问:
Some relation about a simple graph
I have seen some relation about the simple graph $G$ which is not directed.

Suppose $G$ has $n=|V|\ge1$ vertices, $m$ edges, $k$ connected components, $p$ odd cycles, $q$ even cycles. Do the following hold?

$p+q\ge 1$ then $m\ge n$

$p+q=1, k=1,$ then $ m=n$

$k\ge n-m$

If $p = 0$, then $G$ is bipartite.

If $q = 0$, then there are $2k$ proper colorings of $G$ using $2$ colors.

Attempts: For question i, I believe it is False as we may have many isolated vertices.
for question 2 i think it is true. $p+q=1$ mean there exist cycle and k=1 mean that it is a connected graph. I believe it is true but then i have no idea how to show m=n.

for Q4, i think that it is true as it may a graph with no cycle is bipartile and also a graph with even with even cycle only seems to be a bipartitle too.

For question 5 i think it is false as for the case that the graph has no cycle, q=0 but we cna actually colour with 2 colours.

Is my interpretation or guess true? and is there any hints about the proof of some details?

答:
It seems the following.

You already answered.

When you delete one edge from the cycle you should obtain a tree. By induction you can show that $|V(T)|=|E(T)|+1$ for each three $T$.

For each connected component $G_i$ of the graph $G$ should be
$|V(G_i)|\ge|E(G_i)|+1$. Adding these inequalities for all components, we obtain the inequality $k\ge n-m$.

For each $i$ fix a vertex $v_i\in G_i$, and let $A_0$ be the set of all $x_i$. Put $A=\{u\in V(G): \exists v\in A_0$ and a path from $v$ to
$u$ with even length $\}$ and $B=V(G)\backslash A$. Show that $V(G)=A\cup B$ is the required partition.

The triangle graph $K_3$ has no even cycles and no vertex 2-colorings.

Title: Some relation about a simple graph


问:
How can we simplify this integral? $\int{x\frac{f(x)}{\int f(x) dx} dx}$
In a machine learning lecture, we encountered the following integral that we needed to solve to calculate the mean of some random variable $x$:

\begin{equation*}
  \int{x\frac{f(x)}{\int f(x) dx} dx}
\end{equation*}

Without really explaining, the professor just simplified it to this:

\begin{equation*}
  \frac{\int x f(x) dx}{\int f(x) dx}
\end{equation*}

I'm not sure how that works. Since the integration is without limits, then the result is a function not a constant, right? It can't be factored out as if it were a constant. Am I missing something? Does integrating on the same variable twice have any special properties that are relevant here?

I'm sorry if the question is lacking in details, if there's anything I can edit to make it clearer, please let me know.

Edit: The problem is solved. The simplification is because the denominator is a definite integral and I didn't understand that at first. Since the result of a definite is just a constant, it can be factored outside the integral.

答: It should read$$\int x\frac{f(x)}{\int_a^bf(y)dy}dx=\frac{\int xf(x)dx}{\int_a^bf(y)dy},$$where the $x$-integral may or may not be definite (though in this context it would be).

ContentLicense: CC BY-SA 4.0


问:
calculate the limit $\lim\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$
I'm working on finding the limit for this equation, and would kindly welcome your support to my solution:

$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$$

These are my steps in hopefully deriving the correct result:

$$\frac{1-\cos \theta}{\theta \sin \theta} = \frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin\theta}=\frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin \theta}=\frac{\sin\theta}{\theta}\cdot\frac{1}{1+\cos\theta}$$

Given that as $\theta$ approaches $0$ then $\dfrac{\sin\theta}{\theta}$ $\approx$ $0$ and $\cos\theta \approx 1$. I thought the answer would be $0$ given $\sin{\theta}$ tends to $0$ but the answer is $\frac{1}{2}$. Can someone kindly explain and show this to me? Maybe I went wrong in my reasoning or calculation?

EDIT:
Given that:

$\cos \theta < \frac{\theta}{\sin \theta} < \frac{1}{cos \theta} = \sec \theta$

because $\frac{\theta}{\sin \theta}$ is between two variables approaching 1, then it must also approach 1.

hence, $\frac{\sin \theta}{\theta}$ approaches 1.

So the answer is $\frac{1}{2}$

答:
Note that $1-\cos\theta=2\sin^2\frac{\theta}2$:
$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}=\lim_{\theta\to0}\frac{2\sin^2\frac{\theta}2}{\theta \sin\theta}$$
Also we have $\sin u\sim u$ when $u\to0$ so:

$$\lim_{\theta\to0}\frac{2\times(\frac{\theta}2)^2}{\theta\times\theta}=\frac12$$


问:
Prove that if $S$ is subspace then there must exist $Ax=0, x \in S$
I would like to show the first direction of the following statement.

$S$ is a subspace if and only if there exists a matrix $A$ s.t. $S = \{x \in \mathbb{R}^n: Ax =0 \}$.

Definition: $S \subset \mathbb{R}^n$ where $S\neq \emptyset$  is called a subspace if $x \in S$ implies $\lambda x \in S, \forall \lambda \in \mathbb{R}$ and $x,y \in S$ implies $x+y \in S$. In addition,

$``\Leftarrow"$If there exists a matrix $A$ s.t. $S = \{x \in \mathbb{R}^n: Ax =0 \}$, then we can select $x, y \in \mathbb{R}^n$ as the solutions for $Ax =0$. It is clear to show that:

$A(x+y) = Ax + Ay = 0$ and $A(\lambda x) = \lambda (Ax) = 0$. Hence, $S$ is a subspace.

How can I show the other direction? If I assume that $S$ is a subspace, how can I show that there must exists a matrix $A$?

答:
Hint.

Suppose $S$ is a subspace of $\mathbb{R}^n$. Then the exists a basis $\beta$ of $S$. Extend $\beta$ to be a basis $\alpha$ of $\mathbb{R}^n$.

Define $A$ to be a linear transformation on $\mathbb{R}^n$ such that $Av=0$ for every $v\in\beta$ and $Av=v$ for every $v\in\alpha\setminus\beta$. Extend $A$ linearly.


问:
Calculate $\iint_D \frac{\sqrt{x^2 + y^2}}{1+x^2+y^2} dx\,dy$
I am attempting to solve the integral: $$\iint_D \frac{\sqrt{x^2 + y^2}}{1+x^2+y^2} dx\,dy\,,$$ where $D$ is bounded by $1<x^2+y^2<9$ and within the sector bounded by lines $$-\frac{y\sqrt{2}}{2}++\frac{x\sqrt{2}}{2}=0$$ and $$\frac{y\sqrt{3}}{2}+\frac{x}{2}=0$$ The illustration of the sector is shown below:

I tried by first calculating the angles between which the sector is bounded and switching to polar coordinates. However that didn't seem to give me the correct answer. I'm not very used to working with polar coordinates so I might have missed something simple along the way. My calculations are shown below:

I'm grateful for any help, thanks in advance!

/Nick

答: Your upper bound of $\theta=-\frac{\pi}{3}$ should instead be $\theta=-\frac{\pi}{6}$. Look how the red ray in the graph you provided makes an angle of $\theta=-30^\circ$ with the positive $x-$ axis. Therefore $$\iint_{D}\frac{\sqrt{x^2+y^2}}{1+x^2+y^2}dA=\int_{-3\pi/4}^{-\pi/6} \int_1^3\frac{r^2}{1+r^2}drd\theta=\frac{7\pi}{12}\Big[2-\arctan(3)+\frac{\pi}{4}\Big]$$


问:
Differentiate $g(t)= {e^t - e^{-t} \over e^t + e^{-t}}$
I'm having some trouble trying to differentiate the function $g(t)= \dfrac{e^t - e^{-t}}{e^t + e^{-t}}$

Can someone help me? Thanks a lot!

答:
You can cancel the terms in the numerator and denominator: the numerator is exactly the same function as is the denominator.

EDIT:  Okay...now it makes more sense! ;-)

Recall the quotient rule: $$g(t) = \frac{f(t)}{h(t)}$$

$$g'(t) = \frac{f'(t)h(t) - h'(t)f(t)}{[h(t)]^2}\tag{1}$$

In your case, $$g(t) = \frac{e^t - e^{-t}}{e^t - e^{-t}} = \frac{f(t)}{h(t)}$$

$f'(t) = e^t + e^{-t},\; h'(t) = e^t - e^{-t}$, so using the quotient rule $(1)$, find $g'(t)$ and simplify.

ContentLicense: CC BY-SA 3.0


答:
call $Q= x f(x) $, then our integral is:

$$ \int \frac{Q}{\int f(x) dx} dx $$

Now integrating by parts:

$$ \frac{ \int Q dx}{\int f(x) dx} +\int \frac{\int Q dx}{ ( \int f(x) dx)^2} f(x) dx$$

For the above to be equal to, $ \frac{ \int x f(x) dx}{\int f(x) dx} i.e: \frac{\int Q dx}{\int f(x) dx}$,

$$ \int \frac{ \int Q dx}{ (\int f(x) dx )^2} f(x) dx= 0$$

So, the equality you wrote is only true for functions $f(x)$ which satisfy above criteria.

Title: How can we simplify this integral? $\int{x\frac{f(x)}{\int f(x) dx} dx}$

Body:
In a machine learning lecture, we encountered the following integral that we needed to solve to calculate the mean of some random variable $x$:

\begin{equation*}
  \int{x\frac{f(x)}{\int f(x) dx} dx}
\end{equation*}

Without really explaining, the professor just simplified it to this:

\begin{equation*}
  \frac{\int x f(x) dx}{\int f(x) dx}
\end{equation*}

I'm not sure how that works. Since the integration is without limits, then the result is a function not a constant, right? It can't be factored out as if it were a constant. Am I missing something? Does integrating on the same variable twice have any special properties that are relevant here?

I'm sorry if the question is lacking in details, if there's anything I can edit to make it clearer, please let me know.

Edit: The problem is solved. The simplification is because the denominator is a definite integral and I didn't understand that at first. Since the result of a definite is just a constant, it can be factored outside the integral.


问:
Is the set S closed?
Let $S = \{(x_1,x_2):(x_1-1)^2+x_2^2=1 \}$

I have to show that S is closed. I searched a lot about how to prove that a set is closed but was not able to understand much. To be more specific, I did not get the intuitive grasp of the method that uses sequences to show that a set is closed. In general, how do we show that sets of similar definition as the one defined above are closed?

答:
The easiest way to show such algebraically defined sets to be closed is to use a function which is continuous.

Define $f: \Bbb R^2 \to \Bbb R$ by $f(x,y)= (x-1)^2 +y^2$. $f$ is continuous as $\Bbb R$ is a topological ring.

And $S=f^{-1}[\{1\}]$ is closed as the inverse image of a closed set under a continuous map.

Title: Is the set S closed?


答: If $f(x_1,x_2)=(x_1-1)^2+x_2^{\,2}$, then $f$ is continuous. Since $\{1\}$ is closed subset of $\Bbb R$, it follows that $S$ is closed, since $S$ is equal to $f^{-1}\big(\{1\}\big)$.

Title: Is the set S closed?

Body:
Let $S = \{(x_1,x_2):(x_1-1)^2+x_2^2=1 \}$

I have to show that S is closed. I searched a lot about how to prove that a set is closed but was not able to understand much. To be more specific, I did not get the intuitive grasp of the method that uses sequences to show that a set is closed. In general, how do we show that sets of similar definition as the one defined above are closed?


问:
A problem with the equivalent form of Frankl conjecture.
The Union Closed Sets Conjecture (or Frankl conjecture) is described in this link: https://en.wikipedia.org/wiki/Union-closed_sets_conjecture.

It is known that this problem has an equivalent form in terms of graphs.

Namely: Frankl conjecture is equivalent to assertion that in any graph $G=(V,\mathcal{E})$, such $|\mathcal{E}|\ge 1$, there exists two adjacent vertices that each of them belongs to at most half of the maximal independent sets.

But here is the problem:

Imagine the following graph $G=(V,\mathcal{E})$. Where :

$V = \{1,2,3\}$ and $\mathcal{E}=\{\{1,2\},\{2,3\}\}$

The only maximal independent set is $\{1,3\}$

Let's take any two vertices, for instance $1,2$. We see that $1\in\{1,3\}$ so "$1$" belongs to more than half of the maximal independent sets. That contradicts the conjecture.

However i strongly believe that i made some mistakes.

Where is mistake?

Thank you in advance.

答: $\{2\}$ is also a maximal independent set - no vertices can be added to it without creating an edge.


问:
Struggling to determine the bounds for this volume by integration problem
I have
$$y=|x|+1, 2x+3y=6\text{ about }x=-20.$$

I need to set up an integral (but not evaluate it) that will give the volume of the solid when the region enclosed by those curves is revolved about the given line. I have to use either the shell or washer's method.

I have been trying to use the shell method (I think) but I am having trouble setting up the integral. How do I figure out the bounds for it? I'm not sure how I get the numbers for "top" and "bottom" either. I solved for y in the $2x+3y=6$ equation and got $y=-(2/3)x+2$ but how do I know where that belongs in the integral?

Thank you for any help

答: Hint:  If you plot the two curves, they form a triangle.  This is the area that you are revolving, so find the coordinates of the three corners of the triangle.  If you use cylindrical shells, the minimum $y$ value is the start of your integration and the maximum $y$ is the end.  Then for a given $y$ in that range, you need to find the minimum $x$ and maximum $x$.  The minimum will always be along $y=1-x$ (why?) but the maximum will shift from one line to the other part way along, so you probably want to split the integral there.</p>


Title: Differentiate $g(t)= {e^t - e^{-t} \over e^t + e^{-t}}$

Body:
I'm having some trouble trying to differentiate the function $g(t)= \dfrac{e^t - e^{-t}}{e^t + e^{-t}}$

Can someone help me? Thanks a lot!

