问:
Absolute continuity of probability measure
Let $([0,1), \mathcal{B}([0,1))$ be a probability space and define the filtration $\mathcal{F}_n=\sigma([0,\frac{1}{2^n}),[\frac{1}{2^n}\frac{2}{2^n}),\ldots,[\frac{2^n-1}{2^n},1))$. Let $\mu_n$ and $\nu_n$ be the restrictions to $\mathcal{F}_n$ of $\mu=$ any probability measure on $(\Omega, \mathcal{F})$ and $\nu=$ Lebesgue measure respectively. Why is it true for all probability measures $\mu$ that $\mu_n \ll \nu_n$?

答: The sigma algebra consists of just unions of the $2^{n}$ intervals. It is obvious that the only set $E$ in this sigma algebra with $\nu_n(E)=0$ is the emprty set. Of course, $\mu_n(\emptyset)=0$.


问:
Is the set of matrices $X \in M_{n\times n}$ satisfying $AX + XA^T = -L$ for a given but arbitrary $A,L \in M_{n\times n}$ a subspace?
Is the set of matrices $X \in M_{n\times n}$ satisfying $AX + XA^T = -L$ for a given but arbitrary $A,L \in M_{n\times n}$ a subspace under usual definitions of matrix addition and scalar multiplication?

Apparently it is not ... For $X_1, X_2\in M_{n\times n}$ and $c \in R$ it is clearly closed by addition and multiplication due to matrix operations. The zero vector is not obvious to me since if $X=[0]$, this would have to make $L=0$ which may not be the case. Is that enough of a justification?

答: You are right. It is a subspace only if $L=0$. Suppose $X_1$ and $X_2$ belong to the set. Then $AX_1+X_1A^T=-L$ and $AX_2+X_2A^T=-L$ and it follows that $A(X_1+X_2)+(X_1+X_2)A^T=-2L$, but $-2L=-L$ only if $L=0$. In general, non-homogeneous linear relations do not lead to (vector) subspaces.

ContentLicense: CC BY-SA 4.0


问:
Is it possible to give a time restriction to a GAP program?
I have written a GAP program that returns "true" or "false" as output.

I would like to modify the program in the following way:

If the computation is not finished after 20 seconds, then its output shall be "false".

Is this possible with GAP?

Example:

I would like to use this program in order to test if an ideal in a family of ideals has a certain property. If the computation does not finish after 20 seconds, then the ideal will most likely not have this property.

I would like to have a probabilistic program in that sense.

Thank you.

答:
In version 4.8 it is possible to call a function with a provided timeout as described in this answer. However, this functionality seems to have serious problems as described in this issue on GitHub and was therefore withdrawn in version 4.9 as stated here.

There is no convenient way (and with convenient I am rereferring to some sort of build in functionality) to achieve what you desire in the current version of plain GAP if we consider the GAP reference as a complete description of its capabilities.

However, using the parallelization options described in HPC-GAP, we could start our calculation in a separate thread, while letting the main thread checking the exceeded time. I have never used it, so I am not sure if this would actually work. I tried it, but failed.

We could also embed the entire GAP system as a C library and call GAP functions in a C program. We could easily monitor the runtime within this program.

Also, and already mentioned by Ethan Bolker in the comments, you can write a GAP program and call it via a shell script. Monitoring the runtime there is also possible. The procedure is described in this answer.

It might also be convenient in some cases to use the GAP interface in SageMath.


问:
Integration by substitution: $\displaystyle\int\frac{dx}{(1+x^2)^2}$
I came across this solution of a tricky integral using implicit substitution (i.e., manipulating differentials without actually declaring $u=\cdots$).
\begin{align*}
                \int \frac{dx}{(1+x^2)^2} &= \int \frac{1 + x^2 - x^2}{(1+x^2)^2}\,dx\\
                &= \int \frac{dx}{1+x^2} - \int \frac{x^2}{(1+x^2)^2}\,dx\\
                &=\tan^{-1}x - \frac{1}{2}\int \frac{x}{(1+x^2)^2}\,d(x^2+1)\\
                &= \tan^{-1}x+\frac 12\int x\, d\Big(\frac 1{1+x^2}\Big)\\
                &=\tan^{-1}x + \frac12\cdot \frac{x}{1+x^2}- \frac12\int \frac{dx}{1+x^2}\\
                &= \frac12\Big(\tan^{-1}x + \frac{x}{1+x^2}\Big)+c.
            \end{align*}
Can someone help me understand how to go from the third line to the fourth?  It doesn't seem like an obvious step. I'm guessing that it's because
$$d\Big(\frac 1{u(x)}\Big) = -\frac{d(u(x))}{u(x)^2},$$
but it doesn't seem like a natural next step, it's just the way I justified it to myself after already having seen the solution. Why should it be the next step?

答:
Observe that $\;d(x^2+1)=2x\,dx\;$, thus

$$x^2\,dx=\frac12\left(x\,d(x^2+1)\right)$$

and in a similar way

$$d\left(\frac1{x^2+1}\right)=-\frac{2x}{(1+x^2)^2}dx$$

Thus, from fourth to fifth line they do integration by parts, as in the fourth one they have $\;\int uv' dx\;$ , with $\;u=x,\,\;\;v'=d\left(\frac1{x^2+1}\right)\;$ ...

Pretty beautiful way...and the way you justified it is the correct one.


答:
What's been done is Integration by Substitution. The full form would explicitly perform the substitution $u = x^2 + 1$, calculate $du$ in the form of $dx$, and substitute everything explicitly.

What the source of this working has done is directly plop the $x^2 + 1$ in place without bothering to go through the workings. It looks a bit like showing off to me, deliberately being a smartypants in order to confuse those weaker.

If you go through the work of making that $u = x^2 + 1$ substitution on the RHS of line 2, you will get to the same answer with a little bit more work, and a much more transparent argument.

If you want another way of doing it, substituting $x = \tan \theta$ generates a surprisingly simple integration, but you do need to use a tangent half-angle substitution at the end because you get an expression in $\sin 2 \theta$ which is fiddlesome to substitute back for $x$.

https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_x_squared_plus_a_squared_squared

Title: Integration by substitution: $\displaystyle\int\frac{dx}{(1+x^2)^2}$

Body:
I came across this solution of a tricky integral using implicit substitution (i.e., manipulating differentials without actually declaring $u=\cdots$).
\begin{align*}
                \int \frac{dx}{(1+x^2)^2} &= \int \frac{1 + x^2 - x^2}{(1+x^2)^2}\,dx\\
                &= \int \frac{dx}{1+x^2} - \int \frac{x^2}{(1+x^2)^2}\,dx\\
                &=\tan^{-1}x - \frac{1}{2}\int \frac{x}{(1+x^2)^2}\,d(x^2+1)\\
                &= \tan^{-1}x+\frac 12\int x\, d\Big(\frac 1{1+x^2}\Big)\\
                &=\tan^{-1}x + \frac12\cdot \frac{x}{1+x^2}- \frac12\int \frac{dx}{1+x^2}\\
                &= \frac12\Big(\tan^{-1}x + \frac{x}{1+x^2}\Big)+c.
            \end{align*}
Can someone help me understand how to go from the third line to the fourth?  It doesn't seem like an obvious step. I'm guessing that it's because
$$d\Big(\frac 1{u(x)}\Big) = -\frac{d(u(x))}{u(x)^2},$$
but it doesn't seem like a natural next step, it's just the way I justified it to myself after already having seen the solution. Why should it be the next step?


问:
Translating a sin curve into one that lingers on its maximums longer
The goal is to create an animation curve mathematically that moves like a sin curve but holds at the extremes. I would also love to be able to parameterize it.

In the graph note how the domain spent at either 1 or -1 increases while the function still transitions smoothly in and out of these extremes.

