问:
Proving and supplying counterexamples for few basic facts regarding open and closed sets
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
I spent a considerable amount of time ($>2$ hours) thinking about the below questions, however my arguments are sloppy. Could you give me some hints on how to proceed with (b),(c) and (d)?

Decide whether the following statements are true or false. Provide counterexamples for those that are false, and supply proofs for those that are true.

(a) An open set that contains every rational number must necessarily be all of $\mathbf{R}$.

(b) The Nested Interval property remains true if the term "closed interval" is replaced by "closed set".

(c) Every nonempty open set contains a rational number.

(d) Every bounded infinite closed set contains a rational number.

(e) The Cantor Set is closed.

My Attempt.

(a) I think that this proposition is true. We proceed by contradiction. Let $O \supseteq \mathbf{Q}$ be an open set containing all the rationals. Assume that this is not the whole of $\mathbf{R}$. Then, $\exists x \in O^C \cap \mathbf{R}$. Since $O^C$ is a closed set, there exists a sequence of rational numbers $(q_n) \in O^C$, such that $\lim q_n = x$, with $q_n \ne x$ for all $n \in\mathbf{N}$. But, this implies $O \not\supseteq Q$. This leads to a contradiction. Hence, our initial assumption is false.

(b) I could not think of counterexamples here. For instance, consider the decreasing sequence of closed sets
\begin{align*}
    A_n = \left\{-\frac{1}{n},\frac{1}{n}\right\}
\end{align*}

We have $A_1 \supseteq A_2 \supseteq A_3 \ldots$ and $\cap_{n=1}^{\infty} A_n = \{0\} \ne \emptyset$.

Another example could be,
\begin{align*}
    A_n = \left\{-1-\frac{1}{n},1+\frac{1}{n}\right\}
\end{align*}

(c) Let $O$ be a non-empty open set and say, $x \in O$. Then, there exists an $\epsilon$-neighbourhood $V_\epsilon(x)$ such that $V_\epsilon(x) \subseteq O$. Since, $(x-\epsilon,x] \subseteq O$, and the set of real numbers $\mathbf{R}$ are dense in $\mathbf{Q}$, there exists a rational number $m/n$ between the two reals $x - \epsilon < \frac{m}{n} < x$. So, every non-empty open set $O$ contains a rational number.

(d) Consider the bounded infinite closed sets
\begin{align*}
    A_n = \left\{1 + \frac{1}{n}:n \in \mathbf{N}\right\} \cup \{1\}
\end{align*}

\begin{align*}
    B_n = \left\{\frac{1}{(\sqrt{2})^{2n-1}}:n \in \mathbf{N}\right\} \cup \{0\}
\end{align*}

\begin{align*}
    C_n = \left\{x_n: x_{n+1} = \frac{1}{2}\left(x_n + \frac{2}{x_n}, x_1 = 1\right)\right\} \cup \{\sqrt{2}\}
\end{align*}

$A_n, B_n, C_n$ contain rational numbers. I couldn't think of a counterexample here.

By AoC, every bounded infinite closed subset of $\mathbf{R}$ has a least upper bound. Let $s = \sup A$. Since, $A$ is closed, $s \in A$. We have a dichotomy: either (i) the supremum is irrational or (ii) the supremum is rational number. If $s$ is rational, then we are done. If $s$ is irrational and then there exists a sequence of rationals in $A$, such that $\absval{q - s} < \epsilon$.

(e) The Cantor set is the limit set formed by recursively removing the open middle one-thirds. So, \begin{align*}
    C = [0,1] - \left[\left(\frac{1}{3},\frac{2}{3}\right) \cup \left(\frac{1}{9},\frac{2}{9}\right) \cup \left(\frac{7}{9},\frac{8}{9}\right) \ldots \right]
\end{align*}

The countable union of open sets is open. Since $C$ is the complement of an open set, it is closed.

答:
(a) This is false: $\Bbb R\setminus\left\{\pi\right\}\ne\Bbb R$.

(b) It is also false: $\displaystyle\bigcap_{n\in\Bbb N}[n,\infty)=\emptyset$.

(c) True, since $\Bbb Q$ is dense in $\Bbb R$. Indeed, that's what “dense” means.

(d) It is false. Take, say, $\displaystyle\{\pi\}\cup\left\{\pi+\frac1n\,\middle|\,n\in\Bbb N\right\}$.

(e) Your answer is correct.


问:
Geometric way to view the truncated braid groups?
This is perhaps a vague question, but hopefully there exists literature on the subject. The question is motivated by an answer I gave to this question. I also asked a related question on MO, although hopefully this question should be easier.

There exists a rather remarkable relationship between the 5 platonic solids, and the factor groups of the $n$-string braid groups $B_n$ by adjoining the relation $\sigma_i^k=1$ (here the $\sigma_i$, $1\leq i\leq n-1$ are the usual generators of $B_n$). We'll call these groups $B_n(k)$ the truncated braid groups of type $(n,k)$ where $B_n(k)=B_n/\langle \sigma_i^k\rangle$.

Theorem *: For $n\geq 3$, the group $B_n(k)$ is finite if and only if $k=2$ or $(n,k)$ is the Schläfli type of one of the 5 platonic solids. For these cases, $$|B_n(k)|=\left(\frac{f(n,k)}{2}\right)^{n-1}n!$$ where $f(n,k)$ is the number of faces of the platonic solid of type $(n,k).$

The 5 platonic solids correspond to the pairs $(n,k)\in\{(3,3),(3,4),(4,3),(3,5),(5,3)\}$. This is equivalent to the pair $(n,k)$ being a solution to the inequality $$\frac{1}{n}+\frac{1}{k}>\frac{1}{2}.$$

*It appears that this theorem was proved by Coxeter in

H. S. M. Coxeter, Factor groups of the braid group, Proceedings of the Fourth Can. Math. Cong., Banff 1957, University of Toronto Press (1959), 95–122.

although it is proving difficult for me to find a copy of this online or in my institution's library. From what I can gather, the proof is rather algebraic/combinatorial, although without access to a copy I can't say for sure. My question is:

Question Can one view the finite truncated braid groups in a geometric way as some action (in the vague, not necessarily strict group-action, sense) on the corresponding platonic solid or related objects?

Some of the approaches I have taken have included:

considering the isometry group on the corresponding solid (group orders don't match up),

considering paths on the surface which 'remember' the side that a face was entered from (relations don't match up),

considering labellings of the edges of the faces of the solid so that no face has a pair of edges with the same label, with group elements being permutations of edges which preserve this property (not sure if order matches up - seems difficult to calculate but may be the best approach so far),

considering 'rolling' the solid along a surface until it reaches its starting point again (no obvious way of making this a group via some kind of homotopy)

It seems that there should be some nice geometric interpretation of these groups, especially as every face has $n$ edges and every vertex is shared by $k$ faces.

It would be especially nice if the usual generators can be realised in a nice geometric way. Ultimately it would be nice to extend such a geometric interpretation of these finite truncated braid groups to the infinite cases as well (which correspond to regular tilings of the hyperbolic plane in most cases, and the complex plane in the case $(n,k)\in\{(3,6),(4,4),(6,3)\}$).

答:
Here is a suggestion towards a geometric interpretation of the group $B_3/(\sigma_1^k)$ with $3\le k\le 5$.

First note that $B_3$ is generated by $\sigma_1$ and $x=\sigma_1\sigma_2$. The braid relation is equivalent to $(\sigma_1x)^2=x^3$, and $\Delta=x^3$ is the Garside element. It generates the center of $B_3$, but I will not use this fact.

Let $K$ be the set of oriented edges of the platonic solid. Each such edge $e$
determines two vertices (source $s(e)$ and target $t(e)$) and two faces ($l(e)$
on the left of $e$ and $r(e)$ on the right of $e$). For any oriented edge e, let
$\sigma_1\cdot e$ and $x\cdot e$ be the oriented edges with
$$ s(\sigma_1\cdot e)=s(e),\quad r(\sigma_1\cdot e)=l(e) $$
and
$$ s(x\cdot e)=t(e),\quad l(x\cdot e)=l(e). $$
It follows that
$$ x^3\cdot e=e,\quad \sigma_1\cdot (x\cdot (\sigma_1\cdot (x\cdot e)))=e,
$$
and $k$-fold action of $\sigma_1$ on $e$ gives $e$ for each oriented edge $e$.
It is not very difficult to show that this defines a free and transitive action of
$B_3/(\sigma_1^k,\Delta)$ on $K$.

Since $\Delta $ is central in $B_3$, we may extend this action to an action of
$B_3/(\sigma_1^k)$ on $K\times C_N$ with $N$ the order of $\Delta $ in
$B_3/(\sigma_1^k)$ and $C_N$ the cyclic group of order $N$.
Note that $x^3=\Delta $. In order to implement this action, let us fix a set $L$
of oriented edges such that each face is adjacent to exactly one edge in $L$ in

counterclockwise orientation. (There may be other edges in $L$ adjacent to the
face in clockwise orientation.) Moreover, we add to each oriented edge an integer $l'(e)$. Now we define
$$ \sigma_1\cdot (e,m)=(\sigma_1\cdot e,m),\quad
x\cdot (e,m)=(x\cdot e,m+l'(e)+\varepsilon_e),$$
where $\varepsilon_e\in \{0,1\}$, and
$$ \varepsilon_e=1 \quad \Leftrightarrow e\in L. $$
It turns out that this defines an action of $B_3/(\sigma_1^k)$ on $K\times C_N$
if and only if for each face $F$,
$$ \sum_{e\in K,l(e)=F}l'(e)=0 $$
and for each oriented edge $e$,
$$ l'(e)+l'(-e)+\varepsilon_e+\varepsilon_{-e}=1. $$
Here $-e$ is the edge $e$ with opposite orientation.
Such a labeling exists if and only if $N|f/2$ (with $f$ the number of faces).

It follows that $N=f/2$ and
$$ |B_3/(\sigma_1^k)|=\frac f2|K|=6\left(\frac f2\right)^2 $$
since $|K|=3f$.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
When is $\mathbb Z[\sqrt{-2}] /p\mathbb Z[\sqrt{-2}]$ a field?
Let $R = \mathbb Z[\sqrt{-2}]$ and p be an odd prime. Show that $R/pR$ is a field if and only if $p \equiv  5$, $7$ ($\textrm{mod}$ $8$).

I've already proved that $R$ is an euclidean domain, thus a DIP and I've already proved that if $p \equiv 1$, $3 (\textrm{mod}$ $8)$, then $R/pR$ is not a field.

答:
Relating square roots to roots of unity is the key to quadratic reciprocity.

For $p$ odd. Let $k = \Bbb{F}_p(i,\sqrt2)$ and $\zeta_8 = \frac{\sqrt2+i\sqrt2}{2}\in k$ which is a primitive $8$th root of unity.

Then $\sqrt{-2}= \zeta_8-\zeta_8^{-1}$.

It is in $\Bbb{F}_p$ iff $$\zeta_8^p-\zeta_8^{-p}=(\zeta_8-\zeta_8^{-1})^p=\sqrt{-2}^p=\sqrt{-2}=\zeta_8-\zeta_8^{-1}$$ which happens iff $\zeta_8^p = \zeta_8$ or $-\zeta_8^{-1}$ iff $p\equiv ... \bmod 8$.

From there we know when $x^2+2\in \Bbb{F}_p[x]$ is irreducible thus when $$\Bbb{Z}[\sqrt{-2}]/(p)=\Bbb{Z}[x]/(x^2+2)/(p)=\Bbb{F}_p[x]/(x^2+2)$$ is a field.


问:
How to find the orthogonal complement and the expression of the image of the orthogonal projection?
Let $I=[-1,1]$, $Q=I\times I=[-1,1]^2$, $B=L^2(I,\mathcal{B_I},\lambda_1)$, $H=L^2(Q,\mathcal{B}_{Q},\lambda_2).$
Consider the set $$V=\{u\in H|\exists v\in B, \text{s.t.}~u(x,y)=v(x)~\text{a.e.}~(x,y)\in Q\}.$$

(1) Find the orthogonal complement $V^\bot$.

(2) Letting $P_V$ denote the orthogonal projection, find the expression of $P_Vu$ for all $u\in H$.

(3) Let $u(x,y)=xy$, find $P_Vu$.

I have found that $V$ is a closed subspace of $H$, but then how to find the expression of the image of the orthogonal projection? I don't even know how to express the form of the orthogonal complement the orthogonal complement?

Thanks a lot.

答:
Hints: First observe that if $\int_Iu(x,y)dy=0$ for almost every $x\in I$, then $u\perp V$.

Then define
$$Q:B\to H\ \quad Q(v)=(x,y)\mapsto v(x)\\
P:H\to B\ \quad P(u)=x\mapsto\int_Iu(x,y)dy$$
And prove that they are adjoint to each other.


问:
Counting minimum elements needed such that their sum covers the whole finite space.
In $\mathbb{F}_p$, how to find a subset with smallest cardinality such that the sum between its pairs cover $\mathbb{F}_p\setminus\{0\}$.
So for a subset with cardinality $n$ there are ${n\choose 2}$ pairs and correspondingly ${n\choose 2}$ sums. I want $n$ to be smallest such that those ${n\choose 2}$ sums cover $\mathbb{F}_p\setminus\{0\}$.

For example in $\mathbb{F}_7$, let $A=\{0,1,2,4\}$,
The sums between its pairs covers $\mathbb{F}_7\setminus\{0\}$ since
\begin{align*}
0+1=1,0+2=2,1+2=3,0+4=4,1+4=5,2+4=6\;.
\end{align*}
I want to know if over large $p$ similar set can be easily constructed and what would be the size of that set (I expect it to be close to $\sqrt{2p}$). Any similar literature is also appreciated.

答:
You can solve the problem via integer linear programming as follows.  For $i\in \{0,\dots,n-1\}$, let binary decision variable $x_i$ indicate whether $i$ is selected.  For $0 \le i < j \le n-1$, let binary decision variable $y_{i,j}$ represent the product $x_i x_j$.  The problem is to minimize $\sum_i x_i$ subject to
\begin{align}
\sum_{\substack{0 \le i < j \le n-1:\\ i + j = k}} y_{i,j} &\ge 1 &&\text{for $k \in \{1,\dots,n-1\}$} \tag1 \\
y_{i,j} &\le x_i &&\text{for $0 \le i < j \le n-1$} \tag2 \\
y_{i,j} &\le x_j &&\text{for $0 \le i < j \le n-1$} \tag3
\end{align}
Constraint $(1)$ forces each sum $k$ to be covered.
Constraints $(2)$ and $(3)$ enforce the logical implication $y_{i,j} \implies (x_i \land x_j)$.