Body:
$g(t) = \dfrac{a(t)}{b(t)}$

$g'(t)= \dfrac{a'(t)b(t)-a(t)b'(t)}{b^2(t)}$

$g'(t)= \dfrac{(e^t+e^{-t})(e^t+e^{-t})-(e^t-e^{-t})(e^t-e^{-t})}{(e^t+e^{-t})^2}$

$g'(t)= \dfrac{(e^{2t}+2+e^{-2t})-(e^{2t}-2+e^{-2t})}{(e^t+e^{-t})^2} = \dfrac{4}{(e^t+e^{-t})^2}$

By the way,  $g(t)=\tanh(t)$, and   $g'(t)=\dfrac{1}{\cosh^2(t)} = \dfrac{1}{\left(\dfrac{e^t+e^{-t}}{2}\right)^2} = \dfrac{4}{(e^t+e^{-t})^2}$.


问:
A conditional probability problem where the next day depends on the last 3 days
For many years, Meteorologists have spent long visits (5 days) at the Bigtown.
They have observed that, for three consecutive days, if there are EXACTLY two sunny days, the next day is a sunny day*, while half of all the cloudy days are followed by another cloudy day. Assuming days are either cloudy or sunny, estimate how many days have been cloudy in their last ten visits. Visits are not necessarily following each other.

''half of all the cloudy days are followed by another cloudy day'' so $P(C\mid C)=50\%$

Is this possible? If not why not?

*i.e.
SSC -> S

SCS -> S

CSS -> S

where -> means followed by

答:
For when you get a mathematical answer, the following simulations suggest that the answer is around 2.22.</p>
```<code>nclo=0
for (k in 1:100) {
ncloud=0
for (j in 1:10000) {
  chain=rep(0, 10)
  for (i in 1:10) {
    if (i==3) {
      if (chain[1]+chain[2]==2) {
        chain=rep(1, 10)
        break
      }
    }
    if (i>3) {
      if (chain[i-1]+chain[i-2]+chain[i-3]>=2) {
        chain[i]=1
      }
    }
    if (runif(1)>.5) {
      chain[i]=1
    }
  }
  ncloud=ncloud+(10-sum(chain))
}
nclo=nclo+ncloud/10000
}
nclo=nclo/100
</code>```
```<code>> nclo
[1] 2.221082
</code>```

Body:
For when you get a mathematical answer, the following simulations suggest that the answer is around 2.22.

nclo=0
for (k in 1:100) {
ncloud=0
for (j in 1:10000) {
  chain=rep(0, 10)
  for (i in 1:10) {
    if (i==3) {
      if (chain[1]+chain[2]==2) {
        chain=rep(1, 10)
        break
      }
    }
    if (i>3) {
      if (chain[i-1]+chain[i-2]+chain[i-3]>=2) {
        chain[i]=1
      }
    }
    if (runif(1)>.5) {
      chain[i]=1
    }
  }
  ncloud=ncloud+(10-sum(chain))
}
nclo=nclo+ncloud/10000
}
nclo=nclo/100

> nclo
[1] 2.221082


问:
Isometric domains are isospectral for the Laplacian
Let

$X$ and $Y$ be separable Hilbert spaces;

$(e_n)_{n\in\mathbb N}$ and $(f_n)_{n\in\mathbb N}$ be orthonormal bases of $X$ and $Y$, respectively;

$(\lambda_n)_{n\in\mathbb N},(\mu_n)_{n\in\mathbb N}\subseteq[0,\infty)$

$$S(t)x:=\sum_{n\in\mathbb N}e^{-\lambda_nt}\langle x,e_n\rangle_Xe_n\;\;\;\text{for }x\in X\text{ and }t\ge0$$ and $$T(t)y:=\sum_{n\in\mathbb N}e^{-\mu_nt}\langle y,f_n\rangle_Hf_n\;\;\;\text{for }y\in Y\text{ and }t\ge0$$

We can easily show that $(S(t))_{t\ge0}$ is a contractive $C^0$-semigroup on $X$ with generator $$Ax:=-\sum_{n\in\mathbb N}\lambda_n\langle x,e_n\rangle_Xe_n\;\;\;\text{for }x\in\mathcal D(A):=\left\{x\in X:\sum_{n\in\mathbb N}\lambda_n^2\lambda_n^2\left|\langle x,e_n\rangle_X\right|^2<\infty\right\}$$ and the same clearly applies to $(T(t))_{t\ge0}$ (let $B$ denote its generator).

I want to know when we can find an unitary $U\in\mathfrak L(X,Y)$ such that $$\forall t>0:US(t)=T(t)U\tag1.$$ Since I'm not sure how specific this question is, the particular situation I'm interested in is $X=L^2(\Omega)$ and $Y=L^2(\tilde\Omega)$, where $\Omega,\tilde\Omega\subseteq\mathbb R^d$ are bounded and open such that there is an isometry $\iota:\Omega\to\tilde\Omega$, and $(e_n)_{n\in\mathbb N}$ and $(f_n)_{n\in\mathbb N}$ are the orthonormal bases consisting of eigenvectors of $-\Delta$ on $\{u\in H_0^1(\Omega):\Delta u\in L^2(\Omega)\}$ and $\{u\in H_0^1(\tilde\Omega):\Delta u\in L^2(\tilde\Omega)\}$, respectively, and the sequences $(\lambda_n)_{n\in\mathbb N}$ and $(\mu_n)_{n\in\mathbb N}$ are the corresponding eigenvalues.

Unless, I'm missing something, we should be always to find an unitary $U\in\mathfrak L(X,Y)$ with $Ue_n=f_n$ for all $n\in\mathbb N$. This should simply follow from the fact that $V:X\to\ell^2(I),x\mapsto(\langle x,e_i\rangle)_{i\in I}$ and $W:Y\to\ell^2(I),y\mapsto(\langle y,f_i\rangle)_{i\in I}$ are unitary isomorphisms so that we can take $U:=W^{-1}\circ V$. However, in generality I'm not able to deduce $(1)$ from that (using this particular $U$.)

In the particular case of the Laplacian, we could consider $Uf:=f\circ\iota^{-1}$ for $f\in L^2(\Omega)$, which is clearly unitary and hence maps ortrhonormal bases to orthonormal bases. But, again, I'm not able to show $(1)$ for this choice of $U$ either ...

答:
The question seems to be motivated by the "Can you here the shape of a drum" question.

First, let me briefly give a general answer to the question and then explain how it is related to the specific setting of Laplace operators.

General situation.

Let $H_1$ and $H_2$ be two Hilbert spaces and let $A_1: H_1 \supseteq D(A_1) \to H_1$ and $A_2: H_2 \supseteq D(A_2) \to H_2$ be self-adjoint operators with compact resolvent which have finite spectral bound and thus generate $C_0$-semigroups $T_1$ and $T_2$.

Proposition 1.
The following assertions are equivalent:

(i) The operators $A_1$ and $A_2$ have the same eigenvalues, including multiplicities - i.e., they are isospectral.

(ii) There exists a unitary operator $U: H_1 \to H_2$ that intertwines $A_1$ and $A_2$ (equivalently, that intertwines the semigroups $T_1$ and $T_2$).

(iii) There exists a bounded linear bijection $U: H_1 \to H_2$ that intertwines $A_1$ and $A_2$ (equivalently, that intertwines the semigroups $T_1$ and $T_2$).

Sketch of proof.
"(i) $\Rightarrow$ (ii)" Let $(\lambda_n)$ denote the sequence of eigenvalues of both $A_1$ and $A_2$, and let $(e_n)$ and $(f_n)$ be corresponding ONBs of eigenvectors.
Define $Ue_n := f_n$ for each $n$. Then $U$ is a unitary that intertwines $A_1$ and $A_2$.

"(ii) $\Rightarrow$ (iii)" Obvious.

"(iii) $\Rightarrow$ (i)" The fact that $U$ intertwines both operators means precisely that $A_1$ and $A_2$ are similar; hence, they have the same eigenvalues. $\square$

The Laplace operator

Now we specialize to the situation where $H_1 = L^2(\Omega_1)$ and $H_2 = L^2(\Omega_2)$ for two bounded domains $\Omega_1$ and $\Omega_2$ in $\mathbb{R}^d$, und where $A_1$ and $A_2$ denote the Dirichlet Laplace operators on both spaces.

Proposition 2. If there is a Euclidean isometric $\iota: \mathbb{R}^d \to \mathbb{R}^d$ which maps $\Omega_2$ to $\Omega_1$, then $A_1$ and $A_2$ are isospectral.

Sketch of proof. Such a $\iota$ is a composition of a translation and a rotation. Hence, the mapping $U: f \mapsto f \circ \iota$ commutes with the distributional Laplace operator. Moreover, the same mapping maps $H^1_0(\Omega_1)$ to $H^1_0(\Omega_2)$, so it actually intertwines the operators $A_1$ and $A_2$. Hence, isospectrality follows from the implication "(ii) $\Rightarrow$ (i)" in Proposition 1. $\square$

Note that the proof of Proposition 2 has nothing to do with the converse implication "(i) $\Rightarrow$ (ii)" in Proposition 1, i.e., the unitary $U$ in the proof of Proposition 2 is constructed by a very different method than the one in the proof of the implication "(i) $\Rightarrow$ (ii)" in Proposition 1. (But, of course, one can show a posteriori that the unitary $U$ from Proposition 2 maps each eigenspace of $A_1$ to the corresponding eigenspace - for the same eigenvalue - of $A_2$).

Hearing the shape of a drum.

This is a famous problem, made popular by Kac, which asks under which circumstances the converse to Proposition 2 holds. I.e., assume that $A_1$ and $A_2$ are isospectral - does it follows that $\Omega_1$ and $\Omega_2$ they are congruent?

In full generality, the answer has been known to be "no" for a long time. In certain special cases, the answer is known to be "yes".

But for large classes of domains, the problem is still open.

Isothermal drums.

If we are interested in the "Can you here the shape of a drum" problem, why should we care about the implication "(i) $\Rightarrow$ (ii)" in Proposition 1?

Here is an intuitive reason: Assume that $\Omega_1$ and $\Omega_2$ are "good" (for instance, have $C^\infty$-boundary) and that $A_1$ and $A_2$ are isospectral - then Proposition 1 tells us that there exists an operator $U$ as in (ii) (or (iii)). If we could prove that $\Omega_1$ and $\Omega_2$ are congruent, we would have a positive answer to Kac's problem - but as mentioned above, this is open.

Now assume, however, that the intertwining operator $U$ does not only exists but can be chosen such that both $U$ and $U^{-1}$ are positive (in the sense of Banach lattice, i.e., $Uf \ge 0$ iff $f \ge 0$). Then it can indeed be shown that $\Omega_1$ and $\Omega_2$ are congruent; see this paper. Hence, we don't have a solution to Kac's problem, but we have a positive result under additional assumptions on $U$.

(Notational remark: If $U$ and $U^{-1}$ can be chosen to be positive, it is tempting to call $A_1$ and $A_2$ isothermal rather than only isospectral - thus the title of this little subsection here.)


问:
Show that normal distribution has 0 mean and unit variance
For a normal distribution,

$$\phi(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\times e^{\frac{-1(x-\mu)^2}{2\sigma^2}}$$

Question:Prove that the standard normal distribution has 0 mean and unit variance?

答:
Lemmas we will use before we begin:

If $f(x) = -f(-x)$ then $\int_a^a f(x) dx = 0 $ for any $a \in \mathbb{R} $

Mean

$\mathbb{E}[X] = \int_{\mathbb{R}} xf_X(x) dx$

Using lemma above and noticing that $xf_X(x)$ satisfies the property above we are done for zero mean.

Variance

$\mathbb{V}[X] = \mathbb{E}[X^2] - \mathbb{E}[X]^2 = \mathbb{E}[X^2]$ as $\mathbb{E}[X] = 0 $

Hence $\mathbb{V}[X] = \int_{\mathbb{R}} x^2f_X(x) dx = 1$


问:
Existence of smooth partitions of unity without open condition
This is a problem from Evans Chapter 5.

We assume a bounded $U$ and $U \subset \subset \cup_{i=1}^N V_i$ and want to prove the existence of $C^{\infty}$ functions $\zeta_i (i=1 \;to \; N)$ such that
\begin{align*}
\begin{cases} 0 \leq \zeta_i \leq 1, \; \text{support of } \zeta_i \subset V_i \; (i=1,...,N) \\
\sum_{i=1}^N \zeta_i = 1 \; \; \text{ on }\;  U.
\end{cases}
\end{align*}

I think I should use the following preceding exercise, which gives the existence of a smooth $\zeta$ uniformly 1 on a smaller open set compactly contained in a bigger one, where $\zeta$ disappears in some neighbourhood of $\partial U.$

Let $U$, $V$ be open sets, with $V\subset\subset U$. Show there exists a smooth function $\zeta$ such that $\zeta\equiv1$ on $V$, $\zeta=0$ near $\partial U$. (Hint: Take $V\subset\subset W\subset\subset U$ and mollify $\chi_W$.)

For the problem in the image, I have a nice space to work with mollifiers and the smooth partition is there since $U, V$ were open. But now I don't have any openness involved in this problem, which is my concern. I can't even imagine how the domain of each partition should look like. How do we restrict a support of a smooth function to $V_i$ and at the same time guarantee that the partition doesn't disappear on any $U \cap \partial V_i$?

I looked through many other problems concerning this existence of smooth partition of unity, but they all needed open domains.

