 3 1
\con 9 0 3 3
\con{15}{1}{2}{7}
\con{12}{2}{5}{2}
\con{18}{5}{13}{1}
\con{123}{0}{41}{3}
\con{16}{7}{9}{1}
\con{24}{3}{7}{3}
\con{2}{4}{2}{(-1)}
\con{5}{15}{5}{(-2)}
\end{align*}

Title: Questions about Euler Theorem


问:
nbd in subspace topology
I'm studying topology and struggling with simple statement :

Let $X$ be a topological space and $A \subset X$ with subspace topology. Let $x \in A$ then "Every nbd of $x$ in $A$ is a nbd of $x$ in $X$" $\iff$ "$A$ is a nbd of $x$ in $X$"

I need to know why this statement is true to understand the proof of "A locally compact Hausdorff space is $T_{3}$". I tried to use "for open set $V$ in A(subspace topology) there is open $U$ in X s.t. $V = A \cap U$". But it does not help me to prove the above statement. I cannot think of anything but this. How can I prove this statement?

答:
$\implies$ is obvious since $A$ is a neighborhood of $x$ in $A$.

For the converse assume that $A$ is a neighborhood of $x$ in $X$. There exist an open set $V$ in $X$ such that $x \in V \subseteq A$. Let $W$ be any neighborhood of $x$ in $A$. There exists an open set $U$ in $A$ such that $x \in U \subseteq W$. We can write $U=A \cap T$ where $T$ is open in $X$. Now $x \in V\cap T$, $V \cap T$ is open in $X$ and $V \cap T \subseteq W$ (because $V \subseteq A$ and $U \subseteq W$). We have proved that $W$ is a neighborhood of $x$ in $X$.

Title: nbd in subspace topology


问:
$u_t+\frac{x}{t}u_x=0$ for $t>1$
$$u_t+\frac{x}{t}u_x=0,\hspace{20mm}t>1,-\infty<x<\infty,$$
where the initial data is given at $t=t_0$ at $t=1$ and we have $u(x,t=1)=u_{in}(x)=e^{-x^2}$.

I used method of characteristics. I find that $\frac{dx}{dt}=\frac{x}{t}\implies x=ct$. Now, I have $c=\frac{x}{t}$, so my solution would be $u(x,t)=u_{in}(\frac{x}{t})=e^{-(\frac{x}{t})^2}$. Could someone please tell me if I am correct?

答:
Why asking if your solution is correct ? You can check yourself :
$$u(x,t)=e^{-(x/t)^2}$$
$$u_x=\frac{-2x}{t^2}e^{-(x/t)^2}$$
$$u_t=\frac{2x^2}{t^3}e^{-(x/t)^2}$$
$$u_t+\frac{x}{t}u_x=\left(\frac{2x^2}{t^3}e^{-(x/t)^2}\right)+\frac{x}{t}\left(\frac{-2x}{t^2}e^{-(x/t)^2}\right)=0$$
Thus $u=e^{-(x/t)^2}$ is solution of the PDE.
$$u(x,1)=e^{-(x/1)^2}=e^{-x^2}$$
Thus $u=e^{-(x/t)^2}$ satisfies the condition.

Title: $u_t+\frac{x}{t}u_x=0$ for $t>1$


问:
The set of continuous map from $X$ to $K$ is separable

$\mathbf QUESTION :$ Show that $C_K$(X) , the set of all continuous functions from $X$ to $K$, with supremum norm is separable. Where $X$ is a compact , Hausdorff space and $K$ is a complete separable metric space.

I have showed that $Y = C_K(X)$ is a Banach space. Using taking a Cauchy sequence I have showed that the sequence is uniformly Cauchy sequence . And as $K$ is complete so the uniform limit is continuous. Hence $Y$ is Banach . But in proof of separability I want to show that $Y$ has a subset which is countable and dense in $Y$.
but how can I prove that the existence on such subset.

Please help me.

答: In case $X$ is a point (which is compact) then the space of continuous functions from $X$ to the metric space $K$, with the sup norm, is isometric (hence homeomorphic) to $K$ itself. In particular, if $K$ is not separable there is no hope to prove your claim. For metric spaces, separability is equivalent to second countability, and there exist complete metric spaces which are not second countables.