The values for $n \in \{1,\dots,50\}$ are
$$
\begin{matrix}
0 &2 &3 &3 &4 &4 &4 &5 &5 &5\\
6 &6 &6 &6 &7 &7 &7 &7 &8 &8 \\
8 &8 &8 &9 &9 &9 &9 &9 &10 &10 \\
10 &10 &10 &10 &11 &11 &11 &11 &11 &11\\
11 &12 &12 &12 &12 &12 &12 &12 &13 &13
\end{matrix}
$$

If you change the condition to $(i+j) \mod n = k$ in constraint $(1)$, the values for $n \in\{1,\dots,50\}$ are instead
\begin{matrix}
0 &2 &3 &3 &4 &4 &4 &5 &5 &5 \\
5 &6 &6 &6 &7 &7 &7 &7 &7 &7 \\
8 &8 &8 &8 &8 &9 &9 &9 &9 &9 \\
9 &10 &10 &10 &10 &10 &10 &11 &11 &11 \\
11 &11 &11 &11 &12 &12 &12 &12 &12 &12
\end{matrix}


问:
Approach ideas for the integral $\int\frac{dx}{(x^4-16)^2}$
Well, the title sums it up pretty well. I'm in search for some smart approach ideas for solving this indefinite integral:
$$\int\frac{dx}{(x^4-16)^2}$$

I know one that would work for sure, namely partial fraction decomposition, but it gets really heavy when regrouping the coefficients for the powers of $x$ and then solving an $8\times8$ system of linear equations. It will eventually work, but I suspect there is something more ingenuine behind this problem.

I also tried all sorts of trigonometric substitutions and formulations, but that added square power really is a bummer to it all.

I'm generally open to any exchange on the topic and would be glad to hear some advice in such situations. Many thanks in advance!

答:
Just integrate by parts

\begin{align}
\int\frac{dx}{(x^4-16)^2}
& =\int \frac1{64x^3} d\left( \frac{-x^4}{x^4-16}\right)
=-\frac1{64} \frac x{x^4-16} -\frac3{64} \int \frac{dx}{x^4-16}\\
&= -\frac1{64} \frac x{x^4-16} -\frac3{64}\cdot\frac18\int \left( \frac1{x^2-4}-\frac1{x^2+4} \right)dx
\end{align}


问:
Combining two identical uniform distributions
Say two random variables, $X$ and $Y$, are such that  $X$ ~ $U(0,a)$ and $Y$ ~ $U(0,a)$.

What will the pdf be for $Z$, where $Z=X-Y$?

答:
Here is how I would start:

$$P(Z \le z) = P(X-Y \le z) = \int_0^a P(X \le z+y) f(y) dy$$


问:
For $f=u+iv$, if $u$ and $v$ are continuously differentiable and $f$ satisfies Cauchy-Riemann equation, then $f$ is holomorphic
I was following the proof by Stein, and wondering where we used the condition of continuity.

So he starts by decoding differentiability of $u$ and $v$, saying that

$$u(x+h_1,y+h_2)-u(x,y)=\frac{du}{dx}h_1 + \frac{du}{dy}h_2+\mid h\mid\psi_1(h)$$ and

$$v(x+h_1,y+h_2)-v(x,y)=\frac{du}{dx}h_1 + \frac{du}{dy}h_2+\mid h\mid\psi_2(h)$$

where $\psi_j (h)\to 0$ $(j=1,2)$ as $\mid h\mid$ tends to $0$ and $h=h_1+ih_2$

Then we use Cauchy-Riemann equation to show that $$f(z+h)-f(z)=(\frac{du}{dx}-i\frac{du}{dy})(h_1+ih_2)+\mid h\mid\psi(h)$$where $\psi(h)=\psi_1(h)+\psi_2(h)\to 0$ as $\mid h\mid\to 0$.

Where did we use the fact that derivative of $u$ and $v$ is continuous?

答:
The continuity of the derivatives is not needed, unless they're using it to infer that $u$ and $v$ are differentiable.

Differentiability is sufficient, see Page 231 of Rudin's Real and Complex Analysis.


问:
Is A-B positive semidefinite if A's eigenvalues are greater than B's
I know that if matrices $A$ and $B$ are positive semi-definite and $A\succeq B$ ($A-B$ is positive semi-definite), then the $i$-th eigenvalue of $A$ is greater than or equal to the i-th eigenvalue of $B$. In other words:
$$\text{If } A \succ 0, B \succ 0, A\succeq B \text{ then } \lambda_i(A) \ge \lambda_i(B), i=1,...,n$$

My question: Is the opposite of this true?
As in, if I know $A$ and $B$ are positive semi-definite and the $i$-th eigenvalue of $A$ is greater than or equal to the $i$-th eigenvalue of $B$, can I conclude that $A\succeq B$? If so, how? Again in other words:
$$\text{If } A \succ 0, B \succ 0, \lambda_i(A) \ge \lambda_i(B), i=1,...,n \text{ then } A\succeq B?$$

答: Certainly not. E.g. $\pmatrix{2\\ &1}-\pmatrix{1\\ &2}$ is indefinite.


问:
Is it possible for $x_{n+1}-x_{n}$ to converge to $0$, but the limit of $x_n$ to not exist?
I was reading about subsequences today and I came across an example which said the if $x_{n+1}-x_{n}$ converges to $0$, it's possible for $x_n$ to not have a finite limit (for example if we take $x_n$ to be the harmonic series). But then I thought "is it possible for the limit of $x_n$ to not exist at all?" and I think the answer is no,because I couldn't find any examples, but I am not sure. Could you please tell me what your opinion is on this?

Thanks in advance!

答:
The answer is yes!

Not only you can find examples were the limit is not finite but is for example $\infty$. You can also have examples were a full interval is a set of limit points of the sequence.

Look at a sequence that will slide back and forth from $0$ to $1$ with steps decreasing to zero for example. Let's for example consider $\{x_n\}$ whose initial terms are
$$
\frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}, \dots$$

Formal definition of $\{x_n\}$ is

$$
x_n=\begin{cases}
\frac{1}{2} &\text{ for } n= 1\\
\frac{n}{k+2} - \frac{k(k+1)}{2(k+2)} &\text{ for } \frac{k(k+1)}{2} \lt n \le \frac{(k+1)(k+2)}{2}
\end{cases}$$

The values of the sequence $\{x_n\}$ are in $(0,1) \cap \mathbb Q$. Moreover, one can notice that $\{x_n\}$ takes each rational number of $(0,1)$ as value an infinite number of times. Indeed for $\frac{p}{q} \in (0,1)$ with $1 \le p \lt q$ and $m \ge 1$ we have $$
\frac{(mq-2)(mq-1)}{2} \lt \frac{(mq-2)(mq-1)}{2} + mp$$ and $$\begin{aligned}
\frac{(mq-2)(mq-1)}{2} + mp &\le \frac{(mq-2)(mq-1)}{2} + m(q-1)\\
&\le \frac{(mq-2)(mq-1)}{2} + mq-1\\
&= \frac{(mq-1)mq}{2}
\end{aligned}$$ Hence $$\begin{aligned}
r_{\frac{(mq-2)(mq-1)}{2} + mp} &= \frac{(mq-2)(mq-1)}{2mq} + \frac{mp}{mq} - \frac{(mq-2)(mq-1)}{2mq}\\
&= \frac{mp}{mq}\\
&= \frac{p}{q}
\end{aligned}$$ proving the desired result. As the rational numbers of the segment $(0,1)$ are dense in $[0,1]$, we can conclude that the set of limit points of $\{x_n\}$ is exactly the interval $[0,1]$.

See here for additional examples.


问:
Martingale and indicator
Exercise comes from "1000 exercices in probability" (12.9.6). Let $X_1, X_2, \dots$ be independent random variables with
$X_n=\begin{cases}
1,  & \text{with probability} & (2n)^{-1}, \\
0,  & \text{with probability} & 1-(n)^{-1}, \\
-1, & \text{with probability} & (2n)^{-1}.
\end{cases}$

Let $Y_{1}=X_{1}$ and for $n\geq2$

$Y_n=\begin{cases}
X_{n},  & \text{if}& Y_{n-1}=0, \\
nY_{n-1}|X_{n}|, & \text{if} & Y_{n-1}\neq0.
\end{cases}$

Show that, $Y_{n}$ is a martingale, with respect to $F_{n}=\sigma(Y_{1},Y_{2},\dots,Y_{n})$.

Solution

With $I_{k}=I_{\{Y_{k}=0\}}$ , where $I$ is indicator function $E(Y_{n}|F_{n-1})=E(X_{n}I_{n-1}+nY_{n-1}|X_n|(1-I_{n-1})\mid F_{n-1})=\dots$

My question is what information this indicator provides and why it is needed. Thanks for help.

答:
Note that $Y_{n-1}$, $I_{n-1}$ and $(1-I_{n-1})$ are $\mathcal{F}_{n-1}$-measurable, thus

$$\mathbb{E}(Y_n \mid \mathcal{F}_{n-1}) = I_{n-1} \cdot \mathbb{E}(X_n \mid \mathcal{F}_{n-1}) + n (1-I_{n-1}) \cdot Y_{n-1} \cdot \mathbb{E}(|X_n| \mid \mathcal{F}_{n-1})$$

Moreover, we have $\mathcal{F}_n = \sigma(Y_1,\ldots,Y_n) \subseteq \sigma(X_1,\ldots,X_n)$. By applying tower property we obtain $$ \begin{align*} \mathbb{E}(|X_n| \mid \mathcal{F}_{n-1}) &= \mathbb{E}(\mathbb{E}(|X_n| \mid \sigma(X_1,\ldots,X_{n-1}) \mid \mathcal{F}_{n-1}) \stackrel{\ast}{=} \mathbb{E}(\mathbb{E}(|X_n|) \mid \mathcal{F}_{n-1}) \\ &= \mathbb{E}(|X_n|) = 2 \cdot \frac{1}{2n} = \frac{1}{n} \end{align*}$$ where we used in  $(\ast)$ the independence of $X_n$ and $\sigma(X_1,\ldots,X_{n-1})$. Similarily, $$\mathbb{E}(X_n \mid \mathcal{F}_{n-1}) = \mathbb{E}X_n = 0$$

Thus $$\begin{align*} \mathbb{E}(Y_n \mid \mathcal{F}_{n-1}) &= 0 + n \cdot (1-I_{n-1}) \cdot Y_{n-1} \cdot \frac{1}{n} \\ &= (1-I_{n-1}) \cdot Y_{n-1} + 0 \cdot I_{n-1} = Y_{n-1} \end{align*}$$

(The last equality follows from the fact that $Y_{n-1} \cdot I_{n-1} = 0 = 0 \cdot I_{n-1}$.)

Title: Martingale and indicator


问:
Find coordinates in 3rd quadrant given distance of $0.08$ from origin where $x$ and $y$ are the same value

"What are the coordinates of the point which has a distance of 0.08m away from the origin. the x and y are both equal to each other and are also negative values?"

I've tried using the distance formula $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and the closest number I got was like $0.05656$ for both $x$ and $y$ but I'm not entirely sure how to do it without just putting random numbers into the calculator as trial and error, is there a certain method that I can use to do this or is it just complete trial and error?

答:
You were almost there!

The point you are looking for is one of the intersections between the circle
$x^2+y^2=0.8^2$ and the line $y=x$.

Substitute and get

$2x^2=0.0064\to x^2= 0.0032\to x=\pm \sqrt{0.0032}\to x\approx\pm 0.0565685$

The point you need is $(-0.0565685,-0.0565685)$


问:
Rank of a matrix vs Dimensions of solution
Let $\bf{A}$ be a matrix acting on $\bf{r}$ $= (x_1,y_1,z_1)^T$ to produce a new vector $\bf{v}$ $=(x_2,y_2,z_2)^T:$
$$\bf{A.r} =
\begin{pmatrix}
1 & 2 & 3\\
2 & 4 & 6 \\
3 & 6 & 9
\end{pmatrix}
\begin{pmatrix}
x_1 \\ y_1 \\z_1
\end{pmatrix}=
\begin{pmatrix}
x_2 \\ y_2 \\z_2
\end{pmatrix}
$$
Clearly this matrix has rank 1 i.e. we can write

$$\begin{pmatrix}
1 & 2 & 3\\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
x_1 \\ y_1 -2x_1 \\z_1 -3x_1
\end{pmatrix}$$
I have a few questions:

What would it mean to "solve" the system of equations (i.e., what are we doing mathematically and what will it tell us at the end)? (If this sounds confused, it's because it is. I can attempt further clarification if it would help.)

The dimension of $\bf{ker\  A}$ is 2 since any vector that we feed in that satisfies $x_1 + 2(y_1-2x_1) + 3(z_1 -3x_1) = 0$ (i.e. a plane of vectors) will give the null vector. What does this imply about the structure of our solution? That is has dimension 1?

答:
A matrix can represent a system of linear equations. In your example, the equations are:

$x\cdot x_1 +2y \cdot y_1 + 3z \cdot z_1=x_2$

$2x\cdot x_1 +4y \cdot y_1 + 6z \cdot z_1=y_2$

$3x\cdot x_1 +6y \cdot y_1 + 9z \cdot z_1=z_2$

After you did gauss elimination you got the equations:

$x\cdot x_1 +2y \cdot (y_1 - 2x_1) + 3z \cdot (z_1-3x_1)=x_2$

$0x\cdot x_1 +0y \cdot (y_1 - 2x_1) + 0z \cdot (z_1-3x_1)=y_2$

$0x\cdot x_1 +0y \cdot (y_1 - 2x_1) + 0z \cdot (z_1-3x_1)=z_2$

Solving the system means get all $(x,y,z)$ satisfying the equations.

$dim(ker A) = 2$ meaning you will have infinite solution (if no contradiction occurs i.e $0\cdot x+0 \cdot y + 0 \cdot z = a \neq 0$), and the degree of freedom is 2.


问:
Is Nested Interval Property $\iff$ Monotone Convergence Property in an ordered field?
Given any ordered field $R$

Can we conclude Monotone convergence Property $\iff$ Nested Interval Property?

Monotone Convergence Property: Any monotone increasing sequence bounded above in $R$ is convergent in $R.$

Nested Interval Property: If $I_1,I_2,...,I_n,...$ be a collection of nested closed intervals,
i.e $I_1 \supseteq I_2 \supseteq ... \supseteq I_n \supseteq ...$ then
$\bigcap_{1}^{\infty} I_i \neq \phi$.

Monotone convergence Property $\implies$ Nested Interval Property.

Let $\{I_n:=[a_n,b_n]:n \in \mathbb{N}\}$ be a nested sequence of closed intervals in $R.$ Then the sequence $\{a_n\}$ is bounded above by $b_1$. Using MCP one concludes $\{a_n\} \to a \in R$.