答:
It seems to me that you can w.l.o.g. assume that $U$ is open and all the $V_i$ are open.
In other words (making this wlog a bit more formal) you can just take the interiors $\mathring{U}$ and $\mathring{V}_i$ and solve the problem with the interiors.

Once you have solved the problem with the $\mathring{U}$ and the $\mathring{V}_i$ instead of $U$ and $V$ you have found functions $\zeta_i \in C^{\infty}$ such that
$$
\begin{cases}
0 \leq \zeta_i \leq 1\\
\text{supp}(\zeta_i) \subset \mathring{V}_i\\
\sum_{i} \zeta_i = 1 \text{ on } \mathring{U}.
\end{cases}
$$
Now you can use continuity of the $\zeta_i$ and $\sum_{i} \zeta_i$ to conclude that you actually solved the problem for the original $U$ and $V_i$.

Solving the problem for the interiors can be done in a few steps (I think!! but it has been some time since I did this proof, but it can be found online probably, but some hints might be helpful if you want figure it out without full solutions first):

First conclude that $U \subset \subset \cup_i V_i \quad$ also holds when you  substitute $U$ and $V_i$ with the interiors.

Use problem 5 in your image to first find $\xi_i$ for every $V_i$ (actually the interior), this can be done because you are solving the problem with the interiors (which are of course open!)

Some tricks to define $\zeta_i$ using a smart division of the form $\zeta_i := \frac{\xi_{i}}{\sum_{j} \xi_j}$.

Edit:
An update after your comment.
Indeed, we have $\cup_i(\text{int}(V_i)) \subset \text{int}(\cup_i(V_i))$, but in general not the other way around, so your comment is valid and I did not think of this beforehand.

Luckily however, things work out in the following way:
Suppose we have a point $x \in  \text{int}(\cup_i(V_i)) - \cup_i(\text{int}(V_i))$ and $x \in \overline{U}$.
This means that  $x \not \in \text{int}V_i$ for all $i \in \{1,\dots,n\}$ and $x \in \partial V_i$ for at least one $i$.

However, in this case a partition of unity subordinate to $\{V_i\}_i$ cannot exist: Indeed for arbitrary $i \in \{1,\dots,n\}$ we have $\zeta_i$ with support in $V_i$ which means that $\zeta_i(x) = 0$, since either $x \in \partial V_i$ or $x \not \in V_i$.
This implies we have $\sum_i \zeta_i(x) = 0.$

However, since $x \in \overline{U}$ we get by continuity $\sum_i \zeta_i(x) = 1$, a contradiction.
So we are not interested in this case, since no interesting partition of unity exists, so we can wlog assume that $x \in \overline{U} \implies x \in int(V_i)$ for some $i$, and I believe this solves the problem you correctly stated in your comment.


问:
Stationary Distribution of a Stochastic Processes
A Markov Chain with states 0,1,... has transition probabilities
$$p_{jk}=e^{-a} \sum_{r=0}^k \left( \begin{matrix} j \\ r \end{matrix} \right) p^r (1-p)^{j-r} a^{k-r} / (k-r)!$$
Show that the limiting distribution $\{ v_k\}$ is Poisson with parameter $a/(1-p)$.

My attempt was using:

$$ \sum_{k=0}^\infty p_{jk} = 1 $$ and

$$ v_k = \sum_{j=0}^\infty v_j p_{jk} $$

But I am really out of order on how to simplify

$$ v_k = \sum_{j=0}^\infty v_j e^{-a} \sum_{r=0}^k \left( \begin{matrix} j \\ r \end{matrix} \right) p^r (1-p)^{j-r} a^{k-r} / (k-r)! $$

答:
The Markov chain, let's call it $Z=(Z_n)_{n\ge 0}$, looks to be a branching process with immigration. The offspring distribution is Bernoulli with mean value $p$ and the immigration distribution is Poisson with mean $a$. That is to say,
$$
Z_{n+1} \stackrel{d}{=} Y_{n+1}+\sum_{k=1}^{Z_n}\xi_k^{(n)},
$$
where $Y_{n+1}$ is Poisson($a$), the $\xi_k^{(n)}$ are are Bernoulli($p$), and all the $Y_n$s and $\xi_k^{(n)}$s are independent.

If there is to be a stationary distribution for this chain, then a  random variable $X$ with that distribution will satisfy the distributional identity
$$
X  \stackrel{d}{=}  Y+\sum_{k=1}^X\xi_k,\qquad\qquad (1)
$$
where on the right, $Y$, $X$, and the $\xi_k$s are independent and have the same distributions as their earlier discussed namesakes.

Using generating functions it's not hard to see that if $X$ is Poisson($a/(1-p)$) then (1) is true.

Because the branching process with Bernoulli($p$) offspring distribution is subcritical and $P(Y=0)>0$, the Markov chain $Z$ is recurrent.
Because the associated invariant measure equation $v=v{\bf P}$ has a solution $v$ with finite total mass (namely, the Poisson($a/(1-p)$) distribution), the process $Z$ is in fact positive recurrent (and clearly aperiodic). It follows from Markov chain theory that
$$
\lim_n P(Z_n=k) = e^{-\mu}{\mu^k\over k!},\qquad k=0,1,2,\ldots,
$$
where $\mu=a/(1-p)$.


答:
Let $(a_n,b_n)$ be a convergent sequence in $S$. Let $(a,b)$ be the limit i.e. $\lim_{n\to\infty} a_n = a$ and $\lim_{n\to\infty} b_n = b$.

Then $(a_n-1)^2 + b_n^2 = 1$ and the function $f(x,y)=(x-1)^2+y^2$ is continuous, thus
$1 = \lim_{n\to\infty} (a_n-1)^2 + b_n^2 = (a-1)^2+b^2$ which shows that $(a,b)\in S$.
We have shown that for any convergent sequence in $S$ the limit is also contained in $S$, thus $S$ is closed.

Title: Is the set S closed?


问:
Hard improper integral $\int_0^\infty \frac{(1+\frac{1}{x})^{-x}}{x^2}dx $
I have been stuck on this problem for a while:
$$\int_0^\infty \frac{(1+\frac{1}{x})^{-x}}{x^2}dx $$

I have toyed around with substitution and integration by parts with no luck. I have also searched for this on the internet but nothing has come up. Any help would be appreciated

答:
As mentioned, the integral as written diverges. It is easy to see that the controlling factor of $\int\limits_a^\infty dx \ x^{-2} (1+1/x)^{-x}$ is $1/a$ as $a \to 0$. By itself this is not a good approximation. I'm going to consider the integrals

$$
I(a)=\int\limits_a^1 dx \ x^{-2} (1+1/x)^{-x} \\
J(b)=\int\limits_1^b dx \ x^{-2} (1+1/x)^{-x}
$$

For $I(a)$, expand the term $(1+1/x)^{-x}$ around $x=0$, and then integrate term by term

$$
I(a)=\int\limits_a^1 dx \ x^{-2} \left(1+x \ln(x)+ \dots \right) \\
I(a) \sim \frac{1}{a}-1-\frac{1}{2}\ln(a)\ln(a) \ \ , \ \ a \to 0
$$

Here is a plot of the two term series versus the numerical result for $I(a)$:

For $J(b)$, I propose replacing $(1+1/x)^{-x}$ with a Pade approximant around $x=1$, where the integrand is maximum. I'm using $P^1_1$. Let $\mathcal{J}$ be our approximation for $J$

$$
\mathcal{J}(b)=\int\limits_1^b dx \ x^{-2} \left( \frac{1+\alpha(x-1)}{2+\beta(x-1)} \right)
$$

Where $\alpha= \frac{\ln(2)(\ln(2)-1)}{\ln(4)-1}$, $\beta=\frac{2\ln(2)\ln(2)}{\ln(4)-1}-1$, I'll give the indefinite integral here, so I only have to write one awful equation instead of two

$$
\mathcal{J}(b)=\frac{-(\alpha-1)(\beta-2) + x(2\alpha-\beta)(\ln(x)-\ln(2+\beta(x-1))}{x(\beta-2)^2}\Bigg\vert_1^b
$$

Here is a plot of $\mathcal{J}$ versus the numeric result:


问:
System of Equations İnvolving Floor Function
The Question

Solve for $(x,y,z)$

$$\left\{\begin{matrix} x+\left \lfloor y \right \rfloor + \left \{ z
> \right \}=1,1\\  z+\left \lfloor x \right \rfloor + \left \{ y \right
> \}=2,2\\  y+\left \lfloor x \right \rfloor + \left \{ z \right \}=3,3
 \end{matrix}\right.$$

My Understanding

From the definition of floor function we know that $n = \left \lfloor n \right \rfloor + \left \{n  \right \}$ and if you play with equation a bit $(1,1+,2,2=3,3)$ and do some cancelling it's easy to see that

$$x+z=0 \Rightarrow x = -z.$$

After that you set $z=-x$ and rewrite the equation. You'll see from $[2(1,1)=2,2]$ and a bit cancelling $\left \{ y \right \} - \left \{ x \right \}=2,2$. Also an important note that $\left \lfloor x \right \rfloor \neq \left \lfloor -x \right \rfloor , \left \{ x \right \} \neq \left \{ -x \right \}$ example: $\left \lfloor 5,35 \right \rfloor = 5$ but $\left \lfloor -5,35 \right \rfloor = -6$. I got this so far and need help.

答:
These equations have no solution since the LH sides of equations (2) and (3) have the same fractional part, formed from the sum of the fractional parts of $y$ and $z$, and yet differ by $1.1$.

N.B. Looking at the comment of @cosmo5 I suspect that the last equation should be as in that comment. Then there is the solution $x=0.1,y=1.2,z=2$.


问:
Linear extensions that preserve norm of linear functional.
$(a)$ Let $\mathbb{X}$ a normed space and $f \in Z^{\prime},$ where $Z$ is subspace of $\mathbb{X}.$ If $f$ have two distinct extensions $f_{1},f_{2}\in\mathbb{X}^{\prime}$ that preserve the norm of $f,$ show that $f$ have infinity linear extensions that preserve norm of $f.$

$(b)$ Consider the space $\mathbb{R}^{2}$ with norm $\|(x, y)\|=\max \{|x|,|y|\} .$ In subspace $Z=\{(t, t): t \in \mathbb{R}\}$ consider the linear functional $f:Z\to\mathbb{R}$ given by $f(t,t)=t.$ Find two distinct linear extensions defined in $\mathbb{R}^{2}$ that preserve norm of $f.$

Can anybody help me? I had no idea

答:
For (a), consider convex combinations.

For (b), the norm of $f$ is 1. The obvious extensions are $f(x,y)=x$ and $f(x,y)=y$. Do these work?


答: Note that $g(t) = \frac{\sinh(t)}{\cosh(t)}$. You can use quotient rule and remember $\sinh'(t) = \cosh(t), \cosh'(t) = \sinh(t), \cosh^2(t) - \sinh^2(t) = 1$.</p>


问:
What is the single equation for a helix?
Is there a way to describe a helix not by its parametric form

$$ x=R\cos(t) ,\ y=R\sin(t) , \ z=ht , $$

but by a single equation like you can for a sphere with $ r^2 = x^2+y^2+z^2 $?

Also the same question for a 3-dimensional tube that follows a helix curve.

答:
We can have asymmetric equation in terms of x, y and z:

To get such equation consider equation of a straight line in three dimension in form of symmetric:

$\frac{x-x_0}m=\frac{y-y_0}n=\frac{z-z_0}l$

To get parametric equation we can write:

$\frac{x-x_0}m=\frac{y-y_0}n=\frac{z-z_0}l=t$

$x=mt+x_0$, $y=nt+y_0$, $z=lt+z_0$

Clearly reverse procedure gives equation of line in terms of x, y and z. We apply this procedure to parametric equation of helix:

$x=R \cos(t)\Rightarrow t=\cos^{-1}\frac xR$

$y=R\sin (t)\Rightarrow t=\sin^{-1}\frac yR$

$z=ht \Rightarrow t=\frac z h$

Therefore symmetric equation for helix can be:

$\cos^{-1}\frac xR=\sin^{-1}\frac yR=\frac zh$


答:
Here it is a slightly different approach from @amirali.

According to the trigonometric identity $\sin(2x) = 2\sin(x)\cos(x)$, we conclude that
\begin{align*}
\lim_{\theta\to0}\frac{1-\cos(\theta)}{\theta\sin(\theta)} & = \lim_{\theta\to 0}\frac{2\sin^{2}(\theta/2)}{2\theta\sin(\theta/2)\cos(\theta/2)}\\\\
& = \lim_{\theta\to 0}\left[\frac{1}{2}\times\frac{\sin(\theta/2)}{\theta/2}\times\frac{1}{\cos(\theta/2)}\right] = \frac{1}{2}
\end{align*}
where we have applied the continuity of $\cos(x)$ at $x = 0$ as well as the following result:
\begin{align*}
\lim_{x\to0}\frac{\sin(x)}{x} = 1
\end{align*}

Hopefully this helps!

Title: calculate the limit $\lim\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$

Body:
I'm working on finding the limit for this equation, and would kindly welcome your support to my solution:

$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$$

These are my steps in hopefully deriving the correct result:

$$\frac{1-\cos \theta}{\theta \sin \theta} = \frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin\theta}=\frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin \theta}=\frac{\sin\theta}{\theta}\cdot\frac{1}{1+\cos\theta}$$

Given that as $\theta$ approaches $0$ then $\dfrac{\sin\theta}{\theta}$ $\approx$ $0$ and $\cos\theta \approx 1$. I thought the answer would be $0$ given $\sin{\theta}$ tends to $0$ but the answer is $\frac{1}{2}$. Can someone kindly explain and show this to me? Maybe I went wrong in my reasoning or calculation?