Thus far I have graphs of the following forms with no luck

```<code>sin(x)*sin(x)^a
sin(x)^a // This sort of works when a is [0, 1] however I do not like that the derivative appears infinite when y =0
sin(sin(x))
sin(x)+sin(x+a)
</code>```

I really have not seen a formula that can do this. Is it even possible to translate a regular sine curve into one that lingers at its extremes?

A function that is not even technically sinusoidal would be ok so long as it is periodic or can be made periodic by modulo-ing x.

答:
Try:

$$f(x) = \frac{\vert\sin x\vert^{2-p}}{\sin x}$$

If $p=0$, you have a regular sine curve:

When $p$ goes from $0$ to $1$, the function smoothly goes between the extremes. $p = \frac{1}{2}$:

Finally, as $p$ goes to $1$ you get the square wave.

The function can also be written as:

$$f(x) = \text{sgn}(\sin x)\cdot\vert\sin x\vert^{1-p}$$

This might give some more intuition as to why it works. For more, see Desmos.


问:
Slight Confusion in proof of this theorem
I asked for the backwards direction on this proof before. I am concerned if I am able to use this theorem in the proof, even though this is a theorem that a collection is a basis for a known topology:

Theorem:

Let $(X,\tau)$ be a topological space, and let $\mathcal{B}$ be a collection of subsets of $X$. Then $\mathcal{B}$ is a basis for $\tau$ if and only if

$(1) \mathcal{B} \subset \tau$ and

$(2)$ for each set $U$ in $\tau$ and point $p$ in $U$ there is a set $V$ in $\mathcal{B}$ such that $p \in V \subset U$.

Theorem(trying to prove):Suppose $X$ is a set and $\mathcal{B}$ is a collection of subsets of $X$. Then $\mathcal{B}$ is a basis for some topology if and only if

$(1)$ each point of $X$ is in some element of $\mathcal{B}$ and

$(2)$ if $U$ and $V$ are sets in $\mathcal{B}$ and $p$ is a point in $U \cap V$, there is a set $W$ in $\mathcal{B}$ such that $p \in W \subset (U \cap V)$.

($\Rightarrow$)

Since $\mathcal{B}$ is a basis, $X$ can be expressed as the union of elements of $\mathcal{B}$, so each point in $X$ is in some basis element. If $U,V \in\mathcal{B}$ since $U,V$ are open, $U \cap V$ is also open.Let $x \in U \cap V$. Since $U \cap V$ is open there exists a basis element $W$ such that $x \in W \subset U \cap V$.

Clearly I used the first theorem given in this proof. My question is, is it justified to use this result? Thanks

答: Yes. Actually what you refer as the first theorem is the definition of a topological basis. So it is okay to use it in the proof of that theorem.</p>


问:
Is a quotient set the same as a partition?
So I've just started trying to teach myself some topology and im really confused on what a partition is and how it is in any way related to an equivalence relation.

My main confusion however is that of the concept of a quotient set I've read that a quotient set is the set of all equivalence classes of a set namely if $S$ is some set with an equivalence relation $R$

Then is the quotient set $S/R=\{[a]  |  a \in S\} $

My question is how is this quotient set the same as a partition of a set?

答:
You need to be acquainted with the https://proofwiki.org/wiki/Fundamental_Theorem_on_Equivalence_Relations
and its converse.

Every equivalence relation partitions the set it is on into disjoint subsets. Each of the elements in a given subset is equivalent to all the other elements in that subset. They divide the population into "bubbles", if you like.

The set of all these subsets is the quotient set induced by that equivalence relation.


问:
How to partition a 2D plane, given a graph that determines neighbourhoods?
It is very common to reduce spatial structures (e.g. city maps, country borders, rooms in a building, etc) to graphs that abstract some kind of connectivity. The problem I am trying to solve is about this process in reverse.

Assume you have been given a graph G with n nodes. Is there a way (and how?) to partition a 2D plane into n cells in such a way that for every edge (u,v) in G, the corresponing cells of u and v share borders. (A harder version of this problem can ask for cells to share borders if and only if there is an edge between their corresponding nodes.)

I have exhausted google by searching with different keywords to no success. I would appreciate any help.

答:
Usually, we solve the dual problem instead: how can we draw the graph in the plane, with vertices as points and edges as curves, so that edges do not cross? This is called finding a plane embedding of the graph. (This problem, and your problem as well, are only possible if the graph is planar.) The algorithms to do it are not the most straightforward ones, but at least you should have some keywords to search for, there.

Once you have a plane embedding, here's how you solve your problem. First, make your diagram "thicker", so that the vertices are no longer idealized points but actually small disks, and the edges are not infinitely thin curves but have some small thickness. Not too much thicker, though: we don't want to create any new overlaps between these points and edges.

Second, chop each edge in the middle. To each vertex, assign the following region: the portion of the diagram containing that vertex, and half of each edge out of the vertex. These regions will have exactly the adjacencies you want, but there will also be gaps between them in the plane.

To actually get a partition of the plane, just arbitrarily assign the gaps to any adjacent region. This may introduce some unwanted adjacencies.


问:
Quadrilateral side finding.
Imagine a quadrilateral $ABCD$ where $AB$ is ∥ to $CD$ and ∠D is 2 times as large than ∠B.

We will add in 2 more variables, $a$ and $b$, where $a$ is the length of $AD$ and $b$ is the length of $CD$.

Our goal is to find the length of $AB$ in terms of $a$ and $b$

This is already very hard to visualise so I will be very happy if someone explains this to me, thanks!

答:
First note that the corners of $ABCD$ rotate round it: the sides are $AB$, $BC$, $CD$ and $DA$. Hence if you know that the angle at $B$ is half the angle at $D$, you can start by drawing a sketch where you start by drawing $AB$, then put $B$ at, say, $60$ degrees to draw $BC$ (don't make $BC$ all that long, by the way), then you draw $CD$ parallel to $AB$ but stop when $D$ is twice $B$ and then join $D$ to $A$.

You get a trapezoidal shape.

When you have drawn it, you will visualize it just fine.

Title: Quadrilateral side finding.


问:
How to solve this system of equation.
$x^2-yz=a^2$

$y^2-zx=b^2$

$z^2-xy=c^2$

How to solve this equation for $x,y,z$. Use elementary methods to solve (elimination, substitution etc.).

Given answer is:$x=\pm\dfrac{a^4-b^2c^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}\,$, $y=\pm\dfrac{b^4-a^2c^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}\,$ ,$z=\pm\dfrac{c^4-b^2a^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}$

答:
Subtract them we get
$$(x^2-y^2)+xz-yz = a^2-b^2 \implies (x-y)(x+y+z) = (a-b)(a+b)$$
$$(y^2-z^2)+yx-zx = b^2-c^2 \implies (y-z)(x+y+z) = (b-c)(b+c)$$
$$(z^2-x^2)+zy-xy = c^2-a^2 \implies (z-x)(x+y+z) = (c-a)(c+a)$$
Now consider the cases $x+y+z=0$ and $x+y+z\neq 0$. I trust you can finish it off from here.</p>

ContentLicense: CC BY-SA 3.0

Title: How to solve this system of equation.


问:
What are all the integral solutions of $n!=m(m^2-1)$?
Observe that:

$3!=2(2^2-1)$

$4!=3(3^2-1)$

$5!=5(5^2-1)$

$6!=9(9^2-1)$

Question: What are all the integral solutions of $n!=m(m^2-1)$?

I guess it is just $(n,m) = (3,2),(4,3),(5,5),(6,9)$, but how to prove that there is no other one?

I checked that there is no other one for $n<20$.

Let reformulate the problem using Cardano's formula:    consider the cubic equation $$x^3+px+q=0,$$
its discriminant is $\Delta = -(4p^3+27p^2)$. Our case corresponds to $(p,q) = (-1,-n!)$, so $\Delta = 4-27(n!)^2 <0$. Thus the cubic has one real root and two non-real complex conjugate roots, and by Cardano's formula, the real root is $$ \left(-\frac q2+\sqrt{\frac{q^2}4+\frac{p^3}{27}} \right)^{1/3} +\left(-\frac q2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}\right)^{1/3} $$
So the problem "reduces" to ask whether there exists an integer $n>6$ such that

$$ \left(\frac{n!}{2} +\sqrt{\frac{(n!)^2}{4} - \frac{1}{27}} \right)^{1/3} +\left(\frac{n!}{2} - \sqrt{\frac{(n!)^2}{4} - \frac{1}{27}} \right)^{1/3}  $$
is also an integer. It is not clear that it helps...

答:
Variations of your problem are very well studied. For example, Brocard's problem asks about whether or not there are any solutions $(n,m)$ to $n!=m^2-1$ other than $(4,5)$, $(5,11)$, and $(7,71)$. This conjecture dates back to Erdos.

An interesting generalization is asking how many solutions $n!=P(m)$ has for any integer polynomial. Under the abc conjecture, there are only ever finitely many solutions if $P(m)$ has degree at least $2$.


问:
Closure of set in lexicographically ordered square
I am struggling to figure out how to prove this result, or if my method is correct. Consider the set $[0,1] \times [0,1]$ equipped with the dictionary order. $(a,b)<(c,d)$ if and only if $a<c$ or $a=c$ and $b<d$ in the order topology. Prove $(0,1)$ is a limit point of $\{(\frac{1}{n},0)|n \in \mathbb{N}\}$. This was an excercise so probably not requiring a proof, however I am curious as to how one would attempt to prove this.

Attempt since $\frac{1}{n} \rightarrow 0$, for any $\epsilon>0$, there is a natural number $N$, such that $n \geq N \implies \frac{1}{n}<\epsilon$, then for each $n \geq N$, the interval $((0,a),(\frac{1}{n},b)),a<1,b>0$, is open, nonempty and contains $(0,1)$ so this point is a limit point.

Also I wrote the $\epsilon$ based on my knowledge of convergent sequences however, I noticed this does not add anything in the proof, so I think this is wrong.

答:
I will write $\preceq$ for the lexicographic order on $X=[0,1]\times[0,1]$. Let $U$ be any open nbhd of $\langle 0,1\rangle$. Then there $p=\langle a,b\rangle,q=\langle c,d\rangle\in X$ such that

$$\langle 0,1\rangle\in(p,q)_\preceq\subseteq U\,,$$

where $(p,q)_\preceq$ denotes the open interval between $p$ and $q$ in the lexicographic order. Then $\langle a,b\rangle\prec\langle 0,1\rangle\prec\langle c,d\rangle$, so $a=0$, $0\le b<1$, and $0<c\le 1$. There is an $m\in\Bbb Z^+$ such that $\frac1m<c$, and clearly $\left\langle\frac1n,0\right\rangle\in(p,q)_\preceq\subseteq U$ for each $n\ge m$. Thus, not only is $\langle 0,1\rangle$ a limit point of the set $\left\{\left\langle\frac1n,0\right\rangle:n\in\Bbb Z^+\right\}$, it is the limit of the sequence $\left\langle\left\langle\frac1n,0\right\rangle:n\in\Bbb Z^+\right\rangle$.


问:
Finding Big Theta of Recurrence Relation $8T(n/3)+n^2/\log_3(n)$
Getting stuck on finding the geometric series after finding the sum for this recurrence. I didn't think master theorem applied to this because I thought n^2/log3(n) was not a polynomial but I could be wrong.</p>

答:
Welcome to MSE!

The master theorem is applicable here. In particular:

We see
$c_{\text{crit}} = \log_3(8) \approx 1.89$,
so we want to compare the term $n^2 / \log_3(n)$ with $n^{1.89}$ (roughly).

Of course, $n^2 / \log_3(n) = \Omega(n^{1.89})$ since

$$
\lim_{n \to \infty} \frac{n^2 / \log_3(n)}{n^{1.89}} =
\lim_{n \to \infty} \frac{n^{.11}}{\log_3(n)} = \infty > 0
$$

So we fall into case $3$ of the master theorem (at least as listed on wikipedia).

Our last check is to show the "regularity condition" holds. That is,

$$8f(n/3) \leq k f(n)$$

for some $0 < k < 1$ and $n$ large.

Thankfully, we expand this to

$$
8 \frac{(n/3)^2}{\log_3(n/3)} \leq k \frac{n^2}{\log_3(n)}
$$

Which happens if and only if

$$
\frac{8}{9} \frac{\log_3(n)}{\log_3(n) - 1} \leq k
$$

But since $\frac{\log_3(n)}{\log_3(n) - 1}$ limits to $1$, if we choose
$8/9 < k < 1$, the desired inequality will hold for large $n$.

This lets us conclude

$$T(n) = \Theta(n^2/\log(n))$$

I hope this helps ^_^


问:
Can the Lebesgue integral be defined so that it can be applied to non-Lebesgue measurable subsets of $\mathbb{R}^d$?
We can define the integral of a function $f$ on any $M \subseteq \mathbb{R}^d$ as $\int_{\mathbb{R}^d} f \cdot \chi_M$, irrespective of whether $M$ is Lebesgue-measurable or not. It works fine for the zero function on the Vitali set $V$, so that $\int_V 0=0$. However, if we follow the measure theory approach, $\int_V 0$ is simply not defined. Is this not a shortcoming of the measure theory approach to the Lebesgue integral? Could there not be some non-trivial function $f$ with $f \cdot \chi_M$ integrable over $\mathbb{R}^d$, even if $M$ is non-measurable?

答:
Of course you can use an "upper integral" or a "lower integral".  But if you do this, then linearity can fail.  That is, the property $\int_M f + \int_N f = \int_{M \cup N} f$  for disjoint sets $M,N$ could fail.

Your case of the $0$ function on the Vitali set $V$, though, is not a problem.