Claim: $a \in \cap_1^\infty I_i$

I must show that $a_n\leq a$ for all $n \in \mathbb{N}$. If not then there exists $p \in \mathbb{N}$ such that $a<a_p \implies$ $a<a_n$ for all $n\geq p.$ Choose $\epsilon=(a_p-a)/2_{N_{R}}<(a_p-a)$. For this $\epsilon$ there is no $k \in \mathbb{N}$ such that $|a_n-a|_R < \epsilon$ for all $n \geq k$ . Contradiction. Hence we can conclude $a_n\leq a$ for all $n \in \mathbb{N}$. i.e $a$ is an upperbound of the sequence $\{a_n\}$.

Want to show: $a\leq b_n$ for all $n \in \mathbb{N}$. If not then there exists $q \in \mathbb{N}$ such that $b_q\implies a_n<b_n\leq b_q<a$ for all $n\geq q.$ i.e There are no elements of $\{a_n\}$ but finitely many in between $b_q$ and $a$, contradicting $(a_n) \to a$. Hence $a\leq b_n$ for all $n \in \mathbb{N}$.

So we conclude that $a \in \cap_1^\infty I_i$ $\implies$ $\cap_1^\infty I_i \neq \phi.$ i.e Nested Interval Property.

My questions are the following:

Please tell me if the above proof is correct. i.e Is the above proof valid in an arbitrary ordered field with MCP?

I am not sure about the converse. If the converse is also true, how to prove it. If not, counterexample? Is the converse true in an archimedean ordered field?

答:
Your proof is fine.

If an ordered field satisfies MCP, then it is Dedekind complete (see page 11 here), thus isomorphic to $\mathbb R$ and hence has every nice property the reals enjoy.


问:
Integral of rational functions.
I want to evaluate this integral:

$$\int{\frac{ax+b}{(x^2+2px+q)^n}}dx$$

The book only says to integrate by parts $\int{\dfrac{1}{(x^2+2px+q)^{n-1}}dx}$,
for simplicity if $n = 2$ I get:
$\int{\dfrac{1}{(x^2+2px+q)}dx}=\dfrac{x}{x^2+2px+q}+\int{\dfrac{2x^2+2px}{(x^2+2px+q)^2}}dx$.

Now I don't know what to do.

答:
General description. A partial fraction of the form $$\dfrac{Bx+C}{\left[ \left( x-r\right) ^{2}+s^{2}\right] ^{n}}$$ is integrable by substitution, using the change of variables $x=r+st$. With this method an integral of the form

$$\displaystyle\int \frac{Bx+C}{\left[\left( x-r\right) ^{2}+s^{2}\right] ^{n}}dx\tag{1}$$

is transformed into an integral of the form

$$\displaystyle\int \frac{Dt+E}{(t^{2}+1)^{n}}dt=D\int \frac{t}{(t^{2}+1)^{n}}
dt+E\int \frac{1}{(t^{2}+1)^{n}}dt\tag{2}.$$

The first integral is a table integral. The second one satisfies a recurrence relation.

Application to the given integral. Rewrite the integrand as $\dfrac{ax+b}{\left[(x+p)^{2}+q-p^{2}\right]^{n}}$ and make the substitution
$$
\begin{equation*}
x=-p+\sqrt{q-p^{2}}\, t.\tag{3}
\end{equation*}
$$
Then
$$
\begin{eqnarray*}
I &=&\int \frac{ax+b}{(x^{2}+2px+q)^{n}}dx=\int \frac{ax+b}{
\left[(x+p)^{2}+q-p^{2}\right]^{n}}dx, \tag{4}\\
&& \\ \\
M &=&a\sqrt{q-p^{2}},\quad N=b-ap,\quad K=\frac{\sqrt{q-p^{2}}}{\left(
q-p^{2}\right) ^{n}} \\ \\
&& \\
I &=&K\int \frac{M}{(t^{2}+1)^{n}}t+\frac{N}{(t^{2}+1)^{n}}dt, \\
&=&KM\int \frac{1}{(t^{2}+1)^{n}}tdt+KN\int \frac{1}{(t^{2}+1)^{n}}dt \\
&=&-\frac{KM}{2(n-1)}\frac{1}{(t^{2}+1)^{n-1}}+KN\int \frac{1}{(t^{2}+1)^{n}}
dt.\tag{5}
\end{eqnarray*}
$$
The last integral can be split into two, the latter being integrable by
parts which generates the recursive relation $(6)$ bellow. For $n\ne 1$, we have:
$$
\begin{eqnarray*}I_n&=&
\int \frac{1}{(t^{2}+1)^{n}}dt =\int \frac{1+t^{2}-t^{2}}{(t^{2}+1)^{n}}
dt\\ &=&\int \frac{1}{(t^{2}+1)^{n-1}}dt-\int \frac{t^{2}}{(t^{2}+1)^{n}}dt \\
&=&\int \frac{1}{(t^{2}+1)^{n-1}}dt+\frac{t}{2(n-1)\left( t^{2}+1\right)
^{n-1}} \\ &\qquad - &\frac{1}{2(n-1)}\int \frac{1}{\left( t^{2}+1\right) ^{n-1}}\,dt \\
&=&\frac{t}{2(n-1)\left( t^{2}+1\right) ^{n-1}}+\frac{2n-3}{2(n-1)}\int
\frac{1}{\left( t^{2}+1\right) ^{n-1}}\,dt\\
&=&\frac{t}{2(n-1)\left( t^{2}+1\right) ^{n-1}}+\frac{2n-3}{2(n-1)}I_{n-1}.\tag{6}
\end{eqnarray*}
$$

For $n=1$
$$
I_1=\int \frac{1}{t^{2}+1}dt=\arctan t+\text{Constant}.\tag{7}
$$

ADDED. For the particular case $n=2$, we get from $(6),(7)$
$$
\begin{eqnarray*}
I_2=\int \frac{1}{(t^{2}+1)^{2}}dt &=&\frac{1}{2}\frac{t}{t^{2}+1}+\frac{1}{2}I_1
 \\
&=&\frac{1}{2}\frac{t}{t^{2}+1}+\frac{1}{2}\arctan t+\text{Constant}.\tag{6$\mathrm{a}$}
\end{eqnarray*}
$$

and from $(5)$

$$
\begin{eqnarray*}
I &=&\int \frac{ax+b}{(x^{2}+2px+q)^{2}}dx \\
&=&-\frac{a}{2}\frac{1}{x^{2}+2px+q}+\frac{b-ap}{2\left( q-p^{2}\right) }
\frac{x+p}{q+x^{2}+2xp}\\ &+& \frac{b-ap}{2\left( q-p^{2}\right) ^{3/2}}\arctan
\frac{x+p}{\sqrt{q-p^{2}}}+\text{Constant}.
\end{eqnarray*}\tag{5$\mathrm{a}$}
$$

Setting $a=0,b=1$ in $(5\mathrm{a})$ we evaluate your last integral for $q>p^2$:

$$
\boxed{\begin{eqnarray*}
\int \frac{1}{(x^{2}+2px+q)^{2}}dx&=&\frac{1}{2\left( q-p^{2}\right) }\frac{x+p}{x^{2}+2px+q}\\&+&\frac{1}{2\left(
q-p^{2}\right) ^{3/2}}\arctan \frac{x+p}{\sqrt{q-p^{2}}}+\text{Constant}.
\end{eqnarray*}}
$$
$$\tag{5$\mathrm{b}$}$$

Title: Integral of rational functions.


问:
Taking the limit on the first variable, or over the first variable, or?
I asked the following question on the English Usage page, and it made people furious why I didn't ask here. So I do: suppose $f(x,y)$ is a function of two variables and I want to say: "take the limit of $f$, as $x$ approches a given variable". Can I say "take the limit of $f$ on the first variable? Or take the limit over the first variable? Suppose from the context it is clear toward which variable $x$ is approaching.

答:
According to google's n-gram viewer, the best phrase is:

'take the limit of '

Do realise that the meaning is the same regardless, but you're better off saying what is most commonly used.

It actually doesn't really matter whether you refer to more than one variable or multiple. This is because it wouldn't make sense logically or semantically for it to change whether you are taking limits of multiple variables or just one; this is due to the fact that contextually the 'verb' is the same, just not the subject.

I can jump 'over' them. I can also jump 'over' him.

The object of the preposition does not affect it. However:

I rammed 'into' them.

I have had to change my preposition such that it matched the context of my verb.

The same applies here


问:
Maximizing the product of numbers after summing certain set
Let we have three baskets A,B,C.

And we also have a set of integers X={$x_1$, ..., $x_n$}.

For each $x_i$, we should put it in one of the baskets.

After this, for each basket, we sum every integer in it.

Let's name them as a, b, c perspectively.

Our goal is maximizing the product: $a*b*c$.

For example... let X = {1,1,2,2}

Then the solution will be as following:

A : {1,1}

B : {2}

C : {2}

I think this problem is somewhat similar to knapsack... but different.

Any idea will be really appreciated.

Extended version : What if the baskets already have integers?

答:
You can solve this via integer linear programming as follows.  Let $I$ be the set of items, and let $B$ be the set of bins. For $i\in I$ and $b \in B$, let binary decision variable $y_{i,b}$ indicate whether item $i$ appears in bin $b$.  The problem is to maximize
$$\prod_b \sum_i x_i y_{i,b}$$
subject to
\begin{align}
\sum_b y_{i,b} &= 1 &&\text{for $i \in I$} \\
\end{align}
The nonlinear polynomial objective can be linearized by replacing each product of binary variables with a new binary variable and linear constraints, as described here.

For your example, $I=\{1,2,3,4\}$, $B=\{1,2,3\}$, $x_1=x_2=1$, and $x_3=x_4=2$.  Your sample optimal solution is $y_{1,1}=y_{2,1}=y_{3,2}=y_{4,3}=1$, with all other variables $0$.


答:
The proof is fine. You can make the proof that $a$ is an upper bound for the sequence of left endpoints very slightly simpler by letting $\epsilon=a_p-a$: then $|a_n-a|_R\ge\epsilon$ for all $n\ge p$, and you get the same contradiction.

The converse is false, but the counterexamples that demonstrate this are not elementary. One is discussed on page $13$ of this PDF, and another is briefly outlined in this answer to an earlier question on the subject.

Title: Is Nested Interval Property $\iff$ Monotone Convergence Property in an ordered field?

Body:
Given any ordered field $R$

Can we conclude Monotone convergence Property $\iff$ Nested Interval Property?

Monotone Convergence Property: Any monotone increasing sequence bounded above in $R$ is convergent in $R.$

Nested Interval Property: If $I_1,I_2,...,I_n,...$ be a collection of nested closed intervals,
i.e $I_1 \supseteq I_2 \supseteq ... \supseteq I_n \supseteq ...$ then
$\bigcap_{1}^{\infty} I_i \neq \phi$.

Monotone convergence Property $\implies$ Nested Interval Property.

Let $\{I_n:=[a_n,b_n]:n \in \mathbb{N}\}$ be a nested sequence of closed intervals in $R.$ Then the sequence $\{a_n\}$ is bounded above by $b_1$. Using MCP one concludes $\{a_n\} \to a \in R$.

Claim: $a \in \cap_1^\infty I_i$

I must show that $a_n\leq a$ for all $n \in \mathbb{N}$. If not then there exists $p \in \mathbb{N}$ such that $a<a_p \implies$ $a<a_n$ for all $n\geq p.$ Choose $\epsilon=(a_p-a)/2_{N_{R}}<(a_p-a)$. For this $\epsilon$ there is no $k \in \mathbb{N}$ such that $|a_n-a|_R < \epsilon$ for all $n \geq k$ . Contradiction. Hence we can conclude $a_n\leq a$ for all $n \in \mathbb{N}$. i.e $a$ is an upperbound of the sequence $\{a_n\}$.

Want to show: $a\leq b_n$ for all $n \in \mathbb{N}$. If not then there exists $q \in \mathbb{N}$ such that $b_q\implies a_n<b_n\leq b_q<a$ for all $n\geq q.$ i.e There are no elements of $\{a_n\}$ but finitely many in between $b_q$ and $a$, contradicting $(a_n) \to a$. Hence $a\leq b_n$ for all $n \in \mathbb{N}$.

So we conclude that $a \in \cap_1^\infty I_i$ $\implies$ $\cap_1^\infty I_i \neq \phi.$ i.e Nested Interval Property.

My questions are the following:

Please tell me if the above proof is correct. i.e Is the above proof valid in an arbitrary ordered field with MCP?

I am not sure about the converse. If the converse is also true, how to prove it. If not, counterexample? Is the converse true in an archimedean ordered field?


问:
How to find the norm of the operator $(Ax)_n = \frac{1}{n} \sum_{k=1}^n \frac{x_k}{\sqrt{k}}$?
How to find the norm of the following operator
$$
A:\ell_p\to\ell_p:(x_n)\mapsto\left(n^{-1}\sum\limits_{k=1}^n k^{-1/2} x_k\right)
$$
Any help is welcome.</p>

答:
Consider diagonal operator
$$
S:\ell_p\to\ell_p:(x_n)\mapsto(n^{-1/2}x_n)
$$
It is bounded and its norm is $\Vert S\Vert=\sup\{|n^{-1/2}|:n\in\mathbb{N}\} =1$

Consider Caesaro operator
$$
T:\ell_p\to\ell_p:(x_n)\mapsto\left(n^{-1}\sum\limits_{k=1}^nx_k\right)
$$
As it was proved earlier its norm is is $\Vert T\Vert\leq p(p-1)^{-1}$. Since $A=T\circ S$, then
$$
\Vert A\Vert\leq\Vert T\Vert\Vert S\Vert=p(p-1)^{-1}
$$
But unfortunately I can''t say this is the best constant.


答:
Take $$x_n=\sin(\sqrt{n})$$

By MVT,

$$\lim_{n\to +\infty}(\sin(\sqrt{n+1})-\sin(\sqrt{n}))=0$$

but

$$\lim_{n\to+\infty}x_n$$
does not exist.

OR

$$x_n=\sum_{k=0}^n\frac{1}{k!}$$
as a sequence of rationnals.

it has no limit in $ \Bbb Q$.

Title: Is it possible for $x_{n+1}-x_{n}$ to converge to $0$, but the limit of $x_n$ to not exist?

Body:
I was reading about subsequences today and I came across an example which said the if $x_{n+1}-x_{n}$ converges to $0$, it's possible for $x_n$ to not have a finite limit (for example if we take $x_n$ to be the harmonic series). But then I thought "is it possible for the limit of $x_n$ to not exist at all?" and I think the answer is no,because I couldn't find any examples, but I am not sure. Could you please tell me what your opinion is on this?