答:
Given $w\in V$, define
$$g_w(z):V\setminus\{w\}\to\mathbb C,\quad z\mapsto \frac{f(z)-f(w)}{z-w}.$$
Since $f$ is holomorphic on $V$ and since $f'$ is not constant, $g$ is a nonconstant holomorphic function on $V\setminus\{w\}$. Noting that $V\setminus\{w\}$ is open, by open mapping theorem, $g_w(V\setminus\{w\})$ is an open subset of $\mathbb C$.
It follows that
$$\Omega=\bigcup_{w\in V}g_w(V\setminus\{w\})$$
is open.</p>

Title: Application of Open Mapping Theorem

Body:
This was stated without proof in the complex analysis text I am reading (Complex Made Simple by Ulrich, page 107). I'm sure it's easy, but I'm tired and need a little help.

Let $f$ be nonconstant and holomorphic in some region $V$ and assume $f'$ is nonconstant. Define

$$\Omega = \left\{ \frac{f(z)-f(w)}{z-w}: z,w\in V, z\neq w \right\}.$$

Why is $\Omega$ open? I don't see how to use the hypothesis that $f'$ is nonconstant.

To be clearer: I tried something like the answers below. But Ulrich specifically mentions that the hypothesis that $f'$ is nonconstant is needed in addition to the hypothesis that $f$ is nonconstant. So I feel I'm missing some subtle point about the derivative.

Edit: After reading the answers below, the key point is $f'$ nonconstant implies the difference quotient is nonconstant.


问:
Best approximation proof question
Here is proof on best approximation , V is a finite-dimensional inner product space, $W$ is a subspace of $V$, $x \in V$ $\|{x-P_wx}\| \leq\|x-y\|$ for $y \in W$ , where $P_wx$ is the orthogonal projection linear transformation from $V$ onto $W$

$\|x-y\|^2=\|x-P_wx+P_wx-y\|^2$

$=\|x-P_wx\|^2+\|P_wx-y\|^2$ since $x-P_wx$ and $P_wx-y$ $ \in W$are orthogonal

$\geq \|x-P_wx\|^2$

My question is, for the last line, how did the $\|P_wx-y\|^2$ term disappear? I know that it is positive but how does it mean $\|{x-P_wx}\| \leq\|x-y\|$?

答:
It disappeared simply because it is nonnegative.

$$x+y \ge x$$

holds whenever $x,y \ge 0$ for that reason. You can see this by working backwards and adding $x$ to both sides:

$$y \ge 0 \implies x+y \ge x$$

In this case, you're doing that with norms, which are nonnegative. From that step in the proof you mentioned, then, taking the square root of each side gives the desired result.

Title: Best approximation proof question


问:
Passing Weak Convergence Through a Function?
Let $f_k,f : \mathbb{R}^d\to \mathbb{R}$. Assume $f_{k}$ converges weakly to $f$ in $L^p(\mathbb{R}^d)$. Let $p:\mathbb{R}\to \mathbb{R}$ be some function.

What can say about the weak convergence of $p(f_k)$ (what are some general assumptions on $p$ that one would expect to need)? The reason I ask is I have two terms in an equation in which I want to evaluate as $k\to\infty$, these are

$$ \int_{\mathbb{R}^d} \varphi_1(x)f_k(x)+ \varphi_2(x)p(f_k(x))~dx  $$

where $\varphi_i\in C^\infty_c(\mathbb{R})$, infinitely differentiable compact support, i.e are test functions. I don't mind if I can evaluate this limit $k\to\infty$ up to a subsequence.

答:
This is not possible unless $p$ is affine.

Indeed, if $p$ is not affine, then there are $a,b \in \mathbb R$ and $\lambda \in (0,1)$ such that
$$
p( \lambda \, a + (1-\lambda) \, b) \ne \lambda \, p(a) + (1-\lambda) \, p(b).$$

Let us denote the support of $\varphi_2$ by $C$ and the Lebesgue measure by $\mu$. For simplicity, I will assume that $\varphi_2 = 1$ on $C$,
otherwise the argument still works but is a little bit more technical.

Now, we can construct a sequence $f_k$, such that each $f_k$ only takes the values $a$ and $b$ on $C$ with
$$ \mu(\{ x \in C \;|\; f_k(x) = a \}) = \lambda \, \mu(C)$$
and such that $f_k \rightharpoonup f$,
where $f$ equals $\lambda \, a + (1-\lambda)\,b$ on the support of $\varphi_2$.
(This is possible by some standard construction of oscillating sequences).