EDIT:
Given that:

$\cos \theta < \frac{\theta}{\sin \theta} < \frac{1}{cos \theta} = \sec \theta$

because $\frac{\theta}{\sin \theta}$ is between two variables approaching 1, then it must also approach 1.

hence, $\frac{\sin \theta}{\theta}$ approaches 1.

So the answer is $\frac{1}{2}$


问:
How is the negation operator a connective?
While going through a lot of notes online on propositional logic, I've come to the conclusion that people generally refer to the $\neg$ (negation) operator as a connective.

However, strictly speaking, I think that a connective is any operator that connects two propositions; and by that means the negation operator is not a connective.

So is it technically wrong to say that the negation operator is a connective or is it correct?

答:
While the etymology for "connective" does suggest two inputs, the term is used more broadly in this technical context: negation is considered a connective despite having a single input, and higer-arity connectives are also considered (see e.g. here).

This happens quite frequently in (and outside as well) mathematics; ultimately, you just need to accept the standard usage, even if it's somewhat (or even extremely) odd.


问:
How can we simplify this integral? $\int{x\frac{f(x)}{\int f(x) dx} dx}$
In a machine learning lecture, we encountered the following integral that we needed to solve to calculate the mean of some random variable $x$:

\begin{equation*}
  \int{x\frac{f(x)}{\int f(x) dx} dx}
\end{equation*}

Without really explaining, the professor just simplified it to this:

\begin{equation*}
  \frac{\int x f(x) dx}{\int f(x) dx}
\end{equation*}

I'm not sure how that works. Since the integration is without limits, then the result is a function not a constant, right? It can't be factored out as if it were a constant. Am I missing something? Does integrating on the same variable twice have any special properties that are relevant here?

I'm sorry if the question is lacking in details, if there's anything I can edit to make it clearer, please let me know.

Edit: The problem is solved. The simplification is because the denominator is a definite integral and I didn't understand that at first. Since the result of a definite is just a constant, it can be factored outside the integral.

答:
I think maybe it is meant to read something like:
$$\int_a^bx\frac{f(x)}{\int_a^bf(x)dx}dx$$
as then the bottom integral is a constant and you can bring it outside:
$$\frac{1}{\int_a^bf(x)dx}\int_a^bxf(x)dx=\frac{\int_a^bxf(x)dx}{\int_a^bf(x)dx}$$

Or maybe if it is:
$$\int x\frac{f(x)}{\int f(x)dx}dx$$
then you can define:
$$F(x)=\int f(x)dx$$ so you get:
$$\int x\frac{F'(x)}{F(x)}dx=\int\frac{F^{-1}(u)}{u}du$$
I don't really understand the notation but its a suggestion


问:
Finding a base of matrices of rank 1
I came up with an interesting question from linear algebra that I couldn't resolve. I am interested to find out an algorithm, a characterization or perhaps a proof that only one case is possible. Of course, if some other branch of mathematics, more specialized than linear algebra, is suited better for this problem, it will useful to tell what is it.

The statement of the problem:

Let $n>1$ be a natural number. Let $\mathbb{F}$ be some algebraically closed field. Let $M_1,M_2,\dots,M_n$ be $n$ matrices over $\mathbb{F}$ of size $n\times n$ that are linearly independent, meaning that the linear closure $L=\mathcal{L}(M_1,\dots,M_n)$ (linear space span over this set of matrices) has dimension $n$. What can we say about dimension $m$ of linear subspace that is generated by matrices of rank $1$?  Namely, the dimension of $\mathcal{L}\{M\in L\mid\text{rank}M=1\}$?

I believe that in the case of $n=2$, this rank has to be always $2$. However, I couldn't prove it. Also, for other values of $n$, does it still hold that $m=n$? What are all the possible pairs $(m,n)$? Is there perhaps an algorithm or method to calculate that dimension $m$ given the matrices $M_1,M_2,\dots,M_n$?

答:
I’m turning my comment into answer. I’ll show that any pair $n \geq m > 0$ is possible. I’m unsure whether $m=0$ is possible (user10354138’s answer requires non-algebraically closed fields). The argument works in any characteristic, but it’s simpler, as a first reading, to assume that we’re in characteristic zero.

So we consider a Frobenius matrix $P$ with characteristic polynomial $\chi$, and take $M_i=P^{i-1}$.

To study the rank of matrices of $L$ (ie the $f(P)$, $f$ being a polynomial of degree less than $n$) it is easier to pretend they’re endomorphisms of some $\bigoplus_{\lambda}{E_{\lambda}}$ where the $\lambda$ are roots of $\chi$, $E_{\lambda}$ is stable under each endomorphism and has dimension $d_{\lambda}$, and $p_{\lambda}=P_{|E_{\lambda}}-\lambda I$ has nilpotency index $d_{\lambda}$.

Let $f$ be a polynomial of degree less than $n$, then the ranks $r_{\lambda}$ of $f(P)_{|E_{\lambda}}$ add up to the rank of $f(P)$. Moreover, if $\chi(\lambda)=0$, then $f_{\lambda}=f(P)_{|E_{\lambda}}=\sum_{k \geq 0}{(\frac{f^{(k)}}{k!})(\lambda)p_{\lambda}^k}$. Thus, as $p_{\lambda}$ is nilpotent, the rank of $f_{\lambda}$ is $d_{\lambda}-k$, where $k$ is the smallest integer such that $f^{(k)}/k!$ doesn’t vanish at $\lambda$, (or more precisely the minimum of such a $k$ and $d_{\lambda}$), ie the vanishing order of $f \wedge \chi$ at $\lambda$.

Thus the rank of $f(P)$ is $\sum_{\chi(\lambda)=0}{d_{\lambda}-v_{\lambda}(f)}$ where $v_{\lambda}(f)$ is the vanishing order of $f \wedge \chi$ at $\lambda$.

Thus, $f(P)$ has rank at most one iff for some root $\lambda$ of $\chi$, $\chi(x)|f(x)(x-\lambda)$, and in that case $f(P)$ is zero on $E_{\mu}$ if $\mu \neq \lambda$, but $f(P)$ acts as $(f^{(d_{\lambda}-1)}/(d_{\lambda}-1)!)(\lambda)p_{\lambda}^{n-1}$ on $E{\lambda}$.

From this, it follows that the subspace of $L$ generated by the matrices of rank one has the $\frac{\chi(x)}{x-\lambda}(P)$ as a basis so has dimension $m$.

Title: Finding a base of matrices of rank 1


问:
Natural Deduction proof for $\forall x \neg A \implies \neg \exists xA$
$\forall x \neg A \implies \neg \exists xA$

I won't ask you to solve this for me, but can you please give some guiding lines on how to approach a proof in NDFOL?

There are many tricks that the TA shows in class, that I could not dream of...

P.S. I managed to proof $\neg \exists xA \implies \forall x \neg A$ but could not get on from there.

Thanks!

After the proposed answer, let me see if I got this correct:

$\exists x A \implies \exists x A$ (axiom assumption)

$\exists x A \implies A$ (from 1)

$\exists x A, \forall x \neg A \implies \forall x \neg A$ (axiom assumption)

$\exists x A, \forall x \neg A \implies \neg A$ ($\forall$ extraction, from 3)

$\forall x \neg A \implies \neg \exists x A$ (from 2,4)

Am I correct?

I could not understand the justification going from (1) to (2)

答:
Stated in a slightly different way from LF...

$\forall x \neg A(x)$  (Assume)

$A(y)$  (Assume)

$\neg A(y)$  (Universal Specification, line 1)

$A(y) \wedge \neg A(y)$ (Join, lines 2, 3)

$\neg\exists x A(x) $ (Conclusion, line 2)

$\forall x \neg A(x) \rightarrow \neg\exists x A(x)$ (Conclusion, line 1)


问:
Prove that there exist integers $k, l \in \mathbb{N}$ such that $\mathrm{ord}(x^ky^l) = \mathrm{lcm}(\mathrm{ord}(x), \mathrm{ord}(y))$
Let $x, y \in G$, where $G$ is a group, $\operatorname{ord}(x), \operatorname{ord}(y) \in \mathbb{N} < \infty$, and $xy=yx$ (seems like it's very important).

I want to prove that there exist integers $k, l \in \mathbb{N}$ such that $\operatorname{ord}(x^ky^l) = \operatorname{lcm}(\operatorname{ord}(x), \operatorname{ord}(y))$.

I know these two facts:

If $\operatorname{gcd}(\operatorname{ord}(x), \operatorname{ord}(y)) = 1 \Rightarrow \operatorname{ord}(xy) = \operatorname{ord}(x)\operatorname{ord}(y)$.

For $g \in G$ and $n \in \mathbb{N}$, $\operatorname{ord}(g^n) = \frac{\operatorname{ord}(g)}{\operatorname{gcd}(\operatorname{ord}(g), n)}$.

Looks like these two facts can help me, but I have no idea how.

And why is $xy=yx$ so important? I tried to consider infinite non-abelian groups (For example $2\times2$ matrices), but I couldn't find any counterexamples. Maybe you can help me.

答:
Let: $ord(x)=m$, $ord(y)=n$, $d=gcd(m,n)$, then:

$$m=\overline{m}d, n=\overline{n}d,$$ with $gcd(\overline{m},\overline{n})=1$.

The idea is to find a partition $d=d_1d_2$, such that $gcd(d_1\overline{m},d_2\overline{n})=1$.

From fact $2)$, there exist $\overline{x} \in G$, with $ord(\overline{x})=d_1\overline{m}$, and $\overline{y} \in G$, with $ord(\overline{y})=d_2\overline{n}$. Due to fact $1)$, it gives us: $$ord(\overline{x}\overline{y})=ord(\overline{x})ord(\overline{y})=d\overline{m}\overline{n}=lcm(m,n).$$

In fact, $\overline{x}=x^{d/d_1}$ and $\overline{y}=y^{d/d_2}$, so we can take $k:=d/d_1$ and $l:=d/d_2$, and we are done.

Now, returning to the multiplicative partition of $d$:

Take $d=p_1^{\alpha_1}...p_h^{\alpha_h}$ its decomposition into primes. For each $p_i^{\alpha_i}$, $1\leq i \leq h$, we have three cases:

$i)$ $p_i\mid \overline{m}$, then $d_1$ will contain $p_i^{\alpha_i}$, whereas $p_i\nmid \overline{n}$;

$ii)$ $p_i\mid \overline{n}$, then $d_2$ will contain $p_i^{\alpha_i}$, whereas $p_i\nmid \overline{m}$;

$iii)$ $p_i\nmid \overline{m}$ and $p_i\nmid \overline{n}$, then either $d_1$ or $d_2$ will contain $p_i^{\alpha_i}$.


问:
What is the probability of randomly generating a tautology?
Suppose we randomly generate a classical Hilbert propositional calculus formula $F$ with $n$ variables, using the following method:

$F = x_i$ for each of $i \leq n$ with probability $\frac{1}{n+2}$.

$F = \neg F_1$, where $F_1$ is generated independently using the same method.

$F = F_1 \to F_2$, where $F_1$ and $F_2$ are generated independently using the same method.

It follows from the extinction criterion for the Galton-Watson branching processes, that the process of generation will terminate with probability $1$ and thus our random formula is well defined.

My question is:

What is the probability that $F$ is a tautology?

It is clearly less, than $\frac{2}{n+2}$. However, it is clearly greater, than $\frac{n}{(n + 2)^3}$ which is the probability of generating a formula of the form $x_i \to x_i$.

答:
The probability that a formula generated that way  is a tautology is

$$\frac{1}{n^2} + O(\frac{1}{n^3})$$

for large $n$.

To demonstrate this, let's call the probability that our formula is always true as $x$ and that our formula is always false as $y$.

Notice the following facts:

$x_i$ is sometimes true and sometimes false.

$\neg A$ is always true iff $A$ is always false.

$\neg A$ is always false iff $A$ is always true.

$x_i \to A$ is always true iff either $A = x_i$ or $A$ is always true.

$x_i \to A$ is true when $x_i$ is false (which sometimes happens)

$A \to x_i$ is always true iff either $A = x_i$ or $A$ is always false.

$A \to x_i$ is true when $x_i$ is true (which sometimes happens)

THe probability of generating any formula of type different from above mentioned ones is $O(\frac{1}{n^3})$

From this we can derive the following system of equations:

$$x = \frac{y}{n+2} + \frac{n}{n+2}(\frac{n}{(n+2)^2} + x + y) + O(\frac{1}{n^3})$$

$$y = \frac{x}{n+2} + O(\frac{1}{n^3})$$

When solving this system of equations we get $x = \frac{1}{n^2} + O(\frac{1}{n^3})$, Q.E.D.


问:
Comparison of degrees of field extensions
Let $M$ be a field, and let $F$ and $K \subset L$ be subfields of $M$. Do we have the inequality $$[L \cap F \colon K \cap F] \leq [L \colon K] \, \text{?}$$ Using the multiplicativity of degrees, it is equivalent to $$[K \colon K \cap F] \leq [L \colon L \cap F] \, \text{.}$$ If the inequality holds, is the divisibility relation also true? If it does not, is the inequality true under some mild assumptions?

答:
In general this inequality doesn't hold. Consider the case where $L=FK$ is the compositum of $F$ and $K$. Then we have $[L:L\cap F]=[FK:F]\leq [K:K\cap F]$, i.e. the reversed inequality! Now take $K,F$ to be fields that are not linearly disjoint over $F\cap K$ and such that all degrees are finite in order to get the strict inequality. For example: $K=\Bbb Q(\sqrt[3]{2}),F=\Bbb Q(\zeta\sqrt[3]{2})$, $L=FK$ where $\zeta\in\Bbb C$ is a primitive third root of unity. Then \begin{align}
[L:L\cap F]&=2\\
[K:K\cap F]&=3
\end{align}
These are probably not 'mild' assumption but: The inequality holds if for example $K,F$ are linearly disjoint over $K\cap F$ and $L,F$ are linearly disjoint over $L\cap F$. In that case we have: $$[K:K\cap F]=[KF:F]\leq[LF:F]=[L:L\cap F]$$
Since $[LF:F]=[LF:KF][KF:F]$, so we even have divisibility  in this case.