Thanks in advance!


问:
Meager set and $\mathrm{MA}$
Martin's axiom. Let $\langle\mathbb P,\leq\rangle$ be a ccc partially ordered set. If $\mathcal D$ is a family of dense subset of $\mathbb P$ such that $|\mathcal D|<\mathfrak c$, then there exists a $\mathcal D$-generic filter $F$ in $\mathbb P.$ it is usually abbreviated by $\mathrm{MA}$. The following theorem is well known.

Theorem: If $\mathrm{MA}$ holds then a union of less than continuum many meager subsets of $\mathbb R$ is meager in $\mathbb R$.

So, it is clear that any subset of $\mathbb R$ with cardinality less than $\mathfrak c$ is meager.

My question is Can the theorem above be hold with a weaker set-theoretical assumption ? or we need the whole $\mathrm{MA}$ to prove it ?

Any help will be appreciated greatly.

答:
Probably the simplest way to build a model in which MA fails but every union of $<\mathfrak c$ meager sets is meager is to start with a model of GCH and adjoin Hechler reals in a finite-support iteration of length $\aleph_2$.

A table of facts of this sort is in Section 11 of my chapter of the Handbook of Set Theory, a preprint of which is available at http://www.math.lsa.umich.edu/~ablass/hbk.pdf .

For a lot more information about MA (and weaker forms of MA), the standard reference is David Fremlin's book, "Consequences of Martin's Axiom".

Title: Meager set and $\mathrm{MA}$


问:
What problems are known to be require superpolynomial time or greater to solve?
I'm having a hard time finding problems that known to require greater than polynomial time to solve (superpolynomial), particularly in graph theory.

So, what problems (or better yet, class of problems) have been proven to require superpolynomial time or more to reach a definite solution (especially graph theory)?

To clarify, I'm aiming for mathematically important problems, like the traveling salesman problem, graph isomorphism, counting problems, decision and satisfyability problems. There are many, many examples in P and NP, but exptime seems harder to turn up (it's possible I'm just bad at googling :-)).

答:
Wiki gives an interesting example here

Consider Sudoku, a game where the player is given a partially filled-in grid of numbers and attempts to complete the grid following certain rules. Given an incomplete Sudoku grid, of any size, is there at least one legal solution? Any proposed solution is easily verified, and the time to check a solution grows slowly (polynomially) as the grid gets bigger. However, all known algorithms for finding solutions take, for difficult examples, time that grows exponentially as the grid gets bigger.

And for EXPTIME, here:

In computability theory, one of the basic undecidable problems is the halting problem: deciding whether a deterministic Turing machine (DTM) halts. One of the most fundamental EXPTIME-complete problems is a simpler version of this, which asks if a DTM halts in at most k steps. It is in EXPTIME because a trivial simulation requires O(k) time, and the input k is encoded using O(log k) bits which causes exponential number of simulations. It is EXPTIME-complete because, roughly speaking, we can use it to determine if a machine solving an EXPTIME problem accepts in an exponential number of steps; it will not use more

For Graph Theory, this article rather recently proved EXPTIME-completeness:

We investigate the computational complexity of deciding whether k cops can capture a robber on a graph G. Goldstein and Reingold (1995) [8] conjectured that the problem is EXPTIME-complete when both G and k are part of the input; we prove this conjecture.

I'm not sure of your mathematical ability so I'll just leave the wiki article as to what cops and robbers are; sorry if this is patronising (this is new to me).


答:
Let
$$x_n-x_{n-1}=\frac{(-1)^{\lceil\log_2n\rceil}}n$$
This is in effect taking the harmonic $x_n=H_n$ example, which has $\infty$ as a limit, and "folding it" into infinitely many intervals where $x_n$ goes up and down alternately by at least $\frac12$ per interval:

From $x_2$ to $x_4$, $\frac13+\frac14>\frac12$ is added

From $x_4$ to $x_8$, $\frac15+\frac16+\frac17+\frac18>\frac12$ is subtracted, etc.

Thus $x_n$ cannot have a limit.


问:
If $|a_{n+1} - a_n| < 2^{-n}$ prove $(a_n)$ converges.
The solutions present a different proof to what I came up with.

I am given $|a_{n+1} - a_n| < 2^{-n}$

Suppose $m$ >> $n$. Now : $$|a_{n+1} - a_m + a_m - a_n| < 2^{-n}$$ $$|a_{n+1} - a_m + a_m - a_n| \ge |a_m - a_n|-|a_{m}-a_{n+1}| \ge|a_m-a_n| < 2^{-n}$$

Now pick $N$ and $\epsilon > 0$ such that $2^{-N} < \epsilon$.

Then for $m > n > N$ I have $|a_m -  a_n|< \epsilon \iff$ Cauchy therefore convergence.

Is this right? Or do I need to tweak some things

答: You proof is wrong because you can't conclude that $|a_m-a_n|<2^{-n}$, you can only say that $2^{-n}+|a_m-a_{n+1}|\geq |a_m-a_n|$. However this can be fixed, if $m> n$, applying your result repeatedly will lead you to $$2^{-n}+2^{-n+1}+\cdots+2^{-(m-1)} \geq |a_m-a_n|$$ from which you can easily show that $a_n$ is Cauchy.


问:
What is an open set in a topological space?
I recently started learning topology to help me understand limits and continuity better for calculus, and I am struggling with some of the definitions.

What I am getting confused with is why is every set in a topology considered to be open and when talking about sets in the topology we always say the set is open.

My intuitive notion of openness from  previous knowledge of mathematics is an interval that does not contain its endpoints, so there is an infinite sequence at the end points, e.g., $(0,1)$ is an open interval.

However, in topology, for example, the singleton $\{1\}$
is considered an open set—how is this so? Why are sets in a topology always open? And what is the definition of an open set in a topological space?

Thanks in advance.

答:
I have read above interesting answers! Here I would like to share a way how I understand the generalized notion open set from the familiar notion open interval.

Imagine that you are a mathematician! LET'S GO!!!

Part 1. What is "Topology"?

Now, you want is to give a definition for "open set" (in a general set, for example, a discrete set), which generalizes the open intervals $(a,b)$ (in real line). What will you do?

...thinking...

Yes, exactly! You have to collect the key properties of open intervals!

To obtain the key properties, the most important thing is to distinguish the open interval $(a,b)$ and the closed interval $[a,b]$. (Actually, we should compare it with half-open intervals $[a,b)$ and $(a,b]$, but we shall ignore them to avoid the unnecessary complexity ^o^).

The first  difference came to our mind is probably the appearance of the "end points" $a, b$ in $[a,b]$ while $(a,b)$ doesn't have. Now, how to generalize the notion of "end points" to a discrete setting? It seems to be challenging!

...Let's think in a different way...Let's think creatively...

Eureka! Ohlala!

Let $\mathcal T$ be the set of all open intervals (loosely speaking!)

It means: a set is open if and only if it belongs to $\mathcal T$! Now, we only need to study the properties of $\mathcal T$, which is a collection of sets. This approach seems to be promising to defined an open in discrete setting as we desired! That's so cool! Right? Hahaha!

Let's give $\mathcal T$ a name!

$\mathcal T$ is called a topology.

Yup, what a super ultra cool name!

Part 2. The characteristics of Topology.

When we study a collection of sets. What we should do is to investigate the intersection and the union . Let's start!

Part 2.1. The intersection.

Now, look at the intersection of two sets
\begin{align}
(a,b)\cap (c,d)& =(b,c)\\
[a,b]\cap [c,d]& =[b,c]
\end{align}
for $a<c<b<d$. It means the intersection of two open intervals (res. closed intervals) is also an open interval (res. closed interval). So what is the difference here? Nothing! Let's consider  an arbitrary intersection
\begin{align}
\bigcap_{n=1}^\infty (0,1+\frac{1}{n})=(0,1]\\
\bigcap_{n=1}^\infty [0,1+\frac{1}{n}]=[0,1].
\end{align}
Oh la la, this is exactly the difference that we are looking for! It means that the arbitrary intersection of open intervals is not an open interval.
Hence, there is a difference between finite intersection and arbitrary intersection! Keep this in mind!
Now, we may consider this fact as a feature of $\mathcal T$.

Intersection of two (or finite) open intervals is again an open interval.

Part 2.2. Union.
Analogously, we find that
\begin{align}
\bigcup_{n=1}^\infty (0,1-\frac{1}{n})=(0,1)\\
\bigcup_{n=1}^\infty [0,1-\frac{1}{n}]=[0,1).
\end{align}
and claim another feature of $\mathcal T$ is

Union of arbitrary open intervals is still an open interval.

Part 2.3. Trivial sets.
Now, consider the empty set and the whole real line.
Since they cannot be written as $(a,b)$ nor $[a,b]$, how can we give them a reasonable property, open or closed? From the intersection part, we expect that intersection of any two open sets is again an open set, and clearly, the intersection may be empty. Thus, it is reasonable to define the empty set as an open set! Similarly, from the union part, it is reasonable to define the whole set as an open set! Overall, we may state that

The empty set and the whole set are open.

Part 3. Every set can be open in some sense.

Let $X$ be a set, it may be a real line or any discrete set.
Now, any subsets of $X$ can be considered as an open set if it belongs to a collection satisfying the three properties above, and we call such collection a "topology"! Conversely, "topology" is just a collection of open sets.

In a nutshell, any set (including the singleton) will be open as soon as you can put it into a collection, named topology, satisfying the three properties.


问:
Simplifying the integral $\int\frac{dx}{(3 + 2\sin x - \cos x)}$ by an easy approach
$$I=\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)}$$

If $$\tan\left(\frac{x}{2}\right)=u$$

or $$x=2\cdot\tan^{-1}(u)$$

Then,

$$\sin{x}=\dfrac{2u}{1+u^2}$$

$$\cos{x}=\dfrac{1-u^2}{1+u^2}$$

$$dx=\dfrac{2}{1+u^2}$$

Substitute $$\tan\left(\dfrac{x}{2}\right)=u$$

Let us simplify the integrand before integrating

$$\dfrac{1}{3+2\sin{x}-\cos{x}}$$

$$=\dfrac{1}{3+2\frac{2u}{1+u^2}-\frac{1-u^2}{1+u^2}}$$

$$=\dfrac{1}{3+\frac{4u-1+u^2}{1+u^2}}$$

$$=\dfrac{1}{\frac{4u-1+u^2+3+3u^2}{1+u^2}}$$

$$=\dfrac{1+u^2}{4u^2+4u+2}$$

$$=\dfrac{1+u^2}{(2u+1)^2+1}$$

$$I=\displaystyle\int\dfrac{1+u^2}{(2u+1)^2+1}\cdot\dfrac{2}{1+u^2}\ du$$

$$=\displaystyle\int\dfrac{1}{(2u+1)^2+1}\ 2\,du$$

Now,

Take :  $$v=2u+1$$

Therefore, $$dv=2\,du$$

$$I=\displaystyle\int\dfrac{1}{v^2+1}\ dv$$

$$I=\tan^{-1}(v)$$

Substitute everything back

$$I=\tan^{-1}(2u+1)$$

$$I=\tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)$$

$$\boxed{\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)} = \tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)+C}$$

I know that my approach is also not so difficult but still I think there must be an relatively easy approach to this integral. I have tried many different things using trigonometric identities but nothing seems to bring to the solution easily. Kindly help me out.

答:
With $a=\tan^{-1}\frac1{\sqrt5}$ and $t=\frac12(x+a)$

\begin{align}
\int\frac{dx}{3 + 2\sin x - \cos x}
& = \int \frac{dx}{3-\sqrt5\cos(x+a)}\\
&= \int \frac{2d(\tan^2t)}{(3-\sqrt5)+(3+\sqrt5)\tan^2t}=\tan^{-1} \frac{2\tan t}{3-\sqrt5} +C
\end{align}


问:
Why do we take the axiom of induction for natural numbers (Peano arithmetic)?
More precisely, when we define the set of natural numbers $\mathbb{N}$ using the Peano axioms, we assume the following:

$0\in\mathbb{N}$

$\forall n\in\mathbb{N} (S(n)\in\mathbb{N})$

$\forall n\in\mathbb{N}(0\neq S(n))$

$\forall m,n (m\neq n\to S(m)\neq S(n))$

If $P(n)$ denotes the fact that $n$ has property $P$, then $\Big(P(0)\wedge \forall n\in\mathbb{N}\big(P(n)\to P(S(n))\big)\Big)\implies \forall n\in \mathbb{N} (P(n))$

I understand that using these axioms we can derive everything about the natural numbers, but I also think it's helpful to know why the axioms were chosen the way they are. So my question is why we choose to accept the axiom of induction ((5.) above), which in a way makes this more of a metamathematical question.

For example in Tao's Analysis I, it says that the axiom of induction keeps unwanted elements (such as half-integers) from entering the set.

Wikipedia says, "Axioms [1], [2], [3] and [4] imply that the set of natural numbers is infinite, because it contains at least the infinite subset $\{ 0, S(0), S(S(0)), \ldots \}$, each element of which differs from the rest. To show that every natural number is included in this set requires an additional axiom, which is sometimes called the axiom of induction. This axiom provides a method for reasoning about the set of all natural numbers."---But I find this tautological: $\mathbb{N}$ is defined as the set of natural numbers so "$n$ is a natural number" means "$n\in\mathbb{N}$", right? So isn't every natural number included in  $\mathbb{N}$ by definition?

Suppose we want to show $\mathbb{N}=\{0,1,2,3,\ldots\}$ using all five of the Peano axioms.

If we let $P(n)$ denote $n\in\{0,1,2,3,\ldots\}$, then $P(0)$ is true. Suppose $n$ is in $\{0,1,2,3,\ldots\}$. Then (informally) the dots indicate that $S(n)$ is in $\{0,1,2,3,\ldots\}$. So $\mathbb{N}\subseteq\{0,1,2,3,\ldots\}$, i.e., our defined set contains no "extra" elements (as in Tao's Analysis I).

Yet I still do not see how to show $\{0,1,2,3,\ldots\}\subseteq\mathbb{N}$ (in order to complete the "proof" that $\mathbb{N}=\{0,1,2,3,\ldots\}$) without just assuming it. (I think this is what the Wikipedia article was doing(?))

Thanks in advance for any help and I apologize if this kind of question is unsuitable for this site.

答:
Note: I'm still learning about the Peano Axioms, so if any part of my answer is inaccurate, please let me know.

First, consider why we even bother thinking about the natural numbers. Of course, it's because they obey unique properties; namely, the natural numbers are "natural" in the sense that they represent how we count real-life objects. Clearly, the set of natural numbers is no ordinary set, so we might say that $\mathbb{N}$ is a "set with structure," where the "structure" encapsulates all the unique properties that make the natural numbers special.