We define $\int_V 0$ to be $\int_{\mathbb R} 0\cdot \chi_V$.  But since $0 \cdot \chi_V$ is $0$ on all of $\mathbb R$, this integral is computed as the integral of the measurable function $0$.


问:
Finding eigenvectors of matrix of matrices

Let $A$ be a matrix with $\lambda_1,...,\lambda_n$ eigenvalues, and $v_1,...,v_n$ the corresponding eigvenvectors. Let $B= \begin{pmatrix} 0& A \\ A& 0 \end{pmatrix}$. It's known that the eigenvalues of $B$ are $\{\pm \lambda_1,..., \pm \lambda_n\}$. Find the eigenvectors of $B$.

I've managed to find $n$ of the $2n$ eigenvectors of $B$, those who correspond to the eigenvalues: $\{ \lambda_1,...,  \lambda_n\}$:
$u_1= \begin{pmatrix} v_1 \\ v_1 \end{pmatrix},..., u_n = \begin{pmatrix} v_n \\ v_n \end{pmatrix}$. I assumed that the new eigenvectors should be built of the original eigenvectors, thus I started playing with combinations of the original eigenvectors, such as: $\begin{pmatrix} v_i \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ v_i \end{pmatrix}$ however those vectors didn't help at all. Eventually, looking at how $B$ is defined, I understood that since $B$ simply works the same as $A$ works on vectors for the first half and then again for the second half, I found that  $u_1= \begin{pmatrix} v_1 \\ v_1 \end{pmatrix},..., u_n = \begin{pmatrix} v_n \\ v_n \end{pmatrix}$ can work.

However, I couldn't  manage to find the corresponding eigenvectors for the $\{-\lambda_1,...,-\lambda_n\}$ eigenvalues. I can't see how applying the original matrix $A$ in any kind of form will get us to the negative of its' eigenvalues.

Any hint/clue?

答:
Let's look at $u_i =\begin{pmatrix} v_i \\ -v_i \end{pmatrix}$:

$B \begin{pmatrix} v_i \\ -v_i \end{pmatrix} = \begin{pmatrix} 0& A \\ A& 0 \end{pmatrix} \begin{pmatrix} v_i \\ -v_i \end{pmatrix}= \begin{pmatrix} A(-v_i) \\ Av_i \end{pmatrix} = \begin{pmatrix} -\lambda_iv_i \\ \lambda_iv_i \end{pmatrix} = -\lambda_i\begin{pmatrix} v_i \\ -v_i \end{pmatrix}$

Thus we get the other $n$ eigenvectors, ending with total of $2n$ eigenvectors.


问:
set of natural numbers
Prove each set of natural numbers is the set of all smaller natural numbers.
IE $ n =$ { $m \in N | m < n$ }

Hint: use induction to prove that all elements of a natural number are natural numbers.

The natural numbers are defined as : 0, 1, 2, ...etc.

The textbook I am using (Hrbacek and Jech) further defines:

$ 1 = \{ 0 \} = \{ \emptyset \}$

I am not sure if I am going about this correctly.

Step ONE: show it is true for the second smallest natural number 1

it is a given that $ 1 \in $ {$0$}

Step TWO: assume it is true for the natural number $k$
  IE: $k =$ { $m \in N | m< k$ }

Step THREE: prove it is true for natural number $ k + 1 $

since $ k = \{ m \in N | m < k \}$

and since $ k < k + 1$

then $m < k + 1 $

therefore $ k + 1 = \{ m \in N | m < k + 1 \}$

am I correct?

答:
I suspect that you’ve actually defined $N$ to be the smallest set such that $0\in N$ and for all $n\in N$, $S(n)\in N$ or something very similar, and that you’ve defined $n+1$ to be $S(n)=n\cup\{n\}$ for $n\in N$. To show by induction that $n=\{k\in N:k<n\}$ for each $n\in N$, you must show two things:

$0=\{k\in N:k<0\}$, and

if $n\in N$ and $n=\{k\in N:k<n\}$, then $n+1=\{k\in N:k<n+1\}$.

Note that for (1) you want to start at $n=0$, not at $n=1$. (And what you have for $n=1$ isn’t correct: $1\notin 0$. In fact $1=0\cup\{0\}=\{0\}$, so $0\in 1$.) In order to prove (1) you must show that there is no $k\in N$ such that $k<0$. How you do this will depend on you you’ve defined the relation $<$; I don’t know your definition, so I can’t help here.

For (2), assume that $n=\{k\in N:k<n\}$. Then

$$n+1=S(n)=n\cup\{n\}=\{k\in N:k<n\}\cup\{n\}\;,$$

so to finish the induction step you must show that

$$\{k\in N:k<n\}\cup\{n\}=\{k\in N:k<n+1\}\;.$$

It’s very easy to see that

$$\{k\in N:k<n\}\cup\{n\}\subseteq\{k\in N:k<n+1\}\;,$$

so all that remains is to show that

$$\{k\in N:k<n+1\}\subseteq\{k\in N:k<n\}\cup\{n\}\;.$$

For this you’ll want the result from this question.

Title: set of natural numbers


问:
Conditional Expectation of X with respect to a $\sigma$-algebra $\mathcal{G}$
I read this definition from my lecture handout:

If X is a random variable on a countable probability space $(\Omega,\mathcal{F},\mathbb{P})$, then the conditional expectation $\mathbb{E}_\mathbb{P}(X|\mathcal{G})$ of X with respect to $\mathcal{G}$ is defined as a random variable which satisfies, for every $\omega\in A_i$,
$$\mathbb{E}_\mathbb{P}(X|\mathcal{G})(\omega)=\sum_{i\in I}\frac{\mathbb{E}_\mathbb{P}(X\mathbb{1}_{A_i})\mathbb{1}_{A_i}}{\mathbb{P}(A_i)}\quad\quad \forall A_i\in \mathcal{P}$$
$\mathcal{P}$ is a countable partition that generates $\mathcal{G}$.

And I have the following questions:

Some of the other materials give this definition as a conditional expectation of X with respect to the whole $\sigma$-field $\mathcal{G}$, which without indicating the small omega, i.e.:$$\mathbb{E}_\mathbb{P}(X|\mathcal{G})=\sum_{i\in I}\frac{\mathbb{E}_\mathbb{P}(X\mathbb{1}_{A_i})\mathbb{1}_{A_i}}{\mathbb{P}(A_i)}.$$
I personally support this definition, because $\mathbb{E}_\mathbb{P}(X|\mathcal{G})(\omega)$ seems to be a partial condition and it should be equivalent to $\mathbb{E}_\mathbb{P}(X|\mathcal{G})\mathbb{1}_{A_i}=\mathbb{E}_\mathbb{P}(X|A_{i})$. So which one is the correct definition?

Why we have two indicator functions over there? As per my logic, $\mathbb{E}_\mathbb{P}(X|\mathcal{G})=\sum_{i\in I}\mathbb{E}_\mathbb{P}(X|A_i)=\sum_{i\in I}\frac{\mathbb{E}_\mathbb{P}(X\mathbb{1}_{A_i})}{\mathbb{P}(A_i)}.$ I don't know where is the other $\mathbb{1}_{A_i}$ comes from.

答:
In what follows

$(\Omega ,\mathcal F, \mathbb P)$ is a probability space,

$X$ is a random variable defined on $\Omega $,

$\mathcal G\subseteq \mathcal F$ is a sub-$\sigma $-algebra generated by a countable partition $\{A_i\}_{i\in I}$
of $\Omega $, formed  by $\mathcal F$-measurable sets.

The first thing to bear in mind is that
the  conditional expectation $\mathbb E_{\mathbb P}(X|\mathcal G)$ is another random
variable (rather than just a number).  Since random variables are nothing but  functions defined on $\Omega $,
the notation
$$
  \mathbb E_{\mathbb P}(X|\mathcal G)(\omega )
  $$
makes perfect sense as it indicates the value taken by the function
$\mathbb E_{\mathbb P}(X|\mathcal G)$ on a given point $\omega $ of $\Omega $.

The second important point is that   $\mathbb E_{\mathbb P}(X|\mathcal G)$ is constant on every set $A_i$
(although its constant value may change among the various  $A_i$).

By the definition of
conditional expectation, the constant value taken on $A_i$ by $\mathbb E_{\mathbb P}(X|\mathcal G)$  is
the expected value of $X$ on $A_i$,  namely
$$
  \mathbb E_{\mathbb P}(X|A_i ) := \frac {\mathbb E_{\mathbb P}(X1_{A_i} )}{\mathbb P(A_i)}.
  $$

Denoting the above expected value simply by $e_i$, the question is how should  we express the function which
is supposed to take on the value $e_i$ on $A_i$.    I guess the best way to do this is simply to write
$$
  \sum_{i\in I}e_i1_{A_i},
  \tag 1
  $$
observing that if $\omega $ lies in some $A_j$, then $1_{A_i}(\omega )$ vanishes for every $i$, except for $i=j$, in
which case $1_{A_i}(\omega )=1$,  so the above sum comes out as $e_j$, which is precisely what we expect.

Notice that there is no $\omega $ in  expression (1) for the same reason  many people consider it
incorrect to say

"consider the function $\sin(x)$"

In fact,   the  function is called simply "$\sin$", whereas  "$\sin(x)$" is meant to denote the value
of the function $\sin$ at the given real number $x$.

According to this we therefore have that
$$
  \mathbb E_{\mathbb P}(X|\mathcal G)=   \sum_{i\in I}e_i1_{A_i},
  \tag 2
  $$
which, upon substituting the appropriate value for $e_i$, is exactly what the OP writes in the first question.

If we want to explicitly indicate the dependency of these functions on a variable $\omega $,  I'd write
$$
  \mathbb E_{\mathbb P}(X|\mathcal G) (\omega ) =   \sum_{i\in I}e_i1_{A_i}(\omega ), \quad\forall \omega \in \Omega ,
  \tag 3
  $$
noticing that now $\omega $ shows up on both sides.

On the down side,   I believe mixing the LHS of (3) with the RHS of (2) is incorrect so I agree 100%
with the OP that their second formula is preferable over the first one.

Regarding the OP's second question, I see a problem in the sense that
$$
  \mathbb{E}_\mathbb{P}(X|\mathcal{G})
  $$
is supposed to be a random variable, as already discussed, while
$$
  \sum_{i\in I}\mathbb{E}_\mathbb{P}(X|A_i)
  $$
strikes me as a number.

Inserting the  missing  $1_{A_i}$ on the  last expression  would make it a function, which is more in line
with the functional nature of the conditional expectation.


问:
Linear recursion with convergent coefficients
Suppose we have a linear recurrence relation

$$x_n = \alpha_n x_{n-1}+ \beta_n x_{n-2} + y_n,$$

where

$(\alpha_n) \to \alpha >0$,

$(\beta_n) \to \beta >0$,

$(y_n) \to y > 0$, and

the limiting auxiliary equation, $m^2 - \alpha m - \beta$, has roots in $(-1,1)$

(all limits are as $n \to \infty$).

My question:

Does this imply that the sequence $(x_n)_{n=1}^\infty$ converges?

This comes from a convergent example. I am simply wondering if the above information suffices to prove it, or if one needs control on the rates of convergence, say.

答:
There is a nice order reduction result for difference inequalities in the following reference which simplifies the reasoning quite a bit:

https://msp.org/involve/2008/1-1/p07.xhtml

For a given $\epsilon > 0$, for sufficiently large N, we have that for $n>N$:

$$(\alpha + \beta -\epsilon)min(x_{n-1}, x_{n-2}) + (\gamma - \epsilon) \leq x_{n} \leq (\alpha + \beta +\epsilon)max(x_{n-1}, x_{n-2}) + (\gamma + \epsilon).$$

The condition on the roots ensures $\alpha + \beta < 1$.

Applying theorem's 2 and 3 from the reference we can bound solutions of the recursion $x_{n}$ between the following two recursions (taking maxes over chains of terms as shown in the article):

$u_{n} = (\alpha + \beta -\epsilon) u_{n-1} + (\gamma - \epsilon)$

and

$v_{n} = (\alpha + \beta +\epsilon) v_{n-1} + (\gamma + \epsilon)$

Since $\alpha + \beta < 1$ both {$u_{n}$} and {$v_{n}$} converge to the unique equilibrium.

Therefore given $\epsilon_{2} > 0$ there exists $M$ so that for $n>M$:

$$\frac{\gamma - \epsilon}{1-\alpha -\beta + \epsilon} - \epsilon_{2} \leq x_{n} \leq \frac{\gamma + \epsilon}{1-\alpha -\beta - \epsilon} + \epsilon_{2}$$

Thus $x_{n} \to \frac{\gamma}{1 - \alpha - \beta}$.


问:
How can I solve this non-linear differential equation?
I'm trying to solve the equation $$y' = 1 - y^2$$
Here is my attempt:
$$y' = 1 - y^2$$
Divide by (1-y^2)
$$\frac{y'}{1-y^2} = 1$$
Integrate both sides:
$$\frac{1}{2}\log|\frac{y+1}{y-1}|=t+c$$
Rearrange
$$y = \frac{ke^{2t}+1}{ke^{2t}-1}$$
I'd have thought that solution was right, but we have to figure out a specific solution with y(0) = 0. But this isn't possible with the above equation.</p>

答:
I wrote it down and solved it in a slightly different way. The first thing you should notice is that $y = 1$ and $y = -1$ are the two constant solutions, which allows you then to divide $y'$ by $1-y^2$, since you want to study it for $y(0) \in (-1,1)$, knowing that any solution starting in $(-1,1)$ stays there (or dies in 1).

Then yes, with some algebra you manage to get

$\left(\log \frac{1+y}{1-y}\right)' = 2$

if I haven't screwed up with the signs; now integrating it from 0 to $t$ you get:

$\log \frac{1+y(t)}{1-y(t)} - \log \frac{1+y(0)}{1-y(0)} = 2t$

without the absolute value since everything in the argument of the logs is non-negative. By imposing $y(0) = 0$ the second term in the left vanishes and you're left with an easy expression that if inverted gives the following:

$y(t) = \frac{e^{2t} - 1}{e^{2t} + 1}$

which is simply

$y(t) = \tanh (t)$

and of course double checking $y(0) = 0$.


问:
How to express by a single equation a region delimited by confocal ellipse and hyperbola
I am trying to find the equation of a shape formed by a conic (ellipse) that excludes its intersection with another conic (hyperbola). The hyperbola and ellipse are assumed to share common foci. But they are general conics  in 2D which means they can be rotated about the z-axis and translated along x and y axes. As it can be seen in the figure, the conics are rotated by an angle theta and not centered at the origin.

In the figure, I want the equation of resulting shape shown in grey shade (formed by the two conics). This new shape can be shown as follows:

Please let me know if it is possible to get an equality representing this new shape in the quadratic form as follows:

$ Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$

答:
I guess you need inequalities describing the region.

Take $a>c>b>0$, for solid of revolution about the $x$-axis:

Internal region bounded by an ellipsoid:

$$A=\left \{(x,y,z) \in \mathbb{R}^3: \frac{x^2}{a^2-c^2}+\frac{y^2+z^2}{c^2} \le 1  \right \}$$

Middle region bounded by branches of a hyperboloid:

$$B=\left \{(x,y,z) \in \mathbb{R}^3: \frac{x^2}{c^2-b^2}-\frac{y^2+z^2}{b^2} \le 1  \right \}$$

The shaded region is $A\cap B$ or equivalently,

$$b^2-\frac{b^2 x^2}{c^2-b^2} \le y^2+z^2 \le c^2-\frac{c^2 x^2}{a^2-c^2}$$

Intersection of two conics (of revolution):

$$\textstyle \partial A \cap \partial B=
\left \{
  (\xi,\eta,\zeta) \in \mathbb{R}^3: (\xi^2,\eta^2+\zeta^2)=
  \left(
    \frac{(a^2-c^2)(c^4-b^4)}{a^2 b^2+c^4-2b^2 c^2},
    \frac{b^2 c^2(a^2+b^2-2c^2)}{a^2 b^2+c^4-2b^2 c^2}
  \right)
\right \}$$

For solid of revolution about the $y$-axis is left as an exercise.


问:
Subgaussian tail bounds for maximum of subgaussian random variables: concentration around the true expectation?
As in this question, it is not too hard to show that if $X_1,\dots,X_n$ are $\sigma^2$-subgaussian (not even necessarily independent) then $\mathbb{E}[\max_{1\leq i\leq n} X_i] \leq \sqrt{2\sigma^2\log n}$ and
$$
\mathbb{P}\!\left\{\max_{1\leq i\leq n} X_i \geq \sqrt{2\sigma^2\log n}+ u \right\} \leq e^{-\frac{u^2}{2\sigma^2}} \tag{$\dagger$}
$$
for all $u>0$. Now, in $(\dagger)$ above we have our upper bound on the expected value, not the expected value itself. Both basically coincide in the case of actual Gaussians, but not in general: for subgaussians r'v's, the expected value $\mathbb{E}[\max_{1\leq i\leq n} X_i]$ could be much smaller than that.

Question:

is it possible to replace $\sqrt{2\sigma^2\log n}$ by $\mathbb{E}[\max_{1\leq i\leq n} X_i]$ in $(\dagger)$, either with or without an independent assumption for $(X_i)_i$? (the proof I had in the linked post does not allow that)

if not, what is a counterexample? (for independent r.v's? arbitrary ones?)

答:
Not a full answer, but something (inspired by Section 2.2 of this paper) that shows that the independence assumption is crucial.

If not, let $Y_1,\dots,Y_n$ be i.i.d. $N(0,\sigma^2)$, and $\xi$ be a Bernoulli$(p)$ random variable, jointly independent of the $Y_i$'s, where say $p:= 1/6$. Set
$$
(X_1,\dots,X_n) = (\xi Y_1,\dots, \xi Y_n)
$$
so that the $X_i$'s are still $\sigma^2$-subgaussian, and we have
$$
\mathbb{E}[\max_{1\leq i\leq n} X_i] =
p\mathbb{E}[\max_{1\leq i\leq n} Y_i] = (1+o(1)) p \sqrt{2\sigma^2\log n} \tag{1}
$$

However, for $u = p \sqrt{2\sigma^2\log n}$ (and ignoring the $o(1)$ for convenience), we have
$$\begin{align}
\mathbb{P}\{\max_{1\leq i\leq n} X_i &\geq \mathbb{E}[\max_{1\leq i\leq n} X_i] + u\}
= \mathbb{P}\{\max_{1\leq i\leq n} X_i \geq 2\mathbb{E}[\max_{1\leq i\leq n} X_i]\} \\
&\geq \mathbb{P}\{\max_{1\leq i\leq n} X_i \geq \frac{1}{2}\mathbb{E}[\max_{1\leq i\leq n} Y_i]\} \approx p \tag{2}
\end{align}$$
the last inequality as $4p < 1$, and the $\approx$ since, conditioned on $\xi =1$, we have $\max_{1\leq i\leq n} Y_i$ which concentrates very well around its expectation. But, for any constant $c>0$,
$$
e^{-c\frac{u^2}{\sigma^2}} = \frac{1}{n^{2cp}} \ll p \tag{3}
$$
so (2) and (3) together rule out an upper tail bound involving the true expectation.


问:
Are these two definitions of base for a topology equivalent?
I have some experience with metric spaces, but I never took a dedicated Topology class. I recently encountered the term "base for a topology", and looking on Wikipedia, there seem to be two definitions. The very first line of the article says:

[Definition 1]: In mathematics, a base or basis for the topology $\tau$ of a topological space $(X, \tau)$ is a family $B$ of open subsets of $X$ such that every open set is equal to a union of some sub-family of $B.$

However, the first thing in the "Definition and basic properties" section is:

[Definition 2] A base is a collection $B$ of open subsets of $X$ satisfying the following properties:

The base elements cover $X$.

Let $B_1$, $B_2$ be base elements and let $I$ be their intersection. Then for each $x$ in $I$, there is a base element $B_3$ containing $x$ such that $B_3$ is subset of $I$.

I believe I have proved these definitions to be equivalent. But I want to make sure I haven't made any hidden assumptions, because I know that unlike metric spaces, General Topology can be very unintuitive for the outsider.

Def 1 $\implies$ Def 2:

Suppose $B$ is a base according to definition 1. Take any two base elements $B_1, B_2$. Their intersection $I$ is an open set, so it is a union of base elements. If $x$ is in $I$, then it is in the union of those base elements, so it is in at least one of them. Therefore $x$ is in a base element which is contained in the intersection.

Def 2 $\implies$ Def 1:

Suppose $B = \{B_i|i \in I\}$ is a base according to definition $2$. Let $U$ be an open set of $X$. The set
$$S = \bigcup_{i \in I}U \cap B_i.$$
is a union of oepn sets, and we claim that $S = B$. It is obvious that $S \subseteq B$. To see the reverse, let $x$ be any element of $B$. Since the base covers $X$, there is a base element $b$ such that $x \in b$. Therefore $x \in U \cap b$, so $x \in S$.

答:
EDIT: It's been pointed out that your definition is actually a bit off; the proof below assumes a corrected definition, which I assume was just a typo.  Tbh I didn't even notice it haha

Your first part is correct, but the second part has problems.  For example, $S$ is a subset of $X$, but you say that it's equal to $B$, a basis, which is a collection of sets.  Also, every element in $S$ is contained in $U$, but $X$ is not necessarily contained in $U$, so you can't get a base.

The way to do $2 \implies 1$ is suppose $B$ is a basis according to the second definition, and let $U$ be any open set in $X$.  We want to show that it's the union of elements of $B$.  By definition $2$ of a basis, if $x \in U = U \cap U$ then there is an open set $B_x \in B$ such that $x \in B_x \subset U$.  Now $U = \cup B_x$, since each $B_x$ is contained in $U$, but also every point of $U$ is in some $B_x$.  I think you had the right idea, just wrote it down wrong.


问:
Circle Sector Area Calculation - I am getting wrong answer
I am trying to calculate the area of a circle segment using MS-Excel. The radius value is stored in variable "rr", the angle is stored in variable "a".

There are 2 equivalent formulas, depending on whether the angle is in digress or radians, that should yield the same results, but in my case, I am getting different results.

Formula 1 gives:$57.345$ - I think this is the correct one...

Formula 2 gives:$303.025$

Where are my mistake in Formula $2$? Thank you.

答: SIN function expects an argument in radians, so even in formula 2 it should still be SIN(RADIANS(a)).


问:
How can we prove that a continuous function with $f(x) = f(x+1000)$ is constant?
The original problem :

if $f(x) = f(x+1000)$ for any x

and $\int_0^3f(x) = 30 $

What is $\int_3^5f(x+2016)$

I only know that if $f(x) = 10$ it will fulfill all of the conditions, but I don't know how to prove it.

Thank you.

答: A periodic function with period 1000 or any other number need not be constant. Im the context of your problem, consider $f, g$ on [0, 1000) given by $f(x) = 10$ and $g(x) = 20x/3$ and let $\hat f, \hat g$ be their periodic continuations on $\mathbb R$ respectively (if you don’t know what periodic continuation is, it is what you think it is). Clearly, both functions satisfy the given conditions. However, $$\int_{2019}^{2021} \hat f = \int_{19}^{21} f = 20 \neq 800/3 = \int_{19}^{21} g = \int_{2019}^{2021} \hat g.$$ Therefore, the answer is not unique. Perhaps you are asked to find all possible values of the integral or there is a condition missing? Nonetheless, the problem is rather strange. In this form, you could rather easily prove that any real number is possible for the integral to attain.


答:
These two definitions are not defining the same thing, so they are not "equivalent".

In definition 1, one is given the set $X$ and the topology $\tau$ on $X$. What is being defined is a basis for the topology $\tau$.

In definition 2, one is given just the set $X$. What is being defined is a basis. Period.

Note the syntactic difference in these two terminologies: definition 1 is relative to a given topology; definition 2 is not relative to anything. (Well... both of them are relative to the given set $X$).

These definitions are connected in the following manner:

Theorem: Given a topology $\tau$, a basis $\mathcal B$ for $\tau$ (as in definition 1) is a basis (definition 2).

Theorem: Given a basis $\mathcal B$ (as in definition 2), there exists a unique topology $\tau$ on $X$ such that $\mathcal B$ is a basis for $\tau$.

Title: Are these two definitions of base for a topology equivalent?

Body:
I have some experience with metric spaces, but I never took a dedicated Topology class. I recently encountered the term "base for a topology", and looking on Wikipedia, there seem to be two definitions. The very first line of the article says:

[Definition 1]: In mathematics, a base or basis for the topology $\tau$ of a topological space $(X, \tau)$ is a family $B$ of open subsets of $X$ such that every open set is equal to a union of some sub-family of $B.$

However, the first thing in the "Definition and basic properties" section is:

[Definition 2] A base is a collection $B$ of open subsets of $X$ satisfying the following properties:

The base elements cover $X$.

Let $B_1$, $B_2$ be base elements and let $I$ be their intersection. Then for each $x$ in $I$, there is a base element $B_3$ containing $x$ such that $B_3$ is subset of $I$.

I believe I have proved these definitions to be equivalent. But I want to make sure I haven't made any hidden assumptions, because I know that unlike metric spaces, General Topology can be very unintuitive for the outsider.

Def 1 $\implies$ Def 2:

Suppose $B$ is a base according to definition 1. Take any two base elements $B_1, B_2$. Their intersection $I$ is an open set, so it is a union of base elements. If $x$ is in $I$, then it is in the union of those base elements, so it is in at least one of them. Therefore $x$ is in a base element which is contained in the intersection.

Def 2 $\implies$ Def 1:

Suppose $B = \{B_i|i \in I\}$ is a base according to definition $2$. Let $U$ be an open set of $X$. The set
$$S = \bigcup_{i \in I}U \cap B_i.$$
is a union of oepn sets, and we claim that $S = B$. It is obvious that $S \subseteq B$. To see the reverse, let $x$ be any element of $B$. Since the base covers $X$, there is a base element $b$ such that $x \in b$. Therefore $x \in U \cap b$, so $x \in S$.


问:
How to prove that End(R) is isomorphic to R?
I tried to prove this:
$R$ as a ring is isomorphic to $End(R)$, where $R$ is considered as a left $R$-module. The map of isomorphism is

$$F:R\to End(R), \quad   F(r)=fr,$$
where $fr(a)=ar$. And I define the multiplication in $End(R)$ by $(.)$, where $h.g=g\circ h$ for $g,h$ in $End(R)$.

Is this true?

答: Hint: Let $F\in End(R)$. Put $a:=F(1)$, show that $F(x)=xa, \forall x\in R$.</p>


问:
Results that came out of nowhere.
Most big results in mathematics are built on years and years of groundwork by the author and other mathematicians, such as Wiles' proof of FLT or the classification of finite simple groups.  Nevertheless, there is this big romantic idea in pop culture where a young genius comes up with a brilliant proof of a major conjecture, based on some creative new idea which flies in the face of the mathematical community.

What are some examples where this has actually happened?  That is, which results stem from independent work by a mathematician who came along "out of nowhere" and solved a huge problem by surprise through nonstandard techniques?

I read that the recent proposed proof of the ABC conjecture comes from years of autonomous theory by Prof Mochizuki, most of which ventures far outside of the current literature; this would seem to qualify as one example provided the proof turns out to be true.

答: I'm not sure this is correct, but I can't think of any precursors to Stephen Cook's invention of NP-completeness.  With a single paper in 1971 he invented the idea of NP-completeness, previously unimagined, and showed that the satisfiability problem was NP-complete; this revolution has dominated the study of algorithms ever since.  The same paper posed the $P=NP$ question that has remained one of the foremost questions of computer science.

Title: Results that came out of nowhere.


问:
How can one calculate $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$?
I am working through a textbook for my Fourier Series class and I came across the series $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$. I know from the context of the question that the answer has to be $-\frac{\pi^2}{12}$, but I am not sure how to arrive at this answer. I tried partial fraction decomposition to see if there was any telescoping behavior, but I had an issue with the $(-1)^n$ part.