问:
Understand this probability argument
I've got a pretty specific question here. I've included an image of a probabilistic expression relating Brownian Motion and the stopping times for when the Brownian crosses each integer value. This text is Steele: Stochastic Calculus and Financial Applications. The question is, how is this last inequality derived?

I am not seeing exactly how we get the connection between $|B_{\tau_n}-B_n| \ge \epsilon\sqrt{n}$ and $\sup_{s:|s-n| \le \delta n} |B_s - B_n| \ge \epsilon\sqrt{n}$. There isn't really any more explanation in the book about this. I guess it should be obvious but I've spent a long time on it and can't figure it out. Thanks!

答:
I am not an expert, but it looks to me that the event:

$$|B_{\tau_n}-B_n|\ge \epsilon\sqrt{n}\text{ and }|\tau_n-n|<\delta n$$

is contained in (i.e. implies) the event:

$$\sup_{s:|s-n|\le\delta n}|B_s-B_n|\ge \epsilon\sqrt{n}$$

whereas the event:

$$|B_{\tau_n}-B_n|\ge \epsilon\sqrt{n}\text{ and }|\tau_n-n|\ge\delta n$$

is contained in (i.e. implies) the event:

$$|\tau_n-n|\ge\delta n$$

so really this is the consequence of monotonicity of probability.

Title: Understand this probability argument


问:
What is the definition of a free abelian group generated by the set $X$?
What is the definition of a free abelian group generated by the set $X$?

If $X=\{A,B,C,D\}$ what are the elements of the free abelian group on this set?  What is the identity of this group?

答:
Other answers have already pointed out concrete models for free Abelian groups. I want to emphasize that they can be characterized by a universal property, which I found helpful to gain some perspective on the construction.

In order to state the universal property, note that every Abelian group $G$ has an underlying set $\underline G$. The passage from $G$ to $\underline G$ simply forgets the group structure. Similarly, a group homomorphism $\alpha\colon G\to H$ has an underlying map of sets $\underline \alpha\colon \underline G\to\underline H$.

Now let $S$ be a set. An Abelian group $F(S)$ together with a map $i\colon S\to \underline{F(S)}$ is called freely generated by $S$ if for every Abelian group $G$ and every map $a\colon S\to \underline G$ there is a unique group homomorphism $\alpha\colon F(S)\to G$ such that $a=\underline \alpha\circ i$. In other words, the map $i$ gives a bijection $$Hom(F(S),G) \cong map(S,\underline G),\quad \alpha\mapsto \underline \alpha\circ i.$$ This property characterizes the pair $(F(S),i)$ uniquely up to unique isomorphisms. It manifests the slogan that "homomorphisms out of $F(S)$ are determined by the values on $S$".

EDIT: By the way, if you erase the word "Abelian" everywhere, you get the universal property of free groups. In fact, I initially missed that word and wrote the answer for groups. Even more generally, you can speak of "free objects" in other categories.


问:
Bezier curves of an iris.
I've got rather limited math background, having never taken a calculus class. I'm trying to program an iris, and I've got little clue where to start. I know I'm dealing with arcs, and that the angle of the center of the iris never changes, but I'm not sure what else needs to be done.

This is Apple's implementation of what I want to accomplish: http://youtu.be/ZX8nnCR4VyM I think I need to put plot control points against the "radius" of the inner octagon, but I'm not sure.

Where can I start?

答: First you need to specify the curve that represents the edges of the iris.  The easiest is straight lines, which is not what is shown in the link.  The selection of the curve is dependent on what your purpose is.  For photographic applications, you want the out of focus elements to look nice.  For many computer applications, straight lines are fine.  Then you can find much on Wikipedia.  The lines are either horizontal, vertical, of of slope $\pm1$ through the points.</p>

Title: Bezier curves of an iris.


问:
When is a map inducing isomorphisms on homology with rational and mod $p$ coefficients a weak equivalence?
Suppose $f \colon X \to Y$ is a map of simply connected topological spaces that induces isomorphisms on homology with rational coefficients as well as homology with $\mathbb{Z}/p$ coefficients for all primes $p$. Under what conditions is $f$ a weak equivalence?

答: It is an exercise in homological algebra to show that your hypothesis implies a homology isomorphism in integral coefficients (Hint: use a short exact sequence involving $\mathbb{Z}/p^\infty$, $\mathbb{Z}$, and $\mathbb{Q}$). Hence, homology Whitehead applies since your spaces are simply connected.


问:
Prove if $n^2$ is even, then $n$ is even.
I am just learning maths, and would like someone to verify my proof.

Suppose $n$ is an integer, and that $n^2$ is even.  If we add $n$ to $n^2$, we have $n^2 + n = n(n+1)$, and it follows that $n(n+1)$ is even.  Since $n^2$ is even, $n$ is even.

Is this valid?

答:
Another simple proof:

$2|n^2$, but $2$ is prime then $2|n$.


答:
This is a typical abuse of notation.

Unless we have bounds on an integral, like $\int_a^b$, the integral $\int f(x)dx$ stands for a primitive function of $f$.
However, according to the argument of your professor he/she really means an integral with bounds.


问:
bayesian logic converse entialment
In formal epistemology, converse entailment condition states that:

For all e,h and k that are cosistent, if h&k implies e and k does not imply e, then E confirm h relative to k.

Is this justified by a bayesian perspective, where e confirms h relative to k iff P(h/e&k)>P(h/k)?

答:
Roughly, yes - the Bayesian approach is a framework for inferring causes from effects. I wouldn't use the word "confirm" unless a posterior probability came out to 0 or 1, though.

If we model E as depending on H so that $P(E|H)>P(E)$, and we model our belief in $H$ as an unknown parameter $p=P(H)$, then observing $E$ will increase our best guess for $p$, in the sense that the posterior distribution for $p$ will be skewed higher than whatever prior distribution we had for it (unless the prior represented no uncertainty to begin with). The same applies when everything is conditioned on some other event $K$.

Title: bayesian logic converse entialment


问:
Limits coincide distance tends to zero?
$\newcommand{\N}{\mathbb{N}}$
$\newcommand{\max}{\operatorname{Max}}$

Let $(X,d)$, $X \not= \emptyset$, a
metric space and $(x_{n})_{n \in \N}$ and $(y_{n})_{n \in \N}$
sequences in $X$. Show that if there exists a $a \in X$ such that:

$$\lim_{n \to \infty}{x_{n}} = a = \lim_{n \to \infty}{y_{n}}$$
Then:
$$\lim_{n \to  \infty}{d(x_{n},y_{n})} = 0$$

Is the converse also true? (proof or counterexample).

Here is my attempt:

Suppose that:

$$\lim_{n \to \infty}{x_{n}} = a = \lim_{n \to \infty}{y_{n}}$$

We want to see that:

$$\lim_{n \to  \infty}{d(x_{n},y_{n})} = 0$$

Let $\varepsilon > 0$, take $\varepsilon = \frac{\varepsilon}{2}$ so in both cases there exists $N_{1},N_{2} > 0$ such that:

$$\forall_{n > N_{1}}\left[ d(x_{n},a) < \frac{\varepsilon}{2}\right]$$

$$\forall_{n > N_{2}}\left[ d(y_{n},a) < \frac{\varepsilon}{2}\right]$$

Now, set $N = \max{N_{1},N_{2}}$. Then, if we add both equations we have that:

$$\forall_{n > N} [d(x_{n},a) + d(y_{n},a) < \varepsilon]$$

Using the triangular inequality:

$$\forall_{n > N} [d(x_{n},y_{n}) < \varepsilon]$$

In particular:

$$\forall_{n > N} [|d(x_{n},y_{n}) - 0 |< \varepsilon ]$$

In fact we find that:

$$\forall_{\varepsilon}\{\exists_{N > 0} \{\forall_{n > N_{1}}[ |d(x_{n},y_{n}) - 0| <\varepsilon]\}\}$$

So by definitiuon of limit:

$$\lim_{n \to \infty}{d(x_{n},y_{n})} = 0$$

What do you think?

答:
For the first part $d(x_n,y_n) \leq d(x_n,a)+d(a,y_n)$ can also be used.

The converse holds true only if one of the sequences $(x_n)_n$ and $(y_n)_n$ is Cauchy and the  space $(X,d)$ is complete. Because: from
$d(x_n,x_m) \leq d(x_n,y_n)+d(y_n,y_m)+d(y_m,x_m)$ and $d(y_n,y_m) \leq d(y_n,x_n)+d(x_n,x_m)+d(x_m,y_m)$ we get $\lim_{n,m \rightarrow \infty}d(x_n, x_m)=\lim_{n,m \rightarrow \infty}d(y_n, y_m)$. So $(x_n)_n$ is Cauchy iff $(y_n)_n$ is so. By completeness of $(X,d)$, $\exists~~~a,b \in X$ such that $\lim_{n \rightarrow \infty}d(x_n, a)=\lim_{n \rightarrow \infty}d(y_n, b)=0$. $d(a,b) \leq d(a,x_n)+d(x_n,y_n)+d(y_n,b)$, taking limit as $n \rightarrow \infty$ we have $d(a,b)=0\Rightarrow a=b$.

Example that shows Cauchy condition can't be dropped: take $x_{2n-1}=2n-1$, $x_{2n}=\frac{1}{2n}$ and $y_{2n-1}=2n-1$, $y_{2n}=\frac{1}{n}$ in $(\mathbb{R},|\cdot|)$.
That the completeness condition can't be dropped can be seen from @Henno Brandsma's example

Title: Limits coincide distance tends to zero?


问:
Axiom of Choice and Completeness of the Reals
I have a question regarding the Completeness axiom of R.
Does It have a relation with the axiom of choice?
I mean, does the axiom of choice implies the Completeness axiom? Or are they independent?</p>

答: They're unrelated. The completeness of the reals is property $4$ here,  stating a nonempty set of reals bounded above has a least upper bound.


问:
Find the equation of the sine function graphed below.
Find the equation of the sine function graphed below.

Write a cosine function for the graph below. Assume the least possible phase shift.

答: Hint:  Your answer should be of the form $f(x)=A \sin (kx+\phi)$.  What is the amplitude of a sine wave before you multiply it by $A$?  What is the wavelength before you multiply $x$ by $k$?  Where does it pass through $0$?</p>


答:
Connor's answer is of course great and way more elementary than what I'm about to say, but here's maybe a second ("global") perspective on why this works: by simple-connectedness and the Hurewicz theorem, we are reduced to showing that it's an integral homology equivalence.

Hence, we are reduced to the following statement:

Let $C,D$ be chain complexes and $f: C\to D$ a map such that $f\otimes \mathbb Q$ and $f\otimes^L\mathbb Z/p$ are quasi-isomorphisms for all $p$; then $f$ itself is a quasi-isomorphism.

To prove this, consider the cone of $f$, $cone(f)$. The hypotheses imply that $cone(f)\otimes^L \mathbb Z/p \simeq 0$, so $p: cone(f)\to cone(f)$ is a quasi-isomorphism for all primes $p$. In particular, $cone(f)\to cone(f) \otimes \mathbb Q$ is a quasi-isomorphism, but the latter is $0$ by the hypothesis on $\mathbb Q$.

Therefore $cone(f) \simeq 0$ and so $f$ is a quasi-isomorphism.

This can be used to derive the arithmetic fracture square, which states that for a chain complex $C$, letting $C_p$ denote its derived $p$-completion, the following square is a homotopy pullback/pushout :
$$\require{AMScd}\begin{CD}C @>>> \prod_p C_p \\
@VVV @VVV \\
C\otimes \mathbb Q @>>> (\prod_p C_p)\otimes \mathbb Q\end{CD}$$

Now this tells you that you can essentially "always" reduce things to computations/verifications mod $p$ (in the derived sense; so $-\otimes \mathbb Z/p$ and $\mathrm{Tor}(-;\mathbb Z/p)$ appear) and rationnally.

Title: When is a map inducing isomorphisms on homology with rational and mod $p$ coefficients a weak equivalence?

Body: Suppose $f \colon X \to Y$ is a map of simply connected topological spaces that induces isomorphisms on homology with rational coefficients as well as homology with $\mathbb{Z}/p$ coefficients for all primes $p$. Under what conditions is $f$ a weak equivalence?


问:
Condition for getting same numbers when multiplying each row and each column in an n * n grid
A friend gave me an interesting problem recently that has me stumped:

How can I fill a 3 X 3 grid with the numbers 1, 2,...9 such that the three numbers obtained on multiplying each row, are the same three numbers we get on multiplying each column? What if I were to fill a 5 X 5 grid with the numbers 1,2,3...25? Is it possible to formulate a condition(s) that guarantee that it is possible/impossible to write the numbers 1,2,3....n^2 on an n x n grid such that the numbers obtained on multiplying each row and each column are the same?

I've gotten one solution to the 3 x 3 grid, but am struggling to generalize it or even just see if there are more than one combinations. The one I have now is given below:

5  ,  3  ,  8  =  120

6  ,  7  ,  1  =  42

4  ,  2  ,  9  =  72

And accordingly, multiplying each column will give 120, 42, 72 as well.

答: It is easy to see that in an $n \times n$ grid, primes $> n^2/2$ can't share a row or column.  So a necessary condition for the $n \times n$ grid to be possible is that there are at most $n$ such primes.  That rules out $n \times n$ grids for $n=9$ and, I think, all $n \ge 11$.