Think of the Peano Axioms as a means of formulating this "structure." Any set obeying this "structure" should either be the natural numbers, or a set isomorphic to the natural numbers. Therefore, it doesn't make much sense to try and prove that $\{0, 1, 2, 3, \dots\} \subseteq \mathbb{N}$ (as you do in your answer); instead, you should be trying to prove that the set $\mathbb{N} = \{0, 1, 2, 3, \dots \}$ paired with the canonical successor function obeys the Peano Axioms, and thus represents the natural numbers.

If you view the axioms in this way, your question can be reframed as, "Why are the first four Peano Axioms insufficient in defining the 'structure' of $\mathbb{N}$?" A simple counterexample suffices to answer this, which Tao's half-integers does beautifully. Tao defines the half-integers to be

$$\mathbb{N} := \{0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, \dots\}.$$

and its successor function $S$ to be

$$
S : \mathbb{N} \to \mathbb{N} \\
S(n) = n + 1
$$

where $+$ is the standard real addition operator. It isn't difficult to check that this obeys the first four Peano Axioms. But this set is not the natural numbers, nor is it isomorphic to them! Therefore, we add a fifth axiom, which (in plain English) states that, "All natural numbers  are some eventual successor of $0$."


问:
Is "Convolution operator" well-defined and injective?
Problem: Let $g_t(x)=\frac{1}{\sqrt{2\pi}} e^{ \frac{- (x-t)^2}{2}}$ be the Gaussian function centered at $t\in \mathbb R$. Let $I$ be a subset of $\mathbb R$. Consider the "convolution operator"
\begin{align}
\mathcal A: L^1(I) & \longrightarrow L^2(I)\\
f & \longmapsto \left(x\longmapsto  \int_{I} g_t(x)f(t) dt \right).
\end{align}
How can I prove (or disprove) that the operator is well-defined and injective?

My attempts: Here are some of my attempts to prove the well-defined property.

I have tried to use Cauchy inequality but with no success:
\begin{align}
\int_{I} \left( \int_{I} g_t(x)f(t)dt\right)^2 dx \leq \int_{I} \int_{I} g_t^2(x)dx dt \int_{I}f^2(t)dt.
\end{align}
as the right-hand-side may be infinite as $f$ does not belong to $L^2(I)$.

I have tried to use the boundedness of Gaussian function but nothing better. We have
\begin{align}
\int_{I} \left( \int_{I} g_t(x)f(t)dt\right)^2 dx \leq
M \left( \int_{I} f(t)dt \right)^2 \int_I dx.
\end{align}
and the right-hand-side may be infinite as $I$ is unbounded.

答:
I'm going to unclutter the situation and leave out some constants. If it works in the simplified case, you can throw constants at it later.

Suppose $f\in L^1(\mathbb R)$ and $g(t) = e^{-t^2}.$ For $x\in \mathbb R,$ define

$$(g*f)(x) = \int_{\mathbb R}g(x-t)f(t)\,dt.$$

That is certainly well defined for any $x,$ since $g\in L^\infty.$ To show $g*f\in L^2,$ note

$$|g*f(x)|^2 \le (\int_{\mathbb R}g(x-t)|f(t)|\,dt)^2$$ $$ = \|f\|_1^2(\int_{\mathbb R}(g(x-t)(|f(t)|/\|f\|_1)\,dt)^2.$$

Use Jensen to see the last expression is bounded above by

$$\|f\|_1^2\int_{\mathbb R}(g(x-t))^2(|f(t)|/\|f\|_1)\,dt.$$

Thus, using Fubini, we have

$$\int |g*f(x)|^2\,dx \le \|f\|_1^2 \int g(x)^2\,dx.$$

This proves $\|g*f\|_2 \le \|f\|_1\|g\|_2.$ Thus $f\to g*f$ is a bounded linear operator from $L^1$ to $L^2.$

For injectivity, I'll use some Fourier analysis. Let $F$ denote the Fourier transform. Then $F$ is an isometry on $L^2,$ and is injective on $L^1.$ Thus $g*f$ is the zero function iff $F(g*f)$ is the zero function. Now

$$F(g*f)(x) =F(g)(x)\cdot F(f)(x).$$

And as is well known, $F(g)>0$ everywhere. (In fact for the Gaussian $g,$ $F(g)$ is essentially $g,$ meaning $F(g)(x)= ag(bx)$ for some nonzero constants $a,b.$) Thus $F(g*f)=0$ iff $F(f)=0.$ And the later happens iff $f=0.$ It follow that $f\to g*f$ is injective.


Title: If $|a_{n+1} - a_n| < 2^{-n}$ prove $(a_n)$ converges.

Body:
The solutions present a different proof to what I came up with.

I am given $|a_{n+1} - a_n| < 2^{-n}$

Suppose $m$ >> $n$. Now : $$|a_{n+1} - a_m + a_m - a_n| < 2^{-n}$$ $$|a_{n+1} - a_m + a_m - a_n| \ge |a_m - a_n|-|a_{m}-a_{n+1}| \ge|a_m-a_n| < 2^{-n}$$

Now pick $N$ and $\epsilon > 0$ such that $2^{-N} < \epsilon$.

Then for $m > n > N$ I have $|a_m -  a_n|< \epsilon \iff$ Cauchy therefore convergence.

Is this right? Or do I need to tweak some things

Body:
Let $ n,p >0$.

$$|a_{n+p}-a_n|=$$
$$|a_{n+p}-a_{n+p-1}+...+a_{n+1}-a_n|\le$$

$$2^{-(n+p-1)}+...+2^{-(n+1)}+2^{-n}=$$

$$2^{-n}(1+2^{-1}+...+2^{-p+1})=$$

$$2^{1-n}(1-2^{-p})\le 2^{1-n}$$

for $ n $ great enough, we will have

$$|a_{n+p}-a_n|\le 2^{1-n}<\epsilon$$

$(a_n) $ is Cauchy, and converges.


问:
How to generalise $(\wedge^2 \chi)(g) = \frac{1}{2}(\chi(g)^2-\chi(g^2))$?
One can decompose $\bigotimes^2 V = \bigvee^2 V \oplus \bigwedge^2 V$, getting a corresponding decomposition for representations, say when $V$ is a module for some finite group $G$. One then has the relation between the characters $\chi$, $\vee^2 \chi$ and $\wedge^2 \chi$ afforded by the $G$-modules $V$, $\bigvee^2 V$ and $\bigwedge^2 V$, given by $(\wedge^2 \chi)(g) = \frac{1}{2} (\chi(g)^2-\chi(g^2))$ and $(\vee^2 \chi)(g) = \frac{1}{2} (\chi(g)^2 + \chi(g^2))$.

More generally, one can decompose $\bigotimes^r V$ by Schur-Weyl duality in terms of the irreducible representations of the symmetric group $S_r$. Can we use this to give similar formulas for the representations coming up in this decomposition?

答: Well, yes, but I'm not sure what Schur-Weyl duality has to do with it. The eigenvalues of $g \in G$ acting on $\Lambda^n V$ are the elementary symmetric polynomials of the eigenvalues of $g$ acting on $V$, so the character value can be read off the characteristic polynomial of $g$ as a linear endomorphism of $V$. Similarly, the eigenvalues of $g$ acting on $\text{Sym}^n V$ are the complete homogeneous symmetric polynomials of the eigenvalues of $g$, and can be related to the character of $\Lambda^n V$. With some work one can rewrite these as linear combinations of $\chi(g^m)$; for example $\displaystyle \Lambda^3 \chi(g) = \frac{1}{6} \left( \chi(g)^3 - 3 \chi(g^2) \chi(g) + 2 \chi(g^3) \right)$.</p>


问:
Prove $\sin x +\frac{1}{(\sin x)^{\sin x}}<2, \quad x \in(0,\pi/2)$
No idea how to prove that. What should I try?
$$\sin x +\frac{1}{(\sin x)^{\sin x}}<2, \quad x \in(0,\pi/2)$$</p>

答:
Let's give it a try though it looks ugly and evil:

Define

$$f(u):=u+u^{-u}-2=u+e^{-u\log u}-2\implies f'(u)=1-u^{-u}\left(\log u+1\right)\implies f'(1)=0 $$

$$f''(u)=u^{-u}\left[\left(\log u+1\right)^2-\frac1u\right]\implies f''(1)=1(1-1)=0$$

$$f'''(u)=-u^{-u}\left((\log u+1)\left[\left(\log u+1\right)^2-\frac1u\right]-2(\log u+1)\frac1u-\frac1{u^2}\right)\implies$$

$$f'''(1)=-1(-2-1)>0$$

and the above means $\,u=1\,$ is the first coordinate of an inflection point for $\,f\,$ and of an extreme point (a minimum) for $\,f'\,$, from which we get that

$$f'(u)\ge 0\;,\;\forall\,u\in(0,1)\implies f(u)\le f(1)= 0\;\;\;\;\text{Q.E.D.}$$


问:
Question about the formulation of Kenneth Kunen's "Primitive Recursion on Ordinals" Theorem
In Kenneth Kunen's The Foundations of Mathematics, the subject of recursion (on ordinals) is introduced and is described using the following "recipe":

$f(\xi)=G(f_{|\xi})$, where $G$ and $f$ are functions and $f_{|\xi}$ is the function $f$ restricted to the set of ordinals less than $\xi$.

To provide an example, consider the Fibonacci sequence, traditionally written as: $f(0)=1, f(1)=1,$ and when $x \gt 1$, $f(x)=f(x-1)+f(x-2)$.

Using Kunen's definition, I believe we could rewrite this as:

$$f(0)=G(f_{|0})=1$$

$$f(1)=G(f_{|1})=1$$

$$f(x)=G(f_{|x})=f_{|x}(x-1)+f_{|x}(x-2) \text{ when } 2 \leq x \lt \omega$$

In order to let $x$ be defined for all ordinals (not just $\omega$), one could add a final rule as:

$$f(\xi)=G(f_{|\xi})=1 \text{ when } \xi \geq \omega$$

(As we will see shortly, it would also be necessary to define $G(s)$ as equaling an arbitrary default value when $s$ is some set that is not a function with a domain of some ordinal)

In the following paragraphs, Kunen seeks to demonstrate that the general form of $f(\xi)=G(f_{|\xi})$ is "legitimate"; this is accomplished by proving the following theorem:

Primitive Recursion on Ordinals
Suppose that $\color{red}{\forall s\exists!y\varphi(s,y)}$, and define $G(s)$ to be the unique $y$ such that $\varphi(s,y)$. Then we can define a formula $\psi$ for which the following are provable:

$\forall x \exists ! y \psi(x,y)$, so $\psi$ defines a function $F$, where $F(x)$ is the $y$ such that $\psi(x,y)$

$\forall \xi \in \text{ ON } [F(\xi)=G(F_{|\xi})]$ - where $\text{ ON }$ is shorthand for "is an ordinal"... a regular convention throughout this book -

In the above statement  ("1.", in particular) Kunen states that $\psi$ defines a function $F$. There are several scattered comments (in the book) that make it clear to the reader that this "function" is not really a function in the technical sense because $F$ is not really a set (my understanding is that it is a proper class).

With the necessary background provided, my question is as follows:

I am having difficulty understanding what exactly $\varphi(s,y)$ looks like in the formula $\forall s \exists ! y \varphi (s,y)$...colored in $\color{red}{\text{red}}$... and am wondering if someone could please use the Fibonacci sequence (or an easier recursive formula of your choice) to demonstrate what its corresponding $\varphi(s,y)$ would formally/pseudo-formally look like.

I'm still pretty inexperienced when it comes to constructing sentences using FOL but, using the Fibonacci sequence as my recursive function of interest, it seems like I need to encode the following 5 sub-formulas:

if $s = \emptyset$, then $y=1$

This represents $G(f_{|0})=1$

if $s=\{\langle 0, 1 \rangle \}$, then $y=1$

This represents $G(f_{|1})=1$

if $s$ is a function that has an ordinal greater than or equal to $\omega$ as its domain, then $y=1$

This represents $G(f_{|\xi})=1 \text{ when } \xi \geq \omega$

if $s$ is not a function with a domain of some ordinal, then $y=1$.

We mentioned this at the beginning of the post

if $s$ is a function whose domain is an ordinal $\geq 2$ and $\lt \omega$, then $y = $ the second component of the maximal element in $s$ $+$ the second component of the next largest element in $s$

This represents $G(f_{|x})=f_{|x}(x-1)+f_{|x}(x-2) \text{ when } 2 \leq x \lt \omega$

(Edit: after playing around with this a little bit, these 5 sub-formulas would be connected via conjunctions, '$\land$')

Am I on the right track? Any help and/or insight is greatly appreciated! Thank you.

答:
The formula $\varphi$ should say what the function $G$ is. In the Fibonacci case, to simplify matters consider the formula $\theta(s)$ given by
$$
s=0\vee s=\{\langle 0,1\rangle\}\vee [\text{fun}(s)\wedge \text{dom}(s)\in \omega\setminus 2]
$$
where $\text{fun(s)}$ is a formula saying "$s$ is a function".

Now let $\varphi(s,y)$ be the formula
$$
[\neg\theta(s)\wedge y=\emptyset] \vee [s=0\wedge  y=1] \vee [s=\{\langle 0,1\rangle \}\wedge y=1]\vee[\text{fun}(s)\wedge \text{dom}(s)\in\omega\setminus 2 \wedge y=s(\text{dom}(s)-1)+s(\text{dom}(s)-2)]
$$
The first quantity in brackets is there to output the default value $\emptyset$ whenever $s$ does not have one of the interesting forms (relative to the Fibonacci sequence). Of course, there are many variations on $\varphi$, I chose this one because it shared some components with what you wrote.

Now, the sentence $\forall s\exists! y\varphi(s,y)$ is provable in ZF (or some weaker theory), and so the recursion theorem kicks in. I'll now use Kunen's choice of notation ($F$ and $G$) to check that this formula does what we want. Remember, $F\upharpoonright \omega$ should be the Fibonacci sequence.

We have:
$$
F(0)=G(F\upharpoonright0)=G(\emptyset)=1
$$

$$
F(1)=G(F\upharpoonright 1)=G(F\upharpoonright \{0\})=G(\{\langle 0,1\rangle\})=1
$$
So far so good.

Now, suppose $2\le n<\omega$. Then $F \upharpoonright n$ is a function with domain $n\in \omega\setminus 2$, and so

$$
F(n)=G(F\upharpoonright n)=(F\upharpoonright n)(n-1)+(F\upharpoonright n)(n-2)=F(n-1)+F(n-2)
$$
which is exactly what we wanted.