Body:
Assuming that you know $S:=\sum_{n\geq 1} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}$, we get (and I'll replace $(-1)^{n}$ with $(-1)^{n-1}$ which seems that you actually intended to be positive)

$$
\frac{1}{1^2} -\frac{1}{2^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + \cdots \\
= \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \cdots \\
- 2 \left(\frac{1}{2^{2}} + \frac{1}{4^{2}} + \frac{1}{6^{2}} + \cdots\right) \\
= S - 2\cdot \frac{1}{2^{2}}S = \frac{1}{2}S = \frac{\pi^{2}}{12}
$$


问:
Conditions for the existence of solutions to a peculiar Diophantine equation
If $a$ and $b$ are even natural numbers, how can I determine when the equation $$a+kb =2^n$$ has at least one solution where $k$ and $n$ are natural numbers? It's necessary for $a$ and $b$ to not have any common prime factors other than $2$. Is that sufficient as well? Any direction pointing would be much appreciated!

答:
Consider the case of $a=2,b=4.$ Here  $a$ and $b$ are even and not divisible by primes other than $2.$ However:

$2+k \cdot 4$ is $2$ mod $4.$ If we require as you do that $k \ge 1$ then note there are no powers of $2$ which are greater than $2$ but are $2$ mod $4.$


问:
How to prove that if $\gcd(x,n)=1$, then there is a positive integer $m<n$ such that $n\mid x^m-1$
Let $x$ and $n$ be integers, such that $\gcd(x,n)=1$. How do I prove that there is a positive integer $m<n$, such that $n\mid x^m-1$?

I'm supposed to prove this using Pigeonhole Principle, and I tried factoring $x^m-1$ and using Diophantine Equation, but I still didn't see how I could apply Pigeonhole here. Any hints would be appreciated.

答:
The $(n+1)$ elements $\{1, x, x^2, \cdots, x^n\}$ can not each be in a separate congruency class, $\pmod{n}$, because there are only $(n)$ such classes.  Furthermore, none of these elements can be congruent to $0 \pmod{n}.$

Therefore, ...


答:
Note $a$ and $b$ just not having any common prime factors other than $2$ is insufficient. For example, $a = 6$ and $b = 14$ gives

$$6 + 14k = 2^n \iff 3 + 7k = 2^{n-1} \tag{1}\label{eq1A}$$

However, $2^n$ modulo $7$ (i.e., when divided by $7$) has only values $1$, $2$ or $4$, so there's no integral solution to \eqref{eq1A}.

There are $2$ main things to check to determine if there's at least one solution for a particular $a$ and $b$. The first is with the number of factors of $2$ of $a$ (call this $i$) and of $b$ (call this $j$). If $i \lt j$, then normally there's no solution since dividing both sides by $2^{i}$ leaves an odd number, with this only being $1$ if $0$ is included in your definition of natural numbers (plus you assign it to $k$) and there's no odd prime factors in $a$, so there's then a solution of $k = 0$ and $n = i$. Otherwise, there's no solution (a basic example of this is given in coffeemath's answer).

With $i \ge j$ instead, the next check starts with dividing both sides by $2^{j}$ to get

$$a_1 = \frac{a}{2^{j}}, \; b_1 = \frac{b}{2^{j}} \tag{2}\label{eq2A}$$

Since $b_1$ is odd, Euler's theorem states $2^{\varphi(b_1)} \equiv 1 \pmod{b_1}$, where $\varphi$ is Euler's totient function. As $2^{0} \equiv 2^{\varphi(b_1)} \equiv 1 \pmod{b_1}$, the remainders repeat for larger powers than $\varphi(b_1)$. Thus, we need only to check the powers of $2$ up to $\varphi(b_1)$ to determine which one, if any, is congruent to $a_1$ modulo $b_1$. If there are none, like with my example of $7$ above, then there is no solution. Otherwise, with this power, call it $m$ (so $2^{m} \equiv a_1 \pmod{b_1}$), use

$$n_1 = q(\varphi(b_1)) + m \tag{3}\label{eq3A}$$

where $q$ is a positive integer large enough for $2^{n_1} \gt a_1$. Since $2^{n_1} \equiv \left(2^{\varphi(b_1)}\right)^{q}2^{m} \equiv 2^{m} \equiv a_1 \pmod{b_1}$, this means $2^{n_1} - a_1$ is a positive integer which is a multiple of $b_1$, so

$$k = \frac{2^{n_1} - a_1}{b_1}, \; n = n_1 + j \tag{4}\label{eq4A}$$

is a solution. Note increasing $q$ in \eqref{eq3A} shows there are an actually an infinite number of solutions in this case.

Checking the powers of $2$ up to $\varphi(b_1)$ may not be simple, or even very easy, for very large values of $b_1$, especially cases where it's required to rule out a solution if $a_1$ doesn't match any congruence. However, I offhand don't know of any easy way to simplify or speed up that process, at least not in general.

Title: Conditions for the existence of solutions to a peculiar Diophantine equation

Body: If $a$ and $b$ are even natural numbers, how can I determine when the equation $$a+kb =2^n$$ has at least one solution where $k$ and $n$ are natural numbers? It's necessary for $a$ and $b$ to not have any common prime factors other than $2$. Is that sufficient as well? Any direction pointing would be much appreciated!


问:
Closed subspace consisting of continuous functions in $L^{2}([0,1])$ is finite-dimensional
I wish to show that any closed subspace $M$ of $L^{2}[0,1]$ consisting of continuous real-valued functions is finite-dimensional (which amazes me quite a bit, I must admit).

My approach proceeds as follows:

Step 0. Note that $L^{2}[0,1]$ is a separable Hilbert space with the usual scalar product on $L^2$. Moreover, since $M$ is closed, it is separable Hilbert, too, with inner product of $L^2([0,1])$ restricted to $M$.

Step 1. Let $I: (M, \| \cdot \|_{\infty}) \to (M, \| \cdot \|_{2})$ be the identity operator on $M$. Since $I$ is bijective, by the open mapping theorem, $I^{-1}$ exists in the space of bounded linear operators on $M$. Hence, for any $f \in M$, we have $$ \| f\|_{\infty} = \| I^{-1} (f)\|_{\infty} \leq C \|f \|_{2} \quad \forall f \in M,$$
where $C := \| I^{-1} \|.$

Step 2. Now, let $t \in [0,1]$ and consider the linear functional $\ell_{t}: M \to \mathbb{R}, f \mapsto f(t).$ As $M$ is Hilbert, we may apply Riesz representation theorem to deduce that $\ell_{t} = \langle \cdot, g_t \rangle =: \ell_{g_t}$ for some $g_t \in M$ and $\|\ell_{g_t}\| = \| g_t\|_{2}$. Moreover, $$ f(t) = l_t (f) =  \langle f, g_t \rangle \quad \forall t \in [0,1].$$
Using step 1, we may further conclude that $\| g_t \|_{2} \leq C$.

Step 3. By seperability of $M$, we may suppose that $S = \{h_{n}: n \in \mathbb{N}\}$ is a orthonormal basis of $M$. Then, Parseval's identity yields

$$
\infty > \|g_t\|_{2} = \sum_{h \in S} | \langle h, g_t \rangle |^{2} = \sum_{h \in S} |h(t)|^{2}
$$

I am stuck in the last line. I want to argue that this sum has to be finite. Can we bound $|h(t)|$ from below? I guess, this suffices to conclude, right? Does this argument work?

答:
From the last line
$$
  C^2≥ \|g_t\|^2_{2} = \sum_{h \in S} | \langle h, g_t \rangle |^{2} = \sum_{h \in S} |h(t)|^{2}
  $$
so
\begin{align}
  \infty &> \int_0^1 C^2 \,  dx\\ &\geq
 \int_0^1 \sum_{h \in S} |h(t)|^{2} \,  dx\\  &=
\sum_{h \in S}  \int_0^1 |h(t)|^{2} \, \\ &= \sum_{h \in S} \|h\|^2 = \#S.
\end{align}

Here is an interesting consequence of the above fact:

It is known that for a bounded operator $T:H\to K$,  between Hilbert spaces $H$ and $K$,  the following are equivalent:

$T$ is compact,

the range of $T$ has no infinite dimensional closed subspaces.

While thinking about the present question I first tried to answer it by applying the above fact to the
inclusion map
$$
  T:C([0,1])\to L^2([0,1]),
  $$
so it would be enough to show that $T$ satisfies (1), above.  However I later noticed that $T$ is NOT compact, the
reason being that the functions
$$
  e_n(x) = \sqrt2\sin(2\pi nx), \quad x\in [0,1],
  $$
form a bounded set in $C([0,1])$,  but they  do not admit a convergent sub-sequence in $L^2([0,1])$,  as they form an
orthonormal set in the latter space.

Of course I was wrong all along, as the equivalence between (1) and (2) is stated above only for Hilbert spaces!

While it is not hard to show that (1) implies (2) for all Banach spaces, the present question and answer provide a
counter example for (2) $\Rightarrow$ (1) in the context of Banach spaces, as we now know that the range of $T$ contains no infinite dimensional closed
subspaces, and yet $T$ is not compact!

Summarizing, the point I want to make is this:

There are non-compact bounded operators between Banach spaces whose range contains no infinite dimensional subspaces, even though no such operators exist between Hilbert spaces.


问:
Product of marginal Gaussian and conditional Gaussian
Suppose I have
$$
\begin{align}
p(x_1) &= N(x_1; 0, 1) \\
p(x_2 \mid x_1) &= N(x_2; x_1, 1)
\end{align}
$$
How do I compute $p(x_1 \mid x_2)$? I know how to compute their product, giving $N\left(\frac{x_1}{2}, \frac{1}{2}\right)$. The answer of the exercise is $N\left(\frac{x_2}{2}, \frac{1}{2}\right)$, so with $x_2$ rather than $x_1$ in the mean.

答:
The following results work for multivariate Gaussians and can be used for univariate Gaussians as well. I came across them from one of the following texts (perhaps both):

Bayesian Filtering and Smoothing by Simo Särkkä

Linear Models in Statistics 2E by Rencher & Schaalje

Result #1:
If random variables $x \in \mathbb R^n$ and $y \in \mathbb R^m$ have the Gaussian distributions
$$x \sim \mathcal N (\mu, \Sigma)$$
$$y \,|\, x \sim \mathcal N (A x + b, \Omega)$$
then the joint distribution of $x, y$
$$\begin{pmatrix} x \\ y \end{pmatrix}
  \sim \mathcal{N} \left(
  \begin{pmatrix} \mu \\ A \mu + b \end{pmatrix},
  \begin{pmatrix} \Sigma & \Sigma A^{\top} \\ A \Sigma & A \Sigma A^{\top} + \Omega \end{pmatrix}
  \right)$$

You can use result #1 to find the marginal distribution of $x_2$. Then you can use the definition of conditional probability to find the desired pdf:
$$p(x_1 \,|\, x_2) = \frac{p (x_1, x_2)}{p (x_2)} = \frac{p(x_2 \,|\, x_1) p(x_1)}{p(x_2)}.$$

Using result #1 above, with $\mu = 0, \Sigma = 1, A = 1, b = 0, \Omega = 1$, we have that
$$\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \sim
\mathcal{N} \left( \begin{pmatrix} 0 \\ 0 \end{pmatrix},
\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \right).$$
Therefore, $x_2 \sim \mathcal{N} (0, 2)$.
Using a bit of algebra, you'll find that
$$p(x_1 \,|\, x_2) = \frac{ \mathcal{N} \left( x_2; x_1, 1 \right) \cdot \mathcal{N} \left( x_1; 0, 1 \right) }{ \mathcal{N} \left( x_2; 0, 2\right) } = \mathcal{N} \left( x_1; \frac{x_2}{2}, \frac12 \right)$$
which implies that
$$x_1 \,|\, x_2 \sim \mathcal{N} \left( \frac{x_2}{2}, \frac12 \right).$$

Below is another result that can be used, along with result #1, to answer your question without working directly with pdfs.

Result #2: If random variables $x \in \mathbb R^n$ and $y \in \mathbb R^m$ have the joint Gaussian distribution
\begin{equation}
  \begin{pmatrix} x \\ y \end{pmatrix} \sim \mathcal N \left(
  \begin{pmatrix} a \\ b \end{pmatrix},
  \begin{pmatrix} A & C \\ C^{\top} & B \end{pmatrix}
  \right),
\end{equation}
then the conditional distributions of $x$ and $y$ are given as
$$x \,|\, y \sim \mathcal N \left( a + C B^{-1} (y - b), A - C B^{-1} C^{\top} \right)$$
$$y \,|\, x \sim \mathcal N \left( b + C^{\top} A^{-1} (x - a), B - C^{\top} A^{-1} C \right)$$
provided $A$ and $B$ are invertible.

From result #1, we know that the joint distribution of $x_1$ and $x_2$ is
$$\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \sim
\mathcal{N} \left( \begin{pmatrix} 0 \\ 0 \end{pmatrix},
\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \right).$$
Using result #2, with $a=0, b=0, A=1, B=2, C=1$, we see that the conditional distribution of $x_1$ given $x_2$ is
$$x_2 \,|\, x_1 \sim \mathcal{N} \left( \frac{x_2}{2}, \frac12 \right)$$


问:
Integration by substitution: $\displaystyle\int\frac{dx}{(1+x^2)^2}$
I came across this solution of a tricky integral using implicit substitution (i.e., manipulating differentials without actually declaring $u=\cdots$).
\begin{align*}
                \int \frac{dx}{(1+x^2)^2} &= \int \frac{1 + x^2 - x^2}{(1+x^2)^2}\,dx\\
                &= \int \frac{dx}{1+x^2} - \int \frac{x^2}{(1+x^2)^2}\,dx\\
                &=\tan^{-1}x - \frac{1}{2}\int \frac{x}{(1+x^2)^2}\,d(x^2+1)\\
                &= \tan^{-1}x+\frac 12\int x\, d\Big(\frac 1{1+x^2}\Big)\\
                &=\tan^{-1}x + \frac12\cdot \frac{x}{1+x^2}- \frac12\int \frac{dx}{1+x^2}\\
                &= \frac12\Big(\tan^{-1}x + \frac{x}{1+x^2}\Big)+c.
            \end{align*}
Can someone help me understand how to go from the third line to the fourth?  It doesn't seem like an obvious step. I'm guessing that it's because
$$d\Big(\frac 1{u(x)}\Big) = -\frac{d(u(x))}{u(x)^2},$$
but it doesn't seem like a natural next step, it's just the way I justified it to myself after already having seen the solution. Why should it be the next step?

答:
Here's something you might like.

Writing the indefinite integral as a definite one and applying Feynman's method