问:
Prove by induction $2^n \le (n+1)!$

Prove by induction that:  $2^n \le (n+1)! \ \ \ \ \ \ \ \ \forall n\in\mathbb{N}$

I tried to solve in this way:
$$P(1): 2 \le 2! \quad\text{(true)}$$
$$P(n) \implies p(n+1), \;\text{ where }\;\; p(n+1): 2^{n+1} \le (n+2)!$$
so
$$2^n \le (n+1)! \\ 2^n \cdot 2\le 2 \cdot (n+1)! \\ (n+1)! \cdot 2 \le 2 \cdot (n+1)! \quad\text{(true)}$$
where tapped that $2^n \le (n+1)!$

Are my steps right?

Title: Prove by induction $2^n \le (n+1)!$

Body:
After you get $2^{n+1}\le 2(n+1)!$.......(1)

Proceed as follows:

$(n+1)!=(n+1)!$

$\implies \color{red}{(1+1)}(n+1)!\le \color{green}{(n+2)}(n+1)!$

$(\because (1+1)<(n+2)\forall n\in\mathbb N$)

$\implies 2(n+1)!\le (n+2)!$.........(2)

From (1) and (2), you have $2^{n+1}\le (n+2)!$


问:
Proving uncountable sum as series using nets.
I saw this question:

The sum of an uncountable number of positive numbers

Asking about a proof of the following:
Let $A = \{a_i\}_{i\in I}$ be a set of positive numbers. If the uncountable sum converges (as a net) to $a$, then only countably many $a_i$ are nonzero.

And was wondering if this proof any good (I have little experience working with nets):

The closure of $A$ is a closed subset of a metric space and $a$ lies in this closure. Thus there is a sequence in $A$ that converges to $a$.
$$ a = \sum_{i=1}^\infty a_i. $$
If $A$ is not countable, we could find a $a_k \in A$ that is not part of the sequence. If $a_k\neq 0$, then
$$ a_k + \sum_{i=1}^\infty a_i > \sum_{a_i\in A}a_i $$
yielding a contradiction.

I did not immediately see this proof being used anywhere. Sorry if I missed something obvious.

Edit: I missed something obvious: $a$ is not in the closure $A$!

It is, however in the closure of the set of finite sums, so we can find a sequence of finite sums converging to $a$, $(s_i)_{i\in\mathbb{N}}$, where each finite sum features each $a_i$ at most once. Each sum has a finite set of summands, so the union of all summands in the entire sequence is a countable union of finite subsets and is countable. Take the sum of all these summands and we have a countable infinite sum
$$ \sum_{i=1}^\infty a_i. $$

The claim is this converges to $a$. Indeed clearly $\sum_{i=1}^\infty a_i \leq \sum_{a_i\in A}a_i = a$. Conversely, for all $s_j$, there is an $n$ such that $\sum_{i=1}^n a_i \geq s_j$, thus $\sum_{i=1}^\infty a_i \geq a$. The rest of the argument then follows.

答:
You have a nice idea, but it is not properly elaborated. Let us first recall

Let $A = (a_i)_{i \in I}$ be an indexed collection of real numbers. Then $\sum_{i\in I}a_i = a$ means that for each number $\varepsilon > 0$ there exists a finite set $F_\varepsilon \subset I$ such that for each finite set  $F \subset I$ with $F_\varepsilon \subset F$ one has $\lvert a-\sum_{i\in F}a_i \rvert <\varepsilon$.

If all $a_i$ are positive and $I$ is infinite, then clearly we have $0 < \sum_{i\in F}a_i < a$ for all finite $F \subset I$: If there would be a finite $F^*$ with $\sum_{i\in F^*}a_i \ge a$, pick $k \notin F^*$ and let $\varepsilon = a_k > 0$. Then $\sum_{i\in F_\varepsilon \cup F^* \cup \{k\}}a_i \ge \sum_{i\in F^*}a_i + a_k \ge a + a_k = a +\varepsilon$, i.e. $\lvert a - \sum_{i\in F_\varepsilon \cup F^* \cup \{k\}}a_i \rvert \ge \varepsilon$ which is a contradiction.

Now define $B$ as the set of all sums $\sum_{i\in F}a_i$ with a finite $F \subset I$. Then by definition $a  \in \overline B$. Thus there is a sequence of finite $F_n \subset I$ such that $\sum_{i\in F_n}a_i \to a$ as $n \to \infty$. If $A$ is uncountable, there exists $k \in I \setminus \bigcup_{n=1}^\infty F_n$. Convergence implies that $ a - \sum_{i\in F_n}a_i  < a_k$ for some $n$. Therefore
$$a < \sum_{i\in F_n}a_i +a_k = \sum_{i\in F_n \cup \{k \}}a_i < a$$
which is a contradiction. Therefore $A$ cannot be uncountable.


问:
Divergence of $\vec{F} = \frac{\hat{\mathrm{r}}}{r^{2}}$
Consider the vector field
$$\vec{\mathrm{F}} = \frac{\hat{\mathrm{r}}}{r^{2}},$$
then the divergence of this field is:
$$\vec{\nabla}\cdot\left(\frac{\hat{\mathrm{r}}}{r^{2}}\right) = 4\pi\delta^{3}(\mathrm{\vec{r}})$$

What is the proof of this relation?

答:
The below is a more detailed version of bob.sacamento's answer. I will use boldface for vector quantities.

Suppose we integrate over a sphere of radius $R$ centered at the origin. The surface
integral is
$$
\begin{aligned}
\oint \mathbf{F} \cdot d \mathbf{a} &=\int\left(\frac{1}{R^{2}} \hat{\mathbf{r}}\right) \cdot\left(R^{2} \sin \theta d \theta d \phi \hat{\mathbf{r}}\right) \\
&=\left(\int_{0}^{\pi} \sin \theta d \theta\right)\left(\int_{0}^{2 \pi} d \phi\right)=4 \pi
\end{aligned}
$$
Using the divergence theorem:
$$
\begin{aligned}
\oint \mathbf{F} \cdot d \mathbf{a} &=\int \nabla\cdot{\bf F}\, d{\mathbf{\tau}}
\end{aligned}
$$
where ${\mathbf{\tau}}$ is the volume differential. The above two equations imply that
$$
\begin{aligned}
\int \nabla\cdot{\bf F}\, d{\mathbf{\tau}} &=4\pi\,
\end{aligned}
$$
Therefore,
$$
\begin{aligned}
\nabla\cdot\left(\frac{1}{r^{2}} \hat{\mathbf{r}}\right) &=4\pi\delta^3({\bf r}) \, .
\end{aligned}
$$


问:
the curve $y=x^2\cos{4x}$ for $0 \leq x \leq \dfrac{\pi}{8}$. The point $P$ is the maximum
Question a) Show that the $x$ coordinates of $P$ satisfies the equation $4x^2\tan{4x}=2x$

b) Show also the $x$ coordinates of $P$ satisfies the equation $x=\dfrac{1}{4}\tan^{-1}(1/2x)$

I have been stuck on this question for the past hour I don't understand it it is the last section of my A level book chapter 6 and I have no professor and the back of my book all that is written is proof. Any help is much appreciated thank you!

答: Hint. The maximum of the given curve occurs wherever $y'=0,$ since $y=0$ at both endpoints of the domain.


问:
Representing Recursion and Primitive Recursion diagrammatically
I'm interested in how Recursion, and Primitive Recursion, could be represented diagrammatically. It occurred to me that this would be a good way of seeing the difference. Also, I'm interested in how Recursive functions in their various different forms could be represented this way.

If anyone knows of a method of drawing diagrams to visualise the behaviour of recursion and recursive functions, I'd love to learn more about it on this thread!

答:
There is a way to visualize in an algebraic manner the recursive operator of the Primitive Recursive Function class: the Natural Numbers Object (NNO) in the context of a cartesian (or even monoidal) category. It allows to define Primitive Recursion by showing the relations diagramatically. You can see an extensive definition and characterizations in ncatlab.org/nlab/show/natural+numbers+object

On the other hand, in this paper you can find a whole description of the class in terms of trees where the vertices are functions labelled by the usual operators.

The latter description makes use of the NNO diagram (and others).


问:
If a binary quadratic form primitively represents $n$ and $n^3$, must it be the identity form?
I've been searching numerically, and whenever a positive definite binary quadratic form can primitively represent an integer $n>1$ and $n^3$, it turns out to be equivalent to the identity form in that class group.

Is there some elementary way to see that this is only possible for identity forms?

Here is what initially gave me hope that it could be possible with other forms.

For any binary quadratic form $A x^2 + B xy + C y^2$, if it can represent integers $r$, $s$, $t$ then it can also represent $rst$.

This can be seen as follows:

For any form with coefficients $(A,B,C)$, in the class group its inverse is $(A,-B,C)$.

Both a form and its inverse can represent the same numbers (if $(x,y)$ with the former, use $(x,-y)$ with the later).

If a form can represent $r$ with $(x_1,y_1)$, and $s$ with $(x_2,y_2)$ then the identity form can represent $rs$ with some integers $(u,v)$ due to form composition:

$$
\begin{aligned}
rs &= (A x_1^2 + B x_1 y_1 + C y_1^2)(A x_2^2 + B x_2 y_2 + C y_2^2) \\
   &= (A x_1^2 + B x_1 y_1 + C y_1^2)(A x_2^2 - B x_2 (-y_2) + C (-y_2)^2) \\
   &= I(u,v) \\
\end{aligned}
$$

Futhermore, if the form can represent $t$ with $(x_3,y_3)$, then it can represent $rst$ by some integers $(x_4,y_4)$ due to form composition.

$$rst = I(u,v)(A x_3^2 + B x_3 y_3 + C y_3^2) = (A x_4^2 + B x_4 y_4 + C y_4^2)$$

It is definitely possible to primitively represent cubes, for example

```<code>a) 2 x^2 - xy + 6 y^2 = 3^3  with x=3,y=-1
b) 2 x^2 - xy + 6 y^2 = 8^3  with x=2,y=-9
</code>```

But in example 'a', it is not possible for that form to represent 3. While in example 'b', the form can represent 8, but not primitively (x=2,y=0).

This seems to be the case for any form not equivalent to an identity form.

While for an identity form, I've been able to find a lot of examples that work.

```<code>    x^2 + xy + 12 y^2 = 18     with x=2, y=1, gcd(x,y)=1
    x^2 + xy + 12 y^2 = 18^3   with x=-76, y=7, gcd(x,y)=1

    x^2 + xy + 12 y^2 = 118    with x=2, y=3, gcd(x,y)=1
    x^2 + xy + 12 y^2 = 118^3  with x=964, y=207, gcd(x,y)=1

    x^2 + xy + 12 y^2 = 63     with x=3, y=2, gcd(x,y)=1
    x^2 + xy + 12 y^2 = 63^3   with x=-501, y=2, gcd(x,y)=1
</code>```

Can one prove that if a form can primitively represent $n$ and $n^3$, that is must be equivalent to the identity form?

UPDATE:

A related question On products of ternary quadratic forms $\prod_{i=1}^3 (ax_i^2+by_i^2+cz_i^2) = ax_0^2+by_0^2+cz_0^2$ shows a possible solution for any form with B=0 (all that is left is to analyze if the result is a primitive representation). Such forms are in the same equivalence class as its inverse in the class group, and are called "ambiguous forms".

Below Will Jagy gave an explicit solution for an ambiguous form which is not equivalent to the identity form, showing that the result can be a primitive representation.

The connection to what I tried is that the identity is always an ambiguous form. So that likely refines the question to:

If a form can primitively represents $n$ and $n^3$, is the form necessarily an ambiguous form?

答:
As Will Jagy said in his answer, it is simplest to consider the case that $n$ is a prime $p$ not dividing the discriminant. Then if $p$ is represented by some form of the given discriminant, this form is unique up to equivalence, and each power $p^k$ is also represented by a unique form up to equivalence. In terms of the class group, if the form $Q$ represents $p$ then the forms representing $p^k$ are precisely $Q^k$ and $Q^{-k}$.  (A form and its inverse in the class group always represent exactly the same numbers. Throughout this answer I am using the term "represent" to mean "primitively represent".)

Thus if $Q$ represents both $p$ and $p^k$ then we must have $Q^k=Q$ or $Q^k=Q^{-1}$ in the class group, or in other words either $Q^{k-1}$ or $Q^{k+1}$ is the identity element in the class group. Taking $k=3$ as in the question, we see that $p$ and $p^3$ are represented by the same element of the class group exactly when this element has order dividing $4$.

There are plenty of examples.  The simplest might be discriminant $-15$ with class group cyclic of order $2$ generated by the form $Q(x,y)=2x^2+xy+2y^2$.  The primes not dividing the discriminant that are represented by $Q$ are the primes congruent to $2$ or $8$ mod $15$.  In particular $Q(1,0)=2$ and $Q(-1,2)=2^3$. The next prime would be $17$, and $Q(-1,3)=17$ but I wouldn't want to solve $Q(x,y)=17^3$ by hand, though the general theory says that a solution exists.

Another example would be discriminant $-56$ as in Will Jagy's answer.  Here the class group is cyclic of order $4$ generated by the form $Q(x,y)=3x^2+2xy+5y^2$. This form represents exactly the primes congruent to $3,5,13,19,27$, or $45$ mod $56$, so primes in these congruence classes give examples. The first two are $Q(1,0)=3$ and $Q(0,1)=5$ with $Q(1,2)=3^3$ and $Q(6,1)=5^3$.

Allen Hatcher

(Stackexchange thinks I am a new user here, which I am not, so it created a new account for me. If someone knows how to combine the new account with my old one, please do so or tell me how to do it.)