Then
$$\int \varphi_2 \, p(f_k) \mathrm d\mu = (\lambda \, p(a) + (1-\lambda) \, p(b)) \, \mu(C),$$
but
$$\int \varphi_2 \, p(f) \mathrm d\mu = p(\lambda \, a + (1-\lambda) \, b) \, \mu(C).$$


问:
Non-empty Intersection of a set with the closure of some other sets implies Non-empty Intersection of a set with the sets
Let $X$ be a topological space.

Let $(x_i)_{i \in I}$ be a net in $X$.
Let $I$ be an directed set.

Let $S_i= \{ x_j \in X : j \geq i,  \forall i\in I \}$.

Suppose $x\in \cap_{i \in I} \bar{S}_i$, where $\bar{S}_i$ is the closure of $S_i$.

Then we have that any neighbourhood $U_x$ of $x$ intersects with each of $\bar{S}_i$.

From the previous line, why do we have that any neighbourhood $U_x$ of $x$ intersects with each of $S_i$?

Thanks!

答:
Fix $i \in I$ for now, arbitrarily. We know $U_x$ is a neighbourhood of $x$ and we also know that $x \in \overline{S_i}$ by definition (Of the intersection). This means that every neighbourhood of $x$, so also $U_x$ (!), intersects the set $S_i$.

So for every $i \in I$ there is some $j \ge i$ such that $x_j \in U_x$ (the witness to $U_x$ intersecting $S_i$). This says that $x$ is an accumulation (or cluster) point of the net $(x_i)_{i \in I}$.


问:
What is the deepest / most interesting known connection between Trigonometry and Statistics?
I'm teaching both at the same time to different classes in high school, so I just wondered about this.

Added by OP on 16.May.2011 (Beijing time)

I mean Statistics only, without Probability. In other words, Descriptive Statistics only. This rules out Buffon’s Needle Problem.

The occurrence of π is counted only as a connection to geometry. By Trigonometry is meant explicit, non-gratuitous, occurrence of the sine, cosine, tangent, or their reciprocals.

Yes, the Law of Cosines fits the bill, but it is on the surface: everyone knows about it. It would be a hugely interesting meta theorem that this were the “deepest” connection between Trigonometry and Descriptive Statistics. My suspicion/hope is that there are deeper connections, somewhat along the line of the surprising use of trig in solving the cubic in closed form, or the use of trigonometric substitutions in evaluating certain integrals. The comment below about the arcsine transformation at first blush seems to be something along this line, but when you follow the link you see that someone is bringing it up only to say how bad it is.

So, I hope the intent of my question is now much clearer.

答:
The normal distribution has a $\pi$ in it. That's fairly deep. </p>

<p>Edit: I don't see how this doesn't count. $\pi$, after all, is $4 \arctan 1$, or half the period of the $\sin$ and $\cos$ functions. It is closely related to trigonometry and the properties of the trigonometric functions, and to call $\pi$ an occurrence of "geometry" as if that were something unrelated to trigonometry is mystifying. </p>

Body:
The normal distribution has a $\pi$ in it. That's fairly deep.

Edit: I don't see how this doesn't count. $\pi$, after all, is $4 \arctan 1$, or half the period of the $\sin$ and $\cos$ functions. It is closely related to trigonometry and the properties of the trigonometric functions, and to call $\pi$ an occurrence of "geometry" as if that were something unrelated to trigonometry is mystifying.


问:
union of a Sequence of closed subsets
Let $F_1,F_2,...$ be closed subsets of some topological space $X$. If for each $x \in X$ there is a neighborhood of $x$ (open set containing) such that

$N_x \cap F_j \neq \emptyset$

for only finitely many $j$. Prove the arbitrary union of the $F_j$ is closed in $X$.