\begin{align}I(a)&=\int_c^x\dfrac{1}{a^2+x^2}\,\mathrm dx=\dfrac1a\arctan\left(\dfrac xa\right)\bigg|_c^x=\dfrac1a\arctan\left(\dfrac xa\right)+C\\I'(a)&=\int_c^x-\dfrac{2a}{(a^2+x^2)^2}\,\mathrm dx=-\dfrac1{a^2}\arctan\left(\dfrac xa\right)-\dfrac x{a^2+x^2}+K\\-\dfrac1{2a}I'(a)&=\int_c^x\dfrac1{(a^2+x^2)^2}\,\mathrm dx=\dfrac1{2a^3}\arctan\left(\dfrac xa\right)+\dfrac1{2a(a^2+x^2)}+K\\-\dfrac12I'(1)&=\int_c^x\dfrac1{(1+x^2)^2}\,\mathrm dx=\dfrac12\arctan x+\dfrac1{2(1+x^2)}+K\end{align}

$$\boxed{\boxed{\int\dfrac1{(1+x^2)^2}\,\mathrm dx=\dfrac12\left[\arctan x+\dfrac1{1+x^2}+K\right]}}$$

where $C=\dfrac1a\arctan\left(\dfrac ca\right)$


问:
Oscillation less than epsilon gives property near x
I'm working through a Real Analysis textbook and came across this problem that I am having trouble making rigorous

Let $f:[a,b]\rightarrow \mathbb{R}$ be bounded, $\epsilon>0$ and $\omega_f(x) < \epsilon$ for all $x \in [a, b]$. Show for each $x_0 \in (a, b)$ there exists $h > 0$ such that $x, y \in B_h(x_0) \implies |f(x) − f(y)| < \epsilon$.

The book defines oscillation ($\omega_f(x_0)$) to be $\limsup_{x\rightarrow x_0}f(x)-\liminf_{x\rightarrow x_0}f(x)$.

I see the general idea that the oscillation being less than $\epsilon$ tells us that near $x_0$, the function cannot change too much, but I'm not seeing how to rigorously make that connection.

答:
We have for all $z \in [a,b]$,

$$\lim_{\delta \to 0+}\sup\{f(x) :x \in B_\delta(z)\cap[a,b]\} -  \lim_{\delta \to 0+}\inf\{f(x) :x \in B_\delta(z)\cap[a,b]\}\\:= \omega_f(z) < \epsilon, $$

and there exists $\delta_z $ such if that $x,y \in B_{2\delta_z}(z)\cap[a,b]$, then $|f(x) - f(y)| < \epsilon$.

Since $[a,b]$ is compact and $\{B_{\delta_z}(z):z \in [a,b]\}$ is an open cover, there is a finite subset of intervals $B_{\delta_k}(z_k)$ where $k = 1, \ldots, n$ and

$$\tag{*}[a,b] \subset \bigcup_{k=1}^n B_{\delta_k}(z_k)$$

For any $x \in [a,b]$ there exists by (*) a $j$ such that $x \in B_{\delta_j}(z_j)$.  Let $\delta = \min(\delta_1,\ldots, \delta_n)$ and suppose for $y \in [a,b]$ we have $|y-x| < \delta$. Then we have $|x - z_j| < \delta_j < 2\delta_j$ and

$$|y-z_j| \leqslant |y-x| + |x- z_j| < \delta + \delta_j < 2\delta_j$$

Thus, $x,y \in  B_{2\delta_z}(z_j)\cap[a,b]$  and it follows that $|f(x) - f(y)| < \epsilon$.

To summarize, for any $\epsilon > 0$ there exists $\delta > 0$ such that for all $x,y \in [a,b]$ such that $|x-y|< \delta$ we have $|f(x) - f(y)| < \epsilon$.

Finally, for any $x_0 \in (a,b)$ there exists $h_0>0$ such that $B_{h_0}(x_0) \in (a,b)$ since $(a,b)$ is open. Let $h = \min (h_0,\delta/2)$. If $x,y \in B_h(x_0)$, then we have both $x,y \in [a,b]$ and $|x-y|< \delta$, and, hence, $|f(x) - f(y)| < \epsilon$.


问:
In how many ways can we enumerate from 1 to 20 the sides of a icosahedral
In how many ways can we enumerate from 1 to 20 the sides of a icosahedral? Consider the sides indistinguishable.

I thought about starting off by fixing 1 to one of the sides and then choosing the number of its opposite side but I can't go much further than that

答:
I am assuming you want to put a unique number on each face. For example, having a 1 on two or more faces is not allowed.

Say you pick up such a icosohedron. You can turn it so that the 1-face is facing you. Once you do this, notice there are three sides touching the 1-face. You can rotate the icosohedron so that the face with the smallest number is directly below the 1-face. The choice for number for the below face must be one of the following: $\{2,3,...20\}$. Lets consider each case separately.

If you pick 2, there are 18 choices for the face to the upper left of the 1-face and 17 for the face to the upper right (because you can't repeat numbers). Once you pick these numbers, we have numbered 4 of the 20 faces, so there are 16 faces left and 16 numbers to put on them. Therefore there are 16! ways to do this. Thus we have $18*17*16!$ total ways in this case.

If you pick 3, there are 17 choices for the face to the upper left of the 1-face and 16 for the face to the upper right. This is because the bottom face had to be the smallest number, so we can't pick 2 for either of the upper faces. Then, just as before, we have 16 faces remaining and 16 numbers for the faces, so we have $16!$ ways to do this. Thus we have $17*16*16!$ total ways in this case.

Hopefully you see the pattern. If we count up all the possible ways we get:

$(18*17+17*16+...2*1)*16!$ total numberings. If you would like a simplified expression, I invite you to try and do so yourself.


问:
Why does the Curry-Howard correspondence exist?
The Curry-Howard correspondence observes that logics correspond to type systems (in the sense of having the "same" inference rules). Is there some deeper explanation/justification for this? E.g. "in category theory, 'interesting' logics and type systems are both [something] and therefore isomorphisms exist between them".</p>

答:
To make a long answer short, the part of the history I am best aware of is that Curry stumbled onto it when devising a system of types for combinatory logic and found - by a most remarkable coincidence - that the types of the basic combinators just so happen to have the same form as the basic axioms of implicational logic; that the rule for forming types for function application in combinatory logic happen to match the modus ponens rule; and that the process of abstracting out a free variable to find the combinator that gives you a lambda expression happens to match the rule for discharging an hypothesis in a proof conditioned on hypotheses.

The deeper this correspondence was delved into, the more that was found to line up; e.g. if combinatory logic is extended to allow for tuples, then one can add conjunctions to its scope.

The best explanation that I can think of is that in discovering that things lined up this way, it's not so much that anyone discovered that one can be accounted for or explained in terms of the other or vice versa, but that one just happened onto two instances of an elaborate structure that later came to be recognized to occur elsewhere in other - often unrelated - contexts, and that this was the structure that eventually came to be known and formalized in Category Theory as various types of categories: Cartesian closed, bi-Cartesian closed, etc.

Good account is provided for the "and" and "if-then" connectives and of the "true" and even "false" predicate constants. To some degree, the "all" quantifier can also be treated within this framework.

But there is always such a thing as pushing an analogy too far; and I think that those who are imbued with the notion that one can be explained or accounted for in terms of the other are guilty of doing this. The weak spot is - and has always been - the dual operators, "or" (the dual of "all"), "unless" (the dual of "if-then"), "some" (the dual of "all"), and the correspondence starts to become a bit forced and awkward ... or non-existent (i.e. there's no Curry-Howard rendition of "unless" that I'm aware of).

Within Category theory, propositions correspond to objects, while the inference relation between propositions corresponds to the arrows of morphisms, with the morphisms themselves being the proofs that bear witness to the inference. This is not just a "correspondence": they are the elements for the structure of an actual category, for logic; since we always have reflexivity and transitivity amongst our basic rules for inferrability (i.e. the relation is a pre-order for predicates) and the ability to chain up proofs made of two or more inferences in sequence.

Inferences are (1,1) sequents: sequents that have one statement on each side. A general sequent $A₀,A₁,⋯ ⊢ B₀,B₁,⋯$ may permit 0, 1, 2 or any finite number of statements on each side. It is always reducible, equivalently, to (1,1) form as $A₀∧A₁∧⋯ ⊢ B₀∨B₁∨⋯$, where $∧$ is the "and" connective, $∨$ is the "or" connective and $⊢$ is the "infers" relation.

When trying to draw analogy to type theory, what we're actually trying to do is extend the correspondence to also be able to account for "assertions": or (0,1) sequents $⊢ B$. There is nothing in category theory that directly corresponds to this, but we'd like to think of a proof that witnesses this sequent $b: ⊢ B$ as somehow being a "member" of a "type" which $B$ corresponds to. Types are then populated by members, in some sense.

I'm always a fan of symmetry, so I feel uneasy when I see this treatment, while a similar treatment remains absent - for the dual notion of a "query" or counter-factual "is it so?" question: the (1,0) sequent $a: A ⊢$. A language, like Prolog, actually has something that corresponds to both, the notations being respectively $b.$ and $?- a$.

The process of starting with a purely category-theoretic starting point and then trying to populate the objects so that they can be treated as types is what we now call creating an "internal language". There is no dual notion of "co-language" that I am aware of.

So, when it comes time to try and relate types and propositions, there is an asymmetry in the Curry-Howard correspondence reflected by the absence of full duality. This speaks more to the "the two analogues are the first two instances of a general as-yet-discovered phenomenon" view that I just relayed, rather than to the more prevalent "one can be explained by or account for in terms of the other" view.

So, I think it just comes down to this: that the correspondence was the discovery of the first two examples of a category-theoretic structure and that it is actually category theory that lies beneath everything.


问:
Why is $\mathbb{Z}[\alpha ]$ not finitely generated as $\mathbb{Z}$-module?

Assume that $\alpha \in \mathbb{C}$ is an algebraic number which is not an algebraic integer. My question is why $\mathbb{Z}[\alpha]$ is not finitely generated as  $\mathbb{Z}$-module.

Clearly there exists $ a_i \in \mathbb{Z}$, which are not all zero and $n>0$ such that $a_n\alpha^n+a_{n-1}\alpha^{n-1}+\dots + a_1\alpha + a_0=0$. Shouldn't this imply that $\mathbb{Z}[\alpha]$ is finitely generated? The motivation for this question is that it may be shown that $\alpha \in \mathbb{C}$ is an algebraic integer if and only if $\mathbb{Z}[\alpha ]$ is finitely generated as a $\mathbb{Z}$-module. Why does the coefficient $a_n$ has to be one for the module to be finitely generated? And, to check that I haven't misunderstood the terminology, $\mathbb{Z}[\alpha ]$ is equal to all polynomials in $\alpha$ over $\mathbb{Z}$, right?

答: Let me modify M Turgeon's answer slightly, so as not to assume that you want to use the powers $\alpha^0$ through $\alpha^{n-1}$ as your generators.  Suppose you had some more complicated or more clever finite set of generators $\{g_1,\dots,g_k\}$ for $\mathbb Z[\alpha]$.  Each $g_i$ is a polynomial in $\alpha$ with integer coefficients, but we know nothing more about these polynomials, not even how many of them there are.  Nevertheless, we can fix a natural number $N$ strictly greater than the degrees of all these $g_i$'s, and, because $\alpha^N$ is in $\mathbb Z[\alpha]$, we must be able to express $\alpha^N$ as a linear combination of the alleged generators $g_i$ with integer coefficients, say $\alpha^N=\sum_{i=1}^kc_ig_i$.  But now, transposing everything to the left side in this equation, you have a monic polynomial equation, with integer coefficients, satisfied by $\alpha$.  (It's monic because $\alpha^N$ has coefficient $1$, unaffected by any of the other terms, as these have lower degree.)  So $\alpha$ is an algebraic integer, contrary to your assumption.</p>


问:
For any cardinal numbers $a\neq b$, is there a cardinal number $c$ such that $c^a\neq c^b$?
For any cardinal numbers $a\neq b$, is there a cardinal number $c$ such that $c^a\neq c^b$?

If there is no (or ZFC undecidable) such cardinal number, then is the following true?
Let $a,b$ be two cardinal numbers, if $c^a=c^b$ for any cardinal number $c$, then $a=b.$

This is motivated by the question here when I consider the category of set.

答:
Yes. Recall the $\beth$ numbers:

$\beth_0=\aleph_0$;

$\beth_{\alpha+1}=2^{\beth_\alpha}$; and

$\beth_\delta=\sup\{\beth_\gamma\mid\gamma<\delta\}$ for a limit ordinal $\delta$.

So if $\delta$ is a limit ordinal, $\beth_\delta$ is a strong limit cardinal, and moreover, if $\lambda$ is the cofinality of $\beth_\delta$ (or that of $\delta$), then $\beth_\delta^\gamma=\beth_\delta$ for all $\gamma<\lambda$. But, recall König's theorem and its consequence: if $\gamma\geq\lambda$, then $\beth_\delta^\gamma>\beth_\delta$.

Now. If $\kappa<\lambda$, then there is a regular cardinal $\mu$ such that $\kappa<\mu\leq\lambda$, e.g. $\kappa^+$. Therefore $\beth_\mu$ is a cardinal such that:

$\beth_\mu^\kappa=\beth_\mu$, and

$\beth_\mu^\lambda\geq\beth_\mu^\mu>\beth_\mu$.


问:
Guillemin and Pollack, Exercise 4.8.7, understanding the hint.