答:
Choice has nothing to do with the completeness of $\mathbb{R}$. To be precise, none of the constructions of an Archimedean complete ordered field or proofs of the uniqueness of such up to isomorphism (that I'm aware of, anyways) invoke the axiom of choice in any way.

Let me clarify the language of the above paragraph a bit, since it may seem odd at first. Since there are multiple reasonable ways to construct $\mathbb{R}$ in the first place (Dedekind cuts, Cauchy sequences, Eudoxus' functions, and many others) the whole notion of "constructing $\mathbb{R}$" in the first place is a bit more slippery than one might expect. In my opinion, the right interpretation of this phrase is: "proving that there is exactly one Archimedean complete ordered field up to isomorphism." We then use the symbol "$\mathbb{R}$," the term "real numbers," and so on to refer to any such field, the point being that uniqueness-up-to-isomorphism makes this ambiguity benign in a precise sense. Meanwhile, notice that completeness is explicitly part of the definition of $\mathbb{R}$ in this approach - so "constructing $\mathbb{R}$" includes "showing that $\mathbb{R}$ is complete" as a necessary step along the way.

Now admittedly a bit of set theory does come into play in constructing the reals - necessarily so, in a precise sense - but it's the powerset axiom, not the axiom of choice. Specifically, the things we use to build $\mathbb{R}$ from $\mathbb{Q}$ are "higher-type" objects - sets of rationals in the construction via Dedekind cuts, sets of sequences of rationals in the construction via Cauchy sequences, and so on - and showing that there is a set of all such objects in the first place requires the powerset axiom.

It's separately worth noting - and this was mentioned by Patrick Stevens in his comment above - that there is a computability theoretic aspect to completeness of $\mathbb{R}$. Specifically, there is a computable increasing bounded sequence of rational numbers whose least upper bound is not a computable real number. Consequently, in a precise (if somewhat weak) sense we have to go "outside the computable universe" to see completeness.

Title: Axiom of Choice and Completeness of the Reals

Body:
I have a question regarding the Completeness axiom of R.
Does It have a relation with the axiom of choice?
I mean, does the axiom of choice implies the Completeness axiom? Or are they independent?</p>


问:
Prove that if $f:M\rightarrow\Bbb R$ is a scalar function over a 1-manifold M without boundary then $\int_M df=0$
Well James Munkres in the text Analysis on Manifolds prove the general Stoke's theorem for $k$-form when $k>1$ and then he proves it for $k=1$ only when the bounary of the Manifold is not empty and he leaves as exercise to show that if $f$ is a scalar function defined over a compact $1$-manifold $M$ then
$$
\int_M df=0
$$

So let's start to try to prove it and precisely we will do to scalar function whose support is covered by a single coordinate chart becasue as Munkres showed in the general proof this is sufficent. Indeed if $\Phi:\{\phi_i:i=1,...,l\}$ is a partition of unity dominated by the coordinate patches of $M$ then the support of the forn $\phi_i\omega$ is contained in a single coordinate patch for each $i=1,..,l$ and thus
$$
\int_Md\omega=\int_M0+\int_Md\omega=\int_M0\curlywedge\omega+\int_M\Biggl(\sum_{i=1}^l\phi_i\Biggl)\curlywedge d\omega=
\\
\int_Md1\curlywedge\omega+\int_M\Biggl(\sum_{i=1}^l\phi_i\curlywedge d\omega\Bigg)=
\\
\int_Md\Biggl(\sum_{i=1}^l\phi_i\Biggl)\curlywedge\omega+\sum_{i=1}^l\Biggl(\int_M\phi_i\curlywedge d\omega\Biggl)=
\\
\sum_{i=1}^l\Biggl(\int_Md\phi_i\curlywedge\omega\Biggl)+\sum_{i=1}^l\Biggl(\int_M(-1)^0\phi_i\curlywedge d\omega\Biggl)=\sum_{i=1}^l\Biggl(\int_Md\phi_i\curlywedge\omega+\int_M(-1)^0\phi_i\curlywedge d\omega\Biggl)
\\
\sum_{i=1}^l\Biggl(\int_Md\phi_i\curlywedge\omega+(-1)^0\phi_i\curlywedge d\omega\Biggl)=\sum_{i=1}^l\int_Md(\phi_i\omega)=0
$$

Well if $M$ is a compact $1$-manifold without boundary it is possible to prove that for any $p\in M$ there exist a coordinate patch $\alpha$ defined in the unitary positive interval $(0,1)$ and thus if $f:M\rightarrow\Bbb R$ is a scalar function defined over M whose support $S_f$ is covered by a single coordinate patch then
$$
\int_Mdf:=\int_{(0,1)}\alpha^*(df)=\int_{(0,1)}d(\alpha^*f)=\int_{(0,1)}d(\alpha\circ f)
$$
So Unfortunately I do not able to prove that
$$
\int_{(0,1)}d(\alpha\circ f)=0
$$
and thus I ask to do it. So could someone help me, please?

答:
You say you know that it is sufficient to prove this for a scalar function $f$ with support $K$ contained in a coordinate open $U$, then $df$ also has support in $U$, say in $K'$. Now, let $\phi:(a,b)\to U$ be a parametrization. The integral in this case is defined by
$$
\int_a^b\phi^*(df)=\int_a^bd(\phi^*f).
$$
I think you also reached this conclusion. Write $g=\phi^*f$. Then $dg=g'(x)dx$. So, we have
$$
\int_a^b\phi^*(df)=\int_a^bd(\phi^*f)=\int_a^bg'(x)dx=g(b)-g(a)
$$
by the fundamental theorem of calculus. Now, as $K\subsetneq U$, $\phi^{-1}(K)$ is a proper subset of $(a,b)$ and hence does not contain $a$ or $b$. So, $g(b)=g(a)=0$. So
$$
\int_a^b\phi^*(df)=\int_a^bd(\phi^*f)=\int_a^bg'(x)dx=g(b)-g(a)=0.
$$


问:
Plim of the expectation of the reciprocal of the sample mean of an autoregressive process
I am trying to prove the following:

Let $X_t$ be a sequence of random variables that follows an autoregressive process; i.e. $X_t=X_{t-1}+e_t$, where $e_t$ is a zero mean i.i.d. sequence. Then $\lim\limits_{n\to \infty} \mathbf E\left(\left(\sum\limits_{i=1}^n (X_t-\bar X_t)^2\right)^{-1}\right)$ converges in probability to 0, where $\bar X_t$ denotes the sample mean.

So far, I have been able to show that $\mathbf E\left(\sum(X_t-\bar X_t)^2\right)$ converges in probability to infinity, but this alone does not imply the expectation of its reciprocal goes to 0. Are there any results I can use, along with this, to complete my proof?

答:
Let $Y_t=X_t-\bar X_t$ and $S_n=\displaystyle\sum_{t=1}^nY_t^2$. Then $Y_1=S_1=0$ and $(S_n)$ is a nondecreasing sequence of nonnegative random variables. If the distribution of the random variables $e_t$ has an atom at zero, then, for every $n$, $S_n=0$ with positive probability hence $1/S_n$ is not integrable. From now on we assume that the distribution of $e_t$ has no atom at zero. Then $Y_2=\frac12(X_2-X_1)=\frac12e_2$ hence $S_2=\frac14e_2^2$ and $S_n>0$ almost surely, for every $n\ge2$. Let us show that $S_n\to+\infty$ almost surely.

For every $t$, $X_t=X_0+e_1+\cdots+e_t$ hence the mean $\bar X_t$ of the values of $X_s$ from $s=1$ to $s=t$ is $\bar X_t=X_0+e_1+(t-1)e_2/t+(t-2)e_3/t+\cdots+e_t/t$, and
$$
Y_t=e_2/t+2e_3/t+\cdots+(t-1)e_t/t.
$$
Assume that $(S_n)$ is bounded, then $Y_t\to0$. But $Y_{t+1}=tY_t/(t+1)+te_{t+1}/(t+1)$ hence $e_{t+1}\to0$, an event which has probability zero.

Hence $S_n\to+\infty$ almost surely, $1/S_n\to0$ almost surely and $E(1/S_n)\to0$ as soon as one of the expectations $E(1/S_n)$ is finite.

To conclude that at least one random variable $1/S_n$ is integrable, some additional hypothesis on the common distribution of the random variables $e_t$ seem required.


问:
When does E[XY] = XE[Y] hold?
I am wondering, when does

$$ \mathbb{E}[XY] = X\mathbb{E}[Y] $$

hold for two random Variables $X$ and $Y$?

Idea: If $XY$ is independent of a sigma-algebra $\mathcal{F}$ we have $\mathbb{E}[XY] = \mathbb{E}[XY|\mathcal{F}]$. And now we have $ \mathbb{E}[XY] = X\mathbb{E}[Y] $ if $X$ is $\mathcal{F}$-measurable. Is that correct? Can this property be described in a different way?

答:
What you wrote can only hold if $X$ is a constant, since the LHS is a constant, and the RHS is a random variable.

Did you mean to write $\mathbb{E}[XY] = \mathbb{E}[X] \cdot \mathbb{E}[Y]$?

Title: When does E[XY] = XE[Y] hold?


问:
Poisson distribution problem including cdf
The Question

The number of cracks in a section of highway that is significant
enough to require repair is assumed to follow a Poisson distribution.

(a) Let $Y$ be the number of cracks in $4$km, sketch the (CDF) Cumulative
Distribution Function and graph up to $ = 4.5$.

(b) If we should order the material to fix the cracks beforehand, how
many packages of the material (One package for one crack) shall we
order to ensure that all the cracks in $4$km can be fixed with at least
$95\%$ chance?

My Understanding

For part (a), I tried to compute the probabilities of $Y=0, 1, 2, 3, 4, 5$ respectively, by using the formula of $Pr(X=k)=\frac{e^{-}\mu^k}{k!}$ however, when I tried to sum these probabilities together , $F(Y)$ turns out to exceed $1$, where probability should not $\gt 1$, what's wrong with that? What is the appropriate way to find the probabilities and thus I can graph up to $y=4.5$?

For part (b), for my understanding, this Poisson distribution has infinite number of cracks, so I am quite doubt that how to ensure that all the cracks in $4$km can be fixed with at least $95\%$ chance?

答:
@callculus is correct that you do not have enough information to know
(or even estimate) the Poisson mean $\mu.$ I will use $\mu = 4.5$ for
illustration. Maybe that will give you enough information to make sense
of the problem in your question.

If the average number $X$ of cracks in $4$km of highway is distributed
$\mathsf{Pois}(\mu = 4.5),$ then $\sum_{k=0}^5 \frac{e^{-4.5}4.5^K}{k!} = 0.7029304 < 1.$ I used R to compute this in two ways: (a) by making a vector pdf of the six probabilities, and (b) by using the R procedure dpois, which is a Poisson PDF. Alternatively, you might use a calculator to sum these six terms.

```<code>k = 0:5;  pdf = exp(-4.5)*4.5^k/factorial(k)
sum(pdf)
[1] 0.7029304
sum(dpois(k, 4.5))
[1] 0.7029304
</code>```

With $\mu = 4.5$ a graph of the CDF of this Poisson distribution is
as follows.

```<code>pdf = ppois(k, 4.5)
plot(k, pdf, type="s")  # 's' for 'stairstep'
 points(k, pdf, pch=19)
</code>```

If you want to be 95% sure to have enough bags of material to
repair all of the cracks in $4$km of this highway, then you
need to find the number $r = 8$ at which the CDF first exceeds $0.95.$
You can use the R quantile function qpois (inverse CDF) for that.
Alternatively, using a calculator, you could sum the nine required terms to
verify this result.

```<code>qpois(.95, 4.5)
[1] 8
ppois(8, 4.5)
[1] 0.9597427
</code>```

Here is an extended CDF plot that goes up to $P(X \le 9)\approx 1.$
The horizontal red line is at $0.95.$

R code for second figure:

```<code>k = 0:9;  pdf = ppois(k, 4.5)
plot(k, pdf, type="s")  # 's' for 'stairstep'
points(k, pdf, pch=19)
abline(h=.95, col="red")
</code>```

Body:
@callculus is correct that you do not have enough information to know
(or even estimate) the Poisson mean $\mu.$ I will use $\mu = 4.5$ for
illustration. Maybe that will give you enough information to make sense
of the problem in your question.

If the average number $X$ of cracks in $4$km of highway is distributed
$\mathsf{Pois}(\mu = 4.5),$ then $\sum_{k=0}^5 \frac{e^{-4.5}4.5^K}{k!} = 0.7029304 < 1.$ I used R to compute this in two ways: (a) by making a vector pdf of the six probabilities, and (b) by using the R procedure dpois, which is a Poisson PDF. Alternatively, you might use a calculator to sum these six terms.

k = 0:5;  pdf = exp(-4.5)*4.5^k/factorial(k)
sum(pdf)
[1] 0.7029304
sum(dpois(k, 4.5))
[1] 0.7029304

With $\mu = 4.5$ a graph of the CDF of this Poisson distribution is
as follows.

pdf = ppois(k, 4.5)
plot(k, pdf, type="s")  # 's' for 'stairstep'
 points(k, pdf, pch=19)

If you want to be 95% sure to have enough bags of material to
repair all of the cracks in $4$km of this highway, then you
need to find the number $r = 8$ at which the CDF first exceeds $0.95.$
You can use the R quantile function qpois (inverse CDF) for that.
Alternatively, using a calculator, you could sum the nine required terms to
verify this result.

qpois(.95, 4.5)
[1] 8
ppois(8, 4.5)
[1] 0.9597427

Here is an extended CDF plot that goes up to $P(X \le 9)\approx 1.$
The horizontal red line is at $0.95.$

R code for second figure:

k = 0:9;  pdf = ppois(k, 4.5)
plot(k, pdf, type="s")  # 's' for 'stairstep'
points(k, pdf, pch=19)
abline(h=.95, col="red")


答: Write $\mathbb{E}[XY] = \mathbb{E}[XY|\sigma(\emptyset)]$, that is - the Sigma algebra generated by the empty set, i.e $\{\Omega,\emptyset\}$, known also as the trivial sigma-algebra. Then it will follows that $\mathbb{E}[XY] = X\mathbb{E}[Y]$ if and only if $X$ is measurable with respect to the trivial sigma-algebra, which in turn means $X$ is in fact constant.