Here is my attempt and all I have so far,

So I would like to show the complement is open in $X$, that is,

$X \backslash \bigcup_{j=1}^\infty F_j \subset X$

is open. Let $x \in X \backslash \bigcup_{j=1}^\infty F_j$ be arbitrary, we wish to put $x$ in a neighborhood that is disjoint from $\bigcup_{j=1}^\infty F_j$

So we are guaranteed the existence of a neighborhood $N_x$ of $x$ such that

$N_x \cap F_1,...,N_x \cap F_n$

all have non-trivial intersection. But for $F_{n+1},F_{n+2},...$ we know the intersection with $N_x$ is empty, that is

$N_x \cap F_{n+1},N_x \cap F_{n+2}$

are all empty. Also arbitrary union of open is open and since the $F_j$ are closed their complements in $X$ are open, so can I take the complements of the infinitely many $F_j$ that is $X \backslash \bigcup_{k=n+1}^\infty F_k$ is the open neighborhood of $x$? a bit lost. SO basically is it saying that those finitely many $F_j$ have elements that can get close to $x$ but after the threshold of $n$ the $F_k$ become disjoint from balls around $x$? and they are complements of closed so each is open and arbitrary union of open is open?

答:
Let $x \notin \bigcup_{i \in I} F_i$. We know that there is an open neighbourhood $N_x$ of $x$ such that $J:= \{i \in I: F_i \cap N_x \neq \emptyset\}$ is finite.
We also know that $x \notin F_j$ for all $j \in J$, so $O=N_x \bigcap_{j \in J}(X\setminus F_j)$ is an open neighbourhood of $x$, being a finite intersection of such open sets. And $O$ does not intersect any $F_i$: the ones with $i \notin J$ are already missed by $N_x$, so certainly by the smaller $O$, and all the $F_j, j \in J$ are missed by construction. So $O$ misses $\bigcup_{i \in I} F_i$ and as $x$ was arbitrary, this union is indeed closed.

Title: union of a Sequence of closed subsets


问:
coordinate tensor with respect to the standard basis vectors
Consider the bilinear form $\phi: \mathbb{R}^3 \times \mathbb{R}^3 \rightarrow \mathbb{R}$ defined by
$$\phi(v,w)=\langle(1,0,0)^T,v \times w\rangle$$
Then what is the coordinate tensor with respect to the standard basic vectors?

答: In the standard basis, $\phi=v_yw_z-v_zw_y=v_iT_{ij}w_j$ (see $r=p_iT_{ij}q_j$ here) with$$T=\left(\begin{array}{ccc}0 & 0 & 0\\0 & 0 & 1\\0 & -1 & 0\end{array}\right).$$


问:
How to derive a Hypergeometric function from the integration of the square of Bessel function
I have a integral function like below. $ J_0 $ is Bessel function and c is a constant.

$$
\int \left( J_0\left(c*x \right) \right)^2 dx
$$

In the Mathematica software, it gives me the equation as the result.

$$
x * HypergeometricPFQ[\{ 1/2, 1/2 \}, \{1, 1, 3/2\}, -c^2 x^2]
$$

This software does not present any steps for derivation.

Could you help me confirm these equation?