I have copied below exercise 4.8.7 from Guillemin and Pollack's Differential Topology

I'm having trouble following the last part of the hint (Exercise 7 of Section 6 refers to the fact that homotopic maps induce the same map on cohomology).

Following the hint, we pick a form $\theta \in \Omega^k(X)$ and an open cover $\{U_1,\ldots,U_n\}$ where $U_i$ is smoothly isotoptic to a coordinate neighbourhood we started with $U$. Then we choose a partition of unity subordinate to this cover i.e. we have smooth functions $\rho_i$'s with $\mathrm{supp}(\rho_i) \subset U_i$ and $\sum_{i=1}^n\rho_i \equiv 1$. Letting $\theta_i = \rho_i\theta$, we have forms supported in $U_i$ such that they add upto to $\theta$. Okay, now, since $U_i$ is smooth isotopic to $U$, we have a isomorphism of cohomology groups $H^k(U_i) \cong H^i(U)$, so $[\theta_i]$ corresponds to some class of a scalar multiple of $[\omega]$. But the hint is saying that we make it so that $\theta_i$ is cohomologous to a scalar multiple of $\omega$, that makes me think I'm not using the fact that $U_i$ and $U$ are isotopic well enough. I suppose I'm also confused on how're we intepreting isotopy in this context.

Any comments will be helpful. I know I'm asking for help in the middle a chain of thought that you may not be privy to, so apologies for that!

Update: After following the comments, we have the following. Let $h_{i,t}$ denote the isotopy between $U$ and $U_i$, then for $h_{i,1}^*\theta_i$, as a compactly supported form on $U$, we have $[h_{i,1}^*\theta_i] = c_i[\omega]$, where $c_i = \int_Uh_{i,1}^*\theta_i = \int_{U_i}\theta_i$, since $h_{i,1}^*$ is a diffeomorphism (this comes from the initial hint and 4.8.6).

Then I suppose letting $H_i$ denote the inverse of $h_{i,1}$, we have $[\theta_i] = c_i[H_i^*\omega]$ and so $[\theta] = \sum_{i=1}^nc_i[H_i^*\omega]$.

答: The key to this problem is how we construct this cover given in the hint, which I initially took as a black box. We started with a $U$, let's fix a $p\in U$. Then for any $x\in X$, by the Isotopy lemma, we can find a diffeomorphism $h_x:X \to X$ which is isotopic to the identity on $X$ such that $h_x(p) = x$. Then $\{h_x(U)\}_{x\in X}$ is an open cover of a compact manifold, so we can choose a finite subcover, say $\{h_1(U),\ldots,h_N(U)\}$. Then $U_i = h_i(U)$ is the cover given in the hint. Since $h_i \sim \mathrm{id}_X$, then $h_i^\sharp = \mathrm{id}_X^\sharp$ (pullback maps on cohomology), and things follow. I won't bother fixing notation in my question and the update I added. But everything comes together to give us a solution. (I refer to and use notations from Guillemin-Pollack.)


问:
For a normal population with a mean ofµ and a standard deviation of 9, the hypotheses to be tested are H0: µ = 50 versus H1: µ < 50.
Which would result in the highest power? (D)

A. µ1 = 49; n =10

B. µ1 = 48; n =10

C. µ1 = 49; n =100

D. µ1 = 48; n =100

E. µ1 = 48.5; n =15

The answer is D.

So far I got

x^= 50 - za 9/sqrt n

I know I have to standardize for beta, but there's not confidence level stated so I'm not sure what to do next.

答:
You don't need to do any serious calculation for this.  You can use process of elimination.  The farther $\mu_1$ is from $\mu_0$, the higher the power.  The larger $n$ is, the higher the power.  So this eliminates all choices except (D):  compared to every other answer choice, (D) has a difference $|\mu_1 - \mu_0|$ that is at least as large and an $n$ that is at least as large.

The only way that the question becomes "difficult" is if, among the answer choices, you could not find one that is unequivocally "better" than the others; for example, if you had to choose between only options (B) and (C), it is no longer clear which of the two has larger power because while $|\mu_1 - \mu_0| = 2$ for (B) and $1$ for (C), the standard error of the test statistic for (C) is substantially smaller than for (B).  In such a case, you would need to do more computation.


问:
Generating Laguerre polynomials using gamma functions
An exercise given by my complex analysis assistant goes as follows:

For $n \in \mathbb{N}$ and $x>0$ we define
  $$P_n(x) = \frac{1}{2\pi i} \oint_\Sigma \frac{\Gamma(t-n)}{\Gamma(t+1)^2}x^tdt$$
  where $\Sigma$ is a closed contour in the $t$-plane that encircles the points $0,1, \dots, n$ once in the positive region.

Now, we have to prove some things like that $P_n(x)$ is a polynomial of degree $n$, which is no problem. However, the assistant claims that $P_n(x)$ "is known as a Laguerre polynomial". However, calculating $P_n(x)$ for certain $n$ and comparing to values of the Laguerre polynomials $L_n(x)$ on the internet, I find that
$$(-1)^n \cdot n! \cdot P_n(x) = L_n(x)$$
Could my assistant have made a typo which explains this missing factor; or could he mean that $P_n(x)$ have similar properties as the Laguerre polynomials, perhaps?

答:
All you need to do is to multiply the integral representation by $(-1)^n n!$, so you will have the right integral representation

$$ P_n(x) = \frac{(-1)^nn!}{2\pi i} \oint_\Sigma \frac{\Gamma(t-n)}{\Gamma(t+1)^2}x^tdt. $$

For instance,

$$ P_5(x) = 1-5\,x+5\,{x}^{2}-\frac{5}{3}\,{x}^{3}+{\frac {5}{24}}\,{x}^{4}-{\frac {1}{120
}}\,{x}^{5},$$

which agrees with the Laguerre polynomial $L_5(x)$.


问:
Cauchy convergence in L2 of empirical CDF implies convergence in probability to a tail sigma-algebra measurable function
I am reading the proof of de Finetti's theorem in Chow and Teicher's book Probability Theory and there is one claim which I am not sure about. Let me quote some context first:

Theorem: Random variables $X_n, n \ge 1$ on $(\Omega, \mathcal{F}, \mathbb{P})$ are interchangeable iff they are conditionally independent and identically distributed given some $\sigma$-algebra $\mathcal{G}$ of events. Moreover, $\mathcal{G}$ can be taken to be either the $\sigma$-algebra $\mathcal{E}$  of permutable events or the tail $\sigma$-algebra $\mathcal{T}$, and $\mathbb{P}(X_1 < x_1 \mid \mathcal{E}) = \mathbb{P}(X_1 < x_1 \mid \mathcal{T})$ a.s..

Proof: Sufficiency is immediate since ... . Apropos of necessity, for $n \ge 1$ and any real $x$, define $$\xi_n(x) = \frac{1}{n} \sum_{i=1}^n I_{\{X_i < x\}}$$ Then $$\mathbb{E} [\xi_n(x) - \xi_m(x)]^2 = \frac{|m-n|}{mn} \left( \mathbb{P}(X_1 < x) - \mathbb{P}(X_1 < x, X_2 < x) \right) \to 0 \tag{1}$$ as $m,n \to \infty$, whence $\xi_n(x) \stackrel{p}{\to}$ some $\mathcal{T}$-measurable r.v. $\xi(x)$.

I know that since $L^2$ is complete, (1) implies $L^2$ convergence of the sequence $\{\xi_n(x)\}$, which in turn implies convergence in probability to $\xi(x)$, say.

Why is $\xi(x)$  $\mathcal{T}$-measurable?

It intuitively makes sense to me but how do I show this formally? I'll need to show that $\{\xi(x) < t\} \in \sigma(\xi_{m+1}(x), \xi_{m+2}(x), \ldots)$ for every $t \in \mathbb{R}$ and $m \ge 1$. Now convergence in probability implies convergence in distribution, which in turn implies $\lim_{n} \{\xi_n(x) < t\} = \{\xi(x) < t\}$. But $\lim_{n} \{\xi_n(x) < t\} = \lim_n \left\{\frac{1}{n-m} \sum_{i=m+1}^n I_{\{X_i < x\}} < t\right\} \in \sigma(\xi_{m+1}(x), \xi_{m+2}(x), \ldots)$. Is this correct?

答:
Convergence in probability implies that a subsequence $\xi_{n(k)}$ converges almost surely and it is very standard that this implies measurability.

https://en.wikipedia.org/wiki/Convergence_in_measure


问:
inequality in inner product
I want to show that if $$(u-\hat u, v-\hat u)\leq 0$$ and also
$$(v-\hat v, u-\hat v)\leq 0$$
then
$$(\hat u-\hat v, u-v)\geq 0$$

Please help me. Maybe it is easy

Thanks

答:
Hint: Euler identity

$$(u-\hat u, v-\hat u)- (v-\hat v, u-\hat v)+ (\hat u-\hat v, u-v)= 0.$$

(So I suppose you've made a typo)

Title: inequality in inner product


问:
If $\{F_n\}$ is a decreasing sequence of compact subsets of $X$ Hausdorff and $f$ is continuous, then $f(\cap F_n) = \cap f(F_n)$
I have to prove the following statement:

Let $X$ and $Y$ be Hausdorff Spaces and $f:X\to Y$ a continuous function. If $\{F_n\}$ is a decreasing sequence of compact subsets of $X$, then $f(\cap F_n) = \cap f(F_n)$.

I already proved it, however, my proof doesn't use the hypothesis that $Y$ is Hausdorff, and that makes me think it might be wrong, but I don't find any mistakes. I would be really thankful if someone could verify it.

Proof. First we see that $f(\cap F_n) \subseteq \cap f(F_n)$. If $x\in f(\cap F_n)$, there must exists $a\in \cap F_n$ such that $f(a)=x$. It is clear that $a\in F_n$, for all $n\in\mathbb{N}$ and in consequence $x\in f(F_n)$, for all $n\in\mathbb{N}$, which implies $x\in\cap f(F_n)$.

Now we are going to prove that $\cap f( F_n) \subseteq f(\cap F_n)$. Let $x\in\cap f( F_n)$, this implies that $x\in f( F_n)$ for all $n\in\mathbb{N}$. From here we get for each $n\in\mathbb{N}$ there exists $y_n\in F_n$ such that $f(y_n)=x$. Since $\{F_n\}$ is a decreasing sequence, we have that $F_n \subseteq F_1$ for all $n\in\mathbb{N}$ and this implies that $y_n\in F_1$ for all $n\in\mathbb{N}$. This way, $\{y_n\}$ is a sequence in $F_1$. Because $F_1$ is compact, $\{y_n\}$ must have a convergent subsequence. Lets say this subsequence is $\{y_{n_k}\}$, and that it converges to $y$. Futhermore, since $X$ is Hausdorff, this limit is unique.

We see that $y\in\cap F_n$. Let $i\in\mathbb{N}$. There exists a $j\in\mathbb{N}$ such that $i<n_j$. Now lets consider the sequence $\{a_k^j\}$ given by $a_k^j=y_{n_{k+j}}$. It is clear that $a_k^j=y_{n_{k+j}}\in F_{n_{k+j}}\subseteq F_{n_{j}}\subseteq F_i$, which implies that the sequence $\{a_k^j\}$ is a sequence in $F_i$. Also, this sequence is just a shift of $\{y_{n_{k}}\}$, and because of this it must also converge to $y$. Because $F_i$ is compact, and $X$ is Hausdorff, $F_i$ is closed, and since $\{a_k^j\}$ is in $F_i$, we have that $y\in F_i$. As $i\in\mathbb{N}$ is arbitrary, we get that $y\in\cap F_n$.

Finally, using the continuity of $f$ and the uniqueness of the limit of $\{y_{n_k}\}$ we get that $f(y)=f(\text{lim}_{k\to\infty}y_{n_k})=\text{lim}_{k\to\infty}f(y_{n_k})=\text{lim}_{k\to\infty}x=x$, from where we get that $x=f(y)\in f(\cap F_n)$.

In conclusion, $f(\cap F_n) = \cap f(F_n)$

答:
It is wrong to say $$f(\lim_{k\to\infty}y_{n_k})=\lim_{k\to\infty}f(y_{n_k})$$ if $Y$ is not Hausdorff.  Indeed, it is not even meaningful to write $\lim_{k\to\infty}f(y_{n_k})$, since the limit may not be unique.  All you can say is that $f(\lim_{k\to\infty}y_{n_k})$ is equal to some limit of the sequence $f(y_{n_k})$, but it might be a different limit from $x$.

(Separately, it is wrong to say that $(y_n)$ has a convergent subsequence.  It is not true in a compact space that a sequence always has a convergent subsequence.  This error is easily corrected by taking a convergent subnet instead, though.)


问:
Find the coordinate of third point of equilateral triangle.
I have two points A and B whose coordinates are $(3,4)$ and $(-2,3)$ The third point is C. We need to calculate its coordinates.

I think there will be two possible answers, as the point C could be on the either side of line joining A and B.

Now I put AB = AC = BC.

We calculate AB by distance formula : $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

= $ \sqrt{5^2 + 1^2} = \sqrt{26}$

I plug this value into the distance of AC and BC and finally equate to get :

$5x + y = 6$

Now what can I do? There are two variables, I am getting equation of a line! How can I solve this?

答:
Calculate slope of AB (m), distance AB = s
Point A is at (h,k)
tan-1 m = angle N
s x sin (60 + angle N) = y
s x cos (60 + angle N) = x
Third point is at x+h, y+k
Use (180 + 60 + angle N) for second answer
and h,k is now at point B</p>


问:
What is the maximal interval for the IVP $x'(t) = x(t)(1-x(t))$, with $ x(0) = x_{0}$?
I am new to ODEs and I'm stuck on this problem. I will write down my attempt.  The IVP in question is:

\begin{cases}
x'(t) = x(t)(1-x(t)), \\
x(0) = x_{0}.
\end{cases}

I applied separation of variables. I defined the function $f: X \mapsto \mathbb{R}$, $X := X_{1} \times X_{2},$
\begin{align}
f(t,x) := g(t)\cdot h(x) = x(t)(1-x(t))
\end{align} such that $g: X_{1} \mapsto \mathbb{R}$ and $h: X_{2}\mapsto \mathbb{R}$, with $g(t) := 1$ and $h(x) := x(1-x), X_{1} := \mathbb{R}$ and $X_{2} := \mathbb{R}$.  Then I defined the solution $x : I \mapsto \mathbb{R}$, defined on an interval $I \subset \mathbb{R}$ such that $0 = t_{0} \in I$ too.

Firstly, I checked when $h(x) = 0$:
\begin{align}
h(x_{0}) = 0 \implies x_{0} = 0 \text{ or } x_{0} = 1.
\end{align} Hence,
\begin{align}
\text{for } x_{0} = 0 \implies x(t) = 0; \\
\text{for } x_{0} = 1 \implies x(t) = 1.
\end{align}

Afterwards, I solved the equation given that $h(\xi) \neq 0$ ($\implies x_{0} \neq 0$ and $x_{0} \neq 1$),

$$
\int_{x_{0}}^{x}\frac{1}{h(\xi)} d\xi = \int_{t_{0}}^{t} g(\eta) d\eta,
$$
i.e.
\begin{align*}
\int_{x_{0}}^{x}\frac{1}{\xi(1-\xi)} d\xi = \int_{0}^{t}  d\eta