I hope this clarifies matters. I'll second Asaf's comment regarding finding transfinite recursion weird on the first (several) encounters. I found that writing out things explicitly (like I've done above) was very helpful in demystifying things. Once you're more comfortable with this stuff, you'll just mention how $F(\xi)$ depends on previous values of $F$ and move on with your day.


问:
Interchanging Taylor series with Maclaurin series
Is it always valid to say that the Taylor series of some function $f(x)$ about a point $x = a$ equivalent to the the Maclaurin series of another function $h(x) = f(x+a)$?

For example, is it correct to state the Taylor series of $\arctan(x)$ around $x=1$ is equivalent to the Taylor series of $\arctan(x+1)$ around $x=0$, even though their forms are different (i.e. the latter has terms of powers of $(x-a)$ rather than just $x$)?

Is this true in general for all functions or are there convergence issues?

答:
Let $ f $ be a function defined arround $ x= a $, let us say at $ (a-\eta,a+\eta)$.

then, the function $ g $ given  by $ g(x)=f(x+a) $ will be well defined at $(-\eta,\eta) $.

If $$ g(x)=P_n(x)+x^n\epsilon_1(x)$$

with $\lim_{x\to 0}\epsilon_1(x)=0$

then

$$f(x)=g(x-a)=$$
$$P_n(x-a)+(x-a)^n\epsilon_2(x)$$
with $ \lim_{x\to a}\epsilon_2(x)=0$.


问:
step function working only for non negative numbers
I want to create a step function $f(x,n)$ that works like this:

$n$ indicates the size of the interval and $x$ is the position in the $x$-axis.

I thought the following way:

Let $k = x\div n$ (this maps the intervals to $[\dots,-4,-3,-2,-1,0,1,2,3,4,\dots]$).

if $k \; mod \; 2 < 1$, return $1$, otherwise return $0$.

This works fine for non-negative numbers, but fails at negatives. How can I fix this function? Is there any better approach?

答: What you want in precise terms is $k=\lfloor x/n\rfloor$. Python's // operator computes exactly that; in C and C++ you will need to import floor().


问:
If $(E[\int_0^T|f_n(s)|^2ds])_n$ converges, does it exists a subsequence $f_{n(j)}$ such that the integral converges for $t<T$?
Suppose $f, f_1, f_2, \dots: [0, T] \times \Omega \to \mathbb R$, and that
$$
\lim_n \mathrm E\left [  \int_0^T \left |f_n(s, \cdot) \right|^2 ds\right ] \to \mathrm E\left [ \int_0^T \left |f(s, \cdot) \right|^2 ds\right ] < \infty.
$$

Does it follow that for any $0 < t < T$, there exists a subsequence $n(j)$ such that
$$
\lim_j \mathrm E\left [ \int_0^t \left |f_{n(j)}(s, \cdot) \right|^2 ds\right ] \to \mathrm E\left [ \int_0^t \left |f(s, \cdot) \right|^2 ds\right ]\quad ?
$$
If so, how would you prove it?

Must grateful for any help provided!

答:
No, it does not!

$T=1$, $f_n = 1$ everywhere and $f(t,\cdot)=t\sqrt{3}$, then
$$ \mathbb{E}[\int_0^1|f_n(s,\cdot)|^2ds] = 1 \stackrel{n \to \infty}\to 1 = \mathbb{E}[\int_0^1|f(s,\cdot)|^2ds]$$

But $$ \mathbb{E}[\int_0^t|f_n(s,\cdot)|^2ds] = t \stackrel{n \to \infty}{\not\to} t^3 = \mathbb{E}[\int_0^t|f(s,\cdot)|^2ds]$$


问:
Confusing math problem
How would I solve this question? I came across it and is really confused.

The payment of Jon was bigger by $960$ than the payment of David.
After the payment of David got increased by $10\%$, Jon and David got the same payment amount.

$\text{A}$. We will mark with $X$ the payment of David in the beginning and using the $X$ find the payment of Jon.

$\text{B}$. Find out the payment of David in the beginning.

How can I solve this? Any hints?

答: I imagine you are expected to use "algebra," but it might be useful to note that we can figure out the answer without algebra. A $10\%$ increase in David's salary let him catch up to Jon. The increase in David's salary was $960$. This is $10\%$ of his old salary, so his old salary must have been $9600$.

Title: Confusing math problem


问:
The Underlying Set of Product Variety in the Classic Language
I'm really not familiar with product variety in classic algebraic geometry language(NOT SCHEME).

For $X, Y$ two prevarieties, we can define their product prevariety $X\times Y$ by gluing affine cases. My questions:

Is the underlying set of $X\times Y$ the same as the Cartesian product as sets? (I know the topology is not)

If $X, Y$ are varieties, is $X\times Y$ also variety?

Does the projection maps $p: X\times Y\rightarrow X$ have any property? For example, is it a  closed map?

Let $i_{y_0}: X\rightarrow X\times Y$ be $x\mapsto (x,y_0)$ for a fixed $y_0\in Y$. Is this map a closed embedding?

I'm also wondering if there is any reference for algebraic geometry in classic language? Thank you in advance.

答:
By "prevarieties" I assume you mean what others may call not-necessarily-separated algebraic varieties.

Yes, as a set it is the same as the product of the underlying sets.

Yes, it is shown in Kempf's "Algebraic Varieties", page 26. Basically it is shown by proving that a product of affine varieties is an affine variety, and then taking an open cover of products of affines from each variety's own affine cover.

It is not always closed, but it is open. This can be shown due to the fact that the projection satisfies the axioms of a quotient morphism, which is open.

It is closed and if I remember correctly it is in fact an isomorphism onto its image.

I share your frusturation in the search for extensive materials in classical algebraic geometry.


问:
$\int x^{2}\sqrt{a^{2}+x^{2}}\,dx$. Is there another way to solve it faster?
I have to calculate this integral:

\begin{align} \int x^{2}\sqrt{a^{2}+x^{2}}\,dx \qquad\text{with}
\quad a \in \mathbb{R} \end{align}

My attempt:

Using, trigonometric substitution

\begin{align}
\tan \theta &= \frac{x}{a}\\ \Longrightarrow  \ x&=a \tan \theta\\ \Longrightarrow  \ dx&=a \sec^{2}\theta\\ \Longrightarrow \ x^{2}&=a^{2}\tan^{2}\theta
\end{align}

Thus,
\begin{align}
\int x^{2}\sqrt{a^{2}+x^{2}}\,dx&=\int a^2 \tan^{2}\theta \sqrt{a^2+a^2\tan^{2}\theta}\ a\sec^{2}\theta\, d \theta\\&=a^{3}\int \tan^{2}\theta \sqrt{a^{2}(1+\tan^{2}\theta)}\sec^{2}\theta\, d\theta\\&=a^{3}\int \tan^{2}\theta \sqrt{a^{2}(\sec^{2}\theta)}\sec^{2}\theta \, d\theta\\&=a^{4}\int (1-\sec^{2}\theta)\sec^{3}\theta \, d\theta\\&=a^{4}\underbrace{\int \sec^{3}\theta \, d\theta}_{\text{solve by parts}}-a^{4}\underbrace{\int \sec^{5}\theta \, d\theta}_{\text{solve by parts}}
\end{align}

My doubt is: Is there any other way to solve it faster? Because by parts is a large process to solve each one. I really appreciate your help

答:
$$x^2 \sqrt{a^2+x^2};\;\, x\to a \sinh u$$
$$dx=\cosh u\,du$$
$$\int x^2 \sqrt{a^2+x^2}\,dx=\int (a^2  \sinh ^2 u)(a \cosh u )\sqrt{a^2 \sinh ^2 u+a^2}\,du=$$
$$=a^4\int \sinh^2 u\cosh^2 u\,du=\frac{a^4}{4}\int\sinh^2 2u\,du=\frac{a^4}{8} \int (\cosh 4 u-1) \, du=$$
$$=\frac{1}{8} a^4 \left(\frac{1}{4} \sinh 4 u-u\right)+C=\frac{1}{8} \left(x \sqrt{a^2+x^2} \left(a^2+2 x^2\right)-a^4 \text{arcsinh}\left(\frac{x}{a}\right)\right)+C$$

Useful formulas

$\cosh^2 u -\sinh^2 u=1$

$\sinh 2u=2\sinh u\cosh u$

$\cosh 4u =\sinh ^2 2u +\cosh ^2 2u=2\sinh^2 2u+1\to \sinh^2 2u = \frac{1}{2}(\cosh4u - 1)$

$x=a\sinh u\to u=\text{arcsinh}\left(\frac{x}{a}\right)$

$\sinh 4u = 2\sinh 2u \cosh 2u = 4\sinh u\cosh u(\cosh^2 u+ \sinh^2 u)$


答:
No.  If we consider an infinitely differentiable function $f(x)$, then we can write down it's Taylor series about the point $x = a$ as
$$\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n.$$  Now, if we consider the translation $h(x)=f(x+a)$ and compute its Maclaurin series, we get $$\sum_{n=0}^{\infty}\frac{h^{(n)}(0)}{n!}x^n = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}x^n.$$  Note that $$\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n \ne \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}x^n$$ because they clearly disagree at $x = a$.  What we can say, however, is that the Taylor series for $f(x)$ is the Maclaurin series for $g(x) = f(x+a)$ shifted to the right by $a$.

Title: Interchanging Taylor series with Maclaurin series

Body:
Is it always valid to say that the Taylor series of some function $f(x)$ about a point $x = a$ equivalent to the the Maclaurin series of another function $h(x) = f(x+a)$?

For example, is it correct to state the Taylor series of $\arctan(x)$ around $x=1$ is equivalent to the Taylor series of $\arctan(x+1)$ around $x=0$, even though their forms are different (i.e. the latter has terms of powers of $(x-a)$ rather than just $x$)?

Is this true in general for all functions or are there convergence issues?


问:
how many empty sets are there?
Would I be correct in saying that in the category of sets, the "class of sets that are isomorphic to the empty set is a proper class"?

In other words, there are LOTS of initial objects in the category of sets, but they're all related to each other via unique isomorphisms?

Similarly, for any category $C$, if $X$ is an object of $C$, then the full subcategory of objects isomorphic to $X$ is a proper class (ie, there are LOTS of them)?

EDIT: Let me elaborate a bit. I'm not entirely sure if I should be considering ZFC or ETCS, etc. I'm asking this question from the point of view of algebraic geometry and its use of category theory. I mean, I know people say that the class of say, all elliptic curves over $\mathbb{C}$ is a proper class, but the class of $\mathbb{C}$-isomorphism classes of elliptic curves is a set. That would seem to imply that every isomorphism class is a proper class, and hence this would seem to say that the isomorphism class (ie, bijection class) of the empty set in the category of sets should also be a proper class, right?

Lana's answer with the extensionality axiom seems to make sense, in that there is only ONE empty set, so that the isomorphism class of the empty set is a singleton set. On the other hand, the isomorphism class of a set with 1 element should be a proper class right?

Generalizing a bit, would it be true that the isomorphism class of any initial (or terminal) object of a category be a singleton set? It seems to me that the isomorphism class of, say $\text{Spec }\mathbb{Z}$ in the category of schemes is a proper class, simply because there are many different presentations of rings isomorphic to $\mathbb{Z}$?

will

答:
Let $A$ and $B$ be empty sets, then by logic (namely when if-statements are true) $A \subset B$ and $B\subset A$, so $A = B$ (since $A\subset B$ iff $x\in A\implies x\in B$, but x is not in A, so it is vacuously true, same for $B\subset A$). So there is only one initial object in $\textbf{Set}$ up to equality, not just isomorphism.

Edit: "On the other hand, the isomorphism class of a set with 1 element should be a proper class right?" Yes because the class of sets is a proper class and you can take all of the sets of the form $\{A\}$ where A is a set.

"Generalizing a bit, would it be true that the isomorphism class of any initial (or terminal) object of a category be a singleton set?" No because the objects in a general category need not be just sets.

Title: how many empty sets are there?


问:
How to show that this polynomial has all it's roots in in the disk $|z|<3$
This question was asked in my analysis assignment and I am unable to solve it.So, I am looking for help here.

Show that the polynomial $z^3 -z^2 +4z+5=0 $ has all its roots in the disk |z|<3.

I thought of using Rouché's theorem but I am not able to choose a $g(z)$ which would work. So, I would be unable to provide anything as attempt as this is the starting point of the answer if question has to be solved using Rouché's theorem

Can you please help me with finding $g(z)$?

I hope Rouché's theorem would work in this question.

答:
This is a similar approach to Kavi's answer.

Assume by contradiction that one root $z$ satisfies $|z| \geq 3$.

Then
$$
1= \frac{1}{z}-\frac{4}{z^2}-\frac{5}{z^3}= \left| \frac{1}{z}-\frac{4}{z^2}-\frac{5}{z^3}\right| \\
\leq  \left| \frac{1}{z}\right| + \left| \frac{4}{z^2}\right| + \left| \frac{5}{z^3}\right| \leq \frac{1}{3}+\frac{4}{9}+\frac{5}{27}=\frac{9+12+5}{27}=\frac{26}{27}
$$
This is a contradiction.


问:
Show a certain vector field $V$ only commutes with others that are collinear with $V$
I found this on an old qualifying exam. I started the problem, but I'm not sure what my next step should be:

Let $T^2$ be the standard 2-dimensional torus with $\mathbb{Z}$-periodic coordinates $(x,y)$. Consider the vector field
$$ V = \sin(2\pi y) \dfrac{\partial}{\partial x}$$

Prove that any vector field $Y$ on $T^2$ which commutes with $V$, i.e., $[Y,V] = 0$, has zero coefficient in front of $\frac{\partial}{\partial y}$.

I started by letting $Y = f(x,y) \frac{\partial}{\partial x} + g(x,y)  \frac{\partial}{\partial y}$ be an arbitrary vector field that commutes with $V$. I then tried calculating the bracket:

$$0 = [Y,V] = 2\pi\cdot g(x,y) \cos(2\pi y)  \frac{\partial}{\partial x} - \sin(2\pi y)  \frac{\partial f}{\partial x} \cdot  \frac{\partial}{\partial x} - \sin(2\pi y)  \frac{\partial g}{\partial x} \cdot  \frac{\partial}{\partial y}$$

I then deduced that since $\sin(2\pi y)  \frac{\partial g}{\partial x} = 0$, and $\sin(2\pi y)$ is not always zero, so $ \frac{\partial g}{\partial x} = 0$. This means that $g$ is a function of only $y$.

However, at this point I am stuck. Any suggestions?