Title: When does E[XY] = XE[Y] hold?

Body:
I am wondering, when does

$$ \mathbb{E}[XY] = X\mathbb{E}[Y] $$

hold for two random Variables $X$ and $Y$?

Idea: If $XY$ is independent of a sigma-algebra $\mathcal{F}$ we have $\mathbb{E}[XY] = \mathbb{E}[XY|\mathcal{F}]$. And now we have $ \mathbb{E}[XY] = X\mathbb{E}[Y] $ if $X$ is $\mathcal{F}$-measurable. Is that correct? Can this property be described in a different way?


答:
Hint for part b: You have to set up the following inequality:

$$P(X\leq x)=\sum_{k=0}^x e^{-2}\cdot \frac{2^k}{k!}\geq 0.95,$$

where $X\sim Poi(2)$

From one of the previous exercises you know that $P(X\leq 5)>0.95$. So go on and try $x=4,3,...$ until $P(X\leq x)<0.95$.

Title: Poisson distribution problem including cdf

Body:
The Question

The number of cracks in a section of highway that is significant
enough to require repair is assumed to follow a Poisson distribution.

(a) Let $Y$ be the number of cracks in $4$km, sketch the (CDF) Cumulative
Distribution Function and graph up to $ = 4.5$.

(b) If we should order the material to fix the cracks beforehand, how
many packages of the material (One package for one crack) shall we
order to ensure that all the cracks in $4$km can be fixed with at least
$95\%$ chance?

My Understanding

For part (a), I tried to compute the probabilities of $Y=0, 1, 2, 3, 4, 5$ respectively, by using the formula of $Pr(X=k)=\frac{e^{-}\mu^k}{k!}$ however, when I tried to sum these probabilities together , $F(Y)$ turns out to exceed $1$, where probability should not $\gt 1$, what's wrong with that? What is the appropriate way to find the probabilities and thus I can graph up to $y=4.5$?

For part (b), for my understanding, this Poisson distribution has infinite number of cracks, so I am quite doubt that how to ensure that all the cracks in $4$km can be fixed with at least $95\%$ chance?


问:
Does the order of the Fibonacci sequence's initial values matter?
I am reading about generating functions in this reputable engineering textbook and the author uses the Fibonacci sequence as an example:

The Fibonacci sequence is defined by the initial conditions $a_0=0$,
$a_{-1}=1$, and the recursion relation $a_n=a_{n-1}+a_{n-2}$ for
$n>0$.

Using this definition, I am able to work through the author's example to get the correct generating function:
$$G(x)=\frac{1}{1-x-x^2}$$

However, everywhere I read online states the initial conditions of the Fibonacci sequence as (0,1) and not (1,0). So I tried that instead, but I get a different result:

$$G(x)=\frac{1+x}{1-x-x^2}$$

This generating function seems to produce the Fibonacci sequence, but starting from the second number instead of the first $(1,2,3,5,8,13,...)$. Where did I go wrong? I would have thought that the sequence should be identical in both cases.

Many thanks for any help!

My calculations:

The author starts with a general expression for a generating function for a sequence $\{a_n\}$:
$$G(x)=\sum_{n=0}^{\infty}a_nx^n$$

He then looks at the special case where $\{a_n\}$ is a linear recurrence of range $r$:
$$a_n=\sum_{i=1}^rc_ia_{n-i}$$

and proves that this leads to:
$$G(x)=\frac{g(x)}{f(x)}=\frac{\sum_{i=1}^rc_ix^i\left(a_{-i}x^{-i}+\cdots+a_{-1}x^{-1}\right)}{1-\sum_{i=1}^rc_ix^i}$$

So, for the Fibonacci sequence, I guess we have $r=2$ and $c_1=c_2=1$. So we get:

$$g(x)=a_{-1}+a_{-2}+a_{-1}x$$

Now I'm a bit confused because the subscripts don't seem to match up. In all the author's discussion, the "initial conditions" have strictly negative subscripts ($a_{-i}$ for $i=1,2,\dots,r$). But just for the Fibonacci example, we have initial conditions $a_0=0$ and $a_{-1}=1$. So I assume these are just off by 1. Therefore, substituting $a_{-1}=0$ and $a_{-2}=1$ gives the author's result above and using $a_{-2}=0$ and $a_{-1}=1$ gives my result above.

答:
In the particular case of the
Fibonacci number
sequence OEIS A000045
(or series) there is some difference of opinion
as amply evidenced by the Wikipedia article and OEIS entry.
The most common convention is that $\,F_0=0\,$
and $\,F_1=1\,$ and this choice has much in its
favor. This choice implies that its
generating function is
$$ \sum_{n=0}^\infty F_n\,x^n = \frac{x}{1-x-x^2}. $$
Notice the $\,x\,$ in the numerator. For this and
a few other reasons, some prefer the choice
$\,F_0=F_1=1\,$ instead, which implies that the
generating function is now
$$ \sum_{n=0}^\infty F_n\,x^n = \frac{1}{1-x-x^2}. $$
Both choices are valid and useful in some cases
and not as useful in other cases. Thus, it is a good
idea to be clear on which choice is being used in any
given context since the choice will usually matter.

The situation with negative index Fibonacci
sequence elements is that the recurrence
relation for the sequence can be used to
uniquely extend the sequence in the negative
index direction. For the common convention
this implies that
$$ F_{-n} = (-1)^{n-1}F_n \quad\text{ for
all integer }n. $$ The result for the other
convention it is that
$$ F_{-n} = (-1)^n F_{n-2} \quad\text{ for
all integer }n. $$

What if we choose arbitrary starting values with
the same recurrence relation? That is, if we
choose $\,F_0=a\,$ and $\,F_1=b,\,$ what do
we get? The recurrence relation is linear and
this implies that the sequence of numbers we get
is a linear combination of the two initial value
choice sequences. The generating function now is
$$ (a) \!+\! (b)x \!+\! (a+b)x^2 \!+\!
 (a+2b)x^3 \!+\! \cdots \!=\!
\frac{a\!+\!(b\!-\!a)x}{1-x-x^2}. $$

Note that initial values of Fibonacci and related
sequences can be given for any two consecutive
index values--not just for $\,a_0\,$ and $\,a_1\,$
since the recursion going forward and backward can
reach any other integer index value. Again, this is
just a matter of convenience and convention. In the
case you cite, specifying $\,a_{-1}\,$ and $\,a_0\,$
is fine. The initial values $\,a_{-1}=1,\,a_0=0\,$
gives the same sequence as $\,a_0=0,\,a_1=1.$ This
special case may be misleading though. The initial
values $\,a_{-1}=u,\,a_0=v\,$ gives the same
sequence as $\,a_0=v,\,a_1=u+v\,$ and $\,u+v\,$ is
equal to $\,u\,$ if and only if $\,v=0.$

Note that the Wikipedia article
Generalizations of Fibonacci numbers
mentions some of the issues such as extension
to negative integers. It mentions a vector space
of number sequences:

Since the set of sequences satisfying the relation
$\,S(n)=S(n-1)+S(n-2)\,$ is closed under termwise addition and under termwise multiplication by a constant, it can be viewed as a vector space. Any such sequence is uniquely determined by a choice of two elements, so the vector space is two-dimensional. If we abbreviate such a sequence as $\,(S(0),S(1)),\,$ the Fibonacci sequence $\,F(n)=(0,1)\,$ and the shifted Fibonacci sequence $\,F(n-1)=(1,0)\,$ are seen to form a canonical basis for this space, yielding the identity:
$$ S(n)=S(0)F(n-1)+S(1)F(n) $$
for all such sequences $\,S$.


问:
A condition that implies (?) linear dependence of linear functionals
Let $F$, $G$, and $H$ be linear functionals on a real vector space $V$. Assume that there is $x \in V$ such that $G(x)<0$ and $F(x) \geq 0$.

Assume further that $F$, $G$, $H$ satisfy the following condition:
for any $x,y \in V$, if $G(x)<0$, $G(y)<0$, and $F(x)/G(x)=F(y)/G(y) \leq 0$, then $H(x) \geq 0$ iff $H(y) \geq 0$.

Does this imply that $H=aF+bG$ for some constants $a$ and $b$?

答: By assumption, there is $x \in V$ such that $G(x) < 0$ and $F(x) \geq 0$. Pick $y \in V$ such that $F(y)=G(y)=0$. Then $F(\lambda x)/G(\lambda x) = F(\lambda x \pm y)/G(\lambda x \pm y) \leq 0$ for any $\lambda >0$. By the condition, we must have $H(\lambda x) \geq 0$ iff $H(\lambda x \pm y) \geq 0$. This relation holds for any $\lambda >0$ if and only if $F(y)=0$. Now the required result follows from Intersection of kernels and linear dependence of functionals.


问:
How do i verify that $272727...2727$ ($100$ digits) can or cannot be written as a perfect square??
I've been stuck on this question.

I tried writing the number as as geometric progression plus $$2((10^{100}-1)/9)+5+5.10^2+5.10^4...$$

Got stuck in there.

答:
Note that this is $27×1010\ldots 1$ [50 $1$s]. So $1010\ldots 1$ is not divisible by $3$. So the largest power of $3$ that divides this number is $27=3^3$ i.e., the power $3$ is odd.

ETA as noted in the comments by Stinking Bishop already


问:
Probability of distributing n elements into k bins
We are distributing $n$ distinguishable elements into $k$ distinguishable bins where $n>k>2$.

What is the probability of having all $n$ elements landing in the first two bins?

I tackled this problem with star and bars: there are $\binom {n+k-1}{k-1}$ ways to distribute the n elements into the k bins total.

Then I restrict the bins to be 2, and have $\binom{n+2-1}{2-1}$ or n+1 ways to distribute n elements into 2 bins, and took the quotient.

However, I noticed that star and bars works when the elements are indistinguishable, so this answer must be wrong.

My second attempt was to say there is $k^{n}$ ways to distribute $n$ distinguishable elements into $k$ distinguishable bins, which leads to the final answer of $\frac{k^{2}}{k^{n}} = \frac{1}{k^{n-2}}$.

But I picked some numbers and simulated with the computer and the experimental result is also not close with my calculated result.

答:
Each element lands in one of the first two bins. So there are two choices for each element and that makes it $2^n$ choices for $n$ distinguishable elements. But this also includes arrangements where all the $n$ elements are in either bin $1$ or bin $2$. We should subtract those. We also note there is only one way for all elements to be in a bin.

That gives us $2^n-2$ arrangements out of total $k^n$ unrestricted arrangements ($k$ choices for each element and $k^n$ choices for $n$ elements).

So the probability is $\displaystyle \frac{2^n-2}{k^n}$.


问:
Möbius function for prime $p$ and gcd of prime $p$ and $d$ where $d$ divides $n$

Let $\mu(p,d)$ denote the value of the Möbius function at the gcd of
$p$ and $d$. Prove that for every prime $p$ we have

$$\sum_{d|n}\mu(d)\mu(p,d) = \begin{cases} 1  & \text{if $n=1,$} \\ 2
 & \text{if $n=p^a,a \ge1,$}\\ 0 & \text{otherwise} \end{cases}$$

I'm not sure where to start on this question at all; I've always had trouble with these kinds of proofs, and working with primes. Could anyone help with this?

答:
Case 1: $n=1$
In this case our sum simplifies to
$$\mu(1)\mu(p,1)=1$$
The gcd of a prime number and $1$ is obviously $1$.

Case 2: $n=p^a,a\ge1$

We have
$$\sum_{d|p^a}\mu(d)\mu(p,d).$$
Here the crutial term is $\mu(d)$. If $d\neq1$ and $d\neq p$, $\mu(d)$ is always equal to zero, since we can rewrite $d$ as $p^2p^b$, where $b\ge0$. If $d=1$ it is trivial to show that the term is equal to $1$. If $d=p$,  $\mu(p)\cdot \mu(p)$ = $1$.

Case 3: $\mathrm{otherwise}$

For non-quadratics:
Let $p_n,n\in\mathbb{N}$ a prime number.
We start with the product
$p_1\cdot p_2$. The divisors are $1,p_1,p_2$ and $p_1\cdot p_2$. We will define $\nu(p,d) := \mu(d)\mu(p,d)$.
$$\nu(p,1)=+1,\quad \nu(p,p_1)=\mp1, \quad \nu(p,p_2) = -1, \quad \nu(p,p_1p_2)=\pm1$$

If we add another prime, then what happens? The number $p_1p_2p_3$ have the same divisors as $p_1p_2$ but also $p_1p_3$, $p_2p_3$, $p_3$ and $p_1p_2p_3$. So everytime we multiply our expression with another prime we have an increase in $\nu=1$ but the same decrease in $\nu=-1$.

For quadratics: Just an example would be the number $p_1^2p_2$. Divisors, $1,p_1,p_2,p_1p_2,p_1^2,p_1^2p_2$.
Same argument as if it was a non-quadratic but $\nu(p,p_1^2)$ etc. equals zero.

Remark:
You can formulate everything more rigorously and better. I had to do it in my head on my phone.