&\implies (\ln\lvert\xi\rvert - \ln\lvert 1-\xi\rvert)\big\rvert_{x_{0}}^{x} = t \\
&\implies \ln\Big\lvert\frac{\xi}{1-\xi}\Big\rvert\Bigg\rvert_{x_{0}}^{x} = t \\
&\implies \ln\Big\lvert\frac{x}{1-x}\Big\rvert - \ln\Big\lvert\frac{x_{0}}{1-x_{0}}\Big\rvert = t \\
&\implies \ln\Big\lvert\frac{x(1-x_{0})}{x_{0}(1-x)}\Big\rvert = t  \\ &\implies \Big\lvert\frac{x(1-x_{0})}{x_{0}(1-x)}\Big\rvert = e^{t}.
\end{align*}
I distinguished two cases:

When $h(x_{0}) > 0 \implies h(x) > 0$, so
\begin{align*}
&\frac{x(1-x_{0})}{x_{0}(1-x)} = e^{t}
\implies &x(t) = \frac{e^{t}x_{0}}{1-x_{0}(1-e^{t})}.
\end{align*}
Here is where I'm having issues with finding the maximal interval. Since, $h(x_{0}) > 0$ then
\begin{align*}
x_{0}(1-x_{0})>0 \implies
\end{align*}
\begin{cases}
x_{0}>0 \\
1-x_{0}>0
\end{cases}
$$ \textbf{or} $$
\begin{cases}
x_{0}<0 \\
1-x_{0}<0
\end{cases}
which results in
$$ 0 < x_{0} < 1 .$$ Additionally, $$x_{0} \neq \frac{1}{1-e^{t}}.$$ However, in this case, $x(t)$ exists for $t \in \mathbb{R} \setminus \{\ln(1- \frac{1}{x_{0}})\}$, since
$$ 1-x_{0}(1-e^{t}) \neq 0 \implies t \neq \ln\Big(1- \frac{1}{x_{0}}\Big) $$
which is a problem because the bounds on $x_{0}$ make it so that $\ln(1- \frac{1}{x_{0}})$ can't exist.
So, I'm wondering what I did wrong and it's confusing me a lot.

For $h(x_{0})<0$ I got the same $x(t)$ but the bounds on $x_{0}$ are
$$ x_{0} < 0 \cup x_{0} > 1 $$
which in this case works with $t \in \mathbb{R} \setminus \{\ln(1- \frac{1}{x_{0}})\}$.
What confuses me even more, is that by differentiating $x(t)$ you get the ODE in the IVP, so it's presumably right (?).
I'm not sure if I get the wrong interval because I used the wrong approach, or I messed up the signs, or something else I'm not noticing. Any help/suggestions would be appreciated.

答:
In the first case, where $\ 0<x_0<1\ $, your conclusion that $\ \ln\left(1-\frac{1}{x_0}\right)\ $ cannot exist isn't really a problem.  It simply tells you that $\ 1-x_0\left(1-e^t\right)=x_0e^t+1-x_0\ $ can never be zero for any real value of $\ t\ $.  In fact, it's easy to show that if $\ 0<x_0<1\ $, then that expression is always strictly positive. In other words, your solution
$$
x(t)=\frac{x_0e^t}{1-x_0\left(1-e^t\right)}
$$
of the differential equation holds good for all $\ t\ $ in the infinite interval $\ (-\infty,\infty)\ $. You appear to have implicitly assumed that this interval had to be finite.  If so, that's probably what has given rise to your confusion.


问:
closure of A in the product space
For each $\alpha\in I=[0,1]$, let $X_{\alpha} = \{0,1\}$ be the discrete topology. For every $\Delta\subseteq I$, define $f_\Delta=(x_\alpha)\in\prod_{\alpha\in I}X_\alpha$, where $x_\alpha=1$ if $\alpha\in\Delta$, and $x_\alpha=0$ if $\alpha\notin\Delta$. Let $A=\{f_\Delta:\Delta\text{ is a finite subset of }I\}$. Show that $f\in\operatorname{cl}A$, where $\operatorname{cl}A$ is the closure of $A$ in the product space $\prod_{\alpha\in I}X_\alpha$.

答:
Consider some arbitrary $f$, along with an open set $U$ that contains $f$ in the product topology. Recall that the product topology has a basis consisting of sets $\prod_{\alpha \in I} U_{\alpha}$, where each $U_\alpha$ is  non-empty, open in $X_\alpha$, and $U_\alpha = X_{\alpha}$ for all but finite $\alpha$. Thus, we will have:

$$f \in \prod_{\alpha \in I} U_{\alpha} \subset U$$

Let $S$ be the finite set of $\alpha$, where  for which $U_\alpha \neq X_\alpha$. Then, we let $T$ be the finite subset of all $\alpha \in S$ such that $U_\alpha$ contains $1$. Consider the sequence $f_T = (x_\alpha)$. Given some $\alpha \in T$, we note that $x_\alpha = 1$, which is in $U_\alpha$, by definition of $T$.

Given some $\alpha \notin T$, we note that $x_\alpha = 0$. In such a case, $U_\alpha$ is either not equal to $X_\alpha$, and doesn't contain $1$, so it must contain $0$, or $U_\alpha = X_\alpha$, so it also contains $0$ in this case. Thus, in either case, $x_\alpha \in U_\alpha$.

It follows that $f_T \in \prod_{\alpha \in I} U_\alpha \subset U$ and is in $A$. Thus, every open neighbourhood of $f$ intersects $A$, so $f \in \text{cl}(A)$. This completes the proof.

Title: closure of A in the product space


问:
Two weak formulations for a time-dependent PDE: are they equivalent?
My question is concerned with the way people tend to define weak formulations for time-dependent PDEs. Suppose I have a wave equation in a simplest form
$u_{tt}-a^{2}\Delta u = f$ in a bounded domain in $\mathbb{R}^{2}$ or $\mathbb{R}^{3}$ for $t\in(0,T)$. To make it simple, suppose $u = 0$ on $\partial\Omega$ and initial conditions are homogeneous.
Typical weak formulation is constructed from integration over entire variable space $[0,T]\times\Omega$ and, if we do not use integration by parts in time, has a form
$\int_{0}^{T}\int_{\Omega}(u_{tt}\phi + a^{2}\nabla u\nabla\phi)dxdt = \int_{0}^{T}\int_{\Omega}f\phi dxdt$ for each $\phi$ from the closure of space $\dot{D}((0,T)\times\Omega)$ which consists of all smooth functions with compact support with respect to $\Omega$ ( but not time dimension ). The closure is done with respect to norm $H^{2}((0,T)\times\Omega)$. Function $u$ is sought from the same. A good and consistent theory exists for this formulation, such as existence and uniqueness results, as well as some regularity results. I may not specify the spaces correctly here, but that's not the point right now.
What troubles me is that people don't use this formulation in practical situations, such as proving a priori estimates for Finite Element method. They instead keep using what I call a stronger form:
for a.a. $t\in(0,T)$ we have $\int_{\Omega}(u_{tt}\phi + a^{2}\nabla u\nabla\phi)dx = \int_{\Omega}f\phi dx$ for all twice-differentiable maps $\phi$ belonging to $L^{2}(0,T; H_{0}^{1}(\Omega))$ ( adjust the space in your way, if you like ). So it's the difference in how the time evolution is treated: you integrate over time in the first formulation and
avoid it in the second one.
 I've never seen a single proof of existence/uniqueness for the second formulation. It's like it's equivalent to the one above. But if so, can anyone provide at least a sketch of a proof?
I already made a thread concerning the difference between spaces $L^{p}(0,T; L^{p}(\Omega))$ and $L^{p}([0,T]\times\Omega)$: $L^{2}(0,T; L^{2}(\Omega))=L^{2}([0,T]\times\Omega)$?
and it is closesly related to my question here.
Any help will be much appreciated!
By the way, the original weak formulation ( the first one ) implies that you also integrate by parts in time, to reduce regularity assumptions in time. But since you cannot do it in the second formulation, I intentionally skipped it, to give a chance of proving that solutions for both formulations are the same.</p>

答:
For the strong formulation, you only need a test function $\phi \in H_0^1(\Omega)$ (no dependence on time).

Assume that a solution of the weak formulation has the regularity $u_{tt} \in L^2(0,T; H^{-1}(\Omega))$, $u \in L^2(0,T; H_0^1(\Omega))$. Now, take a arbitrary $\phi \in H_0^1(\Omega)$ and some $\psi \in C^\infty(0,T)$. The weak formulation (tested with $\phi(x)\,\psi(t)$) implies $$\int_{0}^{T}\int_{\Omega}(u_{tt}\phi(x) + a^{2}\nabla u\nabla\phi(x) - f \phi(x))dx \, \psi(t) \, dt = 0.$$
Since $\psi$ was arbitrary, the fundamental theorem of calculus of variations implies
$$\int_{\Omega}(u_{tt}\phi(x) + a^{2}\nabla u\nabla\phi(x) - f \phi(x))dx = 0$$
a.e. in $(0,T)$.

Hence, both formulations are equivalent if $u$ has the above mentioned regularity.


问:
Expectation of a sum of equally distributed random variables
I'm studying the following term:

$$W_{ij} = \frac{1}{n(n-1)E[f(X_1,X_2)]} \sum_{i\neq j}^n f(X_i,X_j)Y_j,$$

where $(X_i, Y_i)$ is a sequence of equally distributed random pairs and $f$ a measurable function.

Question: Does $E(W_{ij})=\frac{E[f(X_1,X_2)Y_2]}{E[f(X_1,X_2)]}$ hold?

Considerations:

To get this result, one has to show that $E[f(X_i,X_j)Y_j] = E[f(X_1,X_2)Y_2], \forall i \neq j$, which I believe is false in general.

Take for example $Y = 1$ and $f(x, y) = xy$. It's not correct to say that $E(X_iX_j) = E(X_1X_2)$ for $i \neq j$, only using the equality in distribution, right? If we add independecy, $E(X_iX_j) = E(X_i)E(X_j) = E(X_1X_2)$.

Being more specific

In my case, I have $f(x,y)=[g(x)]^2h(x)h(y)-g(x)h(x)g(y)h(y)$, with g,f measurable. For simplicity's sake, denote $g(X_k):=g_k$ and $h(X_k)=h_k$.
If I assume $(X_i,Y_i)_{i\in\mathbb{N}}$ is also independent, then each $X_i$ is independent of $X_j$ and $Y_j$, for $i\neq j$.

That said, we have

\begin{align}
E(f(X_i,X_j)Y_j)&=E(g_i^2h_ih_jY_j)-E(g_ih_ig_jh_jY_j)\\
&\overset{indep.}{=}E(g_i^2h_i)E(h_jY_j)-E(g_ih_i)E(g_jh_jY_j)\\
&\overset{i.d.}{=}E(g_1^2h_1)E(h_2Y_2)-E(g_1h_1)E(g_2h_2Y_2)\\
&=E(f(X_1,X_2)Y_2).
\end{align}

But when $(X_i,Y_i)_{i\in\mathbb{N}}$ is dependent, the above equality does not hold in general.

Are my thoughts right?

答:
Yes, I also think that an assumption of independence is needed.

Suppose $g(x) = x$, $h(x) = x$ and that for some $i \neq j$ we have $X_i = X_j = Y_j$ (which is allowed under dependence).  Then at line 1 the second term is $E(Y_j^5)$.  But in general, $E(Y_j^5) \neq E(Y_j^2)E(Y_j^3)$, versus the step to line 2 which is valid under assumption of independence.

(Strictly, this example answer doesn’t prove that an assumption of independence is necessary for the original equation proposed for $E(W_{ij})$.  But it does indicate that without such an assumption, the equality isn’t obvious.)


问:
Test convergence of alternating series for real alpha smaller or equal to 1
Test convergence of following series for $\alpha\in \mathbb{R}$:
$\sum_{n=2}^{\infty} (-1)^n \frac{e^{\frac{1}{n}}}{n*\ln^{\alpha}(n)}$

I proved that series converge absolutely for $\alpha>1$ but I can't find anything for all other alphas. I am looking for the solution for $\alpha\in[-\infty,1]$

答:
The alternating series test tells us this is convergent for all $\alpha\in\mathbb{R}$. Note that for negative $\alpha$, the sequence $a_n=\frac{e^{1/n}}{n\ln^{\alpha}n}$ might not be decreasing from the first value ($n=2$) but it will be decreasing for sufficiently large values of $n$, i.e. for $n\geq N$ where $N$ is some constant.

As for absolute convergence,
$$\frac{1}{n\ln^{\alpha}n}<\frac{e^{1/n}}{n\ln^{\alpha}n}<\frac{3}{n\ln^{\alpha}n} $$
and since $\sum_{n=2}^{\infty}\frac{1}{n\ln^{\alpha}n}$ converges iff $\alpha>1$, the given series is absolutely convergent only for $\alpha>1$.


问:
What is the probability of drawing 6 unique cards from 6 different decks?
I am working through some problems while studying and would like to validate my thinking with this problem.

There are six distinct deck of cards and one card is drawn from each deck. What is the probability that each card drawn is different?

I approached it as 1/52* 1/51 *... *1/47 because each time a card is drawn there is one less that is available to choose from if we don't want to match. I figured this is similar to 52 choose 1 times 51 choose 1 etc

Is this the correct way to approach it?

答:
Your answer to this problem is not correct. In fact, the correct answer should be $$ \frac{52}{52} * \frac{51}{52} * \frac{50}{52} * \frac{49}{52} * \frac{48}{52} * \frac{47}{52} $$
This is how I got this answer - Consider event A to be the set of all outcomes in which all of the six cards are pairwise-distinct. Using combinatorics, this is equivalent to $$ 52*51*50*49*48*47 $$
The total number of ways in which 6 cards can be drawn from 6 decks is $$ 52^{6} $$
Thus by probability theory, the answer should be equal to the number of favourable outcomes divided by total outcomes. Hence $$ \mathbb{P}(A) = \frac{52}{52} * \frac{51}{52} * \frac{50}{52} * \frac{49}{52} * \frac{48}{52} * \frac{47}{52} $$
Hope it helps you!!!


问:
In how many ways can we enumerate from 1 to 20 the sides of a icosahedral
In how many ways can we enumerate from 1 to 20 the sides of a icosahedral? Consider the sides indistinguishable.

I thought about starting off by fixing 1 to one of the sides and then choosing the number of its opposite side but I can't go much further than that

答:
Suppose we start by ignoring symmetry and count all $20!$ labelings. In doing so, we've counted each distinct labeling in exactly 60 different orientations:

Some (arbitrary) base orientation $x$

$x$ rotated about an axis that goes through a pair of opposite vertices (6 axes $\times$ 4 nontrivial rotations = 24 rotations)

$x$ rotated about an axis that goes through the centers of two opposite faces (10 axes $\times$ 2 nontrivial rotations = 20 rotations)

$x$ rotated about an axis that goes through the centers of two opposite edges (15 axes $\times$ 1 nontrivial rotation = 15 rotations)

So the number of distinct labelings is $20!/60$. This agrees with Mathew's answer.

We can think of this as a simple application of Burnside's Lemma.