答:
Then you need to have
$$\frac{\partial f}{\partial x} = 2\pi g(y)\cot(2\pi y)\,,$$
and there is no $1$-periodic solution $f$ unless $g=0$.</p>


答:
Integrate by parts

\begin{align}
I & =\int x^{2}\sqrt{a^{2}+x^{2}}\,dx
=\int \frac x3d(a^2+x^2)^{3/2} \\
&=\frac13x( a^2+x^2)^{3/2}-\frac13I-\frac{a^2}3\underset{=J}{\int \sqrt{a^2+x^2}dx} \\
 &=\frac14x( a^2+x^2)^{3/2}-\frac{a^2}4J\tag 1
\end{align}
Integrate $J$ by parts again
\begin{align}
J & =\int \sqrt{a^{2}+x^{2}}\,dx=x\sqrt{ a^2+x^2}-J-\int \frac{a^2}{\sqrt{a^2+x^2}}dx\\
 &=\frac12  x\sqrt{ a^2+x^2}-\frac{a^2}2\sinh^{-1}\frac xa \tag2
\end{align}
Plug (2) into (1) to obtain
$$I= \frac14x( a^2+x^2)^{3/2}- \frac{a^2}8 x\sqrt{a^2+x^2} - \frac{a^4}8\sinh^{-1}\frac xa +C
$$

Title: $\int x^{2}\sqrt{a^{2}+x^{2}}\,dx$. Is there another way to solve it faster?

Body:
I have to calculate this integral:

\begin{align} \int x^{2}\sqrt{a^{2}+x^{2}}\,dx \qquad\text{with}
\quad a \in \mathbb{R} \end{align}

My attempt:

Using, trigonometric substitution

\begin{align}
\tan \theta &= \frac{x}{a}\\ \Longrightarrow  \ x&=a \tan \theta\\ \Longrightarrow  \ dx&=a \sec^{2}\theta\\ \Longrightarrow \ x^{2}&=a^{2}\tan^{2}\theta
\end{align}

Thus,
\begin{align}
\int x^{2}\sqrt{a^{2}+x^{2}}\,dx&=\int a^2 \tan^{2}\theta \sqrt{a^2+a^2\tan^{2}\theta}\ a\sec^{2}\theta\, d \theta\\&=a^{3}\int \tan^{2}\theta \sqrt{a^{2}(1+\tan^{2}\theta)}\sec^{2}\theta\, d\theta\\&=a^{3}\int \tan^{2}\theta \sqrt{a^{2}(\sec^{2}\theta)}\sec^{2}\theta \, d\theta\\&=a^{4}\int (1-\sec^{2}\theta)\sec^{3}\theta \, d\theta\\&=a^{4}\underbrace{\int \sec^{3}\theta \, d\theta}_{\text{solve by parts}}-a^{4}\underbrace{\int \sec^{5}\theta \, d\theta}_{\text{solve by parts}}
\end{align}

My doubt is: Is there any other way to solve it faster? Because by parts is a large process to solve each one. I really appreciate your help


问:
What is my operator norm (cannot get good enough bounds).
Given a space of square integrable functions $x(t)$ over the interval $[0;1]$ one can introduce a norm

$$\|x(t)\|= \sqrt{\int_0^1 (x(t))^2 \, dt};$$

Then what is a norm of the transformation below (result is also square integrable function, same space, same norm) assuming the definition.

$$\operatorname{A}x(t)= \int_0^ts \cdot x(s) \, ds;$$

Well according to one definition $\|\operatorname{A}\| = \sup_{\|x\|=1}\|\operatorname{A}x(t)\|\ge \|\operatorname{A}1\| =\sqrt{\frac{1}{20}}$, I tried quite a few different functions and ain't getting any better lower bound than this. To be specific I tried $x(t) = t^\alpha/\| t^\alpha \|$ (for $\alpha \in (-0.5;+ \infty)$) - result is $\sqrt{\frac{1}{20}}$ or lower.

According to other equivalent definition I look for infimum of bounding constants
$$\|\operatorname{A}x(t)\| = \sqrt{\int_0^1 \left(\int_0^t s \cdot x(s) \, ds \right)^2 \, dt} \le \|x(t) \| \cdot \sqrt{\int_0^1 \int_0^t s^2 \, ds \, dt} = \sqrt{\frac{1}{12}} \cdot \|x(t) \|;$$

As a result $\sqrt{\frac{1}{20}} \le \|x(t) \| \le \sqrt{\frac{1}{12}}$. How can I improve?

答:
This is only a partial answer.

Consider ${\rm A}$ as a Hilbert–Schmidt integral operator, i.e. for every $x\in L^2[0,1]$,
$${\rm A}x(t)=\int_0^1k(t,s)x(s)ds,\quad t\in[0,1],$$
where the kernel $k$ of $A$ is given by:
$$k(t,s)=s\cdot 1_{[0,t]}(s),\quad t,s\in[0,1].$$
As a result,
$$\|{\rm A}\|\le \|k\|_{L^2([0,1]^2)}=\frac{1}{2\sqrt{3}}.$$
Let ${\rm A}^*$ be the adjoint operator ${\rm A}^*$ of ${\rm A}$, i.e. $\langle{\rm A}x,y\rangle=\langle x,{\rm A}^*y\rangle$ for any $x,y\in L^2[0,1]$. By definition,
$${\rm A}^* x(t)=\int_0^1k(s,t)x(s)ds.$$
It follows that for ${\rm B}:={\rm A}^*{\rm A}$,
$${\rm B}x(t)=\int_0^1K(t,s)x(s)ds,\quad t\in[0,1],$$
where the kernel $K$ of $B$ is given by:
$$K(t,s)=\int_0^1k(r,t)k(r,s)dr=(1-\max(t,s))ts,\quad t,s\in[0,1].$$
By definition,
$$\|{\rm A} x\|^2=\langle{\rm B}x,x\rangle.$$
Note that ${\rm B}$ is a compact self-adjoint operator and ${\rm B}$ is positive definite, so from the spectral theorem we know that
$$\|{\rm A}\|^2=\sup_{\|x\|=1}\langle{\rm B}x,x\rangle=\|{\rm B}\|=\text{ maximal eigenvalue of } {\rm B}.$$
Let $\lambda_1\ge\lambda_2\ge\cdots\ge\lambda_i\ge\cdots>0$ be all the all eigenvalues of $B$(counting multiplicity). Then
$$\|{\rm B}^n\|=\lambda_1^n,\quad n\ge 1.$$
Also note that $B^n$ is a Hilbert–Schmidt integral operator with kernel $K_n$, where
$K_1=K$ and
$$K_{n+1}(t,s)=\int_0^1K_n(t,r)K(r,s)dr,\quad t,s\in[0,1],~ n\ge 1.$$
It follows that
$$\|K_n\|_{L^2([0,1]^2)}={\rm tr}{\rm B}_n=\sum_{i=1}^\infty \lambda_i^n,\quad n\ge 1. $$
This provides a practical way to estimate the upper bound of $\|{\rm A}\|$:
$$\|{\rm A}\|=\|{\rm B}^n\|^{\frac{1}{2n}}\le\|K_n\|_{L^2([0,1]^2)}^{\frac{1}{2n}},\quad n\ge 1,$$
and
$$\|{\rm A}\|=\lim_{n\to\infty}\|K_n\|_{L^2([0,1]^2)}^{\frac{1}{2n}}.$$
In particular, for $n=1$, we have a rough upper bound
$$\|{\rm A}\|= \|B\|^{\frac{1}{2}}\le\|K\|_{L^2([0,1]^2)}^{\frac{1}{2}}=6^{-\frac{1}{2}}\cdot7^{-\frac{1}{4}}<\frac{1}{2\sqrt{3}}.$$

Another way to estimate $\lambda_1=\|{\rm B}\|=\|{\rm A}\|^2$ is to notice that the eigenspace
$$E_1:=\{x\in L^2[0,1]\mid {\rm B}x=\lambda_1 x\}$$
contains some positive function. It can be seen as follows. If $x\in E_1\setminus\{0\}$, then $x=x^+-x^-$, where $x^\pm:=\max(\pm x,0)$. We may assume that $x^+\ne 0$. Since $y\ge 0\Rightarrow {\rm B}y\ge 0$,
$$Bx^+\ge \max(Bx,0) =(Bx)^+=\lambda_1x^+\Rightarrow \langle Bx^+,x^+\rangle\ge \lambda_1\langle x^+,x^+\rangle.$$
Recall that $\lambda_1=\sup_{\|y\|=1}\langle{\rm B}y,y\rangle$. Therefore,
$$0<\langle Bx^+,x^+\rangle=\lambda_1\langle x^+,x^+\rangle \Rightarrow x^+\in E_1\setminus\{0\}.$$
Now let $x\in E_1\setminus\{0\}$ with $x\ge 0$. Since  $\langle x,1\rangle>0$, we can write $1$ as
$$1=x_1+x_2,\quad x_1\in E_1\setminus\{0\},~ x_2\perp E_1.$$
It follows that
$$\lim_{n\to \infty}\lambda_1^{-n}\cdot{\rm B}^n 1=x_1,$$
i.e. we an use the formula
$$\lambda_1=\lim_{n\to \infty}\|{\rm B}^n 1\|^{\frac{1}{n}}$$
to estimate $\|{\rm A}\|$.


问:
Find y in terms of $t$
So I am given this equation for the rate of a reaction:
$$\frac{dy}{dt}=y^2-8y+15$$ when $t=0, y=0$ [what does this mean?]

How can I proceed to find $y$ in terms of $t$?

I checked out this site but I still don't get how you're supposed to find $y$ in terms of $t$ from it. Some guidance would be appreciated.

答:
''IF dy/dx is the rate of reaction, then y must be the concentration of product formed'' .

Now, at t=0, y=0 means that when the reaction started( at t=0), the concentration of the products must be zero (y=0)

For rest of the calculation you can refer to @lordcommander 's answer.

That is THE BEST METHOD if you have just got into integrations and differentiations.

Title: Find y in terms of $t$


问:
Unfamiliar Notation in Sum; Encountered in AdaBoost description
While trying to understand the relation between AdaBoost and Forward Stagewise Additive Modeling, I came across the following notation:

$$
G_m = arg \min_{b \in H} \sum_{i=1}^n w_i^{(t)} I_{y_i \neq b(x_i)}
$$

I do not understand what $$ I_{y_i \neq b(x_i)} $$ looks like. Is it a square matrix with ones on the diagonal except where $y_i = b(x_i)$?

I hope someone who has seen this notation before can explain this. Thanks in advance.

答: Found it. It's the indicator function.</p>

Body: Found it. It's the indicator function.


问:
Trouble with $I(\alpha) = \int_0^{\infty} \frac{\cos (\alpha x)}{x^2 + 1} dx$
I'm ultimately trying to solve
$$I(\alpha) = \int_0^{\infty} \dfrac{\cos (\alpha x)}{x^2 + 1} dx$$

by using differentiation under the integral. I realize that this is most easily done using residues but I'm intending this problem to introduce my advanced calculus 2/differential equations students to some interesting techniques before they take real analysis.

Differentiating under the integral a first time leads to

$$I'(\alpha) = \int_0^{\infty} \dfrac{-x \sin (\alpha x)}{x^2 + 1} dx = - \dfrac{\pi}{2} + \int_0^{\infty} \dfrac{\sin (\alpha x)}{x(x^2 + 1)}dx$$

by making use of the Dirichlet integral and again to

$$I''(\alpha) = \int_0^{\infty} \dfrac{\cos (\alpha x)}{x^2 + 1} = I(\alpha)$$

To solve this second-order ODE we'll need two initial conditions. The integral for $I'(\alpha)$ leads to the incorrect result $I'(0) = 0$ but the rewritten version leads to the correct result of $I'(0) = -\dfrac{\pi}{2}$. I'm having trouble justifying this.

Any help or guidance is appreciated. I'll also settle for simpler arguments as to why $I'(0) \neq 0$.

答:
You are assuming that
$$
\int_{0}^{\infty}\frac{\sin(\alpha x)}{x}dx= \frac{\pi}{2}
$$
but if $\alpha=0$, then
$$
\int_{0}^{\infty}\frac{\sin(\alpha x)}{x}dx=0
$$
So, the equality
$$
\int_{0}^{\infty}\frac{−x\sin(αx)}{x^2+1}dx=-\frac{π}{2}+\int_{0}^{\infty}\frac{\sin(αx)}{x(x^2+1)}dx
$$
is true iff $\alpha>0$.


问:
What does $\mathcal{C}^{\infty}_{\mathcal{C}}(Z)$ mean?
In my course material I have the following notation
$$
\varphi \in \mathcal{C}^{\infty}_{\mathcal{C}}(Z)
$$
where $Z⊂R^{n}$ is a bounded open set. I was wondering what does this notation mean? What is the set $\varphi$ belongs into? Please, help me :)

答: $\mathcal{C}^{\infty}_{\mathcal{C}}(Z)$ is the set of infinitely differentiable functions defined on $Z$ with compact support, i.e. they are zero outside some compact set.


问:
Is there a reason the word "collection" is used in Heine-Borel theorem to describe an open covering instead of "set"?
The Heine-Borel theorem is stated as follows:

Suppose $\mathcal{H}$ is an open covering of a compact set $S \subseteq \mathbb{R}$. Then $S$ is of an open covering $\tilde{H}$ consisting of finitely many open sets belonging to $\mathcal{H}$.

Here, an open covering $\mathcal{H}$ of $S$ is defined as a collection of open sets such that every point $s \in S$ can be found in a set $H \in \mathcal{H}$. I am wondering if there is a specific reason why $\mathcal{H}$ is called a collection instead of a set. If this $\mathcal{H}$ cannot be constructed using rules from the set theory, how to interpret the use of it?

答:
This is just for clarity. One of the common stumbling blocks in interpreting a piece of mathematics is the issue of type: keeping track e.g. of what's a set of points versus what's a set of sets of points, and so on. It seems to be the case that using varied terminology often makes things a bit easier to read.

As to formal content, there is none: there are no set theoretic issues with constructing sets of open sets.


问:
Why the root of this tree has to be "1"?
Arrange $2^{n-1}-1$ zeroes and $2^{n-1}$ ones in a balanced full binary tree of depth $n$. If we want the number of edges that connect the same (and respectively different) digits are the same, then one claims that the root of this tree has to be a one. Why is that?

For example, if $n=2$, then we need to arrange 1 zero and 2 ones. One arrangement that makes the number of edges that connect the same digits (which in this case is only one: the edge with a "+" who connects 2 ones,) and the number of edges that connect two different digits (which in this case is also one: the edge with a "-") are the same. Note that the root of this tree is unexceptionally 1.

```<code>      1
    /   \
 + /     \ -
  /       \
 1         0
</code>```

And here is a case where $n=3$.