Here is the code

```<code>Integrate[BesselJ[0, c*x]^2, x]
</code>```

Thank you

Body:
You have
$$J_0(x)=\sum_{n=0}^\infty (-1)^n \frac{x^{2 n} }{2^{2 n}\,\Gamma (n+1)^2 }$$
$$\big[J_0(x)\big]^2=\frac 1 {\sqrt \pi}\sum_{n=0}^\infty (-1)^n\frac{ \Gamma \left(n+\frac{1}{2} \right)}{ \Gamma (n+1)^3}x^{2n}$$
$$\int \big[J_0(x)\big]^2\,dx=\frac 1 {\sqrt \pi}\sum_{n=0}^\infty (-1)^n\frac{ \Gamma \left(n+\frac{1}{2}\right)}{(2 n+1) \Gamma (n+1)^3} x^{2 n+1}$$
$$\int \big[J_0(x)\big]^2\,dx=x \, _2F_3\left(\frac{1}{2},\frac{1}{2};1,1,\frac{3}{2};-x^2\right)$$


问:
Arrange points in a high-dimensional space when you only have distances (no coordinates)
I'm working on an algorithm to determine how similar two files are. You feed $2$ files into the algorithm and the algorithm returns a natural number. If this number is $0$, the files are identical. If they have similar contents, the number is small, if they are different, the result is a larger number (of the order of the length of the larger file).

I want to use this algorithm to compare many files (e.g. thousands of log files) to group them into clusters based on their similarities. But most clustering algorithms work with the coordinates of points in a multidimensional space. But I do not have coordinates. I have distances only.

1st question:

If the triangle inequality applies for each triple of distances (Let $a,b,c$ be the three distances, $a \le c \land b \le c$, then $a+b \ge c$): Is it then guaranteed that the $n$ points between which the distances are known can always be arranged in a ($n-1$)-dimensional space?

I would use the following algorithm:

Sort the files (points) in any order, so you have Points $p_1, p_2, p_3$ etc.

Set the point $p_1$ to the coordinate origin

Set the point $p_2$ on the positive $x_1$ axis, corresponding to the distance $p_1-p_2$

Place point $p_3$ in the plane defined by axes $x_1$ and $x_2$ so that the distances to the points $p_1$ and $p_2$ are correct and $p_3$ has a non-negative value for its $x_2$ coordinate.

Continue with all further points so that the new point correctly takes all distances to the previous points and that it has a non-negative value for the newly added coordinate.

With my imagination I can understand that I can always use this algorithm to insert $4$ points with the correct distances into a $3$-dimensional space, and my gut feeling tells me that this is also true for all higher dimensions, but gut feeling is not a proof.

Is my assumption correct?

2nd question:

Same problem, but now the triangle inequality is violated. For example, among many thousands of points there are 3 points $A, B, C$ with these distances: $AB = 1, BC = 1000, AC = 1100$. (These are rounded results of a real test run.)

What possibilities do you have to arrange these points in a high-dimensional space? (The purpose is still to find clusters. The purpose is not accurate positioning.)

Addendum

I just posted an answer to my own question in which I showed that my assumption from question 1 was wrong.

But question 2 is still left unanswered, so let me ask it again with other words:

Given a complete graph with $n$ vertices and weights for all of its $\frac{n^2-n}{2}$ edges: What is the best way to embed this graph in an Euclidean space with $n-1$ dimensions, so that each vertex becomes a point and each pair of $2$ points has a distance as close as possible to the weight of the edge that connects both vertices?

答:
My assumption was wrong.

If you have a complete graph with more than 3 vertices, with positive weights for each edge, and if the triangle inequality applies for each triple of edges, then it is not always possible to embed this graph in Euclidean space, so that each vertex is a point and the weight of each edge is a distance.

Let $A,B,C,D$ be the vertices of the graph (points in Euclidean space) with these weights (distances):

$$
\overline{AB} = 5\\
\overline{AC} = 8\\
\overline{AD} = 5\\
\overline{BC} = 5\\
\overline{BD} = 8\\
\overline{CD} = 5\\
$$
Since the largest distance is less than twice as large as the smallest distance, the triangle inequality applies to all 4 triangles $ABC, ABD, ACD, BCD$.
Let's begin to create the triangles $ABC$ and $ACD$. They are equal and they have a common edge ($AC$). The height of both triangles over their common edge is $3$. If you embed those triangles in $3D$ space (or any Euclidean space with more dimensions), the points $B$ and $D$ can freely rotate around the side $AC$ as long as they aren't connected. But their distance can never be greater that the sum of both heights, which is $6$. But $\overline{BD}$ should be $8$.