```<code>          1
        /   \
     + /     \ -
      /       \
     1         0
  + / \ -   + / \ -
   /   \     /   \
  1     0   0     1
</code>```

Again, we see that the root is a one. If we change the root to a zero, however, then we can never find an arrangement that makes the number of "+" edges equals to the number of "-" edges. Why is that?

答:
Let's say you have $k$ edge incidences at $1$s and $l$ edge incidences at $0$s. Then if $x$ edges connect a $0$ and a $1$, that leaves $k-x$ incidences at $1$s, so there are $(k-x)/2$ edges connecting a $1$ and a $1$, and likewise $(l-x)/2$ edges connecting a $0$ and a $0$. Then the condition that there are the same number of $+$ and $-$ edges is

$$
\frac{k-x}2+\frac{l-x}2=x\;,
$$

and thus $x=(k+l)/4$. Thus there are

$$
\frac{k-x}2=\frac{k-(k+l)/4}2=\frac38k-\frac18l
$$

edges connecting a $1$ and a $1$. Swapping a $0$ and a $1$ between a branch and a leaf changes this quantity by $\pm1$, so the only way to change whether it's an integer is to change the root, and it turns out that it's an integer if the root is a $1$ and not an integer if it isn't.


问:
Is the complexification of $\mathfrak{sl}(n, \mathbb{C})$ itself?
More precisely if I complexify the algebra, is the following isomorphism true? $$\mathfrak{sl}(n, \mathbb{C})_\mathbb{C} \cong \mathfrak{sl}(n, \mathbb{C})$$
Given that the complex dimension is already $n^2-1$, and each element of the matrix can already be complex, do we gain anything by complexifying here?

答:
I would just like to add to the very good existing answers one basic subtlety which, I think, needs to be pointed out more strongly and explicitly.

Namely, note that if $\mathfrak g$ is a Lie algebra over $\mathbb C$, then of course we have

$$\mathbb C \otimes_{\mathbb C} \mathfrak g \simeq \mathfrak g.$$

But when we say we "complexify" such a Lie algebra $\mathfrak g$ (which already is a complex Lie algebra), we are actually not doing that, at all. Rather, we are doing two things: We first apply scalar restriction, and then scalar extension to it. The shorthand notation for that is

$$\mathbb C \otimes_{\color{red}{\mathbb R}} \mathfrak g \qquad \qquad (1)$$

and what is hidden in there is that the right hand side is no longer the complex Lie algebra $\mathfrak g$, but rather "$\mathfrak g$ viewed as a real Lie algebra". I have found it enlightening (although, admittedly, tedious) to keep track of that with explicit notation, e.g. $\mathrm{Res}_{\mathbb C\vert \mathbb R} ( \mathfrak g)$ (for "restriction"). So if $\mathfrak g$ is an $n$-dimensional complex Lie algebra, then $\mathrm{Res}_{\mathbb C\vert \mathbb R} (\mathfrak g)$ is a $\color{red}{2}n$-dimensional real Lie algebra. And we should properly denote $(1)$ as

$$\mathbb C \otimes_{\color{red}{\mathbb R}} \color{red}{\mathrm{Res}_{\mathbb C\vert \mathbb R}} ( \mathfrak g)$$

Of course this looks horribly cumbersome, but when looked at it this way, it will hopefully become clearer why this does not just give back $\mathfrak g$. Also, it is the restriction step which is responsible for doubling the dimension.

I wrote much more about extension and restriction in my answer here. It is a bit verbose, but it also points out that in general, we cannot even expect that restriction followed by extension (or, for that matter, the other way around, which can give something very different again) would just give several copies of the original Lie algebra, like it does here. In a way, we are still in a good case.


问:
Normal Function, Limit Ordinals and Composition of functions
(ZF) We say that the formula operation F is normal if:

• for each ordinal α, F(α) is an ordinal as well,

• F is strictly increasing over the ordinals:
α < β ⇒ F(α) < F(β),

• F is continious over the limit ordinals:
Lim(λ) ⇒ F(λ) = sup{F(β) | β < λ}.

а) Let F is a normal function. Prove that for every ordinal α the following statements are true:

α ≤ F(α);

for each limit ordinal λ, F(λ) is a limit ordinal as well;

for every non empty set A of ordinals is true that:

sup{F(α) | α ∈ A} = F(sup A), i.e.:
∪{F(α) | α ∈ A} = F(∪A);

each time when F(0) < β it follows that there exists a largest ordinal α which satisfies: F(α) ≤ β.

b) Prove that the composition of normal functional operations is normal as well,i.e. if the formula operations F и G are normal,then the formula operation H, defined as x с H(x) = G(F(x)),
is normal as well.

My ideas are:
a)

1.To use that if  α, β ordinals and α < β ↔ S(α) ≤ β, where S(α) is the successor of the ordinal number α. But S(α) is defined as α∪{α}, hence α∪{α}≤ β. Also, I claim that between an ordinal α and its successor S(α) there is no other ordinal. Hence let S(α):=F(α) which gives us α<F(α).

Every non zero  ordinal which is not a successor ordinal is a limit ordinal:
Limit(λ) ↔ (λ != 0 & (∀ β)[λ!= S(β)]).Let now x ∈ λ is an arbitrary. Hence ord(x) and ord(S(x)).Since the successor of x is the smallest ordinal bigger than x, we have that S(x) ≤ λ. But λ is not a successor, so S(x) < λ.Then Lim(λ) ⇒ F(λ) = sup{F(x) | x < λ}, so F maps all ordinals bigger than x into limit ordinals.

3.We say that a is an upper bound for B in  if for (∀ x )[x ∈ B → <x , a> ∈ R] where  is a partial ordered set with B ⊆ A and a ∈ A; As such F will map a to sup in (B). About unity, i think that induction will solve it for two or more ordinals.

Since 0 is not an ordinal, our function F can be initialized as F(α) = 1 + α. Hence F(0) = 1 < β. But for every successor α of 1, we have that α < α∪{α}=F(α) <β;

For b): It seems obvious that the composition of normal functions is a normal functuons since Dom and Rng are the ordinal numbers, but can I use somehow the Bachmann hierarchy here?

I am sorry for this long post, but in my opinion if I have divided it into two separated posts, the whole idea wouldn't have been fulfilled.
I would be extremely thankful for any help you give me!

答:
This has quite a few problems. For (a) you need to show that every normal function satisfies (1), (2), (3), and (4). Thus, you cannot ‘let $S(\alpha):=F(\alpha)$, as you do in your argument for (1): your $S$ is the successor function, so $S(\alpha)$ is already defined, and there is no reason to think that it is $F(\alpha)$. (And in any case $S$ is not a normal function.)

Your conclusion for (2) is a non sequitur: it doesn’t follow from your argument. It’s also false in general: a normal function need not map all sufficiently large ordinals to limit ordinals. You have no argument for (3), and for (4) you are once again assuming a particular $F$ (which is not in fact normal) rather than proving something about all normal functions. And $0$ is an ordinal.

Let $F$ be normal, and suppose that $F(\alpha)<\alpha$ for some $\alpha$. Let $$A=\{\xi\le\alpha:F(\xi)<\xi\}\,;$$ $A$ is a non-empty set of ordinals, so it has a least element $\mu$. Let $\beta=F(\mu)$; $\beta<\mu$, so $F(\beta)<F(\mu)=\beta$, and therefore $\beta\in A$. But this is impossible: $\mu$ is the smallest member of $A$, and $\beta<\mu$. This contradiction shows that $\alpha\le F(\alpha)$ for all $\alpha$.

Let $\lambda$ be a limit ordinal; since $F$ is normal, $F(\lambda)=\sup\{F(\beta):\beta<\lambda\}$. Suppose that this is not a limit ordinal. Clearly $F(\lambda)\ne 0$ (why?), so $F(\lambda)\ne 0$, and therefore $F(\lambda)=\alpha+1$ for some ordinal $\alpha$. (I write $\alpha+1$ for the immediate successor of $\alpha$.) Thus, $\sup\{F(\beta):\beta<\lambda\}=\alpha+1$. If it were the case that $F(\beta)\le\alpha$ for all $\beta<\lambda$, then $\sup\{F(\beta):\beta<\lambda\}$ would be at most $\alpha$, so there must be a $\gamma<\lambda$ such that $F(\gamma)=\alpha+1$. But $\lambda$ is a limit ordinal, so $\gamma+1<\lambda$, and $\gamma<\gamma+1$, so $$\alpha+1=F(\gamma)<F(\gamma+1)\le\sup\{F(\beta):\beta<\lambda\}=\alpha\,,$$ which is absurd. This contradiction shows that $F(\lambda)$ cannot be a successor ordinal, and since it’s not $0$, it must be a limit ordinal.

Now let $A$ be a non-empty set of ordinals; we want to show that $$\sup\{F(\alpha):\alpha\in A\}=F(\sup A)\,.\tag{1}$$ Let $\beta=\sup A$; then $\alpha\le\beta$ for each $\alpha\in A$, so $F(\alpha)\le F(\beta)$ for each $\alpha\in A$. That is, $F(\beta)$ is an upper bound for $\{F(\alpha):\alpha\in A\}$, so $$\sup\{F(\alpha):\alpha\in A\}\le F(\beta)\,.$$ There are now two possibilities. If $\beta\in A$, then $$F(\beta)\le\sup\{F(\alpha):\alpha\in A\}\le F(\beta)\,,$$ and $(1)$ holds. If $\beta\notin A$, then $\beta$ must be a limit ordinal (why?), and therefore

$$\begin{align*}
F(\beta)&=\sup\{F(\alpha):\alpha<\beta\}\\
&=\bigcup\{F(\alpha):\alpha\in\beta\}\\
&\overset{*}=\bigcup\{F(\alpha):\alpha\in A\}\\
&=\sup\{F(\alpha):\alpha\in A\}\,,
\end{align*}$$

giving us $(1)$ in this case as well. I leave it to you to prove the starred equality; prove and use the fact that for each $\gamma<\beta$ there is an $\alpha\in A$ such that $\gamma<\alpha$.

Finally, suppose that $F(0)<\beta$. Let $A=\{\alpha\le\beta:F(\alpha)<\beta\}$. From (1) we know that if $\alpha$ is an ordinal such that $F(\alpha)\le\beta$, then $\alpha\le\beta$, so the set $A$ contains every ordinal $\alpha$ such that $F(\alpha)\le\beta$, and we want to show that $A$ has a largest element. Let $\gamma=\sup A$. If $\gamma$ is $0$ or a successor ordinal, then $\gamma\in A$. If $\gamma$ is a limit ordinal, then by (3) we know that $$F(\gamma)=\sup\{F(\alpha):\alpha\in A\}\le\beta\,,$$ so again $\gamma\in A$. Thus, in all cases $A$ has a largest element, namely, $\gamma=\sup A$.

For (b) it is trivial that $H(\alpha)$ is an ordinal for every $\alpha$: $F$ is normal, so $F(\alpha)$ is an ordinal, and therefore $G\big(F(\alpha)\big)$ is defined and, because $G$ is normal, is an ordinal. It is also very easy to see that $H$ is strictly increasing: if $\alpha<\beta$, then $F(\alpha)<F(\beta)$, so $H(\alpha)=G\big(F(\alpha)\big)<G\big(F(\beta)\big)=H(\beta)$.

Finally, suppose that $\lambda$ is a limit ordinal. Then $F(\lambda)=\sup\{F(\alpha):\alpha<\lambda\}$. Now apply (4) to $G$:

$$\begin{align*}
H(\lambda)&=G\big(F(\lambda)\big)\\
&=\sup\left\{G\big(F(\alpha)\big):\alpha<\lambda\right\}\\
&=\sup\{H(\alpha):\alpha<\lambda\}\,.
\end{align*}$$

Thus, $H$ is normal.


问:
Need Help Interpreting the Imaginary Component of the Dot Product for Complex Valued Vectors
I've been looking into a copy of Schaum's Outlines for linear algebra and came across the extension of the real-valued dot product into complex-valued vectors. For two vectors $\vec{u}$ and $\vec{v}$ in $\mathbb{C}^{n}$, $$ \vec{u} \cdot \vec{v} = u_1 \bar{v}_1 + u_2 \bar{v}_2 + \cdots + u_n \bar{v}_n $$
Where $\bar{v}_j$ is the complex conjugate of $v_j$.

I understand that using the conjugate of the second argument, $\vec{v}$ in this case, allows this definition to reduce to the real-valued dot product in the real case, and allows the real component, Re($\vec{u} \cdot \vec{v}$), to reflect the familiar idea of projecting and scaling two 'equivalent' vectors in $\mathbb{R}^{2n}$
(that is, if $ \vec{u}_\text{complex} =\: <2+4i,1+3i>,$ then the 'equivalent' real vector is $\vec{u}_\text{real}=\:<2,4,1,3>$).

This gives me a fine understanding for the real component of this complex dot product, but I can't find a similar familiarity for the imaginary component. I'm not sure exactly how to word my question as generally as I would like to, so for a concrete example:

Given two vectors in $\mathbb{C}^3$, $\vec{z}=\:<5+2i,-3-7i,-6+8i>$ and $\vec{w}=\:<-9+2i,1+i,7-2i>$,$$\vec{z}\cdot\vec{w}=-109+12i$$$$\vec{w}\cdot\vec{z}=-109-12i$$

How can I interpret the numbers 12i and -12i? Is there a geometric interpretation (perhaps one relating to rotation?) that I can attribute to those values? Should I not be looking at the imaginary component alone? What does that number tell me, and what can I use it to do?

Another question to go with this: is the choice of taking the complex conjugate of the first argument as opposed to the second an agreed upon definition? Looking into this on my own, I've found many online sources disagree on which terms to take the conjugate of. The two definitions I'm seeing are:$$ \vec{u} \cdot \vec{v} = u_1 \bar{v}_1 + u_2 \bar{v}_2 + \cdots + u_n \bar{v}_n $$$$ \vec{u} \cdot \vec{v} = \bar{u}_1 v_1 + \bar{u}_2 v_2 + \cdots + \bar{u}_n v_n $$Both of these definitions agree that $\vec{u} \cdot \vec{v} = \overline{\vec{v} \cdot \vec{u}}$, but they do give noticeably different results. Is one of these definitions preferable to the other? And, if so, why?

Using these definitions without understanding the justifications behind them really damages my ability to work with the material, so I appreciate any help I can get with this.

答: The imaginary part of the complex dot product is a symplectic form. I have never understood the significance of this, and I recommend that you don't worry about it. Because $\langle v, v \rangle = \| v \|^2$ is still real you can mostly ignore it in practice. The polarization identity guarantees that the inner product is just a convenient way to work with the norm (which is just an ordinary Euclidean norm); this is how I think about things anyway.