So, it is not possible to embed this graph with its weights in Euclidean space. Any larger complete graph might contain such an impossible tetrahedron as described here, and therefore for all complete graphs with more than 3 edges the triangle inequality is a criterion that is to weak to guarantee that the graph might be embedded in Euclidean space.

qed.


问:
generating function for k-combinations
Let $S$ be a set with $n$ elements. I want to compute the following generating function:

$$G(x)=\sum_{A\subseteq S}x^{|A|}.$$ I want to understand the following equality: $G(x)=(1+x)^n$.

In a book, it says "each of the $n$ elements in the set contributes the term $(1+x)$ to the generating function."

Is there a better explanation?

答:
Well by the binomial expansion $$G(x)=(1+x)^n=\sum_{k=0}^n \binom{n}{k} x^k$$
but note that
$$
\#\{A\subseteq X:|A|=k\} =\binom{n}{k},
$$
so we get
$$
G(x)=\sum_{A\subseteq S} x^{|A|}.
$$

Title: generating function for k-combinations


问:
An Interesting Number Theory Problem: Solve $2520=(x+y+xy)^2+2xy+2y-3x, x, y \in \mathbb Z$
Solve:

$2520=(x+y+xy)^2+2xy+2y-3x$

Please help, I tried Solving SFFT and Quadratic, but couldn't end up with a result.

Edit: I solved this much.I factorized it
$((x+1)(y+1)-1)^2+2((x+1)(y+1)-1)-5x$
Then substituted $t=((x+1)(y+1)-1)$
so $t^2+2t-5x=2520$
and after this i am stuck
Edit 2: Solve for Integers

Body:
$$2520=x^2+y^2+(xy)^2+2x^2y+2y^2x+2xy+2xy+2y-3x$$ so, if we group in terms of $y$ $$2520=y^2(x^2+2x+1)+y(2x^2+4x+2)+x^2-3x$$ so $$0=y^2\cdot(x+1)^2+2y(x+1)^2+x^2-3x-2520$$ so solving the quadratic equation in the usual way:

$$\Delta=4(x+1)^4-4(x+1)^2(x^2-3x)=4(x+1)^2(x^2+2x+1-x^2+3x+2520)$$ so we have found out that $$\Delta=(2x+2)^2\cdot(5x+2521)$$

but we know

$$y=\frac{-2(x+1)^2\pm\sqrt{\Delta}}{2(x+1)^2}=\pm\frac{\sqrt{\Delta}}{2(x+1)^2}-1=\pm\frac{2(x+1)\sqrt{5x+2521}}{2(x+1)^2}-1=\pm\frac{\sqrt{5x+2521}}{(x+1)}-1$$ so for $y$ to be an integer, $x+1$ must divide $\sqrt{5x+2521}$ and  $5x+2521$ must be a perfect square. Let $5x+2521=k^2$ so $x=\frac{k^2-2521}{5}$ so $$\big(\frac{k^2-2521}{5}+1\big)^2\text{ divides } k$$ so $$(k^2-2516)^2\text{ divides } 25k$$ so $$(k^2-2516)^2\leq 25|k|$$ which gives us $k=\pm50,\pm51$. If $k=\pm 50$, then because $x=\frac{k^2-2521}{5}$, x wouldn't be an integer so contradiction. So $k=51$, so $x=\frac{51^2-2521}{5}=16$

Because $x=16$ wecan now see that because $y=\pm\frac{\sqrt{5x+2521}}{(x+1)}-1=\pm3-1$ so $y$ is $2$ or $-4$.

So the solutions are $(x,y)=(16,-4)$ and $(16,2)$


问:
A problem involving handshakes and probability with combinatorial elements to prove

My approach in this problem was to use combinatorial arguments. To admit, I cannot understand leave alone solve it. I need to start off...I am actually confused between the probability and the combinatorial part....

答:
Hint:

There are $(n-2)$ people at the party other than $i, j \,$ where $i \ne j$.

A) If $i$ and $j$ meet each other

$i$ and $j$ need to meet $(k-1)$ and $(l-1)$ people respectively from rest $(n-2)$ people.

Number of ways of $i$ meeting $(k-1)$ of $(n-2)$ people $ = {n-2 \choose k-1}$

Similarly for $j$.

$i$ meets any other person with probability $p$ and does not meet with probability $(1-p)$. Same for $j$. Please also consider that $i$ and $j$ meet each other with probability $p$.

Probability that both events occur ($i$ meeting $k$ people and $j$ meeting $l$ people) will be multiplication of both probabilities.

B) If $i$ and $j$ do not meet each other.

Number of ways of $i$ meeting $k$ of $(n-2)$ people $ = {n-2 \choose k}$. Similarly for $j$.

You go on to calculate the probability similar to case $A$.


问:
the Monotonicity of functions
Let  $$g(x)=-x \frac{\cos(\frac{A}{x})}{\sin(\frac{\pi}{x})}+x.\cot(\frac{\pi}{x})$$ be a function with respect to $x$  defined on $(\frac{3}{2},+\infty)$

and $A\in \mathbb{R}$ a constant, such that $$\frac{-\pi}{2}.x\le A \le \frac{\pi}{2}.x$$

shows  that $g(x)$ is increasing on $(\frac{3}{2},+\infty)$

the derivative gives me a bit difficult term to work with

Thanks

答:
$$f=x \cot \left(\frac{\pi }{x}\right)-x \csc \left(\frac{\pi }{x}\right) \cos
   \left(\frac{A}{x}\right)$$
Let $x=\frac 1y$ to make
$$f=\frac{\cot (\pi  y)-\csc (\pi  y) \cos (A y)}{y}$$ Compute $f'_y$ and
$$f'_x=f'_y \times \frac{dy}{dx}$$

Title: the Monotonicity of functions